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 |
|---|---|---|---|---|
745,613 | <p>I've been pondering this since yesterday. I</p>
<blockquote>
<p>Is it true that given two irreducible polynomials <span class="math-container">$f(x)$</span> and <span class="math-container">$ g(x)$</span> will <span class="math-container">$f(g(x))$</span> or <span class="math-container">$g(f(x))$</span> be irreducible?</p>
</blockquote>
| M.darwish | 111,150 | <p>I am extending @Pipicito's answer for the case when both polynomials have degree greater than 1. The answer is also no. For example, consider <span class="math-container">$f(x)=x^2-\frac{4}{3}$</span>; it is irreducible over <span class="math-container">$\mathbb{Q}$</span>. However, <span class="math-container">$f(f(x))={(x^2-\frac{4}{3})}^2-\frac{4}{3}=(x^2-2x+\frac{2}{3})(x^2+2x+\frac{2}{3})$</span> and it is reducible!</p>
<p>Moreover, there is Cappelli's lemma about this question. It says the following:</p>
<p>Let <span class="math-container">$K$</span> be a field, <span class="math-container">$f$</span> and <span class="math-container">$g$</span> be irreducible polynomials over <span class="math-container">$K$</span>. Then, <span class="math-container">$f(g(x))$</span> is irreducible over <span class="math-container">$K$</span> iff <span class="math-container">$f$</span> is irreducible over <span class="math-container">$K$</span> and <span class="math-container">$g-\alpha$</span> is irreducible over <span class="math-container">$K(\alpha)$</span> for every root <span class="math-container">$\alpha$</span> of <span class="math-container">$f$</span>.</p>
<p>In the example, the roots of <span class="math-container">$x^2-\frac{3}{4}$</span> are <span class="math-container">$\pm \frac{\sqrt{3}}{2}$</span>.
So, to show <span class="math-container">$f(f(x))$</span> is reducible, just show that one of <span class="math-container">$x^2-\frac{3}{4}\pm \frac{\sqrt{3}}{2}$</span> has roots in <span class="math-container">$\mathbb{Q}(\sqrt{3})$</span></p>
|
1,666,396 | <p>I can show the convergence of the following infinite product and some bounds for it:</p>
<p>$$\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}=\sqrt{1+\frac{1}{2}} \sqrt[3]{1+\frac{1}{3}} \sqrt[4]{1+\frac{1}{4}} \cdots<$$</p>
<p>$$<\left(1+\frac{1}{4} \right)\left(1+\frac{1}{9} \right)\left(1+\frac{1}{16} \right)\cdots=\prod_{k \geq 2} \left(1+\frac{1}{k^2} \right)=\frac{\sinh \pi}{2 \pi}=1.83804$$</p>
<p>Here I used Euler's product for $\frac{\sin x}{x}$.</p>
<p>The next upper bound is not as easy to evaluate, but still possible, taking two more terms in Taylor's series for $\sqrt[k]{1+\frac{1}{k} }$:</p>
<p>$$\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}<\prod_{k \geq 2} \left(1+\frac{1}{k^2}-\frac{k-1}{2k^4}+\frac{2k^2-3k+1}{6k^6} \right)=$$</p>
<p>$$=\prod_{k \geq 2} \left(1+\frac{1}{k^2}-\frac{1}{2k^3}+\frac{5}{6k^4}-\frac{1}{2k^5}+\frac{1}{6k^6} \right)<$$</p>
<p>$$<\exp \left( \frac{\pi^2}{6}+\frac{\pi^4}{108}+\frac{\pi^6}{5670}-1-\frac{\zeta (3)}{2}-\frac{\zeta (5)}{2} \right)=1.81654$$</p>
<p>The numerical value of the infinite product is approximately:</p>
<p>$$\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}=1.758743628$$</p>
<p>The ISC found no closed from for this number.</p>
<blockquote>
<p>Is there some way to evaluate this product or find better bounds in closed form?</p>
</blockquote>
<hr>
<p><strong>Edit</strong></p>
<p>Clement C suggested taking logarithm and it was a very useful suggestion, since I get the series:</p>
<p>$$\ln \prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}= \frac{1}{2} \ln \left(1+\frac{1}{2} \right)+\frac{1}{3} \ln \left(1+\frac{1}{3} \right)+\dots$$</p>
<p>I don't know how to find the closed form, but I can certainly use it to find the boundaries (since the series for logarithm are very simple).</p>
<p>$$\frac{1}{2} \ln \left(1+\frac{1}{2} \right)+\frac{1}{3} \ln \left(1+\frac{1}{3} \right)+\dots>\sum^{\infty}_{k=2} \frac{1}{k^2}-\frac{1}{2}\sum^{\infty}_{k=2} \frac{1}{k^3}$$</p>
<p>$$\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}>\exp \left( \frac{\pi^2}{6}-\frac{1}{2}-\frac{\zeta (3)}{2} \right)=1.72272$$</p>
<p>$$\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}<\exp \left( \frac{\pi^2}{6}+\frac{\pi^4}{270}-\frac{5}{6}-\frac{\zeta (3)}{2} \right)=1.77065$$</p>
<p>$$\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}>\exp \left( \frac{\pi^2}{6}+\frac{\pi^4}{270}-\frac{7}{12}-\frac{\zeta (3)}{2} -\frac{\zeta (5)}{4}\right)=1.75438$$</p>
<p>$$\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}<\exp \left( \frac{\pi^2}{6}+\frac{\pi^4}{270}+\frac{\pi^6}{4725}-\frac{47}{60}-\frac{\zeta (3)}{2} -\frac{\zeta (5)}{4}\right)=1.76048$$</p>
<blockquote>
<p>$$\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}>\exp \left( \frac{\pi^2}{6}+\frac{\pi^4}{270}+\frac{\pi^6}{4725}-\frac{37}{60}-\frac{\zeta (3)}{2} -\frac{\zeta (5)}{4} -\frac{\zeta (7)}{6}\right)=1.75803$$</p>
</blockquote>
<p>This method generates much better bounds than my first idea. The last two are very good approximations.</p>
<hr>
<p><strong>Edit 2</strong></p>
<p>Actually, would it be correct to write (it gives the correct value of the product):</p>
<blockquote>
<p>$$\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}=\frac{1}{2} \exp \left( \sum_{k \geq 2} \frac{(-1)^k \zeta(k)}{k-1} \right)$$</p>
</blockquote>
| robjohn | 13,854 | <p>I don't know if a closed form exists, but to get geometric convergence, we can use the following.</p>
<p>Since the product starts at $k=2$, we compute the sum
$$
\begin{align}
\sum_{k=2}^\infty\frac1k\log\left(1+\frac1k\right)
&=\sum_{k=2}^\infty\frac1k\sum_{n=1}^\infty\frac{(-1)^{n-1}}{nk^n}\\
&=\sum_{n=1}^\infty\frac{(-1)^{n-1}}n\sum_{k=2}^\infty\frac1{k^{n+1}}\\
&=\sum_{n=1}^\infty(-1)^{n-1}\frac{\zeta(n+1)-1}{n}\\[6pt]
&=0.564599706384424320592667709038
\end{align}
$$
Note that $\frac{\zeta(n+1)-1}{n}\sim\frac1{n2^{n+1}}$. This gives better than geometric convergence.</p>
<p>Applying $e^x$, we get
$$
\prod_{k=2}^\infty\left(1+\frac1k\right)^{1/k}=1.75874362795118482469989684966
$$</p>
|
4,128,050 | <p><a href="https://i.stack.imgur.com/ybNTh.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ybNTh.jpg" alt="enter image description here" /></a></p>
<p>I think we should use the corollary to solve this problem.</p>
<p>For the first question, I think we can use the property that <span class="math-container">$J$</span> is a free <span class="math-container">$\mathbb{Z}[G]$</span> module with a basis <span class="math-container">${s-1,t-1}$</span>. Hence <span class="math-container">$J\otimes \mathbb{Z^{'}}= 2\mathbb{Z}[G]\otimes \mathbb{Z^{'}}$</span>.</p>
<p>But I don't know how to solve the second question.</p>
<p>Thank you for your help.</p>
| J. Darné | 611,408 | <p>Since <span class="math-container">$J$</span> is a free <span class="math-container">$\mathbb Z[G]$</span>-module of rank two (the two generators being <span class="math-container">$(s-1)$</span> and <span class="math-container">$(t-1)$</span>), the tensor product <span class="math-container">$J \otimes_{\mathbb Z[G]} \mathbb Z'$</span> is the direct sum of two copies of <span class="math-container">$\mathbb Z [G] \otimes_{\mathbb Z[G]} \mathbb Z' \cong \mathbb Z$</span>. So the tensor product of the resolution of the trivial representation <span class="math-container">$0 \rightarrow J \rightarrow \mathbb Z[G] \rightarrow 0$</span> with <span class="math-container">$\mathbb Z'$</span> is:
<span class="math-container">$$0 \rightarrow \mathbb Z \oplus \mathbb Z \rightarrow \mathbb Z \rightarrow 0.$$</span>
Then you need to understand the middle map. The calculations that you have already done in your question (which are in fact just a reformulation of the definition of <span class="math-container">$\mathbb Z'$</span>) give you that the two elements <span class="math-container">$(s-1) \otimes 1$</span> and <span class="math-container">$(t-1) \otimes 1$</span>, which form a basis of the source, are both sent to <span class="math-container">$1 \otimes 2 = 2 (1 \otimes 1)$</span> in the target, which means that this map identifies with <span class="math-container">$(m,n) \mapsto 2(m+n)$</span> (it sends both <span class="math-container">$(1,0)$</span> and <span class="math-container">$(0,1)$</span> to <span class="math-container">$2$</span>). Its cokernel is thus <span class="math-container">$\mathbb Z/2$</span>, and its kernel is <span class="math-container">$\mathbb Z$</span>, as expected (and you can write these generators as the classes of explicit cycles if you like).</p>
<p>For computing the homology <span class="math-container">$H_*(T, \mathbb Z')$</span>, you have the same resolution (since <span class="math-container">$T$</span> is free on one generator), and you can use exactly the same reasoning.</p>
|
4,103,366 | <p>I have <span class="math-container">$A \in \mathbb{R}^{q\times n }, B \in \mathbb{R}^{n \times p} $</span> with <span class="math-container">$\text{rank}(A)=q~$</span> and <span class="math-container">$~\text{rank}(B)=p$</span>.</p>
<p>Additionally there is the condition: <span class="math-container">$n\geq p \geq q$</span>.</p>
<p>I know that <span class="math-container">$\text{rank}(AB)\leq \min\{\text{rank}(A), \text{rank}(B)\}=q$</span>.</p>
<p>I want to know if equality (<span class="math-container">$\text{rank}(AB)=q$</span>) always holds with the additional condition <span class="math-container">$n\geq p \geq q~$</span>, or are there some constraints?</p>
| Ben Grossmann | 81,360 | <p>No, the equality does not necessarily hold. Consider the case of <span class="math-container">$n = 3,p=2,q=1$</span>. Take
<span class="math-container">$$
A = \pmatrix{1&0&0},\quad B = \pmatrix{0&0\\1&0\\0&1}.
$$</span>
We have <span class="math-container">$\operatorname{rank}(AB) = 0 < q = 1$</span>.</p>
<hr />
<p>One possibly helpful additional condition: if <span class="math-container">$p = n$</span> (so that <span class="math-container">$B$</span> is square) and both <span class="math-container">$A$</span> and <span class="math-container">$B$</span> have full rank, then it will always hold that <span class="math-container">$\operatorname{rank}(AB) = \operatorname{rank}(A) = q$</span>.</p>
|
3,429,489 | <blockquote>
<p>Let <span class="math-container">$\mathcal F = \{f \mid f : \mathbb R \rightarrow \mathbb R\}$</span> and
define relationship <span class="math-container">$R$</span> on <span class="math-container">$\mathcal F$</span> as follows:</p>
<p><span class="math-container">$$R = \{(f,g) \in \mathcal F \times \mathcal F \mid \exists h \in \mathcal F (f = h \circ g)\}$$</span></p>
<p>Prove that for all <span class="math-container">$f \in \mathcal F$</span>, <span class="math-container">$i_{\mathbb R}Rf$</span> iff <span class="math-container">$f$</span> is
one-to-one.</p>
</blockquote>
<p><span class="math-container">$$i_{\mathbb R} = \{(x,x) \mid x \in \mathbb R\}$$</span></p>
<p>My attempt:</p>
<p><span class="math-container">$(\rightarrow)$</span></p>
<p>Suppose <span class="math-container">$i_{\mathbb R}Rf$</span>. Exists function <span class="math-container">$h \in F$</span>, such that </p>
<p><span class="math-container">$$i_{\mathbb R} = h \circ f $$</span></p>
<p>Suppose <span class="math-container">$(x,a_1) \in f$</span> and <span class="math-container">$(x,a_2) \in f$</span></p>
<p>Since <span class="math-container">$a_1,a_2 \in \mathbb R$</span> and <span class="math-container">$h: \mathbb R \rightarrow \mathbb R$</span>, there exist <span class="math-container">$b_1,b_2 \in \mathbb R$</span> such that <span class="math-container">$(a_1,b_1) \in h$</span> and <span class="math-container">$(a_2,b_2) \in h$</span></p>
<p>Then we have <span class="math-container">$(x,b_1) \in i_{\mathbb R}$</span> and <span class="math-container">$(x,b_2) \in i_{\mathbb R}$</span>, which means that <span class="math-container">$b_1 = b_2$</span> </p>
<p>Since <span class="math-container">$h$</span> is a function, <span class="math-container">$a_1 = a_2$</span></p>
<p><span class="math-container">$(\leftarrow)$</span></p>
<p>Suppose <span class="math-container">$f$</span> is one-to-one. Define <span class="math-container">$f^{-1}$</span> as </p>
<p><span class="math-container">$$f^{-1} = \{(y,x) \mid (x,y) \in f\} $$</span></p>
<p>Then <span class="math-container">$f^{-1} \circ f = i_{\mathbb R}$</span>, and thus <span class="math-container">$i_{\mathbb R}Rf$</span></p>
<p><span class="math-container">$\Box$</span></p>
<hr>
<p>Is it correct?</p>
| Community | -1 | <p>The only problem that I see is that you have assumed that <span class="math-container">$f^{-1}$</span> has a domain of <span class="math-container">$\mathbb R$</span> in the last part. That is only true if <span class="math-container">$f$</span> is also surjective. It is true that every injective function has a left-inverse, but you have to define such a function for the entire domain with a little more care. Not that it's a difficult task, but it's part of the proof.</p>
|
4,008,488 | <p>While looking for the answer for my question I came across <a href="https://math.stackexchange.com/questions/1840801/why-is-ata-invertible-if-a-has-independent-columns?rq=1">this</a> post. It may be a silly idea, but if <span class="math-container">$A^{t}$</span> has independent rows can I just transpose it and get <span class="math-container">$A$</span> with independent columns and proceed as is shown in the linked solution? Or it is not working this way?</p>
| Ali | 251,307 | <p>For every matrix <span class="math-container">$A$</span> with independent rows, we have <span class="math-container">$det A \neq 0$</span> also <span class="math-container">$det A= det A^t$</span>so both <span class="math-container">$A$</span> and <span class="math-container">$A^t$</span> are invertible. Then <span class="math-container">$AA^t$</span> is invertible if and only if <span class="math-container">$A$</span> is.</p>
|
4,351,990 | <p>I have just finished my undergrad and while I haven't studied much in representation theory I find it a very fascinating subject. My current interest is in differential equations, and I am wondering is there any ongoing research that combines these two areas?</p>
| paul garrett | 12,291 | <p>The wide variation in interpretation of your question should be a convincing indicator that your question is not going to get a useful answer. At extremes, the answer is "obviously yes" or "obviously no", depending on interpretations of many of the words... :)</p>
<p>That is, on one hand, if people in one field, with a self-chosen label, want to de-legitimize other people who claim the same label... well, does that really mean anything about the mathematics itself? So the answer to your question can easily be made to be "no", by fiat. :)</p>
<p>On another hand, especially in my experience in "number theory in the broad sense", almost every bit of mathematics has some significant (if not profound) connection to every other. So, the answer to your question is trivially "yes". :)</p>
<p>Still, I would tend to be sympathetic to @markvs's answer: the modern theory of automorphic forms uses a lot of representation theory. The representation theory itself uses some very serious PDE business: in the rank-one case, it's really just the classical theory of asymptotics of ODEs. But in the higher-rank case, the corresponding subquotient theorem of Harish-Chandra, and then the subrepresentation theorem of Casselman (and Milicic), are significantly subtler. (The Casselman-Milicic paper has a wonderful appendix expanding a paper of Deligne's on PDE with regular singular points...)</p>
<p>Not so elementary, though.</p>
|
4,722 | <p>Is there a necessary and sufficient condition for the boundary of a planar region to be a finite union of Jordan curves?</p>
| Harald Hanche-Olsen | 802 | <p>I am going to throw caution to the wind and suggest an Answer, based on the previous comments: No, there is no useful characterization of the regions you seek, other than the requirement stated. On the one hand, there is the region $\{(x,y):0\lt x\lt 1, -2\lt y\lt \sin x^{-1}\}$, and on the other, you can make part of the boundary an Osgood curve, by which I mean a Jordan arc of positive area. (See, e.g., W. F. Osgood, <em>A Jordan curve of positive area</em>. Trans. Amer. Math. Soc. <strong>4</strong> (1903) 107–112). Between these two examples, I think you'll be hard pressed to find an easily checkable local condition on the boundary that will guarantee the “Jordan-ness” of the boundary curve.</p>
|
4,722 | <p>Is there a necessary and sufficient condition for the boundary of a planar region to be a finite union of Jordan curves?</p>
| Clinton Curry | 1,743 | <p>Local connectivity of the boundary provides much of the topological structure that such a domain would have. If you specify the following two conditions, you have that the boundary of a domain $U$ is a finite union of simple closed curves, I think.</p>
<ol>
<li>$\partial U$ is locally connected.</li>
<li>$\overline U$ has finitely many complementary components.</li>
</ol>
<p>The thrust is this: if $V$ is a complementary component of $\overline U$, then $\partial V$ is the common boundary of two simply connected domains ($V$ and the component of the complement of $\overline V$ containing $U$). A locally connected plane continuum which is the common boundary of two connected open sets is always a circle.</p>
|
2,368,827 | <p>I would like to know how a piecewise function and its derivative would look like under these circumstances. Suppose that the function is continuous (and also nice like poly, trig etc) but defined differently for points $\le a$ and point $\gt a $</p>
<p>1) The function is differentiable at $a$. Then the derivative would be continuous at $a$, but would it be differentiable at $a$?</p>
<p>2) The function is continuous at $a$ but not differentiable at $a$. Then the derivative would not be defined at $a$ but defined elsewhere. Is this correct? Also would the left and right limits of the derivative be equal at $a$?</p>
| Emilio Novati | 187,568 | <p>Consider the function:
$$
y= \begin{cases}
x^2 \quad \mbox{for} \quad x\ge0\\
x^3 \quad \mbox{for} \quad x<0\\
\end{cases}
$$
this function is differentiable at $x=0$ ( and for any $x\in \mathbb{R}$) but its derivative is continuous but not differentiable at $x=0$</p>
|
14,712 | <p>I have matrix <code>in</code> as shown, consisting of real numbers and 0. How can I sort it to become <code>out</code> as shown?</p>
<pre><code>in ={
{0, 0, 3.411, 0, 1.343},
{0, 0, 4.655, 2.555, 3.676},
{0, 3.888, 0, 3.867, 1.666}
};
out ={
{1.343, 3.411, 0, 0, 0},
{2.555, 3.676, 4.655, 0, 0},
{1.666, 3.867, 3.888, 0, 0}
};
</code></pre>
<p>This is related to a <a href="https://mathematica.stackexchange.com/questions/14663/">question I asked</a>. It is much easier to add the columns by sorting it this way than in previous question, and easier to visualize than trying to take the first non-zero value in a row.</p>
| Rojo | 109 | <pre><code>in = {{0, 0, 3.411, 0, 1.343}, {0, 0, 4.655, 2.555, 3.676}, {0, 3.888,
0, 3.867, 1.666}};
</code></pre>
<p>A possible solution</p>
<pre><code>Sort /@ (in I - Unitize[in]) // Im
</code></pre>
|
14,712 | <p>I have matrix <code>in</code> as shown, consisting of real numbers and 0. How can I sort it to become <code>out</code> as shown?</p>
<pre><code>in ={
{0, 0, 3.411, 0, 1.343},
{0, 0, 4.655, 2.555, 3.676},
{0, 3.888, 0, 3.867, 1.666}
};
out ={
{1.343, 3.411, 0, 0, 0},
{2.555, 3.676, 4.655, 0, 0},
{1.666, 3.867, 3.888, 0, 0}
};
</code></pre>
<p>This is related to a <a href="https://mathematica.stackexchange.com/questions/14663/">question I asked</a>. It is much easier to add the columns by sorting it this way than in previous question, and easier to visualize than trying to take the first non-zero value in a row.</p>
| Mr.Wizard | 121 | <p>You might use:</p>
<pre><code>SortBy[#, # /. 0 -> {} &] & /@ in
</code></pre>
<p>This works because <code>{}</code> will be placed after atomic elements such as real numbers. If your zeros may not always be precise (head Integer) you may use:</p>
<pre><code>SortBy[#, # /. x_ /; x == 0 -> {} &] & /@ in
</code></pre>
|
2,262,167 | <p>The question in title has been considered for finite groups $G$ and $H$, but I do not know its situation, how far it is known whether $G$ and $H$ could be isomorphic. I have two simple questions regarding it.</p>
<p><strong>Q.0</strong> If $\mathbb{Z}[G]\cong \mathbb{Z}[H]$ then $|G|$ should be $|H|$; because, $G$ is a free basis for the <em>free additive abelian group $\mathbb{Z}[G]$</em>, am I right? (I am asking this because in Isaac's character theory, I saw something different argument, not too lengthy, but I was thinking for the above natural arguments.)</p>
<p><strong>Q.1</strong> Are there known examples of finite groups $G\ncong H$ with $\mathbb{Z}[G]\cong \mathbb{Z}[H]$? (In the book of character theory by Isaacs, he stated for metablelian groups $G,H$, $\mathbb{Z}[G]\cong \mathbb{Z}[H]$ implies $G\cong H$; it was proved by Whitcomb, in $1970$; but book has not been further revised, I don't known what is done after $1970$).</p>
| Jeremy Rickard | 88,262 | <p>You're right about the zeroth question.</p>
<p>For the first, a counterexample was constructed by Hertweck in 2001. There are two non-isomorphic groups $G$ and $H$ of order $2^{21}97^{28}$ with $\mathbb{Z}[G]\cong\mathbb{Z}[H]$.</p>
|
