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3,914,365 |
<p>I'm pretty stuck with this exercise. Hope somebody can help me:</p>
<p>show that
<span class="math-container">$$
\sum_{n=2^{k}+1}^{2^{k+1}} \frac{1}{n} \geq\left(2^{k+1}-2^{k}\right) \frac{1}{2^{k+1}}=\frac{1}{2}
$$</span>
and use this to show that the harmonic series is divergent.</p>
<p>First I can't figure out how to show the inequality.
Second I'm not quite sure how to use the inequality to show divergence. Because it doesn't really help letting k go to infinity.</p>
<hr />
<p>Writing out the series gave me this, but where do I go from here:
<span class="math-container">$$
\sum_{n=2^{k}+1}^{2^{k+1}} \frac{1}{n}=\frac{1}{2^{k}+1}+\frac{1}{2^{k}+2}+\cdots \frac{1}{2^{k}+2^{k}}=\frac{1}{2^{k}+1}+\frac{1}{2^{k}+2}+\cdots+\frac{1}{2^{k+1}}
$$</span></p>
<hr />
<p><strong>Edit:</strong>
So I understand how to show the inequality now. Thanks!
But I'm still lost with how to use the inequality for the divergence proof.</p>
|
gt6989b
| 16,192 |
<p><strong>HINT</strong>
<span class="math-container">$$
\begin{split}
\sum_{n=2^k+1}^{2^{k+1}} \frac1n
\ge \sum_{n=1}^{2^k} \frac{1}{2^k+n}
\ge \sum_{n=1}^{2^k} \frac{1}{2^{k+1}}
= \frac{1}{2^{k+1}} \sum_{n=1}^{2^k} 1
\end{split}
$$</span>
Can you finish?</p>
|
283,816 |
<p>$ 2 \ln (5x) = 16$</p>
<p>$ \ln (5x) = 8 $</p>
<p>$ 5x = e^8 $</p>
<p>$ x = \dfrac {1}{5}e^8$</p>
<p>But why can't we do it like this:</p>
<p>$ \ln(5x)^2 = 16$</p>
<p>I thought that was a possibilty with logaritms?</p>
|
Guest 86
| 58,804 |
<p>$$\ln ((5x)^2) = 16,\qquad x>0$$</p>
<p>$$25x^2 = e^{16}$$</p>
<p>$$x^2 = \frac{e^{16}}{25}$$</p>
<p>$$x = \pm\sqrt{\frac{e^{16}}{25}} =\pm\frac{\sqrt{e^{16}}}{5}=\pm\frac{e^{16/2}}{5}=\pm \frac{e^8}{5}$$</p>
<p>But since you've introduced the square you have to go back and check the answers - The negative one doesn't fit. So they're equivalent.</p>
|
262,773 |
<p>In the curve obtained with following</p>
<pre><code> Plot[{1/(3 x*Sqrt[1 - x^2])}, {x, 0, 1}, PlotRange -> {0, 1},
GridLines -> {{0.35, 0.94}, {}}]
</code></pre>
<p>how can one fill the top and bottom with different colors or patterns such that two regions are perfectly visible in a black-n-white printout?</p>
|
cvgmt
| 72,111 |
<pre><code>Clear[plot, pts];
plot = Plot[{1/(3 x*Sqrt[1 - x^2])}, {x, 0, 1}, PlotRange -> {0, 1}];
pts = Cases[plot, Line[a_] :> a, All] // First;
Graphics[{{GrayLevel[.6], Polygon[pts]}, {EdgeForm[Dashed], LightGray,
Polygon[Join[{{pts[[1, 1]], 0}}, pts, {{pts[[-1, 1]], 0}}]]}},
PlotRange -> {{0, 1}, {0, 1}}, Axes -> True]
</code></pre>
<p><a href="https://i.stack.imgur.com/QHJZK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QHJZK.png" alt="enter image description here" /></a></p>
|
2,901,971 |
<p>The question is as follows: </p>
<blockquote>
<p>Prove that for every $\beta \in \mathbb{R}$, $\sup(-\infty, \beta) = \beta$.</p>
</blockquote>
<p>The goal of the problem is to prove this without using the $\epsilon-\delta$ method. My professor gave us an idea in class, but for some reason, it isn't really making any sense to me. </p>
<p>Call the set $S.$ We need to show that: </p>
<p>1) $\beta$ is an upper bound of $S.$ </p>
<p>2) $\beta \leq B \ \forall $ upper bounds $B$ of $S.$ </p>
<p>The first condition is clearly satisfied by the fact that $x \leq \beta \ \forall x \in S$. </p>
<p>The second condition is what we did in class and the part that I am most confused about. </p>
<p>If $B$ is an upper bound of $S$, then $\beta \leq B.$ </p>
<p>Now assume that $B < \beta.$</p>
<p>If $B < \beta, \exists x$ such that $B < x < \beta$. But then $x < \beta$ and $B<x$, contradicing the assumption that $B$ is an upper bound of $S$, thus proving that $\sup(-\infty, \beta) = \beta$. </p>
<p>Q.E.D. </p>
<p>This proof does not make much sense to me, so I am asking for a clarification. If this prove is indeed wrong, I would appreciate if it someone could provide me with a clear and concise way of proving this. </p>
<p>Thank you so much!</p>
|
Misha Lavrov
| 383,078 |
<p>Here are a few conditions that are equivalent to having an $X$-saturating matching. (This is stronger than what the question asks for, but is still an answer.)</p>
<ol>
<li><p>$G$ has no vertex cover smaller than $X$. </p></li>
<li><p>$G$ has no independent set bigger than $Y$. (This follows from 1 because the complement of a vertex cover is an independent set.)</p></li>
<li><p>For all $A \subseteq Y$, $|\Gamma(A)| \ge |A|+|X|-|Y|$. (A Hall-like condition for $Y$ rather than $X$.)</p></li>
</ol>
|
494,338 |
<p>This is gre preparation question of data interpretation,
Distribution of test score among students(Score range -> total % of students)</p>
<blockquote>
<p>0-65 -> 16<br/>
65-69 -> 37<br/>
70-79 -> 25<br/>
80-89 -> 14 <br/>
90-100 -> 8<br/></p>
</blockquote>
<hr>
<p>Question is that</p>
<p>Which of the following point ranges includes the median reading test score for ninth grade students in School District X for 1993 ? </p>
<hr>
<p>From my understanding, median is the middle value of the dataset</p>
<blockquote>
<p>1,2,3,4,5 <strong>median</strong>=3<br/>
1,2,3,4,5,6 <strong>median</strong>=3.5</p>
</blockquote>
<p>From this logic answers is 0-65 as 16 comes in middle (8,14,16,25,37), However that is not right, Correct answer is 65-69.</p>
<p>Can anyone explain to me what I am doing wrong? </p>
|
Daniel R
| 83,553 |
<p>The right column shows the percentage of students for each range. Adding the percentages, starting from the top (or bottom for that matter), the range where you cross the 50 % limit will be the your answer. </p>
<p>What you're doing wrong is that you use the percentages themselves as the dataset. The dataset is the <em>scores</em>, which have been represented by ranges. </p>
|
347,009 |
<p>I am looking for a generalisation of a modular form that transforms as something like:</p>
<p><span class="math-container">$f(\frac{a \tau+b}{c \tau+d}) = (c \tau+d)^k c^k f(\tau)$</span></p>
<p>I understand this cannot be literally true, as the multiplier c^k is not a root of unity, but does something like this arise in the context of modular forms? (or generalisations of those). Thanks!</p>
|
Yoav Kallus
| 20,186 |
<p>Let A be the measure on R^2 invariant wrt rotations about the origin with total weight at each radius equal to the frequency of that distance in your histogram. This measure is obtained by rotationally averaging the autocorrelation measure. The Fourier transform of the autocorrelation measure is nonnegative. Since the Fourier transform is equivariant under rotation, I believe this means that the Fourier transform of A must also be nonnegative. This yields some strong constraints on the histogram of distances, which are basically the entire theory behind LP bounds.</p>
|
3,379,447 |
<p>I am currently studying for an exam in Mathematical research tools. One concept that has come up again and again throughout the course is continuity and although the definition of Cauchy Continuity or Delta Epsilon continuity was discussed, we were never shown how to actually prove that a function is continuous. Therefore I am very interested if there is a. general approach that one can take if one wants to prove continuity of simple one variable functions like X^2 or the square root of x etc. Any sort of help would be greatly appreciated. </p>
|
fleablood
| 280,126 |
<p>To prove <span class="math-container">$f(x) $</span> is continuous at <span class="math-container">$x = c$</span> we must prove the limit of <span class="math-container">$\lim_{x\to c}f(x) = c$</span>.</p>
<p>So for any <span class="math-container">$\epsilon > 0$</span> there is a <span class="math-container">$\delta_\epsilon > 0$</span> so that whenever <span class="math-container">$|x-c|< \delta_\epsilon$</span> it will follow that <span class="math-container">$|f(x)-f(c)|< \delta$</span>.</p>
<p>There are two strategies</p>
<p>1) Start with <span class="math-container">$|x-c| < \delta$</span></p>
<p><span class="math-container">$-\delta < x-c < \delta$</span></p>
<p><span class="math-container">$c-\delta < x <c+\delta$</span></p>
<p>do series of manipulations on <span class="math-container">$x$</span> to get</p>
<p><span class="math-container">$SOMETHINGTODOWITHDELTA < f(x) - f(c) <SOMETHINGELSETODOWITHDELTA$</span></p>
<p>And find a way to have a <span class="math-container">$\epsilon_{\delta}$</span> based on <span class="math-container">$\delta$</span> so that </p>
<p><span class="math-container">$-\epsilon_{\delta} \le SOMETHINGTODOWITHDELTA < f(x) - f(c) <SOMETHINGELSETODOWITHDELTA +\epsilon_\delta$</span></p>
<p>And then express <span class="math-container">$\delta$</span> in terms <span class="math-container">$\epsilon$</span></p>
<p>Example: Prove <span class="math-container">$f(x) = \sqrt{4x +1}$</span> is continuous at <span class="math-container">$x = 12$</span>.</p>
<p>That's a matter of proving <span class="math-container">$\lim_{x\to 12} f(x)=\sqrt{4x +1} = f(12)=7$</span>.</p>
<p>So if <span class="math-container">$|x -12| <\delta$</span> then</p>
<p><span class="math-container">$-\delta < x-12 < \delta$</span></p>
<p><span class="math-container">$49 -4\delta < 4x +1 < 49 +4\delta$</span></p>
<p>Now we can assume <span class="math-container">$\delta < \frac{49}4$</span>. If <span class="math-container">$|x-c| < \delta \implies something$</span> then <span class="math-container">$|x-c|< smaller\delta < \delta \implies something$</span> so we are allow to assume <span class="math-container">$\delta$</span> is as small as we need it to be.</p>
<p><span class="math-container">$\sqrt{49-4\delta} < \sqrt{4x+1} < \sqrt{49+4\delta}$</span>.</p>
<p>Now if we express <span class="math-container">$7-\epsilon \le \sqrt{49-4\delta}$</span> and <span class="math-container">$\sqrt{49+4\delta} \le 7+\epsilon$</span> we will be done.</p>
<p>We can assume <span class="math-container">$\epsilon$</span> is as small as we like. We need to prove that there is a <span class="math-container">$\delta$</span> that shows <span class="math-container">$|f(x)-f(c)|< \epsilon$</span> and if instead we find a <span class="math-container">$\delta$</span> shows <span class="math-container">$|f(x)-f(c)|< smaller\epsilon < \epsilon$</span> that will be good enough. So we can assume <span class="math-container">$\epsilon < 7$</span></p>
<p>So we need </p>
<p><span class="math-container">$0 < 7-\epsilon \le \sqrt{49-4\delta}< \sqrt{49+4\delta} \le 7+4\epsilon$</span></p>
<p>Or </p>
<p><span class="math-container">$49- 14\epsilon + \epsilon^2 \le {49-4\delta}< 49+4\delta \le 49 -14\epsilon + \epsilon^2$</span></p>
<p><span class="math-container">$-14\epsilon + \epsilon^2 \le -4\delta \le 14\epsilon +\epsilon^2$</span>.</p>
<p>If we pick <span class="math-container">$\epsilon < 1$</span> then <span class="math-container">$\epsilon^2 < \epsilon$</span> so</p>
<p><span class="math-container">$-14\epsilon + \epsilon^2 < -13\epsilon \le 14\epsilon + \epsilon^2 < 15\epsilon$</span>.</p>
<p>If we can force <span class="math-container">$|4\delta| < 13\epsilon$</span> we are done i.e. if</p>
<p><span class="math-container">$\delta < \frac {13}4\epsilon$</span> and <span class="math-container">$\epsilon <1$</span> then </p>
<p><span class="math-container">$|x-7|<\delta \implies$</span></p>
<p><span class="math-container">$\sqrt{49-4\delta} < \sqrt{4x + 1} < \sqrt{49+\delta}$</span></p>
<p><span class="math-container">$\sqrt{49-13\epsilon} < \sqrt{49-4\delta} < \sqrt{4x + 1} < \sqrt{49+\delta} < \sqrt{49+13\epsilon}$</span> so</p>
<p><span class="math-container">$\sqrt{49-14\epsilon + \epsilon^2}<\sqrt{49-13\epsilon} < \sqrt{4x + 1} < \sqrt{49+13\epsilon} < \sqrt{49 + 14\epsilon + \epsilon^2}$</span></p>
<p>so <span class="math-container">$7-\epsilon < \sqrt{4x+1} < 7 +\epsilon$</span></p>
<p><span class="math-container">$-\epsilon < \sqrt{4x+1} -7 < \epsilon$</span></p>
<p><span class="math-container">$|\sqrt{4x+1}-7|<\epsilon$</span>.</p>
<p>So <span class="math-container">$\lim_{x\to 12}\sqrt{4x + 1}=7=\sqrt{4*12+1}$</span> so <span class="math-container">$\sqrt{4x+1}$</span> is continuous at <span class="math-container">$x = 12$</span>.</p>
<p>Method 2:</p>
<p>We start with <span class="math-container">$|f(x)-f(c)| < \epsilon$</span></p>
<p><span class="math-container">$f(c) -\epsilon < f(x) < f(c) +\epsilon$</span>.</p>
<p>Manipulate <em>with steps that are reversable</em> to get </p>
<p><span class="math-container">$SOMETHINGTODOWITHEPSILON < x < SOMETHINGELSETDOITHEPSILON$</span></p>
<p>And try to find some <span class="math-container">$\delta_\epsilon$</span> in terms of <span class="math-container">$\epsilon$</span> where </p>
<p><span class="math-container">$SOMETHINGTODOWITHEPSILON\le c+ \delta_\epsilon < x <c-\delta_\epsilon \le SOMETHINGELSETDOITHEPSILON$</span></p>
<p>Then you have (assuming ever step is reversable) <span class="math-container">$|x-c|<\delta_\epsilon \implies |f(x) - f(c) | <\epsilon$</span>.</p>
<p>Example:</p>
<p>Example: Prove <span class="math-container">$f(x) = \sqrt{4x +1}$</span> is continuous at <span class="math-container">$x = 12$</span>.</p>
<p>That's a matter of proving <span class="math-container">$\lim_{x\to 12} f(x)=\sqrt{4x +1} = f(12)=7$</span>.</p>
<p>So to conclude <span class="math-container">$|\sqrt{4x+1} - 7| < \epsilon$</span> we can show that from</p>
<p><span class="math-container">$0<7-\epsilon < \sqrt{4x+1} < 7+\epsilon$</span>.</p>
<p>And that will follow from <span class="math-container">$\epsilon < 1$</span> and </p>
<p><span class="math-container">$0< 49 -14\epsilon+\epsilon^2 < 4x+1 < 49 + 14 \epsilon +\epsilon^2$</span></p>
<p>If we assume <span class="math-container">$\epsilon < 1$</span> then <span class="math-container">$\epsilon^2 < \epsilon$</span> and we will have the above result if we have</p>
<p><span class="math-container">$48 -14\epsilon+\epsilon^2<48 - 13\epsilon < 4x < 48 + 13\epsilon < 48-+14\epsilon + \epsilon^2$</span>.</p>
<p>or equivalently <span class="math-container">$12 -\frac{13}4\epsilon < x < 12 + \frac{13}4 \epsilon$</span></p>
<p>or equivalently if <span class="math-container">$|x-12|< \frac{13}4\epsilon$</span>.</p>
<p>So if we are given and <span class="math-container">$\epsilon$</span> and we relabel <span class="math-container">$\epsilon = \min(\epsilon, 1)$</span> and we set <span class="math-container">$\delta =\frac{13}4\min(\epsilon, 1)$</span> then we will get</p>
<p><span class="math-container">$|x-12|<\delta \implies$</span></p>
<p><span class="math-container">$|f(x) -7|<\epsilon$</span>.</p>
<p>So <span class="math-container">$\lim_{x\to 12}\sqrt{4x+1} = 7=\sqrt{4(12)+1}$</span> so</p>
<p><span class="math-container">$\sqrt{4x+1}$</span> is continuous at <span class="math-container">$x = 12$</span>.</p>
|
116,201 |
<p>I am curious if the "<a href="http://neumann.math.tufts.edu/~mduchin/UCD/111/readings/architecture.pdf" rel="nofollow noreferrer">Bourbaki's approach</a>" to mathematics is still a viable point of view in modern mathematics, despite the fact that Bourbaki is vilified by many.</p>
<p>Even more specifically, does anyone actively approach mathematics from the more "yielding" <a href="http://www.math.jussieu.fr/~leila/grothendieckcircle/chap1.pdf" rel="nofollow noreferrer">point of view</a> famously practiced by Grothendieck? Which, or what type of, research areas are welcoming to (or practicing) Grothendieck's approach to mathematics?</p>
<p><strong>Motivation:</strong></p>
<p>To me, there is a deep question regarding motivation of mathematicians over time which is addressed by this viewpoint. An emphasis on resolving hard technical problems is quite depressing, generally, whereas the idea of finding a general framework which presents a natural and explanatory solution through the development of a vast theory seems very motivating. In such a view, the open problem only serves to motivate a better development of the general theory surrounding the core difficulty, bringing into focus a clearer picture of the essential issue at hand.</p>
<p>It seems to me that carefully developing a general (sometimes axiomatic) theory is analogous to performing scientific experiment. One is not looking to be clever, but instead is filling in data which may, when examined later, reveal clear and natural answers to mathematical questions. Obviously such an approach can be exhausting, in that one must spend much more time to fill in an entire picture than to, at some point, jump to a resolution of a particular question. On the other hand, It may be possible to persevere longer at such a task, as one is not so sensitive to one's loss of quickness or cleverness and can simply engage the task at hand.</p>
<p>Is this viewpoint valid?</p>
<hr>
<p>[<strong>Edited (Dec. 17, 2012) by A. Caicedo</strong>, following suggestions <a href="http://mathoverflow.tqft.net/discussion/1490/bourbaki-thread/" rel="nofollow noreferrer">here</a>. Question originally asked by user
<a href="https://mathoverflow.net/users/29891/curious1">curious1</a>.]</p>
|
AAK
| 2,503 |
<p>The following appears in "<a href="http://www.ams.org/notices/201009/rtx100901106p.pdf">Reminiscences of Grothendieck and his school</a>", published in Notices of the AMS:</p>
<blockquote>
<p>Bloch: I wonder whether today such a style of mathematics could exist.</p>
<p>Illusie: Voevodsky’s work is fairly general. Several people tried to imitate Grothendieck, but I’m afraid that what they did never reached that “oily” character dear to Grothendieck.</p>
</blockquote>
<p>I am not completely sure what Illusie meant by "oily", but this seems to be a hint:</p>
<blockquote>
<p>Illusie: To him no statement was ever the best one. He could always find something better, more general or more flexible. Working on a problem, he said he had to sleep with it for some time. He liked mechanisms that had oil in them. For this you had to do scales, exercises (like a pianist), consider special cases, functoriality. At the end you obtained a formalism amenable to dévissage.</p>
</blockquote>
|
267,753 |
<p>A number of seemingly unrelated elementary notions can be defined uniformly with help of (iterated) Quillen lifting property
(a category-theoretic construction I define below) "starting" to a single (counter)example or a simple class of morphisms,
for example a finite group being nilpotent, solvable, p-group, a topological space being compact, discrete, T4 (normal).</p>
<p>I would like to see more examplees, to help me understand if there is a bigger picture behind. </p>
<p>Let me give the definitions.</p>
<p>For a property $C$ of arrows (morphisms) in a category, define
its {\em left and right orthogonals} as</p>
<p>$$ C^\perp := \{ f :\text{ for each }g \in C\ f \,\rightthreetimes\, g \} $$
$$ {}^\perp C := \{ g :\text{ for each }f \in C\ f \,\rightthreetimes\, g \} $$</p>
<p>here $f \,\rightthreetimes\, g$ reads " $f$ has the left lifting property wrt $g$ ",
" $f$ is (left) orthogonal to $g$ ",
i.e. for $f:A\longrightarrow B$, $g:X\longrightarrow Y$,
$f \,\rightthreetimes\, g$ iff for each $i:A\longrightarrow X$, $j:B\longrightarrow Y$ such that $ig=fj$ ("the square commutes"),
there is $j':B\longrightarrow X$ such that $fj'=i$ and $j'g=j$ ("there is a diagonal
making the diagram commute").</p>
<p>Examples:</p>
<p>In the category Sets of sets the right orthogonal
${}^\perp \{\emptyset \longrightarrow \{*\}\}$ of the simplest non-surjection
$\emptyset \longrightarrow \{*\}$ is the class of surjections.
The triple left orthogonal $ ((\{\emptyset \longrightarrow \{*\}\}^\perp)^\perp)^\perp$ is the class of functions which split.</p>
<p>The left and right orthogonals of $ \{x_{1},x_{2}\}\longrightarrow \{*\} $, the simplest non-injection, are both precisely the class of injections.</p>
<p>A finite group $H$ is nilpotent iff $H\longrightarrow H\times H$ is in ${}^\perp(\{ 0\longrightarrow G : G\text{ arbitrary} \}^\perp)$ </p>
<p>A Hausdorff space $K$ is compact iff $K\longrightarrow \{*\}$ is in ${}^\perp({}^\perp(\{a\}\longrightarrow \{a{<}b\})_{<5})^{\perp})$;
here $^\perp(\{a\}\longrightarrow \{a{<}b\})_{<5}$ denotes maps in $^\perp(\{a\}\longrightarrow \{a{<}b\})$
between spaces of size less than 5.</p>
<p>I give more examples in the answers to my own question I posted,
as they require some notation.</p>
|
user108780
| 108,780 |
<p>Properties of topological spaces. </p>
<p>Here I need to use some notation for finite topological spaces. I use the fact that a finite topological space
may be thought of as a category such that $card Hom(x,y)\leq 1$ for any objects $x,y$.
