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 |
|---|---|---|---|---|
27,089 | <p>Off hand, does anyone know of some useful conditions for checking if a ring (or more generally a semiring) has non-trivial derivations? (By non-trivial, I mean they do not squish everything down to the additive identity.) Part of the motivation for this is that I was thinking about it the other day, and had trouble finding any good example of a semiring with an interesting derivation. </p>
<p>For example, the multiplicative Banach algebra of positive functions is an algebra of the semifield of nonnegative reals. However, the usual definition for derivative breaks down due to the fact that you can have positive functions with negative slope. So, this leads me to wonder if there are any semirings with derivations at all?</p>
<p>As a related question, is there a known classification of all the derivations for an algebra? It feels like this should be a pretty standard thing, but I don't think I've ever encountered it in one of my courses and my initial googling around was not too successful at finding references.</p>
| Hailong Dao | 2,083 | <p>For your question in the second paragraph, <a href="http://books.google.com/books?id=roEx6lkAp10C&lpg=PA37&ots=jI7AVZExLT&dq=any%20semirings%20can%20be%20embeded%20in%20one%20with%20non%20trivial%20derivations&pg=PA37#v=onepage&q&f=false" rel="nofollow">any semiring can be embedded in one with non-trivial derivation</a>. </p>
|
2,290,395 | <p>What if in Graham’s Number every “3” was replaced by “tree(3)” instead? How big is this number? Greater than Rayo’s number? Greater than every current named number?</p>
| Owen Wei | 676,645 | <p>The numbers themselves are already so big that doing that would barely change it at all.
It would still be ZERO compared to a number like Rayo's number.
Obviously, if doing so really did make it the largest number ever created, why wouldn't people do it earlier? </p>
|
3,145,973 | <p>Show that for every integer <span class="math-container">$n ≥ 3$</span>, the number <span class="math-container">$n!e$</span> is not an integer.</p>
<p>I have shown the inequality <span class="math-container">$\displaystyle0< \sum_{m=n+1} \frac{1}{m!} < \frac{1}{n!}$</span> for <span class="math-container">$n>3$</span></p>
<p>and I know <span class="math-container">$\displaystyle e = \sum_{k=0}^{\infty} \frac{1}{k!}$</span>. Therefore <span class="math-container">$\displaystyle n!e = \sum_{k=0}^{\infty} \frac{n}{k!}$</span>. How do I continue here. I tried to prove this by induction but I wasn't even sure how to show that <span class="math-container">$3e$</span> isn't an integer.</p>
<p>Lastly, how may I incorporate this result in the process of proving that e is irrational?</p>
| Ari Royce Hidayat | 435,467 | <p>I understand your frustration, I also read several proofs but hard to understand them because somehow and I don't know why they all seem to omit the most important part of the explanation.</p>
<p>First about restriction notation, it's easy to understand and a neat thing write. So instead of writing:</p>
<p><span class="math-container">$\phi(N) = N$</span> <span class="math-container">$\forall \phi \in Aut(K)$</span></p>
<p>It could be written as:</p>
<p><span class="math-container">$\phi|_K(N) = N$</span></p>
<p>So we are going to use this notation to prove that if <span class="math-container">$N \: char \: K \: char \: G$</span> then <span class="math-container">$N \: char \: G$</span>.</p>
<p>Let <span class="math-container">$\phi \in Aut(G)$</span>, so <span class="math-container">$\phi(K) = K$</span> since <span class="math-container">$K$</span> is characteristic in <span class="math-container">$G$</span>.</p>
<p>Now please note the most important part of the proof. If we have <span class="math-container">$\varphi \in Aut(K)$</span> we will also have <span class="math-container">$\varphi(K) = K$</span>, so it's no difference than <span class="math-container">$\phi(K) = K$</span> as there is all the elements of <span class="math-container">$Aut(K)$</span> in <span class="math-container">$Aut(G)$</span>. In other words:</p>
<blockquote>
<p><span class="math-container">$\varphi(K) = \phi(K) = K$</span>.</p>
</blockquote>
<p>Note that <span class="math-container">$\varphi$</span> could be written in restriction notation as <span class="math-container">$\phi|_K(K) = K$</span>, it's the same thing because <span class="math-container">$\phi|_K \in Aut(K)$</span>. So we could rewrite the above equality as:</p>
<blockquote>
<p><span class="math-container">$\phi|_K(K) = \phi(K) = K$</span>.</p>
</blockquote>
<p>( So we don't have to define a new symbol such <span class="math-container">$\varphi$</span> every time. )</p>
<p>As <span class="math-container">$N \: char \: K$</span>, we have the same relationship between <span class="math-container">$N$</span> and <span class="math-container">$K$</span> as following:</p>
<blockquote>
<p><span class="math-container">$\phi|_N(N) = \phi|_K(N) = N$</span>.</p>
</blockquote>
<p>But we have had that <span class="math-container">$\phi|_K(K) = \phi(K)$</span>, or is all there elements of <span class="math-container">$Aut(K)$</span> in <span class="math-container">$Aut(G)$</span>, including <span class="math-container">$N < K$</span>, so we have <span class="math-container">$\phi|_K(N) = \phi(N)$</span>. Combine them all, we have:</p>
<blockquote>
<p><span class="math-container">$\phi|_N(N) = \phi|_K(N) = \phi(N) = N$</span>.</p>
</blockquote>
<p>Which is by definition <span class="math-container">$N \: char \: G$</span>.</p>
|
3,073,361 | <p>I think I understood 1-forms fairly well with the help of these two sources. They are dual to vectors, so they measure them which can be visualized with planes the vectors pierce.</p>
<ul>
<li><a href="https://the-eye.eu/public/WorldTracker.org/Physics/Misner%20-%20Gravitation%20%28Freeman%2C%201973%29.pdf" rel="nofollow noreferrer">Gravitation 1973</a><a href="https://i.stack.imgur.com/oU5Cd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oU5Cd.png" alt="1-forms as dual vectorspace to vectors" /></a></li>
<li><a href="http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.126.1099&rep=rep1&type=pdf" rel="nofollow noreferrer">On the Visualisation of Differential Forms - Dan Piponi</a><a href="https://i.stack.imgur.com/zREF5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zREF5.png" alt="enter image description here" /></a></li>
</ul>
<p>But I struggle with the explanations for higher order forms.</p>
<p>The goal is to answer and understand these questions with drawings:</p>
<ol>
<li>How can I visualize the wedge between two 1-forms <span class="math-container">$\alpha\wedge\beta$</span>?
I think I understood the wedge between two vectors, as the parallelogram created by the two in a "area sense". The determinant comes in to make it only about the area which is why <span class="math-container">$v\wedge w = \frac{1}{2}v\wedge 2w$</span> since stretching the parallelogram by two in the w direction is compensated by squishing it in the v direction, so the area stays constant.
So the wedge between two vectors is the area it spans with its vectors.
But how does that translate to the dual space? Where 1-forms measure the length of the component of its dual vector. And can be visualized as planes the vectors pierce through.
What is the visualization between two of these 1-forms as a wedge?</li>
</ol>
<p>Gravitation has this picture:</p>
<p><a href="https://i.stack.imgur.com/62SfS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/62SfS.png" alt="2-forms" /></a>
Dan-Piponi drew it like this:</p>
<p><a href="https://i.stack.imgur.com/MsczU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MsczU.png" alt="2-forms dan" /></a></p>
<p>Now these pictures make some sense as they are generated by intersecting the 1-forms. But I am not quite getting how the result is evaluated. The result (2-form) should map two vectors as input to a number. And I don't see how these intersections do that. While for 1-forms you count the numbers of planes a vector pierced.</p>
<ol start="2">
<li>Why does it make sense that <span class="math-container">$d(d\alpha)=0$</span> for every differential form <span class="math-container">$\alpha$</span></li>
<li>What does Dan Piponi mean by saying: "exterior derivative is none other than finding the boundary of the picture" (4 Exterior Derivatives)</li>
<li>Understand part 5 about Stokes' theorem from Dan Piponi's paper</li>
</ol>
<p>Note: I should maybe add that I have no background in physics, so I didn't understand a lot of the stuff in Gravitation. I just tried to read it after I couldn't quite understand other source since it was cited there.</p>
<p>Similar questions:</p>
<ul>
<li><p><a href="https://math.stackexchange.com/q/548131/445105">What's the geometrical intuition behind differential forms?</a></p>
<p>Edit (since someone voted "close", based on this question as a duplicate):
This question indicates not grasping how 1-forms work ("families of surfaces [...] Why do this interpretation makes any sense?") not only does that invite explanations for 1-forms and hand-waving away the rest with "it works similarly in higher dimensions" but it in particular does not mention specific visualizations for 2-forms which kind of show that there <em>should</em> be an intuition for 2-forms (and maybe higher). And while this question accepted an answer already, this answer does not help to answer the (enumerated) questions above. So this is absolutely not a duplicate.</p>
</li>
<li><p><a href="https://math.stackexchange.com/q/440816/445105">Geometric understanding of differential forms.</a></p>
</li>
<li><p><a href="https://math.stackexchange.com/q/206074/445105">Visualizing Exterior Derivative</a></p>
</li>
</ul>
| Deane | 10,584 | <p>Below, a tensor is an alternating multilinear function on a vector space, and a form is an assignment of a tensor to the tangent space at each point of a manifold.</p>
<p>A finite dimensional vector space can be viewed as a space with an origin and arrows. Even though the dual space is also a vector space, I don't find it useful to view it the same way. Instead, I view a nonzero 1-tensor as a measuring instrument. It defines a way of measuring the spacing between two parallel (affine) hyperplanes in the vector space itself with respect to a unit of distance. An abstract vector space along has no units of distance. A choice of a nonzero 1-tensor defines both the set of parallel hyperplanes, as well as the units used for measuring the spacing.</p>
<p>As for a exterior <span class="math-container">$2$</span>-tensor (<span class="math-container">$2$</span>-form) that is decomposable (i.e., it can be written as a wedge product of <span class="math-container">$1$</span>-tensors), I view it as a way to measure the (oriented) area of a parallelogram defined by two vectors in the space. It captures the fundamental properties of area. One then observes that such tensors can be extended to an abstract vector space. Again, rescaling a <span class="math-container">$2$</span>-tensor corresponds to changing the units with respect to how the area is measured.</p>
<p>The higher order exterior tensors are similar.</p>
<p>From this point of view, the wedge product is simply a way to take a <span class="math-container">$k$</span>-form (i.e., a a way to measure the <span class="math-container">$k$</span>-dimensional volume of a <span class="math-container">$k$</span>-dimensional parallelotope) and an <span class="math-container">$l$</span>-form (i.e., a a way to measure the <span class="math-container">$l$</span>-dimensional volume of a <span class="math-container">$l$</span>-dimensional parallelotope) and use the two together to measure <span class="math-container">$(k+l)$</span>-dimensional volumes. This construction satisfies many remarkable properties that are all captured algebraically in the definition of the wedge product and the properties implied by it.</p>
<p>I view the exterior derivative of a differential form to be the "right way" to extend the fundamental theorem of calculus to higher dimensions. If you write the obvious extension of the FTC to a rectangular region in a way that does not favor any particular direction of integration and that captures orientation properly, the formula for exterior differentiation arises naturally. That <span class="math-container">$d^2 = 0$</span> indicates that if you apply the fundamental theorem of calculus twice, you get nothing because the boundary of the boundary of a region is empty.</p>
<p>As you can see, I do not try to visualize forms as geometric objects. They are instead practical ways to perform measurements on the geometric objects in the vector space.</p>
|
1,154,592 | <p>I was doing some basic Number Theory problems and came across this problem :</p>
<blockquote>
<p>Show that if $a$ and $n$ are positive integers with $n\gt 1$ and $a^{n} - 1$ is prime, then $a = 2$ and $n$ is prime</p>
</blockquote>
<p><strong>My Solution : (Sloppy)</strong></p>
<blockquote>
<ul>
<li>$a^{n}-1$ = $(a-1)$ . $(a^{n-1} + a^{n-2} + ... + a + 1)$</li>
<li>This means that $(a-1)$ | $(a^{n}-1)$ </li>
<li>But $(a^{n}-1)$ is prime</li>
<li>So , $(a-1)$ = 1 $\Rightarrow$ $a = 2$</li>
<li>Now , let $n$ be composite</li>
<li>$n = kl$ , where $1 \lt k \lt n$ and $1 \lt l \lt n$</li>
<li>Now , $a^{kl}-1$ = $(a^{k}-1)$ . $(a^{k.{(l-1)}} + a^{k.{(l-2)}} + ... + a^k + 1)$</li>
<li>This means that $2^{n} -1$ is composite </li>
<li>Hence , we have achieved a contradiction</li>
</ul>
</blockquote>
<p><strong>My Question :</strong>
Am I correct ?</p>
| Community | -1 | <p>I'm a little late to the party, but here is a way to justify this using a contrapositive statement.</p>
<p>Let <span class="math-container">$a,n\in\mathbb{Z}^+$</span>, where <span class="math-container">$n>1$</span>, and assume <span class="math-container">$a^n-1$</span> is prime. Show <span class="math-container">$n$</span> is prime by the contrapostive, i.e., show if <span class="math-container">$n$</span> is composite, then <span class="math-container">$a^n-1$</span> is composite. Since <span class="math-container">$n$</span> is composite, <span class="math-container">$n=ks,$</span> where <span class="math-container">$k,s,\in\mathbb{Z}$</span>.
Now, <span class="math-container">$a^n-1=a^{ks}-1=(a^k)^s-1 =(a^k-1)[(a^k)^{s-1}+(a^k)^{s-2}+\cdots+a+1)$</span>. Clearly <span class="math-container">$a^n-1$</span> is composite.</p>
<p>We can directly show <span class="math-container">$a=2$</span>. Since <span class="math-container">$a^n-1$</span> is prime, <span class="math-container">$a^n-1=(a-1)(a^{n-1}+a^{n-2}+\cdots+a+1)$</span> is also prime. Since the R.H.S. expression is prime, <span class="math-container">$a-1$</span> must equal <span class="math-container">$1$</span>, showing <span class="math-container">$a=2$</span>.</p>
<p><span class="math-container">$\therefore$</span> if <span class="math-container">$a^n-1$</span> is prime, <span class="math-container">$n>1$</span> is prime and <span class="math-container">$a=2$</span>.</p>
|
1,684,741 | <p>I'm able to show it isn't absolutely convergent as the sequence $\{1^n\}$ clearly doesn't converge to $0$ as it is just an infinite sequence of $1$'s. How do I prove the series isn't conditionally convergent to prove divergence!</p>
| Lubin | 17,760 | <p>The only <em>searching</em> that needs to be done is to find where $4$ sits in $\Bbb F_7^\times$ with respect to a generator of this cyclic group of order six.</p>
<p>Now, $\Bbb F_7^\times$ has only the two generators $3$ and $5$, and $4$ is the square of $5$ modulo $7$. That is, calling $5=g$, we have $4\equiv g^2\pmod7$.</p>
<p>Equations in a cyclic group of order six can be reinterpreted as equations in the standard additive cyclic group $\Bbb Z/6\Bbb Z$. So, $4$, being the <em>square</em> of the generator, can be reinterpreted as <em>twice</em> the generator $1$ of $\Bbb Z/6\Bbb Z$. So our original equation $4\equiv y^4\pmod7$ to be solved for $y$ gets reinterpreted as $2\equiv4x\pmod6$. This is equivalent to $1\equiv2x\pmod3$, which you know has the single congruence-solution $x\equiv2\pmod3$. This is your solution: $y^4\equiv4\pmod7$ if and only if $y=g^x$ with $x\equiv2\pmod3$, in other words $x\equiv2,5\pmod6$, in other words $y\equiv5^2\text{ or }5^5\pmod7$, in other words $y\equiv4,3\pmod7$.</p>
|
4,050,893 | <p>Given a linear transformation <span class="math-container">$T: V \rightarrow W$</span> where <span class="math-container">$V$</span> and <span class="math-container">$W$</span> are finite dimensional, then is it true that nullity(<span class="math-container">$T$</span>) = nullity(<span class="math-container">$[T]_\beta$</span>) for any basis <span class="math-container">$\beta$</span> of <span class="math-container">$V$</span>?</p>
<p>A little embarrassing that I'm 20 weeks into studying linear algebra and I've forgotten how to prove this (if I ever knew how). I think this is true, and if it is, I feel like the reasoning is extremely simple, but I can't seem to rigorously spell it out.</p>
<p>Is there a rigorous (and I'm expecting pretty simple) explanation for why this is true?</p>
<p>Thank you!</p>
| user1551 | 1,551 | <p>Here are two simplifications:</p>
<ol>
<li>Up to similarity, there are only six choices of <span class="math-container">$D$</span>. Note that the fourth equation is a Sylvester equation of the form <span class="math-container">$DX+XD=I$</span>. By considering the Jordan form of <span class="math-container">$D$</span> over the algebraic closure of <span class="math-container">$\mathbb Z_2=GF(2)$</span>, we see that <span class="math-container">$DX+XD=I$</span> is solvable only if all Jordan blocks of <span class="math-container">$D$</span> have even sizes. It follows that the characteristic polynomial <span class="math-container">$p(x)$</span> of <span class="math-container">$D$</span> can always be factorised as
<span class="math-container">$$
p(x)=(x+\lambda)^2(x+\mu)^2
=(x^2+\lambda^2)(x^2+\mu^2)
=x^4+(\lambda^2+\mu^2)x^2+\lambda^2\mu^2.\tag{$\ast$}
$$</span>
Since <span class="math-container">$p$</span> has coefficients in <span class="math-container">$GF(2)$</span>, there are only four possible choices of <span class="math-container">$p$</span>, namely <span class="math-container">$x^4,\,x^2(x+1)^2,\,(x+1)^4$</span> or <span class="math-container">$x^4+x^2+1$</span>. In other words, up to similarity, we can always assume that <span class="math-container">$D$</span> is equal to one of the five Jordan forms
<span class="math-container">$$
J_4(0),\,\pmatrix{J_2(0)\\ &J_2(0)},\,\pmatrix{J_2(0)\\ &J_2(1)}\,\pmatrix{J_2(1)\\ &J_2(1)}\, J_4(1)
$$</span>
(where <span class="math-container">$J_m(\lambda)$</span> denotes the <span class="math-container">$m\times m$</span> Jordan block with eigenvalue <span class="math-container">$\lambda$</span>) or the companion matrix
<span class="math-container">$$
\pmatrix{0&0&0&1\\ 1&0&0&0\\ 0&1&0&1\\ 0&0&1&0}.
$$</span></li>
<li>You may replace <span class="math-container">$D$</span> by <span class="math-container">$D+I$</span>. With the new <span class="math-container">$D$</span>, the first and the forth equations then remain unchanged, but the second and the third equations become
<span class="math-container">\begin{cases}
BCB+ABD=0,\\
DCA+CBC=0,\\
\end{cases}</span>
which involve fewer summands.</li>
</ol>
<p>There are also three observations of which I don't know the usefulness. First, (with the old <span class="math-container">$D$</span> or the new <span class="math-container">$D$</span>) the second and the third equations together give the Sylvester equation <span class="math-container">$D(CAB)+(CAB)D=0$</span>. Second, the second or the third equation implies that at least one of <span class="math-container">$B$</span> or <span class="math-container">$C$</span> is necessarily singular when one of <span class="math-container">$A$</span> or <span class="math-container">$D$</span> is singular. Finally, the LHS of the first equation can be rewritten as
<span class="math-container">$$
(I+A)A+A(I+BC)+(I+BC)(I+A)=0,
$$</span>
where the LHS is a cyclic sum.</p>
|
1,357,638 | <p>Here is the problem:</p>
<p>Suppose $n$ people are at a party, and some number of them shake hands. At the end of the party, each guest $G_i$, $1 \leq i \leq n$ shares that they shook hands $x_i$ times. Assume there were a total of $h \geq 0$ handshakes at the party. Use induction on $h$ to prove that:</p>
<p>$x_i + \cdots + x_n = 2h$</p>
<p>I am a little confused about the best way to solve this problem. Performing induction on $h$, means that I will assign my base case for $h=1$. The base case then suggests:</p>
<p>$x_i + \cdots + x_n = 2h$</p>
<p>$x_i + \cdots + x_n = 2$</p>
<p>I am I correct to assume that this means that for my base case for $1$ handshake, that there are 2 ($x_1$ + $x_2$) people who shook hands?</p>
<p>Based on this assumption, my next step would be to assume $h=k, \ \forall \ k \in \mathbb{Z} $. Performing induction on $h$, I would then let $h=(k+1)$ and I want to show that the left side of the equation is equivalent to $2(k+1)$:</p>
<p>$x_i + \cdots + x_n + 2 = 2(k+1)$</p>
<p>^This is the step I'm not really certain of. Since I know that when I add one additional hand shake, that means I am adding $2$ people to the situation... so, would adding $2$ to the left side be a fair way to show this induction? If I do this, then I easily come up with:</p>
<p>$2k + 2 = 2(k+1)$</p>
<p>$ 2k + 2 = 2k + 2$</p>
<p>Which is a true statement, so then I would have (hopefully...) proved this by induction.</p>
<p>What do you guys think? I have traditionally only performed induction on more straight forward math-y problems, so I'm not really sure if I am headed down the right rabbit hole on this one. I am currently in an Introduction to Proofs course at my University. This is a homework problem- I am obviously not looking for someone to do my homework for me (since that wouldn't bode well for my next exam...) but I just want to seek out some advice for my approach to this problem.</p>
<p>Thanks for looking!</p>
| Kopper | 5,218 | <p>This is the right idea. Your base case is correct, and your argument is the right one, modulo some technical points:</p>
<ol>
<li>You don't want to assume $h=k$ $\forall k \in \mathbb{Z}$. That doesn't really make sense and isn't what you are really doing. You need to assume the claim holds for $1\leq h \leq k$, i.e. that $x_1 + \cdots + x_n = 2h$ when there are $h$ handshakes for all $1 \leq h \leq k$. The point of induction is to show that this holds for $h=k+1$, i.e.
$$x_1 + \cdots + x_n = 2(k+1)$$
when there are $k+1$ handshakes.</li>
<li>For clarity you might say, for the inductive step, <em>to add a handshake, two people must shake hands with each other.</em> Say person 1 and person 2 are this new handshake. Then we consider the sum
$$(x_1 + 1) + (x_2 + 1) + x_3 + \cdots + x_n.$$
This is obviously $2+x_1 + \cdots + x_n = 2 + 2k$ by the inductive hypothesis, and $2+2k = 2(k+1)$ as you have observed.</li>
</ol>
|
51,246 | <p>In undergraduate courses we compute the sum $S$ of some series
of the form $\frac{1}{P(n)}$ where $P(x)$ is some simple
polynomial of degree $2$ with integer coefficients, by the following procedure:</p>
<p>(sketch)</p>
<p>(a) Choose an appropriate periodic function $f(x)$ defined over a domain $D.$</p>
<p>(b) Compute the Fourier series $S(x)$ of $f(x).$</p>
<p>(c) Choose a suitable $x$ in $D$ so that we obtain a linear equation for $S.$</p>
<p>(d) Solve the equation to get $S.$</p>
<p>Example: </p>
<p>When $P(x)=x^2+1$ we can take:</p>
<p>$f(x)= \exp(x),$
$D= [-\pi,\pi[$,
and $x=\pi.$ </p>
<p>$S$ is the sum from $n=1$ to infinity
of $\frac{1}{n^2+1}.$</p>
<p>We get the equation: </p>
<p>$$
ch(\pi) = S(\pi) = 2\frac{sh(\pi)}{\pi}(\frac{1}{2}+S)
$$
that gives
$$
S=\frac{1}{2}(\frac{\pi}{th(\pi)}-1).
$$</p>
<p>($ch,sh,th$ denote the classic hyperbolic functions)</p>
<p>Question:</p>
<p>Why this fails (in general) for polynomials $P(x)$ of degree $3.$ ? </p>
<p>Why this fails for the polynomial $P(x)=x^3.$ ?</p>
| Gerry Myerson | 3,684 | <p>Perhaps it fails because if it worked it would give an answer to a question that doesn't have one. </p>
<p>To be a little less cryptic, if there isn't any evaluation of the sum of the reciprocal cubes in terms of, say, the functions of 1st year calculus, then no method that yields only that kind of function is going to succeed in evaluating the sum. </p>
<p>And if you want to know why there should be no evaluation of that sum in those terms, well, when you find out, please let the rest of us know!</p>
|
43,611 | <p>I posted this on Stack Exchange and got a lot of interest, but no answer.</p>
<p>A recent <a href="http://people.missouristate.edu/lesreid/POW12_0910.html" rel="nofollow">Missouri State problem</a> stated that it is easy to decompose the plane into half-open intervals and asked us to do so with intervals pointing in every direction. That got me trying to decompose the plane into closed or open intervals. The best I could do was to make a square with two sides missing (which you can do out of either type) and form a checkerboard with the white squares missing the top and bottom and the black squares missing the left and right. That gets the whole plane except the lattice points. This seems like it must be a standard problem, but I couldn't find it on the web. <strong>Question:</strong> So can the plane be decomposed into unit open intervals? closed intervals?</p>
| Jeff Strom | 3,634 | <p>Conway and Croft show it can be done for closed intervals and cannot
be done for open intervals in the paper:</p>
<p><a href="https://doi.org/10.1017/S0305004100038263" rel="nofollow noreferrer">Covering a sphere with congruent great-circle arcs.
