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 |
|---|---|---|---|---|
481,952 | <p>Why is a union of infinitely many bounded sets not necessarily bounded, please? In addition, what condition can we add to make this union bounded, please?</p>
| Davide Giraudo | 9,849 | <p>Each subset of a metric space is a union of bounded sets (the singletons) but not every subset is bounded. </p>
|
2,143,510 | <p>NOTE: There are some other similar questions, but I got a negative answer to this question from my proof. Please find out the errors in my reasoning. </p>
<p>$\mathbf {Claim:}$ Is every point of every open set $E \subset R^2$ a limit point of E? Answer the same question for closed sets in $E \subset R^2$</p>
<p>From "Baby Rudin"</p>
<p>$\mathbf {Proof:}$ $\emptyset$ is both open and closed in every topological space. $R^2$ is a metric space, which is a kind of topological space, so $\emptyset$ is both open and closed in it. $\emptyset$ has no limit point because its neighborhood has no other point to include, so we can get a negative answer to both questions.</p>
<p>If possible, please have a look at the two questions that I got while reading the answer to this similiar question:
<a href="https://math.stackexchange.com/questions/1446589/proof-that-every-point-of-every-open-set-e">Proof that every point of every open set E⊂ℝ^2 is a limit point of E?</a>⊂ℝ2-is-a-limit-point-of-e</p>
<ol>
<li>"$q_s=(x_1+s,x_2)$" should be $q_s=(x_1+s,y_1)$, right?</li>
<li>I still don't understand why there should be $\epsilon$. Why can't r complete the proof?</li>
</ol>
<p>I don't have the right to comment on the original post led by the above link, so I ask the two questions here.
Finally, I find this forum very active, responsive and helpful, but not quite friendly to newcomers.</p>
| fleablood | 280,126 | <p>Every k is A $\iff $ there is no k that is not A.</p>
<p>So every point of the empty set is a pink alligator that eats square circles... because the empty set does not have any points that are not pink alligators that eat square circles... because the empty has no points at all so none of them can avoid being a pink alligator that eat square circles.</p>
<p>To the statement is true for the empty set. Every point in $\emptyset$ has property $X$ because there are no points in $\emptyset$ that <em>don't</em> have property $X$. </p>
<p>Every point <em>is</em> a limit point. And every point is not a limit point. And every point is green. And every point is not green. This is because there are zero points all "all of zero" is .... zero. And zero of the points are limit points. So all points are limit points because zero is all the points there are and zero are limit points.</p>
<p>So that is not a counter example.</p>
<p>1) Yes, that was a typo.</p>
<p>2)Because $r$ defines <em>one</em> neighborhood. We must prove this is true for <em>all</em> neighborhoods.</p>
|
1,649,194 | <p>Let $S^n\subset\mathbb{R}^{n+1}$ denote the standard unit sphere with normal bundle $\nu$, let $N=(0,\dots,0,1)$ and $S=(0,\dots,0,-1)$. Then there are two sterographic projections
$$\sigma_+\colon S^n-S\to\mathbb{R}^n $$
and
$$\sigma_-\colon S^n-N\to\mathbb{R}^n$$
Both of these maps are homeomorphisms and they form an atlas for the standard smooth structure on $S^n$. I'm interested in how they should induce the standard orientation on $S^n$. (Here I orient $S^n$ by putting the standard orientation on $T S^n \oplus \nu\cong T\mathbb{R}^{n+1}|_{S^n}$ and orient $\nu$ by declaring that the outward-pointing direction is positive.)</p>
<p>Let $v\in S^{n-1}\subset S^n$ so that $\sigma_+^{-1}\circ \sigma_-(v)=v$, and let $B=\{b_1,\dots,b_n\}$ be a positively-oriented orthonormal basis for $T_vS^n$ where $b_1$ points along the great circle from $N$ to $S$. Geometrically, it seems like $D_v (\sigma_+^{-1}\circ \sigma_-) (b_i)=-b_1$ if $i= 1$ and $b_i$ otherwise, suggesting that the transition function is orientation reversing, and hence exactly one of $\sigma_+$ and $\sigma_-$ is orientation reversing. My question is which one reverses orientation and which preserves?</p>
<p>I haven't succeeded in computing anything in terms of the formulas for stereographic projection. The Jacobean matrix I get in the 2-dimensional case for $\sigma_+$ or $\sigma_-$ is $2\times 3$ so I don't know how I'm supposed to interpret its "determinant". </p>
<p>(I should point out that in general my experiences with differential geometry have been very, very, very bad, and I'm much more topologically minded. In particular my definition for an orientation of a vector bundle is a Thom class.)</p>
| user43883 | 114,453 | <p>$\def\ip {\, \lrcorner \, }$
I get a different result than the previous answer: $\sigma^-$ preserves orientation iff $n$ is even:</p>
<p>Consider $\Bbb S^{n-1}$ embedded in $\Bbb R^n$.The orientation form on $\Bbb R^n$ is $dx_1 \wedge \ldots \wedge dx_n$.</p>
<p>If $v(x)$ is a non-vanishing vector field on $\Bbb S^{n-1}$ then the induced orientation form on $\Bbb S^{n-1}$ is $v \ip (dx_1 \wedge \ldots dx_n)$. In particular the standard orientation on $\Bbb S^{n-1}$ is given by using the outward normal to the sphere.</p>
<p>If $p \in \Bbb R^n$, and we can identify $T_p\Bbb R^n$ with $\Bbb R^n$. If $p \in \Bbb S^{n-1}$ then of course the outward normal is just $p$. Hence the orientation form on $\Bbb S^{n-1}$ is $\hat \omega = p \ip (dx_1 \wedge \ldots \wedge dx_n)$.</p>
<p>The stereographic projection map $\sigma^- : \Bbb S^{n-1} \rightarrow \Bbb R^{n-1}$ is defined by:</p>
<p>$$(u_1, \ldots u_{n-1}) = \sigma^-(x_1, \ldots, x_n) = \left({ x_1, \ldots , x_{n-1} \over 1 - x_n} \right)$$</p>
<p>It should be clear that at $-e_n = (0, \ldots, 0, -1)$ we have:</p>
<p>$$d \sigma^-_{e_n}\left({\partial \over \partial x_i}\right) = {\partial \over \partial u_i} \text{ for }i < n\\
d \sigma^-_{e_n}\left({\partial \over \partial x_n}\right) = 0.$$</p>
<p>If we identify tangent vectors in $\Bbb R^{n-1}$ with $\Bbb R^{n-1}$, then set of tangent vectors $(e_1, \ldots, e_{n-1})$ is trivially mapped to $(e_1, \ldots, e_{n-1})$.</p>
<p>In $\Bbb R^{-1}$, $(e_1, \ldots, e_{n-1})$ is positively orientated, but on $\Bbb S^{n-1}$,</p>
<p>$$\begin{align}\hat \omega (e_1, \ldots, e_{n-1}) &= -e_n \ip (dx_1 \wedge \ldots \wedge dx_n)(e_1, \ldots, e_{n-1}) \\
&= (dx_1 \wedge \ldots \wedge dx_n) (-e_n, e_1, \ldots, e_{n-1})\\
&= (-1)^{n-1}(dx_n \wedge dx_1 \wedge \ldots \wedge dx_{n-1}) (-e_n, e_1, \ldots, e_{n-1})\\
&= (-1)^{n}(dx_n \wedge dx_1 \wedge \ldots \wedge dx_{n-1}) (e_n, e_1, \ldots, e_{n-1})\\
&= (-1)^{n} \\\end{align} $$</p>
<p>Thus at the point $-e_n$, $\sigma^-$ is orientation reversing iff $n$ is odd. Since $\sigma^-$ is diffeomorphism, it has to have the same orientation everywhere, so $\sigma^-$ preserves orientation iff $n$ is even.</p>
<p>Am I mistaken somehow?</p>
|
1,649,194 | <p>Let $S^n\subset\mathbb{R}^{n+1}$ denote the standard unit sphere with normal bundle $\nu$, let $N=(0,\dots,0,1)$ and $S=(0,\dots,0,-1)$. Then there are two sterographic projections
$$\sigma_+\colon S^n-S\to\mathbb{R}^n $$
and
$$\sigma_-\colon S^n-N\to\mathbb{R}^n$$
Both of these maps are homeomorphisms and they form an atlas for the standard smooth structure on $S^n$. I'm interested in how they should induce the standard orientation on $S^n$. (Here I orient $S^n$ by putting the standard orientation on $T S^n \oplus \nu\cong T\mathbb{R}^{n+1}|_{S^n}$ and orient $\nu$ by declaring that the outward-pointing direction is positive.)</p>
<p>Let $v\in S^{n-1}\subset S^n$ so that $\sigma_+^{-1}\circ \sigma_-(v)=v$, and let $B=\{b_1,\dots,b_n\}$ be a positively-oriented orthonormal basis for $T_vS^n$ where $b_1$ points along the great circle from $N$ to $S$. Geometrically, it seems like $D_v (\sigma_+^{-1}\circ \sigma_-) (b_i)=-b_1$ if $i= 1$ and $b_i$ otherwise, suggesting that the transition function is orientation reversing, and hence exactly one of $\sigma_+$ and $\sigma_-$ is orientation reversing. My question is which one reverses orientation and which preserves?</p>
<p>I haven't succeeded in computing anything in terms of the formulas for stereographic projection. The Jacobean matrix I get in the 2-dimensional case for $\sigma_+$ or $\sigma_-$ is $2\times 3$ so I don't know how I'm supposed to interpret its "determinant". </p>
<p>(I should point out that in general my experiences with differential geometry have been very, very, very bad, and I'm much more topologically minded. In particular my definition for an orientation of a vector bundle is a Thom class.)</p>
| Narasimham | 95,860 | <p><em>Inversions</em> in the plane and likewise stereographic projection from $\mathbb R^2 $ (having an inversion component) reverse sense in projections of a closed loop of the surface. </p>
|
3,403,255 | <p>I am trying to follow wikipedia's page about matrix rotation and having a hard time understanding where the formula comes from.</p>
<p><a href="https://en.wikipedia.org/wiki/Rotation_matrix" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Rotation_matrix</a> Wiki page about it.</p>
<p>what i have so far:</p>
<p>y<sub>2</sub>=sin(<em>a<sub>1</sub>+a<sub>2</sub></em>)R -> where R is hypotenuse, a1 is current angle and a2 is the angle by which something must rotate.</p>
<p>this how i used to calculate my rotation, but it takes long time to compute and uses up a lot of cpu time for square roots and other heavy stuff due tot he need of finding the initial angle.</p>
<p>So i decided to reduce computation time and found that sin(a1+a2) could be writen as <code>sin(a1)cos(a2)+cos(a1)sin(a2)</code> and from there i got to the point where it is:</p>
<p>y<sub>2</sub>=y<sub>1</sub>cos(<em>a<sub>2</sub></em>)+xsin(<em>a<sub>2</sub></em>)sin(<em>a<sub>1</sub></em>)</p>
<p>But wiki page says that it must b:</p>
<p>y<sub>2</sub>=y<sub>1</sub>cos(<em>a<sub>2</sub></em>)+xsin(<em>a<sub>2</sub></em>)</p>
<p><a href="https://i.stack.imgur.com/vRzo7.png" rel="nofollow noreferrer">My work book</a></p>
| John Alexiou | 3,301 | <p>Consider a rotated rectangle with sides <span class="math-container">$a$</span> and <span class="math-container">$b$</span></p>
<p><a href="https://i.stack.imgur.com/o7w5e.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/o7w5e.png" alt="sketch"></a></p>
<p>Do the trigonometry to find the <em>x</em> and <em>y</em> coordinates of point <strong>P</strong>.</p>
<p>The red triangle contributes <span class="math-container">$a \cos \theta$</span> horizontally, and <span class="math-container">$a \sin \theta$</span> vertically. The blue triangle contributes <span class="math-container">$-b \sin\theta$</span> horizontally, and <span class="math-container">$b\cos \theta$</span> vertically.</p>
<p>Add them up for</p>
<p><span class="math-container">$$ \pmatrix{Px \\ Py} = \pmatrix{ a \cos \theta - b \sin \theta \\ a \sin \theta + b \cos \theta} $$</span></p>
<p>Now factor the rotation matrix <span class="math-container">$\mathbf{R}$</span></p>
<p><span class="math-container">$$ \pmatrix{Px \\ Py} = \pmatrix{\cos\theta & -\sin \theta \\ \sin\theta & \cos\theta} \pmatrix{a \\ b} $$</span></p>
<p>The vector <span class="math-container">$\pmatrix{Px & Py}$</span> is the rotated vector <span class="math-container">$\pmatrix{a & b}$</span> by an angle <span class="math-container">$\theta$</span> and from geometry you derive the rotation matrix.</p>
|
2,634,701 | <p>Let $ f: {{\mathbb{R^n}} \rightarrow {{\mathbb{R}} }}$ be continuous and let $a$ and $b$ be points in $ {{\mathbb{R} }} $
Let the function $g: {\mathbb{R}} \rightarrow {\mathbb{R}}$ be defined as:
$$ g(t) = f(ta+(1-t)b) $$
Show that $g$ is continuous .</p>
<p>If I define a function $ h(t)=ta+(1-t)b$, then I have that $g(t)=f(h(t))$
I know that $f$ is continuous, so I have to prove that $h(t)$ is continuous as a compound function of two continuous function is also continuous. </p>
<p>How do I prove that $h(t)$ is continuous in ${{\mathbb{R^n}}}$? </p>
| ChoMedit | 527,854 | <p>Just use $\varepsilon-\delta$ argument to solve the problem.
Choose arbitrary point $t_0$ in $\mathbb{R}$.
$$
\forall \varepsilon>0 \exists \delta>0 : |t-t_0|<\delta \Rightarrow |h(t)-h(t_0)|<\varepsilon
$$</p>
<p>We could derive the equality, $|h(t)-h(t_0)|=|(t-t_0)(a-b)|=|t-t_0||a-b|$. Note that it is because $|\cdot|$ is norm of $\mathbb{R}^n$.</p>
<p>So if we took $\delta$ as $\frac{\varepsilon}{|a-b|}$, then the argument holds.</p>
<p>This argument is independent of choosing the point $t_0$. So, $h(t)$is continuous function $\mathbb{R} \to \mathbb{R}^n$.</p>
|
2,084,624 | <p>The question is : </p>
<p>Is $\sum_{k=1}^\infty \frac{(-3)^k(k!)}{k^k}$ convergent? </p>
<p>Note : I can't find the limit of its main term. I know the answer must be related to some test about convergence of series ... I don't know which one and i can't find the limit.</p>
| Jan Eerland | 226,665 | <p>Using the ratio test:</p>
<p>$$\lim_{\text{k}\to\infty}\left|\frac{\left(\frac{\left(-\text{n}\right)^{\text{k}+1}\cdot\left(\left(\text{k}+1\right)!\right)}{\left(\text{k}+1\right)^{\text{k}+1}}\right)}{\left(\frac{\left(-\text{n}\right)^\text{k}\cdot\left(\text{k}!\right)}{\text{k}^\text{k}}\right)}\right|=\lim_{\text{k}\to\infty}\left|-\text{n}\left(\frac{\text{k}}{1+\text{k}}\right)^\text{k}\right|=\left|\text{n}\right|\lim_{\text{k}\to\infty}\left|\left(\frac{\text{k}}{1+\text{k}}\right)^\text{k}\right|=\frac{\left|\text{n}\right|}{e}>1$$</p>
<p>Which leads to:</p>
<p>$$\left|\text{n}\right|>e$$</p>
|
31,767 | <p>I am looking for "low-complexity" indexing methods to enumerate binary sequences of a given length and a given weight. </p>
<p>Formally, let $T_k^n = \{x_1^n \in \{0,1\}^n: \sum_{i=1}^n x_i = k\}$. How to construct a bijective mapping $f: T_k^n \to \{1, 2, \ldots, \binom{n}{k}\}$ such that computing each $f(x_1^n)$ needs small number of operations?</p>
<p>For example, one could do <em>lexicographical ordering</em>, that is, e.g., $0110 < 1010$. Then this gives the following scheme:</p>
<p>$f(x_1^n) = \sum_{k=1}^n x_k \binom{n-k}{w_k}$</p>
<p>where $w_k=\sum_{i=k}^n x_i$. Computing $n$ binomial coefficients can be quite demanding. Any other ideas? Or is it impossible to avoid?</p>
| H A Helfgott | 398 | <p>To answer a remark above: yes, I think it would be a very good idea to get together group of interested people to build a commentary. What non-interested people (or people who haven't read and will not read Vinogradov) can do is suggest what current technical tools would be most appropriate for such a collaborative project. I have no idea about that myself.</p>
|
1,117,592 | <blockquote>
<p>Let <span class="math-container">$k$</span> be a finite field and <span class="math-container">$V$</span> a finite-dimensional vector space over <span class="math-container">$k$</span>.</p>
<p>Let <span class="math-container">$d$</span> be the dimension of <span class="math-container">$V$</span> and <span class="math-container">$q$</span> the cardinal of <span class="math-container">$k$</span>.</p>
<p>Construct <span class="math-container">$q+1$</span> hyperplanes <span class="math-container">$V_1,\ldots,V_{q+1}$</span> such that <span class="math-container">$V=\bigcup_{i=1}^{q+1}V_i$</span></p>
</blockquote>
<p>I tried induction on <span class="math-container">$d$</span>, to no avail.</p>
<p>There has been discussion about this topic (<a href="http://alpha.math.uga.edu/%7Epete/coveringnumbersv2.pdf" rel="nofollow noreferrer">here</a>), but it doesn't answer my question.</p>
| xxxxxxxxx | 252,194 | <p>It is useful in this case to consider the dual. In the dual space, your collection of hyperplanes covering every vector corresponds to a set of <span class="math-container">$(q+1)$</span> <span class="math-container">$1$</span>-dimensional subspaces such that each hyperplane contains at least one of these <span class="math-container">$1$</span>-spaces. If we fix a <span class="math-container">$2$</span>-dimensional subspace <span class="math-container">$\pi$</span>, then <span class="math-container">$\pi$</span> contains <span class="math-container">$q+1$</span> <span class="math-container">$1$</span>-dimensional subspaces, and every hyperplane meets <span class="math-container">$\pi$</span> in a <span class="math-container">$1$</span>-dimensional subspace.</p>
<p>This shows that by taking all of the <span class="math-container">$1$</span>-dimensional subspaces in a common plane, we have the desired property. Then we can replace each of these <span class="math-container">$1$</span>-dimensional subspaces with its dual (through any mapping, sending them to the orthogonal complement under the dot product will work), and we obtain a set of <span class="math-container">$q+1$</span> hyperplanes covering all of the points of <span class="math-container">$V$</span>.</p>
|
3,397,548 | <p>For a sequence <span class="math-container">$\{x_n\}_{n=1}^{\infty}$</span>, define <span class="math-container">$$\Delta x_n:=x_{n+1}-x_n,~\Delta^2 x_n:=\Delta x_{n+1}-\Delta x_n,~(n=1,2,\ldots)$$</span> which are named <strong>1-order</strong> and <strong>2-order difference</strong>, respectively. </p>
<p>The problem is stated as follows:</p>
<blockquote>
<p>Let <span class="math-container">$\{x_n\}_{n=1}^{\infty}$</span> be <strong>bounded</strong> , and satisfy
<span class="math-container">$\lim\limits_{n \to \infty}\Delta^2 x_n=0$</span>. Prove or disprove
<span class="math-container">$\lim\limits_{n \to \infty}\Delta x_n=0.$</span></p>
</blockquote>
<p>By intuiton, the conclusion is likely to be true. According to <span class="math-container">$\lim\limits_{n \to \infty}\Delta^2 x_n=0,$</span> we can estimate <span class="math-container">$\Delta x_n$</span> almost equal with an increasing <span class="math-container">$n$</span>. Thus, <span class="math-container">$\{x_n\}$</span> looks like an <strong>arithmetic sequence</strong>. If <span class="math-container">$\lim\limits_{n \to \infty}\Delta x_n \neq 0$</span>, then <span class="math-container">$\{x_n\}$</span> can not be bounded.</p>
