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603,469 |
<p>Ok, so I'm not very good with these proving by induction thingies. Need a little help please.</p>
<p>How do I prove </p>
<p>$$\sum_{j=0}^n\left(-\frac12\right)^j=\frac{2^{n+1}+(-1)^n}{3\cdot2^n}$$</p>
<p>when $n$ is a non-negative integer.</p>
<p>I got the basis step and such down, but I'm pretty bad with exponents so I am having a difficult time in the induction step. </p>
|
Brian M. Scott
| 12,042 |
<p>Your induction hypothesis is that</p>
<p>$$\sum_{j=0}^n\left(-\frac12\right)^j=\frac{2^{n+1}+(-1)^n}{3\cdot2^n}\;,\tag{1}$$</p>
<p>and you want to prove that</p>
<p>$$\sum_{j=0}^{n+1}\left(-\frac12\right)^j=\frac{2^{n+1}+(-1)^{n+1}}{3\cdot2^{n+1}}\;.\tag{2}$$</p>
<p>The natural thing to do is to split the sum in $(2)$ into the sum in $(1)$ and the new, extra term and apply the induction hypothesis to get</p>
<p>$$\begin{align*}
\sum_{j=0}^{n+1}\left(-\frac12\right)^j&=\sum_{j=0}^n\left(-\frac12\right)^j+\left(-\frac12\right)^{n+1}\\\\
&=\frac{2^{n+1}+(-1)^n}{3\cdot2^n}+\left(-\frac12\right)^{n+1}\\\\
&=\frac{2^{n+1}+(-1)^n}{3\cdot2^n}+\frac{(-1)^{n+1}}{2^{n+1}}\;.
\end{align*}$$</p>
<p>Now you clearly want to put the two fractions over a common denominator to combine them:</p>
<p>$$\begin{align*}
\frac{2^{n+1}+(-1)^n}{3\cdot2^n}+\frac{(-1)^{n+1}}{2^{n+1}}&=\frac{2\left(2^{n+1}+(-1)^n\right)+3(-1)^{n+1}}{3\cdot2^{n+1}}\\\\
&=\frac{2^{n+2}+2(-1)^n+3(-1)^{n+1}}{3\cdot2^{n+1}}\;.
\end{align*}$$</p>
<p>Now there’s just one more small step, which I’ll leave to you; remember that $(-1)^n=-(-1)^{n+1}$. (Why?)</p>
|
3,460,204 |
<p>If <span class="math-container">$X\subseteq \mathbb R^n$</span> and <span class="math-container">$x\in X$</span> is an isolated point, then <span class="math-container">$x$</span> is a cluster point of <span class="math-container">$X^c$</span>.</p>
<p>I don't know if that is true or not, can someone help me, please?</p>
|
Todd
| 730,431 |
<p>If <span class="math-container">$x$</span> is a cluster point, then, there is an open set <span class="math-container">$U\subset X^c\cup\{x\}$</span> s.t. <span class="math-container">$U\cap \{x\}=\{x\}$</span>. Let <span class="math-container">$\delta >0$</span> s.t. <span class="math-container">$]x,x+\delta [\subset U$</span>. Let <span class="math-container">$n$</span> big enough to have <span class="math-container">$\frac{1}{N}<\delta $</span>. Then <span class="math-container">$x_n=x+\frac{1}{n}$</span> where <span class="math-container">$n\geq N$</span> is a sequence of <span class="math-container">$X^c$</span> that converges to <span class="math-container">$x$</span>.</p>
|
2,553,853 |
<p>I know that I am to use the thm. R/A. is a integral domain iff A is prime.
So I want an case where for some a,b in R and ab in A we have a nor b in A.</p>
<p>I thought it had something to do with the conjugate, but I'm lost.</p>
|
Ashwin Iyengar
| 95,668 |
<p>The ring is an integral domain if and only if the ideal is prime. The ideal is prime if and only if the polynomial is irredicible, and you can see that $x^7 +1$ is not either by finding an explicit factorization into smaller degree polynomials, or by noting that if it were irreducible, then $\mathbb R[x]/(x^7+1)$ would generate a degree $7$ field extension of $\mathbb R$ which is impossible, because the only finite field extension of $\mathbb R$ with degree $> 1$ is $\mathbb C$, which has degree $2$. This uses the fundamental theorem of algebra.</p>
|
155,237 |
<p>The axiom of constructibility $V=L$ leads to some very interesting consequences, one of which is that it becomes possible to give explicit constructions of some of the "weird" results of AC. For instance, in $L$, there is a definable well-ordering of the real numbers (since there is a definable well-ordering of the universe).</p>
<p>Since AC holds true in $L$, the ultrafilter lemma must be true. Does this mean that a definable non-principal ultrafilter on $\mathbb{N}$ exists in $L$, given by an explicit formula?</p>
<p>If so, what is the formula?</p>
|
Andreas Blass
| 6,794 |
<p>Let me emphasize what Noah de-emphasized (by putting it in parentheses) in his answer: Instead of the "really cheap" idea of the first free ultrafilter in $L$-order, one can use the $L$-order of subsets of $\mathbb N$ to greedily construct an ultrafilter. The $L$-ordering of the subsets of $\mathbb N$ is a $\Delta^1_2$ relation with some additional properties (I believe Moschovakis uses the terminology $\Delta^1_2$-good well-ordering). One can then deduce that the greedily constructed ultrafilter is a $\Delta^1_2$ set of subsets of $\mathbb N$.</p>
<p>Since free ultrafilters on $\mathbb N$ never have the Baire property and are never Lebesgue measurable, this shows that the classical results, that analytic and coanalytic sets have the Baire property and are measurable, cannot be extended higher in the projective hierarchy in ZFC (without additional hypotheses like large cardinals or determinacy).</p>
<p>Similar greedy constructions based on the $L$-ordering of the reals produce $\Delta^1_2$ examples of lots of other sets of reals that, in the absence of $V=L$, would be obtained by applying the axiom of choice and thus might not be definable at all. Examples include Hamel bases over $\mathbb Q$, Vitali sets, and Bernstein sets.</p>
|
2,492,191 |
<p>Let $f(x)=11x^3+15x^2+9x-2$. I want to calculate the remainder of $f(97)/11$.
I know that $97\pmod{11}=9$. But it also doesn't help with calculation of remainder with the help of modular arithmetic. I also cant get to the answer with $f(9)$ with mod of $11$.</p>
<p>What should I do?</p>
|
José Carlos Santos
| 446,262 |
<p>Since $11\equiv0\pmod{11}$, that $15\equiv4\pmod{11}$, and that $9\equiv-2\pmod{11}$, then for any integer $x$, $f(x)\equiv4x^2-2x-2\pmod{11}$. Since $97\equiv-2\pmod{11}$,\begin{align}f(97)&\equiv f(-2)\pmod{11}\\&\equiv4\times(-2)^2-2\times(-2)-2\pmod{11}\\&\equiv7\pmod{11}.\end{align}</p>
|
2,492,191 |
<p>Let $f(x)=11x^3+15x^2+9x-2$. I want to calculate the remainder of $f(97)/11$.
I know that $97\pmod{11}=9$. But it also doesn't help with calculation of remainder with the help of modular arithmetic. I also cant get to the answer with $f(9)$ with mod of $11$.</p>
<p>What should I do?</p>
|
Shaun
| 104,041 |
<p>Well, $$
\begin{align}11(97)^3+15(97)^2+9(97)-2 & =4(-2)^2-2(-2)-2 \\
&= 5+2\pmod{11} \\
&=7\pmod{11}\end{align} $$ because $97=-2\pmod{11}$.</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>
|
Community
| -1 |
<p>Yes, it is okay to not accept answers. Acceptance is a form of vote: this site has upvotes, downvotes, accept-votes, close-votes, delete-votes, etc (availability depends on reputation). Some users choose to not use some or all of categories of votes: e.g., never downvote, or never accept, or never vote at all. They are free to do so. </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>
|
Git Gud
| 55,235 |
<p>I think this question can be interpreted in two ways.</p>
<p>One is <code>Is it morally acceptable/ethical/polite/whatever to rarely accept answers?</code></p>
<p>The other is <code>Do the rules allow people to rarely accept answers?</code></p>
<p>The answer to the second question is 'yes, people are not forced to accept answers' and moderators shouldn't be able -with the current rules - to punish them if they choose not to accept answers.</p>
<p><em>My</em> answer to the first question is 'no, it's not acceptable'. Surely when the number of answered questions isn't small, there must be some instances in which the asker can pinpoint an answer which helped him more than all others. The acceptance feature exists precisely for this occasions. Users aren't obligated to accept. You're not obligated to thank someone who holds the door for you either, but you most likely do.</p>
<hr>
<p><a href="https://math.meta.stackexchange.com/users/8297/martin-sleziak">Martin</a> <a href="https://math.meta.stackexchange.com/questions/4925/list-of-comment-templates/4945#4945">linked</a> linked the following template.</p>
<blockquote>
<p>After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark ✓ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: <a href="https://math.meta.stackexchange.com/questions/3286/">How do I accept an answer?</a>, <a href="https://math.meta.stackexchange.com/questions/3399/">Why should we accept answers?</a>.</p>
</blockquote>
<p>I choose not to use it because of the word 'should' which may be perceived as entailing rule-wise obligation, which isn't the case.</p>
<p>Instead I use the following which simply forwards the reader to informative aspects of answer acceptance. (I frequently replace the first sentence with "I see you're a new user", when it is appropriate).</p>
<blockquote>
<p>You have a low answer-acceptance rate. Please read about accepting answers <a href="https://math.stackexchange.com/help/accepted-answer">here</a> and <a href="https://meta.stackoverflow.com/questions/5234/how-does-accepting-an-answer-work/5235#5235">here</a>.</p>
</blockquote>
<p><code>You have a low answer-acceptance rate. Please read about accepting answers [here](https://math.stackexchange.com/help/accepted-answer) and [here](https://meta.stackoverflow.com/questions/5234/how-does-accepting-an-answer-work/5235#5235).</code></p>
<p>The reader can then decide for himself whether to accept answers or not and similarly I can then decide whether to hold back answers or not. In the past I've rewarded users who, after being told about answer acceptance, accepted answers to their previous questions, by giving a full answer.</p>
<hr>
<p>I would even like to see the answer acceptance rate back and restrictions imposed on users who have a low acceptance rate. Those rare cases of users who genuinely can't choose what answers to accept in a large number of questions, for whatever reason, can be dealt with (I mean forgiven by) the moderators.</p>
|
2,441,660 |
<p>Suppose I have $n$ observations, which are all normally distributed with the same mean (which is unknown) but each has a different variance (the different variances could be called $v_1,...,v_n$ for example, which are all assumed to be known).</p>
<p>What is the best estimate of the mean? And further, how does one compute the variance of the mean?</p>
<p>Clearly, the sample average would be the best estimate if iid, but intuitively, an observation which has a relatively low variance would provide more information about the mean than the others, suggesting that the sample mean would not be the best estimate. My problem is that I don't know how to 'quantify' this intuition to generate an estimate of the mean and it's variance. Any ideas on how to do this? Thanks.</p>
|
Ennar
| 122,131 |
<p>What you did also counts solutions where $x,\,y,\,z$ can be $0$ as well, but that can't happen with dice.</p>
<p>You should do this for $$(x-1)+(y-1) + (z-1) = 9 -3,$$ i.e. $$x'+y'+z' =6$$ because it is now ok to have $x', y'$ or $z'$ to be $0$. </p>
<p>Now we get $\binom{6+3-1}{3-1} = 28.$ But, this is wrong as well, because $x',y',z'\leq 5$, so if we discard the three solutions where one of the $x', y', z'$ is equal to $6$, we get exactly $25$ ways.</p>
|
2,938,372 |
<p>Seems to me like it is. There are only finitely many distinct powers of <span class="math-container">$x$</span> modulo <span class="math-container">$p$</span>, by Fermat's Little Theorem (they are <span class="math-container">$\{1, x, x^2, ..., x^{p-2}\}$</span>), and the coefficient that I choose for each of these powers can only be taken from <span class="math-container">$\{0,1,2,..., p-1\}$</span>. So essentially I'm choosing amongst <span class="math-container">$p$</span> things <span class="math-container">$p-1$</span> many times, resulting in at most <span class="math-container">$p^{p-1}$</span> distinct polynomials. </p>
<p>Yet an assignment claims that <span class="math-container">$\mathbb{Z}_p[x]$</span> is infinite. </p>
|
Franklin Pezzuti Dyer
| 438,055 |
<p>Here's the flaw in your reasoning: when you consider <span class="math-container">$x$</span>, the variable of your polynomial, <span class="math-container">$x$</span> is <em>not</em> taken to be a number in <span class="math-container">$\mathbb Z_p$</span>. In fact, <span class="math-container">$x$</span> is not really a number of any sort, and the polynomial is not meant to be interpreted as a function, but rather a sort of "algebraic object." Think of it as a tuple with a number of entries equal to the degree of the polynomial plus one.</p>
<p>You're thinking of <span class="math-container">$\mathbb Z_p[x]$</span> as the set of polynomial <em>functions</em> <span class="math-container">$f:\mathbb Z_p\mapsto \mathbb Z_p$</span>. But the polynomials are not defined this way. They are defined only by their coefficients.</p>
<p>For example,
<span class="math-container">$$x^{p+1}+1= x+1$$</span>
if <span class="math-container">$x\in\mathbb Z_p$</span>, but
<span class="math-container">$$x^{p+1}+1\ne x+1$$</span>
if the two above polynomials are regarded as elements of <span class="math-container">$\mathbb Z_p[x]$</span>.</p>
<p>One last clarification:
<span class="math-container">$$\mathbb Z_p[x]=\{a_0+a_1 x+...+a_n x^{n}: a_i\in\mathbb Z_p, n\in\mathbb N\}$$</span>
However, you might have thought that
<span class="math-container">$$\mathbb Z_p[x]=\{a_0+a_1 x+...+a_n x^{n}: a_i,x\in\mathbb Z_p, n\in\mathbb N\}$$</span></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>
|
levap
| 32,262 |
<p>A useful theorem states that if $f_n \rightarrow f$ uniformly, then $f$ must be continuous. Since your pointwise limit function is not continuous, the convergence cannot be uniform.</p>
<p>If you don't want to use the theorem, you can check whether $\mathrm{sup}_{x \in \mathbb{R}} |f_n(x) - f(x)| \rightarrow 0$. </p>
<p>Since </p>
<p>$$ \mathrm{sup}_{x \in \mathbb{R}} |f_n(x) - f(x)| \geq \mathrm{sup}_{|x| < 1} |f_n(x) - f(x)| = \mathrm{max}_{|x| \leq 1} \frac{x^{4n}}{4 + x^{4n}} = \frac{1}{5} $$</p>
<p>we see that the convergence cannot be uniform (even on the interval $(-1,1)$ and surely not on $\mathbb{R}$).</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>
|
James Weigandt
| 4,872 |
<p>The short answer to your question is no. </p>
<p>One of the central open problems related to BSD rank conjecture is to find a way to canonically construct points of infinite order on elliptic curves with analytic rank at least 2. This is difficult because it requires passing from analytic properties of the $L$-function to algebraic points on the elliptic curve.</p>
<p>For analytic rank 1, Heegner points do the trick but they turn out to be torsion points when the analytic rank is higher.</p>
<p>There is no theorem proven that says that it can never happen that the analytic rank is 2 and the algebraic rank is 0 (of course if BSD is true, this cannot happen).</p>
<p>That being said, if you have a particular elliptic curve and you know the algebraic rank is 0, after a finite amount of computation you should be able to show that the analytic rank is 0, i.e. you should be able to prove the BSD rank conjecture for this particular curve.</p>
<p>If you have a particular elliptic curve and you know that the analytic rank is 2, you probably proved this by observing three things:</p>
<ol>
<li>The analytic rank is even by the sign of the functional equation.</li>
<li>The analytic rank is $\leq 2$ by some calculation with the L-function.</li>
<li>The algebraic rank is not equal to 0, because you found a point of infinite order, so the analytic rank cannot be 0.</li>
</ol>
<p>So you conclude the analytic rank must be 2. </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>By the following arrangement we have $2,3,4$ points on each side of the triangle. There are a total of $9$ points. Select any $3$ from them. Total number of ways are: $\binom{9}{3} = 84$. Number of collinear point selection cases: $\binom{2}{3} +\binom{3}{3} +\binom{4}{3} = 5$ ways. Totally $84-5 = 79$ ways. Hope it helps.</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>
|
glebovg
| 36,367 |
<p>To be honest I was never good at high school geometry and geometry in general. I am also interested in mathematics and want to pursue a PhD in pure math. I always thought there was something wrong with geometric arguments -- and it turns out there is. Many philosophers and mathematicians explained why they are not sound using examples. You can use geometry to prove false statements. I remember when I took real analysis, I understood most of it, but never felt confident about it. But know it all makes sense. The same goes for complex analysis. It all makes sense when you use it in, say, number theory. Trust me, no mathematician, however great, could prove every known theorem in her field without using a book or two. Of course, after proving a few basic results you will be able to distinguish between one line proofs and one paragraph proofs. If you feel like you cannot prove something, just go back to the definitions and familiar results. Write everything in full i.e. the definitions, theorems, etc. and after a while you will skip those steps. I know some people who claim to love math, but they are also majoring in something like english or chemistry. If you are truly passionate about math, you should not major in anything else, and you should never do math for a potential employer. If you finally realized that math in the only thing that you are interested in, then you should just go for it.</p>
|
2,908,186 |
<p>There's an old-school pocket calculator trick to calculate <span class="math-container">$x^n$</span> on a pocket calculator, where both, <span class="math-container">$x$</span> and <span class="math-container">$n$</span> are real numbers. So, things like <span class="math-container">$\,0.751^{3.2131}$</span> can be calculated, which is awesome.</p>
<p>This provides endless possibilities, including calculating nth roots on a simple pocket calculator.</p>
<p><strong>The trick goes like this:</strong></p>
<ol>
<li>Type <span class="math-container">$x$</span> in the calculator</li>
<li>Take the square root twelve times</li>
<li>Subtract one</li>
<li>Multiply by <span class="math-container">$n$</span></li>
<li>Add one</li>
<li>Raise the number to the 2nd power twelve times (press <code>*</code> and <code>=</code> key eleven times)</li>
</ol>
<p><strong>Example:</strong></p>
<p>I want to calculate <span class="math-container">$\sqrt[3]{20}$</span> which is the same as <span class="math-container">$20^{1/3}$</span>. So <span class="math-container">$x=20$</span> and <span class="math-container">$n=0.3333333$</span>. After each of the six steps, the display on the calculator will look like this:</p>
<ol>
<li><span class="math-container">$\;\;\;20$</span></li>
<li><span class="math-container">$\;\;\;1.0007315$</span></li>
<li><span class="math-container">$\;\;\;0.0007315$</span></li>
<li><span class="math-container">$\;\;\;0.0002438$</span></li>
<li><span class="math-container">$\;\;\;1.0002438$</span></li>
<li><span class="math-container">$\;\;\;2.7136203$</span></li>
</ol>
<p>The actual answer is <span class="math-container">$20^{1/3}\approx2.7144176$</span>. So, our trick worked to three significant figures. It's not perfect, because of the errors accumulated from the calculator's 8 digit limit, but it's good enough for most situations.</p>
<blockquote>
<p><strong>Question:</strong></p>
<p>So the question is now, why does this trick work? More specifically, how do we prove that:
<span class="math-container">$$x^n\approx \Big(n(x^{1/4096}-1)+1\Big)^{4096}$$</span></p>
</blockquote>
<p>Note: <span class="math-container">$4096=2^{12}$</span>.</p>
<p>I sat in front of a piece of paper trying to manipulate the expression in different ways, but got nowhere.</p>
<p>I also noticed that if we take the square root in step 1 more than twelve times, but on a better-precision calculator, and respectively square the number more than twelve times in the sixth step, then the result tends to the actual value we are trying to get, i.e.:</p>
<blockquote>
<p><span class="math-container">$$\lim_{a\to\infty}\Big(n(x^{1/2^a}-1)+1\Big)^{(2^a)}=x^n$$</span></p>
</blockquote>
<p>This, of course doesn't mean that doing this more times is encouraged on a pocket calculator, because the error from the limited precision propagates with every operation. <span class="math-container">$a=12$</span> is found to be the optimal value for most calculations of this type i.e. the best possible answer taking all errors into consideration. Even though 12 is the optimal value on a pocket calculator, taking the limit with <span class="math-container">$a\to\infty$</span> can be useful in proving why this trick works, however I still can't of think a formal proof for this.</p>
<p><strong>Thank you for your time :)</strong></p>
|
Cheerful Parsnip
| 2,941 |
<p>A standard trick is to calculate the natural logarithm first to get the exponent under control:</p>
<p>$$\log(\lim_{a\to\infty}(n(x^{1/a}-1)+1)^a)=\lim_{a\to\infty}a\log(nx^{1/a}-n+1)$$
Set $u=1/a$.
