qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
3,752,402 | <p>I want to find out the existence of the solutions in diophantine equations of the style:</p>
<p><span class="math-container">$$-259y ^2+ 2400yx + 1817y + 2122x =
$$</span>
<span class="math-container">$$1057364602723981500371957207036553770637547302056514367123547565680640946707606178926389130616$$</span></p>
<p>The point is that solving it with current methods (elliptic curves) takes a long time, so I want to find out whether or not it has solutions.</p>
<p>What I have already tried:</p>
<ol>
<li><p>If the GCD of the coefficients does not divide the independent term, then it has no solutions. Even if I divided it, it may be that the equation has solutions or it may not have them</p>
</li>
<li><p>Shaping the <span class="math-container">$x$</span> and the <span class="math-container">$y$</span>: for example <span class="math-container">$x,y$</span> pairs and come to a contradiction (sometimes it works, sometimes it doesn't)</p>
</li>
</ol>
<p>My questions are:</p>
<p>-Do you know of any procedure that allows deciding whether or not it has solutions given any second-degree diophantine equation with two unknowns?</p>
<p>-The equation above is of a hyperbolic type, is it possible to modify it to make it an elliptical type? If yes, how do I do it?</p>
| Dmitry Ezhov | 602,207 | <p><span class="math-container">$$-259y ^2+ 2400yx + 1817y + 2122x =
$$</span>
<span class="math-container">$$1057364602723981500371957207036553770637547302056514367123547565680640946707606178926389130616$$</span></p>
<p><span class="math-container">$$\implies \Bigl(1440000 x + 1364999\Bigr)^2-\Bigl(600(2400 x - 518 y + 1817)\Bigr)^2=$$</span></p>
<p><span class="math-container">$$394354702231936140378725159936353094296979641774997598362398300096251847484068800492386090829229590001$$</span></p>
<p>For solving need factorize RHS.</p>
<p>Any quadratic diophantine equations with 2 unknowns can simplify to 3 type:</p>
<ol>
<li><p>Pell equation like <span class="math-container">$x^2-dy^2=\pm c$</span></p>
</li>
<li><p>difference of squares like <span class="math-container">$x^2-y^2=c$</span></p>
</li>
<li><p>"quadratic Thue" equation like <span class="math-container">$x^2+dy^2=c$</span></p>
</li>
</ol>
<p>where <span class="math-container">$d,c\in\mathbb{N}$</span>.</p>
|
4,520,388 | <p>I'm stuck on this multivariable equation:</p>
<p><span class="math-container">$$
\frac{d}{dx}\left(\int^x_af(g(b,t),t)dt\right)
$$</span></p>
<p>where a and b are just constants.</p>
<p>If this involved a single variable, it looks like one would just apply the fundamental theorem of calculus. Is there an equivalent for multiple variables.</p>
<p>I know that the answer should just be</p>
<p><span class="math-container">$$
f(g(b,x),x)
$$</span></p>
<p>but I'm hoping someone can explain / walk me through. Is there maybe some rule that lets me pass the <span class="math-container">$\frac{d}{dx}$</span> into the integral?</p>
<p>Thanks</p>
| Steven | 606,584 | <p>The function under the integral can be seen as a function of one variable <span class="math-container">$t$</span>, and then the fundamental theorem applies directly. That is, if we define <span class="math-container">$h(t) = f(g(b,t),t)$</span>, then <span class="math-container">$$\frac{d}{dx}\int_a^x h(t) \, dt = h(x),$$</span> and if we unwind the definition this is says that <span class="math-container">$$\frac{d}{dx}\int_a^x f(g(b,t),t) \, dt = f(g(b,x),x).$$</span></p>
|
4,473,543 | <p>For example, given <span class="math-container">$\color{green}{l_1:5x-2y-8=0}$</span> and <span class="math-container">$\color{blue}{l_2:3x+8y-8=0}$</span>,</p>
<p><a href="https://i.stack.imgur.com/sxZJ1m.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sxZJ1m.jpg" alt="img1" /></a></p>
<p>We can compute the set of lines that pass through the intersection of <span class="math-container">$l_1$</span> and <span class="math-container">$l_2$</span></p>
<p><span class="math-container">$$5x-2y-8+\lambda(3x+8y-8)=0$$</span>
<span class="math-container">$$(5+3\lambda)x+(-2+8\lambda)y+(-8-8\lambda)=0$$</span> for any <span class="math-container">$\lambda\in\Bbb{R}$</span></p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th><span class="math-container">$$\lambda$$</span></th>
<th><span class="math-container">$$(5+3\lambda)x+(-2+8\lambda)y+(-8-8\lambda)=0$$</span></th>
</tr>
</thead>
<tbody>
<tr>
<td><span class="math-container">$$-1$$</span></td>
<td><span class="math-container">$$x-5y=0$$</span></td>
</tr>
<tr>
<td><span class="math-container">$$1$$</span></td>
<td><span class="math-container">$$4x+3y-8=0$$</span></td>
</tr>
<tr>
<td><span class="math-container">$$2$$</span></td>
<td><span class="math-container">$$11x+14y-24=0$$</span></td>
</tr>
</tbody>
</table>
</div>
<p><a href="https://i.stack.imgur.com/5w1Kom.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5w1Kom.jpg" alt="img2" /></a></p>
<p>Why does this method work?</p>
<hr />
<p>The closest thread I could find is, <a href="https://math.stackexchange.com/questions/3743591/a-general-circle-through-the-intersection-points-of-line-l-and-circle-s-1-ha">A general circle through the intersection points of line <span class="math-container">$l$</span> and circle <span class="math-container">$S_1$</span> has the form <span class="math-container">$S_1+\lambda L$</span>. What is the significance of <span class="math-container">$\lambda$</span>?</a>.</p>
<blockquote>
<p>For example if we want to find lines through the point of intersection of 3x+4y+5=0 and 2x+y+4=0 . The required lines would be obtained by substituting different values of λ in 3x+4y+5+λ(2x+y+4)=0</p>
</blockquote>
<p>The accepted answer is,</p>
<blockquote>
<p>Let us take up the case of lines first.
Let <span class="math-container">$L_1(x,y)$</span> and <span class="math-container">$L_2(x,y)$</span> be two lines which intersect at <span class="math-container">$(a,b)\\$</span>.
Thus <span class="math-container">$$L_1(a,b)=0\\L_2(a,b)=0$$</span>Now let <span class="math-container">$L_3(x,y)$</span> be another line such that
<span class="math-container">$$L_3(x,y)=L_1(x,y)+\lambda L_2(x,y)$$</span>Now, if we are able to show that <span class="math-container">$L_3$</span> passes through <span class="math-container">$(a,b)$</span>,i.e. the intersection point of <span class="math-container">$L_1$</span> and <span class="math-container">$L_2$</span>, our job will be complete.To do this we put <span class="math-container">$(a,b)$</span> in our expression for <span class="math-container">$L_3$</span>
<span class="math-container">$$L_3(a,b)=L_1(a,b)+\lambda L_2(a,b)$$</span>
<span class="math-container">$$\Rightarrow L_3(x,y)=0+\lambda .0$$</span>
<span class="math-container">$$\Rightarrow L_3(x,y)=0$$</span>So as you can see, for any value of <span class="math-container">$\lambda$</span>, our line <span class="math-container">$L_3$</span> always passes through the intersection of lines <span class="math-container">$L_1$</span> and <span class="math-container">$L_2\\$</span>.
You can the same with any two curves(e.g. two circles) .</p>
</blockquote>
<p>What I understood from this answer is,</p>
<ul>
<li>Consider the lines <span class="math-container">$L_1(x, y)$</span> and <span class="math-container">$L_2(x, y)$</span> which intersept at <span class="math-container">$(a, b)$</span> such that <span class="math-container">$L_1(a, b)=0$</span> and <span class="math-container">$L_2(a, b)=0$</span></li>
<li>Assume that <span class="math-container">$L_3(x,y)=L_1(x,y)+\lambda L_2(x,y)$</span></li>
<li>Then this means that <span class="math-container">$L_3$</span> passes through <span class="math-container">$(a, b)$</span></li>
</ul>
<p>I do not understand how this proves that <span class="math-container">$L_3(x,y)=L_1(x,y)+\lambda L_2(x,y)$</span> spans a set of <em>distinct</em> lines that pass through <span class="math-container">$(a, b)$</span>.</p>
<p>In the case of circles of the form <span class="math-container">$x^2+y^2+Dx+Ey+F=0$</span> and <span class="math-container">$x^2+y^2+D'x+E'y+F'=0$</span> that intersect and are not concentric, we can't have <span class="math-container">$\lambda=-1$</span> because we would get their radical axis, not another circle.</p>
<p>I'm looking for a proof by deduction if possible (as opposed to assumption).</p>
| aerile | 282,796 | <p>One explanation is to show that for any point <span class="math-container">$(p,q)$</span> outside the line <span class="math-container">$L_2(x,y)=0$</span>, we can set <span class="math-container">$\lambda$</span> so that the the point is on the line <span class="math-container">$L_1(x,y)+\lambda L_2(x,y)=0$</span>.</p>
<p>This is possible because by the assumption <span class="math-container">$L_2(p,q)\neq 0$</span> so we can set <span class="math-container">$\lambda=-L_1(p,q)/L_2(p,q).$</span></p>
<p>Edit:
I mean that we can obtain the line <span class="math-container">$L_1(x,y)+\lambda L_2(x,y)=0$</span> passing through <span class="math-container">$(p,q)$</span> by setting <span class="math-container">$\lambda=-L_1(p,q)/L_2(p,q).$</span></p>
|
23,911 | <p>I am teaching a course on Riemann Surfaces next term, and would <strong>like a list of facts illustrating the difference between the theory of real (differentiable) manifolds and the theory non-singular varieties</strong> (over, say, $\mathbb{C}$). I am looking for examples that would be meaningful to 2nd year US graduate students who has taken 1 year of topology and 1 semester of complex analysis.</p>
<p>Here are some examples that I thought of:</p>
<p><strong>1.</strong> Every $n$-dimensional real manifold embeds in $\mathbb{R}^{2n}$. By contrast, a projective variety does not embed in $\mathbb{A}^n$ for any $n$. Every $n$-dimensional non-singular, projective variety embeds in $\mathbb{P}^{2n+1}$, but there are non-singular, proper varieties that do not embed in any projective space.</p>
<p><strong>2.</strong> Suppose that $X$ is a real manifold and $f$ is a smooth function on an open subset $U$. Given $V \subset U$ compactly contained in $U$, there exists a global function $\tilde{g}$ that
agrees with $f$ on $V$ and is identically zero outside of $U$.</p>
<p>By contrast, consider the same set-up when $X$ is a non-singular variety and $f$ is a regular function. It may be impossible find a global regular function $g$ that agrees with $f$ on $V$. When $g$ exists, it is unique and (when $f$ is non-zero) is not identically zero on outside of $U$.</p>
<p><strong>3.</strong> If $X$ is a real manifold and $p \in X$ is a point, then the ring of germs at $p$ is non-noetherian. The local ring of a variety at a point is always noetherian. </p>
<p><em><strong>What are some more examples?</em></strong></p>
<p>Answers illustrating the difference between real manifolds and complex manifolds are also welcome.</p>
| roy smith | 9,449 | <p>Some of these properties are local, and distinguish analytic and algebraic functions, from smooth functions.</p>
<p>Others are global and distinguish compact manifolds from projective manifolds by their difference in containment of many subvarieties.</p>
<p>some are as simple as contrasting the dimension, and homology of say projective Real space from that of projective Complex space, as noted.</p>
<p>As a deep contrast between smooth and analytic structure I like the answer pointing out that analytic riemann surfaces can have many non isomorphic structures on the same smooth manifold. this is interesting already for genus one manifolds. and it is not trivial to show that the sphere has only one complex analytic structure.</p>
<p>that might be a fun challenge to a class, to show two Riemann surfaces both diffeomorphic to the 2-sphere, are holomorphically isomorphic.</p>
<p>You might also be interested in some of the articles by Kolla'r on the Nash conjecture contrasting real varieties and real manifolds. such as "What are the simplest varieties?", Bulletin, vol 38. I like the pair of theorems 54, 51, subtitled respectively: "the Nash conjecture is true in dim 3", and "The Nash conjecture is false in dim 3".</p>
|
683,513 | <p>There is much discussion both in the education community and the mathematics community concerning the challenge of (epsilon, delta) type definitions in real analysis and the student reception of it. My impression has been that the mathematical community often holds an upbeat opinion on the success of student reception of this, whereas the education community often stresses difficulties and their "baffling" and "inhibitive" effect (see below). A typical educational perspective on this was recently expressed by Paul Dawkins in the following terms: </p>
<p><em>2.3. Student difficulties with real analysis definitions. The concepts of limit and continuity have posed well-documented difficulties for students both at the calculus and analysis level of instructions (e.g. Cornu, 1991; Cottrill et al., 1996; Ferrini-Mundy & Graham, 1994; Tall & Vinner, 1981; Williams, 1991). Researchers identified difficulties stemming from a number of issues: the language of limits (Cornu, 1991; Williams, 1991), multiple quantification in the formal definition (Dubinsky, Elderman, & Gong, 1988; Dubinsky & Yiparaki, 2000; Swinyard & Lockwood, 2007), implicit dependencies among quantities in the definition (Roh & Lee, 2011a, 2011b), and persistent notions pertaining to the existence of infinitesimal quantities (Ely, 2010). Limits and continuity are often couched as formalizations of approaching and connectedness respectively. However, the standard, formal definitions display much more subtlety and complexity. That complexity often baffles students who cannot perceive the necessity for so many moving parts. Thus learning the concepts and formal definitions in real analysis are fraught both with need to acquire proficiency with conceptual tools such as quantification and to help students perceive conceptual necessity for these tools. This means students often cannot coordinate their concept image with the concept definition, inhibiting their acculturation to advanced mathematical practice, which emphasizes concept definitions.</em> </p>
<p>See <a href="http://dx.doi.org/10.1016/j.jmathb.2013.10.002" rel="nofollow noreferrer">http://dx.doi.org/10.1016/j.jmathb.2013.10.002</a> for the entire article (note that the online article provides links to the papers cited above).</p>
<p>To summarize, in the field of education, researchers decidedly have <em>not</em> come to the conclusion that epsilon, delta definitions are either "simple", "clear", or "common sense". Meanwhile, mathematicians often express contrary sentiments. Two examples are given below. </p>
<p><em>...one cannot teach the concept of limit without using the epsilon-delta definition. Teaching such ideas intuitively does not make it easier for the student it makes it harder to understand. Bertrand Russell has called the rigorous definition of limit and convergence the greatest achievement of the human intellect in 2000 years! The Greeks were puzzled by paradoxes involving motion; now they all become clear, because we have complete understanding of limits and convergence. Without the proper definition, things are difficult. With the definition, they are simple and clear.</em> (see Kleinfeld, Margaret; Calculus: Reformed or Deformed? Amer. Math. Monthly 103 (1996), no. 3, 230-232.) </p>
<p><em>I always tell my calculus students that mathematics is not esoteric: It is common sense. (Even the notorious epsilon, delta definition of limit is common sense, and moreover is central to the important practical problems of approximation and estimation.)</em> (see Bishop, Errett; Book Review: Elementary calculus. Bull. Amer. Math. Soc. 83 (1977), no. 2, 205--208.)</p>
<p>When one compares the upbeat assessment common in the mathematics community and the somber assessments common in the education community, sometimes one wonders whether they are talking about the same thing. How does one bridge the gap between the two assessments? Are they perhaps dealing with distinct student populations? Are there perhaps education studies providing more upbeat assessments than Dawkins' article would suggest? </p>
<p>Note 1. See also <a href="https://mathoverflow.net/questions/158145/assessing-effectiveness-of-epsilon-delta-definitions">https://mathoverflow.net/questions/158145/assessing-effectiveness-of-epsilon-delta-definitions</a></p>
<p>Note 2. Two approaches have been proposed to account for this difference of perception between the education community and the math community: (a) sample bias: mathematicians tend to base their appraisal of the effectiveness of these definitions in terms of the most active students in their classes, which are often the best students; (b) student/professor gap: mathematicians base their appraisal on their own scientific appreciation of these definitions as the "right" ones, arrived at after a considerable investment of time and removed from the original experience of actually learning those definitions. Both of these sound plausible, but it would be instructive to have field research in support of these approaches.</p>
<p>We recently published <a href="http://dx.doi.org/10.5642/jhummath.201701.07" rel="nofollow noreferrer">an article</a> reporting the result of student polling concerning the comparative educational merits of epsilon-delta definitions and infinitesimal definitions of key concepts like continuity and convergence, with students favoring the infinitesimal definitions by large margins.</p>
| Ellya | 135,305 | <p>I'm really just putting forth my opinions on $\epsilon$-$\delta$, and how I think it should be introduced to people who have not seen it before.</p>
<p>I believe the entire difficulty that the $\epsilon-\delta$ approach puts forth is the idea of a constant that depends on something, since before undergrad level (i.e. A level) constants are constants, and we don't look at situations where they may change (Due to a change in another constant).</p>
<p>The only times things seem to change (at a first glance) is when we look a functions, where the variable changes and the function changes, and this seems natural.</p>
<p>But then when we are approached with an idea that some constant changes, and then another one <strong>must</strong> it seems very foreign.</p>
<p>I am just finishing my BSc and I feel very well versed in $\epsilon-\delta$ type arguments, but to understand it I had to get there on my own, and I feel that this is the way to go, you can't completely understand things just from someone telling you. You need to explore it yourself.</p>
<p>But the part in the explanations that was <strong>always</strong> lacking, is the fact that $\delta$ depends on $\epsilon$, and I think the definitions should be written in the form $\forall\epsilon\gt 0,\exists \delta(\epsilon)$...</p>
<p>The whole scenario could be argued in a challenge formulation, so someone can really see how $\delta$ must change if $\epsilon$ changes.</p>
<p>For instance, say we want to prove that the function $f(x)$ is continuous at $x_0$, we must show that $\forall\epsilon\gt 0,\exists\delta(\epsilon):|x-x_0|\lt\delta(\epsilon)\Rightarrow|f(x)-f(x_0)|\lt\epsilon$.</p>
<p>Then we argue in this sense: someone gives us the challenge of $\epsilon=\epsilon^*$</p>
<p>Then we find a value of $\delta$ <strong>which depends on $\epsilon^*$</strong> so that <strong>for this $\epsilon^*$</strong>, if $x\in(x_0-\delta(\epsilon^*),x_0+\delta(\epsilon^*))$</p>
<p>Then $f(x)\in(f(x_0)-\epsilon^*,f(x_0)+\epsilon^*)$.</p>
<p>Now summarising in non mathematical terms. We have one "$\delta$" that holds for our one "challenge" ($\epsilon^*$) that makes our condition hold, by taking that choice of "$\delta$".</p>
<p>So up until this point we have been very clear that $\delta$ depends on $\epsilon$.</p>
<p>Now we go a step further, and we ask what happens if any $\epsilon$ is given to us as a challenge? Well, clearly in many cases, the $\delta$ we found in the first instance wont work every time, so we must have to find a new $\delta$.</p>
<p>And in this sense again <strong>$\delta$ depends on $\epsilon$</strong>.</p>
<p>Now we realise that we must find a relationship between $\delta$ and $\epsilon$.</p>
<p>So in common sense $\delta$ is a function of $\epsilon$, but the "variable" $\epsilon$, only changes when we are handed a new "challenge" or situation, rather than over a general domain i.e. something like an interval which is much easier to imagine.</p>
|
3,943,199 | <p>How to put <span class="math-container">$(-\sqrt{3}-i)^{\frac{5}{7}}$</span> into polar form and find all roots.</p>
<p>What I tried:</p>
<p><span class="math-container">$$w = -\sqrt{3}-i$$</span>
<span class="math-container">$$\arg(w)=\arctan(\frac{1}{\sqrt 3})-\pi = \frac{\pi}{6} - \pi = \frac{-5\pi}{6}$$</span>
<span class="math-container">$$w = 2(\cos(\frac{-5\pi}{6})+ i\sin(\frac{-5\pi}{6})) $$</span>
<span class="math-container">$$w^5 = 2^5(\cos(-\frac{5\pi}{6}*5)+i\sin(-\frac{5\pi}{6}*5)) $$</span>
<span class="math-container">$$z^7 = w^5$$</span>
<span class="math-container">$$z = w^{\frac{5}{7}} $$</span>
<span class="math-container">$$z = 32^{\frac{1}{7}}(\cos(\frac{-5\pi*5}{6*7}+\frac{2k\pi}{7})+i\sin(\frac{-5\pi*5}{6*7}+\frac{2k\pi}{7})) $$</span>
Here I get stuck and I don't know how to continue...</p>
| Christian Blatter | 1,303 | <p>It seems you have a complicated definition <span class="math-container">$F(\phi)$</span> of <span class="math-container">$\sin\phi$</span> in terms of simpler things defined before, and in the same way you have a complicated definition <span class="math-container">$G(\phi)$</span> of <span class="math-container">$e^{i\phi}$</span> in terms of simpler things defined before. You are told to prove
<span class="math-container">$$F(\phi)=G(\phi)\qquad\forall\,\phi\ ,$$</span>
using the facts valid in the "simpler" world. As "<span class="math-container">$=$</span>" is a symmetric relation such a proof can go in the direction
<span class="math-container">$$F(\phi)= \ldots =\quad\ldots\quad=\ldots=G(\phi)$$</span>
as well as in the direction
<span class="math-container">$$G(\phi)= \ldots =\quad\ldots\quad=\ldots=F(\phi)\ .$$</span></p>
|
2,355,852 | <p>Given are $m$ bins with equal probability of choosing one of them. Unknown number of balls $n$ is placed into the bins, and, at the end of placement, we observe number of empty bins $m_e$ and non-empty bins $m_{n}$.</p>
<p>Given $m$, $m_e$, $m_n$, what is the most likely number of balls $n$, which have been placed into bins?</p>
<p>(UPD) possible additional information: number of bins with <em>exactly</em> one ball can be also known.</p>
| Thanassis | 8,153 | <p>Let's say we have $m$ bins out of which $m_e$ are empty (and the non-empty bins are $m_n=m-m_e$). Let's also assume that $m_e > 0$, otherwise there is very little information we can get, as you discussed in the comments. </p>
<p>If we could compute the probability of having $i$ balls, given this observation, then we can compute the expected value of balls, or the maximum likelihood estimator, or other questions about estimating the number of balls. So we are looking for $P(\text{balls} = i | m_e,m )$. </p>
<p>Applying Bayes' rule we get:</p>
<p>$$P(\text{balls} = i | m_e,m ) = \frac{P(m_e,m|\text{balls}=i)\cdot P(\text{balls}=i)}{\sum_{i=m_n}^\infty P(m_e,m|\text{balls}=i)\cdot P(\text{balls}=i)}$$</p>
<p>Notice that the balls have to be at least as many as the non-empty bins. </p>
<p>We also notice the importance of knowing the prior probability of $P(\text{balls} =i)$. Without any other information on this, let's assume that the all values of $i$ are equiprobable, so that these terms can cancel out in the formula. </p>
<p>What is $P(m_e,m|\text{balls}=i)$? In other words, given that we distributed $i$ balls in the bins, what is the probability that we have exactly $m_e$ empty bins, out of $m$ total bins? If the distribution was done in a uniformly random way, then this probability is equal to $\dfrac{S_2(i,m_n)\cdot m!}{m^n \cdot m_e!}$.</p>
<p><strong>Edit</strong>: Intially I thought that this was a simpler formula, that was more analytically tractable. After seeing Henry's answer I realised my mistake. I am afraid my analysis stops here. You can computationally calculate these probabilities for small number of balls and bins. You can also compute $P(\text{balls} = i | m_e,m )$, especially if you assume a maximum number of possible balls (so that the summation in the denominator of Bayes' rule becomes tractable).</p>
|
2,853,989 | <p>I'm trying to demonstrate that $\left( 1+\frac1 n \right)^n$ is bigger than $2$. I have tried to prove that $\left( 1+\frac1 n \right)^n$ is smaller than $\left( 1+\frac1{n+1} \right)^{n+1}$ by expanding
$\left( 1+\frac1n \right)^n = \sum\limits_{i=0}^n \left( \frac{n}{k} \right) \frac{1}{n^k}$ and $\left( 1+\frac1{n+1} \right)^{n+1} = \sum\limits_{i=0}^{n+1} \left( \frac{(n+1)}{k} \right) \frac{1}{(n+1)^k}$ but it doesn't seem to work.</p>
<p>What am I missing? Also, is there a method to demonstrate that without induction?</p>
| Donald Splutterwit | 404,247 | <p>You just need to consider the first $2$ terms in the binomial expansion
\begin{eqnarray*}
\left( 1+ \frac{1}{n} \right)^{n} = \sum_{i=0}^{n}\binom{n}{i}\frac{1}{n^i}=1+n \frac{1}{n} +\cdots \geq 2.