<p>But how to prove it rigidly?</p>
| robjohn | 13,854 | <p>Since <span class="math-container">$x_n$</span> is bounded, choose <span class="math-container">$M$</span> so that <span class="math-container">$|x_n|\le M$</span>.</p>
<p>Suppose that <span class="math-container">$\lim\limits_{n\to\infty}\Delta x_n\ne0$</span>. Then <span class="math-container">$\exists\epsilon\gt0:\forall n_0,\exists n\ge n_0:|\Delta x_n|\ge\epsilon$</span>.</p>
<p>Let <span class="math-container">$\delta=\frac{\epsilon^2}{6M}$</span>. Since <span class="math-container">$\lim\limits_{n\to\infty}\Delta^2x_n=0$</span>, choose <span class="math-container">$n_0$</span> so that if <span class="math-container">$n\ge n_0$</span>, we have <span class="math-container">$\left|\Delta^2x_n\right|\le\delta$</span>.</p>
<p>Choose <span class="math-container">$n\ge n_0$</span> so that <span class="math-container">$|\Delta x_n|\ge\epsilon$</span>.</p>
<p>Let <span class="math-container">$k=\left\lceil\frac{6M}\epsilon\right\rceil$</span>. Note that <span class="math-container">$(k-1)\delta\lt\epsilon$</span> and <span class="math-container">$k\epsilon\ge6M$</span>.</p>
<p>By the choice of <span class="math-container">$n_0$</span> and <span class="math-container">$n$</span>,
<span class="math-container">$$
|\Delta x_{n+j}|\ge\epsilon-j\delta
$$</span>
Therefore,
<span class="math-container">$$
\begin{align}
|x_{n+k}-x_n|
&\ge\sum_{j=0}^{k-1}(\epsilon-j\delta)\\
&=k\epsilon-\frac{k(k-1)}2\delta\\[3pt]
&\gt\frac{k\epsilon}2\\[9pt]
&\ge3M
\end{align}
$$</span>
which contradicts the choice of <span class="math-container">$M$</span>. Thus,
<span class="math-container">$$
\lim_{n\to\infty}\Delta x_n=0
$$</span></p>
|
3,811,753 | <p>Show that the equation:</p>
<p><span class="math-container">$$
y’ = \frac{2-xy^3}{3x^2y^2}
$$</span></p>
<p>Has an integration factor that depends on <span class="math-container">$x$</span> And solve it that way.</p>
<hr />
<p>Already we got to:</p>
<p><span class="math-container">$$
y’ + \frac{xy^3}{3x^2y^2} = \frac{2}{3x^2y^2}
$$</span></p>
<p>Therefore:</p>
<p><span class="math-container">$$
y’ + \frac{1}{3x}y = \frac{2}{3x^2}y^{-2}
$$</span></p>
<p>But, in order to get an integration factor, shouldn’t we have a linear equation? Of the form:</p>
<p><span class="math-container">$$
y‘ + p(x)y = g(x)
$$</span></p>
<p>That way getting the integration factor:</p>
<p><span class="math-container">$$
\mu = ke^{\int p(x)}, k \in R
$$</span></p>
<p>But what we have is a non-linear equation, so how could an integration factor exists?</p>
<p>Thanks.</p>
| Claude Leibovici | 82,404 | <p>Make life easier letting
<span class="math-container">$$y=\frac z {\sqrt[3]x}\implies 3 x z'(x)=\frac{2}{z(x)^2}$$</span> which is simpler</p>
|
3,413,364 | <blockquote>
<p>Consider the set of points <span class="math-container">$$O = \{ x \in P \mid \alpha^* = C^T x \}$$</span> where <span class="math-container">$P \subseteq \mathbb R^n$</span> is a closed convex set, <span class="math-container">$C \in \mathbb R^n$</span> and <span class="math-container">$\alpha^* = \min \{ C^Tx \}$</span>. Then, <span class="math-container">$O$</span> is closed convex set.</p>
</blockquote>
<p>This seems a pretty simple statement in my linear programming class but I am unsure how to show it formally. I can easily show it is a convex set but I am not sure how to show it is a closed set.</p>
| celtschk | 34,930 | <p>First, the function <span class="math-container">$f(x)=C^T x$</span> is a finite-dimensional linear function, and therefore continuous.</p>
<p>Also, in <span class="math-container">$\mathbb R^n$</span> single-element subsets are always closed; <span class="math-container">$\{\alpha^*\}$</span> is such a set.</p>
<p>Now the preimage of closed sets under continuous functions is closed. The preimage of <span class="math-container">$\{\alpha^*\}$</span> is <span class="math-container">$f^{-1}[\{\alpha^*\}]=\{x\in\mathbb R^n|C^Tx=\alpha^*\}$</span>. Therefore this set also is closed.</p>
<p>Finally, <span class="math-container">$P$</span> is closed by assumption, and the intersection of two closed sets is closed. But <span class="math-container">$P\cap f^{-1}[\{\alpha^*\}] = \{x\in P|C^Tx=\alpha^*\}$</span>, which is exactly the set you asked about.</p>
|
367,364 | <p>How to calculate the following integral:</p>
<p>$\int^{R}_{0}[2 \cos^{-1}(\frac{r}{2R}) -\sin(2 \cos^{-1}(\frac{r}{2R}) ) ] dr$.</p>
<p>This is a part of a complex formula.</p>
| vonbrand | 43,946 | <p>You have $a_{n + 1} = 2 a_n - (-1)^n$, and thus:
$$
\frac{A(z) - a_0}{z} = 2 A(z) - \frac{1}{1 + z}
$$</p>
|
367,364 | <p>How to calculate the following integral:</p>
<p>$\int^{R}_{0}[2 \cos^{-1}(\frac{r}{2R}) -\sin(2 \cos^{-1}(\frac{r}{2R}) ) ] dr$.</p>
<p>This is a part of a complex formula.</p>
| Matt L. | 70,664 | <p>You can also try to see the pattern:</p>
<p>$$a_1 = 2a_0 - 1\\
a_2 = 2\cdot 2 a_0 - 2 + 1\\
a_3 = 2\cdot 2\cdot 2a_0 - 2\cdot 2 + 2 - 1\\\vdots\\
a_n = 2^na_0 + (-1)^n\sum_{k=0}^{n-1}(-2)^k$$
The sum can of course be written in closed form.</p>
|
892,114 | <p>i have three number
1 2 3 which will always be in this order {123}, i want to find out number of cases can be made,
like {1},{2},{23},{13},{12},{123}{3},{}. but each number has two states like "a" "b", i.e, each one will become different entity,like 2a,2b,3a,3b,1a,
with only exception i.e. 1 will have only one state 1a.</p>
<p>please tel me step wise using formulas, so that i can understand, also, any link will be helpfull.
yours sincerly</p>
| amWhy | 9,003 | <p>$$(x-2)^2 = (x - 2)(x-2) = x^2 - 2x -2x + (-2)(-2) = x^2 - 4x + 4$$</p>
<p>This is called <em>expanding</em> $(x-2)^2$. We factor $x^2 - 4x + 4$ when we write it as the product of its factors, in this case $(x-2)(x-2) = (x-2)^2$.</p>
<p>Now,
$$(x-2)^2-12=(x^2-4x+4)-12=x^2-4x-8$$
You can find the zeros of the quadratic by setting $x^2 - 4x - 8 = 0 $, then using the quadratic formula, which will yield $x = 2\pm 2\sqrt 3$.</p>
|
187,975 | <p>Let $\mu$ be a finite nonatomic measure on a measurable space $(X,\Sigma)$, and for simplicity assume that $\mu(X) = 1$. There is a well-known "intermediate value theorem" of Sierpiński that states that for every $t \in [0,1]$, there exists a set $S \in \Sigma$ with $\mu(S) = t$.</p>
<p>I would like to use the following stronger conclusion for such a measure: </p>
<blockquote>
<p>There exists a chain of sets $\{S_t \mid t \in [0,1]\}$ in $\Sigma$,
with $S_t \subseteq S_r$ whenever $0 \leq s \leq r \leq 1$, such that
$\mu(S_t) = t$ for all $t \in [0,1]$.</p>
</blockquote>
<p>(One can view this as the existence a right inverse to the map $\mu \colon \Sigma \to [0,1]$ in the category of partially ordered sets.)</p>
<p>This statement appears (albeit hidden within a proof) on the Wikipedia page for "<a href="http://en.wikipedia.org/wiki/Atom_%28measure_theory%29#Non-atomic_measures" rel="noreferrer">Atom (measure theory)</a>," and even includes a sketch for the proof! However, I would like to see some mention of this in the literature. I've checked the Wiki references and they both seem to prove the weaker statement. I looked in Fremiln's <em>Measure Theory</em>, vol. 2, and again found the weaker version but not the stronger. </p>
<p><strong>Question:</strong> Can anyone provide me with such a reference?</p>
<hr>
<p><strong>A proof.</strong> In case anyone stumbles to this page and wants to see a proof, I'll sketch one that is more constructive than the one that I linked to above. Set $S_0 = \varnothing$ and $S_1 = X$. By Sierpiński, there exists $S_{1/2} \in \Sigma$ of measure $1/2$. For each Dyadic rational $q = m/2^n \in [0,1]$ ($1 \leq m \leq 2^n$), we may proceed by induction on $n$ to construct each $S_q$. Now given $r \in [0,1]$, set $S_r = \bigcup_{q \leq r} S_q$. (This is essentially the same method of proof as the one in the reference provided in Ramiro de la Vega's answer.)</p>
| Henrique de Oliveira | 18,474 | <p>There's a stronger version of that (basic) theorem due to Lyapunov. It is stronger because it concerns vectors of measures, and not only a single measure. It states that given a non-atomic vector measure (a collection of $n$ measures $\mu_1,\ldots, \mu_n$ where each measure is non-atomic) always has an image which is convex (in $\mathbb{R}^n$).</p>
<p>Unfortunately, I could never find a translation of his paper, so I can only link the <a href="http://www.mathnet.ru/links/e8bebeff1522d3398898112145091ef2/im3907.pdf" rel="nofollow">version in russian</a>. The main statements can be found in French at the end of the paper. There's also a <a href="http://www.ams.org/journals/bull/1948-54-04/S0002-9904-1948-09020-6/S0002-9904-1948-09020-6.pdf" rel="nofollow">paper of Halmos</a> that proves the result.</p>
<p>Maybe looking at the proof method or subsequent papers you can find the chain statement that you seek.</p>
|
3,294,123 | <p>Define <span class="math-container">$f(x)=x^{-1}(\log x)^{-2}$</span> if <span class="math-container">$0<x<\frac{1}{2}$</span>, <span class="math-container">$f(x)=0$</span> on the rest of <span class="math-container">$R$</span>. Then <span class="math-container">$f \in L^1(R)$</span>. Show that
<span class="math-container">$$(Mf)(x) \ge |2x \log(2x)|^{-1} \;\;(0<x<1/4)$$</span> so that <span class="math-container">$\int_0^1 (Mf)(x)dx=\infty.$</span></p>
<p><span class="math-container">$$
(Mf)(x) = \sup_{0<r<\infty} \frac{1}{m(B_r)}\int_{B_r(x)}|f(y)|\mathrm{d}y
$$</span>
where <span class="math-container">$B_r(z)$</span> is a ball centered around <span class="math-container">$x$</span> with radius <span class="math-container">$r$</span>. </p>
<p>I am stuck with proving this lower bound for the maximality function. I would greatly appreciate any help.</p>
| zhw. | 228,045 | <p>A simpler way to show <span class="math-container">$Mf\notin L^1$</span> is to note there is a constant <span class="math-container">$c>0$</span> such that <span class="math-container">$|f|>c$</span> on a set of positive measure <span class="math-container">$E\subset (0,1/2).$</span> Then for <span class="math-container">$x > 1/2,$</span></p>
<p><span class="math-container">$$Mf(x) \ge \frac{1}{B(x,x)}\int_{B(x,x)} |f| \ge \frac{1}{2x}\cdot c\cdot m(E),$$</span></p>
<p>which is not in <span class="math-container">$L^1.$</span></p>
|
1,621,363 | <p>Integrate: $$\int \frac{\sin(x)}{9+16\sin(2x)}\,\text{d}x.$$</p>
<p>I tried the substitution method ($\sin(x) = t$) and ended up getting $\int \frac{t}{9+32t-32t^3}\,\text{d}t$. Don't know how to proceed further. </p>
<p>Also tried adding and substracting $\cos(x)$ in the numerator which led me to get $$\sin(2x) = t^2-1$$ by taking $\sin(x)+\cos(x) = t$. </p>
<p>Can't figure out any other method now. Any suggestions or tips? </p>
| Jack Tiger Lam | 186,030 | <p>To attack this integral, we will need to make use of the following facts:</p>
<p>$$(\sin{x} + \cos{x})^2 = 1+\sin{2x}$$</p>
<p>$$(\sin{x} - \cos{x})^2 = 1-\sin{2x}$$</p>
<p>$$\text{d}(\sin{x}+\cos{x}) = (\cos{x}-\sin{x})\text{d}x$$</p>
<p>$$\text{d}(\sin{x}-\cos{x}) = (\cos{x}+\sin{x})\text{d}x$$</p>
<p>Now, consider the denominator.</p>
<p>It can be rewritten in two different ways as hinted by the above information.</p>
<p>$$9+16\sin{2x} = 25 - 16(1-\sin{2x}) = 16(1+\sin{2x})-7$$</p>
<p>$$9+16\sin{2x} = 25 - 16(\sin{x}-\cos{x})^2 = 16(\sin{x}+\cos{x})^2-7$$</p>
<p>Also note that </p>
<p>$$\text{d}(\sin{x}-\cos{x})-\text{d}(\sin{x}+\cos{x}) = 2\sin{x}\text{d}x$$</p>
<p>By making the substitutions </p>
<p>$$u = \sin{x}+\cos{x}, v = \sin{x}-\cos{x}$$</p>
<p>The integral is transformed into two separate integrals which can be evaluated independently.</p>
<p>$$2I = \int \frac{\text{d}v - \text{d}u}{9+16\sin{2x}} = \int \frac{\text{d}v}{25-16v^2} + \int \frac{\text{d}u}{7-16u^2}$$</p>
<p>The remainder of this evaluation is left as an exercise to the reader.</p>
|
2,887,440 | <p>We were asked in our Calculus class to prove that,</p>
<blockquote>
<p>$f(x+y) - f(x) = \frac {\sec^2(x) \tan(y)} {1 - \tan(x) \tan(y)}$ given that $f(x) = \tan(x)$</p>
</blockquote>
<p>I have gotten so far as:</p>
<p>$$f(x+y) - f(x)$$</p>
<p>$$\tan(x+y) - \tan(x)$$</p>
<p>$$\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)} - \tan(x)$$</p>
<p>$$\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)} + \frac{-\tan(x)+\tan^2(x)\tan(y)}{1-\tan(x)\tan(y)}$$</p>
<p>$$\frac{\tan(y) + \tan^2(x)\tan(y)}{1-\tan(x)\tan(y)}$$</p>
<p>$$\frac{\tan(y) [1+\tan^2(x)]}{1-\tan(x)\tan(y)}$$</p>
<blockquote>
<p>Substituting the pythagorean identity, $$1+\tan^2(x) = \sec^2(x)$$</p>
</blockquote>
<p>$$\frac{\tan(y) \sec^2(x)}{1-\tan(x)\tan(y)} = \boxed{\frac{\sec^2(x)\tan(y)}{1-\tan(x)\tan(y)}}$$ </p>
<p>I don't quite understand how $f(x+y)$ became $\tan(x+y)$. I've had a few search results stating that $f(x+y) = f(x)+f(y)$ but it does not quite fit the bill. </p>
<p>I got the idea for my solution above because of a textbook example I've read, where:</p>
<blockquote>
<p>Given $f(x)=x^2-4x+7$, find $\frac {f(x+h)-f(x)}{h}$</p>
<blockquote>
<p>$\frac{[(x+h)^2 - 4(x+h) + 7] - (x^2 - 4x + 7)}{h} = \frac{h(2x+h-4)}{h} = 2x+h-4$</p>
</blockquote>
</blockquote>
<p>...but the book did not describe what property was used in order to 'insert' the value of $f(x)$ into $f(x+h)$, and by extension the $f(x)$ into the $f(x+y)$ of my problem. They feel... similar.</p>
<p>Is there a name for this mathematical property? Thank you very much.</p>
| Henrik supports the community | 193,386 | <p>It's simply substitution of the argument to $f$.</p>
<p>On the other hand: $f(x+y)=f(x)+f(y)$ that you claim to have found several sources for, is <em>not</em> generally true.</p>
|
233,618 | <p>I want to be able to take a polynomial and take the 1st 5 derivatives, then add at least one root of each derivative to a list using a loop. However, each attempt I try only ends up outputting the roots of the 5th derivative, not the rest. So far I have:</p>
<pre><code>rootderivs[n_]:=(
p[x_]:= x^8-3x^5+x-1;
rootlist={};
Do[
AppendTo[rootlist, NSolve[D[p[x],{x,n}]==0]],1];
Print[rootlist])
</code></pre>
<p>Which gives an ouput of :</p>
<pre><code>rootderivs[5]
{{{x->-0.188487-0.326469 I},{x->-0.188487+0.326469 I},{x->0.376974}}}
</code></pre>
<p>Any help would be appreciated!!</p>
| Bob Hanlon | 9,362 | <pre><code>Clear["Global`*"]
p[x_] := x^8 - 3 x^5 + x - 1
</code></pre>
<p>For the real roots</p>
<pre><code>Table[NSolve[D[p[x], {x, n}] == 0, x, Reals] // Union, {n, 5}] //
Grid[#, Frame -> All] &
</code></pre>
<p><a href="https://i.stack.imgur.com/5I0iB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5I0iB.png" alt="enter image description here" /></a></p>
<pre><code>Plot[Evaluate@Table[
Tooltip[D[p[x], {x, n}], Derivative[n]["p"][x]],
{n, 5}],
{x, -0.6, 1.25},
PlotRange -> 10,
PlotLegends -> Table[Derivative[n]["p"][x], {n, 5}],
Frame -> True]
</code></pre>
<p><a href="https://i.stack.imgur.com/KwiEC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KwiEC.png" alt="enter image description here" /></a></p>
|
596,374 | <p>I solved this , but I am not sure if I did in the right way.</p>
<p>$$2^{2x + 1} - 2^{x + 2} + 8 = 0$$</p>
<p>$$2^{x + 2} - 2^{2x + 2} = 8$$</p>
<p>$$\log_22^{x + 2} - \log_22^{2x + 2} = \log_28$$</p>
<p>$$x + 2- 2x - 2 = 3$$</p>
<p>solving for $x$:</p>
<p>$$x = -2$$</p>
<p>any feedback would be appreciated.</p>
| Suraj M S | 85,213 | <p>let $2^{x+2}=t$ then $2^{2x+1}=\frac{1}{8}t^2$</p>
<p>substituting in the equation ,you get
$$\frac{1}{8}t^2-t+8=0$$
which is a quadratic equation and can be easily solved.</p>
|
4,297,051 | <p>Let <span class="math-container">$G$</span> be a non-abelian group of order <span class="math-container">$p^3$</span>, <span class="math-container">$p$</span> prime. Show that <span class="math-container">$Z(G)$</span> is a group of order <span class="math-container">$p$</span>. Deduce that <span class="math-container">$G/Z(G)$</span> is abelian.</p>
<p>From the class equation I can get that <span class="math-container">$p| |Z(G)|$</span>, so <span class="math-container">$|Z(G)|=\{p,p^2,p^3\}$</span>
it can't be of the order <span class="math-container">$p^3 $</span> because that would give us that <span class="math-container">$G$</span> is abelian.</p>
<p>If <span class="math-container">$|Z(G)|=p^2$</span> how to continue ? I don't see how a contradiction can appear.</p>
<p>Any hints how to continue ?</p>
| Marcos | 962,125 | <p>Let us use that <span class="math-container">$G/Z(G)$</span> is cyclic iff <span class="math-container">$G$</span> is abelian. Using this, let us prove the result:</p>
<p>We know that <span class="math-container">$| Z(G)|\in\{p,p^2,p^3\}$</span> (<span class="math-container">$Z(G)\neq 0$</span> since <span class="math-container">$p$</span>-groups have non trivial center). By hypothesis <span class="math-container">$G$</span> is not abelian, so <span class="math-container">$|Z(G)|\neq p^3$</span>. If <span class="math-container">$|Z(G)|= p^2$</span> then <span class="math-container">$|G/Z(G)|=p $</span>, and so <span class="math-container">$G/Z(G)$</span> is cyclic (a group of prime order is always cyclic), which is a contradiction since <span class="math-container">$G$</span> is not abelian. Then <span class="math-container">$|Z(G)|=p$</span></p>
<p>For the second part notice that <span class="math-container">$|G/Z(G)|=p^2 $</span>. Since there are only two groups of order <span class="math-container">$p^2$</span>, which are <span class="math-container">$C_{p^2}$</span> and <span class="math-container">$C_p\times C_p$</span>, both of them abelian, then <span class="math-container">$G/Z(G)$</span> must be abelian.</p>