We get
$$\lim_{u\to 0}\frac{\log (nx^u-n+1)}{u}$$
Use L'Hopital:
$$\lim_{u\to 0}\frac{nx^u\log x}{nx^u-n+1}=n\log x=\log x^n$$
Here we just plugged in $u=0$ to calculate the limit!</p>
<p>So the original limit goes to $x^n$ as desired.</p>
|
2,908,186 |
<p>There's an old-school pocket calculator trick to calculate <span class="math-container">$x^n$</span> on a pocket calculator, where both, <span class="math-container">$x$</span> and <span class="math-container">$n$</span> are real numbers. So, things like <span class="math-container">$\,0.751^{3.2131}$</span> can be calculated, which is awesome.</p>
<p>This provides endless possibilities, including calculating nth roots on a simple pocket calculator.</p>
<p><strong>The trick goes like this:</strong></p>
<ol>
<li>Type <span class="math-container">$x$</span> in the calculator</li>
<li>Take the square root twelve times</li>
<li>Subtract one</li>
<li>Multiply by <span class="math-container">$n$</span></li>
<li>Add one</li>
<li>Raise the number to the 2nd power twelve times (press <code>*</code> and <code>=</code> key eleven times)</li>
</ol>
<p><strong>Example:</strong></p>
<p>I want to calculate <span class="math-container">$\sqrt[3]{20}$</span> which is the same as <span class="math-container">$20^{1/3}$</span>. So <span class="math-container">$x=20$</span> and <span class="math-container">$n=0.3333333$</span>. After each of the six steps, the display on the calculator will look like this:</p>
<ol>
<li><span class="math-container">$\;\;\;20$</span></li>
<li><span class="math-container">$\;\;\;1.0007315$</span></li>
<li><span class="math-container">$\;\;\;0.0007315$</span></li>
<li><span class="math-container">$\;\;\;0.0002438$</span></li>
<li><span class="math-container">$\;\;\;1.0002438$</span></li>
<li><span class="math-container">$\;\;\;2.7136203$</span></li>
</ol>
<p>The actual answer is <span class="math-container">$20^{1/3}\approx2.7144176$</span>. So, our trick worked to three significant figures. It's not perfect, because of the errors accumulated from the calculator's 8 digit limit, but it's good enough for most situations.</p>
<blockquote>
<p><strong>Question:</strong></p>
<p>So the question is now, why does this trick work? More specifically, how do we prove that:
<span class="math-container">$$x^n\approx \Big(n(x^{1/4096}-1)+1\Big)^{4096}$$</span></p>
</blockquote>
<p>Note: <span class="math-container">$4096=2^{12}$</span>.</p>
<p>I sat in front of a piece of paper trying to manipulate the expression in different ways, but got nowhere.</p>
<p>I also noticed that if we take the square root in step 1 more than twelve times, but on a better-precision calculator, and respectively square the number more than twelve times in the sixth step, then the result tends to the actual value we are trying to get, i.e.:</p>
<blockquote>
<p><span class="math-container">$$\lim_{a\to\infty}\Big(n(x^{1/2^a}-1)+1\Big)^{(2^a)}=x^n$$</span></p>
</blockquote>
<p>This, of course doesn't mean that doing this more times is encouraged on a pocket calculator, because the error from the limited precision propagates with every operation. <span class="math-container">$a=12$</span> is found to be the optimal value for most calculations of this type i.e. the best possible answer taking all errors into consideration. Even though 12 is the optimal value on a pocket calculator, taking the limit with <span class="math-container">$a\to\infty$</span> can be useful in proving why this trick works, however I still can't of think a formal proof for this.</p>
<p><strong>Thank you for your time :)</strong></p>
|
Barry Cipra
| 86,747 |
<p>If $x$ (actually $\ln x$) is relatively small, then $x^{1/4096}=e^{(\ln x)/4096}\approx1+(\ln x)/4096$, in which case</p>
<p>$$n(x^{1/4096}-1)+1)\approx1+{n\ln x\over4096}$$</p>
<p>If $n$ is also relatively small, then</p>
<p>$$(n(x^{1/4096}-1)+1)^{4096}\approx\left(1+{n\ln x\over4096}\right)^{4096}\approx e^{n\ln x}=x^n$$</p>
<p>Remark: When I carried out the OP's procedure on a pocket calculator, I got the same approximation as the the OP, $2.7136203$, which is less than the exact value, $20^{1/3}=2.7144176\ldots$. Curiously, the exact value (according to Wolfram Alpha) for the approximating formula,</p>
<p>$$\left({1\over3}(20^{1/4096}-1)+1\right)^{4096}=2.7150785662\ldots$$</p>
<p>is <em>more</em> than the exact value. On the other hand, if you take the square root of $x=20$ <em>eleven</em> times instead of twelve -- i.e., if you use $2048$ instead of $4096$ -- the calculator gives $2.7152613$ while WA gives $2.715739784\ldots$, both of which are too large. Very curiously, if you <em>average</em> the two calculator results, you get</p>
<p>$${2.7136203+2.7152613\over2}=2.7144408$$</p>
<p>which is quite close to the true value!</p>
|
256,138 |
<p>I need to generate four positive random values in the range [.1, .6] with (at most) two significant digits to the right of the decimal, and which sum to exactly 1. Here are three attempts that do not work.</p>
<pre><code>x = {.15, .35, .1, .4}; While[Total[x] != 1,
x = Table[Round[RandomReal[{.1, .6}], .010], 4]];
x = {.25, .25, .25, .25}; While[Total[x] == 1,
x = Table[Round[RandomReal[{.1, .6}], .010], 4]];
NestWhileList[Total[x],
x = Table[Round[RandomReal[{.1, .6}], .010], 4],
Plus @@ x == 1][[1]]
</code></pre>
|
creidhne
| 41,569 |
<p>Use <a href="https://reference.wolfram.com/language/ref/Normalize.html" rel="nofollow noreferrer">Normalize</a> to find a list of random values between 0.1 and 0.6, which are rounded to two decimal places, and that total to 1.</p>
<pre><code>rr = 0;
While[Total[rr] != 1,
rr = Round[Normalize[RandomReal[{.1, .6}, 4], Total], .01]]
</code></pre>
<p>The random values are returned as <code>rr</code>. For example:</p>
<pre><code>SeedRandom[1234];
rr = 0;
While[Total[rr] != 1,
rr = Round[Normalize[RandomReal[{.1, .6}, 4], Total], .01]]
{Total[rr], rr}
</code></pre>
<blockquote>
<p><code>{1., {0.4, 0.27, 0.11, 0.22}}</code></p>
</blockquote>
|
256,138 |
<p>I need to generate four positive random values in the range [.1, .6] with (at most) two significant digits to the right of the decimal, and which sum to exactly 1. Here are three attempts that do not work.</p>
<pre><code>x = {.15, .35, .1, .4}; While[Total[x] != 1,
x = Table[Round[RandomReal[{.1, .6}], .010], 4]];
x = {.25, .25, .25, .25}; While[Total[x] == 1,
x = Table[Round[RandomReal[{.1, .6}], .010], 4]];
NestWhileList[Total[x],
x = Table[Round[RandomReal[{.1, .6}], .010], 4],
Plus @@ x == 1][[1]]
</code></pre>
|
martin
| 9,923 |
<pre><code>RandomSample[IntegerPartitions[100,{4},Range[10,60]]/100.,1]
</code></pre>
|
239,387 |
<p>Someone could explain how to build the smallest field containing to $\sqrt[3]{2}$.</p>
|
Haskell Curry
| 39,362 |
<p>Consider the set $ F $ of expressions of the form
$$
\left\{ a + b \cdot \sqrt[3]{2} + c \cdot \left( \sqrt[3]{2} \right)^{2} \,\Big|\, (a,b,c) \in \mathbb{Q}^{3} \right\}.
$$
This is the smallest subfield of $ \mathbb{R} $ containing $ \mathbb{Q} $ and $ \sqrt[3]{2} $.</p>
<p>Clearly, $ \mathbb{Q} \cup \{ \sqrt[3]{2} \} \subseteq F $. Next, observe that $ F $ is closed under addition and multiplication. The only non-trivial thing that needs to be verified is that every non-zero element of $ F $ has a multiplicative inverse in $ F $. After having established that $ F $ is a field, notice that it contains precisely all numbers that <em>should be</em> in any subfield of $ \mathbb{R} $ containing $ \mathbb{Q} $ and $ \sqrt[3]{2} $. This proves that $ F $ is the <em>smallest</em> subfield of $ \mathbb{R} $ with such a property.</p>
<p>Notice that $ F $ is a vector space of dimension $ 3 $ over $ \mathbb{Q} $, which agrees nicely with the fact that the degree of the minimal polynomial of $ \sqrt[3]{2} $ over $ \mathbb{Q} $ is $ 3 $ (the minimal polynomial is $ X^{3} - 2 $).</p>
|
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>
|
juantheron
| 14,311 |
<p>Given $\displaystyle \alpha,\beta$ are the roots of $2x^2+8x+4=0.$ </p>
<p>So$\displaystyle \alpha+\beta = -\frac{8}{2}=-4$ and $\displaystyle \alpha\cdot \beta = \frac{1}{2}.$</p>
<p>Now for Second part, Using $\bullet\; \bf{x^2-(sum \; of \; roots)x+(product\; of \; roots) =0}$</p>
<p>So here $\displaystyle \bf{sum\; of \; roots } = 2\alpha^4+\frac{1}{\beta^2}+2\beta^4+\frac{1}{\alpha^2} = 2\left[\alpha^4+\beta^4\right]+\frac{1}{\alpha^2}+\frac{1}{\beta^2}$</p>
<p>So we get $\displaystyle = 2\left[(\alpha^2+\beta^2)^2-2(\alpha\cdot \beta)^2\right]+\frac{(\alpha+\beta)^2-2\alpha\cdot \beta}{(\alpha\cdot \beta)^2}$</p>
<p>$\displaystyle = 2\left[\left\{(\alpha+\beta)^2-2\alpha\cdot \beta\right\}^2-2(\alpha\cdot \beta)^2\right]+\frac{(\alpha+\beta)^2-2\alpha\cdot \beta}{(\alpha\cdot \beta)^2} = $</p>
<p>and $\displaystyle \bf{product\; of roots} = \left(2\alpha^4+\frac{1}{\beta^2}\right)\times \left(2\beta^4+\frac{1}{\alpha^2}\right)$</p>
<p>$\displaystyle = 4(\alpha\cdot \beta)^4+2\left[(\alpha+\beta)^2-2\alpha\cdot \beta\right]+\frac{1}{(\alpha\cdot \beta)^2}=$</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>
|
lab bhattacharjee
| 33,337 |
<p>As $54\cdot5\equiv90\pmod{180},$</p>
<p>if $\tan5x=\infty,5x=180^\circ n+90^\circ$ where $n$ is any integer</p>
<p>$x=36^\circ n+18^\circ$ where $n\equiv0,\pm1,\pm2\pmod5$</p>
<p>Using $\tan5x$ <a href="http://mathworld.wolfram.com/Tangent.html" rel="nofollow">expansion</a>, the roots of $$5t^4-10t^2+1=0$$ are $\tan(36^\circ n+18^\circ)$ where $n\equiv0,\pm1,2\pmod5$ </p>
<p>as $n=2\implies\tan(36^\circ n+18^\circ)=?$</p>
<p>$$t^2=\dfrac{10\pm4\sqrt5}{2\cdot5}=\dfrac{5\pm2\sqrt5}5$$</p>
<p>As $\tan54^\circ>\tan18^\circ>0,\tan^254^\circ>\tan^218^\circ,$</p>
<p>$\tan^254^\circ=\dfrac{5+2\sqrt5}5=\dfrac{25+10\sqrt5}{5^2}$
and $\tan^218^\circ=\dfrac{25-10\sqrt5}{5^2}$</p>
<p>For $\tan36^\circ,$ can you use the same idea?</p>
|
3,363,810 |
<p>I;m not sure how to go about proving this. I just started learning about it and would appreciate some help.</p>
|
Da Mike
| 346,084 |
<p>When dealing with equalities about sets, a good technique is to show that the left side of the equation is always a subset of the right side, and that the right side is also a subset of the left. Thus, the sets are equal. In your case, you would want to prove that </p>
<p><span class="math-container">$$
X \cup (X \cap Y) \subseteq X\ \ \ (1)
$$</span></p>
<p>and</p>
<p><span class="math-container">$$
X \subseteq X \cup (X \cap Y)\ \ \ (2).
$$</span></p>
<p>Now, for <span class="math-container">$(1)$</span>, if <span class="math-container">$x \in X \cup (X \cap Y)$</span> then either <span class="math-container">$x$</span> belongs in <span class="math-container">$X$</span> or in <span class="math-container">$(X \cap Y)$</span> (the latter meaning that <span class="math-container">$x$</span> is both in <span class="math-container">$X$</span> and <span class="math-container">$Y$</span>). So either way, you have that <span class="math-container">$x \in X$</span>. For <span class="math-container">$(2)$</span>, if <span class="math-container">$x \in X$</span> then <span class="math-container">$x$</span> belongs in <span class="math-container">$X \cup (X \cap Y)$</span> since <span class="math-container">$x \in X$</span> and you just need to prove that it belongs to just one of <span class="math-container">$X$</span> or <span class="math-container">$(X \cap Y)$</span>. </p>
<p>Since both <span class="math-container">$(1)$</span> and <span class="math-container">$(2)$</span> hold, we have that every element of the left side also belongs to the right, and every element of the right also belongs to the left. The only way this can be true is when <span class="math-container">$ X = X \cup (X \cap Y)$</span>.</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>
|
Michael Hoppe
| 93,935 |
<p>Why not try logarithmic differentiation:</p>
<p>$$\ln\bigl((g(x)\bigr)=x\ln(t)+t\ln(x),$$
thus
$$\frac{g'(x)}{g(x)}=\ln(t)+\frac{t}{x},$$
from which $g'$ is easily onbtained.</p>
<p>Michael</p>
|
931,748 |
<p>I do not understand this solution and this formula and why we are using (1+1)^n...
I need some help to get an idea of what is going on here
Thanks</p>
<p><img src="https://i.stack.imgur.com/S9IL5.png" alt="enter image description here"></p>
|
Rene Schipperus
| 149,912 |
<p>One way is that the assumption implies that $x$ must be odd, so $x=2n+1$ and thus $$x^2=4n^2+4n+1$$</p>
|
931,748 |
<p>I do not understand this solution and this formula and why we are using (1+1)^n...
I need some help to get an idea of what is going on here
Thanks</p>
<p><img src="https://i.stack.imgur.com/S9IL5.png" alt="enter image description here"></p>
|
lab bhattacharjee
| 33,337 |
<p>$x^2\equiv1\pmod2\iff x\equiv1\pmod2,x=2a+1$ where $a$ is any integer</p>
<p>$$\implies x^2=(2a+1)^2=8\frac{a(a+1)}2+1\equiv1\pmod8$$</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 $</span></p>
<p>I do not understand what I have to do.</p>
|
PrincessEev
| 597,568 |
<p>If you want to show that <span class="math-container">$a$</span> is a root of <span class="math-container">$f$</span>, just plug in <span class="math-container">$a=2/5$</span> for <span class="math-container">$x$</span> in the definition of <span class="math-container">$f$</span>. That is to say, find <span class="math-container">$f(2/5)$</span>.</p>
<p>If <span class="math-container">$f(2/5) = 0$</span>, then <span class="math-container">$2/5$</span> is a root of <span class="math-container">$f$</span>. Similarly, if it's not zero, then it's not a root.</p>
<p>If you want know how you can conclude that <span class="math-container">$2/5$</span> <em>might</em> be a root, the <a href="https://en.wikipedia.org/wiki/Rational_root_theorem" rel="nofollow noreferrer">rational root theorem</a> might be worth looking at. You still have to test the potential roots the theorem gives you, but it's better than trying to mess with that cubic right off the bat.</p>
|
1,901,302 |
<p>Let $R$ be a principal ideal ring, meaning that every ideal is generated by one element. Given a subset $A\subseteq R$, it is generally not possible to choose one element $x\in A$ so that $I(\{x\})=I(A)$, i.e. the ideal generated by $A$ need not be generated by a single element of $A$.</p>
<p>As an example one can consider $R=\mathbb Z$ and $A=\{2,3\}$, then $I(A)=\mathbb Z$.</p>
<p>My question is whether one can choose a finite subset $X$ of $A$ so that $I(A)=I(X)$.</p>
<p>In $\mathbb Z$ for example this is always possible:</p>
<p>The ideal generated by a set is always the ideal generated by the GCD of that set. If $X\subseteq Y$ then $\mathrm{GCD}(Y)≤\mathrm{GCD}(X)$. If we have a finite sbuset $X$ of $A$ and the GCDs are not equal, there must exist an element of $A$ so that appending it to $X$ will decrease the GCD. This can only happen a finite amount of times since we are decreasing a positive number by at least $1$ each time.</p>
<p>I don't expect this argument to work in general principal ideal rings, since it is using a total ordering structure on the ideal space for which every ideal has only finitely many "lesser" ideals.</p>
|
Elle Najt
| 54,092 |
<p>PIDs are Noetherian, and Noetherian rings have the acc on ideals. So order some countable subset of A and consider the ideals generated by the first n. </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>
|
Magdiragdag
| 35,584 |
<p>The answer depends on how you interpret "adjoining an element $\alpha$ to $F$ satisfying $f(\alpha) = 0$".</p>
<p>The standard interpretation is: consider the algebraic closure $\bar F$ of $F$, take an element $\alpha \in \bar F$ satisfying $f(\alpha) = 0$ and look at the smallest field containing both $F$ and $\alpha$, i.e, $F(\alpha)$. If $f$ is irreducible, this is isomorphic to $F[x]/(f)$. This makes the answer to (a) ${\Bbb F}_4$ and to (b) and (c) both ${\Bbb F}_2$.</p>
<p>Since Artin says "the <em>ring</em> obtained from ${\Bbb F}_2$ $\dots$", you could also interpret it as: "add a new <em>free</em> element $\alpha$ to $F$, subject <em>only</em> to the constraint $f(\alpha) = 0$". That is, by definition, consider $F[x]/(f)$. With this interpretation, the answer to (a) is still ${\Bbb F}_4$, (b) is (isomorphic to) ${\Bbb F}_2[x]/(x^2)$ and (c) is ${\Bbb F}_2^2$.</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>
|
Angina Seng
| 436,618 |
<p>For small theta
<span class="math-container">$$\frac{\sin\theta}\theta\approx1-\frac{\theta^2}6.$$</span>
So we get a <span class="math-container">$10\%$</span> error about where <span class="math-container">$\theta^2/6\approx 0.1$</span>, that is
<span class="math-container">$\theta\approx\sqrt{0.6}\approx0.8$</span>. A radian is about <span class="math-container">$57$</span> degrees, so that's
about <span class="math-container">$50$</span> degrees or so.</p>
|
592,805 |
<p>How to prove $$\limsup\limits_{n \to \infty} \frac{1}{\sqrt n}B(n) = +\infty$$ using Blumenthal zero-one law, where $(B(t))_{t \geq 0}$ is a Brownian motion?</p>
|
saz
| 36,150 |
<p>The idea of the proof is rather similar to the hints provided by AlexR., but using the time inversion it is easier to see how to apply Blumenthal's 0-1 law.</p>
<p><strong>Hints</strong>:</p>
<ol>
<li>It is widely known that $W_t := t \cdot B_{1/t}$ is a Brownian motion. This implies in particular that $$\limsup_{n \to \infty} \frac{1}{\sqrt{n}} \cdot B_n = \infty \tag{1}$$ is equivalent to $$\limsup_{n \to \infty} \frac{1}{\sqrt{\frac{1}{n}}} W_{\frac{1}{n}} = \infty \tag{2} $$ which means that it suffices to show $(2)$ for any Brownian motion $(W_t)_{t \geq 0}$.</li>
<li>Show that $\Omega(a) := \{\omega; \limsup_{n \to \infty} \frac{1}{\sqrt{\frac{1}{n}}} W_{\frac{1}{n}}(\omega) >a \} \in \mathcal{F}_{0+}^W$ for any $a>0$. By Blumenthal's 0-1 law, $\mathbb{P}(\Omega(a)) \in \{0,1\}$. Mind that $$\mathbb{P} \left( \limsup_{n \to \infty} \frac{1}{\sqrt{\frac{1}{n}}} W_{\frac{1}{n}} = \infty \right) = \lim_{a \to \infty} \mathbb{P}(\Omega(a))$$</li>
<li>We have $$\mathbb{P}(\Omega(a))\geq \limsup_{n \to \infty} \mathbb{P} \left( \frac{1}{\sqrt{\frac{1}{n}} } W_{\frac{1}{n}} > a \right)$$ Now use the scaling property, i.e. $W_{\frac{1}{n}} \sim \sqrt{\frac{1}{n}} \cdot W_1$, to deduce $\mathbb{P}(\Omega(a))>0$; hence $\mathbb{P}(\Omega(a))=1$ by step 2.</li>
</ol>
|
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 |
<p>Let's stick with the Databin idea for now. It will give us the timestamp functionality for free. We can play with the form locally before deploying it.</p>
<ul>
<li><p>Create the databin</p>
<pre><code>exerciseDataBin = CreateDatabin["exercise"]
</code></pre>
</li>
<li><p>Define a function to update the databin and produce the desired plot</p>
<pre><code>UpdateAndDisplayExerciseHistory =
Function[
data,
DatabinAdd[
exerciseDataBin,
<|"pushups" -> data["Pushups"], "water" -> data["Water"]|>];
DateListPlot[TimeSeries[exerciseDataBin]]]
</code></pre>
<p>This can probably be cleaned up, but I wanted to be explicit while testing things out.</p>
</li>
<li><p>Now we can create the form page</p>
<pre><code>FormPage[
FormObject[{"Pushups" -> {0, 5, 10, 15, 20, 25, 30}, "Water" -> {0, .5, 1, 2}}],
UpdateAndDisplayExerciseHistory]
</code></pre>
</li>
</ul>
<p>Summary:</p>
<ul>
<li>The plot below the submit button is just whatever the result is of executing the submit function. You can play with the exact plotting functionality you want--I just chose something easy to test with. Aslo, there's probably a way to get it to display in the initial view before the first submit, but I didn't explore that yet.</li>
<li>It's easier (in my opinion) to have explicit named functions that do the work so you can test them outside of the form context.</li>
<li>If you use Databin, you don't need to do your own timestamping (like you were doing with Now).</li>
<li>To test your databin, it's useful to apply Dataset, e.g. <code>Dataset[exerciseDataBin]</code></li>
<li>Once you're happy with your local testing, you can CloudDeploy it.</li>
</ul>
<p>I think this gets you your basic functionality. I'll try to grab some time later to look at your bonus questions.</p>
|
4,508,840 |
<p>I need to find the summation
<span class="math-container">$$S=\sum_{r=0}^{1010} \binom{1010}r \sum_{k=2r+1}^{2021}\binom{2021}k$$</span></p>
<p>I tried various things like replacing <span class="math-container">$k$</span> by <span class="math-container">$2021-k$</span> and trying to add the 2 summations to a pattern but was unable to find a solution. For more insights into the question, this was essential to solve a probability question wherein there were 2 players, A and B. A rolls a dice <span class="math-container">$2021$</span> times, and B rolls it <span class="math-container">$1010$</span> times. We had to find the probability of A having number of odd numbers more than twice of B. So if B had <span class="math-container">$r$</span> odd numbers, A could have odd numbers from <span class="math-container">$2r+1$</span> to <span class="math-container">$2021$</span>, hence the summation. I can get the required probability by dividing this by <span class="math-container">$2^{2021+1010}$</span>.</p>
|
R. J. Mathar
| 805,678 |
<p>For <span class="math-container">$N=1010$</span> or others
<span class="math-container">\begin{equation}
S_N\equiv \sum_{r=0}^N\binom{N}{r}\sum_{k=2r+1}^{2N+1}\binom{2N+1}{k}
=
\end{equation}</span>
<span class="math-container">\begin{equation}
\sum_{r=0}^N\binom{N}{r}\sum_{k=2r+1}^{2N+1}\binom{2N+1}{k}
+
\sum_{r=0}^N\binom{N}{r}\sum_{k=0}^{2r}\binom{2N+1}{k}
-
\sum_{r=0}^N\binom{N}{r}\sum_{k=0}^{2r}\binom{2N+1}{k}
\end{equation}</span>
<span class="math-container">\begin{equation}
=
\sum_{r=0}^N\binom{N}{r}\sum_{k=0}^{2N+1}\binom{2N+1}{k}
-
\sum_{r=0}^N\binom{N}{r}\sum_{k=0}^{2r}\binom{2N+1}{k}
\end{equation}</span>
<span class="math-container">\begin{equation}
=
\sum_{r=0}^N\binom{N}{r}2^{2N+1}
-
\sum_{r=0}^N\binom{N}{r}\sum_{k=0}^{2r}\binom{2N+1}{k}
\end{equation}</span>
<span class="math-container">\begin{equation}
=
2^N 2^{2N+1}
-
\sum_{r=0}^N\binom{N}{r}\sum_{k=0}^{2r}\binom{2N+1}{k}
\end{equation}</span>
<span class="math-container">\begin{equation}
=
2^{3N+1}
-
% \sum_{k=0}^{2N}\sum_{r=k}^{N}
\sum_{r=0}^{N}\sum_{k=0}^{2r}
\binom{N}{r}\binom{2N+1}{k}
\end{equation}</span>
substituing <span class="math-container">$r'=N-r$</span>
<span class="math-container">\begin{equation}
=
2^{3N+1}
-
\sum_{r'=N}^{0}\sum_{k=0}^{2N-2r'}
\binom{N}{N-r'}\binom{2N+1}{k}
\end{equation}</span>
and substituting <span class="math-container">$k'=2N+1-k$</span>
<span class="math-container">\begin{equation}
=
2^{3N+1}
-
\sum_{r'=N}^{0}\sum_{k'=2N+1}^{2N+1-(2N-2r')}
\binom{N}{N-r'}\binom{2N+1}{2N+1-k'}
\end{equation}</span>
<span class="math-container">\begin{equation}
=
2^{3N+1}
-
\sum_{r'=N}^{0}\sum_{k'=2N+1}^{1+2r'}
\binom{N}{r'}\binom{2N+1}{k'}
\end{equation}</span>
<span class="math-container">\begin{equation}
=
2^{3N+1}
-
\sum_{r'=0}^{N}\sum_{k'=2N+1}^{1+2r'}
\binom{N}{r'}\binom{2N+1}{k'}
=
2^{3N+1}-S.