\end{eqnarray*}</p>
|
1,365,384 | <p>How can the sum of factorials $(n+1)!+n!$ be simplified?</p>
| izœc | 83,639 | <p><strong>HINT:</strong> $(n+1)! = (n+1)\cdot n!$. So...
$$
(n+1)! + n! = (n+1)\cdot n! + n! = ...
$$</p>
|
1,365,384 | <p>How can the sum of factorials $(n+1)!+n!$ be simplified?</p>
| alkabary | 96,332 | <p>$$(n+1)! = (n+1) \times n!$$</p>
<p>and so $$(n+1)! + n! = (n+1) \times n! + n!$$</p>
<p>and $$(n+1) \times n! + n! = n! ((n+1) +1) = n!(n+2)$$</p>
<p>and hence $$\color{blue}{(n+1)! +n! = n!(n+2)}$$</p>
|
3,246,244 | <p>Consider the action of <span class="math-container">$G$</span> on <span class="math-container">$X$</span>.</p>
<p>Let it be a property of <span class="math-container">$G,X$</span> that <span class="math-container">$\forall x,y,\exists g:g\cdot x=g\cdot y$</span>. This is not quite a transitive action - it describes for example a sequence of inclusions. <strong>What is the name for this type of action?</strong> I can't pair it with an appropriate definition from <a href="https://en.wikipedia.org/wiki/Group_action_(mathematics)#Types_of_actions" rel="nofollow noreferrer">here</a>.</p>
<p>My attempt? There seem to be several things going on here, none of which I can associate with documented group theory at the moment.</p>
<p><span class="math-container">$G$</span> seems to define a "contracting epimorphism"</p>
<p><span class="math-container">$G$</span> seems to define the identity function on the trivial group having the powerset of <span class="math-container">$X$</span> as its element.</p>
| Wojowu | 127,263 | <p>No group action (on a set with more than one element) satisfies this. If <span class="math-container">$g\cdot x=g\cdot y$</span>, then <span class="math-container">$x=g^{-1}\cdot(g\cdot x)=g^{-1}\cdot(g\cdot y)=y$</span>, so your condition implies <span class="math-container">$x=y$</span> for all <span class="math-container">$x,y\in X$</span>.</p>
|
4,178,548 | <p>I'm starting a Linear Algebra course and I'm a bit confused.</p>
<p>Say we have a vector
<span class="math-container">$x = \begin{pmatrix}
x_1\\
x_2\\
\end{pmatrix}$</span>, and another vector <span class="math-container">$y = \begin{pmatrix}
y_1\\
y_2\\
\end{pmatrix}$</span></p>
<p>When we have a matrix composed of the two vectors, instead of saying <span class="math-container">$\begin{pmatrix}
x_1&y_1\\
x_2&y_2\\
\end{pmatrix}$</span></p>
<p>we say that the first column corresponds to <span class="math-container">$x_1$</span>, and the second corresponds to <span class="math-container">$x_2$</span> so
<span class="math-container">$\begin{pmatrix}
x_1&x_2\\
y_1&y_2\\
\end{pmatrix}$</span></p>
<p>Maybe I misunderstood something, but I feel like this is a bit confusing to me, may someone explain to me how this works?</p>
<p>Thanks in advance!</p>
| Eric | 909,833 | <p>This just depends on context and definitions. Both are feasible in the right context:
<a href="https://en.m.wikipedia.org/wiki/Row_and_column_spaces" rel="nofollow noreferrer">https://en.m.wikipedia.org/wiki/Row_and_column_spaces</a></p>
|
1,804,042 | <p><strong>Edit:</strong> Here is the original problem; it is possible that my recurrence for the stationary distribution $\pi$ is incorrect.</p>
<blockquote>
<p>Consider a single server queue where customers arrive according to a Poisson process with intensity $\lambda$ and request i.i.d. $\mathsf{Exp}(\mu)$ service times. The server is subject to failures and repairs. The lifetime of a working server is an $\mathsf{Exp}(\theta)$ random variable, while the repair time is an $\mathsf{Exp}(\alpha)$ random variable. Successive lifetimes and repair times are independent, and are independent of the number of customers in the queue. When the server fails, all the customers in the queue are forced to leave, and while the server is under repair no new customers are allowed to join.</p>
</blockquote>
<p><strong>Edit:</strong> I have revised the recurrence.</p>
<p>In a problem on queueing theory I've derived the following recurrence:
\begin{align}
\pi_1 &=\left(\frac{\lambda+\theta}\mu\right)\pi_0 - \frac{\alpha\theta}{\mu(\alpha+\theta)}\\
\pi_{n+1} &= \left(1+\frac{\lambda+\theta}\mu\right)\pi_n - \frac\lambda\mu\pi_{n-1},\ n\geqslant1.
\end{align}
where $\lambda$, $\mu$, $\theta$, and $\alpha$ are positive constants and $$\sum_{i=0}^\infty \pi_i = \frac\alpha{\alpha+\theta}. $$ </p>
<p>After a lot of tedious algebra, I found that
$$\scriptsize\pi_n = \left(\frac{\alpha \theta \left(2 \theta (\lambda +\mu )+(\lambda -\mu )^2\right) \left(\theta +\lambda +\mu+ \sqrt{\theta ^2+2 \theta (\lambda +\mu )+(\lambda -\mu )^2}\right)^n}{(\alpha +\theta ) (2 \mu )^n \sqrt{\theta ^2+2 \theta (\lambda +\mu )+(\lambda -\mu )^2}}\right)(1+\pi_0) $$ for $n\geqslant 1$. To save space, let $$\mathcal C:=\sqrt{\theta ^2+2 \theta (\lambda +\mu )+(\lambda -\mu )^2}. $$</p>
<p>Summing over $n$ and solving for $\pi_0$, I found
$$\pi_0 =\frac{\alpha \mu \left(2 \theta (\lambda +\mu )+(\lambda -\mu )^2\right) \left(\lambda -\mu-\theta-\mathcal C \right)}{2 \theta (\alpha +\theta ) \mathcal C}, $$</p>
<p>and so
$$
\pi_n=\left(\frac{ \left(2 \theta (\lambda +\mu )+(\lambda -\mu )^2\right) \left(\lambda -\mu-\theta-\mathcal C \right)+2 \theta (\alpha +\theta ) \mathcal C }{2(\alpha +\theta )^2\mathcal C^2\left(\alpha^2\mu \left(2 \theta (\lambda +\mu )+(\lambda -\mu )^2\right)\right)^{-1}} \right)\left(\frac{\lambda+\mu+\theta+\mathcal C }{2\mu}\right)^n.
$$</p>
<p>If you see any errors let me know...</p>
<p>I'm also wondering what conditions on $\lambda,\mu,\theta$, and $\alpha$ are necessary for $\sum_{i=0}^\infty \pi_i$ to converge. For context, this is a $M/M/1$ queue with arrival rate $\lambda$, service rate $\mu$, but with an added state $D$ with transitions of rate $\theta$ from each state $n$ to $D$ and a transition of rate $\alpha$ from $D$ to $0$.</p>
| Semiclassical | 137,524 | <p>Note: The present form of the answer reflects an earlier version of the problem. I plan to modify it to reflect the changes, hopefully sooner rather than later...</p>
<p>Let $P(x)=\sum_{n=0}^\infty \pi_n x^n$. The first few lines of this recurrence are
\begin{align}
\pi_1+\frac{1}{\mu}\left(\frac{\alpha\theta}{\alpha+\theta}\right) &= \left(\frac{\lambda+\mu\theta}{\mu}\right)\pi_0,\\
\pi_2+\frac{1}{\mu}\left(\frac{\alpha\theta}{\alpha+\theta}\right) &= \left(\frac{\lambda+\mu\theta}{\mu}\right)\pi_1+\frac{\theta}{\mu}\pi_0,\\
\pi_3+\frac{1}{\mu}\left(\frac{\alpha\theta}{\alpha+\theta}\right) &= \left(\frac{\lambda+\mu\theta}{\mu}\right)\pi_2+\frac{\theta}{\mu}\pi_1+\frac{\theta}{\mu}\pi_0,\\
\end{align}
and so on. Multiplying each line by powers of $x$ (starting with $x^1$) and summing yields</p>
<p>$$P(x)-\pi_0 +\frac{1}{\mu}\left(\frac{\alpha\theta}{\alpha+\theta}\right)(x+x^2+x^3+\cdots) = \left(\frac{\lambda+\mu\theta}{\mu}\right)xP(x)+\frac{\theta}{\lambda}\left(x^2+x^3+\cdots\right)P(x).$$
We clean this up by multiplying both sides by $(1-x)$, yielding $$(1-x)(P(x)-\pi_0)+\frac{1}{\mu}\left(\frac{\alpha\theta}{\alpha+\theta}\right)x=\left(\frac{\lambda+\mu\theta}{\mu}\right)x(1-x)P(x)+\frac{\theta}{\mu}x^2 P(x).$$
As a check, for $x=1$ this implies
$$\frac{1}{\mu}\left(\frac{\alpha\theta}{\alpha+\theta}\right)=\frac{\theta}{\mu} P(1)\implies P(1)=\sum_{n=0}^\infty \pi_n =\frac{\alpha}{\alpha+\theta}$$
which is the desired normalization condition. What remains is to solve for $P(x)$ and then expand to obtain the coefficients $\{\pi_n\}$, a task I leave to the interested reader.</p>
|
818,169 | <p>The variational distance is defined by,
$$
V(P,Q)=\sum _{i}|p_{i} -q_{i} |
$$
where $P=(p_{1} ,...,p_{n})$ and $Q=(q_{1} ,...,q_{n} )$ are discrete distributions.</p>
<p>It is fairly easy to see that $V$ is a metric, and in particular that it satisfies the triangle inequality. I now introduce class prior, $0<\alpha<1$, which represents the relative 'size' of the distributions,
$$
V(\alpha ,P,Q)=\sum _{i}|\alpha p_{i} -(1-\alpha )q_{i} |
$$
and would like to prove that it again satisfies the triangle inequality,</p>
<p>$$V(\alpha ,P,Q)+V(\beta ,Q,R)\ge V(\gamma ,P,R) $$
where $R$ is another distribution, $\beta$ and $\gamma$ are respective priors, and $\gamma$ is naturally,
$$\gamma =\frac{\alpha \beta }{\alpha \beta +(1-\alpha )(1-\beta )}
$$</p>
<p>I have confidence (numerically) that this triangle inequality indeed holds, but would like a formal proof. Any ideas?</p>
| Johannes Schindelin | 166,166 | <p>But $V(\alpha, P, Q) != V(\alpha, Q, P)$ in general, therefore it would not be a metric, right? If that is true, what is the point of proving or disproving the triangle inequality?</p>
|
462,921 | <p>I simply don't get the following question answered:</p>
<p>How can i proof the equality $\lim_{a\to 0}\sup_{z\in\mathbb{Z}}2-2\cos(2\pi a z)=0$?</p>
<p>Or is it even false?</p>
<p>Thanks in advance!</p>
| Supriyo | 70,120 | <p>Think little bit differently.</p>
<p>$2 - 2\cos(2\pi a z) = 2(1-\cos(2 \pi a z)) = 4\sin^2(\pi a z)$$</p>
<p>Now in complex plain $\sin(z)$ is an unbounded function having a zero at origin. So the value of the expression should be $0$.</p>
|
1,977,306 | <p>This is from a math competition so it must not be something really long
If a parabola touches the lines $y=x$ and $y=-x$ at $A(3,3) $ and $b(1,-1)$ respectively, then </p>
<p>(A) equation of axis of parabola is $2x+y=0$ </p>
<p>(B)slope of tangent at vertex is $1/2$</p>
<p>(C) Focus is $(6/5,-3/5)$</p>
<p>(D) Directrix passes through $(1,-2)$ </p>
<p>I thought the axis would be the angle bisector of the tangents passing through the focus but it turns out that is not the case in a parabola so how can I find anything..</p>
| bubba | 31,744 | <p>It's well known that any rational quadratic Bézier curve is a conic section (ellipse, hyperbola, or parabola), and any (polynomial) quadratic Bézier curve is a parabola. There are fairly simple formulae for obtaining the geometric characteristics (directrix, focus, vertex) directly from the control points of the Bézier curve. For example, see the following paper and the earlier ones they cite:</p>
<p>Geometric Characteristics of Conics in Bézier Form<br />
Cantóna, Fernández-Jambrina, Rosado María<br />
Computer-Aided Design 43 (2011) 1413–1421</p>
<p>Here, we have a quadratic Bézier curve with control points <span class="math-container">$\mathbf{P}_0 =(3,3)$</span>, <span class="math-container">$\mathbf{P}_1 = (0,0)$</span>, <span class="math-container">$\mathbf{P}_2 = (1,-1)$</span>. Several of the papers tell you that the axis of symmetry is in the direction of the vector
<span class="math-container">$\tfrac12(\mathbf{P}_0 + \mathbf{P}_2) - \mathbf{P}_1$</span>, which is <span class="math-container">$(2,1)$</span>. Even without the papers, you can see this by letting <span class="math-container">$t\to\infty$</span> in the parametric equations of the curve. This shows that proposition (A) is false, and it follows immediately that (B) is also false.</p>
<p>Now that we know the axis direction, we can use the reflection property of a parabola to construct lines at <span class="math-container">$A$</span> and <span class="math-container">$B$</span> that pass through the focus. Intersecting these two lines gives us the focus. Or, use the formulae from the papers to get the focus.</p>
<p>I didn't write this up as a complete answer because I'm sure this is not how the problem was meant to be solved.</p>
|
730,198 | <blockquote>
<p>Show $f(x)=\sqrt{x^4+1} - \sqrt{x^4+x^2} \rightarrow -1/2$ for $x \rightarrow \infty$, $x \in \mathbb R$.</p>
</blockquote>
<p>I've tried $$\frac {(\sqrt{x^4+1} - \sqrt{x^4+x^2})(\sqrt{x^4+1} + \sqrt{x^4+x^2})}{\sqrt{x^4+1} + \sqrt{x^4+x^2} } = \frac {1-x^2} {\sqrt{x^4+1} + \sqrt{x^4+x^2}}$$</p>
<p>and $$\frac {1} {\sqrt{x^4+1} + \sqrt{x^4+x^2}} \rightarrow 0$$ so it's enough to verify $$\frac {-x^2} {\sqrt{x^4+1} + \sqrt{x^4+x^2}} \rightarrow -1/2$$.</p>
<p>However I'm having trouble showing this. </p>
| Amir Parvardi | 6,715 | <p>In this kind of problems which want you to calculate the limit when $x \to \infty$, constants are not very important and you have to pay attention to powers of $x$ (or anything related to $x$). That makes sense because, e.g., when $x=10$, you have $x^4 = 10000$, which is way greater than $1$.</p>
<p>Try to convert $x^4+x^2$ to a perfect square, that is:
$$x^4+x^2 +\frac 14 - \frac 14 = (x^2+\frac 12)^2 - \frac 14.$$
Now, take perfect squares out of radical. Your limit becomes
$$x^2\sqrt{1+\frac{1}{x^4}} - (x^2+\frac 12)\sqrt{1+\frac{1}{(x^2+\frac 12)^2}}, \quad x \to \infty$$
You see that the phrases under radical go to $1$ as $x$ goes to $\infty$, and the answer is simply $-\frac 12$.</p>
|
878,961 | <p>I'm asked to prove a theorem (if that is the right word) about double derivatives. I'm still struggling with understanding Leibniz notation and I could use a push in the right direction. It's easy enough for me to differentiate the function when I write it down as $f(g(x))$ but not so much with Leibniz notation.</p>
<p>The problem is as follows: If $y = f(u)$ and $u = g(x)$, where $f$ and $g$ are twice differentiable functions, show that:</p>
<p>$$\frac{d^2y}{dx^2} = \frac{d^2y}{du^2}\frac{du}{dx}^2 + \frac{dy}{du}\frac{d^2u}{dx^2}$$</p>
<p>Could someone fill me in on the details of each part of this equation? For example, why is the second derivative of $\frac{dy}{dx}$ written as $\frac{d^2y}{dx^2}$ instead of $\frac{d^2y}{d^2x}$?</p>
<p>Thanks!</p>
| Paul Sundheim | 88,038 | <p>The answer to your last question is that $\frac{dy}{dx}=\frac{d}{dx}y$ so the second derivative is $\frac{d}{dx}\frac{d}{dx}y=\frac{d^2}{dx^2}y$. This should answer the first question as well.</p>
|
1,853,464 | <p>I am using the Lorentz Force Equation and the electric-cross-magnetic field velocity equation] to solve for the $E$ and $B$ fields given the known path of a particle moving in 3D. </p>
<p>So with that I have the following equations where a and v are known:
<a href="https://i.stack.imgur.com/oL7fg.gif" rel="nofollow noreferrer">Lorentz Form</a> and the
<a href="https://i.stack.imgur.com/xuxHP.gif" rel="nofollow noreferrer">E-cross-B Form</a></p>
<p>My question: Are these equations enough to solve for the $x, y, z$ components of $B$ and $E$?</p>
<p>----------Edit---------------</p>
<p>So this is actually being used as an analogy for the propagation of nano-scale self replicating cracks in 3D. In this analogy, the incoming tensile force is represented by the electric force, and the delamination is represented by the magnetic force. </p>
<p>So I have a parabaloid spiral shaped crack which will represent the motion of a charged particle. Since I know the shape/path I can directly get the position, velocity, and acceleration functions in each direction.</p>
<p>With that said, is there a way to use the two equations linked to find all components of the electric and magnetic fields?</p>
| Ng Chung Tak | 299,599 | <p>One interesting case when a charged particles is at rest initially in the crossed electric and magnetic fields, it moves in a cycloid.</p>
<p>See the video <a href="https://www.youtube.com/watch?v=1eOn5K9qZu4" rel="nofollow">here</a>.</p>
<p>See also a detailed discussion in Chapter 2 of <strong>Fundamentals of Plasma Physics</strong> by <em>Bittencourt, J. A.</em> <a href="https://www.google.com.hk/url?sa=t&rct=j&q=&esrc=s&source=web&cd=6&cad=rja&uact=8&ved=0ahUKEwizqZaS4uTNAhWBqZQKHdOIDF0QFgg5MAU&url=http%3A%2F%2Fwww.springer.com%2Fcda%2Fcontent%2Fdocument%2Fcda_downloaddocument%2F9780387209753-c1.pdf%3FSGWID%3D0-0-45-110333-p25145278&usg=AFQjCNEuZT5BZvHNOikGh_UvyP4GNa6Ysw" rel="nofollow">here</a></p>
|
1,369,641 | <p>I do not know how to set this problem up. Any insight as to how to get the equation would be great. </p>
<p>It is John's birthday and his parents want to make him a cake in the shape of a rectangular box. The height of the cake will be $15$ centimeters, and $2$ times the width plus $2$ times the length will be $180$ centimeters. Find the largest possible volume of cake that John can receive. Round your answer to $5$ decimal places.</p>
| DeepSea | 101,504 | <p>$L + W = 90 \to V =f(L)= L\times W\times H = 15L(90-L)\to f'(L) = 15(90-2L)=0 \iff L = 45 \to W = 45$. Thus $V_{max} = 15\times 45\times 45 = ....$</p>
|
1,369,641 | <p>I do not know how to set this problem up. Any insight as to how to get the equation would be great. </p>
<p>It is John's birthday and his parents want to make him a cake in the shape of a rectangular box. The height of the cake will be $15$ centimeters, and $2$ times the width plus $2$ times the length will be $180$ centimeters. Find the largest possible volume of cake that John can receive. Round your answer to $5$ decimal places.</p>
| JMP | 210,189 | <p>We have $W+L=90$. Consider $W=L=45$ and then consider $W=45-k, L=45+k$. It is well known that $(45-k)(45+k)$ is maximum when $k=0$. In fact, this justifies the initial assumption.</p>
|
456,892 | <p>Find all solutions of $4\cos^2(x)-4\sin(x)-5=0$ in the interval $(6\pi, 8\pi)$.</p>
<p>I tried to work it out and got: $4y^2-4y -9 = 0$, but I can't figure out what $\cos x = $from there to finish the problem.</p>
| amWhy | 9,003 | <p>$$\begin{align} 4\cos^2 x - 4\sin x - 5 & = 0 \\ \\\iff 4(1 - \sin^2 x) & = 4\sin x + 5 \\ \\ \iff 4\sin^2 x + 4\sin x + 1 &= 0\end{align}$$</p>
<hr>
<p>$$\begin{align} y = \sin x \implies 4y^2 + 4y + 1 & = 0 \\ \\
(2y+1)^2 & = 0 \\ \\
\implies y & = -1/2\end{align}$$</p>
<hr>
<p>Therefore $y = \sin x = -1/2 \implies x = \sin^{-1}(-1/2) = 7\pi/6, 11\pi/6$. That's in the interval $x \in (0, 2\pi).$ </p>
<p>Adding $6\pi$ to each for the appropriate solutions in your given interval, we have solutions $$x \in \left\{\dfrac{43\pi}{6}, \dfrac{47\pi}{6}\right\}$$</p>
|
796,262 | <p>So, I am computing something seemingly simple involving complex gaussians and constants, but I am getting a big contradiction in my calculations. </p>
<p><strong>The setup:</strong></p>
<ul>
<li>Let $C$ be a complex constant, that is, $C = c_r + jc_i$. </li>
<li>Let $G$ be a complex gaussian variable, $G = g_r + jg_i$, where $g_r$ and $g_i$ are both uncorrelated, and where each are $\sim\mathcal{N}(0,\sigma^2)$.</li>
</ul>
<p>I am computing $z = |C + G|^2$. </p>
<p><strong>The problem:</strong> Now, before I go on, it is obvious that the variable $z$, must always be greater than or equal to $0$, owning to the $| \cdot |^2$ operation. However when I open up and compute $z$, I get an expression that seems like it CAN be less than $0$. </p>
<p>Opening up $z$, I get</p>
<p>$$
z = (c_r^2 + c_i^2) + 2\Big[c_rg_r + c_ig_i \Big] + (g_r^2 + g_i^2)
$$</p>
<p>The first term is a constant, and will always be greater than or equal to zero. The last term is has a gamma distribution, and by definition, will also always be greater than or equal to zero. However, the <em>middle</em> term is simply a summation of two gaussians, but this means that there is a finite probability that they take on a value of less than zero, meaning that $z$ can also be less than zero! </p>
<p>But $z$ can never be less than zero. This is the contradiction... I am not sure where I am making a mistake in my reasoning...</p>
<p>Thank you.</p>
| N. S. | 9,176 | <p>Any spanning tree has $6$ edges. Thus, the sum of degrees of the $3$ respectively $4$ edges is $6$.</p>
<p>Thus, the degrees of the $4$ vertices can be $(3,1,1,1)$ or $(2,2,1,1)$. Count them from here.</p>
|
204,365 | <p>Consider a positive matrix <code>M</code> and a positive vector <code>b</code>, e.g.</p>
<pre><code>nn = 1000;
M = Table[RandomReal[{0, 100}], {i, 1, nn}, {j, 1, nn}];
b = Table[RandomReal[{0, 100}], {i, 1, nn}];
</code></pre>
<p>I would like to find a positive vector <code>X</code></p>
<pre><code>X = Array[x,nn];
</code></pre>
<p>(each <code>x[i]>0</code>) such that given</p>
<pre><code>expr = M.X-b;
</code></pre>
<p>the quantity <code>expr.expr</code> is minimized. Is it possible to do that in Mathematica efficiently (so that it finishes within a few seconds/minutes)?</p>
| KennyColnago | 3,246 | <p>Perhaps you could use the non-negative least-squares algorithm (NNLS) of Lawson and Hanson, discussed <a href="https://mathematica.stackexchange.com/questions/26336/how-to-perform-a-multi-peak-fitting/26338#26338">here</a>. See the links in the comments.</p>
<p>NNLS was ported to Mathematica by Michael Woodhams, and used here as <code>NNLS[M,b]</code>. I am running Mma 11.3.0 for 64-bit Linux.</p>
<pre><code>nn = 1000;
SeedRandom[10];
M = Table[RandomReal[{0, 100}], {i, 1, nn}, {j, 1, nn}];
b = Table[RandomReal[{0, 100}], {i, 1, nn}];
ls = NNLS[M,b];
</code></pre>
<p>Timing was about 10 seconds for <code>nn=1000</code>.</p>
<p><a href="https://i.stack.imgur.com/4Kkp2.png" rel="noreferrer"><img src="https://i.stack.imgur.com/4Kkp2.png" alt="non-negative least-squares solution"></a></p>
|
2,304,379 | <p>My textbook give the following definition.</p>
<blockquote>
<p>Let $G$ be any topological group. A representation of $G$ on a nonzero complex Hilbert space $V$ is a group homomorphism $\phi$ of $G$ into the group of bounded linear operators on $V$ with bounded inverses, such that the resulting map $ G\times V\to V$ is continuous. </p>
</blockquote>
<p>And it says that to have the continuity property it is enough to have that the map $g\mapsto \phi(g)v$ from $G$ to $V$ is continuous at $g=1$ for all $v\in V$ and a uniform bound for $\|\phi(g)\|$ in some neighbourhood of $1$.</p>
<p>I want to show that the map $G\times V\to V$ given by $(g,v)\mapsto \phi(g)v$ is continuous under these assumptions.</p>
<p><strong>My attempt</strong></p>
<p>Let $(g_0,v_0)\in G\times V$. Let $\epsilon>0$. We need to find an open neighbourhood $U\subseteq G\times V$ of $(g_0,v_0)$ such that $\|\phi(g)v-\phi(g_0)v_0\|<\epsilon$ for all $(g,v)\in U$. By the continuity of the map $G\to V$ by $g\mapsto \phi(g)v_0$ at $g=1$, there is an open neighbourhood $N$ of $1$ such that $\|\phi(g)v_0-v_0\|<\epsilon$ for all $g\in N$. On the other hand, for any $g\in N$ we have
\begin{align*}
\|\phi(g)v-\phi(g_0)v_0\|&=\|\phi(g_0^{-1}g)v-v_0\| \qquad \text{by the unitarity} \\&\leq \|\phi(g_0^{-1}g)v-\phi(g_0^{-1}g)v_0\|+\|\phi(g_0^{-1}g)v_0-v_0\| \\& \leq \|\phi(g_0^{-1}g)\|_{op} \|v-v_0\|+\underbrace{\|\phi(g_0^{-1}g)v_0-v_0\|}_\text{$(*)$}
\end{align*}</p>
<p>Here I could not find an upper bound for the term $(*)$. Could anyone help me? Thanks.</p>
<p>Update: I have realized that we have no unitarity assumption. So I will think on this. </p>
| bof | 111,012 | <p>The empty set is closed and does not contain any nonempty open set. Probably you meant to ask for a <strong>nonempty</strong> closed set which does not contain any nonempty open set. The following theorem answers this question in a general topological space X.</p>
<p><strong>Theorem.</strong> If in the space $X$ there is a closed set which is not open, then there is a nonempty closed set which does not contain any nonempty open set.</p>
<p><strong>Proof.</strong> If there is a closed set which is not open, then its complement, call it $U,$ is an open set which is not closed. Of course $U\ne\emptyset,$ since $\emptyset$ is closed. Assuming the axiom of choice, we can extend $\{U\}$ to a maximal collection $\mathcal U$ of pairwise disjoint nonempty open sets. Then the set $A=X\setminus\bigcup\mathcal U$ is a closed set which does not contain any nonempty open set. I claim that $A\ne\emptyset.$</p>
<p>If we had $A=\emptyset,$ it would follow that $\bigcup\mathcal U=X$ and that $\bigcup(\mathcal U\setminus\{U\})=X\setminus U,$ so that $X\setminus U$ would be an open set, contradicting the fact that $U$ is not closed.</p>
|
3,065,818 | <blockquote>
<p>If <span class="math-container">$$z=\dfrac{\sqrt{3}-i}{2}$$</span> then <span class="math-container">$$(z^{95}+i^{67})^{94}=z^n$$</span> then, <span class="math-container">$\text{find the smallest positive integral value of}$</span> <span class="math-container">$n$</span> <span class="math-container">$\text{where}$</span> <span class="math-container">$i=\sqrt{-1}$</span></p>
</blockquote>
<p><span class="math-container">$\text{My Attempt:}$</span> First of all I tried to convert <span class="math-container">$z$</span> into <span class="math-container">$\text{Euler's Form}$</span> so, <span class="math-container">$z=e^{-i(\frac{π}{6})}$</span>
Then, I raised <span class="math-container">$z$</span> to the <span class="math-container">$\text{95th}$</span> power. Then I'm getting stuck. And, not being able to proceed. Help. </p>
| N. S. | 9,176 | <p><strong>Hint</strong> You can find some <span class="math-container">$k,l \in \mathbb Z$</span> such that
<span class="math-container">$$ka+lb=1$$</span></p>
<p>Then
<span class="math-container">$$n=n^{ka+lb}=(n^a)^k\cdot(n^b)^l$$</span></p>
<p>Now, if you want <span class="math-container">$x = n^b, y=n^a$</span> then the only possible choice for <span class="math-container">$n$</span> is
<span class="math-container">$$n=??$$</span></p>
<p>Just prove that <strong>this is an integer</strong> and that this choice works. </p>
|
3,362,115 | <blockquote>
<p>Find the maximum value of <span class="math-container">$y/x$</span> if it satisfies <span class="math-container">$(x-5)^2+(y-4)^2=6$</span>.</p>
</blockquote>
<p>Geometrically, this is finding the slope of the tangent from the origin to the circle. Other than solving this equation with <span class="math-container">$x^2+y^2=35$</span>, I cannot see any synthetic geometry solution. Thanks!</p>
| albert chan | 696,342 | <p>Let a line be <span class="math-container">$y = mx$</span></p>
<p><span class="math-container">$$(x-5)^2 + (mx-4)^2 - 6 = 0$$</span>
<span class="math-container">$$(m^2+1)x^2 + (-8m-10)x + 35 = 0$$</span></p>
<p>If the line does not touch the circle, above have no solution.<br>
In other words, discriminant is negative.</p>
<p>Line is a tangent if discriminant is zero</p>
<p><span class="math-container">$$(-8m-10)^2 - 4(m^2+1)(35) = 0$$</span>
<span class="math-container">$$35(m^2+1)-(4m+5)^2 = 0$$</span>
<span class="math-container">$$19m^2 -40m + 10 = 0$$</span>
<span class="math-container">$$m = {20 ± \sqrt{210} \over 19}$$</span></p>
<p>For maximum slope, pick the + sign case.</p>
|
3,554,188 | <blockquote>
<p>I was given <span class="math-container">$$|ax - 11| = 4x - 10$$</span> has a positive integral solution and <span class="math-container">$a$</span> is a positive integer.</p>
<p>I was asked what was <span class="math-container">$x, a$</span></p>
</blockquote>
<p><span class="math-container">$$ax > 11 $$</span> then we have <span class="math-container">$x=\frac{1}{a-4}$</span>
<span class="math-container">$$ax < 11$$</span> then we have <span class="math-container">$x=\frac{22}{a+4}$</span></p>
<p>What should I do now? I don't understand what does the "one solution" mean because all I know there are some possibilities for <span class="math-container">$x,a$</span> to be the solutions. The answer keys are <span class="math-container">$x, a =3$</span></p>
| VIVID | 752,069 | <p>I think you made a little error because you had to have <span class="math-container">$x=\frac{21}{a+4}$</span> when <span class="math-container">$ax<11$</span>.