|
4,297,051 | <p>Let <span class="math-container">$G$</span> be a non-abelian group of order <span class="math-container">$p^3$</span>, <span class="math-container">$p$</span> prime. Show that <span class="math-container">$Z(G)$</span> is a group of order <span class="math-container">$p$</span>. Deduce that <span class="math-container">$G/Z(G)$</span> is abelian.</p>
<p>From the class equation I can get that <span class="math-container">$p| |Z(G)|$</span>, so <span class="math-container">$|Z(G)|=\{p,p^2,p^3\}$</span>
it can't be of the order <span class="math-container">$p^3 $</span> because that would give us that <span class="math-container">$G$</span> is abelian.</p>
<p>If <span class="math-container">$|Z(G)|=p^2$</span> how to continue ? I don't see how a contradiction can appear.</p>
<p>Any hints how to continue ?</p>
| Arturo Magidin | 742 | <p>While one can prove this using the oft-quoted result that if <span class="math-container">$G/Z(G)$</span> is cyclic then <span class="math-container">$G$</span> is abelian, this fact is not necessary to prove the result.</p>
<p>Assume, for the sake of contradiction, that the order of <span class="math-container">$Z(G)$</span> is exactly <span class="math-container">$p^2$</span>. Since <span class="math-container">$G\neq Z(G)$</span>, let <span class="math-container">$x\in G$</span> be an element with <span class="math-container">$x\notin Z(G)$</span>.</p>
<p>Then <span class="math-container">$\langle Z(G),x\rangle$</span> is strictly larger than <span class="math-container">$Z(G)$</span>, and hence must have order <span class="math-container">$p^3$</span> (since it divides <span class="math-container">$p^3$</span> but is strictly larger than <span class="math-container">$p^2$</span>). Therefore, <span class="math-container">$\langle Z(G),x\rangle = G$</span>.</p>
<p>But now let <span class="math-container">$a,b\in Z(G)\cup\{x\}$</span> be two generators. If at least one of them lies in <span class="math-container">$Z(G)$</span>, then <span class="math-container">$ab=ba$</span>. And if this is not the case, then they are both equal to <span class="math-container">$x$</span>, so again <span class="math-container">$ab=xx=ba$</span>. Thus, any two elements in the generating set commute, and that implies that <span class="math-container">$G$</span> is abelian.</p>
<p>But that yields that <span class="math-container">$Z(G)=G$</span> is of order <span class="math-container">$p^3$</span>, a contradiction. Thus, the order cannot be exacty <span class="math-container">$p^2$</span>.</p>
|
778,605 | <blockquote>
<p>Let $f,g,h : D \to \mathbb{R}$ be functions, $D \subset \mathbb{R}$. Let c be an accumulation point of $D$. Suppose that $$f(x) \le g(x) \le h(x)$$
for all $x \in D$ with $x \neq c$ and suppose
$$\lim_{x \to c} f(x) = \lim_{x \to c} h(x) = L \in \mathbb{R}$$
Prove that $\lim_{x \to c}g(x) = L$</p>
</blockquote>
<p>I'm not really sure how to start this exercise. If someone could help me start this or give ideas, that would help a lot.</p>
| drhab | 75,923 | <p>To be calculated is integral:</p>
<p>$\frac{1}{4}\int_{0}^{2}\int_{0}^{2}1_A\left(a,b\right)dadb$ where
$1_A\left(a,b\right)=1$ if $a>\frac{1}{4}\wedge\left|a-b\right|>\frac{1}{4}$
and $1_A\left(a,b\right)=0$ otherwise. </p>
<p>Here $a$ corresponds with Alice and $b$ with Bob.</p>
<p>It comes to determining $\frac{1}{4}A$ where $A$ denotes area: $\left\{ \left(a,b\right)\in\left[0,2\right]^{2}\mid a>\frac{1}{4}\wedge\left|a-b\right|>\frac{1}{4}\right\} $.</p>
|
3,176,629 | <p>In the evening, pizza was ordered nine people sat around a round table, 50 slices of pizza were served to these nine people. Prove that there were two people sitting next to each other who ate at least 12 pizza slices.</p>
<p>I used the pigeon hole principle to determine 50/9 = 5.5 => 6</p>
<p>Therefore, at least one person ate 6 slice of pizza. </p>
<p>I just don't know how to prove that two people ate at least 12 slices..</p>
<p>Help would be greatly appreciated!</p>
| awkward | 76,172 | <p>Assume the contrary, i.e., every pair of adjacent persons ate no more than <span class="math-container">$11$</span> slices. </p>
<p>There are <span class="math-container">$9$</span> pairs of adjacent persons, where we count persons 1 and 9 as adjacent. So if we sum up the slices eaten by each pair (persons 1 and 2, persons 2 and 3, etc.), the total is at most <span class="math-container">$9 \times 11 = 99$</span>. But in so doing, we have counted each slice exactly twice, so the total must be <span class="math-container">$2 \times 50 = 100$</span>.</p>
<p>This contradiction shows our initial assumption must be false.</p>
|
1,138,789 | <p><a href="http://en.wikipedia.org/wiki/Free_object" rel="nofollow noreferrer">Wikipedia</a> defines free objects as follows:</p>
<blockquote>
<p>Let <span class="math-container">$(\mathcal{C},F)$</span> be a concrete category (i.e. <span class="math-container">$F : \mathcal{C} \to {\rm \bf{Set}}$</span> is a faithful functor), let <span class="math-container">$X$</span> be a set (called <em>basis</em>), <span class="math-container">$A \in \mathcal{C}$</span> an object, and <span class="math-container">$i: X \to F(A)$</span> a map between sets (called <em>canonical injection</em>). We say that <span class="math-container">$A$</span> is the <em>free object</em> on <span class="math-container">$X$</span> (with respect to <span class="math-container">$i$</span>) if and only if they satisfy this universal property:</p>
<p>for any object <span class="math-container">$B$</span> and any map between sets <span class="math-container">$f : X \to F(B)$</span>, there exists a unique morphism <span class="math-container">$\tilde{f} : A \to B$</span> such that <span class="math-container">$f = F(\tilde{f})\circ i$</span>. That is, the following diagram commutes:</p>
<p><img src="https://i.stack.imgur.com/F53eE.png" alt="I took this image from Wikipedia." /></p>
</blockquote>
<p>I just can't seem to understand the categorical definition of free objects in terms of commutative diagrams, so I would appreciate any help in grasping the intuition behind the way free objects are defined in categorical setting.</p>
| Will Jagy | 10,400 | <p>Lucian, I think you can do this yourself (for primes) with some hints and some formalism, from Leonard Eugene Dickson, <em>Introduction to the Theory of Numbers</em>. If a prime $q$ is represented as $q = x^2 + xy+ 15 y^2,$ then $3 x^2 + xy + 5 y^2$ does not represent it or its square or cube. If a prime $p$ is represented as $p = 3x^2 + xy+ 5 y^2,$ then $ x^2 + xy + 15 y^2$ does not <strong>primitively</strong> represent $p^2.$ Meanwhile, $ 3x^2 + xy+ 5 y^2$ does primitively represent $p^2,$ and does represent $p^3$ but not primitively. For primitively, Dickson says properly.</p>
<p>Start with problem 4 at the bottom of page 93, continues onto page 94. </p>
<p>Oh, for $2$ and any prime $r$ such that $(-59|r) = -1,$ no form of discriminant $-59$ represents $r$ or $r^3.$ </p>
<p>Not needed: the primes represented by $x^2 + xy + 15 y^2$ are $59$ along with those $p$ for which
$$ z^3 + 2 z + 1 \equiv 0 \pmod p $$ has three distinct solutions. The form $3 x^2 + xy + 5 y^2$ represents all other primes with $(-59|p)=1.$</p>
<p><img src="https://i.stack.imgur.com/JIkoX.jpg" alt="enter image description here"></p>
<p><img src="https://i.stack.imgur.com/Txipw.jpg" alt="enter image description here"></p>
|
18 | <p>Some teachers make memorizing formulas, definitions and others things obligatory, and forbid "aids" in any form during tests and exams. Other allow for writing down more complicated expressions, sometimes anything on paper (books, tables, solutions to previously solved problems) and in yet another setting students are expected to take questions home, study the problems in any way they want and then submit solutions a few days later.</p>
<p>Naturally, the memory-oriented problem sets are relatively easier (modulo time limit), encourage less understanding and more proficiency (in the sense that the student has to be efficient in his approach). As the mathematics is in big part thinking, I think that it is beneficial to students to let them focus on problem solving rather than recalling and calculating (i.e. designing a solution rather than modifying a known one). There is a huge difference between work in time-constrained environment (e.g. medical teams, lawyers during trials, etc.) where the cost of "external knowledge" is much higher and good memory is essential.
However, math is, in general (things like high-frequency trading are only a small part math-related professions), slow.</p>
<p>On the other hand, memory-oriented teaching is far from being a relic of the past. Why is this so? As this is a broad topic, I will make it more specific:</p>
<p><strong>What are the advantages of memory-oriented teaching?</strong></p>
<p><strong>What are the disadvantages of allowing aids during tests/exams?</strong></p>
| Mark Meckes | 129 | <p>I'd like to expand on points that Thomas made. Learning math is like learning a language, and a certain amount of memorization (note: not necessarily drilling!) is a necessary component of that. To use a language, you need to have immediate mental access to the basic vocabulary and grammar. (In math, that means not only the definitions, but the fundamental results of a given field.) This is not just an issue when solving exam problems, but for further learning. A student in a linear algebra class isn't going to get much understanding of the proof of the spectral theorem, for example, if they're still struggling to remember what a linear map is.</p>
|
2,257,365 | <p>In a semicircle of diameter $CD$ there's a chord $AB$ of length 7, and it's parallel to the diameter. There's also a small semicircle that is tangent to $AB$ and its diameter is a segment in $CD$ . Find the area of the semicircle without the small semicircle.</p>
<p>I'm pretty curious about this problem, i've tried many things, like drawing triangles that are similar and also rectangles, so i tries with pithagorean theorem but i got nothing. I'd really like to know how to solve it, thanks.</p>
| Abhinav Dhawan | 346,937 | <p>You can calculate tan $ \angle AMD$ which equals 2. Also tan $ \angle BMC $ is 2 .</p>
<p>You can take AMD=BMC=$\theta$ and then can easily compute tan ( 180 - $\theta$) = tan $\alpha$ using identity tan (a-b)</p>
<p>Hope it helps and it's easier as you asked . Your solution have certain problems as mentioned in comments. </p>
|
2,926,270 | <p>The base step is pretty obvious: <span class="math-container">$1 \geq \frac{2}{3}$</span>.</p>
<p>Then we assume that <span class="math-container">$P(k)$</span> is true for some <span class="math-container">$k \in \mathbb{Z}^{+}$</span> and try to prove <span class="math-container">$P(k+1)$</span>. So I have</p>
<p><span class="math-container">$ \sqrt{1}+\sqrt{2}+...+\sqrt{k} + \sqrt{k+1} \geq \frac{2}{3}k\sqrt{k}+\sqrt{k+1}$</span> </p>
<p>by the induction hypothesis. But I'm not too sure how to proceed to prove that this is also greater than <span class="math-container">$\frac{2}{3}(k+1)\sqrt{k+1}$</span>.</p>
<p>Would appreciate any help!</p>
| For the love of maths | 510,854 | <p>For <span class="math-container">$P(k+1)$</span>:<br>
<span class="math-container">$\sqrt{1}+\sqrt{2}+...+\sqrt{k} + \sqrt{k+1} \geq \frac{2}{3}(k+1)\sqrt{k+1}$</span><br>
<span class="math-container">$\frac 23k\sqrt k+\sqrt {k+1}\geq \frac 23k\sqrt k+\frac 23\sqrt {k+1}$</span><br>
<span class="math-container">$\frac 13\sqrt{k+1}\geq 0$</span> </p>
<p>Does this help?</p>
|
2,926,270 | <p>The base step is pretty obvious: <span class="math-container">$1 \geq \frac{2}{3}$</span>.</p>
<p>Then we assume that <span class="math-container">$P(k)$</span> is true for some <span class="math-container">$k \in \mathbb{Z}^{+}$</span> and try to prove <span class="math-container">$P(k+1)$</span>. So I have</p>
<p><span class="math-container">$ \sqrt{1}+\sqrt{2}+...+\sqrt{k} + \sqrt{k+1} \geq \frac{2}{3}k\sqrt{k}+\sqrt{k+1}$</span> </p>
<p>by the induction hypothesis. But I'm not too sure how to proceed to prove that this is also greater than <span class="math-container">$\frac{2}{3}(k+1)\sqrt{k+1}$</span>.</p>
<p>Would appreciate any help!</p>
| user581912 | 581,912 | <p>For the induction step, you want to show that:
$$
\frac{2k\sqrt{k} + 3\sqrt{k+1}}{3} \geq \frac{2(k+1)\sqrt{k+1}}{3} \\
2k\sqrt{k} + 3\sqrt{k+1} \geq 2k\sqrt{k+1} + 2\sqrt{k+1}\\
$$
Working backwards:
$$
2k\sqrt{k} + \sqrt{k+1} \geq 2k\sqrt{k+1} \\
4k^2 \times k \geq (4k^2 - 4k+1)(k+1) = 4k^3 - 4k^2 + k + 4k^2 - 4k+ 1 = 4k^3 - 3k + 1
$$
The rest should follow since $k \geq \frac1{3}$, since the induction is over the positive integers.</p>
|
541,644 | <p>I want to know why $p \leftrightarrow q$ is equivalent to $(p \wedge q) \vee (\neg p \wedge \neg q)$? Without using the truth table.</p>
<p>Thanks all</p>
| drhab | 75,923 | <p>If $p$ is true then $p\iff q$ tells us that $q$ is true as well. Also if $q$ is true then $p\iff q$ tells us that $p$ is true as well. So it cannot be that exactly one of them is true. They are both true or both not true.</p>
|
3,313,697 | <p>To calculate the <span class="math-container">$n$</span>-period payment <span class="math-container">$A$</span> on a loan of size <span class="math-container">$P$</span> at an interest rate of <span class="math-container">$r$</span>, the formula is:</p>
<p><span class="math-container">$A=\dfrac{Pr(1+r)^n}{(1+r)^n-1}$</span> </p>
<p>Source: <a href="https://en.wikipedia.org/wiki/Amortization_calculator#The_formula" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Amortization_calculator#The_formula</a></p>
<p>And so the total amount paid over those n-periods is simply:</p>
<p><span class="math-container">$n*A=\dfrac{nPr(1+r)^n}{(1+r)^n-1}$</span></p>
<p>For example, to full amortize a 10-year loan of $10,000 with 5.00% annual interest would require annual payments (principal + interest) of:</p>
<p><span class="math-container">$A=\dfrac{10000*0.05(1.05)^{10}}{(1.05)^{10}-1}\approx1295$</span> per year</p>
<p>And over those 10 years then, the person would have paid a total of: <span class="math-container">$n*A=10*1295=12950$</span>.</p>
<p>This is the underlying formula for most "amortizing" loans with <span class="math-container">$n$</span> equal installment payments (e.g. car loans, mortgages, student loans). As principal balance is being paid off over time, the interest payments that are based on that decreasing principal balance are decreasing too -- allowing more of the fixed <span class="math-container">$n$</span>-period payment <span class="math-container">$A$</span> to go toward paying off principal. In the end it all balances out (i.e. the increasing portion of <span class="math-container">$A$</span> going toward paying principal offsets the a decreasing portion of <span class="math-container">$A$</span> going toward paying interest payments).</p>
<p>Investing on the other hand works differently with the idea of "compound interest" being earned. The total amount <span class="math-container">$B$</span> you will have after investing <span class="math-container">$P$</span> at rate <span class="math-container">$r$</span>
over <span class="math-container">$n$</span> periods is simply:</p>
<p><span class="math-container">$B=P(1+i)^n$</span></p>
<p>For instance, if one invests $10,000 at 5.00%/year for 10 years, the compound interest results in:</p>
<p><span class="math-container">$B=P(1+i)^n=10000*1.05^{10}=16289$</span>.</p>
<p>Comparing investing rate (<span class="math-container">$i$</span>) to borrowing rate (<span class="math-container">$r$</span>), the break-even analysis for <span class="math-container">$B=nA$</span> should result in <span class="math-container">$0<i<r$</span>.</p>
<p>Computing this explicitly, assume <span class="math-container">$B=nA$</span>:</p>
<p><span class="math-container">$B=nA$</span> </p>
<p><span class="math-container">$P(1+i)^n=\dfrac{nPr(1+r)^n}{(1+r)^n-1}$</span> </p>
<p><span class="math-container">$(1+i)^n=\dfrac{nr(1+r)^n}{(1+r)^n-1}$</span> </p>
<p><span class="math-container">$i=\bigg(\dfrac{nr(1+r)^n}{(1+r)^n-1}\bigg)^{(\frac{1}{n})}-1$</span> </p>
<p>Thus <span class="math-container">$0<i<r$</span> (I couldn't come up with a more simplified formula above, sorry, but the graph plot checks out). </p>
<p>Using the example above borrowing at <span class="math-container">$r=5\%$</span>, if we invest at <span class="math-container">$i\approx2.619\%$</span> then <span class="math-container">$nA=B$</span>. Notice how much smaller <span class="math-container">$i$</span> is than <span class="math-container">$r$</span> to simply break even... amazing!</p>
<p>In fact, for typical <span class="math-container">$r$</span> like what we would see for common long-term loans, say <span class="math-container">$2\%<r<8\%$</span>, the formula is approximately:</p>
<p><span class="math-container">$i\approx\dfrac{r}{2}+0.1\%$</span> (where <span class="math-container">$2\%<i<r<8\%$</span>) (based on regression approximation)</p>
<blockquote>
<blockquote>
<p>Question: Is this true or not? So many people have told me "Only say yes to an X% loan if you think you can beat that same X% investing in the market!" This math makes it seem like actually, you should say "Yes" to loans at X% rates if you can simply beat at least <em>half</em> of that rate investing in the market over the same period.</p>
</blockquote>
</blockquote>
| Micah | 30,836 | <p>Let's be more explicit about what's happening. If you borrow <span class="math-container">$\$10000$</span> at <span class="math-container">$5\%$</span> over a <span class="math-container">$10$</span>-year term, you must pay</p>