Proc. Cambridge Philos. Soc. 60, 1964, pp787–800</a>.</p>
|
4,031,476 | <p>I recently completed a variation of a problem I found from a mathematical olympiad which is as follows:</p>
<p>Prove that, for all <span class="math-container">$n \in \mathbb{Z}^+$</span>, <span class="math-container">$n \geq 1$</span>, <span class="math-container">$$\sum_{k=1}^n \frac{k}{2^k} < 2 $$</span></p>
<p>I would like to hear comments on my method of proving the inequality above as well as ways of improving it or making it more rigorous. Thank you!</p>
<p>Proof: We begin by noticing that the series above is a Riemann sum (or approximation by areas of rectangles) for the integral <span class="math-container">$$ \int_1^n \frac{x}{2^x} dx$$</span> where the right endpoints are taken, hence <span class="math-container">$$\sum_{k=1}^n \frac{k}{2^k} < \int_1^n \frac{x}{2^x} dx$$</span> for all <span class="math-container">$n$</span>.</p>
<p>By evaluating the original integral, it can be shown that <span class="math-container">$$ \int_1^n \frac{x}{2^x} dx = \frac{n \ln 2 + 1}{2^n (\ln 2)^2} + \frac{\ln 2 + 1}{2 (\ln 2)^2}$$</span></p>
<p>As <span class="math-container">$n$</span> approaches infinity, the integral approaches the value <span class="math-container">$\frac{\ln 2 + 1}{2 (\ln 2)^2}$</span>, which is less than <span class="math-container">$2$</span>. Hence, it can be deduced that <span class="math-container">$$ \int_1^n \frac{x}{2^x} dx < 2$$</span> for a large enough <span class="math-container">$n$</span>.</p>
<p>By inspection and calculation, it is obvious that <span class="math-container">$$ \int_1^n \frac{x}{2^x} dx < 2$$</span> when <span class="math-container">$n > 5, n \in \mathbb{Z}^+$</span>. Hence, when <span class="math-container">$1 \leq n \leq 5$</span>, the integral has a value that is greater than <span class="math-container">$2$</span>.</p>
<p>However, it can be verified by calculation that for <span class="math-container">$1 \leq n \leq 5$</span>, <span class="math-container">$$\sum_{k=1}^n \frac{k}{2^k} < 2 $$</span> Since the statement holds true for <span class="math-container">$1 \leq n \leq 5$</span> as well as for <span class="math-container">$n > 5$</span>, where <span class="math-container">$n \in Z^+$</span>, the statement must be true for all <span class="math-container">$n \geq 1$</span>. Hence, the proof is complete.</p>
<p>Hope to hear your thoughts on this!</p>
| student | 11,211 | <p>As commented by Albus, the values can be found.</p>
<p>In fact, let <span class="math-container">$p \in (0,1)$</span>, then</p>
<p><span class="math-container">$$\sum_{k=1}^{n}kp^k\le \sum_{k=1}^{\infty}kp^k=p\sum_{k=1}^{\infty}kp^{k-1}=p\frac{d}{dp}\left(\sum_{k=1}^{\infty}p^k\right)=p\frac{d}{dp}\frac{p}{1-p}=\frac{p}{(1-p)^2}.$$</span></p>
<p>Therefore, when <span class="math-container">$p=1/2$</span>, the R.H.S. of the above is <span class="math-container">$2$</span>, which proves your inequality.</p>
<p>In general, you can use integration to estimate the partial sum, as you did in your post.</p>
|
10,977 | <p>When I taught calculus, I posted my notes after the lecture. Then I had the students fill out a mid-quarter evaluation, and a lot of them wanted me to post my notes before class.</p>
<p>What I started doing was printing and handing out the notes to them, leaving the examples blank so they can fill those in. Many of them expressed to me that they liked it, so that they can concentrate on the problem instead of trying to write everything down. </p>
<p>I just read my course evaluations and I got mixed reviews about the notes. Some students said they liked them and found them helpful, and an equal amount said that they didn't like it and preferred to take their own notes, and learned better that way. I did tell them that they don't have to use me notes if they don't want them, and to let me know if they don't want to use them so I can save paper.</p>
<p><strong>In your experience, what has worked well, to provide notes, let the students make their own notes, or give them the option by posting it before class to they can print out notes if they want to use it?</strong> I usually print out the notes, a copy for each student, and pass it out to all of them individually, but I am wondering if I should change this.</p>
| Marian Minar | 6,845 | <p><strong>My answer</strong></p>
<p>In my experience, I have used guided notes with success, as long as students straddled the "sweet spot" of giving them ample opportunities to interact with the topic and avoiding the creation of mindless, note-taking robots. With guided notes, my guess is that your goal is to accelerate students by saving them time from writing. I think that this is a valuable tool for long paragraphs of text, and you should continue using it as long as it serves a purpose in your plan. I personally prepare the notes before hand, but this may lead to wastage in some cases.</p>
<p><strong>Extended Answer</strong></p>
<p>I think it's helpful to understand the stages of learning for a particular concept and to <strong>understand what role your lecture (and guided notes) plays in this development</strong>. There is an anxiety that wells up in us to concentrate and distill our lessons into the shortest time possible. You only have students for approximately 45 to 90 minutes, but successful learning of a concept could take somewhere between 120 to 300 minutes (my own estimation). In short, students will probably go through the following stages (read: Bloom's taxonomy):</p>
<ol>
<li>Basic Knowledge, including definitions and new symbolic representation</li>
<li>Understanding and Application</li>
<li>Connections to other fields of study (synthesizing knowledge)</li>
<li>Analyzing and critiquing</li>
</ol>
<p>Consider: <strong>in designing your lecture or lesson</strong>, to what part of these stages will you apply the time that you have with them? How much preparation will (can) be done before the lecture to prepare students? Maybe one of your lectures will focus only on one of the stages, having already given students an assignment to understand the definitions and foundations of your lecture (i.e. the whole hour can be devoted to application and analysis).</p>
<p><strong>Learning can be empirically improved</strong> (search "effect of feedback on learning") when timely and appropriate feedback is given to students. An idea: stop your lecture, and give students several multiple choice questions to answer on the screen; connect it to an online polling system so that they can see their own (anonymous) results. Then, you can make the best use of your time as a lecturer in the remaining amount of time, either focusing on clarification or continuing on with the lesson in confidence that everyone has the prerequisite understanding. Use this with your guided notes and you'll have a lecture to be envied.</p>
|
1,579,528 | <p>You decide to play a holiday drinking game. You start with 100 containers of eggnog in a row. The 1st container contains 1 liter of eggnog, the 2nd contains 2 liters, all the way until the 100th, which contains 100 liters. You select a container uniformly at random and take a one liter sip from it. If the container is empty after taking this sip, you remove it from the row and select only from the remaining bottles. You continue this process until there is only 1 bottle remaining. What is the expected number of liters of eggnog in this last bottle? What is this as this as a function of n, the number of starting bottles?</p>
<p>I came up with this problem myself recently, and I'm not really sure how to approach it. I can find the conditional expectation of a bottle given that it is the last one remaining using linearity of expectations, but it's not clear to me how to use this to get the overall expectation. </p>
| BGM | 297,308 | <p>You may think there is a group of numbers, with $k$ number $k$, $k = 1, 2, \ldots, n$, a total of $\frac {n(n+1)} {2}$ numbers in the group. Each permutation of numbers corresponding to a sequence of taking the bottles.</p>
<p>The total number of permutation is given by the multinomial coeffcient:</p>
<p>$$ \frac {\displaystyle \left(\frac {n(n+1)}{2}\right)!} {\displaystyle \prod_{k=1}^n k!}$$</p>
<p>The number of permutation given that the number $i$ is put at the end
$$ \frac {\displaystyle \left(\frac {n(n+1)}{2}-1\right)!} {\displaystyle \prod_{\substack{k=1 \\ k \neq i}}^n k! (i-1)!}$$</p>
<p>So the probability of having number $k$ at the end is</p>
<p>$$ \frac {\displaystyle 2k} {n(n+1)}, k = 1, 2, \ldots, n$$</p>
<p>Note that this has a more intuitive way to interpret: You may also consider fixing the first number - the number of permutation will be the same. The probability is equal to the number of the $k$th bottle divided by the total number of bottles.</p>
<p>Then the expectation is given by</p>
<p>$$ \sum_{k = 1}^n \frac {2k^2} {n(n+1)} = \frac {2} {n(n+1)} \times \frac {n(n+1)(2n+1) } {6} = \frac {2n+1} {3}$$</p>
|
438,263 | <p>Is there a concrete example of a <span class="math-container">$4$</span> tensor <span class="math-container">$R_{ijkl}$</span> with the same symmetries as the Riemannian curvature tensor, i.e.
<span class="math-container">\begin{gather*}
R_{ijkl} = - R_{ijlk},\quad R_{ijkl} = R_{jikl},\quad R_{ijkl} = R_{klij}, \\
R_{ijkl} + R_{iklj} + R_{iljk} = 0.
\end{gather*}</span>
for which there is no metric for which it is the Riemannian curvature tensor?</p>
<p>The existence of such a curvature was already shown by <a href="https://mathoverflow.net/questions/202211/equations-satisfied-by-the-riemann-curvature-tensor">Robert Bryant</a>, however, I'm looking for a concrete example.</p>
| Denis Serre | 8,799 | <p>If you have half-a-dozen open problems in some definite area, why not write an article where you explain them, why there are important, what are the difficulties ? I did that once (<em>Five open problems in compressible mathematical fluid dynamics</em>, Methods and Applications in Analysis, 20 (2013) pp 197-210). Of course, you must agree with the editor. This is often appropriate in proceedings of conferences.</p>
|
438,263 | <p>Is there a concrete example of a <span class="math-container">$4$</span> tensor <span class="math-container">$R_{ijkl}$</span> with the same symmetries as the Riemannian curvature tensor, i.e.
<span class="math-container">\begin{gather*}
R_{ijkl} = - R_{ijlk},\quad R_{ijkl} = R_{jikl},\quad R_{ijkl} = R_{klij}, \\
R_{ijkl} + R_{iklj} + R_{iljk} = 0.
\end{gather*}</span>
for which there is no metric for which it is the Riemannian curvature tensor?</p>
<p>The existence of such a curvature was already shown by <a href="https://mathoverflow.net/questions/202211/equations-satisfied-by-the-riemann-curvature-tensor">Robert Bryant</a>, however, I'm looking for a concrete example.</p>
| Fedor Petrov | 4,312 | <p>Arnold Mathematical Journal has a problems section.</p>
|
438,263 | <p>Is there a concrete example of a <span class="math-container">$4$</span> tensor <span class="math-container">$R_{ijkl}$</span> with the same symmetries as the Riemannian curvature tensor, i.e.
<span class="math-container">\begin{gather*}
R_{ijkl} = - R_{ijlk},\quad R_{ijkl} = R_{jikl},\quad R_{ijkl} = R_{klij}, \\
R_{ijkl} + R_{iklj} + R_{iljk} = 0.
\end{gather*}</span>
for which there is no metric for which it is the Riemannian curvature tensor?</p>
<p>The existence of such a curvature was already shown by <a href="https://mathoverflow.net/questions/202211/equations-satisfied-by-the-riemann-curvature-tensor">Robert Bryant</a>, however, I'm looking for a concrete example.</p>
| Joseph O'Rourke | 6,094 | <p><a href="https://topp.openproblem.net/" rel="nofollow noreferrer">The Open Problems Project (TOPP)</a> is
focussed on discrete and computational geometry.
We (Erik Demaine, Joe Mitchell, and I) started it in 2001 but its <span class="math-container">$78$</span> problems are now only sporadically updated or augmented. I believe my last update was 2020.
It now has a <a href="https://github.com/edemaine/topp" rel="nofollow noreferrer">GitHub Repository</a>.</p>
<p>The annual <a href="https://cccg.ca/" rel="nofollow noreferrer">Canadian Conf. on Computational Geometry</a>
has an open problems session, often feeding into TOPP.</p>
|
143,070 | <p>Suppose whole square and the left square in the diagram below are pullbacks, then we may wonder whether the right square is a pullback. It is usually not the case. </p>
<p><img src="https://i.stack.imgur.com/yhrcd.jpg" alt="square"></p>
<p>Now we seek some addition condition on $X\to Y$ that forces the right square is a pullback too. </p>
<p>My question: is epic a sufficient condition? (If the category is Sets, then yes.)</p>
<p><strong>Added</strong>: Let $P$ be the pullback of the right square, then there exists $B\to P$, and the square $A\to P \to Y$ // $A\to X \to Y$ is a pullback, so we have the following diagram in which the bottom and the whole squares are pullback, so is the upper square. If the category is Sets, $X\to Y$ is surjective then $A\to P $ is also surjective. Since the pullback of $B\to P$ along a surjective map is an bijection, $B\to P$ must be a bijection. This shows the right square of the original diagram is a pullback. We can also see why we consider some nice condition on $X\to Y$.</p>
<p><img src="https://i.stack.imgur.com/Uepcb.jpg" alt="2nd square"></p>
| Eric Wofsey | 75 | <p>Consider the category consisting of that diagram, together with an extra object $P$ with maps $Y\leftarrow P\to C$ that commute with the maps to $Z$. Then in this category, the whole diagram and the left square are pullbacks, $X\to Y$ is epic, but the right square is not a pullback.</p>
|
754,583 | <p>Write <span class="math-container">$$\phi_n\stackrel{(1)}{=}n+\cfrac{n}{n+\cfrac{n}{\ddots}}$$</span> so that <span class="math-container">$\phi_n=n+\frac{n}{\phi_n},$</span> which gives <span class="math-container">$\phi_n=\frac{n\pm\sqrt{n^2+4n}}{2}.$</span> We know <span class="math-container">$\phi_1=\phi$</span>, the <a href="http://en.wikipedia.org/wiki/Golden_ratio" rel="nofollow noreferrer">Golden Ratio</a>, so let's take <span class="math-container">$\phi_n\stackrel{(2)}{=}\frac{n+\sqrt{n(n+4)}}{2}$</span>. (Is that justified?)</p>
<p><a href="http://m.wolframalpha.com/input/?i=%28n%2B%E2%88%9A%28n%28n%2B4%29%29%29%2F2&x=0&y=0" rel="nofollow noreferrer">Wolfram Alpha</a> states that, with <span class="math-container">$(2)$</span>, <span class="math-container">$$\lim\limits_{n\to -\infty}\phi_n=-1.$$</span> Why? Can I infer that this is true for <span class="math-container">$(1)$</span> and, if so, <em>why</em>?</p>
<p><strong>I wonder what happens in <span class="math-container">$(1)$</span> for <span class="math-container">$n\in\mathbb{C}\backslash\mathbb{Z}$</span> too</strong>. I got something horrendous looking in <span class="math-container">$(2)$</span> for <span class="math-container">$n=i$</span>.</p>
<hr>
<p><em>Clarification:</em> I'm trying to <strong>find <span class="math-container">$\phi_n$</span> in terms of <span class="math-container">$n$</span></strong>. See the comments below.</p>
| Steven Stadnicki | 785 | <p>Let $f_x(t) = x+\frac{x}{t}$ (I'm using $x$ rather than $n$ since we're interested in continuous behavior now, and $f_n$ is a little confusing as a family of functions rather than a sequence); the continued fraction thus corresponds to the sequence $\{t_0 = x, t_n = f_x(t_{n-1})\}$. Then $\frac{df}{dt} = -\frac{x}{t^2}$; since the absolute value of this is larger than $1$ in a neighborhood of $t=-1$ once $x$ gets large — and in fact will be larger than $1$ around <em>any</em> finite value of $t$ once $x$ is large enough — then no fixed point of $f_x$ that's bounded as $x\to-\infty$ can be stable. Since $\{t_n\}$, if it converges, must converged to a fixed point of $f_x$, this implies that the sequence can't converge as $x\to -\infty$.</p>
|
418,724 | <p>This question arises in STEP 2011 Paper III, question 2. The paper can be found <a href="http://www.admissionstestingservice.org/our-services/subject-specific/step/preparing-for-step/" rel="nofollow">here</a>. </p>
<p>The first part of the question requires us to prove the result that if the polynomial
$$x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\ldots+a_{0}$$
where each of the $a_{n}$ are integers, has a rational roots if and only if that root is an integer. It does not give a name for this result.<br>
EDIT: user69810 has pointed out that this is in fact the rational root theorem. </p>
<p>We are then to use this result to prove that the polynomial $$x^{n}-5x+7=0$$
has no rational solutions for $n\ge 2$. My argument was the following:<br>
If there exists a rational root, then it must be an integer.<br>
If there exists an integer root, then $$x^{n}=5x-7$$
for some integers $x$ and $n\ge 2$. Then the LHS is divisible by $x$, meaning that $7$ must be divisible by $x$. Therefore $x \in \{-7,-1,1,7\}$. Checking each of these shows that there is no rational root. </p>
<p>Is this solution correct? It isn't the one in the solutions, but if it's right then I think it's more elegant than theirs.</p>
| Simone | 77,622 | <p>I think it's not correct, because you are assuming that if $x|5x-7$ then $x|7$. This is not true in general.Think for example to $2+4$. Of course $6|2+4$ but $6$ doesn't divide $2$ nor $4$. What you can say is that $x(x^{n-1}-5)=7$, but $7$ is prime, then you must have $x=1$ and $x^{n-1}-5=7$ or $x=7$ and $x^{n-1}-5=1$. But both are absurd because $x=1$ and $x=7$ doesn't solve the equation. As you can see the result is quite the same, but the procedure is quite different for what you assumed at the beginning. Just a remark, what you called Eisenstein criterion is Gauss Lemma, Eisenstein criterion is about irriducibility in $\mathbb{Q}[x]$. </p>
|
445,816 | <p>I have to show that</p>
<blockquote>
<blockquote>
<p>$\mathbb{C}=\overline{\mathbb{C}\setminus\left\{0\right\}}$,</p>
</blockquote>
</blockquote>
<p>what is very probably an easy task; nevertheless I have some problems.</p>
<p>In words this means: $\mathbb{C}$ is the smallest closed superset of $\mathbb{C}\setminus\left\{0\right\}$.</p>
| JLA | 30,952 | <p>${\mathbb{C}\setminus\left\{0\right\}}$ is open, $\mathbb{C}$ is closed, and since $\overline{\mathbb{C}\setminus\left\{0\right\}}$ is the smallest closed set containing ${\mathbb{C}\setminus\left\{0\right\}}$, it must be equal to $\mathbb{C}$ (since the only set containing but not equal to ${\mathbb{C}\setminus\left\{0\right\}}$ is $\mathbb{C}$).</p>
|
161,616 | <p>There is a well known result that every one dimensional topological manifold without boundary is homeomorphic either to the circle or to the whole real line. However there is one detail hidden: manifold is understood to be second countable (or paracompact). If we drop this assumption it becomes possible to construct different example, so called <em>open long line</em> or <em>Aleksandroff line</em>. It is defined as $\omega_1 \times [0,1) \setminus \{(0,0)\}$ with suitable order topology. What might be surprising, is that replacing $\omega_1$ by bigger ordinal does not produce manifold anymore (this would produce points with uncountable neighbourhood system). There is also a variant of long line "in both directions". So the natural question is: if we drop the assumption for (one dimensional) manifolds to be second countable, is it possible to characterise all of them?$
Edit: what about two dimensional case? </p>
| Benoît Kloeckner | 4,961 | <p>The one-dimensional case is well known: you have the circle, the line $\mathbb{R}$, the long line $L$ and the long ray $R$. The proof is not that easy to find in the literature since non-metrizable manifolds are (in my opinion) underestimated. It is described in a previous answer, so let me give a few pointers for the two-dimensional case. Before that, let me stress that there is another hypothesis in the usual definition of manifolds that is often overlooked: being Hausdorff. Non-Hausdorff manifolds are interesting too because they appear naturally, e.g. space of leaves of a foliation. I do not know if anyone has studied non-Hausdorff non-metrizable manifolds. That would really be looking for trouble.</p>
<p>Back to the point. There is very, very little hope to classify non-metrizable surfaces: there are many of them, and of various kind. </p>
<p>First one think about easy examples: product of two one-dimensional non-metrizable manifolds; that makes 4 examples. </p>
<p><em>But there are more</em>: take the first octant of the product of two long lines (i.e. $\{(x,y)\in R\times R | y \le x \}$). Or glue a bunch of octants together along their edges. This already makes you a (small) bunch of examples. At that point, I should stress that in the product of two long lines, the diagonal is very different from the coordinate axes. Each axis $A$ is homotopic to a copy of $L$ that is disjoint from $A$, while any embedding of $L$ which is homotopic to the diagonal must meet it on an unbounded set. This comes from the homotopy theory of $L$ and $R$: e.g. there are two homotopy classes of maps $R\to R$: the ones homotopic to a constant, which are exactly the bounded maps, and the ones homotopic to identity, which have an unbounded set of fixed points.</p>
<p><em>But there are more</em>: one can produce many different "long pipes", which are obtained as increasing unions of annuli indexed by $\omega_1$, where at each non-limit ordinal the inclusion is as trivial as one might want, but at each limit ordinal the lower annuli can accumulate to a circle with a segment attached, or to something worse. By choosing the shape of such singularities, and at which limit ordinal they do appear gives you a very large range of long pipes. </p>
<p>The good news is that classifying long pipes is sufficient to get a classification of $\omega$-bounded surfaces (i.e. surfaces in which any countable sequence has an accumulation point; e.g. the long line is $\omega$-bounded but the long ray is not). This is thank to the beautiful "bagpipe theorem" of Nyikos (The theory of non-metrizable manifolds, in K. Kunen and J. Vaughan, eds, “Handbook of Set-Theoretic Topology” (Elsevier, 1984), 633–684) which says that any $\omega$-bounded surface is obtained by gluing finitely many long pipes (the pipes, obviously) to a compact surface with some disks removed (the bag). The bad news is that even a classification of long pipes seems out of reach. If I remember well, it is an open question whether every long pipe contains an embedded long line. </p>
<p>The worse news is: <em>but there are more</em>. $\omega$-bounded surfaces are a very particular kind of surfaces. A non-metrizable surface which is very different from everything above is the <a href="http://en.wikipedia.org/wiki/Pr%C3%BCfer_manifold">Prüfer manifold</a>. Basically, you glue a bunch (i.e., one for each real number) of planes to a half plane in a way that maps half infinite strips to cones, so that the different planes do not interfere two much one with the other. This is a huge, weird space.</p>
<p>But I guess that <em>there are more</em> (if I remember well, it has been proved that there are $2^{\aleph_1}$ pairwise non-homeomorphic non-metrizable surfaces, but I do not know in which axiom system it holds).</p>
|
1,523,392 | <p>This is question 2.4 in Hartshorne. Let $A$ be a ring and $(X,\mathcal{O}_X)$ a scheme. We have the associated map of sheaves $f^\#: \mathcal{O}_{\text{Spec } A} \rightarrow f_* \mathcal{O}_X$. Taking global sections we obtain a homomorphism $A \rightarrow \Gamma(X,\mathcal{O}_X)$. Thus there is a natural map $\alpha : \text{Hom}(X,\text{Spec} A) \rightarrow \text{Hom}(A,\Gamma(X,\mathcal{O}_X))$. Show $\alpha$ is bijective.</p>
<p>I figure we need to start off with the fact that we can cover $X$ with affine open $U_i$, and that a homomorphism $A \rightarrow \Gamma(X,\mathcal{O}_X)$ induces a morphism of schemes from each $U_i$ to $\text{Spec} A$ and some how glue them together. But I have no idea how to show that the induced morphisms agree on intersections. How does this work? </p>
| hm2020 | 858,083 | <p><strong>Answer:</strong> Let <span class="math-container">$A(Y)$</span> be a commutative unital ring and assume you are given a map of commutative unital rings</p>
<p><span class="math-container">$$f: A(Y) \rightarrow \Gamma(X,\mathcal{O}_X).$$</span></p>
<p>Let <span class="math-container">$U:=Spec(A) \subseteq X$</span> be an open affine subscheme. There is an equality
<span class="math-container">$\Gamma(U, \mathcal{O}_X)=A$</span> and you get a restriction map</p>
<p><span class="math-container">$$\rho_U: \Gamma(X, \mathcal{O}_X) \rightarrow \Gamma(U, \mathcal{O}_X)=A$$</span></p>
<p>and a composed map <span class="math-container">$ f_A:=\rho_U \circ f: A(Y) \rightarrow A$</span> and a corresponding map of affine schemes</p>
<p><span class="math-container">$$F_U: Spec(A):=U \rightarrow Y:=Spec(A(Y)).$$</span></p>
<p>Hence for any open affine subscheme <span class="math-container">$U\subseteq X$</span> you get a canonical map <span class="math-container">$F_U: U \rightarrow Y$</span> and it follows (since all maps are canonical, induced by the restriction map <span class="math-container">$\rho_U$</span>) that these maps glue to a well defined map <span class="math-container">$F: X \rightarrow Y$</span>: Given any open affine scheme <span class="math-container">$V:=Spec(B)$</span> and any open affine subscheme <span class="math-container">$W:=Spec(R) \subseteq U \cap V$</span> you get canonical maps</p>
<p><span class="math-container">$$F_U: U \rightarrow Y, F_W:W \rightarrow Y$$</span></p>
<p>and since "all diagrams commute", it follows</p>
<p><span class="math-container">$$F_U \circ i_U = F_W = F_V \circ i_V$$</span></p>
<p>where <span class="math-container">$i_U: W \rightarrow U$</span> is the inclusion <span class="math-container">$W \subseteq U$</span>. Hence the maps <span class="math-container">$F_U$</span> glue to a map <span class="math-container">$F: X \rightarrow Y$</span>.</p>
|
2,416,071 | <p>I have this integral: $\displaystyle \int^{\infty}_0 kx e^{-kx} dx$.</p>
<p>I tried integrating it by parts:</p>
<p>$\dfrac{1}{k}\displaystyle \int^{\infty}_0 kx e^{-kx} dx = ... $. But I'm stuck </p>
<p>now. Can you help me please?</p>
| Ant | 241,144 | <p>Try explicitly writing out what $u$ and $dv$ are in your integration by parts. In this case, you should have $u=kx$, and $dv=e^{-kx}\,dx$. Thus, $du=k\,dx$, and $v=-\frac{1}{k}e^{-kx}$. Now apply the integration-by-parts formula:</p>
<p>$$\int_{0}^{\infty}{u\,dv}={\left.uv\right|}_0^{\infty}-\int_{0}^{\infty}{v\,du}$$</p>
|
3,238,563 | <p>I have a question about a proof I saw in a book about basic algeba rules. The rule to prove is:
<span class="math-container">\begin{eqnarray*}
\frac{1}{\frac{1}{a}} = a, \quad a \in \mathbb{R}_{\ne 0}
\end{eqnarray*}</span></p>
<p>And the proof: </p>
<p><span class="math-container">\begin{eqnarray*}
1 = a \frac{1}{a} \Longrightarrow 1 = \frac{1}{a} \frac{1}{\frac{1}{a}} \Longrightarrow a = a \frac{1}{a} \frac{1}{\frac{1}{a}} \Longrightarrow \frac{1}{\frac{1}{a}} = a
\end{eqnarray*}</span></p>
<p>Why is it allowed to just replace <span class="math-container">$a$</span> with <span class="math-container">$1/a$</span>? What's the explanation behind it? </p>
| Peter Szilas | 408,605 | <p><span class="math-container">$1/a$</span> is the multiplicative inverse <span class="math-container">$a^{-1}$</span> of </p>
<p><span class="math-container">$a (\not =0)$</span>, i .e. <span class="math-container">$a^{-1}a=1$</span>.</p>
<p>Need to show:</p>
<p><span class="math-container">$(a^{-1})^{-1} =a;$</span></p>
<p>Since</p>
<p><span class="math-container">$(a^{-1})^{-1}(a^{-1})=1$</span>;</p>
<p><span class="math-container">$(a^{-1})^{-1}(a^{-1})a=1a=a$</span>;</p>
<p><span class="math-container">$(a^{-1})^{-1}(a^{-1}a)=a;$</span></p>
<p><span class="math-container">$(a^{-1})^{-1}1=(a^{-1})^{-1}= a$</span>.</p>
|
4,452,885 | <p>Let <span class="math-container">$z = e^{i\theta}, \theta \in \mathbb{R}$</span>. Then, does there exist <span class="math-container">$n \in \mathbb{N}$</span> such that:</p>
<p><span class="math-container">$$1 - z^n = re^{2 \pi i \tau}$$</span></p>
<p>for some <span class="math-container">$\tau \in \mathbb{Q}$</span>?</p>
<p>Naturally, this exists if <span class="math-container">$\theta$</span> is a rational multiple of <span class="math-container">$\pi$</span>. However, does this hold for any <span class="math-container">$\theta$</span>?</p>
<p>Although this question appears quite simple, I have no idea how I would approach it, and I suspect that its proof or disproof would be very difficult.</p>
<p>Link to <a href="https://math.stackexchange.com/questions/4452775/does-this-property-of-certain-fields-have-a-better-description?noredirect=1#comment9325845_4452775"><em>motivation</em></a> (it may appear entirely unrelated (it almost is); I wish to show that <span class="math-container">$\mathbb{C}$</span> has a certain property that I defined on fields with the addition of some analysis.)</p>
| aschepler | 2,236 | <p>No, on basis of set cardinalities. The possible pairings of <span class="math-container">$n \in \mathbb{N}$</span> and <span class="math-container">$\tau \in \mathbb{Q}$</span> are countable. Each of those equations has at most <span class="math-container">$n$</span> solutions for <span class="math-container">$z$</span>, so a finite count. So the set of <span class="math-container">$\theta$</span> where the problem has any solution is countable. <span class="math-container">$[0, 2\pi] \subset \mathbb{R}$</span> is uncountable, so it contains many values of <span class="math-container">$\theta$</span> with no solution.</p>
|
4,095,248 | <blockquote>
<p>Suppose that given any <span class="math-container">$\epsilon>0$</span>, <span class="math-container">$$ \sum_{n=1}^\infty P [|X| > n
\epsilon ]< \infty. $$</span> Does this imply that <span class="math-container">$$ E| X| < \infty \quad ?$$</span></p>
</blockquote>
<p>I made the obvious attempt
<span class="math-container">\begin{align*}
E| X| &= \sum_{n=1}^\infty \int 1_{\{|X| \in (\epsilon(n-1), \epsilon n)\}}|X | dP\\
&\le \sum_{n=0} ^\infty \epsilon n P\{|X| \in (\epsilon(n-1), \epsilon n)\} \\
&\le \epsilon \sum_{n=1}^\infty n P\{ |X | \ge \epsilon n \}.