Hence <span class="math-container">$(x,a)=(3,3)$</span>.</p>
|
4,326,073 | <p><a href="https://i.stack.imgur.com/RTyOy.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RTyOy.jpg" alt="enter image description here" /></a>
I came across questions in the free module section of my abstract algebra text. In the text, the notation <span class="math-container">$End_{R}(V)$</span> denotes the set of all <span class="math-container">$R$</span>-module endomorphisms of <span class="math-container">$M$</span>. <a href="http://homepage.math.uiowa.edu/%7Egoodman/algebrabook.dir/book.2.6.pdf" rel="nofollow noreferrer">Algebra: Abstract and Concrete, exercise 8.1.9 on p358 in the attached picture of the linked text</a> Onto the question:</p>
<blockquote>
<p>Let <span class="math-container">$V$</span> be a finite dimensional vector space over a field <span class="math-container">$K$</span>. Let <span class="math-container">$T\in End_{K}(V)$</span>. Give <span class="math-container">$V$</span> the corresponding <span class="math-container">$K[x]$</span>-module structure defined by <span class="math-container">$\sum_{i} \alpha_{i}x^{i}v=\sum_{i}\alpha_{i}T^{i}(v).$</span> Show that <span class="math-container">$V$</span> is not free as a <span class="math-container">$K[x]$</span>-module.</p>
</blockquote>
<p>In this question, I don't understand how in <span class="math-container">$\sum_{i} \alpha_{i}x^{i}v=\sum_{i}\alpha_{i}T^{i}(v).$</span> affects whether <span class="math-container">$V$</span> can be a free <span class="math-container">$K[x]$</span>-module. If I take a finite basis <span class="math-container">$B$</span> for <span class="math-container">$V$</span>, where <span class="math-container">$B=\{v_1, v_2,...v_n\}$</span>, with coefficients <span class="math-container">$\alpha_i \in K$</span>, then would it be that each of the <span class="math-container">$\alpha_i x^i v_{i}$</span> term in the identity <span class="math-container">$\alpha_i x^i v_{i} = \alpha_{i}T^{i}(v_i)$</span>, the coefficients <span class="math-container">$\alpha_{i}$</span> or <span class="math-container">$\alpha_{i}x^{i}$</span> might not equal to zero? Actually, I am assuming the <span class="math-container">$v$</span> in the definition <span class="math-container">$\alpha_i x^i v = \alpha_{i}T^{i}(v)$</span> refers to basis elements from <span class="math-container">$B$</span>, but I am not sure where the <span class="math-container">$x^{i}$</span> is suppose to come from. Is it from the vector space <span class="math-container">$V$</span> or from the field <span class="math-container">$K$</span> along with the <span class="math-container">$\alpha_i$</span>.</p>
<p>Thank you in advance.</p>
| user284331 | 284,331 | <p>A few comments to the answers provided by @QC_QAOA and @Zieac.</p>
<p>Personally I think one should go through by <span class="math-container">$\limsup$</span> and <span class="math-container">$\liminf$</span> argument.</p>
<p>Note that
<span class="math-container">\begin{align*}
\liminf(u_{n}+v_{n})=\liminf u_{n}+\lim v_{n}
\end{align*}</span>
provided that <span class="math-container">$\lim v_{n}$</span> exists. Similar statements hold for <span class="math-container">$\limsup$</span>.</p>
<p>Then,
<span class="math-container">\begin{align*}
\dfrac{a-\epsilon}{2}&=\lim\cdots\\
&\leq\liminf(b_{n}+\cdots)\\
&=\liminf b_{n}+\lim\cdots\\
&=\liminf b_{n}+0\\
&\leq\limsup b_{n}+0\\
&=\limsup b_{n}+\lim\cdots\\
&=\limsup(b_{n}+\cdots)\\
&\leq\lim\cdots\\
&=\dfrac{a+\epsilon}{2}
\end{align*}</span>
Lastly, since <span class="math-container">$\epsilon>0$</span> is arbitrary, we conclude that <span class="math-container">$\liminf b_{n}=\limsup b_{n}$</span> and hence the existence of the limit of <span class="math-container">$b_{n}$</span>.</p>
|
1,938,253 | <p>Consider the following limit:$$\lim_{x \to\infty}\frac{2+2x+\sin(2x)}{(2x+\sin(2x))e^{\sin(x)}}$$</p>
<p>If we apply L'hospital's rule then we get:</p>
<blockquote>
<p><a href="https://i.stack.imgur.com/slrYm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/slrYm.png" alt="Blockquote"></a></p>
</blockquote>
<p>Since the function $e^{-\sin(x)}$ is bounded and $\frac{4\cos(x)}{2x+4\cos(x)+\sin(2x)}e^{-\sin(x)}$ tends to $0$. </p>
<p>However, it is also stated that: </p>
<blockquote>
<p><a href="https://i.stack.imgur.com/gqxqU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gqxqU.png" alt="Blockquote"></a></p>
</blockquote>
<p>Furthermore, (referring to the application of the L'hospital's Rule) it is stated in conclusion that:</p>
<blockquote>
<p><a href="https://i.stack.imgur.com/M98KC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/M98KC.png" alt="Blockquote"></a></p>
</blockquote>
| zxcvber | 329,909 | <p>Since</p>
<p>$$-1 \leq -\sin x \leq 1, \ \ \ 1/e\leq e^{-\sin x}\leq e$$</p>
<p>Thus,</p>
<p>$$\frac{4\cos{x}}{2x+4\cos{x}+\sin{2x}}\cdot\frac{1}{e}\leq \frac{4\cos{x}}{2x+4\cos{x}+\sin{2x}}e^{-\sin{x}}\leq\frac{4\cos{x}}{2x+4\cos{x}+\sin{2x}}e$$</p>
<p>We know that
$$\lim_{x\to\infty}\frac{4\cos{x}}{2x+4\cos{x}+\sin{2x}}=0$$</p>
<p>Thus by Squeeze Theorem,
$$\lim_{x\to\infty}\frac{4\cos{x}}{2x+4\cos{x}+\sin{2x}}e^{-\sin{x}}=0$$</p>
<hr>
<p>Nonexistence of the limit of $e^{-\sin{x}}$ does not imply that the limit of the whole expression doesn't exist.</p>
<p>For instance, limit of $(-1)^n$ as $n$ goes to $\infty$ does not exist.
But, limit of $\frac{(-1)^n}{n}$ does exist, and it is equal to $0$, which can also be proven by using the Squeeze Theorem.</p>
<p>There seems to be nothing wrong with the original question.</p>
<hr>
<p><strong>EDIT</strong>
After reading the edited post, I was very surprised to find an example where L'Hopital's rule doesn't really seem to work!
I looked up on Wikipedia, and it states that the derivative of the denominator must not be zero.</p>
<p>I guess this is not a counterexample, just that L'Hopital's rule cannot be applied here.</p>
|
1,938,253 | <p>Consider the following limit:$$\lim_{x \to\infty}\frac{2+2x+\sin(2x)}{(2x+\sin(2x))e^{\sin(x)}}$$</p>
<p>If we apply L'hospital's rule then we get:</p>
<blockquote>
<p><a href="https://i.stack.imgur.com/slrYm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/slrYm.png" alt="Blockquote"></a></p>
</blockquote>
<p>Since the function $e^{-\sin(x)}$ is bounded and $\frac{4\cos(x)}{2x+4\cos(x)+\sin(2x)}e^{-\sin(x)}$ tends to $0$. </p>
<p>However, it is also stated that: </p>
<blockquote>
<p><a href="https://i.stack.imgur.com/gqxqU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gqxqU.png" alt="Blockquote"></a></p>
</blockquote>
<p>Furthermore, (referring to the application of the L'hospital's Rule) it is stated in conclusion that:</p>
<blockquote>
<p><a href="https://i.stack.imgur.com/M98KC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/M98KC.png" alt="Blockquote"></a></p>
</blockquote>
| Paramanand Singh | 72,031 | <p>First of all +1 for coming up with an example which seems to violate the L'Hospital's Rule. However note that if $f(x)/g(x)$ is the original expression then $$\frac{f'(x)}{g'(x)} = \frac{4\cos^{2}x}{4\cos^{2}x + (2x + \sin 2x)\cos x}\cdot e^{-\sin x}$$ The problem now is that the limit of $f'(x)/g'(x)$ does not exist as $x \to \infty$ because the function $f'(x)/g'(x)$ is not defined in any interval of type $(a, \infty)$ precisely because the denominator $g'(x)$ vanishes for $x = (2n + 1)\pi/2$ for all $n \in \mathbb{Z}$.</p>
<p>As I have mentioned <a href="https://math.stackexchange.com/a/1798950/72031">elsewhere on MSE</a> it is important to understand the conditions under which L'Hospital's Rule is applicable. We must ensure that the ratio $f'(x)/g'(x)$ tends to a limit (or diverges to $\pm\infty$) and only then we can conclude the same behavior for ratio $f(x)/g(x)$. Apart from this we need the ratio $f(x)/g(x)$ to be one of the indeterminate forms "$0/0$" and "$\text{anything}/\pm\infty$".</p>
|
2,653,726 | <p>Prove that $1, r, r^2,\ldots, r^{n-1}, t, tr, tr^2,\ldots, tr^{n-1}$ are $2n$ distinct symmetries of the regular n-gon.</p>
<p>It is clear that there are always $2n$ symmetries for any regular $n$-gon, with half of them being rotations and the other half being their reflections. Therefore for an $n$-gon with an even number of sides, every nontrivial symmetry in $D_n$, we shall call it, is either a reflection or rotation. Therefore $|D_n| = 2n$, as we have $2n$ elements, and each nontrivial element of $D_n$ is a reflection or rotation. </p>
<p>That was my thought, but I think I'm missing the mark. I haven't meddled with proofs for some time and am struggling with the seeming simplicity of this question.</p>
| MCT | 92,774 | <p>You have showed that there are at least $2n$ symmetries, but there could be more. Note that these symmetries are distance preserving, in that if the distance between vertices $a$ and $b$ is $D$, then after applying the symmetry $s$ the vertices $s(a)$ and $s(b)$ will still have distance $D$. </p>
<p>Ok so now pick an arbitrary vertex $P$; the symmetry sends this to $Q$ ($n$ choices). But now the two vertices around it must be the two vertices adjacent to $P$, and we have two choices in where to put each. Fill in the details that after this, we no longer have any choice in our symmetry!</p>
|
2,653,726 | <p>Prove that $1, r, r^2,\ldots, r^{n-1}, t, tr, tr^2,\ldots, tr^{n-1}$ are $2n$ distinct symmetries of the regular n-gon.</p>
<p>It is clear that there are always $2n$ symmetries for any regular $n$-gon, with half of them being rotations and the other half being their reflections. Therefore for an $n$-gon with an even number of sides, every nontrivial symmetry in $D_n$, we shall call it, is either a reflection or rotation. Therefore $|D_n| = 2n$, as we have $2n$ elements, and each nontrivial element of $D_n$ is a reflection or rotation. </p>
<p>That was my thought, but I think I'm missing the mark. I haven't meddled with proofs for some time and am struggling with the seeming simplicity of this question.</p>
| Siong Thye Goh | 306,553 | <p>Prove that $1,r,r^2,\ldots,r^{n−1},t,tr,tr^2, \ldots,tr^{n−1}$, are $2n$ <strong>distinct</strong> symmetries of the regular n-gon.</p>
<p>I would focus more on proving that they are distinct. </p>
<p>If $r^i=r^j, j> i, \{i,j\} \subseteq \{ 0,n-1\}$, then $r^{j-i}=1$ but the order of $r$ is $n$.</p>
<p>We also check that we do not have $tr^i=r^j, \{i,j\} \subseteq \{ 0,n-1\}$, then we have $t=r^{j-i}$ but we know that $t$ is not a rotation. </p>
<p>The case to rule out $tr^i = tr^j$ should be trivial.</p>
<p>Also, I am not sure if you are required to prove that these are the only elements.</p>
|
27,271 | <p>A few days ago I recall finding a visual calendar of my sign-in history on Math Stack Exchange. It looked like an ordinary monthly calendar, where the days were white if I hadn't signed in, and green if I had. (I found it when I wondering how close I've been in the past to getting the "fanatic" badge.)</p>
<p>However, I can no longer find this page/pop-up.</p>
<p>I thought it might have been a feature of Stack Exchange as a whole, but I was unable to find an answer on Meta Stack Exchange.</p>
| Martin Sleziak | 8,297 | <p>A brief answer is that you should go to <a href="https://math.stackexchange.com/users/current?tab=profile">your profile</a> and click on the number of consecutive days. (Remember that you'll find this in the profile tab, not in the activity tab.)</p>
<p>You can find more details here:</p>
<ul>
<li><a href="https://math.meta.stackexchange.com/q/24876">Statistics of visits on M.SE</a></li>
<li><a href="https://meta.stackexchange.com/q/68956">Implementation date for the consecutive days calendar</a> </li>
<li><a href="https://meta.stackexchange.com/q/128605">How can I see which days I visited site?</a></li>
<li>The <a href="https://meta.stackexchange.com/tags/calendar/info">tag-info for the calendar tag</a> is also a reasonable place to find more information.</li>
</ul>
<p>(Originally I considered that this question could be closed as a duplicate of the first one in this list. But maybe they are slightly different after all - the other one asks also whether it is possible to see this information for other users, not just your own visits. That's why I decided to convert my comments to an answer.)</p>
|
2,186,076 | <p><strong>Question.</strong></p>
<blockquote>
<p>If $p$ is a polynomial of degree $n$ with $p(\alpha)=0$, what do we know of the polynomial $q$ (with degree $n-1$) such that the numbers $(q^k(\alpha))_{k=1}^n$ contain all of the zeroes of $p$?</p>
</blockquote>
<p>Here I denote $q(q(\cdots q(\alpha)))=q^k(\alpha)$.</p>
<p><hr>
<strong>Notes.</strong></p>
<p>We know for a fact such $q$ exists, since there always exists a polynomial of degree $n-1$ through $n$ given points. $q$ is not unique, however, since there are multiple permutations we can put the zeroes in.
<hr>
<strong>Examples.</strong></p>
<p>For linear $p$ (write $p(x)=a_1x+a_0$), this is obvious; $q(x)=\tfrac{-a_0}{a_1}=\alpha$ suffices. If $p(\alpha)=0$, then $q(\alpha)=\alpha$ indeed are all the zeroes of $p$.</p>
<p>If $p$ is quadratic, write $p(x)=a_2x^2+a_1x+a_0$, and have $p(\alpha)=0$ again; now $q(x)=\frac{-a_1}{a_2}-x$.</p>
<p>If $p$ is cubic, write $p(x)=a_3x^3+a_2x^2+a_1x+a_0$. This is where I get stuck, since the roots of cubic equations aren't expressions that are easy to work with.
<hr>
<strong>Attempts.</strong></p>
<p>First I see (denote the (not necessarily real) zeroes by $z_1,z_2,\cdots,z_n$) that $z_1+\cdots+z_n=\frac{-a_{n-1}}{a_n}$ and $z_1z_2\cdots z_n=\frac{(-1)^na_0}{a_n}$. We can produce similar expressions for the other coefficients, but I doubt this is useful; they're not even solvable for $n>4$. We also have (given $z_1$)
$$-a_nz_1^n=a_{n-1}z_1^{n-1}+\cdots+a_1x+a_0$$
with which we can reduce every expression of degree $n$ or larger in $z_1$ to an expression of degree $n-1$ or smaller.</p>
<p>For $n=3$ (let's do some specific examples), we could write $q(x)=b_2x^2+b_1x+b_0$, and take for example $p(x)=x^3-x-1$. Then, if $\alpha$ is a zero of $p$, then $\alpha^3=\alpha+1$, and so $q(\alpha)^3=q(\alpha)+1$, which is</p>
<p>$$(b_2\alpha^2+b_1\alpha+b_0)^3=b_2\alpha^2+b_1\alpha+b_0+1$$</p>
<p>working out the constant terms gives $b_2^3+b_1^3+b_0^3+6b_0b_1b_2=b_0+1$ which isn't very useful either.</p>
<p>Please, enlighten me. Has there been done work on this subject, am I missing something obvious, or perhaps you see something that I missed?</p>
| mercio | 17,445 | <p>For any sequence of $n$ distinct (complex) numbers $\alpha_1 \ldots \alpha_n$, there is a unique interpolating polynomial $Q$ of degree $n-1$ with $P(\alpha_i) = \alpha_{i+1}$.</p>
<p>In the generic case, the coefficients of $Q$ are in $\Bbb Q[\alpha_i, \delta^{-1}]$, where $\delta = \prod_{i<j} (\alpha_i - \alpha_j)$ (to be more precise, $\delta Q \in \Bbb Z[\alpha_i]$)</p>
<p>In the case where the $\alpha_i$ are roots of a polynomial $P$ with no repeated roots, then for any $n$-cycle $\sigma$, $Q_\sigma$ is defined over the splitting field of $P$, but sometimes it is defined over a smaller field.</p>
<p>If $G$ is the Galois group of $P$, the action of $G$ on the $Q_\sigma$ corresponds to its action on the set of $n$-cycles by conjugation.</p>
<p>If $K = \Bbb Q(a_i)$ is the field of definition of $P$ and $L_\sigma$ is the field of definition of $Q_\sigma$ for an $n$-cycle $\sigma$, then $[ L_\sigma : K ]$ is exactly the size of the orbit of $\sigma$ under this conjugation action. Since the normalizer of an $n$-cycle in $S_n$ is the subgroup of $S_n$ generated by the $n$-cycle, we have that the subgroup of $G$ fixing the $n$-cycle is $G_\sigma = G \cap \langle \sigma \rangle$, and then $L_\sigma = \Bbb Q(\alpha_i)^{G_\sigma}$, and finally $[L_\sigma : K] = |G| / |G_\sigma|$</p>
<p>For example,
if $n=2$, there is only one $2$-cycle, so there is only one $Q$, and $L= K$ which means that there are formulas for the coefficients of $Q$ in terms of the $a_i$.</p>
<p>If $n=3$, there are two $3$-cycles. If $G \not \subset A_3$ then $G$ will switch them so you will get an orbit of size $2$, so the coefficients of $Q$ will be in a degree $2$ extension of $K$. If $G \subset A_3$ then $G$ won't switch them, which tells you that in this case $L_\sigma = K$. In fact you have formulas for the two $Q$ in terms of the $a_i$ and $\delta = \pm \sqrt \Delta$ (and you switch between the two $Q$ by switching the sign of $\delta$)</p>
<p>If $n=4$, there are $3! = 6$ $4$-cycles, living in $6/\phi(4) = 3$ different cyclic subgroups of order $4$, so lots of stuff can happen.
In the worst case, $G = S_4$ and they are all defined in $3$ different (non-normal) extensions of order $6$ of $K$. $S_4$ is still solvable so if you really want to you can get formulas for $Q$ with square roots and cube roots.</p>
<p>If $n \ge 5$ well then $G$ may not even be solvable anymore so you won't have those formulas anymore.</p>
<p>The only way you can have $L_\sigma = K$ is when $|G_\sigma| = |G|$, which means $G \subset \langle \sigma \rangle$, and for this to happen $G$ has to be cyclic too (not necessarily of order $n$ if $P$ is not irreducible, though).</p>
|
1,425,907 | <p>Show that $$1+e^{-j\theta} =2e^{-j\theta/2}*\cos{\frac{\theta}2}$$</p>
<p>I know of the Euler equation: $e^{j\theta}=\cos(\theta)+j\sin(\theta)$ but am unsure how to simply show that the above are equal.</p>
| ajotatxe | 132,456 | <p>Substitute the expresion for $a_n$ in the formula of $S_n$:</p>
<p>$$363=\frac n2\big(-12+[-12+(n-1)9]\big)$$</p>
|
3,662,286 | <p>Can you raise the imaginary number i to a power that is an irrational number?</p>
| Community | -1 | <p>Short answer:</p>
<p><span class="math-container">$$(e^{i(\pi/2+2k\pi)})^x=e^{ix(\pi/2+2k\pi)}.$$</span></p>
|
1,161,631 | <p>Let $z \in \mathbb{C}$ be a root of real polynomial $p(x)=\sum_{k=0}^{n} a_k x^k$ ,$a_k\in \mathbb{R} \forall k$.</p>
<p>How to proove, that $\overline{z}$ is also a root of given polynomial? Is that true for complex polynomials, too?</p>
| quid | 85,306 | <p>Recall that complex conjugation can be interchanged with products and sums. Plug in $\overline{z}$ an use this repeatedly. </p>
<p>Note that $a \overline{z}= \overline{az}$ when $a$ is real yet not in general. </p>
|
1,161,631 | <p>Let $z \in \mathbb{C}$ be a root of real polynomial $p(x)=\sum_{k=0}^{n} a_k x^k$ ,$a_k\in \mathbb{R} \forall k$.</p>
<p>How to proove, that $\overline{z}$ is also a root of given polynomial? Is that true for complex polynomials, too?</p>
| kryomaxim | 212,743 | <p>May be $z_0=x+iy$ a root of the polynomial $p$. Then, if $\bar{z_0} = x-iy$ is also a root then the polynomial can be divided by $(z-z_0)(z-\bar{z_0}) = (z-x)^2+y^2 \in R$ because the numbers $x,y$ are real numbers.</p>
<p>For complex polynomials, this relation does not hold. </p>
|
1,161,631 | <p>Let $z \in \mathbb{C}$ be a root of real polynomial $p(x)=\sum_{k=0}^{n} a_k x^k$ ,$a_k\in \mathbb{R} \forall k$.</p>
<p>How to proove, that $\overline{z}$ is also a root of given polynomial? Is that true for complex polynomials, too?</p>
| Bumblebee | 156,886 | <p>Suppose $z=re^{i\theta}.$ </p>
<blockquote>
<p><strong>If you set $b_k=ra_k$ then the polynomial
$$q(x)=\sum_{k=0}^{n} b_k x^k$$ has the zero $e^{i\theta}$ if and only if $$p(x)=\sum_{k=0}^{n} a_k x^k$$ has the zero $z=re^{i\theta}.$</strong> </p>
</blockquote>
<p>If $b_k\in\mathbb{R},$ then all $a_k\in\mathbb{R},$ and by <a href="http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0CCAQFjAA&url=http%3A%2F%2Fen.wikipedia.org%2Fwiki%2FDe_Moivre%2527s_formula&ei=yC_rVNrbBcTZ8gXKzID4Ag&usg=AFQjCNG4RQ4D8o0S1x_I4fLQQb2zCGol1g&bvm=bv.86475890,d.dGY" rel="nofollow">D'Moivers theorem</a><br>
$$q(e^{i\theta})=0\iff\sum^n_{k=0}b_k\left(\cos k\theta+i\sin k\theta\right)=0$$
$$q(e^{i\theta})=0\iff\sum^n_{k=0}b_k\cos k\theta+i\sum^n_{k=0}b_k\sin k\theta=0$$
$$q(e^{i\theta})=0\iff\sum^n_{k=0}b_k\cos k\theta=0 ,\,\,\text{and} \sum^n_{k=0}b_k\sin k\theta=0$$
$$q(e^{i\theta})=0\iff\sum^n_{k=0}b_k\cos k\theta-i\sum^n_{k=0}b_k\sin k\theta=0$$
$$q(e^{i\theta})=0\iff q(e^{-i\theta})=0.$$
Hence $$p(re^{i\theta})=0\iff p(re^{-i\theta})=0.$$</p>
|
3,735,798 | <blockquote>
<p><strong>QUESTION:</strong> Given a square <span class="math-container">$ABCD$</span> with two consecutive vertices, say <span class="math-container">$A$</span> and <span class="math-container">$B$</span> on the positive <span class="math-container">$x$</span>-axis and positive <span class="math-container">$y$</span>-axis respectively. Suppose the other vertex <span class="math-container">$C$</span> lying in the first quadrant has coordinates <span class="math-container">$(u , v)$</span>. Then find the area of the square <span class="math-container">$ABCD$</span> in terms of <span class="math-container">$u$</span> and <span class="math-container">$v$</span>.</p>
</blockquote>
<hr>
<p><strong>MY APPROACH:</strong> I was trying to solve it out using complex numbers, but I need a minor help. I have assumed <span class="math-container">$A$</span> to be <span class="math-container">$(x_1+0i)$</span>, <span class="math-container">$B$</span> to be <span class="math-container">$(0+y_2i)$</span> and <span class="math-container">$C$</span> is <span class="math-container">$(u+vi)$</span>. We know that multiplying a point by <span class="math-container">$i$</span> basically rotates it by <span class="math-container">$90°$</span>, <strong>about the origin</strong>. Here, <span class="math-container">$C$</span> is nothing but the reflection of <span class="math-container">$A$</span> about the line <span class="math-container">$BD$</span>. So if I can somehow rotate <span class="math-container">$A$</span> about <span class="math-container">$B$</span> by <span class="math-container">$90°$</span> then we will get <span class="math-container">$x_1$</span> and <span class="math-container">$y_2$</span> in terms of <span class="math-container">$u$</span> and <span class="math-container">$v$</span>.