By convention, $ \{a{\rightarrow}b\}$ denote the space where $a$ is open and $b$ is closed; more generally, a subset is closed iff there are no arrows going outside the subset.
in maps, each point goes to "itself". The arrow $(\{a{\rightarrow}b\}\longrightarrow \{a=b\}$ denotes
the map to a singleton gluing together points $a$ and $b$.</p>
<p>$(\{a{\rightarrow}b\})^r_{<5}$ denotes maps in the class $(\{a{\rightarrow}b\})^r$ between spaces of size less than $5$.</p>
<p>a Hausdorff space $K$ is compact iff $K\longrightarrow \{*\}$ is in $((\{a\}\longrightarrow \{a{\rightarrow}b\})^r_{<5})^{lr}$</p>
<p>a Hausdorff space $K$ is compact iff $K\longrightarrow \{*\}$ is in $$
\{\, \{a\leftrightarrow b\}\longrightarrow \{a=b\},\, \{a{\rightarrow}b\}\longrightarrow \{a=b\},\,
\{b\}\longrightarrow \{a{\rightarrow}b\},\,\{a{\leftarrow}o{\rightarrow}b\}\longrightarrow \{a=o=b\}\,\,\}^{lr}$$</p>
<p>a space $D$ is discrete iff $ \emptyset \longrightarrow D$ is in $ (\emptyset\longrightarrow \{*\})^{rl} $ </p>
<p>a space $D$ is antidiscrete iff $ {D} \longrightarrow \{*\} $ is in
$(\{a,b\}\longrightarrow \{a=b\})^{rr}= (\{a\leftrightarrow b\}\longrightarrow \{a=b\})^{lr} $ </p>
<p>a space $K$ is connected or empty iff $K\longrightarrow \{*\}$ is in $(\{a,b\}\longrightarrow \{a=b\})^l $</p>
<p>a space $K$ is connected and non-empty iff for some arrow $\{*\}\longrightarrow K$ it holds that
$ \{*\}\longrightarrow K$ is in<br>
$ (\emptyset\longrightarrow \{*\})^{rll} = (\{a\}\longrightarrow \{a,b\})^l$ </p>
<p>a space $K$ is non-empty iff $K\longrightarrow \{*\}$ is in $ (\emptyset\longrightarrow \{*\})^l$ </p>
<p>a space $K$ is empty iff $K \longrightarrow \{*\}$ is in $ (\emptyset\longrightarrow \{*\})^{ll}$ </p>
<p>a space $K$ is $T_0$ iff $K \longrightarrow \{*\}$ is in $ (\{a\leftrightarrow b\}\longrightarrow \{a=b\})^r$ </p>
<p>a space $K$ is $T_1$ iff $K \longrightarrow \{*\}$ is in $ (\{a{\rightarrow}b\}\longrightarrow \{a=b\})^r$ </p>
<p>a space $X$ is Hausdorff iff $\{x,y\} \longrightarrow {X} \,\rightthreetimes\, \{ {x} {\rightarrow} {o} {\leftarrow} {y} \} \longrightarrow \{ x=o=y \}$</p>
<p>a non-empty space $X$ is regular (T3) iff for each arrow $ \{x\} \longrightarrow X$ it holds
$ \{x\} \longrightarrow {X} \,\rightthreetimes\, \{x{\rightarrow}X{\leftarrow}U{\rightarrow}F\} \longrightarrow \{x=X=U{\rightarrow}F\}$</p>
<p>a space $X$ is normal (T4) iff $\emptyset \longrightarrow {X} \,\rightthreetimes\, \{a{\leftarrow}U{\rightarrow}x{\leftarrow}V{\rightarrow}b\}\longrightarrow \{a{\leftarrow}U=x=V{\rightarrow}b\}$ </p>
<p>a space $X$ is completely normal iff $\emptyset\longrightarrow {X} \,\rightthreetimes\, [0,1]\longrightarrow \{0{\leftarrow}x{\rightarrow}1\}$
where the map $[0,1]\longrightarrow \{0{\leftarrow}x{\rightarrow}1\}$ sends $0$ to $0$, $1$ to $1$, and the rest $(0,1)$ to $x$
a space $X$ is path-connected iff $\{0,1\} \longrightarrow [0,1] \,\rightthreetimes\, {X} \longrightarrow \{*\}$ </p>
<p>a space $X$ is path-connected iff for each Hausdorff compact space $K$ and each injective map $\{x,y\} \hookrightarrow K$ it holds
$\{x,y\} \hookrightarrow {K} \,\rightthreetimes\, {X} \longrightarrow \{*\}$</p>
<p>$(\emptyset\longrightarrow \{*\})^r$ is the class of surjections</p>
<p>$(\emptyset\longrightarrow \{*\})^{rr}$ is the class of subsets, i.e. injective maps $A\hookrightarrow B$ where the topology on $A$ is induced from $B$</p>
<p>$(\{b\}\longrightarrow \{a{\rightarrow}b\})^l$ is the class of maps with dense image</p>
<p>$(\{b\}\longrightarrow \{a{\rightarrow}b\})^{lr}$ is the class of closed subsets $A \subset X$, $A$ a closed subset of $X$</p>
<p>$((\{a\}\longrightarrow \{a{\rightarrow}b\})^r_{<5})^{lr}$ is roughly the class of proper maps, i.e. a map between T4 spaces is in the class iff it is proper</p>
|
1,972,252 |
<blockquote>
<p>If $v^TAv = v^TBv$ for all constant vectors $v$ and $A,B$ are matrices of size $n$ by $n$, is it true that $A=B$? </p>
</blockquote>
<p>I have thought about using basis vectors but cannot get system of equations uniquely. Is there a trick here?</p>
|
edm
| 356,114 |
<p>It is not true. Take $A$ to be the zero matrix, and $B$ to be any nonzero skew-symmetric matrix. You can prove that both sides are always zero, but $A$ and $B$ are not equal.</p>
<p>If you assume that both $A$ and $B$ are symmetric real matrices, then they are equal. You can see that $v^TAv = v^TBv$ is equivalent to $v^T(A-B)v=0$. We know that $A-B$ is also a symmetric matrix. When the last equality holds for all vectors $v$, $A-B$ must be a zero matrix. This is due to the spectral theorem, as the existence of eigenvectors of $A-B$ is justified by the spectral theorem. See the details in <a href="https://math.stackexchange.com/questions/1177632/existence-of-nonzero-eigenvalues-in-a-symmetric-matrix">another post</a>. Now get any eigenvector $v$ of $A-B$ to see that $v^T(A-B)v=\lambda\left\lVert v \right\rVert^2$, where $\lambda$ is the eigenvalue of $A-B$ corresponding to the eigenvector $v$. Since eigenvectors are by definition nonzero, it must be the eigenvalue that is zero. So all eigenvalues of $A-B$ are zero. $A-B$ is similar to the zero matrix, so that $A-B$ itself is also the zero matrix.</p>
|
4,417,325 |
<p>When I say "divisibility trick" I mean "a recursive algorithm designed to show that, after multiple iterations, if the final output is a multiple of the desired number, then the original was also a multiple of the same number." Here's an example for a divisibility trick for 17.</p>
<blockquote>
<p>Rewrite <span class="math-container">$n$</span> as <span class="math-container">$10q+r$</span>, with <span class="math-container">$r<10$</span>. Then, evaluate <span class="math-container">$|q-5r|$</span>. Repeat this process until left with an easily factorable number.</p>
</blockquote>
<p>Here's another.</p>
<blockquote>
<p>Rewrite <span class="math-container">$n$</span> as <span class="math-container">$100q+r$</span>, with <span class="math-container">$r<100$</span>. Then, evaluate <span class="math-container">$|r-2q|$</span>. Repeat this process until left with an easily factorable number.</p>
</blockquote>
<p>Just to show these both work (or, at least, work for one particular number, let's try both on <span class="math-container">$31382$</span>.</p>
<blockquote>
<p>METHOD ONE:
<span class="math-container">$31382\rightarrow3128\rightarrow272\rightarrow17$</span>, ergo <span class="math-container">$17\ |\ 31382$</span>.</p>
</blockquote>
<blockquote>
<p>METHOD TWO:
<span class="math-container">$31382\rightarrow544\rightarrow34$</span>, ergo <span class="math-container">$17\ |\ 31382$</span>.</p>
</blockquote>
<p>These divisibility tricks rely on breaking the number down into groups of digits and applying some linear operation to them. However, when we try to use a <em>non-linear</em> function, things seem to break down. For example, much as how method one here draws off the fact that <span class="math-container">$17\ |\ 51$</span> and the second relies off of <span class="math-container">$17\ |\ 119$</span>, let's try to do something with <span class="math-container">$17\ |\ 34$</span>. Namely:</p>
<blockquote>
<p>Rewrite <span class="math-container">$n$</span> as <span class="math-container">$10q+r$</span>, with <span class="math-container">$r<10$</span>. Then, evaluate <span class="math-container">$|6q^2-5r|$</span>. Repeat this process until left with an easily factorable number.</p>
</blockquote>
<p>We can try this with <span class="math-container">$34$</span> and see, yes, <span class="math-container">$6(9)-5(4)=34$</span>, so <span class="math-container">$17\ |\ 34$</span>. But this fails for most numbers. For <span class="math-container">$51$</span>, we have <span class="math-container">$51\rightarrow145$</span>, and it diverges from there (also note that <span class="math-container">$17\not|\ 145$</span>). Even with <span class="math-container">$17$</span>, which is obviously a multiple of <span class="math-container">$17$</span>, we have <span class="math-container">$17\rightarrow29$</span>.</p>
<p>What separates the wheat from the chaff here, so to speak? Why is it that if we break down the digits of the multiple of some prime and make a linear relation around it, it seems to be true for all other multiples of the prime, but the same doesn't work for, say, a quadratic relation?</p>
|
Misha Lavrov
| 383,078 |
<p>We can invent "divisibility tricks" that don't have this property. For example:</p>
<p>"To test if <span class="math-container">$n$</span> is divisible by <span class="math-container">$3$</span>, write <span class="math-container">$n$</span> as <span class="math-container">$10q+r$</span>. Then, compute <span class="math-container">$q^3 + r^3$</span> and see if this is divisible by <span class="math-container">$3$</span>."</p>
<p>This works because of Fermat's little theorem: for all integers <span class="math-container">$a$</span> and primes <span class="math-container">$p$</span>, <span class="math-container">$a^p \equiv a \pmod p$</span>. In particular, <span class="math-container">$q^3 \equiv q \pmod3$</span> and <span class="math-container">$r^3 \equiv r\pmod3$</span>, so <span class="math-container">$q^3+r^3 \equiv q + r \equiv 10q+r \pmod 3$</span>.</p>
<p>This is not a good trick! Not because it doesn't work, but because checking <span class="math-container">$q^3+r^3$</span> for divisibility by <span class="math-container">$3$</span> is probably harder than checking <span class="math-container">$n$</span> was.</p>
<p>In highly specific cases, we might be able to take advantage of this kind of trick, though. For example, if you are faced with trying to figure out if <span class="math-container">$10000256$</span> is divisible by <span class="math-container">$7$</span>, it might help to realize that <span class="math-container">$10000256 = 10^7 + 2 \cdot 2^7$</span>, so it is congruent to <span class="math-container">$10 + 2 \cdot 2 = 14$</span> modulo <span class="math-container">$7$</span>.</p>
|
2,375,023 |
<p>So I have to find an interval (in the real numbers) such that it contains all roots of the following function:
$$f(x)=x^5+x^4+x^3+x^2+1$$</p>
<p>I've tried to work with the derivatives of the function but it doesn't give any information about the interval, only how many possible roots the function might have.</p>
|
Michael Rozenberg
| 190,319 |
<p>Let $f(x)=x^5+x^4+x^3+x^2+1$.</p>
<p>Hence, $f'(x)=x(5x^3+4x^2+3x+2)$ and since $(5x^3+4x^2+3x+2)'=15x^2+8x+3>0$,
we see that the polynomial $5x^3+4x^2+3x+2$ has one real root $x_1$ and this root is negative.</p>
<p>Thus, $x_{min}=0$ and $x_{max}=x_1$. </p>
<p>But $f(0)>0$, which say that $f$ has an unique real root and since
$$f(-1.25)f(-1.24)<0,$$
we get that this root placed in $(-1.25,-1.24)$.</p>
|
1,109,918 |
<p>Is it always possible to add terms into limits, like in the following example? (Or must certain conditions be fulfilled first, such as for example the numerator by itself must converge etc)</p>
<p>$\lim_{h \to 0} {f(x)} = \lim_{h \to 0} \frac{e^xf(x)}{e^x}$</p>
|
Vim
| 191,404 |
<p>It is very simple if you apply Taylor's series with Peano's residue:<br/>
When $h\to 0$, we have
$$f(x+h)=f(x)+hf'(x)+\frac{1}{2}h^2 f''(x)+o(h^2)$$
$$f(x-h)=f(x)-hf'(x)+\frac{1}{2}h^2 f''(x)+o(h^2)$$
where $o(h^2)$ denotes any infinitesimal amount that converges to zero faster than $h^2$.<br/>
Put them back to the limit, the result should be
$$f''(x)+\frac{o(h^2)}{h^2}$$
and as $h\to 0$, it is obvious that the residue $\frac{o(h^2)}{h^2} \to 0$, thus the final result should be
$$f''(x)$$</p>
|
1,423,456 |
<p>I am completely stuck on this problem: $C[0,1] = \{f: f\text{ is continuous function on } [0,1] \}$ with metric $d_1$ defined as follows:</p>
<p>$d_1(f,g) = \int_{0}^{1} |f(x) - g(x)|dx $.</p>
<p>Let the sequence $\{f_n\}_{n =1}^{\infty}\subseteq C[0,1]$ be defined as follows:</p>
<p>$
f_n(x) = \left\{
\begin{array}{l l}
min\{1, {1\over 2}-n(x -{1\over 2} )\} & \quad \text{ $x\in [0, 1/2]$}\\
max\{0, {1\over 2}-n(x -{1\over 2} )\} & \quad \text{$x\in [1/2, 1]$}\\
\
\end{array} \right.
$<br>
Using this function to show that $C[0,1]$ is not complete.</p>
<p>I need help to solve this problem.</p>
|
Squirtle
| 29,507 |
<p>Essentially the idea is that you start with functions that look like a flat line, a continuous elevation (or decline) and then a flat line. You then make the elevation (or decline) more and more steep until it disappears and you are left with a noncontinuous step function.</p>
<p>EDIT: The point is that in a complete space every Cauchy sequence is convergent. It's trivial to see that your sequence is Cauchy, but by the above argument it converges to a not continuous function (hence it does not converge in the set of continuous functions) and hence the space $(C[0,1],d_1)$ is not complete.</p>
|
2,077,275 |
<p>Let $U$ be the set $U$ of quaternions of unit length. I know that $U\times S^1$ is compact, connected and is a $2n$ manifold in a $2n+1$ dimensional vector space $V$.</p>
<blockquote>
<p>How can I construct a differentiable tangent vector field on $U\times S^1$ that has no zeroes?</p>
</blockquote>
<p>What I know:<br>
A tangent vector field is a $C^k$ map $F:U\times S^1\rightarrow \mathbb{R}^7$ s.t. $F(x)\in T_x(U\times S^1)$ for all $x\in U\times S^1$. But how can I construct one? </p>
|
Community
| -1 |
<p>Assuming you're trying to solve for $(x, y, z)^{T}$ then</p>
<p>Hint: Rewrite
$$
\begin{pmatrix}
\sin a & \sin 2a & \sin 3a\\
\sin b & \sin 2b & \sin 3b\\
\sin c & \sin 2c & \sin 3c\\
\end{pmatrix}\begin{pmatrix}
x\\
y\\
z
\end{pmatrix}=\begin{pmatrix}
\sin 4a \\\sin 4b \\ \sin 4c
\end{pmatrix}$$
Thus
$$
\begin{pmatrix}
x\\
y\\
z
\end{pmatrix}=\begin{pmatrix}
\sin a & \sin 2a & \sin 3a\\
\sin b & \sin 2b & \sin 3b\\
\sin c & \sin 2c & \sin 3c\\
\end{pmatrix}^{-1}\begin{pmatrix}
\sin 4a \\\sin 4b \\ \sin 4c
\end{pmatrix}$$
The next step is to compute the inverse, using double angle formula for $\sin$.</p>
|
1,879,440 |
<p>I am reading the definition of associative <span class="math-container">$R$</span>-algebra, and am confused about the following definition from Wiki:</p>
<p><a href="https://en.wikipedia.org/wiki/Associative_algebra" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Associative_algebra</a></p>
<p><a href="https://i.stack.imgur.com/M6HMf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/M6HMf.png" alt="enter image description here" /></a></p>
<p>The first equality comes from the associativity of being an algebra.</p>
<p>However, how to obtain the second equality?</p>
<p>I believe <span class="math-container">$\mathcal{A}$</span> does noe have to be a commutative ring though <span class="math-container">$\mathcal R$</span> is a commutative ring.</p>
|
Berci
| 41,488 |
<p>There are two operations here: one is 'multiplication' of 'scalars' with algebra elements, denoted by dot, and the other one is the multiplication of the algebra, between two of its elements.</p>
<p>The axiom says:
$$r\cdot(xy)=(r\cdot x)\,y=x\,(r\cdot y),\\
r\in R,\quad x,y\in A\,.$$</p>
<p>Since $R$ is commutative, its left action on $A$ can be mirrored to the right side by defining $a\cdot r:=r\cdot a$. This way you can also read it as
$$r\cdot(xy)=(r\cdot x)y=(x\cdot r)y=x(r\cdot y)=x(y\cdot r)=(xy)\cdot r\,.$$</p>
|
877,683 |
<p><img src="https://www.anonimg.com/img/c0423915f5f5fae8f9790506c36393f8.jpg" alt="">
Based on my last question I learned that this is an envelope of a parabola</p>
<p><a href="https://math.stackexchange.com/questions/874478/what-is-this-geometric-pattern-called">What is this geometric pattern called?</a></p>
<p>But how can I prove it ?</p>
|
mvw
| 86,776 |
<p>We try to model the family of lines and then try to infer the envelope.</p>
<p>The guiding lines (left and right arms of the V shape) are
$$
g(t) = u_g \, (1-t) + v_g \, t \quad h(t) = u_h \, (1-t) + v_h \, t
$$
for $t \in [0, 1]$, where $u$ is the start point and $v$ the end point of that line.</p>
<p>A line $f_r$ of the family starts on $g$ and ends on $h$:
$$
f_r(t) = g(r) \, (1-t) + h(r) \, t
$$
again with the parameter $t \in [0, 1]$.</p>
<p>The envelope is approximated by points where two of the family lines intersect:
$$
\begin{align}
f_r(t) &= f_s(t') \iff \\
g(r) \, (1-t) + h(r) \, t &= g(s) \, (1-t') + h(s) \,t'
\end{align}
$$</p>
<p>This gives the matrix equation
$$
\left[
\begin{matrix}
h(r) - g(r) & -(h(s) - g(s))
\end{matrix}
\right]
\left[
\begin{matrix}
t \\
t'
\end{matrix}
\right]
=
g(s) - g(r) \quad (*)
$$</p>
<p><strong>Example:</strong>
$$
u_g =
\left[
\begin{matrix}
-1 \\
1
\end{matrix}
\right]
\quad
v_g =
\left[
\begin{matrix}
0 \\
-1
\end{matrix}
\right]
\quad
u_h =
\left[
\begin{matrix}
0 \\
-1
\end{matrix}
\right]
\quad
v_h =
\left[
\begin{matrix}
1 \\
1
\end{matrix}
\right]
$$</p>
<p>This gives
$$
g(r) =
\left[
\begin{matrix}
r - 1 \\
1 - 2r
\end{matrix}
\right]
\quad
h(r) =
\left[
\begin{matrix}
r \\
2r - 1
\end{matrix}
\right]
\quad
g(s) =
\left[
\begin{matrix}
s - 1 \\
1 - 2s
\end{matrix}
\right]
\quad
h(s) =
\left[
\begin{matrix}
s \\
2s - 1
\end{matrix}
\right]
$$
and thus for two different family lines ($r \ne s$):
$$
\left[
\begin{matrix}
1 & -1 \\
4r-2 & 2 - 4s
\end{matrix}
\right]
\left[
\begin{matrix}
t \\
t'
\end{matrix}
\right]
=
\left[
\begin{matrix}
s - r \\
2 (r - s)
\end{matrix}
\right]
$$
The first row gives $t' = t + r - s$ and the second row
$$
2(r-s) = (4r-2)t + (2-4s)(t+r-s) = 4(r-s) t + (2-4s)(r-s) \iff \\
4s(r-s) = 4(r-s) t \iff \\
t = s
$$</p>
<p>and thus $t' = r$. The intersection point results to
$$
p(r,s) =
f_r(s) =
f_s(r)
$$
however we want $r \to s$ and thus:
$$
p(r) =
f_r(r) =
\left[
\begin{matrix}
(r-1)(1-r) + r^2 \\
(1-2r)(1-r) + (2r-1) r
\end{matrix}
\right]
=
\left[
\begin{matrix}
2r - 1 \\
4r^2 -4r + 1
\end{matrix}
\right]
$$
Substituting $x = 2 r - 1 \iff r = (x + 1) / 2$ and remembering $r \in [0, 1]$ we get
$$
p(x) =
\left[
\begin{matrix}
x \\
4\left(\frac{x+1}{2}\right)^2 -4 \frac{x+1}{2} + 1
\end{matrix}
\right]
=
\left[
\begin{matrix}
x \\
x^2
\end{matrix}
\right]
$$
which is the graph of a parabola.</p>
<p><img src="https://i.stack.imgur.com/POrFq.png" alt="Graph of the parabola as parametric plot"></p>
<p><strong>Note:</strong></p>
<p>I picked the example to correspond to a nice symmetric V shape.
Equation (*) holds for general situations, e.g. a V with a longer and a shorter arm. In those cases, other envelopes, if at all, arise!</p>
<p>E.g. it is possible to arrange the guiding lines as a | | shape, where the result should look like a butterfly or a (boring) stripes pattern, depending on the relative orientations of the guiding lines. </p>
|
2,289,813 |
<p>Let$\ f(x,y,z)=4xz -y^2 +z^2$ be a differentiable function, let$\ P=(0,1,1)$ and$\ Q=(1,3,2)$. Find$\ T$ such that $\ f(P)-f(Q) = \nabla f(T)(P-Q)$</p>
<p>What I did:</p>
<p>$\ f(P)=0$, $\ f(Q)=3$, $\ (P-Q)=(-1,-2,-1)$, $\nabla f(P-Q)=(-4z, 4y, -4x+2z)$</p>
<p>Let $\ T=(a,b,c)$ then$\ f(P)-f(Q) = \nabla f(T)(P-Q)$ if and only if </p>
<p>$\ -3 = (-4c, 4b, -4a-2c)$, I'm stuck at this point, how can I determine this? </p>
<p>Looking for some advice, thanks!</p>
|
Manuel Guillen
| 416,103 |
<p>The function $f$ takes a vector in $\mathbb{R}^3$ and returns a scalar in $\mathbb{R}$. The left hand side of the expression is a scalar, and so is the right hand side, because the product is a dot product:
\begin{align*}
f(P)-f(Q) &= \nabla f(T) \cdot (P-Q) \\
0 - 3 &= \langle 4z, -2y, 4x + 2z \rangle \cdot \langle 0-1, 1-3, 1-2 \rangle \\
-3 &= \langle 4z, -2y, 4x + 2z \rangle \cdot \langle -1, -2, -1 \rangle \\
&= -4z+2y-4x-2z \\
3 &= 4x-2y+6z
\end{align*}
This defines a plane of points, any of which will satisfy the relation, so we can choose $(0,0,1/2)$</p>
|
1,783,837 |
<p>I'm studying how to use Laplace transform to solve ODEs. <br/><br/>
I have thought to use this very simple example: $$y'(t)=t+1 \qquad y(0)=0$$</p>
<p>I can use integration to find $y(t)$: $$y(t)=\int (t+1) \ \ dt=\frac{1}{2} t^2+t+C$$</p>
<p>$C \in \mathbb{R} $, $C=0$ for the initial condition, so:</p>
<p>$$y(t)=\frac{1}{2} t^2+t$$</p>
<p><br /></p>
<p>I consider $F(s)$ the Laplace transform of $f(t)=t+1$:
$$F(s)=\frac{1}{s}+\frac{1}{s^2}=\frac{1+s}{s^2}$$</p>
<p>I consider, now, the coefficient of the linear ODE:
$$H(s)=\frac{1}{s}$$</p>
<p>So: $$Y(s)=H(s) \ F(s)=\frac{1+s}{s^3}$$</p>
<p>Partial Fraction Decomposition of $Y(s)$:</p>
<p>$$Y(s)=\frac{1}{s^2}+\frac{1}{s^3}$$</p>
<p>Antitransform:
$$y(t)=t^2+t \ne \frac{1}{2} t^2+t $$</p>
<p>Where is the mistake?</p>
<p>Thanks!</p>
|
kennytm
| 171 |
<p>Your inverse Laplace transform of $\frac1{s^3}$ is wrong, it should be $\frac{t^2}2$ as you expect.</p>
<p>$$\mathcal L^{-1} \left\{ \frac1{s^{n+1}} \right\} = \frac{t^n}{\color{red}{n!}} $$</p>
|
1,783,837 |
<p>I'm studying how to use Laplace transform to solve ODEs. <br/><br/>
I have thought to use this very simple example: $$y'(t)=t+1 \qquad y(0)=0$$</p>
<p>I can use integration to find $y(t)$: $$y(t)=\int (t+1) \ \ dt=\frac{1}{2} t^2+t+C$$</p>
<p>$C \in \mathbb{R} $, $C=0$ for the initial condition, so:</p>
<p>$$y(t)=\frac{1}{2} t^2+t$$</p>
<p><br /></p>
<p>I consider $F(s)$ the Laplace transform of $f(t)=t+1$:
$$F(s)=\frac{1}{s}+\frac{1}{s^2}=\frac{1+s}{s^2}$$</p>
<p>I consider, now, the coefficient of the linear ODE:
$$H(s)=\frac{1}{s}$$</p>
<p>So: $$Y(s)=H(s) \ F(s)=\frac{1+s}{s^3}$$</p>
<p>Partial Fraction Decomposition of $Y(s)$:</p>
<p>$$Y(s)=\frac{1}{s^2}+\frac{1}{s^3}$$</p>
<p>Antitransform:
$$y(t)=t^2+t \ne \frac{1}{2} t^2+t $$</p>
<p>Where is the mistake?</p>
<p>Thanks!</p>
|
Ángel Mario Gallegos
| 67,622 |
<p>$$\mathscr{L}^{-1}\left\{\frac1{s^2}+\frac1{s^3}\right\}=\mathscr{L}^{-1}\left\{\frac1{s^2}\right\}+\mathscr{L}^{-1}\left\{\frac1{s^3}\right\}=\mathscr{L}^{-1}\left\{\frac1{s^2}\right\}+\color{blue}{\frac{1}{2}}\mathscr{L}^{-1}\left\{\frac{\color{blue}{2}}{s^3}\right\}=t+\frac12t^2$$</p>
|
3,371,694 |
<p>Let <span class="math-container">$ \forall n \in \mathbf{N}, ~ u_{n+1} = \frac{1}{\tanh^2(u_n)} - \frac{1}{u^2_n} $</span> with <span class="math-container">$ u_0 = a > 0 $</span>.</p>
<p>What is the limit of <span class="math-container">$(u_n)$</span> ?</p>
<p>I tried to find a fixed point of this sequence but the equation is impossible to solve algebraically.</p>
|
A.J.