\end{equation}</span>
Adding <span class="math-container">$S$</span> to both sides gives <span class="math-container">$2S=2^{3N+1}$</span>, therefore
<span class="math-container">\begin{equation}
S=2^{3N}.
\end{equation}</span></p>
|
1,266,525 |
<p>I'm shown part of the function $g(x) = \sqrt{4x-3}$. Is it injective? I said yes as per definition if $f(x) = f(y)$, then $x =y$. Is this right? </p>
<p>Under what criteria is $g(x)$ bijective? For what domain and co domain does $g(x)$ meet this criteria? I'm totally lost on this one! </p>
<p>Find the inverse of $g(x)$, sketch the graph and explain how $g(x)$ and $g^{-1}(x)$ relate geometrically. </p>
<p>Any help on this is greatly appreciated. </p>
|
wythagoras
| 236,048 |
<p>A function is bijective if it is surjective and injective. A function is surjective if the range is the codomain, i.e. if every value in the codomain is the output of the function. </p>
<p>An inverse function is the function reflected trough $y=x$. </p>
|
598,838 |
<p><span class="math-container">$11$</span> out of <span class="math-container">$36$</span>? I got this by writing down the number of possible outcomes (<span class="math-container">$36$</span>) and then counting how many of the pairs had a <span class="math-container">$6$</span> in them: <span class="math-container">$(1,6)$</span>, <span class="math-container">$(2,6)$</span>, <span class="math-container">$(3,6)$</span>, <span class="math-container">$(4,6)$</span>, <span class="math-container">$(5,6)$</span>, <span class="math-container">$(6,6)$</span>, <span class="math-container">$(6,5)$</span>, <span class="math-container">$(6,4)$</span>, <span class="math-container">$(6,3)$</span>, <span class="math-container">$(6,2)$</span>, <span class="math-container">$(6,1)$</span>. Is this correct?</p>
|
amWhy
| 9,003 |
<p>Yes indeed, you've got them all. So counting them, we get $11$ of the possible $36$ outcomes of which at least one $6$ is rolled. Now simply express this probability as a fraction!</p>
|
598,838 |
<p><span class="math-container">$11$</span> out of <span class="math-container">$36$</span>? I got this by writing down the number of possible outcomes (<span class="math-container">$36$</span>) and then counting how many of the pairs had a <span class="math-container">$6$</span> in them: <span class="math-container">$(1,6)$</span>, <span class="math-container">$(2,6)$</span>, <span class="math-container">$(3,6)$</span>, <span class="math-container">$(4,6)$</span>, <span class="math-container">$(5,6)$</span>, <span class="math-container">$(6,6)$</span>, <span class="math-container">$(6,5)$</span>, <span class="math-container">$(6,4)$</span>, <span class="math-container">$(6,3)$</span>, <span class="math-container">$(6,2)$</span>, <span class="math-container">$(6,1)$</span>. Is this correct?</p>
|
constructor
| 158,202 |
<p><a href="http://www.wolframalpha.com/input/?i=chance+of+throwing+1+6%27s+with+2+dice" rel="nofollow">http://www.wolframalpha.com/input/?i=chance+of+throwing+1+6%27s+with+2+dice</a></p>
<p>X = # of occurrences of 6 with 2 dice</p>
<p>P(X=1) = P(the first dice has 6, the second hasn't) + P(the second dice has 6, the first one hasn't) = 1/6 * 5/6 + 5/6 * 1/6 = 5/ 36 + 5/36 = 5/18 = 0.2778.</p>
<p>Which is also what Wolfram Alpha computes.</p>
<p>Then P(X=2) = 1/6 * 1/6 = 1/36.</p>
<p>5/18 + 1/36 = 11/36, which is indeed the answer.</p>
|
3,851,650 |
<p>We want to prove that the number of sequences <span class="math-container">$(a_1,...,a_{2n})$</span> such that
<span class="math-container">$$
• \text{ every } a_i \text{ is equal to} ±1;\\• a_1 + ··· + a_{2n} = 0;\\• \text{ every partial sum satisfies } a_1 + ··· + a_i > −2
$$</span>
is a Catalan number.</p>
<p>I've been trying to form a bijection between this and ballot sequences of size 2(n+1) by showing that you can remove any one +1 and any one -1 to yield our new sequences, but I am not sure if this is the best method. Any help would be super helpful!</p>
|
pipiyun
| 825,006 |
<p>This question is similar to the paths (0,2) to (2n,2) that do not touch X axis. The number of paths touches at least one point on the axis should be <span class="math-container">$2n \choose n+2$</span>.
The total number of paths should be <span class="math-container">$2n \choose n$</span>.
Then you can get the final answer, <span class="math-container">$2n \choose n+2$</span> - <span class="math-container">$2n \choose n$</span></p>
|
2,389,324 |
<p>For a continuous Function $f$, prove that:</p>
<p>$$\lim_{x\to0^+}\int_{x}^{2x} \frac{1}{t} f(t) dt = ln(2)f(0)$$ </p>
<p>I have already concluded that since f is continuous, it's therefore integrable.Moreover I assumed there is a function $F$
$$F(x)=\int_{x}^{2x} f(s)ds$$
in order to simplify the limit expression by partial integration. Unfortunately that gave me no solution and I am stuck.</p>
|
hamam_Abdallah
| 369,188 |
<p><strong>hint</strong> </p>
<p>the substitition $$t=ux $$ makes it easier.</p>
<p>$$I=\int_1^2 f (ux)\frac {du}{u} $$</p>
<p>use the second Mean formula and continuity at $x=0$.</p>
<p>$$I=f (c_x)\int_1^2\frac {du}{u} $$</p>
<p>Done.</p>
|
4,563,725 |
<p>Is it possible for two non real complex numbers a and b that are squares of each other? (<span class="math-container">$a^2=b$</span> and <span class="math-container">$b^2=a$</span>)?</p>
<p>My answer is not possible because for <span class="math-container">$a^2$</span> to be equal to <span class="math-container">$b$</span> means that the argument of <span class="math-container">$b$</span> is twice of arg(a) and for <span class="math-container">$b^2$</span> to be equal to <span class="math-container">$a$</span> means that arg(a) = 2.arg(b) but the answer is it is possible.</p>
<p>How is it possible when arg(b) = 2.arg(a) and arg(a) = 2.arg(b) contradict each other?</p>
|
Deepak
| 151,732 |
<p>It is entirely possible. Consider the complex conjugate cube roots of one. <span class="math-container">$\omega = -\frac 12 + \frac{\sqrt 3}{2}i$</span> and <span class="math-container">$\overline \omega = -\frac 12 - \frac{\sqrt 3}{2}i$</span></p>
<p>Note that <span class="math-container">$\omega = (\overline \omega)^2$</span>, but also <span class="math-container">$\overline \omega = \omega^2$</span>.</p>
<p>There is no contradiction here, because from the two equations, you get <span class="math-container">$\omega = \omega^4 = \omega^3\omega = 1\cdot \omega$</span>.</p>
<p>Going by arguments, <span class="math-container">$\mathrm{arg}(\omega) = \frac{2\pi}{3}$</span> while <span class="math-container">$\mathrm{arg}(\overline \omega) = \frac{4\pi}{3}$</span>. Clearly, <span class="math-container">$2\cdot \frac{2\pi}{3} = \frac{4\pi}{3}$</span> but also <span class="math-container">$2\cdot \frac{4\pi}{3} = \frac{8\pi}{3} \pmod{2\pi} = \frac{2\pi}{3}$</span>. Essentially, you're forgetting that arguments work modulo <span class="math-container">$2\pi$</span>.</p>
<p>Note that that pair is the only possible solution. If you write <span class="math-container">$x = y^2$</span> and <span class="math-container">$y = x^2$</span>, then <span class="math-container">$x = (x^2)^2 = x^4 \implies x^4 - x = 0 \implies x(x^3-1)=0 \implies x = 0$</span> or <span class="math-container">$x^3 - 1 = 0$</span>. The latter implies <span class="math-container">$(x-1)(x^2 + x + 1)=0 \implies x=1$</span> or <span class="math-container">$x^2 + x + 1=0$</span>. Solving the quadratic gives you the complex conjugate roots of one. Therefore, <span class="math-container">$x = 0, 1, \omega, \overline \omega$</span> are the only solutions. If you impose the additional constraint that <span class="math-container">$x \neq y$</span>, you're left with the only solutions being <span class="math-container">$\omega$</span> and <span class="math-container">$\overline \omega$</span>, as already stated.</p>
|
3,190,601 |
<p>Suppose <span class="math-container">$f$</span> is continuous on <span class="math-container">$[-1,1]$</span> and differentiable on <span class="math-container">$(-1,1)$</span>. Find:</p>
<p><span class="math-container">$$\lim_{x\rightarrow{0^{+}}}{\bigg(x\int_x^1{\frac{f(t)}{\sin^2(t)}}\: dt\bigg)}$$</span></p>
<p>I am trying to use l'hopital's rule with:</p>
<p><span class="math-container">$$\lim_{x\rightarrow{0^{+}}}{\bigg(\frac{\int_x^1{\frac{f(t)}{\sin^2(t)}}\: dt}{\frac{1}{x}}\bigg)}$$</span></p>
<p>However this only works if the integral evaluates to <span class="math-container">$\pm$</span>infinity. when <span class="math-container">$x=0$</span>. Since <span class="math-container">$f(t)$</span> is general I'm not really sure what to do next. </p>
<p>Help would be appreciated :)</p>
|
zhw.
| 228,045 |
<p>Hint: L'Hopital is applicable in the case where the denominator <span class="math-container">$\to \infty,$</span> no matter what the numerator is doing.</p>
|
2,523,570 |
<p>$15x^2-4x-4$,
I factored it out to this:
$$5x(3x-2)+2(3x+2).$$
But I don’t know what to do next since the twos in the brackets have opposite signs, or is it still possible to factor them out?</p>
|
T C Molenaar
| 497,768 |
<p>It has to be $-2$:
$$15x^2-4x-4 = 5x(3x-2)+2(3x-2) = (5x+2)(3x-2)$$</p>
|
2,523,570 |
<p>$15x^2-4x-4$,
I factored it out to this:
$$5x(3x-2)+2(3x+2).$$
But I don’t know what to do next since the twos in the brackets have opposite signs, or is it still possible to factor them out?</p>
|
Vassilis Markos
| 460,287 |
<p>Note that:
$$15x^2-4x-4=15x^2-10x+6x-4=5x(3x-2)+2(3x-2)=(5x+2)(3x-2)$$</p>
<p>You probably missed a sign! ;)</p>
|
1,461,311 |
<p>In axiomatic approach to real numbers, that is by defining them to be the complete ordered field, one is expected to prove every theorem and solve every problem by using ultimately only the axioms. I was trying to solve a Spivak's calculus problem that asked to show that the sum of any number of real numbers is still meaningful without using any parentheses. I think we have to use induction principle to solve this problem. But I don't know how to prove the induction principle using axioms of the reals.</p>
|
Will R
| 254,628 |
<p>The Principle of Induction is a property of the natural number system; see, for example, <a href="https://en.wikipedia.org/wiki/Peano_axioms" rel="nofollow">Peano's axioms</a>. Hence, there isn't really any way to "prove" the Principle of Induction from the ordered field axioms, i.e., $(\text{P}1)$-$(\text{P}12)$ in Spivak. As Spivak himself says in the text of Chapter 2:</p>
<blockquote>
<p>It is also possible to prove the principle of induction from the well-ordering principle (Problem 9). Either principle may be considered <strong>a basic assumption about the natural numbers</strong>.</p>
</blockquote>
<p>I have added the emphasis, here.</p>
<p>However, as I recall, there's a neat exercise towards the end of Chapter 2 (Problem 2-23 in mine) in which, starting with the reals $\Bbb{R}$, you define the natural numbers $\Bbb{N}$, which you do, basically, "by induction" (or rather, you define the notion of an "inductive set" of real numbers, and then the natural numbers are those numbers in every inductive set). In this sense, you can start with the reals, and <em>construct</em> the natural numbers in such a way that you are <em>guaranteed</em> that the Principle of Induction applies; probably this is as close as you're going to get.</p>
|
2,965,993 |
<p>Suppose you have the surface <span class="math-container">$\xi$</span> defined in <span class="math-container">$\mathbb{R}^3$</span> by the equation:
<span class="math-container">$$ \xi :\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 $$</span>
For <span class="math-container">$ x \geq 0$</span> , <span class="math-container">$ y \geq 0$</span> and <span class="math-container">$ z \geq 0$</span>. Now take any point <span class="math-container">$P \in \xi$</span> and consider the tangent plane (<span class="math-container">$\pi_t)$</span> to <span class="math-container">$\xi$</span> at <span class="math-container">$P$</span>. Calculate the minimum volume of the region determined by the <span class="math-container">$xy$</span>, <span class="math-container">$yz$</span>, <span class="math-container">$xz$</span> planes and <span class="math-container">$\pi_t$</span>.</p>
<p><img src="https://i.stack.imgur.com/6zwo0.png" alt="Ellipsoid and tetrahedron."></p>
|
Christian Blatter
| 1,303 |
<p>Assume for the moment <span class="math-container">$a=b=c=1$</span>. The surface <span class="math-container">$\xi$</span> then is the unit sphere <span class="math-container">$S^2$</span>. A typical point <span class="math-container">${\bf n}$</span> of <span class="math-container">$S^2$</span> in the first octant then has coordinates <span class="math-container">$(n_1,n_2,n_3)$</span> with <span class="math-container">$n_i>0$</span> and <span class="math-container">$\sum_i n_i^2=1$</span>. Furthermore the tangent plane to <span class="math-container">$S^2$</span> at <span class="math-container">${\bf n}$</span> is given by the equation <span class="math-container">${\bf n}\cdot{\bf x}=1$</span>, i.e., <span class="math-container">$n_1x_1+n_2x_2+n_3x_3=1$</span>. Intersecting this plane with the three coordinate axes gives the points <span class="math-container">$\bigl({1\over n_1},0,0\bigr)$</span>, <span class="math-container">$\bigl(0,{1\over n_2},0\bigr)$</span>, <span class="math-container">$\bigl(0,0,{1\over n_3}\bigr)$</span>. It follows that the simplex <span class="math-container">$S$</span> in question has volume
<span class="math-container">$${\rm vol}(S)={1\over 6 n_1n_2n_3}\ .$$</span>
This volume is minimal when its reciprocal is maximal. By the AMG inequality
<span class="math-container">$$\root3\of {n_1^2 n_2^2n_3^2}\leq{1\over3}(n_1^2+n_2^2+n_3^2)={1\over3}\ ,$$</span>
with equality iff <span class="math-container">$n_i=3^{-1/2}$</span> for all <span class="math-container">$i\in[3]$</span>. It follows that
<span class="math-container">$${\rm vol}(S)\geq{1\over6} 3^{3/2}={\sqrt{3}\over2}\ .$$</span>
In the case of arbitrary <span class="math-container">$a$</span>, <span class="math-container">$b$</span>, <span class="math-container">$c>0$</span> we therefore have
<span class="math-container">$${\rm vol}(S)\geq{\sqrt{3}\over2}abc\ ,$$</span>
by standard geometric principles.</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>
|
Thusle Gadelankz
| 840,795 |
<p>I recently revisited this problem and stumbled into another way around this, using a standard linear analysis trick. For the sake of completeness for future readers, I will add it in this answer.</p>
<p>It actually suffices to show that <span class="math-container">$$| \text{Re} \langle x,y \rangle | \leq \lVert x \rVert \lVert y \rVert$$</span> for all <span class="math-container">$x,y$</span>, which one gets from using the discriminant trick on the expansion of <span class="math-container">$\lVert \lambda x + y \rVert^2$</span>. To see this, consider vectors <span class="math-container">$x,y$</span>. If <span class="math-container">$\langle x, y \rangle = 0$</span>, there is nothing to show. Hence, suppose otherwise and let <span class="math-container">$\alpha = \frac{\overline{ \langle x, y \rangle}}{| \langle x, y \rangle| }$</span>. Then <span class="math-container">$| \alpha | = 1$</span> and we have</p>
<p><span class="math-container">\begin{align*}
0 \leq | \langle x, y \rangle | &= \alpha \langle x, y \rangle = \langle \alpha x, y \rangle \\
&= |\text{Re} \langle \alpha x, y \rangle| \leq \lVert \alpha x \rVert \lVert y \rVert \\
&= \lVert x \rVert \lVert y \rVert.