This is where I am stuck. How to rotate a point with respect to another?</p>
<blockquote>
<p>Note that this question has been asked before. But I want to know how to solve it using complex numbers..</p>
</blockquote>
<p>Any answers, possibly with a diagram will be much helpful..</p>
<p>Thank you so much..</p>
| atul ganju | 788,520 | <p>You can calculate it like this:</p>
<p><span class="math-container">$$(u-v)^2+u^2$$</span></p>
<p>To understand why, lets define two variables: <span class="math-container">$x, y$</span>. Given the fact that the square is "leaning" on the y-axis, <span class="math-container">$x$</span> is the distance from the origin to the lowest vertex of the square that touches the y-axis and <span class="math-container">$y$</span> is the distance from the origin to the furthest vertex of the square that touches the x-axis. Now, we can see that <span class="math-container">$(u, v)$</span> is just <span class="math-container">$(y, x+y)$</span>. Identically, we can see the fourth vertex is at <span class="math-container">$(x+y, x)$</span> or <span class="math-container">$(v, u-v)$</span>. The distance between these is a side length of the square so this distance squared is the area of the square. Using the distance formula we get:</p>
<p><span class="math-container">$A = \left(\sqrt{(u-v)^2+(v-(u-v))^2}\right)^2 = \left(\sqrt{(u-v)^2+v^2} \right)^2 = (u-v)^2+u^2$</span></p>
|
433,816 | <p>While I was studying the measurements of pressure at earth's atmosphere,I found the barometric formula which is more complex equation ($P'=Pe^{-mgh/kT}$) than what I used so far ($p=h\rho g$).</p>
<p>So I want to know how this complex formula build up? I could reach at the point of
$${dP \over dh}=-{mgP \over kT}$$
From this how can I obtain that equation. Please give me a Mathematical explanation.</p>
| Robert Lewis | 67,071 | <p>If </p>
<p>$\frac{dP}{dh} = (-\frac{mgP}{kT})$,</p>
<p>then</p>
<p>$\frac{1}{P} \frac{dP}{dh} = (-\frac{mg}{kT})$,</p>
<p>or</p>
<p>$\frac{d(ln P)}{dh} = (-\frac{mg}{kT})$.</p>
<p>Integrating with respect to $h$ over the interval $[h_0, h]$ yields</p>
<p>$ln(P(h)) - ln(P(h_0)) = (-\frac{mg}{kT})(h - h_0)$,</p>
<p>or</p>
<p>$ln(\frac{P(h)}{P(h_0)}) = (-\frac{mg}{kT})(h - h_0)$,</p>
<p>or</p>
<p>$\frac{P(h)}{P(h_0)} = exp(-\frac{mg}{kT}(h - h_0))$,</p>
<p>or</p>
<p>$P(h) = P(h_0)exp(-\frac{mg}{kT}(h - h_0))$.</p>
<p>If we now take $h_0 = 0$, and set $P = P(0)$, $P' = P(h)$, the desired formula is had.</p>
|
3,817,340 | <p>I'm trying to draw the bio-hazard symbol for <a href="https://codegolf.stackexchange.com/questions/191294/draw-the-biohazard-symbol">a codegolf challenge</a> in Java, for which I've been given the following picture (later referred to as unit diagram):</p>
<p><a href="https://i.stack.imgur.com/fIsNl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fIsNl.png" alt="enter image description here" /></a></p>
<p>Most existing answers in other programming languages use an approach involving a loop of 3, in which they rotate by 120 degrees and draw the circle again. In Java however, drawing each shape one by one from a fixed position would be shorter (and the shorter the better in <a href="https://codegolf.stackexchange.com/tags/code-golf/info">code-golf</a> challenges).<br />
I want to draw the shapes in the following order:</p>
<ol>
<li>Three big circles in black</li>
<li>Three inner circles in white</li>
<li>The small center circle in white</li>
<li>The three gaps at the center in white</li>
<li>The three gaps at the outer parts in white</li>
<li>A black ring in the middle, with three white rings along the circles we've drawn in step 2; which will create three arcs</li>
</ol>
<p>I won't go too deep into detail of what each Java method does, but in general, most of the methods are given an <span class="math-container">$x,y$</span>-coordinate of the top-left corner of the rectangle surrounding the oval, and a <span class="math-container">$width$</span> and <span class="math-container">$height$</span>. Because of this, I want to calculate all <span class="math-container">$x,y$</span>-coordinates of the circle given the unit diagram, while I only assume the coordinates of the very center of the screen.</p>
<p>Here a more visual representation of the steps and what I want to calculate (quickly made in paint, so excuse any inaccuracies):</p>
<p><a href="https://i.stack.imgur.com/h3j5n.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/h3j5n.png" alt="enter image description here" /></a></p>
<p>So to use the Java methods, I need to know the <span class="math-container">$x,y$</span>-coordinates of all red dots; the width/height of the purple lines; and the angles of the blue lines (for the arcs of step 6).</p>
<p>Assumption: the pink dot at the very center is at <span class="math-container">$x,y$</span>-position <span class="math-container">$[300,300]$</span>; and the units in the first picture are multiplied by 10 for my output.</p>
<p>Here the ones I've been able to figure out myself thus far:</p>
<ol>
<li>Width/height (purple line): This is <span class="math-container">$H$</span> in the unit diagram, thus <span class="math-container">$300$</span>.
<ol>
<li>The first <span class="math-container">$x,y$</span>-coordinate (first red dot): we know that from the very center of the screen (pink dot) to the center of the large circles (yellow dot) is unit <span class="math-container">$E=110$</span> (green line). The yellow dot therefore is at position <span class="math-container">$[300, 300-E] → [300,190]$</span>. From there, we can subtract halve of <span class="math-container">$H$</span> from both the <span class="math-container">$x$</span> and <span class="math-container">$y$</span> positions to get to coordinates of the red dot: <span class="math-container">$[300-\frac{H}{2}, 300-E-\frac{H}{2}] → [150,40]$</span>.</li>
<li>The second <span class="math-container">$x,y$</span>-coordinate (second red dot): <span class="math-container">$\color{red}?$</span></li>
<li>The third <span class="math-container">$x,y$</span>-coordinate (third red dot): <span class="math-container">$\color{red}?$</span></li>
</ol>
</li>
<li>Width/height (purple line): This is <span class="math-container">$G$</span> in the unit diagram, thus <span class="math-container">$210$</span>.
<ol>
<li>The first <span class="math-container">$x,y$</span>-coordinate (first red dot): <span class="math-container">$\color{red}?$</span></li>
<li>The second <span class="math-container">$x,y$</span>-coordinate (second red dot): <span class="math-container">$\color{red}?$</span></li>
<li>The third <span class="math-container">$x,y$</span>-coordinate (third red dot): <span class="math-container">$\color{red}?$</span></li>
</ol>
</li>
<li>Width/height (purple line): This is <span class="math-container">$D$</span> in the unit diagram, thus <span class="math-container">$60$</span>.
<ol>
<li><span class="math-container">$x,y$</span>-coordinate (red dot): This is the position of the pink dot, minus halve its width/height for both the <span class="math-container">$x$</span> and <span class="math-container">$y$</span> coordinates: <span class="math-container">$[300-\frac{D}{2}, 300-\frac{D}{2}] → [270,270]$</span>.</li>
</ol>
</li>
<li>Width/height (purple lines): The width is <span class="math-container">$A$</span> in the unit diagram, thus <span class="math-container">$10$</span>. The height doesn't really matter in this case, as long as it's large enough to create the entire gap, but also not too large. Although it doesn't reflect my paint drawing, we could for example use <span class="math-container">$D$</span> as height and draw up to the pink dot.
<ol>
<li>The first <span class="math-container">$x,y$</span>-coordinate (first red dot): Assuming the height is <span class="math-container">$D$</span> and we draw up to the pink dot, we know the <span class="math-container">$x,y$</span> coordinate is at position <span class="math-container">$[300-\frac{A}{2}, 300-D] → [295,240]$</span>.</li>
<li>The second/third/fourth/fifth <span class="math-container">$x,y$</span>-coordinates / red dots (the Java method to draw irregular oriented rectangles requires all four <span class="math-container">$x,y$</span>-coordinates of the corners): <span class="math-container">$\color{red}?$</span></li>
<li>The sixth/seventh/eight/ninth <span class="math-container">$x,y$</span>-coordinates / red dots (the Java method to draw irregular oriented rectangles requires all four <span class="math-container">$x,y$</span>-coordinates of the corners): <span class="math-container">$\color{red}?$</span></li>
</ol>
</li>
<li>Width/height (purple lines): The width is <span class="math-container">$C$</span> in the unit diagram, thus <span class="math-container">$40$</span>. The height is just like with step 4 not really important, so let's just use twice the <span class="math-container">$x$</span> coordinate of the very top, which we've calculated in step 1.1 and was <span class="math-container">$40$</span>, so we'll use a height of <span class="math-container">$80$</span> here.
<ol>
<li>The first <span class="math-container">$x,y$</span>-coordinate (first red dot): Assuming the height <span class="math-container">$80$</span> and we draw from <span class="math-container">$y=0$</span>, we know the <span class="math-container">$x,y$</span>-coordinate is at position <span class="math-container">$[300-\frac{C}{2}, 0] → [280,0]$</span>.</li>
<li>The second/third/fourth/fifth <span class="math-container">$x,y$</span>-coordinates / red dots (the Java method to draw irregular oriented rectangles requires all four <span class="math-container">$x,y$</span>-coordinates of the corners): <span class="math-container">$\color{red}?$</span></li>
<li>The sixth/seventh/eight/ninth <span class="math-container">$x,y$</span>-coordinates / red dots (the Java method to draw irregular oriented rectangles requires all four <span class="math-container">$x,y$</span>-coordinates of the corners): <span class="math-container">$\color{red}?$</span></li>
</ol>
</li>
<li>Width/height (purple line): Unlike the other circles, the height of the circle along which the ring is drawn isn't known in the unit diagram. We know the thickness of the ring (orange line) is <span class="math-container">$B=35$</span>. In the unit diagram we also see that from the very center (pink dot) to the center of the circles we've drawn in step 1, the unit is <span class="math-container">$E=110$</span>. And from the center of this circle of step 1 to the bottom of the arc is unit <span class="math-container">$A=10$</span>. We can therefore deduct that the width/height (purple line) is <span class="math-container">$2(E-A+B)→270$</span>.
<ol>
<li>The <span class="math-container">$x,y$</span>-coordinate (red dot): Since we know the circle is in the center and we also know it's width/height, we can easily calculate the <span class="math-container">$x,y$</span>-coordinate as: <span class="math-container">$[300-(E-A+B), 300-(E-A+B)] → [165,165]$</span>.</li>
<li>We also know the thickness of the last three white rings we draw on top is <span class="math-container">$A=10$</span>, and their width/height and <span class="math-container">$x,y$</span>-coordinates are the exact same as the three circles we've drawn in step 2.</li>
</ol>
</li>
</ol>
<p>Can anyone help me determine the <span class="math-container">$\color{red}?$</span> above. Thus the unknown <span class="math-container">$x,y$</span> coordinates in the steps 1, 2, 4 and 5? Just general information on how I could go about calculating these is fine as well, but right now I don't know where to even begin. Also, sorry if asking all steps at once is too much for a single question. I could split it up into the unknowns of each individual step in separated questions if that's preferable.</p>
| Kevin Cruijssen | 368,863 | <p>I've been able to figure out all calculations. As I mentioned earlier, I've used 10 times the units of the picture in the challenge description, so those sizes are: <span class="math-container">$A=10, B=35, C=40, D=60, E=110, F=150, G=210, H=300$</span>. I've also assumed the very center is at coordinate <span class="math-container">$[300,300]$</span>. Using just this information alone, I had to calculate all the other sizes and coordinates, which I will go over down below. (<strong>NOTE</strong>: the Paint image I created when I asked this challenge is outdated and irrelevant for this answer; I've also split up step 6 into steps 6 and 7.)</p>
<p><em>1a) Top black circle:</em></p>
<p>Width/height: this is mentioned in the diagram: <span class="math-container">$H=300$</span>.<br />
<span class="math-container">$x,y$</span>-coordinate top-left square corner: line <span class="math-container">$E$</span> goes from the center of the bio-hazard symbol (<span class="math-container">$[300,300]$</span>) to the center of the black circle. So the coordinate at the center of this circle is therefore <span class="math-container">$[300, 300-E]$</span>. From there, we can subtract halve the width/height from both the <span class="math-container">$x$</span> and <span class="math-container">$y$</span> coordinate of this center to get the coordinate of the top-left corner of the square surrounding the circle: <span class="math-container">$[300-\frac{H}{2}, 300-E-\frac{H}{2}] → [150, 40]$</span>.</p>
<p><em>1b) Bottom-left black circle:</em></p>
<p>Width/height: again <span class="math-container">$H=300$</span>.<br />
<span class="math-container">$x,y$</span>-coordinate top-left square corner: we again know the length of line <span class="math-container">$E$</span>. We also know that the angle is at 330°. If we draw a triangle with <span class="math-container">$E$</span> as long side, and with the three corners as angles <span class="math-container">$90,60,30$</span>, we can calculate the other two sides:</p>
<p><a href="https://i.stack.imgur.com/HwxH2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HwxH2.png" alt="enter image description here" /></a></p>
<p>Here <span class="math-container">$a=\frac{E}{2}$</span> and <span class="math-container">$c=\frac{E}{2}\sqrt{3}$</span>. So the center coordinates of this black circle is therefore <span class="math-container">$[300-\frac{E}{2}\sqrt{3}, 300+\frac{E}{2}]$</span>. From there, we can again subtract halve the width/height from both to get the coordinate of the top-left corner of the square surrounding the circle: <span class="math-container">$[300-\frac{E}{2}\sqrt{3}-\frac{H}{2}, 300+\frac{E}{2}-\frac{H}{2}] → [54.737, 205]$</span></p>
<p><em>1c) Bottom-right black circle:</em></p>
<p>Width/height: again <span class="math-container">$H=300$</span>.<br />
<span class="math-container">$x,y$</span>-coordinate top-left square corner: we do something similar as above, but in the other direction: <span class="math-container">$[300+\frac{E}{2}\sqrt{3}-\frac{H}{2}, 300+\frac{E}{2}-\frac{H}{2}] → [245.262, 205]$</span></p>
<p><em>2a) Top inner white circle:</em></p>
<p>Width/height: this is mentioned in the diagram: <span class="math-container">$G=210$</span>.<br />
<span class="math-container">$x,y$</span>-coordinate top-left square corner: line <span class="math-container">$F$</span> goes from the center of the bio-hazard symbol (<span class="math-container">$[300,300]$</span>) to the center of the inner white circle. So the coordinate at the center of this circle is therefore <span class="math-container">$[300, 300-F]$</span>. From there, we can subtract halve the width/height from both the <span class="math-container">$x$</span> and <span class="math-container">$y$</span> coordinate of this center to get the coordinate of the top-left corner of the square surrounding the circle: <span class="math-container">$[300-\frac{G}{2}, 300-F-\frac{G}{2}] → [195, 45]$</span>.</p>
<p><em>2b) Bottom-left inner white circle:</em></p>
<p>Width/height: again <span class="math-container">$G=210$</span>.<br />
<span class="math-container">$x,y$</span>-coordinate top-left square corner: similar as what we did in step 1b: <span class="math-container">$[300-\frac{F}{2}\sqrt{3}-\frac{G}{2}, 300+\frac{F}{2}-\frac{G}{2}] → [65.096, 270]$</span></p>
<p><em>2c) Bottom-right inner white circle:</em></p>
<p>Width/height: again <span class="math-container">$G=210$</span>.<br />
<span class="math-container">$x,y$</span>-coordinate top-left square corner: similar as what we did in step 1c: <span class="math-container">$[300+\frac{F}{2}\sqrt{3}-\frac{G}{2}, 300+\frac{F}{2}-\frac{G}{2}] → [324.903, 270]$</span></p>
<p><em>3) Center white circle:</em></p>
<p>Width/height: this is mentioned in the diagram: <span class="math-container">$D=60$</span>.<br />
<span class="math-container">$x,y$</span>-coordinate top-left square corner: subtracting halve this width/height from the center coordinate is enough: <span class="math-container">$[300-\frac{D}{2}, 300-\frac{D}{2}] → [270, 270]$</span></p>
<p><em>4a) Top white rectangle gap at the center of the bio-hazard symbol:</em></p>
<p>Width: this is mentioned in the diagram: <span class="math-container">$A=10$</span>.<br />
Height: Not too irrelevant, as long as it's large enough to create the gap, and not too large to go over other thing that should remain black. So I've just used <span class="math-container">$D=60$</span> here.<br />
<span class="math-container">$x,y$</span>-coordinate top-left corner: <span class="math-container">$[300-\frac{A}{2}, 300-D] → [295, 240]$</span></p>
<p><em>4b) Bottom-left rectangle gap at the center of the bio-hazard symbol:</em></p>
<p>Single the rectangle is angled, the Java method <code>fillPolygon(int[] xPoints, int[] yPoint, int amountOfPoints)</code> doesn't need the width/height, but instead needs the four individual coordinates of the corners of this rectangle. By again creating multiple triangles with corner-angles at 90, 60, and 30 degrees with the long side known, we can calculate the other sides. The calculations of the four points in the order I've used them in the Java method are:<br />
<span class="math-container">$[300-\frac{D}{2}\sqrt{3}-\frac{A}{4}, 300+\frac{D}{2}-\frac{A}{4}\sqrt(3)] → [245.528, 325.669]$</span><br />
<span class="math-container">$[300-\frac{D}{2}\sqrt{3}+\frac{A}{4}, 300+\frac{D}{2}+\frac{A}{4}\sqrt(3)] → [250.538, 334.330]$</span><br />
<span class="math-container">$[300+\frac{A}{4}, 300+\frac{A}{4}\sqrt{3}] → [302.5, 304.330]$</span><br />
<span class="math-container">$[300-\frac{A}{4}, 300-\frac{A}{4}\sqrt{3}] → [297.5, 295.669]$</span></p>
<p><em>4c) Bottom-right rectangle gap at the center of the bio-hazard symbol:</em></p>
<p>Likewise as step 4b:<br />
<span class="math-container">$[300-\frac{A}{4}, 300+\frac{A}{4}\sqrt{3}] → [297.5, 304.220]$</span><br />
<span class="math-container">$[300+\frac{D}{2}\sqrt{3}-\frac{A}{4}, 300+\frac{D}{2}+\frac{A}{4}\sqrt{3}] → [349.461, 334.330]$</span><br />
<span class="math-container">$[300+\frac{D}{2}\sqrt{3}+\frac{A}{4}, 300+\frac{D}{2}-\frac{A}{4}\sqrt{3}] → [354.461, 325.669]$</span><br />
<span class="math-container">$[300+\frac{A}{4}, 300-\frac{A}{4}\sqrt{3}] → [302.5, 295.669]$</span></p>
<p><em>5a) Top big white gap:</em></p>
<p>Width: this is mentioned in the diagram: <span class="math-container">$C=40$</span>.<br />
Height: Not too irrelevant, as long as it's large enough to create the gap, and not too large to go over other thing that should remain black. So I've just used <span class="math-container">$2\times\text{1a.y}=80$</span> here.<br />
<span class="math-container">$x,y$</span>-coordinate top-left corner: <span class="math-container">$[300-\frac{C}{2}, 0] → [280, 0]$</span> The <span class="math-container">$0$</span> isn't calculated, it was just easier to use (as mentioned earlier, the height is mostly irrelevant).</p>
<p><em>5b) Bottom-left big rectangle gap:</em></p>
<p>Similar as step 4b for the first two points:<br />
<span class="math-container">$[300-\frac{H}{2}\sqrt{3}-\frac{C}{4}, 300+\frac{H}{2}-\frac{C}{4}\sqrt{3}] → [30.192, 432.679]$</span><br />
<span class="math-container">$[300-\frac{H}{2}\sqrt{3}+\frac{C}{4}, 300+\frac{H}{2}+\frac{C}{4}\sqrt{3}] → [50.192, 467.320]$</span></p>
<p>For the other two we can't base it on the center of the screen like we did in step 4b, but instead we'll calculate it based on the two points we've just calculated:</p>
<p><span class="math-container">$[300-\frac{H}{2}\sqrt{3}+\frac{C}{4}+\frac{80}{2}\sqrt{3}, 300+\frac{H}{2}+\frac{C}{4}\sqrt{3}-\frac{80}{2}] → [119.474, 427.320]$</span>
<span class="math-container">$[300-\frac{H}{2}\sqrt{3}-\frac{C}{4}+\frac{80}{2}\sqrt{3}, 300+\frac{H}{2}-\frac{C}{4}\sqrt{3}-\frac{80}{2}] → [99.474, 392.679]$</span><br />
(where the <span class="math-container">$80$</span> is the <span class="math-container">$2\times\text{1a.y}$</span> mentioned in step 5a)</p>
<p><em>5c) Bottom-right big rectangle gap:</em></p>
<p>Likewise as step 5b:<br />
<span class="math-container">$[300+\frac{H}{2}\sqrt{3}-\frac{C}{4}, 300+\frac{H}{2}+\frac{C}{4}\sqrt{3}] → [549.807, 467.320]$</span><br />
<span class="math-container">$[300+\frac{H}{2}\sqrt{3}+\frac{C}{4}, 300+\frac{H}{2}-\frac{C}{4}\sqrt{3}] → [569.807, 432,679]$</span><br />
<span class="math-container">$[300+\frac{H}{2}\sqrt{3}+\frac{C}{4}-\frac{80}{2}\sqrt{3}, 300+\frac{H}{2}-\frac{C}{4}\sqrt{3}-\frac{80}{2}] → [500.525, 392.679]$</span><br />
<span class="math-container">$[300+\frac{H}{2}\sqrt{3}-\frac{C}{4}-\frac{80}{2}\sqrt{3}, 300+\frac{H}{2}+\frac{C}{4}\sqrt{3}-\frac{80}{2}] → [480.525, 427.320]$</span></p>
<p><em>6) Black ring that will form the arcs:</em></p>
<p>Thickness: this is mentioned in the diagram: <span class="math-container">$B=35$</span>.<br />
Width/height: this can be calculated with the units in the diagram: <span class="math-container">$2(E-A+B) → 270$</span>, after which we'll remove the thickness: <span class="math-container">$2(E-A+B)-B → 235$</span> (halve the thickness at both sides)<br />
<span class="math-container">$x,y$</span>-coordinate top-left corner: we simply subtract halve the width/height from the center coordinate: <span class="math-container">$[300-\frac{2(E-A+B)-B}{2}, 300-\frac{2(E-A+B)-B}{2}] → [182.5, 182.5]$</span></p>
<p><em>7) White ring inside the inner circles to form the arcs:</em></p>
<p>Thickness: this is mentioned in the diagram: <span class="math-container">$A=10$</span>.<br />
Width/height: this is the same as step 2a: <span class="math-container">$G=210$</span>, but with this thickness removed: <span class="math-container">$G-A → 200$</span><br />
<span class="math-container">$x,y$</span>-coordinate top-left corner: these are the same calculations as in step 2a, but with the adjusted width/height <span class="math-container">$G-A$</span> instead of <span class="math-container">$G$</span>:<br />
<span class="math-container">$[300-\frac{G-A}{2}, 300-F-\frac{G-A}{2}] → [200, 50]$</span><br />
<span class="math-container">$[300-\frac{F}{2}\sqrt{3}-\frac{G-A}{2}, 300+\frac{F}{2}-\frac{G-A}{2}] → [65.096, 270] → [70.096, 275]$</span><br />
<span class="math-container">$[300+\frac{F}{2}\sqrt{3}-\frac{G-A}{2}, 300+\frac{F}{2}-\frac{G-A}{2}] → [324.903, 270] → [329.903, 275]$</span></p>
<p>Rounding all those values we've calculated to integers ('half up') we get the code seen in <a href="https://codegolf.stackexchange.com/a/191350/52210">this codegolf answer of mine</a>, with the following output:</p>
<p><a href="https://i.stack.imgur.com/kAaaf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kAaaf.png" alt="https://i.stack.imgur.com/kAaaf.png" /></a></p>
<p>Or with each step a different color:</p>
<p><a href="https://i.stack.imgur.com/jxRsj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jxRsj.png" alt="enter image description here" /></a></p>
|
1,675 | <p>This is a follow-up to <a href="https://mathoverflow.net/questions/1039/explicit-direct-summands-in-the-decomposition-theorem">this post</a> on the Decomposition Theorem. Hopefully, this will also invite some discussion about the theorem and perverse sheaves in general.</p>
<p>My question is how does one use the Decomposition Theorem in practice? Is there any way to pin down the subvarieties and local systems that appear in the decomposition. For example, how do you compute intesection homology complexes using this theorem? Does anyone have a link to a source with worked out examples?</p>
<p>Another related question: What is the deep part of the theorem? Is it the fact that the pushforward of a perverse sheaf is isomorphic to its perverse hypercohomology? Is it the fact that these pieces are semisimple? Or are these both hard statements? And what is so special about algebraic varieties?</p>
| Geordie Williamson | 919 | <p>To supplement Ben's answer, basically every aspect of the decomposition theorem is hard.</p>
<p>To give you a simple example of something which is implied by the decomposition theorem but is far from trivial is the following statement: given a proper smooth map of smooth varieties f : X -> Y the direct image of the constant sheaf splits as a direct sum of local systems. Note that this implies (but is stronger than) the degeneration of the Leray-Serre spectral sequence for the fibration. This answers to some extent your question "what is so special about algebraic varieties" because Leray-Serre just doesn't degenerate in general.</p>
<p>I think the situation has been cleared up considerably by the work of de Cataldo and Migliorini which (IMHO) is the first genuinely geometric proof of the decomposition theorem.</p>
<p>One might think of the "smooth map" case above as the "easiest case" (and indeed it does have an easier proof). However de Cataldo and Migliorini point out that in fact the "easiest case" is the case of a semi-small map, for which the decomposition theorem can be deduced from the non-degeneracy of certain bilinear forms. In a difficult work, they deduce the general case by reducing to this case by induction on the "defect of semi-smallness" (how far away a map is from being semi-small) and by taking hyperplane sections to reduce this defect.</p>