<p><span class="math-container">$$
A=\frac{\$10000(0.05)(1+0.05)^{10}}{(1+0.05)^{10}-1}\approx\$1295.045
$$</span></p>
<p>per year, which means that the total sum you pay is <span class="math-container">$\$12950.45$</span>.</p>
<p>If you invest at your interest rate of <span class="math-container">$i \approx 2.619\%$</span>, then you'll have <span class="math-container">$\$10000(1+i)^{10} \approx \$12950.24$</span> after <span class="math-container">$10$</span> years, which is basically equal to the above sum up to rounding error.</p>
<p>The problem is, <em>you won't have the money when you need it</em>. When you're borrowing the money, you have to start paying it back after a year; with the investment you've set up, you're assuming you won't make any withdrawals until the end of the full <span class="math-container">$10$</span>-year period. If you work out what happens if you withdraw enough money every year to make the payment you owe in that particular year, you'll discover that you'd need your investment to return the full <span class="math-container">$5\%$</span> in order to cover the payments on your loan.</p>
|
2,654,538 | <p>If $2\tan^2x - 5\sec x = 1$ has exactly $7$ distinct solutions for $x\in[0,\frac{n\pi}{2}]$, $n\in N$, then the greatest value of $n$ is?</p>
<p>My attempt:</p>
<p>Solving the above quadratic equation, we get $\cos x = \frac{1}{3}$</p>
<p>The general solution of the equation is given by $\cos x = 2n\pi \pm \cos^{-1}\frac{1}{3}$</p>
<p>For having $7$ distinct solutions, $n$ can have value = 0,1,2,3</p>
<p>So, from here we can conclude that $n$ is anything but greater than $6$. So, according to the options given in the questions, the greatest value of $n$ should be $13$. But the answer given is $14$. Can anyone justify?</p>
| Mostafa Ayaz | 518,023 | <p>By replacing $\tan^2 x=\sec^2x-1$ we have$$4\sec^2x-5\sec x=5$$which yields to $$\sec x=\dfrac{5\pm\sqrt{25+80}}{8}=\dfrac{5\pm\sqrt{105}}{8}$$where only $\sec x=\dfrac{5+\sqrt{105}}{8}$ is acceptable. Also the equation $\sec x=p>1$ has exactly two roots in $[0,2\pi)$ so for having 7 distinct roots we must be in the interval $[0,3\times 2\pi+\pi=14\dfrac{\pi}{2})$ which yields to $n<14$ or $n=13$. An illusion is as following
<a href="https://i.stack.imgur.com/tkOlH.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tkOlH.jpg" alt="enter image description here"></a></p>
|
2,205,950 | <p>If $f:\mathbb{R} \rightarrow \mathbb{R}$ satisfies $f'(a) \neq 0$ for all $a \in \mathbb{R}$, show that $f$ is one-to-one for all $a\in \mathbb{R}$.</p>
<h2>My attempt</h2>
<p>We know that $f(a)$ is not a constant because $f'(a)\neq 0$.Define $f$ by $f(a)=bx$. $f'(a)=x\neq 0$</p>
<p>If $f(x)=f(v)$ then $$bx=bv$$</p>
<p>$$x=v$$</p>
<p>Thus, $f$ is one-to-one.</p>
| Jonas Meyer | 1,424 | <p>"$f(a)$ is not a constant" means what? If $a$ is a number, then $f(a)$ is a number. You mean $f$ is not constant. And you can't say $f(a)=bx$, and it isn't clear what that means; what are $b$ and $x$? And this wouldn't imply $f'(a) = x$. </p>
<p>If you are going to prove that the hypothesis implies the conclusion, you can't assume what form the function has. The above is hard to follow, but it is more along the lines of showing that an example exists that satisfies the hypothesis and the conclusion, rather than proving anything.</p>
<hr>
<p>In a proof by contraposition, we can show that if $f:\mathbb R\to\mathbb R$ is a differentiable function that is not one-to-one, then there exists $a\in\mathbb R$ such that $f'(a)=0$. </p>
<p>So suppose that $f$ is differentiable and not one-to-one. Then there exist $x_1$ and $x_2$ not equal such that $f(x_1)=f(x_2)$. By the Mean Value Theorem, or the special case called Rolle's Theorem, there is a number $a$ between $x_1$ and $x_2$ such that $f'(a) = 0$. </p>
|
37,252 | <p>Let $V$ be a vector space with a finite Dimension above $\mathbb{C}$ or $\mathbb{R}$.</p>
<p>How does one prove that if $\langle\cdot,\cdot\rangle_{1}$ and $\langle \cdot, \cdot \rangle_{2}$ are two Inner products</p>
<p>and for every $v\in V$ $\langle v,v\rangle_{1}$ = $\langle v,v\rangle_{2}$ so $\langle\cdot,\cdot \rangle_{1} = \langle\cdot,\cdot \rangle_{2}$</p>
<p>The idea is clear to me, I just can't understand how to formalize it.</p>
<p>Thank you.</p>
| Calle | 5,966 | <p>You can use the <a href="http://en.wikipedia.org/wiki/Polarization_identity" rel="noreferrer">polarization identity</a>.</p>
<p>$\langle \cdot, \cdot \rangle_1$ and $\langle \cdot, \cdot \rangle_2$ induces the norms $\| \cdot \|_1$ and $\| \cdot \|_2$ respectively, i.e.:</p>
<p>$$\begin{align}
\| v \|_1 = \sqrt{\langle v, v \rangle_1} \\
\| v \|_2 = \sqrt{\langle v, v \rangle_2}
\end{align}$$</p>
<p>From this it is obvious that $\|v\|_1 = \|v\|_2$ for all $v \in V$, so we can write $\| \cdot \|_1 = \| \cdot \|_2 = \| \cdot \|$.</p>
<p>By the polarization identity we get (for complex spaces):
$$\begin{align}
\langle x, y \rangle_1 &=\frac{1}{4} \left(\|x + y \|^2 - \|x-y\|^2 +i\|x+iy\|^2 - i\|x-iy\|^2\right) \ \forall\ x,y \in V \ \\
\langle x, y \rangle_2 &=\frac{1}{4} \left(\|x + y \|^2 - \|x-y\|^2 +i\|x+iy\|^2 - i\|x-iy\|^2\right) \ \forall\ x,y \in V
\end{align}$$
since these expressions are equal, the inner products are equal.</p>
|
1,753,620 | <p>How do I find the matrix exponential $e^{tA}$ with </p>
<p>$$A = \left(\begin{matrix} 2 & 8 \\ 0 & 2\end{matrix}\right)$$</p>
<p>The eigenvalue is 2 with multiplicity 2, but it yields only 1 eigenvector {${1, 0}$}, so the matrix isn't diagonalizable. I'm confused what to do. One option is to convert it into a Jordan form, but how do I do that?</p>
<p>Any help is appreciated. Seems like a simple problem, but it's been bugging me for a while. </p>
| Olivier Oloa | 118,798 | <p>The given series <strong>converges uniformly</strong> on $(-\infty,\infty)$ to $f$, since each term is a continuous function on $(-\infty,\infty)$ thus $f$ is a <strong>continuous</strong> function on $(-\infty,\infty)$. The uniform convergence of the series may be obtained from the normal convergence:
$$
\left|\sum_{n=1}^{\infty} \frac{\sin(nx)}{1+n^3}\right| \leq \sum_{n=1}^{\infty} \frac{\left|\sin(nx)\right|}{1+n^3} \leq \sum_{n=1}^{\infty} \frac1{1+n^3}<\infty.
$$ On the other hand, from
$$
\left|\sum_{n=1}^{\infty} \frac{\sin(nx)}{1+n^3}-\sum_{n=1}^{\infty} \frac{\sin(ny)}{1+n^3}\right| \leq \sum_{n=1}^{\infty} \frac{\left|2 \cos \left( {{nx+ny} \over 2} \right) \sin \left( {{nx-ny} \over 2} \right)\right|}{1+n^3} \leq \underbrace{\left(\sum_{n=1}^{\infty} \frac{n}{1+n^3}\right)}_{<\infty}|x-y|,
$$ where we have used $|\sin x|\leq |x|$, we deduce that $f$ is <strong>uniformly continuous</strong> on $(-\infty,\infty)$.</p>
|
3,862,408 | <p>This is the second example of 1. in <a href="http://www-personal.umich.edu/%7Ebhattb/teaching/mat679w17/lectures.pdf" rel="nofollow noreferrer">Ex. 2.0.3 </a> of Bhatt's notes in perfectoid space.</p>
<p>We define <span class="math-container">$R^{perf}:= \varprojlim ( \cdots R \xrightarrow{\phi} R)$</span> where <span class="math-container">$\phi$</span> is the Frobenius map.</p>
<p>He claims that <span class="math-container">$R=\Bbb F_p[t]$</span> we have <span class="math-container">$R^{perf}=\Bbb F_p$</span>.</p>
<p><strong>I don't see why.</strong></p>
<hr />
<p>My thoughts:
From the map <span class="math-container">$\Bbb F_p[t] \rightarrow \Bbb F_p$</span>, <span class="math-container">$t \mapsto 0$</span>. We get an induced map on limits. Since <span class="math-container">$\Bbb F_p$</span> is perfect, <span class="math-container">$$g:\varprojlim \Bbb F_p[t] \rightarrow \Bbb F_p$$</span></p>
<p>There is also a canonical <span class="math-container">$\Bbb F_p \rightarrow \Bbb F_p[t]$</span>, inducing <span class="math-container">$$h:\Bbb F_p \rightarrow \varprojlim \Bbb F_p[t]$$</span></p>
<p>I'm guess that these two maps are inverses. Its clear that <span class="math-container">$hg=id$</span>. However, it is less clear to me why <span class="math-container">$gh=id$</span>.</p>
| Eric Wofsey | 86,856 | <p>Suppose <span class="math-container">$g\in N_G(H)$</span>, <span class="math-container">$h\in H$</span>, and <span class="math-container">$x\in X^H$</span>. Then since <span class="math-container">$g$</span> normalizes <span class="math-container">$H$</span>, <span class="math-container">$hg=gh'$</span> for some <span class="math-container">$h'\in H$</span>. We thus have <span class="math-container">$hgx=gh'x=gx$</span>. So, since <span class="math-container">$h\in H$</span> was arbitrary, <span class="math-container">$gx\in X^H$</span>.</p>
<p>Thus every element of <span class="math-container">$N_G(H)$</span> actually maps <span class="math-container">$X^H$</span> to itself, and so <span class="math-container">$N_G(H)$</span> acts on <span class="math-container">$X^H$</span>. But every element of <span class="math-container">$H$</span> fixes <span class="math-container">$X^H$</span>, so <span class="math-container">$H$</span> is contained in the kernel of this action, and so it induces an action of the quotient group <span class="math-container">$N_G(H)/H$</span>.</p>
|
681,543 | <p>So I have a function </p>
<p>$$r= ( x^2 + y^2)^{1/2}$$</p>
<p>and I want to show that </p>
<p>$$\operatorname{grad} f(r) = f'(r)(\operatorname{grad} r).$$</p>
<p>I don't really know where to begin do you say that $f(r) = (f \circ r)(x,y)$ and then use the definition of gradient to work it out. Please give a relatively basic answer as I'm new to multi-variable calculus, thanks. </p>
| Mark Fantini | 88,052 | <p>To compute the gradient of $f(r)$ you need to compute the partial derivatives. How to do it?</p>
<p>Use the chain rule: $f(r)$ stands for $f[(x^2+y^2)^{1/2}]$ since $r=(x^2+y^2)^{1/2}$, therefore</p>
<p>$$
\begin{align}
\frac{\partial}{\partial x} f(r) & = \frac{\partial}{\partial x} f[(x^2+y^2)^{1/2}] \\
& = \frac{d}{dr} f (r) \cdot \frac{\partial}{\partial x} (x^2+y^2)^{1/2} \\
& = f'(r) \cdot \frac{x}{(x^2+y^2)^{1/2}}.
\end{align}
$$</p>
<p>Likewise you obtain</p>
<p>$$\frac{\partial}{\partial y} f(r) = f'(r) \cdot \frac{y}{(x^2+y^2)^{1/2}}.$$</p>
<p>Notice that</p>
<p>$$\operatorname{grad}r = \left( \frac{x}{(x^2+y^2)^{1/2}}, \frac{y}{(x^2+y^2)^{1/2}} \right),$$</p>
<p>therefore</p>
<p>$$\operatorname{grad} f(r) = f'(r) \left( \frac{x}{(x^2+y^2)^{1/2}}, \frac{y}{(x^2+y^2)^{1/2}} \right) = f'(r) \cdot \operatorname{grad} r.$$</p>
|
1,375,085 | <p>It is the first time I met such a question:</p>
<blockquote>
<p>Which is greater as $n$ gets larger, $f(n)=2^{2^{2^n}}$ or $g(n)=100^{100^n}$?</p>
</blockquote>
<p>Intuitively I think $f(n)$ would gradually become larger as $n$ gets larger, but I find it hard to produce an argument.
Is there any trick to use for this type of question?</p>
| ccorn | 75,794 | <p>Compare $f(n)$ with $G(n)=256^{256^n}>g(n)$. You arrive at
$G(n) = 2^{2^{8n+3}}$, so you just need to compare $2^n$ vs. $8n+3$.</p>
|
3,043,846 | <p>I want to rewrite a question not so well written on this site and clarified by Mr. Lahtonen (thank you again).</p>
<p>So here the question:</p>
<blockquote>
<p>Let the extention <span class="math-container">$GF(p^m) \supset GF(p)$</span> that contains roots of
<span class="math-container">$p(x)=x^{p^{m}}-1$</span>. Show that those roots are distinct and that forms
a field</p>
</blockquote>
<p>I know that the roots of <span class="math-container">$p(x)=x^{p^{}}-1$</span> are contained in <span class="math-container">$p(x)=x^{p^{m}}-1$</span>, but then?</p>
<p>edit: probably the correct exercise was <span class="math-container">$p(x)=x^{p^{m}-1}$</span></p>
| C Monsour | 552,399 | <p>You can even use non-isomorphic finite groups of the same order. The smallest example is <span class="math-container">$Aut(C_4\times C_2)\cong Aut(D_8)\cong D_8$</span>.</p>
|
2,699,621 | <p>To show $1 + \frac12 x - \frac18 x^2 < \sqrt{1+x}$ is it enough to tell that the taylor series expansion of $\sqrt{1+x}$ around $0$ has more positive terms?</p>
| DeepSea | 101,504 | <p>If $x > 4$, then $LHS < 1 < \sqrt{x+1} = RHS$ and inequality holds. Assume $0 < x \le 4\implies LHS > 0$, square both sides and simplify:$$1+\dfrac{x^2}{4}+\dfrac{x^4}{64}+x- \dfrac{x^2}{4}- \dfrac{x^3}{8}< 1+x\iff \dfrac{x^4}{64}< \dfrac{x^3}{8}\iff\dfrac{x^3}{8}\left(\dfrac{x}{8}-1\right)< 0$$ which is true since $0 < x < 8$. The claim is verified...</p>
|
2,699,621 | <p>To show $1 + \frac12 x - \frac18 x^2 < \sqrt{1+x}$ is it enough to tell that the taylor series expansion of $\sqrt{1+x}$ around $0$ has more positive terms?</p>
| K B Dave | 534,616 | <p>From the <a href="https://en.wikipedia.org/wiki/Taylor%27s_theorem#Explicit_formulas_for_the_remainder" rel="nofollow noreferrer">integral form of the remainder</a> for Taylor's theorem,
$$\begin{split}(1+x)^{1/2}&=1+\tfrac{x}{2} -\tfrac{x^2}{8} +\tfrac{x^3}{16} \int_0^1 (1+tx)^{-5/2}3(1-t)^2\mathrm{d}t\\
&>1+\tfrac{x}{2} -\tfrac{x^2}{8}
\end{split}$$
whenever $x>0$.</p>
|
2,669,277 | <p>In his textbook <em>Calculus</em>, Spivak presents integration by parts as follows: </p>
<p>If $f'$ and $g'$ are continuous then
\begin{align*}
\int fg'&=fg-\int f'g\\
\int f(x)g'(x)\,dx&=f(x)g(x)-\int f'(x)g(x)\,dx\\
\int_a^b f(x)g'(x)\,dx&=f(x)g(x)\bigg|_a^b-\int_a^b f'(x)g(x)\,dx\\
\end{align*}
I understand that without the continuity requirement, $fg'$ and $gf'$ may not be integrable, but why isn't it enough to have $f'$ and $g'$ be integrable functions? Isn't the product of two Riemann-integrable functions necessarily Riemann-integrable?</p>
| RRL | 148,510 | <blockquote>
<p>Spivak's statement for integration-by-parts (in the context of Riemann
integration) holds when $fg’$ and $gf’$, individually, are
Riemann integrable, and it is enough just that $f’$ and $g'$ be Riemann
integrable when $f$ and $g$ are continuous.</p>
</blockquote>
<p>The "counterexample" in the linked paper is relevant for improper integrals. </p>
<p>In the example, $f(x) = x^2 \sin(x^{-4})$ and $g(x) = x^2 \cos(x^{-4})$ on $[0,1]$ with $f(0), g(0) := 0$, and</p>
<p>$$\sin(1)\cos(1) = \int_0^1 (fg’ + gf’) \neq \int_0^1fg’ + \int_0^1 g f’,$$</p>
<p>since the integrals on the RHS do not exist as Riemann integrals (nor as finite Lebesgue integrals).</p>
<blockquote>
<p>Rudin's theorem is correct in that $f', \, g' \in \mathcal{R}([a,b])$
requires $f'$ and $g'$ to be bounded. This is not the case in the
counterexample.</p>
</blockquote>
<p>For more details, note that $f$ and $g$ are differentiable with $f'(0) = g'(0) = 0.$</p>
<p>On $(0,1]$ we have,</p>
<p>$$f'(x) = 2x \sin(x^{-4}) - 4 x^{-3}\cos(x^{-4}), \\ g'(x) = 2x \cos(x^{-4}) + 4 x^{-3}\sin(x^{-4}),\\ f(x)g(x) = \frac{1}{2}x^4 \sin(2 x^{-4}), \\ (fg)'(x) = 2x^2 \sin(2x^{-4}) - 4 x^{-1}\cos(2x^{-4}) $$</p>
<p>Notice that $f(x)g'(x) = 2x^3\sin(x^{-4})\cos(x^{-4})- 4x^{-1} \sin^2(x^{-4})$ where the second term does not have a convergent improper integral over $[0,1]$.</p>
|
895,759 | <p>Is there some sorts of Krull's theorem (that every ring has maximal ideal) for rings that do not have multiplicative identity (unit)? So I know that non-unital rings do not satisfy Krull's theorem, but for some types of non-unital rings, theorem does get satisfied. So what is it?</p>
<p>Edit: Wikipedia seems to mention the case with regular ideal, but does not explain it.</p>
| rschwieb | 29,335 | <p>As you noted, there is no version of Krull's theorem that asserts the existence of maximal ideals in all rngs. The best you can hope for is truth in certain subclasses of rngs. One obvious example is if you ask for the maximal condition on right/left or two-sided ideals.</p>
<p>Here's the result that Wikipedia is referring to in Jacobson's <em>Structure of rings</em> page 6.</p>
<p><img src="https://i.stack.imgur.com/CyXVX.png" alt="enter image description here"></p>
<p>Of course, we've got no guarantee a proper modular right ideal exists in a given rng. Not even $\{0\}$ is guaranteed to be modular. In fact, $\{0\}$ being a modular right and left ideal is exactly saying that the ring has an identity.</p>
|
50,227 | <p>The problem I'm having is mapping a 3D triangle into 2 dimensions. I have three points in $(x,y,z)$ form, and want to map them onto the plane described by the normal of the triangle, such that I end up with three points in $(x,y)$ form.</p>
<p>My guess would be it'd assign an arbitrary up vector and then doing something? Finding the distance traveled along the plane from one vertex to another? What do I do, and how do I do it?</p>
| marty cohen | 13,079 | <p>My take on this is that you want to find a mapping of the form
<span class="math-container">$(x, y, z) \mapsto (ax+by+cz, dx+ey+fz) $</span>
so the resulting triangle in the plane is the same shape and size as the original triangle.
Let <span class="math-container">$L_{ij}$</span> be the distance between points <span class="math-container">$i$</span> and <span class="math-container">$j$</span>.
Since we have 6 unknowns, we need 6 equations.</p>
<p>Let’s map <span class="math-container">$(x_1, y_1, z_1)$</span> into <span class="math-container">$(0, 0)$</span>. We get
<span class="math-container">$ax_1+by_1+cz_1 = 0$</span> and <span class="math-container">$dx_1+ey_1+fz_1 = 0$</span>.<br />
Let’s map <span class="math-container">$(x_2, y_2, z_2)$</span> into <span class="math-container">$(L_{12}, 0)$</span>. We get
<span class="math-container">$ax_2+by_2+cz_2 = L_{12}$</span> and <span class="math-container">$dx_2+ey_2+fz_2 = 0$</span>.</p>
<p>Compute the point <span class="math-container">$(u, v)$</span> which is <span class="math-container">$L_{13}$</span> from <span class="math-container">$(0, 0)$</span> and <span class="math-container">$L_{23}$</span> from <span class="math-container">$(L_{12}, 0)$</span>. The sine and cosine laws are your friends here.