\end{align*}</span>
But this does not seem to leed to the desired conclusion.</p>
<p>Most grateful for any help provided!</p>
| Kavi Rama Murthy | 142,385 | <p>This is false. <span class="math-container">$(0,1)$</span> is totally bounded w.r.t. the usual metric. An equivalent metric is <span class="math-container">$|\frac 1 x-\frac 1 y|$</span> and <span class="math-container">$(0,1)$</span> is not bounded in this metric.</p>
<p>[Definition of equivalent metrics I am using: two metrics are equivalent if they have the same convergent sequeneces with the same limits; equivalently, they have the same open sets].</p>
|
4,095,248 | <blockquote>
<p>Suppose that given any <span class="math-container">$\epsilon>0$</span>, <span class="math-container">$$ \sum_{n=1}^\infty P [|X| > n
\epsilon ]< \infty. $$</span> Does this imply that <span class="math-container">$$ E| X| < \infty \quad ?$$</span></p>
</blockquote>
<p>I made the obvious attempt
<span class="math-container">\begin{align*}
E| X| &= \sum_{n=1}^\infty \int 1_{\{|X| \in (\epsilon(n-1), \epsilon n)\}}|X | dP\\
&\le \sum_{n=0} ^\infty \epsilon n P\{|X| \in (\epsilon(n-1), \epsilon n)\} \\
&\le \epsilon \sum_{n=1}^\infty n P\{ |X | \ge \epsilon n \}.
\end{align*}</span>
But this does not seem to leed to the desired conclusion.</p>
<p>Most grateful for any help provided!</p>
| Moishe Kohan | 84,907 | <p>My guess about this problem is that there are two issues:</p>
<ol>
<li>There was a typo and the problem should read</li>
</ol>
<blockquote>
<p>Prove a metric space is totally bounded iff it is totally bounded in every equivalent metric.</p>
</blockquote>
<p><strong>and</strong></p>
<ol start="2">
<li>The definition of "equivalent" in the problem was not completely standard, the author of the problem was assuming <a href="https://en.wikipedia.org/wiki/Equivalence_of_metrics#:%7E:text=Strong%20equivalence%20of%20two%20metrics,under%20a%20topologically%20equivalent%20metric." rel="nofollow noreferrer">strong equivalence</a> of metrics. (Or, more generally, that the identity map between the two metric spaces <span class="math-container">$(X,d_1)\to (X,d_2)$</span> is a uniform homeomorphism.)</li>
</ol>
<p>With these two corrections, the problem becomes a pleasant exercise.</p>
<p><strong>Addendum.</strong> Another possibility is that the question was meant to be:</p>
<blockquote>
<p>Prove that a metric space is totally bounded if and only if it is bounded in every uniformly equivalent metric.</p>
</blockquote>
<p>Here two metrics <span class="math-container">$d_1, d_2$</span> on a set <span class="math-container">$X$</span> are called uniformly equivalent if both identity maps
<span class="math-container">$$
id: (X, d_1)\to (X,d_2), id: (X, d_2)\to (X,d_1)
$$</span>
are uniformly continuous.</p>
<p>This question again has a positive answer but a proof requires much more work. Since the question is likely used as a homework or an exam problem, I will refrain from writing a proof.</p>
|
187,432 | <p>Can we evaluate the integral using <a href="http://en.wikipedia.org/wiki/Jordan%27s_lemma#Application_of_Jordan.27s_lemma">Jordan lemma</a>?
$$ \int_{-\infty}^{\infty} {\sin ^2 (x) \over x^2 (x^2 + 1)}\:dx$$</p>
<p>What de we do if removeable singularity occurs at the path of integration?</p>
| DonAntonio | 31,254 | <p>Taking
$$C_R:=[-R,-\epsilon]\cup\left(\gamma_\epsilon:=\{z=\epsilon e^{it}\;\;|\;\;0\leq t\leq \pi\}\right)\cup [\epsilon,R]\cup\left(\gamma_R:=\{z=Re^{it}\;\;|\;\;0\leq t\leq \pi\}\right)$$
$$f(z)=\frac{1-e^{2iz}}{z^2(z^2+1)}$$<br>
we get
$$\oint_{C_R}\frac{1-e^{2iz}\,dz}{z^2(z^2+1)}=2\pi i\,Res_{z=i}(f)=2\pi i\frac{1-e^{-2}}{i^2(2i)}=-\pi\left(1-e^{-2}\right)$$
But
$$\oint_{C_R}f\,dz=\int_{-R}^{-\epsilon} f\,dx\,-\int_{\gamma_\epsilon}f\,dz+\int_\epsilon^Rf\,dx+\int_{\gamma_R}f\,dz$$
Using now the nice lemma in the answer <a href="https://math.stackexchange.com/questions/83828/definite-integral-calculation-with-0-pole-and/184874#184874">here</a> , we get
$$\lim_{\epsilon\to 0}\int_{\gamma_\epsilon}f(z)dz=-i\pi\,Res_{z=0}(f)=-i\pi(-2i)=-2\pi$$
And either using Jordan's lemma or directly:
$$\left|\int_{\gamma_R}f(z)\,dz\right|\leq \frac{1+e^{-2R\sin t}}{R^2(R^2-1)}\pi R\xrightarrow [R\to\infty]{}0$$
So passing to the limit when $\epsilon\to 0\,\,,\,\,R\to\infty\,$ and using Cauchy's Integral Theorem, we get:
$$\int_{-\infty}^\infty \frac{2\sin^2x\,dx}{x^2(x^2+1)}=2\pi-\pi(1-e^{-2})=\pi\left(1+\frac{1}{e^2}\right)\Longrightarrow $$
$$\Longrightarrow\int_{-\infty}^\infty\frac{\sin^2 x\,dx}{x^2(x^2+1)}=\frac{\pi}{2}\left(1+\frac{1}{e^2}\right)$$</p>
|
3,700,440 | <p>As stated in the title, it is requested to define a linear transformation <span class="math-container">$T:\Bbb R^3 \to \Bbb R^3$</span> such that the null space of <span class="math-container">$T$</span> is the <span class="math-container">$z$</span>-axis, and the range of <span class="math-container">$T$</span> is the plane: <span class="math-container">$x+y+z=0$</span> </p>
<hr>
<p>I don't really know how to begin with the solution of the exercise, I think that I should try to get a matrix using the standard base, but after that, I don't have any concrete ideas.</p>
| Davide Motta | 405,131 | <p>You have to find a basis of that plane: <span class="math-container">$x+y+z=0$</span> then <span class="math-container">$x=-y-z$</span> so you can pick <span class="math-container">$v_1=(1,-1,0), v_2=(1,0,-1)$</span>. The <span class="math-container">$z$</span>-axis is the vector <span class="math-container">$e_3=(0,0,1)$</span>.</p>
<p>If you define your map in a basis you are done. In particular take the standard basis, then</p>
<p><span class="math-container">$f(e_1)=v_1, f(e_2)=v_2, f(e_3)=(0,0,0)$</span></p>
|
696,511 | <p>Find the absolute maximum and minimum values of the function:</p>
<p>$$f(x,y)=2x^3+2xy^2-x-y^2$$</p>
<p>on the unit disk $D=\{(x,y):x^2+y^2\leq 1\}$.</p>
| Evgeny | 87,697 | <p><strong>Hint</strong>: for the interior of disk you may use usual condition $\nabla f(x,y) = 0$. For the boundary of the disk you may use parametrisation $x = \cos \phi$, $y = \sin \phi$ and minimize it w.r.t. to $\phi$ (remember, it's $2\pi$-periodic, so you should investigate it only on segment $\lbrack 0, 2\pi )$ really). Or you may use Lagrange multipliers method.</p>
|
5,528 | <p>Let H be a subgroup of G. (We can assume G finite if it helps.) A complement of H in G is a subgroup K of G such that HK = G and |H∩K|=1. Equivalently, a complement is a transversal of H (a set containing one representative from each coset of H) that happens to be a group.</p>
<p>Contrary to my initial naive expectation, it is neither necessary nor sufficient that one of H and K be normal. I ran across both of the following counterexamples in Dummit and Foote:</p>
<ul>
<li><p>It is not necessary that H or K be normal. An example is S<sub>4</sub> which can be written as the product of H=⟨(1234), (12)(34)⟩≅D<sub>8</sub> and K=⟨(123)⟩≅ℤ<sub>3</sub>, neither of which is normal in S<sub>4</sub>.</p></li>
<li><p>It is not sufficient that one of H or K be normal. An example is Q<sub>8</sub> which has a normal subgroup isomorphic to Z<sub>4</sub> (generated by i, say), but which cannot be written as the product of that subgroup and a subgroup of order 2.</p></li>
</ul>
<p>Are there any general statements about when a subgroup has a complement? The <A href="http://en.wikipedia.org/wiki/Complement_%28group_theory%29">Wikipedia page</A> doesn't have much to say. In practice, there are many situations where one wants to work with a transversal of a subgroup, and it's nice when one can find a transversal that is also a group. Failing that, one can ask for the smallest subgroup of G containing a transversal of H.</p>
| Alex Collins | 1,713 | <p>A partial answer to this question is known as the Schur-Zassenhaus lemma (or theorem). If N is a normal subgroup of a finite group G whose order is prime to its index (such a subgroup is called a (normal) Hall subgroup of G) then N has a complement in G.</p>
<p>Check the Wikipedia page <a href="http://en.wikipedia.org/wiki/Schur-Zassenhaus_theorem">http://en.wikipedia.org/wiki/Schur-Zassenhaus_theorem</a>. I also think that Rotman's coverage of this in his `Introduction to the Theory of Groups' is particularly good; the chapter on group cohomology IIRC.</p>
|
5,528 | <p>Let H be a subgroup of G. (We can assume G finite if it helps.) A complement of H in G is a subgroup K of G such that HK = G and |H∩K|=1. Equivalently, a complement is a transversal of H (a set containing one representative from each coset of H) that happens to be a group.</p>
<p>Contrary to my initial naive expectation, it is neither necessary nor sufficient that one of H and K be normal. I ran across both of the following counterexamples in Dummit and Foote:</p>
<ul>
<li><p>It is not necessary that H or K be normal. An example is S<sub>4</sub> which can be written as the product of H=⟨(1234), (12)(34)⟩≅D<sub>8</sub> and K=⟨(123)⟩≅ℤ<sub>3</sub>, neither of which is normal in S<sub>4</sub>.</p></li>
<li><p>It is not sufficient that one of H or K be normal. An example is Q<sub>8</sub> which has a normal subgroup isomorphic to Z<sub>4</sub> (generated by i, say), but which cannot be written as the product of that subgroup and a subgroup of order 2.</p></li>
</ul>
<p>Are there any general statements about when a subgroup has a complement? The <A href="http://en.wikipedia.org/wiki/Complement_%28group_theory%29">Wikipedia page</A> doesn't have much to say. In practice, there are many situations where one wants to work with a transversal of a subgroup, and it's nice when one can find a transversal that is also a group. Failing that, one can ask for the smallest subgroup of G containing a transversal of H.</p>
| Marty Isaacs | 9,694 | <p>Given $H \subseteq G$, there are a number of conditions sufficient to guarantee that there exists a $normal$ complement for $G$. One of the more interesting of these is due to Frobenius: Assume that $H \cap H^g = 1$ for all elements $g \in G - H$. Then $H$ has a normal complement in $G$. As yet, there is no proof known that does not use characters.</p>
|
5,528 | <p>Let H be a subgroup of G. (We can assume G finite if it helps.) A complement of H in G is a subgroup K of G such that HK = G and |H∩K|=1. Equivalently, a complement is a transversal of H (a set containing one representative from each coset of H) that happens to be a group.</p>
<p>Contrary to my initial naive expectation, it is neither necessary nor sufficient that one of H and K be normal. I ran across both of the following counterexamples in Dummit and Foote:</p>
<ul>
<li><p>It is not necessary that H or K be normal. An example is S<sub>4</sub> which can be written as the product of H=⟨(1234), (12)(34)⟩≅D<sub>8</sub> and K=⟨(123)⟩≅ℤ<sub>3</sub>, neither of which is normal in S<sub>4</sub>.</p></li>
<li><p>It is not sufficient that one of H or K be normal. An example is Q<sub>8</sub> which has a normal subgroup isomorphic to Z<sub>4</sub> (generated by i, say), but which cannot be written as the product of that subgroup and a subgroup of order 2.</p></li>
</ul>
<p>Are there any general statements about when a subgroup has a complement? The <A href="http://en.wikipedia.org/wiki/Complement_%28group_theory%29">Wikipedia page</A> doesn't have much to say. In practice, there are many situations where one wants to work with a transversal of a subgroup, and it's nice when one can find a transversal that is also a group. Failing that, one can ask for the smallest subgroup of G containing a transversal of H.</p>
| Gabe Conant | 21,240 | <p>A lot can be said in the finitely generated abelian case, just by using the structure theorem. </p>
<p>Call a group <strong>transversal</strong> if every subgroup has a complement; <strong>non-transversal</strong> of it has proper, nontrivial subgroups, none of which has a complement; and <strong>semi-transversal</strong> if it is not transversal but some proper, nontrivial subgroup has a complement.</p>
<p>Take a finite abelian group $G=\mathbb{Z}/p_1^{e_1}\mathbb{Z}\times\ldots\times\mathbb{Z}/p_k^{e_k}\mathbb{Z}$ with $r\geq 0$, distinct primes $p_1,\ldots,p_k$ and $e_i\geq 1$. </p>
<p>Then $G$ is transversal when $e_i=1$ for all $i$, semi-transversal when $k>1$ and $e_i>1$ for some $i$, and non-transversal when $k=1$ and $e_1>1$.</p>
<p>For an infinite finitely generated abelian group $G$, $G$ is non-transversal if $G=\mathbb{Z}$, and semi-transversal otherwise.</p>
<p>Given a finitely generated abelian group $G$ and $S\subseteq G$, call $S$ <strong>independent</strong> if $0\not\in S$ and for all $x_1,\ldots,x_k\in S$, $r_1,\ldots, r_k\in\mathbb{Z}$, we have that $\sum_{i=1}^k r_ix_i=0$ implies $r_ix_i=0$ for all $i$. Call $S$ a <strong>basis</strong> of $G$ if it is independent and $G=\left< S\right>$.</p>
<p>Then if $H\leq G$, $H$ has a complement if and only if $H$ has a basis that can be expanded to a basis of $G$. If $S$ is a basis for $H$ and $S\subseteq T$ for some basis $T$ of $G$, then the complement of $H$ is $\left< T\backslash S\right>$.</p>
<p>So if $G$ is a vector space over $\mathbb{Z}/p\mathbb{Z}$ then any subgroup has a complement, since any subspace has a basis that can be completed to the whole space. My definition of independence is the same as linear independence.</p>
<p>If $G$ is a free module over $\mathbb{Z}/p^n\mathbb{Z}$ then my definition of independence is weaker than linear independence. I would like to say that $H\leq G$ has a complement if and only if it has a module basis, but I can't prove the reverse direction of this.</p>
<p>I know less about the divisible abelian case, except that if $G$ is a divisible abelian group then saying that $H\leq G$ has a complement is the same as saying that $H$ is divisible. In particular $\mathbb{Q}$ and the Prufer $p$-group $\mathbb{Z}(p^\infty)$ are both non-transversal.</p>
|
508,104 | <p>I want to understand more about this proof from Lang's Algebra:</p>
<p>Let $B$ be a subgroup of a free abelian group $A$ with basis $(x_i)_{i=1...n}$. It has already been shown that $B$ has a basis of cardinality $\leq n$.</p>
<blockquote>
<p>...
We also observe that our proof shows that there exists at least one basis
of $B$ whose cardinality is $\leq n$. We shall therefore be finished when we prove
the last statement, that any two bases of $B$ have the same cardinality. Let $S$
be one basis, with a finite number of elements $m$. Let $T$ be another basis, and
suppose that $T$ has at least $r$ elements. It will suffice to prove that $r \leq m$ (one can then use symmetry). </p>
<p>Let $p$ be a prime number. Then <strong>$B/pB$ is a direct sum of cyclic groups of order $p$, with $m$ terms in the sum</strong>. Hence its order is $p^m$. Using the basis $T$ instead of $S$, we conclude that $B/pB$ contains an <strong>$r$-fold</strong> product of cyclic groups of order $p$, whence $p^r \leq p^m$ and $r \leq m$, as was to be shown. (Note that we did not assume a priori that T was finite.) </p>
</blockquote>
<p>I've bolded the parts I'm having trouble with. How do I show the first part and what's an $r$-fold product?</p>
<p>Alternative proofs to the problem welcome.</p>
<p>I know that $pB = \{ \sum_{i} p k_i x_i\ | \sum_{i} k_i x_i \in B\}$ and that it forms a normal subgroup, $A$ being abelian.</p>
| Daniel Fischer | 83,702 | <p>That $S = \{s_1,\,\dotsc,\, s_m\}$ is a basis of $B$ means that every $b \in B$ can be written in a unique way as $b = \sum\limits_{i=1}^m k_i\cdot s_i$ with all $k_i \in \mathbb{z}$. Thus $B$ is the direct sum of $m$ copies of $\mathbb{Z}$,</p>
<p>$$B = \bigoplus_{i=1}^m \mathbb{Z}\cdot s_i.$$</p>
<p>Then we have</p>
<p>$$pB = \bigoplus_{i=1}^m p\mathbb{Z}\cdot s_i,$$</p>
<p>and that yields</p>
<p>$$B/pB \cong \bigoplus_{i=1}^m (\mathbb{Z}/p\mathbb{Z})\cdot s_i,$$</p>
<p>that $B/pB$ is the direct sum of $m$ cyclic groups of order $p$.</p>
<p>Now $T$ is by assumption also a basis of $B$, so we also have</p>
<p>$$B = \bigoplus_{\tau \in T} \mathbb{Z}\cdot \tau,$$</p>
<p>and</p>
<p>$$B/pB \cong \bigoplus_{\tau \in T} (\mathbb{Z}/p\mathbb{Z})\cdot\tau.$$</p>
<p>If $T$ contains at least $r$ elements, say we have $t_1,\,\dotsc,\,t_r \in T$, then</p>
<p>$$\bigoplus_{j=1}^r (\mathbb{Z}/p\mathbb{Z})\cdot t_j$$</p>
<p>is a subgroup of $B/pB$, and thus $B/pB$ contains the direct sum of $r$ cyclic groups of order $p$. Since for finitely many summands/factors the direct sum and direct product of (abelian) groups are isomorphic, it contains a product of $r$ cyclic groups of order $p$, an $r$-fold product of $\mathbb{Z}/p\mathbb{Z}$.</p>
|
3,165,460 | <p>I am reading a survey on Frankl's Conjecture. It is stated without commentary that the set of complements of a union-closed family is intersection-closed. I need some clearer indication of why this is true, though I guess it is supposed to be obvious. </p>
| A. Kriegman | 649,089 | <p>An uncountable set does not have to be dense. A simple example would be to take the half interval <span class="math-container">$[0,\frac{1}{2}]$</span> which is uncountable but not dense in <span class="math-container">$[0,1]$</span>. But we can do even better, because we can find a set that is uncountable and not dense in any open interval. The Cantor set is uncountable and dense nowhere. <a href="https://en.wikipedia.org/wiki/Cantor_set" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Cantor_set</a></p>
|
1,482,644 | <p>I am trying to find the shortest equivalent expression of the following:</p>
<p>((C → D) $\wedge$ (D → C)) $↔$ (C $\wedge$ D ∨ ¬C $\wedge$ ¬D)</p>
<p>I have "simplified" the expression into the following:</p>
<p>(($\neg$C $\vee$ D) $\wedge$ ($\neg$D $\vee$ C)) $↔$ ((C $\wedge$ D) ∨ (¬C $\wedge$ ¬D))</p>
<p>I am able to figure our that the expression on the left hand side of the $↔$ operator requires either D or C to be true to evaluate to true. However, the expression on the right hand side requires both to be either true or both to be false.</p>
<p>However I am not sure on how to find an even shorter equivalent expression than I already have.