2,905,022 | <p>I recently stumbled upon the problem $3\sqrt{x-1}+\sqrt{3x+1}=2$, where I am supposed to solve the equation for x. My problem with this equation though, is that I do not know where to start in order to be able to solve it. Could you please give me a hint (or two) on what I should try first in order to solve this equation?</p>
<p><strong>Note</strong> that I only want hints.</p>
<p>Thanks for the help!</p>
| Rakibul Islam Prince | 551,644 | <p><strong>Hint :</strong></p>
<p>Let,$$x=t^2+1$$
So, $$3t+\sqrt{3t^2+4}=2$$
$$\implies 3t^2+4=(2-3t)^2$$</p>
<p>Now,it's your turn to go on...</p>
<p>The answer should be $$x=1,if~t=0$$</p>
<p>$$and$$ $$x=5,if~t=2$$</p>
|
2,905,022 | <p>I recently stumbled upon the problem $3\sqrt{x-1}+\sqrt{3x+1}=2$, where I am supposed to solve the equation for x. My problem with this equation though, is that I do not know where to start in order to be able to solve it. Could you please give me a hint (or two) on what I should try first in order to solve this equation?</p>
<p><strong>Note</strong> that I only want hints.</p>
<p>Thanks for the help!</p>
| user579102 | 579,102 | <p><span class="math-container">$$3\sqrt{x-1}+\sqrt{3x+1}=2$$</span>
this equation is defined for <span class="math-container">$$x\geq 1$$</span>
<span class="math-container">$$9(x-1)+6\sqrt{(x-1)(3x+1)}+3x+1=4$$</span>
<span class="math-container">$$12x+6\sqrt{(x-1)(3x+1)}=12$$</span>
<span class="math-container">$$2x+\sqrt{(x-1)(3x+1)}=2$$</span>
<span class="math-container">$$\sqrt{(x-1)(3x+1)}=2-2x ; 0\leq x\leq1$$</span>
<span class="math-container">$$(x-1)(3x+1)=4(1-x)^2$$</span>
<span class="math-container">$$(x-1)(3x+1+4-4x)=0$$</span>
<span class="math-container">$$(x-1)(-x+5)=0$$</span>
<span class="math-container">$$x=1$$</span>
I know there is something wrong with my solution, but I hope someone will help to make it clear and correct any mistake.</p>
|
99,506 | <p>I am trying to show that how the binary expansion of a given positive integer is unique.</p>
<p>According to this link, <a href="http://www.math.fsu.edu/~pkirby/mad2104/SlideShow/s5_3.pdf" rel="nofollow">http://www.math.fsu.edu/~pkirby/mad2104/SlideShow/s5_3.pdf</a>, All I see is that I can recopy theorem 3-1's proof?</p>
<p>Is this polished enough of an argument. Thanks</p>
| N. S. | 9,176 | <p>Assume by contradiction that a number $n$ has two different binary expansions.</p>
<p>Then $n=a_m...a_0=b_k...b_0$.</p>
<p>Let $l$ be the smallest index so that $a_l \neq b_l$. It follows that the binary numbers $a_{l-1}...a_1a_0=b_{l-1}...b_1b_0$. By subtracting these from $n$ we get</p>
<p>$$a_m...a_l00000..0=b_k...b_l0000...0 \,.$$</p>
<p>Now $a_l\neq b_l$ means that one of these is $0$ and the other is $1$. But then, one side ends in at least $l+1$ zeroes, meaning is divisible by $2^{l+1}$, while the other ends in exactly $l$ zeroes, meaning is not divisible by $2^{l+1}$.</p>
<p>This contradicts the fact that they are equal....</p>
|
2,971,313 | <blockquote>
<p>Prove that every even degree polynomial function <span class="math-container">$f$</span> has maximum or minimum in <span class="math-container">$\mathbb{R}$</span>. (without direct using of derivative and making <span class="math-container">$f'$</span>)</p>
</blockquote>
<p>The problem seems very easy and obvious but I don't know how to write it in a mathematical way.</p>
<p>For example if the largest coefficient is positive, it seem obvious to me that from a point <span class="math-container">$x=a$</span> to <span class="math-container">$+\infty$</span> the function must be completely ascending. And from <span class="math-container">$-\infty$</span> to a point <span class="math-container">$x=b$</span> the function must be completely descending. If it is not like that, its limit will not be <span class="math-container">$+\infty$</span> at <span class="math-container">$\pm\infty$</span>. Now, because it is continuous, it will have a maximum and minimum in <span class="math-container">$[b,a]$</span> so it will have a minimum (because every other <span class="math-container">$f(x)$</span> where <span class="math-container">$x$</span> is outside <span class="math-container">$[b,a]$</span> is larger than <span class="math-container">$f(a)$</span> or <span class="math-container">$f(b)$</span> and we get the minimum that is less than or equal to both of them) . We can also the same for negative coefficient.</p>
<p>But I can't write this in a formal mathematical way.</p>
| José Carlos Santos | 446,262 | <p>Let <span class="math-container">$f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$</span>. Let us assume that <span class="math-container">$a_n>0$</span> (the case in which <span class="math-container">$a_n<0$</span> is similar). Then<span class="math-container">\begin{align}\lim_{x\to\pm\infty}f(x)&=\lim_{x\to\pm\infty}a_nx^n\left(1+\frac{a_{n-1}}{a_nx}+\frac{a_{n-2}}{a_nx^2}+\cdots+\frac{a_0}{a_nx^n}\right)\\&=+\infty\times1\\&=+\infty.\end{align}</span>Therefore, there is a <span class="math-container">$R>0$</span> such that <span class="math-container">$\lvert x\rvert>R\implies f(x)\geqslant f(0)$</span>. So, consider the restriction of <span class="math-container">$f$</span> to <span class="math-container">$[-R,R]$</span>. Since <span class="math-container">$f$</span> is continuous and since <span class="math-container">$[-R,R]$</span> is closed and bounded, <span class="math-container">$f|_{[-R,R]}$</span> attains a minimum at some point <span class="math-container">$x_0\in[-R,R]$</span> and, of course, <span class="math-container">$f(x_0)\leqslant f(0)$</span>. Since outside <span class="math-container">$[-R,R]$</span> you always have <span class="math-container">$f(x)\geqslant f(0)$</span>, <span class="math-container">$f$</span> attains its absolute minimum at <span class="math-container">$x_0$</span>.</p>
|
2,971,313 | <blockquote>
<p>Prove that every even degree polynomial function <span class="math-container">$f$</span> has maximum or minimum in <span class="math-container">$\mathbb{R}$</span>. (without direct using of derivative and making <span class="math-container">$f'$</span>)</p>
</blockquote>
<p>The problem seems very easy and obvious but I don't know how to write it in a mathematical way.</p>
<p>For example if the largest coefficient is positive, it seem obvious to me that from a point <span class="math-container">$x=a$</span> to <span class="math-container">$+\infty$</span> the function must be completely ascending. And from <span class="math-container">$-\infty$</span> to a point <span class="math-container">$x=b$</span> the function must be completely descending. If it is not like that, its limit will not be <span class="math-container">$+\infty$</span> at <span class="math-container">$\pm\infty$</span>. Now, because it is continuous, it will have a maximum and minimum in <span class="math-container">$[b,a]$</span> so it will have a minimum (because every other <span class="math-container">$f(x)$</span> where <span class="math-container">$x$</span> is outside <span class="math-container">$[b,a]$</span> is larger than <span class="math-container">$f(a)$</span> or <span class="math-container">$f(b)$</span> and we get the minimum that is less than or equal to both of them) . We can also the same for negative coefficient.</p>
<p>But I can't write this in a formal mathematical way.</p>
| G Cab | 317,234 | <p>In an even degree polynomial <span class="math-container">$p_{\,2n}(x) \quad |\; 0<n$</span>, we have that
<span class="math-container">$$
\mathop {\lim }\limits_{x\; \to \, + \infty } p_{\,2n} (x) = \mathop {\lim }\limits_{x\; \to \, - \infty } p_{\,2n} (x) = \pm \infty
$$</span>
and it does not have other <em>poles</em> other than those.</p>
<p>Assume that the limit is <span class="math-container">$+ \infty$</span>.</p>
<p>Since a polynomial is a continuous surjective function, its co-domain will be limited below, and this limit will correspond to one or more
values of <span class="math-container">$x$</span>.</p>
<p>The same applies, flipped over, if the limit is <span class="math-container">$-\infty$</span>.</p>
|
547,050 | <p>Which trigonometric formulas are used for these problems?
<img src="https://i.stack.imgur.com/TVBCx.png" alt="enter image description here"></p>
| lab bhattacharjee | 33,337 | <p>HINT:</p>
<p>Use $$\sin C+\sin D=2\sin\frac{C+D}2\cos\frac{C-D}2$$</p>
<p>and $$\cos C+\cos D=2\cos\frac{C+D}2\cos\frac{C-D}2$$</p>
|
2,195,197 | <blockquote>
<p>A circle goes through $(5,1)$ and is tangent to $x-2y+6=0$ and $x-2y-4=0$. What is the circle's equation?</p>
</blockquote>
<p>All I know is that the tangents are parallel, which means I can calculate the radius as half the distance between them: $\sqrt5$. So my equation is
$$(x-p)^2+(y-q)^2=5$$
How can I get the locations of the centre? (I think there are 2 solutions.)</p>
| lab bhattacharjee | 33,337 | <p>As the radius $=$ the perpendicular distance of a tangent from the center.</p>
<p>If $(h,k)$ is the center, radius $r=\dfrac{|h-2k+6|}{\sqrt{1^2+2^2}}=\dfrac{|h-2k-4|}{\sqrt{1^2+2^2}}$</p>
<p>Squaring we get, $$(h-2k+6)^2=(h-2k-4)^2\iff h=2k-1$$</p>
<p>$r=\dfrac{|-1+6|}{\sqrt{1^2+2^2}}=\sqrt5$</p>
<p>Finally $$(\sqrt5)^2=(h-5)^2+(k-1)^2=(2k-1-5)^2+(k-1)^2$$</p>
<p>$k=?$</p>
|
360,663 | <p>Let <span class="math-container">$\mathcal{D}$</span> be a triangulated category and a <span class="math-container">$t$</span>-structure <span class="math-container">$(\mathcal{D}^{\leq 0},\mathcal{D}^{\geq 0})$</span> on <span class="math-container">$\mathcal{D}$</span>. The heart of the <span class="math-container">$t$</span>-structure, <span class="math-container">$\mathcal{A}=\mathcal{D}^{\leq 0} \cap \mathcal{D}^{\geq 0}$</span>, is an abelian category. </p>
<p>I know that in general there is not a natural functor from the derived category of the heart and the triangulated category .</p>
<p>But it's seem to be very linked ...</p>
<p>But if you consider stable infinite category or Grothendieck derivator is there such functor ?</p>
| Maxime Ramzi | 102,343 | <p>Assume <span class="math-container">$\mathcal D$</span> is a presentable stable <span class="math-container">$\infty$</span>-category with a <span class="math-container">$\mathrm t$</span>-structure (which is accessible and compatible with filtered colimits), and let <span class="math-container">$\mathcal A$</span> be its heart, <span class="math-container">$\mathcal{D(A)}$</span> its derived <span class="math-container">$\infty$</span>-category.</p>
<p>Note that under those hypotheses, <span class="math-container">$\mathcal A$</span> is Grothendieck abelian (<em>Higher Algebra</em>, 1.3.5.23.).</p>
<p>You get a natural inclusion functor <span class="math-container">$\mathcal A\to \mathcal D$</span>, which you can extend to <span class="math-container">$Fun(\Delta^{op},\mathcal A)\to \mathcal D$</span> (by geometric realization).</p>
<p>This functor preserves weak equivalences : indeed, (see <em>HA</em>, 1.2.4.4. and 1.2.4.5.), if <span class="math-container">$X$</span> is a simplicial object of <span class="math-container">$\mathcal A$</span> (and therefore <span class="math-container">$\mathcal D_{\geq 0}$</span>), there is a spectral sequence with <span class="math-container">$E^2_{p,q}=\pi_p\pi_q(X)$</span> converging to <span class="math-container">$\pi_{p+q}(|X|)$</span> in <span class="math-container">$\mathcal A$</span>.</p>
<p>Since <span class="math-container">$\pi_q(X) = 0$</span> for <span class="math-container">$q\neq 0$</span> in our situation (as <span class="math-container">$X$</span> takes values in <span class="math-container">$\mathcal A$</span>), this spectral sequence degenerates, and <span class="math-container">$\pi_p(|X|) = \pi_p(X_\bullet)$</span> (where the latter are homotopy groups as computed in <span class="math-container">$\mathcal A$</span> via the classical Dold-Kan correspondance)</p>
<p>It follows that this functor yields a (unique up to a contractible space of choices) functor <span class="math-container">$\mathcal D_{\geq 0}(\mathcal A)\to \mathcal D$</span> (via, again the Dold-Kan correspondance).</p>
<p>Now <span class="math-container">$\mathcal D$</span> is presentable and stable, and this functor <span class="math-container">$\mathcal D_{\geq 0}(\mathcal A)\to \mathcal D$</span> preserves colimits, so it extends (again, uniquely) to a functor <span class="math-container">$\mathcal{D(A) \to D}$</span> which preserves colimits and preserves <span class="math-container">$\mathcal A$</span> (<em>HA</em> 1.3.5.21. : <span class="math-container">$\mathcal{D(A)}$</span> is right complete with its classical <span class="math-container">$\mathrm t$</span>-structure, which implies in particular that <span class="math-container">$\mathcal{D(A)} = \lim(\dots \overset{\Omega}\to \mathcal D_{\geq 0}(\mathcal A)\overset{\Omega}\to \mathcal D_{\geq 0}(\mathcal A))$</span>, and then we use 1.4.4.5. which says that this precisely has the universal property of "presentable stabilization")</p>
<p>Now these hypotheses on <span class="math-container">$\mathcal D$</span> may look pretty strong but they're reasonable ; and in fact you can't really hope for much better : if <span class="math-container">$\mathcal D$</span> isn't presentable, then it could be something like <span class="math-container">$\mathcal D^{-}(\mathcal A)$</span> and then there's no hope to get a sensible functor <span class="math-container">$\mathcal{D(A)}\to \mathcal D^{-}(\mathcal A)$</span> (this is something that won't change whether you're in an <span class="math-container">$\infty$</span>-categorical setting or not). </p>
<p>This is probably already somewhere in <em>HA</em> but I couldn't find it written down completely. </p>
<p>As I pointed out in the comments, under different hypotheses (which may look weaker), you can get away with a natural functor <span class="math-container">$\mathcal D^{-}(\mathcal A)\to \mathcal D$</span> (morally, this is because <span class="math-container">$\mathcal D^{-}(\mathcal A) \subset \mathcal{D(A)}$</span> is <span class="math-container">$\bigcup_n \mathcal D_{\geq n}(\mathcal A)$</span>, and so it is determined by <span class="math-container">$\mathcal D_{\geq 0}(\mathcal A)$</span> via finite limits, so you don't need a presentability hypothesis- you do, however need some hypothesis on <span class="math-container">$\mathcal D$</span> to be able to replace the first step, since you can't take colimits as easily).</p>
<p>I guess there may also be a more general statement about <span class="math-container">$\mathcal D^b(\mathcal A)$</span>, the bounded derived <span class="math-container">$\infty$</span>-category, which should require less hypotheses, since "everything is finite" but I couldn't tell you on the top of my head. </p>
|
360,663 | <p>Let <span class="math-container">$\mathcal{D}$</span> be a triangulated category and a <span class="math-container">$t$</span>-structure <span class="math-container">$(\mathcal{D}^{\leq 0},\mathcal{D}^{\geq 0})$</span> on <span class="math-container">$\mathcal{D}$</span>. The heart of the <span class="math-container">$t$</span>-structure, <span class="math-container">$\mathcal{A}=\mathcal{D}^{\leq 0} \cap \mathcal{D}^{\geq 0}$</span>, is an abelian category. </p>
<p>I know that in general there is not a natural functor from the derived category of the heart and the triangulated category .</p>
<p>But it's seem to be very linked ...</p>
<p>But if you consider stable infinite category or Grothendieck derivator is there such functor ?</p>
| Dan Petersen | 1,310 | <p>I had reason to think about this a few years ago. When <span class="math-container">$\mathcal D$</span> arises as the derived category of an abelian category (with a possibly exotic <span class="math-container">$t$</span>-structure), a construction of a realization functor <span class="math-container">$D^b(A) \to \mathcal D$</span> can be found already in Beilinson-Bernstein-Deligne-Gabber. (For them <span class="math-container">$\mathcal D$</span> is the derived category of constructible sheaves, equipped with the perverse <span class="math-container">$t$</span>-structure, so <span class="math-container">$A$</span> is the category of perverse sheaves.)
Their construction of the realization functor can in fact be imitated in the <span class="math-container">$\infty$</span>-categorical setting, too, and in this case it works more generally for an arbitrary stable <span class="math-container">$\infty$</span>-category <span class="math-container">$\mathcal D$</span>. The way B-B-D-G construct the functor is they use the <em>filtered derived category</em> <span class="math-container">$\mathcal{D}F$</span>. If <span class="math-container">$\mathcal D$</span> is the derived category of an abelian category <span class="math-container">$B$</span>, then <span class="math-container">$\mathcal D F$</span> is the category of complexes in <span class="math-container">$B$</span> with a bounded filtration, localized at filtered quasi-isomorphisms. There is an induced <span class="math-container">$t$</span>-structure on <span class="math-container">$\mathcal D F$</span> from a <span class="math-container">$t$</span>-structure on <span class="math-container">$\mathcal D$</span>, such that if the <span class="math-container">$t$</span>-structure on <span class="math-container">$\mathcal D$</span> has heart <span class="math-container">$A$</span> then the heart of the <span class="math-container">$t$</span>-structure on <span class="math-container">$\mathcal D F$</span> is isomorphic to the abelian category <span class="math-container">$\mathrm{Ch}^b(A)$</span> of bounded chain complexes in <span class="math-container">$A$</span>. Thus we can consider the composition <span class="math-container">$\mathrm{Ch}^b(A) \to \mathcal D F \to \mathcal D$</span> where the first is the inclusion of the heart and the second forgets the filtration. A spectral sequence argument shows that this composition takes quasi-isomorphisms in <span class="math-container">$\mathrm{Ch}^b(A)$</span> to equivalences in <span class="math-container">$\mathcal D$</span>, so there is an induced functor <span class="math-container">$D^b(A) \to \mathcal D$</span> which is the one we want. </p>
<p>The point of the above is that for a general triangulated category <span class="math-container">$\mathcal T$</span> there is no sensible triangulated category <span class="math-container">$\mathcal T F$</span> of filtered objects in <span class="math-container">$\mathcal T$</span>, but if <span class="math-container">$\mathcal T$</span> happens to be the derived category of an abelian category we can write down the filtered derived category by hand. This is not a problem in the world of stable <span class="math-container">$\infty$</span>-categories.</p>
<p>I believe that an analogous argument works also for complexes that are not necessarily bounded (using instead filtrations that are unbounded to the left or right or both). But if we want there to exist a functor <span class="math-container">$\mathcal DF \to \mathcal D$</span> that forgets the filtration then we need <span class="math-container">$\mathcal D$</span> to have sequential limits or colimits, and one should be careful with the spectral sequence argument.</p>
|
633,858 | <p>If G is cyclic group of 24 order, then how many element of 4 order in G?
I can't understand how to find it, step by step. </p>
| DonAntonio | 31,254 | <p>Hints:</p>
<p>1) A cyclic group of order $\;n\;$ has exactly $\;\varphi(n)\;$ generators.</p>
<p>(2) A cyclic group of order $\;n\;$ has exactly one unique subgroup of order $\;d\;$ for <strong>any</strong> divisor $\;d\;$ of $\;n\;$</p>
|
3,298,445 | <p>A random variable is defined by it distribution function. The density function is the derivative of the distribution function. Thus the density function exisst iff the distribution function is absolutely continuous. However, can we construct a distribution function without a density function, except for the finite discrete random variable(i.e. the distribution function is a step function)?</p>
<p>I think the hardness is that the distribution function is non-decreasing and upper-bounded by 1.</p>
| Masacroso | 173,262 | <p>Of course we can, there are the mixed distributions. </p>
<p>In short: any increasing and right continuous function <span class="math-container">$F$</span> with <span class="math-container">$\lim_{x\to -\infty} F(x)=0$</span> and <span class="math-container">$\lim_{x\to+\infty}F(x)=1$</span> defines a probability distribution on <span class="math-container">$\Bbb R$</span>. When <span class="math-container">$F$</span> is differentiable then the random variable represented by <span class="math-container">$F$</span> have a density. If <span class="math-container">$F$</span> is a step function then it represent a discrete random variable. All the other cases are mixed cases.</p>
<p>Moreover: if <span class="math-container">$X:\Omega\to\Bbb R$</span> is a random variable then <span class="math-container">$X$</span> and the probability measure <span class="math-container">$P$</span> in <span class="math-container">$\Omega$</span> induces a probability measure <span class="math-container">$\mu$</span> on <span class="math-container">$\Bbb R$</span>. Then the distribution of <span class="math-container">$X$</span> is the function <span class="math-container">$F$</span> such that <span class="math-container">$\mu((a,b])=F(b)-F(a)$</span>, and it can be shown that <span class="math-container">$F$</span> have the properties above described.</p>
|
321,230 | <p>Suppose $Z$ is a topological space; and $X$ is dense in $Z$. Then do we have $W(X)= W(Z)$?