<p>An excellent informal survey about the decomposition theorem, with lots of wonderful examples can be found in <em><a href="http://arxiv.org/abs/0712.0349">The decomposition theorem, perverse sheaves and the topology of algebraic maps</a></em> by de Cataldo and Migliorini.</p>
<p>Note that there are really three statements in the decomposition theorem, all of which are hard:</p>
<ol>
<li>the direct image is the sum of its perverse cohomology groups;</li>
<li>each perverse cohomology is a direct sum of IC extensions of a local system;</li>
<li>each local system is semi-simple.</li>
</ol>
<p>As is often the case in mathematics, a nice way to learn why the decomposition theorem is hard is to go to situations when it fails. This occurs when one takes perverse sheaves with coefficients in positive characteristic (or even Z). Daniel Juteau, Carl Mautner and I have written a survey called "Perverse sheaves and modular representation theory" which contains lots of examples of the failure of the decomposition theorem. (Note that all of 1), 2) and 3) above can fail!)</p>
|
2,793,384 | <p>I am taking a course in Algebraic Topology next semester, so I thought of starting to read about it on my own.</p>
<p><strong>My concern:</strong> I have read <em>Topology</em> by Munkres, and read the first chapter on algebraic topology from that book, so I have an idea about the fundamental group and covering spaces. Later I read the book <em>Topology</em> by Klaus Janich till the chapter on CW complexes. I really liked the book: it was not rigorous (somewhat unclear at some places) but enough to draw my interest in the subject. I was planning to start reading algebraic topology more seriously, so a senior suggested me the book <em>Algebraic Topology: A First Course</em> by William Fulton. I have read the first chapter, but I am having some problems. I actually don't have a good background in multivariable calculus. The author gives an overview of calculus in the plane and he introduces some basic concepts about multivariable calculus. I really like algebra; I have a good background in fields and Galois theory, ring and group theory (don't know free abelian groups yet). I am reading commutative algebra side by side. So I would really like to learn algebraic topology in an 'algebraic way', and I am concerned that Fulton in this book does this by another way (a differential approach or something). But at the same time I want to learn algebraic topology in an intuitive way (with pictures) and I've been told that this is the most expository book available on algebraic topology. Am I right about this book, I mean does it approach algebraic topology using category theory and other algebraic tools or in some other way? If so please tell me is it more important for intuition, so that I can read more on multivariable calculus and then start again?</p>
<p>I know this kind of question is not supported in this community, but I do need help about this. My course will be starting from next month, and I want to get a good concept about this subject. And I want to take up a course about algebraic geometry too. That's why I don't want to spend time experimenting. Thanks in advance.</p>
| Osama Ghani | 341,911 | <p>A cursory glance through the book suggests it is has a strong differential flavour. If you’re familiar with multivariable calculus especially as formulated through differential forms and exterior derivatives, this isn’t necessarily a bad thing, and does help in understand what it means for two cycles to be homologous. The book presents de Rham cohomology even before homology, and restricts itself to the 0th and 1st (co)homology groups. It also briefly discusses vector fields which are more likely to be covered in a differential topology course. It includes Čech cohomology which is not something you meet in a first algebraic topology course usually. </p>
<p>So, I wouldn’t think that this book matches your background or aims. I’d recommend Topology & Geometry by Bredon as one that I liked, but there are multiple books out there that would serve you better! Feel free to reach out to me for more suggestions or recommendations on books.</p>
|
1,560,050 | <p>I want to solve the homogenous part of a stretched string problem where $y=y(x)$.</p>
<p>$$y'' + y = 0$$</p>
<p>with the boundary conditions such that: $y(0)=y(\pi/2)=0$</p>
<p>The differential equation gives rise to a solution on the form:
$$y = a \cos(x) + b \sin(x)$$</p>
<p>But when applying the boundary conditions I end up with only trivial solution ($a=b=0$).</p>
<p>Have I made a mistake or does these B.C only lead to $a=b=0$?</p>
| Martin Argerami | 22,857 | <p>Yes, the only solution to that boundary value problem is $y(x)=0$. </p>
|
1,585,408 | <p>I have the equation: (1-x<sup>2</sup>)u<sup>''</sup> -xu<sup>'</sup>+ku=0,
where ' represents differentiation with respect to x and k is a constant.</p>
<p>I am asked to show that cos(k<sup>1/2</sup>cos<sup>-1</sup>x) is a solution to this equation.</p>
<p>I assumed to show this you need to set u=cos(k<sup>1/2</sup>cos<sup>-1</sup>x) and substitute it into the the Tchebycheff equation and show it equals zero. However, when doing this I get quite a lot of messy differentiation. </p>
<p>The question before this asked to put the equation into Sturm Liouville form, so I am unsure if that is meant to be used to solve the question.</p>
<p>Any hints on how to advance would be greatly appreciated.</p>
| David Holden | 79,543 | <p>if you make the substitution $x=\cos t$ then
$$
u_t=-u_x\sin t
$$
and
$$
u_{tt} = -u_x\cos t +u_{xx}\sin^2 t = u_{xx}(1-x^2) - x u_x
$$
the equation becomes:
$$
u_{tt} +ku =0
$$</p>
|
514 | <p>I just came back from my Number Theory course, and during the lecture there was mention of the Collatz Conjecture.</p>
<p>I'm sure that everyone here is familiar with it; it describes an operation on a natural number – <span class="math-container">$n/2$</span> if it is even, <span class="math-container">$3n+1$</span> if it is odd.</p>
<p>The conjecture states that if this operation is repeated, all numbers will eventually wind up at <span class="math-container">$1$</span> (or rather, in an infinite loop of <span class="math-container">$1-4-2-1-4-2-1$</span>).</p>
<p>I fired up Python and ran a quick test on this for all numbers up to <span class="math-container">$5.76 \times 10^{18}$</span> (using the powers of cloud computing and dynamic programming magic). Which is millions of millions of millions. And all of them eventually ended up at <span class="math-container">$1$</span>.</p>
<p>Surely I am close to testing every natural number? How many natural numbers could there be? Surely not much more than millions of millions of millions. (I kid.)</p>
<p>I explained this to my friend, who told me, "Why would numbers suddenly get different at a certain point? Wouldn't they all be expected to behave the same?"</p>
<p>To which I said, "No, you are wrong! In fact, I am sure there are many conjectures which have been disproved by counterexamples that are extremely large!"</p>
<p>And he said, "It is my conjecture that there are none! (and if any, they are rare)".</p>
<p>Please help me, smart math people. Can you provide a counterexample to his conjecture? Perhaps, more convincingly, several? I've only managed to find one! (Polya's conjecture). One, out of the many thousands (I presume) of conjectures. It's also one that is hard to explain the finer points to the layman. Are there any more famous or accessible examples?</p>
| ShreevatsaR | 205 | <p>Another example: <a href="http://en.wikipedia.org/wiki/Euler%27s_sum_of_powers_conjecture" rel="noreferrer">Euler's sum of powers conjecture</a>, a generalization of Fermat's Last Theorem. It states:<br>
If the equation <span class="math-container">$\sum_{i=1}^kx_i^n=z^n$</span> has a solution in positive integers, then <span class="math-container">$n \leq k$</span> (unless <span class="math-container">$k=1$</span>). Fermat's Last Theorem is the <span class="math-container">$k=2$</span> case of this conjecture.</p>
<p>A counterexample for <span class="math-container">$n=5$</span> was found in 1966: it's<br>
<span class="math-container">$$
61917364224=27^5+84^5+110^5+133^5=144^5
$$</span>
The smallest counterexample for <span class="math-container">$n=4$</span> was found in 1988:<br>
<span class="math-container">$$
31858749840007945920321 = 95800^4+217519^4+414560^4=422481^4
$$</span>
This example used to be even more useful in the days before FLT was proved, as an answer to the question "Why do we need to prove FLT if it has been verified for thousands of numbers?" :-)</p>
|
514 | <p>I just came back from my Number Theory course, and during the lecture there was mention of the Collatz Conjecture.</p>
<p>I'm sure that everyone here is familiar with it; it describes an operation on a natural number – <span class="math-container">$n/2$</span> if it is even, <span class="math-container">$3n+1$</span> if it is odd.</p>
<p>The conjecture states that if this operation is repeated, all numbers will eventually wind up at <span class="math-container">$1$</span> (or rather, in an infinite loop of <span class="math-container">$1-4-2-1-4-2-1$</span>).</p>
<p>I fired up Python and ran a quick test on this for all numbers up to <span class="math-container">$5.76 \times 10^{18}$</span> (using the powers of cloud computing and dynamic programming magic). Which is millions of millions of millions. And all of them eventually ended up at <span class="math-container">$1$</span>.</p>
<p>Surely I am close to testing every natural number? How many natural numbers could there be? Surely not much more than millions of millions of millions. (I kid.)</p>
<p>I explained this to my friend, who told me, "Why would numbers suddenly get different at a certain point? Wouldn't they all be expected to behave the same?"</p>
<p>To which I said, "No, you are wrong! In fact, I am sure there are many conjectures which have been disproved by counterexamples that are extremely large!"</p>
<p>And he said, "It is my conjecture that there are none! (and if any, they are rare)".</p>
<p>Please help me, smart math people. Can you provide a counterexample to his conjecture? Perhaps, more convincingly, several? I've only managed to find one! (Polya's conjecture). One, out of the many thousands (I presume) of conjectures. It's also one that is hard to explain the finer points to the layman. Are there any more famous or accessible examples?</p>
| Bill Dubuque | 242 | <p>Another class of examples arise from diophantine equations with huge minimal solutions. Thus the conjecture that such an equation is unsolvable in integers has only huge counterexamples. Well-known examples arise from Pell equations, e.g. the smallest solution to the classic <a href="http://en.wikipedia.org/wiki/Archimedes%27_cattle_problem" rel="noreferrer">Archimedes Cattle problem</a> has <strong>206545 decimal digits</strong>, namely 77602714 ... 55081800.</p>
|
514 | <p>I just came back from my Number Theory course, and during the lecture there was mention of the Collatz Conjecture.</p>
<p>I'm sure that everyone here is familiar with it; it describes an operation on a natural number – <span class="math-container">$n/2$</span> if it is even, <span class="math-container">$3n+1$</span> if it is odd.</p>
<p>The conjecture states that if this operation is repeated, all numbers will eventually wind up at <span class="math-container">$1$</span> (or rather, in an infinite loop of <span class="math-container">$1-4-2-1-4-2-1$</span>).</p>
<p>I fired up Python and ran a quick test on this for all numbers up to <span class="math-container">$5.76 \times 10^{18}$</span> (using the powers of cloud computing and dynamic programming magic). Which is millions of millions of millions. And all of them eventually ended up at <span class="math-container">$1$</span>.</p>
<p>Surely I am close to testing every natural number? How many natural numbers could there be? Surely not much more than millions of millions of millions. (I kid.)</p>
<p>I explained this to my friend, who told me, "Why would numbers suddenly get different at a certain point? Wouldn't they all be expected to behave the same?"</p>
<p>To which I said, "No, you are wrong! In fact, I am sure there are many conjectures which have been disproved by counterexamples that are extremely large!"</p>
<p>And he said, "It is my conjecture that there are none! (and if any, they are rare)".</p>
<p>Please help me, smart math people. Can you provide a counterexample to his conjecture? Perhaps, more convincingly, several? I've only managed to find one! (Polya's conjecture). One, out of the many thousands (I presume) of conjectures. It's also one that is hard to explain the finer points to the layman. Are there any more famous or accessible examples?</p>
| Oscar Lanzi | 248,217 | <p>A case where you can "dial in" a large counterexample involves this theorem:</p>
<p>"A natural number is square if and only if it is a <span class="math-container">$p$</span>-adic square for all primes <span class="math-container">$p$</span>."</p>
<p>A <span class="math-container">$p$</span>-adic square is a number <span class="math-container">$n$</span> for which <span class="math-container">$x^2\equiv n\bmod p^k$</span> has a solution for any positive <span class="math-container">$k$</span>.</p>
<p>If we include all primes <span class="math-container">$p$</span> we have no counterexamples, but suppose we are computationally testing for squares and we have only space and time to include finitely many primes. How high we can go before we get a non-square number that slips through the sieve depends on how many primes we include in our test.</p>
<p>With <span class="math-container">$p=2$</span> as the only prime base, the first non-square we miss is <span class="math-container">$17$</span>. Putting <span class="math-container">$p=3$</span> in addition to <span class="math-container">$p=2$</span> raises that threshold to <span class="math-container">$73$</span>. Using <span class="math-container">$2,3,5,7$</span> gives <span class="math-container">$1009$</span> as the first "false positive". The numbers appear to be growing fast enough to make the first counterexample large with a fairly modest number of primes. See <a href="http://oeis.org/A002189" rel="nofollow noreferrer">http://oeis.org/A002189</a> for more details.</p>
|
514 | <p>I just came back from my Number Theory course, and during the lecture there was mention of the Collatz Conjecture.</p>
<p>I'm sure that everyone here is familiar with it; it describes an operation on a natural number – <span class="math-container">$n/2$</span> if it is even, <span class="math-container">$3n+1$</span> if it is odd.</p>
<p>The conjecture states that if this operation is repeated, all numbers will eventually wind up at <span class="math-container">$1$</span> (or rather, in an infinite loop of <span class="math-container">$1-4-2-1-4-2-1$</span>).</p>
<p>I fired up Python and ran a quick test on this for all numbers up to <span class="math-container">$5.76 \times 10^{18}$</span> (using the powers of cloud computing and dynamic programming magic). Which is millions of millions of millions. And all of them eventually ended up at <span class="math-container">$1$</span>.</p>
<p>Surely I am close to testing every natural number? How many natural numbers could there be? Surely not much more than millions of millions of millions. (I kid.)</p>
<p>I explained this to my friend, who told me, "Why would numbers suddenly get different at a certain point? Wouldn't they all be expected to behave the same?"</p>
<p>To which I said, "No, you are wrong! In fact, I am sure there are many conjectures which have been disproved by counterexamples that are extremely large!"</p>
<p>And he said, "It is my conjecture that there are none! (and if any, they are rare)".</p>
<p>Please help me, smart math people. Can you provide a counterexample to his conjecture? Perhaps, more convincingly, several? I've only managed to find one! (Polya's conjecture). One, out of the many thousands (I presume) of conjectures. It's also one that is hard to explain the finer points to the layman. Are there any more famous or accessible examples?</p>
| BlueRaja - Danny Pflughoeft | 136 | <p>Here's a recent one I didn't see on either page. The following are true statements:</p>
<p><span class="math-container">$\displaystyle{ \int_0^\infty \frac{\sin t}{t} \, dt = \frac{\pi}{2} }$</span><br>
<span class="math-container">$\displaystyle{ \int_0^\infty \frac{\sin t}{t} \, \frac{\sin \left(\frac{t}{101}\right)}{\frac{t}{101}} \, dt = \frac{\pi}{2} }$</span><br>
<span class="math-container">$\displaystyle{ \int_0^\infty \frac{\sin t}{t} \, \frac{\sin \left(\frac{t}{101}\right)}{\frac{t}{101}} \, \frac{\sin \left(\frac{t}{201}\right)}{\frac{t}{201}} \, dt = \frac{\pi}{2} }$</span><br>
<span class="math-container">$\displaystyle{ \int_0^\infty \frac{\sin t}{t} \, \frac{\sin \left(\frac{t}{101}\right)}{\frac{t}{101}} \, \frac{\sin \left(\frac{t}{201}\right)}{\frac{t}{201}} \, \frac{\sin \left(\frac{t}{301}\right)}{\frac{t}{301}} \, dt = \frac{\pi}{2} }$</span></p>
<p>Is it true that</p>
<p><span class="math-container">$\forall n,\displaystyle{ \int_0^\infty \prod_0^n\frac{\sin \left(\frac{t}{100n+1}\right)}{\frac{t}{100n+1}} \, dt = \frac{\pi}{2} }$</span></p>
<p>?</p>
<hr>
<p>No, it's not. However, you could have a computer calculating this for the rest of your life and never find a counter-example. <strong>The first counter-example for n has 43 digits.</strong></p>
<p>I found this example <a href="https://johncarlosbaez.wordpress.com/2018/09/20/patterns-that-eventually-fail/" rel="noreferrer">here</a>. It was specially constructed to have a large counter-example - the page gives a way to construct similar statements with arbitrarily large counter-examples.</p>
|
1,246,522 | <p>What are the prime ideals of $\mathbb F_p[x]/(x^2)$? I have been told that the only one is $(x)$, but I would like a proof of this. I want to say that a prime ideal of $\mathbb F_p[x]/(x^2)$ corresponds to a <strong>prime</strong> ideal $P$ of $\mathbb F_p[x]$ containing $(x^2)$. And then $P$ contains $(x)$ since it is prime. But I don't know if prime ideals correspond to prime ideals under the correspondence theorem, and I still can't seem to prove that if they do, $P$ can't be some non-principal ideal properly larger than $(x)$.</p>
<p>Some context: I'm considering why the prime ideals $\mathfrak p$ of $\mathcal O_K$, (with $K=\mathbb Q(\sqrt d)$ and $\textrm{Norm}(\mathfrak p)=p$, a ramified prime) are unique. My definition of a ramified prime is that $\mathcal O_K/(p) \cong \mathbb F_p[x]/(x^2)$ and I know nothing else about these primes.</p>
| Spencer Leslie | 129,418 | <p>Prime ideals do correspond under the correspondence theorem, so your argument suffices.</p>
<p>To see this,</p>
<p>Let $I\subset P\subset R$ be any prime in $R$ containing an ideal $I$, then
$R/P \cong \frac{R/I}{P/I}$ by the 3rd isomorphism theorem. Since $P$ is prime, $R/P$ is an integral domain, hence so is $\frac{R/I}{P/I}$.
Thus, $P/I$ is a prime ideal in $R/I$.</p>
<p>Now start with a prime ideal $Q\subset R/I$, lift it to an ideal containing $I\subset Q'\subset R$, and apply the same argument to see that $Q'$ is a prime ideal in $R$.</p>
<p>For uniqueness, you are correct in saying that if $(x^2)\subset P$, then $(x)\subset P$, as $P$ is prime. But $(x)$ is maximal, so this forces $(x) = P$ by definition of a maximal ideal. To see that it is maximal, note $F[x]/(x)\cong F$ is a field.</p>
|
40,116 | <p>I'm building a program that calculates the cost of an item based on it's size (let's say a bamboo pole). As the customer requests a longer pole, it gets hard to find a bamboo, plus requires more resources to grow, therefore, I would want to charge more per inch for the piece of bamboo based on it's length approaching a particular length. Then after that point the cost per inch would really escalate.</p>
<p>I believe that log would be the function that I need to use, but I just can't figure out how to make my formula. I've tried log(-x), log(x-1), log(y-x), I can't figure out how to get the log to shoot up to infinity, nor target a specific point.</p>
<p>Referencing the example above, I would want the cost/inch of the bamboo to stay reasonable up to 72", but after that, the cost/inch should rapidly increase, until it gets to 100", where it would become ridiculously expensive. Before 72", it should rise in cost/inch, but at a slow rate (it costs a little bit more per inch to grow a 72" stick than a 6" stick). I'm looking for a uniform growth, not a split formula. No f(x) where x<72, g(x) where x>72.</p>
<p>I'm not necessarily looking for the formula to solve the above question. I am looking for the HOW to research and solve the above question.</p>
<p>Many Thanks,
Matt</p>
| zmaril | 11,124 | <p>Try looking into statistics distributions. By messing with the degrees of freedom you can get something that could do what you want. If you take the inverse of the probability of something happening, then you can get your arbitrarily large number. Bonus: you can probably set 75" as the average parameter for the distro and not have to worry about it. </p>
|
3,506,659 | <p>Let <span class="math-container">$\sigma_i$</span> denote the <a href="https://en.wikipedia.org/wiki/Pauli_matrices" rel="nofollow noreferrer">Pauli matrices</a>:
<span class="math-container">$$
\sigma_1\equiv \begin{pmatrix}0&1\\1&0\end{pmatrix}, \quad
\sigma_2\equiv \begin{pmatrix}0&-i\\i&0\end{pmatrix}, \quad
\sigma_3\equiv \begin{pmatrix}1&0\\0&-1\end{pmatrix}.
$$</span>
It isn't hard to see that any <span class="math-container">$2\times 2$</span> unitary <span class="math-container">$U$</span> can be written in terms of these matrices as
<span class="math-container">$$ U = c_0 I + \sum_{k=1}^3 ic_k \sigma_k, $$</span>
for some real coefficients <span class="math-container">$c_j$</span> normalised to one: <span class="math-container">$\mathbf c\equiv(c_0,c_1,c_2,c_3)\in S^3$</span>.</p>
<p>It turns out to be the case that
<span class="math-container">$$ U\sigma_i U^\dagger = \sum_{j=1}^3 B_{ij} \sigma_j, \tag A$$</span>
for any <span class="math-container">$i\in\{1,2,3\}$</span>, with <span class="math-container">$B$</span> a unitary matrix.
I can see why this must be the case by direct analysis on <span class="math-container">$U\sigma_i U^\dagger$</span>: expanding <span class="math-container">$U$</span> in terms of Pauli matrices and using the known expressions for products of Pauli matrices to get to a final expression for <span class="math-container">$B_{ij}$</span>. My problem with this is that it's a somewhat tedious procedure, and the final expression doesn't make it particularly obvious that <span class="math-container">$B$</span> is always unitary.</p>
<p>I am looking for a better way to prove (A), especially because the expression seems to lend itself to be understood on more abstract grounds (I don't know much about Lie theory, but it seems to be saying something on the lines of <span class="math-container">$U(2)$</span> acting on its Lie algebra unitarily via the adjoint representation... if that makes sense).</p>
| Ben Grossmann | 81,360 | <p>We define an inner product over <span class="math-container">$\Bbb C^{n \times n}$</span> by <span class="math-container">$\frac 1n \langle A,B \rangle = \operatorname{tr}(A^\dagger B)$</span>; this is (a normalized version of what is) known as the <a href="https://en.wikipedia.org/wiki/Frobenius_inner_product" rel="nofollow noreferrer">"Frobenius" or "Hilbert-Schmidt"</a> inner-product.</p>
<p>Note that for any <span class="math-container">$U$</span>, the matrices <span class="math-container">$U\sigma_j U^\dagger$</span> form an orthonormal basis for the space of trace free <span class="math-container">$2 \times 2$</span> matrices (if you like, the orthogonal complement of the span of <span class="math-container">$I$</span>). That is, we have
<span class="math-container">$$
\langle U\sigma_jU^\dagger,U\sigma_kU^\dagger \rangle = \delta_{jk}
$$</span>
where <span class="math-container">$\delta_{jk}$</span> is a Kronecker-delta, and every trace-zero matrix can be written as a linear combination of these matrices.</p>
<p>The matrix <span class="math-container">$B_{ij}$</span> that you describe is the <a href="https://en.wikipedia.org/wiki/Change_of_basis" rel="nofollow noreferrer">change-of-basis</a> matrix that takes us from a coordinate-vector relative to the basis <span class="math-container">$\{U\sigma_jU^\dagger: j =1,2,3\}$</span> to a coordinate vector relative to the basis <span class="math-container">$\{\sigma_j: j = 1,2,3\}$</span>. Because where are changing between two orthonormal bases, the resulting change-of-basis matrix is unitary.</p>
|
3,753,474 | <p><strong>Question:</strong></p>
<blockquote>
<p>If <span class="math-container">$\alpha,\beta,\gamma$</span> are the roots of the equation, <span class="math-container">$x^3+x+1=0$</span>, then find the equation whose roots are: <span class="math-container">$({\alpha}-{\beta})^2,({\beta}-{\gamma})^2,({\gamma}-{\alpha})^2$</span></p>
</blockquote>
<p>Now, the normal way to solve this question would be to use the theory of equations and find the sum of roots taken one at a time, two at a time and three at a time. Using this approach, we get the answer as <span class="math-container">$(x+1)^3+3(x+1)^2+27=0$</span>. However, I feel that this is a very lengthy approach to this problem. Is there an easier way of doing it?</p>
| Z Ahmed | 671,540 | <p>Let <span class="math-container">$a,b,c$</span> be the roots of <span class="math-container">$x^3+x+1=0$</span> so we have <span class="math-container">$a+b+c=0, ab+bc+ca=1,abc=-1$</span>, so <span class="math-container">$a^2+b^2+c^2=-2$</span> and <span class="math-container">$c^3=-c-1$</span></p>
<p>We would explore a transformation from <span class="math-container">$x$</span> to <span class="math-container">$y$</span> to get the required cubic equation of <span class="math-container">$y$</span>.