Map <span class="math-container">$(x_3, y_3, z_3)$</span> into <span class="math-container">$(u, v)$</span> via
<span class="math-container">$ax_3+bx_3+cz_3 = u$</span> and <span class="math-container">$dx_3+ey_3+fz_3 = v$</span>.</p>
<p>Solve these two sets of <span class="math-container">$3\times 3$</span> equations, and there’s your mapping.</p>
|
2,174,061 | <p>in $\Delta ABC$ if the $AD\perp BC$,$D\in BC$,and such $$|BC|=2|AD|$$
show that
$$\dfrac{|AB|}{|AC|}\le\sqrt{2}+1$$
<a href="https://i.stack.imgur.com/SXDvI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SXDvI.png" alt="enter image description here"></a></p>
<p>since
$$\cot{B}+\cot{C}=\dfrac{BD}{AD}+\dfrac{CD}{AD}=2$$
so
$$\dfrac{AB}{AC}=\dfrac{\sin{C}}{\sin{B}}$$</p>
| tehjh | 101,039 | <p>This is more of an algebra problem in fact, and I misled myself into thinking of Stewart's theorem at first. In fact if we let the length of $AD$ be 1 and $BC$ be 2 it becomes equivalent to trying to maximise an expression like $\frac{x^2+1}{(2-x)^2+1}$ where $x$ varies between $0$ and $2$ and since the numerator increases while the denominator decreases as $x$ increases it suffices to let $x=2$ giving a bound of $\sqrt{5}$ which is smaller than the given bound. I do not know if there is another method that gives $\sqrt{2}+1$ though.</p>
|
4,170,940 | <blockquote>
<p><a href="https://www.isical.ac.in/%7Eadmission/IsiAdmission2017/PreviousQuestion/BStat-BMath-UGA-2016.pdf" rel="nofollow noreferrer">Question 36</a>: Finding graph corresponding to <span class="math-container">$\int_0^{\sqrt{x} } e^{ -\frac{u^2}{x} } du$</span>
<a href="https://i.stack.imgur.com/KIVRA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KIVRA.png" alt="" /></a>
<span class="math-container">$x>0$</span> and <span class="math-container">$f(0)=0$</span></p>
</blockquote>
<p>Clearly we can't say the function is increasing or decreasing just by inspection because the bound and both integrand is variable. To make statements there we have to consider it's derivative, which is done by the <a href="https://en.wikipedia.org/wiki/Leibniz_integral_rule" rel="nofollow noreferrer">Leibniz integral rule</a>:</p>
<p><span class="math-container">$$ \frac{d}{dx}( \int_0^{\sqrt{x} }e^{ - \frac{u^2}{x} } du) = F(x)= \frac{1}{2 \sqrt{x}} e^{ - \frac{u^2}{\sqrt{x}} } du+ \frac{u^2}{x^2} \int_0^{\sqrt{x} }e^{ - \frac{u^2}{x} } du$$</span></p>
<p>Now... ummm... we still have the integral again, so it's still not easily possible to make statement if it's increasing or decreasing again.</p>
<p>I thought that I could rule out option D by checking if there is a extrema point on the function by checking roots for <span class="math-container">$F(x)=0$</span> but that too seems too difficult.</p>
<p>Is there any trick which I am not seeing? This question was meant for higher schooler's who are entering undergraduate, so methods in that level would be best (other methods are still fine)</p>
| saulspatz | 235,128 | <p>You haven't done the differentiation correctly, but that's not the best way to do the problem, anyway.</p>
<p>Write <span class="math-container">$$e^{-u^2/x}=\sum_{n=0}^\infty\frac{(-1)^n}{n!}\left(\frac{u^2}{x}\right)^n$$</span></p>
<p>Term-by-term integration gives <span class="math-container">$$\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)n!}\frac{u^{2n+1}}{x^n}$$</span> When we substitute <span class="math-container">$\sqrt x$</span> for <span class="math-container">$u$</span> in any of the terms, we <span class="math-container">$\frac{x^{n+1/2}}{x^n}=\sqrt x$</span> so
<span class="math-container">$$\int_0^{\sqrt x}e^{-u^2/x}\,\mathrm du=\sqrt x\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)n!}$$</span> and the function is a constant multiple of <span class="math-container">$\sqrt x$</span>.</p>
<p><strong>EDIT</strong></p>
<p>As to the differentiation, the wiki page you linked gives the formula
<span class="math-container">$$\frac{d}{dx}\int_0^{b(x)}f(x,u)\,\mathrm{d}u=f(x,b(x))\cdot\frac{d}{dx}b(x)+\int_0^{b(x)}\frac{\partial}{\partial x}f(x,u)\,\mathrm du$$</span> where I've substituted <span class="math-container">$0$</span> for <span class="math-container">$a(x)$</span> and <span class="math-container">$u$</span> for <span class="math-container">$t$</span>. Here we have <span class="math-container">$f(x,u)=e^{-u^2/x}$</span>, and <span class="math-container">$b(x)=\sqrt x$</span>, so
<span class="math-container">$$
f(x,b(x))= e^{-1}\\
\frac{d}{dx}b(x)=\frac1{2\sqrt x}\\
\frac{\partial}{\partial x}f(x,u)=\frac{u^2e^{-u^2/x}}{x^2}
$$</span></p>
|
2,302,067 | <p>I'm trying to prove that if ${\kappa}$ is an infinite cardinal, then there are $2^{\kappa}$ bijective functions from ${\kappa}$ to ${\kappa}$. I would greatly appreciate any tips. Thank you. </p>
| Asinomás | 33,907 | <p>I would do it differently.</p>
<p>The number of bijections is clearly not more than $\kappa^\kappa=2^\kappa$.</p>
<p>Let $S$ be the set of subsets of $\kappa$ that leave more than $1$ element in the complement. Clearly $S$ has the same cardinality as $2^\kappa$.</p>
<p>To show that there are at least $2^\kappa$ bijections we give an injection from $S$ to the set of bijections of $\kappa$. To do this we send each $s\in S$ to a bijection $\sigma$ of $\kappa$ such that $s$ is the set of fixed points of $\sigma$.</p>
|
2,250,638 | <p>I feel I am doing the problem correctly however my answers are not following the solution.</p>
<p>My attempt: </p>
<p>$y^{2}+2y+12x-23=0$</p>
<p>$(y^{2}+2y+1) +12x = 23-1$</p>
<p>$(y+1)^{2}+12x=22$ </p>
<p>$\dfrac{(y+1)^{2}}{22}+\dfrac{6x}{11}=1$</p>
<p>Note $a > b$</p>
<p>$a^{2}=22$</p>
<p>$b^{2}=11$</p>
<p>$c^{2}= 22-11$ </p>
<p>$c= ± \sqrt{11}$</p>
<p>I'm not sure what I'm doing incorrectly, but the answers are supposed to be:</p>
<p>vertex: $(2,-1)$</p>
<p>focus: $(-1,-1)$</p>
<p>directrix: $x=5$</p>
| Mick | 42,351 | <p>This is not a solution. The following figure is just my guess on the meaning of the question.</p>
<p><a href="https://i.stack.imgur.com/Pwh6T.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Pwh6T.png" alt="enter image description here"></a></p>
|
281,714 | <p>The absolute maximum value of $f\left(x\right) = x^3-3x^2+12$ on closed interval $\left[-2,4\right]$ occurs at $x = $ </p>
<p>Confused what does <em>absolute maximum value</em> means. </p>
<p>Does it mean </p>
<ol>
<li>The largest of the large values? $\max \{f\left(x\right)\mid x\in [-2,4]\}$</li>
<li>The largest absolute value of $\max \{\vert f\left(x\right)\vert : x\in [-2,4]\}$</li>
</ol>
<p>I have figured out that both values are the same when $x=4$, I can see that just from the graph of the function </p>
<p><img src="https://i.stack.imgur.com/UGz3p.gif" alt="function"></p>
| Rustyn | 53,783 | <p>It means your first assumption. </p>
<p>Setting the derivative equal to $0$, we obtain:<br></p>
<p>$3x^2 - 6x = 0 \Rightarrow$<br>
$x(3x - 6)=0 \Rightarrow$ <br></p>
<p>$x=0,$ or $x=2$</p>
<p>$f(2) = 8$, $f(0) = 12$</p>
<p>Now we test end points,</p>
<p>$f(4) = 64 - 48 + 12 = 28$ <br>
$f(-2)= -8 -12 + 12 = -8$</p>
<p>Hence $f(4)=28$ = $\max \{ f(x): x \in [-2,4]\}$ </p>
|
281,714 | <p>The absolute maximum value of $f\left(x\right) = x^3-3x^2+12$ on closed interval $\left[-2,4\right]$ occurs at $x = $ </p>
<p>Confused what does <em>absolute maximum value</em> means. </p>
<p>Does it mean </p>
<ol>
<li>The largest of the large values? $\max \{f\left(x\right)\mid x\in [-2,4]\}$</li>
<li>The largest absolute value of $\max \{\vert f\left(x\right)\vert : x\in [-2,4]\}$</li>
</ol>
<p>I have figured out that both values are the same when $x=4$, I can see that just from the graph of the function </p>
<p><img src="https://i.stack.imgur.com/UGz3p.gif" alt="function"></p>
| Martin Argerami | 22,857 | <p>I would guess it is your first option. A usual terminology in calculus is about absolute and relative (or local) maxima and minima. </p>
<p>The absolute maximum would be then $\max\{f(x):\ x\in[-2,4]\}$. </p>
<p>The phrase "absolute maximum <em>value</em>" probably has to do with the fact that when looking at extrema of functions, one usually focus on where they are (i.e. $x=\ldots$) rather than what they are (i.e. $f(x)=\ldots$). The latter is the value, so saying "absolute maximum value" one wants the answer "$28$" as opposed "$x=4$". </p>
|
1,581,545 | <p>I am looking for a proof of Euclid's Lemma, i.e if a prime number divides a product of two numbers then it must at least divide one of them.</p>
<p>I am coding this proof in Coq, and i'm doing it over <em>natural numbers</em>. I aim to prove the uniqueness of prime factorization (So I cannot use this lemma!). However, I can use the existence of a prime factorization, which I already proved.</p>
<p>I do not want to use the gcd algorithm as that would involve coding it in Coq and proving it is correct which may be difficult. The idea is to use this proof in a computer science course, so I do not want to overcomplicate things.</p>
<p>Is there any proof of this lemma that does not use gcd, or Bezout's lemma, or the uniqueness of prime factorization? Maybe something using induction?</p>
<p>Thank you in advance.</p>
<p>EDIT: The Proof should be on NATURAL NUMBERS. No answer did the proof in N.</p>
| robjohn | 13,854 | <p>Claim 2 below should answer the question.</p>
<p>Since the only <a href="https://en.wikipedia.org/wiki/Unit_%28ring_theory%29" rel="nofollow">unit</a> in $\mathbb{N}$ is $1$, we have</p>
<p>$p$ is <a href="https://en.wikipedia.org/wiki/Prime_element" rel="nofollow">prime</a> iff $p\mid ab\implies p\mid a\lor p\mid b$</p>
<p>$p$ is <a href="https://en.wikipedia.org/wiki/Irreducible_element" rel="nofollow">irreducible</a> iff $a\mid p\implies a=1\lor a=p$</p>
<blockquote>
<p><strong>Claim 1:</strong> $p$ is prime $\implies$ $p$ is irreducible</p>
</blockquote>
<p><strong>Proof:</strong> Assume that $a\mid p$. For some $b$, we have
$$
p=ab\tag{1}
$$ Since $p\mid ab$, we know that $p\mid a$ or $p\mid b$.</p>
<p>Case $p\mid a$: for some $c$, we have $a=pc$. Therefore, $(1)$ implies that $abc=a$. Since the only unit in $\mathbb{N}$ is $1$, we have that $b=c=1$. Therefore $a=p$.</p>
<p>Case $p\mid b$: for some $c$, we have $b=pc$. Therefore, $(1)$ implies that $abc = b$. Since the only unit in $\mathbb{N}$ is $1$, we have that $a=c=1$. Therefore, $a=1$.</p>
<p>Thus, assuming that $p$ is prime and $a\mid p$, we have shown that $a=1$ or $a=p$.</p>
<p><strong>QED</strong></p>
<blockquote>
<p><strong>Claim 2:</strong> $p$ is irreducible $\implies$ $p$ is prime</p>
</blockquote>
<p><strong>Proof:</strong> Assume that $p$ is irreducible, $p\mid ab$, and $p\nmid a$. Let $g$ be the smallest positive element of the set
$$
S=\{\,ax+py:x,y\in\mathbb{Z}\,\}\tag{2}
$$
If $g\nmid p$, then there is an $r$ so that $0\lt r\lt g$ and $qg+r=p$. However, then
$$
r=p-q(ax+py)=a(-qx)+p(1-qy)\in S\tag{3}
$$
but $g$ is the smallest positive element of $S$. Therefore, $g\mid p$. Similarly, $g\mid a$.</p>
<p>Since $p$ is irreducible, $g=1$ or $g=p$. Since $p\nmid a$ and $g\mid a$, we must have $g=1$. Therefore, we have $x,y$ so that
$$
1=ax+py\tag{4}
$$
Since $p\mid ab$, for some $c$, we have $ab=pc$. Multiply $(4)$ by $b$ to get
$$
\begin{align}
b
&=abx+pby\\
&=p(cx+by)\tag{5}
\end{align}
$$
Equation $(5)$ says that $p\mid b$.</p>
<p>Thus, assuming that $p$ is irreducible and $p\mid ab$, we have shown that if $p\nmid a$, then $p\mid b$.</p>
<p><strong>QED</strong></p>
|
4,317,945 | <p>A function <span class="math-container">$h : A → \mathbb{R}$</span> is Lipschitz continuous if <span class="math-container">$\exists K$</span> s.t.</p>
<p><span class="math-container">$$|h(x) - h(y)| \leq K \cdot |x - y|, \forall x, y \in A$$</span></p>
<p>Suppose that <span class="math-container">$I = [a, b]$</span> is a closed, bounded interval; and <span class="math-container">$g : I → \mathbb{R}$</span> is differentiable on <span class="math-container">$I$</span> and the function <span class="math-container">$G = Dg = g' : I → \mathbb{R}$</span> is continuous. Prove that <span class="math-container">$g$</span> is Lipschitz continuous on <span class="math-container">$I$</span>.</p>
| heropup | 118,193 | <p>It's really difficult to read your writing; for me to try to detect where you might have made a calculation error would probably take longer than to just write my own step-by-step solution.</p>
<p>Let <span class="math-container">$f(x) = x^2$</span>, <span class="math-container">$g(x) = (x+1)^n$</span>, and <span class="math-container">$h(x) = f(x)g(x) = x^2 (x+1)^n$</span>. Then we want
<span class="math-container">$$h^{(n)}(x) = (fg)^{(n)} = \sum_{k=0}^n \binom{n}{k} f^{(k)}(x) g^{(n-k)}(x). \tag{1}$$</span></p>
<p>As you observed, <span class="math-container">$f^{(k)}(x) = 0$</span> for all <span class="math-container">$k > 2$</span>, so the sum on the RHS in Equation <span class="math-container">$(1)$</span> only contains <span class="math-container">$3$</span> terms. These require the evaluation of <span class="math-container">$g^{(n-k)}(x)$</span> for <span class="math-container">$k \in \{0, 1, 2\}$</span>. To this end, we observe
<span class="math-container">$$\begin{align}
g'(x) &= n(x+1)^{n-1} , \\
g''(x) &= n(n-1)(x+1)^{n-2} , \\
&\vdots \\
g^{(m)}(x) &= n(n-1)\cdots(n-m+1)(x+1)^{n-m} = \frac{n!}{(n-m)!} (x+1)^{n-m}, \quad m \in \{1, \ldots, n\}. \tag{2}
\end{align}$$</span> So now letting <span class="math-container">$m \in \{n-2, n-1, n\}$</span> in Equation <span class="math-container">$(2)$</span> gives in particular
<span class="math-container">$$\begin{align}
g^{(n-2)}(x) &= \frac{n!}{2!} (x+1)^2 = \frac{n!}{2} (x+1)^2, \\
g^{(n-1)}(x) &= \frac{n!}{1!} (x+1)^1 = n! (x+1), \\
g^{(n)}(x) &= \frac{n!}{0!} (x+1)^0 = n!.
\end{align}$$</span>
It follows that
<span class="math-container">$$\begin{align}
h^{(n)}(x) &= \binom{n}{0} f(x) g^{(n)}(x) + \binom{n}{1} f'(x) g^{(n-1)}(x) + \binom{n}{2} f''(x) g^{(n-2)}(x) \\
&= (x^2) n! + n (2x) n!(x+1) + \frac{n(n-1)}{2} (2) \frac{n!}{2}(x+1)^2 \\
&= n! \left( x^2 + 2nx(x+1) + \frac{n(n-1)}{2}(x+1)^2 \right).
\end{align}$$</span></p>
<hr />
<p>Upon examination of your handwritten notes, I think I have found your error. You wrote
<span class="math-container">$$h^{(n)}(x) = \binom{\color{red}{2}}{0} f(x) g^{(n)}(x) + \binom{\color{red}{2}}{1} f'(x) g^{(n-1)}(x) + \binom{\color{red}{2}}{2} f''(x) g^{(n-2)}(x)$$</span> where the red text is incorrect; the upper index of the binomial coefficients must be <span class="math-container">$n$</span>, not <span class="math-container">$2$</span>, even though there are only three terms in the sum.</p>
|
163,465 | <p>I am interested in the quantity $$f(X,t) = \int_t^\infty\negthinspace x\ p(x)\ dx,$$ where $p$ is a probability distribution for a positive variable $X$.</p>
<p>1) <strong>Does this quantity $f(X,t)$ have a name</strong>? As the title question suggests, it is a tail of the integral that is cut out when computing the trimmed mean.</p>
<p>2) <strong>Are there any bounds</strong> on $f(X,t)\ / \ f(X,0)$ in terms of $t$ and properties only of $X$ (e.g. moments or quantiles of $X$)? Note $f(X,0)=E[X]$.</p>
<p>Chebyshev's inequality has the right form for a bound on a different quantity, $$\int_t^\infty\negthinspace p(x)\ dx \le \frac{\text{Var}(X)}{(t - E[X])^2}.$$ I am looking for both upper and lower bounds on $f(X,t)$ and Chebyshev's inequality doesn't seem to provide either.</p>
<p>I am especially interested in the case where $X=|Y_1-Y_2|$ where the $Y$'s are i.i.d. variables. Results for the general case might be better-known, and I would appreciate either.</p>
| eisit | 48,584 | <p>I think results from extreme value theory will be helpful here. The standard condition to put on $p(x)$ in these kinds of situations is that $p$ is regularly varying in $x$, and that there exists a constant $\gamma > 0$ such that
$$ \lim_{x \rightarrow \infty} \frac{1 - F(tx)}{1 - F(x)} = t^{-1/\gamma} \text{ for all } t > 0, $$
where $F$ is the CDF of $p$. The constant $\gamma$ is called the tail index. (There also exists a more general theory with $\gamma \leq 0$ that applies to thin-tailed distributions like the Gaussian.) </p>
<p>Once you assume that your density is regularly varying, the quantity $f(X, \, t)$ becomes easier to analyze. For example, you can show that $f(X, \, t)$ scales with $t$.</p>
<p>A good reference for getting started is Chapter 4.3 of Coles (2001) "An Introduction to Statistical Modeling of Extreme Values." For a more theoretical approach, the textbook by de Haan and Ferreira (2006) is excellent.</p>
|
189,689 | <pre><code>CountryData[ "UnitedStates", {"Population", 2014} ]
</code></pre>
<blockquote>
<p>322 422 965 people</p>
</blockquote>
<pre><code>CountryData[ "UnitedStates", {"Population", 2015} ]
</code></pre>
<blockquote>
<p>Missing[ "NotAvailable" ]</p>
</blockquote>
<p>How can I update it?