\end{align*}</span></p>
|
2,174,270 |
<p>I have an elliptic curve $E$ over $\mathbb{F}_{7}$ defined by $y^2=x^3+2$ with the point at infinity $\mathcal{O}$</p>
<p>I am given the point $(3,6)$ and need to find the line which intersects with $E$ at <strong>only this point</strong></p>
<p>I am told that this line is $y\equiv (4x+1)\mod 7$</p>
<p>I have verified that this is the case, however my question is, how would I go about finding that equation in the first place?</p>
<hr>
<p>What I'm asking for is a method, which would take in the elliptic curve $E$ and a point on the curve $P$ and output the line which intersects with $E$ at <strong>only this point</strong></p>
<p>In the example above, we would have the following</p>
<p>\begin{align}\text{input} &= \begin{cases}y^2=x^3 + 2\\
P=(3,6)\end{cases}\\
\text{output} &= y\equiv (4x+1)\mod 7\end{align}</p>
|
Travis Willse
| 155,629 |
<p>If the line $L$ intersects $E$ only at the given point $P$ (and the point $\mathcal O$ at infinity), then the intersection of $L$ and $E$ at $P$ must have multiplicity $2$, and so $L$ is the tangent line to $E$ at $P$. For an algebraic plane curve $\{p(x, y) = 0\}$, the tangent line at $P = (a, b)$ is
$$p_x(a, b)(x - a) + p_y(a, b)(y - b) = 0 ,$$
where $p_x$ and $p_y$ are the (formal) partial derivatives of $p$.</p>
<p>In our case, $E$ is the zero set of $p(x, y) := x^3 - y^2 + 2$, so $p_x(x, y) = 3 x^2$, and thus $p_x(3, 6) = -1$. Likewise we get $p_y(3, 6) = 2$, and substituting in the above equation gives the equation
$$-(x - 3) + 2(y - 6) = 0$$
for $L$. Multiplying both sides by $-3$ to eliminate the coefficient of $y$ and rearranging gives
$$y = 4 x + 1.$$</p>
|
1,900,788 |
<blockquote>
<p>The probability of rain in Greg's area on Tuesday is $0.3$. The probability that Greg's teacher will give him a pop quiz in Tuesday is $0.2$. The events occur independently of each other. What is the probability of neither events occur?</p>
</blockquote>
<p>My approach: </p>
<p>Probability of rain or quiz or both = $0.3+0.2= 0.5$</p>
<p>So, probability of neither= $1-0.5$ = $0.5$</p>
<p>Question: Actual probability of this problem is $0.7\cdot0.8$ = $0.56$. But, I don't understand what is the mistake in above approach? </p>
|
JMoravitz
| 179,297 |
<p>Note the following two things:</p>
<blockquote>
<p><strong>Principle of inclusion-exclusion (2-set case)</strong>
<span class="math-container">$$Pr(A\cup B) = Pr(A)+Pr(B)-Pr(A\cap B)$$</span></p>
<p><strong>Definition of independent events</strong></p>
<p>The following are equivalent statements</p>
<ul>
<li><p><span class="math-container">$A$</span> and <span class="math-container">$B$</span> are independent events</p>
</li>
<li><p><span class="math-container">$Pr(A\mid B) = Pr(A)$</span></p>
</li>
<li><p><span class="math-container">$Pr(B\mid A) = Pr(B)$</span></p>
</li>
<li><p><span class="math-container">$Pr(A\cap B)=Pr(A)\cdot Pr(B)$</span></p>
</li>
</ul>
</blockquote>
<p>For your problem: you know <span class="math-container">$Pr(A)=0.3, Pr(B)=0.2$</span> and they are independent, so</p>
<p><span class="math-container">$$Pr(A\cup B) = Pr(A)+Pr(B)-Pr(A\cap B) = 0.3+0.2-0.3\cdot 0.2 = 0.5-0.06 = 0.44$$</span></p>
<p>So the probability of at least one of the events occurring is <span class="math-container">$0.44$</span>. The probability of no events occurring is then one minus that, i.e. <span class="math-container">$1-0.44=0.56$</span>.</p>
<hr />
<p>Alternatively, if <span class="math-container">$A$</span> and <span class="math-container">$B$</span> are independent, then <span class="math-container">$A^c$</span> and <span class="math-container">$B^c$</span> are also independent. We have then <span class="math-container">$Pr(A^c\cap B^c)=Pr(A^c)Pr(B^c)=(1-0.3)(1-0.2)=0.7\cdot 0.8 = 0.56$</span></p>
|
2,596,457 |
<ul>
<li>If $\lim _{n\rightarrow \infty }a_{n}=a$ then $\left\{a_{n}:n\in\mathbb{N}\right\} \cup \left\{ a\right\}$ is compact.</li>
</ul>
<p><strong>I couldn't do anything. Can you give a hint?</strong></p>
<p>Note: in the question $a_n\in\mathbb{R}$.</p>
|
Gödel
| 467,121 |
<p><strong>Hint:</strong> Prove that $\{a_n:n\in\mathbb{N}\}\cup\{a\}$ is closed and bounded</p>
|
2,596,457 |
<ul>
<li>If $\lim _{n\rightarrow \infty }a_{n}=a$ then $\left\{a_{n}:n\in\mathbb{N}\right\} \cup \left\{ a\right\}$ is compact.</li>
</ul>
<p><strong>I couldn't do anything. Can you give a hint?</strong></p>
<p>Note: in the question $a_n\in\mathbb{R}$.</p>
|
Kenny Lau
| 328,173 |
<p>Let $X$ be an open cover of $\{ a_n \mid n \in \Bbb N \} \cup \{a\}$.</p>
<p>Then, there is an open set containing $a$, called $X_a$.</p>
<p>Since $\displaystyle \lim_{n \to \infty} a_n = a$, $X_a$ must contain all but finitely many members of $\{ a_n \mid n \in \Bbb N \}$.</p>
<p>Then, pick the open sets containing those finitely many members, and together with $X_a$ you obtain a finite subcover of $\{ a_n \mid n \in \Bbb N \} \cup \{a\}$.</p>
|
2,896,780 |
<p>With a trusted 3rd party, running Secret Santa is easy: The 3rd party labels each person $1,\dotsc,n$, and then randomly chooses a derangement from among all possible derangements of $n$ numbers. Person $i$ will then give a gift to the number in position $i$ of the derangement. The trusted 3rd party is responsible for keeping the derangement secure, and for telling each person whom to give a gift to.</p>
<p>The question is: Is there an algorithm that would allow Secret Santa to be played without a trusted 3rd party?</p>
<p>I thought perhaps a clever use of secret keys and a one way hash function could accomplish it, but I've failed to find an algorithm so far.</p>
<p>So I'm looking for a description of a valid algorithm or a (informal) proof that one does not exist.</p>
<h2>EDIT</h2>
<p>I believe my problem is different from the possible duplicate. I want a solution that will work for a distributed group of players. That is, you cannot assume the players are in the same room and have the ability to shuffle envelopes or notecards or anything like that. </p>
<p>To make this concrete, a valid solution must work across a group instant messenger or over group emails.</p>
<p>Also, to clarify again, it must be a random derangement and not merely a random n-cycle.</p>
|
Misha Lavrov
| 383,078 |
<p>Here's one possible strategy. Number the people $1, 2, \dots, n$. They do the following things, in order. (Whenever someone chooses a random number, I assume it's uniformly random from some fixed range $[1,N]$, say $[1,2n]$, excluding all values they've seen before.)</p>
<ol>
<li>Person $1$ picks a random number $x_1$ and passes it to person $2$.</li>
<li>Person $2$ picks a random number $x_2$, scrambles the list $(x_1,x_2)$, and passes it to person $3$.</li>
<li>Person $3$ picks a random number $x_3$, scrambles the list $(x_1,x_2,x_3)$, and passes it to person $4$.</li>
<li>And so on, with person $n$ receiving a permutation of the list $(x_1, x_2, \dots, x_{n-1})$.</li>
<li>Person $n$ picks a random number $x_n$, scrambles the list $(x_1, \dots, x_n)$, and passes that to person $1$.</li>
<li>Person $1$ records the position of $x_1$ (that's their secret Santa target), replaces $x_1$ by a new random number $y_1$, and passes the resulting list to person $2$.</li>
<li>Person $2$ records the position of $x_2$ (that's their secret Santa target), replaces $x_2$ by a new random number $y_2$, and passes the resulting list to person $3$.</li>
<li>And so on, until we get to person $n$, whose target is the position of $x_n$.</li>
</ol>
<p>This does not guarantee that the permutation is a derangement. But everyone can check if they got themselves as a target; if they complain, we can start over. (On average, we'll have to start over $e$ times.)</p>
<p>No information is shared about other people's targets: in steps 1 through 4, person $k$ sees the values $x_1, x_2, \dots, x_{k-1}$, which are not helpful, because in steps 5 through 8, person $k$ is given a permutation of the values $y_1, y_2, \dots, y_{k-1}, x_k, \dots, x_n$: from their perspective, these are uniformly chosen from the complement of $\{x_1,\dots, x_{k-1}\}$.</p>
<p>Another awkward moment, though, is that person $n$ picks the final permutation, so they get to choose their secret Santa target if they cheat and don't do it randomly. However, we can have the other people collaboratively choose a random seed for person $n$ to use (say, they send them numbers $r_1, \dots, r_{n-1}$, and then $r_1 + \dots + r_{n-1}$ has to be the seed) and then this cheating can be exposed after everyone opens their presents.</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>
|
Infinity_hunter
| 826,797 |
<p>This is regarding your second edit. First of all denoting <span class="math-container">$A_1 \subseteq A_2 .... \subseteq A$</span> is not correct when the index set <span class="math-container">$I$</span> is uncountable and moreover you don't know the elements of <span class="math-container">$I$</span>.</p>
<p>Take <span class="math-container">$A_i =(0,i)$</span> where <span class="math-container">$i\in (0,1)$</span> and <span class="math-container">$A = [0,1]$</span>. These are nested in the sense that <span class="math-container">$i \leq j$</span> implies that <span class="math-container">$A_i \subseteq A_j$</span> and observe that for <span class="math-container">$i$</span> in index set <span class="math-container">$A_i \subset A$</span>. Clearly union of all <span class="math-container">$A_i$</span> is the interval <span class="math-container">$(0,1)$</span> which is not <span class="math-container">$A$</span>.</p>
|
934,353 |
<p>I am a high school student in Calculus, and we are finishing learning basic limits. I am reviewing for a big test tomorrow, and I could do all of the problems correctly except this one.</p>
<p>I have no idea how to solve the problem this problem correctly. I looked up the answer online, but I can't figure out how they got their answer. All of the online tools show the steps using L'Hospital's rule or derivation, but I haven't learned either yet.</p>
<p>This is the problem:</p>
<p>$$\large\lim_{x\rightarrow 0}{\left(\frac{\frac{1}{\sqrt{1+x}}-1}{x}\right)}$$</p>
<p>This is the problem that I did incorrectly. I converted the $-1$ to $\frac{\sqrt{1+x}}{\sqrt{1+x}}$, then subtracted the fraction, and multiplied the result by $\frac{1}{x}$ to remove the double division.</p>
<p>$$\large\frac{1-\sqrt{1+x}}{x\sqrt{1+x}}$$</p>
<p>When I substitute $x$, I get $0$, but the answer is $-\frac{1}{2}$. I am doing something simple incorrectly, but I really cannot figure it out.</p>
|
Clinton Bradford
| 167,198 |
<p>As you did, convert to $\dfrac{1-\sqrt{1+x}}{x\sqrt{1+x}}$. (If you substitute, you still get $0$ in the denominator, so there is a little more work to be done.)</p>
<p>Then use the method of "algebraic conjugates":</p>
<p>$$\dfrac{1-\sqrt{1+x}}{x\sqrt{1+x}} \cdot \dfrac {1+\sqrt{1+x}}{1+\sqrt{1+x}} = \dfrac{1-(1+x)}{{x \sqrt{1+x}(1+\sqrt{1+x}})}$$</p>
<p>and simplify, then substitute, to your answer.</p>
<p>If you want to learn the formatting here, it's done in $\LaTeX$, between \$s. To do a fraction, write something like <code>$\frac{\sqrt{2}}{2}$</code>- there are lots of tutorials online.</p>
|
732,334 |
<p>How can we solve this equation?
$x^4-8x^3+24x^2-32x+16=0.$ </p>
|
wendy.krieger
| 78,024 |
<p>You could factorise it, in the manner of $(x-2)^4=0$. I saw those factors immediately.</p>
<p>One process is to note that $16$ has divisors, and one can try various combinations of this such that the sum gives eight.</p>
<p>Possibilities include $2, 2, 2, 2$ and $4, 4, 1, -1$. However, one can not produce the second set to give +16, so trying $(x-2)(x-2)(x-2)(x-2)$ is more likely than $(x-4)(x-4)(x-1)(x+1)$.</p>
|
803,800 |
<p>Define a sequence $(a_n)_{n = 1}^{\infty}$ of real numbers by $a_n = \sin(n)$. This sequence is bounded (by $\pm1$), and so by the <a href="http://en.wikipedia.org/wiki/Bolzano%E2%80%93Weierstrass_theorem" rel="noreferrer">Bolzano-Weierstrass Theorem</a>, there exists a convergent subsequence.</p>
<p><strong>My question is this:</strong> Is there any way to construct an explicit convergent subsequence?</p>
<p>Naïvely, I tried considering the subsequence $$\sin(3), \,\sin(31), \,\sin(314), \,\sin(3141),\, \sin(31415),\, \dots$$ hoping it would converge to $0$, but I was unable to prove this (and I think it's probably not even true).</p>
|
Dominic Michaelis
| 62,278 |
<p>I am not very confident that it converges, because this is how well you can approximate $10^k \cdot \pi$ with an integer. But as the difference will be infinite many times greater than $0.9$ and as you get bounds you see that this sequence is divergent.</p>
<p>I think this is more a question what is explicit for you, because you need to know how well you can approximate a multiple of $\pi$ with an Integer.</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>
|
Johann Cigler
| 5,585 |
<p>A simple proof is in the paper <a href="http://arxiv.org/abs/math/0609283" rel="noreferrer">"Fibonacci numbers and orthogonal polynomials"</a>
by Christian Berg.</p>
|
1,091,653 |
<p>Is this correct ?</p>
<p>$$
\frac{d}{dt} \left( \int_0^t \phi(t)dt \right) = \phi(t)
$$</p>
<p>If not, how can I recover $$ \phi(t) $$ knowing only $$ \int_0^t \phi(t)dt $$ ?</p>
|
drhab
| 75,923 |
<p>The expression $\int_{0}^{t}\phi\left(t\right){\rm d}t$ makes no
sense to me. It cannot be that $t$ is a variable and a constant at
the same time. I presume that you mean $\int_{0}^{t}\phi\left(x\right){\rm d}x$ </p>
<p>For $t>0$ prescribe $\phi(t)$ by $t\mapsto 0$ if $t\neq 1$ and $t\mapsto c$ otherwise, where $c$ is some constant. </p>
<p>Then $\Phi\left(t\right)=\int_{0}^{t}\phi\left(x\right){\rm d}x=0$ for
each $t$ so that $\Phi'\left(t\right)=0$ for each $t$. </p>
<p>However,
if $c\neq0$ then $\Phi'\left(1\right)=0\neq c=\phi\left(1\right)$. </p>
<p>This can be done with every $c$ so apparantly $\phi(1)$ cannot be recovered.
Things get better if you demand $\phi(t)$ to be continuous.</p>
|
164,328 |
<p>What are good introductory textbooks available on Cohomology of Groups?</p>
|
Zurab Silagadze
| 32,389 |
<p>At the undergraduate level, some elements of group cohomology is explained in John Moody's book "Groups for Undergraduates" <a href="http://rads.stackoverflow.com/amzn/click/9810221053" rel="nofollow">http://www.amazon.com/Groups-Undergraduates-John-Atwell-Moody/dp/9810221053</a> At more advanced (but still undergraduate) level, see Maxim Stykow's Honours B.Sc. thesis "Introduction to
Group Cohomology" <a href="http://www.math.ubc.ca/~maxim/stykow_group_cohomology.pdf" rel="nofollow">http://www.math.ubc.ca/~maxim/stykow_group_cohomology.pdf</a></p>
<p>At the graduate level, a good choice seems to be the above mentioned Ken Brown's book "Cohomology of Groups" <a href="http://rads.stackoverflow.com/amzn/click/0387906886" rel="nofollow">http://www.amazon.com/Cohomology-Groups-Graduate-Texts-Mathematics/dp/0387906886</a> </p>
|
1,553,440 |
<p>Let , $g(x)=f(x)+f(1-x)$ and $f'(x)<0$ for all $x\in (0,1)$. Then , $g$ is monotone increasing in </p>
<p>(A) $(1/2,1)$.</p>
<p>(B) $(0,1/2)$</p>
<p>(C) $(0,1/2)\cup (1/2,1)$.</p>
<p>(D) none.</p>
<p>We have , $g'(x)=f'(x)-f'(1-x)$. Now $g'(x)>0$ if $f'(x)>f'(1-x)$. As $f'(x)<0$ so , $f$ is monotone decreasing in $(0,1)$. From here how I can conclude ?</p>
|
Hamit
| 277,958 |
<p>Can u conclude this inequality???
$$f(\frac{2}{8})+f(\frac{6}{8})<f(\frac{1}{8})+f(\frac{7}{8}) $$
So $D$ is the answer.</p>
|
2,965,756 |
<p>I want to intuitively say tht the answer is yes, but if it so happens that <span class="math-container">$|\mathbf{a}|=|\mathbf{b}|cos(\theta)$</span>, where <span class="math-container">$\theta$</span> is the angle between the two vectors, then the equation will be satisfied without the two vectors being the same.</p>
<p>However, my friend keeps telling me that I'm wrong and that this would contradict the given result in our homework question anyway, which tells us that <span class="math-container">$\mathbf{a}\times\mathbf{b} = \mathbf{a}-\mathbf{b}$</span> and then asks us to prove <span class="math-container">$\mathbf{a}=\mathbf{b}$</span> (the equation in the question was obtained by dotting both sides with <span class="math-container">$\mathbf{a}$</span>. </p>
<p>Which one of us is wrong, and why?</p>
|
Kavi Rama Murthy
| 142,385 |
<p>If <span class="math-container">$a$</span> is the zero vector and <span class="math-container">$b$</span> is not zero you get a contradiction to your statement. </p>
|
263,983 |
<p>Is there a method to measure <strong>theoretic time load</strong> (hope +theoretic memory load) of an evaluation?</p>
<p>They should not change - <strong>always generate same value after pressing Enter key</strong>(=evaluation).</p>
<p>Hope they are even same in different environment(=different computers).</p>
<p>For example</p>
<pre><code>Timing[1000000!;]
</code></pre>
<p>gives sometimes <code>0.125</code>, sometimes <code>0.140625</code>, sometimes <code>0.15625</code>.</p>
<p>I want an imaginary function <code>TheoreticTiming</code></p>
<pre><code>TheoreticTiming[1000000!;]
</code></pre>
<p>which outputs always <code>0.125</code></p>
<p>There are built-in function <code>AbsoluteTiming</code> but it also changes for each evaluation, just like <code>Timing</code>.</p>
|
Syed
| 81,355 |
<p>Mathematica runs on general-purpose operating systems (Windows/Linux/MacOS). On such systems, a multitude of services (daemons) are constantly running in the background. These services load the system differently at different times. The OS alone determines when programs get access to the requested resource (CPU, Memory, Disk, Network etc).</p>
<p>A CPU is a resource. You may run multiple kernels through Mathematica, but access to the CPU core(s) is granted by the OS depending on other tasks awaiting execution in the queue. The answer to your question is "No" as it is not the design goal of a general purpose OS to provide the same (or pre-determined) execution trajectory (resource assignment) to any program or service on first/successive runs (let alone a calculation).</p>
<hr />
<p>(Not relevant to the question but) On the CPU level itself, the instruction pipelines are not deterministically enqueued meaning that the CPU prefetches instructions based on a heuristic algorithm. You can look up "branch predictor". Furthermore, the CPU may or may not execute instructions in the same order as submitted (lookup "Out of Order Execution"). Add to all this a cache scheduling algorithm trying to minimize memory access.</p>
<hr />
<p>There is a hint of an xy-problem in your question. If you want guaranteed execution time for a real-time application, you will have to resort to other (more deterministic) platforms.</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>
|
DanielWainfleet
| 254,665 |
<p>Let $A=\{n: x_n\in X\}$ and $B=\{n: x_n\in Y\}.$ At least one of $A,B$ is an infinite set. </p>
<p>If $A$ is infinite let $A=\{f(n):N\in\Bbb N\}$ where $f(n)<f(n+1).$ The sequence $\sigma=(x_{f(n)})_{n\in \Bbb N}$ is a sub-sequence of $(x_n)_{n\in\Bbb N}.$ </p>
<p>All of the entries of $\sigma$ belong to $X$ so $\sigma$ has a sub-sequence $\tau$ converging to some $x\in X.$ Now $\tau$ is also a sub-sequence of $(x_n)_{n\in \Bbb N}$ and of course $x\in X\cup Y.$ </p>
<p>Similarly if $B$ is infinite.</p>
|
774,332 |
<p>I have a radial Schrödinger equation for a particle in Coulomb potential:</p>
<p>$$i\partial_t f(r,t)=-\frac1{r^2}\partial_r\left(r^2\partial_r f(r,t)\right)-\frac2rf(r,t)$$</p>
<p>with initial condition</p>
<p>$$f(r,0)=e^{-r^2}$$</p>
<p>and boundary conditions</p>
<p>$$\begin{cases}
|f(0,t)|<\infty\\
|f(\infty,t)|<\infty.