Let <span class="math-container">$$y=(a-b)^2=a^2+b^2-2ab\implies y=-2-c^2+2/c \implies c=\frac{3}{1+y}$$</span>
Replacing <span class="math-container">$c$</span> by <span class="math-container">$x$</span> we get the required transformation <span class="math-container">$x=\frac{3}{1+y}$</span>, putting it in the given <span class="math-container">$x$</span> equation, we get:
<span class="math-container">$$\frac{27}{(1+y)^3}+\frac{3}{(1+y)}+1=0 \implies y^3+6y^2+9y+31=0,$$</span>
which is the required cubic equation.</p>
|
3,653,212 | <p>Due to Covid -19 , in our university quizzes are held online and it's hard to ask questions. </p>
<p>3 Days back in my Combinatorics quiz this question was asked on which I am struck. I couldn't solve it in the time alloted and struggled to find a proper strategy. </p>
<p>Question is ->Determine the number of non equivalent colourings of the corners of regular tetrahedron with k different colours. </p>
<blockquote>
<p>My attempt -> I am trying to solve it by Burnside Theorem ( Number of non equivalent colourings in C are given by N(G, C) = 1/ |G| <span class="math-container">$\sum_{f \epsilon G } | C(f) | $</span>. [C(f) = set of all colourings in C that are fixed by f ]</p>
</blockquote>
<p>Group of permutations is <span class="math-container">$S_4$</span> and all<span class="math-container">$ (k^4)$</span> will be fixed by identity . But I am not able to think how to find colourings fixed by each permutation caused due to rotations and reflections. I have done it for pentagon which was easy. </p>
<blockquote>
<p>Can someone please tell a way on how to efficiently and elegentally compute the value of C(f) in case of rotations and reflections. </p>
</blockquote>
<p>I will be really thankful for the ideas. </p>
| Robert Z | 299,698 | <p>The symmetric group is <span class="math-container">$S_4$</span> with <span class="math-container">$4!=24$</span> elements:</p>
<ul>
<li>6 with cycle structure <span class="math-container">$(abcd)$</span> then <span class="math-container">$C(f)=k$</span>;</li>
<li>8 with cycle structure <span class="math-container">$(abc)(d)$</span> then <span class="math-container">$C(f)=k^2$</span>;</li>
<li>3 with cycle structure <span class="math-container">$(ab)(cd)$</span> then <span class="math-container">$C(f)=k^2$</span>;</li>
<li>6 with cycle structure <span class="math-container">$(ab)(c)(d)$</span> then <span class="math-container">$C(f)=k^3$</span>;</li>
<li>1 with cycle structure <span class="math-container">$(a)(b)(c)(d)$</span> then <span class="math-container">$C(f)=k^4$</span>.</li>
</ul>
<p>It follows that
<span class="math-container">$$N(G, C)=\frac{6k+(8+3)k^2+6k^3+k^4}{24}=\frac{(k+3)(k+2)(k+1)k}{4!}=\binom{k+3}{4}.$$</span></p>
|
897,043 | <p>I'm having issues getting my head around cartesian products and their cardinalities.</p>
<p>$A = \{0, 1, \{2, 3, 4\}\}$<br>
$B = \{1,5\}$<br>
$D = B \times N$ (where $N$ is the set of natural numbers)</p>
<p><strong>The first problem:</strong> What is the cardinality of:</p>
<p>(a) $A \times B$ (cartesian product)</p>
<p>(b) $A \times D$</p>
<p><strong>Part 2:</strong> true/false
(a) $N$ is a subset of $D$</p>
<p>for (a) I used $|A \times B|$ = $|A| * |B|$
and got $3*2 = 6$ </p>
<p>is this the correct way to do this?</p>
<p>for (b) I assumed that the cardinality was infinite since it involved the set of natural numbers, am I correct in assuming this?</p>
<p>for part 2 (a) I assumed that it was true since $D$ contains the natural set so presumably the natural set is a subset of $D$, am I correct in assuming this?</p>
| AlexR | 86,940 | <p>$\mathbb N$ is <strong>not</strong> a subset of $D$. Subsets of $D$ alike to $\mathbb N$ are, for example
$$\{1\} \times \mathbb N, \{5\} \times \mathbb N$$
The difference is that elements from $D$ look like $(1, a)$ or $(5, a)$ with $a\in\mathbb N$ whereas elements from $\mathbb N$ are just natural numbers (no tuples). </p>
<p>As for the cardinalities, you are right;
$$|A\times B| = 6, |A\times D| = |D| = |\mathbb N| = \aleph_0 \quad \text{("countable infinity")}$$</p>
<hr>
<p>More generally spoken, there are subsets of $A\times B$ looking like $A$ or $B$, namely sets of the form $A\times \{b\}, \{a\} \times B$ with $a\in A, b\in B$, but $A,B$ are no subsets of $A\times B$.</p>
|
125,610 | <p>I have question about sets. I need to prove that: $$X \cap (Y - Z) = (X \cap Y) - (X \cap Z)$$</p>
<p>Now, I tried to prove that from both sides of the equation but had no luck.</p>
<p>For example, I tried to do something like this: $$X \cap (Y - Z) = X \cap (Y \cap Z')$$ but now I don't know how to continue.</p>
<p>From the other side of the equation I tried to do something like this: $$(X \cap Y) - (X \cap Z) = (X \cap Y) \cap (X \cap Z)' = (X \cap Y) \cap (X' \cup Z')$$ and from here I don't know what to do again.</p>
<p>I will be glad to hear how should I continue from here and what I did wrong. Thanks in advance.</p>
| MarnixKlooster ReinstateMonica | 11,994 | <p>Here is a third approach, which first translates to the element level, and then uses logic manipulation.</p>
<p>Starting at the most complex side, we have for any $\;a\;$
\begin{align}
& a \in (X \cap Y) - (X \cap Z) \\
\equiv & \;\;\;\;\;\text{"expand definitions of $\;-\;$ and of $\;\cap\;$ (twice)"} \\
& a \in X \land a \in Y \;\land\; \lnot (a \in X \land a \in Z) \\
\equiv & \;\;\;\;\;\text{"logic: use $\;a \in X\;$ on other side of $\;\land\;$"} \\
& a \in X \land a \in Y \;\land\; \lnot (\text{true} \land a \in Z) \\
\equiv & \;\;\;\;\;\text{"logic: simplify"} \\
& a \in X \;\land\; a \in Y \land \lnot(a \in Z) \\
\equiv & \;\;\;\;\;\text{"reintroduce $\;-\;$ and $\;\cap\;$ using their definitions"} \\
& a \in X \cap (Y - Z) \\
\end{align}
Using set extensionality, this proves the statement in question.</p>
|
1,661,986 | <p>I was out all last week sick with the flu and am trying to get caught up in my Discrete Mathematics course. One set of questions in my book goes as follows:</p>
<p>Find the least integer $n$ such that $f(x) \in O(x^n)$ for each of the functions:</p>
<p>(a). $f(x) = 8x+4$<br>
(b). $f(x) = x\sqrt(x)$<br>
(c). $f(x) = log_3 9$<br>
(d). $f(n) = n!$</p>
<p>Can someone walk me through this? I have no idea where to even begin on these or what it is I am attempting to do in the end.</p>
| user137481 | 137,481 | <p>grand_chat has already explained the concept of Big-O to you.
However, for your exercise, it might be easier to understand if you use the following definition:</p>
<p>$f(x) \in O(g(x))$ if there exists positive constants $C$ and $x_0$ such that: </p>
<p>$\quad\quad 0 \le f(x) \le Cg(x)$ for all $x \gt x_0$</p>
<p>a) $f(x) = 8x + 4 \le 8x + 8x = 16x$ for $x \ge 1$<br>
$\quad$So, $C = 16$, $x_0=1$ and $g(x) = x$<br>
$\quad$So, $f(x)=O(g(x))=O(x^1)$</p>
<p>b) $f(x) = x\sqrt(x) = x^{\frac {3}{2}} \le x^2 $ for $x \gt 0$<br>
$\quad$So, $C = 1$, $x_0=1$ and $g(x) = x$<br>
$\quad$So, $f(x)=O(g(x))=O(x^1)$</p>
<p>c) $f(x)= log_39 = 2 \le 3(x^0)$ for $x \ge 1$<br>
$\quad$So, $C = 3$, $x_0=1$ and $g(x) = x^0$<br>
$\quad$So, $f(x)=O(g(x))=O(x^0)$ </p>
<p>d) $f(n) = n! = n(n-1)...1 \le n^n$<br>
$\quad$So, $C = 1, g(n) = n^n$ </p>
|
1,264,195 | <p>So there is a way of extending the set $N$ of natural numbers with 0, equipped with ordinary multiplication, to its Grothendieck group, the group of integers with respect to addition.</p>
<p>This group, $Z$, is the <em>best</em> extension of $N$ in the sense that ...? (question)</p>
<p>So, here are my questions:</p>
<ol>
<li>I know the field of fractions $F$ of a commutative ring $R$ (without a zero divisor) is the <em>best</em> extension of $R$ into a field in the sense that given a natural inclusion map $\iota:R\to F$ (by sending $r\in R$ to $r/1\in F$), and any other field $K$ and an in injective homomorphism $\phi:R\to K$, there exists a (unique?) injective homomorphism $\overline{\phi}:F\to K$ such that the diagram commutes. So in a sense $F$, the field of fraction $F$ is the <em>smallest</em> extension of $R$.</li>
</ol>
<p>How does this idea then apply to $Z$, the Grothendieck group of $N$? Wikipedia has something like what I said, but slightly different (in that the homomorphisms are not necessarily injective, so one cannot talk about "inclusion relation").</p>
<ol start="2">
<li>Okay, the integers are not only a group with respect to addition, but it is also a ring with respect to addition and multiplication. How do you then extend a commutative monoid (such as $N$) to a commutative ring (such as $Z$)? Do we do so simply by defining one more binary operation on $(Z,+)$? If s0, how do we guarantee that it is indeed a very <em>special</em> structure (in the sense that it is the <em>smallest</em> ring extension of $N$)?</li>
</ol>
<p>Much appreciated in advance!</p>
| Michael Albanese | 39,599 | <p>Not necessarily. Consider the <a href="http://en.wikipedia.org/wiki/Peano_curve" rel="nofollow">Peano curve</a>, a continuous surjective map $f : [0, 1] \to [0, 1]^2$ (the codomain can be identified with a subset of $\mathbb{C}$ is you prefer). Not every element of $[0, 1]^2$ is a boundary point.</p>
|
3,755,288 | <p>I'm trying to solve this:</p>
<blockquote>
<p>Which of the following is the closest to the value of this integral?</p>
<p><span class="math-container">$$\int_{0}^{1}\sqrt {1 + \frac{1}{3x}} \ dx$$</span></p>
<p>(A) 1</p>
<p>(B) 1.2</p>
<p>(C) 1.6</p>
<p>(D) 2</p>
<p>(E) The integral doesn't converge.</p>
</blockquote>
<p>I've found a lower bound by manually calculating <span class="math-container">$\int_{0}^{1} \sqrt{1+\frac{1}{3}} \ dx \approx 1.1547$</span>. This eliminates option (A). I also see no reason why the integral shouldn't converge. However, to pick an option out of (B), (C) and (D) I need to find an upper bound too. Ideas? Please note that I'm not supposed to use a calculator to solve this.</p>
<p>From <strong>GRE problem sets by UChicago</strong></p>
| Community | -1 | <p><span class="math-container">$$\int_0^1\sqrt{1+\dfrac1{3x}}dx=2\int_0^1\sqrt{t^2+\dfrac13}dt$$</span> proves convergence.</p>
<p>Then</p>
<p><span class="math-container">$$\frac1{\sqrt 3}\le\sqrt{t^2+\frac13}\le t+\frac1{\sqrt3}$$</span>
implies</p>
<p><span class="math-container">$$\frac2{\sqrt 3}\approx 1.155\le I\le1+\frac2{\sqrt 3}\approx2.155$$</span></p>
<p>A tighter upper bound is obtained by noting that the function is convex and</p>
<p><span class="math-container">$$\sqrt{t^2+\frac13}\le \frac1{\sqrt3}+t\left(\sqrt{\frac 43}-\frac1{\sqrt3}\right),$$</span> giving <span class="math-container">$$I\le\sqrt3\approx1.732$$</span>
A tighter lower bound could be found by considering the tangents at both endpoints up to their intersection, but we can already conclude C.</p>
<p><a href="https://i.stack.imgur.com/0eiJJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0eiJJ.png" alt="enter image description here" /></a></p>
<p>The exact value is <span class="math-container">$$1.5936865\cdots$$</span> The bounds can be computed by hand, by squaring to avoid square roots.</p>
|
3,755,288 | <p>I'm trying to solve this:</p>
<blockquote>
<p>Which of the following is the closest to the value of this integral?</p>
<p><span class="math-container">$$\int_{0}^{1}\sqrt {1 + \frac{1}{3x}} \ dx$$</span></p>
<p>(A) 1</p>
<p>(B) 1.2</p>
<p>(C) 1.6</p>
<p>(D) 2</p>
<p>(E) The integral doesn't converge.</p>
</blockquote>
<p>I've found a lower bound by manually calculating <span class="math-container">$\int_{0}^{1} \sqrt{1+\frac{1}{3}} \ dx \approx 1.1547$</span>. This eliminates option (A). I also see no reason why the integral shouldn't converge. However, to pick an option out of (B), (C) and (D) I need to find an upper bound too. Ideas? Please note that I'm not supposed to use a calculator to solve this.</p>
<p>From <strong>GRE problem sets by UChicago</strong></p>
| Anonymous | 807,989 | <p>We know that,</p>
<p><span class="math-container">${\displaystyle\int}\sqrt{\dfrac{1}{3x}+1}\,\mathrm{d}x$</span>=<span class="math-container">$=\class{steps-node}{\cssId{steps-node-1}{\dfrac{1}{\sqrt{3}}}}{\displaystyle\int}\sqrt{\dfrac{1}{x}+3}\,\mathrm{d}x$</span></p>
<p>Substitute <span class="math-container">$u=\sqrt{\dfrac{1}{x}+3}$</span> and <span class="math-container">$\dfrac{\mathrm{d}u}{\mathrm{d}x} = -\dfrac{1}{2\sqrt{\frac{1}{x}+3}x^2}$</span>,i.e <span class="math-container">$\mathrm{d}x=-2\sqrt{\dfrac{1}{x}+3}x^2\,\mathrm{d}u$</span></p>
<p><span class="math-container">${\displaystyle\int}\sqrt{\dfrac{1}{x}+3}\,\mathrm{d}x$</span>=<span class="math-container">$-\class{steps-node}{\cssId{steps-node-2}{2}}{\displaystyle\int}\dfrac{u^2}{\left(u^2-3\right)^2}\,\mathrm{d}u$</span>
<span class="math-container">$={\displaystyle\int}\left(\dfrac{\class{steps-node}{\cssId{steps-node-5}{u^2-3}}}{\left(u^2-3\right)^2}+\dfrac{\class{steps-node}{\cssId{steps-node-6}{3}}}{\left(u^2-3\right)^2}\right)\mathrm{d}u$</span></p>
<p><span class="math-container">$={\displaystyle\int}\dfrac{1}{u^2-3}\,\mathrm{d}u+\class{steps-node}{\cssId{steps-node-7}{3}}{\displaystyle\int}\dfrac{1}{\left(u^2-3\right)^2}\,\mathrm{d}u$</span></p>
<p>Perform partial fraction decomposition:</p>
<p><span class="math-container">$={\displaystyle\int}\left(\dfrac{1}{2\sqrt{3}\left(u-\sqrt{3}\right)}-\dfrac{1}{2\sqrt{3}\left(u+\sqrt{3}\right)}\right)\mathrm{d}u$</span> + <span class="math-container">${\displaystyle\int}\dfrac{1}{u+\sqrt{3}}\,\mathrm{d}u$</span></p>
<p>On solving this further we get</p>
<p><span class="math-container">$\dfrac{\sqrt{\frac{1}{3x}+1}}{3\left(\frac{1}{3x}+1\right)-3}+\dfrac{\ln\left(\sqrt{\frac{1}{x}+3}+\sqrt{3}\right)}{6}-\dfrac{\ln\left(\left|\sqrt{\frac{1}{x}+3}-\sqrt{3}\right|\right)}{6}+C$</span></p>
<p>I.e,</p>
<p><span class="math-container">$\dfrac{6\sqrt{\frac{1}{3x}+1}x+\ln\left(\sqrt{\frac{1}{x}+3}+\sqrt{3}\right)-\ln\left(\left|\sqrt{\frac{1}{x}+3}-\sqrt{3}\right|\right)}{6}+C$</span></p>
<p><span class="math-container">$\boldsymbol{\int\limits^{1}_{0}{f(x)}\,\mathrm{d}x =}$</span>=<span class="math-container">${1\over6}[4 \sqrt(3) + \ln(7 + 4 \sqrt(3)]$</span></p>
<p>Approximation:
<span class="math-container">$1.593686504020857$</span>,</p>
<p>i.e <span class="math-container">$1.6$</span>.</p>
<p>The answer is option <span class="math-container">$(C)$</span></p>
|
4,095,831 | <p>If <span class="math-container">$\frac{dy}{dx}=y\sec^2x$</span> and <span class="math-container">$y=5$</span> when <span class="math-container">$x=0$</span>, then <span class="math-container">$y=$</span>?
A) <span class="math-container">$e^{\tan x} + 4$</span></p>
<p>B) <span class="math-container">$e^{\tan x} + 4$</span></p>
<p>C) <span class="math-container">$5e^{\tan x}$</span></p>
<p>D) <span class="math-container">$\tan x +5$</span></p>
<p>E) <span class="math-container">$\tan x + 5e^x$</span></p>
<p>This is how I solved it:</p>
<p><span class="math-container">$\frac{dy}{y}=\sec^2x dx$</span></p>
<p><span class="math-container">$\int \frac{dy}{y}= \int \sec^2x dx$</span></p>
<p><span class="math-container">$\ln|y|= \tan x +C$</span></p>
<p><span class="math-container">$e^{\ln|y|}=e^{\tan x}+C$</span></p>
<p><span class="math-container">$|y|=e^{\tan x} + C$</span></p>
<p><span class="math-container">$y=\pm (e^{\tan x} + C)$</span></p>
<p>Since <span class="math-container">$y=5$</span> we choose <span class="math-container">$y=e^{\tan x} + C$</span></p>
<p>But the answer is C)</p>
<p>I know that I can find <span class="math-container">$dy/dx$</span> for each answer choice and C) is the only one that fits.. however I don't know what I did wrong in my solution.</p>
| GReyes | 633,848 | <p>When you take the exponential function, you actually get
<span class="math-container">$$
|y|=e^{\tan x}e^C
$$</span>
or
<span class="math-container">$$
y=De^{\tan x}
$$</span>
where <span class="math-container">$D=\pm e^C\ne 0$</span>.Since <span class="math-container">$y=0$</span> is also a solution, you can include <span class="math-container">$D=0$</span> to get the general solution. Finally, you impose the given initial condition.</p>
|
61,509 | <p>I'm trying to read Elias Stein's "Singular Integrals" book, and in the beginning of the second chapter, he states two results classifying bounded linear operators that commute (on $L^1$ and $L^2$ respectively).</p>
<p>The first one reads:</p>
<p>Let $T: L^1(\mathbb{R}^n) \to L^1(\mathbb{R}^n)$ be a bounded linear transformation. Then $T$ commutes with translations if and only if there exists a measure $\mu$ in the dual of $C_0(\mathbb{R}^n)$ (continuous functions vanishing at infinity), s.t. $T(f) = f \ast \mu$ for every $f \in L^1(\mathbb{R}^n)$. It is also true that $\|T\|=\|\mu\|$. </p>
<p>The second one says:</p>
<p>Let $T:L^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n)$ be bounded and linear. Then $T$ commutes with translation if and only if there exists a bounded measurable function $m(y)$ so that $(T\hat{f})(y) = m(y) \hat{f}(y)$ for all $f \in L^2(\mathbb{R}^n)$. It is also true that $\|T\|=\|m\|_\infty$. </p>
<p>I was wondering if anyone had a reference to a proof of these two results or could explain why they are true. </p>
| Willie Wong | 1,543 | <p>Let me just give a literature reference, in case that's what you are interested in. The classical paper on translation invariant operators on Lebesgue spaces is Hormander's 1960 <em>Acta</em> paper "Estimates for translation invariant operators in $L^p$ spaces". If you have access to MathSciNet, <a href="http://www.ams.org/mathscinet-getitem?mr=121655" rel="nofollow">here's the mref</a>. </p>
|
2,361,602 | <p>The "Heine–Cantor theorem" states: If $f : M → N$ is a continuous function between two metric spaces, and $M$ is compact, then $f$ is uniformly continuous.</p>
<p>I do not doubt its validity, of course, just trying to understand <strong>why</strong> it is valid.</p>
<p>If we, say, take the function: $y = x^4$.
It rises very quickly with the rising value of the argument. How is it that according to Heine–Cantor theorem just because we enclose the argument of the function, say, between $[0, 10]$ it automatically becomes "<strong>uniformly</strong> continuous" (considering of course that it is "continuous")? </p>
<p>There are areas of function values inside the argument segment where function will rise quicker than in some other areas.</p>
<p>Does the reason have to do with the fact that we could in worst case choose $\delta=10$ (length of the segment in example) and thus cover all possible cases for $\varepsilon$?</p>
| AbstractNonsense | 429,931 | <p>you can also consider the following idea(take for simplicty $N = \mathbb{R})$: since $f$ is continouus and $M$ is compact, the function $f$ obtains its <strong>maximal value</strong>, which is some real number, inside $M$ (see also <a href="https://en.wikipedia.org/wiki/Extreme_value_theorem" rel="nofollow noreferrer">here</a>).