| 654,406 |
<p>You just have to consider the symmetry of the situation. Since the circle has center <span class="math-container">$(0,0)$</span>, the fact that <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> have opposite <span class="math-container">$x$</span>-coordinates and are in opposite quadrants implies that they are endpoints of a diameter of the circle, so <span class="math-container">$XY=2b$</span>.</p>
<p>If you must have an algebraic solution, find the <span class="math-container">$y$</span>-coordinates of the two points and then use the distance formula.</p>
<p>For <span class="math-container">$X$</span>, we have <span class="math-container">$a^2+y^2=b^2$</span>, so <span class="math-container">$y=\sqrt{b^2-a^2}$</span>. Similarly, for <span class="math-container">$Y, y=-\sqrt{b^2-a^2}$</span>.<br>
(Note the signs for y are determined by the quadrants.)</p>
<p>Then, using <span class="math-container">$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$</span>, one gets</p>
<p><span class="math-container">$\begin{align}
XY&=\sqrt{(-a-a)^2+(\sqrt{b^2-a^2}-(-\sqrt{b^2-a^2}))^2}\\
&=\sqrt{(-2a)^2+(2\sqrt{b^2-a^2})^2}\\
&=\sqrt{4a^2+4(b^2-a^2)}\\
&=\sqrt{4b^2}\\
&=2b\\
\end{align}$</span></p>
|
2,108,352 |
<p>It is a question in functional analysis by writer Erwin Kryzic </p>
|
Franck Panche
| 615,159 |
<p>Ok. The problem is the triangle inequality, then: </p>
<p>Let <span class="math-container">$x,y,z \in \mathbb{R}$</span>, so <span class="math-container">$$d(x,z) = (x+y-y+z)^2$$</span> <span class="math-container">$$d(x,z) = ((x-y)+(y-z))^2$$</span> <span class="math-container">$$\qquad d(x,z) = d(x,y)+2(x-y)(y-z)+d(y,z)$$</span> <span class="math-container">$$d(x,z) \geq d(x,y)+d(y,z)\quad iff\quad (x-y)(y-z)\geq0$$</span></p>
<p>Then, suppose what <span class="math-container">$x=3$</span>, <span class="math-container">$z=1$</span>, <span class="math-container">$y=2$</span>, that which does not meet the triangular inequality, therefore, <span class="math-container">$d$</span> is not a metric</p>
|
183,100 |
<p>Let $\phi: M_2(\mathbb{C})\times M_2(\mathbb{C})\rightarrow M_2(\mathbb{C})$ be the map
$$(B_1,B_2)\mapsto [B_1,B_2]$$ which takes two $2\times 2$ matrices to its Lie bracket. </p>
<p>Then why does $d\phi_{(B_1,B_2)}:M_2(\mathbb{C})\times M_2(\mathbb{C})\rightarrow M_2(\mathbb{C})$ send
$$(D_1,D_2)\mapsto [B_1,D_2]+[D_1,B_2]?$$ </p>
<p>$$
$$
Taking $B_1=(g_{ij})$ and $B_2=(h_{ij})$, I do not think taking the partials of the Lie bracket $[B_1,B_2]=$
$$
\left[ \begin{array}{cc}
g_{12} h_{21} - g_{21} h_{12} & -g_{12} h_{11} + g_{11} h_{12} - g_{22} h_{12} +
g_{12} h_{22} \\
g_{21} h_{11} - g_{11} h_{21} + g_{22} h_{21} - g_{21} h_{22} & g_{21} h_{12} - g_{12} h_{21} \\
\end{array}\right]
$$ is a clever way to figure out the map. </p>
|
Georges Elencwajg
| 3,217 |
<p>This has actually nothing to do with Lie brackets nor Lie algebras! </p>
<p>Given finite-dimensional complex (or real) vector spaces $E,F,G$ and a bilinear map $f:E\times F\to G$, the differential of $f$ at $(a,b)\in E\times F$ is given by the formula (whose proof follows directly from the definition)<br>
$$ df_{(a,b)} (x,y)=f(a,y)+f(x,b) $$ </p>
<p>The only thing you have to check in your case is that the bracket is bilinear, which is obvious in the concrete case of matrices and in the abstract theory of Lie algebras is an axiom . </p>
|
539,457 |
<p>Suppose we have a natural number $N$ with decimal representation $A_kA_{k-1}\ldots A_0$. How do I prove that if the $\sum\limits_{i=0}^kA_i$ is divisible by $9$ then $N$ is divisible by $9$ too?</p>
|
Carlos Eugenio Thompson Pinzón
| 99,344 |
<p>If you are familiar with modularity you have:</p>
<ol>
<li>If $x \equiv a\mod{m}$ and $y \equiv b\mod{m}$ then $x+y\equiv a+b\mod{m}$.</li>
<li>If $x \equiv a\mod{m}$ and $y \equiv b\mod{m}$ then $x\cdot y\equiv a\cdot b\mod{m}$.</li>
<li>If $x \equiv 1\mod{m}$ then $x\cdot b\equiv b\mod{m}$.</li>
<li>If $x \equiv 1\mod{m}$ then $x^i\equiv 1\mod{m}$.</li>
<li>if $x\equiv0\mod m$ then $m$ divides $x$.</li>
</ol>
<p>So if $A_kA_{k-1}\ldots A_0$ is the decimal representation of $N$, then
$$N=\sum_{i=0}^k10^iA_i.$$</p>
<p>First we have that $10\equiv1\mod9$, and therefore (by 4.) $10^i\equiv1\mod9$</p>
<p>Let's evaluate it $\operatorname{mod}9$:
$$N=\sum_{i=0}^k10^iA_i \equiv\sum_{i=1}^k1\cdot A_i,\qquad\text{by applying 1. 3. and 4.}$$</p>
<p>I let you continue from here.</p>
|
320,019 |
<p>Assume $Y$ is non negative random variable. Prove that $X+Y$ is stochastically greater than $X$ for any random variable $X$.</p>
<p>We have to prove there that $\Pr(X+Y > x) \geq \Pr(X>x) $ for all $x$</p>
|
Reid
| 64,872 |
<p>One approach:</p>
<p>Would be glad if anyone points out any mistake if there is one.(Or comment to say if there is no mistake).</p>
<p>$P(X+Y \leq x) = \int_{-\infty}^{\infty} F_X(x-y)dG_Y(y)dy$ where $F$ and $G$ are cdf of $X,Y$ respectively.</p>
<p>$\leq \int_{-\infty}^{\infty} F_X(x)dG_Y(y)dy$. As $F(x-t) \leq F(x) \forall t\geq 0$ ($F$ is non decreasing.)</p>
<p>$= F_X(x) \int_{-\infty}^{\infty}dG_Y(y)$
$=F_X(x)$ as $\int_{-\infty}^{\infty}dG_Y(y)=1$</p>
<p>First step used convolution</p>
|
2,591,976 |
<p>If $f(x-1)=2x^2-10x+3$, find $f(x)$. I tried the problem and received the answer $f(x)=2x^2-14x-5$, is this right? </p>
|
Enrico M.
| 266,764 |
<p>Use $x-1 = y$, hence $x = y+1$, thence</p>
<p>$$f(y) = 2(y+1)^2 - 10(y+1) + 3 = 2y^2 + 2 + 4y - 10y - 10 + 3$$</p>
<p>$$f(y) = 2y^2 - 6y - 5$$</p>
<p>Now just call $y = x$ and you're done.</p>
|
3,644,823 |
<p>Let <span class="math-container">$f$</span> be a non constant holomorphic function on and inside of the unit circle <span class="math-container">$C:=\{z\in \mathbb C~:~|z|=1\}$</span>. Suppose <span class="math-container">$|f(z)|=1$</span> on <span class="math-container">$C$</span>, then for <span class="math-container">$D=\{z\in \mathbb C~:~|z|\leq 1\}$</span>, prove that <span class="math-container">$f: \bar{D}\rightarrow \bar{D}$</span> is onto.</p>
<p>Since <span class="math-container">$f$</span> is non constant, maximum modulus gives that <span class="math-container">$|f(z)| <1$</span> on <span class="math-container">$D$</span>. But it is not enough to say that <span class="math-container">$f$</span> is onto.</p>
|
Conrad
| 298,272 |
<p><span class="math-container">$f$</span> has finitely many zeroes in <span class="math-container">$\mathbb D$</span> (why?); number them <span class="math-container">$a_1,...a_n$</span> repeating them if of higher multiplicities and take the finite Blaschke product <span class="math-container">$B(z)=\Pi_{k=1}^n \frac{z-a_k}{1-\bar a_k z}, B(z)=1$</span> if no zeroes. </p>
<p>Then <span class="math-container">$g=f/B$</span> is analytic satisfies the same properties as <span class="math-container">$f$</span> (why?) but has no zeroes inside the unit disc, so <span class="math-container">$\frac{1}{g}$</span> satisfies same properties so both <span class="math-container">$|g| \le 1, |\frac {1}{g}| \le 1$</span> on the unit disc, hence <span class="math-container">$g=\alpha,|\alpha|=1$</span>, hence <span class="math-container">$f=\alpha B$</span>. Since <span class="math-container">$f$</span> nonconstant, <span class="math-container">$n \ge 1$</span> so <span class="math-container">$f$</span> is a nonconstant rational function of a special type.</p>
<p>Now show that <span class="math-container">$B$</span> is a <span class="math-container">$n:1$</span> map of the unit disc to itself (hint: argument principle or fundamental theorem of algebra plus <span class="math-container">$B(\mathbb D) \subset \mathbb D, B(\mathbb D_e) \subset \mathbb D_e$</span>, whhere <span class="math-container">$\mathbb D_e$</span> is the exterior of the closed unit disc)</p>
|
17,270 |
<p>I just joined MathSE and it's beautiful here, except for the fact that some unregistered users ask a question and never come back. Most of the time these questions are trivial, though they still consume answerers' (valuable) time which never gets rewarded. I thought it was okay until I saw someone's profile with the following statistics: active $1$ year $7$ Months, $0$ Answers , $72$ Questions, $0$ accept votes. Yes, I agree that the answers are up-voted in this case but is it really okay to never accept any answers?</p>
|
Anonymous Computer
| 128,641 |
<p>Seventy-two questions without any accepted answers? Guess I'm not answering any of your questions....</p>
<p>Being serious, accepting answers shows gratitude for the answerer that helped the person asking the question the most. If people don't accept answers, the answerers may be discouraged from answering because they think that their efforts will be for naught. They do not have any sign that the person asking the question has even read their answers. Accepting answers essentially tells answerers, "I have read your answer, and I have found it to be the most useful. Thank you for your answer."</p>
<p>I also hope that you have been upvoting/downvoting as well, as this also provides feedback for answerers.</p>
|
17,270 |
<p>I just joined MathSE and it's beautiful here, except for the fact that some unregistered users ask a question and never come back. Most of the time these questions are trivial, though they still consume answerers' (valuable) time which never gets rewarded. I thought it was okay until I saw someone's profile with the following statistics: active $1$ year $7$ Months, $0$ Answers , $72$ Questions, $0$ accept votes. Yes, I agree that the answers are up-voted in this case but is it really okay to never accept any answers?</p>
|
Dunka
| 159,628 |
<p>It is okay to not accept any answers if they haven't really clarified what you were trying to solve. You're under no obligation to give people participation points. If the answers lead you to understanding the problem, you should always accept. In this particular case I think the fact that they have asked 72 questions implies they have gotten useful answers or else why would they come back, and thus they should be accepting more answers. I suggest you send them a message</p>
|
2,559,350 |
<p>Consider a circumference centered in the origin and with radius $r$. Let $C$ be a point on the circumference, and let $A,B$ be its projections on the axes, respectively $x$-axis and $y$-axis. What is the length of $AB$?</p>
<p>I tried applying the laws of sines and cosines on $ABC$, but I only got tautologies....</p>
|
heropup
| 118,193 |
<p>Think about what $|z-i|^4 = 1$ <em>means</em>. It means that the magnitude of the complex number $z-i$, when raised to the fourth power, equals $1$. Since magnitude is always a nonnegative real number, it follows that the magnitude itself must equal $1$, since the only nonnegative real solution to the equation $$x^4 = 1$$ is $x = 1$. Therefore, the set of $z$ that satisfy $|z - i|^4 = 1$ is the same as the set that satisfy $|z-i| = 1$. This is obviously a circle of radius $1$ centered at $i$.</p>
<p>Now that we understand the desired locus, all that remains is to express it algebraically. To this end, we recall that the Cartesian equation of a circle centered at $(0,1)$ with radius $1$ is simply $$(x-0)^2 + (y-1)^2 = 1$$ or $$x^2 + y^2 - 2y = 0.$$ So in rectangular form, $z$ is characterized by $$x + iy = \pm \sqrt{2y - y^2} + iy, \quad y \in [0,1],$$ and here $y$ represents the imaginary part of $z$, geometrically the vertical distance from the real axis.</p>
<p>In polar form, let $z = re^{i\theta}$. We seek a relationship between $r$ and $\theta$ that describes such a circle. But since $e^{i\theta} = \cos\theta + i \sin \theta$, we simply have $$(r \cos \theta)^2 + (r \sin \theta)^2 - 2(r \sin \theta) = 0,$$ or $$r^2 - 2r \sin \theta = 0,$$ or $$r = 2 \sin \theta.$$ It follows that the desired locus is $$z = 2e^{i\theta} \sin\theta, \quad \theta \in [0,\pi),$$ noting that the circle is swept out twice if we were to use $\theta \in [0,2\pi)$. Here, $\theta$ represents the counterclockwise angle $z$ makes with the positive real axis.</p>
|
132,879 |
<p>This question is motivated by the physical description of magnetic monopoles. I will give the motivation, but you can also jump to the last section.</p>
<p>Let us recall Maxwell’s equations: Given a semi-riemannian 4-manifold and a 3-form $j$. We describe the field-strength differential form $F$ as a solution of the equations</p>
<p>$\mathrm{d}F=0$</p>
<p>$\mathrm{d}\star F=j$ (where $\star$ denotes the Hodge star).</p>
<p>If the second de-Rham-cohomology vanishes (for example in Minkowski space), $F$ is exact and we can write it as $F=\mathrm{d}A$, where $A$ denotes a 1-form.</p>
<p>Now let us consider monopoles: We use two 3-forms $j_m$ (magnetic current) and $j_e$ (electric current) and consider the equations</p>
<p>$\mathrm{d}F=j_m$</p>
<p>$\mathrm{d}\star F=j_e$.</p>
<p>Essentially, it is described in <a href="http://arxiv.org/pdf/math-ph/0203043v1.pdf">this paper</a>, but the author Frédéric Moulin (a physicist) uses coordinates. Now he assumes that (in Minkowski space) $F$ can be decomposed using two potentials — into an exact (in the image of the derivative) and a coexact (in the image of the coderivative) form: $F=\mathrm{d}A-\star\mathrm{d}C$. Is there a mathematical justification for this assumption (maybe it is just very pragmatic)?</p>
<h3>The actual question</h3>
<p>Given a 2-form $F$ on 4-dimensional Minkowski space (more generally: semi-riemannian manifolds)—are there any known conditions such that $F$ decomposes into an exact and a coexact form: $F=\mathrm{d}A+\star\mathrm{d}C$)?</p>
<p>For <em>compact riemannian</em> manifolds there is the well-known Hodge decomposition: There is always a decomposition into an exact, a coexact and a harmonic form. In the non-compact case you might be able to get rid of the harmonic form by only considering “rapidly decaying” forms (<a href="http://en.wikipedia.org/wiki/Helmholtz_decomposition#Differential_forms">Wikipedia suggests that</a>, but I do not have a good reference, in euclidean space there is the Helmholtz decomposition, and non-trivial (smooth) harmonic 1-forms do not vanish at infinity).</p>
<p>That is why I also ask: Are there “rapidly decaying” harmonic 2-forms in Minkowski space? Any references where I could see what is known about harmonic forms and Hodge theory in the semi-riemannian case are also welcome.</p>
|
Igor Khavkine
| 2,622 |
<p>Willie's answer is of course correct. But the part that I think is most interesting in the physical context is only briefly mentioned in point 3. of the last paragraph. Let me expand on that.</p>
<p>The relevant condition that make everything work nicely is that the background Lorentzian spacetime be globally hyperbolic. There are two important spaces of forms, $\Omega^p_0$ and $\Omega^p_{SC}$, $p$-forms with compact and <em>spacelike compact</em> supports, respectively. A set $X$ is <em>spacelike compact</em> if it is contained in the causal influence set of a compact set $Y$, $X\subseteq J(Y)$.</p>
<p>On Lorentzian manifolds, the D'Alembertian operator $\square = d\delta + \delta d$ is the analog of the Laplacian operator $\Delta$ in Riemannian signature. Note that I'm using the notation $\delta = (-1)^p{\star d \star}$. It is not elliptic, but it is hyperbolic with very nice properties. Instead of the nice analytical properties of the elliptic Laplacian, we have the following <em>exact sequence</em> (the image of each map is equal to the kernel of the next one):
$$ 0 \to \Omega^p_0 \stackrel{\square}{\to} \Omega^p_0 \stackrel{G}{\to} \Omega^p_{SC} \stackrel{\square}{\to} \Omega^p_{SC} \to 0 , $$
where $G=G^+-G^-$ is the <em>causal Green function</em>, defined as the difference of the retarded ($G^+$) and advanced ($G^-$) Green functions. The retarded and advanced Green functions of the D'Alembertian always exist on Globally hyperbolic manifolds (see, for instance, <a href="http://arxiv.org/abs/0806.1036" rel="nofollow noreferrer">here</a> or <a href="http://arxiv.org/abs/1208.4706" rel="nofollow noreferrer">here</a>).</p>
<p>A version of the result that you want is this: if $F\in \Omega^2_{SC}$, $d F = 0$ and $\delta F = 0$, then there exist two forms $A\in \Omega^1_{SC}$ and $B\in \Omega^3_{SC}$, with $\delta A = 0$ and $d B = 0$, such that $F = d A + \delta B$. Moreover, there exist $\alpha \in \Omega^1_0$ and $\beta \in \Omega^3_0$ such that $A=G\alpha$ and $B=G\beta$. This result can be found, for instance, as Prop.2.2 <a href="http://arxiv.org/abs/1104.1374" rel="nofollow noreferrer">here</a>. I think you can straight forwardly adapt the proof to the case with non-vanishing currents $j_e$ or $j_m$.</p>
|
4,624,695 |
<p>I have an exercise with text: With lines</p>
<p><span class="math-container">$$
p_1 \ldots \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-4}{1}, \quad p_2 \ldots \frac{x-5}{2}=\frac{y-1}{-1}=\frac{z-2}{1} .
$$</span>
Determine one plane <span class="math-container">$\pi$</span> with respect to which the lines <span class="math-container">$p_1$</span> and <span class="math-container">$p_2$</span> are symmetric.</p>
<p>How to approach this? I should somehow conclude that lines go through dots A1(1,-1,4) A2(5,1,2).</p>
|
Sammy Black
| 6,509 |
<p>Shift your coordinates so that <span class="math-container">$x = a$</span> is the origin. In other words, define
<span class="math-container">$u = x - a$</span> so that <span class="math-container">$x = a + u$</span>. Geometrically, this measures <span class="math-container">$u$</span> units to the right of <span class="math-container">$a$</span>. Where is the reflected point <span class="math-container">$u$</span> units to the left of <span class="math-container">$a$</span>?
<span class="math-container">$$
a - u = a - (x - a) = 2a - x.
$$</span>
Thus, the condition <span class="math-container">$-f(x) = f(2a - x)$</span> for all <span class="math-container">$x$</span> corresponds in the shifted coordinates to the symmetry
<span class="math-container">$$
-f(a + u) = f(a - u).
$$</span>
Since any integral with respect to <span class="math-container">$x$</span> over the domain <span class="math-container">$[0, 2a] = [0, a] \cup [a, 2a]$</span> is equivalent to an integral with respect to <span class="math-container">$u$</span> over the domain
<span class="math-container">$[-a, a] = [-a, 0] \cup [0, a]$</span>. In either case, each value on the left has a corresponding value on the right of equal magnitude <em>but opposite sign</em>, so the net integral must be zero.</p>
|
1,554,871 |
<p>The mean value theorem tells</p>
<p>" If $f:[a,b]\rightarrow \mathbb R$ is continuous and $g$ is an integrable function that does not change sign on $[a, b]$, then there exists $c$ in $(a, b)$ such that</p>
<p>$\int_a^b f(x)g(x)dx=f(c)\int_a^b g(x)dx$."</p>
<p>If $g$ changes its sign, is this theorem still true?</p>
|
npatrat
| 220,440 |
<p>Well, $f_n$ are continuous functions. If $(f_n(x))$ is uniformly convergent , then $\lim_{n \to \infty}f_n(x)$ is continuous, which is false.</p>
<p>So, our sequence is not uniformly convergent over $\mathbb{R}$. </p>
|
123,813 |
<p>Let $E$ be an elliptic curve over $\mathbb{Q}$. It is known from Gross and Zagier that if $\textrm{rank}_{\textrm{an}}(E) \leq 1$, then</p>
<p>$$\textrm{rank}(E) \geq \textrm{rank}_{\textrm{an}}(E).$$ </p>
<p>Instead, suppose I know that $\textrm{rank}(E) \leq 1$, are there any (unconditional) inequalities that relate rank and analytic rank of $E$?</p>
|
user30035
| 30,035 |
<p>This is not really the right question. I think it's easier to just list the two things that are known unconditionally and then you can figure out if this answers your question.</p>
<p>1) If analytic rank is zero then algebraic rank is zero.</p>
<p>2) If analytic rank is 1 then algebraic rank is 1.</p>
<p>Proof: see quid's comment.</p>
<p>One now deduces easily that if the algebraic rank is at most 1 then the analytic rank is at least the algebraic rank.</p>
|
2,040,175 |
<p>The following diagram shows 9 distinct points chosen from the sides of a triangle.</p>
<p><a href="https://i.stack.imgur.com/3DUhm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3DUhm.png" alt="enter image description here" /></a></p>
<p><span class="math-container">$(1)$</span> How many line segments are there joining any two points on different sides?</p>
<p><span class="math-container">$(2)$</span> How many triangles can be formed from these points?</p>
<p>I manage to solve <span class="math-container">$(1)$</span>, by considering the following:</p>
<p>Fix one end of a line at the hypotenuse. Then it has <span class="math-container">$5$</span> choices for another end. Since there are <span class="math-container">$4$</span> points on hypotenuse, we have <span class="math-container">$5 \times 4$</span>.</p>
<p>Fix one end of a line at the bottom length of the triangle. Then we have <span class="math-container">$2 + 2 + 2 = 6$</span> lines formed between bottom and left side of the triangle.</p>
<p>By the addition principle, we have <span class="math-container">$20+6=26$</span> lines formed.</p>
<p>But I don't know how to start <span class="math-container">$(2)$</span>, as we can have two vertices of a triangle lie on the same side. This extra situation confuses me.</p>
<p>Answer to <span class="math-container">$(2)$</span> is <span class="math-container">$79$</span>.</p>
|
Community
| -1 |
<p>With all vertices on different sides, you form $2\cdot3\cdot4$ triangles.</p>
<p>With two vertices on the same side, you form $1,3$ and $6$ pairs respectively; then you can match every pair with a vertex on another side, giving</p>
<p>$$1\cdot(3+4)+3\cdot(2+4)+6\cdot(2+3).$$</p>
<p>Total $79$.</p>
<p>(I assume that we can rule out the $0+1+4$ flat triangles obtained with three vertices on the same side.)</p>
<hr>
<p>The approach by @Rohan is better: from all $9\cdot8\cdot7/6$ possible triangles, discard the $0+1+4$ degenerate ones. </p>
|
216,473 |
<p>Let me begin with some background: I used to enjoy mathematics immensely in school, and wanted to pursue higher studies. However, everyone around me at that time told me it was a stupid area (that I should focus on earning as soon as possible), and so instead I opted for an engineering degree. While in college, I used to find myself fascinated by higher math and other stuff, and used to score near-perfect all the time (but only in mathematics!). I must make it clear that I don't have an extraordinary talent for math; it's just that I enjoy exploring it very much, and then describing it to others in interesting ways. Anyway, it so happened that I went on changing job after job and was never satisfied. Now at the age of 27, I'm sitting home and realizing that I should have given mathematics a thought. So I'm thinking of taking up a graduate course and picking up where I left off.</p>
<p>HOWEVER . . .</p>
<p>I dream of becoming a teacher cum researcher some day, even if it's at school-level only. Before my graduate course begins (in 2-3 months), I've started reviewing math from my school books. Now the point is that I expect myself to be much more mature and smarter by now. That means I should be able to solve any problem and prove any theorem given in school math. But I can't, and it's shattering me. I mean, if I can't even prove basic theorems related to Euclid's geometry, how can I ever hope to do some authentic research like the mathematicians I so admire? This makes me wonder if I’ll ever be fit for teaching. If, at the age of 27, I can’t even master basic school mathematics with certainty, how can I ever hope to tackle problems in calculus and polynomials that students bring me tomorrow? It’s as if I’m cheating myself and those who’ll come to me for instruction.</p>
<p>Am I being too hard on myself? Am I expecting too much too soon? Does there come a point in a person’s math studies when he is able to discern properties and theorems all on his own? Or are all students of mathematics struggling and hiding their weaknesses? Or I really have no talent for math and it’s just an idle indulgence of mine. I mean, I’m not sure “how good” I’m supposed to be in order to feel confident that I can pull it off. In general, I find myself wondering how much do the others know math. Do all the teachers have perfect knowledge of, say, geometry, and can tackle every problem? If not, what gives them the right to call themselves teachers?</p>
<p>I have a feeling most of these questions are absurd, but I’ll be very thankful if someone can put me out of my misery.</p>
|
zyx
| 14,120 |
<blockquote>
<p>"I mean, if I can't even prove basic theorems related to Euclid's geometry, how can I ever hope to do some authentic research like the mathematicians I so admire? "</p>
</blockquote>
<p>These are different skills. There are many researchers who (as students) could solve all the problems in the calculus or geometry books, and now cannot as easily find the answer when students come asking questions. Hence the reliance on graduate student assistants, solution books, and permanent calculus lecturers. </p>
<blockquote>
<p>"This makes me wonder if I’ll ever be fit for teaching. If, at the age of 27, I can’t even master basic school mathematics with certainty, how can I ever hope to tackle problems in calculus and polynomials that students bring me tomorrow"?</p>
</blockquote>
<p>An overestimate of the requirements for being a school instructor! Only a small fraction of teachers can explain the mathematics credibly + know the theory behind it + can solve hard problems, and fewer have all these skills before working as a teacher. Some of the most effective teachers begin with relatively little advanced knowledge, but love the subject and constantly deepen their understanding by reading, solving problems, using math software, taking classes, assisting with student research projects and competitions ... or participating in online math sites. </p>
|
216,473 |
<p>Let me begin with some background: I used to enjoy mathematics immensely in school, and wanted to pursue higher studies. However, everyone around me at that time told me it was a stupid area (that I should focus on earning as soon as possible), and so instead I opted for an engineering degree. While in college, I used to find myself fascinated by higher math and other stuff, and used to score near-perfect all the time (but only in mathematics!). I must make it clear that I don't have an extraordinary talent for math; it's just that I enjoy exploring it very much, and then describing it to others in interesting ways. Anyway, it so happened that I went on changing job after job and was never satisfied. Now at the age of 27, I'm sitting home and realizing that I should have given mathematics a thought. So I'm thinking of taking up a graduate course and picking up where I left off.</p>
<p>HOWEVER . . .</p>
<p>I dream of becoming a teacher cum researcher some day, even if it's at school-level only. Before my graduate course begins (in 2-3 months), I've started reviewing math from my school books. Now the point is that I expect myself to be much more mature and smarter by now. That means I should be able to solve any problem and prove any theorem given in school math. But I can't, and it's shattering me. I mean, if I can't even prove basic theorems related to Euclid's geometry, how can I ever hope to do some authentic research like the mathematicians I so admire? This makes me wonder if I’ll ever be fit for teaching. If, at the age of 27, I can’t even master basic school mathematics with certainty, how can I ever hope to tackle problems in calculus and polynomials that students bring me tomorrow? It’s as if I’m cheating myself and those who’ll come to me for instruction.</p>
<p>Am I being too hard on myself? Am I expecting too much too soon? Does there come a point in a person’s math studies when he is able to discern properties and theorems all on his own? Or are all students of mathematics struggling and hiding their weaknesses? Or I really have no talent for math and it’s just an idle indulgence of mine. I mean, I’m not sure “how good” I’m supposed to be in order to feel confident that I can pull it off. In general, I find myself wondering how much do the others know math. Do all the teachers have perfect knowledge of, say, geometry, and can tackle every problem? If not, what gives them the right to call themselves teachers?</p>
<p>I have a feeling most of these questions are absurd, but I’ll be very thankful if someone can put me out of my misery.</p>
|
Wouter Stekelenburg
| 27,375 |
<p>I don't have an answer, but I have a suggestion for finding one.</p>
<p>Most mathematics professors will allow you to visit lectures and exercise classes, and even welcome you if you show more enthusiasm than the average student (trust me: not hard to do). You won't get a degree that way, but you will get a better idea if studying mathematics is something for you.</p>
|
4,221,530 |
<p>Assume terms <span class="math-container">$a_n$</span> and <span class="math-container">$a_{n+1}$</span> share some common factor <span class="math-container">$x$</span> so that <span class="math-container">$a_n = xm$</span> for some integer <span class="math-container">$m$</span> and <span class="math-container">$a_{n+1} = xk$</span> for some integer <span class="math-container">$k$</span>.</p>
<p>Since <span class="math-container">$a_{n-1}$</span> = <span class="math-container">$a_{n+1} - a_n = xk - xm = x(k-m)$</span>, <span class="math-container">$a_{n-1}$</span> is a multiple of <span class="math-container">$x$</span> as well. Then <span class="math-container">$a_{n-2}$</span> must also be a multiple of <span class="math-container">$x$</span> because <span class="math-container">$a_{n-2} = a_{n} - a_{n-1} = xm - (xk - xm) = x(2m-k)$</span>. Since each lower or upper term can be found by subtracting or adding one multiple of <span class="math-container">$x$</span> to another, we can say that every term in the sequence must then be a multiple of <span class="math-container">$x$</span>.</p>
<p>Therefore, if any two consecutive terms share a common factor, then all terms must share a common factor. However, if <span class="math-container">$a_1=1$</span> we can see the counterexample of <span class="math-container">$a_3 = 2, a_4=3$</span> where <span class="math-container">$2$</span> and <span class="math-container">$3$</span> are consecutive terms with no common factors. Since not all terms then share a common factor, no two consecutive terms can share a common factor.</p>
|
Math Lover
| 801,574 |
<p><span class="math-container">$\triangle ABM$</span> is isosceles so <span class="math-container">$\angle AMB = \angle ABM = \cfrac{180^ \circ - 30^\circ}{2} = 75^ \circ$</span></p>
<p>Say side of hexagon is <span class="math-container">$a$</span> and perp from <span class="math-container">$T$</span> to <span class="math-container">$FN$</span> is <span class="math-container">$TH$</span> and to <span class="math-container">$AM$</span> is <span class="math-container">$TG$</span>, then</p>
<p><span class="math-container">$TH = FT \sin 60^ \circ = \cfrac{\sqrt3 a}{2}$</span></p>
<p><span class="math-container">$TG = a - \cfrac{\sqrt3 a}{2}$</span></p>
<p><span class="math-container">$\tan (\angle TAM) = \cfrac{TG}{AM/2} = 2 - \sqrt3$</span></p>
<p><span class="math-container">$\implies \angle TAM = 15^ \circ$</span></p>
<p>So, <span class="math-container">$x = \angle AIB = 180^ \circ - 75^ \circ - 45^ \circ = 60^ \circ$</span></p>
<hr />
<p>Without trigonometry, note that <span class="math-container">$\triangle AFT$</span> is isosceles with <span class="math-container">$\angle FAT = 75^0$</span> so <span class="math-container">$\angle MAT = 15^0$</span>. That leads to the answer.</p>
|
636,359 |
<p>How do you proof that $\angle CTP = \angle CBP+ \angle BCN$ ?