I am new to Mathematica.</p>
| Edmund | 19,542 | <p>You should report to WRI. As a workaround you may use the entity functions instead; it is a bit verbose.</p>
<p>First you can get a list of qualifier values of an <code>EntityProperty</code> by</p>
<pre><code>EntityValue[EntityProperty["Country", "Population"], "QualifierValues"]
</code></pre>
<blockquote>
<pre><code>{Age->{Adult,MiddleAge,PreSchool,SchoolAge,Senior,Young,YoungAdult},
CitizenshipStatus->{BornInPuertoRico,BornInUS,BornToAmericanParents,NaturalizedCitizen,NotCitizen,TotalCitizens},
Date->{},
Gender->{Female,Male},
HispanicOrigin->{Argentinean,Bolivian,CentralAmerican,Chilean,Colombian,CostaRican,Cuban,Dominican,Ecuadorian,Guatemalan,Hispanic,HispanicOrLatinoAllOther,Honduran,Mexican,Nicaraguan,NotHispanic,OtherCentralAmerican,OtherHispanicOrLatino,OtherSouthAmerican,Panamanian,Paraguayan,Peruvian,PuertoRican,Salvadoran,SouthAmerican,Spaniard,Spanish,SpanishAmerican,Uruguayan,Venezuelan},
MarginOfError->{MarginOfError,StandardError},
Percent->{Main},
Race->{AmericanIndian,Asian,Black,NativeHawaiian,Other,TwoOrMore,White,{All,Hispanic}},
TwoOrMore->{ThreeOrMore,TwoIncludingOther},
UrbanRural->{Rural,Urban}}
</code></pre>
</blockquote>
<p><code>"Date"</code> is a qualifier. This post gives details on how to specify a <code>"Date"</code> qualifier (<a href="https://mathematica.stackexchange.com/questions/156014">156014</a>); an area where the docs can improve.</p>
<pre><code>EntityValue[
Entity["Country", "UnitedStates"],
EntityProperty["Country", "Population", {"Date" -> DateObject[{2015}]}]
]
</code></pre>
<blockquote>
<pre><code>319,929,162 people
</code></pre>
</blockquote>
<p>However, you do have a bit more flexibility with this syntax.</p>
<pre><code>EntityValue[
Entity["Country", "UnitedStates"],
EntityProperty["Country",
"Population", {"Date" -> Interval[{DateObject[{2015}], DateObject[{2019}]}]}]
]
</code></pre>
<blockquote>
<p><img src="https://i.stack.imgur.com/YWecL.png" alt="Mathematica graphics"></p>
</blockquote>
<p>and</p>
<pre><code>EntityValue[
Entity["Country", "UnitedStates"],
EntityProperty["Country", "Population", {"Date" -> All}]
]
</code></pre>
<blockquote>
<p><img src="https://i.stack.imgur.com/YmYLu.png" alt="Mathematica graphics"></p>
</blockquote>
<p>Hope this helps.</p>
|
1,369,482 | <p>I'm have problem proving: Law for Scalar Multiplication :</p>
<p>Vector spaces possess a collection of specific characteristics and properties. Use the definitions in the attached “Definitions” to complete this task.</p>
<p>Define the elements belonging to $\mathbb{R}^2$ as $\{(a, b) | a, b \in \mathbb{ R}\}$. Combining elements within this set under the operations of vector addition and scalar multiplication should use the following notation:</p>
<p>Vector Addition Example: $(–2, 10) + (–5, 0) = (–2 – 5, 10 + 0) = (–7, 10)$</p>
<p>Scalar Multiplication Example: $–10 × (1, –7) = (–10 × 1, –10 × –7) = (–10, 70)$, where –10 is a scalar.</p>
<p>Under these definitions for the operations, it can be rigorously proven that R2 is a vector space.</p>
<p>Prove Closure under Scalar Multiplication - **i need help with this law ** </p>
<p>Can someone put it in a proof form?</p>
| Patrick Da Silva | 10,704 | <p>If you're asking if vector spaces are closed under multiplication by a scalar, then yes, it is true. If you're asking why, it's because it's written in the definition of a vector space that it must be true ; there is nothing to prove here. It's true because we assume it is when we speak of a vector space.</p>
<p>EDIT : So if I understand this correctly, you need to show that $\mathbb R^2$ is a vector space and you need help showing that $\mathbb R^2$ is closed under scalar multiplication. Scalar multiplication is defined for $\lambda \in \mathbb R$ and $(a,b) \in \mathbb R^2$ via
$$
\lambda \cdot (a,b) \overset{def}= (\lambda a, \lambda b)
$$
where $\lambda a$ is the usual multiplication of real numbers. What you want to show is that
$$
\forall \lambda,a,b \in \mathbb R, \quad \lambda \cdot (a,b) \in \mathbb R^2.
$$
Is it obvious now? </p>
<p>Hope that helps,</p>
|
1,369,482 | <p>I'm have problem proving: Law for Scalar Multiplication :</p>
<p>Vector spaces possess a collection of specific characteristics and properties. Use the definitions in the attached “Definitions” to complete this task.</p>
<p>Define the elements belonging to $\mathbb{R}^2$ as $\{(a, b) | a, b \in \mathbb{ R}\}$. Combining elements within this set under the operations of vector addition and scalar multiplication should use the following notation:</p>
<p>Vector Addition Example: $(–2, 10) + (–5, 0) = (–2 – 5, 10 + 0) = (–7, 10)$</p>
<p>Scalar Multiplication Example: $–10 × (1, –7) = (–10 × 1, –10 × –7) = (–10, 70)$, where –10 is a scalar.</p>
<p>Under these definitions for the operations, it can be rigorously proven that R2 is a vector space.</p>
<p>Prove Closure under Scalar Multiplication - **i need help with this law ** </p>
<p>Can someone put it in a proof form?</p>
| Rescy_ | 248,257 | <p>If $V$ is a vector space over the field $\mathbf F$, then it must satisfy two properties, namely closure under addition and closure under multiplication.</p>
<p>For closure under multiplication, we demand that if $u \in V$, $a \in \mathbf F$, then $a \mathbf F \in V$. Note that the 'multiplication' needs to be defined beforehand. </p>
<p>Specifically to your example, perhaps you are having trouble with what's meant by closure. According to the definition, you want to prove that $S:\{(a,b)|a,b \in \mathscr R\}$ is a vector space.</p>
<p>You have actually done it. You have shown it is closed under addition as well as multiplication.</p>
<p>Let's take your example,$–10 \times (1, –7) = (–10 \times 1, –10 \times –7) = (–10, 70)$, the original vector$(1,-7) \in S$,right? Also$(-10,70) \in S$ because let $a=-10,b=70$, then you see $(-10,70)$ satisfy the requirement $a,b \in R$. Thus you have proved closure under multiplication.</p>
<p><strong>Edit: I will try to prove the statement</strong></p>
<blockquote>
<p>$\mathbb R^2$ is a vector space over the field $\mathbb R$</p>
</blockquote>
<p>Proof:
if $ a,b \in \mathbb R$, then $(a,b) \in \mathbb R^2$. Let $k \in \mathbb R$ be a scalar, then $k(a,b)=(ka,kb)$. As $k,a,b \in \mathbb R$, $ka,kb \in \mathbb R$, thus $(ka,kb) \in \mathbb R^2$. Therefore scalar multiplication on $\mathbb R^2$ is closed.</p>
|
3,854,446 | <p>I am reading a textbook on representation theory which says the following.</p>
<p><span class="math-container">$G$</span> is a finite group with irreducible representation <span class="math-container">$\rho:G\to GL(V)$</span> over field <span class="math-container">$k$</span> (possibly algebraically closed, there's an assumption that all fields are algebraically closed which I'm not certain extends to this page in the book). <span class="math-container">$\phi$</span> is a class function from <span class="math-container">$G$</span> to <span class="math-container">$k$</span> satisfying <span class="math-container">$(\phi,\chi_\rho)=0$</span>. Define
<span class="math-container">$$T=\frac{1}{\#G}\sum\limits_{g\in G}\phi(g^{-1})\rho_g.$$</span>
The text claims that <span class="math-container">$T=0$</span>. I am not sure how to see this. I see why <span class="math-container">$T\in End_GV$</span>, so if <span class="math-container">$k$</span> is algebraically closed then we can identify it with some element of <span class="math-container">$k$</span> (and regardless <span class="math-container">$End_GV$</span> is a division ring, by Schur's lemma), but I don't see why <span class="math-container">$T$</span> must be <span class="math-container">$0$</span>.</p>
<p>Is there something that I'm missing here? Thanks in advance.</p>
| runway44 | 681,431 | <p>Suppose <span class="math-container">$T=aI$</span> is a scalar multiple of the identity map.</p>
<p>Taking traces yields <span class="math-container">$\mathrm{tr}(T)=(\dim V)a=(\phi,\chi)=0$</span>, so <span class="math-container">$a=0$</span>.</p>
<p>(I assume <span class="math-container">$\dim V$</span> is invertible in the base field.)</p>
|
2,021,557 | <p>I'm not really sure how to do this, I guessed it had something to do with Vector Functions but overall couldn't find a way to do it. Can you please help?</p>
<p>The equations are:</p>
<p>$$f(x,y) = x^2 + y^2
\ g(x,y) = xy + 10 $$</p>
<p>and I need a Vectorial equation.
Thank you in advance!</p>
| mvw | 86,776 | <p>$f$ is a paraboloid (red colour), $g$ is a hyperbolic paraboloid (green colour):</p>
<p><a href="https://i.stack.imgur.com/JAc4mm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JAc4mm.png" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/J0c8Rm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/J0c8Rm.png" alt="enter image description here"></a>
(Large versions: <a href="https://i.stack.imgur.com/JAc4m.png" rel="nofollow noreferrer">link</a> and <a href="https://i.stack.imgur.com/J0c8R.png" rel="nofollow noreferrer">link</a>)</p>
<p>The intersection reminds me of a Lissajous curve.</p>
<p>For the intersection points we have
$$
x^2 + y^2 = xy + 10 \iff \\
0 = x^2 + y^2 - xy - 10
$$
This seems to be a cylinder with elliptic cross section.
$$
x = \frac{1}{\sqrt{2}}(\xi + \eta) \\
y = \frac{1}{\sqrt{2}}(\xi - \eta)
$$
So we get
\begin{align}
0 &= x^2 + y^2 - xy - 10 \\
&= \frac{1}{2}(\xi^2 + \eta^2 + 2\xi\eta)
+ \frac{1}{2}(\xi^2 + \eta^2 - 2\xi\eta)
- \frac{1}{2}(\xi^2 - \eta^2) - 10 \\
&= \frac{1}{2} \xi^2 + \frac{3}{2} \eta^2 - 10 \iff \\
\end{align}
$$
\left( \frac{\xi}{\sqrt{20}} \right)^2 +
\left( \frac{\eta}{\sqrt{20/3}} \right)^2 = 1
$$
We can parameterize
$$
\xi(t) = \sqrt{20} \cos(2\pi t) \\
\eta(t) = \sqrt{20/3} \sin(2\pi t)
$$
for $t \in [0, 2\pi)$ and thus with $z = x^2 + y^2 = \xi^2 + \eta^2$ get
\begin{align}
x(t) &= \sqrt{10} \cos(2\pi t) + \sqrt{10/3} \sin(2\pi t) \\
y(t) &= \sqrt{10} \cos(2\pi t) - \sqrt{10/3} \sin(2\pi t) \\
z(t) &= 20 \cos(2\pi t)^2 + (20/3) \sin(2\pi t)^2 \\
&= (40/3) \cos(2\pi t)^2 + (20/3)
\end{align}
The left image shows the ellipse in $\xi$-$\eta$-coordinates (light green), in $x$-$y$-coordinates (pink) and the scene in $3D$ (right image).</p>
<p><a href="https://i.stack.imgur.com/Clq7xm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Clq7xm.png" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/JlFbNm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JlFbNm.png" alt="enter image description here"></a></p>
<p>(Large versions: <a href="https://i.stack.imgur.com/Clq7x.png" rel="nofollow noreferrer">link</a> and <a href="https://i.stack.imgur.com/JlFbN.png" rel="nofollow noreferrer">link</a>)</p>
<p>You can fiddle with an interactive version <a href="https://ggbm.at/GHVhsv7D" rel="nofollow noreferrer">here</a>: </p>
<ul>
<li>adjust the $t$ slider in the left window and see the points on the ellipses ($A(t)$, $B(t)$) and the intersection curve ($C(t)$) change.</li>
<li>drag the upper part of the right 3D view to change the 3D view</li>
</ul>
|
1,103,624 | <p>Ashamed to admit that I cannot aid my friend's niece with her second grade homework problem. So much for that college education, eh? Here's the problem.</p>
<p>Using only the natural numbers 1 through 9 without repeating any of them (natural because there cannot occur any rationals anywhere in this process, i.e. the first number cannot be prime, since otherwise there would be a fraction after the first operation), place the numbers in the boxes so that the following statement is true.</p>
<p>$\square \div \square \times \square + \square \times \square \times \square \div \square + \square \times \square = 100$</p>
<p>I am aware that I can use brute force to work this problem out, but this is going to be tedious at best, and might require a computer at worst. Furthermore, the problem doesn't specify anything about using the order of operations. On one hand, that would make sense, but on the other, this is a second grade problem in an American school, and I know that order of operations wasn't on my second grade curriculum.</p>
<p>In either case, does anyone know how to approach this in a somewhat elegant way? (Also apologies over tagging, someone please retag this appropriately if they think it isn't).</p>
| Fizz | 173,347 | <p>I do have a suggestion for an algorithm that would reduce the cases to try. If I understand correctly that there isn't an operator precedence, but all operations are simply done left to right, then last thing you do is a multiplication. But $100 = 2 \times 2 \times 5 \times 5$. So the last box can only be 1, 2, 4 or 5. And so forth (handwaving!), i.e. use <a href="http://en.wikipedia.org/wiki/Backtracking" rel="nofollow">backtracking</a>. But it's better than brute force for sure.</p>
<p>Ok, here are a few steps...</p>
<p>Assume $x_9 = 5$, for the last unknown (which is highest available value given what I said above about the factorization of 100). Then you want $y_8 = 20$, where $y_8$ is the result of all operations up to $x_8$, inclusively. So assume $x_8 = 9$ (again, the highest available value). Now you want $y_7 = 11$. This is where it gets a bit more "intuitive", or rather where you can eliminate some more candidates. You want a quotient of $11 = y_6 / x_7$. Note that you have 3 numbers that you need to multiply to produce $y_6$. So it needs at least three different factors in its factorization including 1. And two of these factors have to be single-digit. What could work? $22/2$, $33/3$? No! Because $22/2$ has already "used up" 2 for the division, so now you can't have it as $11 \times 2 \times 1$ because 2 would be duplicated. Likewise for $33/3$. So let's go for $44/4$, i.e. set $x_7$ to 4. Since $44=22 \times 2\times 1$, set $x_6 = 2$ and set $x_5 = 1$. Now $y_4=22$. Since 9 is used up, set $x_4=8$ so $y_3=22-8=14$. Then set $x_3=7$ as the next available number. All you need now is $y_2=14/7=2$. And luckily we can get that with the two numbers left, 6 and 3 by their division. </p>
<p>This is not entirely computer-driven backtracking, but close enough I hope (for one paragraph that I had the patience to write). So whoever downvoted this answer better explain yourself now why you did that.</p>
<p>N.B. I have "cheated" a little bit above, in that I actually forgot when I got to the quotient business that until there I had picked the default value (i.e. when I have no immediate clue what to choose) as the highest available digit. For the quotient I used the lowest and got to a solution. So what happens if I use the highest? Neither $88/8$ nor $77/7$ lead to a solution, although showing this is more involved for $88/8$ (there are several branches to backtrack from), but for $77/7$ it is as easy as it was for $22/2$ or $33/3$. So next try $66/6$. This ultimately leads to a solution but you have more cases to try as $33 \times 2 \times 1$ does not yield a solution, but $22 \times 3 \times 1$ does so again because 8 is free and so is 7, yielding $(22-8)/7=2$ and we've used up 6 and 3 this time, but 4 and 2 are available and their quotient is again 2.</p>
<p>So here you have two solutions obtained this way.</p>
<p>Looking at the other answer, its computer-generated list does show that there are no solutions for the last digit set to 1 (but there are some for 2). I can't think off the top of my head how to quickly eliminate the last-digit-1 branch, so if you instead start by using the lowest available digit as the default value, you'll probably be in a world of frustration for quite a while. There are also only two solution for the last digit 2, while the rest of 30 solutions are for 4 or 5 as last digit. So maybe there is some room for rational improvement to my method... like somehow inferring that 4 or 5 as last digit could/would yield a lot more solutions than 1 or 2 do, but I can't say right now how to infer that without actually trying-cases/backtracking; a proof of that without using backtracking might involve Ramsey theory... which would obviously be way inaccessible for the 2nd grade.</p>
|
1,103,624 | <p>Ashamed to admit that I cannot aid my friend's niece with her second grade homework problem. So much for that college education, eh? Here's the problem.</p>
<p>Using only the natural numbers 1 through 9 without repeating any of them (natural because there cannot occur any rationals anywhere in this process, i.e. the first number cannot be prime, since otherwise there would be a fraction after the first operation), place the numbers in the boxes so that the following statement is true.</p>
<p>$\square \div \square \times \square + \square \times \square \times \square \div \square + \square \times \square = 100$</p>
<p>I am aware that I can use brute force to work this problem out, but this is going to be tedious at best, and might require a computer at worst. Furthermore, the problem doesn't specify anything about using the order of operations. On one hand, that would make sense, but on the other, this is a second grade problem in an American school, and I know that order of operations wasn't on my second grade curriculum.</p>
<p>In either case, does anyone know how to approach this in a somewhat elegant way? (Also apologies over tagging, someone please retag this appropriately if they think it isn't).</p>
| epi163sqrt | 132,007 | <p>As it is a second grade homework problem, I don't think the problem is stated with the intention to find <em>all</em> solutions.</p>
<p>Note, the problem does <em>not ask</em> to find all solutions. To me it seems more of being a stimulus to <em>play</em> with numbers and fundamental operations in order to increase experience and familiarity with basic arithmetic.</p>
<p>It's more a task to arise interest in playing with numbers, to train not giving up, even when many trials might fail and also to have fun and a good feeling when a solution could finally be found. </p>
<p>Since the solution is not unique there will most probably be different answers which could additionally stimulate discussions how specific solutions were found.</p>
|
2,781,801 | <p>When asked to evaluate $g$ at the point specified above we would get $\dfrac{1}{e} \cdot \log_e(\frac{1}{\sqrt e})$ and that evaluates to some -0.18393... but the correct answer is -1/2e. How does it get simplified to that?</p>
| max_zorn | 506,961 | <p>You did well. The "correct" answer is just written sloppily. It should be
$$-1/(2e)$$
which evaluates to what you found. </p>
|
3,306,571 | <p>I know that the function <span class="math-container">$f(x) = \frac{x}{x}$</span> is not differentiable at <span class="math-container">$x = 0$</span>, but according to the definition of differentiable functions:</p>
<blockquote>
<p>A differentiable function of one real variable is a function whose derivative exists at each point in its domain</p>
</blockquote>
<p>since <span class="math-container">$x = 0$</span> is not in the domain of <span class="math-container">$f$</span>, it doesn't have to be differentiable at that point for the function to be differentiable. This suggests that <span class="math-container">$f$</span> is differentiable as every other points in the domain has a derivative of <span class="math-container">$0$</span>.</p>
<p>However, some say that a function must be continuous if it's differentiable. This disproves the fact that <span class="math-container">$f$</span> is differentiable since it's not a continuous function.</p>
<p>Then is it really a differentiable function?</p>
| mlchristians | 681,917 | <p>To see why the function <span class="math-container">$f(x) = \frac{x}{x}$</span> is differentiable everywhere except at <span class="math-container">$0$</span>, and has derivative equal to <span class="math-container">$0$</span> where it is differentiable, consider the following: </p>
<p>The graph of <span class="math-container">$f(x) = \frac{x}{x}$</span> is the graph of <span class="math-container">$y=1$</span> with the point <span class="math-container">$(0,1)$</span> removed. Hence, <span class="math-container">$f'(x) = 0$</span> over <span class="math-container">$(-\infty, 0) \cup (0, \infty)$</span> and <span class="math-container">$f(x)$</span> is not differentiable at <span class="math-container">$x=0$</span>. </p>
|
3,699,105 | <p>If <span class="math-container">$T$</span> is normal operator and <span class="math-container">$T^3=T^2$</span>,then show that <span class="math-container">$T$</span> is idempotent .</p>
<ol>
<li><span class="math-container">$TT*=T*T$</span> </li>
<li><span class="math-container">$T^3=T^2$</span></li>
<li>We are to prove that <span class="math-container">$T^2=T$</span></li>
</ol>
<p>I have tried it many times by operating <span class="math-container">$T$</span> in both sides of <span class="math-container">$1$</span> and <span class="math-container">$2$</span> please tell me what will be the proper way</p>
| fleablood | 280,126 | <p>For a counter example just remove an essential item from R but keep it in S. And maybe (but not necessarily) vice versa</p>
<p>For example if <span class="math-container">$T= R\cup S = \mathbb Z \times \mathbb Z$</span> be the equivalence relationship on <span class="math-container">$\mathbb Z$</span> that <em>everything</em> is equivalent to <em>everything</em> (can't get more equivalent than that! <span class="math-container">$a T b$</span> for all <span class="math-container">$a, b \in \mathbb Z$</span>. So <span class="math-container">$T$</span> is reflexive because <span class="math-container">$r T r$</span> for all <span class="math-container">$r\in \mathbb Z$</span>). And <span class="math-container">$T$</span> is symmetric because <span class="math-container">$r T s$</span> and <span class="math-container">$s T r$</span> for all <span class="math-container">$r,s \in \mathbb Z$</span>. And <span class="math-container">$T$</span> is transitive because <span class="math-container">$r Ts; sTt; rTt$</span> for all <span class="math-container">$r,s,t\in \mathbb Z$</span>.)</p>
<p>Now let <span class="math-container">$R = \{(5, 7), (7,2), (2,2)\}$</span> and <span class="math-container">$S = (\mathbb Z \times \mathbb Z)\setminus R$</span>.</p>
<p><span class="math-container">$R$</span> is not reflexive because if <span class="math-container">$r \ne 2$</span> then <span class="math-container">$(r,r) \not \in \mathbb R$</span> (although <span class="math-container">$(r,r) \in S$</span>). And <span class="math-container">$S$</span> is not reflexive because <span class="math-container">$(2,2) \not \in S$</span>.</p>
<p>And <span class="math-container">$R$</span> is not symmetric because <span class="math-container">$(5,7) \in R$</span> but <span class="math-container">$(7,5)\not \in R$</span>. <span class="math-container">$S$</span> is not symmetric because <span class="math-container">$(7,5) \in S$</span> but <span class="math-container">$(5, 7) \not \in S$</span>.</p>
<p><span class="math-container">$R$</span> is not transitive because <span class="math-container">$(5, 7), (7,2) \in R$</span> but <span class="math-container">$(5,2) \not \in R$</span>. ANd <span class="math-container">$S$</span> is not transitive because <span class="math-container">$(5,2), (2,7) \in S$</span> but <span class="math-container">$(5,7) \not \in R$</span>.</p>
|
2,567,332 | <p>A Greek urn contains a red, blue, yellow, and orange ball. A ball is drawn from the urn at random and then replaced. If one does this $4$ times, what is the probability that all $4$ colors were selected?</p>
<p>I approached this questions by doing $(1/4)^4$ because there's always a $1/4$ chance of selected a specific color ball if it's replaced. I also tried doing if not the correct ball was selected; so I did $(3/4)^4$ but that didn't work either. What am I doing wrong?</p>
| Remy | 325,426 | <p>The first ball can be any of the four with probability $\frac{4}{4}$</p>
<p>The second ball must be any of the other three with probability $\frac{3}{4}$</p>
<p>The third ball must be any of the other two with probability $\frac{2}{4}$</p>
<p>The fourth ball must be the ball that hasn't been selected yet with probability $\frac{1}{4}$</p>
<p>All together,</p>
<p>$$\frac{4}{4}\cdot \frac{3}{4}\cdot \frac{2}{4}\cdot\frac{1}{4}=0.09375$$</p>
|
2,567,332 | <p>A Greek urn contains a red, blue, yellow, and orange ball. A ball is drawn from the urn at random and then replaced. If one does this $4$ times, what is the probability that all $4$ colors were selected?</p>
<p>I approached this questions by doing $(1/4)^4$ because there's always a $1/4$ chance of selected a specific color ball if it's replaced. I also tried doing if not the correct ball was selected; so I did $(3/4)^4$ but that didn't work either. What am I doing wrong?</p>
| Benji Altman | 398,014 | <p>We could do this by counting the number of ways to draw four balls and the number of ways to draw four balls without getting any duplicates.