In other words, you can control the values in $N$ taken by $f$ and in particular, $f:M \to \mathbb{R}$ is a bounded function (because $f$ is always less or equal than its maximum). That example makes it easier to understand why you can control $f$ on the entire compact set $M$, as you questioned.</p>
|
20,942 | <p>If I had a recursive function (<code>f(n) = f(n-1) + 2*f(n-2)</code> for example), how would I derive a formula to solve this? For example, with the Fibonacci sequence, Binet's Formula can be used to find the nth term.</p>
| Community | -1 | <p>If the recurrence relation is linear, homogeneous and has constant coefficients, here is the way to solve it. First obtain the characteristic equation. To do this, assume $f(n) = m^n$. Plug it in to get a quadratic in $m$. Solve for $m$. Get the two roots say $m_1$ and $m_2$. Now the general solution is given by the linear combination namely $f(n) = a_1 m_1^n + a_2 m_2^n$. Solve for $a_1$ and $a_2$ using the initial conditions.</p>
<p>For your recurrence, the corresponding equation becomes, $m^2-m-2 = 0 \Rightarrow (m-2)(m+1) = 0$. So $f(n) = a 2^n + b (-1)^n$. You need to specify two initial conditions to get $a$ and $b$.</p>
<p>This methodology is analogous to plugging in $y=e^{mx}$ when you want to solve a linear, homogeneous ODE with constant coefficients. Here you have a difference equation instead of a differential equation.</p>
|
1,719,840 | <p>If there is a group $G$ with order $a$, having a subgroup $H_1$ with order $b$, and $H_2$ with order $c$, and $bc=a$, $H_1 \cap H_2 = e $. Is $H_1 H_2 =G$? </p>
| Math1000 | 38,584 | <p>Since $H_1$ and $H_2$ are subgroups of $G$, we have $$|H_1H_2|=\frac{|H_1||H_2|}{|H_1\cap H_2|}=|H_1||H_2|=|G|,\tag1 $$ so $H_1H_2=G$.</p>
<p>(See @Bungo's answer for a proof of the identity in (1))</p>
|
1,303,485 | <p>Evaluate the integral $$\int_{C}\frac{z^2}{z^2+9}dz$$
where C is the circle $|z|=4$</p>
<p>I know that if f is analytic in simply connected domain $D$, $C$ a simple closed positively oriented contour that lies in D and $z_o$ lies interior to $C$, then
$$\int_{C}\frac{f(z)}{z-z_o}dz=2\pi i f(z_o)$$</p>
<p>But for this problem, the circle contains both interior points which is $3i$ and $-3i$. And I found that reducing the fraction into partial fraction seems to be useless in solving the problem. So what is the $f(z)$ here?</p>
| Kaj Hansen | 138,538 | <p><strong>Hint</strong>:</p>
<p>Note that $z^2+9 = (z+3i)(z-3i)$. So with partial fraction decomposition, we get:</p>
<p>$$\frac{z^2}{(z^2+9)} = \frac{iz^2}{6(z+3i)} - \frac{iz^2}{6(z-3i)}$$</p>
<p>Therefore:</p>
<p>$$\int_C \frac{z^2}{z^2 + 9} \ dz = \int_C \frac{iz^2}{6(z+3i)} \ dz - \int_C \frac{iz^2}{6(z-3i)} \ dz$$</p>
<p>And you can apply Cauchy's integral formula to each of the two pieces on the right hand side.</p>
<hr>
<p><strong>Alternative</strong>:</p>
<p>You can split the contour into two pieces $C_1$ and $C_2$, where $C_1$ is the upper half of the circle $|z| = 4$ together with the segment $[-2, 2]$, and likewise $C_2$ is the lower half of the circle $|z|=4$ together with the same segment. If both are oriented counterclockwise, then $C = C_1 + C_2$, and so the integral splits accordingly, and you'll notice that each of $C_1$ and $C_2$ contains only one point where analyticity fails. I won't write out all the details, but you can refer to my previous post <a href="https://math.stackexchange.com/questions/1299127/cauchy-integral-formula/1299143#1299143"><strong>here</strong></a> for more information.</p>
|
188,492 | <p>$A$ is an $n\times n$ matrix (not symmetric). If $\rho(A)$, spectral radius of $A$, is less than or equal to 1, can we say that $x^TAx\leq x^Tx$? </p>
<p>In another word,</p>
<p>if $\rho(A)\leq 1$, then $\frac{1}{2}\rho(A+A^T)\leq 1$?</p>
| copper.hat | 27,978 | <p>No. Choose $A$ to be a matrix of all zeroes except $[A]_{1n} = n+1$. Let $x = (1,...,1)^T$. Then $x^T A x = n+1$, but $x^T x = n$. The spectral radius is $\rho (A) = 0$.</p>
|
2,809,090 | <p>The stochastic vector $(X,Y)$ has a continuous distribution with pdf:
$$f(x,y) =
\begin{cases}
xe^{-x(y+1)} & \text{if $x,y>0$} \\[2ex]
0 & \text{otherwise}
\end{cases}$$<br>
Define $Z:=XY$.<br>
I would like to know what exactly $XY$ is. It seems to me that the function $f(x,y)$ has but one output, so what does $XY$ mean here? I feel like I'm missing something obvious. Thanks in advance.</p>
| Mythomorphic | 152,277 | <p><strong>Hint:</strong>
$$\cos A\cos B+\sin A\sin B=\cos(A-B)$$
$$\cos^2 A-\sin^2 A=\cos2A$$</p>
|
2,878,553 | <p>I was trying to think about what $\pi$ actually is. There are a lot of ways to get $\pi$ for example $4(1-\frac{1}{3}+\frac{1}{5}-\cdots)$.</p>
<p>But there is no one way to define it. </p>
<p>On the other hand a fraction like $\frac{1}{2}$ also has multiple definitions e.g. $\frac{2}{4}$ or $\frac{3}{6}$. Although we generally take the simplest case as the definition. But it is really the class of all pairs $(n,2n)$. We might say that a tuple of the form $(n,2n)$ has the property of half-ness. Two fractions that both have this half-ness property can be set equal. </p>
<p>In the same way, there is not one algorithm to define $\pi$. So $\pi$ must be the class of all algorithms that define it(?). Or the class of all mathematical expressions that define it.</p>
<p>We might say that an expression $4\sum_{n=0}^\infty (-1)^n\frac{1}{2n+1}$ has the property of $\pi$-ness and if two expressions both have this property then they can be set equal.</p>
<p>The difficultly I see is that some expressions, it might be unknown if they have the property of $\pi$-ness. </p>
<p>Also, if this definition was true we could not write that anything is equal to $\pi$ as $\pi$ is just a property. How would we write this? Perhaps we would write that the expression belongs to the set of expressions that have the $\pi$-ness property. Perhaps:</p>
<p>$$\text{Expression}\left\{4\sum_{n=0}^\infty (-1)^n\frac{1}{2n+1} \right\}\subset \pi$$</p>
<p>And then this would work for other irrational numbers too:</p>
<p>$$ \text{Expression}\left\{\sum_{n=0}^\infty \frac{1}{n!}\right\} \subset e$$</p>
<p>But then this is not very convenient if we want to express the numerical value of one of these algorithms we would have to write something like:</p>
<p>$$\text{Expression}\left\{x\right\} \subset \pi \implies x \approx 3.14159$$</p>
<p>The alternative would simply be to have $\pi$ set equal to one of the expressions with the $\pi$-ness property but this seems a bit like cheating.</p>
| YeatsL | 563,841 | <p>Just as rationals are defined by the equivalence class of ordered pairs of integers under the equivalence relation $\langle a, b\rangle \sim \langle c, d\rangle$ iff $a \cdot d = b \cdot c$, reals are defined as the equivalence class of Cauchy sequences of rational numbers under the equivalence relation $f : \mathbb{N} \rightarrow \mathbb{Q} \sim g : \mathbb{N} \rightarrow \mathbb{Q}$ iff $h : \mathbb{N} \rightarrow \mathbb{Q}$ defined by $h(2n) = f(n)$ and $h(2n + 1) = g(n)$ is also a Cauchy sequence.</p>
<p>For the definition of a Cauchy sequence see here: <a href="https://en.wikipedia.org/wiki/Cauchy_sequence" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Cauchy_sequence</a></p>
<p>Thus $\pi$ being a real number is defined to be the equivalence class of Cauchy sequences of rational numbers under the above equivalence relation that converges to the ratio of the circumference of a circle to its diameter.</p>
|
2,878,553 | <p>I was trying to think about what $\pi$ actually is. There are a lot of ways to get $\pi$ for example $4(1-\frac{1}{3}+\frac{1}{5}-\cdots)$.</p>
<p>But there is no one way to define it. </p>
<p>On the other hand a fraction like $\frac{1}{2}$ also has multiple definitions e.g. $\frac{2}{4}$ or $\frac{3}{6}$. Although we generally take the simplest case as the definition. But it is really the class of all pairs $(n,2n)$. We might say that a tuple of the form $(n,2n)$ has the property of half-ness. Two fractions that both have this half-ness property can be set equal. </p>
<p>In the same way, there is not one algorithm to define $\pi$. So $\pi$ must be the class of all algorithms that define it(?). Or the class of all mathematical expressions that define it.</p>
<p>We might say that an expression $4\sum_{n=0}^\infty (-1)^n\frac{1}{2n+1}$ has the property of $\pi$-ness and if two expressions both have this property then they can be set equal.</p>
<p>The difficultly I see is that some expressions, it might be unknown if they have the property of $\pi$-ness. </p>
<p>Also, if this definition was true we could not write that anything is equal to $\pi$ as $\pi$ is just a property. How would we write this? Perhaps we would write that the expression belongs to the set of expressions that have the $\pi$-ness property. Perhaps:</p>
<p>$$\text{Expression}\left\{4\sum_{n=0}^\infty (-1)^n\frac{1}{2n+1} \right\}\subset \pi$$</p>
<p>And then this would work for other irrational numbers too:</p>
<p>$$ \text{Expression}\left\{\sum_{n=0}^\infty \frac{1}{n!}\right\} \subset e$$</p>
<p>But then this is not very convenient if we want to express the numerical value of one of these algorithms we would have to write something like:</p>
<p>$$\text{Expression}\left\{x\right\} \subset \pi \implies x \approx 3.14159$$</p>
<p>The alternative would simply be to have $\pi$ set equal to one of the expressions with the $\pi$-ness property but this seems a bit like cheating.</p>
| Michael Hardy | 11,667 | <p>The answer is that with $\pi$ as with many mathematical concepts, there is no objective reason to single out one characterization as being "the definition". One establishes a definition within a context; for example one may write "For the purposes of this paper [or this textbook, or this web page, etc.] we take $\pi$ to be the ratio of the circumference of a circle to the diameter", but in another context, one might define it to be the ratio of the area of a circle to the area of the square on its radius, and in another as the sum of a certain infinite series, and another as the value of a certain integral, and in another (in Baby Rudin, I think?) as $2$ times the smallest positive zero of the solution of $y''=-y$ and $y(0)= 1$ and $y'(0)=0.$</p>
<p>Likewise a parabola can be characterized as the graph of $y=x^2,$ or by means of a focus and a directrix, or as a section of a cone by a plane, or as what's left of an ellipse in a projective plane when you deleted one of its tangent lines from the projective plane to get an affine plane, etc. Which characterization you take to be the definition within a context is just a matter of which one best serves the purpose that you have on the particular occasion.</p>
|
306,788 | <p>I understand determinants but I cannot understand the following question, can someone explain it to me ?</p>
<p>Suppose that a $4 x 4$ matrix with rows $v_1,v_2,v_3$ and $v_4$ has determinant det A = -1. Find the following determinants: </p>
<p>$$det \begin{bmatrix}v_1\\6v_2\\v_3\\v_4 \end{bmatrix}=$$
$$det \begin{bmatrix}v_2\\v_1\\v_4\\v_3 \end{bmatrix}=$$
$$det \begin{bmatrix}v_1\\v_2+7v_1\\v_3\\v_4 \end{bmatrix}=$$</p>
<p>I know how to find the determinate of a matrix but do not understand how this works.</p>
| Julien | 38,053 | <p>Hints:</p>
<ul>
<li>the determinant is multilinear on the rows.</li>
<li>the sign changes when you swap two rows.</li>
<li>the determinant is zero when two rows are the same</li>
</ul>
<p>With this, you can compute all three determinants in your question</p>
<p>Note: multilinear means $\det(ar_1+br_1',r_2,r_3,r_4)=a\det(r_1,r_2,r_3,r_4)+b\det(r_1',r_2,r_3,r_4)$ in the first. </p>
<p>And likewise for other entries.</p>
|
2,883,370 | <p>If I want to determine whether a sequence, ${a_n}$, is bounded above $\forall n \in \Bbb{N} $, is it enough to find a sequence that is larger than $a_n$, and show that it converges and is therefore bounded? For example:</p>
<p>$\forall n \in \Bbb{N}, let,$</p>
<p>$$
a_n = \frac{1}{n+1} + \frac{1}{n+2} + ...+\frac{1}{2n}\\
\frac{1}{n+1} + \frac{1}{n+2} + ...+\frac{1}{2n} \leq \frac{1}{n} + \frac{1}{n} + ...+\frac{1}{n} = n\cdot\frac{1}{n}=1
$$
and since $\lim\limits_{n\to\infty}1 = 1$, then 1 must be an upper bound for $a_n$. Is this correct? Thanks.</p>
| miracle173 | 11,206 | <p>I think you mess up some ideas.</p>
<p>You say "and since $\lim\limits_{n\to\infty}1 = 1$", but you never showed that $\lim\limits_{n\to\infty}1 = 1$. And if you check the <a href="https://math.stackexchange.com/users/6460/henry">comment of Henry</a> this seems to be wrong. But you don't need the limes.</p>
<p>You showed that</p>
<p>$$a_n = \frac{1}{n+1} + \frac{1}{n+2} + ...+\frac{1}{2n}\\
\frac{1}{n+1} + \frac{1}{n+2} + ...+\frac{1}{2n} \leq \frac{1}{n} + \frac{1}{n} + ...+\frac{1}{n} = n\cdot\frac{1}{n}=1$$
this means</p>
<p>$$a_n\le 1,\; \forall n \in \Bbb{N}$$</p>
<p>And this means that $a_n$ is bounded above by $1$. There is nothing else to show.</p>
<p><strong>Remark 1:</strong>
<em>An increasing sequence that is bounded above is convergent</em></p>
<p>We have
$$a_{n+1}=a_n+\frac{1}{(2n+1)(2n+2)}$$
This means $$a_{n+1}>a_n$$ and so $a_n$ is monotone increasing. If a sequence is increasing and bounded above then the sequence is convergent.</p>
<p><strong>Remark 2:</strong>
<em>An convergent sequence is bounded</em></p>
<p>If a sequence $a_n$ converges to $a$ then there exists a number $N$ such that
$$\mid a_n-a\mid\le 1,\; \forall n>N$$
and so we have $$a_n\le a+1,\; \forall n>N$$ and
$$a_n\le\max\{a_1,\ldots,a_N\},\; \forall n\le N$$
and therefore the sequence $a_n$ is bounded by $$\max\{N,a_1,\ldots,a_N\}$$</p>
|
4,435,088 | <p>In an <span class="math-container">$8×8$</span> table one of the square is colored black and all the others are white . Prove that one cannot make all the boxes white by recoloring the rows and columns . "Recoloring" is the operation of changing the color of all boxes in a row or in a column .</p>
<p>This is a problem taken from Mathematical Circles by Dimitri Fomin , Sergey Genkin and IIa Itenberg.</p>
<p>My solution goes like this:</p>
<blockquote>
<p>We have a square colored black while all other squares are white. So, in order to make it white we must first recolor its row or its column in which the black colored square is present . Now , if we recolor the row or the column the number black squares will be increased and it's parity will still be odd as we now have <span class="math-container">$7$</span> black squares in that particular row or column . Now, if we again recolor that row or column it will again get reverted back to its initial configuration. So, we must now color each of the row or column in which we have one of those <span class="math-container">$7$</span> black sqaures . But this will also result in an odd number of black squares as we have then <span class="math-container">$7-1+7=13$</span> black squares in total. So, after any transformations we are left with an odd number of black squares. If all the squares are colored black then we have <span class="math-container">$0$</span> black squares . This has an even parity. So, this is not possible to have all squares colored white.</p>
</blockquote>
<p>I want to verify my proof whether it is alright or not? Is this valid? Of course there is a duplicate link about this question but that asks for a different thing. I am asking whether thus proof is valid or not...that link asks probably for a verification of a different proof ...but I want to know whether this proof is valid or not?....</p>
<p>Also if we have a <span class="math-container">$3×3$</span> square like the one given in the figure can we do the do the same thing.<a href="https://i.stack.imgur.com/yfXmxm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yfXmxm.jpg" alt="enter image description here" /></a></p>
<p>Can we solve this using the same above reasoning?</p>
| user2661923 | 464,411 | <p>Your analysis for the <span class="math-container">$8 \times 8$</span> square amounts to saying that each operation leaves the parity of the total number of black squares unchanged. This assertion is accurate, and therefore, the analysis is valid. The assertion critically depends on the idea that <span class="math-container">$(8-1)$</span> has the same parity as <span class="math-container">$(1)$</span>.</p>
<p>For the <span class="math-container">$3 \times 3$</span> square, this specific approach is invalid, because (for example) <span class="math-container">$(3-1)$</span> does not have the same parity as <span class="math-container">$(1)$</span>.</p>
<p>At this point, it is unclear whether the same conclusion is accurate in the <span class="math-container">$3 \times 3$</span> square. That is, there may exist a similar but different valid argument that involves <span class="math-container">$\pmod{n}$</span>, where <span class="math-container">$n$</span> is some positive integer other than <span class="math-container">$(2)$</span>.</p>
<hr />
<p>For what it's worth, an alternative (convoluted and therefore inferior) approach for the <span class="math-container">$8 \times 8$</span> square is to recognize that the computation of <span class="math-container">$(W - B)$</span> (i.e. # of white squares - number of black squares) is unchanged <span class="math-container">$\pmod{4}$</span> because (for example) <span class="math-container">$[(+6) - (-6)] \equiv 0\pmod{4}$</span>.</p>
<p>This yields a similar contradiction, as your analysis of the <span class="math-container">$8 \times 8$</span> square, because <span class="math-container">$(63 - 1) \equiv 2\pmod{4}.$</span></p>
|
2,418,916 | <blockquote>
<p>Find how many terms there are in this geometric sequence:</p>
<p><span class="math-container">$-1, 2, -4, 8, ..., -16777216$</span></p>
</blockquote>
<p>My attempt:</p>
<p><span class="math-container">$a_k=a.r^{k-1}$</span></p>
<p>And in this sequence:</p>
<p><span class="math-container">$a=-1$</span>, <span class="math-container">$r=-2$</span></p>
<p>So</p>
<p><span class="math-container">$a_k=(-1){(-2)}^{k-1}$</span></p>
<p><span class="math-container">$-16777216=(-1){(-2)}^{k-1}$</span></p>
<p><span class="math-container">$16777216={(-2)}^{k-1}$</span></p>
<p><span class="math-container">$log(16777216)=log({(-2)}^{k-1})$</span></p>
<p><span class="math-container">$log(16777216)=(k-1)log{(-2)}$</span></p>
<p><span class="math-container">$k-1={{log(16777216)} \over {log{(-2)}}}$</span></p>
<p>But <span class="math-container">$-2$</span> is negative, and logarithm not defined for negative numbers?,
So what can I do ?</p>
<p>Thanks</p>
| farruhota | 425,072 | <p>It is:
$$-1,2,-4,8,...,(-1)\cdot 2^{24}$$</p>
|
3,896,817 | <blockquote>
<p>Solve <span class="math-container">$x^2 \equiv 12 \pmod {13}$</span></p>
</blockquote>
<p>By guessing I can say that the solutions are <span class="math-container">$5$</span> and <span class="math-container">$8$</span>, but is there another way to find the solution besides guessing?</p>
| Nathanael Skrepek | 423,961 | <p>As cosmo5 mentioned in his comment. You can do the following which is a little bit better than guessing since you probably can instantly recognise square numbers.
For every natural number <span class="math-container">$n$</span> we have
<span class="math-container">$$
x^2 \equiv 12 \mod 13
\quad\Leftrightarrow\quad
x^2 \equiv 12+13n \mod 13.
$$</span>
Hence, you can start
<span class="math-container">$$
\begin{array}{c|c}
n & 12+13n \\
\hline
1 & 25 \\
2 & 38 \\
3 & 51 \\
4 & 64
\end{array}
$$</span>
Then you will see that for <span class="math-container">$n=1$</span> you have <span class="math-container">$5^2$</span> and for <span class="math-container">$n=4$</span> you have <span class="math-container">$8^2$</span>, which gives you the solutions <span class="math-container">$x_1=5$</span> and <span class="math-container">$x_2=8$</span>.</p>
<p>Alternative you can make a polynomial division after you found one solution, because
<span class="math-container">$$
x^2-12 \equiv (x-x_1)(x-x_2) \mod 13.
$$</span>
However, this might take longer than guessing for small numbers.</p>
|
2,732,562 | <p>I'm trying to figure out how to find the number of ternary strings of length $n$ that have 3 or more consecutive 2's.
So far I've been able to establish that there is $n(2^{n-1})$ with a single 2.
And I think (but am not certain) that this can be extrapolated to the number of strings with a single group of 2's of length $x$ by:
$$\bigl(n-(x-1)\bigr)(2^{n-x})$$
What I'm getting caught on is the 'or more part', any help would be greatly appreciated.</p>
| epi163sqrt | 132,007 | <blockquote>
<p>We consider the alphabet $V=\{0,1,2\}$ and we are looking for the number $c_n$ of strings of length $n$ having runs of $2$ less than three. The wanted number is $$a_n=3^n-c_n$$</p>
<p>We derive a generating function $C(z)=\sum_{n=0}^\infty c_nz^n$ from which the number $a_n$ can be obtained easily since
\begin{align*}
a_n=[z^n]\left(\frac{1}{1-3z}-C(z)\right)\tag{1}
\end{align*}
with $[z^n]$ denoting the coefficient of $z^n$ of a series.</p>
</blockquote>
<p>Strings with no consecutive equal characters at all are called Smirnov or Carlitz words. See example III.24 <em>Smirnov words</em> from <em><a href="http://algo.inria.fr/flajolet/Publications/books.html" rel="nofollow noreferrer">Analytic Combinatorics</a></em> by Philippe Flajolet and Robert Sedgewick for more information. </p>
<p>A generating function for the number of Smirnov words over a ternary alphabet $V$ is given by
\begin{align*}
\left(1-\frac{3z}{1+z}\right)^{-1}\tag{2}
\end{align*}</p>
<p>Replacing occurrences of $0$ in a Smirnov word by one or more zeros generates words having runs of $0$ with length $\geq 1$.
\begin{align*}
z\longrightarrow z+z^2+z^3+\cdots=\frac{z}{1-z}
\end{align*}</p>
<p>The same can be done with the digit $1$.</p>
<p>Replacing occurrences of $2$ in a Smirnov word by one or two $2$'s generates words with runs of $2$ of length less than three.
\begin{align*}
z\longrightarrow z+z^2=z(1+z)
\end{align*}</p>
<p>The resulting generating function is according to (2)
\begin{align*}
\color{blue}{C(z)}=\left(1- 2\cdot\frac{\frac{z}{1-z}}{1+\frac{z}{1-z}}-\frac{z(1+z)}{1+z(1+z)}\right)^{-1}
&\color{blue}{=\frac{1+z+z^3}{1-2z-2z^2-2z^3}}\tag{3}
\end{align*}</p>
<blockquote>
<p>We obtain for $n\geq 3$ the number of wanted words of length $n$ according to (1) and (3) as
\begin{align*}
\color{blue}{a_n}&=3^n-c_n\\
&=[z^n]\left(\frac{1}{1-3z}-\frac{1+z+z^3}{1-2z-2z^2-2z^3}\right)\\
&=[z^n]\left(\color{blue}{1}z^3+\color{blue}{5}z^4+\color{blue}{21}z^5+\color{blue}{81}z^6+\color{blue}{295}z^7+\color{blue}{1\,037}z^8+\color{blue}{3\,555}z^9+\cdots\right)
\end{align*}</p>
<p>whereby the last line was obtained with some help of Wolfram Alpha.</p>
</blockquote>
|
264,745 | <p>When I was learning statistics I noticed that a lot of things in the textbook I was using were phrased in vague terms of "this is a function of that" e.g. a statistic is a function of a sample from a distribution. I realized that while I know the definition of a function as a relation and I have an intuitive notion of what "function of" means, it's unclear to me how you transform this into a rigorous definition of "function of". So what is the actual definition of "function of"?</p>
| Michael Hardy | 11,667 | <p>"$y$ is a function of $x$" means the value of $y$ is determined by that of $x$. For example, to say that the area of a circle is a function of the radius implies that all circles with the same radius have the same area.</p>
|
264,745 | <p>When I was learning statistics I noticed that a lot of things in the textbook I was using were phrased in vague terms of "this is a function of that" e.g. a statistic is a function of a sample from a distribution. I realized that while I know the definition of a function as a relation and I have an intuitive notion of what "function of" means, it's unclear to me how you transform this into a rigorous definition of "function of". So what is the actual definition of "function of"?</p>
| Michael Bächtold | 1,984 | <p>Very much the <a href="https://mathoverflow.net/questions/307947">same question</a> was asked several years later on mathoverflow and received a few interesting answers, including one from a fields medalist.</p>
|
2,203,008 | <p>Consider the quadratic program</p>
<blockquote>
<p><span class="math-container">$ \min$</span> <span class="math-container">$ f(x) $</span></p>
<p><span class="math-container">$ \text{s.t.} \space Ax=c$</span></p>
</blockquote>
<p>Prove that <span class="math-container">$ x^* $</span> is a local minimum if and only if it is a global minimum. No convexity is assumed. The back direction is trivial.</p>
<p>Could anyone help with proof? Thanks!</p>
| copper.hat | 27,978 | <p>If there is a feasible point $x_0$ then the feasible set is given by
$\{x_0\} + \ker A$ which can be written as $\{x_0 + Ly \}_y$ for some $L$ with ${\cal R} L = \ker A$.</p>
<p>Hence we can define $\phi(y) = f(x_0+Ly)$, note that $\phi$ is also
quadratic and the problem is to show that $y_0$ is a local $\min$
<strong>iff</strong> $y_0$ is a global $\min$.</p>
<p>Suppose $y_0$ is a local $\min$, then we can write $
\phi(y_0+h) = \phi(y_0) + { 1\over 2}\langle h, Q h \rangle $ and
hence it follows that $Q \ge 0$, otherwise $y_0$ would not be
an unconstrained local $\min$. (That is, the 'restricted' Hessian of
$f$ is positive semi definite). Hence $\phi$ is convex and $y_0$
is a global solution.</p>
<p>Note that the above does <strong>not</strong> imply that $f$ is convex, for example we could take $f(x) = x_1^2-x_2^2$ subject to $x_2 = 0$.</p>
|
389,460 | <blockquote>
<p>For $n>0$ let $A(n) = \underbrace{111 \ldots 11}_{n}$. Prove that if $A(n)$ is divisible by a prime number $p>3$, then $\gcd(n, p-1) > 1$.</p>
</blockquote>
<p>It is no huge discovery that if $n$ is even, then $2$ is a common divisor of $n$ and $p-1$, thus the implication holds. I don't know how to justify the general case though, so I would appreciate some hints.</p>
| lsr314 | 66,964 | <p>If $10^n\equiv 1 \pmod p,(n,p-1)=1,(p,10)=1$,then there exist integer $x,y$, satisfies $xn+y(p-1)=1,$ hence
$$10^1=10^{xn+y(p-1)}=(10^n)^x\cdot(10^{p-1})^y\equiv1^x\cdot 1^y=1 \pmod p$$
We get $p=3.$</p>
|
3,191,402 | <p>I have tried to answer by taking change the variable <span class="math-container">$\theta$</span> to <span class="math-container">$\theta/2$</span>, so the integration is now over unit circle, then I have taken <span class="math-container">$z=e^{i\theta}$</span>. Now I tried to use residue formula for integration, but I failed.</p>
| eyeballfrog | 395,748 | <p>Winther notes in the comments that integration by parts gives
<span class="math-container">$$
(n+1)I_{n+1} = (2n+1)a I_n - n(a^2-1)I_{n-1}
$$</span>
This is suspiciously close to the recursion relation for the Legendre Polynomials,
<span class="math-container">$$
(n+1)P_{n+1}(x) = (2n+1)x P_n(x) - nP_{n-1}(x).
$$</span>
To get it into the same form, divide through by <span class="math-container">$(a^2-1)^{(n+1)/2}$</span>
<span class="math-container">$$
(n+1)\left[\frac{I_{n+1}}{(a^2-1)^{(n+1)/2}}\right] = (2n+1)\frac{a}{\sqrt{a^2-1}} \left[\frac{I_n}{(a^2-1)^{n/2}}\right] - n\left[\frac{I_{n-1}}{(a^2-1)^{(n-1)/2}}\right],
$$</span>
which means that <span class="math-container">$I_n = C(a^2-1)^{n/2}P_n(a/\sqrt{a^2-1})$</span> for some constant <span class="math-container">$C$</span>. Since <span class="math-container">$I_0 = \pi$</span>, we must have <span class="math-container">$C = \pi$</span>, and
<span class="math-container">$$
\int_0^\pi(a+\cos\theta)^nd\theta = \pi(a^2-1)^{n/2}P_n\left(\frac{a}{\sqrt{a^2-1}}\right).