Where $W(X)$, $W(Z)$ denote the weight of the $X$ and $Z$ respectively. </p>
<p><strong>What I've tried:</strong> On one hand, $W(X)\le W(Z)$, clearly; On the other hand, for any open set $U$ of $Z$, we have $U\cap X$, an open set in $X$, because $X$ is dense in $Z$. So $W(X)= W(Z)$. Am I right?</p>
<p>Thanks ahead.</p>
| user642796 | 8,348 | <p>The weight of a space does not necessarily equal the weight of a dense subspace. </p>
<p>As an example, note that $\mathbb N$ is clearly second-countable ($w(\mathbb{N}) = \aleph_0$), but its Stone–Čech compactification $\beta \mathbb{N}$ has weight $2^{\aleph_0}$. This can be generalised for the Stone–Čech compactification of any infinite discrete space (see, <em>e.g.</em>, Engelking, <em>General Topology</em>, Theorem 3.6.11, pp.174-5).</p>
|
831,472 | <p>I am learning about Karnaugh maps to simplify boolean algebra expressions. I have this:</p>
<p>$$\begin{bmatrix}
& bc & b'c & bc' & b'c' \\
a & 0 & 1 & 1 & 0\\
a' & 1 & 1 & 0 & 1
\end{bmatrix}$$</p>
<p>There are no groups of four, so I am now looking for groups of two. I have highlighted the groups of two that I chose:
$$\begin{bmatrix}
& bc & b'c & bc' & b'c' \\
a & 0 & \color{red}1 & 1 & 0\\
a' & \color{blue}1 & \color{red}1 & 0 & \color{blue}1
\end{bmatrix}$$</p>
<p>One red, and another blue.</p>
<p>Now, there is one $1$ hanging over there. Normally, I would say that it will belong to a third group (of size one) and be done with it.</p>
<p>However, I remember the professor doing an example in which he was in a similar situation, but he actually joined the $1$ with another $1$ that was <strong>already</strong> grouped. I cannot recall his reasoning though.</p>
<p>What should I do?</p>
| thewillix | 156,545 | <p>Karnaugh maps require a particular ordering of the variables different from a normal truth table. Your K-map is ordered like a truth table: bc b'c bc' b'c' (or 11 01 10 00) whereas it has to be ordered such that only one variable changes going from one column (or row) to the next, and it is usually written with 00 on the left, namely 00 01 11 10 (or b'c' b'c bc bc').
When the K-map is arranged like this, any adjacent pair (or 4 or 8) allows the elimination of one (or 2 or 3) variables. Take as an example the two adjacent red "ones" in your table above (we can use the vertical axis of your table since the "a" variable is ordered correctly. The red "ones" mean (ab'c + a'b'c) which reduces to b'c(a+a'). a+a' is always true, ie equals 1, thus the expression reduces to b'c. So putting a ring around two adjacent "ones" eliminates the variable which has both its true and complement present.
To proceed, we have to rewrite your table:</p>
<p>\begin{array}{ccccc}
& 00 & 01 & 11 & 10\\
\hline
0 & 1 & 1 & 1 & 0\\
1 & 0 & 1 & 0 & 1\\
\end{array}
We see here three adjacent ones on the a' (a=$0$) line and two adjacent ones in the b'c column ($01$). We can make two pairs horizontally and one pair vertically. The centre "one" is shared by all three pairs. These three pairs represent a'b' + b'c + a'c. The remaining, single "one" (bottom right) represents abc'. Thus the minimised function is a'b' + b'c + a'c + abc'.</p>
|
1,617,462 | <p>Is this a line or a plane, I thought it would be a plane where z=0 always so it will be the xy plane.</p>
<p>Also: what will be the normal vector for this if it is a plane?</p>
| David Quinn | 187,299 | <p>In $\mathbb{R}^3$, this represents a plane with normal vector $$\left(\begin{matrix}2\\-1\\0\end{matrix}\right)$$ and is therefore perpendicular to the $z$ plane</p>
|
2,136,937 | <p>Find $$\lim_{z \to \exp(i \pi/3)} \dfrac{z^3+8}{z^4+4z+16}$$</p>
<p>Note that
$$z=\exp(\pi i/3)=\cos(\pi/3)+i\sin(\pi/3)=\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}$$
$$z^2=\exp(2\pi i/3)=\cos(2\pi/3)+i\sin(2\pi/3)=-\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}$$
$$z^3=\exp(3\pi i/3)=\cos(\pi)+i\sin(\pi)=1$$
$$z^4=\exp(4\pi i/3)=\cos(4\pi/3)+i\sin(4\pi/3)=-\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}$$</p>
<p>So,
\begin{equation*}
\begin{aligned}
\lim_{z \to \exp(i \pi/3)} \dfrac{z^3+8}{z^4+4z+16} & = \dfrac{1+8}{-\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}+4\left(-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\right)+16} \\
& = \dfrac{9}{\dfrac{27}{2}+i\frac{3\sqrt{3}}{2}} \\
& = \dfrac{6}{9+i\sqrt{3}} \\
& = \dfrac{9}{14}-i\dfrac{\sqrt{3}}{2} \\
\end{aligned}
\end{equation*}</p>
<p>But, when I check my answer on wolframalpha, their answer is $$\dfrac{245}{626}-i\dfrac{21\sqrt{3}}{626}.$$</p>
<p>Can someone tell me what I am doing wrong?</p>
| A. Fenzry | 295,901 | <p>I studied Physics and would definetely recommend you Apostol's. It is a good book, rigorous enough and complete, while at the same time keeps a casual layout with a lot of explanative text, which makes it a good transition between High School and college.</p>
<p>As a personal reference I can tell you it was the recommended book for my first Calculus course on Physics, I bought it, and I still occasionally consult it as a TA to see how it explains some concepts.</p>
<p>Also keep in mind that you may want to buy both volumes, the second is not a more advanced book but simply covering other topics on the same basic level (sequences, integrals,etc.)</p>
|
94,440 | <p>In Sean Carroll's <em>Spacetime and Geometry</em>, a formula is given as
$${\nabla _\mu }{\nabla _\sigma }{K^\rho } = {R^\rho }_{\sigma \mu \nu }{K^\nu },$$</p>
<p>where $K^\mu$ is a Killing vector satisfying Killing's equation ${\nabla _\mu }{K_\nu } +{\nabla _\nu }{K_\mu }=0$ and the convention of Riemann curvature tensor is</p>
<p>$$\left[\nabla_{\mu},\nabla_{\nu}\right]V^{\rho}={R^\rho}_{\sigma\mu\nu}V^{\sigma}.$$</p>
<p>So how to prove the this formula (the connection is Levi-Civita)?</p>
| AstoundingJB | 93,907 | <p>@C.R. This is my 'simpler proof'; I'm pretty sure it's correct, and simpler than Zhen's one as well.</p>
<p>From the <em>first Bianchi identity</em> [Carroll, (3.132)]
$$R_{\mu\nu\rho\sigma}+R_{\mu\rho\sigma\nu}+R_{\mu\sigma\nu\rho}=0$$
we have that, for every vector $V^\rho$,
$$\nabla_{[\mu}\nabla_\nu V_{\rho]}=\tfrac{1}{6}\big(R_{\rho\alpha\mu\nu}+R_{\mu\alpha\nu\rho}+R_{\rho\mu\nu\alpha}\big)V^\alpha=0,$$
where in the last equation I used the symmetry properties of the indices in the Riemann tensor to reduce it to the Bianchi identity. <em>This is a very useful formula!</em> I'm quite sure about the index placement thanks to the <em>metric compatibility</em>, $\nabla_\mu g_{\nu\rho}=0$ [Carroll, (1.32)], which implies that the metric $g_{\nu\rho}$ commutes with the covariant derivative $\nabla_\mu$.
Now, we take the Killing vector $K^\mu$ and we expand the antisymmetrization in the previous equation:
$$0=6\nabla_{[\mu}\nabla_\nu K_{\rho]}=\nabla_\mu\nabla_\nu K_\rho +
\nabla_\nu \nabla_\rho K_\mu +\nabla_\rho\nabla_\mu K_\nu-
\nabla_\nu\nabla_\mu K_\rho-\nabla_\mu\nabla_\rho K_\nu-\nabla_\rho\nabla_\mu K_\nu.$$
Now we use the <em>Killing's equation</em> [Carroll, (3.174)], $\nabla_{(\nu}K_{\rho)}=0$ or $\nabla_\nu K_\rho =-\nabla_\rho K_\nu$, to simplify the previous equation:
$$\nabla_\mu\,\nabla_\nu K_\rho=-\nabla_\mu\,\nabla_\rho K_\nu\quad \Rightarrow\quad 0=2\nabla_\mu\nabla_\nu K_\rho+(\nabla_\nu\nabla_\rho-\nabla_\rho\nabla_\nu)K_\mu+\nabla_\rho\nabla_\mu K_\nu-\nabla_\nu\nabla_\mu K_\rho.$$
Again, using Killing's equations $\nabla_\rho\nabla_\mu K_\nu=-\nabla_\rho\nabla_\nu K_\mu$ and $\nabla_\nu\nabla_\mu K_\rho=-\nabla_\nu\nabla_\rho K_\mu$ we get:
$$0=2\big(\nabla_\mu\nabla_\nu K_\rho+(\nabla_\nu\nabla_\rho-\nabla_\rho\nabla_\nu)K_\mu\big)=2\big(\nabla_\mu\nabla_\nu K_\rho+R_{\mu\alpha\nu\rho}K^\alpha\big)$$
$$\nabla_\mu\nabla_\nu K_\rho=-R_{\mu\alpha\nu\rho}K^\alpha=R_{\rho\nu\mu\alpha}K^\alpha.$$
Then, we can rise the index $\rho$ multiplying with the metric $g^{\rho\sigma}$ (and using the metric compatibility):
$$\nabla_\mu\nabla_\nu K^\sigma=R^\sigma_{\phantom{\sigma}\nu\mu\alpha}K^\alpha.$$
That's all folks! I hope it would be helpful (and correct)!</p>
|
1,677,868 | <p>The sequence is:</p>
<p>$$a_n = \frac {2^{2n} \cdot1\cdot3\cdot5\cdot...\cdot(2n+1)} {(2n!)\cdot2\cdot4\cdot6\cdot...\cdot(2n)} $$</p>
| parsiad | 64,601 | <p>A direct approach is to try to show that is the same as $a_n=(2n+1)!/(2(n!)^{3})$, which goes to zero as $n\rightarrow\infty$. Alternatively, use the ratio test as suggested in the comments above.</p>
|
2,468,326 | <p>I want to read <a href="https://www.amazon.co.uk/Introduction-Cyclotomic-Fields-Graduate-Mathematics/dp/0387947620" rel="noreferrer">Lawrence Washington's An <em>Introduction to Cyclotomic Fields</em></a>. However, my knowledge of algebraic number theory doesn't extend farther than what's found in <a href="https://www.amazon.co.uk/Classical-Introduction-Modern-Number-Theory/dp/038797329X/ref=sr_1_1?s=books&ie=UTF8&qid=1507759261&sr=1-1&keywords=ireland%20and%20rosen%20number%20theory" rel="noreferrer"><em>A Classical Introduction to Modern Number Theory</em> - Ireland and Rosen</a>. Does anyone know whether this will be sufficient or whether I'll have to learn more about algebraic number theory before I can get to it? </p>
| paul garrett | 12,291 | <p>I think you'd be happier to have a full introduction to algebraic number theory, like S. Lang's, or equivalent. Then you'll be able to see many of the features of cyclotomic fields as special cases of what would happen more generally, rather than having those special cases appear as novelties.</p>
<p>That is, I think it is useful to see the basic features of cyclotomic fields appear as especially accessible examples of general algebraic number theory, rather than as first-encounter extensions of the number theory of $\mathbb Z$ and quadratic extensions, for example.</p>
|
330,991 | <p>Many things in math can be formulated quite differently; see the list of statements equivalent to RH <a href="https://mathoverflow.net/questions/39944/collection-of-equivalent-forms-of-riemann-hypothesis">here</a>, for example, with RH formulated as a bound on lcm of consecutive integers, as an integral equality, etc.</p>
<p>I am wondering about equivalent formulations of the P vs. NP problem. Formulations that are very much different from the questions such "Is TSP in P?", formulation that may seem unrelated to complexity theory.</p>
| Sam Hopkins | 25,028 | <p>I think "Geometric Complexity Theory" is roughly speaking an attempt to do what you're talking about: formulate P vs. NP in very different language. See <a href="https://en.wikipedia.org/wiki/Geometric_complexity_theory" rel="noreferrer">https://en.wikipedia.org/wiki/Geometric_complexity_theory</a>. I think that technically it may be dealing with "VP vs. VNP" rather than "P vs. NP" but in spirit it fits your request.</p>
|
330,991 | <p>Many things in math can be formulated quite differently; see the list of statements equivalent to RH <a href="https://mathoverflow.net/questions/39944/collection-of-equivalent-forms-of-riemann-hypothesis">here</a>, for example, with RH formulated as a bound on lcm of consecutive integers, as an integral equality, etc.</p>
<p>I am wondering about equivalent formulations of the P vs. NP problem. Formulations that are very much different from the questions such "Is TSP in P?", formulation that may seem unrelated to complexity theory.</p>
| Ryan O'Donnell | 658 | <p>There is the descriptive complexity formulation: </p>
<p>P = NP is equivalent to the statement that every property expressible by a second order existential statement is also expressible in first order logic with a least fixed point operator.</p>
<p>See, e.g., Immerman's survey here: <a href="https://people.cs.umass.edu/~immerman/pub/capture.pdf" rel="noreferrer">https://people.cs.umass.edu/~immerman/pub/capture.pdf</a></p>
|
330,991 | <p>Many things in math can be formulated quite differently; see the list of statements equivalent to RH <a href="https://mathoverflow.net/questions/39944/collection-of-equivalent-forms-of-riemann-hypothesis">here</a>, for example, with RH formulated as a bound on lcm of consecutive integers, as an integral equality, etc.</p>
<p>I am wondering about equivalent formulations of the P vs. NP problem. Formulations that are very much different from the questions such "Is TSP in P?", formulation that may seem unrelated to complexity theory.</p>
| none | 140,370 | <p>"<a href="https://en.wikipedia.org/wiki/P_versus_NP_problem#Polynomial-time_algorithms" rel="nofollow noreferrer">This version of</a> Levin's universal search algorithm solves SUBSET-SUM in polynomial time" is equivalent to P=NP.</p>
|
1,499,583 | <p>If $a=b\log b$, how does $b$ grow asymptotically in terms of $a$?</p>
<p>I think the answer should be $b=\Theta\left(\frac{a}{\log a}\right)$. I tried taking logs to get $\log a=\log b+\log\log b$, but it's not clear how to separate $b$.</p>
| Deepak | 151,732 | <p>There is no contradiction, only a misconception. When working with complex numbers, $1^{\frac{25}{4}}$ has multiple values, not just $1$. (Or, to put it more generally, real positive numbers to fractional exponents can return multiple complex values).</p>
<p>A correct manipulation would be $i^{25} = ({i^4})^6 \cdot i = 1^6 \cdot i = i$.</p>
|
1,499,583 | <p>If $a=b\log b$, how does $b$ grow asymptotically in terms of $a$?</p>
<p>I think the answer should be $b=\Theta\left(\frac{a}{\log a}\right)$. I tried taking logs to get $\log a=\log b+\log\log b$, but it's not clear how to separate $b$.</p>
| Paul Sinclair | 258,282 | <p>When dealing with complex functions, you often run into multi-valued functions, because when you circle around a singularity, it will pick up some constant value. There are three ways in which this is often handled:</p>
<ol>
<li>You can restrict the function to domains, called branches, that do not completely circle a singularity.</li>
<li>You can live with the concept of a "multi-valued" function, treating the function value as being a set of complex numbers instead of a single complex number, or</li>
<li>You can build a layered extension of the complex numbers, called a Riemann surface, to act as the domain, so that the function picks up each of its different values on a different layer.</li>
</ol>
<p>The answer your question depends on which approach you take:</p>
<ol>
<li>Multiplication of exponents only works in general if the combined exponent lies in the same branch.</li>
<li>Multiplication of exponents always works, and your $i = 1$ example is not a problem because the actual value is a set that contains both.</li>
<li>Multiplication of exponents always works, but your example is wrong because you are mixing layers.</li>
</ol>
|
275,371 | <p>I was wondering if it is possible to decompose any symmetric matrix into a positive definite and a negative definite component. I can't seem to think of a counterexample if the statement is false.</p>
| Arin Chaudhuri | 404 | <p>If $X$ is symmetric then $X = (X + \lambda I) - \lambda I$. Since the eigenvalues of $X + \lambda I $ are $ \lambda_i + \lambda$ where $\lambda_i$'s are the eigenvalues of X we can find a positive $\lambda$ such that $(X + \lambda I)$ is positive definite.</p>
|
364,278 | <p>Let <span class="math-container">$X$</span> be a variety over a number field <span class="math-container">$K$</span>. Then it is known that for any topological covering <span class="math-container">$X' \to X(\mathbb{C})$</span>, the topological space <span class="math-container">$X'$</span> can be given the structure of a <span class="math-container">$\overline{K}$</span>-variety in such a way so that the morphism <span class="math-container">$f: X' \to X$</span> inducing the topological map is a finite etale morphism over <span class="math-container">$\overline{K}$</span>. However, the variety <span class="math-container">$X'$</span> and the morphism <span class="math-container">$f$</span> may not descend to <span class="math-container">$K$</span>.</p>
<p>My question is as follows: does there always exist a further finite etale covering <span class="math-container">$f' : X'' \to X'$</span> such that the composition <span class="math-container">$X'' \to X$</span> may be defined over <span class="math-container">$K$</span>?</p>
<p>EDIT: Just to be clear, I'd like all the covers involved to be geometrically connected to avoid trivial solutions.</p>
| SashaP | 39,304 | <p>Let's assume that <span class="math-container">$X$</span> admits a <span class="math-container">$K$</span>-point <span class="math-container">$x$</span> and use the corresponding geometric point as the base point. The existence of a rational point is in fact necessary for a positive answer, as explained by S. carmeli.</p>
<p>In terms of etale fundamental groups the question can be paraphrased as follows: given an open subgroup <span class="math-container">$H\subset \pi_1(X_{\overline{K}},x)$</span> does there exist an open subgroup <span class="math-container">$H'\subset H$</span> such that the action of the Galois group <span class="math-container">$G_K$</span> on <span class="math-container">$\pi_1(X_{\overline{K}},x)$</span> preserves <span class="math-container">$H'$</span>.</p>
<p>This is true and follows from <span class="math-container">$\pi_1(X_{\overline{K}},x)$</span> being topologically finitely generated. Consider the subgroup <span class="math-container">$\Gamma_H\subset G_K$</span> consisting of elements <span class="math-container">$\gamma\in G_K$</span> such that <span class="math-container">$\gamma(H)=H$</span>. Let <span class="math-container">$h_1,\dots, h_n$</span> be a set of topological generators of <span class="math-container">$H$</span> (<span class="math-container">$H$</span> is topologically finitely generated because it has finite index in <span class="math-container">$\pi_1(X_{\overline{K}})$</span>). Then <span class="math-container">$\Gamma_H$</span> can be expressed as <span class="math-container">$\{\gamma\in G_K|\gamma(h_i)\in H\}$</span> so <span class="math-container">$\Gamma_H$</span> is an intersection of finitely many open subset, hence is an open subgroup. In particular, <span class="math-container">$\Gamma_H$</span> has finite index in <span class="math-container">$G_K$</span>. Take <span class="math-container">$\Gamma\subset \Gamma_H$</span> to be an open subgroup which is moreover normal in <span class="math-container">$G_K$</span>.</p>
<p>Let <span class="math-container">$g_1,\dots, g_m$</span> be a set of representatives of cosets of <span class="math-container">$\Gamma$</span> in <span class="math-container">$G_K$</span>. Then <span class="math-container">$H'=\bigcap g_i(H)$</span> is an open subgroup with the desired property. Indeed, suppose that <span class="math-container">$x\in H'$</span> and <span class="math-container">$\gamma g_i\in G_K$</span> are arbitrary elements where <span class="math-container">$\gamma\in \Gamma$</span> and <span class="math-container">$i\in\{1,\dots, m\}$</span>. The result of the action <span class="math-container">$\gamma \circ g_i(x)$</span> lies in <span class="math-container">$H'$</span> because for each <span class="math-container">$k=1,\dots, m$</span> we have <span class="math-container">$g_k^{-1}\gamma g_i=\gamma'g_j^{-1}$</span> for some <span class="math-container">$\gamma'\in \Gamma$</span> and <span class="math-container">$j\in\{1,\dots, m\}$</span> so <span class="math-container">$\gamma g_i(x)\in \gamma g_ig_j(H)=g_k\gamma'(H)=g_k(H)$</span>.</p>
<p>We can think of this argument as of a generalization of the proof that a compact group acting on a finite-dimensional <span class="math-container">$\mathbb{Q}_p$</span>-vector space always preserves some <span class="math-container">$\mathbb{Z}_p$</span>-lattice.</p>
|
3,453,408 | <p>I'm reading through some lecture notes and see this in the context of solving ODEs:
<span class="math-container">$$\int\frac{dy}{y}=\int\frac{dx}{x} \rightarrow \ln{|y|}=\ln{|x|}+\ln{|C|}$$</span> why is the constant of integration natural logged here?</p>
| Quantum_Magnet | 383,722 | <p>See this:</p>
<p><span class="math-container">$\int\frac{dy}{y}=\int\frac{dx}{x} \Rightarrow \int\frac{dy}{y}=\int\frac{dx}{x} + c$</span></p>
<p><span class="math-container">$\ln |y| = \ln |x| + c \Rightarrow \ln |y/x| = c \Rightarrow y/x = \pm e^c \Rightarrow y= \pm e^c x \Rightarrow y = C x $</span> [Where C is non zero constant.]</p>
<p><span class="math-container">$\Rightarrow \ln |y| = \ln |x| + \ln |C|$</span></p>
<p>Instead of doing this much: </p>
<p>One prefers to write in short as: <span class="math-container">$ \ln |y| = \ln |x| + \ln |C|$</span></p>
|
238,547 | <p>I have a PDE like</p>
<pre><code>D[h[x1, x2], x1]*a[x1,x2]+D[h[x1,x2], x2]*b[x1,x2] + c[x1,x2] == h[x1,x2]
s.t. gradient(h(0,0))==0
</code></pre>
<p>where a,b,c are known functions of x1 and x2, and h are the function to be solved. x1 and x2 are both in [-2, 2].