Also, would the resulting expression be a tautology or contradiction?</p>
| Community | -1 | <p>Really, the shortest you can really come up with to preserve meaning is:</p>
<p>$$(c \iff d)\iff ((c\ \wedge d) \vee (\neg c \wedge \neg d))$$</p>
<p>Or if you want to have something that is just equivalent to the above:</p>
<p>$$(c \iff d) \iff (c \iff d)$$</p>
<p>Although this losing some of the intended meaning :)</p>
<p>If you want to simplify even further, you can just simplify this to:</p>
<p>$$\mathrm{true}$$</p>
|
1,652,846 | <p>Let $s$ be any complex number, $t = e^s$ and $z = t^{1/t}$. Define the sequence $(a_n)_{n\in\mathbb{N}}$ by $a_0 = z $ and $a_{n+1} = z^{a_n} $ for $n \geq 0$, that is to say $a_n$ is the sequence $z$, $z^z$, $z^{z^z}$, $z^{z^{z^{z}}}$ and so on.</p>
<p>I want to show that the sequence $(a_n)_{n\in\mathbb{N}}$ converges to $t$ <em>if and only if</em> $s$ lies in the unit disk. I know that when the sequence converges the limit is $\frac{W(-\ln(z))}{-\ln(z)}$ where $W$ is the Lambert W function.</p>
<p>I have verified the above statements numerically for several thousand values of $z$ but I have no idea how to actually prove it.</p>
<p>I graphed the natural logs of the limits on my computer. I got what appeared to be the unit disk. </p>
<p>If we take the log of the limit we get $ln(\frac{W(−ln(z)}{−ln(z)})=-ln(−ln(z))−W(−ln(z))+ln(−ln(z))=−W(−ln(z))$</p>
<p>So what I want to prove is equivalent to showing the sequence $a_n$ is convergent if and only if $|W(−ln(z))| \leq 1$ </p>
<p>I was reading the Wikipedia article on the Lambert W Function and I found this proof that the limit $c$, when it exists, is $c= \frac{W(-ln(z))}{-ln(z)}$</p>
<p>$z^c = c\implies z = c^{1/c} \implies z^{-1} = c^{-1/c} \implies 1/z = (1/c)^{1/c} \implies -ln(z) = \frac{ln(1/c)}{c} \implies -ln(z) = e^{ln(1/c)}ln(1/c) \implies ln(1/c) = W(-ln(z)) \implies 1/c = e^{W(-ln(z))} \implies \frac{1}{c} = \frac{-ln(z)}{W(-ln(z)} \implies c = \frac{W(-ln(z)}{-ln(z)}$</p>
<p>I can only assume that at least 1 step is not justified when $|W(-ln(z)| > 1$ though I am not sure which one. I think that part of the problem is the equation $z^c=c$ has a solution for every non-zero complex number $c$ while the sequence $a_n$ only converges for certain special values of z. In other words the convergence of the $a_n$ is a sufficient but not necessary condition for the existence of a solution to the equation.</p>
| cpiegore | 268,070 | <p>I was reading the Wikipedia article on the Lambert W Function and I found this proof that the limit $c$, when it exists, is $c= \frac{w(-ln(z))}{-ln(z)}$</p>
<p>$z^c = c\implies z = c^{1/c} \implies z^{-1} = c^{-1/c} \implies 1/z = (1/c)^{1/c} \implies -ln(z) = \frac{ln(1/c)}{c} \implies -ln(z) = e^{ln(1/c)}ln(1/c) \implies ln(1/c) = W(-ln(z)) \implies 1/c = e^{W(-ln(z))} \implies \frac{1}{c} = \frac{-ln(z)}{W(-ln(z)} \implies c = \frac{W(-ln(z)}{-ln(z)}$</p>
<p>I can only assume that at least 1 step is not justified when $|W(-ln(z)| > 1$ though I am not sure which one.</p>
|
1,119,634 | <p>Find the point on the curve $y=x^2+2$ where the tangent is parallel to the line $2x+y-1=0$</p>
<p>I understand the answer is $(-1,3)$ but I can't find a way to get there... Thanks </p>
| turkeyhundt | 115,823 | <p>Do you know how to find the derivative of a function? The derivative of the curve function will tell you the slope at any point $x$. </p>
<p>So, figure out the slope of $2x+y=1$ and then find the $x$ value of the curve's derivative that returns that same slope. </p>
<p>That will be the $x$ value for your point. You will have to plug that into your curve function to get the $y$ value.</p>
|
316,601 | <p>Can anyone tell me what I am doing wrong? need to prove for $k\ge2$
$$(5-\frac5k )(1+\frac{1}{(k+1)^2}) \le 5 - \frac{5}{k+1}$$$$(5-\frac5k )(1+\frac{1}{(k+1)^2})= 5(1-\frac1k)(1+\frac1{(k+1)^2})$$
$$=5(1+\frac1{k+1)^2}-\frac1k-\frac1{k(k+1)^2})$$
$$= 5(1-\frac{k^2+k+2}{k(k+1)^2})$$
$$=5(1-\frac{k(k+1)}{k(k+1)^2}+\frac2{k(k+1)^2})$$
$$=5(1-\frac{1}{k+1}+\frac2{k(k+1)^2})$$
$$= 5 - \frac5{k+1}+\frac{10}{k(k+1)^2}\le5-\frac5{k+1}$$
which doesn't look true.</p>
| copper.hat | 27,978 | <p>Let $f(x) = (5-\frac{5}{x})(1+\frac{1}{(1+x)^2})-5+\frac{5}{1+x}$. After a little algebra, this gives $f(x) = -\frac{10}{x(x+1)^2}$. Hence $f(x) \leq 0$ when $x>0$.</p>
<p>Hence $f(k) \leq 0$ for $k \geq 2$.</p>
|
316,601 | <p>Can anyone tell me what I am doing wrong? need to prove for $k\ge2$
$$(5-\frac5k )(1+\frac{1}{(k+1)^2}) \le 5 - \frac{5}{k+1}$$$$(5-\frac5k )(1+\frac{1}{(k+1)^2})= 5(1-\frac1k)(1+\frac1{(k+1)^2})$$
$$=5(1+\frac1{k+1)^2}-\frac1k-\frac1{k(k+1)^2})$$
$$= 5(1-\frac{k^2+k+2}{k(k+1)^2})$$
$$=5(1-\frac{k(k+1)}{k(k+1)^2}+\frac2{k(k+1)^2})$$
$$=5(1-\frac{1}{k+1}+\frac2{k(k+1)^2})$$
$$= 5 - \frac5{k+1}+\frac{10}{k(k+1)^2}\le5-\frac5{k+1}$$
which doesn't look true.</p>
| Gigili | 181,853 | <p>$$=5(1+\frac1{(k+1)^2}-\frac1k-\frac1{k(k+1)^2})$$
$$=5(1+\frac{k}{k(k+1)^2}-\frac{(k+1)^2}{k(k+1)^2}-\frac1{k(k+1)^2})$$
$$= 5(1-\frac{k^2+k+2}{k(k+1)^2})$$
$$=5(1-\frac{k(k+1)}{k(k+1)^2}\color{red}{-}\frac2{k(k+1)^2})$$
$$=5(1-\frac{1}{k+1}\color{red}{-}\frac2{k(k+1)^2})$$
$$= 5 - \frac5{k+1}\color{red}{-10}\frac1{k(k+1)^2}$$</p>
<hr>
<p>Where $k \geq 2$ , so $$\frac{-10}{k(k+1)^2} < 0$$
Therefore</p>
<p>$$5 - \frac5{k+1}-\frac{10}{k(k+1)^2} \leq 5 - \frac5{k+1}$$</p>
<hr>
<p>$$(5-\frac5k )(1+\frac{1}{(k+1)^2}) \le 5 - \frac{5}{k+1}$$ Which is the desired result.</p>
|
2,083,460 | <p>While trying to answer <a href="https://stackoverflow.com/questions/41464753/generate-random-numbers-from-lognormal-distribution-in-python/41465013#41465013">this SO question</a> I got stuck on a messy bit of algebra: given</p>
<p>$$
\log m = \log n + \frac32 \, \log \biggl( 1 + \frac{v}{m^2} \biggr)
$$</p>
<p>I need to solve for $m$. I no longer remember enough logarithmic identities to attempt to do this by hand. Maxima can’t do it at all, and Wolfram Alpha <a href="http://www.wolframalpha.com/input/?i=solve+[log+m+%3D+log+n+%2B+%283%2F2%29*log%281+%2B+v%2F%28m%5E2%29%29]+for+m" rel="nofollow noreferrer">coughs up a hairball</a> that appears to be the zeroes of a quartic, with no obvious relationship to the original equation.</p>
<p>Is there a short, tidy solution? Failing that, an explanation of how WA managed to turn this into a quartic, and the quartic itself, would be ok.</p>
| Cleisthenes | 393,542 | <p>If by solve you mean isolate $m$ in terms of $n$ and $v$ you can use the one-to-one property of logarithms:
\begin{align*}
\log m & = \log n + \frac{3}{2} \log \left(1 + \frac{v}{m^2}\right) \\
\log\left(\frac{m}{n}\right) & = \log\left(\left(1+\frac{v}{m^2}\right)^\frac{3}{2}\right) \\
\frac{m}{n} & = \left(1+\frac{v}{m^2}\right)^\frac{3}{2} \\
\frac{m^2}{n^2} & = \left(1+\frac{v}{m^2}\right)^3
\end{align*}</p>
<p>Now the problem is simple algebra in solving a cubic in $m^2$.</p>
|
2,162,452 | <p>Question: Find the slope of the tangent line to the graph of $r = e^\theta - 4$ at $\theta = \frac{\pi}{4}$.</p>
<p>$$x = r\cos \theta = (e^\theta - 4)\cos\theta$$</p>
<p>$$y = r\sin \theta = (e^\theta - 4)\sin\theta$$</p>
<p>$$\frac{dx}{d\theta} = -e^\theta\sin\theta + e^\theta\cos\theta + 4\sin\theta$$
$$\frac{dy}{d\theta} = e^\theta\cos\theta + e^\theta\sin\theta - 4\cos\theta$$</p>
<p>$$\frac{dy}{dx} = \frac{e^\theta(\cos\theta + \sin\theta) - 4\cos\theta}{e^\theta(\cos\theta - \sin\theta) + 4\sin\theta}$$</p>
<p>$$\frac{dy}{dx} = \frac{\sqrt{2}(e^{\frac{\pi}{4}}-2)}{2\sqrt{2}} = \frac{1}{2}e^{\frac{\pi}{4}} - 1$$</p>
<p>When I plugged this problem into Wolfram Alpha (<a href="http://www.wolframalpha.com/input/?i=slope+of+the+tangent+line+to+r+%3D+e%5E(theta)-4+at+theta%3D(pi%2F4)" rel="nofollow noreferrer">http://www.wolframalpha.com/input/?i=slope+of+the+tangent+line+to+r+%3D+e%5E(theta)-4+at+theta%3D(pi%2F4)</a>), it said that the answer was just $e^{\frac{\pi}{4}}$, so I'm confused where I went wrong in my steps. I tried looking over the arithmetic a couple of times but couldn't find an incorrect step. </p>
<p>Any pointers or help would be appreciated - thank you very much!</p>
| mrnovice | 416,020 | <p>Since $r$ and $s$ are roots of the equation, we have that $(x-r)(x-s)=0$</p>
<p>Then $x^{2} - (r+s)x +rs=0$</p>
<p>Comparing coefficients:</p>
<p>$r+s = 2m$</p>
<p>$rs = m^{2} + 2m + 3$</p>
<p>$\Rightarrow (r+s)^{2} = 4m^{2}$</p>
<p>$\Rightarrow r^{2}+ s^{2} = 4m^{2} - 2rs = 4m^{2} -2m^{2} -4m -6$ </p>
<p>$r^{2}+ s^{2} = 2m^{2} -4m -6$</p>
<p>So we now just need to find the minimum value of this function of $m$:</p>
<p>$r^{2}+ s^{2} = 2[(m-1)^{2} -4]$</p>
<p>Which we can clearly see has a minimum of -$8$ at $m=1$</p>
<p>Edit:</p>
<p>But then we also require the initial equation to have real roots </p>
<p>$\Rightarrow 4m^{2} - 4m^{2} -8m - 12 \geq 0$</p>
<p>$\Rightarrow m \leq -\frac{3}{2}$</p>
<p>It is then simple to see by considering the graph of $m$ that $m= -\frac{3}{2}$ would provide the minimum value of $r^{2} + s^{2} = \frac{9}{2}$</p>
|
233,238 | <p>I am just practicing making some new designs with Mathematica and I thought of this recently. I want to make a tear drop shape (doesn't matter the orientation) constructed of mini cubes. I am familiar with the preliminary material, I am just having some difficulty getting it to work.</p>
| Ulrich Neumann | 53,677 | <p>Try</p>
<pre><code>list={{Position,{Code}},{1,{0000,0001}},{2,{0100,0011}},{3,{0110,0111}},{4,{1000,1001}},{5,{1100,1011}},{6,{1110,1111}}}
list /. {a_ , b_List } -> Join[{a}, b]
</code></pre>
|
1,873,180 | <p>The final result should be $C(n) = \frac{1}{n+1}\binom{2n}{n}$, for reference.</p>
<p>I've worked my way down to this expression in my derivation:</p>
<p>$$C(n) = \frac{(1)(3)(5)(7)...(2n-1)}{(n+1)!} 2^n$$</p>
<p>And I can see that if I multiply the numerator by $2n!$ I can convert that product chain into $(2n)!$ like so, also taking care to multiply the denominator all the same:</p>
<p>$$C(n) = \frac{(2n)!}{(n+1)!(2n!)} 2^n =\frac{1}{n+1} \cdot \frac{(2n)!}{n!n!} 2^{n-1} = \frac{1}{n+1} \binom{2n}{n} 2^{n-1}$$</p>
<p>Where did I go wrong? Why can't I get rid of that $2^{n-1}$?</p>
| Ovi | 64,460 | <p>You can combine to get $\int (f(x)-f(x))dx= \int 0 dx=C$.</p>
|
630,838 | <p>I was woundering if anyone knows any good references about Kähler and complex manifolds? I'm studying supergravity theories and for the simpelest N=1 supergravity we'll get these. Now in the course-notes the're quite short about these complex manifolds. I was hoping someone of you guys might know a good (quite complete book) about the subject ?</p>
<p><strong>edit</strong></p>
<blockquote>
<p>I've also posted this on the <a href="https://physics.stackexchange.com/questions/92776/kahler-and-complex-manifolds">physics stackexchange</a>-site, hoping that maybe a string-theorist of supergravity-specialist might be able to provide some information. But from what I'm seeing in the posts here the answers are already very nice! A big thanks in advance, I'll be going trough the sources somwhere in the end of this week or the beginning of next week somewhere!</p>
</blockquote>
| Michael Albanese | 39,599 | <p>Here are some references that I have used in the past for various reasons. They are listed in no particular order.</p>
<ul>
<li>Huybrechts - <em>Complex Geometry: An Introduction</em></li>
<li>Moroianu - <em>Lectures in Kähler Geometry</em> (pdf version available <a href="http://moroianu.perso.math.cnrs.fr/tex/kg.pdf" rel="noreferrer">here</a>, but I believe the book has more details)</li>
<li>Ballman - <em>Lectures on Kähler Manifolds</em> (pdf available <a href="http://people.mpim-bonn.mpg.de/hwbllmnn/archiv/kaehler0609.pdf" rel="noreferrer">here</a>)</li>
<li>Griffiths and Harris - <em>Principles of Algebraic Geometry</em></li>
<li>Demailly - <em>Complex Algebraic and Analytic Geometry</em> (pdf available <a href="http://www-fourier.ujf-grenoble.fr/~demailly/manuscripts/agbook.pdf" rel="noreferrer">here</a>)</li>
<li>Wells - <em>Differential Analysis on Complex Manifolds</em></li>
</ul>
<p>If I had to recommend a single book for you to consult for complex and Kähler geometry, I'd select Huybrechts' book. Having said that, complex and Kähler geometry are incredibly diverse areas, so it is hard to know exactly what it is you are looking for. I know that Huybrechts has at least a comment about SUSY and how it relates to the Kähler identities.</p>
<p>A one-dimensional complex manifold is called a Riemann surface. All Riemann surfaces are Kähler manifolds (you should try to learn why). For this reason, learning about Riemann surfaces is usually considered a good introduction to complex and Kähler geometry. Here are some books on Riemann surfaces, again, listed in no particular order.</p>
<ul>
<li>Forster - <em>Lectures on Riemann Surfaces</em></li>
<li>Donaldson - <em>Riemann Surfaces</em> (pdf available <a href="http://www2.imperial.ac.uk/~skdona/RSPREF.PDF" rel="noreferrer">here</a>, but I believe the book has more details)</li>
<li>Ahlfors & Sario - <em>Riemann Surfaces</em></li>
</ul>
|
4,624,058 | <p>Godsil&Royle <a href="https://doi.org/10.1007/978-1-4613-0163-9" rel="nofollow noreferrer">Algebraic Graph Theory</a> section 2.5 states (slightly paraphrased):</p>
<blockquote>
<p>Let <span class="math-container">$G$</span> be a transitive group acting on a set <span class="math-container">$V$</span>. A nonempty subset <span class="math-container">$S$</span> of <span class="math-container">$V$</span> is a <em>block of imprimitivity</em> for G if for any element <span class="math-container">$g\in G$</span>, either <span class="math-container">$g(S) = S$</span> or <span class="math-container">$g(S)\cap S = \emptyset$</span>. Because <span class="math-container">$G$</span> is transitive, it is clear that the translates of <span class="math-container">$S$</span> form a partition on <span class="math-container">$V$</span>.</p>
</blockquote>
<p>Assuming "translates of <span class="math-container">$S$</span>" refers to all the <span class="math-container">$g(S)$</span>, I understand why those <em>cover</em> <span class="math-container">$V$</span>.</p>
<p><strong>Question:</strong> But how does it follow that any two <span class="math-container">$g(S)$</span> and <span class="math-container">$h(S)$</span>, if not identical, have an empty intersection?</p>
<p>Put another way: why is the situation in this diagram not possible?</p>
<p><a href="https://i.stack.imgur.com/mlqU8m.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mlqU8m.jpg" alt="overlapping g(S) and h(S)" /></a></p>
<p>The diagram shall describe <span class="math-container">$x_1,x_2\in S$</span> both mapped to some <span class="math-container">$x\notin S$</span> by <span class="math-container">$g$</span> and <span class="math-container">$h$</span> respectively where otherwise <span class="math-container">$g(S)\neq h(S)$</span>.
I tried to derive a contradiction from this situation, but failed.</p>
| colt_browning | 446,709 | <p>Consider <span class="math-container">$h^{-1}(g(S))$</span>. On one hand, its intersection with <span class="math-container">$S$</span> is nonempty because <span class="math-container">$g(S)\cap h(S)$</span> is nonempty. On the other hand, it is not <span class="math-container">$S$</span>: consider <span class="math-container">$y$</span> which is in <span class="math-container">$g(S)$</span> but not <span class="math-container">$h(S)$</span>; then <span class="math-container">$h^{-1}(y)$</span> is not in <span class="math-container">$S$</span>. (If such <span class="math-container">$y$</span> doesn't exist, then use the same reasoning with <span class="math-container">$g\leftrightarrow h$</span>, which is possible unless <span class="math-container">$g(S)=h(S)$</span>.)</p>
|
674,448 | <p>Prove $F: \mathbb{R}\to\mathbb{R}$ where $F(x) = \int_a^x f(t)\, dt$ ($a<x$) is surjective. </p>
<p>$f$ is continuous and bounded below by $m>0$. Also $a$ belongs to $\mathbb{R}$ (reals).</p>
| Paramanand Singh | 72,031 | <p>I suppose it is easier to handle it using derivatives. Clearly $F'(x) = f(x) \geq m > 0$ for all $x$ so that $F(x)$ is strictly increasing. We need to show that range of $F(x)$ is whole of $\mathbb{R}$. Since $F$ is increasing it follows that either $F(x) \to L$ or $F(x) \to \infty$ as $x \to \infty$. If $F(x) \to L$ then it is obvious (by mean value theorem) that $$F(x) - F(x/2) = (x/2)F'(c) = (x/2)f(c) \geq mx/2$$ Then as $x \to \infty$ we get LHS as $L - L = 0$ and RHS as $\infty$. Hence it follows that $F(x) \to \infty$ as $x \to \infty$. Similarly it can be proved that $F(x) \to -\infty$ as $x \to -\infty$. By continuity of $F(x)$ and intermediate value theorem it follows that $F(x)$ takes all values between $-\infty$ and $\infty$ so that the range of $F$ is $\mathbb{R}$ and hence $F(x)$ is surjective.</p>
|
2,991,366 | <blockquote>
<p>Consider a point <span class="math-container">$Q$</span> inside the <span class="math-container">$\triangle ABC$</span> triangle, and <span class="math-container">$M$</span>, <span class="math-container">$N$</span>, <span class="math-container">$P$</span> the intersections of <span class="math-container">$\overleftrightarrow{AQ}$</span>, <span class="math-container">$\overleftrightarrow{BQ}$</span>, <span class="math-container">$\overleftrightarrow{CQ}$</span> with respective sides <span class="math-container">$\overline{BC}$</span>, <span class="math-container">$\overline{CA}$</span>, <span class="math-container">$\overline{AB}$</span>.</p>
<p>What are the triangles <span class="math-container">$\triangle ABC$</span>, respectively, which are the positions of <span class="math-container">$Q$</span>, for which <span class="math-container">$Q$</span> is the intersection of the heights (aka, the orthocenter) in <span class="math-container">$\triangle MNP$</span>?</p>
</blockquote>
<p>I have not been able to solve the problem using synthetic geometry elements. In order to solve the problem with methods of analytical geometry, we considered <span class="math-container">$A(0, a)$</span>, <span class="math-container">$B(b, 0)$</span>, <span class="math-container">$C(c, 0)$</span>, and <span class="math-container">$Q(m, n)$</span>, where <span class="math-container">$0<n<a,b<0<c, b<m<c$</span>. We calculated the coordinates of <span class="math-container">$M$</span>, <span class="math-container">$N$</span>, <span class="math-container">$P$</span> and then we set the conditions that the <span class="math-container">$\overleftrightarrow{AQ}$</span>, <span class="math-container">$\overleftrightarrow{BQ}$</span>, <span class="math-container">$\overleftrightarrow{CQ}$</span> are perpendicular to <span class="math-container">$\overline{NP}$</span>, <span class="math-container">$\overline{PM}$</span>, <span class="math-container">$\overline{MN}$</span>. But the calculations became complicated, and I quit.</p>
<p>If I could have continued, I would have found three conditions (equalities) that I would have to satisfy <span class="math-container">$m$</span> and <span class="math-container">$n$</span> simultaneously. It follows that the <span class="math-container">$\triangle ABC$</span> triangle must be a particular triangle; and, within this triangle, <span class="math-container">$Q$</span> must have a particular position. One such case is the equilateral triangle and <span class="math-container">$Q$</span> is its center.</p>
<p>How can triangles be characterized with this property?</p>
| Blue | 409 | <p>Changing notation a bit, we'll consider a triangle with circumdiameter <span class="math-container">$1$</span> and angles <span class="math-container">$\alpha$</span>, <span class="math-container">$\beta$</span>, <span class="math-container">$\gamma$</span>. Duplicating OP's configuration with <span class="math-container">$A$</span> on the <span class="math-container">$y$</span>-axis, we can write
<span class="math-container">$$A = (0,\sin\beta\sin\gamma) \qquad B = (\cos\beta\sin\gamma,0) \qquad C = (-\sin\beta\cos\gamma,0) \tag{1}$$</span>
Now, define our variable point, <span class="math-container">$P$</span>, using barycentric coordinates:
<span class="math-container">$$P = \frac{p A + q B + r C}{p + q + r} \qquad(p, q, r \neq 0) \tag{2}$$</span>
Let <span class="math-container">$\overleftrightarrow{AP}$</span>, <span class="math-container">$\overleftrightarrow{BP}$</span>, <span class="math-container">$\overleftrightarrow{CP}$</span> meet <span class="math-container">$\overleftrightarrow{BC}$</span>, <span class="math-container">$\overleftrightarrow{CA}$</span>, <span class="math-container">$\overleftrightarrow{AB}$</span> at <span class="math-container">$D$</span>, <span class="math-container">$E$</span>, <span class="math-container">$F$</span>. We find that
<span class="math-container">$$
D = \frac{qB+rC}{q+r} \qquad
E = \frac{rC+pA}{r+p} \qquad
F = \frac{pA+qB}{p+q}
\tag{3}$$</span>
For <span class="math-container">$P$</span> to be the orthocenter of <span class="math-container">$\triangle DEF$</span>, we must have <span class="math-container">$\overleftrightarrow{AP}\perp\overleftrightarrow{EF}$</span>, <span class="math-container">$\overleftrightarrow{BP}\perp\overleftrightarrow{FD}$</span>, <span class="math-container">$\overleftrightarrow{CP}\perp\overleftrightarrow{DE}$</span>, where any two guarantee the third. Considering the last two gives equations that reduce to
<span class="math-container">$$\begin{align}
(B-P)\cdot(F-D) &= 0 \quad\to\quad
\overline{p}^2\left(\overline{q} + \overline{r}\cos\alpha \right)
= \overline{r}^2\left(\overline{q} + \overline{p}\cos\gamma \right)
\tag{4}\\
(C-P)\cdot(D-E) &= 0 \quad\to\quad
\overline{q}^2 \left(\overline{r} + \overline{p} \cos\beta \right)
= \overline{p}^2 \left(\overline{r} + \overline{q} \cos\alpha \right)
\tag{5}
\end{align}$$</span>
where <span class="math-container">$\overline{p}:= p/\sin\alpha$</span>, <span class="math-container">$\overline{q}:=q/\sin\beta$</span>, <span class="math-container">$\overline{r}:=r/\sin\gamma$</span>.</p>
<p>Using the method of resultants (that is, the <code>Resultant[]</code> function in <em>Mathematica</em>), we can eliminate <span class="math-container">$\overline{p}$</span> from <span class="math-container">$(4)$</span> and <span class="math-container">$(5)$</span>. The relation reduces to</p>
<blockquote>
<p><span class="math-container">$$\begin{align}
0
&= \phantom{2} \overline{q}^4\phantom{\overline{r}^4} \sin^2\beta \\[4pt]
&+ \phantom{2} \overline{q}^3 \overline{r} \phantom{^4} \cos\beta \left(\cos\gamma - \cos\alpha \cos\beta \right) \\[4pt]
&- 2 \overline{q}^2 \overline{r}^2 \left(1 - \cos\alpha \cos\beta \cos\gamma\right) \\[4pt]
&+ \phantom{2}\overline{q}\phantom{^4} \overline{r}^3 \cos\gamma \left(\cos\beta - \cos\alpha\cos\gamma \right) \\[4pt]
&+ \phantom{2}\phantom{\overline{q}^4}\overline{r}^4 \sin^2\gamma
\end{align} \tag{$\star$}$$</span></p>
</blockquote>
<p>To get an orthocenter configuration, we can take <span class="math-container">$r=1$</span> (so that <span class="math-container">$\overline{r}=1/\sin\gamma$</span>), solve <span class="math-container">$(\star)$</span> for <span class="math-container">$\overline{q}$</span>, then return to <span class="math-container">$(4)$</span> to solve for <span class="math-container">$\overline{p}$</span>. These values in turn give <span class="math-container">$p$</span> and <span class="math-container">$q$</span>, which determine the desired <span class="math-container">$D$</span>, <span class="math-container">$E$</span>, <span class="math-container">$F$</span>. <span class="math-container">$\square$</span></p>
