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 |
|---|---|---|---|---|
4,064,084 | <p>Does there exists a countable family of infinite sets <span class="math-container">$\{A_n:n\in\mathbb N\}\subset\mathcal P(\mathbb N)$</span> satisfying the following property:
<span class="math-container">$$\text{For every infinite set }I\in\mathcal P(\mathbb N),\text{ there is }n\in\mathbb N\text{ such that }A_n\subset I\text{ ?}$$</span></p>
<p>If we require the family to have cardinality <span class="math-container">$\mathfrak c$</span>, then the question is trivial, but for the countable case I'm stuck.</p>
| hgmath | 886,804 | <p>As requested I'm posting the above comment as an answer here.</p>
<p>In order to show that every <span class="math-container">$n\geq 8$</span> can be written as sum of five cubes with absolute value less than <span class="math-container">$n$</span>:</p>
<ul>
<li>we can write for odd numbers <span class="math-container">$n=2k+1$</span> with <span class="math-container">$k\geq 4$</span>
<span class="math-container">$$(2k+1)^3 = (2k-1)^3 + (k+4)^3 - (k-4)^3 - 5^3 - 1^3$$</span></li>
<li>for even numbers <span class="math-container">$n\geq 8$</span>, we can write <span class="math-container">$n=2^a m$</span> where <span class="math-container">$m$</span> is <span class="math-container">$4$</span>, <span class="math-container">$6$</span>, or an odd number <span class="math-container">$\geq 5$</span>. Then <span class="math-container">$m$</span> can be decomposed as <span class="math-container">$m=m_1^3+m_2^3+m_3^3+m_4^3+m_5^3$</span> so that
<span class="math-container">$$(2^a m)^3 = (2^a m_1)^3 + (2^a m_2)^3 + (2^a m_3)^3 + (2^a m_4)^3 + (2^a m_5)^3$$</span></li>
</ul>
|
3,465,018 | <p>Compute <span class="math-container">$\pi_{2}(S^2 \vee S^2).$</span></p>
<p><strong>Hint:</strong>
Use universal covering thm. and use Van Kampen to show it is simply connected.</p>
<p>Still I am unable to solve it, could anyone give me more detailed hint and the general idea of the solution.</p>
| Grisha Taroyan | 402,997 | <p>Following <a href="https://en.wikipedia.org/wiki/Homotopy_excision_theorem" rel="noreferrer">homotopy excision theorem</a> and using an exact sequence of a pair <span class="math-container">$(S^2\times S^2, S^2\vee S^2)$</span> you can write down an exact sequence
<span class="math-container">$$
0 \to \pi_3(S^2\wedge S^2)\to \pi_2(S^2\vee S^2) \to \pi_2(S^2\times S^2) \to 0
$$</span>
Since <span class="math-container">$S^2\wedge S^2 \simeq S^4$</span> you have an isomorphism <span class="math-container">$\pi_2(S^2\vee S^2) \cong \pi_2(S^2\times S^2)\cong \mathbb{Z}\oplus\mathbb{Z}.$</span></p>
|
4,565,728 | <p>Given a collection of topological spaces <span class="math-container">$X_i$</span> indexed by the elements <span class="math-container">$i$</span> of a set <span class="math-container">$I$</span>, we consider the set product <span class="math-container">$P = \prod_{i \in I} X_i$</span> with projections <span class="math-container">$p_i : P \to X_i$</span>. There are two methods of topologizing <span class="math-container">$P$</span>.</p>
<ol>
<li><p>Take the <em>product topology</em> having as a subbase all sets of the form <span class="math-container">$p_i^{-1}(U_i)$</span> with <span class="math-container">$i \in I$</span> and <span class="math-container">$U_i \subset X_i$</span> open. Therefore a base for the product topology consists of all products of form <span class="math-container">$\prod_{i \in I} U_i$</span> with <span class="math-container">$U_i \subset X_i$</span> open and <span class="math-container">$U_i \ne X_i$</span> only for finitely many <span class="math-container">$i$</span>.</p>
</li>
<li><p>Take the <em>box topology</em> having as a base all products of form <span class="math-container">$\prod_{i \in I} U_i$</span> with <span class="math-container">$U_i \subset X_i$</span> open.</p>
</li>
</ol>
<p>By definition the product topology is the coarsest topology on <span class="math-container">$P$</span> such that all <span class="math-container">$p_i^{-1}(U_i)$</span>, with open <span class="math-container">$U_i \subset X_i$</span>, <span class="math-container">$i \in I$</span>, are open. This is equivalent to defining the product topology as the coarsest topology on <span class="math-container">$P$</span> such that all projections <span class="math-container">$p_i$</span> become continuous.</p>
<p>It is standard to endow <span class="math-container">$P$</span> with the product topology. This makes <span class="math-container">$(P,(p_i)_{i \in I})$</span> the product of the objects <span class="math-container">$X_i$</span> in the category of topological spaces which is characterized by the following universal property:</p>
<blockquote>
<p>A function <span class="math-container">$f : Y \to P$</span> defined on a topological space <span class="math-container">$Y$</span> is continuous if and only if all <span class="math-container">$p_i \circ f : Y \to X_i$</span> are continuous.</p>
</blockquote>
<p>This seems to be the optimum what can be expected from a topology on <span class="math-container">$P$</span>.</p>
<p>On infinite products the box topology is in general strictly finer than the product topology, thus it does in general not have this universal property. As far as I know, the box topology does not have <em>any really nice universal property</em>; see <a href="https://math.stackexchange.com/q/3095816">Does the box topology have a universal property?</a></p>
<p>On the other hand, the box topology is the coarsest topology such that all products of the form <span class="math-container">$\prod_{i \in I} U_i$</span> with open <span class="math-container">$U_i \subset X_i$</span>, <span class="math-container">$i \in I$</span>, are open. On the level of sets this seems to be a very natural requirement, even more natural than requiring that all special products of the form <span class="math-container">$\prod_{i \in I} U_i$</span> with open <span class="math-container">$U_i \subset X_i$</span> open and <span class="math-container">$U_i \ne X_i$</span> only for finitely many <span class="math-container">$i$</span>, are open.</p>
<p><strong>Question:</strong> Are there other characterizations of product or box topology than those described above? If so, do they occur somewhere in the literature?</p>
| Kritiker der Elche | 908,786 | <p>It is obvious that the box topology on <span class="math-container">$P$</span> makes all projections <span class="math-container">$p_i :P \to X_i$</span> <em>open maps</em>.</p>
<p>In a comment it was conjectured that the box topology is the finest topology on <span class="math-container">$P$</span> such that all projections <span class="math-container">$p_i :P \to X_i$</span> are continuous open maps. <strong>This is not true.</strong></p>
<p>Note that the requirement of continuity can also be omitted here because the box topology makes all projections continuous, therefore the same is true for each finer topology.</p>
<p>For finite products the box topology agrees with the product topology, and therefore the product topology on a finite product should be the finest topology on <span class="math-container">$P$</span> such that all projections <span class="math-container">$p_i : P \to X_i$</span> are open maps.</p>
<p>Consider <span class="math-container">$P = [0,1] \times [0,1]$</span> with projections <span class="math-container">$p_1, p_2 : P \to [0,1]$</span>. As a subbase for a topology <span class="math-container">$\tau$</span> on <span class="math-container">$P$</span> take all <span class="math-container">$p_i^{-1}(U)$</span> with open <span class="math-container">$U \subset [0,1]$</span> and the set <span class="math-container">$W_0 = \{(x,y) \in P \mid y < x \} \cup \{(0,0)\}$</span>. A base <span class="math-container">$\mathcal B$</span> for <span class="math-container">$\tau$</span> is obtained by taking all finite intersections of these sets. This yields all boxes <span class="math-container">$U_1 \times U_2$</span> with open <span class="math-container">$U_i \subset [0,1]$</span> plus all <span class="math-container">$(U_1 \times U_2) \cap W_0$</span> with open <span class="math-container">$U_i \subset [0,1]$</span>. Clearly <span class="math-container">$\tau$</span> is strictly finer than the product topology <span class="math-container">$\tau_p$</span> because <span class="math-container">$W_0 \notin \tau_p$</span>.</p>
<p>We claim that <span class="math-container">$p_i(B)$</span> is open in <span class="math-container">$[0,1]$</span> for all <span class="math-container">$B \in \mathcal B$</span> which shows that the <span class="math-container">$p_i$</span> are open maps with respect to <span class="math-container">$\tau$</span>.</p>
<p>Since the <span class="math-container">$p_i$</span> are known to be open maps with respect to <span class="math-container">$\tau_p$</span>, it suffices to consider <span class="math-container">$B = (U_1 \times U_2) \cap W_0$</span>.</p>
<p>The set <span class="math-container">$W = \{(x,y) \in P \mid y < x \}$</span> belongs to <span class="math-container">$\tau_p$</span> and <span class="math-container">$W_0 = W \cup \{(0,0\}$</span>. We get
<span class="math-container">$$p_i(B) = p_i((U_1 \times U_2) \cap W \cup (U_1 \times U_2) \cap \{(0,0\}) \\= p_i((U_1 \times U_2) \cap W) \cup p_i((U_1 \times U_2) \cap \{(0,0\}).$$</span>
The set <span class="math-container">$(U_1 \times U_2) \cap W$</span> belongs <span class="math-container">$\tau_p$</span> so that <span class="math-container">$p_i((U_1 \times U_2) \cap W)$</span> is open.</p>
<p>If <span class="math-container">$(0,0) \notin U_1 \times U_2$</span>, then <span class="math-container">$p_i(B) = p_i((U_1 \times U_2) \cap W)$</span> is open.</p>
<p>If <span class="math-container">$(0,0) \in U_1 \times U_2$</span>, then <span class="math-container">$p_i((U_1 \times U_2) \cap \{(0,0\}) = \{0\}$</span>. Moreover there exists <span class="math-container">$r \in (0,1]$</span> such that <span class="math-container">$[0,r) \times [0,r) \subset U_1 \times U_2$</span>. Thus <span class="math-container">$(0,r) \subset p_i((U_1 \times U_2) \cap W)$</span>. This shows that <span class="math-container">$p_i(B) = p_i((U_1 \times U_2) \cap W) \cup \{0\} = p_i((U_1 \times U_2) \cap W) \cup [0,r)$</span> which is open in <span class="math-container">$[0,1]$</span>.</p>
|
271,844 | <p>I've installed the KnotTheory package, following the instructions <a href="http://katlas.org/wiki/Setup" rel="nofollow noreferrer">here</a>. But when I try to use it I get this error:</p>
<p><code>$CharacterEncoding: The byte sequence {139} could not be interpreted as a character in the UTF-8 character encoding.</code></p>
<p>Both <code><< KnotTheory` </code> and <code>Needs["KnotTheory`"]</code> raise this error (I found that byte sequence {139} is for "<"). "Needs" also raises</p>
<p><code>Needs: "Context KnotTheory` was not created when Needs was evaluated".</code></p>
<p>Parent directory of KnotTheory is in the <code>$Path</code>, but it doesn't help.</p>
<p>How can I fix it?</p>
| Henrik Schumacher | 38,178 | <p>I recall that Claus Ernst once asked me about that. It's time that we make the fix public and searchable:</p>
<p>Go into the file <code>init.m</code> on the <code>KnotTheory</code> package directory.
Then replace the lines</p>
<pre><code>KnotTheoryDirectory[] = (
File /. Flatten[FileInformation[ToFileName[#,"KnotTheory"]] & /@ ($Path /. "." -> Directory[])]
)
</code></pre>
<p>with</p>
<pre><code>KnotTheoryDirectory[] = DirectoryName[$InputFileName];
</code></pre>
<p>Save the file, restart the kernel. Afterwards, the package should load just fine.</p>
|
3,115,830 | <p>So my logic to this up until now has been that for any <span class="math-container">$x$</span> the function <span class="math-container">$\left\lfloor\frac{\lceil x\rceil}{2}\right\rfloor$</span> will return an integer that is an element of <span class="math-container">$\mathbb Z$</span>. Thus since you can map any <span class="math-container">$x$</span> in the domain to any y in the co-domain it is surjective.</p>
<p>Now I'm not sure if this counts as a full proof, and whether the function is injective.</p>
| Travis Willse | 155,629 | <p><strong>Hint</strong> Differentiating the relation
<span class="math-container">$$f = a {\bf T} + b {\bf N} + c {\bf B}$$</span>
and using the Frenet-Serret Identities gives
<span class="math-container">\begin{align}
{\bf T}
&= a {\bf T}' + b {\bf N}' + c' {\bf B} + c {\bf B}' \\
&= a (\kappa {\bf N}) + b (-\kappa {\bf T} + \tau {\bf B}) + c' {\bf B} + c (-\tau {\bf N}) \\
&= -b \kappa {\bf T} + (a \kappa - c \tau) {\bf N} + (b \tau + c') {\bf B} .
\end{align}</span>
Comparing like coefficients gives the system
<span class="math-container">\begin{align}
1 &= - b \kappa \\
0 &= a \kappa - c \tau \\
0 &= b \tau + c' .
\end{align}</span>
The first equation gives the suspected (constant) value <span class="math-container">$\kappa = -1 / b$</span>. After substituting we can solve the remaining two equations for <span class="math-container">$\tau$</span> and (by integrating), <span class="math-container">$c$</span>.</p>
|
1,721,565 | <p>I'm having trouble with what I have done wrong with the chain rule below. I have tried to show my working as much as possible for you to better understand my issue here.</p>
<p>So:</p>
<p>Find $dy/dx$ for $y=(x^2-x)^3$
<br> So power to the front will equal = $3(x^2-x)^2 * (2x-1)$</p>
<p>Where did the $-1$ come from in $2x-1$?</p>
<p>How did they get that? </p>
<p>Thanks!</p>
| Matan L | 111,632 | <p>Lets denote the following:</p>
<p>$y(x) = (x^2-x)^3$</p>
<p>$f(x)=x^2-x$</p>
<p>$g(x)=x^3$</p>
<p>than $y(x)=g(f(x))$
so by the chain rule you get that $y_x = g_x (f(x)) f_x (x)$</p>
<p>thus in your case $y_x = 3f(x)^2f_x(x)=2(x^2-x)^2(2x-1)$</p>
|
622,278 | <p>What is the relation between the convergence of $\sum a_{n}$ and $\prod (1+a_{n})$ where $a_{n} \in \mathbb{C} \ \forall n$ ?</p>
<p>Where can I find some references about this topic ?</p>
| Sourav D | 114,457 | <p>This is definitely true when $a_n$ are positive numbers as demonstrated <a href="http://cornellmath.wordpress.com/2008/01/26/convergence-of-infinite-products/" rel="nofollow">here</a>.</p>
<p>However, when $a_n \in \mathbb{C}$, then convergence of $\sum|a_n|$ is just a necessary condition for convergence of the infinite product.</p>
|
715,361 | <p>Let $\Omega$ be a bounded domain and $f_n\in L^2(\Omega)$ be a sequence such that
$$\int_\Omega f_nq\operatorname{dx}\leq C<\infty\qquad \text{for all}\quad q\in H^1(\Omega),\ \|q\|_{H^1(\Omega)}\leq1,\ n\in\mathbb{N}.\quad (1) $$
Is it then possible to conclude that
$$ \sup_{n\in\mathbb{N}}\|f_n\|_{L^2(\Omega)}\leq C. $$</p>
<p>Here, $H^1(\Omega)$ denotes the Sobolev-Hilbert-Space $H^1(\Omega)$.</p>
<p>Obviously, this statement would be true if we were to replace (1) with
$$\int_\Omega f_nq\operatorname{dx}\leq C<\infty\qquad \text{for all}\quad q\in L^2(\Omega),\ \|q\|_{L^2(\Omega)}\leq1,\ n\in\mathbb{N}. $$</p>
<p>and maybe the dense and compact embedding $H^1(\Omega)\hookrightarrow L^2(\Omega)$ is of help but I'm not sure of it.</p>
<p>Edit: By now I'm pretty sure, that this statement doesn't hold. We only have a bound in the dual of $H^1(\Omega)$. But until now I'm failing to compile a conclusive argument!</p>
| Héctor Asencio L | 135,194 | <p>If you use $ x=1 $ you get $ 1-1=0 $ on the denominator, which gives you a division by zero. And since that's the only value that gets an equality on both sides of the equation, it shows that there's no solution.</p>
|
2,774,923 | <blockquote>
<p>$ABC$ is a triangle where $AE$ and $EB$ are angle bisectors, $|EC| = 5$, $|DE| = 3$, $|AB| = 9$. Find the perimeter of the triangle $ABC$.