\end{cases}$$</p>
<p>Trying to solve it, I couldn't come up to any analytical solution, so I tried to solve it numerically.</p>
<p>Simplest what I thought of was finite differences method. But while I can limit domain to make it finite, imposing zero Dirichlet condition at $r=A$ for some $A$, I would need to somehow impose the condition of regularity at $r=0$, where $f$ doesn't have to be neither zero, nor non-zero in general. I don't know how to do this with finite differences.</p>
<p>Another approach I tried was to expand the initial condition in eigenfunctions of the Hamiltonian operator (the RHS of the PDE) and then approximate the solution taking finite number of eigenstates into account. But the expansion appeared to converge extremely slowly, and after taking 200 bound eigenstates (simple because of closed-form solution in terms of Laguerre polynomials) and 70 eigenstates of continuous spectrum (taking those of them which vanish at $r=A$, so as to impose zero boundary condition at $A$; had $A=10$), I still got nothing similar to initial function.</p>
<p>So, the question is: is there any explicit solution for this IBVP (be it closed-form one or in some sort of series, but necessarily explicit and fast convergent)? If no, how can it be solved numerically?</p>
|
Dmoreno
| 121,008 |
<p>Regarding the regularity condition when applying finite differences, some authors make the following approach:</p>
<p>Consider the RHS of the equation at $r \to 0 $ and compute the limit with the help of L'Hôpital's rule:</p>
<p>$$\lim_{r\to 0} \left(- \frac{1}{r^2} \partial_r (r^2 f_r) - \frac{2}{r} f \right) = \lim_{r\to 0} \left( - f_{rr} - \frac{2}{r} f_r - \frac{2}{r} f \right) = \lim_{r\to 0} \left( -f_{rr} - 2 f_{rr} - 2 f_r \right), $$</p>
<p>and hence you have:</p>
<p>$$ - \frac{1}{r^2} \partial_r (r^2 f_r) - \frac{2}{r} f \approx -3 f_{rr} - 2 f_r.$$</p>
<p>Now, you may discretize your equation at $i = 0$, if $\, r_i = r_0 + i \Delta r$, using this formula, avoiding the geometrical singularity.</p>
<p>Although this is a pretty "rough" way to approximate both the Laplacian and the source term, it's quite well known when applying a finite difference scheme.</p>
<p>I hope this helps you.</p>
<p>Cheers.</p>
|
774,332 |
<p>I have a radial Schrödinger equation for a particle in Coulomb potential:</p>
<p>$$i\partial_t f(r,t)=-\frac1{r^2}\partial_r\left(r^2\partial_r f(r,t)\right)-\frac2rf(r,t)$$</p>
<p>with initial condition</p>
<p>$$f(r,0)=e^{-r^2}$$</p>
<p>and boundary conditions</p>
<p>$$\begin{cases}
|f(0,t)|<\infty\\
|f(\infty,t)|<\infty.
\end{cases}$$</p>
<p>Trying to solve it, I couldn't come up to any analytical solution, so I tried to solve it numerically.</p>
<p>Simplest what I thought of was finite differences method. But while I can limit domain to make it finite, imposing zero Dirichlet condition at $r=A$ for some $A$, I would need to somehow impose the condition of regularity at $r=0$, where $f$ doesn't have to be neither zero, nor non-zero in general. I don't know how to do this with finite differences.</p>
<p>Another approach I tried was to expand the initial condition in eigenfunctions of the Hamiltonian operator (the RHS of the PDE) and then approximate the solution taking finite number of eigenstates into account. But the expansion appeared to converge extremely slowly, and after taking 200 bound eigenstates (simple because of closed-form solution in terms of Laguerre polynomials) and 70 eigenstates of continuous spectrum (taking those of them which vanish at $r=A$, so as to impose zero boundary condition at $A$; had $A=10$), I still got nothing similar to initial function.</p>
<p>So, the question is: is there any explicit solution for this IBVP (be it closed-form one or in some sort of series, but necessarily explicit and fast convergent)? If no, how can it be solved numerically?</p>
|
Ruslan
| 64,206 |
<p>Another approach to set a regularity condition is to make a change of variables to switch the type of boundary condition. For this equation, if $f(r)$ is regular at $r=0$, then $\frac{\text{d}}{\text{d}\rho}f(\rho^2)=0$ at $\rho=0$.</p>
<p>We can switch to $g(\rho)=f(\rho^2)$ and rewrite the equation for $g(\rho)$:</p>
<p>$$i\partial_tg(\rho,t)=-\frac1{4\rho^2}\partial_{\rho\rho}g(\rho,t)-\frac3{4\rho^3}\partial_\rho g(\rho,t)-\frac2{\rho^2}g(\rho,t)$$</p>
<p>with boundary conditions:</p>
<p>\begin{cases}
g^{(1,0)}(0,t)=0\\
|g(\infty,t)|<\infty,
\end{cases}</p>
<p>i.e. we switched from regularity condition at $r=0$ to homogeneous Neumann boundary condition at $\rho=0$. Now this is trivial to setup for finite differences.</p>
<hr>
<p>Yet simpler approach is to use $q(r,t)=rf(r,t)$. Then the equation transforms to:</p>
<p>$$i\partial_t q(r,t)=-\partial_{rr}q(r)-\frac2rq(r,t),$$</p>
<p>and the regularity condition becomes homogeneous Dirichlet condition (because finite $f$ is multiplied by zero $r$). Now we can't set good boundary condition at infinity though, but for numerical treatment this is not a problem. For finite domain $r\in[0,L]$ we have the following boundary conditions:</p>
<p>\begin{cases}
q(0,t)=0\\
q(L,t)=0,
\end{cases}</p>
<p>and this problem is also trivial to setup for finite differences.</p>
|
3,343,550 |
<p>Show that if <span class="math-container">$(a,15)=1$</span>, then <span class="math-container">$a^4\equiv1 \mod 15$</span>, so that we do not have primitive roots of <span class="math-container">$15$</span>
Please help me with this problem.</p>
|
Bernard
| 202,857 |
<p><strong>Hint</strong>:</p>
<p>Use the <em>Chinese remainder theorem</em> to check that if <span class="math-container">$a$</span> is coprime with <span class="math-container">$15$</span>, i.e. prime to <span class="math-container">$3$</span> and to <span class="math-container">$5$</span>, the order of <span class="math-container">$a\bmod 15$</span> is the l.c.m. of the orders of <span class="math-container">$a\bmod 3$</span> and <span class="math-container">$\bmod 5$</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>
|
John
| 7,163 |
<p>Here's a hint on applying <a href="https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)" rel="noreferrer">stars and bars</a> to your problem.</p>
<p>The guy has $14$ animals of some combination of type. The animals are the "stars":</p>
<p>$$\star\star\star\star\star\star\star\star\star\star\star\star\star\star$$</p>
<p>You can divide the starts into three groups using $3-1=2$ "bars." Here's one way:</p>
<p>$$\star\star\star | \star\star\star\star\star\star | \star\star\star\star\star$$</p>
<p>If we say that the first, second, and third groups are cats, dogs, and guinea pigs, respectively, then the above situation would count the case of $3$ cats, $6$ dogs, and $5$ guinea pigs.</p>
<p>You can also have something like this:</p>
<p>$$\star\star\star\star\star\star\star || \star\star\star\star\star\star\star$$</p>
<p>There are no stars between the bars, so therefore no dogs! (Which is sad, but I digress.) In other words, he has $7$ cats and $7$ guinea pigs.</p>
<p>Or, a much happier situation:</p>
<p>$$|\star\star\star\star\star\star\star\star\star\star\star\star\star\star|$$</p>
<p>$14$ dogs, and no other types.</p>
<p><strong>It boils down to figuring out the number of places you can put the bars.</strong></p>
<p>Can you take it from here?</p>
|
3,029,585 |
<p>I have been stumped for a few days on this...It would be great if anyone can point me to enlightenment :)</p>
<p>Here's what I have tried. Let <span class="math-container">$Y = X^3$</span>, where X is a standard normal distribution with mean 0 and variance 1. Then</p>
<p><span class="math-container">$P(Y \leq y) = P(X^3 \leq y) = P(X \leq y^{1/3})$</span>, for both <span class="math-container">$y$</span> positive or negative.</p>
<p>The above is the CDF of <span class="math-container">$X^3$</span>; differentiating w.r.t. to <span class="math-container">$y$</span> should give me the PDF, which is</p>
<p><span class="math-container">$f_X(y^{1/3})\frac{1}{3}y^{-2/3}$</span>, where <span class="math-container">$f_X$</span> is the PDF of the standard normal. </p>
<hr>
<p>But THAT cannot be the answer, because it would take on negative values for <span class="math-container">$y < 0$</span>. Whereas a PDF should always be non-negative. I have seen a "correct" answer in a math paper, where it changes <span class="math-container">$y$</span> to <span class="math-container">$|y|$</span> and results in an integral that converges (in Mathematica). </p>
<p>But where can I introduce the absolute terms!? Much thanks :D</p>
|
Hagen von Eitzen
| 39,174 |
<p>Any element <span class="math-container">$\alpha$</span> of <span class="math-container">$F$</span> is a rational expression in the numbers adjoined. As such, it can involve only <em>finitely</em> many of the <span class="math-container">$\sqrt[2^k]3$</span>. If in such an expression, <span class="math-container">$\sqrt[2^n]3$</span> is one with maximal <span class="math-container">$k$</span>, then all other <span class="math-container">$\sqrt[2^k]3$</span> are powers of <span class="math-container">$\sqrt[2 k]3$</span>. It follows that <span class="math-container">$\alpha\in\Bbb Q(\sqrt[2^n]3)$</span> and <span class="math-container">$\alpha $</span> is algebraic.</p>
<p>A different approach for the infinity part: By Eisenstein, the polynomial <span class="math-container">$X^{2^n}-3$</span> is irreducible. Hence <span class="math-container">$[F:\Bbb Q]\ge[\Bbb Q(\sqrt[2^n]3):\Bbb Q]\ge 2^n$</span>, where <span class="math-container">$n$</span> is arbitrariy. It follows that <span class="math-container">$[F:\Bbb Q]$</span> is inifnite.</p>
|
3,481,531 |
<p>I was solving a problem where there are 4 points and we need to check if they form a square or not. Now if all 4 points are the same, is it square or not. My initial thought was it is a zero-sided square but wanted to clarify if my understanding is right or not. </p>
|
Z Ahmed
| 671,540 |
<p>A point say (h,k) is a point circle:of zero radius.
<span class="math-container">$$(x-h)^2+(y-k)^2=0$$</span> or also a point-square <span class="math-container">$$|x-h|+|y-k|=0.$$</span></p>
|
3,518,232 |
<p>I am given with a Triangle <span class="math-container">$ABC$</span> which is inscribed in circle <span class="math-container">$\omega$</span>. The tangent lines to <span class="math-container">$\omega$</span> at <span class="math-container">$B$</span> and <span class="math-container">$C$</span> meet at <span class="math-container">$T$</span>. Point <span class="math-container">$S$</span> lies on ray <span class="math-container">$BC$</span> such that <span class="math-container">$AS$</span> is perpendicular to <span class="math-container">$AT$</span>. Points <span class="math-container">$B_1$</span> and <span class="math-container">$C_1$</span> lies on ray <span class="math-container">$ST$</span> (with <span class="math-container">$C_1$</span> in between <span class="math-container">$B_1$</span> and <span class="math-container">$S$</span>) such that <span class="math-container">$B_{1}T=BT=C_{1}T$</span>. I need to prove that the triangles <span class="math-container">$ABC$</span> and <span class="math-container">$AB_1C_1$</span> are similar to each other. </p>
|
dan_fulea
| 550,003 |
<p>Here is a rewritten version of the OP with a picture. The order of "inventing" the points (in the solution using inversion) is the same as the order of the "inverted" points in a lemma. The lemma is simple, extracts the essence and gives the bridge for the solution. (Writing the lemma is not so simple, points have to be constructed in the right order.)</p>
<blockquote>
<p><strong>Proposition</strong>:
There are given two circles with centers <span class="math-container">$O$</span> and <span class="math-container">$T$</span>, which intersect orthogonally in two points, <span class="math-container">$B$</span> and <span class="math-container">$C$</span>. The two circles are denoted by <span class="math-container">$(O)$</span>, <span class="math-container">$(T)$</span>, using their centers, if there is no danger of confusion.</p>
<p>Let <span class="math-container">$U=OT\cap BC$</span> be the mid point of the segment <span class="math-container">$BC$</span>.</p>
<p>Let <span class="math-container">$S$</span> be a point on the <strong>ray</strong> = half-line <span class="math-container">$BC$</span>, so that the circle with diameter <span class="math-container">$ST$</span> intersects <span class="math-container">$(O)$</span> in two points, <span class="math-container">$A$</span> and <span class="math-container">$a$</span>. Here <span class="math-container">$A$</span> is on the bigger arc <span class="math-container">$\overset\frown{BC}$</span>.</p>
<p>Let <span class="math-container">$B_1,C_1$</span> be the intersections of the <strong>line</strong> <span class="math-container">$ST$</span> with the circle <span class="math-container">$(T)$</span> , so that the points <span class="math-container">$B_1,T,C_1,S$</span> appear in this order on the line.</p>
<p><a href="https://i.stack.imgur.com/lchVV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lchVV.png" alt="math stackexchange problem, dan_fulea, 3518232"></a></p>
<p><strong>Then we have the following properties:</strong></p>
<p><span class="math-container">$(1)$</span> The points <span class="math-container">$B,T,C_1,A$</span> are on a circle.</p>
<p><span class="math-container">$(2)$</span> The points <span class="math-container">$B,T,B_1,a$</span> are on a circle.</p>
<p><span class="math-container">$(3)$</span> The points <span class="math-container">$A,B,B_1,S$</span> are on a circle.</p>
<p><span class="math-container">$(1')$</span> The points <span class="math-container">$C,T,B_1,A$</span> are on a circle.</p>
<p><span class="math-container">$(2')$</span> The points <span class="math-container">$C,T,C_1,a$</span> are on a circle.</p>
<p><span class="math-container">$(3')$</span> The points <span class="math-container">$A,C,C_1,S$</span> are on a circle.</p>
<p><span class="math-container">$(4)$</span> The triangles <span class="math-container">$\Delta ABB_1$</span>, <span class="math-container">$\Delta ACC_1$</span> are similar.</p>
<p><span class="math-container">$(5)$</span> The triangles <span class="math-container">$\Delta ABC$</span>, <span class="math-container">$\Delta AB_1C_1$</span> are similar.</p>
</blockquote>
<hr>
<p>Before we proceed, we consider the situation obtained by applying an inversion <span class="math-container">$X\to X'$</span> centered in <span class="math-container">$B$</span>. Let <span class="math-container">$D$</span> be the intersection of <span class="math-container">$BT$</span> with the circle <span class="math-container">$(T)$</span>, so that <span class="math-container">$BD$</span> is a diameter of <span class="math-container">$(T)$</span>. By inversion, circles through <span class="math-container">$B$</span> are mapped into projective lines, so that <span class="math-container">$B\to B'=\infty$</span>, and <span class="math-container">$\infty\to\infty'=B$</span>. A line through <span class="math-container">$B$</span> is mapped in the same line. The inversion picture is then constructed using points from the following lemma.</p>
<hr>
<blockquote>
<p><strong>Lemma:</strong> Let <span class="math-container">$B$</span> be a point in the plane. Consider two perpendicular lines passing through <span class="math-container">$B$</span>, and two points <span class="math-container">$A', T'$</span> on them respectively, so that we have
<span class="math-container">$$BA'\perp BT'\ .$$</span> </p>
<p>Let <span class="math-container">$D'$</span> be the mid point of segment <span class="math-container">$BT'$</span>.</p>
<p>The perpendiculars in <span class="math-container">$D'$</span> and <span class="math-container">$A'$</span> on the lines <span class="math-container">$BD'T'$</span> and respectively <span class="math-container">$BA'$</span> intersect in a point <span class="math-container">$C'$</span>. </p>
<p>Let <span class="math-container">$U'$</span> be the reflection of <span class="math-container">$B$</span> w.r.t. <span class="math-container">$C'$</span>. (So <span class="math-container">$C'$</span> is the mid point of <span class="math-container">$BU'$</span>.)</p>
<p>The circle <span class="math-container">$(A'U'T')$</span> intersects the lines <span class="math-container">$A'C'$</span> in <span class="math-container">$a'$</span>, so that <span class="math-container">$A'a'$</span> is the diameter of this circle.</p>
<p>The line <span class="math-container">$A'T'$</span> intersects <span class="math-container">$C'D'$</span> in a point <span class="math-container">$B_1'$</span>.</p>
<p>We consider the heights in triangle <span class="math-container">$\Delta A'a'B_1'$</span>, and let its orthocenter be <span class="math-container">$C_1'$</span>. Let <span class="math-container">$S'\in B_1'A'$</span> be the foot of the height from <span class="math-container">$a'$</span>, <span class="math-container">$a'S'\perp B_1'A'$</span>.
and <span class="math-container">$BC'U'$</span> in <span class="math-container">$S'$</span>.</p>
<p>The circle <span class="math-container">$(BS'T')$</span> intersects <span class="math-container">$D'C'$</span> in two points, <span class="math-container">$B_1', C_1'$</span>, chosen so that <span class="math-container">$C_1'$</span> is between <span class="math-container">$D'$</span> and <span class="math-container">$C'$</span>.</p>
<p><a href="https://i.stack.imgur.com/ZSpPi.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZSpPi.jpg" alt="math stackexchange, inversion solution, problem 3518232, dan_fulea"></a></p>
<p><strong>Then we have:</strong></p>
<ul>
<li><span class="math-container">$C_1'$</span> is the intersection of the diagonals in the parallelogram <span class="math-container">$A'D'T'C'$</span>, so it is the mid point of <span class="math-container">$D'C'$</span>, and of <span class="math-container">$A'T'$</span>.</li>
<li>The points <span class="math-container">$B, S', C_1', T', B_1'$</span> are on the circle.</li>
<li>The points <span class="math-container">$B,S',C',U'$</span> are colinear.</li>
</ul>
</blockquote>
<hr>
<p><strong>Proof of the lemma:</strong> Consider the four rectangles with a vertex in <span class="math-container">$C'$</span> which are congruent to the rectangle <span class="math-container">$A'C'D'B$</span>, and the other vertices are among <span class="math-container">$A',B,D',T',U'$</span> or reflections of them w.r.t. <span class="math-container">$C'$</span>.