Please check the image below:</p>
<p><img src="https://i.stack.imgur.com/SqeXa.png" alt="enter the great image description here"></p>
|
MikhaelM
| 247,415 |
<p>$\angle CTP=\angle BTN=180^\circ-\angle PTN=180^\circ-\angle CTB=\angle CBT+\angle TCB = \angle CBP + \angle BCN$</p>
|
4,057,959 |
<p>What is the number of N step random walks starting from a point (x0,y0) to a point (x1,y1) assuming each direction (right,left,up,down) has equal probability. I know the expression for the one dimension case is:</p>
<p><span class="math-container">$$ \binom{N}{\frac{N+(y1-y0)}{2}} $$</span></p>
<p>is there a similar expression for two dimensions?
I saw this post (<a href="https://math.stackexchange.com/questions/2157448/probability-distribution-of-a-2d-lattice-random-walk">Probability Distribution of a 2D lattice random walk</a>) but the answer given is clearly wrong.</p>
|
grand_chat
| 215,011 |
<p>Here's another approach, less slick than that of @MikeEarnest (+1):</p>
<p>Wlog we are taking <span class="math-container">$N$</span> steps from <span class="math-container">$(0,0)$</span> to <span class="math-container">$(x,y)$</span>. Suppose <span class="math-container">$k$</span> of these steps are horizontal (either left or right). Use the one-dimensional formula to count the number of ways to make <span class="math-container">$k$</span> (horizontal) steps starting at <span class="math-container">$0$</span> and ending at <span class="math-container">$x$</span>. Next, multiply by the number of ways to embed a given sequence of <span class="math-container">$k$</span> horizontal steps into your total <span class="math-container">$N$</span> steps. Then given that embedding, count the ways to make <span class="math-container">$N-k$</span> (vertical) steps starting at <span class="math-container">$0$</span> and ending at <span class="math-container">$y$</span>. This yields the sum
<span class="math-container">$$\sum_k{k\choose \frac12(k+x)}{N\choose k}{N-k\choose\frac12(N-k+y)}.$$</span>
Change index of summation to <span class="math-container">$a:=\frac12(k-x)$</span> and define <span class="math-container">$m:=\frac12(N-x-y)$</span>. This gives
<span class="math-container">$$
\sum_a\frac{N!}{(x+a)!a!(m+y-a)!(m-a)!}={N\choose m}\sum_a{m\choose a}{N-m\choose m+y-a}$$</span> after some rearranging along with the identity <span class="math-container">$N-m=(x+a)+(m+y-a)$</span>. The rightmost sum <a href="https://math.stackexchange.com/q/337923/215011">evaluates to</a> <span class="math-container">${N\choose m+y}$</span>, yielding
<span class="math-container">$${N\choose m}{N\choose m+y}={N\choose \frac12(N-x-y)}{N\choose\frac12(N-x+y)}.$$</span></p>
|
3,612,916 |
<p>Find a function <span class="math-container">$f$</span> such that <span class="math-container">$f(x)=0$</span> for <span class="math-container">$x\leq0$</span>, <span class="math-container">$f(x)=1$</span> for <span class="math-container">$x\geq1$</span>, and <span class="math-container">$f$</span> is infinitely differentiable. </p>
<p>I've tried cojoining two quarters of a circle to get an S-shaped graph of the function and tried with different combinations of <span class="math-container">$e^{\frac{-1}{x}}$</span> and <span class="math-container">$e^{\frac{1}{x-1}}$</span> but so far nothing has worked. </p>
<p>Any help would be appreciated.</p>
<p>Edit: Not allowed to use integrals</p>
|
David Kraemer
| 224,759 |
<p>Try the function</p>
<p><span class="math-container">$$
f(x)= \begin{cases}
0 & x < 0 \\
\frac{1}{1+\exp(-g(x))} & 0 \leq x \leq 1 \\
1 & 1 < x
\end{cases}
$$</span>
where <span class="math-container">$g(x)= (1-x)^{-1} - x^{-1}$</span>.</p>
<p>Here's a <a href="https://www.desmos.com/calculator/yvgcxqyztb" rel="nofollow noreferrer">link</a> to Desmos where you can try it out.</p>
|
1,916,232 |
<p>I have massive problems with questions like these:</p>
<p>Let $\{v_1, . . . , v_r\}$ be a set of linearly independent vectors in $\mathbb{R}^n$
(with
$r < n$), and let $w\in\mathbb{R}^n$ be a vector such that $w \in \mathrm{span}\{v_1, . . . , v_r\}$.
Prove that $\{v_1, . . . , v_r, w\}$ is a linearly independent set.</p>
<p>Let $U$ and $V$ be subspaces of $\mathbb{R}^n$
Define the set $U + V = \{u + v|u ∈ U, v ∈ V \}$. Prove that $U + V$
is a subspace of $\mathbb{R}^n$.</p>
<p>I'm not looking for the answers to these 2 questions but instead I want to know how do I learn to approach these problems. These proving problems are my absolute Achilles' heel. I can't get the solution at all. What can I do to learn to solve these problems? Any online resources you guys can recommend? I usually learn how to do real questions by following examples but I get nothing from watching people talk about these theories and principles behind how it's done...</p>
|
Siong Thye Goh
| 306,553 |
<p>In general, there is no sure work formula to successfully prove something and there could be multiple ways to solve the same thing . The advices in the comments are very useful. </p>
<p>Reading mathematics does help if you can see the thought process.</p>
<p>Mathematical proofs is really like solving real life problem.
I usually ask myself questions to guide myself to solve problems:</p>
<ul>
<li>what is my goal? </li>
<li>what tools do I have? </li>
<li>how do i use my tools to reach my goal?</li>
<li>if I get stucked, can I use tricks like contradiction or contrapositive to prove it. Do I have other lemmas or theorems that can help me achieve my goal.</li>
<li>Sometimes, you really get stucked, ask for help/ hint from tutor/ friends. Take a walk around and come back to the question later. Try to solve special cases or look at similar questions.</li>
</ul>
<p>For example, in the first question:</p>
<p>Let $\{v_1, . . . , v_r\}$ be a set of linearly independent vectors in $\mathbb{R}^n$
(with
$r < n$), and let $w\in\mathbb{R}^n$ be a vector such that $w \notin \mathrm{span}\{v_1, . . . , v_r\}$.
Prove that $\{v_1, . . . , v_r, w\}$ is a linearly independent set.</p>
<p>Goal: prove that $\{v_1, . . . , v_r, w\}$ is a linearly independent set.</p>
<p>hmmm... what does linearly independent set mean? Let me check the definition and rewrite it.</p>
<p>New goal: If $\sum_{i=1}^r c_iv_i+c_{r+1}w = 0$, show that $c_i=0, \forall i=1, \ldots, r+1.$</p>
<p>Now let's see what do we know about $v_i$ and $w$.</p>
<p>Tool $1$: Let $\{v_1, . . . , v_r\}$ be a set of linearly independent vectors in $\mathbb{R}^n$
(with
$r < n$). Meaning, whenever I form a linear combination of $v_i$ and equate it to $0$, the coefficients must be $0$.</p>
<p>Tool $2$: $w \notin \mathrm{span}\{v_1, . . . , v_r\}$, that is $w$ cannot be written as linear combination of $v_i$.</p>
<p>The second tool motivates me to try to isolate $w$.Hence, it is natural to consider whether :</p>
<ul>
<li>case 1: $c_{r+1}=0$: If $\sum_{i=1}^r c_iv_i+c_{r+1}w = 0$ and $c_{r+1}=0$, we have $\sum_{i=1}^r c_iv_i = 0$ and we can check our tool list and see that we can use tool $1$ to conclude that $c_i=0, \forall i=1,\ldots, r$.</li>
<li>case 2: $c_{r+1}\neq0$: We better get a contradiction for this case, as we really want $c_{r+1}$ to be zero. Suppose it is not, $w =-\frac{1}{c_{r+1}}\sum_{i=1}^r c_iv_i$, check our tool list and we can see that tool $2$ says that we can't have this case.</li>
</ul>
|
1,438,512 |
<p>I was given this problem at school to look at home as a challenge, after spending a good 2 hours on this I can't seem to get further than the last part of the equation. I'd love to see the way to get through 2) before tomorrow's lesson as a head start.</p>
<p>So the problem is as follows:</p>
<p>1) Quadratic Equation $$2x^2 + 8x + 1 = 0$$ </p>
<p>i. Find roots $$\alpha + \beta$$</p>
<p>ii. Find roots $$\alpha\beta$$</p>
<p>2) Find an Equation with integer coefficients who's roots are:</p>
<p>$$2\alpha^4+\frac{1}{\beta^2}$$$$2\beta^4+\frac{1}{\alpha^2}$$</p>
<p>I'm completely puzzled on the second part of the question and I've tried following the method I was taught. Sorry if formatting is a bit off, first time posting here :) </p>
<p>Thanks in advance for any help!</p>
|
R.N
| 253,742 |
<p>1) Quadratic Equation $$2x^2 + 8x + 1 = 0$$ </p>
<p>i. $$\alpha + \beta=\frac{-b}{a}=\frac{-8}{2}=-4$$</p>
<p>ii. $$\alpha\beta=\frac{c}{a}=\frac{1}{2}$$
For remain just find roots and use this fact that if $x_1+x_2=s$ and $x_1 x_2=p$ equation will be $x^2 -sx+p=0$. </p>
|
175,535 |
<p>I have <code>data1</code> and a target point <code>targetPts</code>, and want to find the closest point from the data. As you can see from below, the 29th point from the data is the closest point and thus the value of <code>data1[[29]]</code> is same as the desired value, which is <code>{0.67033, 0.84245}</code>.</p>
<pre><code>data1 = BlockRandom[SeedRandom[7]; RandomReal[1, {30, 2}]];
targetPts = {0.68, 0.75};
data1[[29]]
Flatten@Nearest[data1, targetPts]
</code></pre>
<p>Now my problem changed a bit. I was given <code>data1data2</code> and I need to use it as the input. My desired output for this case is <code>{0.67033, 0.84245, 0.177696, 0.49873}</code>. The first 2 elements are used in the <code>Nearest</code> searching, while all 4 elements are the desired output. It just like a lookup function.</p>
<pre><code>data2 = BlockRandom[SeedRandom[8]; RandomReal[1, {30, 2}]];
data1data2 = ArrayFlatten[{{data1, data2}}];
data1data2[[29]]
</code></pre>
<p>How can I get it done? Many thanks!</p>
|
kglr
| 125 |
<pre><code>First @ Nearest[data1data2[[All, ;; 2]] -> "Index", targetPts]
</code></pre>
<blockquote>
<p>29</p>
</blockquote>
<pre><code>data1data2[[First @ Nearest[data1data2[[All, ;; 2]] -> "Index", targetPts]]]
</code></pre>
<blockquote>
<p> {0.67033, 0.84245, 0.177696, 0.49873} </p>
</blockquote>
<p>For versions prior to version 11.1 replace <code>"Index"</code> above with <code>Automatic</code> (thanks: Henrik Schumacher).</p>
|
1,791,849 |
<p>On the <a href="https://en.wikipedia.org/wiki/Pentagon" rel="nofollow">Wikipedia Page about Pentagons</a>, I noticed a statement in their work saying that $\sqrt{25+10\sqrt{5}}=5\tan(54^{\circ})$ and $\sqrt{5-2\sqrt{5}}=\tan(\frac {\pi}{5})$</p>
<p>My question is: How would you justify that? My goal is to simplify $\sqrt{25+10\sqrt{5}}$ and $\sqrt{5-2\sqrt{5}}$ without knowledge on the detested radical!</p>
|
egreg
| 62,967 |
<p>You surely know that $z=\cos\frac{2\pi}{5}+i\sin\frac{2\pi}{5}$ is the root in the first quadrant of $z^5-1=0$ so it satisfies
$$
z^4+z^3+z^2+z+1=0
$$
that can also be written as
$$
z^2+\frac{1}{z^2}+z+\frac{1}{z}+1=0
$$
or, by noting that $z^2+1/z^2=(z+1/z)^2-2$,
$$
\left(z+\frac{1}{z}\right)^2+\left(z+\frac{1}{z}\right)-1=0
$$
Therefore,
$$
z+\frac{1}{z}=2\cos\frac{2\pi}{5}=\frac{\sqrt{5}-1}{2}
$$
Thus
$$
\cos\frac{2\pi}{5}=\frac{\sqrt{5}-1}{4},
\qquad
\sin\frac{2\pi}{5}=\sqrt{1-\frac{5-2\sqrt{5}+1}{16}}=
\frac{\sqrt{10+2\sqrt{5}}}{4}
$$
Now use that
$$
\tan\frac{x}{2}=\frac{\sin x}{1+\cos x}
$$
and you get
$$
\tan\frac{\pi}{5}=\frac{\sqrt{10+2\sqrt{5}}}{3+\sqrt{5}}=
\frac{\sqrt{(10+2\sqrt{5})(3-\sqrt{5})^2}}{4}=
\frac{\sqrt{80-32\sqrt{5}}}{4}=\sqrt{5-2\sqrt{5}}
$$</p>
<p>Since $54^\circ=3\pi/10=\pi/2-\pi/5$, we have
$$
\tan\frac{3\pi}{10}=\cot\frac{\pi}{5}=\frac{1}{\sqrt{5-2\sqrt{5}}}=
\sqrt{\frac{5+2\sqrt{5}}{25-20}}=\frac{\sqrt{25+10\sqrt{5}}}{5}
$$</p>
|
393,280 |
<p>I am reading about the construction of the Affine Grassmannian in Dennis Gaitsgory's seminar <a href="http://www.math.harvard.edu/%7Egaitsgde/grad_2009/SeminarNotes/Oct13(AffGr).pdf" rel="noreferrer">notes</a>
and there are some commutative algebra facts that I am not able to figure out by myself apparently, like the following:</p>
<blockquote>
<p>Let <span class="math-container">$k$</span> be an algebraically closed field, <span class="math-container">$A$</span> a finite type <span class="math-container">$k$</span>-algebra and <span class="math-container">$A\subseteq B$</span> some (non finite type) extension. Let <span class="math-container">$M$</span> be a finitely generated <span class="math-container">$A[[t]]$</span>-module which is flat over <span class="math-container">$A$</span> (actually finite free over <span class="math-container">$A$</span>) and <span class="math-container">$t$</span> acts nilpotently on <span class="math-container">$M$</span>. Then, <span class="math-container">$\operatorname{Tor}^{A[[t]]}_n(M,B[[t]])=0$</span> for all <span class="math-container">$n>0$</span>.</p>
</blockquote>
<p>Intuitively, since we get <span class="math-container">$B[[t]]$</span> from <span class="math-container">$A[[t]]$</span> by extending only the coefficient ring <span class="math-container">$A$</span> in a "free" way, flatness over <span class="math-container">$A$</span> of <span class="math-container">$M$</span> should suffice, but I can't make it into a proof. Notice that <span class="math-container">$A[[t]]\otimes _A B\ne B[[t]]$</span> in general. Am I missing something obvious?</p>
<h3>Edit:</h3>
<p>After reading it again, it seems that the only additional hypothesis I missed is that <span class="math-container">$t$</span> acts nilpotently on <span class="math-container">$M$</span> which I think does not follow from what I have written. The relevant place in the notes is the first line on page 7 and a bit before that.</p>
|
Dylan Wilson
| 423 |
<p>Here's an attempt (could be mistakes so be wary!).</p>
<p>We have
$$
M \otimes_{A[[t]]}^{\mathbb{L}} B[[t]] \cong (M \otimes_{A[[t]]}^{\mathbb{L}} A[[t]]/t^n) \otimes_{A[[t]]}^{\mathbb{L}} B[[t]].
$$</p>
<p>By associativity this is (quasi-isomorphic) to
$$
M \otimes_{A[[t]]}^{\mathbb{L}} B[[t]]/t^n.
$$
Since $A[[t]] \rightarrow A[[t]]/t^n$ is surjective, this is the same as
$$
M \otimes_{A[[t]]/t^n}^{\mathbb{L}} B[[t]]/t^n.$$</p>
<p>And this satisfies the conclusion you want because of your conditions on $M$. </p>
|
1,905,186 |
<blockquote>
<p>Let <span class="math-container">$R$</span> be a commutative Noetherian ring (with unity), and let <span class="math-container">$I$</span> be an ideal of <span class="math-container">$R$</span> such that <span class="math-container">$R/I \cong R$</span>. Then is it true that <span class="math-container">$I=(0)$</span> ?</p>
</blockquote>
<p>I know that a surjective ring endomorphism of a Noetherian ring is also injective, and since there is a natural surjection from <span class="math-container">$R$</span> onto <span class="math-container">$R/I$</span> we get a surjection from <span class="math-container">$R$</span> onto <span class="math-container">$R$</span>, but the problem is I can not determine the map explicitly and I am not sure about the statement. Please help. Thanks in advance.</p>
|
Community
| -1 |
<p>Assume $f:R/I\to R$ is an isomorphism and $I \ne (0)$. Let $\overline{J} = f^{-1}I\subset R/I$ and $J\subset R$ be the preimage of $\overline J$ in $R$. Now $I \ne (0)$ implies $J$ strictly contains $I$; but $R/J \cong (R/I)/\overline J$ which is isomorphic to $R/I$ via $f$, and so isomorphic to $R$ by hypothesis. Now you can repeat for $J$; you will find a never-ending sequence of ever-larger ideals, contradicting the Noetherian property.</p>
<p>Note that the problem statement is somewhat ambiguous; you could interpret the isomorphism to be "isomorphic as $R$-modules" (in which case the problem would be trivial).</p>
|
496,178 |
<p>Let $t$ be a positive real number. Differentiate the function</p>
<blockquote>
<p>$$g(x)=t^x x^t.$$</p>
</blockquote>
<p>Your answer should be an expression in $x$ and $t$.</p>
<p>came up with the answer </p>
<blockquote>
<p>$$(x/t)+(t/x)\ln(t^x)(x^t)=\ln(t^x)+\ln(x^t)=x\ln t+t\ln x .$$</p>
</blockquote>
<p>and the derivative to that is $(x/t)+(t/x)$. Not sure if I've done it right.</p>
|
user94529
| 94,529 |
<p>$g(x)= t^xx^t = e^{ln(t^xx^t)}
=e^{xln(t) + tln(x)}$</p>
<p>Now,</p>
<p>$g'(x)= \frac{d}{dx}e^{ln(t^xx^t)}$</p>
<p>$=\frac{d}{dx}e^{xln(t) + tln(x)}$</p>
<p>$= e^{xln(t) + tln(x)}.\frac{d}{dx}({xln(t) + tln(x)})$</p>
<p>$= e^{xln(t) + tln(x)}.(ln(t)+\frac{t}{x})$</p>
<p>$=t^xx^t.(ln(t)+\frac{t}{x})$</p>
|
2,861,362 |
<p>I have the following question:
- We proved that if $T$ is 1-1 and $\{v_1...v_n\}$ is linearly independent then $\{T(v_1)...T(v_n)\}$ is linearly independent! I understood the proof! But can’t $\{T(v_1)...T(v_n)\}$ be linearly independent without having $T$ 1-1?
The image I uploaded shows my work on proving that the set of the images is linearly independent without having $T$ 1-1.
Can someone help me?