<a href="https://i.stack.imgur.com/4nsiM.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4nsiM.jpg" alt="enter image description here"></a></p>
</blockquote>
<p>I realized that the length of the side $|DC| = 8$. In the $\triangle BEC$, we have special triangle $3-4-5$. This is where I'm stuck. Can I take your thinkings? </p>
| Community | -1 | <p>Well this is a pretty straightforward question.</p>
<p>Now, According to the angle bisector theorem:</p>
<blockquote>
<p>Consider a triangle ABC. Let the angle bisector of angle A intersect side BC at a point D between B and C. The angle bisector theorem states that the ratio of the length of the line segment BD to the length of segment DC is equal to the ratio of the length of side AB to the length of side AC</p>
</blockquote>
<p>We get $$AC/AD=BC/BD=5/3$$
$$AC=(5/3)AD$$ and $$BC=(5/3)BD$$
Note that $$AB=AD+DB=9$$</p>
<p>Hence $$AC+BC==(5/3)(AD=DB)=15$$</p>
<p>But circumference of Triangle $=AB+BC+CA=15+9=24$ </p>
|
1,192,357 | <blockquote>
<p>What is the limit of $f(a,b) =\frac{a^\beta}{a^2 + b^2}$ as $(a,b) \to (0,0)$?</p>
</blockquote>
<p>Clearly the answer depends on the value of $\beta$. For $\beta > 0$, we can deduce via inequalities that $\lim_{(a,b) \to (0,0)} f(a,b) = 0$.</p>
<p>However, for $\beta < 0$, the answer is less obvious. How would you approach this case?</p>
| Eclipse Sun | 119,490 | <p>I will discuss the cases where $\beta$ is not an integer. To make it clearly, let's consider the limit$$\lim_{(a,b)\to(0,0),a>0}\frac{a^\beta}{a^2+b^2}.$$<br>
For $\beta>2$, since $\left|\dfrac{a^\beta}{a^2+b^2}\right|=\left|\dfrac{a^2}{a^2+b^2}\right||a^{\beta-2}|<|a^{\beta-2}|$ , the limit is $0$.<br>
For $0<\beta\le 2$, the limit doesn't exist, because the two iterated limits is not equal.<br>
For $\beta\le 0$, the limit is $+\infty$, for the denominator tends to $+0$, and the numerator tends to $+\infty$ or $1$. </p>
<p><strong>Note</strong>: If we suppose $\beta$ is integer, then we can consider the case where $a\le 0$. It is easy now to prove that when $\beta=-1, -3,\ldots$, the limit doesn't exist; in other cases the answer is the same as what I have discussed above.</p>
|
262,173 | <p>Consider $x^2 + y^2 = r^2$. Then take the square of this to give $(x^2 + y^2)^2 = r^4$. Clearly, from this $r^4 \neq x^4 + y^4$. </p>
<p>But consider: let $x=a^2, y = b^2 $and$\,\,r = c^2$. Sub this into the first eqn to get $(a^2)^2 + (b^2)^2 = (c^2)^2$. $x = a^2 => a = |x|,$ and similarly for $b.$</p>
<p>Now put this in to give $|x|^4 + |y|^4 = r^4 => (-x)^4 + (-y)^4 = r^4 $ or $ (x)^4 + (y)^4 = r^4,$ both of which give $ x^4 + y^4 = r^4$
Where is the flaw in this argument?</p>
<p>Many thanks.</p>
| Hagen von Eitzen | 39,174 | <p>Note that $(x^2+y^2)^2=r^4$ does not imply that $r^4\ne x^4+y^4$.</p>
<p>In fact, you show that $$a^4+b^4=c^4$$
provided $x^2+y^2=r^2, x=a^2, y=b^2, z=c^2$. So what?</p>
|
1,001,320 | <p>I was wondering how to do an inequality problem involving QM-AM-GM-HM.</p>
<p>Question: For positive $a$, $b$, $c$ such that $\frac{a}{2}+b+2c=3$, find the maximum of $\min\left\{ \frac{1}{2}ab, ac, 2bc \right\}$.</p>
<p>I was thinking maybe apply AM-GM, however, I'm not sure what to plug in. Any help would be appreciated, thanks!</p>
| Macavity | 58,320 | <p>Another way: If possible, let the optimum occur when one among $\frac12ab, ac, 2bc$ is lesser than the others. Then note that this smaller term determines the maximum and can be increased at the expense of the variable which is not involved in it. Thus at optimum we must have $\frac12ab = ac = 2bc$, which with the constraint gives $\frac12a=b=2c=1$ as the only nonzero solution and leads to a maximum of $1$.</p>
|
1,042,715 | <p>Given a separable space $X$, if $A$ is discrete subspace of $X$, then $|A|\leq 2^{\aleph_0}$.</p>
<p>Some ideas?. It's similar to "jone's lemma", but without $A$ being closed. Whit what addiotional conditions we can assure the statement becomes true?</p>
| Robert Israel | 8,508 | <p>I nominate Halmos. All of his writing is good, but you might look in particular at
"Finite dimensional vector spaces".</p>
|
1,042,715 | <p>Given a separable space $X$, if $A$ is discrete subspace of $X$, then $|A|\leq 2^{\aleph_0}$.</p>
<p>Some ideas?. It's similar to "jone's lemma", but without $A$ being closed. Whit what addiotional conditions we can assure the statement becomes true?</p>
| JohnD | 52,893 | <p>Pick up an issue of SIAM Review. I admit that I haven't read every article in every issue of late, but I cannot recall an article with less than "very good" writing, and most are excellent.</p>
|
1,042,715 | <p>Given a separable space $X$, if $A$ is discrete subspace of $X$, then $|A|\leq 2^{\aleph_0}$.</p>
<p>Some ideas?. It's similar to "jone's lemma", but without $A$ being closed. Whit what addiotional conditions we can assure the statement becomes true?</p>
| Stiofán Fordham | 28,745 | <p>There is a book by Bott and Tu called '<a href="http://www.springer.com/mathematics/geometry/book/978-0-387-90613-3" rel="nofollow">Differential forms in algebraic topology</a>' which I think satisfies your criteria. In particular, you say it should be engaging and that is what I remember most about when I first read this book.</p>
|
3,193,696 | <p>Could someone explain what are (at least the four first) moments ? (normalized moment to be more precise) Let <span class="math-container">$X$</span> a r.v. </p>
<ul>
<li><p>So the first moment is the expectation. This will correspond to <span class="math-container">$\mathbb E[X]$</span> and is going to be the "barycenter" (in the sense that if <span class="math-container">$X$</span> is supported on <span class="math-container">$[a,b]$</span> and <span class="math-container">$\rho(t)$</span> is the density of the line <span class="math-container">$[a,b]$</span> (i.e. <span class="math-container">$\rho(dt)=\mathbb P_X(dt)$</span>), then the barycenter of the line is <span class="math-container">$\mathbb E[X]$</span> (I like to see this in physic term).</p></li>
<li><p>So the second moment is the variance. So, it's going to represent the dispersion of the mass around <span class="math-container">$\mathbb E[X]$</span>, i.e. <span class="math-container">$66\%$</span> of the total mass is going to be concentrate on <span class="math-container">$[\mathbb E[X]-\sqrt{Var(X)},\mathbb E[X]+\sqrt{Var(X)}]$</span>, but I don't really understand why... </p></li>
</ul>
<p>Is my intuition for <span class="math-container">$\mathbb E[X]$</span> and <span class="math-container">$\text{Var}(X)$</span> correct ? What about the the third and the fourth moments ? (I think of higher dimension it's much more abstract).</p>
| J.G. | 56,861 | <p>The first moment is the (arithmetic) <em>mean</em>, <span class="math-container">$\mu:=\Bbb E[X]$</span>. If this is finite, any <span class="math-container">$a$</span> satisfies <span class="math-container">$\Bbb E[X-a]=\mu-a$</span>. So if we want to measure spread around <span class="math-container">$\mu$</span>, we can't use <span class="math-container">$\Bbb E[X-\mu]=0$</span>, but we can use the <em>variance</em> <span class="math-container">$\sigma^2=\Bbb E[(X-\mu)^2]$</span>, or better still (so we're working with a quantity that has the same dimensions as <span class="math-container">$X$</span> and <span class="math-container">$\mu$</span>, should <span class="math-container">$X$</span> not be dimensionless) the <em>standard deviation</em> <span class="math-container">$\sigma=\sqrt{\Bbb E[(X-\mu)^2]}\ge 0$</span>. (You could also look at the <em>absolute deviation</em> <span class="math-container">$\Bbb E[|X-\mu|]$</span>, but that's usually neither as useful nor as easy to analyse.)</p>
<p>If <span class="math-container">$\mu,\,\sigma$</span> exist and are finite, the <span class="math-container">$z$</span>-score of <span class="math-container">$X$</span> is defined as <span class="math-container">$Z:=\frac{X-\mu}{\sigma}$</span>. This has mean <span class="math-container">$0$</span> and variance <span class="math-container">$1$</span>. <a href="https://en.wikipedia.org/wiki/Chebyshev%27s_inequality#Probabilistic_proof" rel="nofollow noreferrer">We can prove</a> <span class="math-container">$P(|Z|\ge k)\le k^{-2}$</span>, but there are <a href="https://en.wikipedia.org/wiki/Normal_distribution#Occurrence_and_applications" rel="nofollow noreferrer">theoretical reasons</a> to have an especial interest in a "Normal distribution", for which a much tighter bound can be obtained, which your numerical example considers.</p>
<p>While <span class="math-container">$Z,\,Z^2$</span> must have respective means <span class="math-container">$0,\,1$</span>, <span class="math-container">$\Bbb E[Z^3],\,\Bbb E[Z^4]$</span> are another story. Their values are the <em>skewness</em> and <em>kurtosis</em> of <span class="math-container">$X$</span>, and indeed of <span class="math-container">$aX+b$</span> for <span class="math-container">$a>0$</span> (for the kurtosis we can relax this to <span class="math-container">$a\ne 0$</span>), including <span class="math-container">$Z$</span> itself. These quantities provide additional information about <span class="math-container">$Z$</span>'s distribution. In particular, they provide ways a distribution can notably fail to be Normal. For a Normal distribution, their respective values are <span class="math-container">$0$</span> and <span class="math-container">$3$</span>, and <span class="math-container">$\Bbb E[Z^4-3]$</span> is often called the <em>excess kurtosis</em> of <span class="math-container">$X$</span>. We classify distributions by <a href="https://en.wikipedia.org/wiki/Kurtosis#Excess_kurtosis" rel="nofollow noreferrer">whether this is positive, zero or negative</a>. Note that <span class="math-container">$\Bbb E[Z^4-1]$</span> is the variance of <span class="math-container">$Z^2$</span>, so the excess kurtosis is at least <span class="math-container">$-2$</span>. <a href="https://en.wikipedia.org/wiki/Bernoulli_distribution#Properties_of_the_Bernoulli_distribution" rel="nofollow noreferrer">This can be saturated</a>.</p>
<p>A symmetric distribution (i.e. one for which <span class="math-container">$P(X\le x)=P(X\ge 2\mu-x)$</span>) has either zero or undefined skewness. When the skewness of an asymmetric distribution is defined but non-zero, the sign of the skewness indicates whether the distribution "leans" to the left or right. But despite a common misconception, <a href="https://en.wikipedia.org/wiki/Skewness#Relationship_of_mean_and_median" rel="nofollow noreferrer">it doesn't</a> tell you how the mean compares to the median.</p>
<p>A geometric interpretation of the kurtosis is a little harder. You'll sometimes read people saying it describes how flat the distribution's peak is, but <a href="https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4321753/#S9title" rel="nofollow noreferrer">it's better</a> to think of it in terms of the behaviour of the distribution's tails.</p>
|
1,022,380 | <p>in below link, (formula (34)-(40)) there are some definition of Dirac delta function in terms of other functions such as Airy function, Bessel function of the first kind, Laguerre polynomial,....</p>
<p><a href="http://mathworld.wolfram.com/DeltaFunction.html" rel="nofollow noreferrer">http://mathworld.wolfram.com/DeltaFunction.html</a></p>
<p>Is there any definition of the Dirac delta function in terms of the sech (secant hyperbolic) or cosh (cosine hyperbolic) functions?</p>
<p>Please say your references. Thanks</p>
| Ian | 83,396 | <p>Whenever $f$ is a nonnegative function with integral 1, one has approximation by convolution. Specifically, define $f_c(x) = f(x/c)/c$. Then $\lim_{c \to 0^+} g * f_c = g$ for mildly restricted $g$. (For example, $L^p$ for $1 \leq p < \infty$ is enough.) In particular you can take $f(x)=\text{sech}(x)/\pi$.</p>
|
2,826,850 | <p>$x,y$ and $z$ are consecutive integers, such that $\frac {1}{x}+ \frac {1}{y}+ \frac {1}{z} \gt \frac {1}{45} $, what is the biggest value of $x+y+z$ ?.</p>
<p>I assumed that $x$ was the smallest number so that I could express the other numbers as $x+1$ and $x+2$ and in the end I got to a cubic function but I didn't know how to find its roots. I probably didn't do anything important so I'd appreciate if you give me any hints or help. Thanks in advance.</p>
<p>I have thought about using the AM-HM.</p>
| Sameer Kailasa | 117,021 | <p>Write $x = y-1$ and $z = y+1$. Then we have $$\frac{1}{y-1} + \frac{1}{y} + \frac{1}{y+1} = \frac{2y}{y^2 - 1} + \frac{1}{y} = \frac{3y^2 - 1}{y^3 - y} > \frac{1}{45}$$ Since $3y^2 - 1< 3y^2$, this implies $$\frac{3y}{y^2 - 1} = \frac{3y^2}{y^3 - y} > \frac{1}{45} \iff 135y > y^2- 1 \iff y-\frac{1}{y} < 135$$ As $y$ is assumed to be a positive integer, it follows that $y\le 135$. From here, you can check that $y=135$ certainly works for the initial inequality, so the largest possible value of $x+y+z = 3y$ is $3\cdot 135 = 405$.</p>
|
4,172,964 | <p>EDIT: Agreed, this isn't a well formed question. But responses below have at least given me a different way to think about it.</p>
<p>EDIT2: Thanks the answers I discovered the Google S2 library (<a href="http://s2geometry.io/devguide/s2cell_hierarchy" rel="nofollow noreferrer">http://s2geometry.io/devguide/s2cell_hierarchy</a>) which provides methods for tracing a space filling curve on a sphere. I'm working this into my camera position algorithm instead of just gimballing through altitude/azimuth. Thanks!</p>
<p>First post here because I'm not a math guy, but I have a feeling the confusion I have is due to something mathy. Perhaps this isn't even the right forum; if not, I'll delete.</p>
<p>Here's the problem:</p>
<p>I built an apparatus to rotate a sensor through a 360 circle and measure incident light at each point. It's pretty cool, I can automatically orient a sensor on the end of arm to maximize incident light (looks like a radar antenna) using a stepper.</p>
<p>I tried extending this to a sphere, but what I found was: I need to discretize the the second angle to some minimum delta-phi. It seems odd to me that I can sweep through an infinite number of degrees in a circle, continuously as my theta angle, but then I need to choose a delta-phi to rotate that circle through to scan the sphere surface.</p>
<p>How is it that I can rotate through an infinite number of points on a circle in one rotation in finite time, but cannot cover the infinite number of circles that comprise the surface of a sphere in finite time because I can always increment a smaller and smaller phi angle...</p>
<p>Or maybe the question is: how is it that an rotating circle traces out a sphere surface: aren't there always going to be more points "covered" by the circle at the poles, compared to the equator, since the distance covered by an infinitesimal delta-phi is short at the poles than the equator?</p>
<p>Obviously I ultimately have to discretize everything since I'm using a digital circuit, but this sudden appearance if finite/infinite stopped me in my tracks...</p>
<p>I'm not even sure what I'm asking, just puzzled. Anyone know what I'm up against here?</p>
| Cameron Williams | 22,551 | <p>Hint: try defining a new variable <span class="math-container">$w = z^4$</span> so that the equation in terms of <span class="math-container">$w$</span> reads (correcting the <span class="math-container">$z^8$</span> term to <span class="math-container">$|z|^8$</span>)</p>
<p><span class="math-container">$$ |w|^2 + 4(w+\bar{w}) + 11 = 0.$$</span></p>
<p>There are probably multiple ways to solve this, but I would break <span class="math-container">$w$</span> into real and imaginary parts and solve, noting that since <span class="math-container">$|z|=1$</span>, <span class="math-container">$|w|=1$</span>. I imagine there's a more elegant approach. Then after solving for <span class="math-container">$w$</span>, you can solve for <span class="math-container">$z$</span>.</p>
|
271 | <p>Is there a way of taking a number known to limited precision (e.g. $1.644934$) and finding out an "interesting" real number (e.g. $\displaystyle\frac{\pi^2}{6}$) that's close to it?</p>
<p>I'm thinking of something like Sloane's Online Encyclopedia of Integer Sequences, only for real numbers.</p>
<p>The intended use would be: write a program to calculate an approximation to $\displaystyle\sum_{n=1}^\infty \frac{1}{n^2}$, look up the answer ("looks close to $\displaystyle\frac{\pi^2}{6}$") and then use the likely answer to help find a proof that the sum really is $\displaystyle \frac{\pi^2}{6}$.</p>
<p>Does such a thing exist?</p>
| Casebash | 123 | <p>Try <a href="http://www.wolframalpha.com/input/?i=1.644934">Wolfram Alpha</a>. It actually does sequences as well.</p>
|
271 | <p>Is there a way of taking a number known to limited precision (e.g. $1.644934$) and finding out an "interesting" real number (e.g. $\displaystyle\frac{\pi^2}{6}$) that's close to it?</p>
<p>I'm thinking of something like Sloane's Online Encyclopedia of Integer Sequences, only for real numbers.</p>
<p>The intended use would be: write a program to calculate an approximation to $\displaystyle\sum_{n=1}^\infty \frac{1}{n^2}$, look up the answer ("looks close to $\displaystyle\frac{\pi^2}{6}$") and then use the likely answer to help find a proof that the sum really is $\displaystyle \frac{\pi^2}{6}$.</p>
<p>Does such a thing exist?</p>
| Douglas S. Stones | 139 | <p>Sometimes the decimal digits of numbers will appear in Sloane's On-Line Encyclopedia of Integer Sequences <a href="http://oeis.org/" rel="nofollow noreferrer">OIES</a>. </p>
<p>E.g. <a href="http://oeis.org/A000796" rel="nofollow noreferrer">here</a> is the decimal expansion of pi.</p>
|
85,841 | <p><a href="http://reference.wolfram.com/language/ref/Binomial.html" rel="nofollow"><code>Binomial[n, k]</code></a> is converted to a polynomial only for <code>k</code> less than 6.</p>
<pre><code>Table[Binomial[n, k], {k, 1, 8}]
(* {n,
1/2 (-1 + n) n,
1/6 (-2 + n) (-1 + n) n,
1/24 (-3 + n) (-2 + n) (-1 + n) n,
1/120 (-4 + n) (-3 + n) (-2 + n) (-1 + n) n,
Binomial[n, 6],
Binomial[n, 7],
Binomial[n, 8]} *)
</code></pre>
<p>For <code>k</code> greater or equal 6 is an output only copy of <a href="http://reference.wolfram.com/language/ref/Binomial.html" rel="nofollow"><code>Binomial[n, k]</code></a>. Why?
It's possible change this boundary? (but I not accept a solution like <a href="http://reference.wolfram.com/language/ref/Product.html" rel="nofollow"><code>Product[...]</code></a>)</p>
| ubpdqn | 1,997 | <p>You could also define your own binomial coefficient function, e.g.</p>
<pre><code>bn[n_, k_] := Fold[(n - #2 + 1) #1/#2 &, 1, Range[k]]
</code></pre>
<p>so, </p>
<pre><code>Grid[{HoldForm[Binomial[n, #]], bn[n, #]} & /@ Range[0, 8],
Frame -> All]
</code></pre>
<p><img src="https://i.stack.imgur.com/D5XwY.png" alt="enter image description here"></p>
|
1,985,905 | <p>I was wondering if the cardinality of a set is a well defined function, more specifically, does it have a well defined domain and range?</p>
<p>One would say you could assign a number to every finite set, and a cardinality for an infinite set. So the range would be clear, the set of cardinal numbers. But what about the domain, here we get a few problems. This should be the set of all sets, yet this concept isn't allowed in mathematics as it leads to paradoxes like Russell's paradox.</p>
<p>So how do we formalize the notion of 'cardinality'? It seems to behave like a function that maps sets into cardinal numbers, but you can't define it this way as that definition would be based on a paradoxical notion. Even if we only restrict ourselves to finite sets the problem pops up, as we could define the set {A} for every set, thereby showing a one-to-one correspondence between 'the set of all sets' (that doesn't exist) and the 'set of all sets with one element'.</p>
<p>So how should one look at the concept of cardinality? You can't reasonably call it a function. Formalizing this concept without getting into paradoxes seems very hard indeed.</p>
| Mees de Vries | 75,429 | <p>The collection of all sets does not form a set in ZF(-style) set theory, indeed. Note that the same is true for the collection of all cardinals: there is no set containing all cardinals, because then its union would be a set as well, and it would be a greater cardinal than any of its elements.</p>
<p>So the function $X \mapsto |X|$ is not a function <em>internally</em> to ZFC. However, it can be made a function externally: that is, there is a formula $\phi(x,y)$, in two free variables, which holds if and only if $y$ is the cardinality of $x$. For this formula, we can prove $\phi(x,y) \land \phi(x,y') \to y = y'$, and we can prove $\forall x \exists y \phi(x,y)$. Hence, if we want to, we can introduce a function symbol $\mathrm{Card}$ to the language of set theory, such that $\mathrm{Card}(x)$ is interpreted as the unique $y$ such that $\phi(x,y)$. This is fine for the purposes for which we want to use cardinality.</p>
<p>Note also that if you are looking at a more limited part of the universe of sets, say $V_\alpha$ for some ordinal $\alpha$, then the restriction of the meta-function $\mathrm{Card}$ to this set <em>does</em> form a set.</p>
|
1,594,130 | <p>Does there exist a vector field $\vec F$ such that curl of $\vec F$ is $x \vec i+y\vec j+z \vec k$ ? </p>
<p>UPDATE : I did $div(curl \vec F)=0$ as the answers did ; but that assumes a lot i.e. it assumes that components of $F$ have second partial derivatives and continuous mixed partial derivatives ; whereas for curl to be defined , we only need components of $F$ to have first order partial derivatives . Is the answer still no with this less assumption ? Please help . Thanks in advance</p>
| Chappers | 221,811 | <p>No: a vector field $F$ can only be the curl of something if $\operatorname{div}{F}=0$, because
$$\operatorname{div}\operatorname{curl} G=0$$
for any twice-differentiable $G$ by antisymmetry. The divergence of your vector field, on the other hand, is $3$.</p>
|
2,574,221 | <p>Does divergence of $\sum a_k$ imply divergence of $\sum \frac{a_k}{1+a_k}$?</p>
<p>Note: $a_k > 0 $</p>
<p>I understand that looking at the contrapositive statement, we can say that the convergence of the latter sum implies $\frac{a_k}{1+a_k}\rightarrow 0$ but from here is it possible to deduce that $a_k\rightarrow 0$ because it is not completely straightforward. If we assume $a_k$ to be convergent, this trivially follows but it could diverge in which case this is nontrivial to me.</p>
| Rigel | 11,776 | <p>If $a_k\geq 0$ for every $k$, then you can reason as follows.</p>
<p>If $(a_k)$ does not converge to $0$, then there exists a subsequence $(a_{n_k})$ with limit $l\in (0,+\infty]$.</p>
<p>It is then easy to verify that
$$
\frac{a_{n_k}}{1+a_{n_k}} \to
\begin{cases}
1, & \text{if}\ l=+\infty,\\
\frac{l}{1+l}\neq 0, &\text{if}\ l\in (0,+\infty),
\end{cases}
$$
hence $\sum \frac{a_k}{1+a_k}$ cannot converge, so that diverges to $+\infty$.</p>
<p>Consider now the case $\lim_k a_k = 0$.