For some selected a,b,c, DSolveValue can give me perfect analytical solutions, like (a=0.1*x1, b=x2-x1^2, c=-x1^2)</p>
<pre><code>eq1 = -x1^2 + D[h1[x1, x2], x1]*(0.1)*x1 +
D[h1[x1, x2], x2]*(x2 - x1^2) == h1[x1, x2];
s1 = DSolveValue[eq1, h1[x1, x2], {x1, x2}]
(*x1^2. (-1.25 + x1^8. C[1][(0.25 (-5. x1^2 + 4. x2))/x1^10])*)
</code></pre>
<p>For this analytical solution, I then use the gradient constraints to figure out how C[1] should look like and finally get h1=-1.25*x1^2, which is the solution I want.
But for other a,b,c, DSolveValue reports that "Inverse functions are being used by Solve, so solutions may not be found". For example (a=x2, b=Sin[x1]+x2, c=0):</p>
<pre><code>eq2 = D[h2[x1, x2], x1]*x2 + D[h2[x1, x2], x2]*(Sin[x1] + x2) ==
h2[x1, x2];
s2 = DSolveValue[eq2, h2[x1, x2], {x1, x2}]
</code></pre>
<p>The output just repeats my cmd and doesn't have a solution.
I'm aware that this is because h2 doesn't have an analytical solution. How to get a numerical solution under this circumstance?</p>
| bbgodfrey | 1,063 | <p>It also is possible to obtain an approximate symbolic solution, valid for small <code>x1</code>. Expand the PDE in <code>x1</code> to obtain</p>
<pre><code>DSolve[(x1 + x2)*D[h[x1, x2] + x2*D[h[x1, x2] == h[x1, x2], h[x1, x2], {x1, x2}]
</code></pre>
<p>Although <code>DSolve</code> cannot solve this PDE directly, it can solve the characteristic ODEs.</p>
<pre><code>{x2'[s] == x1[s] + x2[s], x1'[s] == x2[s], h'[s] == h[s]}
</code></pre>
<p>Focus first on the ODEs for {x1, x2}</p>
<pre><code>DSolve[{x1'[s] == x2[s], x2'[s] == x1[s] + x2[s]}, {x1[s], x2[s]}, s] // Flatten
sol = ((Simplify[% /. C[n_] -> C[n] Exp[-s/2]] // ExpToTrig // Simplify)
/. C[n_] -> C[n] Exp[s/2]) // Simplify
(* {x1[s] -> 1/5 E^(s/2) (5 C[1] Cosh[(Sqrt[5] s)/2] -
Sqrt[5] (C[1] - 2 C[2]) Sinh[(Sqrt[5] s)/2]),
x2[s] -> 1/5 E^(s/2) (5 C[2] Cosh[(Sqrt[5] s)/2] +
Sqrt[5] (2 C[1] + C[2]) Sinh[(Sqrt[5] s)/2])} *)
</code></pre>
<p>which can be plotted as</p>
<pre><code>Show[ParametricPlot[Table[{x1[s], x2[s]} /. sol /. {C[1] -> n, C[2] -> 0},
{n, -2, 2, .1}], {s, -3, 3}, PlotRange -> {{-2, 2}, {-2, 2}},
ImageSize -> Large, LabelStyle -> {15, Bold, Black}],
ParametricPlot[Table[{x1[s], x2[s]} /. solt /. {C[1] -> 0, C[2] -> n}, {n, -3, 3, .1}],
{s, -3, 3}]]
</code></pre>
<p><a href="https://i.stack.imgur.com/wR1zo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wR1zo.png" alt="enter image description here" /></a></p>
<p>Compare this to the first plot in my strictly numerical answer above. Clearly, they are about the same for small <code>x1</code>. Next, obtain a symbolic solution for <code>h</code>. Consider the quadratic expression, <code>(x1[s] + x2[s]/2)^2 - 5/4 x2[s]^2)</code>, which reduces to</p>
<pre><code>Simplify[((x1[s] + x2[s]/2)^2 - 5/4 x2[s]^2) /. sol /. C[2] -> 0]
(* E^s C[1]^2 *)
</code></pre>
<p>and</p>
<pre><code>DSolve[h'[s] == h[s], h[s], s] /. C[1] -> C[3]
(* {{h[s] -> E^s C[3]}} *)
</code></pre>
<p><code>E^s</code> can be eliminated between these two results to yield</p>
<pre><code>h[x1, x2] == Simplify[(x1 + x2/2)^2 - 5/4 x2^2] C[3]/C[1]^2
(* h[x1, x2] == ((x1^2 + x1 x2 - x2^2) C[3])/C[1]^2 *)
</code></pre>
<p>Similarly, assuming <code>C[2] -> 0</code> leads to</p>
<pre><code>(* h[x1, x2] == ((-x1^2 - x1 x2 + x2^2) C[3])/C[2]^2 *)
</code></pre>
<p>Choose as boundary conditions <code>h[x1,x2] = (x1^2 +x2^2)/4</code> along the <code>x1</code> and <code>X2</code> axes, as done in the numerical solution above. Then, <code>C[3]/C[2]^2</code> reduces to <code>1/4</code> on the <code>x1</code> axis, and <code>C[3]/C[1]^2</code> reduces to <code>1/4</code> on the <code>x2</code> axis. Combining these expressions gives for <code>h</code></p>
<pre><code>h[x1, x2] = Abs[(x1^2 + x1 x2 - x2^2)/4];
</code></pre>
<p>which can be plotted as</p>
<pre><code>Plot3D[Abs[(x1^2 + x1 x2 - x2^2)/4], {x1, -2, 2}, {x2, -2, 2},
PlotRange -> {0, 1.5}, AxesLabel -> {x1, x2, h}, ImageSize -> Large,
LabelStyle -> {15, Bold, Black}, PlotPoints -> 50]
</code></pre>
<p><a href="https://i.stack.imgur.com/aygyw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aygyw.png" alt="enter image description here" /></a></p>
<p>which agrees well for small <code>x1</code> with the final plot in the numerical answer above.</p>
|
2,631,220 | <p>If $z$ is a variable complex number , and $a$ is a fixed complex number , is it true that if $z$ , $a$ satisfy the following condition </p>
<p>$|z+a| = |z-a|$ </p>
<p>Then the locus of $z$ is the perpendicular bisector of $a$ and $-a$ ?</p>
| egreg | 62,967 | <p>Of course, you have to assume $a\ne0$. Write $a=ru$, with $|u|=1$ and $r>0$; then, writing $z=wu$, the equation becomes
$$
|w-r|=|w+r|
$$
Notice that this corresponds to a rotation around the origin by the negative of the angle determined by $u$.</p>
<p>By squaring,
$$
(w-r)(\bar{w}-r)=(w+r)(\bar{w}+r)
$$
that simplifies to
$$
w+\bar{w}=0
$$
that is, $w$ is purely imaginary. The locus is therefore the $y$-axis, the perpendicular bisector of the segment joining $-r$ and $r$. Multiplying back by $u$, which is a rotation, we finish.</p>
|
4,021,994 | <p>I was taught in high school algebra to translate word problems into algebraic expressions. So when I encountered <a href="https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_3" rel="nofollow noreferrer">this</a> problem I tried to reason out an algebra formula for it</p>
<blockquote>
<p>For every dollar Ben spent on bagels, David spent 25 cents less. Ben
paid $12.50 more than David. How much did they spend in the bagel store
together?</p>
</blockquote>
<p>To solve this I imagined a series of comparisons when Ben spends <span class="math-container">$x$</span>, David spends <span class="math-container">$.75x$</span>. Loop this relationship until <span class="math-container">$x - .75x \approx 12.50$</span>. Good. Done. <span class="math-container">$x = 50$</span>, then add David's for the answer. Coming from computers, I would have set this up in code where a loop (recursion) would increase <span class="math-container">$x$</span> until the condition <span class="math-container">$x - .75x = 12.50$</span> was met, then the "loop counter/accumulator" would be how much Ben spent, i.e., <span class="math-container">$50$</span>, etc.</p>
<p>I'm a beginner with math, but it seems like there should be a better approach, something with series and sequences or even calculus derivatives, something better than my brute-force computer algorithm. Can someone enlighten? The "answer" given at the site (see link) is its own brute-force and hardly satisfying. I'm thinking there should be something more formal -- at least for the first part that derives <span class="math-container">$50$</span>.</p>
<p><strong>Update</strong></p>
<p>I think everyone so far has missed my point. Many of you simply re-did the problem again. I'm wondering if there is a more <em>formal</em> way to do this other than just "figuring it out" (FIO). The whole FIO routine is murky. It looks like a limit problem; it looks like a system of equations, but I'm not experienced enough to know exactly. If there isn't, then let's call it a day....</p>
| fleablood | 280,126 | <p>I have to admit I'm having a hard time seeing how you "imagine a series of comparisons " and how you "loop" them.</p>
<p>As a mathematician, I simply view this as Ben did a <em>single</em> transaction where in one go he spent <span class="math-container">$x$</span> dollars on bagels.</p>
<p>Therefore David spend <span class="math-container">$0.75x$</span> on bagels (in one go).</p>
<p>Set <em>one</em> equation (no series of loops) to get <span class="math-container">$x - 0.75x = 12.50$</span> and solve:</p>
<blockquote>
<p><span class="math-container">$x-0.75x = 12.50$</span><br />
<span class="math-container">$0.25x = 12.50$</span><br />
<span class="math-container">$\frac {0.25}{0.25} x = \frac {12.50}{0.25}$</span><br />
<span class="math-container">$x = 50$</span></p>
</blockquote>
<p>So Ben spent <span class="math-container">$50$</span> dollars on bagels.</p>
<p>Now add Davids answer. David spent <span class="math-container">$0.75x$</span> and <span class="math-container">$x = 50$</span> so David spent <span class="math-container">$0.75 \times 50 = 37.50$</span>. Adding that they added <span class="math-container">$50+37.50 = 87.50$</span>.</p>
<p>Or.... Ben spent <span class="math-container">$x$</span> and David spent <span class="math-container">$0.75x$</span> so together they spent <span class="math-container">$x + 0.75x = 1.75x$</span> and as <span class="math-container">$x = 50$</span> together they spent <span class="math-container">$1.75\times 50 = 87.50$</span>.</p>
|
319,262 | <p>If the first 10 positive integer is placed around a circle, in any order, there exists 3 integer in consecutive locations around the circle that have a sum greater than or equal to 17? </p>
<p>This was from a textbook called "Discrete math and its application", however it does not provide solution for this question. </p>
<p>May I know how to tackle this question. </p>
<p>Edit: I relook at the actual question and realize it is sum greater or equal to 17. My apologies.</p>
| Gerry Myerson | 8,269 | <p>EDIT: it has been pointed out that this answer only gives $\ge17$, while the question asks for $\gt17$. More work is needed. </p>
<p>Let $A_1=a_1+a_2+a_3$, $A_2=a_2+a_3+a_4$, and so on, $A_{10}=a_{10}+a_1+a_2$. Then $A_1+A_2+\cdots+A_{10}=3(a_1+a_2+\cdots+a_{10})=(3)(55)=165$, so some $A_i\ge165/10=16.5$, so some $A_i\ge17$. </p>
|
319,262 | <p>If the first 10 positive integer is placed around a circle, in any order, there exists 3 integer in consecutive locations around the circle that have a sum greater than or equal to 17? </p>
<p>This was from a textbook called "Discrete math and its application", however it does not provide solution for this question. </p>
<p>May I know how to tackle this question. </p>
<p>Edit: I relook at the actual question and realize it is sum greater or equal to 17. My apologies.</p>
| joriki | 6,622 | <p>Original answer:</p>
<blockquote>
<p>To have all sums $\le17$, all four numbers from $7$ to $10$ would have to be separated by at least two numbers; but it would take at least $12$ slots to space them like that.</p>
</blockquote>
<p>As has been pointed out in the comments, this is wrong, but Gerry showed how to complete it.</p>
<p>The numbers from $8$ to $10$ must be separated by at least two numbers. That leaves two number to separate the $7$ from them. If it's not adjacent to any of them, it must be separated from both $8$ and $9$ by just one number, and those have to be $2$ and $1$, respectively; that leaves only $3$ to $6$ in the four slots around $10$, which leads to at least one sum of at least $18$.</p>
<p>So the $7$ must be next to one of the numbers from $8$ to $10$. It can't be next to the $9$ or $10$ because that would lead to a sum of at least $18$ even with $1$ and $2$ adjacent. So it must be next to the $8$, and Gerry's argument completes the proof.</p>
<p>That's rather inelegant case work; a more systematic proof would be nice.</p>
|
1,092,665 | <p>My question is really simple, how can I write symbolically this phrase: </p>
<blockquote>
<p>$x=\sum a_mx^m$ where $m$ range over
$\{1,\ldots,g\}\setminus\{t_1,\ldots,t_u\}$</p>
</blockquote>
<p>Being more specific, I would like to know how to write with mathematical symbols this part: "range over $\{1,\ldots,g\}\setminus\{t_1,\ldots,t_u\}$"</p>
<p>Thanks</p>
| Kez | 201,782 | <p>I'd suggest
$$\Large x=\sum_{\substack{m=1\\[0.1cm] m\,\notin\, \{t_1,\,\ldots\,,\,t_u\}}}^g a_mx^m$$</p>
|
599,126 | <p>Question is to check which of the following holds (only one option is correct) for a continuous bounded function $f:\mathbb{R}\rightarrow \mathbb{R}$.</p>
<ul>
<li>$f$ has to be uniformly continuous.</li>
<li>there exists a $x\in \mathbb{R}$ such that $f(x)=x$.</li>
<li>$f$ can not be increasing.</li>
<li>$\lim_{x\rightarrow \infty}f(x)$ exists.</li>
</ul>
<p>What all i have done is :</p>
<ul>
<li>$f(x)=\sin(x^3)$ is a continuous function which is bounded by $1$ which is not uniformly continuous.</li>
<li>suppose $f$ is bounded by $M>0$ then restrict $f: [-M,M]\rightarrow [-M,M]$ this function is bounded ad continuous so has fixed point.</li>
<li>I could not say much about the third option "$f$ can not be increasing". I think this is also true as for an increasing function $f$ can not be bounded but i am not sure.</li>
<li>I also believe that $\lim_{x\rightarrow \infty}f(x)$ exists as $f$ is bounded it should have limit at infinity.But then I feel the function can be so fluctuating so limit need not exists. I am not so sure.</li>
</ul>
<p>So, I am sure second option is correct and fourth option may probably wrong but i am not so sure about third option.</p>
<p>Please help me to clear this.</p>
<p>Thank You. :)</p>
| Dan | 79,007 | <p>For the third point, consider $f(x) = \arctan(x)$. For the fourth point, you've already found a counterexample in one of your other points!</p>
|
599,126 | <p>Question is to check which of the following holds (only one option is correct) for a continuous bounded function $f:\mathbb{R}\rightarrow \mathbb{R}$.</p>
<ul>
<li>$f$ has to be uniformly continuous.</li>
<li>there exists a $x\in \mathbb{R}$ such that $f(x)=x$.</li>
<li>$f$ can not be increasing.</li>
<li>$\lim_{x\rightarrow \infty}f(x)$ exists.</li>
</ul>
<p>What all i have done is :</p>
<ul>
<li>$f(x)=\sin(x^3)$ is a continuous function which is bounded by $1$ which is not uniformly continuous.</li>
<li>suppose $f$ is bounded by $M>0$ then restrict $f: [-M,M]\rightarrow [-M,M]$ this function is bounded ad continuous so has fixed point.</li>
<li>I could not say much about the third option "$f$ can not be increasing". I think this is also true as for an increasing function $f$ can not be bounded but i am not sure.</li>
<li>I also believe that $\lim_{x\rightarrow \infty}f(x)$ exists as $f$ is bounded it should have limit at infinity.But then I feel the function can be so fluctuating so limit need not exists. I am not so sure.</li>
</ul>
<p>So, I am sure second option is correct and fourth option may probably wrong but i am not so sure about third option.</p>
<p>Please help me to clear this.</p>
<p>Thank You. :)</p>
| Hayden | 27,496 | <p>Here is an incredibly non-interesting trivial example: $f(x)=a$ for $a$ being some real number.</p>
|
1,924,033 | <blockquote>
<p><strong>Question.</strong> Let $\mathfrak{g}$ be a real semisimple Lie algebra admitting an invariant inner-product. Is every connected Lie group with Lie algebra $\mathfrak{g}$ compact?</p>
</blockquote>
<p>I know that the converse is true: If $G$ is a compact connected Lie group, then the Haar measure may be used to give an invariant inner-product on $\mathrm{Lie}(G)$. Also, semisimplicity is necessary since $\mathrm{Lie}(\mathbb{R})=\mathbb{R}$ trivially admits an invariant inner-product.</p>
| Community | -1 | <p>I will assume that gluing data is also meant to include the condition $U_{ii} = U_i$. </p>
<p>I'm too tired to organize this all in a narrative, so this will be fairly disjointed.</p>
<hr>
<p>It is fairly common in category theory to consider two families of objects $X_k$ and $X_{ij}$, families of maps $f_{ij} : X_{ij} \to X_i$ and $g_{ij} : X_{ij} \to X_j$, and the corresponding coequalizer</p>
<p>$$ \coprod_{ij} X_{ij} \overset{f}{\underset{g}{\rightrightarrows}} \coprod_k X_k \xrightarrow{\rho} X$$</p>
<p>or, if we include 'redundant' $X_{ii} = X_i$, a pushout square</p>
<p>$$ \begin{matrix} \coprod_{ij} X_{ij} &\xrightarrow{f}& \coprod_k X_k
\\ \!g\!\downarrow & & \downarrow\!\rho\!