We see that <span class="math-container">$A'C'$</span> goes through the mid point of <span class="math-container">$T'U'$</span>. So <span class="math-container">$A'C'$</span> is the side bisector of <span class="math-container">$T'U'$</span> (in the isosceles triangle <span class="math-container">$\Delta A'T'U'$</span>), this implies that <span class="math-container">$A'a'$</span> is a diameter in the circumcircle
<span class="math-container">$(A'S'T'a'U')$</span> of this triangle. So <span class="math-container">$A'T'\perp T'a'$</span>.</p>
<p>By construction, <span class="math-container">$C_1'=A'T'\cap B_1'C'$</span> (intersection of two heights), so <span class="math-container">$C_1'=
D'C'\cap A'T'$</span> is the intersection of the diagonals in the parallelogram <span class="math-container">$A'C'T'D'$</span>.</p>
<p>The triangles <span class="math-container">$\Delta B_1'C_1'T'$</span>, <span class="math-container">$\Delta B_1'C_1'S'$</span> have both a right angle opposite to the side <span class="math-container">$B_1'C_1'$</span>, so <span class="math-container">$B_1'C_1'$</span> is the diameter of the circle <span class="math-container">$B_1'T'C_1'S'$</span>, and on this circle there is also <span class="math-container">$B$</span>, the reflection of <span class="math-container">$T'$</span> w.r.t. the same diameter. </p>
<p>We need now to show that <span class="math-container">$B',S',C'$</span> are colinear. For this we compute the angle
<span class="math-container">$$
\begin{aligned}
\widehat{BS'C'}
&=
\widehat{BS'B_1'}
+\widehat{B_1'S'C_1'}
+\widehat{C_1'S'C'}
\\
&=
\widehat{BT'B_1'}
+90^\circ
+\widehat{C_1'A'C'}
\\
&=
\widehat{A'a'T'}
+90^\circ
+\widehat{T'A'C'}
\\
&=
90^\circ
+90^\circ
=180^\circ\ .
\end{aligned}
$$</span>
This finishes the lemma.</p>
<p><span class="math-container">$\square$</span></p>
<hr>
<p><strong>Proof of the proposition:</strong> Lines of colinearity containing points <span class="math-container">$X',Y',Z',\dots$</span> in the lemma correspond to circles through <span class="math-container">$B$</span> containing points inverted points <span class="math-container">$X,Y,Z$</span>. So we have the claimed propositions
<span class="math-container">$(1)$</span>, <span class="math-container">$(2)$</span>, <span class="math-container">$(3)$</span>;
<span class="math-container">$(1')$</span>, <span class="math-container">$(2')$</span>, <span class="math-container">$(3')$</span>. Let us use these circles to obtain equalities of angles.
We have:
<span class="math-container">$$
\begin{aligned}
\widehat{ABS} &= \widehat{AB_1S} &&\text{ since $(ABB_1S)$ cyclic.}\\
\widehat{ACS} &= \widehat{AC_1S} &&\text{ since $(ACC_1S)$ cyclic. Passing to suplements:}\\
\widehat{ACB} &= \widehat{AC_1B_1}\ .
\\
\widehat{CAC_1}
&= \widehat{CSC_1} &&\text{ since $(ACC_1S)$ cyclic,}\\
&= \widehat{BSB_1} \\
&= \widehat{BAB_1} &&\text{ since $(ABB_1S)$ cyclic. This implies}\\
\widehat{BAC} &= \widehat{B_1AC_1}
\ .
\end{aligned}
$$</span></p>
<p><span class="math-container">$\square$</span></p>
|
1,059,989 |
<p>I mean are there examples of problems that have been proven to be undecidable, in the sense that it would not be possible to devise a deterministic computer program that outputs a solution for an instance of the problem. And yet human mathematicians have come up with such a solution.</p>
|
Matt Samuel
| 187,867 |
<p>It is common to find that undecidable problems often have a solution in certain special cases even though there is no general solution. The halting problem is an excellent example; certainly it is possible to prove that (most) programs written by humans intended for a practical purpose do eventually halt.</p>
|
2,893,388 |
<p>My textbook is confusing me a little. Here is a worked example from my textbook:</p>
<blockquote>
<p>Line <span class="math-container">$l$</span> has the equation <span class="math-container">$\begin{pmatrix}3\\ -1\\ 0\end{pmatrix}+\lambda \begin{pmatrix}1\\ -1\\ 1\end{pmatrix}$</span> and point <span class="math-container">$A$</span> has co-ordinates <span class="math-container">$(3, 9, -2)$</span>.</p>
<p>Find the coordinates of point <span class="math-container">$B$</span> on <span class="math-container">$l$</span> so that <span class="math-container">$AB$</span> is perpendicular to <span class="math-container">$l$</span>.</p>
<p><span class="math-container">$\vec{AB\cdot }\begin{pmatrix}1\\ -1\\ 1\end{pmatrix}=0$</span></p>
<p><span class="math-container">$\vec{OB}=r=\begin{pmatrix}3+\lambda \\ -1-\lambda \\ \lambda \end{pmatrix}$</span></p>
<p><span class="math-container">$\vec{AB}=\begin{pmatrix}\lambda \\ -10-\lambda \\ \lambda +2\end{pmatrix}$</span></p>
<p><span class="math-container">$\begin{pmatrix}\lambda \\ -10-\lambda \\ \lambda +2\end{pmatrix}\cdot \begin{pmatrix}1\\ -1\\ 1\end{pmatrix}=0$</span></p>
<p><span class="math-container">$3\lambda= -12, \lambda = -4$</span></p>
<p>Coordinates of <span class="math-container">$B$</span>: <span class="math-container">$(-1, 3, -4)$</span></p>
</blockquote>
<p>The thing I don't understand is why they found the dot product of the line AB and the direction vector of line l. My textbook does mention that to check whether two vectors are perpendicular, <span class="math-container">$a\cdot b = 0 $</span> and for lines, the dot product of their direction vectors = 0. So why did they mix both here? Didn't they use the entire line <span class="math-container">$AB$</span> and then just the direction vector of line l? Or am I missing something as usual?</p>
|
Simon Terrington
| 302,396 |
<p>OK so if $(X, \tau)$ has a countable basis then there may still be some bases that are not countable. If we let $X$ be the real line and $\tau$ the usual set of open sets then $\tau$ is a basis for the reals, as is the set of all open intervals, but the set of all open intervals with rational end points is a countable basis. </p>
<p>In a general topological space, every set can be written as a union of members of the basis. This need not be a countable union. In the case of the basis given above for the real line (the set of all open intervals with rational end points) then is so happens that every set can be written as a countable union but that is a happy accident.</p>
<p>And answering your last point, you are right, the converse is not true. If we take the set of all open sets as the basis then every set can be written as the union of a single set (itself) but the basis is not countable.</p>
|
52,677 |
<p>Why people have to find quadratic formula,isn't that the formula cannot solve a polynomial with 2 and 1/2 degree?
and just curious, how many roots does a polynomial with 2 and 1/2 degree have and how to solve them all(by formula)?</p>
|
André Nicolas
| 6,312 |
<p>A <strong>polynomial</strong> $P(x)$ in one indeterminate with coefficients in a ring $A$ is an expression of the form
$$a_nx^n+a_1x^{n-1}+ \cdots + a_0$$
where $n$ is an integer $\ge 0$ and the <em>coefficients</em> $a_0$, $a_1$, $\dots$, $a_n$ are elements of the ring $A$.</p>
<p>Familiar examples are polynomials with integer coefficients (the ring $A$ is then the ring $\mathbb{Z}$ of integers), polynomials with rational coefficients (the ring $A$ is then the ring $\mathbb{Q}$ of rational numbers), and polynomials with real coefficients.</p>
<p>There are much more "exotic" examples, of great theoretical and practical importance. For example, polynomials with coefficients in the set $\{0,1\}$, with addition and multiplication defined modulo $2$, are important in coding theory.
The arithmetic here is very simple, $0+0=0\;$, $0+1=1+0=1\;$, $1+1=0\;$, $0\cdot 0=0\;$, $0\cdot 1=1\cdot 0=0\;$, $1\cdot 1=1$.</p>
<p>In addition to polynomials with one indeterminate, we can also consider polynomials in two indeterminates, such as $3x^2y^3-7xy+17$, three indeterminates, and so on.</p>
<p>The expression
$$6x^{5/2} -4x^{3/2} -x+17$$
is therefore technically <em>not</em> a
polynomial. However, in this example, let $y=x^{1/2}$. Then we can rewrite the above expression as
$$6y^5 -4y^3-y^2 +17$$
which is definitely a polynomial in $y$. We ordinarily say that
$6x^{5/2} -4x^{3/2} -x+17$ is not a polynomial "in" $x$, but is a polynomial
in $x^{1/2}$. </p>
<p>Not that this helps if we want to find roots, since solving even polynomial equations "by formula" is hopeless by degree $5$.</p>
<p>The Quadratic Formula can be adapted to solve some equations that strictly speaking are not polynomial equations. For example, look at the equation
$$x-x^{1/2}-1=0.$$</p>
<p>Make the substitution $y=x^{1/2}$.
We obtain the equation
$$y^2-y-1=0$$
which has the solutions
$$y=\frac{1\pm\sqrt{5}}{2}.$$</p>
<p>So now we know the possible values of $x^{1/2}$, and we can square them to get the possible values of $x$.</p>
<p>But please remember that when the plain word "polynomial" is used, the exponents (powers) used are non-negative integers.</p>
<p>About your question on the number of solutions of a polynomial equation of "degree" $5/2$, the first thing I would say is that it is not a polynomial equation. In certain cases, you would be able to make a substitution, like our earlier $y=x^{1/2}$, and turn the equation into a polynomial equation in $y$. But depending on the values of other exponents, the number of solutions could be large.</p>
<p>And if you look for example at the equation $x^{5/2}-x^{\pi/2}+1=0$, no substitution will turn this into a polynomial equation.</p>
<p>I hope this was not overly elaborate and technical!</p>
|
1,250,459 |
<p>Let $A\in M_{m \times n}(\mathbb{R})$, $x\in \mathbb{R}^n$ and $b,y\in \mathbb{R}^m$. Show that if $Ax=b$ and $A^ty=0_{\mathbb{R}^m}$, then $\langle b,y\rangle=0$. Also make a geometric interpretation.</p>
<p>I think I may have something to do with overdetermined / underdetermined system, but do not know how to prove it.</p>
<p>Take this opportunity to ask for the appointment of a linear algebra and matrix reference, I'm able to do the numerical exercises mostly, but those involving statements are a big problem.</p>
|
TravisJ
| 212,738 |
<p>I (personally) think of closed sets the way I explained here: <a href="https://math.stackexchange.com/questions/1185719/how-would-i-explain-an-open-set/1185749#1185749">How would I explain an 'open set'</a></p>
<p>One way to think of closed sets is they are the complement of an open set. (In many instances this may even be the definition of a closed set.)</p>
<p>The way I think of closed sets is in terms of convergence of sequences... if you take any sequence $s_{n}\to s$ where each $s_{i}\in S$ then if $S$ is closed you are guaranteed that $s\in S$ too. The "intuitive" way to think of this is that if you are moving around inside a set but slowing down and coming to a stop (eventually a stop... after $\infty$-ly many steps). Once you've slowed down enough, you can predict where you will stop. If it must be the case that your prediction is always a stopping point inside the set, then it is closed. Of course in practice, a sequence could be very bouncy and not come to a stop evenly at all... but this is how I think about it.</p>
|
1,434,486 |
<p>In case the question didn't display in the title correctly: How do we bound the function $\frac{-1}{2}\sum_{m=n}^{\infty}\frac{1}{m^{2}}$?</p>
<p>I think a way I can do this is to show that $$\sum_{m=n}^{\infty}\frac{1}{m^{2}} < \frac{1}{n^{2}} + \int_{n}^{\infty}\frac{1}{x^{2}}$$</p>
<p>But I'm a little unsure of the process it takes to get there.</p>
|
André Nicolas
| 6,312 |
<p>Estimating via the integral works well. Another way is to note that the tail (if we forget about the $-\frac{1}{2}$ in front) is less than
$$\frac{1}{(n-1)(n)}+\frac{1}{(n)(n+1)}+\frac{1}{(n+1)(n+2)}+\cdots.$$</p>
<p>But $\frac{1}{(n-1)(n)}=\frac{1}{n-1}-\frac{1}{n}$ and the next term is $\frac{1}{n}-\frac{1}{n+1}$, and the next term is $\frac{1}{n+1}-\frac{1}{n+2}$, and so on. Now note the massive cancellation (telescoping). We get $\frac{1}{n-1}$.</p>
<p>We conclude that the tail is less than $\frac{1}{n-1}$. A similar argument shows that the tail is greater than $\frac{1}{n}$, so we have obtained reasonably tight bounds for the tail.</p>
<p><strong>Remark:</strong> Since the post expresses uncertainty about the estimation with the integral, we give a brief explanation. The sum $\sum_{n+1}^\infty \frac{1}{k^2}$ can be thought of as the combined area of a bunch of rectangles, all of base $1$. The base of the first rectangle is $[n,n+1]$ and its height is $\frac{1}{(n+1)^2}$. The base of the second rectangle is $[n+1,n+2]$ and its height is $\frac{1}{(n+2)^2}$, and so on.</p>
<p>Draw the curve $y=\frac{1}{x^2}$. In the interval $[n,n+1]$ this curve is above the first rectangle, so the first rectangle has area less than $\int_n^{n+1}\frac{dx}{x^2}$. In the same way, the second rectangle has area less than $\int_{n+1}^{n+2}\frac{dx}{x^2}$, and so on. So the sum of the areas of all the rectangles is less than $\int_n^\infty \frac{dx}{x^2}$. </p>
|
1,434,486 |
<p>In case the question didn't display in the title correctly: How do we bound the function $\frac{-1}{2}\sum_{m=n}^{\infty}\frac{1}{m^{2}}$?</p>
<p>I think a way I can do this is to show that $$\sum_{m=n}^{\infty}\frac{1}{m^{2}} < \frac{1}{n^{2}} + \int_{n}^{\infty}\frac{1}{x^{2}}$$</p>
<p>But I'm a little unsure of the process it takes to get there.</p>
|
DanielWainfleet
| 254,665 |
<p>The process is this: For every $m>n$ and for $m-1 \leq x <m$ let $f(x)=1/m$. And let $g(x)=1/x$ for all $x >0$. Then for all $m>n$ we have $f(x) \le g(x)$ with equality only when $x$ is an integer greater than $n $.Therefore $$\sum_{m=n}^{\infty} 1/m^2=1/n^2 + \sum_{m=n+1}^{\infty} 1/m^2=$$ $$=1/n^2 +\sum_{m=n+1}^{\infty} \int_{x=m-1}^{x=m} f(x) dx =$$= $$=1/n^2+\int_{x=n}^{\infty} f(x) dx $$ $$ <1/n^2+\int_{x=n}^{\infty} g(x) dx =$$ $$= 1/n^2 +\int_{x=n}^{\infty}( 1/x) dx=1/n^2 +1/n.$$</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>
|
user
| 505,767 |
<p>We have that</p>
<p><span class="math-container">$$\frac{z-1}{z+1}=\sqrt 2 e^{i\theta} \implies z=\frac{1+\sqrt 2 e^{i\theta}}{1-\sqrt 2 e^{i\theta}}$$</span></p>
<p>and <span class="math-container">$$z+3= \frac{4-2\sqrt 2 e^{i\theta}}{1-\sqrt 2 e^{i\theta}} \implies |z+3|^2=\frac{4-2\sqrt 2 e^{i\theta}}{1-\sqrt 2 e^{i\theta}} \frac{4-2\sqrt 2 e^{-i\theta}}{1-\sqrt 2 e^{-i\theta}} =\frac{24-16\sqrt 2 \cos \theta}{3-2\sqrt 2 \cos \theta}=8$$</span></p>
|
398,409 |
<p>I was asked to help someone with this problem, and I don't really know the answer why. But I thought I'd still try.</p>
<p>$$\lim_{t \to 10} \frac{t^2 - 100}{t+1} \cos\left( \frac{1}{10-t} \right)+ 100$$</p>
<p>The problem lies with the cos term. What can I do with the cos term to remove divide by 0 ? </p>
<p>I found the answer to be $100$ (Google), but I do not know what they did to the $\cos$ term. Is that even the answer ?</p>
<p>Thanks!</p>
|
Benjamin Dickman
| 37,122 |
<p><strong>Incomplete Answer:</strong>
At least for the cube root case, this - like the square root case - follows from a theorem of Katok (2007). You can also get nice approximations in both the square root case and cube root case.</p>
<p>For more details on the above, see (<a href="http://www.ccsenet.org/journal/index.php/jmr/article/download/24259/15867" rel="nofollow">link</a>):</p>
<p>Ignacio, P. S., Addawe, J., Alangui, W., & Nable, J. (2013). Computation of Square and Cube Roots of $p$-Adic Numbers via Newton-Raphson Method. Journal of Mathematics Research, 5(2), p. 31.</p>
|
398,409 |
<p>I was asked to help someone with this problem, and I don't really know the answer why. But I thought I'd still try.</p>
<p>$$\lim_{t \to 10} \frac{t^2 - 100}{t+1} \cos\left( \frac{1}{10-t} \right)+ 100$$</p>
<p>The problem lies with the cos term. What can I do with the cos term to remove divide by 0 ? </p>
<p>I found the answer to be $100$ (Google), but I do not know what they did to the $\cos$ term. Is that even the answer ?</p>
<p>Thanks!</p>
|
Lubin
| 17,760 |
<p>Let me start from @Cantlog’s response, which points out that the difficulty is in knowing when $1+pa$ is an $m$-th power. The answer depends strongly on the divisibility of $m$ by $p$. In particular, if $m$ is indivisible by $p$, then $1+pa$ is <em>always</em> an $m$-th power. Indeed, as @Jyrki points out, $(1+pa)^z$ has a well-defined meaning for any $p$-adic integer $z$. For, if $\{n_i\}$ is a sequence of positive integers with $p$-adic limit equal to $z$, one shows (it’s a nice exercise) that $\{(1+pa)^{n_i}\}$ is Cauchy, and you define $(1+pa)^z$ to be the limit.</p>
<p>When $m$ is indivisible by $p$, $1/m$ is a $p$-adic integer, and by the remarks above, you’re good to go. So the question reduces to what the $pa$’s are such that $1+pa$ is a $p$-th power. When $p>2$, you see that $(1+pb)^p=1+p^2bu$, where $u$ is a $p$-adic unit. This also is an easy consequence of the Binomial Theorem. So, as long as $a$ itself is divisible by $p$, $1+pa$ is a $p$-th power. The story is just a little more complicated for $p=2$, because the expansion of $(1+z)^2$ involves so few terms. But if $a$ is a $p$-adic unit, then $1+pa$ is definitely not a $p$-th power.</p>
|
1,108,918 |
<p>Given two functions $f$ and $g$ whose derivatives $f'$ and $g'$ satisfy : $f'(x)=g(x), g'(x)=-f(x),f(0)=0, g(0)=1$ for all $x$ in an interval $J$. $~~~\cdots(A)$</p>
<p>(a) Prove that $f^2(x)+g^2(x)=1 ~\forall~x \in J $</p>
<p>(b) Let $F$ and $G$ be another pair of functions in $J$ which satisfy the given conditions $(A)$. Prove that $f(x)=F(x)$ and $g(x)=G(x)$.</p>
<p><strong>Attempt:</strong> $(a)$</p>
<p>$f^2(x)+g^2(x)+c= 2[ ~\int f(x) ~d ~(f(x)) + ~\int g(x) ~d ~(g(x))~]$</p>
<p>$=2 [ ~\int f(x) ~ g(x)~ dx - ~\int g(x) ~f(x)~ dx~]$</p>
<p>$ = 0$</p>
<p>Applying the initial conditions, we get :$f^2(x)+g^2(x)=1$</p>
<p>$(b)$</p>
<p>Consider : $h(x)=[f(x)-F(x)]^2+[g(x)-G(x)]^2$</p>
<p>If we prove that $h(x)=0~\forall~x \in J$, then $f(x)=F(x), g(x)=G(x)$</p>
<p>Expanding : $h(x)=f^2(x)+F^2(x)-2f(x)F(x)+g^2(x)+G^2(x)-2g(x)G(x)$</p>
<p>$=2[1-f(x)F(x)-g(x)G(x)]$</p>
<blockquote>
<p>How do I prove that $f(x)F(x)+g(x)G(x) =1$?</p>
</blockquote>
<p>Thank you for your help.</p>
|
Martin R
| 42,969 |
<p>You have already defined
$$
h(x)=[f(x)-F(x)]^2+[g(x)-G(x)]^2
$$
You only need to differentiate this equation:</p>
<p>$$
\begin{align}
h'(x) &= 2 [f(x)-F(x)][f'(x)-F'(x)] + 2 [g(x)-G(x)][g'(x)-G'(x)] \\
&= 2 [f(x)-F(x)][g(x)-G(x)] - 2 [g(x)-G(x)][f(x)-F(x)] \\
&= 0
\end{align}
$$</p>
<p>so that $h$ is constant and therefore identical to $h(0) = 0$.</p>
|
3,652,879 |
<p>If <span class="math-container">$\langle x_n\rangle $</span> is a sequence of positive real numbers such that <span class="math-container">$$x_{(n+2)}=\frac{(x_{n+1}+ x_{n})}{2}$$</span> for all <span class="math-container">$n \in \mathbb{N},\ $</span> let <span class="math-container">$x_1 <x_2$</span></p>
<p>then subsequence of odd terms is increasing and subsequence of even terms is decreasing .But how to prove it mathematically? We have </p>