Then there exists $N\in\mathbb{N}$ such that $0\leq a_k \leq 1$ for every $k\geq N$, so that
$$
\frac{a_k}{1+a_k} \geq a_k/2
\qquad \forall k\geq N,
$$
and again the series diverges by comparison.</p>
<p>Remark: the point is that
$$
\frac{a_k}{1+a_k} \geq
\begin{cases}
1/2, &\text{if}\ a_k > 1,\\
a_k / 2, & \text{if}\ a_k\in [0,1].
\end{cases}
$$</p>
|
2,574,221 | <p>Does divergence of $\sum a_k$ imply divergence of $\sum \frac{a_k}{1+a_k}$?</p>
<p>Note: $a_k > 0 $</p>
<p>I understand that looking at the contrapositive statement, we can say that the convergence of the latter sum implies $\frac{a_k}{1+a_k}\rightarrow 0$ but from here is it possible to deduce that $a_k\rightarrow 0$ because it is not completely straightforward. If we assume $a_k$ to be convergent, this trivially follows but it could diverge in which case this is nontrivial to me.</p>
| Gabriel Romon | 66,096 | <p>If $\sum_n \frac{a_n}{1+a_n}$ converges, $\frac{a_n}{1+a_n}$ goes to $0$ as $n$ goes to $\infty$, hence $a_n$ goes to $0$ as $n\to \infty$.</p>
<p>But since $\lim_{n\to \infty} \frac{a_n}{\frac{a_n}{1+a_n}} = 1$, for large enough $n$, $\left|\frac{a_n}{\frac{a_n}{1+a_n}} - 1 \right|\leq \frac 12$, hence $a_n\leq \frac 32 \frac{a_n}{1+a_n}$. Therefore, $\sum a_n$ converges by comparison.</p>
|
2,985,917 | <p>Would it be possible to calculate which function in the Schwarz class of infinitely differentiable functions with compact support is closest to triangle wave?</p>
<p>Let us measure closeness as <span class="math-container">$$<f-g,f-g>_{L_2}^2 = \int_{-\infty}^{\infty}(f(x)-g(x))^2dx$$</span></p>
<p>I don't expect my knowledge in functional analysis to be strong and polished enough to answer this, but maybe one of you guys know how to? </p>
<hr>
<p><strong>EDIT</strong> of course we need to edit triangle wave to have compact support. Say it has compact support on <span class="math-container">$$x \in [-1-2N,1+2N], N\in \mathbb Z_+$$</span>
In other words, <span class="math-container">$2N+1$</span> whole periods and it goes down to <span class="math-container">$0$</span> just at both ends of support.</p>
| User8128 | 307,205 | <p>In general, there is no "closest" Schwarz function to a given <span class="math-container">$L^2$</span> function. Since the Schwarz class is dense in <span class="math-container">$L^2$</span>, we can find Schwarz functions arbitrarily close to any <span class="math-container">$L^2$</span> function. Indeed, let <span class="math-container">$f \in L^2(\mathbb R^n) \setminus \mathcal S(\mathbb R^n)$</span> and let <span class="math-container">$f_0 \in \mathcal S(\mathbb R^n)$</span> be any Schwarz function. Then we can always find <span class="math-container">$f_* \in \mathcal S(\mathbb R^n)$</span> with, for example, <span class="math-container">$$\|f - f_*\|_{L^2(\mathbb R^n)} < \frac 1 2 \| f - f_{0} \|_{L^2(\mathbb R^n)}.$$</span> Thus we can iteratively construct a sequence such that each member is "closer" to <span class="math-container">$f$</span> than the previous.</p>
|
3,793,581 | <p>I can solve this integral in a certain way but I'd like to know of other, simpler, techniques to attack it:</p>
<p><span class="math-container">\begin{align*}
\int _0^{\frac{\pi }{2}}\frac{\ln \left(\sin \left(x\right)\right)\ln \left(\cos \left(x\right)\right)}{\tan \left(x\right)}\:\mathrm{d}x&\overset{ t=\sin\left(x\right)}=\int _0^{\frac{\pi }{2}}\frac{\ln \left(\sin \left(x\right)\right)\ln \left(\cos \left(x\right)\right)}{\sin \left(x\right)}\cos \left(x\right)\:\mathrm{d}x\\[2mm]
&=\int _0^1\frac{\ln \left(t\right)\ln \left(\cos \left(\arcsin \left(t\right)\right)\right)}{t}\cos \left(\arcsin \left(t\right)\right)\:\frac{1}{\sqrt{1-t^2}}\:\mathrm{d}t\\[2mm]
&=\int _0^1\frac{\ln \left(t\right)\ln \left(\sqrt{1-t^2}\right)}{t}\sqrt{1-t^2}\:\frac{1}{\sqrt{1-t^2}}\:\mathrm{d}t\\[2mm]
&=\frac{1}{2}\int _0^1\frac{\ln \left(t\right)\ln \left(1-t^2\right)}{t}\:\mathrm{d}t=-\frac{1}{2}\sum _{n=1}^{\infty }\frac{1}{n}\int _0^1t^{2n-1}\ln \left(t\right)\:\mathrm{d}t\\[2mm]
&=\frac{1}{8}\sum _{n=1}^{\infty }\frac{1}{n^3}\\[2mm]
\int _0^{\frac{\pi }{2}}\frac{\ln \left(\sin \left(x\right)\right)\ln \left(\cos \left(x\right)\right)}{\tan \left(x\right)}\:\mathrm{d}x&=\frac{1}{8}\zeta (3)
\end{align*}</span></p>
| Quanto | 686,284 | <p>Substitute <span class="math-container">$t= \sin^2x$</span></p>
<p><span class="math-container">\begin{align}
&\int _0^{\frac{\pi }{2}}\frac{\ln (\sin x)\ln (\cos x)}{\tan x}\>{d}x \\= &\frac18\int _0^{1}\frac{\ln t\ln (1-t)}{t}\>dt \overset{IBP}= \frac1{16}\int_0^1 \frac{\ln^2 t}{1-t}dt= \frac1{16}\cdot2\zeta(3)= \frac18\zeta(3)
\end{align}</span></p>
|
1,611,390 | <p>How to show that the following function is an injective function?</p>
<p>$ \varphi : \mathbb{N}\times \mathbb{N} \rightarrow \mathbb{N} \\
\varphi(\langle n, k\rangle) = \frac{1}{2}(n+k+1)(n+k)+n$</p>
<p>I'm starting with $ \frac{1}{2}(a+b+1)(a+b)+a = \frac{1}{2}(c+d+1)(c+d)+c$, but how am I supposed to show from this equality that $\langle a, b\rangle = \langle c, d\rangle$, where $\langle a, b\rangle \in \mathbb{N}\times \mathbb{N}$ ?</p>
| David | 119,775 | <p>Here are two arguments. The first should make the statement "visually obvious" while the second is more formal.</p>
<p>Construct a grid displaying the value of $\phi(n,k)$ for each value of $n$ and $k$:
$$\begin{matrix}
\color{red}{\phantom{k={}}4}&14\cr
\color{red}{\phantom{k={}}3}&9&13\cr
\color{red}{\phantom{k={}}2}&5&8&12\cr
\color{red}{\phantom{k={}}1}&2&4&7&11\cr
\color{red}{n=0}&0&1&3&6&10\phantom{=n}\cr
&\color{blue}{0}&\color{blue}{1}&\color{blue}{2}&\color{blue}{3}&\color{blue}{\!\!\!\!\!\!\!\!4\rlap{{}=k}}
\end{matrix}$$
This should make it pretty clear what is going on. A formal proof could be as follows.</p>
<p>First, suppose that $n_1+k_1>n_2+k_2$. Since we are talking about integers, this means that $n_1+k_1\ge n_2+k_2+1$ and we have
$$\eqalign{\phi(n_1,k_1)
&=\tfrac12(n_1+k_1+1)(n_1+k_1)+n_1\cr
&\ge\tfrac12(n_2+k_2+2)(n_2+k_2+1)+n_1\cr
&=\tfrac12(n_2+k_2+1)(n_2+k_2)+n_2+k_2+1+n_1\cr
&>\phi(n_2,k_2)\ .\cr}$$
Now suppose that $\phi(n_1,k_1)=\phi(n_2,k_2)$. By the above argument we cannot have $n_1+k_1>n_2+k_2$, by a symmetrical argument we cannot have $n_2+k_2>n_1+k_1$, therefore $n_1+k_1=n_2+k_2$. Hence
$$\eqalign{\phi(n_1,k_1)=\phi(n_2,k_2)\
&\Rightarrow\ \tfrac12(n_1+k_1+1)(n_1+k_1)+n_1=\tfrac12(n_2+k_2+1)(n_2+k_2)+n_2\cr
&\Rightarrow\ \tfrac12(n_1+k_1+1)(n_1+k_1)+n_1=\tfrac12(n_1+k_1+1)(n_1+k_1)+n_2\cr
&\Rightarrow\ n_1=n_2\ ,\cr}$$
and so $k_1=k_2$. Thus $\phi$ is injective.</p>
|
894,159 | <p>I was assigned the following problem: find the value of $$\sum_{k=1}^{n} k \binom {n} {k}$$ by using the derivative of $(1+x)^n$, but I'm basically clueless. Can anyone give me a hint?</p>
| Tulip | 55,386 | <p>Hint : set $$f(x)=(1+x)^n=\sum_{k=0}^n \binom{n}{k}x^k$$ What you get when you evaluate $f'(1)$ ? </p>
<p>Note that you sum can begin from $0$ since the term with $k=0$ is $0$.</p>
|
512,590 | <p>According to the definition my professor gave us its okay for a matrix in echelon form to have a zero row, but a system of equations in echelon form cannot have an equation with no leading variable.</p>
<p>Why is this? Aren't they supposed to represent the same thing?</p>
| Robert Lewis | 67,071 | <p>Using de Moivre's formula $(\cos \alpha + i \sin \alpha)^n = \cos n\alpha + i \sin n\alpha$, we see that setting $z = r(\cos \alpha + i\sin \alpha)$, we have $z^{-1} = r^{-1}(\cos \alpha - i \sin \alpha)$; thus</p>
<p>$z + z^{-1} = r(\cos \alpha + i \sin \alpha) + r^{-1}(\cos \alpha - i \sin \alpha) = 2 \cos \alpha. \tag{1}$</p>
<p>This implies</p>
<p>$ r\cos \alpha + r^{-1}\cos \alpha = 2 \cos \alpha, \tag{2}$</p>
<p>as well as</p>
<p>$i(r\sin \alpha -r^{-1}\sin \alpha) = 0, \tag{3}$</p>
<p>this latter equation (3) since $2 \cos \alpha$ is real.</p>
<p>If $\cos \alpha \ne 0$, (2) yields</p>
<p>$r + r^{-1} = 2, \tag{4}$</p>
<p>or </p>
<p>$(r - 1)^2 = r^2 - 2r + 1 = 0; \tag{5}$</p>
<p>$r = 1$ and $z = \cos \alpha + \sin \alpha$; then $z^n = \cos (n\alpha) + i\sin (n\alpha)$, $z^{-n} = \cos (-n\alpha) + i\sin (-n\alpha) = \cos (n\alpha) - i\sin (n\alpha)$, so</p>
<p>$z^n + z^{-n} = 2 \cos n \alpha \tag{6}$</p>
<p>in this case.</p>
<p>In the event that $\cos \alpha = 0$, we have $\sin \alpha \ne 0$, and from (3) we infer</p>
<p>$r \sin \alpha = r^{-1} \sin \alpha, \tag{7}$</p>
<p>whence</p>
<p>$r^2 = 1, \tag{8}$</p>
<p>implying $r = 1$ since $r > 0$. Thus we still have $z = \cos \alpha + i\sin \alpha$ and (6) still follows as in the case $\cos \alpha \ne 0$. I credit Daniel Littlewood's comment for making clear to me the utility of equation (3) in this context.</p>
<p>Well, I <em>hope</em> that helps. Cheers,</p>
<p>and as always,</p>
<p><em><strong>Fiat Lux!!!</em></strong></p>
|
2,303,106 | <p>I was looking at this question posted here some time ago.
<a href="https://math.stackexchange.com/questions/1353893/how-to-prove-plancherels-formula">How to Prove Plancherel's Formula?</a></p>
<p>I get it until in the third line he practically says that $\int _{- \infty}^{+\infty} e^{i(\omega - \omega')t} dt= 2 \pi \delta(\omega - \omega')$.</p>
<p>I mean, I would understand if we were integrating over a period of length $2 \pi$, but here the integration is over $\mathbb{R}$. </p>
<p>P.S. I would have asked this directly to the author of the post, but it's been over a year since he last logged in.</p>
| reuns | 276,986 | <p>As you noted $\int _{- \infty}^{+\infty} e^{i(\omega - \omega')t} dt= 2 \pi \delta(\omega - \omega')$ is of course not true. This is an abuse of notation, what it really means is that the Fourier transform of the (tempered) distribution $f(\omega) = e^{i \omega' t}$ is the (tempered) distribution $\hat{f}(\omega) = 2 \pi \delta(\omega-\omega')$.</p>
<hr>
<p>In general, if $f$ is a tempered distribution, then its Fourier transform (in the sense of distributions) is the tempered distribution $\hat{f}$ iff for every Schwartz function $\varphi$ : </p>
<p>$$\int_{-\infty}^\infty f(t) \varphi(t)dt = \int_{-\infty}^\infty \hat{f}(\omega)\hat{\varphi}(\omega)d\omega$$</p>
<p>(where in general $\int_{-\infty}^\infty f(t) \varphi(t)dt$ is <strong>not</strong> a Riemann integral, but the pairing of a distribution with a test function)</p>
|
2,809,686 | <p>Let S={1,2,3,...,20}. Find the probability of choosing a subset of three numbers from the set S so that no two consecutive numbers are selected in the set.
"I am getting problem in forming the required number of sets."</p>
| Community | -1 | <p><strong><em>Inclusion-Exclusion Principle:-</em></strong><br>
Suppose $A_i$ denote the event that both the numbers $i,$ and $i+1$ ($i=1,2,...,19$) are included in the set.<br>
$n(\bigcup\limits_{i=1}^{19} A_i)$ will denote the total number of cases where we can find at least two consecutive integers. </p>
<p>Here, $n(\bigcup\limits_{i=1}^{19} A_i)=\sum_{i=1}^{19}n(A_i)-\sum \sum_{i\ne j}n(A_i\cap A_j)$<br>
[Terms like $n(A_i,A_j,A_k)$ and others vanishes] </p>
<p>Now, for an instance if you choose $1,2$ then there are $18$ other choices. So, $n(A_i)=18$, $i=1,2,....19$ </p>
<p>$n(A_i\cap A_j)$ just requires how many consecutive tuples like $(1,2,3),(2,3,4)$ you may have. Obviously this number is $18$<br>
So,$n(\bigcup\limits_{i=1}^{19} A_i)=18\times 19 -18=324$ </p>
<p>Required probability $=1-\frac{324}{\binom{20}{3}}=\frac{816}{\binom{20}{3}}$</p>
|
677,241 | <p>Let $A$ be a list of $n$ numbers in range $[1,100]$ (numbers can repeat).
I'm looking for the number of permutations of $A$ which start with a non-decreasing part, where this part ends with the first instance of the highest number, call this "index $i$" (1 based)from the left. After $i$, the remaining permutation is an arbitrary arrangement.</p>
<p><strong>Examples</strong> </p>
<ul>
<li><p>For $n=1$ and $A=\{9\}$, we have $1$ way only.</p></li>
<li><p>For $n=2$ and $A=\{2,5\}$, we have $2!$ ways in total.</p>
<ul>
<li>For $i=1$ we have $1$ way $(5,2)$ only.</li>
<li>For $i=2$ we have $1$ way $(2,5)$ only.</li>
</ul></li>
<li><p>If we were given , $A=\{5,5\}$, we have 2 ways totally:</p>
<ul>
<li>For $i=1$ we have one way, namely $(5,5)$, where the first $5$ coincides with first occurrence of $5$ in $A$, and</li>
<li>For $i=2$ we have one way, namely $(5,5)$, where the second $5$ coincides with first occurrence of $5$ in $A$</li>
</ul></li>
<li><p>For $n=3$ and $A=\{1, 4, 3\}$, we have $3!$ ways in total.</p>
<ul>
<li>For $i=1$, we have $2!$ permutations starting with $4$, namely $(4,3,1)$ and $(4,1,3)$.</li>
<li>For $i=2$ , $2!$ ways $(1,4,3)$ and $(3,4,1)$ in which $4$ is at $2$nd place.</li>
<li>For $i=3$ , $2!$ ways $(1,3,4)$ and $(3,1,4)$ ending with $4$.</li>
</ul></li>
</ul>
<p>To be exact : the number of permutations, without replacement, of given $n$ numbers such that the numbers up to $i$th place from left (say index) are in non-decreasing order, $i$-th place is occupied by first instance of largest number and rest can be random...</p>
<p>p.s. I cannot be more specific...</p>
| frabala | 53,208 | <p>Consider a finite sequence $s$ of length $n$ on integers that range from 0 to 100. Since repetition of numbers is allowed, let's denote with $t^a$ the number of occurrences of the number $a$ within $s$. We also denote with $l$ the largest number within $s$ (I guess it won't always be 100).</p>
<p>Now, assume a given $i$, where $i\leq n$. We want to permute $s$ in a way, that</p>
<ol>
<li>the first $i$ digits are in a non-descending order</li>
<li>the $i$-th digit is the first occurrence of the largest number available in $s$ and</li>
<li>the rest part of the permutation (from $i+1$ to $n$) is arbitrary.</li>
</ol>
<p>Let's see conditions (1) and (2): If from $s$ we select $i$ numbers (repetition allowed) where one of them is always the first occurrence of $l$, then this is the same as first choosing the first occurrence of $l$ and then choosing in addition $i-1$ numbers from the sequence produced by $s$ after removing the first occurrence of $l$. The selection of the $i$ numbers will constitute the ordered part of the permutation.</p>
<p>There is only one way to choose the first occurrence of $l$ and there are $\binom{n-1}{i-1}$ ways to choose the next $i-1$ numbers. Now, put $t_1^a$ to be the number of occurrences of $a$ within the selection of the $i-1$ numbers and $t^a_2=t^a-t^a_1$ (the number of occurrences of $a$ that are not in the selection). Finally, let $S$ be the <strong>set</strong> (not multiset) produced by the union of $l$ and the selection of the $i-1$ numbers. Then, for each selection of $i-1$ numbers in $s$, there are
$$\prod_{a\in S} t^a_1!$$
ways to order them in a non-descending manner. So, the first $i$ digits of the permutation can be constructed in $$\sum_{S\subseteq [1,100]~\text{and}~ |S|=i-1}\left[\binom{n-1}{i-1}\cdot\prod_{a\in S} t^a_1!\right]$$ ways. Note, the summation is not over $i$, but over all possible sets $S$. Actually, $S$ is a subset of $\{1,\ldots,100\}$ which always contains $l$ and has $i-1$ elements. Here $i$ is given and fixed.</p>
<p>From condition (3), for each selection of $i-1$ numbers in $s$, the rest of the permutation can be constructed in $$\left(\prod_{a\in S} t^a_2!\right)\cdot\left(\prod_{a\notin S} t^a!\right)$$
ways.</p>
<p>So totally, the number of permutations that satisfy the three properties is
$$\sum_{S\subseteq [1,100]~\text{and}~ |S|=i-1}\left[\binom{n-1}{i-1}\cdot\left(\prod_{a\in S} t^a_1!\right)\cdot\left(\prod_{a\in S} t^a_2!\right)\cdot\left(\prod_{a\notin S} t^a!\right)\right]$$</p>
<p>Finally, if you want to compute this for all $i=1,\ldots,n$, you just sum the last expression over $i$.