\\ \coprod_k X_k &\xrightarrow{\rho}& X \end{matrix} $$</p>
<p>The picture here is that the $X_k$'s are a description of an object of interest, and the $X_{ij}$'s describe relations between the descriptions.</p>
<p>Geometrically, we might think of the $X_k$'s as a cover, and the $X_{ij}$'s describe the overlap between them. Algebraically, we might think of $X_k$ as being generators, and the $X_{ij}$ as being relations.</p>
<p>Either way, it's clear that in the "nicest" arrangement, we want each $X_{ij}$ to to be the pullback of $X_i \to X \leftarrow X_{j}$, so that it truly does describe <em>all</em> relations between $X_i$ and $X_j$, and you only really want one $X_{ij}$ per pair of indices.</p>
<p>Gluing data is in this nice situation, but not all such diagrams are: e.g. we might only have a more minimal description of the relations, or the covers might actually self-intersect nontrivially (e.g. taking the interval $X_1 = [0,2\pi]$ as a cover of the circle, with $X_{11}$ being a single point mapped to both ends) or other deficiencies may apply.</p>
<hr>
<p>The gluing data gives an example of this sort of diagram: on each $U_{ij}$, the two maps into $\coprod_k U_k$ come from</p>
<ul>
<li>$ U_{ij} \hookrightarrow U_i $</li>
<li>$ U_{ij} \xrightarrow{\varphi_{ji}} U_{ji} \hookrightarrow U_j $</li>
</ul>
<p>Furthermore, the data does assert there is "just one" in the sense that the $\varphi_{ji}$ gives a homeomorphism $U_{ij} \to U_{ji}$, and they do so coherently in the sense that $\varphi_{ij} = \varphi_{ji}^{-1}$ and $\varphi_{ii} = 1_{U_i}$, so different "paths" $U_{ij} \to U_{ji}$ (e.g $\varphi_{ji}$ versus $\varphi_{ji} \circ \varphi_{ii} \circ \varphi_{ij} \circ \varphi_{ji}$) all give the same map.</p>
<hr>
<p>The property that <strong>Top</strong> has that makes this setup convenient to work with is that it is <a href="https://ncatlab.org/nlab/show/extensive+category" rel="nofollow">infinitary extensive</a> — i.e. that coproducts really do act like disjoint unions. I do not know if this is actually <em>required</em>, but all the ways I want to reason about gluing data rely on it.</p>
<hr>
<p>Sometimes, we also want to consider another family of $X_{ijk}$, this time with three maps down to the various $X_{mn}$. We have this in gluing data too: we can define $U_{ijk} = U_{ij} \cap U_{ik}$. And again we have "just one" per triple of indices, because we again have coherent homeomorphisms between the different permutations. It is enough to check</p>
<ul>
<li>$U_{ijk} = U_{ikj}$</li>
<li>$\varphi_{ji} : U_{ijk} \to U_{jik}$ is a homeomorphism</li>
<li>The two homeomorphism $U_{ijk} \to U_{kji}$ given by $\varphi_{ki}$ and $\varphi_{kj} \circ \varphi_{ji}$ are the same.</li>
</ul>
<p>We could go further. This leads to a simple example of a simplicial object.</p>
<hr>
<p><strong>Top</strong> is nice enough to talk about relations. The colimit defining $X$ can be viewed as taking the quotient of $\coprod_k U_k$ by the relation that the two maps $\coprod_{ij} U_{ij} \rightrightarrows \coprod_k U_k$ give equivalent outputs for each input.</p>
<p>Normally, this relation is not an equivalence relation, and so the quotient is by the equivalence relation <em>generated</em> by this relation.</p>
<p>However, the neat thing about having the transition maps $\varphi$ is that the relation really is an equivalence relation, so the colimit is much, much easier to work with. The interesting part is that it is a transitive relation, which you can check by noting that</p>
<p>$$ x \sim \varphi_{ji}(x) \quad \text{and} \quad \varphi_{ji}(x) \sim \varphi_{kj}(\varphi_{ji}(x)) $$</p>
<p>only makes sense when $x \in U_{ijk}$, and that the transitive property requires $x \sim \varphi_{ki}(x)$, which we have.</p>
<hr>
<p>In <strong>Top</strong>, having the transition maps implies that the $\psi_i : U_i \to X$ are monic. I think you can even argue it's regular monic. Off hand I don't know what you want from the category to say such things.</p>
<hr>
<p>Finally, last feature is about <em>open subspaces</em>. While seemingly the part most topological in flavor, it too has an abstract analog.</p>
<p><strong>Top</strong> has a <em>open subspace classifier</em>. Let $S = \{0,1\}$ be the Sierpinski space, with topology $\{\varnothing, \{1\}, S\}$. Then there is a natural bijection between open subspaces of $X$ and continuous maps $X \to S$:</p>
<ul>
<li>For every open subset $U \subseteq X$, the characteristic function $\chi_U : X \to S$ is continuous</li>
<li>For any continuous map $\chi : X \to S$, the inverse image $f^{-1}(1)$ is an open subset</li>
</ul>
<p>You can thus show that $\psi_i(U_i) \subseteq X$ is an open subspace by showing that the maps
$$ \chi_{ij} : \coprod_i U_j \to S : x \mapsto \begin{cases}
1 & x \in U_{ji} \\ 0 & x \notin U_{ji} \end{cases} $$
induce a well-defined map $\chi_i : X \to S$, and that $\chi_i^{-1}(1) = \psi_i(U_i)$.</p>
<p>Off hand, I'm not sure what additional properties you need (if any) to be able to make this argument in <strong>Top</strong>.</p>
|
370,058 | <p>How can I take this integral?</p>
<p>$$\int_{0}^{x} (z- u)_+^2 du $$</p>
<p>which <code>+</code> means If $z$ is bigger than u its equal $z - u$ and else it's equal zero.</p>
| Ron Gordon | 53,268 | <p>Wouldn't this integral just be</p>
<p>$$\int_{\tau}^t du \: (\tau-u)^2 = \frac{1}{3} (t-\tau)^3$$</p>
|
627,871 | <p>Let $\mathbf{A}$ be an algebra (in the sense of universal algebra) of some signature $\Sigma$. By <em>quasi-identity</em> I mean the formula of the form</p>
<p>$$(\forall x_1) (\forall x_2) \dots (\forall x_n) \left(\left[\bigwedge_{i=1}^{k}t_i(x_1, \dots, x_n)=s_i(x_1, \dots, x_n)\right]\rightarrow t(x_1, \dots, x_n)=s(x_1, \dots, x_n) \right) \;, $$</p>
<p>where $t_i(x_1, \dots, x_n), s_i(x_1, \dots, x_n), t(x_1, \dots, x_n), s(x_1, \dots, x_n)$ are terms (using the algebra operations) with all its variables among $x_1, \dots, x_n$.</p>
<p>Since the class of all algebras (of the considered signature) satisfying some quasi-identity is (allegedly) not a variety in general and clearly such class is closed under taking subalgebras and products, it follows that it is not closed under taking quotients in general.</p>
<p>So the question is:</p>
<p><strong>Is there some (possibly elementary) example of an algebra satisfying some quasi-identity and its quotient where the quasi-identity does not hold?</strong></p>
<p>My only idea was to show this for the cancellation law in some monoid, but I do not see any example (mainly because I do not see how the quotients look like).</p>
<p>Thanks in advance for any help.</p>
| bof | 111,012 | <p>The quasi-identity $\forall x(x+x=0\to x=0)$ holds in $\mathbb Z$ but not in $\mathbb Z/2\mathbb Z$.</p>
|
2,574,117 | <p>For a matrix $A$, define the operator $\ell_p$-norm of $A$ to be
$$
\|A\|_p = \sup_{x \neq 0} \frac{\|Ax\|_p}{\|x\|_p}.
$$
Here $\|x\|_p$ denotes the $\ell_p$ norm of the vector $x$.</p>
<p>For $1 \le p \le q \le 2$ and $x \in \mathbb{R}^n$, we know that $\|x\|_q \le \|x\|_p \le n^{1 / p - 1 / q} \|x\|_q$. </p>
<p>Is there any similar conclusion for the operator $\ell_p$-norm of a given matrix $A \in \mathbb{R}^{n \times m}$?</p>
<p>Or a more concrete problem: if we know $\|A\|_1$ and $\|A\|_2$, what is the best possible upper bound we can achieve for $\|A\|_p$ if $1 \le p \le 2$?</p>
<p>E.g.,
$$\|Ax\|_p \le n^{1 / p - 1 /2} \|Ax\|_2 \le n^{1 / p - 1 / 2} \|A\|_2 \cdot \|x\|_2 \le n^{1 / p - 1 / 2} \|A\|_2 \cdot \|x\|_p,$$
which implies
$$
\|A\|_p \le n^{1 / p - 1 / 2} \|A\|_2.
$$</p>
<p>Similarly,
$$\|Ax\|_p \le \|Ax\|_1 \le \|A\|_1 \cdot \|x\|_1 \le m^{1 - 1 / p} \|A\|_1 \cdot \|x\|_p,$$
which implies
$$
\|A\|_p \le m^{1 - 1 / p} \|A\|_1.
$$
Combine them we have $\|A\|_p \le \min\{n^{1 / p - 1 / 2} \|A\|_2, m^{1 - 1 / p} \|A\|_1\}$, which is quite naive.
Are there any tighter upper bounds?</p>
| daw | 136,544 | <p>Using H\"older's inequality, we can prove that these norms are equivalent. Let $1\le p\le q\le +\infty$. Then
$$
\|Ax\|_p \le n^{1/p-1/q} \|Ax\|_q \le n^{1/p-1/q} \|A\|_q\|x\|_q\le n^{1/p-1/q} \|A\|_q\|x\|_p,
$$
hence
$$
\|A\|_p \le n^{1/p-1/q}\|A\|_q.
$$
Similarly,
$$
\|A\|_q \le m^{1/p-1/q}\|A\|_p.
$$
Equality holds in one of the inequalities if $A$ contains a column (row) of all ones, with the remaining entries zero:
$$
A=\pmatrix{1& 0 & \dots & 0 \\ \vdots &\vdots&&\vdots\\1& 0 & \dots & 0}.
$$
Take $x\ne0$, then $\|A\|_p = |x_1| n^{1/p}$. Taking the supremum over $\|x\|_p\le 1$ yields $\|A\|_p = n^{1/p}$. And the first inequality holds with equality.</p>
<p>Equality holds in both inequalities for $p\ne q$ only if $n=m$ for the same matrix $A$ as above.</p>
|
313,794 | <p>I am not able to calculate extremums for the given function:
$u = 3x^2 - 3xy + 3x +y^2 + 5y$</p>
<p>I am able to calculate
$
u_x = 6x - 3y+ 3
$</p>
<p>$
u_{xx} = 6
$</p>
<p>$
u_{xy} = -3 = u_{yx}
$</p>
<p>$
u_y = -3x + 2y + 5
$</p>
<p>$
u_{yy} = 2
$</p>
<p>But what is next? Where are extremum points?</p>
| wj32 | 35,914 | <p>Since $u$ is differentiable, any local min/max point $(x,y)$ must satisfy $Du(x,y)=0$. You've already done the calculations, so just solve \begin{align}
6x-3y+3&=0 \\
-3x+2y+5&=0
\end{align} to get a single point $(x_0,y_0)$. The matrix representing $D^2u(x_0,y_0)$ (i.e. the "Hessian matrix") is $$H=\begin{bmatrix}6 & -3 \\ -3 & 2\end{bmatrix},$$ as you've calculated. Clearly $\det(H)>0$, so $D^2u(x_0,y_0)$ is positive definite. Then apply the higher derivative test to deduce that $u$ has a local minimum at $(x,y)$.</p>
<p>(In fact, this is a global minimum, since the fact that $D^2u$ is positive definite everywhere implies that $u$ is convex.)</p>
|
313,794 | <p>I am not able to calculate extremums for the given function:
$u = 3x^2 - 3xy + 3x +y^2 + 5y$</p>
<p>I am able to calculate
$
u_x = 6x - 3y+ 3
$</p>
<p>$
u_{xx} = 6
$</p>
<p>$
u_{xy} = -3 = u_{yx}
$</p>
<p>$
u_y = -3x + 2y + 5
$</p>
<p>$
u_{yy} = 2
$</p>
<p>But what is next? Where are extremum points?</p>
| Bob | 48,443 | <p>You're almost there.</p>
<p>$$u_x=6x-3y+3=0 \to y= 2x+1$$
$$u_y=-3x+2y+5=0$$</p>
<p>Substitution leads to
$$-3x+2(2x+1)+5=0 \to x=-7, y=-13$$</p>
<p>Which is the global minimum since both $u_{xx},u_{yy}>0$</p>
|
2,096,408 | <p>My Physics book has many graphs. Some are straight lines, some parabolas while others are hyperbolas. I have not studied these curves (conic sections) yet and to me parabola and hyperbola look just the same. Is there any way of knowing whether a line is a parabola or a hyperbola just by seeing the graph of the line.</p>
| Edward Porcella | 403,946 | <p>As you go out from the vertex (turning-point) of a parabola, tangents to opposite sides of the curve approach parallelism. With a hyperbola there's a limit to how small an angle the tangents can make with each other--the angle of the "asymptotes". Parabolas are more u-shaped, hyperbolas more v-shaped.</p>
|
2,780,403 | <p>I've tried to solve this but I don't seem to get anywhere.</p>
<p>The question states:</p>
<blockquote>
<p>Tom's home is $1800$ m from his school. One morning he walked part of the way to school and then ran the rest. If it took him $20$ mins or less to get to school, and he walks at $70$ m/min and runs at $210$ m/min, how far did he run?</p>
</blockquote>
<p>My attempts to solve it got me stuck here:</p>
<blockquote>
<p>$$\frac{x}{70}+\frac{y}{210}≥\frac{1800}{20}$$</p>
</blockquote>
<p>Any help would be appreciated! </p>
| Dylan | 135,643 | <p>The answer you were given has a several mistakes. First, the gradient should be</p>
<p>$$ \nabla T(1,1,1) = \langle 2,4,4 \rangle $$</p>
<p>The direction vector is in the wrong direction
$$ \hat{u} = -\frac{\nabla T}{|\nabla T|} = -\left\langle \frac13, \frac23, \frac23 \right\rangle $$</p>
<p>And the directional derivative is off by a factor of $2$</p>
<p>$$ \nabla T \cdot\hat{u} = -|\nabla T| = -6 $$</p>
|
20,726 | <p>The following situation is ubiquitous in mathematical physics. Let
$\Lambda_N$
be a finite-size lattice with linear size
$N$. An typical example would be the subset of
$\mathbb{Z}\times\mathbb{Z}$
given by those pairs of integers
$(j,k)$ such that
$j,k \in$ {
$0,\ldots,N-1$}. On each vertex
$j$ of the lattice place a copy of the vector space
$\mathbb{C}^d$. The total space will be the tensor product of all of these spaces. Then define a Hamiltonian acting on this total space as follows:
$$ H = \sum_{k \in \Lambda_N} h_k$$
for some Hermitian matrices
$h_k$ which act like the identity everywhere except on the vector spaces located on site
$k$ and in the neighborhood surrounding
$k$. Typically, one is interested in the case where there is a translational symmetry (except at the boundary) in the definition of the
$h_k$. Denote the eigenvalues of
$H$ in increasing order by
$\lambda_1 \le \lambda_2 \le \ldots \le \lambda_M$. </p>
<blockquote>
<p>For an arbitrary fixed family of Hamiltonians
$H$, what proof techniques exist for computing an upper and a lower bound on
$\Delta = \lambda_2 - \lambda_1$
as a function of
$N$? In particular, we want to know if
$\Delta$ decays to zero as a function of
$N$, or if it is lower-bounded by some constant independent of
$N$.</p>
</blockquote>
<p>The gap
$\Delta$ is the energy gap between the ground state and the first excited state of an interacting quantum system. Understanding this quantity tremendously impacts our understanding of the different phases of matter, but it is extremely difficult to compute or even bound for all but the simplest cases (like when all the
$h_k$ commute). This difficulty persists even when there is significant additional (physically motivated) structure in the problem, such as considering only
$h_k$ which are projectors, and where there is a unique zero-energy eigenstate (all others having positive energy for any finite
$N$).</p>
<p>More general formulations of this question also have applications to expansion properties of graphs, mixing times of Markov chains, and many other things. I’m happy to hear answers related to these as well, but I’m hoping to find answers that are useful for the structure of local Hamiltonians, as defined above.</p>
| Steve Flammia | 1,171 | <p><em>Bounding angles in projector-valued Hamiltonians</em></p>
<p>Suppose that each of the
<span class="math-container">$h_k$</span> are projectors, and suppose that we shift the total energy so that
<span class="math-container">$\lambda_1 = 0$</span>. Further suppose that this eigenspace is known to be non-degenerate for every finite
<span class="math-container">$N$</span>. Then one proof technique that sometimes works (at least in one dimension, but it can be generalized) to prove an
<span class="math-container">$N$</span>-independent lower bound is the following, which to the best of my knowledge was first discussed in</p>
<blockquote>
<p>M. Fannes, B. Nachtergaele, R. F. Werner, <i>Finitely correlated states on quantum spin chains</i>, Comm. Math. Phys. Vol. 144, Num. 3 (1992), 443-490. <a href="https://mathscinet.ams.org/mathscinet-getitem?mr=1158756" rel="nofollow noreferrer">MathSciNet:MR1158756</a></p>
</blockquote>
<p>The bound is as follows. Let
<span class="math-container">$\theta_{j,k}$</span> be the smallest non-zero angle between the pair of projectors
<span class="math-container">$h_j$</span> and
<span class="math-container">$h_k$</span>. That is,
<span class="math-container">$\cos^2(\theta_{j,k})$</span> is the largest eigenvalue not equal to one of
<span class="math-container">$h_j h_k h_j$</span>. Define
<span class="math-container">$\theta = \min_{j,k} \theta_{j,k}$</span>. Then FNW show that
<span class="math-container">$$ \Delta \ge 1-2 \cos(\theta) .$$</span>
As long as
<span class="math-container">$\cos(\theta)$</span> is less than 1/2, there is an
<span class="math-container">$N$</span>-independent lower bound on
<span class="math-container">$\Delta$</span>, since this involves only local information. The proof proceeds by squaring
<span class="math-container">$H$</span> and then bounding how negative the anti-commutators can be by using this angle
<span class="math-container">$\theta$</span>.</p>
|
2,129,830 | <p>I am wondering if this is generally true for any topology. I think there might be counter examples, but I am having trouble generating them. </p>
| user404961 | 404,961 | <p>The set $(0,1) \cup (1,2)$ is a counterexample.</p>
|
2,129,830 | <p>I am wondering if this is generally true for any topology. I think there might be counter examples, but I am having trouble generating them. </p>
| Eric Wofsey | 86,856 | <p>No. A quick way to verify counterexamples is the following observation: if $U$ is the interior of a closed set $C$, then $U$ is also the interior of $\overline{U}$. Indeed, since $C$ is closed and $U\subseteq C$, $\overline{U}\subseteq C$, so the interior of $\overline{U}$ is contained in the interior of $C$. But $U$ is contained in the interior of $\overline{U}$ (since it is open and contained in $\overline{U}$) and equal to the interior of $C$, so $U$ must be equal to the interior of $\overline{U}$.</p>
<p>We can now give an example: take $U=(0,1)\cup(1,2)$ as a subset of $\mathbb{R}$. Since $\overline{U}=[0,2]$ has interior $(0,2)\neq U$, $U$ cannot be the interior of any closed set. In general, a set which is the interior of a closed set (equivalently, of its closure) is called a <em>regular open set</em>.</p>
|
2,129,830 | <p>I am wondering if this is generally true for any topology. I think there might be counter examples, but I am having trouble generating them. </p>
| gerw | 58,577 | <p>Every open, dense set is a counterexample. In $\mathbb{R}$ you can use
$$\bigcup_{n \in \mathbb{N}} (q_n - 1/2^n, q_n + 1/2^n),$$
where $\mathbb{Q} = \{q_0, q_1, \ldots\}$ are the rationals.</p>
|
1,034,335 | <p>I'm preparing for my calculus exam and I'm unsure how to approach the question: "Explain the difference between convergence of a sequence and convergence of a series?" </p>
<p>I understand the following:</p>
<p>Let the sequence $a_n$ exist such that $a_n =\frac{1}{n^2}$ </p>
<p>Then $\lim_{n\to\infty} a_n=\lim_{n\to\infty} \frac{1}{n^2}=0$ therefore $a_n$ converges to $0$.</p>
<p>And the series $\sum_{i=1}^{n}a_n=1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + ... +\frac{1}{n^2}$</p>
<p>And by the $n$-th term test, this series converges. But, I don't understand <em>why</em> or <em>how</em> the convergence between the series and the sequence is different.</p>
<p>I looked online and I find a lot of answers on how to determine convergence or divergence, but the only difference I've found is that you use limits to test sequences and series have more complex testing requirements. Please help!</p>
| Gyu Eun Lee | 52,450 | <p>If we are talking about sequences and series of real or complex numbers, or of vectors in a real (or complex) normed vector space, then convergence of sequences and series are equivalent concepts.</p>
<p>Convergence of a series $\sum_{n=1}^\infty a_n$ is simply the convergence of the sequence of partial sums $S_N = \sum_{n=1}^N a_n$.</p>
<p>Convergence of a sequence $(a_n)$ of numbers is equivalent to the convergence of the series $a_1 + \sum_{n=1}^\infty (a_{n+1} - a_n)$; note the $N$-th partial sum is
$$
a_1 + \sum_{n=1}^N (a_{n+1} - a_n) = a_1 + \sum_{n=2}^{N+1} a_n - \sum_{n=1}^N a_n = a_1 + a_N - a_1 = a_N.
$$</p>
<p>However, in the more general setting of topology the notion of a sequence is more general than the notion of a series, because in topological spaces without a vector space structure the notion of a series makes no sense (how will you define a series if you can't add things?). Therefore it may be convenient to think of sequences as the more natural idea, and series as a special case of sequences.</p>
|
50,002 | <p>a general version: connected sums of closed manifold is orientable iff both are orientable.
I think this can be prove by using homology theory, but I don't know how.Thanks.</p>
| gary | 6,595 | <p>Consider the perspective of simplicial homology, for manifolds M,M'. Assume WOLG that M,M' are both connected: if an m-manifold M is orientable (I think that there is a result that all manifolds can be made into simplicial complexes), this means that the top cycle --call it m'-- can be assigned a coherent orientation, so that m' is a cycle --that does not bound, since m is the highest dimension--, i.e., the net boundary of m cancels out , e.g., in the simplest case of a loop with boundary a-a=0. This means m', which represents M itself, is a non-trivial cycle, which generates the top homology class. If your coefficient ring is $\mathbb Z$, then the top homology will be $\mathbb Z$; consider going n-times about the cycle.