There are $4!$ ways to not get a duplicate as every drawing can be thought of as an ordering and if we don't allow duplicates then we have a permutation. There are $4^4$ different possible drawings as replacement is allowed, this gives us $$\frac{4!}{4^4} = \frac{3}{32}$$</p>
|
1,102,885 | <p>I have exams in Machine Learning coming up and I need help answering this question.</p>
<blockquote>
<p>There are a million identical fish in a lake, one of which has
swallowed the One True Ring. You must get it back! After months of
effort, you catch another random fish and pass your metal detector
over it, and the detector beeps! It is the best metal detector money
can buy, and has a very low error rate: it fails to beep when near the
ring only one in a billion times, and it beeps incorrectly only one in
ten thousand times. What is the probability that, at long last, you’ve
found your precious ring?</p>
</blockquote>
<p>This is my answer I worked out using Bayes rule:</p>
<p><img src="https://i.stack.imgur.com/76WjZ.gif" alt="enter image description here"></p>
<p>Is this the right way to work out this type of question and is that somewhat the correct answer?</p>
| ChocolateBar | 161,284 | <p>You have $2$ cases in which the detector will beep:</p>
<p>$1:$ You have found the fish with the ring and the detector beeps. Probability: </p>
<p>$$\frac{1}{10^6} \cdot (1-\frac{1}{10^9}) = 9.99999999 \cdot 10^{-7}$$</p>
<p>$2:$ You have not found the fish with the ring but the detector beeps anyway. Probability:</p>
<p>$$\frac{10^6 -1}{10^6} \cdot \frac{1}{10^4} = 0.0000999999$$</p>
<p>To get the probability that you catched the correct fish under the condition that your detector beeped is then</p>
<p>$$\frac{\frac{1}{10^6} \cdot (1-\frac{1}{10^9})}{\frac{1}{10^6} \cdot (1-\frac{1}{10^9})+\frac{10^6 -1}{10^6} \cdot \frac{1}{10^4}}= \frac{1001001}{101101001} \approx 0.9901 \%,$$ so slightly less than $1\%$.</p>
|
1,566,471 | <p>Hi can someone please help?</p>
<p>I need to evaluate this indefinite integral:</p>
<p>$$\int \frac{(\ln x)^5}x dx$$</p>
<p>I know I need to use substitution, so if I let <em>u= x</em> but I can't figure out the antiderivative for the top portion.</p>
<p>Thank you!</p>
| Hagen von Eitzen | 39,174 | <p>Actually, that is the <em>definition</em> of independent. Then again, $P(B\mid A)$ is not $=1$ ("if $A$ happens, $B$ must happen"), it is not even defined.</p>
|
874,300 | <p>I'm having trouble grasping how to set these types of problems. There are a lot of related questions but it's difficult to abstract a general procedure on finding constants that give the given function bounding constraints to make it big-theta(general function). </p>
<p>so $\frac{x^4 +7x^3+5}{4x+1}$ is $ \Theta (x^3) $</p>
<p>to show this, we need to find constants such that.</p>
<p>$$ |c_1|(x^3) \leq \frac{x^4 +7x^3+5}{4x+1} \leq |c_2|(x^3)$$
In addition, there also has to be a $k$ such that for all values $x >k $ the argument holds.</p>
<p>start with one inequality
$$ |c_1|(x^3) \leq \frac{x^4 +7x^3+5}{4x+1}$$
$$ = |c_1| \leq \frac{x^4 +7x^3+5}{4x^4+x^3}$$
$$ = |c_1| \leq \frac{x^4}{x^3(4x+1)} + \frac{7x^3}{x^3(4x+1)} + \frac{5}{x^3(4x+1)}$$
so basically for $x > 0$, $$ |c_1| \leq \frac{1}{4} + 0 + 0$$
I'm assuming after I take the limit as x goes to infinity, i could choose any $c_1$ less than or equal to $\frac{1}{4}$? The other way would then have the same procedure? What would I set $k$ to?</p>
| David | 119,775 | <p>Here is a nice simple method.</p>
<p>If $x>1$ then
$$\frac{x^4 +7x^3+5}{4x+1}<\frac{x^4+7x^4+5x^4}{4x}=\frac{13}{4}x^3$$
and
$$\frac{x^4 +7x^3+5}{4x+1}>\frac{x^4}{4x+x}=\frac{1}{5}x^3\ .$$
That is, we have shown that if $x>1$ then
$$\frac{1}{5}x^3<f(x)<\frac{13}{4}x^3\ .$$</p>
|
373,958 | <p>Is $\sum_{n=1}^\infty(2^{\frac1{n}}-1)$ convergent or divergent?
$$\lim_{n\to\infty}(2^{\frac1{n}}-1) = 0$$
I can't think of anything to compare it against. The integral looks too hard:
$$\int_1^\infty(2^{\frac1{n}}-1)dn = ?$$
Root test seems useless as $\left(2^{\frac1{n}}\right)^{\frac1{n}}$ is probably even harder to find a limit for. Ratio test also seems useless because $2^{\frac1{n+1}}$ can't cancel out with ${2^{\frac1{n}}}$. It seems like the best bet is comparison/limit comparison, but what can it be compared against?</p>
| Aryabhata | 1,102 | <p>Elementary method: use AM $\ge$ GM!</p>
<p>$n-2$ copies of $1$, two copies of $\frac{1}{\sqrt{2}}$ gives</p>
<p>$$ \frac{n-2 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}}{n} \ge \left(\frac{1}{2}\right)^{1/n}$$</p>
<p>$$\frac{n-c}{n} \ge \left(\frac{1}{2}\right)^{1/n}$$</p>
<p>for some $c \gt 0$ ($ c= 2 - \sqrt{2}$)</p>
<p>$$ 2^{1/n} \ge \frac{n}{n-c} = 1 + \frac{c}{n-c}$$</p>
<p>Thus $$2^{1/n} - 1 \ge \frac{c}{n-c}$$</p>
<p>and so the series diverges.</p>
<p>But more simply:</p>
<p>$$2^{1/n} = e^{\log 2/n} \ge 1 + \frac{\log 2}{n}$$</p>
<p>(using $e^x \ge 1 + x$).</p>
|
91,590 | <p>So I'm reviewing old homeworks for an upcoming comp sci test and I came across this question:</p>
<p>Say whether the following statement is True, False or Unknown: </p>
<blockquote>
<p>The problem of checking whether a given Boolean formula has exactly
one satisfying assignment, is NP-complete</p>
</blockquote>
<p>My original answer to this was True because it seems to me that you can reduce SAT to this. Here's my solution:</p>
<p>Let's call this problem EX_SAT. Given a boolean formula s, we can construct a TM M where L(M) = SAT using EX_SAT. Assume that we have a NTM P that decides EX_SAT, and a NTM Q that decides DOUBLE_SAT (the problem of determining whether a Boolean formula has two or more satisfying assignments). We that DOUBLE_SAT is NP-complete because we reduced SAT to it in an earlier homework problem.</p>
<pre><code>M = on input s
1. Run P on s.
2. If P accepts, then accept.
3. If P rejects then run Q on s.
4. If Q accepts then accept.
5. If Q rejects then reject.
</code></pre>
<p>I see that EX_SAT doesn't have a polynomial time verifier, and I also see the one flaw in this proof is that I also have to use DOUBLE_SAT to complete it - which probably doesn't allow us to conclude that EX_SAT is NP-complete, but I thought I would ask this here because it might aid in my understanding of the topic.</p>
<p>Any thoughts would be much appreciated :) </p>
| Kyle Jones | 21,376 | <p>For the problem to be NP-complete it has to be in NP, which means it has to have a polynomial time verifier for all "yes" answers. If the exactly-1-satisfying-assignment question is in NP, that means there must be "yes" answers and verifiers for the following two questions:</p>
<p>Does this formula have a satisfying assignment?
Does this formula have no more satisfying assignments?</p>
<p>There's a polynomial time verifier for the first question, but not the second unless NP = coNP. Since the status of NP = coNP is unknown, the status of the main question also has to be "unknown."</p>
|
37,325 | <p>Suppose Bob has 50% chance to stand at each of two points $p_1, p_2$ on a unit circle. If one tries to naïvely answer the question "what is Bob's average position on the circle", an ambiguity shows up: the answer can be either the midpoint along the shortest arc between $p_{1,2}$ or along the longest. </p>
<p>More generally, any probability distribution for a position on the circle will seem to have an ambiguous mean. This will seem to also lead to ambiguity in the definition of the variance and higher moments. </p>
<p>The question is: is there a systematic way of dealing or getting rid of this ambiguity? (Note: I'm not interested in answers that assume the circle is embedded in a higher-dimensional space such as a plane, and give as an answer a point outside it. I want a point on the circle itself).</p>
| Alon Amit | 308 | <p>If you wish to have a specific, unambiguous "average position" for any distribution then, in particular, you expect to have one for the uniform distribution. This implies that whatever your definition is, it must give preference to some point on the circle.</p>
<p>Granting that, you can parametrize the circle by the angle $\theta$ relative to some fixed radius. With that, the average and all other moments are unambiguously defined as $\int \theta^k dp(\theta)$. Of course this definition breaks the symmetry of the circle but, as we've seen, this is unavoidable.</p>
|
37,325 | <p>Suppose Bob has 50% chance to stand at each of two points $p_1, p_2$ on a unit circle. If one tries to naïvely answer the question "what is Bob's average position on the circle", an ambiguity shows up: the answer can be either the midpoint along the shortest arc between $p_{1,2}$ or along the longest. </p>
<p>More generally, any probability distribution for a position on the circle will seem to have an ambiguous mean. This will seem to also lead to ambiguity in the definition of the variance and higher moments. </p>
<p>The question is: is there a systematic way of dealing or getting rid of this ambiguity? (Note: I'm not interested in answers that assume the circle is embedded in a higher-dimensional space such as a plane, and give as an answer a point outside it. I want a point on the circle itself).</p>
| joriki | 6,622 | <p>It depends on what you mean by "systematic". You can arbitrarily place a cut somewhere and parametrize the circle using angles $\phi\in[0,2\pi]$. The moments of these angles are unambiguous, but also arbitrary. There can't be a non-arbitrary way to do this because of the symmetry of the circle. If you have two points opposite each other, or more generally $n$ points at angles $2\pi/n$, there's no reason to pick any particular point as the average.</p>
<p>If you have additional structure on the circle, i.e. the circle is $U(1)$ or $\{z\in\mathbb C\mid \lvert z\rvert=1\}$, then you can use that for a less arbitrary definition of angles measured from the identity, but I doubt that the result will be particularly useful.</p>
|
3,760,594 | <p>Is there a proper notation to <em>compose</em> sets and produce a set of sets? (<em>I am referring to this as compose due to ignorance of a proper manner to call it</em>)</p>
<p>To illustrate what I want, let me <em>suppose</em> that <span class="math-container">$\otimes$</span> does the job, so that</p>
<p><span class="math-container">\begin{align}
\{1\} \otimes \{2\} &\rightarrow \{ \{1\} , \{2\} \}\\
\{1\} \otimes \{1,2\} &\rightarrow \{ \{1\} , \{1,2\} \}
\end{align}</span></p>
<p>Also, how can we write a <em>composition</em> for a finite number of sets? Say that <span class="math-container">$U_i = \{i\}$</span> (trivial example) is there something that can make (again using <span class="math-container">$\otimes$</span>):
<span class="math-container">\begin{equation}
\bigotimes_{i=1}^N U_i = \{ \{1\} , \{2\}, \ldots , \{N\} \}
\end{equation}</span></p>
| Tuvasbien | 702,179 | <p>If <span class="math-container">$\nabla f=\nabla g$</span>, then <span class="math-container">$\frac{\partial f}{\partial x_k}=\frac{\partial g}{\partial x_k}$</span> for all <span class="math-container">$k\in\{1,\ldots,n\}$</span>. Thus there exists <span class="math-container">$c_k(x_1,\ldots,x_{k-1},x_{k+1},\ldots,x_n)$</span> such that <span class="math-container">$f=g+c_k$</span>. And, for <span class="math-container">$l\neq k$</span>, we have
<span class="math-container">$$ \frac{\partial c_k}{\partial x_l}=0 $$</span>
and thus <span class="math-container">$dc_k=0$</span> and <span class="math-container">$c_k$</span> is a constant. The value of <span class="math-container">$c_k$</span> then does not depend of <span class="math-container">$k$</span> since for all <span class="math-container">$k,l$</span>, <span class="math-container">$f-g=c_k=c_l$</span>. Thus there exists a constant <span class="math-container">$c$</span> such that <span class="math-container">$f=g+c$</span>.</p>
|
804,882 | <p>If both $L:V\rightarrow W$ and $M:W\rightarrow U$ are linear transformations that are invertible, how can you prove that the composition $(M\circ L):V\rightarrow U$ is also invertible.</p>
| Kevin Sheng | 150,297 | <p>Hint: Suppose $L$ and $M$ are two invertible linear transformations. This means that the standard matrix $[L]$ and $[M]$ are invertible. The composition of two linear transformations is just the product of their standard matrices. Go from there.</p>
|
1,560,209 | <p>Prove that $f(x):\mathbb{R}\to\mathbb{R}$ , $x \mapsto x^3$ is injective.</p>
<hr>
<p>I want to prove this claim is true. </p>
<p>Here is my outline so far:</p>
<hr>
<p>We want to show that $f(a)=f(b)$ implies that $a=b$, for all $a,b \in \mathbb{R}$</p>
<p>We have $f(a)=a^3$, and $f(b)=b^3$</p>
<p>So, if $f(a)=f(b)$, we have $a^3=b^3$, or $a^3-b^3=0$, or $(a-b)(a^2+ab+b^2)=0$</p>
<p>Consider two cases:</p>
<p>Case 1: $(a-b)=0$, then $a=b$, and we are done.</p>
<p>Case 2: $(a^2+ab+b^2)=0$ We want to show that $a=b$.</p>
<hr>
<p>I am having trouble with Case 2. I need to figure out how to show that $a=b$. I think I can use this inequality some how: $a^2+2ab+b^2>0$ </p>
<hr>
<p>Any help would be appreciated.</p>
| Alekos Robotis | 252,284 | <p>Here is an alternative suggestion for a proof. First, show that $f$ is strictly increasing. Consider $x\in\mathbb{R}$, and define some $\epsilon>0$.