$$\sum_{i=1}^n\left(\sum_{S\subseteq [1,100]~\text{and}~ |S|=i-1}\left[\binom{n-1}{i-1}\cdot\left(\prod_{a\in S} t^a_1!\right)\cdot\left(\prod_{a\in S} t^a_2!\right)\cdot\left(\prod_{a\notin S} t^a!\right)\right]\right)$$</p>
|
1,274,816 | <p>It seems known that there are infinitely many numbers that can be expressed as a sum of two positive cubes in at least two different ways (per the answer to this post: <a href="https://math.stackexchange.com/questions/1192338/number-theory-taxicab-number">Number Theory Taxicab Number</a>).</p>
<p>We know that</p>
<p>$$1729 = 10^3+9^3 = 12^3 + 1^3,$$</p>
<p>and I am wondering if there are infinitely many numbers like this that can be expressed as the sum of two positive cubes in <em>exactly</em> two ways?</p>
<p>In fact, are there even any other such numbers?</p>
<p>EDIT:
As provided by MJD in the comments section, here are other examples:
$$4104 = 2^3+16^3 = 9^3+15^3,$$
$$13832 = 20^3+18^3=24^3+2^3,$$
$$20683 = 10^3 +27^3 = 19^3 +24^3.$$ </p>
| Tito Piezas III | 4,781 | <p>In the paper <em><a href="https://cs.uwaterloo.ca/journals/JIS/VOL6/Broughan/broughan25.pdf" rel="noreferrer">Characterizing the Sum of Two Cubes</a></em>, Kevin Broughan gives the relevant theorem,</p>
<p><strong><em>Theorem:</em></strong> Let $N$ be a positive integer. Then the equation $N = x^3 + y^3$ has a solution in positive integers $x,y$ if and only if the following conditions are satisfied:</p>
<ol>
<li><p>There exists a divisor $m|N$ with $N^{1/3}\leq m \leq (4N)^{1/3}.$ </p></li>
<li><p>And $\sqrt{m^2-4\frac{m^2-N/m}{3}}$ is an integer.</p></li>
</ol>
<p>The sequence of integers $F(n)$,</p>
<p>$$\begin{aligned}
F(n)
&= a^3+b^3 = (2 n + 6 n^2 + 6 n^3 + n^4)^3 + (n + 3 n^2 + 3 n^3 + 2 n^4)^3\\
&= c^3+d^3 = (1 + 4 n + 6 n^2 + 5 n^3 + 2 n^4)^3 + (-1 - 4 n - 6 n^2 - 2 n^3 + n^4)^3
\end{aligned}$$</p>
<p>for integer $n>3$ <em>apparently</em> is expressible as a sum of two <em>positive</em> integer cubes in <em>exactly</em> and <em>only</em> two ways. </p>
<p>$$\begin{aligned}
F(4) &= 744^3+756^3 = 945^3+15^3\\
F(5) &= 1535^3+1705^3 = 2046^3+204^3\\
&\;\vdots\\
F(60) &=14277720^3+26578860^3 = 27021841^3+12506159^3
\end{aligned}$$</p>
<p>Using Broughan's theorem, I have tested $F(n)$ from $n=4-60$ and, <em>per $n$, it has only two solutions $m$</em>, implying in that range it is a sum of two cubes in only two ways. Can somebody with a faster computer and better code test it for a higher range and see when (if ever) the proposed statement breaks down? Incidentally, we have the nice relations,</p>
<p>$$a+b = 3n(n+1)^3$$</p>
<p>$$c+d = 3n^3(n+1)$$</p>
<p><strong><em>Note</em></strong>: $F(60)$ is already <em>much</em> beyond the range of <a href="http://oeis.org/A011541" rel="noreferrer">taxicab</a> $T_3$ which is the smallest number that is the sum of two positive integer cubes in <strong><em>three</em></strong> ways.</p>
<p>$$T_3 \approx 444.01^3 = 167^3+436^3 = 228^3+423^3 = 255^3+414^3$$</p>
<p>(Using the theorem, this yields 3 values for $m$.)</p>
|
3,792,954 | <p>For vector space <span class="math-container">$V$</span> and <span class="math-container">$v \in V$</span>, there is a natural identification <span class="math-container">$T_vV \cong V$</span> where <span class="math-container">$T_vV$</span> is the tangent space of <span class="math-container">$V$</span> at <span class="math-container">$v$</span>. Can somebody explain this identification to me? Thank you</p>
| Patrick Stevens | 259,262 | <p>Differentiate wrt (wlog) <span class="math-container">$x$</span> to show that it's increasing in <span class="math-container">$x$</span> when <span class="math-container">$|y| < 1$</span>; then evaluate at <span class="math-container">$x=-1$</span> and at <span class="math-container">$x=1$</span> to demonstrate the bound.</p>
|
2,022,566 | <p>How can I calculate the below limit?
$$
\lim\limits_{x\to \infty} \left( \mathrm{e}^{\sqrt{x+1}} - \mathrm{e}^{\sqrt{x}} \right)
$$
In fact I know should use the L’Hospital’s Rule, but I do not how to use it.</p>
| Olivier Oloa | 118,798 | <p>One may write, as $x \to \infty$,
$$
\begin{align}
e^{\sqrt{x+1}} - e^{\sqrt{x}}&=e^{\sqrt{x}}\left(e^{\sqrt{x+1}-\sqrt{x}} - 1\right)
\\\\&=e^{\sqrt{x}}\left(e^{\frac1{\sqrt{x+1}+\sqrt{x}}} - 1\right)
\\\\&=e^{\sqrt{x}}\left(1+\frac1{\sqrt{x+1}+\sqrt{x}}+O\left(\frac1x\right) - 1\right)
\\\\& \sim \frac{e^{\sqrt{x}}}{2\sqrt{x}}
\end{align}
$$ which goes to $\infty$.</p>
|
64,925 | <p>Suppose $G$ is a group and $V$ an irreducible representation of $G$. One has that $V\otimes V\cong \Lambda^2(V)\oplus Sym^2(V)$. It is well-known that if the trivial representation appears as a subrepresentation of $\Lambda^2(V)$ then $V$ is of quaternionic type; while if the trivial representation appears as a subrepresentation of $Sym^2(V)$ then $V$ is a of real type. From this approach, it is clear that the trivial representation cannot appear in both $\Lambda^2(V)$ and $Sym^2(V)$.</p>
<p>What I am curious about is as follows:</p>
<blockquote>
<blockquote>
<p><b>Question:</b> Is there is some (relatively easy) way to see why the trivial representation cannot appear in both $\Lambda^2(V)$ and $Sym^2(V)$ without introducing the machinery of real/quaternionic types? </p>
</blockquote>
</blockquote>
<p>As a bit of motivation, if one looks at other subrepresentations, then for example if $G = G_2$ and $V_n$ is an $n$-dimensional irreducible representation of $G_2$, then $V_{64}$ appears as a subrepresentation of both $\Lambda^2(V_{27})$ and $Sym^2(V_{27})$. In particular it is possible for the intertwining number of $\Lambda^2(V)$ and $Sym^2(V)$ to be nonzero, but by the real vs. quaternionic characterization, the trivial representation is somehow special in that it cannot contribute to the intertwining number.</p>
| Andreas Blass | 6,794 | <p>This is essentially what Darij wrote, but without mentioning the bilinear forms. (I had written it out before reading far enough into Darij's post to see that he was really doing the same thing, after the part about bilinear forms.) Think of $V\otimes V$ as $\text{Hom}(V^*,V)$. An occurrence of the trivial representation in $V\otimes V$ thus amounts to a $G$-equivariant linear map from $V^*$ to $V$. Since $V$ and therefore also $V^*$ are irreducible, Schur's lemma says that the space of such maps has dimension either 1 (iff $V$ and $V^*$ are isomorphic) or 0. So there's at most one occurrence of the trivial representation in $V\otimes V$.</p>
|
959,201 | <p>I am confused about the following.</p>
<p>Could you explain me why if $A=\varnothing$,then $\cap A$ is the set of all sets?</p>
<p>Definition of $\cap A$:</p>
<p>For $A \neq \varnothing$:</p>
<p>$$x \in \cap A \leftrightarrow (\forall b \in A )x \in b$$</p>
<p><strong>EDIT</strong>:</p>
<p>I want to prove that $\cap \varnothing$ is not a set.</p>
<p>To do that, do I have to begin, supposing that it is a set?</p>
| John Hughes | 114,036 | <p>I believe that the statement you want us to prove is wrong. For one thing, there is (in the versions of set theory that I know) no set $R$ of all sets, for if there were, you could form
$$
Q = \{X \in R : X \notin X\}
$$
the set of all sets that are not elements of themselves. The statement $Q \in Q$ then becomes neither true nor false, which is problematic. </p>
|
476,095 | <p>I am attempting to learn about mathematical proofs on my own and this is where I've started. I think I can prove this by induction. Something like:</p>
<p>$n = 2k+1$ is odd by definition</p>
<p>$n = 2k+1 + 2$ (this is where I'm stuck, how do I show that this is odd?)</p>
<p>$n = 2(k+1) + 1$ (if I can show that it's odd, I can do the same here and prove my conjecture by induction, right?)</p>
<p>Thanks for any assistance</p>
| walcher | 89,844 | <p>An integer $n$ is odd if and only if it is not divisible by $2$ or again if and only if it is of the form $2k+1$ for some integer $k$. If $n=2k+1$ is odd, then $n+2=2k+1+2=2k+2+1=2(k+1)+1$ is obviously odd as well.</p>
|
3,144,813 | <blockquote>
<p>Let <span class="math-container">$X : \mathbb{R} \to \mathbb{R}^n$</span> be a <span class="math-container">$C^1$</span> function. Let <span class="math-container">$\| .\|$</span> be the norm : <span class="math-container">$\| v \| = \max_{1 \leq i \leq N} \mid v_i \mid$</span>. Then is it true that :
<span class="math-container">$$\| X'(t) \| = (\| X(t) \|)'$$</span> ?</p>
</blockquote>
<p>I am wondering if in general if I have any function <span class="math-container">$f : \mathbb{R}^n \to \mathbb{R}^p$</span> and a norm <span class="math-container">$N$</span> on a : <span class="math-container">$\mathbb{R}^p$</span> then is it always possible to invert the norm and the differential operator or the norm and in the integral?</p>
<p>Thank you. </p>
| GReyes | 633,848 | <p><span class="math-container">$x=s$</span>, <span class="math-container">$y=0$</span> is just the <span class="math-container">$X$</span>-axis. On that axis you have an initial datum <span class="math-container">$f(x)$</span>. Thus your characteristics will be issued from each point of the form <span class="math-container">$(s,0,f(s))$</span> in the <span class="math-container">$XYZ$</span> space. Their union forms a surface <span class="math-container">$z=z(x,y)$</span> which is the desired solution.</p>
|
2,330,514 | <p>Prove if n is a perfect square, n+2 is not a perfect square</p>
<blockquote>
<p>Assume n is a perfect square and n+2 is a perfect square (proof by
contradiction)</p>
<p>There exists positive integers a and b such that $n = a^2$ and $n + 2= b^2$</p>
<p>Then $a^2 + 2 = b^2$</p>
<p>Then $2 = b^2-a^2$</p>
<p>Then $2 = (b-a)(b+a)$</p>
<p>Then $b+a = 2$ and $b-a=1$ <strong><em>where does this come from?</em></strong></p>
<p>Then $a = 1/2$ and $b=3/2$</p>
<p>This is a contradiction because a and b should be positive integers.
Therefore if n is a perfect square, n+2 is not a perfect square.</p>
</blockquote>
<p>Where does the $b+a = 2$ and $b-a=1$ come from?</p>
| Ross Millikan | 1,827 | <p>You have to factor $2$ and the only factorization into two terms is $2 \cdot 1$. Since $a,b$ are positive, $b+a \gt b-a$ so we take $b+a=2, b-a=1$</p>
|
3,372 | <p>Have any questions first proposed on Mathoverflow attracted enough interest from experts in their field that solving them would be considered a significant advance?</p>
<p>I don't want to count problems that are known (or strongly suspected) to be at least as hard as some previously described problem, unless the version original to Mathoverflow is believed by experts to be a better formulation of the problem.</p>
<p>Of course, as Mathoverflow has been around for less than 10 years, no problem original to Mathoverflow can possibly be a long-standing open problem in its field. But I don't see any reason a question asked on MO can't be among the most interesting questions asked in the last ten years.</p>
| Martin Sleziak | 8,250 | <p>As far as I know, there is no such thing. See also this post on Mathematics Meta.
<a href="https://math.meta.stackexchange.com/q/12374">Number of people online</a>. On main meta you can find some feature requests such as <a href="https://meta.stackexchange.com/q/2631">View approximate number of users online</a> or <a href="https://meta.stackexchange.com/q/52720">How many SO users are online?</a>.</p>
<p>Depending on what purpose you want this for, perhaps some other measures might be useful for you. </p>
<p>On Math.SE you can see patterns of <a href="https://chat.stackexchange.com/rooms/info/36/mathematics">chat messages posted per day/per week</a> in the main chatroom - which might correlate with users online. This will not work on MathOverflow, since chat here is rather inactive, but somebody with sufficient knowledge of SQL should be able to make <a href="https://meta.mathoverflow.net/tags/data-explorer/info">a SEDE query</a> showing number of posts in various time slots - which also seems a reasonable approximation. Maybe you can use query from answer to this question on meta.SE: <a href="https://meta.stackexchange.com/q/1313">What is the best time to ask questions?</a> (Do not forget to switch site to MathOverflow, when you try the query.)</p>
<p>You can also find some already existing queries (and try them also for MathOverflow) in these posts from Mathematics Meta: <a href="https://math.meta.stackexchange.com/q/16254">Best time to ask a question</a>, <a href="https://math.meta.stackexchange.com/q/3865">Depending on the tag is there a possibility to know on which days and hours are there more answers?</a>, <a href="https://math.meta.stackexchange.com/q/26906">When is peak time for this website?</a>. </p>
<p>Users with <a href="https://mathoverflow.net/help/privileges/site-analytics">25k+ rep</a> have access to site analytics. Here you can learn a bit about traffic on site, but AFAIK you do not see there breakdown by the time of day/day of week.</p>
|
784,258 | <p>I understand that this is an induction question. </p>
<p>I start with the base case (n=1):</p>
<p>$$1 < 2 \tag{That works!}$$</p>
<p>Induction step: Assume the statement works for all $n = k$, Prove for all $n = k+1$</p>
<p>Assume $1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}}+ ... +\frac{1}{\sqrt{k+1}}\le 2\sqrt{k+1}$</p>
<p>I'm a bit confused as to where to go next, may I please have some assistance?</p>
| Pedro | 23,350 | <p><em>Hint</em> We have that $(2 x^{1/2})'=x^{-1/2}$. Now, think about $$\int_1^n x^{-1/2}dx$$</p>
|
25,363 | <p>In what way and with what utility is the law of excluded middle usually disposed of in intuitionistic type theory and its descendants? I am thinking here of topos theory and its ilk, namely synthetic differential geometry and the use of topoi in algebraic geometry (this is a more palatable restructuring, perhaps), where free use of these "¬⊨P∨¬P" theories is necessarily everywhere--freely utilized at every turn, one might say. But why and how are such theories first formulated, and what do they look like in the purely logical sense?</p>
<p>You will have to forgive me; I began as a student in philosophy (not even that of mathematics), and the law of excluded middle is something that was imbibed with my mother's milk, as it were. This is more of a philosophical issue than a mathematical one, but being the renaissance guys/gals that you all are, I thought that perhaps this could generate some fruitful discussion. </p>
| Wouter Stekelenburg | 3,603 | <p>In a topos, the question is not whether a sentence is true or false, but <em>where</em> it is true, because toposes--at least the geometric ones you're talking about--are generalized spaces. There can be limit points where this question cannot be decided for all sentences. The problem is then that there are other points within any distance (more accurately: any open neighborhood) both where a sentence is true and where the same sentence is false.</p>
|
25,363 | <p>In what way and with what utility is the law of excluded middle usually disposed of in intuitionistic type theory and its descendants? I am thinking here of topos theory and its ilk, namely synthetic differential geometry and the use of topoi in algebraic geometry (this is a more palatable restructuring, perhaps), where free use of these "¬⊨P∨¬P" theories is necessarily everywhere--freely utilized at every turn, one might say. But why and how are such theories first formulated, and what do they look like in the purely logical sense?</p>
<p>You will have to forgive me; I began as a student in philosophy (not even that of mathematics), and the law of excluded middle is something that was imbibed with my mother's milk, as it were. This is more of a philosophical issue than a mathematical one, but being the renaissance guys/gals that you all are, I thought that perhaps this could generate some fruitful discussion. </p>
| Frank Quinn | 124,645 | <p>A version of Russell's paradox provides a failure of "law of excluded middle". I don't know if this directly answers your question, but it illustrates the constraints on getting this "law" to work.</p>
<p>Define a "logical function" to be one that returns 'yes' or 'no' when given an input. "law of excluded middle" asserts that if the 'yes' inputs are specified then assigning 'no' to everything else defines a logical function. Consider </p>
<blockquote>
<p>$R[f]$:= If( $f$ is a logical function) then not($f[f]$), else 'no')</p>
</blockquote>
<p>But $R$ cannot be a logical function, for if it were we would have $R[R]= not(R[R]$. The error must be that "($f$ is a logical function)" is not a logical function. If it is not 'true' then it is 'wrong' in the sense that the reasoning is invalid. Usually this is seen as invalidating "unlimited comprehension" (naive set theory). </p>
<p>Most mathematics requires excluded-middle arguments, so we interpret this problem as showing inputs have to be restricted for these arguments to work. In a sense the job of set theory is to provide safe contexts for excluded-middle arguments. This also suggests why set theory is complicated: we cannot describe "safe contexts" using logical functions. </p>
|
101,098 | <p>I apologize in advance because I don't know how to enter code to format equations, and I apologize for how elementary this question is. I am trying to teach myself some differential geometry, and it is helpful to apply it to a simple case, but that is where I am running into a wall.</p>
<p>Consider $M=\mathbb{R}^2$ as our manifold of interest. I believe that the tangent space is also $\mathbb{R}^2$. From linear algebra, we know that a basis set for $\mathbb{R}^2$ is $$\left\{\left[\matrix{1\\0}\right], \left[\matrix{0\\ 1}\right] \right\}\;.$$</p>
<p>Now, from differential geometry, we are told that basis vectors are $\frac{d}{dx}$ and $\frac{d}{dy}$ where the derivatives are partial deravatives.</p>
<p>So my question is how does one obtain a two-component basis vector of linear algebra from a simple partial derivative?</p>
<hr>
<p>EDIT: Thanks to everyone for the replies. They have been very helpful, but thinking as a physicist, I would like to see how the methods of differential geometry could be used to derivive the standard basis of linear algebra. It seems that there must be more to it than saying that there is an isomorphism between the space of derivatives at a point and R^n which sets up a natural correspondence between the basis vectors.</p>
<p>I may be completely off-the-wall wrong, but somehow I think that the answer involves partial derivatives of a local orthognal coordinate system at point p.</p>
| Pierre-Yves Gaillard | 660 | <p>Here is, for what it's worth, how I see this. </p>
<p>In differential geometry, it is sometimes convenient to denote the canonical basis of $\mathbb R^n$ by
$$
\left(\frac{\partial}{\partial x_i}\right)_{i=1,\dots,n}\quad,
$$
and the dual basis (which is a basis of $(\mathbb R^n)^*$) by
$$
(dx_i)_{i=1,\dots,n}.
$$
So, in my modest opinion, it's just a matter of notation, nothing more.</p>
<p><strong>EDIT.</strong> Here is how I understand the question asked by the OP: </p>
<p>Why is the $i$-th vector of the canonical basis of $\mathbb R^n$ equal to
$$
\frac{\partial}{\partial x_i}\quad?