<p><span class="math-container">$$x_{(n+2)}-x_n=\frac{(x_{n+1}- x_{n})}{2}= \frac{(x_{n}- x_{n-2})}{4}$$</span> How to proceed from here? Any hint please.</p>
|
Hrishabh
| 776,474 |
<p>Suppose that <span class="math-container">$x_n <= x_2$</span> for all n less than or equal to k then <span class="math-container">$x_{k+1}=\frac{{x_k}+x_{k-1}}{2} <= \frac{2x_2}{2}=x_2$</span> so by induction sequence is bounded.Now suppose <span class="math-container">$x_{n-1} <= x_{n}$</span> for all n less than or equal to k,then <span class="math-container">$x_{k+1}=\frac{{x_k}+x_{k-1}}{2} => \frac{2x_k}{2}=x_k$</span> so by induction sequence is increasing.As it is bounded and increasing it is convergent.</p>
|
1,540,412 |
<p>Let x be a (right) eigenvector of A corresponding to an eigenvalue λ and let y be a left eigenvector of A corresponding to a different eigenvalue µ, where λ $\neq$ µ. Show that x∗y = 0. Hint : Ax = λx and y'A = µy'</p>
|
Aaron
| 9,863 |
<p><strong>Hint:</strong> Calculate $y'Ax$ two different ways, and relate the answer to $y'x$.</p>
|
2,834,219 |
<p>I'd like to numerically evaluate the following integral using computer software but it has a singularity at $x=1$:</p>
<p>\begin{equation}
\int_1^{\infty} \frac{x}{1-x^4} dx
\end{equation}</p>
<p>I was thinking of a variable transformation, rewriting the expression, or something of a kind. One of my attempts was to do differentiation under the integral sign, but I only got so far:</p>
<p>\begin{equation}
I(b) = \int_1^{\infty} \frac{x}{1-x^4}e^{-bx} dx
\end{equation}</p>
<p>where setting $b=0$ gives the original expression. Differentiating w.r.t. $b$ gives</p>
<p>\begin{equation}
I'(b) = -\int_1^{\infty} \frac{x^2}{1-x^4}e^{-bx} dx = \int_1^{\infty} \frac{1}{1+x^2}e^{-bx} dx - \int_1^{\infty} \frac{1}{1-x^4}e^{-bx} dx
\end{equation}</p>
<p>It remains to integrate the two terms w.r.t. $x$ and then $b$, but they are not standard integrals.</p>
<p>Is there a better, faster, or easier way that I'm not thinking of?</p>
|
José Carlos Santos
| 446,262 |
<p>You can't do that, because$$\frac x{x-x^4}=\frac1{1-x}\times\frac x{1+x+x^2+x^3}$$and therefore near $1$ this functions behaves as $\frac1{4(1-x)}$. And if $a>1$, the integral$$\int_1^a\frac1{4(1-x)}\,\mathrm dx$$diverges.</p>
|
323,971 |
<p>I know basic things about cardinality (I'm only in High School) like that since $\mathbb{Q}$ is countable, its cardinality is $\aleph_0$. Also that the cardinality of $\mathbb{R}$ is $2^{\aleph_0}$.</p>
<blockquote>
<p>Are there any direct applications of these numbers outside of theoretical math?</p>
</blockquote>
<p>I know this can be convenient for certain proofs and help understanding sets of numbers, but are there any applications of this?</p>
|
Asaf Karagila
| 622 |
<p>The answer is an obvious <em>no</em>. For two main reasons:</p>
<ol>
<li><p>With the exception of occasional naive approach to sets, there is little to no use of set theory outside theoretical mathematics. So any application would be indirect and purely coincidental.</p></li>
<li><p>Applied mathematics is not concerned with infinite objects. Let alone "<em>vastly huge beyond any reasonable visualization and imagination of a human being</em>" sizes of infinity.</p></li>
</ol>
<p>It is important to understand that mathematics is not "merely a tool for engineers" (or physicists). It is a world filled with magic and mind boggling ideas which have absolutely nothing to do with this physical reality. Infinite sets is one of them. These ideas trickle slowly and some of them eventually get to the point where they have some use, but these uses are far from being "direct" in any sense of the word.</p>
<p>For example, by plain cardinality arguments it is easy to see that almost any function from $\Bbb R$ to itself is not continuous, or even Borel measurable. Almost any continuous function is nowhere differentiable, and almost all the differentiable functions are not continuously differentiable, and so on and so forth (although some of these arguments require more than sheer cardinality).</p>
<p>But have you ever seen someone "applying" everywhere-discontinuous functions to a real world situation? I can't recall anything like that (although it might be in some quantum theory sort of application I am unaware of).</p>
<p>As long as mankind is limited by a finite powers of perception we cannot even distinguish between $100^{100^{100^{100^{100}}}}$ and $\aleph_0$.</p>
<p><strong>Might also be relevant:</strong> <a href="https://math.stackexchange.com/questions/154234/can-we-distinguish-aleph-0-from-aleph-1-in-nature">Can we distinguish $\aleph_0$ from $\aleph_1$ in Nature?</a></p>
|
368,800 |
<p>Let <span class="math-container">$G$</span> be a finite group, let <span class="math-container">$X$</span> be a locally compact Hausdorff space, and let <span class="math-container">$G$</span> act freely on <span class="math-container">$X$</span>. It is well-known that the canonical quotient map <span class="math-container">$\pi\colon X\to X/G$</span> onto the orbit space <span class="math-container">$X/G$</span> admits local cross-sections. More precisely, for every <span class="math-container">$z\in X/G$</span> there are an open set <span class="math-container">$U$</span> in <span class="math-container">$X/G$</span> containing <span class="math-container">$z$</span>, and a continuous function <span class="math-container">$s\colon U\to X$</span> such that <span class="math-container">$\pi\circ s$</span> is the identity on <span class="math-container">$U$</span>. In particular, there is an open cover of <span class="math-container">$X/G$</span> consisting of sets where a local cross-section can be defined.</p>
<p><strong>Question</strong>: is there a <em>finite</em> open cover of <span class="math-container">$X/G$</span> consisting of sets where a local cross-section can be defined?</p>
<p>(This is the same as asking whether the Schwarz genus of the fiber map <span class="math-container">$X\to X/G$</span> is finite.)</p>
<p>The answer is "yes" if <span class="math-container">$X$</span> (or at least <span class="math-container">$X/G$</span>) is finitistic, so in particular whenever <span class="math-container">$X$</span> has finite covering dimension, and clearly also whenever <span class="math-container">$X$</span> is compact. I wonder if it is true in general.</p>
|
Mark Grant
| 8,103 |
<p>There is a general cohomological lower bound for the Schwarz genus of a map <span class="math-container">$p:E\to B$</span>. Namely, if there are cohomology classes <span class="math-container">$x_1,\ldots , x_k\in H^*(B)$</span> such that <span class="math-container">$0=p^*(x_i)\in H^*(E)$</span> for all <span class="math-container">$i=1,\ldots , k$</span> and <span class="math-container">$x_1\cup\cdots \cup x_k \neq 0$</span>, then the genus of <span class="math-container">$p$</span> is greater than <span class="math-container">$k$</span>. Here the coefficients are completely arbitrary, in particular can be twisted. (This is a generalisation of the cup-length lower bound for Lusternik-Schnirelmann category, since the LS-category of a space <span class="math-container">$X$</span> is equal to the genus of any fibration over <span class="math-container">$X$</span> with contractible total space.)</p>
<p>So you can get many counter-examples using this cohomological criterion. In fact, whenever <span class="math-container">$X$</span> is a contractible CW-complex then it is a model for <span class="math-container">$EG$</span>, and <span class="math-container">$X/G$</span> is a model for <span class="math-container">$BG$</span>. The cup-length of <span class="math-container">$BG$</span> is always infinite for a finite group <span class="math-container">$G$</span> (with appropriately chosen, possibly twisted coefficients). This generalises the example in Hannes Thiel's answer.</p>
|
1,344,883 |
<p>I had an idea that passes by declaring a new type of <strong>computer variable</strong> (like Integer, Double, etc.) <strong>that represents a statistical probability distribution (PDF)</strong>, for that I would need to define the basic operations; sum, multiplication, inverse and negation.</p>
<p>The problem is that I have no idea of how to define such operations or if that is even possible. (I'm electrical engineer...)</p>
<p>So the question is; How to <code>sum</code>, <code>multiply</code>, <code>inverse</code> and <code>negate</code> probability distributions, or where can I learn how to do it. A numerical approach would be sufficient.</p>
<p>Thanks.</p>
|
Tryss
| 216,059 |
<p>This can be rewritten as</p>
<p>$$ xe^x = 1$$</p>
<p>And here, there is no "simple" answer, you need to introduce the
<a href="https://en.wikipedia.org/wiki/Lambert_W_function#Applications" rel="nofollow">Lambert W function</a>.</p>
|
1,752,506 |
<p>Question: $ \sqrt{x^2 + 1} + \frac{8}{\sqrt{x^2 + 1}} = \sqrt{x^2 + 9}$</p>
<p>My solution: $(x^2 + 1) + 8 = \sqrt{x^2 + 9} \sqrt{x^2 + 1}$</p>
<p>$=> (x^2 + 9) = \sqrt{x^2 + 9} \sqrt{x^2 + 1}$</p>
<p>$=> (x^2 + 9) - \sqrt{x^2 + 9} \sqrt{x^2 + 1} = 0$</p>
<p>$=> \sqrt{x^2 + 9} (\sqrt{x^2 + 9} - \sqrt{x^2 + 1}) = 0$</p>
<p>So, either $\sqrt{x^2 + 9} = 0$ or $(\sqrt{x^2 + 9} - \sqrt{x^2 + 1}) = 0$</p>
<p>From the first expression, I get $x = \pm 3 i$ and from the second expression, I get nothing.</p>
<p>Now, notice how in the 2nd step, I could've divided both the sides by $\sqrt{x^2 + 9}$, but I didn't because I learned here that we must never do that and that we should always factor: <a href="https://math.stackexchange.com/questions/1751410/how-to-square-both-the-sides-of-an-equation#comment3574205_1751435">Why one should never divide by an expression that contains a variable.</a></p>
<p>So, my question is: is the solution above correct? Would it have been any harm had I divided both the sides by $\sqrt{x^2 + 9}$?</p>
|
Community
| -1 |
<p>The equation says$$ \sqrt{x^2 + 1} + \frac{8}{\sqrt{x^2 + 1}}=\frac{x^2 + 9}{\sqrt{x^2 + 1}} = \sqrt{x^2 + 9}.$$</p>
<p>So either $x^2+9=0$ and $x=\pm3i$, or $\sqrt{x^2+9}=\sqrt{x^2+1}$, which is impossible (by squaring).</p>
|
246,571 |
<p>How can I calculate the following limit epsilon-delta definition?</p>
<p>$$\lim_{x \to 0} \left(\frac{\sin(ax)}{x}\right)$$</p>
<p>Edited the equation, sorry...</p>
|
Brusko651
| 50,608 |
<p>This is easy if you know the power series of the sine function.</p>
<p><span class="math-container">$$\sin(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{6} + \frac{x^5}{120} - \frac{x^7}{5040} +\cdots$$</span></p>
<p>Then <span class="math-container">$$\sin(ax) = a x - \frac{a^3 x^3}{6} + \frac{a^5 x^5}{120} -\cdots$$</span></p>
<p>and <span class="math-container">$\frac{\sin(ax)}{x} = a - \frac{a^3 }{6}x^2 + \frac{a^5 }{120}x^4 -\cdots$</span> for all <span class="math-container">$x \neq 0$</span>. Since power series are continuous you can find the limit for <span class="math-container">$x \rightarrow 0$</span> by simply setting <span class="math-container">$x=0$</span> in this expression, so the limit is <span class="math-container">$a$</span>.</p>
|
2,906,797 |
<p>I want to express this polynomial as a product of linear factors:</p>
<p>$x^5 + x^3 + 8x^2 + 8$</p>
<p>I noticed that $\pm$i were roots just looking at it, so two factors must be $(x- i)$ and $(x + i)$, but I'm not sure how I would know what the remaining polynomial would be. For real roots, I would usually just do use long division but it turns out a little messy in this instance (for me at least) and was wondering if there was a simpler method of finding the remaining polynomial. </p>
<p>Apologies for the basic question!</p>
|
Mohammad Riazi-Kermani
| 514,496 |
<p>If you divide $$ x^5 + x^3 + 8x^2 + 8$$ by $$(x-i)(x+i) = x^2+1$$ you will get $$x^3+8$$ which factors as $$(x^3+8) = (x+2)(x^2-2x+4)$$ which has a solution of $x=-2$</p>
<p>Now use quadratic formula to solve $x^2-2x+4=0$ to find other roots and factor if you wish.</p>
|
1,187,142 |
<p>Let a(n) be an arithmetic sequence and, as usual let $A(n)=a(1) + a(2) + \dots + a(n).$ If $A(5) = 50$ and $A(20) = 650$, find $A(15).$</p>
<p>I'm not so sure how to solve that.
So </p>
|
Joffan
| 206,402 |
<p>Using $d$ to denote the common difference between terms, we can see that the terms from $a_6$ to $a_{10}$ are each exactly $5d$ more than the terms from $a_1$ to $a_5$. </p>
<p>This means that
$$\begin{align}A(10) &= 2A(5)+25d \\
\text{Similarly, }A(15) &= 3A(5)+75d \\
\text{ and } A(20) &=4A(5) + 150d
\end{align}$$
Then it's apparent that $A(15)= A(20)/2 + A(5) = 325+50=375$.</p>
<p>Of course you can also compute $d$ from this if you want to, but it isn't needed here.</p>
|
890,988 |
<p>My book states that: </p>
<p>$f$ is a quasiconcave function on $U$ if for all $x,y \in U $ and $t \in [0,1]$: </p>
<p>$f(x) \geq f(y) \implies f(tx + (1 - t)y) \geq f(y)$ </p>
<p>$f$ is a quasiconvex function on $U$ if for all $x,y \in U $ and $t \in [0,1]$: </p>
<p>$f(y) \leq f(x) \implies f(tx + (1 - t)y) \leq f(x)$</p>
<p>Now take a look to the graph below $f(x)=xy$. I know it should be quasi-concave, but given the definitions above I seems to be both quasi-convave and quasi-convex. What am I doing wrong?</p>
<p><img src="https://i.stack.imgur.com/mzj4p.png" alt="enter image description here"> </p>
<p>Thanks!</p>
|
Community
| -1 |
<p>Nothing is wrong. Every monotone function is both quasiconvex and quasiconcave.</p>
<p>Indeed, the definition of quasiconvexity amounts to saying that on every closed interval, the function attains its maximum at an endpoint. Same for quasiconcavity, except replace maximum with minimum. Every monotone function has both of these properties. </p>
|
890,988 |
<p>My book states that: </p>
<p>$f$ is a quasiconcave function on $U$ if for all $x,y \in U $ and $t \in [0,1]$: </p>
<p>$f(x) \geq f(y) \implies f(tx + (1 - t)y) \geq f(y)$ </p>
<p>$f$ is a quasiconvex function on $U$ if for all $x,y \in U $ and $t \in [0,1]$: </p>
<p>$f(y) \leq f(x) \implies f(tx + (1 - t)y) \leq f(x)$</p>
<p>Now take a look to the graph below $f(x)=xy$. I know it should be quasi-concave, but given the definitions above I seems to be both quasi-convave and quasi-convex. What am I doing wrong?</p>
<p><img src="https://i.stack.imgur.com/mzj4p.png" alt="enter image description here"> </p>
<p>Thanks!</p>
|
user160518
| 451,930 |
<p>I see incorrect answers on this topic everywhere and nobody is correcting them.
In the case of z= f(x,y)= xy, the graph that you draw ( rectangular hyperbola) is a level curve and not the function of the graph. The graph is originally in 3-D. To check its properties ( mainly quasi-concavity and quasi- convexity), level curves are used. I see people getting confused among the two.</p>
<p>Property of monotonicity is a property of functions and not of level curves. </p>
<p>F(x,y) = xy is therefore not quasiconvex and only quasiconcave ( because the upper contour set is a convex set).</p>
|
890,988 |
<p>My book states that: </p>
<p>$f$ is a quasiconcave function on $U$ if for all $x,y \in U $ and $t \in [0,1]$: </p>
<p>$f(x) \geq f(y) \implies f(tx + (1 - t)y) \geq f(y)$ </p>
<p>$f$ is a quasiconvex function on $U$ if for all $x,y \in U $ and $t \in [0,1]$: </p>
<p>$f(y) \leq f(x) \implies f(tx + (1 - t)y) \leq f(x)$</p>
<p>Now take a look to the graph below $f(x)=xy$. I know it should be quasi-concave, but given the definitions above I seems to be both quasi-convave and quasi-convex. What am I doing wrong?</p>
<p><img src="https://i.stack.imgur.com/mzj4p.png" alt="enter image description here"> </p>
<p>Thanks!</p>
|
nrazzak
| 737,658 |
<p>The function <span class="math-container">$f(x,y) = xy$</span> is neither quasiconcave nor quasiconvex and you can show that by solving the bordered Hessian.</p>
<p>For a formal definition, let <span class="math-container">$f(x,y)$</span> be a function with continuous partial derivatives and continuous cross partial derivatives in a convex set <span class="math-container">$S$</span>. Let <span class="math-container">$D_r$</span> be the determinant of its <span class="math-container">$rth$</span> order bordered Hessian. If <span class="math-container">$f$</span> is quasiconcave then <span class="math-container">$D_1 \leq 0 $</span> , <span class="math-container">$D_2 \geq 0$</span> and <span class="math-container">$D_n \leq 0$</span> if <span class="math-container">$n$</span> is odd and <span class="math-container">$D_n \geq 0$</span> if <span class="math-container">$n$</span> is even for all <span class="math-container">$x$</span> in <span class="math-container">$S$</span>.</p>
|
760,654 |
<p>If $\lambda$ is the eigenvalue of matrix $A$,what is the eigenvalue of $A^TA$?I have no clue about it. Can anyone help with that?</p>
|
Klaas van Aarsen
| 134,550 |
<p>Generally, you won't be able to say much about them.</p>
<p>However, if $A$ is for instance real symmetric, it is diagonalizable with real eigenvalues, meaning there is an orthogonal matrix B (that is with $B^{-1}=B^T$) and a diagonal matrix D such that:
$$A = B D B^T$$
$$A^TA = (B D B^T)^T B D B^T = B D^T B^T B D B^T = BD^2B^T$$
In other words, the eigenvalues of $A^TA$ are the squares of the eigenvalues of $A$.</p>
|
410,663 |
<p>Can anyone in detail explain the procedure for finding the harmonic conjugate of the function $$u(x,y) = x^{2} - y\cdot (y+1)$$</p>
<p>I am new to this and I would like to know. </p>
|
Pedro
| 23,350 |
<p>OK, so $$u_x(x,y)=2x$$ and $$u_y(x,y)=-2y-1$$</p>
<p>This means that $v_y(x,y)=2x$ and $v_x(x,y)=2y+1$. Integrating w.r.t to $y$ and $x$ respectively you get that $$v(x,y)=2xy+h(x)+C$$ $$v(x,y)=2xy+x+f(y)$$ thus $v(x,y)=2xy+x+C$, $C$ some constant.</p>
|
182,301 |
<p>I have asked a question <a href="https://mathematica.stackexchange.com/questions/182247/constructing-possibilities-of-a-generating-function">here</a>.
I want to reproduce the coefficients of a generating function of the form:
<span class="math-container">$$(1 + x)^2 (1 + x + x^2 + x^3+\cdots+x^n)^{n-1}$$</span>
It is important that <span class="math-container">$x_0$</span> and <span class="math-container">$x_n$</span> to be strictly 0 or 1 and <span class="math-container">$x_1$</span> to <span class="math-container">$x_2$</span> can be any number within 0 and n (nothing higher). Here are two examples:
For <span class="math-container">$n=3$</span> we have:</p>
<pre><code>In: (1 + x)^2 (1 + x + x^2 + x^3)^2 // Expand
Out: 1 + 4 x + 8 x^2 + 12 x^3 + 14 x^4 + 12 x^5 + 8 x^6 + 4 x^7 + x^8
</code></pre>
<p>We can produce the coefficients as: </p>
<pre><code>n = 3;
m = n + 1;
tabel = Table[
v = Array[x, m, 0];
eqn = Total[v] == t;
constraints =
And[0 <= v[[1]] <= 1, 0 <= v[[2]] <= n, 0 <= v[[3]] <= n,
0 <= v[[4]] <= 1];
v /. Solve[{eqn, constraints}, v, Integers], {t, 0, 8}];
Table[Length[tabel[[i]]], {i, Length[tabel]}]
</code></pre>
<p>which gives: </p>
<pre><code>{1, 4, 8, 12, 14, 12, 8, 4, 1}
</code></pre>
<p>as desired. For <span class="math-container">$n=4$</span> one had to add extra constraint and change the <span class="math-container">$t$</span> range.