<hr>
<p>There's no bias in <span class="math-container">$(\star)$</span> for isosceles triangles. Indeed, after admittedly scant experimentation, it <em>appears</em> that the four roots are valid for any triangle (barring unanticipated degeneracies), and that they correspond to one "internal" orthocenter and three "external" orthocenters (one per side). For example:</p>
<p><a href="https://i.stack.imgur.com/2KXGP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2KXGP.png" alt="enter image description here"></a></p>
|
4,399,371 | <p>According to my textbook, the formula for the distance between 2 parallel lines has been given as below:</p>
<p><a href="https://i.stack.imgur.com/ZQtQk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZQtQk.png" alt="enter image description here" /></a></p>
<p>Where PT is a vector from the first line that makes a perpendicular on the second line, vector B is a vector to which both the lines are parallel to and vector (a2 - a1) is a vector that joins one arbitrary point on the second line, to yet another arbitrary point on the other</p>
<p>This is what I am confused by. The book, along with the numerous threads I've scoured through already provide similar diagrams for the proof:</p>
<p><a href="https://i.stack.imgur.com/qoua3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qoua3.png" alt="enter image description here" /></a></p>
<p>From what I understand, the crossing of ST with B should yield us a vector pointing OUT of the plane to which the lines (and in conjunction, ST) belong</p>
<p>How would that yield us TP/PT? TP/PT belongs to the same plane to which the lines and ST belong as well, so how'd crossing ST and B yield us PT?</p>
<p>I understand the end goal is to calculate the MAGNITUDE of the shortest vector joining both the lines, but I can't seem to understand how d is the magnitude of PT as opposed to being the magnitude of the vector jutting OUT of the plane</p>
| bubba | 31,744 | <p>The formula uses only the <em>magnitude</em> of the the cross product. And the magnitude of <span class="math-container">$u \times v$</span> is <span class="math-container">$\|u\|$</span> times <span class="math-container">$\|v\|$</span> times the sine of the angle between <span class="math-container">$u$</span> and <span class="math-container">$v$</span>. So, using the cross product is just a fancy way of getting the sine of an angle. And, if you do a bit of trig, you'll see that the sine of an angle is exactly what you need to calculate the desired distance. The direction of the cross product is irrelevant in all of this, so you don't need to worry about the fact that it's perpendicular to the plane in which all the action occurs.</p>
|
119 | <p>One thing I could never wrap my head around is how <code>Flatten</code> works when provided with a matrix as the second argument, and the <em>Mathematica</em> help isn't particularly good on this one.</p>
<p>Taken from the <a href="http://reference.wolfram.com/mathematica/ref/Flatten.html"><code>Flatten</code></a> <em>Mathematica</em> documentation:</p>
<pre><code>Flatten[list, {{s11, s12, ...}, {s21, s22, ...}, ...}]
</code></pre>
<blockquote>
<p>Flattens <code>list</code> by combining all levels $s_{ij}$ to make each level $i$ in the result.</p>
</blockquote>
<p>Could someone elaborate on what this actually means/does?</p>
| WReach | 142 | <p>A second list argument to <code>Flatten</code> serves two purposes. First, it specifies the order in which indices will be iterated when gathering elements. Second, it describes list flattening in the final result. Let's look at each of these capabilities in turn.</p>
<p><strong>Iteration Order</strong></p>
<p>Consider the following matrix:</p>
<pre><code>$m = Array[Subscript[m, Row[{##}]]&, {4, 3, 2}];
$m // MatrixForm
</code></pre>
<p><img src="https://i.stack.imgur.com/ZjUuL.png" alt="matrix result"></p>
<p>We can use a <code>Table</code> expression to create a copy of the matrix by iterating over all of its elements:</p>
<pre><code>$m === Table[$m[[i, j, k]], {i, 1, 4}, {j, 1, 3}, {k, 1, 2}]
(* True *)
</code></pre>
<p>This identity operation is uninteresting, but we can transform the array by swapping the order of the iteration variables. For example, we can swap <code>i</code> and <code>j</code> iterators. This amounts to swapping the level 1 and level 2 indices and their corresponding elements:</p>
<pre><code>$r = Table[$m[[i, j, k]], {j, 1, 3}, {i, 1, 4}, {k, 1, 2}];
$r // MatrixForm
</code></pre>
<p><img src="https://i.stack.imgur.com/eK2Ws.png" alt="matrix result"></p>
<p>If we look carefully, we can see that each original element <code>$m[[i, j, k]]</code> will be found to correspond to the resulting element <code>$r[[j, i, k]]</code> -- the first two indices have been "swapped".</p>
<p><code>Flatten</code> allows us to express an equivalent operation to this <code>Table</code> expression more succintly:</p>
<pre><code>$r === Flatten[$m, {{2}, {1}, {3}}]
(* True *)
</code></pre>
<p>The second argument of the <code>Flatten</code> expression explicitly specifies the desired index order: indices 1, 2, 3 are altered to become indices 2, 1, 3. Note how we did not need to specify a range for each dimension of the array -- a significant notational convenience.</p>
<p>The following <code>Flatten</code> is an identity operation since it specifies no change to index order:</p>
<pre><code>$m === Flatten[$m, {{1}, {2}, {3}}]
(* True *)
</code></pre>
<p>Whereas the following expression re-arranges all three indices: 1, 2, 3 -> 3, 2, 1</p>
<pre><code>Flatten[$m, {{3}, {2}, {1}}] // MatrixForm
</code></pre>
<p><img src="https://i.stack.imgur.com/VNms8.png" alt="matrix result"></p>
<p>Again, we can verify that an original element found at the index <code>[[i, j, k]]</code> will now be found at <code>[[k, j, i]]</code> in the result.</p>
<p>If any indices are omitted from a <code>Flatten</code> expression, they are treated as if they had been specified last and in their natural order:</p>
<pre><code>Flatten[$m, {{3}}] === Flatten[$m, {{3}, {1}, {2}}]
(* True *)
</code></pre>
<p>This last example can be abbreviated even further:</p>
<pre><code>Flatten[$m, {3}] === Flatten[$m, {{3}}]
(* True *)
</code></pre>
<p>An empty index list results in the identity operation:</p>
<pre><code>$m === Flatten[$m, {}] === Flatten[$m, {1}] === Flatten[$m, {{1}, {2}, {3}}]
(* True *)
</code></pre>
<p>That takes care of iteration order and index swapping. Now, let's look at...</p>
<p><strong>List Flattening</strong></p>
<p>One might wonder why we had to specify each index in a sublist in the previous examples. The reason is that each sublist in the index specification specifies which indices are to be flattened together in the result. Consider again the following identity operation:</p>
<pre><code>Flatten[$m, {{1}, {2}, {3}}] // MatrixForm
</code></pre>
<p><img src="https://i.stack.imgur.com/ZjUuL.png" alt="matrix result"></p>
<p>What happens if we combine the first two indices into the same sublist?</p>
<pre><code>Flatten[$m, {{1, 2}, {3}}] // MatrixForm
</code></pre>
<p><img src="https://i.stack.imgur.com/XzPfw.png" alt="matrix result"></p>
<p>We can see that the original result was a 4 x 3 grid of pairs, but the second result is a simple list of pairs. The deepest structure, the pairs, were left untouched. The first two levels have been flattened into a single level. The pairs in the third level of the source matrix remained unflattened.</p>
<p>We could combine the second two indices instead:</p>
<pre><code>Flatten[$m, {{1}, {2, 3}}] // MatrixForm
</code></pre>
<p><img src="https://i.stack.imgur.com/O0TZ2.png" alt="matrix result"></p>
<p>This result has the same number of rows as the original matrix, meaning that the first level was left untouched. But each result row has a flat list of six elements taken from the corresponding original row of three pairs. Thus, the lower two levels have been flattened.</p>
<p>We can also combine all three indices to get a completely flattened result:</p>
<pre><code>Flatten[$m, {{1, 2, 3}}]
</code></pre>
<p><img src="https://i.stack.imgur.com/tJkj2.png" alt="matrix result"></p>
<p>This can be abbreviated:</p>
<pre><code>Flatten[$m, {{1, 2, 3}}] === Flatten[$m, {1, 2, 3}] === Flatten[$m]
(* True *)
</code></pre>
<p><code>Flatten</code> also offers a shorthand notation when no index swapping is to take place:</p>
<pre><code>$n = Array[n[##]&, {2, 2, 2, 2, 2}];
Flatten[$n, {{1}, {2}, {3}, {4}, {5}}] === Flatten[$n, 0]
(* True *)
Flatten[$n, {{1, 2}, {3}, {4}, {5}}] === Flatten[$n, 1]
(* True *)
Flatten[$n, {{1, 2, 3}, {4}, {5}}] === Flatten[$n, 2]
(* True *)
Flatten[$n, {{1, 2, 3, 4}, {5}}] === Flatten[$n, 3]
(* True *)
</code></pre>
<p><strong>"Ragged" Arrays</strong></p>
<p>All of the examples so far have used matrices of various dimensions. <code>Flatten</code> offers a very powerful feature that makes it more than just an abbreviation for a <code>Table</code> expression. <code>Flatten</code> will gracefully handle the case where sublists at any given level have differing lengths. Missing elements will be quietly ignored. For example, a triangular array can be flipped:</p>
<pre><code>$t = Array[# Range[#]&, {5}];
$t // TableForm
(*
1
2 4
3 6 9
4 8 12 16
5 10 15 20 25
*)
Flatten[$t, {{2}, {1}}] // TableForm
(*
1 2 3 4 5
4 6 8 10
9 12 15
16 20
25
*)
</code></pre>
<p>... or flipped and flattened:</p>
<pre><code>Flatten[$t, {{2, 1}}]
(* {1,2,3,4,5,4,6,8,10,9,12,15,16,20,25} *)
</code></pre>
|
3,783,878 | <p>Hey everyone can anyone help me in simplifying the following boolean expression with explanation?</p>
<p><span class="math-container">\begin{equation}[((p\land q)\implies r)\implies((q\land r')\implies r')]\land[(p \land q)\implies(q\iff p)]\end{equation}</span></p>
| Beyond Infinity | 683,570 | <p>Note that <span class="math-container">$(q\land r')\implies r'$</span> is always true. Thus first part of the statement becomes trivial.
Consider the statement <span class="math-container">$(p \land q)\implies(q\iff p)$</span>. If <span class="math-container">$p=q=1$</span>, it is true, otherwise it is vacuous. Hence, both parts of the to-be-simplified statement are always true and hence it is a tautology.</p>
|
286,798 | <blockquote>
<p>Find the limit $$\lim_{n \to \infty}\left[\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdots\left(1-\frac{1}{n^2}\right)\right]$$</p>
</blockquote>
<p>I take log and get $$\lim_{n \to \infty}\sum_{k=2}^{n} \log\left(1-\frac{1}{k^2}\right)$$</p>
| user 1591719 | 32,016 | <p>Your way works nice if you employ the Euler's infinite product for the sine function. Then </p>
<p>$$\lim_{n \to \infty}\sum_{k=2}^{n} \ln\left(1-\frac{1}{k^2}\right)=\lim_{x\to\pi}\ln\left(\frac{\pi^2\sin x}{x(\pi^2-x^2)}\right)=\lim_{y\to0}\ln\left(\frac{\pi^2\sin y}{y(2\pi-y)(\pi-y)}\right)=\ln(1/2)$$
Thus, your product is $1/2$.</p>
|
1,103,533 | <p>I am reading John Lee's book <em>Riemannian Manifolds</em>. On page 91, he begins a chapter called "Geodesics and Distance," which is I think the first chapter that seriously addresses geodesics. </p>
<p>I was very surprised when I came across the following sentence: </p>
<p><em>Most of the results of this chapter do <strong>not</strong> apply to pseudo-Riemmanian metrics, at least not without substantial modification.</em></p>
<p>I thought the only real difference between the two was about the positive-definite constraint. But this makes it sound like there's a whole host of properties that don't apply to pseudo-Riemannian metrics but that do apply to Riemannian metrics. </p>
<p>Can someone clarify this for me? Are the things we can do on Riemannian manifolds that can't be done on pseudo-Riemannian ones? </p>
| Andrew D. Hwang | 86,418 | <p>For one thing, a Lorentz-signature metric on a compact manifold can fail to be geodesically complete. If memory serves, Chapter 3 of <em>Einstein Manifolds</em> by Besse contains an example of a metric on a torus where a finite-length geodesic "winds" infinitely many times.</p>
<p>Generally, the "unit sphere" in a tangent space is non-compact for a metric of indefinite signature (e.g., it's a hyperbola on a Lorentz-signature surface), which can cause all manner of fun.</p>
|
1,103,533 | <p>I am reading John Lee's book <em>Riemannian Manifolds</em>. On page 91, he begins a chapter called "Geodesics and Distance," which is I think the first chapter that seriously addresses geodesics. </p>
<p>I was very surprised when I came across the following sentence: </p>
<p><em>Most of the results of this chapter do <strong>not</strong> apply to pseudo-Riemmanian metrics, at least not without substantial modification.</em></p>
<p>I thought the only real difference between the two was about the positive-definite constraint. But this makes it sound like there's a whole host of properties that don't apply to pseudo-Riemannian metrics but that do apply to Riemannian metrics. </p>
<p>Can someone clarify this for me? Are the things we can do on Riemannian manifolds that can't be done on pseudo-Riemannian ones? </p>
| Jack Lee | 1,421 | <p>OK, I'll accept the challenge...</p>
<p>The biggest difference in the pseudo-Riemannian case is that curves can have zero length, and the "Riemannian distance function" (the supremum of the lengths of curves between two points) is not a metric in the sense of metric spaces. Thus most of the results in Chapter 6 of my book don't make sense if the metric isn't positive definite.</p>
<p>Nonetheless, there is still a lot that can be said, at least in the Lorentz case (pseudo-Riemannian metrics of index $1$). In that case, one has to distinguish curves according to the nature of their velocity vectors: a curve $\gamma$ is <strong><em>spacelike</em></strong> if $g(\dot\gamma,\dot\gamma)>0$, <strong><em>lightlike</em></strong> if $g(\dot\gamma,\dot\gamma)\equiv 0$, and <strong><em>timelike</em></strong> if $g(\dot\gamma,\dot\gamma)<0$. The resulting geometric properties, as you might imagine, are intimately connected with the physics of space-time.</p>
<p>But this is a subject that is quite different from Riemannian geometry, which is why I didn't treat it in my book.</p>
|
4,058,600 | <p>Please pardon the elementary question, for some reason I'm not grocking why all possible poker hand combinations are equally probable, as all textbooks and websites say. Just intuitively I would think getting 4 of a number is much more improbably than getting 1 of each number, if I were to draw 4 cards. For example, ignoring order, to get 4 of a single number there are only <span class="math-container">$4 \choose 4$</span> distinct possibilities, whereas for 1 of each number I would have <span class="math-container">${4 \choose 1}^4$</span> distinct possibilities.</p>
| Neel Sandell | 405,304 | <p>I can try deriving it. Imagine a classic case of a dealer drawing 5 cards from the top of a shuffled deck one at a time.</p>
<p>Assume that there exists an ordering in a hand, so JQK12 is different from 1JQK2.</p>
<p>This means that the probability of choosing a hand is <span class="math-container">$\frac{1}{52} \times \frac{1}{51} \times \frac{1}{50} \times \frac{1}{49} \times \frac{1}{48}$</span>. However, the multiplication operation assumes an ordering. There are <span class="math-container">$5!$</span> ways of drawing the "same" hand. This means that we can multiply our previous product by <span class="math-container">$5!$</span>. After some simplification, you will notice that it is equivalent to <span class="math-container">$\frac{1}{\binom{52}{5}}$</span>.</p>
|
1,534,724 | <p>Let $U$ be a unitary matrix, show that $r(x,y) := x^*Uy$ is an inner product satisfying </p>
<ol>
<li><p>$(u,v) = \overline{(v,u)}$</p></li>
<li><p>$(u,u)> 0$ for $u\neq0$; $(u,u)=0$ for $u= 0$</p></li>
<li><p>$(u+sv,w)=(u,w)+s(v,w)$</p></li>
</ol>
<p>for a complex vector space $V$</p>
<p>Explain why this would not work if $U$ is simply invertible</p>
<p>Note: A matrix $U$ is unitary if $U^*U = I$</p>
<p>Don't know how to start this question... Hope someone can help. Thank you very much!</p>
| Lutz Lehmann | 115,115 | <p>This should not work, matrices to generate scalar products are self-adjoint positive definite (SPD).</p>
<p>Indeed, property 1) translates to $U^*=U$, property 2) to $U>0$ and 3) is generally true for this type of construction.</p>
|
1,534,724 | <p>Let $U$ be a unitary matrix, show that $r(x,y) := x^*Uy$ is an inner product satisfying </p>
<ol>
<li><p>$(u,v) = \overline{(v,u)}$</p></li>
<li><p>$(u,u)> 0$ for $u\neq0$; $(u,u)=0$ for $u= 0$</p></li>
<li><p>$(u+sv,w)=(u,w)+s(v,w)$</p></li>
</ol>
<p>for a complex vector space $V$</p>
<p>Explain why this would not work if $U$ is simply invertible</p>
<p>Note: A matrix $U$ is unitary if $U^*U = I$</p>
<p>Don't know how to start this question... Hope someone can help. Thank you very much!</p>
| egreg | 62,967 | <p>Let's examine your claim, where I guess you define $x^*$ as the conjugate transpose. Now
$$
r(v,u)=v^*Uu=\overline{u^*U^*v}=\overline{r(u,v)}
$$
if and only if
$$
u^*U^*v=u^*Uv
$$
for all $u$ and $v$. This is the same as requiring that $U^*=U$, so $U$ must be Hermitian, not unitary.</p>
<p>However a Hermitian unitary matrix is not really interesting: the eigenvalues of a unitary matrix have modulus $1$, those of a Hermitian matrix are real. Thus a Hermitian unitary matrix can only have $1$ and $-1$ as eigenvalues. If you require it to define an inner product, the eigenvalues must be positive, so you just get the identity.</p>
|
1,677,035 | <p>I'm new to this website so I apologize in advance if what I'm going to ask isn't meant to be posted here.</p>
<p>A bit of background though: I haven't been to school in 6 years and the last level I've graduated was Grade 7 due to financial problems, as well as my mom frequently being in and out of the hospital. I am now 18 and I wish to go to college as soon as I can, but I need to be caught up on all the math I've missed (I have been studying these past few years with what's available on the internet, but I don't think it's enough).</p>
<p>So my question is, are there any good, easy to understand, high school math books suited for my situation? I learn better with a teacher who can explain the lesson, but since I don't have one I'd prefer books that aren't too difficult, but at the same time provide everything necessary for high school level math and more. I used to be a bright student so I'm sure I can do this on my own with the right material.</p>
<p>Again, if this question isn't meant to be on this site I'd be more than willing to delete it asap! That's all. Thank you for reading. :)</p>
| Live Free or π Hard | 126,067 | <p>Good complimentary books to go with whatever resources you chose are the Schaum’s Outline series. You can find some high school level math books in these series with a lot of practice problems and full solutions. Good luck! <a href="https://www.mhprofessional.com/schaum-s" rel="nofollow noreferrer">https://www.mhprofessional.com/schaum-s</a></p>
|
1,643,201 | <p>The spectrum-functor
$$
\operatorname{Spec}: \mathbf{cRng}^{op}\to \mathbf{Set}
$$
sends a (commutative unital) ring $R$ to the set $\operatorname{Spec}(R)=\{\mathfrak{p}\mid \mathfrak{p} \mbox{ is a prime ideal of R}\}$ and a morpshim $f:S\to R$ to the map $\operatorname{Spec}(R)\to \operatorname{Spec}(S)$ with $\mathfrak{p}\mapsto f^{-1}(\mathfrak{p})$. Does this functor send pullback squares
\begin{eqnarray}
S\times_R T&\to& T\\
\downarrow && \downarrow\\
S&\to& R
\end{eqnarray}
of (commutative unital) rings to pushout squares
\begin{eqnarray}
\operatorname{Spec}(R)&\to& \operatorname{Spec}(T)\\
\downarrow && \downarrow\\
\operatorname{Spec}(S)&\to& \operatorname{Spec}(S\times_R T)
\end{eqnarray}
of sets? Put in other words, does the functor $\operatorname{Spec}$ from above preserve pushouts?</p>
| John Molokach | 90,422 | <p>I would count the number of three digits numbers (which is 900) and then count how many of them are a multiple of 7 (which is the greatest integer less than or equal to 900/7).</p>
|
3,357,841 | <p>In the diagram (which is not drawn to scale) the small triangles each have the area shown. Find the area of the shaded quadrilateral.</p>
<p><a href="https://i.stack.imgur.com/DK8sn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DK8sn.png" alt="enter image description here"></a></p>
| dfnu | 480,425 | <p><strong>A POSSIBLE PATH</strong></p>
<p>Consider the Figure below and let <span class="math-container">$x$</span> be the desired area.</p>
<p><a href="https://i.stack.imgur.com/D9b5B.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/D9b5B.png" alt="enter image description here"></a></p>
<ol>
<li>By Menelaus's Theorem on <span class="math-container">$\triangle BCE$</span> cut by <span class="math-container">$AD$</span> you get
<span class="math-container">\begin{equation}\frac{\overline{EP}}{\overline{PB}}\cdot\frac{\overline{BD}}{\overline{DC}}\cdot\frac{\overline{AC}}{\overline{AE}}=1.\tag{*}\label{eq1}\end{equation}</span></li>
<li>Observe that the ratio <span class="math-container">$\frac{\overline{EP}}{\overline{PB}} = \frac12$</span>. Why?</li>
<li>Similarly, you can find <span class="math-container">$\frac{\overline{BD}}{\overline{DC}}=\frac{18}{7+x}$</span>, and <span class="math-container">$\frac{\overline{AC}}{\overline{AE}}=\frac{25+x}{21}$</span>.</li>
<li>This info in \eqref{eq1} will give the equation in <span class="math-container">$x$</span> <span class="math-container">$$7(7+x) = 3(25+x),$$</span>and the final result <span class="math-container">$x=\frac{13}2$</span>.</li>
</ol>
|
3,357,841 | <p>In the diagram (which is not drawn to scale) the small triangles each have the area shown. Find the area of the shaded quadrilateral.</p>
<p><a href="https://i.stack.imgur.com/DK8sn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DK8sn.png" alt="enter image description here"></a></p>
| albert chan | 696,342 | <p>I tried to build a general formula for the shaded area.<br>
To simplify, <strong>assume base area = 1</strong>. We can scale it back later.</p>
<p>Let left and right triangle area be <span class="math-container">$x, y$</span>. Solving for shaded area, <span class="math-container">$z$</span><br>
Using @dfnu setup, with symbols instead of actual numbers:<br>
<span class="math-container">$${EP \over PB}{BD \over DC}{AC \over AE} = {x \over 1}{1+y \over x+z}{1+x+y+z \over 1+x} = 1$$</span>
<span class="math-container">$$\large z = {x + y + 2 \over {1 \over x\,y} - 1}$$</span></p>
<p>Note: the formula implied <span class="math-container">$x\,y < 1$</span></p>
<p>For this example, <span class="math-container">$\large x={7\over14}, y={4\over14}$</span>, we have <span class="math-container">$\large z={13\over28}$</span> </p>
<p>Scaled back with actual base area of 14, we have shaded area = <span class="math-container">$\large{13\over2}$</span> </p>
<p>Lets get the area of ΔABC (still, assume base area = 1):</p>
<p><span class="math-container">$$ΔABC = 1+x+y+z = {(1+x)(1+y) \over 1-x\,y}$$</span></p>
|
2,317,625 | <p>How do you compare $6-2\sqrt{3}$ and $3\sqrt{2}-2$? (no calculator)</p>
<p>Look simple but I have tried many ways and fail miserably.
Both are positive, so we cannot find which one is bigger than $0$ and the other smaller than $0$.