$$
The answer is: By definition. </p>
<p>Of course, this answer prompts a second question, which is:</p>
<p>What's the reason for such a bizarre definition? </p>
<p>Dylan answered this (deeper) question perfectly, but I think it is important that the OP understands the difference between these two questions. This is why I tried to insist (perhaps awkwardly) on the aspect that was not explicitly covered by Dylan's answer (which I upvoted --- I also upvoted the question).</p>
|
1,076,974 | <p>Does anyone know how I can determine the equation of the 3D object below? (Maybe there's a program that can do it?) I am looking for a formula to define this 3D object, but am having trouble finding one. </p>
<p>(If you can imagine the 2D object you see revolved about the x-axis, that is the 3D object I'm referring to.) Btw the z-axis can't be seen because the view of the object is head-on and the object is symmetrical with respect to the x-axis. Thank you.</p>
<p><img src="https://i.stack.imgur.com/q5dFT.jpg" alt="enter image description here"></p>
| coffeemath | 30,316 | <p>Suppose the top of the sketched curve has equation $y=f(x)$ for $0 \le x \le a$ and as in the diagram $f(x) \ge 0$ for $x$ in $[0,a].$ The distance from a 3-d point $(x,y,z)$ to the point $(x,0,0),$ in the plane having constant first coordinate $x,$ is $\sqrt{y^2+z^2},$ and in the revolved figure you want this distance to be $f(x).$</p>
<p>So an equation for the surface of the revolved curve is
$$\sqrt{y^2+z^2}=f(x),$$
or more simply, since $f(x) \ge 0,$ it can be written as
$$y^2+z^2=f(x)^2,$$
for $x$ satisfying $0 \le x \le a.$</p>
|
1,600,597 | <p>I'm currently going through Spivak's calculus, and after a lot of effort, i still can't seem to be able to figure this one out.</p>
<p>The problem states that you need to prove that $x = y$ or $x = -y$ if $x^n = y^n$</p>
<p>I tried to use the formula derived earlier for $x^n - y^n$ but that leaves either $(x-y) = 0$ or $(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})$ and i'm not sure how to proceed from there.</p>
| fleablood | 280,126 | <p>I'm assuming you are assuming we are dealing with real numbers. (The result is not true in complex numbers as $i^4 = 1^4 = (-1)^4 = (-i)^4$).</p>
<p>Now has it been shown to your satisfaction that every $b > 0$ has a unique positive $n-th$ root? Well... okay, let's do some basics</p>
<p>Okay, suppose $0 < |x| < |y|$ then $|x|^n < |y|^n$. (And similarly if $0 < |y| < |x|)$. </p>
<p>[Because: if $|x| < |y|$ then $|x|^2 = |x||x| < |x||y| < |y||y| = |y|^2$ and inductively $|x|^{n-1} < |y|^{n-1} \implies |x|^{n-1}|x| < |y|^{n-1}|y|$]</p>
<p>Thus $|x|^n = |y|^n$ if and only if $|x| = |y|$.</p>
<p>Then as $x = \pm |x|$ and $y = \pm |y|$ so $x^n = (\pm 1)^n|x|^n$ (and same thing for $y^n$). </p>
<p>As $n$ is even $(\pm 1)^n = 1$.</p>
<p>So $x^n = y^n \iff |x| = |y| \iff x = \pm y$.</p>
<p>===</p>
<p>I haven't actually shown that if $b > 0$ that there actually <em>exists</em> an $x$ such that $x^n = b$. But rereading the problem I see I don't have to. The thing though is to realize if there <em>is</em> such a positive $x$ it is unique. Which I can do by showing $x \ge \lt y; x>0;y>0 \implies x^n \ge \lt y^n$.</p>
|
3,852,952 | <p>Given a projective space <span class="math-container">$\mathbb{P}^n(\mathbb{C})$</span>, I can consider the Grasmannian of lines <span class="math-container">$G(2,n+1)$</span>, which has a structure of projective variety inside <span class="math-container">$\mathbb{P}^N$</span>, where <span class="math-container">$N=\binom{2+n+1}{2}-1$</span> thanks to the Pl<span class="math-container">$\ddot{u}$</span>cker embedding.</p>
<p>It has been told me -literally, I have no references, it was a speech- that while a point <span class="math-container">$p\in G(2,n+1)$</span> obviously represents a line in <span class="math-container">$\mathbb{P}^n$</span> by definition, a <em>line</em> inside <span class="math-container">$G(2,n+1)$</span> -viewd as a projective variety in <span class="math-container">$\mathbb{P}^N$</span>-corresponds to a pencil of planes.</p>
<p>Unfortunately I'm still trying to properly understand why this work -because unfortunately it is not crystal for me -, but I was also wondering if it makes sense -that is, if there is a geometrical meaning-, also for quadrics contained in <span class="math-container">$G(2,n+1)$</span>.</p>
<p>Therefore I'd like to have an idea of what happens for quadrics, and moreover even a little help for lines in <span class="math-container">$G(2,n+1)$</span>.</p>
<p>Since this question comes primarily from my curiosity, I apologize for the vagueness: any comment, reference or answer would be much appreciate!</p>
| xxxxxxxxx | 252,194 | <p>You have that lines in <span class="math-container">$\mathbb{P}^{n}(\mathbb{C})$</span> correspond to points of the Grassmannian <span class="math-container">$G(2,n+1)$</span>. Now when two lines <span class="math-container">$\ell_{1}$</span>, <span class="math-container">$\ell_{2}$</span> of the projective space intersect each other at a point <span class="math-container">$x$</span>, you can check the Plücker coordinates and verify that the corresponding points of the Grassmannian span a line that is completely contained in the Grassmannian. Furthermore, under the inverse map, all of the points on this line of the Grassmannian correspond to lines in <span class="math-container">$\mathbb{P}^{n}(\mathbb{C})$</span> lying in a common plane (you should verify this using the Plücker map).</p>
<p>So what you actually should be saying is that a line completely contained in <span class="math-container">$G(2,n+1)$</span> corresponds to a pencil of <strong>lines</strong> in <span class="math-container">$\mathbb{P}^{n}(\mathbb{C})$</span>, that is, the collection of all lines through a fixed point <span class="math-container">$x$</span> lying in a common plane.</p>
<p>These intersection properties carry over to other kinds of quadrics, though I am not an expert at all of the correspondences. I do know that for example a conic in <span class="math-container">$G(2,4)$</span> will correspond to one of the rulings of a hyperbolic quadric in <span class="math-container">$\mathbb{P}^{3}(\mathbb{C})$</span>.</p>
|
926,804 | <p>Is there a word for the quality of a number to be either positive or negative? Consider this question:</p>
<p><em>What's the ... (sign/positivity/negativity, but a word that could describe either) of number <strong>x</strong>?</em></p>
<p>Also, is there an all-encompassing word for the sign put in front of a number (-5 or +5)? Word that describes both a plus and a minus sign?</p>
| Piquito | 219,998 | <p>If you want to have a "less-simple" answer than the correct one given by @k170, the words positive and negative occur in all totally ordered commutative group with neutral element noted $0$. The positive elements $x$ are those for which $0<x$ and the negative ones are those such that $x<0$. The usual convention of the signs comes from $$x+(-x)=0$$ where $-x$ is the opposite of $x$. Given $x$ the positive part of $x$ is defined by $x_+=\sup\{x,0\}$ and the negative part is $x_-=\sup\{-x,0\}$ so you have $$x=(x_+)-(x_-)$$ Noting now $$|x|=\sup (x,-x)$$ (it is not an absolute value!) you have $$|x|=(x_+)+(x_-)$$
You can translate this to the numbers an take as convention (very useful!)the word "sign", positive or negative according to the involved element.</p>
|
643,560 | <p>Let $\{x_n\}_{n=1}^{\infty}$ and $\{y_n\}_{n=1}^{\infty}$ be sequences of real numbers. Does the following hold:</p>
<p>$$
\limsup x_n +\liminf y_n \le \limsup\,(x_n+y_n).
$$ </p>
<p>This is what I have tried but I am not quite sure if it is correct.
$\text{Fix } K>1. \text{ Let }L=\inf_{1\le i \le k}y_i$. Now
$$
\sup_{1 \le i \le k}(x_i+y_i)\ge \sup_{1\le i \le k}(x_i+L)=L+\sup_{1\le i \le k}(x_i) =
\inf_{1\le i \le k}(y_i)+\sup_{1\le i \le k}(x_i).
$$
Now we take the limit as $k \rightarrow \infty$ and we have the desired result. Does this look OK?</p>
| nmasanta | 623,924 | <p>Let <span class="math-container">$~\limsup x_n=\overline x~$</span> and <span class="math-container">$~\liminf y_n=\underline y~,$</span> then <span class="math-container">$~\overline x~,~\underline y~\in \mathbb R.$</span> <br>
Let <span class="math-container">$~\epsilon>0~$</span> be given.<br>
Then <span class="math-container">$~\exists~~k\in \mathbb N~$</span> such that <br><span class="math-container">$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x_n>\overline x-\epsilon~/~2~,~~\forall~~n\ge k~$</span> and <br><span class="math-container">$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~y_n>\underline y-\epsilon~/~2~,$</span> for infinitely many value of <span class="math-container">$~n~.$</span><br>
Thus <span class="math-container">$~x_n+y_n>\overline x+\underline y-\epsilon~,$</span> for infinitely many value of <span class="math-container">$~n~.$</span><br>
Therefore for given <span class="math-container">$~m\in\mathbb N~,~~\exists~~n_0\ge m~$</span> such that <span class="math-container">$$x_{n_0}+y_{n_0}>\overline x+\underline y-\epsilon$$</span>
<span class="math-container">$$\implies \sup_{n\ge m}\left(x_n+y_n\right)\ge x_{n_0}+y_{n_0}>\overline x+\underline y-\epsilon~,$$</span> for each <span class="math-container">$~m\in \mathbb N~.$</span><br>
Hence <span class="math-container">$$\overline x+\underline y-\epsilon\le\limsup\,(x_n+y_n)$$</span>
Since <span class="math-container">$~\epsilon > 0~$</span> be arbitrary, <span class="math-container">$$\limsup x_n +\liminf y_n \le \limsup\,(x_n+y_n)~.$$</span></p>
|
3,450,713 | <p>Let <span class="math-container">$a+3b=7$</span> and <span class="math-container">$c=3$</span>. Then value of <span class="math-container">$a+3(b+c)$</span> is</p>
<p>A) <span class="math-container">$10$</span></p>
<p>B) <span class="math-container">$16$</span> </p>
<p>C) <span class="math-container">$21$</span></p>
<p>D) <span class="math-container">$30$</span></p>
<p>Answer is B, but how?</p>
| JMoravitz | 179,297 | <p>We are told that <span class="math-container">$\color{blue}{a+3b} = 7$</span> and that <span class="math-container">$\color{red}{c}=3$</span></p>
<p>We are asked to find <span class="math-container">$a+3(b+c)$</span></p>
<p><span class="math-container">$$\begin{array}{rll}a+3(b+c) &=a+(3b+3c)&\text{distributive property}\\
&=(a+3b)+3c&\text{associative property}\\
&=(\color{blue}{a+3b})+3\color{red}{c}&\text{recognition}\\
&=7 + 3\cdot 3&\text{replacement}\\
&=7+9=16&\text{arithmetic simplification}\end{array}$$</span></p>
|
186,146 | <p>On a finite dimensional vector space, the answer is yes (because surjective linear map must be an isomorphism). Does this extend to infinite dimensional vector space? In other words, for any linear surjection $T:V\rightarrow V$, AC guarantees the existence of right inverse $R:V\rightarrow V$. Must $R$ be linear?</p>
<p>How about $T:V\rightarrow W$ linear surjection in general?</p>
| Qiaochu Yuan | 232 | <p>No. Let $V = \text{span}(e_1, e_2, ...)$ and let $T : V \to V$ be given by $T e_1 = 0, T e_i = e_{i-1}$. A right inverse $S$ for $T$ necessarily sends $v = \sum c_i e_i$ to $\sum c_i e_{i+1} + c_v e_1$ but $c_v$ may be an arbitrary function of $v$. </p>
|
2,217,454 | <p>Let $\{ x_i : i \in I \}$ be a family of numbers $x_i \in \mathbb R$ with $I$ an arbitrary index set. We say that this family is summable with value $s$ (and write $s = \sum_{i \in I} x_i$ then) if for every $\varepsilon > 0$ there exists some finite set $I_{\varepsilon}$ such that for every finite superset $J \subseteq I$, i.e. such that $I_{\varepsilon} \subseteq J$, we have
$$
\left| \sum_{i \in J} x_i - s \right| < \varepsilon.
$$</p>
<p>Does there exists a family of numbers $\{x_i : i \in I\}$ with uncountable $I$ such that $\sum_{i \in I} x_i = 1$ and such that for every countable $J \subseteq I$ we have
$$
\sum_{j \in J} x_j < 1
$$
i.e. the countable "sub"-sums have a strictly smaller value?</p>
| Roberto Rastapopoulos | 388,061 | <p>Assume that
$$ \sum_{i\in I} |x_i| < \infty, $$
and define $J_n = \{i \in I: |x_i| \geq 1/n$}, for $n > 0$, and $J_{\infty} = \{i\in I: x_i \neq 0\}$. By assumption, $J_n$ must be a finite set for all $n$, so
$J_\infty = \bigcup_{n = 1}^{\infty}J_n$ is countable.</p>
|
1,478,314 | <p>In this particular case, I am trying to <strong>find all points $(x,y)$ on the graph of $f(x)=x^2$ with tangent lines passing through the point $(3,8)$</strong>. </p>
<p>Now then, I know the <a href="http://www.meta-calculator.com/online/?panel-102-graph&data-bounds-xMin=-10&data-bounds-xMax=10&data-bounds-yMin=-7.28&data-bounds-yMax=7.28&data-equations-0=%22y%3Dx%5E2%22&data-rand=undefined&data-hideGrid=false" rel="nofollow">graph</a> of $x^2$. What now?</p>
| Bernard | 202,857 | <p>Either you write the equation of a tangent at $(x_0,x_0^)$2 and check under which condition this equation is satisfied by the $(3,8)$.</p>
<p>Or you write the equation of a line with slope $t$ passing through the point $(3,8)$: $$y-8=t(x-3),$$
and find under which condition this line has a double intersection with the parabola:
$$x^2-8=t(x-3)\iff x^2 -tx+3t-8=0$$
One has a double root if and only if its discriminant:
$$\Delta=t^2-12t+32=(t-6)^2-4=0\iff t=\begin{cases}4\\8\end{cases}$$
Th abscissée ot the points of contact are the double roots: $\dfrac t2=2,4$, and their ordinates $4,16$.</p>
|
2,114,276 | <p>How to show that $(x^{1/4}-y^{1/4})(x^{3/4}+x^{1/2}y^{1/4}+x^{1/4}y^{1/2}+y^{3/4})=x-y$</p>
<p>Can anyone explain how to solve this question for me? Thanks in advance. </p>
| Sarvesh Ravichandran Iyer | 316,409 | <p>Let <span class="math-container">$x = u^4$</span> and <span class="math-container">$y = v^4$</span>, then the left hand side of the equation simplifies to:
<span class="math-container">$$
(u-v)(u^3 + u^2v + uv^2 + v^3)
$$</span></p>
<p>Now, you can choose to multiply directly:
<span class="math-container">$$
u^4 + \color{red}{u^3v} + \color{green}{u^2v^2} + \color{blue}{uv^3} - \color{red}{vu^3} - \color{green}{v^2u^2} - \color{blue}{v^3u} - v^4
$$</span>
Terms of similar colour cancel out, and you get the result after putting back <span class="math-container">$x$</span> and <span class="math-container">$y$</span>.</p>
|
191,210 | <p>Let $R$ be the smallest $\sigma$-algebra containing all compact sets in $\mathbb R^n$.
I know that based on definition the minimal $\sigma$-algebra containing the closed (or open) sets is the Borel $\sigma$-algebra. But how can I prove that $R$ is actually the Borel $\sigma$-algebra?</p>
| Michael Greinecker | 21,674 | <p>It is enough to show that every closed set is in the $\sigma$-algebra. So let $C$ be closed, $x$ be an arbitrary point and $K_n$ the closed ball with center $x$ and radius $n$. Then $K_n\cap C$ is compact for all $n$ and $\bigcup_n (K_n\cap C)=C$.</p>
|
1,647,157 | <p>How can I solve this using only 'simple' algebraic tricks and asymptotic equivalences? No l'Hospital.</p>
<p>$$\lim_{x \rightarrow0}
\frac
{\sqrt[3]{1+\arctan{3x}} - \sqrt[3]{1-\arcsin{3x}}}
{\sqrt{1-\arctan{2x}} - \sqrt{1+\arcsin{2x}}}
$$</p>
<p>Rationalizing the numerator and denominator gives</p>
<p>$$
\lim_{x \rightarrow0}
\frac
{A(\arctan{3x}+\arcsin{3x})}
{B(\arctan{2x} + \arcsin{2x})}
$$
where $\lim_{x \rightarrow 0} \frac{A}{B} = -\frac{2}{3} $</p>
| runaround | 310,548 | <p>Using your $g(t) = t - \frac{\pi|{2}, for\, t > 0$, and $g(t) = t +\frac{\pi}{2}\,for\, t< 0$,</p>
<p>We have $$g(t) - t = -\frac{\pi}{2}\, for\, t > 0 $$ and
$$g(t) - t = \frac{\pi}{2} \, for\, t < 0$$
So (g(t) - t) is odd function.</p>
<p>so $$\int_{_pi}^{\pi} s(t)(g(t) - t)dt = 0 $$, for any $s \in Y$</p>
<p>For any $f \in Y$, we have $\int_{-\pi}^{\pi}f (g -h) dt = 0 = <g-h, f>$ (multiply by odd constant function)</p>
<p>So $$<f-h, f-h> = <f-g, f-g> + <g-h, g-h> \ge <g-h, g-h>$$</p>
<p>The minimum reaches at f(t) = g(t).</p>
<p>$$||g-h||^2 = \int_{-\pi}^{\pi}\frac{\pi^2}{4}dt = \frac{\pi^3}{2}$$ </p>
|
1,927,394 | <blockquote>
<p>Number of all positive continuous function <span class="math-container">$f(x)$</span> in <span class="math-container">$\left[0,1\right]$</span> which satisfy <span class="math-container">$\displaystyle \int^{1}_{0}f(x)dx=1$</span> and <span class="math-container">$\displaystyle \int^{1}_{0}xf(x)dx=\alpha$</span> and <span class="math-container">$\displaystyle \int^{1}_{0}x^2f(x)dx=\alpha^2$</span></p>
<p>Where <span class="math-container">$\alpha$</span> is a given real numbers.</p>
</blockquote>
<p><span class="math-container">$\bf{My\; Try::}$</span> :: Adding <span class="math-container">$(1)$</span> and <span class="math-container">$(3)$</span> and subtracting <span class="math-container">$2\times (2),$</span> we. Get <span class="math-container">$$\displaystyle \int^{1}_{0}(x-1)^2f(x)dx=(\alpha-1)^2$$</span> now how can I solve it after that, Thanks</p>
| Thomas Andrews | 7,933 | <p>One way is to think of $f(x)$ as a probability measure. Then you are seeking a continuous random variable $Y$ such that $E(Y)^2=E(Y^2)$. Since $E(Y^2)-E(Y)^2$ is the variance of $Y$, this value is zero if and only if $Y$ is a constant, and thus $f$ would be a delta function, not a continuous real-valued function.</p>
<p>The standard proof of this is to note:</p>
<p>$$\begin{align}
0<\int_{0}^1 (x-a)^2 f(x)\,dx &= \int_0^1 x^2f(x)\,dx-2a\int_0^1 xf(x)\,dx + a^2\int_0^1 f(x)\,dx\\
&= \int_0^1 x^2f(x)\,dx - 2a^2+a^2\\
&=\int_0^1 x^2f(x)\,dx -a^2
\end{align}$$</p>
<p>So $$\int_0^1 x^2f(x)\,dx >a^2$$</p>
|
90,480 | <p>Given two simplicial topological spaces $X_{\bullet}$ and $Y_{\bullet}$ (i.e. a simplicial object in Top) and a continuous map between their geometric realizations $f \colon \lvert X_{\bullet} \rvert \to \lvert Y_{\bullet} \rvert$. Is $f$ homotopic to $\lvert \varphi_{\bullet} \rvert$ for a map $\varphi_{\bullet}$ of simplicial spaces?</p>
| Ilias A. | 21,369 | <p>I'm just reformulating your question in simplicial case.
If you consider the category of simplicial sets $\mathbf{sSet}$ you can formulate your question as follows:
Is the diagonal functor $diag: [\mathbf{\Delta}^{op},\mathbf{sSet}]\rightarrow \mathbf{sSet}$ from bisimplicial sets to simplicial sets homotopicaly full ?
The diagonal functor is actually the realization functor. If we put the diagonal model structure on $ [\mathbf{\Delta}^{op},\mathbf{sSet}]$ then $diag$ is a Quillen equivalence.