<a href="https://i.stack.imgur.com/IRqZV.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IRqZV.jpg" alt="enter image description here"></a></p>
|
Jack D'Aurizio
| 44,121 |
<p>In a elementary fashion we have
$$\begin{eqnarray*}\log\left(\frac{n+1}{n}\right)&=&\int_{n}^{n+1}\frac{dx}{x}=\int_{0}^{1}\frac{dx}{n+x}=\int_{-1/2}^{1/2}\frac{dx}{n+\frac{1}{2}+x}\\&=&\int_{0}^{1/2}\left(\frac{1}{n+\frac{1}{2}+x}+\frac{1}{n+\frac{1}{2}-x}\right)\,dx\\&=&\int_{0}^{1}\frac{dz}{\left(n+\frac{1}{2}\right)-\frac{z^2}{4n+2}}\end{eqnarray*}$$
so for any $n\geq 1$
$$ \log\left(\frac{n+1}{n}\right)-\frac{1}{\left(n+\frac{1}{2}\right)}=\frac{1}{4}\int_{0}^{1}\frac{z^2}{\left(n+\frac{1}{2}\right)^3-\frac{z^2}{4}\left(n+\frac{1}{2}\right)}\,dz$$
is bounded by $\frac{1}{12}\cdot\frac{1}{n\left(n+\frac{1}{2}\right)(n+1)}$ and it is decreasing towards zero as $n\to +\infty$.<br>
This essentially is a naive implementation of the <a href="https://en.wikipedia.org/wiki/Hermite%E2%80%93Hadamard_inequality" rel="nofollow noreferrer">Hermite-Hadamard inequality</a>.<br>
Since $\log\left(\frac{n+1}{n}\right)$ and $\frac{1}{\left(n+\frac{1}{2}\right)}$ are pretty close (with the former being a telescopic term) and $\sum_{n\geq 1}\frac{1}{n\left(n+\frac{1}{2}\right)(n+1)}$ is finite (equal to $\frac{3-4\log 2}{6}$) your sequence is convergent.</p>
<p>The Euler-Mascheroni constant $\gamma$ is exactly defined as $\lim_{n\to +\infty}\left(H_n-\log n\right)$. From
$$ H_n = \log(n) + \gamma + O\left(\frac{1}{n}\right) $$
we have
$$ H_{2n}-\frac{1}{2}H_n-\frac{1}{2}\log(n) = \underbrace{\color{red}{\frac{\gamma}{2}+\log 2}}_{\approx 0.981755}+O\left(\frac{1}{n}\right).$$</p>
<p>Long story short, your limit is just a bit less than one.</p>
|
3,221,652 |
<p><span class="math-container">$ a= \frac{2}{5}; f(x) = 10x^3 +6x^2 + x -2 $</span></p>
<p>Have difficulties with advanced polynomials of this type. I can not factor.To find the right root. Here is my solution:</p>
<p><span class="math-container">$f(x) = 10x^3 +6x^2 + x -2 = 10x^3 +6x^2 +1x^2 -1x^2 +x -2 = (5x^2+x)(2x+1) -x^2 -2 $</span></p>
<p>I do not understand what I have to do.</p>
|
Community
| -1 |
<p><span class="math-container">$f(\dfrac25)=\dfrac {80}{125}+\dfrac {24}{25}+\dfrac25-2=\dfrac{80}{125}+\dfrac{120}{125}+\dfrac{50}{125}-\dfrac{250}{125}=0$</span>. Thus it is a root.</p>
|
885,478 |
<p>Let $f$ be a function defined on an interval $I$ differentiable at a point $x_o$ in the interior of $I$.</p>
<p>Prove that if $\exists a>0$ $ \ [x_o -a, x_o+a] \subset I$ and $ \ \forall x \in [x_o -a, x_o+a] \ \ f(x) \leq f(x_o)$, then $f'(x_o)=0$.</p>
<p>I did it as follows:</p>
<p>Let b>0.</p>
<p>Since $f$ is differentiable at $x_o$,
$$ \exists a_o>0 \ \ \text{s.t} \ \ \forall x \in I \ \ \ \ \ 0<|x-x_o|<a_o \implies \left| \frac{f(x)-f(x_o)}{x-x_0} - f'(x_o)\right| <b$$
Let $x_1 \in (x_o,x_o+a) \forall x \in I; f(x_1) \leq f(x_o)$
$$ \left| \frac{f(x_1)-f(x_o)}{x_1-x_0} - f'(x_o)\right| <b \\
-b < f'(x_o)-\frac{f(x_1)-f(x_o)}{x_1-x_0} <b \\
f'(x_o) < b+ \frac{f(x_1)-f(x_o)}{x_1-x_0} < b$$
$$f'(x_o) < b \tag{1} $$
Similarly Let $x_2 \in (x_o-a,x_o) \forall x \in I; f(x_2) \leq f(x_o)$</p>
<p>$$ \left| \frac{f(x_2)-f(x_o)}{x_2-x_0} - f'(x_o)\right| <b \\
-b < \frac{f(x_2)-f(x_o)}{x_2-x_0} - f'(x_o) <b \\
-b< -b + \frac{f(x_2)-f(x_o)}{x_2-x_0} < f'(x_o)$$
$$-b<f'(x_o) \tag{2} $$</p>
<p>From $(1)$ and $(2)$,
$$ -b < f'(x_o) <b \\
|f'(x_o)|<b $$</p>
<p>I'm stuck here, how can I go to $f'(x_o)=0$ from here?</p>
<p>Any help?</p>
|
JC574
| 165,562 |
<p>It seems you're pretty much there. There are a couple of things you can do to make the argument clearer, both in the early line using the property of differentiable:</p>
<p>1.Differentiation talks about a limit, make sure you say $\forall b > 0 $, as this is what gets you the final step.</p>
<p>2.You want to include the given property here, because it seems you have assumed this. What it should say is something like:</p>
<p>By differentiability at $x_0$ and the property given (about $f$ being max at $x_0$ ) we have:</p>
<p>$$ \forall b> 0 \ \ \exists a > 0 \ \ \text{s.t} \ \ \forall x \in (x_0-a,x_0+a) \subset I \ \ \left| \frac{f(x)-f(x_0)}{x-x_0} - f'(x_0) \right| < b \ \ \text{and} \ \ f(x) \le f(x_0)$$</p>
|
885,478 |
<p>Let $f$ be a function defined on an interval $I$ differentiable at a point $x_o$ in the interior of $I$.</p>
<p>Prove that if $\exists a>0$ $ \ [x_o -a, x_o+a] \subset I$ and $ \ \forall x \in [x_o -a, x_o+a] \ \ f(x) \leq f(x_o)$, then $f'(x_o)=0$.</p>
<p>I did it as follows:</p>
<p>Let b>0.</p>
<p>Since $f$ is differentiable at $x_o$,
$$ \exists a_o>0 \ \ \text{s.t} \ \ \forall x \in I \ \ \ \ \ 0<|x-x_o|<a_o \implies \left| \frac{f(x)-f(x_o)}{x-x_0} - f'(x_o)\right| <b$$
Let $x_1 \in (x_o,x_o+a) \forall x \in I; f(x_1) \leq f(x_o)$
$$ \left| \frac{f(x_1)-f(x_o)}{x_1-x_0} - f'(x_o)\right| <b \\
-b < f'(x_o)-\frac{f(x_1)-f(x_o)}{x_1-x_0} <b \\
f'(x_o) < b+ \frac{f(x_1)-f(x_o)}{x_1-x_0} < b$$
$$f'(x_o) < b \tag{1} $$
Similarly Let $x_2 \in (x_o-a,x_o) \forall x \in I; f(x_2) \leq f(x_o)$</p>
<p>$$ \left| \frac{f(x_2)-f(x_o)}{x_2-x_0} - f'(x_o)\right| <b \\
-b < \frac{f(x_2)-f(x_o)}{x_2-x_0} - f'(x_o) <b \\
-b< -b + \frac{f(x_2)-f(x_o)}{x_2-x_0} < f'(x_o)$$
$$-b<f'(x_o) \tag{2} $$</p>
<p>From $(1)$ and $(2)$,
$$ -b < f'(x_o) <b \\
|f'(x_o)|<b $$</p>
<p>I'm stuck here, how can I go to $f'(x_o)=0$ from here?</p>
<p>Any help?</p>
|
copper.hat
| 27,978 |
<p>This is not an answer, but is too long for the comments.</p>
<p>Another approach is to show that if $f'(x_0) > 0$, where $x_0$ is in the interior of $I$, then for some $\delta>0$, then if $x \in (x_0-\delta,x_0)$ we have $f(x) <f(x_0)$ and for $x \in (x_0, x_0+\delta)$ we have $f(x) > f(x_0)$.</p>
<p>A corresponding result will hold for $f'(x_0)<0$, of course.</p>
<p>The proof is straightforward, since if $f'(x_0)>0$, we can find some $\delta>0$ such that ${ f(x)-f(x_0) \over x-x_0 } \ge {f'(x_0) \over 2}$ for all $x$ such that $|x-x_0|< \delta$. For $x>x_0$ this gives $f(x) \ge f(x_0) + {1 \over 2} (x-x_0) f'(x_0)$, and similarly for $x<x_0$ this gives $f(x) \le f(x_0) + {1 \over 2} (x-x_0) f'(x_0)$.</p>
<p>Hence if $f$ has a local maximum (or indeed a local minimum) at $x_0$, we must have $f'(x_0) = 0$ (otherwise we could find nearby points that violate the assumption).</p>
|
2,657,632 |
<p>My question involves part (b) of Chapter 11 problem 6.4 in Artin's Algebra textbook.</p>
<blockquote>
<p>In each case, describe the ring obtained from <span class="math-container">$\mathbb{F_2}$</span> by adjoining an element <span class="math-container">$α$</span> satisfying the given relation:</p>
<p>(a) <span class="math-container">$α^2+α+1=0$</span></p>
<p>(b) <span class="math-container">$α^2+1=0$</span></p>
<p>(c) <span class="math-container">$α^2+α=0$</span></p>
</blockquote>
<p>Now, I obtained that the ring in part (a) is isomorphic to <span class="math-container">$\mathbb{F_4}$</span> and that the ring in part (c) is isomorphic to <span class="math-container">$\mathbb{F_2}\times\mathbb{F_2}$</span>.</p>
<p>It seems to me that the ring in part (b) would be isomorphic to <span class="math-container">$\mathbb{F}_2[x]/(x^2+1)$</span>, but my teacher doesn't agree.</p>
<p>He said,</p>
<blockquote>
<p>"Be careful: notice the polynomial <span class="math-container">$x^2+1$</span> is not irreducible over <span class="math-container">$\mathbb{F}_2$</span>. Adjoining a root of a reducible polynomial is not the same as taking the quotient <span class="math-container">$\mathbb{F}_2[x]/(x^2+1)$</span>"</p>
</blockquote>
<p>So, is my teacher right, or am I? And why?</p>
|
Tsemo Aristide
| 280,301 |
<p>Yes, $x^2+1=(x+1)^2$ in $\mathbb{F_2}[x]$ you deduce that $\mathbb{F_2}[x]/(x^2+1)$ is a $2$ dimensiobal vector space over $\mathbb{F_2}$.</p>
<p>Adjoining an element (of the algebraic closure) which satisfies $\alpha^2+1=0$ is adjoining $1$ and the result is $\mathbb{F_2}$.</p>
|
2,983,370 |
<p><strong>question:</strong></p>
<p>maximum value of <span class="math-container">$\theta$</span> untill which the approximation <span class="math-container">$\sin\theta\approx \theta$</span> holds to within <span class="math-container">$10\%$</span> error is </p>
<p><span class="math-container">$(a)10^{\circ}$</span></p>
<p><span class="math-container">$(b)18^{\circ}$</span></p>
<p><span class="math-container">$(c)50^{\circ}$</span></p>
<p><span class="math-container">$(d)90^{\circ}$</span></p>
<p><strong>my attempt:</strong></p>
<p>i calculated percentage error for each of 4 options and got <span class="math-container">$\theta = 50 $</span>degree</p>
<p>but is there any quick method to arrive at answer without verifying all options </p>
<p>one by one . because it is MCQ there will be very less time availaible per question to solve it.</p>
<p>thank you</p>
|
Jam
| 161,490 |
<p>This is similar to the other two answers but another way of thinking about it. By definition, the true <span class="math-container">$\theta$</span> is at <span class="math-container">$\left|\frac{\sin(\theta)-\theta}{\sin(\theta)}\right|=\frac{10\%}{100\%}$</span>. This would give us <span class="math-container">$\theta=0.749=42.9^\circ$</span>. We can use Taylor series to approximate the LHS as <span class="math-container">$\left|\frac{\theta-\frac{\theta^3}{6}-\theta}{\theta-\frac{\theta^3}{6}}\right|=\left|\frac{\theta^2}{6-\theta^2}\right|=0.1$</span>. This solves for <span class="math-container">$\theta\approx42^\circ$</span>, which is around <span class="math-container">$50^\circ$</span>.</p>
|
273,086 |
<p>I'd like to deploy a website to display and increment counters that track daily activities.</p>
<p>Here's a minimal working example:</p>
<pre><code>bin=CreateDatabin[]
CloudDeploy[FormPage[{"Pushups" -> {0, 5, 10, 15, 20, 25, 30},
"Water" -> {0, .5, 1, 2}},
DatabinAdd[bin, {{"Pushups", Now, #Pushups}, {"Water", Now, #Water}}]; &], Permissions->"Public"]
</code></pre>
<p><a href="https://i.stack.imgur.com/A6ddn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/A6ddn.png" alt="enter image description here" /></a></p>
<p>I'm unsure how to customize the content of a <code>FormPage</code>. I would like to add this plot on the page below the submit button:</p>
<pre><code>DateListPlot@Cases[Flatten[Normal@bin,1],{"Pushups",r__}:>{r}]
</code></pre>
<p><a href="https://i.stack.imgur.com/uAoba.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uAoba.png" alt="enter image description here" /></a></p>
<p>Along with a few tweaks for bonus points:</p>
<ul>
<li>replace drop-downs with buttons</li>
<li>display the current daily counter values</li>
<li>custom image banner at top of page</li>
<li>custom fonts and colors</li>
</ul>
|
lericr
| 84,894 |
<ul>
<li><p><strong>Replace drop-downs with buttons.</strong> You can add your own controls for each form element. Not sure if SetterBar is what you want, and it seems to behave a bit strangely when it's deployed to the cloud, so you might have to experiment with other controls.</p>
<pre><code>exerciseForm =
FormPage[
FormObject[
{"Pushups" -> <|"Interpreter" -> {0, 5, 10, 15, 20, 25, 30}, "Control" -> SetterBar|>,
"Water" -> <|"Interpreter" -> {0, .5, 1, 2}, "Control" -> SetterBar|>}],
UpdateAndDisplayExerciseHistory]
</code></pre>
</li>
<li><p><strong>Custom image banner at top of page.</strong> You could probably use AppearanceRules for this.</p>
<pre><code>title = ImageResize[ExampleData[{"TestImage", "Volubilis"}], 400];
exerciseForm =
FormPage[
FormObject[
{"Pushups" -> <|"Interpreter" -> {0, 5, 10, 15, 20, 25, 30}, "Control" -> SetterBar|>,
"Water" -> <|"Interpreter" -> {0, .5, 1, 2}, "Control" -> SetterBar|>}],
UpdateAndDisplayExerciseHistory,
AppearanceRules -> <|"Title" -> title|>]
</code></pre>
</li>
<li><p><strong>Custom fonts and colors</strong></p>
<p>You can add your own labels. For example, the pushup form element specification might look like this:</p>
<pre><code>"Pushups" ->
<|"Interpreter" -> {0, 5, 10, 15, 20, 25, 30},
"Control" -> SetterBar,
"Label" -> Style["Pushups", Blue]|>
</code></pre>
</li>
</ul>
<p>I'm not sure what "display the current daily counter values" means.</p>
|
3,753,143 |
<p>We spend a lot of time learning different theories (for instance, theory of differential forms, sobolev spaces, homology groups, distributions). Although (at least most parts of) these theories are very natural and understandable when we read them from books, they are very difficult to create at the first place: it could take tens of years of effort of a large number of excellent mathematicians.</p>
<p>After learning those theories, we do exercises or solve problems, but most of the time, we are just using the tools stated in the book. Even the chance that we come up with a "new" definition ourselves is rare. (By "new", I mean "have not learnt", even if someone else have created it before.) <strong>So here is my question:</strong></p>
<p><strong>What are some problems which prompts the creation of a new theory?</strong></p>
<p><strong>EDIT:</strong> Just to clarify, I am looking for some problems which give everybody a chance to experience the process of creating new mathematics; so the problem need not be as difficult as Riemann conjecture.</p>
<p>By "new theory", I just mean something that help us formulate the problem in a different way. For example, this <a href="https://www.youtube.com/watch?v=wTJI_WuZSwE" rel="nofollow noreferrer">video on a chess board puzzle</a> has the idea of creating new theories, because <strong>unlike</strong> other less interesting puzzles about chess board which can be solved by just carefully counting the squares, this video mentions a new way of looking at the problem, namely the vertices of a hypercube.</p>
<p>I have also seen other similar puzzles like this. Apparently, almost all of them are on discrete mathematics, so <strong>it would be really interesting if anyone could provide such a "theory creating" problem in other areas of mathematics (e.g. analysis).</strong></p>
<p>Of course, not all theories are created to tackle specific problems, so other ways of experiencing inventing new maths could also be suggested.</p>
|
Martin Hansen
| 646,413 |
<p>Well, here is my h'penny's worth :</p>
<p>In trying and eventually succeeding in proving Fermat's last Theorem, Andrew Wiles developed vast swaths of new mathematics. (As had others before him even without managing a proof). The problems are all around; The Riemann Hypothesis - with the tantalising "video's" of the zeros in the complex plane - and other "millennium problems" for example. The worthwhile problems tend to not be easy though !</p>
<p>The Millenium Prize problems : <a href="https://en.wikipedia.org/wiki/Millennium_Prize_Problems" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Millennium_Prize_Problems</a></p>
<p>The Riemnann Hypothesis :
<a href="https://en.wikipedia.org/wiki/Riemann_hypothesis" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Riemann_hypothesis</a></p>
<p>Andrew Wiles :
<a href="https://en.wikipedia.org/wiki/Andrew_Wiles" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Andrew_Wiles</a></p>
<p>Kummer Theory :
<a href="https://en.wikipedia.org/wiki/Kummer_theory" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Kummer_theory</a></p>
<p>(Thanks for the lead to Kummer Theory, Hagen von Eitzen !)</p>
|
3,267,216 |
<p>An urn contains equal number of green and red balls. Suppose you are playing the following game. You draw one ball at random from the urn and note its colour. The ball is then placed back in the urn, and the selection process is repeated. Each time a green ball is picked you get 1 Rupee. The first time you pick a red ball, you pay 1 Rupee and the game ends. Your expected income from this game is..</p>
<p>The answer given is 0 but shouldn't it be positive as the as the income rises if you consecutively draw the green ball? Please help me clarify this doubt</p>
|
Anurag A
| 68,092 |
<p>Assuming <span class="math-container">$ab>0$</span>.
<span class="math-container">\begin{align*}
\frac{a+b}{2\sqrt{ab}} & = \frac{m}{n}\\
n^2(a+b)^2& = 4m^2ab\\
n^2\left(\frac{a}{b}+\frac{b}{a}+2\right) &=4m^2.
\end{align*}</span>
Let <span class="math-container">$\frac{a}{b}=t$</span>, then you have a quadratic equation to solve
<span class="math-container">$$n^2t^2+(2n^2-4m^2)t+n^2=0$$</span></p>
|
3,120,090 |
<blockquote>
<p>Find all the values of <span class="math-container">$\theta$</span> that satisfy the equation
<span class="math-container">$$\cos(x \theta ) + \cos( (x+2) \theta ) = \cos( \theta )$$</span></p>
</blockquote>
<p>I've tried simplifying with factor formulae and a combo of compound angle formulae, and I'm still stuck. I get to <span class="math-container">$\theta = 180^\circ$</span> and <span class="math-container">$\theta = \frac{60^\circ}{x+1}$</span>, but I'm unsure if that's correct. </p>
<p>It seems to work for <span class="math-container">$\theta=180^\circ$</span>, but I can't verify the other solution. I feel as though it should be a numerical solution, but I'm unsure. </p>
|
Claude Leibovici
| 82,404 |
<p>Use
<span class="math-container">$$\cos(p)+\cos(q)=2 \cos \left(\frac{p+q}{2}\right)\cos \left(\frac{p-q}{2}\right)$$</span> Make <span class="math-container">$p=(x+2)\theta$</span> and <span class="math-container">$q=x\theta$</span> making your problem to be
<span class="math-container">$$2\cos((x+1)\theta) \cos(\theta)= \cos(\theta)$$</span> and then the two cases.</p>
|
227,296 |
<p>I happened to ponder about the differentiation of the following function:
$$f(x)=x^{2x^{3x^{4x^{5x^{6x^{7x^{.{^{.^{.}}}}}}}}}}$$
Now, while I do know how to manipulate power towers to a certain extent, and know the general formula to differentiate $g(x)$ wrt $x$, where $$g(x)=f(x)^{f(x)^{f(x)^{f(x)^{f(x)^{f(x)^{f(x)^{.{^{.^{.}}}}}}}}}}$$
I'm still unable to figure out as to how I can adequately manipulate the function to differentiate it within its domain of convergence, if it exists.</p>
<p><strong>Sub-Q: What is the domain over which the function converges? Does it have only a finite domain (meaning there's no necessity to discuss its derivative)?</strong></p>
<hr>
<p>General formula (making domanial assumptions for f(x) of course): $$g'(x)=\frac{g^2(x)f'(x)}{f(x)\left[1-g(x)\ln(f(x))\right]}$$</p>
<hr>
<p><strong>Note</strong>: This has been posted in stackexchange; a look at the (incomplete) answers and their respective comments' sections will give you an insight into what has been looked into so far - [<a href="https://math.stackexchange.com/questions/1592377/what-is-the-derivative-of-fx-x2x3x4x5x6x7x">https://math.stackexchange.com/questions/1592377/what-is-the-derivative-of-fx-x2x3x4x5x6x7x</a> ]</p>
|
Fedor Petrov
| 4,312 |
<p>Update: it answers a different question.</p>
<p>If $1/(N-1)>x> 1/N$, then $p_N:=(Nx)^{(N+1)x^{\dots}}=\infty$, $p_{N-1}=0$, $p_{N-2}=1$, and so we have a tower of height $N-3$. Thus your function has a jump in points $1/N$ and is smooth between consecutive jumps (values at points of jumps is difficult to define, since $p_N$ for $x=1/N$ is $1^{\infty}$).</p>
<p>How to differentiate a tower? Simply fix all but one appearings of $x$, differentiate with respect to this $x$, and then sum up.</p>
|
2,227,135 |
<p>I have this information from my notes:<span class="math-container">$\def\rk{\operatorname{rank}}$</span></p>
<p>Let <span class="math-container">$A ∈ \mathbb{R}^{m\times n}$</span>. Then </p>
<ul>
<li><span class="math-container">$\rk(A) = n$</span></li>
<li><span class="math-container">$\rk(A^TA) = n$</span></li>
<li><span class="math-container">$A^TA$</span> is invertible.</li>
</ul>
<p>In my case, <span class="math-container">$n = 1$</span>, so I would need to show <span class="math-container">$\rk(vv^T) = \rk(v^Tv) = \rk(v) = 1$</span>. Suppose <span class="math-container">$A^TAx = 0$</span>. Because <span class="math-container">$A^TA$</span> is invertible, I can multiply both sides by its inverse to get <span class="math-container">$x = 0$</span>, meaning the nullity of <span class="math-container">$A^TA$</span> is <span class="math-container">$0$</span>. Can I apply the same logic to <span class="math-container">$AA^T$</span>? i.e. I have some matrix <span class="math-container">$B = A^T$</span>, so <span class="math-container">$B^TB$</span> = <span class="math-container">$AA^T $</span> has a nullity of <span class="math-container">$0$</span> (and therefore they have the same rank by the rank-nullity theorem)?</p>
|
Marc van Leeuwen
| 18,880 |
<p><span class="math-container">$\def\rk{\operatorname{rank}}$</span>No you cannot apply the same logic to <span class="math-container">$vv^T$</span>, in other words for <span class="math-container">$AA^T$</span> where <span class="math-container">$A$</span> is (still) a <span class="math-container">$1\times m$</span> matrix, even though the conclusion that it has rank<span class="math-container">$~1$</span> is true. The point is that you are trying to apply the implications in your notes with the roles of <span class="math-container">$m$</span> and <span class="math-container">$n$</span> interchanged with respect to your initial application, in other words with <span class="math-container">$m=1$</span> and <span class="math-container">$n>1$</span>, but the first two points are still talking about <span class="math-container">$n$</span> (and necessarily so: if you replace them by <span class="math-container">$m$</span> the implication is not valid any more). The first point (which is the hypothesis) could be more clearly stated as "<span class="math-container">$A$</span> has full column rank" (meaning all its columns are linearly independent) to emphasise the asymmetry with respect to rows and columns of the statements.</p>
<p>Assuming that you know (or have been told) that the first dotted point implies the two other dotted points (which is true) then substituting <span class="math-container">$v^T$</span> for <span class="math-container">$A$</span> it says: "if <span class="math-container">$v^T\in\Bbb R^{1\times n}$</span> and <span class="math-container">$\rk(v^T)=n$</span>, then <span class="math-container">$\rk(vv^T)=n$</span> and <span class="math-container">$vv^T$</span> is invertible". This is valid, but unhelpful since the hypothesis <span class="math-container">$\rk(v^T)=n$</span> can never be satisfied if <span class="math-container">$n>1$</span>, since the rank of an <span class="math-container">$1\times n$</span> matrix is at most <span class="math-container">$1$</span>. A <em>row</em> vector cannot have linearly independent <em>columns</em> (if it has more than one of them). Note by the way that the conclusions that <span class="math-container">$\rk(vv^T)=n$</span> and <span class="math-container">$vv^T$</span> is invertible would <em>contradict</em> what you are trying so prove (namely <span class="math-container">$\rk(vv^T)=1$</span>), so it is fortunate that you cannot deduce them in this situation.</p>
<p>The conclusion <span class="math-container">$\rk(vv^T)=1$</span> is easy to obtain since <span class="math-container">$\rk(vv^T)\leq1$</span> follows from the general fact that the rank of a product of matrices cannot exceed the rank of any factor in the product. All that remains to show is that <span class="math-container">$vv^T\neq0$</span>, which can be easily done in many ways. I would like to point out that while <span class="math-container">$\rk(vv^T)=1=\rk(v^Tv)$</span> in this case, it is <em>not true</em> that <span class="math-container">$AB$</span> and <span class="math-container">$BA$</span> always have the same rank.</p>
|
699,933 |
<p>There is a proof of the real case of Cauchy-Schwarz inequality that expands $\|\lambda v - w\|^2 \geq 0 $, gets a quadratic in $\lambda$, and takes the discriminant to get the Cauchy-Schwarz inequality. In trying to do the same thing in the complex case, I ran into some trouble. First, there are proofs <a href="http://www.artofproblemsolving.com/Wiki/index.php/Cauchy-Schwarz_Inequality#Complex_Form" rel="nofollow noreferrer">here</a>, <a href="https://math.stackexchange.com/questions/446148/cauchy-schwarz-for-complex-numbers">here</a>, <a href="https://math.stackexchange.com/questions/202406/proof-of-cauchy-schwarz-inequality">here</a>, and <a href="http://ckrao.wordpress.com/2011/02/18/two-interesting-proofs-of-the-cauchy-schwarz-inequality-complex-case/" rel="nofollow noreferrer">here</a>, but none of them do it the way I'm thinking of.</p>
<p>If I similarly expand $\|\lambda v - w\|^2_{\mathbb{C}},$ I get $|\lambda|^2\|v\|^2 - 2 \text{Re}(\lambda \langle v,w\rangle) + \|w\|^2$. How can I manipulate this to get Cauchy-Schwarz using the discriminant? My problem is that $$|\lambda|^2\|v\|^2 - 2 \text{Re}(\lambda \langle v,w\rangle) + \|w\|^2 \geq |\lambda|^2\|v\|^2 - 2 | \lambda ||\langle v,w\rangle | + \|w\|^2,$$ so I can't be sure that the latter term is $\geq 0$.</p>
|
Eric Auld
| 76,333 |
<p>I just thought of $$|\lambda|^2\|v\|^2 - 2 \text{Re}(\lambda \langle v,w\rangle) + \|w\|^2 \leq |\lambda|^2\|v\|^2 + 2 | \lambda ||\langle v,w\rangle | + \|w\|^2,$$ which yields the same discriminant. This seems to work!</p>
|
3,960,189 |
<p>An auto insurance company is implementing a new bonus system. In each month, if a policyholder does not have an accident, he or she will receive a $5 cashback bonus from the insurer. Among the 1000 policyholders, 400 are classified as low-risk drivers and 600 are classified as high-risk drivers. In each month, the probability of zero accidents for high-risk drivers is 0.8 and that for low-risk drivers is 0.9.</p>
<p>Calculate the expected bonus payment from the insurer to the 1000 policyholders in one year.</p>
<p><strong>The question is what's wrong with my (following) solution?</strong></p>
<p>Probability of zero accidents <span class="math-container">$= (0.8\times 0.6) + (0.90 \times 0.4) = 0.84$</span></p>
<p>Then, the expected payment is given by <span class="math-container">$$\sum_{x=1}^{12} (0.84^x) \times 1000 \times 5x$$</span></p>
|
Ross Millikan
| 1,827 |
<p>Your first line computes that the average number of accident free drivers in a month is <span class="math-container">$840$</span>. Those people receive <span class="math-container">$840\cdot 5=4200$</span> for the month. Each month is the same, so the annual total is <span class="math-container">$12\cdot 4200=50400$</span></p>
|
2,727,974 |
<blockquote>
<p>I am in need of this important inequality $$\log(x+1)\leqslant x$$.</p>
</blockquote>
<p>I understand that $\log(x)\leqslant x$. For $c\in\mathbb{R}$. However is it true that $\log(x+c)\leqslant x$?</p>
<p>It is hard to accept because it seems like $c$ cannot be arbitrary.