$$</span></p>
|
374,209 | <p>Need to show that if $f$ is a glide reflection then there is only one line $L$
such that $f(L) = L$</p>
<p>What I know is that a glide reflection is an isometry </p>
<p>$$f(z)=a\bar{z}+b,$$ such that $|a|=1$ and $a\bar{b}+b\neq0$.</p>
<p>Now assume that two lines $L_1$ and $L_2$ such that are axes for this glide reflection. Take $x_1 \in L_1$ and $x_2 \in L_2$. Since glide reflection maps given point to some new "location", then</p>
<p>$$f(x_1)=a\bar{x_1}+b=a\bar{x_2}+b=f(x_2)$$</p>
<p>But then $\bar{x_1}=\bar{x_2}$ and consequently $x_1=x_2$ and two lines are the same.</p>
<p>Could the suffice for a proof? Thanks!</p>
| MvG | 35,416 | <p>What the proof idea in your question really shows is that $f$ is injective.</p>
<p>One problem here is the fact that it isn't easy to describe a “line” in the complex number plane. At least not as easy as say the real Euclidean plane. Most formalisms don't provide a nice way to compare lines for equality.</p>
<p>One possible approach would be the following:</p>
<ol>
<li>Take an arbitrary point $z\in\mathbb C$.</li>
<li>Apply the transformation once, which gives you $f(z)$</li>
<li>Apply the transformation a second time, resulting in $f(f(z))$</li>
<li>Compute the differences $f(z)-z$ and $f(f(z))-z$</li>
<li>If all three points lie on a single line, then the second difference will be in the same direction as the first</li>
<li>You can express this by requiring that the fraction between the two differences is a real number (and will in fact be $\frac12$ in this case)</li>
</ol>
<p>Solving the resulting equation for z should result in a condition which restricts $z$ to lie on a single line. The step 5 above is only a necessary condition for $f(L)=L$, not a sufficient one, so on the whole this will proove that there can be no more than a single fixed line. Showing that the axis of reflection expressed in this way actually is a fixed line should be rather easy by comparison.</p>
<p><em>Edit:</em><br>
I attempted to provide more details on the computation. The computation is quite lengthy, and turns to using real variables fairly quickly, but I hope it will be possible to understand my idea from this.</p>
<p>$$
\mathbb R \ni r =
\frac{f(z)-z}{f(f(z))-z} =
\frac{a\bar z + b - z}{a(\bar a z + \bar b) + b - z} =
\frac{a\bar z + b - z}{a\bar b + b} \\
r(a\bar b + b) = a\bar z + b - z
$$</p>
<p>Now let's move to real-valued variables: $a = a_x+a_yi, b = b_x+b_yi$ and $z = z_x + z_yi$. Splitting the resulting equation into real and imaginary part results in the following two real-valued equations (transformed with the help of a computer algebra system):</p>
<p>$$
\begin{pmatrix}
(-a_xb_x - a_yb_y - b_x) & (a_x - 1) & a_y \\
(a_xb_y - a_yb_x - b_y) & a_y & (-a_x-1)
\end{pmatrix}
\cdot\begin{pmatrix}r \\ z_x \\ z_y\end{pmatrix}
= \begin{pmatrix}b_x \\ b_y\end{pmatrix}
$$</p>
<p>This system is underdefined, and will in general have a 1-dimensional solution space, corresponding to your 1-dimensional fixed line $L$. When I used <a href="http://www.sagemath.org/" rel="nofollow">sage</a> to solve the above, I got the following solution, depending on an arbitrarily chosen parameter $r$:</p>
<p>\begin{align*}
z_x &= \frac{(2r - 1)a_yb_y + (a_x^2r + a_y^2r + (2r - 1)a_x + r - 1)b_x}{a_x^2 + a_y^2 - 1}
\\
z_y &= \frac{(2r - 1)a_yb_x + (a_x^2r + a_y^2r - (2r - 1)a_x + r - 1)b_y}{a_x^2 + a_y^2 - 1}
\end{align*}</p>
<p>If you look closely, you'll notice however that the denominator is zero since $\lvert a\rvert=1$. For this reason, you cannot simply choose $r$ arbitrarily.</p>
<p>But there <em>might</em> still be solutions, for which the numerator of these must be zero as well. Solving for that numerator, with the added knowledge of $a_x^2+a_y^2=1$, gives the condition $r=\frac12$. So you know that if there is any chance for a solution, you'll have to try $r=\frac12$. That gives the system of equations</p>
<p>$$\begin{pmatrix}
a_x - 1 & a_y \\
a_y & -a_x-1
\end{pmatrix}
\cdot\begin{pmatrix} z_x \\ z_y \end{pmatrix}
=\frac12\begin{pmatrix}
- a_xb_x - a_yb_y + b_x \\
a_xb_y - a_yb_x + b_y
\end{pmatrix}$$</p>
<p>There is a unique solution to this:</p>
<p>\begin{align*}
z_x &= \tfrac12 b_x & z_y &= \tfrac12 b_y
\end{align*}</p>
<p>But a single fixed solution isn't appropriate here, so let's compute the determinant of the matrix to see when there will be more than a single solution.</p>
<p>$$\begin{vmatrix}
a_x - 1 & a_y \\
a_y & -a_x-1
\end{vmatrix}
= 1-a_x^2-a_y^2 = 0$$</p>
<p>So for $\lvert a\rvert=1$ the two equations are linearily dependent. There are two possible cases. Eithere there is no solution at all, or you have one real degree of freedom in solving one, and then the other one will be solved automatically. We can show that the second is the case here, by combining the two rows of the right hand side:</p>
<p>$$
a_y(- a_xb_x - a_yb_y + b_x) - (a_x-1)(a_xb_y - a_yb_x + b_y) = 0
$$</p>
<p>Strictly speaking we should have handled the case $a=1$ separately, as in that case both factors in the above comparison would be equal to $0$. But for $a=1$ the system becomes a lot simpler, so showing that a commn solution is still possible is left as an excercise.</p>
<p>So by now we basically have a single equation describing our line:</p>
<p>$$ L = \left\{
z_x + z_yi \in\mathbb C
\;\middle\vert\;
z_x,z_y\in\mathbb R,
(a_x - 1)z_x + a_yz_y = \frac12(- a_xb_x - a_yb_y + b_x)
\right\}$$</p>
<p>So far we have shown that any point on this line will be collinear with its first and second image under $f$. What remains to be shown is that these images do in fact lie on the same line. You can achieve this by plugging $z''=f(f(z))$ into the equation:</p>
<p>\begin{gather*}
(a_x - 1)z''_x + a_yz''_y = \\ =
(a_x - 1)(a_x^2z_x + a_y^2z_x + a_xb_x + a_yb_y + b_x) +
a_y(a_x^2z_y + ay^2*z_y - a_xb_y + a_yb_x + b_y) = \\ =
(a_x - 1)z_x + a_yz_y =
\frac12(- a_xb_x - a_yb_y + b_x)
\end{gather*}</p>
<p>So now you know that if $z$ lies on $L$, then $f(f(z))$ will lie on $L$ as well, and $f(z)$ will lie on the same line as $z$ and $f(f(z))$, so it will lie on $L$ as well. Therefore $L$ is a fixed line. It is the only such fixed line, since the above computations did not provide any alternatives.</p>
|
4,350,015 | <p>I'm trying to learn calculus through self study and I happened upon the following exercise:</p>
<p><span class="math-container">$$ \int_{0}^{2} \sqrt{4 - x^2}\cdot\operatorname{sgn}(x-1) \,dx $$</span></p>
<p>Seeing the sgn I thought: well this is easy and concluded that since <span class="math-container">$ \int_{0}^{2} \operatorname{sgn}(x-1) \,dx = 0$</span>, the answer should be zero. But of course then I remembered that <span class="math-container">$ \int f(x)g(x) \,dx \neq \int f(x) \,dx \int g(x) \,dx$</span> and got lost.</p>
<p>I know how to solve <span class="math-container">$ \int_{0}^{2} \sqrt{4 - x^2}dx$</span>, since that geometrically corresponds to a quarter circle of radius <span class="math-container">$2$</span> it's just <span class="math-container">$\pi$</span>, but how do I approach the product of these things?</p>
| José Carlos Santos | 446,262 | <p>That integral is equal to<span class="math-container">$$-\int_0^1\sqrt{4-x^2}\,\mathrm dx+\int_1^2\sqrt{4-x^2}\,\mathrm dx.$$</span>So, all you have to do is to compute those two integrals. In order to do that, it is usful to know that<span class="math-container">$$\int\sqrt{4-x^2}\,\mathrm dx=\frac12x\sqrt{4-x^2}+2\arcsin\left(\frac{x}{2}\right).$$</span></p>
|
1,685,523 | <p>What is the domain of $f(x)=x^x$ ? </p>
<p>I used Wolfram alpha where it is said that the domain is all positive real numbers. Isn't $(-1)^{(-1)} = -1$ ? Why does the domain not include negative real numbers as well?</p>
<p>I also checked graph and its visible for only $x>0$ . Can someone help me clarify this?</p>
| John Bentin | 875 | <p>The expression $x^y$ can be assigned a reasonable meaning for all real $x$ and all rational numbers of the form $y=m/n$, where $m$ is even and $n$ is odd and positive. Thus $x^y=(x^m)^{1/n}$, interpreted as the unique real $n$th root of $x^m$ (define $0^0$ to be $1$). Since every real number can be arbitrarily well approximated by such "even/odd" rationals, by continuity, a synthetic definition of $x^y$ can be obtained for all real $x$ and $y$. For example, using this definition, a graph can be plotted for the relation $y^y=x^x$, which runs smoothly as a loop through all four quadrants (along with the obvious line $y=x$).</p>
|
4,198,805 | <p>A <a href="https://en.wikipedia.org/wiki/Sober_space" rel="nofollow noreferrer">sober space</a> is a topological space such that every irreducible closed subset is the closure of exactly one point. Looking for examples I convinced myself that the following is true.</p>
<blockquote>
<p>Every finite <span class="math-container">$T_0$</span> topological space is sober.</p>
</blockquote>
<p>As I could not find this mentioned anywhere, can someone provide a proof to have it as a reference?</p>
| Eric Wofsey | 86,856 | <p>Suppose <span class="math-container">$X$</span> is a finite <span class="math-container">$T_0$</span> space and <span class="math-container">$A\subseteq X$</span> is an irreducible closed subset. Let <span class="math-container">$B\subset A$</span> be a maximal closed proper subset (which exists by finiteness), and let <span class="math-container">$x\in A\setminus B$</span>. By maximality of <span class="math-container">$B$</span>, <span class="math-container">$B\cup\overline{\{x\}}$</span> must be all of <span class="math-container">$A$</span>. By irreducibility of <span class="math-container">$A$</span>, this means either <span class="math-container">$B$</span> or <span class="math-container">$\overline{\{x\}}$</span> must be all of <span class="math-container">$A$</span>, and thus <span class="math-container">$\overline{\{x\}}=A$</span> and <span class="math-container">$x$</span> is a generic point of <span class="math-container">$A$</span>.</p>
|
1,627,713 | <p>This is maybe math $101$ question:</p>
<p>Let $z_1=1+i$.</p>
<p>I know that $r=\sqrt 2$ and $\theta=\arctan(1/1)=\pi/4$ so $$z_1=\color{blue}{\sqrt 2e^{i\pi/4}} .$$</p>
<p>But now if I take a look at</p>
<p>$z_2=-1-i$,</p>
<p>I know that $r=\sqrt 2$ and $\theta=\arctan(-1/-1)=\pi/4$ so $$z_1=\color{blue}{\sqrt 2e^{i\pi/4}}.$$</p>
<p>But $z_2$ should be equal to $$\color{red}{\sqrt 2 e^{5i\pi/4}} .$$</p>
<p>Why in $z_2$ should I add $\pi$ in the power of the exponent?</p>
| Balarka Sen | 117,002 | <p>Path-connectedness, by definition, means for every pair of points $a, b$ in $\Bbb R^2 \setminus E$, there is a continuous map $f : [0, 1] \to \Bbb R^2 \setminus E$ with $f(0) = a$, $f(1) = b$. So what you said is not wrong.</p>
<p>To prove that such paths exist, note that there are uncountably many lines passing through $a$ in $\Bbb R^2$. As $E$ is countable, there are uncounably many lines through $a$ not intersecting $E$. Similarly, there are uncountably many lines passing through $b$ which does not hit $E$. You can pick two such lines passing through $a$ and $b$ respectively so that they are not parallel, i.e., so that they intersect.</p>
<p>This gives you a path between $a, b$ in $\Bbb R^2$ not hitting $E$, as desired.</p>
|
163,873 | <p>With the exception of a few miscellaneous cases, the axioms (and/or schemeta) of ZFC can roughly be divided into two kinds:</p>
<ol>
<li><p>Those that guarantee the existence of more complicated sets, given that simpler sets are already around (e.g. separation, replacement schema). Also, uniqueness of these entities immediately follows, which is very satisfying.</p></li>
<li><p>Those that guarantee the existence of larger sets, given that smaller sets are already around (e.g. powerset, union). These also tend to satisfy uniqueness properties; in particular, powersets and unions are indeed unique.</p></li>
</ol>
<p>Of course, this is a gross oversimpification (e.g. replacement is needed to prove the existence of $\beth_\omega,$ a kind of "large" cardinal, albeit a very small one). However, the point is that we can also apply the above categorization scheme to $\in$-sentences that aren't theorems of ZFC, especially to proposed axioms for set theory. In particular, large cardinal axioms are (by definition) of the latter variety.</p>
<blockquote>
<p><strong>Question.</strong> Have any axioms or axiom schemata of the former variety (i.e. those guaranteeing the existence of more complicated sets) been proposed or otherwise considered?</p>
</blockquote>
<p>I'm especially interested in:</p>
<ul>
<li><p>axioms and/or schemata that legitimize non-mainstream ways of defining and/or constructing things. For example, I'd be interested to hear of a schema asserting that certain definable (proper-class) functions always have greatest and/or least fixed points. Or an axiom asserting that a particular class of self-referential definitions do indeed define unique functions. etc.</p></li>
<li><p>axioms or axiom schemata that guarantee the existence of entities whose uniqueness can then be proved (or which assert not only existence but also uniqueness). For example, Martin's axiom does not have this property.</p></li>
</ul>
| Monroe Eskew | 11,145 | <p>Axiom: $0^\sharp$ exists.</p>
<p>$0^\sharp$ is a pivotal principle in the large cardinal hierarchy, but it is actually a set of natural numbers. If it exists, it is unique.</p>
|
2,793,983 | <p>For example I find myself wanting to write $x$ is an element of the integers from $1$ to $50$,</p>
<p>Is this the quickest way? </p>
<p>$x\in \left[ 1,50\right] \cap \mathbb{N} $</p>
<p>Also is this standard on here? $\mathbb{N} = \{0, 1, 2,\dotsc \}$,
$\mathbb{ℤ}_+ = \{1, 2, \dotsc \}$.</p>
| Tesla | 338,187 | <p>It depends on your own preference on how to write things down, there are countless variations, for example</p>
<p>$x \in \{ n \in \mathbb N : 1 ≤ n ≤ 50\}$</p>
<p>$x \in \{1,2,...,50\}$</p>
<p>$x \in \mathbb N_1^{50}$ </p>
|
1,627,619 | <p>Could anyone please check my solution to the following problem?</p>
<blockquote>
<p><strong>Problem:</strong> Let $f(x, y) = (x^2 + y^2)e^{-(x^2 + y^2)}$. Find global extrema of $f$ on $M = {\mathbf R}^2$.</p>
</blockquote>
<p><strong>Proposed solution:</strong> Taking partial derivatives of $f$, we conclude that critical points are $[0,0]$ and points of the unit circle $C = \{[x,y] \in {\mathbf R}^2:\ x^2 + y^2 = 1\big\}$.</p>
<p>We can reason immediately that the global minimum is attained at $[0,0]$ as the function is nonnegative. We observe that the value of $f$ on $C$ is $e^{-1}$.</p>
<p>To prove that $f$ attains global maximum on $C$, we let $r := x^2 + y^2$. We observe that for any two points $[x_1,y_1]$ and $[x_2,y_2]$, $r_1 = r_2$ (ie. the function is constant on circles). Now let $r \to \infty$. Then
$re^{-r} \to 0$. From the definition of limit, it follows that for any $\varepsilon > 0$, we find $\delta > 0$ such that</p>
<p>$$\forall r \in P(\infty, \delta) = ({1 \over \delta}, \infty): re^{-r} < \varepsilon.$$</p>
<p>Let $\varepsilon = (2e)^{-1}$. Then there's $\delta$ from the definition above and we know that for $r \in ({1 \over \delta}, \infty)$, the value of f is less than the value of f on $C$. Restricting ourselves to the compact set</p>
<p>$$C' = \big\{[x, y] \in {\mathbf R}^2:\ x^2 + y^2 \le {1 \over \delta}\big\},$$</p>
<p>we can now argue that $f$ on $C'$ is indeed maximized on $C$, as it is a continuous function on a compact set, it's value around the boundary is at most $(2e)^{-1}$ and all critical points have been considered.</p>
<p>Therefore, the maximum value of $f$ is attained on $C$ with respect to $M$ as well.</p>
| levap | 32,262 | <p>Looks good. Note that you can also start by translating the problem into a one-dimensional problem by letting $g(r) := re^{-r}$ and finding the global maximum and minimum of $g$ on $[0, \infty)$ and then note that $f(x,y) = g(x^2 + y^2)$ and conclude.</p>
|
60,152 | <p>What is the motivation behind topology?</p>
<p>For instance, in real analysis, we are interested in rigorously studying about limits so that we can use them appropriately. Similarly, in number theory, we are interested in patterns and structure possessed by algebraic integers and algebraic prime numbers.</p>
<p>Some googling and wiki-ing gave me that topology studies about deformation of objects in some space i.e. how an object in some space behaves under a continuous map. However, when I started reading the subject it starts of by defining what a topology is i.e. a set of subsets of a set with certain properties. I fail to see the connection immediately. I would appreciate if someone could give a short bird's eye view of topology.</p>
<p>Thanks,
Adhvaitha</p>
| Matt E | 221 | <p><em>Topology</em> can mean different things in mathematics, depending on the context. </p>
<p>If a mathematician describes themselves as a <em>topologist</em>, this likely means that they study various kinds of shapes (technically, <em>manifolds</em>, or related kinds of spaces), with an eye perhaps to classifying them (a typical example being the classification of closed orientable surfaces via the number of handles attached to a sphere), or understanding certain invariants or other aspects of their structure (e.g. the Poincare conjecture, which gives a characterization of the three-dimensional sphere in terms of a certain invariant, namely its fundamental group).</p>
<p>On the other hand, the basic axioms of topology that you are asking about (a topology on a space is a collection of subsets, called open subsets, satisfying the following axioms ...) are much more general. There is a branch of mathematics that focuses on studying (more or less) these axioms and their consequences (appropriately known as <em>general topology</em>, where <em>general</em> refers to the generality of the axioms), but this investigation is rather different in flavour to the mathematics described in the preceding paragraph --- it is perhaps closer to set theory than it is to the investigation of the topology of manifolds and so on.</p>
<p>But thinking in terms of the subject of general topology is not a good way to understand the point of the axioms of topology. By analogy, consider the group axioms: there is a well-developed area of mathematics called <em>group theory</em> which studies (more or less) these axioms and their consequences, but the notion of group is ubiquitous in mathematics, and plays a fundamental role in many areas of mathematics besides group theory proper (including number theory, topology, and geometry). Thus the group axioms capture a concept which is of fundamental mathematical importance, and this is why we isolate the notion of group and study it.</p>
<p>Similary, the axioms of a topology, and the basic definitions associated to them (such as continuity of maps, and connectedness and compactness of subsets) were formulated as the consequence of a rather long attempt during the 19th and early 20th centuries to isolate and abstract the basic notions related to continuity and limits. These concepts are again ubiquitous in mathematics, and so the notion of <em>topology</em> (in the sense of the axioms you asked about) are applied in an enormous number of different contexts throughout mathematics (not just in those particular areas studied by topologists or general topologists, but in geometry, analysis, number theory, parts of algebra, logic, ...). Although the axioms may seem unusual, and unrelated to the notions of shape and position that you see discussed in an intuitive account of topology, they are in fact carefully crafted to capture, in very general language, the notions of continuity, nearness, limits, and so on. </p>
<p>In order to see the truth of my last claim, you will have to invest some time studying the axioms (i.e. learning some basic general topology), and then see how it is applied in various contexts. At least in the U.S., this typically happens in advanced undergraduate and beginning graduate courses.</p>
|
2,567,607 | <p>$$\arctan 2x +\arctan 3x = \left(\frac{\pi}{4}\right)$$
$$\arctan \left(\frac{2x+3x}{1-2x*3x}\right)=\frac {\pi}{4}$$
$$\frac {5x}{1-6x^2}=\tan \frac{\pi}{4}=1$$
$$6x^2 + 5x -1 = 0$$
$$(6x-1)(x+1)=0$$
$$x=-1, \frac{1}{6}$$</p>
<p>The answer however rejects the solution $x=-1$ saying that it makes the L.H.S of the equation negative. I don't understand this, I don't see how $x=-1$ makes the L.H.S. negative.</p>
| Vivek | 546,817 | <p>Concep: Let $f$ be a function, which means $$x_1=x_2\Rightarrow f(x_1)=f(x_2)$$
But note that if $f(x_1)=f(x_2)$, then $x_1=x_2$ is one of the possibility, among perhaps many possibilities if the function is not injective</p>
<p>Now Given $\tan^{-1}(2x)+\tan^{-1}(3x)=\frac{\pi}{4}$, taking $\tan$ of both sides(since $\tan$ is a function),we get
$$\frac{2x+3x}{1-2x.3x}=1$$</p>
<p>which after simplification gives $x=\frac16,-1$. Taking $x=\frac16$, we want to find what is $\tan^{-1}\frac13+\tan^{-1}\frac12$. For that let us take $\tan^{-1}\frac13=\alpha,\tan^{-1}\frac12=\beta$, which is equal to $\tan\alpha=\frac13,and \tan\beta=\frac12$, where $\alpha,\beta$ being principal values belongs to $(-\frac{\pi}2,\frac{\pi}2)$, which means $\alpha+\beta\in(-\pi,\pi)$. Now we know $$\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$$ which implies $$\tan(\alpha+\beta)=\frac{\frac13+\frac12}{1-\frac13\frac12}=1$$ Hence $$\alpha+\beta=\tan^{-1}1=\frac{\pi}4$$ or $$\tan^{-1}\frac13+tan^{-1}\frac12=\frac{\pi}4$$ and $x=\frac16$ becomes a root of the above equation</p>
<p>Now take $x=-1$, then the lhs of the equation becomes $$\tan^{-1}(-2)+\tan^{-1}(-3)$$ which is a negative angle $-\frac{3\pi}{4}$ , which can not be a positive number $\frac{\pi}{4}$. </p>
|
1,307,269 | <p>In my textbook, before introducing the epsilon delta definition, they gave a working definition of what a limit is. The definition sounded something like this "$\lim \limits_{x \to a}f(x) = L$, if when $x$ gets closer to $a$, $f(x)$ gets closer to $L$" </p>
<hr>
<p>But is that always the case with limits? What if $f(x) = 4,$ then we have $\lim \limits_{x \to 2}f(x) = 4$, but it is never true that when x gets closer to 2, f(x) gets closer to 4. Maybe instead we should say: "$\lim \limits_{x \to a}f(x) = L$, if when $x$ gets closer to $a$, $ f(x)$ gets closer to or equals $L$".</p>
<hr>
<p>Please correct me if I'm wrong. I'm pretty new to this stuff. Btw, i understand that the epsilon delta definition has the constant function limit case covered, but I'm more interested in the working definition.</p>
| Community | -1 | <p>Probably a better intuitive definition is $f(x)$ can be made arbitrarily close to $L$ by making $x$ close enough to $a$.</p>
<p>You avoid the awkwardness about the constant case. Additionally this emphasizes that it is for ALL $\epsilon$. It's not just that $f(x)$ gets "closer", it's that it can be made as close as you'd like.</p>
<p>For a counterexample of "gets closer" is $\lim\limits_{x\to 0} -x^2 $. I propose that $\lim\limits_{x\to 0} -x^2 =1$ because as $x$ gets closer to $0$ then $-x^2$ gets closer to $1$. However it doesn't get within $\frac12$ of $1$.</p>
|
1,740,458 | <p>I saw somewhere on Math Stack that there was a way of finding integrals in the form $$\int \frac{dx}{a+b \cos x}$$ <strong>without</strong> using Weierstrass substitution, which is the usual technique. </p>
<p>When $a,b=1$ we can just multiply the numerator and denominator by $1-\cos x$ and that solves the problem nicely. </p>
<p>But I remember that the technique I saw was a nice way of evaluating these even when $a,b\neq 1$.</p>
| C. Dubussy | 310,801 | <p>One usual trick is the substitution $x=2y$. Note that $$\frac{1}{a+b\cos(2y)}=\frac{1}{a+b(2\cos^2(y)-1)}=\frac{\sec^2(y)}{2b+(a-b)\sec^2(y)}=\frac{\sec^2(y)}{(a+b)+(a-b)\tan^2(y)}.$$ Hence $$\int \frac{dx}{a+b\cos(x)}=\int \frac{\sec^2(y)}{(a+b)+(a-b)\tan^2(y)} \, dy.$$ Now conclude with the substitution $t=\tan(y).$</p>
|
19,325 | <p>I'm looking for a simple way to define mathematics to primary/elementary school teachers and explain some of the confusion children have.</p>
<p>I'm hoping some Algebraist could help me properly state the following:</p>
<blockquote>
<p>A number in and of itself has no true meaning. True in the sense that it relates to an existing object within our world. The question we need to ask is how do we teach children meaning if that meaning is not at first grounded within something concrete.</p>
</blockquote>
<blockquote>
<p>Numbers in and of themselves represent abstract notions and in the pure study of mathematics we study mostly patterns: the various patterns that emerge from these abstract notions and the various means through which some relation is developed or expressed between them. Meaning that between the value 1 and 2 for instance there is no relation except when explicitly defined for example as some additive operation, in general an addition of multiples of the unit element.</p>
</blockquote>
| user52817 | 1,680 | <p>I highly recommend the book <em>The Number Sense: How the Mind Creates Mathematics</em>, by Stanislas Dehaene and published by Oxford University Press. Another book that comes to mind is <em>The Language Instinct: How The Mind Creates Language</em>, by Steven Pinker. Both books have had broad impact on scholarship related to your question.</p>
<p>As you try to refine your question and find an answer that satisfies you, it might help to think about how artificial intelligence "learns." Let's take the sentence from your OP "<em>The question we need to ask is how do we teach children meaning if that meaning is not at first grounded within something concrete.</em>" and replace "<em>children</em>" by "<em>machines</em>" or "<em>artificial intelligence.</em>" To be sure, there is much controversy about all of this, related to meaning and understanding.</p>
|
56,847 | <p>What are the angles formed at the center of a tetrahedron if you draw lines to the vertices?</p>
<p>I'm trying to make these:</p>
<p><img src="https://i.stack.imgur.com/FRUi8.jpg" alt="caltrop"> </p>
<p>I need to know what angles to bend the metal.</p>
| Nick Alger | 3,060 | <p>The angle is
<span class="math-container">$$\theta = \arccos \left(-\frac{1}{3}\right) \approx 109.47.$$</span>
You can find this angle by writing the volume of a regular tetrahedron in two different ways.</p>
<p>(this is an adaptation of a blog post I wrote for a website that has since been deleted)</p>
<hr>
<p>For setup, here is a picture of the regular tetrahedron in question:</p>
<p><a href="https://i.stack.imgur.com/Ta30v.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ta30v.png" alt="enter image description here"></a></p>
<p>If we knew the vertical distance, <span class="math-container">$h$</span>, from the base of the tetraheron to the center, then we could find the angle to be
<span class="math-container">$$\theta = \arccos \left(-h\right).$$</span>
This follows from basic 2D trigonmetry, as shown in the following diagram which views the tetrahedron from the side:</p>
<p><a href="https://i.stack.imgur.com/bXUfG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bXUfG.png" alt="enter image description here"></a></p>
<p>We can find <span class="math-container">$h$</span> in a roundabout way by, taking advantage of the symmetry of the tetrahedron. Break the big tetrahedron into 4 smaller tetrahedra:</p>
<p><a href="https://i.stack.imgur.com/KlGUj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KlGUj.png" alt="enter image description here"></a></p>
<p>Now we may use the formula <span class="math-container">$\text{volume} = \frac{1}{3} \times \text{base} \times \text{height}$</span> on the smaller tetrahedra, and the larger tetrahedra, to come up with two different expressions for the volume. Then, we can set these two expressions equal, solve for <span class="math-container">$h$</span>, and plug the result into the previous formula to get the desired angle:</p>
<p><a href="https://i.stack.imgur.com/SzF29.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SzF29.png" alt="enter image description here"></a></p>
<hr>
<p>One may trivially extend this argument to regular simplices of arbitrary dimension. There one would break an <span class="math-container">$n$</span>-simplex into <span class="math-container">$n+1$</span> equal pieces, therefore yielding the angle
<span class="math-container">$$\theta = \arccos \left(-\frac{1}{n}\right).$$</span></p>
|
1,822,336 | <p>My friend asked me what the roots of $y=x^3+x^2-2x-1$ was.</p>
<p>I didn't really know and when I graphed it, it had no integer solutions. So I asked him what the answer was, and he said that the $3$ roots were $2\cos\left(\frac {2\pi}{7}\right), 2\cos\left(\frac {4\pi}{7}\right)$ and $2\cos\left(\frac {8\pi}{7}\right)$.</p>
<blockquote>
<p><strong>Question:</strong> How would you get the roots without using a computer such as Mathematica? Can other equations have roots in Trigonometric forms? </p>
</blockquote>
<p>Anything helps!</p>
| David Quinn | 187,299 | <p>Consider the equation $$\cos4\theta=\cos3\theta$$ whose roots are $$\theta=n\cdot\frac{2\pi}{7}$$</p>
<p>Representing this as a polynomial in $c=\cos\theta$, we have $$8c^4-4c^3-8c^2+3c+1=0$$
$$\Rightarrow (c-1)(8c^3+4c^2-4c-1)=0$$</p>
<p>Now write $x=2c$ and we see that the polynomial equation $$x^3+x^2-2x-1=0$$ has roots as stated in your question. Note that $$2\cos\frac{6\pi}{7}=2\cos\frac{8\pi}{7}$$</p>
|
3,001,335 | <p>I know little formal math terminology and don't understand much of anything about complex analysis. Also, if this isn't a good starting point for complex integration feel free to say (I'm learning about it partly for Cauchy's residue theorem). </p>
<p>My first and intuitive idea of residue has to do with remainder, subtraction, division, etc., But more generally something that's leftover, extra, or unused by an operation or something. Do these ideas tie together? </p>
| Nodt Greenish | 614,664 | <p>The residue (latin residuere - remain) is named that way because
<span class="math-container">$\frac{1}{2\pi i}\int_{|w-z_0|=r}f(w)dw=\sum_{n=-\infty}^{\infty} a_n\int_{|w-z_0|=r}(w-z_0)^n\,dw= a_{-1}$</span> is what remains after integration.</p>
|
254,030 | <p>I am following a course in basic algebra, and we have covered rings & groups in class, but I am having trouble visualising them. Are there applications of group &/or ring theory that can be more easily visualized than the abstract object? For instance, are there objects, or properties of objects, that behave as elements of a group in physics, chemistry, or other fields?</p>
| Bill Dubuque | 242 | <p>Group theory may be viewed roughly as a general study of symmetry. In chemistry this applies to crystals via the study of crystallographic groups, and in art via <a href="http://en.wikipedia.org/wiki/Wallpaper_group" rel="nofollow">wallpaper groups.</a> For an example in physics, the Lie symmetry groups of partial differential equations play fundamental roles, e.g, governing conservation laws and separation of variables. See for example Weyl's <em>Symmetry</em> and Budden's <em>Fascination of Groups</em>.</p>
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.