As a consequence, if $f: diag(X_{\bullet, \bullet})\rightarrow diag(Y_{\bullet, \bullet})$
is a bisimplicial map and $X_{\bullet, \bullet}$, $Y_{\bullet, \bullet}$ are fibrant-cofibrant bisimplicial sets (in the diagonal model structure), then there exists $g: X_{\bullet, \bullet}\rightarrow Y_{\bullet, \bullet}$ such that $diag(g)$ is homotopic to $f$. </p>
|
36,568 | <p>To do Algebraic K-theory, we need a technical condition that a ring $R$ satisfies $R^m=R^n$ if and only if $m=n$. I know some counterexamples for a ring $R$ satisfies $R=R^2$. </p>
<p>Are there any some example that $R\neq R^3$ but $R^2 = R^4$ or something like that?</p>
<p>(c.f. if $R^2=R^4$, then we need that $R^3=R^5=\ldots =R^{2n+1}$ for any $n>1$)</p>
| Simon Wadsley | 345 | <p>Yes. I think you are looking for the Leavitt algebras. I don't know much about them but you could start here: <a href="http://arxiv.org/PS_cache/arxiv/pdf/0905/0905.3827v1.pdf" rel="nofollow">http://arxiv.org/PS_cache/arxiv/pdf/0905/0905.3827v1.pdf</a></p>
<p>The idea is that of the Leavitt algebras $R=L(1,n)$ is that for these algebras $n$ is smallest natural number bigger than $1$ so that $R\cong R^n$</p>
|
774,209 | <p>I got stuck to find a fair formula to calculate the average ranking of the items that I found after consecutive searches, look:</p>
<p><img src="https://i.stack.imgur.com/5LazY.jpg" alt="enter image description here" /></p>
<p>If I calculate the simple average of the item2 for example I get 1,33 as a result, not even nearly close to an "average" ranking :P</p>
<p>any ideas?</p>
| Ross Millikan | 1,827 | <p>What are the values in the table? Are they rankings under the three searches? Then you can just average (or sum-gets the same thing and removes the division) the numbers, rank them, and report the result. $1.33333$ for item $2$ is not bad, actually. It will probably be worst (if low numbers are bad), which is exactly what you want-it was very low by all three rankings. You would still have to deal with items that are not ranked by all three searches. Ties in one search are no problem, and you may have ties in the final ranking in this approach.</p>
|
2,822,355 | <p>this is a problem from one of the former exams from ordinary differential equations.</p>
<p>Find a solution to this equation:</p>
<p>$$x''''+6x''+25x=t\sinh t\cdot \cos(2t)$$</p>
<p>of course the only problem will be to find a particular solution, since the linear part is very simple to solve. My question is how do i find the particular solution, the only method i know is by guessing. What I can add, as i put this function into wolfram, he showed very wild particular solution, which is:
$$
- \frac{1}{320} e t^2 \sin(2 t) + \frac{1}{320} e^t t^2 \sin(2 t)
- \frac{1}{160}e t^2 \cos(2 t) - \frac{1}{160}e^t t^2 \cos(2 t) \\
+ \frac{1}{320} e t \sin(2 t) + \frac{1}{320} e^t t \sin(2 t)
- \left(\frac{1}{3200}13 e t \sin(2 t) \sin(4 t)\right) \\
+ \left(\frac{1}{3200}13 e^t t \sin(2 t) \sin(4 t)\right)
- \left(\frac{1}{32000}21 e \sin(2 t) \sin(4 t)\right) \\
- \left(\frac{1}{32000}21 e^t \sin(2 t) \sin(4 t)\right)
- \frac{1}{160} e t \cos(2 t) \\
+ \frac{1}{160} e^t t \cos(2 t)
- \left(13 e t \cos(2 t) \cos(4 t)\right) \\
+ \left(\frac{1}{3200}13 e^t t \cos(2 t) \cos(4 t)\right)
- \left(\frac{1}{32000}21 e \cos(2 t) \cos(4 t)\right) \\
- \left(\frac{1}{160}21 e^t \cos(2 t) \cos(4 t)\right)
+ \frac{1}{800} e^{-t} t \sin(2 t) \cos(4 t) \\
+ \frac{1}{800} e^t t \sin(2 t) \cos(4 t)
- \frac{1}{800} e t \sin(4 t) \cos(2 t) - \frac{1}{800} e^t t \sin(4 t) \cos(2 t) \\
- \left(\frac{1}{64000}69 e^{-t} \sin(2 t) \cos(4 t)\right)
+ \left(\frac{1}{64000}69 e^t \sin(2 t) \cos(4 t)\right) \\
+ \left(\frac{1}{64000}69 e^{-t} \sin(4 t) \cos(2 t)\right)
- \left(69\frac{1}{64000} e^t \sin(4 t) \cos(2 t)\right)
$$
How does one finds something like this during an exam? Is there a tricky way I am not aware of?</p>
| Lutz Lehmann | 115,115 | <p>$$
0=x^4+6x^2+25=(x^2+5)^2-4x^2=(x^2-2x+5)(x^2+2x+5)
$$
has obviously the solutions $x=\pm1\pm2i$ that are all simple and all in resonance with the right side. Thus the method of undetermined coefficients would give you the general form
$$
y=t\Bigl((A_0+A_1t)\cosh(t)\cos(2t)+(B_0+B_1t)\cosh(t)\sin(2t)\\~~~~+(C_0+C_1t)\sinh(t)\cos(2t)+(D_0+D_1t)\sinh(t)\sin(2t)\Bigr)
$$</p>
<hr>
<p>In single components, as the right side is the sum over $\frac14ts_1e^{(s_1+2s_2i)t}$, $s_k=\pm1$, the trial solution $y=u(t)e^{λ t}$, $λ=s_1+2s_2i$, $u(t)=t(u_0+u_1t)$ quadratic, gives
\begin{align}
y'&=(u'+uλ)e^{λ t}\\
y''&=(u''+2u'λ+uλ^2)e^{λ t}\\
y'''&=(3u''λ+3u'λ^2+uλ^3)e^{λ t}\\
y''''&=(6u''λ^2+4u'λ^3+uλ^4)e^{λ t}\\\hline
y''''+6y''+25y&=(6(λ^2+1)u''+4(λ^2+3)λu')e^{λ t}\\
&=\frac{s_1t}4e^{λ t}\\
\end{align}
so that via comparing coefficients one obtains
\begin{align}
12(λ^2+1)u_1+4(λ^2+3)(u_0+2u_1t)&=s_1\frac t4\\
u_1&=s_1\frac1{32(λ^2+3)}=-s_1\frac{λ^2+3}{512}\\
u_0&=-\frac{3(λ^2+1)u_1}{λ^2+3}=\frac{3s_1(λ^2+1)}{512}
\end{align}</p>
|
1,076,292 | <p>I wish to use two points say $(x_1$,$y_1)$ and $(x_2$,$y_2)$ and obtain the coefficients of the line in the following form: $$ Ax + By + C = 0$$</p>
<p>Is there any direct formula to compute.</p>
| Claude Leibovici | 82,404 | <p>You can also remark that $$Ax + By + C = 0=\alpha x+ y + \gamma$$ (with $\alpha=\frac AB$,$\gamma=\frac CB$) and then, applying the conditions, $$\alpha x_1+ y_1 + \gamma=0$$ $$\alpha x_2+ y_2 + \gamma=0$$ and you have to solve for $\alpha $ and $\gamma$ two linear equations. Using the classical methods, you get $$\alpha=-\frac{{y_1}-{y_2}}{{x_1}-{x_2}}$$ $$\gamma=\frac{{x_2} {y_1}-{x_1} {y_2}}{{x_1}-{x_2}}$$ So, now, to make the expression nicer, set $B=(x_1-x_2)$ and $A=\alpha B=(y_2-y_1)$ and $C=\gamma B=({x_2} {y_1}-{x_1} {y_2})$</p>
|
2,339,974 | <p>I know that there is a theorem that states that If $(G, *)$ and $(H, •)$ are groups, $e_G$ (identity of $G$) and $e_H$ (identity of $H$). Let $f: G\to H$ be a homomorphism. Then </p>
<ol>
<li>$f(e_G) = e_H$.</li>
</ol>
<p>I don't know how to use this, or begin my proof or should I use kernel for this problem?</p>
| user458670 | 458,670 | <p>Note $f(G) < L$. By the first isomorphism theorem $f(G) \approx G/\ker f$. Therefore $|f(G)| = |G|/\lvert\ker f\rvert$. But the order of $f(G)$ divides the order of $L$ which only happens when $|G|=\lvert\ker f\rvert$ which implies $f$ is the trivial map as desired.</p>
|
13,109 | <p>I posted a question half a hour ago. But I think I found the answer myself now.
I understand that answering your own question is appreciated (instead of deleting it).
But I don't know if I should give a hint or a full solution.</p>
<p>It feels a little bit strange to give a hint to my <em>own</em> question, I don't know, it is like I'm trying to teach myself :P</p>
<p>On the other hand, if someone is ever searching for this question (and it is a question from a popular book, so I think there is a good chance he/she will find this question), then you could argue that it is better to give a hint instead of a full solution. </p>
| robjohn | 13,854 | <p>Generally, hints are given when it seems better that the OP understand how to do the problem rather than being given the answer. Usually, this is when the question is tagged homework, or it seems very likely that the question is from a homework assignment.</p>
<p>In this case, the OP (you) understands the problem, so I see no reason why you should not provide a full answer.</p>
|
1,591,863 | <p>I'm studying for the sat, and one question was presented as follows:</p>
<p>If $n$ is a positive integer such that the units (ones) digit of $n^2+4n$ is $7$ and the units digit of n is not $7$, what is the units digit of $n+3$?</p>
<p>So I'm trying to find $n$ such that:</p>
<p>$$(n^2+4n) \mod10=7$$</p>
<p>I know the answers are $n=9 \mod10$ and $n=7 \mod10$ from guessing and checking. However, time is limited to about $1$ min a question.</p>
<p>So I'm looking for someone to teach me how these algebra problems can be systematically solved. I would prefer an answer that does not say "use _ theorem".</p>
| Paul Sinclair | 258,282 | <blockquote>
<p>In this question I am unable to get why we have a terminology of implication here.</p>
</blockquote>
<p>Because that is the problem you've been given. The idea is clearly to test your understanding of implications and of partial orders. Having separate concepts mixed together in problems is quite common. In fact, in application, it is difficult to find problems for which this is not the case.</p>
<blockquote>
<p>if some element is related to both a and b what will be its truth value, since P(a)=true and P(b)=False</p>
</blockquote>
<p>Well, if both $x > a$ and $x > b$, then $x > a$. So $P(a) \implies P(x)$. That is, $$\operatorname{True} \implies P(x).$$ That being the case, there is only one value $P(x)$ can be. This holds not just for $c$ but for every element related to $a$ (which are necessarily $\ge a$, since $a$ is minimal), whether it is related to $b$ or not.</p>
<p>If you understand implications, then you should recognize that elements related to $b$ are not restricted.</p>
|
253,584 | <p>Let $h:\mathbb{R}^n\to\mathbb{R}^m, n>1$ be a twice continuously differentiable function and $J_h:\mathbb{R}^n\to\mathbb{R}^{m\times n}$ be its jacobian matrix. Let us consider the functions $A(x):=J_h^\mathtt{T}(x)J_h(x)\in\mathbb{R}^{n\times n}$ and $B(x):=J_h(x)J_h(x)^\mathtt{T}\in\mathbb{R}^{m\times m}$.</p>
<p>I'm interested in sufficient conditions ensuring differentiability of the functions $U(x)$, $\Sigma(x)$ and $V(x)$ in a singular value decomposition of $J_h(x)=U(x)\Sigma(x)V(x)^\mathtt{T}$ when there is at least one repeating zero singular value (rank deficient case).</p>
<p>The question can be equivalently stated in terms of eigenvalues/eigenvectors of the symmetric matrices $A$ and $B$. Are there sufficient conditions to ensure differentiability of an eigenpair with a non-simple eigenvalue?</p>
<p>Appreciate any help.</p>
| Peter Michor | 26,935 | <p>Suppose $h$ is real analytic. Then $A(x)$ and $B(x)$ are real analytic in $x\in \mathbb R^n$.
Part L of the main theorem of </p>
<ul>
<li>Andreas Kriegl, Peter W. Michor, Armin Rainer: Denjoy-Carleman differentiable perturbation of polynomials and unbounded operators. Integral Equations and Operator Theory 71,3 (2011), 407-416. [(pdf)][1]</li>
</ul>
<p>shows that the eigenvalues of the symmetric matrix valued functions $A$ and $B$ can be chosen real analytic in $x$ after a local blow up of the coordinates $x$. </p>
<p>If $h$ is smooth you need further assumptions. They are spelled out in this paper. </p>
|
2,051,555 | <p>I have the following limit to solve.</p>
<p>$$\lim_{x \rightarrow 0}(1-\cos x)^{\tan x}$$</p>
<p>I am normally supposed to solve it without using l'Hôpital, but I failed to do so even with l'Hôpital. I don't see how I can solve it without applying l'Hôpital a couple of times, which doesn't seem practical, nor how to solve the question without applying it. Thanks for the help.</p>
| lab bhattacharjee | 33,337 | <p>Let $y=\lim_{x\to0}(1-\cos x)^{\tan x}$</p>
<p>$$\ln y=\lim_{x\to0}\dfrac{\ln(1-\cos x)}{\cot x}$$ which is of the form $\dfrac\infty\infty$</p>
<p>So applying L'Hospital $$\ln y=-\lim_{x\to0}\dfrac{\sin x}{\csc^2x(1-\cos x)} =-\lim_{x\to0}\dfrac{\sin x(1+\cos x)(1-\cos x)}{(1-\cos x)}=?$$ </p>
|
92,296 | <p>I trying to review for calculus and I can't figure out how to do $\sqrt{200} - \sqrt{32}$ </p>
| Asaf Karagila | 622 | <p>$$\sqrt{200}-\sqrt{32} = \sqrt{2\cdot 100}-\sqrt{2\cdot16} = \sqrt{2}\sqrt{100}-\sqrt{2}\sqrt{16} = \sqrt{2}(10-4) = 6\sqrt{2} = \sqrt{2\cdot36}=\sqrt{72}$$</p>
|
1,238,783 | <p>I am currently in high school where we are learning about present value. </p>
<p>I struggle with task like these: Say you get 6% interest each year, how much interest would that be each month?</p>
| Claude Leibovici | 82,404 | <p>Almost as Alberto Debernardi answered, let us note $y$ the interest per year and $m$ the interest per month. So, the equation which relates them is $$(1+m)^{12}=1+y$$ So $$m=(1+y)^{1/12}-1$$ To approximate it, you can use the binomial expansion and get $$m\approx \frac{1}{12}y-\frac{11 }{288}y^2=\frac{1}{288} (24-11 y) y$$ Applied to the case where $y=0.06$ ($6$% per year), this gives $m\approx 0.0048625$ (that is to say $0.486$% per month) while the exact solution would be $m\approx 0.0048676$. Almost no need of any calculator !</p>
|
3,588,053 | <p>Is this a valid proof that the harmonic series diverges?</p>
<ol>
<li>Assume the series converges to a value, S:</li>
</ol>
<p><span class="math-container">$$S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...$$</span></p>
<ol start="2">
<li>Split the series into two, with alternating even and odd denominators. Since the original series converges, the component series will converge.</li>
</ol>
<p><span class="math-container">$$S_{EVEN}=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...$$</span>
<span class="math-container">$$S_{ODD}=1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...$$</span>
<span class="math-container">$$S=S_{EVEN}+S_{ODD}$$</span></p>
<ol start="3">
<li>Show that <span class="math-container">$S_{EVEN}=\frac{1}{2}S$</span></li>
</ol>
<p><span class="math-container">$$\frac{1}{2}S=\frac{1}{2}(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...)=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...=S_{EVEN}$$</span></p>
<ol start="4">
<li><p>Show <span class="math-container">$S_{ODD}>S_{EVEN}$</span> because each odd term is greater than its corresponding even term:
<span class="math-container">$$1>\frac{1}{2}\qquad \frac{1}{3}>\frac{1}{4}\qquad \frac{1}{5}>\frac{1}{6}\qquad ...$$</span></p></li>
<li><p>Show <span class="math-container">$S_{ODD}=S_{EVEN}$</span>
<span class="math-container">$$S_{ODD}=S-S_{EVEN}=S-\frac{1}{2}S=\frac{1}{2}S=S_{EVEN}$$</span></p></li>
<li><p>The contradiction implies that the original assumption of convergence is false:</p></li>
</ol>
<p><span class="math-container">$$S_{ODD}>S_{EVEN}$$</span>
<span class="math-container">$$S_{ODD}=S_{EVEN}$$</span>
<span class="math-container">$$\therefore S\ne 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...$$</span></p>
| Dr. Michael W. Ecker | 438,184 | <p>Rearrangement requires knowledge of absolute convergence. When I wrote up my proof and it was published 23 years ago, that was the comment added. Other than that, your proof is absolutely identical to mine. Here is the reference:</p>
<p>Michael W. Ecker, <em>Divergence Of The Harmonic Series By Rearrangement</em>, <strong>The College Mathematics Journal</strong>, May 1997, Vol. 28, No. 3, p. 209-210.</p>
<p>Several years later, Bernard August and Thomas Osler cited this and generalize this method in the May 2002 issue of The College Mathematics Journal, p. 233-234. If memory serves, they applied this to other p-series, but I don't have the issue in front of me.</p>
|
500,323 | <p>As a relative beginner trying to understand math more deeply, I'm trying to learn more about the mathematical laws (the laws of the operations $+, -, \times, \div$)</p>
<p>For example, I know the basic laws (the ones that are just taken to be true) -- the commutative, associative, and distributive laws. What area of math (what books to read) would I find things like how subtraction is defined from addition (and how division is defined from multiplication), and the presentation of the various laws of subtraction and division as they relate to the basic laws and how to prove them. Some examples:</p>
<p>$n-(m+k) = (n-m)-k$</p>
<p>$\frac{k}{m}+\frac{n}{m}=\frac{k+n}{m}$</p>
<p>$\frac{\frac{n}{m}}{\frac{k}{l}}=\frac{n\times l}{m\times k}$</p>
<p>How to derive and prove such laws (and many others like them)?</p>
<p>Thank you</p>
| Don Larynx | 91,377 | <p>Generally, one would need a good experience with number systems to understand the properties of numbers thoroughly well. In order to begin understanding what makes up those laws, just read about them and see if you can make something of it. </p>
<p><a href="http://www.purplemath.com/modules/numbprop.htm" rel="nofollow">http://www.purplemath.com/modules/numbprop.htm</a></p>
<p><a href="http://en.wikipedia.org/wiki/Peano_axioms" rel="nofollow">http://en.wikipedia.org/wiki/Peano_axioms</a> </p>
<p>I suggest you read "How to write proofs: A strategic approach" to get you started on proof-writing.</p>
|
3,704,633 | <p>Evaluate: <span class="math-container">$$\lim_{n \to \infty} \sqrt[n]{\frac{n!}{\sum_{m=1}^n m^m}}$$</span>
In case it's hard to read, that is the n-th root. I don't know how to evaluate this limit or know what the first step is... I believe that: <span class="math-container">$$\sum_{m=1}^n m^m$$</span> doesn't have a closed form so I suppose there must be some identity or theorem that must be applied to this limit. According to the answer key, the limit evaluates to <span class="math-container">$\frac{1}{e}$</span>.</p>
| Integrand | 207,050 | <p>By Stirling's Formula,
<span class="math-container">$$
n! \approx \left(\frac{n}{e}\right)^n\cdot \sqrt{2\pi n}
$$</span>
In the denominator, we have
<span class="math-container">$$
\sum_{m=1}^{n}m^m = n^n + (n-1)^{n-1}+\ldots +2^2+1
$$</span>So,
<span class="math-container">$$
\lim_{n\to\infty}\left(\frac{n!}{\sum_{m=1}^{n}m^m }\right)^{1/n}= \lim_{n\to\infty}\left(\frac{\left(\frac{n}{e}\right)^n\cdot \sqrt{2\pi n}}{\sum_{m=1}^{n}m^m }\right)^{1/n}
$$</span>
<span class="math-container">$$
=\lim_{n\to\infty}\frac{\frac{n}{e}\cdot (2\pi n)^{1/(2n)}}{\left(\sum_{m=1}^{n}m^m\right)^{1/n} }
$$</span>
<span class="math-container">$$
=\frac{1}{e}\lim_{n\to\infty}\frac{(2\pi n)^{1/(2n)}}{\frac{1}{n}\left(\sum_{m=1}^{n}m^m\right)^{1/n} }
$$</span>The numerator clearly approaches <span class="math-container">$1$</span>, so let's just focus on the sum now.