Now, the key is that the two orientable manifolds can be glued so that, at the circle of gluing, the total boundary cancels out, and the resulting manifold M#M' is still orientable. As a specific example, consider a square a,b,c,d, with arrows going all in the same direction, so that the net boundary is : (b-a)+(c-b)+(d-c)+(a-d)=(b-b)+(a-a)+(c-c)+(d-d)=0 . Now glue a second square a',b',c',d' along a common edge, (say (b,c) with (b',c')), but reverse the orientation of the edge (b',c') in M when gluing, and notice how the simplex resulting from the gluing also has net boundary zero. </p>
<p>Now, the key general point is that , at the cycle C where we collapse M with M', we change the orientation of C in either M or M', so that, along the common cycle, where you are doing the gluing, the respective boundaries cancel each other out, and the remaining orientations of M-C and M-C' remain the same, so that M#M' is orientable.</p>
|
942,470 | <p>I am trying to count how many functions there are from a set $A$ to a set $B$. The answer to this (and many textbook explanations) are readily available and accessible; I am <strong>not</strong> looking for the answer to that question and <strong>please do not post it</strong>. Instead I want to know what fundamental mistake(s) I am making in counting the number of these functions. My reasoning is below, which I know is wrong after checking this question: <a href="https://math.stackexchange.com/questions/639326/how-many-functions-there-is-from-3-element-set-to-2-element-set">How many functions there is from 3 element set to 2 element set?</a>.</p>
<hr>
<p>For an example case, I consider counting how many functions there are from set $A = \{0,1\}$ to set $B = \{a,b\}$. My understanding of the term <em>function</em> is that it is any possible mapping between elements of set $A$ to elements of set $B$. Thus, a possible function $F: A \times B$ is the function that maps each element of $A$ to no element of $B$, i.e. $f_0(0) = \emptyset, f_0(1) = \emptyset$. Another possible function is $f_1(0) = a, f_1(1) = \{a, b\}$. </p>
<p>I notice a pattern here: for each element of the set $A$, there are $|\mathcal P (B)|$ unique combinations of elements that it can map to. In this case, $\mathcal P(B) = \{\{a,b\}, \{a\}, \{b\}, \emptyset\}$. To count these functions, then, we can use the product rule, since the choice of what each element of $A$ maps to does not affect what another element of $A$ can map to (since we consider all functions). </p>
<p>There are $4$ choices for $0$ and $4$ choices for $1$. Therefore there are $16$ unique functions $F: A \times B$. For a sanity check, I've listed out all <strong>16</strong> possible functions.</p>
<p>$f_0(0) = \emptyset, f_0(1) = \emptyset$</p>
<p>$f_1(0) = \emptyset, f_1(1) = \{a\}$</p>
<p>$f_2(0) = \emptyset, f_2(1) = \{b\}$</p>
<p>$f_3(0) = \emptyset, f_3(1) = \{a, b\}$</p>
<p>$f_4(0) = \{a\}, f_4(1) = \emptyset$</p>
<p>$f_5(0) = \{a\}, f_5(1) = \{a\}$</p>
<p>$f_6(0) = \{a\}, f_6(1) = \{b\}$</p>
<p>$f_7(0) = \{a\}, f_7(1) = \{a, b\}$</p>
<p>$f_8(0) = \{b\}, f_8(1) = \emptyset$</p>
<p>$f_9(0) = \{b\}, f_9(1) = \{a\}$</p>
<p>$f_{10}(0) = \{b\}, f_{10}(1) = \{b\}$</p>
<p>$f_{11}(0) = \{b\}, f_{11}(1) = \{a, b\}$</p>
<p>$f_{12}(0) = \{a,b\}, f_{12}(1) = \emptyset$</p>
<p>$f_{13}(0) = \{a,b\}, f_{13}(1) = \{a\}$</p>
<p>$f_{14}(0) = \{a,b\}, f_{14}(1) = \{b\}$</p>
<p>$f_{15}(0) = \{a,b\}, f_{15}(1) = \{a, b\}$</p>
<p>The generalization: The number of functions $F: A \times B$ is $|\mathcal P(B)|^{|A|}$.</p>
<hr>
<p>Now I know my reasoning is completely wrong, but why? Am I double counting? Do I misunderstand the definition of a function? </p>
| Stijn Hanson | 145,554 | <p>A function requires assignment so you cannot just not return anything. What you're calling a function is, in fact, a partial function.</p>
|
2,049,207 | <p>The question is this: <strong><em>How many ways are there to put 5 different balls into 3 different boxes so that none of the boxes is empty?</em></strong> </p>
<p>The correct answer as per my lecturer's notes is <strong>150</strong>, and I would like to know where I am going wrong in my approach.</p>
<p><strong>Here is how I approached it (wrongly):</strong></p>
<p>Separated into three tasks: </p>
<p>1) Picking three balls from 5 to put in the boxes</p>
<pre><code> {5 \choose 3} ways
</code></pre>
<p>2) Permuting those balls</p>
<pre><code> 3! ways
</code></pre>
<p>3) Placing the other two balls can be done in two ways:</p>
<p>Either put both in one of the boxes</p>
<pre><code> 3 ways
</code></pre>
<p>Or put both each in a different box</p>
<pre><code> P(3,2) ways to pick two boxes and arrange the balls in them
</code></pre>
<p>Task 3 has a total of </p>
<pre><code> 6 +3 = 9 ways
</code></pre>
<p>Total using product rule is (as per my approach, which is incorrect):</p>
<pre><code> 60 * (9) = 540
</code></pre>
<p>I have seen other approaches to getting the correct answer(including using Stirling numbers of the second kind followed by permutation, inclusion-exclusion), but would like to know what is the correct way to split this into tasks and use the product rule (without using Stirling numbers).</p>
| mesel | 106,102 | <p>A subgroup generated by conjugacy classes is always normal.</p>
<p><strong>Hint:</strong> Let $N$ be a subgroup generated by $<g^x|x\in G>$. Take a $n\in N$.</p>
<p>Then $n=h_1h_2...h_k$ for $h_i=g^x$ for some $x\in G$. Now what can you say about $n^y$ for some $y\in G$ ?</p>
|
2,542,056 | <p>Baire's Category Theorem states that a meager subset of a complete metric space has empty interior. </p>
<p>Are there examples of meager subsets of non-complete metric spaces which do not have empty interior?<br>
In particular, are the rationals numbers as a subset of themselves an example?</p>
| Asinomás | 33,907 | <p>Yes they do because $\mathbb Q \setminus \{a\}$ is open and dense for each $a\in \mathbb Q$. When you take the intersection of these sets for every $a$ you get $\varnothing$, notice that this is a countable intersection since there ate countably many rationals.</p>
|
2,651,537 | <p>Prove that $(5m+3)(3m+1)=n^2$ is not satisfied by any <strong>positive</strong> integers $m,n$.</p>
<p>I have been staring at this for some time (it's the difficult part of a competition problem, I won't spoil it by naming the problem). I tried looking at it modulo 3,4,5,7,8,16 for a contradiction, as well as looking at the discriminant with respect to $m$. I couldn't finish either way. I wonder whether it can somehow be used that $(-1,2)$ is a solution.</p>
<p>There is a very small chance that there are actually positive solutions but I've ruled out the first few millions for $m$ with a little Python script and, as I said, this was set in a competition so presumably any solutions would be reasonably small.</p>
<p>I've heard somewhere that all diophantine quadratics can be solved by some method and would be interested to read more (though not an entire book, if possible).</p>
| Macavity | 58,320 | <p>You can rewrite that equation as $(15m+7)^2-15n^2=4$. Looking at this $\pmod 4$, we can find both $15m+7$ and $n$ need to be even. Hence let $15m+7=2a, \,n=2b$, so that we have the more familiar Pell equation $a^2-15b^2=1$ to solve. </p>
<p>Note the smallest positive solution to the Pell equation is $(a_1, b_1) = (4, 1)$, and all others are generated by $a_n+\sqrt{15}\,b_n = (4+\sqrt{15})^n$, or more particularly of our interest, as a recursion, $a_{n+1} = 8a_n-a_{n-1}$ with $a_1=4, a_2=31$.</p>
<p>Further, note we need $2a=15m+7 \implies 2a \equiv 7 \pmod {15} \implies a \equiv 11 \pmod {15}$. However, the Pell recursion solution above seen $\mod {15}$ generates the sequence $a_i = 4, 1, 4, 1, 4, 1...$ which repeats without ever containing an $11$. Hence there can be no positive solution for $(m, n)$.</p>
|
3,255,654 | <p>In a multiple choice question, there are five different answers, of which only one is correct. The probability that a student will know the correct answer is 0.6. If a student does not know the answer, he guesses an answer at random.</p>
<p>a) What is the probability that the student gives the correct answer?</p>
<p>b) If the student gives the correct answer, what is the probability that he guessed?</p>
<p>Let <strong>A<span class="math-container">$_1$</span></strong> be: student knows the answer, <strong>A<span class="math-container">$_2$</span></strong>: student doesn't know the answer, <strong>B</strong>: student gives the correct answer</p>
<p>P(A<span class="math-container">$_1$</span>) = 0.6 and P(A<span class="math-container">$_2$</span>) = 0.4. How do I find P(B|A<span class="math-container">$_1$</span>) and P(B|A<span class="math-container">$_2$</span>)?</p>
| yurnero | 178,464 | <p>Slightly different notation. Let <span class="math-container">$A$</span> be the event that the student knows the answer and <span class="math-container">$B$</span> is as you defined.
<span class="math-container">$$
P(B)=P(B|A)P(A)+P(B|\neg A)P(\neg A)=1\times 0.6+0.2\times0.4=0.68.
$$</span>
Next,
<span class="math-container">$$
P(\neg A|B)=\frac{P(\neg A\text{ and }B)}{P(B)}=\frac{P(B|\neg A)P(\neg A)}{P(B)}=\frac{0.08}{0.68}=\frac{2}{17}.
$$</span></p>
|
73,410 | <p>Gromov proved that if $$
f,g:\left[ {a,b} \right] \to R
$$
are integrable functions, such that the function $$
t \to \frac{{f\left( t \right)}}
{{g\left( t \right)}}
$$
is also integrable, and decreasing. Then the function $$
r \to \frac{{\int\limits_a^r {f\left( t \right)dt} }}
{{\int\limits_a^r {g\left( t \right)dt} }}
$$
is decreasing.
I could not proved, and I could not find a proof )=</p>
| AD - Stop Putin - | 1,154 | <p><em>Extra assumption 1: $g$ is non-negative or non-negative. (Thanks robjohn)</em></p>
<p><em>Extra assumption 2: $f$ and $g$ are absolute continuous (e.g. they are strictly increasing/decreasing). (Thanks Mariano Suárez-Alvarez and t.b.)</em></p>
<p>Fix $r$. Since $f/g$ is decreasing we have
$$\frac{f(x)}{g(x)}\ge\frac{f(r)}{g(r)}$$
for all $a\le x\le r$. Hence (by Extra assumption 1)
$$f(x)g(r)\ge f(r)g(x)$$
next we integrate this with respect to $x$ over $[a,r]$ which leads to
$$g(r)\int_a^rf(x)dx\ge f(r)\int_a^rg(x)dx$$
(recall $r$ was fixed). But then (by Extra assumption 2)
$$\left(\frac{\int_a^rf(x)dx}{\int_a^rg(x)dx}\right)'= \frac{f(r)\int_a^rg(x)dx-g(r)\int_a^rf(x)dx}{\left(\int_a^rg(x)dx\right)^2}\le0.$$</p>
|
4,368,464 | <p>How to solve <span class="math-container">$\sum_{i=1}^{n} \frac{P_i}{1+(d_i-d_1)x/365} = 0$</span> in spreadsheet?</p>
<p>We have already known that in Excel,</p>
<p>XIRR() find the root of the equation:
<span class="math-container">$\sum_{i=1}^{n} \frac{P_i}{(1+x)^{(d_i-d_1)/365}} = 0$</span>, which is the IRR (Internal Rate of Return) of a series of compounding cash flow.</p>
<p>However, what if the interest rate is flat rather than compounded?
that is, solve <span class="math-container">$\sum_{i=1}^{n} \frac{P_i}{1+(d_i-d_1)x/365} = 0$</span></p>
<p>I have used approximations like</p>
<p><span class="math-container">$1/(1+\tau x)=1-\tau x$</span></p>
<p>or</p>
<p><span class="math-container">$1/(1+\tau x)=(1+x)^{\tau}$</span></p>
<p>where <span class="math-container">$\tau=(d_i-d_1)/365$</span></p>
<p>But I am not sure the error is negligible.</p>
<p>I hope that I could have a nice numerical result in excel.</p>
<p>Thanks!</p>
| Greg Martin | 16,078 | <p>Yes, the formula is the same, since
<span class="math-container">$$
\sum_{i=0}^{n} i = 0+1+\cdots+n = 0+(1+\cdots+n) = 0 + \sum_{i=1}^n i.
$$</span></p>
|
3,112,043 | <p>The first part of the problem is:</p>
<p>Prove that for all integers <span class="math-container">$n \ge 1$</span> and real numbers <span class="math-container">$t>1$</span>,
<span class="math-container">$$ (n+1)t^n(t-1)>t^{n+1}-1>(n+1)(t-1)$$</span></p>
<p>I have done the first part by induction on <span class="math-container">$n$</span> for any real <span class="math-container">$t>1$</span>.</p>
<p>However, I don't know how to do the second part, which is:</p>
<p>Use this to prove that if <span class="math-container">$m$</span> and <span class="math-container">$n$</span> are positive integers,</p>
<p><span class="math-container">$$ \frac{m^{n+1}}{n+1}<1^n+2^n+\dots+m^n<\Big(1+\frac{1}{m}\Big)^{n+1}\frac{m^{n+1}}{n+1} $$</span></p>
<p>I factorized <span class="math-container">$t^{n+1}-1$</span> into <span class="math-container">$(t-1)(1+t+t^2+\dots+t^n)$</span> and cancelled <span class="math-container">$t-1$</span> to obtain <span class="math-container">$(n+1)t^n>1+t+t^2+\dots+t^n>n+1$</span>. And have no idea what to do next. This is the only way I could think of in order to get a "sum", but couldn't see any relation between the two sum (if there are any...).</p>
<p>The last part of the question is:</p>
<p>Find
<span class="math-container">$$ \lim_{m\rightarrow \infty} \frac{1^n+2^n+\dots+m^n}{m^{n+1}}$$</span></p>
<p>I think I know how to do the last part. It should be divide the inequality by <span class="math-container">$m^{n+1}$</span>, then apply Squeeze Theorem. I believe the answer is <span class="math-container">$\frac{1}{n+1}$</span>.</p>
| G Cab | 317,234 | <p>Hint:</p>
<p>Note that <span class="math-container">$x^n$</span> is convex when <span class="math-container">$1 \le n$</span>.<br>
Thus you can profitably demonstrate the inequality through a Riemann sum of the
integral of <span class="math-container">$x^n$</span> (histogram below and above the continuous curve).</p>
|
3,466,870 | <p>Suppose </p>
<p><span class="math-container">$$a^2 = \sum_{i=1}^k b_i^2$$</span> </p>
<p>where <span class="math-container">$a, b_i \in \mathbb{Z}$</span>, <span class="math-container">$a>0, b_i > 0$</span> (and <span class="math-container">$b_i$</span> are not necessarily distinct).</p>
<p>Can any positive integer be the value of <span class="math-container">$k$</span>?</p>
<hr>
<p>The reason I am interested in this: in a irreptile tiling where the smallest piece has area <span class="math-container">$A$</span>, we have <span class="math-container">$a^2A = \sum_{i=1}^k b_i^2A$</span>, where we have <span class="math-container">$k$</span> pieces scaled by <span class="math-container">$b_i$</span> to tile the big figure, which is scaled by <span class="math-container">$a$</span>. I am wondering what constraints there are on the number of pieces.</p>
<p>Here is an example tiling that realizes <span class="math-container">$4^2 = 3^2 + 7 \cdot 1^2$</span>, so <span class="math-container">$k = 8$</span>.</p>
<p><a href="https://i.stack.imgur.com/LdwQy.png" rel="noreferrer"><img src="https://i.stack.imgur.com/LdwQy.png" alt="enter image description here"></a></p>
| Keith Backman | 29,783 | <p>"Is any <span class="math-container">$k$</span> possible?" An easy route to "Yes": You know from the Pythagorean theorem that two squares can add to a perfect square. <span class="math-container">$$c^2=a^2+b^2$$</span></p>
<p><span class="math-container">$c^2$</span> must be either odd or even. If odd, it is the difference between two consecutive squares. <span class="math-container">$$c^2=2n-1=n^2-(n-1)^2$$</span>
If even, <span class="math-container">$c^2$</span> is divisible by <span class="math-container">$4$</span> and is also the difference between two squares. <span class="math-container">$$c^2=4n=(n+1)^2-(n-1)^2$$</span>
So in either case, <span class="math-container">$c^2$</span> equals the difference between two squares. <span class="math-container">$$c^2=r^2-s^2 \\ r^2=c^2+s^2=a^2+b^2+s^2$$</span>
Here, <span class="math-container">$r^2$</span> is the sum of three squares. </p>
<p>This can be repeated indefinitely, increasing by one the number of squares in the sum which adds to a square. There is no limit to the number of squares that can be accumulated in the sum.</p>
|
3,466,870 | <p>Suppose </p>
<p><span class="math-container">$$a^2 = \sum_{i=1}^k b_i^2$$</span> </p>
<p>where <span class="math-container">$a, b_i \in \mathbb{Z}$</span>, <span class="math-container">$a>0, b_i > 0$</span> (and <span class="math-container">$b_i$</span> are not necessarily distinct).</p>
<p>Can any positive integer be the value of <span class="math-container">$k$</span>?</p>
<hr>
<p>The reason I am interested in this: in a irreptile tiling where the smallest piece has area <span class="math-container">$A$</span>, we have <span class="math-container">$a^2A = \sum_{i=1}^k b_i^2A$</span>, where we have <span class="math-container">$k$</span> pieces scaled by <span class="math-container">$b_i$</span> to tile the big figure, which is scaled by <span class="math-container">$a$</span>. I am wondering what constraints there are on the number of pieces.</p>
<p>Here is an example tiling that realizes <span class="math-container">$4^2 = 3^2 + 7 \cdot 1^2$</span>, so <span class="math-container">$k = 8$</span>.</p>
<p><a href="https://i.stack.imgur.com/LdwQy.png" rel="noreferrer"><img src="https://i.stack.imgur.com/LdwQy.png" alt="enter image description here"></a></p>
| poetasis | 546,655 | <p>Yes, any number of squares can be a perfect square. Take the example of Pythagorean triples generated by the formula with a table of triples shown below it where <span class="math-container">$n$</span> is a set and where <span class="math-container">$k$</span> is a set member.</p>
<p><span class="math-container">\begin{equation}
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
\end{equation}</span>
<span class="math-container">\begin{array}{c|c|c|c|c|c|c|}
n & k=1 & k=2 & k=3 & k=4 & k=5 & k=6 \\ \hline
Set_1 & 3,4,5 & 5,12,13& 7,24,25& 9,40,41& 11,60,61 & 13,84,85 \\ \hline
Set_2 & 15,8,17 & 21,20,29 &27,36,45 &33,56,65 & 39,80,89 & 45,108,117 \\ \hline
Set_3 & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 & 75,100,125 & 85,132,157 \\ \hline
Set_{4} &63,16,65 &77,36,85 &91,60,109 &105,88,137 &119,120,169 & 133,156,205 \\ \hline
Set_{5} &99,20,101 &117,44,125 &135,72,153 &153,104,185 &171,140,221 & 189,180,261 \\ \hline
\end{array}</span>
Since <span class="math-container">$3^2+4^2=5^2$</span>, we can replace the <span class="math-container">$5$</span> in <span class="math-container">$5,12,13$</span> with <span class="math-container">$3,4$</span> so <span class="math-container">$3^3+4^2+12^2=13^2$</span>. The value of <span class="math-container">$A$</span> can be any odd number greater than <span class="math-container">$1$</span> so there is always a triple where side-A of one equals side-C of another. Extending to a couple more squares, <span class="math-container">$3^2+4^2+12^2+84^2+132^2+12324=12325$</span>. The process can be extended to infinite numbers of squares summed.</p>
<p>Let me know if you would like to know how to find "extensions".</p>
|
3,466,870 | <p>Suppose </p>
<p><span class="math-container">$$a^2 = \sum_{i=1}^k b_i^2$$</span> </p>
<p>where <span class="math-container">$a, b_i \in \mathbb{Z}$</span>, <span class="math-container">$a>0, b_i > 0$</span> (and <span class="math-container">$b_i$</span> are not necessarily distinct).</p>
<p>Can any positive integer be the value of <span class="math-container">$k$</span>?</p>
<hr>
<p>The reason I am interested in this: in a irreptile tiling where the smallest piece has area <span class="math-container">$A$</span>, we have <span class="math-container">$a^2A = \sum_{i=1}^k b_i^2A$</span>, where we have <span class="math-container">$k$</span> pieces scaled by <span class="math-container">$b_i$</span> to tile the big figure, which is scaled by <span class="math-container">$a$</span>. I am wondering what constraints there are on the number of pieces.</p>
<p>Here is an example tiling that realizes <span class="math-container">$4^2 = 3^2 + 7 \cdot 1^2$</span>, so <span class="math-container">$k = 8$</span>.</p>
<p><a href="https://i.stack.imgur.com/LdwQy.png" rel="noreferrer"><img src="https://i.stack.imgur.com/LdwQy.png" alt="enter image description here"></a></p>
| Erick Wong | 30,402 | <p>Many of the other answers (based on extending smaller values of <span class="math-container">$k$</span>) yield a fairly large value for the LHS. It’s worth pointing out that we actually know a great deal about the set of integers representable as the sum of <span class="math-container">$k$</span> nonzero squares.</p>
<p>See, for instance, <a href="https://math.stackexchange.com/a/1419292/30402">this other answer</a>, where it is shown that every integer <span class="math-container">$\ge 34$</span> is the sum of five positive squares. Immediately, by padding with <span class="math-container">$1$</span>s, we can extend this to</p>
<blockquote>
<p>For <span class="math-container">$k\ge 5$</span>, every integer <span class="math-container">$> k+28$</span> can be written as the sum of <span class="math-container">$k$</span> nonnegative squares.</p>
</blockquote>
<p>Therefore, we can just choose <span class="math-container">$a^2$</span> to be the first square greater than <span class="math-container">$k+28$</span>, which makes <span class="math-container">$a^2 \le k + 2\sqrt k + O(1)$</span>, which makes <span class="math-container">$$a \le \sqrt k + 1 + O(k^{-1/2}).$$</span> This basically optimal as clearly <span class="math-container">$a \ge \lceil \sqrt k\rceil^2$</span>.</p>
<p>We might be able to shave down the constant <span class="math-container">$28$</span> somewhat by increasing the lower bound on <span class="math-container">$k$</span>, but it can’t be reduced all the way to <span class="math-container">$0$</span> because of the gap between the first two nonzero squares <span class="math-container">$1$</span> and <span class="math-container">$4$</span>. There must be some constant <span class="math-container">$C>0$</span> that is the optimal replacement for <span class="math-container">$28$</span> for all <span class="math-container">$k\ge k_0$</span> (I wouldn’t be surprised if it is already optimal).</p>
|
649,502 | <p>What do we mean when we talk about a topological <em>space</em> or a metric <em>space</em>? I see some people calling metric topologies metric spaces and I wonder if there is some synonymity between a topology and a space? What is it that the word means, and if there are multiple meanings how can one distinguish them?</p>
| Thomas Andrews | 7,933 | <p>A metric space is a set with a metric.</p>
<p>A topological space is a set with a topology.</p>
<p>Both metric spaces and topologies are useful for defining "continuity." </p>
<p>Every metric space can be made a topological in a useful way, so that the notion of continuity in metric spaces agrees with the notion of continuity in topological spaces. The reverse is not true - the notion of topology is more general, so not every topological space "comes from" a metric space.</p>
<p>Also, two different metrics on the same set can result in the same topology.</p>
<p>A topology is then a "metric topology" if it arises from converting some metric on the set to a topology.</p>
<p>One concept you have in metric spaces that you don't have in topologies is the notion of "uniform continuity."</p>
|
2,461,506 | <p>I am trying to derive / prove the fourth order accurate formula for the second derivative:</p>
<p>$f''(x) = \frac{-f(x + 2h) + 16f(x + h) - 30f(x) + 16f(x - h) - f(x -2h)}{12h^2}$.</p>
<p>I know that in order to do this I need to take some linear combination for the Taylor expansions of $f(x + 2h)$, $f(x + h)$, $f(x - h)$, $f(x -2h)$. For example, when deriving the the centered-difference formula for the first derivative, the Taylor expansion of $f(x + h)$ minus $f(x-h)$ can be computed to give the desired result of $f'(x)$, in that case.</p>
<p>In what way would I have to combine these Taylor expansions above to obtain the required result?</p>
| Paul Aljabar | 435,819 | <p>$$
f(x+h) =
f(x) + h f'(x) + \frac{h^2}{2} f''(x)
+ \frac{h^3}{6} f'''(x)
+ O(h^4)
$$</p>
<p>$$
f(x-h) =
f(x) - h f'(x) + \frac{h^2}{2} f''(x)
- \frac{h^3}{6} f'''(x)
+ O(h^4)
$$</p>
<p>$$
f(x+2h) =
f(x) + 2h f'(x) + 2 h^2 f''(x)
+ \frac{4 h^3}{3} f'''(x) + O(h^4)
$$</p>
<p>$$
f(x-2h) =
f(x) - 2h f'(x) + 2 h^2 f''(x)
- \frac{4 h^3}{3} f'''(x)
+ O(h^4)
$$</p>
<p>Calculate:
$$
-f(x + 2h) + 16f(x + h) - 30f(x) + 16f(x - h) - f(x -2h)
$$</p>
<p>Which is
$$
\begin{aligned}
&
- \left[
f(x) + 2h f'(x) + 2 h^2 f''(x)
+ \frac{4 h^3}{3} f'''(x)
\right]
\\
&
+16
\left[
f(x) + h f'(x) + \frac{h^2}{2} f''(x)
+ \frac{h^3}{6} f'''(x)
\right]
\\
&
-30
f(x)
\\
&
+16
\left[
f(x) - h f'(x) + \frac{h^2}{2} f''(x)
- \frac{h^3}{6} f'''(x)
\right]
\\
&
-
\left[
f(x) - 2h f'(x) + 2 h^2 f''(x)
- \frac{4 h^3}{3} f'''(x)
\right]
\\
&
+ O(h^4)
\end{aligned}
$$</p>
<p>Which evaluates to
$
12 h^2
$
to give the required result.</p>
|
2,087,107 | <p>In the following integral</p>
<p>$$\int \frac {1}{\sec x+ \mathrm {cosec} x} dx $$</p>
<p><strong>My try</strong>: Multiplied and divided by $\cos x$ and Substituting $\sin x =t$. But by this got no result.</p>
| DXT | 372,201 | <p>$\displaystyle I = \int\frac{1}{\sec x+\csc x}dx = \frac{1}{2}\int\frac{\sin 2x}{\sin x+\cos x}dx = \frac{1}{2}\int\frac{(\sin x+\cos x)^2-1}{\sin x+\cos x}dx$</p>
<p>$\displaystyle I = \frac{1}{2}\int (\sin x+\cos x)dx - \frac{1}{2}\int \frac{1}{\sin x+\cos x}dx$</p>
<p>$\displaystyle I =\frac{1}{2}(\sin x-\cos x)-\frac{1}{2\sqrt{2}}\int \frac{1}{\sin (x+45^0)}dx$</p>
<p>$\displaystyle I = \frac{1}{2}(\sin x-\cos x)-\frac{1}{2\sqrt{2}}\int \csc (x+45^0)dx$</p>
<p>Use $\displaystyle \int \csc xdx = \ln |\csc x-\cot x|+\mathcal{C} = \ln\tan \frac{x}{2}+\mathcal{C}$</p>
|
2,087,107 | <p>In the following integral</p>
<p>$$\int \frac {1}{\sec x+ \mathrm {cosec} x} dx $$</p>
<p><strong>My try</strong>: Multiplied and divided by $\cos x$ and Substituting $\sin x =t$. But by this got no result.</p>
| Animesh Ashish | 401,839 | <p>Since I don't know LaTex, I am going to add a <a href="http://m.meritnation.com/ask-answer/question/what-is-the-integration-of-1-secx-cosecx/integrals/489125" rel="nofollow noreferrer">link</a> so that you can understand with your best of abilities. Let me know if you find a problem/bug.</p>
<p>Since, I could only come up with a link. Let me know if there is any problem in it if you come across. I'll always have your back ! </p>
|
350,747 | <p>Base case: $n=1$. Picking $2n+1$ random numbers 5,6,7 we get $5+6+7=18$. So, $2(1)+1=3$ which indeed does divide 18. The base case holds.