I have tried to prove this inequality:</p>
<p>$\log(x+c)\leqslant x\iff x+c\leqslant e^x$</p>
<p>It is true that $f(x)=x$ grows much faster than $g(x)=\log(x+c)$, since the $\frac{df(x)}{dx}=1\geqslant \frac{1}{x+c}=\frac{dg(x)}{dx}$ </p>
<p><strong>Question:</strong></p>
<p>Is the derivative argument enough to prove the more general inequality $\log(x+c)\leqslant x$?</p>
<p>Thanks in advance!</p>
|
fleablood
| 280,126 |
<p>"However is it true that log(x+c)⩽x?"</p>
<p>Let $x =1$ and $c = 500,000,000$ then is $\log 500,000,001 \le 1$?</p>
<p>It is true that if $\log (x+c) \le x\iff x+c \le e^x$ and it is true that $\frac {d(x+c)}{dx} = 1 \le \frac{d e^x}{dx} = e^x$ if $x > 0$. So, for positive $x$ we have that $e^x$ increases faster than $x + c$.</p>
<p>But it's not enough that a function increases (has a greater derivative) faster than another function to ensure it is always greater. The function must also have a greater initial value.</p>
<p>If $0 + c > e^0 = 1$ then there will but some $(b,d)$ where $x \in (b,d)$ will imply $x + c > e^x$. ($b,d$ will be the two solutions to $e^x - x = c$. It's easy to convince ourselves that if $c > 1$ then there are exactly two such points where $b < 0 < d$. And that if $x \ge d$ then $\log (x + c) \le x$.)</p>
|
2,029,538 |
<blockquote>
<p>A function $f(x)$, where $x$ is a real number , is defined implicitly by the following formula:
$$f(x)=x-\int^{\frac{\pi}{2}}_0f(x)\sin(x)dx$$
Find the explicit function for $f(x)$ in its simplest form.</p>
</blockquote>
<p>This question appeared in the recent New Zealand Qualifications Authority 2016 Scholarship Calculus examination.</p>
<hr>
<p>What I have done</p>
<p>Let $f(x)=y$</p>
<p>$$f(x)=x-\int^{\frac{\pi}{2}}_0f(x)\sin(x)dx \Rightarrow y=x-\int^{\frac{\pi}{2}}_0y\sin(x)dx$$</p>
<p>$$ y=x-\int^{\frac{\pi}{2}}_0y\sin(x)dx \Leftrightarrow \int^{\frac{\pi}{2}}_0y\sin(x)dx=x-y$$</p>
<p>Consider the integral</p>
<p>$$ \int^{\frac{\pi}{2}}_0y\sin(x)dx$$</p>
<p>Let $u=y\Rightarrow du=dy$ and $dv= \sin(x) \Rightarrow v=-\cos(x)$</p>
<p>$$ \int^{\frac{\pi}{2}}_0y\sin(x)dx = \left[y\sin(x) \right]^{\frac{\pi}{2}}_0+\int^{\frac{\pi}{2}}_0\cos(x) dydx$$ </p>
<p>How can I continue?</p>
|
David
| 119,775 |
<p><strong>Hint</strong>. The integral in your equation is a <strong>definite</strong> integral and evaluates to a constant. Therefore
$$f(x)=x-c$$
where $c$ is a constant satisfying
$$c=\int_0^{\pi/2} (x-c)\sin x\,dx\ .$$
Evaluate the integral, then solve the equation to find $c$.</p>
<p>Answer:</p>
<blockquote class="spoiler">
<p> $c=\frac12$.</p>
</blockquote>
|
4,233,619 |
<p>Consider all natural numbers whose decimal expansion has only the even digits <span class="math-container">$0,2,4,6,8$</span>. Suppose these are arranged in increasing order. If <span class="math-container">$a_n$</span> denotes the <span class="math-container">$n$</span>-th number in this sequence then the value of the limit: <span class="math-container">$\displaystyle \lim_{n\to \infty}\frac{\log a_n}{\log n}=$</span></p>
<p>(a) <span class="math-container">$0$</span>.</p>
<p>(b) <span class="math-container">$\log_5 10$</span></p>
<p>(c) <span class="math-container">$\log_210$</span></p>
<p>(d) <span class="math-container">$2$</span></p>
<p>I observe that the sequence <span class="math-container">$\{a_n\}=\{2,4,6,8,20,22,24,26,28,40,42,44,46,48,60,62,64,68,80,82,84,86,88,200,202,204,206,208,\cdots\}$</span>.</p>
<p>I am not getting the explicit formula for <span class="math-container">$a_n$</span>, but it shows that the growth is exponential. How to find the exact value of the limit ?</p>
<p>Also I found that <span class="math-container">$a_n \ge n$</span> always, from which we get the required limit is <span class="math-container">$\ge 1$</span>. So option (a) is incorrect.</p>
<p>Any hint. please</p>
|
Lalit Tolani
| 913,165 |
<p>Note that <span class="math-container">$y^z$</span> is one number. To avoid confusion you can let it equal to <span class="math-container">$a$</span></p>
<p>Therefore <span class="math-container">$ \displaystyle\log x^{(y^{z})}=\log x^a=a\log x=y^z\log x$</span></p>
<p>Also <span class="math-container">$yz\log x=\log x^{yz}\neq \log x^{y^{z}}$</span></p>
|
2,049,777 |
<p>$2.$ Find the dimensions of </p>
<p>(a) the space of all vectors in $R^n$ whose components add to zero;</p>
<p>(c) the space of all solutions to $\frac{d^2y(t)}{dt^2} −3 \frac{dy(t)}{dt} +2y(t) = 0$. </p>
<p>for (a) Im pretty sure that the dimension is $n-1$ but people seem to differ, I thought it is $n-1$ since there is a condition of components adding to zero, but I can't make up with the basis could someone help??
and for (c) its a differential equation, Ive dealt with polynomial matrixes, but this is my first time to encounter one that involves differential factor does I really am curious about the basis of (c) since the dimension will equal to the rank of the basis right??</p>
|
Community
| -1 |
<p>For first choose (0,0,...,0,1,0,...0,-1) where 1 is in the i th place i=1 to n-1.
For the second the general solution is a.e^2x +b e^x. So the dimension is 2. Since e^2x and e^x are linearly independent functions.</p>
|
181,321 |
<p>Assuming that <code>a.b=z</code>,where <code>b={b1,b2,b3}</code> and <code>z=x b1+y b2+z b3</code>, obviously <code>a={x,y,z}</code>, how to realize it by mathematica?</p>
|
kglr
| 125 |
<p>If <code>a.{b1,b2,b3} == x b1+y b2+z b3</code> <em><strong>for all</strong></em> <code>{b1,b2,b3}</code>:</p>
<pre><code>Reduce[ForAll[{b1, b2, b3}, Dot[{a1, a2, a3}, {b1, b2, b3}] == x b1 + y b2 + z b3]]
</code></pre>
<blockquote>
<p>a3 == z && a2 == y && a1 == x </p>
</blockquote>
<p>or</p>
<pre><code>SolveAlways[ Dot[{a1, a2, a3}, {b1, b2, b3}] == x b1 + y b2 + z b3, {b1, b2, b3}]
</code></pre>
<blockquote>
<p>{{a1 -> x, a2 -> y, a3 -> z}}</p>
</blockquote>
|
2,348,811 |
<p>Whenever I go through the big pile of socks that just went through the laundry, and have to find the matching pairs, I usually do this like I am a simple automaton:</p>
<p>I randomly pick a sock, and see if it matches any of the single socks I picked out earlier and that haven't found a match yet. If there is a match, I will fold the two socks together and put them in the 'done' pile, otherwise I will add the single sock to the 'no match yet' pile of single socks, and pick out another random sock.</p>
<p>So, as I was doing this last night, I started thinking about this, and figured that the following would be true: The 'no match yet' pile can be expected to slowly grow, up to some point somewhere in the 'middle' of the process, after which the pile will gradually shrink, and eventually go down back to $0$. In fact, my intuition is that the expected number of loose socks as a function of the number of socks picked so far, is a symmetric function, with the maximum being when I have picked half of the socks.</p>
<p>So, my questions are:</p>
<p>With $n$ pairs of socks, what is the expected number of loose socks that are in my 'no match yet' pile after having picked $k$ socks?</p>
<p>Is it true that this function is a symmetric function, and that the maximum is for $k=n$? (if so, I figure there must be a conceptual way of looking at the problem that makes this <em>immediately</em> clear, without using any formulas ... what is that way? Is it just that I can think of reversing the process?)</p>
<p>Of course, this is all assuming there are $n$ pairs of socks total, and that there are no single socks in the original pile, and while this is something that <em>never</em> seems to apply to the pile of socks coming through my actual laundry, let's assume for the sake of mathematical simplicity that there really just are $n$ pairs of socks.</p>
|
lulu
| 252,071 |
<p>The expected number can be computed via Linearity of Expectation. Let $E[n,k]$ denote the answer and let $\{X_i\}_{i=1}^n$ denote the indicator variable for the $i^{th}$ pair. Thus $X_i=1$ if exactly one member of the $i^{th}$ pair has been chosen in your $k$ trials, and $X_i=0$ otherwise. It is easy to see that $$E[X_i]=2\times \frac k{2n}\times \left(1-\frac {k-1}{2n-1}\right)$$ from which it follows that $$E[n,k]=E\left[\sum X_i\right] =\sum E[X_i]= k\times \left(1-\frac {k-1}{2n-1}\right)$$</p>
<p>Sanity check: $k=1\implies E[n,1]=1$ as it should. Also $k=2n\implies E[n,2n]=0$ as it should.</p>
<p>Remark: it is easily seen that this function is maximized with $k=n$, confirming your intuition. Also the expression can be written as $$E[n,k]=\frac {k(2n-k)}{2n-1}$$ which is symmetric under the exchange of $k,2n - k$ also in line with your expectations.</p>
<p>Remark: more strongly, it is clear that at any time the number of unmatched socks in one pile is the same as the number in the other pile (indeed it's exactly the same pairs of socks which are split between the piles). That provides clear justification for the symmetry.</p>
|
2,348,811 |
<p>Whenever I go through the big pile of socks that just went through the laundry, and have to find the matching pairs, I usually do this like I am a simple automaton:</p>
<p>I randomly pick a sock, and see if it matches any of the single socks I picked out earlier and that haven't found a match yet. If there is a match, I will fold the two socks together and put them in the 'done' pile, otherwise I will add the single sock to the 'no match yet' pile of single socks, and pick out another random sock.</p>
<p>So, as I was doing this last night, I started thinking about this, and figured that the following would be true: The 'no match yet' pile can be expected to slowly grow, up to some point somewhere in the 'middle' of the process, after which the pile will gradually shrink, and eventually go down back to $0$. In fact, my intuition is that the expected number of loose socks as a function of the number of socks picked so far, is a symmetric function, with the maximum being when I have picked half of the socks.</p>
<p>So, my questions are:</p>
<p>With $n$ pairs of socks, what is the expected number of loose socks that are in my 'no match yet' pile after having picked $k$ socks?</p>
<p>Is it true that this function is a symmetric function, and that the maximum is for $k=n$? (if so, I figure there must be a conceptual way of looking at the problem that makes this <em>immediately</em> clear, without using any formulas ... what is that way? Is it just that I can think of reversing the process?)</p>
<p>Of course, this is all assuming there are $n$ pairs of socks total, and that there are no single socks in the original pile, and while this is something that <em>never</em> seems to apply to the pile of socks coming through my actual laundry, let's assume for the sake of mathematical simplicity that there really just are $n$ pairs of socks.</p>
|
HelloGoodbye
| 119,068 |
<p>If you let your function that gives the expected number of socks in your 'no match yet' pile take $k$ and $2n-k$ as arguments, i.e.</p>
<p>$$f\,=\,f(k,\,2n-k),$$</p>
<p>it will be <a href="https://en.wikipedia.org/wiki/Symmetric_function" rel="nofollow noreferrer">symmetric</a>. The 'no match yet' pile is just the subset of all $k$ socks that have been picked that have their matching sock among all $2n-k$ socks that have not been picked yet.</p>
<p>Therefore, we can make the following reinterpretation of $f$:</p>
<blockquote>
<p>Out of $n$ pairs of socks, $f(a,\,b)$, where $a+b=2n$, is the expected number of pairs of socks for which the two socks in the pair end up in different piles when all $2n$ socks are randomly divided into two piles of sizes $a$ and $b$, respectively.</p>
</blockquote>
<p>Since the piles don't have any order, the order of the arguments to $f$, which are just the sizes of the piles, doesn't matter. Hence, $f$ is symmetric.</p>
|
3,946,974 |
<blockquote>
<p>To prove:<span class="math-container">$$\text{If } A_1\subseteq A_2 \subseteq ... \subseteq A_n\text{ ,then } \bigcup_{i=1}^n A_i = A_n$$</span> using the axioms of ZFC Set Theory.</p>
</blockquote>
<p>Honestly, this statement is <strong>very obvious</strong>, but I do not want to take that for granted. How can I prove this from the axioms, especially the axiom of unions?</p>
<p>I'm fairly new to ZFC Set Theory, so I don't know where to start. I do know that the Axiom of Unions (and Specification) gives rise to the definition of <span class="math-container">$\bigcup_{i=1}^n A_i$</span> for sets <span class="math-container">$A_1,...,A_n$</span> though. The uniqueness follows from the Axiom of Extensionality.</p>
<p><em>My thoughts:</em><br>
We already know that the union exists, and it is defined such that for every element <span class="math-container">$x\in \bigcup_{i=1}^n A_i$</span>, there exists at least one set <span class="math-container">$A_j$</span> such that <span class="math-container">$x\in A_j$</span>. What's next?</p>
<p>Thank you!</p>
<p><strong>Edit:</strong><br>
Induction works if the chain is finite or countably infinite. What if the chain length is <strong>uncountable</strong>? Does this result still hold, and if yes, how do we prove it?</p>
<p><strong>Edit 2:</strong><br>
So now I wish to prove (or disprove) the following using the axioms of ZFC Set Theory:</p>
<blockquote>
<p>Consider an infinite set <span class="math-container">$A$</span>, and an index set <span class="math-container">$I$</span> such that <span class="math-container">$$A_1\subseteq A_2\subseteq ... \subseteq A_i \subseteq ... \subseteq A$$</span>
where <span class="math-container">$i\in I$</span>. The index set <span class="math-container">$I$</span> is uncountable. We have <span class="math-container">$$\bigcup_{i\in I}A_i = A$$</span></p>
</blockquote>
|
Luiz Cordeiro
| 58,818 |
<p>To be completely formal, you'd could first prove by induction that <span class="math-container">$A_i\subseteq A_n$</span> for every <span class="math-container">$i\leq n$</span>. Then prove that <span class="math-container">$\bigcup_{i=1}^n A_i\subseteq A_n$</span> and that <span class="math-container">$A_n\subseteq\bigcup_{i=1}^n A_i$</span>. This last one should be trivial by definition of the union.</p>
<p>As for the first one, if <span class="math-container">$x\in \bigcup_{i=1}^nA_i$</span>, then <span class="math-container">$x\in A_j$</span> for some <span class="math-container">$j$</span>, and then we know (from the proof by induction) that <span class="math-container">$x\in A_n$</span>.</p>
|
1,691,825 |
<p>My textbook goes from</p>
<p>$$\frac{\left( \frac{6\ln^22x}{2x} \right)}{\left(\frac{3}{2\sqrt{x}}\right)}$$</p>
<p>to:</p>
<p>$$\frac{6\ln^22x}{3\sqrt{x}}$$</p>
<p>I don't see how this is right. Could anyone explain?</p>
|
Martigan
| 146,393 |
<p>Well...</p>
<p>$\dfrac ab=a\times \dfrac 1b$</p>
<p>In your case $a=\frac{6\ln^22x}{2x}$ and $b=\frac{3}{2\sqrt{x}}$</p>
<p>Then... $\dfrac 1b=\dfrac{2\sqrt{x}}{3}$.</p>
<p>Can you go from here?</p>
|
2,033,790 |
<p>How do you prove the sequence $x_n = (\frac{n}{2})^n$ diverges?</p>
<p>Here is my attempt: </p>
<p>Suppose $x_n \to L$. This means $(\forall \epsilon > 0) (\exists N \in \mathbb{N}) (\forall n>N)|x_n-L| < \epsilon$</p>
<p>Assume $n > N$</p>
<p>Then $|x_n-L| < \epsilon$</p>
<p>$|(\frac{n}{2})^n-L| < \epsilon$</p>
<p>$|(\frac{n}{2})^n-\frac{2L}{2}| < \epsilon$</p>
<p>At this point I am stuck. I'm trying to arrive to a contradiction, but I'm unable to.</p>
|
Retired account
| 372,313 |
<p>Any sequence which is unbounded is divergent. Note that $(n/2)^{n}>n/2$ for $n>2$. Then if we pick any M, $n = 2M$ will have $(n/2)^{n}>n/2=M$. Therefore, the sequence is unbounded and divergent.</p>
|
33,303 |
<p>Is it possible to attach certain pieces of code to certain controls in a Manipulate? For example, consider the following Manipulate</p>
<pre><code>Manipulate[
data = Table[function[x], {x, -Pi*10, Pi*10, Pi/1000}];
ListPlot[{x, data}, PlotRange -> {{start, stop}, Automatic}]
, {function, {Sin, Cos, Tan}}
, {start, 1, Length[data]}
, {{stop, 300}, 1, Length[data]}
]
</code></pre>
<p>Generation of the data is expensive but it only needs to be done if I change function. So, I'd like the line</p>
<pre><code>data = Table[function[x], {x, -Pi*10, Pi*10, Pi/1000}];
</code></pre>
<p>to only run when I change the function control...i.e. I want to attach that line of code to the 'function' control. As it stands, the data is generated when I move the plot range too which is not what I want. </p>
<p><img src="https://i.stack.imgur.com/OHJeN.png" alt="enter image description here"></p>
|
Nasser
| 70 |
<p>This is good reason to use the second argument of dynamics.</p>
<p><img src="https://i.stack.imgur.com/ezhK3.gif" alt="enter image description here"></p>
<pre><code>Manipulate[
function; (*just to allow tracking, since not explicity in the command*)
ListLinePlot[data, PlotRange -> {{start, stop}, Automatic},
ImageSize -> 300, ImagePadding -> 30, Frame -> True,
FrameLabel -> {{None, None}, {"x", function}}
],
Grid[
{
{"function", SetterBar[Dynamic[function, {function = #;
data = Table[function[x], {x, -Pi*10, Pi*10, Pi/10}]} &], {Sin, Cos, Tan}]
},
{"start", Manipulator[Dynamic[start, {start = #} &], {1, Dynamic@Length@data, 1}],
Dynamic[start]
},
{"stop",Manipulator[Dynamic[stop, {stop = #} &], {300, Dynamic@Length@data, 1}],
Dynamic[stop]
}
}],
{{start, 1}, None},
{{stop, 300}, None},
{{data, Table[Sin[x], {x, -Pi*10, Pi*10, Pi/10}]}, None},
{{function, Sin}, None},
{{x, Symbol}, None},
TrackedSymbols :> {function, start, stop}
]
</code></pre>
|
2,082,815 |
<p>Find the $100^{th}$ power of the matrix $\left( \begin{matrix} 1& 1\\ -2& 4\end{matrix} \right)$.</p>
<p>Can you give a hint/method?</p>
|
kub0x
| 309,863 |
<p>Since $A=PDP^{-1}$ then $A^{100}=PD^{100}P^{-1}$</p>
<p>Then we know that $M=\begin{bmatrix} 1& 1\\ -2& 4\end{bmatrix}$ has eigenvalues $\lambda=(3,2)$ and eigenvector matrix P=$\begin{bmatrix} 1&1\\2&1\end{bmatrix}$ therefore $D=\begin{bmatrix} 3&0\\0&2\end{bmatrix}$</p>
<p>Now apply $A^{100}=PD^{100}P^{-1}$</p>
<p>$D^{100}=\begin{bmatrix} 3^{100}&0\\0&2^{100}\end{bmatrix}$</p>
<p>$P^{-1}=\begin{bmatrix} -1&1\\2&-1 \end{bmatrix}$</p>
<p>Then $M^{100}=(-515377520732011328501159929309162469708701111249,
515377520732011329768810529537391871205404316625,
-1030755041464022659537621059074783742410808633250,
1030755041464022660805271659303013143907511838626)$</p>
|
2,082,815 |
<p>Find the $100^{th}$ power of the matrix $\left( \begin{matrix} 1& 1\\ -2& 4\end{matrix} \right)$.</p>
<p>Can you give a hint/method?</p>
|
Community
| -1 |
<p>Let $$P= \begin {pmatrix} 1& 1\\ -2 & 4 \end {pmatrix} $$The key is to write the matrix $P $ in terms of a diagonal matrix $D $. Thus we wish to write $$P=ADA^{-1} $$ where $D $ is a diagonal matrix. Thus, $$P^n =AD^n A^{-1} $$ $D^n $ being very easy to compute, we find the values of $D $ and $A $ using eigenvalues and eigenvectors. </p>
<hr>
<p>For more, you can see <a href="https://math.stackexchange.com/questions/735532/how-to-compute-the-n-textrmth-power-of-a-matrix">here</a>. Hope it helps.</p>
|
47,561 |
<p>The Hilbert matrix is the square matrix given by</p>
<p>$$H_{ij}=\frac{1}{i+j-1}$$</p>
<p>Wikipedia states that its inverse is given by</p>
<p>$$(H^{-1})_{ij} = (-1)^{i+j}(i+j-1) {{n+i-1}\choose{n-j}}{{n+j-1}\choose{n-i}}{{i+j-2}\choose{i-1}}^2$$</p>
<p>It follows that the entries in the inverse matrix are all integers.</p>
<p>I was wondering if there is a way to prove that its inverse is an integer matrix without using the formula above.</p>
<p>Also, how would one go about proving the explicit formula for the inverse? Wikipedia refers me to a paper by Choi, but it only includes a brief sketch of the proof.</p>
|
Henricus V.