We have:</p>
<pre><code>In: (1 + x)^2 (1 + x + x^2 + x^3 + x^4)^3 // Expand
Out: 1 + 5 x + 13 x^2 + 25 x^3 + 41 x^4 + 58 x^5 + 70 x^6 + 74 x^7 +
70 x^8 + 58 x^9 + 41 x^10 + 25 x^11 + 13 x^12 + 5 x^13 + x^14
</code></pre>
<p>thus <span class="math-container">$t$</span> should be from 0 to 14,
so we have: </p>
<pre><code>n = 4;
m = n + 1;
tabel = Table[
v = Array[x, m, 0];
eqn = Total[v] == t;
constraints =
And[0 <= v[[1]] <= 1, 0 <= v[[2]] <= n, 0 <= v[[3]] <= n,
0 <= v[[4]] <= n, 0 <= v[[5]] <= 1];
v /. Solve[{eqn, constraints}, v, Integers], {t, 0, 14}];
Table[Length[tabel[[i]]], {i, Length[tabel]}]
</code></pre>
<p>I wonder if these modifications can be done automatically so that one doesn't have to add a constraint by hand and change the range.</p>
<p><strong>Note: I want the output <code>table</code> to be in a format so that I can see all of the possibilities of the sums.</strong>
For instance, for n=3, <code>table[[2]]</code> should give all of the possibilities such that the total is equal to 1.</p>
<pre><code>{{0, 0, 0, 1}, {0, 0, 1, 0}, {0, 1, 0, 0}, {1, 0, 0, 0}}
</code></pre>
|
Ulrich Neumann
| 53,677 |
<p>Perhaps I didn't understand your question, but </p>
<pre><code>Table[ CoefficientList[(1 + x)^2 (Sum[x^i, {i, 0, n}])^(n - 1),x] , {n, 1, 5}]
</code></pre>
<p>evaluates the list of coefficients you're looking for.</p>
|
182,301 |
<p>I have asked a question <a href="https://mathematica.stackexchange.com/questions/182247/constructing-possibilities-of-a-generating-function">here</a>.
I want to reproduce the coefficients of a generating function of the form:
<span class="math-container">$$(1 + x)^2 (1 + x + x^2 + x^3+\cdots+x^n)^{n-1}$$</span>
It is important that <span class="math-container">$x_0$</span> and <span class="math-container">$x_n$</span> to be strictly 0 or 1 and <span class="math-container">$x_1$</span> to <span class="math-container">$x_2$</span> can be any number within 0 and n (nothing higher). Here are two examples:
For <span class="math-container">$n=3$</span> we have:</p>
<pre><code>In: (1 + x)^2 (1 + x + x^2 + x^3)^2 // Expand
Out: 1 + 4 x + 8 x^2 + 12 x^3 + 14 x^4 + 12 x^5 + 8 x^6 + 4 x^7 + x^8
</code></pre>
<p>We can produce the coefficients as: </p>
<pre><code>n = 3;
m = n + 1;
tabel = Table[
v = Array[x, m, 0];
eqn = Total[v] == t;
constraints =
And[0 <= v[[1]] <= 1, 0 <= v[[2]] <= n, 0 <= v[[3]] <= n,
0 <= v[[4]] <= 1];
v /. Solve[{eqn, constraints}, v, Integers], {t, 0, 8}];
Table[Length[tabel[[i]]], {i, Length[tabel]}]
</code></pre>
<p>which gives: </p>
<pre><code>{1, 4, 8, 12, 14, 12, 8, 4, 1}
</code></pre>
<p>as desired. For <span class="math-container">$n=4$</span> one had to add extra constraint and change the <span class="math-container">$t$</span> range.
We have:</p>
<pre><code>In: (1 + x)^2 (1 + x + x^2 + x^3 + x^4)^3 // Expand
Out: 1 + 5 x + 13 x^2 + 25 x^3 + 41 x^4 + 58 x^5 + 70 x^6 + 74 x^7 +
70 x^8 + 58 x^9 + 41 x^10 + 25 x^11 + 13 x^12 + 5 x^13 + x^14
</code></pre>
<p>thus <span class="math-container">$t$</span> should be from 0 to 14,
so we have: </p>
<pre><code>n = 4;
m = n + 1;
tabel = Table[
v = Array[x, m, 0];
eqn = Total[v] == t;
constraints =
And[0 <= v[[1]] <= 1, 0 <= v[[2]] <= n, 0 <= v[[3]] <= n,
0 <= v[[4]] <= n, 0 <= v[[5]] <= 1];
v /. Solve[{eqn, constraints}, v, Integers], {t, 0, 14}];
Table[Length[tabel[[i]]], {i, Length[tabel]}]
</code></pre>
<p>I wonder if these modifications can be done automatically so that one doesn't have to add a constraint by hand and change the range.</p>
<p><strong>Note: I want the output <code>table</code> to be in a format so that I can see all of the possibilities of the sums.</strong>
For instance, for n=3, <code>table[[2]]</code> should give all of the possibilities such that the total is equal to 1.</p>
<pre><code>{{0, 0, 0, 1}, {0, 0, 1, 0}, {0, 1, 0, 0}, {1, 0, 0, 0}}
</code></pre>
|
Daniel Lichtblau
| 51 |
<p>One can automate the generating construction readily enough using <code>Product</code> and <code>Sum</code>. Getting total degrees and then the subsets of balues that give them is a bit more work. I show one method below.</p>
<pre><code>gen[n_] := Module[
{vars, x, t, genFunc, coeffs},
vars = Array[x, n + 1, 0];
genFunc = (1 + First[vars])*(1 + Last[vars])*
Product[Sum[vars[[j]]^k, {k, 0, n}], {j, 2, n}] /.
Thread[vars -> t*vars];
coeffs = GroebnerBasis`DistributedTermsList[genFunc, t][[1]];
Map[{#[[1, 1]],
GroebnerBasis`DistributedTermsList[#[[2]], vars][[1, All, 1]]} &,
coeffs]
]
</code></pre>
<p>Example:</p>
<pre><code>gen[3]
(* Out[59]= {{8, {{1, 3, 3, 1}}}, {7, {{1, 3, 3, 0}, {1, 3, 2, 1}, {1, 2,
3, 1}, {0, 3, 3, 1}}}, {6, {{1, 3, 2, 0}, {1, 3, 1, 1}, {1, 2, 3,
0}, {1, 2, 2, 1}, {1, 1, 3, 1}, {0, 3, 3, 0}, {0, 3, 2, 1}, {0,
2, 3, 1}}}, {5, {{1, 3, 1, 0}, {1, 3, 0, 1}, {1, 2, 2, 0}, {1, 2,
1, 1}, {1, 1, 3, 0}, {1, 1, 2, 1}, {1, 0, 3, 1}, {0, 3, 2, 0}, {0,
3, 1, 1}, {0, 2, 3, 0}, {0, 2, 2, 1}, {0, 1, 3, 1}}}, {4, {{1, 3,
0, 0}, {1, 2, 1, 0}, {1, 2, 0, 1}, {1, 1, 2, 0}, {1, 1, 1,
1}, {1, 0, 3, 0}, {1, 0, 2, 1}, {0, 3, 1, 0}, {0, 3, 0, 1}, {0, 2,
2, 0}, {0, 2, 1, 1}, {0, 1, 3, 0}, {0, 1, 2, 1}, {0, 0, 3,
1}}}, {3, {{1, 2, 0, 0}, {1, 1, 1, 0}, {1, 1, 0, 1}, {1, 0, 2,
0}, {1, 0, 1, 1}, {0, 3, 0, 0}, {0, 2, 1, 0}, {0, 2, 0, 1}, {0, 1,
2, 0}, {0, 1, 1, 1}, {0, 0, 3, 0}, {0, 0, 2, 1}}}, {2, {{1, 1, 0,
0}, {1, 0, 1, 0}, {1, 0, 0, 1}, {0, 2, 0, 0}, {0, 1, 1, 0}, {0,
1, 0, 1}, {0, 0, 2, 0}, {0, 0, 1, 1}}}, {1, {{1, 0, 0, 0}, {0, 1,
0, 0}, {0, 0, 1, 0}, {0, 0, 0, 1}}}, {0, {{0, 0, 0, 0}}}} *)
</code></pre>
<p>Looks spiffier if one uses <code>TableForm</code> or similar.</p>
|
2,973,825 |
<p>when you are checking to see if a sum of say <span class="math-container">$k^2$</span> from <span class="math-container">$k=1$</span> to to <span class="math-container">$k=n$</span> is equal to a sum of <span class="math-container">$(k+1)^2$</span> from <span class="math-container">$k=0$</span> to <span class="math-container">$n−1$</span> can someone explain what is going on here. THanks</p>
<p>(looking for a fairly simple way to work the problem without writing out the sums which may help me understand what is going on, )</p>
|
LeoDucas
| 214,422 |
<p>It's a cute one. Let's just assume the cards are numbered <span class="math-container">$\{1 \dots 52\}$</span>. We draw cards <span class="math-container">$a_1 < a_2 < a_3 < a_4 < a_5$</span>, and hand out in this order <span class="math-container">$b_1, b_2, b_3, b_4$</span>. The way we order this set <span class="math-container">$B={b_1, b_2, b_3, b_4}$</span> already allows to transmit quite some information. Indeed, choosing the orders paramounts to choosing a permutation on the 4 elements of <span class="math-container">$B$</span>. There are <span class="math-container">$4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24$</span> such permutations.</p>
<p>This is not enough still to transmit the <span class="math-container">$5$</span>th which is in a set of size <span class="math-container">$52$</span>. Well, we also know that the 5th card is different from those <span class="math-container">$4$</span> firsts ones: there is only <span class="math-container">$52-4 = 48$</span> possibilities. </p>
<p>I'll leave it to you to complete the strategy. Note that <span class="math-container">$48/24 = 2$</span>, so we just need to find a trick to only half the search space. Note also that we have still not fixed how we choose the secret card among the set <span class="math-container">$A$</span>. At last, note at least one of the two inequalities:<br>
- <span class="math-container">$a_2 \leq 25$</span><br>
- <span class="math-container">$a_4 \geq 58-25$</span>.</p>
|
869,358 |
<p>This is a seemingly simple induction question that has me confused about perhaps my understanding of how to apply induction</p>
<p>the question;</p>
<p>$$\frac{1}{1^2}+ \cdots+\frac{1}{n^2}\ \le\ 2-\frac{1}{n},\ \forall\ n \ge1.$$</p>
<p>this true for $n=1$, so assume the expression is true for $n\le k$. which produces the expression, </p>
<p>$$\frac{1}{1^2} + \cdots + \frac{1}{k^2} \le\ 2-\frac{1}{k}.$$ now to show the expression is true for $k+1$,</p>
<p>$$\frac{1}{1^2}+\cdots+ \frac{1}{k^2} + \frac{1}{(k+1)^2} \le\ 2-\frac{1}{k}+\frac{1}{(k+1)^2}.$$</p>
<p>this the part I am troubled by, because after some mathemagical algebraic massaging, I should be able to equate,</p>
<p>$$2-\frac{1}{k}+\frac{1}{(k+1)^2}=2-\frac{1}{(k+1)},$$</p>
<p>which would prove the expression is true for $k+1$ and I'd be done. right? but these two are not equivalent for even $k=1$, because setting $k=1$ you wind up with $\frac{5}{4}=\frac{3}{2}$, so somewhere i am slipping up and I'm not sure how else to show this if someone has some insight into this induction that I'm not getting. thanks.</p>
|
Marco Flores
| 164,575 |
<p>What you really need is $2 − \frac{1}{k} + \frac{1}{(k+1)^2} \leq 2 − \frac{1}{(k+1)}$,</p>
|
1,545,945 |
<p>Here my question :Perform the following arithmetic operations for the numbers in the bases indicated and write out answers in base 5 notation:
a) 244 + 132 (base 5).
b) 11101 × 111 (base 2).
c) F7 – B6 (base 16).
this is the first time I hear of(Base 5 notation)
can anyone help please ?</p>
|
Lubin
| 17,760 |
<p>It’s valuable to have a little experience and skill in such computations. The idea is to think carefully of what you’re doing when you do decimal computations with paper and pencil, then carry the principles over to the unfamiliar situation.</p>
<p>To add $244_5$ and $132_5$, you start with the units digit, and add $4+2=6_{10}=11_5$. So your units digit is $1$, with a carry of $1$ to the next column. Now you’re adding $4+3+1=8_{10}=13_5$, so the second place gets a $3$, with a carry of $1$ again to the next column. And so it goes.</p>
<p>If you want to do multiplication, you should prepare yourself with a $5\times5$ multiplication table of numbers written in base $5$: for instance $3\times4=12_{10}=22_5$. Again the procedure is the same as in the familiar situation, but you just have to be careful and think of what you’re doing.</p>
|
1,545,945 |
<p>Here my question :Perform the following arithmetic operations for the numbers in the bases indicated and write out answers in base 5 notation:
a) 244 + 132 (base 5).
b) 11101 × 111 (base 2).
c) F7 – B6 (base 16).
this is the first time I hear of(Base 5 notation)
can anyone help please ?</p>
|
John Molokach
| 90,422 |
<p>You can think of different number bases like a variation of the base 10 decimal system we already use. I'll explain by working out the first problem you list and then leave the others to you.</p>
<p>In base 5, you would count starting at zero and have the following numbers:</p>
<p>$$\{0,1,2,3,4,10,11,12,13,14,20,21,\dots\}.$$</p>
<p>In this system, the 'ones column' tells how many ones, or multiples of $5^0$ you have, the next column to the left of a number is not the tens place, but the fives, so it tells what multiple of $5^1$ you have. So for example, $14_5$ is $9_{10}$, meaning 14 in base 5 is the same as 9 in base 10 because you have 1 $5^1$ and 4 ones.</p>
<p>With this in mind, you can carry and borrow with addition and subtraction as usual, so</p>
<p>$$244_5+132_5=3(5^2)+7(5^1)+6(5^0).$$ </p>
<p>Now we need to carry any multiple that is 5 or more to the column to the left, so we rewrite this expression as</p>
<p>$$3(5^2)+8(5^1)+1(5^0)=4(5^2)=3(5^1)+1(5^0)=431_5.$$</p>
|
378,228 |
<p>This is probably known, but I have not located a reference.</p>
<p>Let <span class="math-container">$P$</span> be the convex hull of <span class="math-container">$k$</span> points in <span class="math-container">$\mathbb R^n$</span> with rational coordinates. Consider the Euclidean square norm function <span class="math-container">$F:P\to\mathbb R$</span>, <span class="math-container">$F(x)=||x||^2$</span>.</p>
<p>Is it true that <span class="math-container">$\mathrm{min}_{x\in P}F(x)$</span> is rational?</p>
|
Fedor Petrov
| 4,312 |
<p>Yes, this is true in general case. Let <span class="math-container">$P={\rm conv} M$</span> for a finite set <span class="math-container">$M\subset \mathbb{Q}^n$</span>, and let <span class="math-container">$x_0$</span> be a minimizer of <span class="math-container">$\|x\|^2$</span> over <span class="math-container">$x\in P$</span>. I am going to prove that <span class="math-container">$x_0\in \mathbb{Q}^n$</span>, that implies <span class="math-container">$\|x_0\|^2\in \mathbb{Q}$</span>.</p>
<p>Consider the plane <span class="math-container">$\alpha:=\{x\colon \langle x,x_0\rangle=\|x_0\|^2\}$</span> passing through <span class="math-container">$x_0$</span> and orthogonal to <span class="math-container">$x_0$</span>, and the half-space <span class="math-container">$\alpha_+:=\{x\colon \langle x,x_0\rangle\geqslant \|x_0\|^2\}$</span>. Note that <span class="math-container">$P\subset \alpha_+$</span>: otherwise, if <span class="math-container">$p\in P$</span> satisfy <span class="math-container">$\langle p,x_0\rangle< \|x_0\|^2$</span>, then <span class="math-container">$$\|x_0+t(p-x_0)\|^2=\|x_0\|^2+2t\langle x_0,p-x_0\rangle+t^2\|p-x_0\|^2<\|x_0\|^2$$</span> if <span class="math-container">$t>0$</span> is small enough, that contradicts to minimality as <span class="math-container">$x_0+t(p-x_0)\in P$</span> for <span class="math-container">$t\in (0,1)$</span>.</p>
<p>Choose the minimal number of points <span class="math-container">$v_1,\ldots,v_r\in M$</span> for which <span class="math-container">$x_0\in {\rm conv}(v_1,\ldots,v_r)$</span>. Then <span class="math-container">$x_0=c_1v_1+\ldots+c_rv_r$</span> for strictly positive <span class="math-container">$c_i\in (0,1)$</span>, <span class="math-container">$\sum c_i=1$</span>. We have <span class="math-container">$v_i\in \alpha$</span> for all <span class="math-container">$i$</span>, otherwise <span class="math-container">$x_0$</span> would lie in the interior of <span class="math-container">$\alpha_+$</span>. Next, the vectors <span class="math-container">$v_r-v_1,v_r-v_2,\ldots,v_r-v_{r-1}$</span> are linearly independent. Indeed, if they were dependent, we would have a linear dependence <span class="math-container">$t_1v_1+\ldots+t_r v_r=0$</span> with <span class="math-container">$\sum t_i=0$</span>. Choosing maximal <span class="math-container">$s>0$</span> for which all numbers <span class="math-container">$c_i+st_i$</span> are nonnegative (thus at least one of them equals to 0), we get a representation <span class="math-container">$x_0=\sum (c_i+st_i)v_i$</span>, i.e., <span class="math-container">$x_0$</span> belongs to a convex hull of less than <span class="math-container">$r$</span> <span class="math-container">$v_i$</span>'s, a contradiction. Then taking the inner product of the representation
<span class="math-container">$$
x_0=v_r+\sum_{i=1}^{r-1} c_i (v_i-v_r)
$$</span>
with vectors <span class="math-container">$v_1-v_r$</span>, <span class="math-container">$v_2-v_r$</span>, <span class="math-container">$\ldots$</span>, <span class="math-container">$v_{r-1}-v_r$</span> we get a linear system of <span class="math-container">$r-1$</span> equations for <span class="math-container">$c_1,\ldots,c_{r-1}$</span> (note that <span class="math-container">$\langle x_0, v_i-v_r\rangle=0$</span> as <span class="math-container">$v_i,v_r\in \alpha$</span>, so the system does not involve <span class="math-container">$x_0$</span> and has rational coefficients). the matrix of this system is Gram matrix of the vectors <span class="math-container">$v_i-v_r$</span>, <span class="math-container">$i=1,\ldots,r-1$</span>. They are linearly independent, thus this matrix is non-singular. So, <span class="math-container">$c_i$</span>'s may be found from this system of equations and they are rational. Thus <span class="math-container">$x_0\in \mathbb{Q}^n$</span>.</p>
|
2,647,868 |
<p>I'm very confused at the following question:</p>
<blockquote>
<p>Find the basis for the image and a basis of the kernel for the following matrix:
$\begin{bmatrix} 7 & 0 & 7 \\ 2 & 3 & 8 \\ 9 & 0 & 9 \\ 5 & 6 & 17 \end{bmatrix}$</p>
</blockquote>
<p>I just don't know how to do any of this. We find the image by doing the following:
$\begin{bmatrix} 7 & 2 & 9 & 5 \\\ 0 & 3 & 0 & 6\\ 7 & 8 & 9 &17 \end{bmatrix}$
Then, after doing RREF, we get:
$\begin{bmatrix} 1 & 0 & \frac{9}{7} & \frac{1}{7} \\\ 0 & 1 & 0 & 2\\ 0&0&0&0 \end{bmatrix}$. This gives us an image of {$\begin{bmatrix} 1 \\ 0 \\ \frac{9}{7} \\ \frac{1}{7} \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \\ 2 \end{bmatrix}$}. However, I don't know how to proceed from here. Please help me?</p>
|
gbox
| 103,441 |
<p>$Im(T)$ is the column span.</p>
<p>For vectors to be a basis they need to be: </p>
<ol>
<li>linear independent </li>
<li>span the subspace</li>
</ol>
<p>So we are left to check for linear dependence $$\begin{pmatrix} 7 & 0 & 7 \\ 2 & 3 & 8 \\ 9 & 0 & 9 \\ 5 & 6 & 17 \end{pmatrix}\approx \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$</p>
<p>We have $2$ pivots in the in the first 2 columns so $$\{\begin{pmatrix} 7 \\ 2 \\ 9 \\ 5 \end{pmatrix},\begin{pmatrix} 0 \\ 3 \\ 0 \\ 6 \end{pmatrix}\}$$ are basis for the image </p>
<p>To find the kernel we need to solve $$\left(\begin{array}{ccc|c}
7 & 0 & 7 & 0\\ 2 & 3 & 8 & 0\\ 9 & 0 & 9 & 0\\ 5 & 6 & 17 & 0
\end{array}\right)$$</p>
|
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