Taking the first minus the second in order to see the result positive or negative get me no where (perhaps I am too dense to see through).</p>
| Peter Green | 278,485 | <p>I'll use >=< to represent the unknown comparison.</p>
<p>$ 6-2\sqrt{3} >=< 3\sqrt{2}-2$</p>
<p>Lets start by adding two to both sides to reduce the number of numbers. This doesn't change the comparison result.</p>
<p>$ 8-2\sqrt{3} >=< 3\sqrt{2}$</p>
<p>Both sides are clearly positive ( $ 2\sqrt{3} < 6 $ ) so we can square both sides without changing the comparison result.</p>
<p>In a more maginal case where we were unsure if the left hand side was positive we could have compared the two terms in the left hand side by squaring both of them and hence determined whether the left hand side was positive or negative.</p>
<p>$ 64 -32\sqrt{3} + 12 >=< 18$</p>
<p>Now lets collect terms.</p>
<p>$ 60 >=< 32\sqrt{3}$</p>
<p>Divide by four.</p>
<p>$ 15 >=< 8\sqrt{3}$</p>
<p>Square again.</p>
<p>$ 225 >=< 64 \times 3 $</p>
<p>$ 225 > 192 $</p>
<p>Therefore </p>
<p>$ 6-2\sqrt{3} > 3\sqrt{2}-2$</p>
|
1,114,007 | <p>How to simplify $$\arctan \left(\frac{1}{2}\tan (2A)\right) + \arctan (\cot (A)) + \arctan (\cot ^{3}(A)) $$ for $0< A< \pi /4$?</p>
<p>This is one of the problems in a book I'm using. It is actually an objective question , with 4 options given , so i just put $A=\pi /4$ (even though technically its disallowed as $0< A< \pi /4$) and got the answer as $\pi $ which was one of the options , so that must be the answer (and it is weirdly written in options as $4 \arctan (1) $ ).</p>
<p>Still , I'm not able to actually solve this problem. I know the formula for sum of three arctans , but it gets just too messy and looks hard to simplify and it is not obvious that the answer will be constant for all $0< A< \pi /4$. And I don't know of any other way to approach such problems.</p>
| Jack D'Aurizio | 44,121 | <p>Over the given interval we have $\arctan\cot A=\frac{\pi}{2}-A$ and, by setting $t=\tan A$:
$$\begin{eqnarray*}&&\tan\left(\arctan\cot^3 A+\arctan\left(\frac{\tan(2A)}{2}\right)\right)=\frac{\cot^3 A+\frac{1}{2}\tan(2A)}{1-\frac{1}{2}\cot^3 A\tan(2A)}\\&=&\frac{\cot^3 A+\frac{1}{2}\tan(2A)}{1-\frac{1}{2}\cot^3 A\tan(2A)}=\frac{\frac{1}{t^3}+\frac{t}{1-t^2}}{1-\frac{1}{t^2(1-t^2)}}=\frac{1-t^2+t^4}{t(t^2-t^4-1)}=-\frac{1}{t}\end{eqnarray*}$$
so:
$$ \arctan\cot^3 A+\arctan\left(\frac{\tan(2A)}{2}\right)=\frac{\pi}{2}+A $$
and the sum of the three arctangents is $\color{red}{\pi}$ as wanted. Another chance is given by differentiating such a sum wrt to $A$ and check that the derivative is zero oven the given interval, so the sum equals its value in the point $A=\frac{\pi}{8}$, for instance.</p>
|
583,030 | <p>I have to show that the following series convergences:</p>
<p>$$\sum_{n=0}^{\infty}(-1)^n \frac{2+(-1)^n}{n+1}$$</p>
<p>I have tried the following:</p>
<ul>
<li>The alternating series test cannot be applied, since $\frac{2+(-1)^n}{n+1}$ is not monotonically decreasing.</li>
<li>I tried splitting up the series in to series $\sum_{n=0}^{\infty}a_n = \sum_{n=0}^{\infty}(-1)^n \frac{2}{n+1}$ and $\sum_{n=0}^{\infty}b_n=\sum_{n=0}^{\infty}(-1)^n \frac{(-1)^n}{n+1}$. I proofed the convergence of the first series using the alternating series test, but then i realized that the second series is divergent.</li>
<li>I also tried using the ratio test: for even $n$ the sequence converges to $\frac{1}{3}$, but for odd $n$ the sequence converges to $3$. Therefore the ratio is also not successful.</li>
</ul>
<p>I ran out of ideas to show the convergence of the series.</p>
<p>Thanks in advance for any help!</p>
| Siméon | 51,594 | <p>It is not convergent. To see this, let
$$
a_n = (-1)^n\frac{2}{n+1},\qquad b_n =\frac{1}{n+1},\qquad c_n = a_n + b_n.
$$
The series $\sum a_n$ is convergent by the alternating test.</p>
<p>We are interested in the convergence of $\sum c_n$. If $\sum c_n$ was convergent, then $\sum b_n = \sum c_n - \sum a_n$ would also be convergent, which is known to be false (divergence of the harmonic series).</p>
|
4,285,143 | <p>Let's suppose I've got a function <span class="math-container">$f(x)$</span> where I'd like to differentiate with respect to <span class="math-container">$t$</span>, but <span class="math-container">$t$</span> depends on <span class="math-container">$x$</span>: <span class="math-container">$t(x)$</span>. Thus the whole linked derivative thing: <span class="math-container">$\dfrac{\mathrm{d}f(x)}{\mathrm{d}t(x)}$</span>. Is this possible at all? Alternatively I had to find <span class="math-container">$t^{-1}(x)$</span>: <span class="math-container">$x(t)$</span> and then calculate the derivative fairly easy: <span class="math-container">$\mathrm{d} f(x(t))/\mathrm{d}t$</span>. But finding the inverse is not always possible. Is there another way? At all?</p>
| Brian Lai | 821,645 | <p>Maybe this is what you're looking for:</p>
<p><span class="math-container">$$ \frac{df}{dt} = \frac{df}{dx} \frac{dx}{dt} = \frac{df}{dx} \frac{1}{\frac{dt}{dx}}. $$</span></p>
<p>For example: Consider <span class="math-container">$f(x) = x^2$</span> and <span class="math-container">$t = e^x$</span>.</p>
<ol>
<li>Direct calculation of <span class="math-container">$\frac{df}{dt}$</span>: Note that <span class="math-container">$x = \ln{t}$</span>, so <span class="math-container">$f = \ln^2t$</span>. Then using chain rule:</li>
</ol>
<p><span class="math-container">$$ \frac{df}{dt} = 2\ln{t} (\frac{1}{t}) = 2x \frac{1}{e^x}.$$</span></p>
<ol start="2">
<li>Using the formula above:</li>
</ol>
<p><span class="math-container">$$\frac{df}{dt} = \frac{df}{dx} \frac{1}{\frac{dt}{dx}} = 2x \frac{1}{e^x}. $$</span></p>
|
4,434,832 | <p>I have taken this question from molodovian national MO 2008
The question is as follows</p>
<p>The sequence <span class="math-container">$(a_p)_p\ge 0$</span> is defined as <span class="math-container">$$a_p=\sum_{i=0}^p (-1)^i\frac{\binom{p}{i}}{(i+2)(i+4)}$$</span></p>
<p>Now let's find the limit</p>
<p><span class="math-container">\begin{align*} \lim_{n \to +\infty} a_0+a_1+a_2+.......a_n \end{align*}</span></p>
<p><a href="https://i.stack.imgur.com/CE73S.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CE73S.png" alt="enter image description here" /></a></p>
<p>Can anyone help me to go further.</p>
| nobody | 1,050,165 | <p>The solution of Stefan Lafon is elegant and shows the usefulness of professional tools. Integral representations (or, moreover, the dominated convergence theorem) are probably not the intended solution of an MO problem, though. So it may be interesting to see how to derive explicit expressions for <span class="math-container">$a_p$</span> and the partial sums without any calculus.</p>
<p>This can be done with finite differences (<a href="https://en.wikipedia.org/wiki/Finite_difference" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Finite_difference</a>):
If we define <span class="math-container">$\Delta_h[f](x)=f(x+h)-f(x)$</span>, the formula for <span class="math-container">$n$</span>th order forward differences will be
<span class="math-container">$$\Delta^n_h[f](x)=\sum^n_{i=0}\,(-1)^{n-i}\binom{n}{i}f(x+i\,h),$$</span> or for the special case of stepsize <span class="math-container">$h=1$</span>,
<span class="math-container">$$\Delta^n[f](x)=\sum^n_{i=0}\,(-1)^{n-i}\binom{n}{i}f(x+i).$$</span> This means that
<span class="math-container">$$a_p=(-1)^p\Delta^n[f](0),\tag1$$</span>
where <span class="math-container">$\displaystyle f(x)=\frac1{(x+2)(x+4)}$</span>. Now it's easy to calculate forward differences for <span class="math-container">$\displaystyle f_a(x)=\frac1{x+a}$</span>:
<span class="math-container">$$\Delta[f_a](x)=\frac1{x+a+1}-\frac1{x+a}=\frac{-1}{(x+a)(x+a+1)},$$</span>
<span class="math-container">$$\Delta^2[f_a](x)=\frac{-1}{(x+a+1)(x+a+2)}-\frac{-1}{(x+a)(x+a+1)}=\frac{(-1)(-2)}{(x+a)(x+a+1)(x+a+2)},$$</span> and by induction
<span class="math-container">$$\Delta^p[f_a](x)=\frac{(-1)^p\,p!}{(x+a)\ldots(x+a+p)}.$$</span> For a positive integer <span class="math-container">$a$</span>, this gives <span class="math-container">$$\Delta^p[f_a](0)=\frac{(-1)^p\,(a-1)!\,p!}{(p+a)!}=(-1)^p\,\frac{(a-1)!}{(p+1)\ldots(p+a)}.\tag2$$</span>
Since <span class="math-container">$$f(x)=\frac1{(x+2)(x+4)}=\frac12\,\left(f_2(x)-f_4(x)\right),$$</span> combining (1) and (2) yields
<span class="math-container">$$a_p=\frac12\,\left(\frac{1!}{(p+1)(p+2)}-\frac{3!}{(p+1)(p+2)(p+3)(p+4)}\right)=\frac{p+6}{2\,(p+2)(p+3)(p+4)},\tag3$$</span> as announced in a comment. Of course, it would be possible to obtain (3) from the integral representation as well, substituting <span class="math-container">$x\to 1-y$</span>. The partial sums <span class="math-container">$\displaystyle\sum^n_{p=0}a_p$</span> can be computed as telescoping sums, using <span class="math-container">$$\frac1{(p+1)(p+2)}=\frac1{p+1}-\frac1{p+2}$$</span> and
<span class="math-container">$$\frac1{(p+1)(p+2)(p+3)(p+4)}=\frac13\left(\frac1{(p+1)(p+2)(p+3)}-\frac1{(p+2)(p+3)(p+4)}\right).$$</span>
So,
<span class="math-container">$$\sum^n_{p=0}a_p=\frac12\left(1-\frac1{n+2}\right)-\frac16+\frac1{(n+2)(n+3)(n+4)}=\frac{2n^2+11n+9}{6\,(n+3)(n+4)}.$$</span></p>
|
351,846 | <p>The following problem was on a math competition that I participated in at my school about a month ago: </p>
<blockquote>
<p>Prove that the equation $\cos(\sin x)=\sin(\cos x)$ has no real solutions.</p>
</blockquote>
<p>I will outline my proof below. I think it has some holes. My approach to the problem was to say that the following equations must have real solution(s) if the above equation has solution(s):</p>
<p>$$
\cos^2(\sin x)=\sin^2(\cos x)\\
1-\cos^2(\sin x)=1-\sin^2(\cos x)\\
\sin^2(\sin x)=\cos^2(\cos x)\\
\sin(\sin x)=\pm\cos(\cos x)\\
$$</p>
<p>I then proceeded to split into cases and use the identity $\cos t = \sin(\frac{\pi}{2} \pm t\pm y2\pi)$ to get </p>
<p>$$
\sin x=\frac{\pi}{2} \pm \cos x\pm y2\pi\\
$$</p>
<p>and the identity $-\cos t = \sin(-\frac{\pi}{2}\pm t\pm y2\pi) $ to get </p>
<p>$$
\sin x=- \frac{\pi}{2}\pm \cos x \pm y2\pi.\\
$$</p>
<p>where $y$ is any integer. I argued that $y=0$ was the only value of $y$ that made any sense (since the values of sine and cosine remain between $-1$ and $1$). Therefore, the above equations become </p>
<p>$$
\sin x=\frac{\pi}{2} \pm \cos x\implies \sin x \pm \cos x=\frac{\pi}{2}\\
$$</p>
<p>and </p>
<p>$$
\sin x=- \frac{\pi}{2}\pm \cos x\implies \cos x\pm\sin x= \frac{\pi}{2}.\\
$$</p>
<p>Then, by a short optimization argument, I showed that these last two equations have no real solutions.</p>
<p>First, does this proof make sense? Second, if my proof makes sense, then I feel that it was not very elegant nor simple. Is my approach the best, or is there a better (i.e., more elegant, shorter, simpler) proof?</p>
| Anupriya | 799,968 | <p>Here's how I did it -</p>
<p><span class="math-container">$\cos( \sin x) = \sin(\cos x)$</span> can be written as,
<span class="math-container">$\sin (\frac{\pi}{2} - \sin x) = \sin (\cos x)$</span>, which implies,</p>
<p><span class="math-container">$\cos x + \sin x = \frac{\pi}{2}$</span>, or,
<span class="math-container">$\cos x + \cos(\frac{\pi}{2}-x) = \frac{\pi}{2}$</span>
Now, using <span class="math-container">$\cos x + \cos y$</span> identity,</p>
<p><span class="math-container">$2 \cos(\frac{\pi}{4})cos (\frac{\pi}{4} - x) = \frac{\pi}{2}$</span>, or,
<span class="math-container">$\cos (\frac{\pi}{4} - x) = \frac{\pi}{2 \sqrt 2} $</span>
Clearly, <span class="math-container">$\frac{\pi}{2\sqrt 2}$</span> is greater than 1 and that's not possible since the range of cosx is from <span class="math-container">$-1$</span> to <span class="math-container">$1$</span> only.</p>
<p>Therefore there are no real solutions to this equation.</p>
|
3,038,965 | <p>Here's the question I'm puzzling over:</p>
<p><span class="math-container">$\textbf{Find the perpendicular distance of the point } (p, q, r) \textbf{ from the plane } \\ax + by + cz = d.$</span></p>
<p>I tried bringing in the idea of a dot product and attempted to get going with solving the problem, but I'm heading nowhere. That is:</p>
<p><span class="math-container">$\text{The direction vector of the normal of the plane } = (a\textbf{i}+b\textbf{j}+c\textbf{k}) \text{, where } \\\textbf{i}, \textbf{ j},\textbf{ k } \text{ are unit vectors.}$</span> </p>
<p>This dotted with the direction vector of the point (position vector, precisely) should equal 0. Am I right? And will this method work even?</p>
| qualcuno | 362,866 | <p>A different approach, by a manual verification,</p>
<p><span class="math-container">$$
\frac{1}{1-x} = \sum_{n \geq 0}X^n,
$$</span></p>
<p>and so</p>
<p><span class="math-container">$$
\frac{1}{(1-x)^2} = \frac{1}{1-x} \cdot \frac{1}{1-x} = \sum_{n \geq 0}X^n \cdot \sum_{m \geq 0}X^m = \sum_{i \geq 0}\left(\sum_{k = 0}^i1\right)X^i = \sum_{i \geq 0}(i+1)X^i
$$</span></p>
|
2,674,217 | <p>Let $\{ a_{n}\}_{n}$ be a sequence and let $a\in \mathbb{R}$. Define $\{ c_{n}\}_{n}$ as:</p>
<p>$$c_{n}=\frac{a_{1}+...+a_{n}}{n}.$$</p>
<p>I want to prove the following claim: if $\lim\limits_{n\to +\infty}a_{n}=+\infty$ then $\lim\limits_{n\to +\infty}c_{n}=+\infty$</p>
<p>Approach: Suppose $\lim\limits_{n\to +\infty}a_{n}=+\infty$</p>
<p>Let $M>0$, $\exists\space n_{0}\in\mathbb{N}$ such that $\forall\space n\ge n_{0}, a_{n}>M$</p>
<p>\begin{align}
c_{n} &=\frac{a_{1}+...+a_{n_{0}-1}+a_{n_{0}}+...+a_{n}}{n} \\
&>\frac{a_{1}+...+a_{n_{0}-1}+(n-n_{0})M}{n}\\
&=\frac{a_{1}+...+a_{n_{0}-1}}{n}+\frac{(n-n_{0})M}{n}
\end{align}</p>
<p>So $\lim\limits_{n\to +\infty}c_{n}\ge\lim\limits_{n\to +\infty}\left[\frac{a_{1}+...+a_{n_{0}-1}}{n}+(1-\frac{n_{0}}{n})M\right]=M$</p>
<p>Since this is true $\forall\space M>0$, then $\lim\limits_{n\to +\infty}c_{n}=+\infty$</p>
<p>Is this approach correct?</p>
| hamam_Abdallah | 369,188 | <p>$$\lim_{\infty}(1-\frac {n_0}{n})M=M $$</p>
<p>$$\implies \exists n_1 \in \Bbb N \;: $$
$$n>n_1\implies (1-\frac {n_0}{n})M>\frac {M}{2} $$
on the other hand</p>
<p>$$\lim_{\infty}\frac {a_1+...a_{n_0}}{n}=0\implies $$</p>
<p>$$\exists \; n_2\in \Bbb N \;:$$
$$ n>n_2\implies \frac {a_1+...a_{n_0}}{n}>\frac {-M}{4} $$
thus for $n>\max (n_0,n_1,n_2) $ we have
$$c_n>\frac {M}{4} $$
and this proves that $$\lim_{\infty}c_n=+\infty. $$</p>
|
279,808 | <p>I was working on a way of calculating the square root of a number by the method of x/y → (x+4y)/(x+y) as shown by bobbym at <a href="https://math.stackexchange.com/questions/861509/">https://math.stackexchange.com/questions/861509/</a></p>
<p>I tried to do it via functions on mathematica, everything seems correct. Why am I not seeing 2.5 as the answer? How can I fix it?
<a href="https://i.stack.imgur.com/TXA8S.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TXA8S.png" alt="code" /></a></p>
| Bob Hanlon | 9,362 | <pre><code>Clear["Global`*"]
a = 1;
b = 4;
</code></pre>
<p>Using <a href="https://reference.wolfram.com/language/ref/RSolve.html" rel="nofollow noreferrer"><code>RSolve</code></a> will provide the general result for arbitrary <code>n</code></p>
<pre><code>Clear[f]; f[n_] = RSolveValue[{
f[n + 1] == (Numerator[f[n]] + b*Denominator[f[n]])/
(Numerator[f[n]] + Denominator[f[n]]), f[1] == a}, f[n], n] //
FullSimplify
(* -2 - 4/(-1 + (-(1/3))^n) *)
</code></pre>
<p><code>f</code> is complex for non-integer <code>n</code></p>
<pre><code>f[1.5]
(* 1.85714 - 0.742307 I *)
</code></pre>
<p>It is real for integer <code>n</code></p>
<pre><code>f /@ Range[10] // Simplify
(* {1, 5/2, 13/7, 41/20, 121/61, 365/182, 1093/547, 3281/1640,
9841/4921, 29525/14762}
Show[
Plot[Re[f[n]], {n, 1, 10}, PlotRange -> All],
DiscretePlot[f[n], {n, 0, 10}, PlotStyle -> Red, Filling -> None]]
</code></pre>
<p><a href="https://i.stack.imgur.com/uzMMB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uzMMB.png" alt="enter image description here" /></a></p>
<p><strong>EDIT:</strong> Alternatively, use <a href="https://reference.wolfram.com/language/ref/RecurrenceTable.html" rel="nofollow noreferrer"><code>RecurrenceTable</code></a> and <a href="https://reference.wolfram.com/language/ref/FindSequenceFunction.html" rel="nofollow noreferrer"><code>FindSequenceFunction</code></a></p>
<pre><code>Clear[f2]; seq = RecurrenceTable[{
f2[n + 1] == (Numerator[f2[n]] + b*Denominator[f2[n]])/
(Numerator[f2[n]] + Denominator[f2[n]]), f2[1] == a}, f2[n], {n, 8}]
(* {1, 5/2, 13/7, 41/20, 121/61, 365/182, 1093/547, 3281/1640} *)
f2[n_] = FindSequenceFunction[seq, n] // FullSimplify
(* 2 + 4/(-1 + (-3)^n) *)
</code></pre>
<p><code>f</code> and <code>f2</code> are equal for integer values of <code>n</code></p>
<pre><code>f[n] == f2[n] // Simplify[#, n ∈ Integers] &
(* True *)
</code></pre>
<p>Their real parts are equal</p>
<pre><code>Re[f[n]] == Re[f2[n]] // ComplexExpand // Simplify
(* True *)
</code></pre>
<p>They are complex conjugates of each other</p>
<pre><code>f[n] == Conjugate[f2[n]] // ComplexExpand // Simplify
(* True *)
</code></pre>
|
279,808 | <p>I was working on a way of calculating the square root of a number by the method of x/y → (x+4y)/(x+y) as shown by bobbym at <a href="https://math.stackexchange.com/questions/861509/">https://math.stackexchange.com/questions/861509/</a></p>
<p>I tried to do it via functions on mathematica, everything seems correct. Why am I not seeing 2.5 as the answer? How can I fix it?
<a href="https://i.stack.imgur.com/TXA8S.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TXA8S.png" alt="code" /></a></p>
| bill s | 1,783 | <p>This can also be approached recursively:</p>
<pre><code>Clear[f]; a = 1; b = 4; f[1] = a;
f[n_] := f[n] = (Numerator[f[n - 1]] + b Denominator[f[n - 1]])
/(Numerator[f[n - 1]] + Denominator[f[n - 1]])
</code></pre>
<p>Then the fist 10 values can be calculated</p>
<pre><code>f /@ Range[10]
</code></pre>
|
3,561,664 | <p>I did part of this question but am stuck and don't know how to continue</p>
<p>I let <span class="math-container">$x= 2k +1$</span></p>
<p>Also noticed that <span class="math-container">$x^3+x = x(x^2+1)$</span></p>
<p>therefore
<span class="math-container">$4m+2 = 2k+1((2k+1)^2+1)$</span></p>
<p>I simplified this and ended up with</p>
<p><span class="math-container">$4m+2 = 8k^3+12k^2+8k+2$</span></p>
<p>I don't know how to continue from and prove that <span class="math-container">$x^3+x$</span> has remainder 2 when divided by 4</p>
| Z Ahmed | 671,540 | <p>Let <span class="math-container">$x=2n+1, then
f(x)=x^3+x=(2n+1)^3+(2n+1)=2+8n^3+12n^2+8n \implies \frac{f(x)}{4}=\frac{2}{4}+2n^3+3n^2+2n.$</span> So the remaner is 2.</p>
|
246,071 | <p>How do I solve the following equation?</p>
<p>$$x^2 + 10 = 15$$</p>
<p>Here's how I think this should be solved.