| 84,447 |
<p>(Note: The matrix elements are indexed from $0$. To avoid confusion, I will not index into a matrix without brackets)</p>
<p>Here is a naive approach to this problem, without hindsight of the already-derived formula.</p>
<blockquote>
<p>Lemma:
If $M \in \mathbb F^{n \times n}$ is an invertible matrix and $(\alpha_i),(\beta_i)$ are non-zero elements of $\mathbb F$, then the matrix formed by $$N := \left[(M)_{i,j}\alpha_i\beta_j\right]$$
is invertible, with its inverse given by
$$N^{-1} = \left[(M^{-1})_{i,j}\beta_i^{-1}\alpha_j^{-1}\right]$$</p>
</blockquote>
<hr>
<blockquote>
<p>Let $H_n$ be the $n$-th Hilbert matrix, given by
$$ H_n = \left[\frac1{i+j+1}\right]_{i,j}
$$
Then $H_n$ is invertible and every element of $H_n^{-1}$ is an integer.</p>
</blockquote>
<p><em>Proof:</em> The $H_n$s are nested within each other:
$$ H_{n+1} = \begin{bmatrix} H_n & u \\ u^\intercal & 1/(2n+1)\end{bmatrix}
$$
Define $V_n := H_n - (2n+1)uu^\intercal$. Then
$$\begin{align*}
V_n &= \left[\frac1{i+j+1} - \frac{2n+1}{(n+1+i)(n+1+j)}\right] \\
&= \left[\frac{(n-i)(n-j)}{(i+j+1)(n+1+i)(n+1+j)}\right] \\
&= \left[(H_n)_{i,j} \frac{n -i}{n+1+i} \frac{n-j}{n+1+j}\right]
\end{align*}$$
It is well-known that if $V_n$ is invertible, then so is $H_{n+1}$, but the invertibility of $V_n$ is intrinsically linked to $H_n$, so an inductive argument gives that $H_n$ is invertible.</p>
<p>The inverse of $H_{n+1}$ is given by Blockwise inversion formula:
$$ H_{n+1}^{-1} = \begin{bmatrix}
V_n^{-1} & -(2n+1)V_n^{-1}u \\ -(2n+1)u^\intercal V_n^{-1} & (2n+1) + (2n+1)^2u^\intercal V_n^{-1} u\end{bmatrix}
$$
By the Lemma above, the inverse of $V_n$ is given by
$$ V_n^{-1} = \left[(H_n^{-1}) \frac{(n+1+i)(n+1+j)}{(n-i)(n-j)} \right]
$$
Since $V_n^{-1}$ is a constituent of $H_{n+1}$, this can be used to telescope the entries of $H_n^{-1}$.</p>
<p>By the symmetry of $H_n,V_n,H_n^{-1}$, it suffices to consider the upper triangular portion of $H_n$. We can see that eventually a "interior" element of $H_n^{-1}$ "lands" on an edge element of some $H_n^{-1}$: Let $i,j<n$ with $i \leq j$. Then
$$\begin{align*}
(H_{n+1}^{-1})_{i,j} &= (V_n^{-1})_{i,j}
= (H_{j+1}^{-1})_{i,j} \frac{(2j+2)(i+j+2)}{1(j+1-i)} \cdots \frac{(n+1+i)(n+1+j)}{(n-i)(n-j)} \\
&= (H_{j+1}^{-1})_{i,j} \frac{(n+1+i)!(n+1+j)!}{(2j+1)!(i+j+1)!} \frac{(j-i)!}{(n-i)!(n-j)!} \\
&= (H_{j+1}^{-1})_{i,j} \frac{(j-i)!}{(2j+1)!(i+j+1)!} \frac{(n+1+i)!(n+1+j)!}{(n-i)!(n-j)!(2i+1)!(2j+1)!} (2i+1)!(2j+1)! \\
&= (H_{j+1}^{-1})_{i,j} \frac{(j-i)!(2i+1)!}{(i+j+1)!} \binom{n+1+i}{n-i}\binom{n+1+j}{n-j} \\
&= (H_{j+1}^{-1})_{i,j} \frac{1}{\binom{i+j+1}{2i+1}} \binom{n+1+i}{n-i}\binom{n+1+j}{n-j}
\end{align*}$$
This is a very strong hint on what the inductive hypothesis should be.</p>
<p><strong>Inductive Hypothesis</strong> P(n):</p>
<blockquote>
<p>Every element of $H_n^{-1}$ is an integer, and $$\binom{n+i}{n-1-i}\binom{n+j}{n-1-j} \mid (H_n^{-1})_{i,j}$$
where $i,j \in \{0,\dots,n-1\}$.</p>
</blockquote>
<p>Define the matrix
$$ A_n := \left[\frac{(H_n^{-1})_{i,j}}{\binom{n+i}{n-1-i} \binom{n+j}{n-1-j}}\right]_{i,j}
$$
$P(n)$ is just the statement that $A_n$ has integer entries. Evidently $P(1)$ holds. Suppose $P(n)$ holds for some $n$. We shall examine the validity of $P(n+1)$.</p>
<p>There are three cases:</p>
<ol>
<li>Suppose $i,j < n$. This corresponds to a "interior" element $(H_{n+1}^{-1})_{i,j}$ of $H_{n+1}^{-1}$.</li>
</ol>
<p>We have that
$$\begin{align*}
(H_{n+1}^{-1})_{i,j} &= (V_n^{-1})_{i,j} = (H_n^{-1})_{i,j} \frac{(n+1+i)(n+1+j)}{(n-i)(n-j)} \\
&= (A_n)_{i,j} \binom{n+i}{n-1-i}\binom{n+j}{n-1-j} \frac{(n+1+i)(n+1+j)}{(n-i)(n-j)} \\
&= (A_n)_{i,j} \binom{n+1+i}{n-i} \binom{n+1+j}{n-j}
\end{align*}$$
Establishing the interior case. (It can also be inferred from this equation that $A_n$'s are nested within each other.)</p>
<ol start="2">
<li><p>The edge case: Consider the edge $-(2n+1)V_n^{-1}u$. Let $i \in \{0,\dots,n-1\}$ The $i$th element $(H_{n+1}^{-1})_{i,n}$ on the edge is given by the dot product
$$\begin{align*}
-(2n+1)(V_n^{-1}u)_i
&= -(2n+1) \sum_{j=0}^{n-1} (V_n^{-1})_{i,j} (u)_j \\
&= -(2n+1) \sum_{j=0}^{n-1} (H_n^{-1})_{i,j} \frac{(n+1+i)(n+1+j)}{(n-i)(n-j)} \frac1{n+1+j} \\
&= -\frac{(n+1+i)(2n+1)}{n-i} \sum_{j=0}^{n-1} (H_n^{-1})_{i,j} \frac1{n-j} \\
&= -\frac{(n+1+i)(2n+1)}{n-i} \sum_{j=0}^{n-1} (A_n)_{i,j} \binom{n+i}{n-1-i}\binom{n+j}{n-1-j} \frac1{n-j} \\
&= -\frac{n+1+i}{n-i}\binom{n+i}{n-1-i} \sum_{j=0}^{n-1} (A_n)_{i,j} \frac{2n+1}{n-j} \binom{n+j}{n-1-j}\\
\end{align*}$$
<a href="https://math.stackexchange.com/questions/2622874/product-of-binomial-coefficient-and-fraction-is-integer">Here</a> people proved that the multiplier in the sum is an integer, so it remains to show that
$$ \frac{n+1+i}{n-i}\binom{n+i}{n-1-i} = \binom{n+i+1}{n-i}
= \binom{n+i+1}{n-i}\binom{n+n+1}{n-n}
$$
is also an integer.</p></li>
<li><p>The corner element $(H_{n+1}^{-1})_{n,n}$ is given by the quadratic form:
$$\begin{align*}
(H_{n+1}^{-1})_{n,n} - (2n+1)
&= (2n+1)^2 \sum_{i=0}^{n-1}\sum_{j=0}^{n-1} \frac{(H_n^{-1})_{i,j}}{(n-i)(n-j)} \\
&= \sum_{i=0}^{n-1}\sum_{j=0}^{n-1} (A_n)_{i,j} \frac{(2n+1)(2n+1)}{(n-i)(n-j)}\binom{n+i}{n-1-i}\binom{n+j}{n-1-j}
\end{align*}$$
The last sum is a sum integers, and hence an integer. Trivially it holds that
$$ \binom{n+1+n}{n-n}\binom{n+1+n}{n-n} \mid (H_{n+1}^{-1})_{n,n}
$$
With all three cases established, the proof is complete by induction.</p></li>
</ol>
|
2,983,199 |
<p>I am looking for an example demonstrating that <span class="math-container">$\lim\inf x_n+\lim \inf y_n<\lim \inf(x_n+y_n)$</span> but for the life of me i can't find one. any suggestions?</p>
|
ZAF
| 609,023 |
<p>Let <span class="math-container">$(x_{n})_{n}$</span> such that <span class="math-container">$x_{2n} = 1$</span> and <span class="math-container">$x_{2n - 1} = -1$</span></p>
<p>Let <span class="math-container">$(y_{n})_{n}$</span> such that <span class="math-container">$y_{2n} = -1$</span> and <span class="math-container">$y_{2n -1 } = 1$</span></p>
<p>Then <span class="math-container">$\liminf x_{n} = -1$</span>, <span class="math-container">$\liminf y_{n} = -1$</span></p>
<p>And <span class="math-container">$\liminf (x_{n} + y_{n}) = 0$</span></p>
|
914,440 |
<p>By the definition of topology, I feel topology is just a principle to define "open sets" on a space(in other words, just a tool to expand the conception of open sets so that we can get some new forms of open sets.) But I think in the practical cases, we just considered Euclidean space most and the traditional form open set in Euclidean space works pretty well. And these new form open sets seem to have no common use. So why we need to expand the conception of the open sets? What's the motivation behind?Thanks.</p>
|
Theon Alexander
| 165,460 |
<p>For starters, we have things defined abstractly (at least in principle) such as manifolds. These are defined through charts, which endow your manifold with a topology.</p>
<p>The important tool of partitions of unity (which reduces Stokes's theorem and others to a toy case) works due to the 2nd Countability Axiom (see e.g. Warner, Foundations of Differential Manifolds and Lie groups).</p>
<p>There are important results in Commutative Algebra such as finiteness of irreducible components of the spectrum of a noetherian commutative ring: topology furnishes a proof of an algebraic result (see Atiyah & Macdonald, Introduction to Commutative Algebra).</p>
<p>Also, the topology is kind of qualitative, i.e. you are not considering a particular metric only, but only those equivalent to it (i.e. those which yield the same topology).</p>
<p>There are many more things one could say, but we'll leave it at that until further comments appear.</p>
<hr>
<p>Where you can see a nice interplay between topology and analysis is in the Gelfand-Naimark theorem: the maximal rings of the ring of (real or complex) continuous functions over a compact topological space $K$ is homeomorphic to $K$ itself.</p>
<p>Again, a good reference is Atiyah-Macdonald (it's in the Exercises section, if I remember rightly).</p>
|
142,507 |
<p>Let $\psi(x) := \sum_{n\leq x} \Lambda(n)$ where $\Lambda(n)$ is the Von-Mangoldt function.
I want to show that if $$ \lim_{x \rightarrow \infty} \frac{\psi(x)}{x} =1 $$ then also $$\lim_{x\rightarrow \infty} \frac{\pi(x) \log x }{x}=1.$$</p>
<p>I tried to play a little bit with $\psi$, what I want to show is that:</p>
<p>$$\left| \frac{\pi(x) \log x}{x} -1 \right| \leq \left| \frac{\psi(x)}{x} -1 \right| \rightarrow 0$$</p>
<p>So I tried to develop $\psi$ a little bit, but I got astray.</p>
<p>So I have
$$ \frac{\psi(x)}{x} -1 = \sum_{p^k \leq x , k \geq 1} \frac{\log p}{x} -1 = \frac{1}{x}\left(\sum_{p\leq x} \log p + \sum_{p^2\leq x} \log p + ...+ \sum_{p^k \leq x, p^{k+1} >x} \log p \right) -1 $$
and I want to estimate its aboslute value from below, but I don't have any idea?</p>
<p>Any hints?</p>
<p>Thanks.</p>
|
Syd
| 27,920 |
<p>Consider each set as {0,1} Map the elements of the product onto the closed unit interval by mapping each to a binary number. For example, [0,1,0,1,0,1,..] maps to .01010101... [I'm assuming the Axiom of Choice.]</p>
<p>The image is uncountable with countably many duplications, hence B must have the same cardinality.</p>
|
2,947,911 |
<p>I have a homework with this equation:
<span class="math-container">$$(x+2y)dx + ydy = 0$$</span></p>
<p>However I have no idea how to solve it.
I tried couple things:</p>
<ol>
<li>Is it linear equation, or does it have "standard form" : <span class="math-container">$\frac{dy}{dx}+\frac{x}{y} + 2 = 0$</span>. Well <span class="math-container">$\frac{1}{y}$</span> is not a standard form.</li>
<li>Is it exact? <span class="math-container">$\frac{∂M}{∂y}=2$</span> and <span class="math-container">$\frac{∂N}{∂x}=0$</span>. I tried using <span class="math-container">$\mu(y)=e^{\int\frac{2}{y}dy}=y^2$</span> to make it exact, but that didn't work</li>
<li>Is it homogeneous? In theory yes as <span class="math-container">$M(tx,ty)=tx+2ty=t(x+2y)=tM()$</span> and <span class="math-container">$N(tx,ty)=ty=tN()$</span> and I reached up to <span class="math-container">$$x(dx)+2ux(dx)+u^2x(dx)+x^2u(du)=0$$</span> But I can't separate it.</li>
</ol>
<p>Tried searching online and going with subsitution (method #3) was suggested, but they never went beyond that point. Maybe it's just too late and I'm not thinking straight.</p>
<p>If you could point me in the right direction, or list steps instead of giving solution if would be great.</p>
|
Joseph Vidal-Rosset
| 62,171 |
<p><a href="https://www.vidal-rosset.net/sequent_calculus_prover_with_antisequents_for_classical_propositional_logic.html" rel="nofollow noreferrer">This prover is helpful to reply to this question. See example 3</a></p>
|
1,038,152 |
<p>Let $B \subseteq \mathbb{R}_{+}$ such that B is non-empty.
consider $B^{-1} = \left \{b^{-1} : b\in B \right \}$.<br>
Show that if $B^{-1}$ is unbounded from above, then $\inf\left(B\right)=0$</p>
<p>How can i prove that? tnx!</p>
|
Matt Samuel
| 187,867 |
<p>The long exact sequence is not really helpful here. What we need to do is look at chain groups. Any two 0-chains in $A$ of the form $1f$ where $f$ is a constant map are homologous and map to a generator when we pass to homology. 0 chains of the form $1(i\cdot f)$ generate the direct summand in $H_0(X)$ corresponding to the path component containing $A$. Therefore a generator of $H_0(A)$ is mapped to a nonzero element in $H_0(X)$, so since the groups are free abelian the map is injective.</p>
|
4,272,214 |
<h2>The Equation</h2>
<p>How can I analytically show that there are <strong>no real solutions</strong> for <span class="math-container">$\sqrt[3]{x-3}+\sqrt[3]{1-x}=1$</span>?</p>
<h2>My attempt</h2>
<p>With <span class="math-container">$u = -x+2$</span></p>
<p><span class="math-container">$\sqrt[3]{u-1}-\sqrt[3]{u+1}=1$</span></p>
<p>Raising to the power of <span class="math-container">$3$</span></p>
<p><span class="math-container">$$(u+1)^{2/3}(u-1)^{1/3} - (u+1)^{1/3}(u-1)^{2/3}=1\\(u+1)^{1/3}(u^2-1)^{1/3} - (u-1)^{1/3}(u^2-1)^{1/3}=1\\(u^2-1)^{1/3}\cdot\boxed{\left[(u+1)^{1/3}-(u-1)^{1/3}\right]}=1$$</span></p>
<p>Raising to the power of <span class="math-container">$3$</span>:</p>
<p><span class="math-container">$$(u^2-1)\cdot\left[3(u+1)^{1/3}(u-1)^{2/3}-3(u+1)^{2/3}(u-1)^{1/3}+2\right]=1\\(u^2-1)\cdot\left[3(u^2-1)^{1/3}(u-1)^{1/3}-3(u+1)^{1/3}(u^2-1)^{1/3}+2\right]=1$$</span></p>
<p>Thus: <span class="math-container">$(u^2-1)\cdot\left[3(u^2-1)^{1/3}\boxed{\left[(u-1)^{1/3}-(u+1)^{1/3}\right]}+2\right]=1$</span></p>
<p>And with <span class="math-container">$y = (u-1)^{1/3}-(u+1)^{1/3}$</span>, we can say that:</p>
<p><span class="math-container">$y^3=-3y(u^2-1)^{1/3}+2$</span></p>
<p>I am stuck... Any tips for this radical equation?</p>
|
Rene Schipperus
| 149,912 |
<p>Use the fact that <span class="math-container">$a^3+b^3+c^3-3abc$</span> is divisible by <span class="math-container">$a+b+c$</span>.</p>
<p>Let</p>
<p><span class="math-container">$$a=\sqrt[3]{x-3}$$</span>
<span class="math-container">$$b=\sqrt[3]{1-x}$$</span>
<span class="math-container">$$c=-1$$</span></p>
<p>Then you have <span class="math-container">$a+b+c=0$</span> so</p>
<p><span class="math-container">$$a^3+b^3+c^3=3abc$$</span> or</p>
<p><span class="math-container">$$x-3+1-x -1=-3\sqrt[3]{(x-3)(1-x)}$$</span></p>
<p>and</p>
<p><span class="math-container">$$1=\sqrt[3]{(x-3)(1-x)}$$</span></p>
<p>So</p>
<p><span class="math-container">$$(x-3)(1-x)=1$$</span></p>
<p>and you get <span class="math-container">$x=2$</span>. If you substitute you find it does not satisfy the equation.
Indeed, this show that <span class="math-container">$x-3=2-x=-1$</span></p>
<p>Now the cube roots of <span class="math-container">$-1$</span> are roots of <span class="math-container">$$x^3+1=(x+1)(x^2-x+1)$$</span> and if <span class="math-container">$\alpha$</span> and <span class="math-container">$\beta$</span> are the two non real roots, then <span class="math-container">$$\alpha+\beta =1.$$</span>
So by taking these non real roots, the equation can be solved.</p>
|
4,272,214 |
<h2>The Equation</h2>
<p>How can I analytically show that there are <strong>no real solutions</strong> for <span class="math-container">$\sqrt[3]{x-3}+\sqrt[3]{1-x}=1$</span>?</p>
<h2>My attempt</h2>
<p>With <span class="math-container">$u = -x+2$</span></p>
<p><span class="math-container">$\sqrt[3]{u-1}-\sqrt[3]{u+1}=1$</span></p>
<p>Raising to the power of <span class="math-container">$3$</span></p>
<p><span class="math-container">$$(u+1)^{2/3}(u-1)^{1/3} - (u+1)^{1/3}(u-1)^{2/3}=1\\(u+1)^{1/3}(u^2-1)^{1/3} - (u-1)^{1/3}(u^2-1)^{1/3}=1\\(u^2-1)^{1/3}\cdot\boxed{\left[(u+1)^{1/3}-(u-1)^{1/3}\right]}=1$$</span></p>
<p>Raising to the power of <span class="math-container">$3$</span>:</p>
<p><span class="math-container">$$(u^2-1)\cdot\left[3(u+1)^{1/3}(u-1)^{2/3}-3(u+1)^{2/3}(u-1)^{1/3}+2\right]=1\\(u^2-1)\cdot\left[3(u^2-1)^{1/3}(u-1)^{1/3}-3(u+1)^{1/3}(u^2-1)^{1/3}+2\right]=1$$</span></p>
<p>Thus: <span class="math-container">$(u^2-1)\cdot\left[3(u^2-1)^{1/3}\boxed{\left[(u-1)^{1/3}-(u+1)^{1/3}\right]}+2\right]=1$</span></p>
<p>And with <span class="math-container">$y = (u-1)^{1/3}-(u+1)^{1/3}$</span>, we can say that:</p>
<p><span class="math-container">$y^3=-3y(u^2-1)^{1/3}+2$</span></p>
<p>I am stuck... Any tips for this radical equation?</p>
|
user
| 505,767 |
<p>We have that by <span class="math-container">$A^3-B^3=(A-B)(A^2+AB+B^2)$</span></p>
<p><span class="math-container">$$\sqrt[3]{x-3}+\sqrt[3]{1-x}=\sqrt[3]{x-3}-\sqrt[3]{x-1}=$$</span></p>
<p><span class="math-container">$$=\frac{-2}{\sqrt[3]{(x-3)^2}+\sqrt[3]{(x-3)(x-1)}+\sqrt[3]{(x-1)^2}}<0$$</span></p>
<p>indeed the latter is true for <span class="math-container">$x= 1,2,3$</span> and for <span class="math-container">$x\neq 1,2,3$</span> by AM-GM</p>
<p><span class="math-container">$$\sqrt[3]{(x-3)^2}+\sqrt[3]{(x-1)^2} > 2\sqrt[3]{|x-3||x-1|}>0$$</span></p>
<p>and therefore</p>
<p><span class="math-container">$$\sqrt[3]{(x-3)^2}+\sqrt[3]{(x-3)(x-1)}+\sqrt[3]{(x-1)^2} >\sqrt[3]{|x-3||x-1|}>0$$</span></p>
|
103,317 |
<p>I have a 3D Plot with 3 separate functions for <code>z</code> in terms of <code>x</code> and <code>y</code>. I would like to convert this to a RegionPlot with the <code>max</code> of the three functions plotted for the variables <code>x</code> and <code>y</code>.</p>
<p>Essentially, this would be the top view of the 3D Plot, as whatever function is visible from the top is the highest and thus the maximum at that point.</p>
<p>However, I also need to add a <code>Manipulate</code> to this so it cannot merely be an image of the top view of the 3D Plot.</p>
<pre><code> Manipulate[
Plot3D[{0.1,
Piecewise[{{α - ((2 α - 1) α)/y + (
x*α^2)/(2 y ), (1 - α) + α^2/(2 y) >=
0.35}}],
Piecewise[{{α + (1 - α)^2/(
2 y), (1 - α) + (((3 α -
1) - ((Abs[1 - x]) (1 - α))) (1 - α))/(
2 y) >= 0.35}}],
Piecewise[{{α - ((2 α - 1) α)/((2 -
a) 2) + ((3 - 2 a - Abs[x - a] ) α^2)/(
2 y*(2 - a)^2), (1 - α) + α^2/(2 y (2 - a)) >=
0.35}}]} /. {a -> ((1 + z)/2), x -> 1} //
Evaluate, {α, 0, 1}, {z, 0, 1},
AxesLabel -> Automatic,
PlotRange -> Automatic,
PlotStyle -> {Black, Green, Orange, Blue}], {{y, 2,
Gamma}, 1, 5}]
</code></pre>
<p><a href="https://i.stack.imgur.com/jgFLS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jgFLS.png" alt=""></a></p>
<p>I would like a 2D Plot from this top view</p>
|
Michael E2
| 4,999 |
<p>Here's my answer <a href="https://mathematica.stackexchange.com/a/58257/4999">here</a> applied to this question:</p>
<pre><code>Manipulate[
With[{fns = {0.1,
Piecewise[{{α - ((2 α - 1) α)/y + (x*α^2)/(2 y), (1 - α) + α^2/(2 y) >= 0.35}}],
Piecewise[{{α + (1 - α)^2/(2 y),
(1 - α) + (((3 α - 1) - ((Abs[1 - x]) (1 - α))) (1 - α))/(2 y) >= 0.35}}],
Piecewise[{{α - ((2 α - 1) α)/
((2 - a) 2) + ((3 - 2 a - Abs[x - a]) α^2)/(2 y*(2 - a)^2),
(1 - α) + α^2/(2 y (2 - a)) >= 0.35}}]} /. {a -> ((1 + z)/2), x -> 1}},
ContourPlot[Ordering[fns, -1], {α, 0, 1}, {z, 0, 1},
Contours -> 1/2 + Range[Length@fns - 1],
ContourShading -> {Black, Green, Orange, Blue},
MaxRecursion -> ControlActive[1, 4]]
],
{{y, 2, Gamma}, 1, 5}]
</code></pre>
<p><img src="https://i.stack.imgur.com/4a3Zv.png" alt="Mathematica graphics"></p>
|
2,491,881 |
<p>I'm trying to factorise $(4 + 3i)z^2 + 26iz + (-4+3i)$.</p>
<p>I tried to use the quadratic formula to factorise $(4 + 3i)z^2 + 26iz + (-4+3i)$, where $b = 26i$, $a = (4 + 3i)$, $c = (-4+3i)$. This gives us the roots $z = \dfrac{-4i - 12}{25}$ and $z =-4i - 3$.</p>
<p>But $\left( z + \dfrac{4i + 12}{25} \right) \left( z + 4i + 3 \right) \not= (4 + 3i)z^2 + 26iz + (-4+3i)$. </p>
<p>So I did what we do when using the quadratic formula with real numbers and have that $a \not= 0, 1$: $(4 + 3i)\left( z + \dfrac{4i + 12}{25} \right) \left( z + 4i + 3 \right)$. But, as far as I can tell, $(4 + 3i)\left( z + \dfrac{4i + 12}{25} \right) \left( z + 4i + 3 \right) \not= (4 + 3i)z^2 + 26iz + (-4+3i)$.</p>
<p>So what is the correct way to go about this?</p>
<p>I would greatly appreciate it if people could please take the time to explain this.</p>
|
Sarvesh Ravichandran Iyer
| 316,409 |
<p>See, the definition is very simple : given a sequence in the set, there exists a convergent subsequence, which converges in the set itself. Then, the set is compact.</p>
<p>Now, what you have to start with is a sequence $x_n \in X \cup Y$. There is nothing to assume about the nature of this sequence at all. It is therefore not assumed convergent, as you have asked in the comments.</p>
<p>Now, as you have said, this sequence $x_{n}$ will have a subsequence $x_{n_k}$ contained entirely either in $X$ or $Y$, because the sequence is infinite, so if you split it in two parts depending upon whether that term is in $X$ or in $Y$, on of the parts will be infinite .Suppose it is contained in $X$ (i.e. there are infinitely many terms of $x_n$ which are contained in $X$, and these we call as $x_{n_k}$). </p>
<p>Now, $x_{n_k}$ is a subsequence of $x_n$, but it is contained in $X$. Since $X$ is compact, and $x_{n_k}$ is a sequence in $X$, there is a convergent subsequence $x_{n_{k_l}}$ of $x_{n_k}$, which converges in $X$ (to some point $x \in X$, if you like). Now, $x_{n_{k_l}}$ is a subsequence of $x_{n_k}$, which is a subsequence of $x_n$. So, $x_{n_{k_l}}$ is a convergent subsequence of $x_n$ (<em>and not $x_{n_k}$, as written in your answer</em>), which converges to a point $x \in X$, and therefore $x \in X \cup Y$. </p>
<p>Now, do the same thing as above, but instead assume that the subsequence $x_{n_k}$ is in $Y$ rather than in $X$. The argument is exactly the same, and you could write it out for practice if you like.</p>
<p>Therefore, $X \cup Y$ is compact.</p>
|
2,971,980 |
<p>Show that if <span class="math-container">$0<b<1$</span> it follows that
<span class="math-container">$$\lim_{n\to\infty}b^n=0$$</span>
I have no idea how to express <span class="math-container">$N$</span> in terms of <span class="math-container">$\varepsilon$</span>. I tried using logarithms but I don't see how to find <span class="math-container">$N$</span> from this.</p>
|
Gâteau-Gallois
| 393,828 |
<p>Welcome to MSE. What you want is, given some <span class="math-container">$\epsilon > 0$</span>, <span class="math-container">$b^N \leq \epsilon$</span>. Note that, by properties of the logarithm, and since both <span class="math-container">$b$</span> and <span class="math-container">$\epsilon$</span> are positive,
<span class="math-container">$$ b^N \leq \epsilon \Leftrightarrow N \ln(b) \leq ln(\epsilon).$$</span> </p>
<p>I'll let you write the conclusion from here (there is a little thing to be careful of). </p>
|
843,763 |
<blockquote>
<p>Let $x\in \mathbb{R}$ an irrational number. Define $X=\{nx-\lfloor nx\rfloor: n\in \mathbb{N}\}$. Prove that $X$ is dense on $[0,1)$. </p>
</blockquote>
<p>Can anyone give some hint to solve this problem? I tried contradiction but could not reach a proof.</p>
<p>I spend part of the day studying this question <a href="https://math.stackexchange.com/questions/450493/multiples-of-an-irrational-mod-1-are-dense">Positive integer multiples of an irrational mod 1 are dense</a>
and its answers. Only one answer is clear and give clues to solve the problem. This answer is the first one. However, this answer does not answer the question nor directly, nor the proof follows from this answer. </p>
<p>This answer has some mistakes, he use that $[(k_1-k_2)\alpha]=[k_1\alpha]-[k_2\alpha]$ which is not true. Consider $k_1=3, k_2=1, \alpha=\sqrt{2}$ we have $[(k_1-k_2)\alpha]=2\not= 3=[k_1\alpha]-[k_2\alpha] $. We only can assure that $[k_2\alpha]-[k_1\alpha]-1\leq [(k_2-k_1)\alpha]\leq[k_2\alpha]-[k_1\alpha]$. </p>
<p>Who answered said something interesting about additive subgroups of $\mathbb{R}$, but unfortunately the set $X=\{nx-[nx] : n\in \mathbb{N} \}$ is not a subgroup. Considering the additive subgroup $G=\langle X \rangle$, if we prove the part (a) of the link, we get that indeed $G$ is dense on $\mathbb{R}$ but we can not conclude that $X$ is dense on $[0,1)$.</p>
<p>I think this problem has not been solved.</p>
<p>Thanks!</p>
|
Adam Rubinson
| 29,156 |
<p>A pictoral representation of mm-aops's proof. I also avoid use of: "the distance from <span class="math-container">$a_n$</span> to <span class="math-container">$a_m$</span> is the same as between <span class="math-container">$a_{n-m}$</span> and <span class="math-container">$a_0 = 0$</span>."</p>
<p>Let <span class="math-container">$\ \varepsilon = \frac{1}{200}.\ $</span> Then, we split the circle up into <span class="math-container">$201$</span> arcs of equal length. Think of these arcs as "boxes" in the sense of the pigeonhole principle: then there must be <span class="math-container">$\ n,m \leq 201\ $</span> such that <span class="math-container">$\ \mid a_{n} - a_m\mid \leq \frac{1}{201} <\varepsilon.$</span></p>
<p>Suppose the first two such <span class="math-container">$\ n,m\ $</span> are <span class="math-container">$\ a_{45}\ $</span> and <span class="math-container">$\ a_{162}\ .$</span></p>
<p>Any point on the circle must lie within <span class="math-container">$\ \varepsilon=\frac{1}{200}\ $</span> distance from <span class="math-container">$\ a_{45}\ $</span> or <span class="math-container">$\ a_{162}\ $</span> or <span class="math-container">$\ a_{279}\ $</span> or <span class="math-container">$\ a_{396}\ $</span> or ...</p>
<p><a href="https://i.stack.imgur.com/T0LFv.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/T0LFv.jpg" alt="enter image description here" /></a></p>
|
2,213,528 |
<p>Combination formula is defined as:</p>
<p>$$\binom{n}{r}=\frac{n!}{(n-r)!\cdot r!}$$</p>
<p>We do $(n-r)!$ because we want the combinations of only $r$ objects given $n$. We do $r!$ again because in combinations, order does not matter so doing this gets rid of repeated results. But why? Why does $r!$ get rid of anything.</p>
<p>Any visual help is appreciated.</p>
<p>thanks</p>
|
jnez71
| 295,791 |
<p>Like others have already pointed out, the conundrum you are facing is because you are confusing probability with probability <em>density</em>.</p>
<p>Recall that we have an underlying <a href="https://en.wikipedia.org/wiki/Probability_space" rel="nofollow noreferrer">probability space</a> $(\Omega, \mathscr{F}, P)$, and $X_2:\Omega \to \mathbb{R}$. For some event $A \in \mathscr{F}$, we can express the integral relationship between probability density and probability itself as,
$$P(\{\omega \in \Omega : X_2(\omega) \in A\})\ \overset{\text{shorthand}}{=}\ P(X_2 \in A) = \int_A f_2(x_2)dx_2$$</p>
<p>For example, let the event be $A=[-10,12.4]$, then
$$P(X_2 \in [-10,12.4]) = \int_{-10}^{12.4} f_2(x_2)dx_2$$</p>
<p>If the event is $A=\{3\}$ then we might say,
$$P(X_2 \in \{3\}) = \int_{3}^{3} f_2(x_2)dx_2 = 0$$
but this does <strong>not</strong> imply that $f_2(x_2) \equiv 0$ or that $f_2(3)=0$.</p>
<p>It may ease confusion if you only put random variables (capital letters) as arguments of $P$ and only "samples" (really dummy variables, lowercase letters) as arguments of $f$. Saying $f_1(X_1)$ is weird, but $f_1(x_1)$ is not.</p>
|
1,104,163 |
<p>I have come up with the equation in the form $${{dy}\over dx} = axe^{by}$$, where a and b are arbitrary real numbers, for a project I am working on. I want to be able to find its integral and differentiation if possible. Does anyone know of a possible solution for $y$ and/or ${d^2 y}\over {dx^2}$?</p>
|
graydad
| 166,967 |
<p>I think there are some scenarios to consider if <span class="math-container">$a$</span> and or <span class="math-container">$b$</span> is equal to zero.</p>
<p><strong>Case 1:</strong> <span class="math-container">$a=0$</span></p>
<p>Then <span class="math-container">$$\frac{dy}{dx} = 0 \implies y = C$$</span></p>
<p><strong>Case 2:</strong> <span class="math-container">$a \neq 0, b=0$</span>.</p>
<p><span class="math-container">$$\frac{dy}{dx} = ax \implies y = \frac{ax^2}{2}+C$$</span></p>
<p><strong>Case 3:</strong> <span class="math-container">$a \neq 0 \neq b$</span>.</p>
<p>Then <span class="math-container">$$\frac{dy}{dx} = axe^{by} \implies e^{-by}dy = axdx \\ \implies \int e^{-by}dy = \int axdx \\ \implies \frac{-1}{b}e^{-by} = \frac{ax^2}{2}+C \\ \implies e^{-by} = \frac{-ax^2}{2b}+\tilde{C} \\ \implies y = \frac{-1}{b}\ln\left(\frac{-ax^2}{2b}+\tilde{C} \right)$$</span>
It would probably be wise to look at the solution of <span class="math-container">$y$</span> in case <span class="math-container">$3$</span> to figure out if there are other constraints on the constants <span class="math-container">$a,b,\tilde{C}$</span>. For example, if <span class="math-container">$a>0, b<0, \tilde{C}<0$</span> then you would be taking the natural log of a negative number.</p>
|
3,220,135 |
<p>How many integer numbers, <span class="math-container">$x$</span>, verify that the following</p>
<p><span class="math-container">\begin{equation*}
\frac{x^3+2x^2+9}{x^2+4x+5}
\end{equation*}</span></p>
<p>is an integer?</p>
<p>I managed to do:</p>
<p><span class="math-container">\begin{equation*}
\frac{x^3+2x^2+9}{x^2+4x+5} = x-2 + \frac{3x+19}{x^2+4x+5}
\end{equation*}</span></p>
<p>but I cannot go forward.</p>
|
nonuser
| 463,553 |
<p>Let <span class="math-container">$t=x+2$</span>, then <span class="math-container">$$t^2+1\mid 3t+13$$</span> and thus <span class="math-container">$$t^2+1\mid t(3t+13)-3(t^2+1)= 13t-3$$</span>
so <span class="math-container">$$t^2+1\mid 13(3t+13)-3(13t-3) = 178$$</span></p>
<p>So <span class="math-container">$$t^2+1\in\{1,2,89,178\}\implies t=\pm 1,0 \implies x\in\{-1,-2,-3\}$$</span> </p>
|
2,319,766 |
<p>Lets say ,I have 100 numbers(1 to 100).I have to create various combinations of 10 numbers out of these 100 numbers such that no two combinations have more than 5 numbers in common given a particular number can be used max three times.