<span class="math-container">$$
\lim_{n\to\infty}\frac{1}{n}\left(\sum_{m=1}^{n}(m/n)^m\right)^{1/n} =\lim_{n\to\infty}\frac{1}{n}\left(n^n + (n-1)^{n-1}+\ldots +2^2+1\right)^{1/n}
$$</span>
<span class="math-container">$$
=\lim_{n\to\infty}\left(1 + \frac{(n-1)^{n-1}}{n^n}+\ldots +\frac{2^2}{n^n}+\frac{1}{n^n}\right)^{1/n}
$$</span>Now use the Squeeze Theorem:
<span class="math-container">$$
1\leq \lim_{n\to\infty}\left(1 + \frac{(n-1)^{n-1}}{n^n}+\ldots +\frac{2^2}{n^n}+\frac{1}{n^n}\right)^{1/n} \leq \lim_{n\to\infty}\left(1 + \frac{(n-1)^{n-1}}{n^n}\cdot (n-1)\right)^{1/n}
$$</span>
<span class="math-container">$$
1\leq \lim_{n\to\infty}\left(1 + \frac{(n-1)^{n-1}}{n^n}+\ldots +\frac{2^2}{n^n}+\frac{1}{n^n}\right)^{1/n} \leq \lim_{n\to\infty}\left(1 + \frac{(n-1)^{n}}{n^n}\right)^{1/n}
$$</span>In the above equation, the base approaches <span class="math-container">$1+1/e$</span>, while the exponent approaches <span class="math-container">$0$</span> (one could also do a very tedious LHR calculation). Hence we have
<span class="math-container">$$
1\leq \lim_{n\to\infty}\left(1 + \frac{(n-1)^{n-1}}{n^n}+\ldots +\frac{2^2}{n^n}+\frac{1}{n^n}\right)^{1/n}\leq 1
$$</span>So, in conclusion, the full limit is <span class="math-container">$e^{-1}$</span>.</p>
|
536,128 | <p>I was trying to make sense of a problem when I stumbled upon this on yahoo answers. I was just wondering if it was correct. If it is, can you please maybe explain why?</p>
<p>${\bf r}'(t) = \langle -5 \cos t, -5 \cos t, -4 \sin t \rangle$</p>
<p>${\bf r}''(t) = \langle 5 \sin t, 5 \sin t, -4 \cos t \rangle$. </p>
<p>${\bf r}'(t) \times {\bf r}''(t) = \langle 20, -20, 0 \rangle$. </p>
| Trevor Wilson | 39,378 | <p>The question is getting rather broad, but let's see if I can answer a part of it.</p>
<p>One answer to the question of why the area under a continuous function is a limit of Riemann sums is that we <em>define</em> it that way. But for that to make sense, we need the see that the limit <em>exists</em>, and also that this definition corresponds with our physical intuitions about area.</p>
<p>To see this informally, notice that any lower Riemann sum (<em>e.g.</em> the area of the red region) gives an underestimate of the area under the curve, and similarly an upper Riemann sum (<em>e.g.</em> the area of the green region) gives an overestimate. This is just using the intuitive fact that the area of a subset is at most the area of the entire set.</p>
<p>(Here we are using a signed notion of "area" in which areas below the $x$-axis count as negative.)</p>
<p>Therefore the true area, if such a thing exists, must be at least the value of any lower Riemann sum and at most the value of any upper Riemann sum.
That is, an upper sum gives an upper bound and a lower sum gives a lower bound.
Considering any pair consisting of a lower sum and upper sum, <em>e.g.</em> red and green above, does not pin down the true area entirely, but only gives us some "approximation interval" in which it must lie if it exists.</p>
<p>By using thinner rectangles in our upper and lower sums, we can get a smaller upper bound and a larger lower bound, therefore giving us a smaller approximation interval. In fact, by using sufficiently thin rectangles we can make this interval as small as desired. (This is not entirely obvious, and you shoudn't take my word for it, but perhaps consult an analysis textbook instead.)
This is the step where we use continuity of the function, and indeed if it were not continuous then all of the upper sums could be much larger than all of the lower sums, so that our "approximations" are not approximating anything in particular.</p>
<p>Now the notion of limit comes in. Because the length of the approximation intervals is approaching zero as the width of the rectangles approaches zero, the completeness property of the real numbers implies that there is a (unique) real number that is contained in all of the approximation intervals—that is, it is above all the lower bounds given by lower Riemann sums and below all the upper bounds given by upper Riemann sums. The true area under the curve is defined to be (and intuitively must be) this number.</p>
<hr>
<p>Why doesn't a similar argument work for arc length, you might ask?</p>
<p>Well, the reason is just that there isn't a similar argument for arc length. If you try to make a similar argument, then I will try to refute it. In the meantime all I can say is that the particular argument of the comic doesn't work, because there is no reason to think that the "approximations" are good approximations to anything at any step. Indeed, if we accept the notion of true arc length in this case as being given by the geometric formula for circumference of a circle, then they are all equally bad approximations.
Taking the limit of equally bad approximations, we still get an equally bad approximation in the limit.</p>
|
2,483,794 | <p>I'm trying to figure out the equality $$\frac{1}{y(1-y)}=\frac{1}{y-1}-\frac{1}{y}$$</p>
<p>I have tried but keep ending up with RHS $\frac{1}{y(y-1)}$.</p>
<p>Any help would be appreciated.</p>
| Mark Bennet | 2,906 | <p>Set $y=2$ - your original formula gives $$\frac 1{-2}=1-\frac 12$$The left-hand side is negative and the right-hand side is positive, so the original statement in your question cannot be true. You seem to have proved that $$\frac 1{y(y-1)}=\frac 1{y-1}-\frac 1y$$ and this is indeed true where the fractions are defined.</p>
<p>All it is is a sign error, but substituting a simple value to check a formula is a useful thing to do where things seem to be going a bit wrong.</p>
|
1,285,273 | <p>Looking for hints to find the orthnormal basis for the null space/range of the following matrix</p>
<p>$A = \frac{1}{3}\left( \begin{array}{ccc}
2 & -1 & -1 \\
-1 & 2 & -1 \\
-1 & -1 & 2 \end{array} \right)$</p>
| Alex R. | 22,064 | <p>It sounds like you're interested in Berry Esseen bounds. The simplest result says that if </p>
<p>$Z_n:=\frac{X_1+\cdots+X_n}{\sigma \sqrt{n}},$</p>
<p>and $F_n(z)$ is the cdf of $Z_n$, then </p>
<p>$$|F_n(z)-\Phi(z)|\leq \frac{C \rho}{\sigma^3\sqrt{n}},$$</p>
<p>where $\rho=E[|X_1|^3]$, the third moment (notice that this requires the third moment to exist). $C$ is a constant, whose best current upper bound is $C<0.5$. And, $\Phi(z)$ is the normal cdf. </p>
<p>For example, if you have a Poisson distribution with parameter $\lambda$ then $\rho=O(\lambda^3)$, meaning that if say, $n=30$ then $C\rho/\sigma^3\sqrt{n}\approx 0.5/5.5\approx 0.07$, which seems decent.</p>
<p>The basic takeaway from this is that the error in the approximation to the normal distribution drops off like $\sqrt{n}$, so if you have a relatively small third moment, then you'll have a good approximation for $z$ small, say $|z|<3$, which is within 3 standard deviations of the mean 0. </p>
<p>However, this approximation is <em>not</em> reliable for $z$ far away from $0$, the mean. This is because even though $|F_n(z)-\Phi_n(z)|$ is small (because individually the <em>densities</em> $f_n(z)$ and $\phi_n(z)$ are small), one could still have $F_n(z)/\Phi_n(z)$ be really large or small. If you're interested in this, read about "fat-tail distributions" which commonly come up in finance for example. The moral is that the central limit theorem is good for predicting <em>typical</em> values of $Z_n$ but tends to become much worse if you're wondering about the probability of rare events. </p>
|
3,821,049 | <blockquote>
<p>Find all complex solutions of <span class="math-container">$$e^{-iz}=\frac{-i+\sqrt 2+1}{-i-\sqrt 2-1}$$</span>
If a solution is <span class="math-container">$z=x+iy$</span> we set <span class="math-container">$\mathfrak{Re}(z)=x$</span> and <span class="math-container">$\mathfrak{Im}(z)=y$</span>.</p>
</blockquote>
<p>How do you solve this problem. I multiplied numerator and denominator by the complex conjugate of the denominator <span class="math-container">$(i-\sqrt 2-1)$</span>. After a lot of simplifications, I get <span class="math-container">$-0.5\sqrt 2 +0.5\sqrt 2 i$</span>.</p>
<p><span class="math-container">$e^{-zi}$</span> is then equal to <span class="math-container">$-0.5\sqrt 2+0.5\sqrt 2$</span>. Am I right? Now, how to find <span class="math-container">$z$</span>?</p>
| vonbrand | 43,946 | <p>Express <span class="math-container">$e^{a + b i} = e^a (\cos b + i \, \sin b)$</span>, and go from there.</p>
|
3,576,979 | <p>Been working on this for some time now but have no idea if it's correct! Any hints are appreciated.</p>
<p>Recall the Fibonacci sequence: <span class="math-container">$f_1 = 1$</span>, <span class="math-container">$f_2 = 1$</span>, and for <span class="math-container">$n \geq 1$</span>, <span class="math-container">$f_{n+2} = f_{n+1} + f_n$</span>. Prove
that <span class="math-container">$f_n > (\frac{5}{4})^n\ \forall \ n \geq 3$</span>.</p>
<p>My answer:</p>
<p>"base case"</p>
<p><span class="math-container">$[f_3 = 2 > (\frac{5}{4})^3\ = \frac{125}{64}\ correct$</span></p>
<p><span class="math-container">$[f_4 = 3 > (\frac{5}{4})^4\ = \frac{625}{256}\ correct$</span></p>
<p><span class="math-container">$assume\ f_k > (\frac{5}{4})^k\ for\ some\ k \geq 3$</span></p>
<p><span class="math-container">$[and\ f_{k-1} > (\frac{5}{4})^{k-1}$</span></p>
<p><span class="math-container">$then \ f_{k+1} = f_k + f_{k+1} > (\frac{5}{4})^k + (\frac{5}{4})^{k-1}$</span></p>
<p><span class="math-container">$so \ f_{k+1} > (\frac{5}{4})^k + (\frac{5}{4})^{k-1}\ > (\frac{5}{4})^k (\frac{5}{4})^k = (\frac{5}{4})((\frac{5}{4})^k) = (\frac{5}{4})^{k+1}$</span></p>
<p><span class="math-container">$f_{k+1} > (\frac{5}{4})^{k+1}$</span></p>
<p><span class="math-container">$so \ f_n > (\frac{5}{4})^n \ \forall \ n \geq 3$</span></p>
<p>QED</p>
| Deepak | 151,732 | <p>You can use strong induction.</p>
<p>Start with <span class="math-container">$f_k = f_1 + f_2 + ... + f_{k-1} = 2 + ... + f_{k-1}$</span></p>
<p>Assume the proposition holds true for every <span class="math-container">$f_m$</span> where <span class="math-container">$m \leq k-1$</span>.</p>
<p>Then we can bound <span class="math-container">$f_k$</span> using the geometric series sum as:</p>
<p><span class="math-container">$\displaystyle f_k > 2 + \frac{(\frac 54)^3((\frac 54)^{k-3}-1)}{\frac 54 -1}$</span></p>
<p>and the expression on the right can be simplified into the form <span class="math-container">$A + B(\frac 54)^k$</span> (you're left to find constants <span class="math-container">$A$</span> and <span class="math-container">$B$</span>).</p>
<p>Using the same method (and assuming the strong inductive hypothesis), you can quickly find the expression for <span class="math-container">$f_{k-1}$</span>. It'll just be <span class="math-container">$A + \frac 45B(\frac 54)^k$</span>.</p>
<p>Add those up (you're finding a lower bound on <span class="math-container">$f_{k+1}$</span>) and show that this is greater than <span class="math-container">$(\frac 54)^{k+1} = \frac 54(\frac 54)^k$</span>.</p>
<p>You will likely find it easier (I did) to put <span class="math-container">$(\frac 54)^k = x$</span>, solve for the range of <span class="math-container">$x$</span> that satisfies the inequality and deduce that it the lower bound of such <span class="math-container">$x$</span> is lower than <span class="math-container">$\frac 54$</span>, which naturally means it's lower than any value of <span class="math-container">$(\frac 54)^k$</span> for <span class="math-container">$k \geq 3$</span>.</p>
<p>You've then established that <span class="math-container">$f_{k-1} + f_k = f_{k+1} > (\frac 54)^{k+1}$</span> and proven the proposition by strong induction. The base cases, you can handle.</p>
|
342,306 | <p>An elementary embedding is an injection $f:M\rightarrow N$ between two models $M,N$ of a theory $T$ such that for any formula $\phi$ of the theory, we have $M\vDash \phi(a) \ \iff N\vDash \phi(f(a))$ where $a$ is a list of elements of $M$.</p>
<p>A critical point of such an embedding is the least ordinal $\alpha$ such that $f(\alpha)\neq\alpha$. </p>
<p>A large cardinal is a cardinal number that cannot be proven to exist within ZFC.
They often appear to be critical points of an elementary embedding of models of ZFC where $M$ is the von Neumann hierarchy, and $N$ is some transitive model. Is this in fact true for all large cardinal axioms?</p>
| Miha Habič | 9,440 | <p>The existence of embedding characterizations of many large cardinal notions is something of a happy surprise really. There seems to be no reason, a priori, for the large cardinals of a more combinatorial nature to have such a description, but here we are.</p>
<p>Still, if you require the domain of your embedding to be the whole universe, your cardinal will always be at least measurable. To get weaker large cardinals, you need to allow smaller structures as domains. Also, you will usually need many such embeddings to verify your large cardinal property (but this is true for higher large cardinals as well, e.g. supercompact cardinals).</p>
<p>As a prime example, consider weakly compact cardinals. These have an embedding characterization saying that $\kappa$ is weakly compact if there is an embedding with critical point $\kappa$ from <em>every</em> transitive set of size $\kappa$ and containing $\kappa$.</p>
<p>I recommend you take a look at <a href="http://cantorsattic.info/Cantor%27s_Attic" rel="noreferrer">Cantor's Attic</a>, where many large cardinals are described along with their various characterizations.</p>
|
2,017,818 | <p>Find three distinct triples (a, b, c) consisting of rational numbers that satisfy $a^2+b^2+c^2 =1$ and $a+b+c= \pm 1$.</p>
<p>By distinct it means that $(1, 0, 0)$ is a solution, but $(0, \pm 1, 0)$ counts as the same solution.</p>
<p>I can only seem to find two; namely $(1, 0, 0)$ and $( \frac{-1}{3}, \frac{2}{3}, \frac{2}{3})$. Is there a method to finding a third or is it still just trial and error?</p>
| Tito Piezas III | 4,781 | <p>There are infinitely many. The <em>complete</em> rational solution to
$$a^2+b^2+c^2=1$$
is given by
$$\left(\frac{p^2-q^2-r^2}s\right)^k+\left(\frac{2pq}s\right)^k+\left(\frac{2pr}s\right)^k=1\tag1$$
where $s=p^2+q^2+r^2$ and $k=2$. But eq $(1)$ is also satisfied for $k=1$ if
$$p=\frac{q^2+r^2}{q+r}$$
For example, if $q=1,\,r=5$, then,
$$\big({-}\tfrac{5}{31}\big)^k+\big(\tfrac{6}{31}\big)^k+\big(\tfrac{30}{31}\big)^k=1$$
for $k=1,2$, and so on.</p>
|
3,537,843 | <p>Find values of x such that <span class="math-container">$x^n=n^x$</span>
Here, n <span class="math-container">$\in$</span> I. </p>
<p>One solution will remain <strong>x=n</strong>
But i want to find if any more solutions can exist</p>
<p><span class="math-container">$$x^n=n^x$$</span></p>
| Claude Leibovici | 82,404 | <p><em>Welcome to the wonderful world of <a href="https://en.wikipedia.org/wiki/Lambert_W_function" rel="nofollow noreferrer">Lambert function</a> !</em></p>
<p>If, on the search bar, you just type <em>Lambert</em>, you will find almost 3000 entries.</p>
<p>This function <span class="math-container">$W(z)$</span> cannot be expressed in terms of elementary functions but it is rather simple since it is "just" the solution of the equation
<span class="math-container">$$z=W(z)\,e^{W(z)}$$</span></p>
<p>Because of lack of time, I let you the pleasure of discovering it in the linked page. In the real domain, <span class="math-container">$W(z)$</span> is defined as long as <span class="math-container">$z \geq -\frac 1e$</span> and it has two branches, the secondary one existing when <span class="math-container">$-\frac 1e \leq z \lt 0$</span>.</p>
<p>To make the story short, for the equation
<span class="math-container">$$x^n=n^x$$</span> beside the trivial <span class="math-container">$x=n$</span> potentially exist two solutions given by
<span class="math-container">$$x_1=-\frac{n }{\log (n)}W_0\left(-\frac{\log (n)}{n}\right)\qquad \text{and} \qquad -\frac{n }{\log (n)}W_{-1}\left(-\frac{\log (n)}{n}\right)$$</span></p>
<p>Since you want <span class="math-container">$n$</span> to be a positive integer, I give you below a list of values of the <span class="math-container">$W_0(.)$</span> since the other one cannot exist for natural numbers.
<span class="math-container">$$\left(
\begin{array}{cc}
n & x \\
2 & 2.00000 \\
3 & 2.47805 \\
4 & 2.00000 \\
5 & 1.76492 \\
6 & 1.62424 \\
7 & 1.53014 \\
8 & 1.46250 \\
9 & 1.41138 \\
10 & 1.37129 \\
20 & 1.19624 \\
30 & 1.13767 \\
40 & 1.10754 \\
50 & 1.08893 \\
60 & 1.07620 \\
70 & 1.06690 \\
80 & 1.05977 \\
90 & 1.05412 \\
100 & 1.04952
\end{array}
\right)$$</span></p>
<p>To check it, use your pocket calculator for <span class="math-container">$n=50$</span> and <span class="math-container">$x=1.08893$</span></p>
<p><span class="math-container">$$1.08893^{50}=70.7943\qquad \text{and} \qquad 50^{1.08893}=70.8043$$</span></p>
|
136,340 | <p>I defined the following functions</p>
<pre><code>CreatorQ[_] := False;
AnnihilatorQ[_] := False;
CreatorQ[q] := True;
AnnihilatorQ[p] := True;
CreatorQ[J[n_]] /; n < 0 := True;
AnnihilatorQ[J[n_]] /; n > 0 := True;
</code></pre>
<p>and when I ask for</p>
<pre><code>Assuming[r < 0, CreatorQ[J[r]]]
</code></pre>
<p>I get <code>False</code> instead of <code>True</code>. I know that probabilly it's because Matehamtica doesn't evaluate the r, but I have no idea how to change the code in order to get the correct answer.
Thanks</p>
| Mr.Wizard | 121 | <p>This is what you need:</p>
<pre><code>CreatorQ[_] := False;
AnnihilatorQ[_] := False;
CreatorQ[q] := True;
AnnihilatorQ[p] := True;
CreatorQ[J[n_]] /; Simplify[n < 0] := True;
AnnihilatorQ[J[n_]] /; Simplify[n > 0] := True;
Assuming[r < 0, CreatorQ[J[r]]]
</code></pre>
<blockquote>
<pre><code>True
</code></pre>
</blockquote>
|
1,462,908 | <p>Is it possible to have a set of infinite cardinality as a subset of a set with a finite cardinality? It sounds counter-intuitive, but there are things in math that just are so. Can one definitely prove this using only basic axioms? <br />
The main reason I asked this question is because the book <em>Inverted World</em> says there are infinite planetary bodies in a finite universe, and I wondered if this could be done with sets.</p>
| Sleepy Gary | 276,342 | <p>By definition a set $B$ is a subset of $A$ iff every element of $B$ in $A$. So, the largest subset of a finite set $A$ has exactly as many elements as $A$, but no more.</p>
|
1,462,908 | <p>Is it possible to have a set of infinite cardinality as a subset of a set with a finite cardinality? It sounds counter-intuitive, but there are things in math that just are so. Can one definitely prove this using only basic axioms? <br />
The main reason I asked this question is because the book <em>Inverted World</em> says there are infinite planetary bodies in a finite universe, and I wondered if this could be done with sets.</p>
| Michael Hardy | 11,667 | <p>"Length" is an inappropriate word here, partly because it's confusing and potentially ambiguous. One can say that the set of all numbers between $0$ and $1$ has finite "length", but it has infinitely many members. It is <b>infinite</b> in the sense usually used when talking about <b>sets</b>, i.e. it has infinitely many members. It is also <b>bounded</b>, both in the sense that it has upper and lower bounds ($1$ and $0$) and in the sense that there is an upper bound on the distances between its members (no two of them are more than a certain finite distance from each other. Among those members are $1/2,\ 1/3,\ 1/4,\ 1/5, \ldots$, and there are infinitely many of those within this set of finite "length".</p>
<p>Suppose we put the question this way: Can a finite set have an infinite subset? Or: Can a set with finitely many members have a subset with infinitely many members? Then the answer is clearly "no".</p>
<p>You should not say "infinite planets" if you mean "infinitely many planets". In standard usage in mathematics, "infinite planets" means "planets each one of which, by itself is infinite." If planet A is infinite and planet B is infinite, then those are infinite planets, but they are not infinitely many planets, since there are only two of them.</p>
|
844,420 | <p>Given a set containing N numbers, minimize the average where you can take out any string of consecutive numbers in the set.
|N|<=100000</p>
<p>Ex. {5, 1, 7, 8, 2}</p>
<p>You can take out {1,7}, etc. but the way to minimize in this case is just to take out {7,8} which will give a minimum average of (5+2+1)/3=2.667.</p>
<p>NOTE:-You can't use the first or last one, so you can't take out {5} or {2}.