Let $n=k>=1$ and let $2k+1$ be true. We want to show $2(k+1)+1$ is true. So, $2(k+1)+1=(2k+2) +1$....</p>
<p>Now I'm stuck. Any ideas?</p>
| robjohn | 13,854 | <p>The sum of a sequence of consecutive integers is the average of the sequence, which is also the average of the largest and smallest terms, times the number of terms in the sequence. If the smallest term of a sequence of length $2n+1$ is $k$, the largest is $k+2n$. The average is $k+n$. Thus, the sum is $(2n+1)(k+n)$, which is easily divisible by $2n+1$.</p>
|
733,675 | <p>A new question has emerged after this one was successfully answered by r9m: <a href="https://math.stackexchange.com/questions/731292/inequality-with-abcd-2/731930#731930">If $a+b+c+d = 2$, then $\frac{a^2}{(a^2+1)^2}+\frac{b^2}{(b^2+1)^2}+\frac{c^2}{(c^2+1)^2}+\frac{d^2}{(d^2+1)^2}\le \frac{16}{25}$</a>. I thought of this generalization. Does it hold?</p>
<p>$$\dfrac{x_1^2}{(x_1^2+1)^2}+\dfrac{x_2^2}{(x_2^2+1)^2}+\cdot\cdot\cdot+\dfrac{x_n^2}{(x_n^2+1)^2}\le \dfrac{n^2}{(n+1)^2}$$ with $$x_1+x_2+\cdot\cdot\cdot+x_n= \sqrt{n}$$ $$x_1,x_2,\cdot\cdot\cdot,x_n \ge0$$ $$ n \in \mathbb{N}$$</p>
| Jimmy R. | 128,037 | <p><em>Alternative way to reach the result, that is already proposed in the other answers:</em></p>
<p>Consider the non-linear maximization problem $$\max_{x_i}\sum_{i=1}^{n}\frac{x_i^2}{(x_i^2+1)^2}$$ subject to $x_1+x_2+\ldots+x_n=\sqrt{n}$ and $x_i\ge0$. In that case the <a href="http://en.wikipedia.org/wiki/Lagrange_multiplier" rel="nofollow">Lagrange function</a> is defined as $$Λ(x, λ)=\sum_{i=1}^{n}\frac{x_i^2}{(x_i^2+1)^2}+λ\left(\sum_{i=1}^{n}x_i-\sqrt{n}\right)$$ The equation $\nablaΛ=0$ yields $$λ=\frac{2x_i(1-x_i^2)}{(1+x_i^2)^3}$$ for all $1\le x_i \le n$. Obviously the point $x^0$ with $$x^0_i=\begin{cases}1, & 1\le i \le \lfloor \sqrt{n}\rfloor \\ \\0, & \lfloor \sqrt{n}\rfloor \le i \le n \end{cases}, \qquad λ=0$$ solves the above system (both sides are equal to zero) and is thus a stationary point. The value of the objective function in $x^0$ is equal to $$\lfloor \sqrt{n}\rfloor\frac14$$
It is easy to see that for $n \ge 16$ this values is greater than $1$ and thus larger than the RHS of the given inequality. For $n<16$ this point does not yield a value larger than the RHS.</p>
<hr>
<p>Perhaps there are other stationary points that yield a higher value for the objective function, even for $n<16$.</p>
|
2,928,196 | <p>I thought that I could take all points with rational coordinates, but this space is not discrete</p>
| Henno Brandsma | 4,280 | <p>In <a href="http://at.yorku.ca/p/a/c/a/26.pdf" rel="nofollow noreferrer">this note</a> a classic fact is shown that in a metric space the boundary of an open set is the closure of a discrete set. So e.g. the circle in the plane, being the boundary of the open unit disk, is such a set.</p>
|
2,476,865 | <p>As the title suggests, I'm trying to establish a good bound on</p>
<p>\begin{equation}
S(n) = \sum_{k = 2}^n (en)^k k^{-Cn/\log{n} - k - 1/2},
\end{equation}</p>
<p>where $C$ is some reasonably large positive constant. In particular I would like to have $S(n) = o(1)$, i.e., </p>
<p>\begin{equation}
\lim_{n \to \infty} S(n) = 0,
\end{equation}</p>
<p>which numerical evaluations suggest to be the case.</p>
<p>Moreover it appears to be the case that the terms in the series are monotonically decreasing; if this were true my claim follows trivially (replace all terms by the one at $k = 2$ and check) but verifying the ''monotone decrease conjecture'' is again a difficult task in itself.</p>
<p>I appreciate any ideas on how to tackle this problem.</p>
<p>EDIT: The sequence terms are unfortunately not monotonically decreasing as can be verified by studying the logarithm of the latter but the conjecture about the term at $k = 2$ being largest still stands.</p>
| Nate River | 279,404 | <p>In spite of my previous comment, I do think your intuition is correct.</p>
<p>Attempted proof (sorry but it is very ugly): Again, let $r(d)=\frac{s_2}{s_{2+d}}$. As you noted,
$$
r(d)=\left(\frac{2+d}{en}\right)^d\left(1+\frac{d}{2}\right)^{Cn/\log n + 2 + 1/2}
$$</p>
<p>Taking logarithm, we get
$$
\log r(d)=d\left(\log\left(\frac{2+d}{en}\right)\right)+ (Cn/\log n+2+1/2)\left(\log\left(\frac{2+d}{2}\right)\right)
$$
where $d\in [0,n-2]$. Now, let $\lambda=\frac{2+d}{n}\in[2/n,1]$, or, equivalently, $d=\lambda n - 2$. Thus,
$$
\log r(d)=(\lambda n - 2)\left(\log\left(\frac{\lambda}{e}\right)\right) + (Cn/\log n+2+1/2)\left(\log\left(\frac{\lambda n}{2}\right)\right):=f(\lambda)
$$</p>
<p>Cleaning this mess a bit, we get
$$
f(\lambda)=(\lambda n -2)(\log\lambda - 1) + (Cn/\log n+2+1/2)(\log\lambda + \log n -\log2)\\
= \lambda n(\log\lambda - 1) -2\log\lambda+2+(Cn/\log n + 2 + 1/2)\log\lambda\\
+ Cn + 2.5\log n-2.5\log 2\\
=n(\lambda \log\lambda - \lambda) + (Cn/\log n + 0.5)\log\lambda + D(n)
$$
where $D(n)$ depends only on $n$. </p>
<p>Taking derivative, we get $f'(\lambda)=n\log\lambda + \frac{1}{\lambda}(Cn/\log n + 0.5)$, and the second derivative is $f''(\lambda)=\frac{n}{\lambda} - \frac{1}{\lambda^2}(Cn/\log n + 0.5)=\frac{n}{\lambda^2}(\lambda-C/\log n + 0.5/n)$</p>
<p>Hence. If $n$ is large enough, $f''(\lambda)>0$ for all $\lambda \ge 2/n$. Ie, $f'$ is increasing. Now, for such $n$,
$$
f'\left(\frac{2}{n}\right)=n\log 2 - n\log n + \frac{n}{2}(Cn/\log n + 0.5)\\
=n\log 2 - n\log n + \frac{Cn^2}{2\log n} + n$$
which is bigger than $0$ for large enough $n$. And since $f'$ increasing, $f'>0$ in $[2/n,1]$. Ie, $f$ is increasing. Ie, for large enough $n$, $f(\lambda)\ge 0$ for $\lambda \ge 2/n$. Hence $r(d)\ge 1$ for $d\in [0,n-2]$. Hence $S(n)=o(1)$.</p>
|
1,955,591 | <p>I have to prove that ' (p ⊃ q) ∨ ( q ⊃ p) ' is a tautology.I have to start by giving assumptions like a1 ⇒ p ⊃ q and then proceed by eliminating my assumptions and at the end i should have something like ⇒(p ⊃ q) ∨ ( q ⊃ p) but could not figure out how to start.</p>
| Hagen von Eitzen | 39,174 | <p>As a quick overview: You should be able to show $p\vdash q\supset p$ as well as $\neg p\vdash p\supset q$, and then by case distinction $p\lor\neg p\vdash (p\supset q)\lor (q\supset p)$.</p>
|
96,468 | <blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="https://math.stackexchange.com/questions/22537/how-many-fixed-points-in-a-permutation">How many fixed points in a permutation</a> </p>
</blockquote>
<p>Suppose we have a collection of n objects, numbered from 1 to n. These objects are placed in a random order.</p>
<p>What is the probability that p of the objects are in the position in the order corresponding to their order. </p>
<p>For instance, </p>
<p>For n=3, 1-2-3 has all objects in the correct position, p = 3, and has probability P(p=3) = 1/3! = 1/6.</p>
<p>However P(p=2) = 0</p>
<p>P(p=1) = 3/3! = 1/2. (1-3-2, 3-2-1, 2-1-3)</p>
<p>P(p=0) = 2/3!. (2-3-1, 3-1-2)</p>
| Elvis | 21,435 | <p>Let’s denote $P$ the random variable defined by $P = {}$ number of objects in correct position. We compute $\mathbb P(P = p)$ by counting the number of permutations "with $p$ fixed points" (just an other way to say "$p$ objects in correct position).</p>
<ul>
<li><p>You have a total of $n!$ permutations ;</p></li>
<li><p>You have ${n \choose p}$ choices for the indices of the objects in correct position ;</p></li>
<li><p>The remaining $n-p$ objects have to be permutated without fixed points. This kind of permutation is called a <a href="http://en.wikipedia.org/wiki/Derangement" rel="nofollow">derangement</a>. The number of derangement is denoted by $!(n-p)$ and is computable via an <a href="http://en.wikipedia.org/wiki/Derangement" rel="nofollow">induction formula</a> or with $!(n-p) = (n-p)!\sum_{i=0}^{n-p} {(-1)^i \over i!}$ (same reference).</p></li>
<li><p>The probability $\mathbb P(P=p)$ is then</p></li>
</ul>
<p>$$ \mathbb P(P=p) = { {n \choose p} \cdot !(n-p) \over n! } = {!(n-p) \over (n-p)! p! }.$$</p>
<p>When $n-p \gg 1$ we have ${!(n-p) \over (n-p)!} \simeq e^{-1}$, so $\mathbb P(P=p) \simeq {1 \over e \cdot p!}$.</p>
|
121,546 | <p>Consider the 2-Wasserstein distance between probability measures $\mu$ and $\nu$ (on $\mathbb{R}^d$), defined as
$$
d_{W_2}(\mu,\nu) = \inf_{\gamma} \Big[\int \|x-y\|^2 d\gamma(x,y)\Big]^{1/2}
$$
where the $\inf$ is over all couplings $\gamma$ of $\mu$ and $\nu$. Can we define a norm (or something norm-like) on the space of signed measures (or a linear subspace of it containing the cone of probability measures) which gives rise to $W_2$ for probability measures. (I suppose not, but why?)</p>
<p>If not, can we approximate $d_{W_2}$ by a norm?</p>
| Vladimir Zolotov | 32,454 | <p>(I guess you missed a square in your definition.)</p>
<p>2-Wasserstein distance doesn't respect the convex structure on measures. Consider two points $x_1 \ne x_2$ and Dirac measures $\delta(x_1), \delta(x_2)$. The measure $\frac{\delta(x_1)+\delta(x_2)}{2}$ is not a midpoint between $\delta(x_1)$ and $\delta(x_2)$. </p>
|
816,088 | <blockquote>
<p>The sum of two variable positive numbers is $200$.
Let $x$ be one of the numbers and let the product of these two numbers be $y$. Find the maximum value of $y$.</p>
</blockquote>
<p><em>NB</em>: I'm currently on the stationary points of the calculus section of a text book. I can work this out in my head as $100 \times 100$ would give the maximum value of $y$. But I need help making this into an equation and differentiating it. Thanks!</p>
| Andrés E. Caicedo | 462 | <p>You call one of your numbers $x$. Write $x$ in the form $100-k$, so the other number is $100+k$ (it does not matter here whether $k$ is zero, or positive, or negative). We have $$z=(100-k)(100+k)=100^2-k^2\le 100^2,$$ with equality happening if and only if $k=0$.</p>
|
606,431 | <p>Can someone explain to me how to solve this using inverse trig and trig sub?</p>
<p>$$\int\frac{x^3}{\sqrt{1+x^2}}\, dx$$</p>
<p>Thank you. </p>
| lab bhattacharjee | 33,337 | <p>$$ \frac{x^3}{\sqrt{1+x^2}}=\frac{x^3+x-x}{\sqrt{1+x^2}}=x\sqrt{1+x^2}-\frac x{\sqrt{1+x^2}}$$</p>
<p>Set $1+x^2=u$ in each case</p>
|
606,431 | <p>Can someone explain to me how to solve this using inverse trig and trig sub?</p>
<p>$$\int\frac{x^3}{\sqrt{1+x^2}}\, dx$$</p>
<p>Thank you. </p>
| Alraxite | 61,039 | <p>Set $u=1+x^2$. Then $\dfrac{du}{dx}=2x$ and so,</p>
<p>$\displaystyle\int \dfrac{x^3}{\sqrt{1+x^2}}dx=\int \dfrac12\dfrac{u-1}{\sqrt{u}}du=\dfrac12\int (u^{1/2}-u^{-1/2})\, du$.</p>
|
17,713 | <p>I am a bit perplexed in trying to find values <span class="math-container">$a,b,c$</span> so that the approximation is as precise as possible:</p>
<p><span class="math-container">$$\sum_{k=n}^{\infty}\frac{(\ln(k))^{2}}{k^{3}} \approx \frac{1}{n^{2}}[a(\ln (n))^{2}+b \ln(n) + c]$$</span></p>
<p>I can see from Wolfram that <span class="math-container">$\ln(x)$</span> could be written in many different <a href="http://www.wolframalpha.com/input/?i=ln(x)+series" rel="nofollow noreferrer">ways</a> and that the sum starting from <span class="math-container">$1$</span> can be written with Riemann zeta function <a href="http://www.wolframalpha.com/input/?i=sum((ln(x)%5E2%2Fx%5E3)" rel="nofollow noreferrer">here</a>. They are probably not something I am looking here. The decreasing exponent hints me that perhaps the term in the sum can be approximated with its derivations, the first derivate <a href="http://www.wolframalpha.com/input/?i=Dervivate((ln(x)%5E2%2Fx%5E3)" rel="nofollow noreferrer">here</a> and the second <a href="http://www.wolframalpha.com/input/?i=Derivate(Dervivate((ln(x)%5E2%2Fx%5E3))" rel="nofollow noreferrer">here</a> but when I try to approximate something goes wrong, perhaps wrong premise. Let <span class="math-container">$z=\frac{\ln(k)^{2}}{k^{3}}$</span> then</p>
<p><span class="math-container">$$\sum_{k=n}^{\infty} z \approx z' z-z' z^3/2!$$</span></p>
<p>by the Taylor approximation for the odd function (<span class="math-container">$\ln(x)$</span> is odd, <span class="math-container">$x^3$</span> is odd and the oddity is preserved after the operations, maybe wrong, so <span class="math-container">$z$</span> is odd like <span class="math-container">$\sin(x)$</span>), I get:</p>
<p><span class="math-container">$$\sum_{k=n}^{\infty} z \approx (\frac{1}{x^3}(2\ln(x) - 3(\ln(x))^2)) -\frac{1}{x^2}(6 \ln(x)^2-7 \ln(x)+1)$$</span></p>
<p>something wrong because the first term has <span class="math-container">$x^3$</span> instead of <span class="math-container">$x^2$</span>. I am sorry if this hard to read but stuck here. So is it the correct way to approximate the sum? Is it really with Riemann zeta function or does Taylor work here, as I am trying to do above?</p>
<p><em>I labeled the question with <code>parity</code> because I think it is central here. I feel I may be misunderstanding how the oddity works over different operations, hopefully shown in my vague explanations, and hence the error.</em></p>
| Aryabhata | 1,102 | <p>You can try using elementary estimates (upper and lower bounds) with integrals (see the end of the answer), or can use the more general and quite useful <a href="http://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula" rel="nofollow">Euler Mclaurin Summation Formula</a>, which gives us</p>
<p>$$\sum_{k=1}^{n} \frac{\log^2 k}{k^3} = \int_{1}^{n} \frac{\log^2 x}{x^3} \ \text{d}x + \frac{\log ^2 n}{2n^3} + C + \mathcal{O}(\frac{1}{n^{3+e}})$$</p>
<p>where $\displaystyle e > 0$.</p>
<p>Now $$\int_{1}^{n} \frac{\log^2 x}{x^3} \ \text{d}x = - \frac{2\log^2 n + 2 \log n + 1}{4n^2} + 1/4$$</p>
<p>Thus</p>
<p>$$\sum_{k=1}^{n} \frac{\log^2 k}{k^3} = K - \frac{2\log^2 n + 2 \log n + 1}{4n^2} + \frac{\log^2 n}{2n^3} + \mathcal{O}\left(\frac{1}{n^{3+e}}\right)$$</p>
<p>Since the LHS converges we have that</p>
<p>$$\sum_{k=1}^{n} \frac{\log^2 k}{k^3} = \sum_{k=1}^{\infty}\frac{\log^2 k}{k^3} - \frac{2\log^2 n + 2 \log n + 1}{4n^2} + \frac{\log^2 n}{2n^3} + \mathcal{O}\left(\frac{1}{n^{3+e}}\right)$$</p>
<p>Thus</p>
<p>$$\sum_{k=n+1}^{\infty}\frac{\log^2 k}{k^3} = \frac{2\log^2 n + 2 \log n + 1}{4n^2} - \frac{\log^2 n}{2n^3} + \mathcal{O}\left(\frac{1}{n^{3+e}}\right)$$</p>
<p>And so</p>
<p>$$\sum_{k=n}^{\infty}\frac{\log^2 k}{k^3} = \frac{2\log^2 n + 2 \log n + 1}{4n^2} + \frac{\log^2 n}{2n^3} + \mathcal{O}\left(\frac{1}{n^{3+e}}\right)$$</p>
<p>Note that the summation formula allows us to figure out the exact terms which make up the $\displaystyle \mathcal{O}\left(\frac{1}{n^{3+e}}\right)$ and so we can make the formula more accurate by including as many lower order terms we need.</p>
<hr>
<p>A more elementary way:</p>
<p>If $\displaystyle f(x) \ge 0$ and is monotonically decreasing then we have that</p>
<p>$$\int_{n}^{M+1} f(x)\ \text{d}x \le \sum_{k=n}^{M} f(k) \le f(n) + \int_{n}^{M} f(x)\ \text{d}x$$</p>
<p>which can be seen by bounding the area under the curve by rectangles of width $1$.</p>
<p>For $\displaystyle f(n) = \frac{\log ^2 n}{n^3}$, we can take limit as $\displaystyle M \to \infty$ to get the asymptotic formula you need.</p>
|
1,871,103 | <blockquote>
<p>Given that $a_{1}=0$, $a_{2}=1$ and
$$a_{n+2}=\frac{(n+2)a_{n+1}-a_{n}}{n+1}$$
prove that $\lim\limits_{n\to\infty} a_n=e$</p>
</blockquote>
<p>What I did:</p>
<p>It was hinted to prove that $a_{n+1}-a_{n}=\frac{1}{n!}$ which I did inductively.