E.g.</p>
<ol>
<li>Combination 1: 1,2,3,4,5,6,7,8,9,10</li>
<li>Combination 2: 1,2,3,4,5,11,12,13,14,15</li>
<li>Combination 3: 1,2,3,4,5,16,17,18,19,20</li>
<li>Combination 4: 6,7,8,9,10,11,12,13,14,15</li>
</ol>
<p>Here Combination 1,2,3 have numbers 1 to 5 in common whereas combination 1 and 4 have numbers 6 to 10 in common.
I am finding it difficult to understand how to approach this problem.
What would be the starting point if I have to apply this logic on N numbers.</p>
|
Community
| -1 |
<p>This is how I will model it: each o (of a total of 14) represents one animal and there are two / to divide them into three group; first group will be cats, second one dogs and the last one pigs.</p>
<p>An example demonstration:
ooo/oo/ooooooooo This is 3 cats, 2 dogs and 9 pigs.</p>
<p>You got the idea. So basically, the question is how many permutations of 14 o's and 2 /'s exist. The answer is $\frac{16!}{14!2!}$, which is "16 choose 2".</p>
<p>Notice that this model allows zero number of animals such as
/o/ooooooooooooo is 0 cats, 1 dog, 13 pigs. If you want at least 1 for each, you then assign 1 to each, leaving $14-3=11$ free assignments and you repeat the idea above as if you have 11 animals in total, in the end you add 1 more to each number because you pre-assigned 1 to each to prevent zero number for each of them.</p>
<p>If the number of animals is $x$ (instead of 14), and number of type of animals is $y$ (instead of 3), the formula reduces to "(x+y-1) choose (y-1)"</p>
|
496,479 |
<p>This symbols are used to describe left recursion : </p>
<blockquote>
<p>$A\to B\,\alpha\,|\,C$<br>$B\to A\,\beta\,|\,D,$</p>
</blockquote>
<p>It is taken from :
<a href="http://en.wikipedia.org/wiki/Left_recursion" rel="nofollow">http://en.wikipedia.org/wiki/Left_recursion</a></p>
<p>How can these symbols be explained to a non-mathematician ? Can provide an explanation for these statements in image ?</p>
|
Eike Schulte
| 11,853 |
<p>This notation is used in <a href="http://en.wikipedia.org/wiki/Formal_language" rel="nofollow" title="Wikipedia on formal languages">formal language theory</a> to describe <a href="http://en.wikipedia.org/wiki/Formal_grammar" rel="nofollow" title="Wikipedia on formal grammars">formal grammars</a>. A formal grammar consists of four things:</p>
<ul>
<li>A collection of non-terminal symbols (usually written as uppercase letters), </li>
<li>a collection of terminal symbols (usually lowercase and sometimes Greek letters), </li>
<li>a start symbol (which has to be one of the non-terminal symbols) and </li>
<li>production rules. </li>
</ul>
<p>It is common to use $S$ as the start symbol and to specify the non-terminal and terminal symbols implicitly as those used in the rules.</p>
<p>The example in the image shows some rules. Each rule consists of a head, written to the right of the arrow, and a tail, written to the left. In this case, each tail specifies two alternatives, divided by the pipe. The symbols are implicitly specified: $A$, $B$, $C$ and $D$ are non-terminal symbols, $\alpha$ and $\beta$ are terminal symbols. There is no start symbol indicated, so it probably meant to be a part of a bigger grammar.</p>
<p>To generate words from a grammar, you usually start with the start symbol. You may then repeatedly replace any sequence of characters in the word you generated so far with one of the alternatives in the tail of a rule, provided the replaced sequence of characters is exactly the head of that rule. When there are no more non-terminal symbols left, the word you generated is part of the formal language.</p>
<p>In your example, taking $A$ as the start symbol, we may replace $A$ by $B\alpha$ (by the first rule). We can then replace the $B$ by $A\beta$ (by the second rule), so our word becomes $A\beta\alpha$. Replacing $A$ by $B\alpha$ again, we get $B\alpha\beta\alpha$ and so on.</p>
<p>You see that we add characters to the left of our already generated word by using the same rules over and over again. Therefore, it is left recursion. If at any point we decide to replace $A$ by $C$ or $B$ by $D$, the generation stops. </p>
<p>(In a real-world example, there would be further rules for $C$ and $D$, but in this example there aren’t, because they are not part of the left recursion.)</p>
|
4,350,781 |
<p><a href="https://i.stack.imgur.com/57qXm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/57qXm.jpg" alt="enter image description here" /></a></p>
<p>I was able to follow most of the proof but I don’t understand how the author concludes that <span class="math-container">$r=0$</span> at the final part. I would appreciate it if someone could clarify. Thanks</p>
|
Miguel Mars
| 226,969 |
<p>First we can state that <span class="math-container">$z$</span> is the "free variable", so it doesn't depend on <span class="math-container">$x$</span> and <span class="math-container">$y$</span> but both of them will depend on <span class="math-container">$z$</span>. We can see in the figure that it goes from <span class="math-container">$0$</span> to <span class="math-container">$1$</span>.</p>
<p>Then, we will see the dependence for <span class="math-container">$y$</span>. You can visualize it projecting the figure on the <span class="math-container">$YZ$</span> plane. To do this, let us make <span class="math-container">$x=0$</span> and we get the line <span class="math-container">$z= 1-y$</span>. If <span class="math-container">$z$</span> varies from <span class="math-container">$0$</span> to <span class="math-container">$1$</span>, then <span class="math-container">$y$</span> goes from <span class="math-container">$0$</span> to <span class="math-container">$1-z$</span>.</p>
<p>Then, from the equation of the surface, we get that <span class="math-container">$x$</span> varies from <span class="math-container">$0$</span> to <span class="math-container">$1-y-z$</span>.</p>
<p>Analogously, we could have chosen other dependences, like <span class="math-container">$x$</span> being from <span class="math-container">$0$</span> to <span class="math-container">$1$</span>, projecting in the <span class="math-container">$XY$</span> plane to get that <span class="math-container">$y$</span> goes from <span class="math-container">$0$</span> to <span class="math-container">$1-x$</span> and obtaining from the equation that <span class="math-container">$z$</span> goes from <span class="math-container">$0$</span> to <span class="math-container">$1-x-y$</span> (it is a very symmetric figure, so no surprise that the equations look the same switching variables).</p>
|
1,258,623 |
<p>Let $a, b, c, d, e$ be distinct positive integers such that $a^4 + b^4 = c^4 + d^4 = e^5$. Show that $ac + bd$ is a composite number.</p>
|
Singh
| 121,735 |
<p>I used <a href="https://en.wikipedia.org/wiki/Beal%27s_conjecture" rel="nofollow">Beal's conjecture</a> to show that $ac+bd$ is composite.</p>
<p>We have $a^4+b^4=e^5$, Using Beal's conjecture we say that $a$ and $b$ will have common prime factor.</p>
<p>Let the common prime factor be $f$.</p>
<p>$a=a_1f$ and $b=b_1f$ for some $a_1,b_1\in \mathbb{Z^+}$</p>
<p>In $ac+bd=a_1fc+b_1fd=f(a_1c+b_1d)$</p>
<p>Therefore $ac+bd$ is composite.</p>
<p><strong>Above was a proof using a conjecture.</strong> I tried of proving it without the use of any conjecture but I failed to prove. </p>
<p><strong>I was able to prove that $ac+bd$ is composite when $e$ is even.</strong></p>
<p>$e$ is even, then $e^5$ will be even, then $a^4+b^4$ will be even, then $a+b$ will be even.</p>
<p>Therefore either both $a$ and $b$ are odd or both are even.</p>
<p>Similarly we will have either $c$ and $d$ odd or both even.</p>
<p>Now with any combination we will get $ac+bd$ as even.</p>
<p>Therefore $ac+bd$ is composite.</p>
<p><strong>When doing the same analysis when taking $e$ as odd</strong></p>
<p>Then we will get that <strong>either $a$ or $b$ is even and either $c$ or $d$ is even.</strong></p>
<p><strong>When $a$ and $c$ are odd and $b$ and $d$ are even then this may be possible that $ac+bd$ is prime or this may be composite.(yet I don't have any proof of this)</strong> This is the reason why I used the conjecture. </p>
|
2,553,610 |
<p>I come across this result:</p>
<blockquote>
<p>Any power is conditionally convergent for at most two values of $x$, the endpoints of its interval of convergence.</p>
</blockquote>
<p>If it is so then why?</p>
|
Fred
| 380,717 |
<p>Let $\sum_{n=0}^\infty a_n(x-x_0)^n$ be a power series with radius of convergence $=r>0$.</p>
<p>If $r= \infty$, then the power series converges absolutely in each $x \in \mathbb R$.</p>
<p>Let $r< \infty$ and $I=(x_0-r,x_0+r)$. Then the power series converges absolutely in each $x \in I$.</p>
<p>Therefore there are at most two points in which the power series converges but not absolutely: $x_0-r$ amd $x_0+r$.</p>
|
2,985,256 |
<blockquote>
<p>Let there are three points <span class="math-container">$(2,5,-3),(5,3,-3),(-2,-3,5)$</span> through which a plane passes. What is the equation of the plane in Cartesian form?</p>
</blockquote>
<p>I know how to find it in using vector form by computing the cross product to get the normal vector and passing through any one of the given points. But I want to do it a bit differently. </p>
<p>We know, the equation of any plane passing through the first point is <span class="math-container">$$a(x-2)+b(y-5)+c(z+3)=0$$</span></p>
<p>This equation must satisfy the other two points. However, this given me two equations with three unknowns <span class="math-container">$a,b,c$</span>. So can I not solve by this method?</p>
|
Josh B.
| 611,981 |
<p>You in fact can solve it by this method, even if it does seem like there will be infinite solutions. This is because the normal vector for a plane is not unique: if <span class="math-container">$\vec{n}$</span> is a normal vector to the plane, then <span class="math-container">$c\vec{n}$</span> is as well, provided that <span class="math-container">$c\neq 0$</span>.</p>
<p>For your example, your two equations for the remaining two points yield</p>
<p><span class="math-container">$$3a-2b=0$$</span></p>
<p><span class="math-container">$$-4a-8b+8c=0$$</span></p>
<p>One way you could solve this is by multiplying the first equation by <span class="math-container">$-4$</span> and then adding them, giving you</p>
<p><span class="math-container">$$-16a+8c=0\implies c=2a$$</span></p>
<p>Rearranging the first equation gives</p>
<p><span class="math-container">$$b=\frac{3}{2}a$$</span></p>
<p>As you pointed out, there are not enough equations for a unique solution, but once we specify a value for <span class="math-container">$a$</span>, we gain a value for <span class="math-container">$b$</span> and <span class="math-container">$c$</span>. Moreover, should we scale <span class="math-container">$a$</span>, this scales <span class="math-container">$b$</span> and <span class="math-container">$c$</span> by the same amount, which is exactly what we would expect. The normal that I would choose here is when <span class="math-container">$a=2$</span>, so <span class="math-container">$b=3$</span> and <span class="math-container">$c=4$</span>. </p>
|
3,860,623 |
<p>I'm trying to prove
<span class="math-container">$$\forall z\in\mathbb C-\{-1\},\ \left|\frac{z-1}{z+1}\right|=\sqrt2\iff\left|z+3\right|=\sqrt8$$</span>
thus showing that the solutions to <span class="math-container">$\left|(z-1)/(z+1)\right|=\sqrt2$</span> form the circle of center <span class="math-container">$-3$</span> and radius <span class="math-container">$\sqrt8$</span>. But my memories of algebra in <span class="math-container">$\mathbb C$</span> fail me. The simplest I get is writing <span class="math-container">$z=x+i\,y$</span> with <span class="math-container">$(x,y)\in\mathbb R^2-\{(-1,0)\}$</span> and doing the rather inelegant
<span class="math-container">$$\begin{align}\left|\frac{z-1}{z+1}\right|=\sqrt2&\iff\left|z-1\right|=\sqrt2\,\left|z+1\right|\\
&\iff\left|z-1\right|^2=2\,\left|z+1\right|^2\\
&\iff(x-1)^2+y^2=2\,((x+1)^2+y^2)\\
&\iff0=x^2+6\,x+y^2+1\\
&\iff(x+3)^2+y^2=8\\
&\iff\left|z+3\right|^2=8\\
&\iff\left|z+3\right|=\sqrt8\\
\end{align}$$</span></p>
<p>How can I avoid the steps with <span class="math-container">$x$</span> and <span class="math-container">$y$</span> ?</p>
|
fgrieu
| 35,016 |
<p>Using thrice that <span class="math-container">$\forall c\in\mathbb C,\, \left|c\right|^2=c\,\bar c$</span>, I got it down to</p>
<p><span class="math-container">$$\begin{align}\left|\frac{z-1}{z+1}\right|=\sqrt2&\iff\left|z-1\right|=\sqrt2\,\left|z+1\right|\\
&\iff\left|z-1\right|^2=2\,\left|z+1\right|^2\\
&\iff(z-1)\,(\bar z-1)=2\,(z+1)\,(\bar z+1)\\
&\iff0=z\,\bar z+3\,z+3\,\bar z+1\\
&\iff(z+3)(\bar z+3)=8\\
&\iff\left|z+3\right|^2=8\\
&\iff\left|z+3\right|=\sqrt8\\
\end{align}$$</span></p>
|
3,291,190 |
<p>Compute</p>
<p><span class="math-container">$$\text{const} \left( \frac{1}{(1-z_1z_2)(1-z_1^2z_2)(1-z_1)(1-z_2)z_1^{3t}z_2^{2t}} \right) $$</span> </p>
<p>The answer should be a polynomial in <span class="math-container">$t$</span>. This is exercise 2.38 from Beck's Computing the Continuous Discretely, and the correct answer is </p>
<p><span class="math-container">$$\frac{7}{4} t^2 + \frac{5}{2}t + \frac{7+(-1)^t}{8} $$</span></p>
<p>I'm honestly at a loss here. Can someone please explain to me how to compute the constant term here? What about a situation where we have a greater number of variables and products of functions? </p>
<p>This is closely related to an earlier post of mine at <a href="https://math.stackexchange.com/questions/3290516/finding-constant-term-in-product-of-series">Finding Constant Term in Product of Series</a>, which still doesn't have an answer. I would greatly appreciate help on either or both of these problems, and would greatly appreciate a general procedure for doing this.
Thank you so much. </p>
|
gt6989b
| 16,192 |
<p><strong>HINT</strong>
Denote the coefficient of <span class="math-container">$x^k$</span> in <span class="math-container">$f(x)$</span> as <span class="math-container">$\left[x^k\right]f(x)$</span>.</p>
<p>You are looking for
<span class="math-container">$$
\left[x^0 y^0 \right] \frac{F(x,y)}{x^{3t} y^{2t}}
= \left[x^{3t} y^{2t} \right] F(x,y)
$$</span></p>
<p>Can you now finish it? <span class="math-container">$F(x,y)$</span> is just a product of geometric series...</p>
|
4,402,613 |
<p>I know that the limits of integration in spherical coordinates are these, but I can't find the reason why that 2 appears,but I can't find a way to go further, and evaluate r in a <span class="math-container">$a \sin \theta$</span>,</p>
<p><span class="math-container">$$V = \int_0^{2\pi}\int_0^{a\sin{\theta}}\int_{-\sqrt{a^{2} - r^{2}}}^{\sqrt{a^2-r^{2}}}r\mathrm{d}z\mathrm{d}r\mathrm{d}\theta = 2\int_0^{2\pi}\int_0^{a\sin{\theta}}r\sqrt{a^2-r^2}\mathrm{d}r\mathrm{d}\theta$$</span></p>
|
eyeballfrog
| 395,748 |
<p>The 2 comes from the limits of integration: <span class="math-container">$\sqrt{a^2-r^2} - (-\sqrt{a^2-r^2}) = 2\sqrt{a^2-r^2}$</span></p>
<p>As for proceeding, here's a hint:
<span class="math-container">$$
2\int_0^{2\pi}\int_0^{a\sin{\theta}}r\sqrt{a^2-r^2}\mathrm{d}r\mathrm{d}\theta = \int_0^{2\pi}\int_0^{a\sin{\theta}}\sqrt{a^2-r^2}(2r\mathrm{d}r)\mathrm{d}\theta
$$</span>
Does a possible <span class="math-container">$u$</span>-substitution come to mind?</p>
|
740,294 |
<p>Let $y_n$ satisfy the nonlinear difference equation:</p>
<p>$$(n+1)y_n=(2n)y_{n-1}+n.$$</p>
<p>Let $u_n=(n+1) y_n$. Show that</p>
<p>$$u_n= 2u_{n-1}+n.$$</p>
<p>Solve the linear difference equation for $u_n$. Hence find $y_n$ subject to the initial condition $y_0=4$.</p>
<p>I have showed that $u_n=2u_{n-1}+n$, but I don't know how to do the next step, can anyone help me with this please?</p>
|
vonbrand
| 43,946 |
<p>This is a linear recurrence of the first order, thus solvable. Write:
$$
y_{n + 1} - \frac{2 n + 2}{n + 2} y_n = \frac{n + 1}{n + 2}
$$
If this is written:
$$
y_{n + 1} - u_n y_n = f_n
$$
you can divide by the summing factor:
$$
s_n = \prod_{0 \le k \le n} u_n
$$
to get the nicely telescoping:
$$
\frac{y_{n + 1}}{s_n} - \frac{y_n}{s_{n - 1}} = \frac{f_n}{s_n}
$$
Here the summing factor is:
$$
s_n
= \prod_{0 \le k \le n} \frac{2 k + 2}{k + 2}
= \frac{2 \cdot 4 \cdot \ldots \cdot (2 n + 2)}
{2 \cdot 3 \cdot \ldots \cdot (n + 2)}
= \frac{2^n (n + 1)!}{(n + 2)!}
= \frac{2^n}{n + 2}
$$
Sure enough:
$$
\frac{(n + 2) y_{n + 1}}{2^n} - \frac{(n + 1) y_n}{2^{n - 1}}
= \frac{n + 1}{2^n}
$$
Sum over $0 \le n \le k - 1$ to get:
$$
\frac{(k + 1) y_k}{2^{k - 1}} - \frac{y_0}{2^{-1}}
= \sum_{0 \le n \le k - 1} \frac{n + 1}{2^{n - 1}}
$$
Still need the sum on the right hand side:
$$
\sum_{0 \le n \le k - 1} \frac{n + 1}{2^{n - 1}}
= 2 \sum_{0 \le n \le k - 1} (n + 1) \cdot 2^{-n}
$$
\begin{align}
\sum_{0 \le n \le k} z^n
&= \frac{1 - z^{k + 1}}{1 - z} \\
\sum_{0 \le n \le k} n z^{n - 1}
&= \sum_{0 \le n \le k - 1} (n + 1) z^n \\
&= \frac{\mathrm{d}}{\mathrm{d} z} \frac{1 - z^{k + 1}}{1 - z}
&= \frac{1 - (k + 1) z^k + k z^{k + 1}}{(1 - z)^2}
\end{align}
Evaluating the last expression at $z = 1/2$ gives:
$$
\sum_{0 \le n \le k - 1} (n + 1) 2^{-n}
= \frac{2^{k + 2} - 2 k - 4}{2^k}
$$
Pulling all together:
$$
y_k = \frac{2^{k + 2} - 2 k - 4 + 2^k y_0}{k + 1}
$$
Phew!</p>
<p><a href="http://maxima.sourceforge.net" rel="nofollow">Maxima</a> did much of the algebra.</p>
|
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