\begin{align*}
x^2 + 10 - 10 & = 15 - 10 \\
x^2 & = 15 - 10 \\
x^2 & = 5 \\
x & = \sqrt{5}
\end{align*}
I was thinking that the square root of 5 is iregular repeating 2.23606797749979 number. 2.236 multipled by itself equals 5ish.</p>
<p>I've also seen another equation like this:
\begin{align*}
x^2 & = 4 \\
x^2 + 4 & = 0 \\
(x - 2)(x + 2) & = 0 \\
x & = 2 \text{ or } -2
\end{align*}
So I guess I could near the end of my equation do the following:</p>
<p>$$x^2 + 5 = 0$$</p>
<p>and then go from there?</p>
<p>Is my first attempt at solving correct?</p>
| Henry | 6,460 | <p>You actually want to show </p>
<p>$$\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{(n+1)+(n+1)}+\frac{1}{(n+1)+(n+2)}\le\frac{5}{6}$$</p>
<p>So you take the inductive hypothesis, and subtract $\frac{1}{n+1}$ from and add $\frac{1}{(n+1)+(n+1)}+\frac{1}{(n+1)+(n+2)}$ to the left hand side. Since you can show that change is less than zero, the overall sum is still less than or equal to $\frac56$.</p>
|
170,240 | <p>I have the following function of ω</p>
<pre><code>f[ω_] := (2 Sqrt[Γ] (4*g2^2 + (κ1 - 2*I*ω) (κ2 - 2*I*ω)))/(4*g2^2 (Γ -
2*I*ω) + (4*
g1^2 + (Γ - 2*I*ω) (κ1 -
2*I*ω)) (κ2 - 2*I*ω))
</code></pre>
<p>And I wish to obtain the poles for the denominator of the function:</p>
<pre><code>wroots1 = x /. Solve[(Denominator[f[ω]] /. {ω -> x}) == 0, x]
</code></pre>
<p>The result is of the following:</p>
<pre><code>{-(1/6) I (Γ + κ1 + κ2) + (I (-16 (\
Γ + κ1 + κ2)^2 +
48 (4 g1^2 +
4 g2^2 + Γ κ1 + Γ...}
</code></pre>
<p>Basically a really long ugly solution. My goal is to obtain the conjugate of wroots1, however, when I do it straightforwardly:</p>
<pre><code>wroots2 =
Simplify[Conjugate[wroots1],
Assumptions -> {Γ ∈
Reals, κ1 ∈ Reals, κ2 ∈ Reals,
g1 ∈ Reals,
g2 ∈ Reals, ω ∈ Reals,
4 (-16 (Γ + κ1 + κ2)^2 +
48 (4 g1^2 +
4 g2^2 + Γ κ1 + (Γ + \
κ1) κ2))^3 +
4096 (-36 g1^2 (Γ + κ1 -
2 κ2) + (2 Γ - κ1 - \
κ2) (36 g2^2 + (Γ + κ1 -
2 κ2) (Γ -
2 κ1 + κ2)))^2 > 0}];
</code></pre>
<p>I am returned with:</p>
<pre><code>{1/48 I (8 (Γ + κ1 + κ2) +
8 2^(1/3) (-12 g1^2 -
12 g2^2 + (Γ + κ1 + κ2)^2 -
3 (κ1 κ2 + Γ (κ1 + \
κ2))) Conjugate[
1/(-36 g1^2 Γ + 72 g2^2 Γ +
2 Γ^3 - 36 g1^2 κ1 -
36 g2^2 κ1 - 3 Γ^2 κ1 -
3 Γ κ1^2 + 2 κ1^3 +
72 g1^2 κ2 - 36 g2^2 κ2 -
3 Γ^2 κ2 +
12 Γ κ1 κ2 -
3 κ1^2 κ2 - 3 Γ κ2^2 -
3 κ1 κ2^2 + 2 κ2^3...}
</code></pre>
<p>Clearly, Conjugate refuses to take the conjugate of said function. I tried doing</p>
<pre><code>wroots2 = wroots1 /. {I -> -I}
</code></pre>
<p>But that doesn't work as well. I'm at lost at what to do here and I could use any help I can get. Thank you very much in advance.</p>
| Roman | 26,598 | <p>If you <code>Conjugate</code> the equation before <code>Solve</code>, you'll get the complex-conjugated solutions (complex analysis is nice!):</p>
<pre><code>wroots1 = w /. Solve[Denominator[f[w]] == 0, w];
wroots1C = w /. Solve[ComplexExpand[Conjugate[Denominator[f[w]]]] == 0, w];
</code></pre>
<p>This way you don't have to complex-conjugate the solutions. The downside is that the order of the solutions in wroots1 and wroots1C may not be the same.</p>
|
3,703,981 | <p>If we consider an equation <span class="math-container">$x=2x^2,$</span> we find that the values of <span class="math-container">$x$</span> that solve this equation are <span class="math-container">$0$</span> and <span class="math-container">$1/2$</span>. Now, if we differentiate this equation on both sides with respect to <span class="math-container">$x,$</span> we get <span class="math-container">$1=4x.$</span> Now, I know that it is wrong to say that the value of <span class="math-container">$x=1/4,$</span> but then, what does <span class="math-container">$x=1/4$</span> signify? Is it related to maxima or minima? Please help me with this.</p>
| J. W. Tanner | 615,567 | <p>There are no solutions, because <span class="math-container">$x^3,y^3\equiv0$</span> or <span class="math-container">$\pm1\pmod7$</span>, but <span class="math-container">$2020\equiv4\bmod7$</span>.</p>
|
194,547 | <p>I know the definition of a linear transformation, but I am not sure how to turn this word problem into a matrix to solve:</p>
<p>$T(x_1, x_2) = (x_1-4x_2, 2x_1+x_2, x_1+2x_2)$</p>
<p><strong>Find the image of the line that passes through the origin and point $(1, -1)$.</strong></p>
| Brian M. Scott | 12,042 | <p>HINT: A linear transformation sends straight lines to straight lines. If $T$ sends the origin and the point $\langle 1,-1\rangle$ to the points $P$ and $Q$, it must send the line through the origin and the point $\langle 1,-1\rangle$ to the line through $P$ and $Q$.</p>
|
2,736,426 | <p>Let's imagine a point in 3D coordinate such that its distance to the origin is <span class="math-container">$1 \text{ unit}$</span>.</p>
<p>The coordinates of that point have been given as <span class="math-container">$x = a$</span>, <span class="math-container">$y = b$</span>, and <span class="math-container">$z = c$</span>.</p>
<p>How can we calculate the angles made by the vector with each of the axes?</p>
| sirous | 346,566 | <p>Suppose angle of vector related to x axis is $\alpha$, related to y axis is $\beta$ and related to z axis is $\gamma$ then we have:</p>
<p>Due to presumption; $\sqrt {a^2+b^2+c^2}=1$ </p>
<p>$1\times \ cos \alpha=a$</p>
<p>$1\times \ cos \beta=b$</p>
<p>$1 \times \ cos \gamma=c$</p>
|
3,757,213 | <blockquote>
<p>Prove that the maximum area of a rectangle inscribed in an ellipse <span class="math-container">$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$</span> is <span class="math-container">$2ab$</span>.</p>
</blockquote>
<p><strong>My attempt:</strong></p>
<p>Equation of ellipse: <span class="math-container">$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$</span>.</p>
<p>Assume that <span class="math-container">$a>b$</span> & let <span class="math-container">$l$</span> & <span class="math-container">$w$</span> are length and width of rectangle then area will be <span class="math-container">$$A=l\times w \tag{1}$$</span></p>
<p>Now, I substituted the point <span class="math-container">$\left(\dfrac {l}{2}, \dfrac {w}{2}\right)$</span> in the equation of ellipse
<span class="math-container">$$\frac{(l/2)^2}{a^2}+\frac{(w/2)^2}{b^2}=1,$$</span>
<span class="math-container">$$b^2l^2+a^2w^2=4a^2b^2.\tag{2}$$</span></p>
<p>I am not sure how to proceed from here.</p>
| Quanto | 686,284 | <p>Let one vertex of the rectangle be <span class="math-container">$(a\cos t, b\sin t)$</span>. Then, the other three are known as well and the area is</p>
<p><span class="math-container">$$A= 4ab|\sin t \cos t |\le 2ab (\cos^2t+\sin^2t)=2ab$$</span></p>
<p>where the inequality <span class="math-container">$2uv\le u^2+v^2 $</span> is used.</p>
|
1,015,498 | <p>I am merely looking for the result of the convolution of a function and a delta function.
I know there is some sort of identity but I can't seem to find it. </p>
<p>$\int_{-\infty}^{\infty} f(u-x)\delta(u-a)du=?$</p>
| JohnD | 52,893 | <p>It's called the <a href="http://en.wikipedia.org/wiki/Dirac_delta_function#Translation" rel="noreferrer">sifting property</a>:</p>
<p>$$
\int_{-\infty}^\infty f(x)\delta(x-a)\,dx=f(a).
$$</p>
<p>Now, if
$$
f(t)*g(t):=\int_0^t f(t-s)g(s)\,ds,
$$
we want to compute
$$
f(t)*\delta(t-a)=\int_0^t f(t-s)\delta(s-a)\,ds.
$$
With an eye on the sifting property above (which requires that we integrate "across the spike" of the Dirac delta, which occurs at $a$, we consider two cases.</p>
<ol>
<li><p>If $t<a$, then $\delta(s-a)\equiv 0$ since $0\le s\le t<a$. Therefore $\int_0^t f(t-s)\delta(s-a)\,ds\equiv 0$.</p></li>
<li><p>If $t\geq a$, then by the sifting property, $\int_0^t f(t-s)\delta(s-a)\,ds=f(t-s)\Big|_{s=a}=f(t-a)$.</p></li>
</ol>
<p>Thus,
\begin{align}
f(t)*\delta(t-a)=\int_0^t f(t-s)\delta(s-a)\,ds&=\begin{cases} 0, &t<a,\\ f(t-a) &t\ge a,\end{cases}=f(t-a)u(t-a),
\end{align}
where $u(t)$ denotes the unit step function.</p>
|
22,839 | <p>Is it possible to have the text generated by <code>PlotLabel</code> (or any other function) aligned to the left side of the plot instead of in the center?</p>
| Verbeia | 8 | <p>You can specify your <code>PlotLabel</code> to be a construct that takes a <code>TextAlignment</code> or <code>Alignment</code> option, such as <code>Pane</code>. For example:</p>
<pre><code>Plot[Sin[x], {x, 0, 5},
PlotLabel -> Framed@Pane["This is the title", Alignment -> Left,
ImageSize -> 270], ImageSize -> 300]
</code></pre>
<p><img src="https://i.stack.imgur.com/E5rYK.png" alt="enter image description here"></p>
<p>The <code>Framed</code> isn't necessary; I just included it so you could see what was going on.</p>
<p>The main trick is that you need to specify the size of both the total graphic and the <code>Pane</code>, so that the text will align to the left edge of the graphic or the axis, as shown here.</p>
<p>If you don't do it this way, the <code>Pane</code> will indeed be left-aligned text, but it will "shrink" to fit the text and still be aligned center, like this:</p>
<pre><code>Plot[Sin[x], {x, 0, 5},
PlotLabel -> Framed@Pane["This is the title", Alignment -> Left]]
</code></pre>
<p><img src="https://i.stack.imgur.com/WmrR6.png" alt="enter image description here"></p>
<p>As an alternative, you can bend <code>AxesLabel</code> to fit, by working out how wide the label will be in pixels, and then adding that amount of spacing to its left in a <code>Row</code> construct to make the argument to <code>AxesLabel</code>. This is similar in spirit to Mr.Wizard's answer but has the advantage of automation of the spacing.</p>
<pre><code>With[{space =
First@ImageDimensions[Rasterize[TraditionalForm[Sinc[x]]]]},
Plot[Sinc[x], {x, 0, 10}, AxesLabel -> Row[{Spacer[space], Sinc[x]}] ]]
</code></pre>
<p><img src="https://i.stack.imgur.com/40IYJ.png" alt="enter image description here"></p>
<p>Obviously if you have styled the label for the plot label in some other way, you will need to adjust the argument to <code>Rasterize</code> accordingly.</p>
|
22,839 | <p>Is it possible to have the text generated by <code>PlotLabel</code> (or any other function) aligned to the left side of the plot instead of in the center?</p>
| Carl Woll | 45,431 | <p>Here is a refinement of @DavidC's approach.</p>
<p>We can use <code>PlotRangeClipping->False</code> and then stick the label outside of the plot range and still have it show up. In order to do this we need to know the <a href="http://reference.wolfram.com/language/ref/ImagePadding" rel="nofollow noreferrer"><code>ImagePadding</code></a> of the plot, and the height of the plot label. Here is the plot:</p>
<pre><code>plot = Plot[Sinc[x], {x, 0, 9}]
</code></pre>
<p><a href="https://i.stack.imgur.com/w4EhF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/w4EhF.png" alt="enter image description here"></a></p>
<p>We can find out the image padding using my <a href="https://mathematica.stackexchange.com/a/138907/45431"><code>GraphicsInformation</code></a> function. Install with:</p>
<pre><code>PacletInstall[
"GraphicsInformation",
"Site" -> "http://raw.githubusercontent.com/carlwoll/GraphicsInformation/master"
];
</code></pre>
<p>Then, load it:</p>
<pre><code><<GraphicsInformation`
</code></pre>
<p><code>GraphicsInformation</code> returns a list of rules:</p>
<pre><code>pad = "ImagePadding" /. GraphicsInformation[plot]
</code></pre>
<blockquote>
<p>{{16.1531, 1.5}, {1.5, 0.5}}</p>
</blockquote>
<p>To get the size of the label, we can use <a href="http://reference.wolfram.com/language/ref/Rasterize" rel="nofollow noreferrer"><code>Rasterize</code></a>:</p>
<pre><code>label = Style["Plot label", "Section"];
Rasterize[label, "BoundingBox"]
</code></pre>
<blockquote>
<p>{105, 33, 27}</p>
</blockquote>
<p>The second element of the list is the height in points. Now, we are ready to add the label to the plot:</p>
<pre><code>Show[
plot,
Graphics @ Text[label, Offset[{-16, 10}, Scaled[{0, 1}]], {-1, -1}],
PlotRangeClipping->False,
ImagePadding -> pad + {{0, 0}, {0, 43}}
]
</code></pre>
<p><a href="https://i.stack.imgur.com/vRSna.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vRSna.png" alt="enter image description here"></a></p>
<p>I used an offset of 10 points in the vertical direction (to create space between the label and the y-axis), so that the total padding needed is actually 33 + 10, and not just 33.</p>
|
336,834 | <p>It is a well known theorem, that every signed measure can be split into its positive and negative parts (Hahn-Jordan-Decomposition). My question is, if something similar is possible for functionals on Sobolev spaces.</p>
<p>To be precise, let $\Omega \subset \mathbb{R}^n$ be some open domain and $\mu \in H^{-1}(\Omega) = H_0^1(\Omega)^*$. Are there $\mu^+, \mu^- \in H^{-1}(\Omega)$, which are positive in the sense that
$$\langle \mu^+, v \rangle \ge 0 \quad\text{for all } v \in H_0^1(\Omega), v \ge 0,$$
(and the same for $\mu^-$) and $\mu = \mu^+ - \mu^-$?</p>
<p>(By duality arguments, one obtains, that the set of all differences $\mu^+ - \mu^-$ of positive functionals $\mu^+, \mu^-$ is dense in $H^{-1}(\Omega)$.)</p>
<p><strong>Edit:</strong> Found a very related question on MO: <a href="https://mathoverflow.net/questions/149151/is-any-order-bounded-continuous-linear-functionals-a-difference-of-positive-cont">https://mathoverflow.net/questions/149151/is-any-order-bounded-continuous-linear-functionals-a-difference-of-positive-cont</a></p>
| gerw | 58,577 | <p>Just for reference, I post a different answer (which doesn't rely on the fact that positive distributions are measures).</p>
<p>Define $\Omega = (-1, 1)$, and
$$
f = \begin{cases} 0 & x \le 0, \\ |x|^{-1/3} & x > 0. \end{cases}
$$
Then, $f \in L^2(\Omega)$ and hence, $\mu$ defined by
$$
\langle \mu, v \rangle := \int_\Omega f \, v' \, \mathrm{d} x
$$
belongs to $H^{-1}(\Omega)$.</p>
<p>Now, assume that $\mu = \mu^+ - \mu^-$ for positive $\mu^+,\mu^- \in H^{-1}(\Omega)$. Let $v \in H_0^1(\Omega)$, $v \ge 0$ be given.
Then, for all $w \in H_0^1(\Omega)$ with $0 \le w \le v$, we have
$$
\langle \mu, w \rangle =
\langle \mu^+, w \rangle -
\langle \mu^-, w \rangle
\le
\langle \mu^+, v \rangle -
\langle \mu^-, 0 \rangle
= \mathrm{const}.
$$
(In books about Banach lattices, this property is called "order bounded". This property is necessary and sufficient for the decomposition into a positive and a negative part.)</p>
<p>Now, we construct $v$ and $w$ which contradicts this boundedness. Let $v \ge 0$, $v \in H_0^1(\Omega)$ be such that $v = 1$ on $(-1/2, 1/2)$.
Define for $n \in \mathbb{N}$, $n \ge 2$,
$$
w = \begin{cases} 0 & |x| \ge 1/n,\\ 1+x\,n & x \in (-1/n, 0), \\ 1-x\,n & x \in (0,1/n).\end{cases}
$$
Then, $w \in H_0^1(\Omega)$ and $0 \le w \le v$.
However,
$$
\langle \mu, w \rangle = \int_0^{1/n} |x|^{-1/3} \, n \mathrm{d} x = \ldots = \frac{n^{1/3}}{2/3} \to \infty \quad\text{for } n \to \infty
$$
This is a contradiction. Hence, $\mu$ cannot be decomposed into a positive and a negative part.</p>
|
92,983 | <p><strong>Does every polyhedron in $\mathbb{R}^3$ with $n$ triangular facets have a <em>topological</em> triangulation with complexity $O(n)$?</strong></p>
<p>Suppose $P$ is a non-convex polyhedron in $\mathbb{R}^3$ with $n$ triangular facets, possibly with positive genus. A <em>topological</em> triangulation of $P$ is a simplicial complex whose underlying space is the closure of the interior of $P$, such that every facet of $P$ is a cell in the complex. These boundary facets are true geometric triangles, but interior simplices may be arbitrarily bent and twisted. In the more standard <em>geometric</em> triangulations, every simplex is the convex hull of its vertices.</p>
<p>Results of <a href="http://www.cs.princeton.edu/~chazelle/pubs/BoundsSizeTetrahedral.pdf" rel="nofollow">Chazelle and Shouraboura</a> imply that every polyhedron has a geometric triangulation with complexity $O(n^2)$. Moreover, a classical construction of <a href="http://www.cs.princeton.edu/~chazelle/pubs/ConvexPartitionPolyhedra.pdf" rel="nofollow">Chazelle</a> implies that the $O(n^2)$ bound is is optimal in the worst case, even when the genus is zero.</p>
<p>But we can get tighter bounds for topological triangulations, at least for genus-zero polyhedra. If $P$ has genus zero, <a href="http://en.wikipedia.org/wiki/Steinitz%27s_theorem" rel="nofollow">Steinitz's theorem</a> implies that there is a <em>convex</em> polyhedron $Q$ that is combinatorially equivalent to $P$. <a href="http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1085499/" rel="nofollow">Alexander's extension of the Schönflies theorem</a> implies that the interiors of $P$ and $Q$ are both homeomorphic to open balls. Thus, applying a suitable homeomorphism to a minimal <em>geometric</em> triangulation of $Q$ gives us a <em>topological</em> triangulation of $P$ with complexity $O(n)$. (Alternatively, we can triangulate $P$ by joining an arbitrary interior point to every facet.)</p>
<p>What makes the question tricky for higher-genus polyhedra is the possibility of knottedness; the topology of the interior of $P$ is not determined by its genus. Intuitively, the question is how knotted the interior of a polyhedron can be, as a function of the number of facets.</p>
<p>The following question may be equivalent: Let $K$ be a closed polygonal chain (or "stick knot") in $S^3$ with $n$ edges. Is there a <em>topological</em> triangulation of $S^3$ with complexity $O(n)$ that includes $K$ in its 1-skeleton? Again, if we insist on <em>geometric</em> triangulations, $\Theta(n^2)$ tetrahedra are always sufficient and sometimes necessary, even if $K$ is unknotted.</p>
<p><strong>Added for bounty (Apr 13):</strong> Partial results, subquadratic upper bounds, or references that imply this problem is open (or the crossing-number problem in my comment on the first answer) would be welcome.</p>
| Sam Nead | 1,650 | <p>I've been thinking about the main question in the original post on and off for a few days. All of my efforts have been in the direction of finding enough examples to prove a super-linear lower bound, following Misha's suggestion to use hyperbolic volume. This hasn't worked yet - the problem appears to be tricky! In any case, here is the most appealing of the constructions. </p>
<p><strong>Stick braids</strong></p>
<p>Let $x, y, z$ be the usual coordinates on $\mathbb{R}^3$. Let $D$ be the unit disk in the plane $z = 0$ and let $E$ be the unit disk in the plane $z = 1$. Suppose that $\{a_i\}_1^n \subset D$ is a collection of points. Let $b_i$ be the point in $E$ with the same $x$ and $y$ coordinates as $a_i$. Suppose that $\sigma \in \Sigma_n$ is a permutation. Let $B = B(a, \sigma)$ be the collection of line segments where the $i$'th segment has endpoints $a_i$ and $b_{\sigma(i)}$. If the segments are pairwise disjoint then we call $B$ a <em>stick braid</em>.</p>
<p>It follows that the braid closure of $B$ is a link with stick number at most $5n$. As a concrete example, the $(p,q)$-torus knot can be obtained by placing the points $a_i$ at the $p$'th roots of unity, taking $\sigma(i) = i + q$, modulo $p$, and taking a braid closure. </p>
<p><strong>Hyperbolic volume</strong></p>
<p>Now we must use the Euclidean geometry of the braid $B$ to draw conclusions about the hyperbolic volume of the braid closure. Consider the unit disk $D_t$ in the plane $z = t$. As $t$ varies from $0$ to $1$, the points of intersection $D_t \cap B = \{a_i^t\}$ move along straight lines at speeds depending on the slope of the $i$'th strand. Here is a lie: when two points $a_i^t$ and $a_j^t$ come much closer to each other than they are to any of the other points, then there is a definite contribution to hyperbolic volume. Making this precise (ie, actually true) and then finding a braid $B$ that arranges superlinearly many such meetings would give the desired lower bound. </p>
<p>One way to do this would be to take $n$ sufficently large, $\epsilon$ correspondingly small, and take the points $a_i$ to be a generic $\epsilon$--net in $D$. Choose $\sigma$ to be a random permutation. Let $B = B(a, \sigma)$. Take the braid closure and plug everything into SnapPy. I've not tried to do this yet, but it would at least give some data...</p>
<p><strong>Edit</strong></p>
<p>Before writing the above, I had the idea of generating a random stick knot using <a href="http://en.wikipedia.org/wiki/Outer_billiard" rel="nofollow noreferrer">outer billiards</a> -- namely, let $P$ and $Q$ be concentric spheres and build a knot by taking segments tangent to $Q$ with endpoints on $P$. This has the virtue that when $Q$ has smallish radius, the expected crossing number will be quadratic. But it seems easier to estimate volume using the braid construction, and it is volume that really matters to us. </p>
<p>And then... after all this thinking and writing, I started poking randomly around the web and found O'Rourke's <a href="https://mathoverflow.net/questions/54412/complexity-of-random-knot-with-vertices-on-sphere">question</a> on our very own MO. O'Rourke gives a very simple model for random stick knots: just bounce around in a sphere. The Thurstons suggest that the expected volume grows as $n^{3/2}$.</p>
|
1,431,464 | <p>Does anyone know a good reference where it is shown that the Schwartz class $\mathcal{S}(\mathbb R)$ is a dense subset of $L^2(\mathbb R)$?</p>
<p>Many thanks</p>
| Dan | 79,007 | <p>The most frequent/easiest way I've seen this proved is to show instead that $C_c^\infty(\mathbb R)$ is dense $L^2(\mathbb R)$ and then just note $C_c^\infty(\mathbb R) \subset S(\mathbb R)$. This can be found in anything from <s>big Rudin to</s> Folland's <em>Real Analysis</em> to Trèves's <em>Topological Vector Spaces, Distributions, and Kernels</em>.</p>
<p>The last reference, along with similar classic texts on locally convex spaces (or really any books emphasizing the role of locally convexity in approximation), would be my (very biased) suggestion if you're at a level where you feel comfortable reading from them.</p>
<p>Edit: According to Silvia's comment, it might not appear in big Rudin (although I'd be surprised not to see it there). Proposition 8.17 in Folland states that $C_c^\infty$ (and therefore $S$) is dense in $L^p$ for $1 \leq p \leq \infty$.</p>
|
183,077 | <p>A complex Lie group may have several real forms.
Are there any duality/trinity... between them?
Maybe a trivial question to ask, is $SL(3,\mathbb{C})$ a real form of $SL(3,\mathbb{C})\times SL(3,\mathbb{C})$ ?</p>
| Amritanshu Prasad | 9,672 | <p>According to <a href="http://dx.doi.org/10.1016/0021-8693(75)90041-1">Djokovic and Maizan</a>, the Specht module $V_{(3, 1, 1)}$ of $S_5$ is monomial. This is a representation of dimension $6$, induced from a representation of dimension $3$ of $A_5$. Since $A_5$ has no subgroup of index $3$ (see <a href="http://groupprops.subwiki.org/wiki/Subgroup_structure_of_alternating_group:A5">here</a> for example), this representation of $A_5$ cannot be monomial.</p>
|
155,455 | <p>I want to find the maximum of a function (f) over a variable (t). The function is huge and it's not possible to maximize f(t) directly. So I want to create f inside a Table and then find the highest value over a small range of t. How can I add the steps to construct f into a Table? It seems "/." is not working. </p>
<p>A simplified version of my problem is:</p>
<pre><code> v = Table[f = TR1 + TR2
/. TR1 = Solve[TR1 - t = 0, {TR1}]
/. TR2 = t + 1
, {t, 1, 3, 1}];
tstar = Max[v]
</code></pre>
<p>f(t) has many components like TR1 and TR2 which makes it so huge.</p>
| danielsmw | 24,976 | <p>You don't actually need to create a variable called <code>f</code> in each element of the table; just return the value you need. You also want rules, rather than assignments. Try</p>
<pre><code>v = Table[ TR1 + TR2 /. Flatten[{
Solve[TR1 - t == 0, TR1],
TR2 -> t + 1}],
{t, 1, 3, 1}];
tstar = Max[v]
</code></pre>
<p>Note that <code>Solve[...]</code> actually returns a list of substitutions for <code>TR1</code>, so we've simply flattened it with the <code>TR2</code> rule.</p>
|
20,314 | <p>Hi all.
I'm looking for english books with a good coverage of distribution theory.
I'm a fan of Folland's Real analysis, but it only gives elementary notions on distributions.
Thanks in advance.</p>
| Anweshi | 2,938 | <p>Lieb and Loss, "Analysis" quickly starts with measure theory and after a short break with Fourier transforms, gets on to Distributions. I would imagine this is the fastest way to learn distributions. </p>
|
20,314 | <p>Hi all.
I'm looking for english books with a good coverage of distribution theory.
I'm a fan of Folland's Real analysis, but it only gives elementary notions on distributions.
Thanks in advance.</p>
| O.R. | 5,506 | <p>I liked Functional Analysis by Kosaku Yosida. It is book on functional analysis but oriented to get the applications of it to differential equations. </p>
|
20,314 | <p>Hi all.
I'm looking for english books with a good coverage of distribution theory.
I'm a fan of Folland's Real analysis, but it only gives elementary notions on distributions.