But then using this information now I get:</p>
<p>$a_{n+1}=a_{1}+\frac{n}{n!}=\frac{1}{(n-1)!}$ so $a_{n}=\frac{n-1}{n!}$.</p>
<p>Now $a_{n}$ is bounded above by $\sum_{k=0}^{n}\frac{1}{k!}$ which converges to $e$. I can't find a lower bound that converges to $e$ as well. I'm starting to think I was supposed to go about this differently.</p>
<p>Thanks in advance.</p>
| user84413 | 84,413 | <p>Now you can use induction to show that $\displaystyle a_n=1+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{(n-1)!}$, </p>
<p>so then $\displaystyle\lim_{n\to\infty}a_n=\sum_{k=1}^{\infty}\frac{1}{k!}=\color{red}{e-1}$</p>
|
3,075,979 | <p>Prove that <span class="math-container">$$\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}$$</span> is an integer using mathematical induction.</p>
<p>I tried using mathematical induction but using binomial formula also it becomes little bit complicated.</p>
<p>Please show me your proof.</p>
<p>Sorry if this question was already asked. Actually i did not found it. In that case only sharing the link will be enough.</p>
| Michael Rozenberg | 190,319 | <p>Because
<span class="math-container">$$\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}=\frac{k^7-k}{7}+\frac{k^5-k}{5}+\frac{2(k^3-k)}{3}+k.$$</span></p>
|
170,830 | <p>I have two circles with the same radius and I want to calculate the points of tangency. </p>
<p>For example, in the picture below, I want to calculate $(x_3, y_3)$ and $(x_4,y_4)$. I have the radius and the distance between the two circles as shown below:</p>
<p><img src="https://i.stack.imgur.com/tQ2qu.png" alt=""></p>
| Barun Dasgupta | 34,858 | <p>The gradient of the tangent to any circle is given by $$-\frac{(x'-a)}{(y'-b)}$$ Where (x',y') is the point of tangency and (a,b) is the center of the circle. Now the gradient of the line joining the centers of the two circles is same as the gradient of the tangent. Hence in this case this essentially translates to the following equation $$ -\frac{(x_3-x_1)}{(y_3-y_1)}=\frac{(y_2-y_1)}{(x_2-x_1)}$$ The other equation is $$ (x_3-x_1)^2+(y_3-y_1)^2= R^2$$ Solving the above two equations you will get two points for $(x_3,y_3)$. This shows the existence of two parallel tangents. Similarly you can solve for $(x_4,y_4)$ </p>
|
1,955,212 | <p>I couldn't find this on the whole internet. My life depends on solving this. Please help.
I must write a formula for this sequence
<span class="math-container">$-8, -14, 8, 14, -8, -14, 8, 14$</span>.</p>
| IntegrateThis | 270,702 | <p>Consider for an indice $n$ that if n is divisible by 2 the absolute value of your sequence is 14, else the absolute value is 8.</p>
<p>Next if an indice $n$ can be represented by $4k-1$ or $4k$ for $k \in \Bbb N$ then the sequence value is positive, else the sequence value is negative.</p>
|
1,368,899 | <p>I am fairly new to statistics and just recently encountered queueing theory.</p>
<p>I have programmed a simulation for a $M/M/1$ queue in which I specify the inter-arrival times and service times. I input say, an exponential distribution with a mean of $1$ for both the inter-arrival and service time.</p>
<p>I also measure for the effective arrival rate, meaning I run for, say, $1000$ time steps and at each time step a random value is drawn from the exponential distribution. I collect these random values in a list and then compute the effective arrival rate, this being the mean of the list. In theory, this value should converge to the mean, yet, in practice I end up with values not so close to the mean.</p>
<p>My question is, how many random values from an exponential distribution should I draw such that the mean of these converges to the mean of the distribution?</p>
| Vinod Kumar Punia | 255,098 | <p>$3\alpha=\frac{-a}{a-b},
2\alpha^2=\frac{1}{a-b}$
divide
$\frac{3\alpha}{2\alpha^2}=-a$,
$a\alpha=\frac{-3}{2}...............(1)$,
As $\alpha$ is a root of $(a-b)x^2+ax+1=0$,
$(a-b)\alpha^2+a\alpha+1=0$
$(a-b)\frac{9}{4a^2}-\frac{3}{2}+1=0$
$2a^2-9a+9b=0$
this is quadratic in $a$ and as $a$ being real
$81-4(2)(9b)\ge0$
$b\le\frac{9}{8}$</p>
|
2,792,770 | <p>I found the following question in a test paper:</p>
<blockquote>
<p>Suppose $G$ is a monoid or a semigroup. $a\in G$ and $a^2=a$. What can we say
about $a$?</p>
</blockquote>
<p>Monoids are associative and have an identity element. Semigroups are just associative. </p>
<p>I'm not sure what we can say about $a$ in this case other than that $a$ could be other things apart from the identity. Any idea if there's a definitive answer to this question?</p>
| BCLC | 140,308 | <p>Let $(\Omega',\mathcal{F}',\mathbb{P}')$ be a probability space where $\Omega'$ is countable. It can be shown that any $X$ in the probability space must be discrete. Thus, for any non-discrete random variable $Y$ on the probability space $(\Omega,\mathcal{F},\mathbb{P})$, $Y$ is not $\mathcal{F}$-measurable but not $\mathcal{F}'$-measurable.</p>
|
2,919,266 | <p>Let $(x_n)$ be a sequence in $(-\infty, \infty]$. </p>
<p>Could we define the sequence $(x_n)$ so that limsup$(x_n) = -\infty$? </p>
<p>My intuitive thought is no, but I’m not 100% sure. </p>
| parsiad | 64,601 | <p>You are interested in the sum
$$
S_{N}\equiv\sum_{n=1}^{N}\frac{n}{\left(n+1\right)\left(n+2\right)}2^{n}.
$$
As Donald suggests, you can use the partial fraction expansion
$$
\frac{n}{\left(n+1\right)\left(n+2\right)}=\frac{2}{n+2}-\frac{1}{n+1}
$$
to get
\begin{multline*}
S_{N}=\sum_{n=1}^{N}\frac{2}{n+2}2^{n}-\sum_{n=1}^{N}\frac{1}{n+1}2^{n}=\sum_{n=1}^{N}\frac{1}{\left(n+1\right)+1}2^{n+1}-\sum_{n=1}^{N}\frac{1}{n+1}2^{n}
=\sum_{n=2}^{N+1}\frac{1}{n+1}2^{n}-\sum_{n=1}^{N}\frac{1}{n+1}2^{n}\\=\sum_{n=2}^{N+1}\frac{1}{n+1}2^{n}-\sum_{n=1}^{N}\frac{1}{n+1}2^{n}=\frac{1}{N+2}2^{N+1}-\frac{1}{2}2^{1}
=\frac{1}{N+2}2^{N+1}-1.
\end{multline*}
From the above, it is clear that $S_{N}\rightarrow\infty$ as $N\rightarrow\infty$.</p>
|
1,831,134 | <p><a href="https://i.stack.imgur.com/cQEeH.jpg" rel="nofollow noreferrer">Worked examples</a></p>
<p>Can somebody please explain to me how the generator matrix is obtained when we are given the codewords of the binary code in the examples attached.</p>
<p>I tried arranging the codes in a matrix with each row being a codeword , I then reduced to row echelon form and hence found basis of the code. I then Tried to construct the generator matrix using the basis but it does not work out. Please help! </p>
| Dietrich Burde | 83,966 | <p>$C_1$ is a $2$-dimensional vector space over the finite field $\mathbb{F}_2$ with basis $e_1=(0,1),e_2=(1,0)$. So we have $C_1=\{\lambda e_1+\mu e_2\mid \lambda,\mu\in \mathbb{F}_2\}=\{(0,0),(1,0),(0,1),(1,1)\}$. Of course the generator matrix $G$ is formed by $e_1$ and $e_2$, which is the canonical basis for the linear code $C_1$.</p>
|
2,285,202 | <p>So I want to find a field extension that has the galois group $Z_{3} \times Z_{3} \times Z_{3} $. Now if the 3's where changed to 2's then I guess for example $(x^2-2)(x^2-3)(x^2-5)$ would suffice but I don't see how any clever way to do it with $Z_{3}$. I tried a bit with cyclotomic extensions but came up empty handed. Any hints & help are appreciated, thanks!</p>
| Chickenmancer | 385,781 | <blockquote>
<p>First: <strong>Your example won't work.</strong> For example take $\theta=\sqrt[3]{2}.$ </p>
</blockquote>
<p>Then $$x^3-2=(x-\theta)(x^2+\theta x+\theta^2).$$</p>
<p>Since You can see that $(x^2+\theta x+\theta^2)$ has roots $$x=\theta\left(\dfrac{1}{2}\pm\dfrac{\sqrt{3}}{2}i\right).$$</p>
<p>Taking an extension over $\Bbb{Q}$ you find that the extension $\Bbb{Q}(\sqrt[3]{2})$ has a group of $\Bbb{Q}$ fixing automorphisms $G=\{1\}$ since the only place you can send $\sqrt[3]{2}$ is itself. So you need a bigger extension: $\Bbb{Q}\subset \Bbb{Q}(\sqrt[3]{2},\sqrt{3}i).$ Now, you have three places to send $\sqrt[3]{2},$ that is, any of the roots of the above polynomial, and that gives you a copy of $\Bbb{Z}/3\Bbb{Z}.$ However, you need to check that the automorphisms fixing $\Bbb{Q}$ don't commute (if you don't believe then use the complex conjugation automorphism), so you get a Galois group isomorphic to $S_3.$ </p>
<blockquote>
<p><strong>Now go read Lord Shark the Unknown's answer.</strong></p>
</blockquote>
|
2,584,688 | <p>Consider normed spaces $X$ and $Y$. You can assume that they are Banach spaces if needed. Let $\mathcal{L}(X, Y)$ denote the spaces of bounded linear operators from $X$ to $Y.$ Now consider the set </p>
<p>$$\Omega=\{T \in L(X,Y): T \textrm{ is onto}\}.$$ Is $\Omega$ open with the norm topology? </p>
| David C. Ullrich | 248,223 | <p>We've seen in another answer that this is false if $X$ is not a Banach space.</p>
<blockquote>
<blockquote>
<p><strong>Lemma</strong> Suppose $X$ is a Banach space, $T:X\to Y$ is linear and bounded, $0<\epsilon<1$, $\delta>0$, and for every $y\in Y$ there exists $x\in X$ with $||x||\le\delta ||y||$ and $||Tx-y||\le\epsilon||y||$. Then $T$ is surjective.</p>
</blockquote>
</blockquote>
<p>Proof. Let $y\in Y$. Choose $x_0\in X$ with $||x_0||\le\delta||y||$ and $||Tx_o-y||\le\epsilon||y||$. Having chosen $x_0,\dots x_n$, let $s_n=x_0+\dots+x_n$ and choose $x_{n+1}\in X$ with $||x_{n+1}||\le\delta||y-Ts_n||$ and $||Tx_{n+1}-(y-Ts_n)||\le\epsilon||y-Ts_n||$.</p>
<p>Then $||y-Ts_n||\to0$, in fact $\sum||y-Ts_n||<\infty$, since $||y-Ts_{n+1}||\le\epsilon||y-Ts_n||$. And so $||x_{n+1}||\le\delta||y-Ts_{n}||$ implies that $\sum||x_n||<\infty$, so $s_n\to s$. It follows that $Ts=y$, since $T$ is continuous. QED.</p>
<p>Now suppose $T:X\to Y$ is surjective. The Open Mapping Theorem says that there exists $\delta>0$ such that for every $y\in Y$ there exists $x\in X$ with $Tx=y$ and $||x||\le\delta||y||$. Hence for any $T'$ we have $$||T'x-y||\le||T'-T||\,||x||\le\delta||T'-T||\,||y||.$$So if $\delta||T'-T||<1$ the lemma shows that $T'$ is surjective.</p>
|
1,849,608 | <p>Given the irreducible fraction $\frac a b$, with $a, b \in \mathbb N$, what is the expression that enumerates all the irreducible fractions of integers that add up to $\frac a b$? Namely, an expression (in terms of $a$ and $b$) for all the $\frac c d$ and $\frac e f$, with $c,d,e,f \in \mathbb N$, such that $\frac c d + \frac e f = \frac a b$.</p>
| G Cab | 317,234 | <p>Every couple of fractions $c/a$ and $d/b$ in the Stern-Brocot tree can be represented (in the inverse notation according to <em>Concrete Mathematics</em>) by the matrix
${\bf M} = \left\| {\,\begin{array}{*{20}c}
a & b \\
c & d \\
\end{array}\,} \right\|$, where, iff the fractions are the generators of another fraction in the tree, then the determinant of the matrix $=1$, so that:
$$
a\,d - c\,b = 1\quad \Rightarrow \quad \left\{ \begin{array}{l}
\frac{1}{{a\,b}} + \frac{c}{a} = \frac{d}{b}\,\quad \Rightarrow \quad \frac{c}{a} < \frac{d}{b} \\
\frac{1}{{a\,c}} + \,\frac{b}{a} = \frac{d}{c}\quad \Rightarrow \quad \frac{b}{a} < \frac{d}{c} \\
\end{array} \right.
$$
So, given $d/b$, it will be equal to a lower-value (left-side) "co-parent"fraction $c/a$ (determinant $= 1$) plus $1/(ab)$.<br>
Iterating along the tree you get infinite couples, unless you fix a limit on the $1/(ab)$ term.
If that is the case, that is that you fix a maximum denominator, than you may work better on the Farey sequences, arranged on a tree.</p>
<hr>
<p>Another, geometrical, approach.<br>
Suppose you may fix a upper limit (say $d$) to the denominator of the fractions that shall sum to $a/b$,
since otherwise their number is infinite, as already seen.
Then, the solutions in the 1st quadrant (octant) to the diophantine linear equation
$$
\frac{x}{d} + \frac{y}{d} = \frac{a}{b}\quad \Leftrightarrow \quad x + y = \frac{a}{b}d\quad \left| {\;{\rm integers}\;a,b,d,x,y} \right.
$$
will provide all the reduced and un-reduced fractions with denominator $d$.<br>
Their number will clearly be null if $d$ is not a multiple of $b$,
otherwise it will be equal to $n=ad/b+1$ including the null one ($\left\lceil {n\backslash 2} \right\rceil $).</p>
<p>When we consider the fractions reduced, they will span all the fractions with denominators being a divisor of $d$
$$
\frac{{x_{\,k} }}{{m_k }}\quad \left| \begin{array}{l}
\;m_k \backslash d \\
\;b\backslash d \\
\end{array} \right.
$$</p>
<p>That is where the connection with Farey sequences comes in.</p>
<p>Now you can plan various choices for $d$, for instance $d=b!$, or $d=b*q!$ …<br>
'--------- </p>
<p><strong>Examples</strong> </p>
<p><em>1st approach)</em><br>
Take the fraction $5/8$: the picture shows a portion of the SB tree around it.
<a href="https://i.stack.imgur.com/gzFHr.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gzFHr.jpg" alt="enter image description here"></a><br>
Thus we can write:
$$
{5 \over 8} = \left\{ \matrix{
{3 \over 5} + {1 \over {40}} = \left( {{4 \over 7} + {1 \over {35}}} \right) + {1 \over {40}} = {4 \over 7} + {3 \over {56}} = \;\left( {{1 \over 2} + {1 \over {14}}} \right) + {3 \over {56}} = \cdots \hfill \cr
{8 \over {13}} + {1 \over {104}} = \; \cdots \hfill \cr} \right.
$$
Note that the fractions that with $5/8$ will generate a new fraction in the tree (are "co-parents" with $5/8$) are those that are adjacent to it (at its level or lower). In our case we shall choose among those adjacent to the left ($3/5$, $8/13$, and lower levels). </p>
<p><em>2nd approach)</em><br>
from
$$
{x \over d} + {y \over d} = {5 \over 8}\quad \Leftrightarrow \quad x + y = {5 \over 8}d
$$
choosing e.g.<br>
$$
d = 16\;\quad \Rightarrow \quad x + y = 10
$$
we get
$$
{5 \over 8} = {1 \over {16}} + {9 \over {16}} = {1 \over 8} + {1 \over 2} = {3 \over {16}} + {7 \over {16}} = {1 \over 4} + {3 \over 8} = {5 \over {16}} + {5 \over {16}}
$$
and these are all the fractions with denominator <= 16.</p>
|
3,796,216 | <p>Use a group-theoretic proof to show that <span class="math-container">$\mathbb{Q}^*$</span> under multiplication is
not isomorphic to <span class="math-container">$\mathbb{R}^*$</span> under multiplication.</p>
<p><strong>I have tried this:</strong></p>
<p>Suppose <span class="math-container">$$ \phi: \mathbb{Q}^*\to \mathbb{R}^* $$</span>
where <span class="math-container">$\phi(x)=x^2$</span></p>
<p>Now for some <span class="math-container">$3 \in \mathbb{R}^*$</span>
there is no mapping in <span class="math-container">$\mathbb{Q}^*$</span> since <span class="math-container">$\sqrt3$</span> does not belong to <span class="math-container">$\mathbb{Q}^*$</span>.</p>
<p>Hence <strong><span class="math-container">$\phi$</span></strong> is <strong>not an onto function.</strong>
Therefore <span class="math-container">$\mathbb{Q}^* \not\cong \mathbb{R}^*$</span>.</p>
<p>But I am not sure if it is correct.</p>
<p>Also, What is a <strong>group-theoretic proof</strong>?</p>
| lhf | 589 | <p>Your idea is in the right direction but <span class="math-container">$x\mapsto x^2$</span> is not onto. However, <span class="math-container">$x\mapsto x^3$</span> is onto and works. More precisely:</p>
<p>The map <span class="math-container">$\mathbb R^* \to \mathbb R^*$</span> given by <span class="math-container">$x\mapsto x^3$</span> is surjective and so every element of <span class="math-container">$\mathbb R^*$</span> is a cube.</p>
<p>The map <span class="math-container">$\mathbb Q^* \to \mathbb Q^*$</span> given by <span class="math-container">$x\mapsto x^3$</span> is not surjective and so not every element of <span class="math-container">$\mathbb Q^*$</span> is a cube of an element of <span class="math-container">$\mathbb Q^*$</span>. For instance, <span class="math-container">$2 \in \mathbb Q^*$</span> is not a cube (of a rational number).</p>
|
265,067 | <p>$$\lim_{n\to\infty}\frac{(2n-1)!}{3^n(n!)^2}$$</p>
<p>How can I associate limit problem with series? And how can i find limits from series?
Can anyone help?</p>
| André Nicolas | 6,312 | <p><strong>Hint:</strong> Let $a_n=\dfrac{(2n-1)!}{3^n(n!)^2}$. </p>
<p>It is useful to look at the ratio $\dfrac{a_{n+1}}{a_n}$ for large $n$. </p>
|
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