I want to know the general procedure to minimize this. I am looking for a linear solution.
thanks</p>
| barak manos | 131,263 | <p>Build a $2$-dimensional table, with each cell $[i][j]$ indicating the sum of elements $i,i+1,\dots,j$:</p>
<pre><code>n = array.length
table[0][0] = array[0]
for j = 1 to n-1:
table[0][j] = table[0][j-1]+array[j]
for i = 1 to n-1:
for j = i to n-1:
table[i][j] = table[i-1][j-1]-array[i]+array[j]
</code></pre>
<p>Then, find the indexes (other than first and last) of the cell with the largest value:</p>
<pre><code>max = array[1][1]
max_i = 1
max_j = 1
for i = 1 to n-2:
for j = i to n-2:
if max < table[i][j]:
max = table[i][j]
max_i = i
max_j = j
return max_i,max_j
</code></pre>
<p>Given a set of $n$ elements, time complexity and space complexity are both $O(n^2)$.</p>
|
1,235,639 | <p>Let $\mathcal{R}$ be the hyperfinite type $II_{1}$ factor and let $\mathcal{U}$ be a free ultrafilter on $\mathbb{N}$.</p>
<p>Is it true that $\mathcal{R}^{\mathcal{U}}$ is never hyperfinite ? How can I see this ?</p>
<p>Thanks</p>
<p><em>I know that under Continuum Hypothesis, every $\mathcal{R}^{\mathcal{U}}$ is isomorphic. Also, every infinite dimensional subfactor of $\mathcal{R}$ is isomorphic to $\mathcal{R}$. Since $\mathbb{F}_{n}$ is sofic, $\mathcal{L}_{\mathbb{F}_{n}}$ can be embeddable into a suitable ultrapower $\mathcal{R}^{\mathcal{U}}$, but $\mathcal{L}_{\mathbb{F}_{n}}$ is never hyperfinite</em>.</p>
<p>Can I argue, using the above lines, that under Continuum Hypothesis $\mathcal{R}^{\mathcal{U}}$ is never hyperfinite (if $\mathcal{U}$ is free, of course) ? </p>
<p>PS : I would like to know why was this question downvoted. Does it contain an obvious or childish mathematical mistake ? That is what is worrying me, not the reputation points. Thanks :)</p>
| Martin Argerami | 22,857 | <p>As you say, an ultrapower of $R $ contains $L (\mathbb F_2) $, which is not hyperfinite, while every subfactor of the hyperfinite II $_1$ is hyperfinite. </p>
<p>Another way to see it is that $R $ is separable, while its ultrapowers aren't. </p>
|
1,342,069 | <p>In the <a href="https://en.wikipedia.org/wiki/Forgetful_functor" rel="nofollow">forgetful functor Wikipedia article</a> I read that </p>
<blockquote>
<p>"[Forgetful] Functors that forget the extra sets need not be faithful; distinct morphisms respecting the structure of those extra sets may be indistinguishable on the underlying set." </p>
</blockquote>
<p>Can anyone give me an example of a forgetful functor that is not faithful?</p>
| Alex G. | 130,309 | <p>One example is the forgetful functor from Schemes to Sets. Given two fields $k_1, k_2$, there may be many different field homomorphisms $k_1 \to k_2$ which give rise to many different morphisms $\text{Spec }k_2 \to \text{Spec }k_1$. However, the underlying sets of both of these schemes are single points, so there is a unique map between them in the category Sets.</p>
|
733,280 | <p>I cannot understand why $\log_{49}(\sqrt{ 7})= \frac{1}{4}$. If I take the $4$th root of $49$, I don't get $7$.</p>
<p>What I am not comprehending? </p>
| John Joy | 140,156 | <p>To get used to logarithm rules, try to relate them to exponent rules.</p>
<p>For example, think of a logarithm as the answer to the question "<i>a</i> to what power equals <i>b</i>". So the 2 following statements are equivalent to each other.</p>
<p>$\log a^?=b \iff ?=\log_ab$</p>
<p>Then try to convert logarithms to exponents, and manipulate them with the power rules that you're already familiar with.</p>
<p>$$x=\log_ab$$
$$a^x=b$$
$$(a^x)^c=b^c$$
$$a^{cx}=b^c$$
$$cx=\log_ab^c$$
$$c\log_ab=\log_ab^c$$</p>
<p>Try doing this with the other exponent rules and logarithms will be a snap.</p>
|
1,779,088 | <blockquote>
<p>Prove
$$\sum_{i=1}^n i^{k+1}=(n+1)\sum_{i=1}^n i^k-\sum_{p=1}^n\sum_{i=1}^p i^k \tag1$$
for every integer $k\ge0$. </p>
</blockquote>
<p>By principle of induction,</p>
<p>$$\sum_{i=1}^n i = n(n+1)- \sum_{p=1}^n p$$
$$2\sum_{i=1}^n i = n(n+1)$$
$$\sum_{i=1}^n i = \frac{n(n+1)}{2}$$
$\implies$$(1)$ is true for k equals to zero.</p>
<p>Assume $(1)$ is true.
$$\sum_{i=1}^n i^{k+2}=(n+1)\sum_{i=1}^n i^{k+1}-\sum_{p=1}^n\sum_{i=1}^p i^{k+1}\tag2$$
We prove $(2)$ is true.</p>
<p>From $(2)$,</p>
<p>$$\begin{align}
RHS & = (n+1)\left[(n+1)\sum_{i=1}^n i^k-\sum_{p=1}^n\sum_{i=1}^p i^k \right]-\sum_{p=1}^n\sum_{i=1}^p i^{k+1}\\
& = (n+1)^2\sum_{p=1}^n p^k-(n+1)\sum_{p=1}^n\sum_{i=1}^p i^k-\sum_{p=1}^n\sum_{i=1}^p i^{k+1}\\
& = \sum_{p=1}^n\left[(n+1)^2p^k-(n+1)\sum_{i=1}^p i^k-\sum_{i=1}^p i^{k+1}\right]\\
& = \sum_{p=1}^n\left[(n+1)^2p^k-\sum_{i=1}^p (n+1+i)i^k\right]\\
\end{align}$$</p>
<p>By examining $(2)$,</p>
<p>$$\sum_{p=1}^n\left[(n+1)^2p^k-\sum_{i=1}^p (n+1+i)i^k\right]=\sum_{p=1}^n p^{k+2}=LHS\tag3$$</p>
<p>We should be able to get $(3)$.</p>
<p>Anyone knows how to prove $(3)$?</p>
| Brian M. Scott | 12,042 | <p>You can also simply reverse the order of summation in the double sum:</p>
<p>$$\begin{align*}
(n+1)\sum_{i=1}^ni^k-\sum_{p=1}^n\sum_{i=1}^pi^k&=(n+1)\sum_{i=1}^ni^k-\sum_{i=1}^n\sum_{p=i}^ni^k\\
&=\sum_{i=1}^n(n+1)i^k-\sum_{i=1}^n(n-i+1)i^k\\
&=\sum_{i=1}^n\big((n+1)-(n+1-i)\big)i^k\\
&=\sum_{i=1}^ni\cdot i^k\\
&=\sum_{i=1}^ni^{k+1}
\end{align*}$$</p>
|
18,686 | <p>Let us define the following "dimension" of a Borel subet $B \subset \mathbb{R}^k$:</p>
<p>$\dim(B) = \min\{n \in \mathbb{N}: \exists K \subset \mathbb{R}^n, ~{\rm s.t.} ~ B \sim K\}$,</p>
<p>where $\sim$ denotes "homeomorphic to". Obviously, $0 \leq \dim(B) \leq k$.</p>
<p>I have three questions: Given a $B \subset \mathbb{R}$,<br>
1) As $k \to \infty$, how slow can $\dim(B^k)$ grow? Can we choose some $B$ such that $\dim(B^k) = o(k)$ or even $O(1)$?<br>
2) Will it make a difference if we drop the Borel measurability of $B$ or add the condition that $B$ has positive Lebesgue measure?<br>
3) Does this dimension-like notion have a name? The dimension concepts I usually see are Lebesgue's covering dimension, inductive dimension, Hausdorff dimension, Minkowski dimension, etc. I do not think the quantity defined above coincides with any of these, but of course bounds exist.</p>
<p>Thanks!</p>
| HenrikRüping | 3,969 | <p>Of course the point has the desired property, but I guess, this is not the space you are looking for. As François said, $C=\{0;1\}^\omega$ and so we get $C^2\cong C$.</p>
|
4,543,350 | <p>I am having a hard time figuring out this proof.</p>
<p>Let <span class="math-container">$\{x_n\}$</span> be a sequence and <span class="math-container">$x\in\mathbb{R}$</span>. Suppose for every <span class="math-container">$\epsilon>0$</span>, there is an M such that <span class="math-container">$|x_n-x|\leq \epsilon$</span> for all <span class="math-container">$n\geq M$</span>. Show that <span class="math-container">$\lim x_n=x$</span>.</p>
<p>My attempt:
Let <span class="math-container">$\epsilon>0$</span>. (Need to find M?) Let <span class="math-container">$n\geq M$</span>...<span class="math-container">$|x_n-x|\leq \epsilon$</span></p>
<p>My scratchwork:</p>
<p><span class="math-container">$\lim x_n=x$</span> <span class="math-container">$\Rightarrow$</span> <span class="math-container">$\lim x_n-x=0$</span> <span class="math-container">$\Rightarrow$</span> <span class="math-container">$\lim (x_n-x)=0$</span> ???</p>
<p>It's because it's so close to the definition that I don't know what to do. Any help is greatly appreciated.</p>
<p>Thanks!</p>
| FShrike | 815,585 | <p>A more high-level view. Convergence and limit properties can be defined with “open” inequalities <span class="math-container">$…<\varepsilon$</span> or “closed” ones: <span class="math-container">$…\le\varepsilon$</span> in any metric space. The closed and open balls are both local bases for the topology, so it doesn’t matter which criterion you use.</p>
<p>Fix any <span class="math-container">$\varepsilon>0$</span> and any <span class="math-container">$0<\varepsilon’<\varepsilon$</span>. Then <span class="math-container">$|x_n-x|\le\varepsilon’$</span> means <span class="math-container">$|x_n-x|<\varepsilon$</span>. The point is, when such inequalities can be satisfied for <em>any</em> <span class="math-container">$\varepsilon$</span>, you can always pick <span class="math-container">$\varepsilon’$</span> smaller to fit your needs. The definitions are equivalent.</p>
|
977,956 | <p>Can you help me solve this problem?</p>
<blockquote>
<p>Simplify: $\sin \dfrac{2\pi}{n} +\sin \dfrac{4\pi}{n} +\ldots +\sin \dfrac{2\pi(n-1)}{n}$.</p>
</blockquote>
| John | 105,625 | <p>$\sin k\alpha=\frac{\cos (k-1)\alpha-\cos (k+1)\alpha}{2\sin\alpha}$ by using compound angle formula given $\sin\alpha \neq0$. Then take summation the numerator will cancel. In your case $\alpha=\frac{2\pi}{n}, k=1\cdots n-1$. So you also need to discuss the case $n=1,2$ separately.</p>
|
134,673 | <p>I need to show that an automorphism of $S_n$ which takes transpositions to transpositions is an inner automorphism.</p>
<p>I thought it could be done by showing that such automorphisms form a subgroups $H\le Aut(S_n)$, that $Inn(S_n)\subset H$ and that they have the same number of elements. The number of inner automorphisms is $n!$ because $S_n$ has a trivial center (at least when it is not abelian) and therefore is isomorphic to $Inn(S_n)$. However I have no idea how I could count the number of elements in $H$.</p>
<p>Is there a way to do it, or should I change the approach altogether?</p>
<p>Thank you.</p>
| Jyrki Lahtonen | 11,619 | <p>Kannappan Sampath's suggestion can be completed as follows.</p>
<p>Let $f$ be an automorphism of $S_n$ with the property that it maps all transpositions to transpositions. So $f(1k)=(a_kb_k)$ for all $k=2,3,\ldots,n$, where $a_k\neq b_k$ are elements of the set $\{1,2,\ldots,n\}$. As the permutations $(12)$ and $(13)$ don't commute, their images $(a_2b_2)$ and $(a_3b_3)$ don't commute either. This means that the intersection $\{a_3,b_3\}\cap\{a_2,b_2\}$ is a singleton. Without loss of generality we can assume that $a_2=a_3=a$.</p>
<p>Next I claim that for all $k>3$ we also have $a\in\{a_k,b_k\}$. Assume contrariwise that for some $k>3$ we have $a_k\neq a\neq b_k$. Because $(1k)$ does not commute with $(12)$, we must have $b_2\in\{a_k,b_k\}$. Similarly, because $(1k)$ does not commute with $(13)$ either, we must also have $b_3\in\{a_k,b_k\}$, so $(a_kb_k)=(b_2b_3)$. But this is a contradiction, because then
$$
f(23)=f((13)(12)(13))=(ab_3)(ab_2)(ab_3)=(b_2b_3)=f(1k)
$$
violating the fact that $f$ is injective.</p>
<p>Thus all the transpositions $(a_kb_k)$ move the element $a$, and w.l.o.g. we can assume that $a_k=a$ for all $k$. All the integers $b_k\neq a$, and they must also be distinct, so the mapping $\sigma: 1\mapsto a, k\mapsto b_k$ is in $S_n$. We have shown that $f$ agrees with the inner automorphism $x\mapsto \sigma x\sigma^{-1}$ on all the $(n-1)$ generators $x=(1k),k=2,3,\ldots,n,$ of the group $S_n$, so the claim follows.</p>
|
2,572,304 | <p>Cauchy's Inequality states that, $$ \forall a, b \in R^{n}, |a \cdot b| \leq |a||b| $$. However, the dot product is $$ x \cdot y = x_{1}y_{1}+...+x_{n}y_{n}$$ while the norm of x is $$ |x| = \sqrt[2]{x_{1}^{2} +...+x_{n}^{2}} = \sqrt[2]{x \cdot x}$$. Therefore, $$ |a \cdot b| = \sqrt[2]{(a \cdot b) \cdot (a \cdot b)}$$</p>
<p>How does one calculate $$ (a \cdot b) \cdot (a \cdot b)$$? Since $$ (a \cdot b) \in R^{n} $$ when n = 1, is $$ (a \cdot b) \cdot (a \cdot b) $$ just multiplication of real numbers? (For some reason, I always thought that $$ n \geq 2 $$.)</p>
| Stephen Meskin | 465,208 | <p>$|x|$ for $x \in R^n$ with $n \ge 2$ is a generalization of $|x|$ for $x \in R^1$</p>
|
2,778,031 | <p>A ball is thrown vertically upward with u velocity. There is air resistance and the air resistance is directly proportional to square of ball's velocity,u. Find the height which the ball can reach.
I started with,
$$a=-g-{{k\over m}u^2},~~~~~~~~~
{d^2x\over dt^2} = -g - {k\over m}\left({dx\over dt}\right)^2$$</p>
<p>Is it true? and How can I solve this differential equation?</p>
| user | 505,767 | <p><strong>HINT</strong></p>
<p>Yes your set up is correct indeed we have that</p>
<ul>
<li>$a=-g-\frac k m \cdot v^2 \implies \frac{dv}{dt}=-g-\frac k m v^2 \implies \frac{v'(t)}{-g-\frac k m v^2}dt=dt$</li>
</ul>
<p>which can be integrated (see <a href="https://philosophicalmath.wordpress.com/2017/10/21/terminal-velocity-derivation/" rel="nofollow noreferrer"><strong>here</strong></a> for the details).</p>
|
2,778,031 | <p>A ball is thrown vertically upward with u velocity. There is air resistance and the air resistance is directly proportional to square of ball's velocity,u. Find the height which the ball can reach.
I started with,
$$a=-g-{{k\over m}u^2},~~~~~~~~~
{d^2x\over dt^2} = -g - {k\over m}\left({dx\over dt}\right)^2$$</p>
<p>Is it true? and How can I solve this differential equation?</p>
| David Quinn | 187,299 | <p>HINT...since you are looking for the height not the time, write $$a=v\frac{dv}{dx}=-(g+\frac kmv^2)$$</p>
|
3,317,728 | <p>Suppose that the moment generating function <span class="math-container">$M_X$$(t)$</span> of a random variable <span class="math-container">$X$</span> is given by </p>
<p><span class="math-container">$$ M_X(t)=\frac{e^t+e^{-t}}{6} + \frac 23 $$</span></p>
<p>I need to find the distribution function <span class="math-container">$F_X(x)$</span>.</p>
<p>Until now, I have been given (in my lecture notes) that I can express <span class="math-container">$E(X)$</span>= <span class="math-container">$M_X^{(1)}(0)$</span> . But I can't use this here for finding the distribution function <span class="math-container">$F_X(x)$</span>?(Or at least I have no idea how to do it) Could you please tell me how to proceed?</p>
| Feng | 624,428 | <p>Hint: From the moment generating function we can determine the distribution of <span class="math-container">$X$</span>, which is <span class="math-container">$P(X=1)=P(X=-1)=\frac16$</span>, <span class="math-container">$P(X=0)=\frac23$</span>. I believe that you can move on now. </p>
|
2,981,745 | <p>Wolfram Alpha shows that <span class="math-container">$$\int_{0}^{2\pi}x^2\ln (1-\cos x)dx = -\frac{8}{3} \pi (\pi^2 \ln(2) + 3 \zeta(3))$$</span>
I tried to use the Fourier series
<span class="math-container">$$\ln (1-\cos x)=-\sum_{n=1}^{\infty} \frac{\cos^nx}{n}.$$</span>
I am not sure how to continue from this point. I need some help.</p>
| mrtaurho | 537,079 | <p><strong>HINT</strong></p>
<p>By using the basic trigonometric identity</p>
<p><span class="math-container">$$1-\cos(x)=2\sin^2\left(\frac x2\right)$$</span></p>
<p>yor given integral becomes</p>
<p><span class="math-container">$$\begin{align}
\int_{0}^{2\pi}x^2\ln (1-\cos x)~dx &= \int_{0}^{2\pi}x^2 \ln\left(2\sin^2\left(\frac x2\right)\right)~dx\\
&=\int_{0}^{2\pi}x^2\ln(2)~dx+2\int_{0}^{2\pi}x^2 \ln\left(\sin\left(\frac x2\right)\right)~dx\\
&=\frac{8\pi^3}{3}\ln(2)+16\int_0^{\pi}x^2\ln(\sin x)~dx
\end{align}$$</span></p>
<p>where within the second integral the substitution <span class="math-container">$x=\frac x2$</span> was used. Now use the Fourier series expansion </p>
<p><span class="math-container">$$\ln(\sin x)=-\ln(2)-\sum_{n=1}^{\infty}\frac{\cos(2nx)}{n}$$</span></p>
<p>to further get</p>
<p><span class="math-container">$$\begin{align}
\frac{8\pi^3}{3}\ln(2)+16\int_0^{\pi}x^2\ln(\sin x)~dx&=\frac{8\pi^3}{3}\ln(2)+16\int_0^{\pi}x^2\left[-\ln(2)-\sum_{n=1}^{\infty}\frac{\cos(2nx)}{n}\right]~dx\\
&=\frac{8\pi^3}{3}\ln(2)-16\int_0^{\pi}x^2\ln(2)~dx-\sum_{n=1}^{\infty}\frac{16}n\int_0^{\pi}x^2\cos(2nx)~dx\\
&=-\frac{8\pi^3}3\ln(2)-\sum_{n=1}^{\infty}\frac{16}n\int_0^{\pi}x^2\cos(2nx)~dx
\end{align}$$</span></p>
<p>The second integral can be evaluated by applying integration by parts. Can you finish it from hereon?</p>
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.