Thanks in advance.</p>
| Anand | 36,814 | <p>Why don't people mention about Rudin's book, <em>Functional Analysis</em>. Chapter 1-8 are pretty good for the theory of distribution. The problem is that this book is quite dry, no much motivations behind. So you might have a difficult time in the beginning. It is good to read the book Strichartz, R. (1994), <em>A Guide to Distribution Theory and Fourier Transforms</em>, besides.</p>
|
842,266 | <p>I have a tiny little doubt related to one proof given in Ahlfors' textbook. I'll copy the statement and the first part of the proof, which is the part where my doubt lies on.</p>
<p><strong>Statement</strong>
The stereographic projection transforms every straight line in the $z$-plane into a circle on $S$ which passes through the pole $(0,0,1)$ and the converse is also true. More generally, any circle on the sphere corresponds to a circle or straight line in the $z$-plane.</p>
<p><strong>Proof</strong></p>
<p>To prove this we observe that a circle on the sphere lies in a plane $\alpha_1x_1+\alpha_2x_2+\alpha_3x_3=\alpha_0$, where we can assume ${\alpha_1}^2+{\alpha_2}^2+{\alpha_3}^2=1$ and $0\leq \alpha_0 <1$</p>
<p>I don't understand why it is always the case that the condition $0\leq \alpha_0 <1$ can be satisfied. I mean, a plane can be described as:</p>
<p>$$\Pi: \space n.(v-v_0)=0 \tag{1}$$ where $v$ and $v_0$ are two vectors with endpoints lying on $\Pi$. I know that $n$ is a perpendicular vector to the plane, and I understand that if $n=(\alpha_1,\alpha_2,\alpha_3)$ doesn't satisfy $||n||=1$, then the vector $n'=\dfrac{n}{||n||}$ is a unit vector which also satisfies equation (1).</p>
<p>Equation (1) is the same as $$\space n.v=n.v_0 \tag{2}$$</p>
<p>In this problem, $\alpha_0=n.v_0$, I don't understand why we can always choose $n$ and $v_0$ such that all the conditions said in my previous lines are satisfied.</p>
<p>I put the title "complex-analysis" but I am not sure if it is the proper tag, if anyone can think of a better tag for this post, tell me and I'll change it.</p>
| Alucard | 167,097 | <p>i was searching right this question for my complex analysis course because i was stuck on the same page ( guess we are studying the same thing) but my doubt was on ${\alpha_1}^2+{\alpha_2}^2+{\alpha_3}^2=1$. Anyway, now that i have understood this equation i think i can give you another answer for why $ 0 <{\alpha_0} <1 $ . my explanation involve the formula that give you the distance between a point and a plane</p>
<p>$$\frac{|{\alpha_1}x_1 + {\alpha_2}y_1 + {\alpha_3}z_1 +{\alpha_0}|}{\sqrt{{\alpha_1}^2+{\alpha_2}^2+{\alpha_3}^2}}$$</p>
<p>remembering that the sphere has centre in (0,0,0) this formula give you </p>
<p>$ 1 > ||{\alpha_0}|| / \sqrt { ({\alpha_1}^2+{\alpha_2}^2+{\alpha_3}^2)} $</p>
<p>but the denominator we know to be 1 .</p>
|
1,048,668 | <p>Let $f\colon (a,b) \to \mathbb{R}$ a non constant differentiable function. </p>
<p>Is the following statement true:</p>
<p>If $f$ has a local maximum <em>and</em> a local minimum then $f$ also does have an inflection point.</p>
<p>If so, how to prove it, if not, what would be a counterexample?</p>
<p><em>Remark</em></p>
<p>If there are $a_0,b_0 \in [a,b]$ with $a_0 < x_0 < b_0$ such that $f|_{(a_0,x_0)}$ is convex and $f|_{(x_0,b_0)}$ concave or vice versa, then $(x_0,f(x_0))$ is called inflection point of $f$ (Amann Escher Analysis I, p349).</p>
<p>Sometimes <em>stricly</em> concave (convex) is used in the definition. Does this change the theorem?</p>
<p>If the theorem is not true, does it hold if one allows only smooth functions or even more restrictive only polynomial ones (excluding linear functions)?</p>
| Narasimham | 95,860 | <p>Definition of inflection point: when $ f''(x)=0. $</p>
<p>Between two extrema with second derivatives of opposite sign there will always be at least one inflection point. The graph of second derivative must pass through zero as consequence of Rolle's theorem.</p>
<p>Non-strict concave/convex situation is when $ f'(x)=0, f''(x)=0. $ It does not change the theorem.</p>
<p>A point of inflection (PI) exists in all the four situations:</p>
<p>maximum,PI,minimum real root,PI,maximum</p>
<p>maximum,PI,minimum imaginary root,PI,maximum</p>
<p>minimum,PI,maximum real root,PI,minimum</p>
<p>minimum,PI,maximum imaginary root,PI,minimum</p>
|
1,048,668 | <p>Let $f\colon (a,b) \to \mathbb{R}$ a non constant differentiable function. </p>
<p>Is the following statement true:</p>
<p>If $f$ has a local maximum <em>and</em> a local minimum then $f$ also does have an inflection point.</p>
<p>If so, how to prove it, if not, what would be a counterexample?</p>
<p><em>Remark</em></p>
<p>If there are $a_0,b_0 \in [a,b]$ with $a_0 < x_0 < b_0$ such that $f|_{(a_0,x_0)}$ is convex and $f|_{(x_0,b_0)}$ concave or vice versa, then $(x_0,f(x_0))$ is called inflection point of $f$ (Amann Escher Analysis I, p349).</p>
<p>Sometimes <em>stricly</em> concave (convex) is used in the definition. Does this change the theorem?</p>
<p>If the theorem is not true, does it hold if one allows only smooth functions or even more restrictive only polynomial ones (excluding linear functions)?</p>
| Nate Eldredge | 822 | <p>Consider the bump function
$$\phi(x) = \begin{cases} e^{-1/(1-x^2)}, & -1 < x < 1 \\ 0, & \text{else.} \end{cases}$$
It's a standard exercise to verify that $\phi$ is $C^\infty$ and has a local maximum at $x=0$. Set $g(x) = \phi(x-2) - \phi(x+2)$, so that $g$ is $C^\infty$, strictly negative on $(-3,-1)$, strictly positive on $(1,3)$, and zero on $[-1,1]$. In particular, there is no point at which $g(x)$ changes from strictly positive to strictly negative or vice versa. </p>
<p>Let $f(x) = \int_0^x g(t)\,dt - c$ for some sufficiently small positive constant $c$ so that $f$ has exactly one zero in $(-3,-1)$ and another in $(1,3)$. Then let $F(x) = \int_0^x f(t)\,dt$. Now $F$ is $C^\infty$ and has a local maximum in $(-3,-1)$ and a local minimum in $(1,3)$. But $F'' = g$ so there is no "strict inflection point" at which $F$ changes from strictly concave to strictly convex or vice versa.</p>
<p>For polynomials this cannot happen. I will try to write more later, but in short the issue is that a polynomial can only be zero at finitely many points.</p>
|
3,752,455 | <blockquote>
<p><strong>Problem.</strong> Show that for <span class="math-container">$n\ge 2$</span> there are no solution <span class="math-container">$$x^n+y^n=z^n$$</span> such that <span class="math-container">$x$</span>, <span class="math-container">$y$</span>, <span class="math-container">$z$</span> are prime numbers.</p>
</blockquote>
<p>Personally I'd consider this a relatively cute problem which can be given to students when talking about <a href="https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem" rel="noreferrer">Fermat's Last Theorem</a> - and which should be relatively easily solvable.
(I can post my solution - but I suppose that the solutions which will be given here are very likely to be cleverer than mine.)</p>
<p>I will stress that we're looking that the solutions where <em>simultaneously</em> all three numbers are primes - unlike a more difficult problem posted here: <a href="https://math.stackexchange.com/q/1317346">Diophantine Equation <span class="math-container">$x^n + y^n =z^n (x<y, n>2)$</span></a>.</p>
<p>I have searched on the site a bit to see whether this problem has been posted here before. I only found this deleted question: <a href="https://math.stackexchange.com/q/1311295">How we can deal with this equation <span class="math-container">$a^n+b^n=c^n$</span> if it was given to have solutions in primes numbers not integers numbers?</a> (Of course, it is quite possible that I might have missed something. After all, searching on this site is not easy.)</p>
| poetasis | 546,655 | <p>Fermat's last theorem, <a href="https://en.wikipedia.org/wiki/Wiles%27s_proof_of_Fermat%27s_Last_Theorem" rel="nofollow noreferrer">now proven</a>, shows that <span class="math-container">$A^x+B^x=C^x$</span> cannot be true for <span class="math-container">$x\ne2$</span> so it is not an issue here.</p>
<p>If we use a formula I developed that generates only the subset of Pythagorean triples where <span class="math-container">$(C-B)$</span> is an odd square, we can see some relationships that show how two numbers of a triple can be prime, but not all three.</p>
<p><span class="math-container">$$A=(2n-1)^2+2(2n-1)k \qquad B=2(2n-1)k+2k^2 \qquad C=(2n-1)^2+2(2n-1)k+2k^2$$</span></p>
<p>The value of <span class="math-container">$C$</span> must be an odd number of the form <span class="math-container">$(C=4n+1)$</span> and and some of these numbers are prime but we can see that <span class="math-container">$A=(2n-1)^2+2(2n-1)k\implies A=(2n-1)(2n-1+2k)$</span> so <span class="math-container">$A$</span> is composite for all <span class="math-container">$(n>1)$</span>–– only <span class="math-container">$(n=1)$</span> can yield prime numbers for both <span class="math-container">$A$</span> and <span class="math-container">$C$</span>.</p>
<p>The value of <span class="math-container">$B$</span> can never be <span class="math-container">$2$</span> because the smallest is <span class="math-container">$4$</span> and it happens the <span class="math-container">$B$</span> is always a multiple of <span class="math-container">$4$</span> so <span class="math-container">$B$</span> cannot be prime.</p>
<p><span class="math-container">$\therefore$</span> No Pythagorean triple, primitive or otherwise, can contain three prime numbers.</p>
|
3,761,689 | <p>I was watching a YouTube video where it showed how length of daylight changes depending on the time of year, and I was curious and wanted to try calculating the value of how long the daylight is in the Tropic of Cancer (23.5 degrees latitude) during the winter solstice, apparently 10 hours and 33 minutes or so according to the video. Here is the <a href="https://www.youtube.com/watch?v=WLRA87TKXLM&t=5m27s" rel="noreferrer">timestamp</a> for reference.</p>
<p>This is my work (the yellow blobs represent 23.5 degrees and the pink blobs 43 degrees):</p>
<p><a href="https://i.stack.imgur.com/yKmZe.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/yKmZe.jpg" alt="enter image description here" /></a></p>
<p><span class="math-container">$\sin(66.5 \text{ degrees}) = (\text{yellow leg + orange leg}) / r$</span> implies <span class="math-container">$0.917060r = \text{yellow leg + orange leg}$</span></p>
<p><span class="math-container">$\cos(66.5 \text{ degrees}) = \text{purple leg} / r$</span> implies <span class="math-container">$0.398749r = \text{purple leg}$</span></p>
<p><span class="math-container">$\tan(23.5 \text{ degrees}) = \text{orange leg / purple leg}$</span> implies <span class="math-container">$0.434812 \cdot \text{ purple leg} = \text{orange leg}$</span></p>
<p>Subbing in the value we already got from the purple leg, we get <span class="math-container">$0.173381r = \text{orange leg}$</span></p>
<p>That means the orange leg is <span class="math-container">$0.173381r/ 0.917060r$</span> fraction of the yellow and orange leg, about <span class="math-container">$0.189061784$</span>. This represents how much extra darkness there is along the line.</p>
<p>Since this darkness is on both sides of the globe, I multiply it by two, to get <span class="math-container">$0.37812$</span>.</p>
<p>So the daylight is about <span class="math-container">$37.81$</span>% shorter, down from <span class="math-container">$12$</span> hours to about <span class="math-container">$7.46$</span> hours. Way off compared to the video's <span class="math-container">$10$</span> hours <span class="math-container">$33$</span> minutes.</p>
<p>Where is my mistake?</p>
| David K | 139,123 | <blockquote>
<p>That means the orange leg is <span class="math-container">$0.173381r/0.917060r$</span> fraction of the yellow and orange leg, about <span class="math-container">$0.189061784.$</span> This represents how much extra darkness there is along the line.</p>
</blockquote>
<p>Yes, that's how much extra darkness there is when you project the picture onto a flat screen and measure it there. But the Earth is not flat.</p>
<p>The distance <em>along the surface of the earth</em> for the part of the globe that you have colored orange is far less than <span class="math-container">$0.18906$</span> of the total distance along the surface of the earth that you have colored yellow and orange.</p>
<p>The path of someone standing on the tropic of Cancer for <span class="math-container">$12$</span> hours is a semicircle.
If we look straight down on the north pole from far away in space, the path looks like the semicircle in the figure below.</p>
<p><a href="https://i.stack.imgur.com/AxOxz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AxOxz.png" alt="enter image description here" /></a></p>
<p>(Note: this figure is <strong>not</strong> to scale; the orange part is drawn much larger than it should be in order to fit the numbers into the figure.)</p>
<p>What you see in the photograph (taken from far away in the direction of the arrows labeled "to the camera") is just the outer arcs of the yellow and orange segments of the semicircle.
Since the radius of the semicircle is <span class="math-container">$0.917060r$</span> and since the projection of the orange segment onto the plane of the camera is <span class="math-container">$0.173381r,$</span>
the angle of the orange segment is approximately <span class="math-container">$10.9^\circ,$</span>
since <span class="math-container">$\sin(10.9^\circ) \approx 0.18906.$</span></p>
<p>That means the distance along the arc of the orange segment is actually only
<span class="math-container">$$ \frac{10.9}{90} = 0.12111 $$</span>
portion of the length of the arcs of the yellow and orange segments combined.</p>
<blockquote>
<p>Since this darkness is on both sides of the globe, I multiply it by two, to get <span class="math-container">$0.37812.$</span></p>
</blockquote>
<p>This is a mistake. Yes, there is a dark segment on the other side of the Earth corresponding to the orange segment on the visible side. But you've forgotten that the <span class="math-container">$12$</span>-hour time period around solar noon also includes yellow and orange segments on the other side of the earth.
When you take those segments into account, the portion of that <span class="math-container">$12$</span>-hour period that is dark is</p>
<p><span class="math-container">$$ \frac{\text{orange} + \text{orange}}
{\text{orange} + \text{yellow} + \text{yellow} + \text{orange}} $$</span></p>
<p>which comes out to <span class="math-container">$0.18906$</span> again, not <span class="math-container">$0.37812$</span> ...
except that as noted above, the orange portion of the path along the Earth's surface is only <span class="math-container">$0.12111$</span> of the total, not <span class="math-container">$0.18906.$</span></p>
<p>So we end up having approximately</p>
<p><span class="math-container">$$ 0.12111 \times 12 = 1.45333 $$</span></p>
<p>hours in darkness, or about <span class="math-container">$87$</span> minutes, leaving <span class="math-container">$10$</span> hours <span class="math-container">$33$</span> minutes of daylight.</p>
<p>Note that in real life, the sun does not blink out of sight exactly when the line from the Sun's center to your position is tangent to the earth's surface. The Sun has a disk whose apparent radius is about <span class="math-container">$1/4$</span> degree, and there is also some refraction in the atmosphere which makes the sun appear to be higher in the sky than the true direction to the sun.
These effects would cause the time between sunrise and sunset to be longer than <span class="math-container">$10$</span> hours <span class="math-container">$33$</span> minutes.</p>
<p>I have no idea where the captions "<span class="math-container">$5$</span> hours <span class="math-container">$31$</span> minutes" and
"<span class="math-container">$18$</span> hours <span class="math-container">$29$</span> minutes" come from.</p>
|
873,992 | <p>I am having problem with the onto part of this problem.</p>
<p>$\mathbb{N}\rightarrow \mathbb{E}$</p>
<p>My function or pattern is </p>
<p>$x \rightarrow f(x)=2x$ </p>
<p>Which take my natural to even.</p>
<p><strong>One to One</strong></p>
<p>$f(x)=f(y)$</p>
<p>$2x=2y$</p>
<p>$x=y$</p>
<p><strong>Onto</strong></p>
<p>Let</p>
<p>$n\in E$</p>
<p>Not I have to make $x$ equal something when plug into $f(x)$ that will give n ,for all n in even.</p>
<p>And I cannot say let $x=\frac{1}{2}n$</p>
<p>Because the domain is a natural one.</p>
| ant11 | 110,047 | <p>Let $Nx$ be "$x$ is naive" and $Bx$ be "$x$ is bad".</p>
<p>Well, if $x$ is naive, $x$ can't be bad. So
$$\forall x(Nx\rightarrow \neg Bx)$$</p>
|
44,562 | <p>The question is motivated from the definition of $C^r(\Omega)$ I learned from S.S.Chern's <em>Lectures on Differential Geometry</em>:</p>
<p>Suppose $f$ is a real-valued function defined on an open set $\Omega\subset{\bf R}^m$. If all the $k$-th order partial derivatives of $f$ exist and are continuous for $k\leq r$, then we say $f\in C^r(\Omega)$. Here $r$ is some positive integer.</p>
<p>While in Folland's <em>Introduction to Partial Differential Equations</em>, $C^r(\Omega)$ denotes the space of functions possessing continuous derivatives up to order $r$ on $\Omega$, where $\Omega$ is an open subset of ${\bf R}^m$ and $k$ is a positive integer.</p>
<p>It's trivial to show that these two definitions are equivalent when $m=1$. So here is my question:</p>
<blockquote>
<p>Are these two definitions equivalent in the higher dimensions? How to prove it?</p>
</blockquote>
<hr>
<p><strong>Edit</strong>: The question was partially answered by Didier <a href="https://math.stackexchange.com/questions/44355/can-being-differentiable-imply-having-continuous-partial-derivatives/44364#44364">here</a>. I do not think it is trivial for me. It boils down to the following one:</p>
<blockquote>
<p>In the higher dimension case, say,
$f:{\bf R}^n\to {\bf R}$($n\geq 2$), what's
the relationship between the higher
order partial derivative of $f$ and
the high order derivative $f^{(k)}$?</p>
</blockquote>
<p>When $1\leq k\leq 2$, $f^{(k)}$ is the <a href="http://en.wikipedia.org/wiki/Gradient" rel="nofollow noreferrer">gradient</a> and the <a href="http://en.wikipedia.org/wiki/Hessian_matrix" rel="nofollow noreferrer">Hessian matrix</a> respectively. The answer to the above questions in these two cases is clear. I have no idea for $k>2$. I don't know much about tensor, I'm not sure if the question is related to the topic of multilinear algebra.</p>
| JDH | 413 | <p>Now that your question has been answered, let me point out that it may be interesting to observe furthermore that all the countable well-orderings are in fact represented by suborders of $\langle\mathbb{R},\lt\rangle$, and even of $\langle\mathbb{Q},\lt\rangle$. Let me give two proofs. </p>
<p>The first proof is an elementary exercise in transfinite induction. One shows that every countable ordinal $\alpha$ embeds into $\mathbb{Q}$. Note that $0$ embeds trivially. If an ordinal $\alpha$ embeds, then by composing with an isomorphism of $\mathbb{Q}$ with an interval in $\mathbb{Q}$, we may suppose the embedding is bounded above, and thereby extend it to an embedding of $\alpha+1$. If $\lambda$ is a countable limit ordinal, with $\lambda=\text{sup}_n\alpha_n$, then by induction we may map $\alpha_n$ into $\mathbb{Q}\cap (n,n+1)$, and this is an embedding of an ordinal at least as large as $\lambda$. QED</p>
<p>The second proof is simply to argue that $\langle\mathbb{Q},\lt\rangle$ is universal for countable linear orders: every countable linear order embeds into $\mathbb{Q}$. This is proved by using just the "forth" part of Cantor's famous back-and-forth argument, namely, given a linear order $L=\langle\{p_n\mid n\in\mathbb{N}\},\lt_L\rangle$, then map $p_n$ to a rational $q_n$ so that one has a order-preserving map at each finite stage. The next element $p_{n+1}$ relates to the previous elements either by being above them all, between two of them, or below them all, and we may choose a corresponding $q_{n+1}$ of the same type. So we get an order-preserving map $L\to \mathbb{Q}$. Thus, in particular, every countable well-ordering embeds into $\mathbb{Q}$. QED</p>
|
4,528,629 | <p>When doing an exercise about linear representations of finite groups I stumbled upon this Isomorphism in the comments of another <a href="https://math.stackexchange.com/questions/308680/basic-identity-of-characters?rq=1">post</a> which I was not aware of.</p>
<p>In this context <span class="math-container">$V$</span> and <span class="math-container">$W$</span> are finite dimensional Vector complex vector spaces.
I was already able to show that the dimension of the two Vector spaces or rather Tensors are the same. As such it should be sufficient to construct an injective or surjective map between them.</p>
<p>However I tried to construct one such as:
<span class="math-container">$$
\phi: (v_1 \oplus w_1) \cdot (v_2\oplus w_2) \in Sym^2(V \oplus W) \mapsto (v_1 \cdot v_2 ) \oplus ((v_1+v_2) \otimes (w_1 + w_2)) \oplus (w_1 \cdot w_2) \in Sym^2(V)\oplus (V\otimes W)\oplus Sym^2(W)
$$</span>
However it seems to me as if this map is not injective nor surjective since <span class="math-container">$\phi((v_1 \oplus w_1) \cdot (v_2\oplus w_2)) = \phi((v_1 \oplus w_2) \cdot (v_2\oplus w_1))$</span>.</p>
<p>Any tips for constructing such a map or should I be trying a different approach?</p>
| bluemaster | 460,565 | <p>Part of the answer to your question is about notation. If <span class="math-container">$\mathbf{Y}$</span> is a matrix <span class="math-container">$n\times m$</span> such that <span class="math-container">$\mathbf{Y}=[Y_{ij}]$</span>, with <span class="math-container">$Y_{ij}$</span> being a r.v. in the position <span class="math-container">$i,j$</span>, <span class="math-container">$$\mathbb{E}(\mathbf{Y})=[\mathbb{E}(Y_{ij})], i=1,\ldots,n,\ \ j=1,\ldots,m$$</span>.</p>
<p>Therefore, for <span class="math-container">$\mathbf{A}$</span> <span class="math-container">$n\times l$</span> and <span class="math-container">$\mathbf{X}$</span> <span class="math-container">$l\times m$</span> if <span class="math-container">$\mathbf{Y=AX}$</span>, <span class="math-container">$Y_{ij}=\sum_{k=1}^l A_{ik} X_{kj}$</span> and because of the linearity property of <span class="math-container">$\mathbb{E}$</span>, <span class="math-container">$$\mathbb{E}(Y_{ij})=\sum_{k=1}^m \mathbb{E}(A_{ik} X_{kj})=\sum_{k=1}^m \sigma_{ik,kj}+\mathbb{E}(A_{ik}) \mathbb{E}(X_{kj}), i=1,\ldots,n,\ \ j=1,\ldots,m.$$</span></p>
<p>In this last expression <span class="math-container">$\sigma_{ik,kj}=\text{cov}(A_{ik},X_{kj})$</span>. If <span class="math-container">$\sigma_{ik,kj}=0$</span> for all possible i,k,j, because of independence or zero covariance,
<span class="math-container">$$\mathbb{E}(Y_{ij})=\sum_{k=1}^m \mathbb{E}(A_{ik}) \mathbb{E}(X_{kj}), i=1,\ldots,n,\ \ j=1,\ldots,m. (\text{zero covariance})$$</span>
That is <span class="math-container">$$\mathbb{E}(\mathbf{A}\mathbf{X})=\mathbb{E}(\mathbf{A})\mathbb{E}(\mathbf{X})\ \text{(with zero covariance or independence)}$$</span>
If <span class="math-container">$\mathbf{A}$</span> is a matrix of constants, this last expression is valid with <span class="math-container">$\mathbb{E}(\mathbf{A})=\mathbf{A}$</span> and, in this case, <span class="math-container">$$\mathbb{E}(\mathbf{A}\mathbf{X})=\mathbf{A}\mathbb{E}(\mathbf{X}) \ \ \ \ (\mathbf{A}\ \text{is a matrix of constants}).$$</span></p>
|
1,663,113 | <p>I'm having a mind wrenching question that I just cannot answer. It's been a while since I was at the school bench so I wonder if anyone can help me out? :)</p>
<p>We have 10 students with 5 cakes each to be shared amongst each other.
The students can give the cakes out, but they can’t give a piece to a person who gives them a piece (and vice versa). They also can't give more than one piece for the same person. </p>
<p>How can they distribute the cakes so that everyone gets as many cakes as possible? </p>
| Steve Kass | 60,500 | <p>The first solution that comes to mind is for no one to give any cakes to anyone, so everyone "gets" (to keep) the five cakes they brought. But I assume you want "gets" to mean "gets from someone else."</p>
<p>If this is the idea, no one can give out all five cakes and get five back, because that would require there to be 11 students. If everyone must get the same number of cakes, the highest that number can be is four.</p>
<p>Sit the students at a round table and have everyone give one cake to each of the four people to their left. Each student will then receive four cakes, one from each of the four people on their right.</p>
|
3,676,284 | <p><a href="https://i.stack.imgur.com/9xfxz.png" rel="nofollow noreferrer">This is link to question</a>
[Here is my attempt, but the answer key is convergent. I dont think I count it wrong.<a href="https://i.stack.imgur.com/nAcEs.jpg" rel="nofollow noreferrer">][1]</a></p>
| Henno Brandsma | 4,280 | <p>Maybe working in the <a href="https://en.wikipedia.org/wiki/Constructible_universe" rel="nofollow noreferrer">constructible universe</a>, or Gödel's model of ZF would be interesting for you. There we restrict our universe (to so-called constructible sets), so that Choice becomes a theorem and we also have a "constructive" well-order of the whole universe. It solves many independent questions in a definite way (GCH holds e.g., and there is a Suslin line and many more interesting objects). </p>
|
3,676,284 | <p><a href="https://i.stack.imgur.com/9xfxz.png" rel="nofollow noreferrer">This is link to question</a>
[Here is my attempt, but the answer key is convergent. I dont think I count it wrong.<a href="https://i.stack.imgur.com/nAcEs.jpg" rel="nofollow noreferrer">][1]</a></p>
| Asaf Karagila | 622 | <p>Well, the question is what is a list. If a list is just a well ordered set, then in principle you are correct.</p>
<p>In some sense this is similar to working with Global Choice, where every set has a distinguished well ordering.</p>
<p>The problem is formalizing lists is somehow more elaborate than formalizing sets. Since lists have an inherent structure to them, you need to incorporate this into your language and into your semantics. It's not impossible, but what do you get that you can't do with ZFC, or ZF+Global Choice? Ultimately, not much. </p>
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.