$$ f(x+\epsilon)=(x+\epsilon)^3=x^3+3x^2\epsilon+3x\epsilon^2+\epsilon^3>x^3=f(x),\forall x\in\mathbb{R}. $$
Now, suppose this function were not injective, that is, there existed some $y,y'\in \mathbb{R}$ such that $f(y)=f(y')$ but $y\ne y'$. Because $y\ne y'$, by trichotomy of the reals, $y>y'$ or $y'>y$. Assume w.l.o.g. that $y'>y$. Then we have $y'>y$, but $f(y')\not>f(y)$. This contradicts $f$ strictly increasing. $f$ is injective. $\square$</p>
|
160,518 | <p>In Mathematics, we know the following is true:</p>
<p>$$\int \frac{1}{x} \space dx = \ln(x)$$</p>
<p>Not only that, this rule works for constants added to x:
$$\int \frac{1}{x + 1}\space dx = \ln(x + 1) + C{3}$$
$$\int \frac{1}{x + 3}\space dx = \ln(x + 3) + C$$
$$\int \frac{1}{x - 37}\space dx = \ln(x - 37) + C$$
$$\int \frac{1}{x - 42}\space dx = \ln(x - 42) + C$$</p>
<p>So its pretty safe to say that $$\int \frac{1}{x + a}\space dx = \ln(x + a) + C$$ But the moment I introduce $x^a$ where $a$ is not equal to 1, the model collapses. The integral of $1/x^a$ is <strong>not</strong> equal to $\ln(x^a)$. The same goes for $\cos(x)$, and $\sin(x)$, and other trig functions. </p>
<p>So when are we allowed or not allowed to use the rule of $\ln(x)$ when integrating functions?</p>
| Nicholas Kirchner | 33,957 | <p>It boils down to $u$-substitution (which if you haven't covered yet, you soon will). You know that
$$ \int \frac{1}{x} dx = \ln x + C $$
(I'll not bother with absolute values here. You should remember them on problems that you do for class, but they aren't the focus of this question.)</p>
<p>Now, to handle
$$ \int \frac{1}{ax+b}dx $$
we do a $u$-substitution with $u = ax+b$. This makes $du=adx$, so the integral becomes
$$ \int \frac{1}{u} \frac{du}{a} = \frac{1}{a} \ln u + C = \frac{1}{a}\ln(ax+b) + C$$</p>
<p>Note the result:
$$ \int \frac{1}{ax+b}dx = \frac{1}{a}\ln(ax+b) + C $$</p>
<p>Check it by taking a derivative of the right-hand side. A chain rule was needed to take that derivative. The $u$-substitution we did was the chain rule reversed.</p>
|
160,518 | <p>In Mathematics, we know the following is true:</p>
<p>$$\int \frac{1}{x} \space dx = \ln(x)$$</p>
<p>Not only that, this rule works for constants added to x:
$$\int \frac{1}{x + 1}\space dx = \ln(x + 1) + C{3}$$
$$\int \frac{1}{x + 3}\space dx = \ln(x + 3) + C$$
$$\int \frac{1}{x - 37}\space dx = \ln(x - 37) + C$$
$$\int \frac{1}{x - 42}\space dx = \ln(x - 42) + C$$</p>
<p>So its pretty safe to say that $$\int \frac{1}{x + a}\space dx = \ln(x + a) + C$$ But the moment I introduce $x^a$ where $a$ is not equal to 1, the model collapses. The integral of $1/x^a$ is <strong>not</strong> equal to $\ln(x^a)$. The same goes for $\cos(x)$, and $\sin(x)$, and other trig functions. </p>
<p>So when are we allowed or not allowed to use the rule of $\ln(x)$ when integrating functions?</p>
| Joe | 24,942 | <p>Generally speaking, "using $\ln (x)$" as a rule or technique is unheard of. When one speaks of techniques, they usually include integration by substitution, integration by parts, trig substitutions, partial fractions, etc. With introductory calculus in mind, $\ln |x|$ is <strong>defined</strong> as $\int \frac{1}{x} \ dx.$ This can be extended to $\ln |u| = \int \frac{1}{u} \ du.$ Note that there are many more definitions for $\ln (x)$, but I felt this best related particularly to your examples.</p>
<p>For your first couple of examples, when choosing your $u$ to be the denominator, the $du$ is simply equal to $dx.$ This is what 'allows' the integrand to be evaluated to just $\ln |u|$ where $u$ is a linear expression. </p>
<p>In regards to $\int \frac{1}{x^2 + a} \ dx$, this can be handled using an inverse tangent and would be evaluated to
$$\dfrac{\operatorname{arctan}(\frac{x}{\sqrt{a}})}{\sqrt{a}} + C$$</p>
<p>For integrals of the form
$$\int \frac{1}{x^n + a} \ dx$$ </p>
<p>where $n \ge 3$, you will have to revert to partial fractions. For more on partial fractions, see <a href="https://math.stackexchange.com/questions/20963/integration-by-partial-fractions-how-and-why-does-it-work/21112#21112">this.</a></p>
|
63,015 | <p>Why does assigning a DownValue using <code>Apply</code>, e.g.,</p>
<pre><code>Remove[a]
index={3,4};
(a @@ index) = 5;
a @@ index
(*Set::write: Tag Apply in a @@ {3, 4} is Protected. >>*)
(*a[3,4]*)
</code></pre>
<p>not work, while an assignment such as</p>
<pre><code>Remove[a]
a[Sequence @@ index] = 5
a @@ index
(*5*)
(*5*)
</code></pre>
<p>does?</p>
| Chris Degnen | 363 | <p>Evaluation happens too late, but you can fix it by forcing evaluation first:</p>
<pre><code>Remove[a]
index = {3, 4};
Evaluate[a @@ index] = 5;
a @@ index
</code></pre>
<blockquote>
<p>5</p>
</blockquote>
<p>This can only be done on the initial definition though. After that</p>
<pre><code>a[Sequence @@ index] = 6;
a @@ index
</code></pre>
<blockquote>
<p>6</p>
</blockquote>
|
3,529,359 | <p>Let <span class="math-container">$\Omega$</span> be a bounded and smooth domain and let <span class="math-container">$J:H^1(\Omega) \times H^1_0(\Omega) \to \mathbb{R}$</span> be defined by</p>
<p><span class="math-container">$$J(u,v) = \int_\Omega f(u)|\nabla v|^2$$</span>
where <span class="math-container">$f\colon \mathbb{R} \to \mathbb{R}$</span> is a smooth function, bounded above and below away from zero (other assumptions can be added as necessary).</p>
<p>Under what conditions on <span class="math-container">$f$</span> do I get that <span class="math-container">$J$</span> is weakly lower semicontinuous? </p>
<p>Obviously if <span class="math-container">$f \equiv 1$</span> then it is true, but what about the more genera case?</p>
| daw | 136,544 | <p>Here is the proof of @JohannesHahn in a more elementary way.
The idea is to show that
<span class="math-container">$$
(u,v) \mapsto \sqrt{ f(u)}\nabla v
$$</span>
is weakly sequentially continuous from <span class="math-container">$H^1(\Omega)\times H^1(\Omega)$</span> to <span class="math-container">$L^2(\Omega)$</span>.</p>
<p>Let me denote <span class="math-container">$g(u):=\sqrt{ f(u)}$</span>. Let <span class="math-container">$(u_n,v_n) \rightharpoonup (u,v)$</span> in <span class="math-container">$H^1(\Omega)\times H^1(\Omega)$</span>.</p>
<p>Then (after possibly extracting subsequences) <span class="math-container">$u_n \to u$</span> in <span class="math-container">$L^2(\Omega)$</span> and <span class="math-container">$u_n(x)\to u(x)$</span> pointwise almost everywhere. </p>
<p>Let <span class="math-container">$w\in L^2(\Omega)$</span>. By dominated convergence,
<span class="math-container">$$
\int_\Omega |g(u_n)-g(u)|^2 w^2 \to 0
$$</span>
Then we have
<span class="math-container">$$
\int_\Omega (g(u_n)\nabla v_n - g(u)\nabla v)w \ dx
=\int_\Omega (g(u_n) - g(u))\nabla v_n w\ dx + \int_\Omega g(u)(\nabla v_n - \nabla v)w \ dx.
$$</span>
The first integral goes to zero, since <span class="math-container">$(g(u_n)-g(u))w\to0$</span> in <span class="math-container">$L^2(\Omega)$</span>. The second integral vanishes due to the weak convergence <span class="math-container">$\nabla v_n - \nabla v\rightharpoonup 0$</span> in <span class="math-container">$L^2(\Omega)$</span>.
It follows
<span class="math-container">$$
g(u_n)\nabla v_n \rightharpoonup g(u)\nabla v
$$</span>
in <span class="math-container">$L^2(\Omega)$</span>
as required. (Here the usual subsequence argument has to be used: Every subsequence of <span class="math-container">$(g(u_n)\nabla v_n)$</span> contains a subsequence converging weakly to <span class="math-container">$g(u)\nabla v$</span>.)</p>
<hr>
<p>Note that a similar argument does not work when applied directly to the functional:
<span class="math-container">$$
\int_\Omega f(u_n)|\nabla v_n|^2 - f(u)|\nabla v|^2\ dx
=
\int_\Omega (f(u_n)-f(u))|\nabla v_n|^2\ dx + \int_\Omega f(u)(|\nabla v_n|^2- |\nabla v|^2)\ dx.
$$</span>
While the <span class="math-container">$\liminf$</span> of the second integral is non-negative, nothing can be said about the first integral: <span class="math-container">$f(u_n)-f(u)$</span> does not converge to zero in <span class="math-container">$L^\infty(\Omega)$</span> if <span class="math-container">$\Omega\subset \mathbb R^d$</span> for <span class="math-container">$d>1$</span>.</p>
|
2,603,239 | <p>(The Cauchy principal value of)
$$
\int_0^{\infty}\frac{\tan x}{x}\mathrm dx
$$</p>
<p>I tried to cut this integral into $$\sum_{k=0}^{\infty}\int_{k\pi}^{(k+1)\pi}\frac{\tan x}{x}\mathrm dx$$
And then
$$\sum_{k=0}^{\infty}\lim_{\epsilon \to 0}\int_{k\pi}^{(k+1/2)\pi-\epsilon}\frac{\tan x}{x}\mathrm dx+\int_{(k+1/2)\pi+\epsilon}^{(k+1)\pi}\frac{\tan x}{x}\mathrm dx$$
$$\sum_{k=0}^{\infty}\int_{k\pi}^{(k+1/2)\pi}\frac{((2k+1)\pi-2x)\tan x}{((2k+1)\pi-x)x}\mathrm dx$$
And I did not know how to continue. I did not know if I was right or not. How to calculate this integral?</p>
| robjohn | 13,854 | <p>This can also be handled using contour integration. Since there are no singularities inside $\gamma$,
$$
\int_\gamma\frac{\tan(z)}z\,\mathrm{d}z=0
$$
where $\gamma$ is the contour</p>
<p><a href="https://i.stack.imgur.com/4HUqG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4HUqG.png" alt="enter image description here"></a></p>
<p>where $k\to\infty$ then $m\to\infty$.</p>
<p>First notice that the integrals along all the small arcs, which are $-\pi i$ times the residues at the singularities of $\frac{\tan(z)}z$, cancel out for each $k$.</p>
<p>Since $\tan(z)=i\tanh(\operatorname{Im}(z))$ for $\operatorname{Re}(z)\in\mathbb{Z}$, the absolute value of the integrals along the vertical lines are at most $\frac mk$. These vanish as $k\to\infty$.</p>
<p>Therefore, we get
$$
\mathrm{PV}\int_{-\infty}^\infty\frac{\tan(x)}x\,\mathrm{d}x
=\int_{-\infty+mi}^{\infty+mi}\frac{\tan(z)}z\,\mathrm{d}z
$$
as $m\to\infty$, $\tan(z)\to i$. Therefore,
$$
\mathrm{PV}\int_{-\infty}^\infty\frac{\tan(x)}x\,\mathrm{d}x=\pi
$$
Since $\frac{\tan(x)}x$ is even, we get that
$$
\mathrm{PV}\int_0^\infty\frac{\tan(x)}x\,\mathrm{d}x=\frac\pi2
$$</p>
|
1,151,653 | <p>How can I express the following as a function sequence? Namely, how can I properly express <span class="math-container">$f_n(x)$</span>?</p>
<p>Here are the following function graphs:</p>
<p><img src="https://i.stack.imgur.com/2GFYj.png" alt="enter image description here" /></p>
<p>Text only (color-coded with image):</p>
<ol>
<li><span class="math-container">$\color{red}{f_1(x)=x^x}$</span></li>
<li><span class="math-container">$\color{blue}{f_2(x)=x^{x^x}}$</span></li>
<li><span class="math-container">$\color{green}{f_3(x)=x^{x^{x^x}}}$</span></li>
<li><span class="math-container">$\color{purple}{f_4(x)=x^{x^{x^{x^x}}}}$</span></li>
<li><span class="math-container">$\color{orange}{f_5(x)=x^{x^{x^{x^{x^x}}}}}$</span></li>
<li><span class="math-container">$f_6(x)=x^{x^{x^{x^{x^{x^x}}}}}$</span></li>
</ol>
<p>So how may I express <span class="math-container">$f_n(x)$</span>? e.g. <span class="math-container">$f_n(x)=x^{x^{x^{.^{.^{.^{x}}}}}}$</span>?</p>
| AvZ | 171,387 | <p>A unit circle will have the equation $$x^2 + y^2=1$$<br>
To find the intersection points of this with the parabola $y=x^2$ we substitute the value of $y$.<br>
We can write this as<br>
$$x^2 + x^4=1$$<br>
This is a quadratic equation in terms of $x^2$.
We can now write $$x^2=\frac{-1+\sqrt{5}}{2}$$
Note that a real number squared cannot be negative.
$$x=\pm\sqrt{\frac{-1+\sqrt{5}}{2}}$$<br>
The negative solution will be eliminated as it will not be in first quadrant. Hence,<br>
$$x=\sqrt{\frac{-1+\sqrt{5}}{2}}$$</p>
|
3,956,292 | <p>Consider the Euclidean ball <span class="math-container">$B^n(x,r)$</span> in <span class="math-container">$\mathbb{R}^n$</span> given by:
<span class="math-container">$$B^n(x,r) = \{z\in\mathbb{R}^n : ||z-x||_2 \leq r\}$$</span> with centre <span class="math-container">$x\in\mathbb{R}^n$</span> and radius <span class="math-container">$r\geq 0$</span>.
Now we <strong>break</strong> this ball by <em><strong>removing an arbitrary subset from the boundary</strong></em> of <span class="math-container">$B^n(x,r)$</span>. Is the resulting set still convex?</p>
<p>My intuition says <em>Yes</em>, and it's easy to visualise this in <span class="math-container">$\mathbb{R}^2$</span> or <span class="math-container">$\mathbb{R}^3$</span>. I'm wondering how I can prove it generally, i.e. for <span class="math-container">$\mathbb{R}^n$</span> by considering an arbitrary subset <span class="math-container">$S\subseteq B^n(x,r)$</span>? If <span class="math-container">$S = B^n(x,r)$</span> then the resulting set is empty, which is convex - so we can ignore that situation. We assume <span class="math-container">$S\subset B^n(x,r)$</span>. What's next?</p>
<p>I consider two points <span class="math-container">$y_1,y_2 \in B^n(x,r)\backslash S$</span>, and want to show that for <span class="math-container">$t\in [0,1]$</span> we have <span class="math-container">$ty_1 + (1-t)y_2 \in B^n(x,r)\backslash S$</span>. How do I take it from here? I'm stuck.</p>
<p>Thanks!</p>
<p><strong>Addendum</strong>:<br>
What if we consider the <span class="math-container">$p$</span>-norm in general, instead of the Euclidean norm? How do things change with <span class="math-container">$p$</span> if the ball is defined as follows?
<span class="math-container">$$B^n(x,r) = \{z\in\mathbb{R}^n : ||z-x||_p \leq r\}$$</span>
<em>It sounds fun breaking balls in higher dimensions under different norms!</em></p>
| Andreas Blass | 48,510 | <p>To simplify notation, I'll assume, without loss of generality, that the center of the ball is at the origin.
Suppose, toward a contradiction, that you have a counterexample in some high dimension. Say your set contains <span class="math-container">$p$</span> and <span class="math-container">$q$</span> but not some point <span class="math-container">$tp+(1-t)q$</span> on the line segment joining <span class="math-container">$p$</span> to <span class="math-container">$q$</span>. Then this is also a counterexample in the <span class="math-container">$2$</span>-dimensional subspace spanned by (the vectors from the origin to) <span class="math-container">$p$</span> and <span class="math-container">$q$</span>. So, once you know the result for <span class="math-container">$2$</span> dimensions, it follows for all higher dimensions.</p>
|
3,973,006 | <p>The question is fully contained in the title.</p>
<p>I tried to prove maximality (if that happens, <span class="math-container">$I$</span> is prime as well) in <span class="math-container">$\mathbb Z[X]$</span>, but I am not able to figure a strategy out for that purpouse. Obviously, if <span class="math-container">$I$</span> is not maximal, I am expected to say whether <span class="math-container">$I$</span> is a prime ideal, which is a problem too.</p>
<p>How would you solve such exercise?</p>
| Bernard | 202,857 | <p><strong>Hint</strong>: <span class="math-container">$I$</span> contains <span class="math-container">$\;7(X^3+2X^2+1)-X^2(7X+14)=7$</span>, hence
<span class="math-container">$$\mathbf Z[X]/I\simeq \mathbf Z/7 \mathbf Z[X]/(I/7 \mathbf Z[X])= \mathbf Z/7 \mathbf Z[X]/(X^3+\bar 2X^2+\bar 1).$$</span>
Can you show that <span class="math-container">$X^3+\bar 2X^2+\bar 1$</span> is irreducible in <span class="math-container">$ \mathbf Z/7 \mathbf Z[X]$</span>?</p>
|
64,544 | <blockquote>
<p>Please let me know what is the standard notation for group action.</p>
</blockquote>
<p>I saw the following three notations for group action.
(All the images obtained as <code>G\acts X</code> for different deinitions of <code>\acts</code>.) </p>
<p>(1) <img src="https://lh5.googleusercontent.com/_7jyZyirE1is/TcM6Q736oVI/AAAAAAAAACU/7li7VA1-FTc/s144/B.png" alt="alt text"> </p>
<p>I saw this one most, but only in handwriting and I like it. But I did not find a better way to write it in LaTeX.</p>
<pre><code>\usepackage{mathabx,epsfig}
\def\acts{\mathrel{\reflectbox{$\righttoleftarrow$}}}
</code></pre>
<p>(2) <img src="https://lh5.googleusercontent.com/_7jyZyirE1is/TcM6XymimxI/AAAAAAAAACY/byg0FDFv7wM/s144/C.png" alt="alt text"> </p>
<p>It is almost as good as 1, but in handwriting this arrow can be taken as $G$.</p>
<pre><code>\usepackage{mathabx}
\def\acts{\lefttorightarrow}
</code></pre>
<p>(3) <img src="https://lh6.googleusercontent.com/_7jyZyirE1is/TcM6JeflvjI/AAAAAAAAACQ/n_dfYCfeQEc/s800/A.png" alt="alt text"> </p>
<p>I saw this one in print, I guess it is used since there is no better symbol in "amssymb".</p>
<pre><code>\usepackage{amssymb}
\def\acts{\curvearrowright}
</code></pre>
| Theo Johnson-Freyd | 78 | <p>It is always a shame, of course, when none of the many LaTeX packages has precisely the symbol that you might use on the chalkboard. In writing, you should try to use words when possible, or at least supplement your symbols with words. Someone just reading the words should be able to pick up the nuances of your notation. Very importantly, use your notation consistently throughout the paper.</p>
<p>When speaking at the chalkboard, you should speak everything you write, and write everything you speak, in the following sense: someone blind should be able to follow the majority of your talk, and also someone deaf should be able to follow the majority of the talk. Even when talking one-on-one with someone else, this is important. So you write "$G \acts X$" for your favorite definition of $\acts$, but you speak "Let $G$ act on $X$." And in the most important statements in the talk, namely the statements of definitions and theorems, you should write all the words: "Let $G$ act on $X$."</p>
<p>Personally, I like some variation on your option 1. I have seen 3, and I use it in some lecture notes. I do not like 2: when people read, they tend to give a lot more visual importance to the tops of words and letters than to the bottoms (this is why almost all letters do not hang below the baseline, but there are many that stick up higher, and you can almost read just from knowing which letters go higher up). So put the arrow at the top.</p>
|
64,544 | <blockquote>
<p>Please let me know what is the standard notation for group action.</p>
</blockquote>
<p>I saw the following three notations for group action.
(All the images obtained as <code>G\acts X</code> for different deinitions of <code>\acts</code>.) </p>
<p>(1) <img src="https://lh5.googleusercontent.com/_7jyZyirE1is/TcM6Q736oVI/AAAAAAAAACU/7li7VA1-FTc/s144/B.png" alt="alt text"> </p>
<p>I saw this one most, but only in handwriting and I like it. But I did not find a better way to write it in LaTeX.</p>
<pre><code>\usepackage{mathabx,epsfig}
\def\acts{\mathrel{\reflectbox{$\righttoleftarrow$}}}
</code></pre>
<p>(2) <img src="https://lh5.googleusercontent.com/_7jyZyirE1is/TcM6XymimxI/AAAAAAAAACY/byg0FDFv7wM/s144/C.png" alt="alt text"> </p>
<p>It is almost as good as 1, but in handwriting this arrow can be taken as $G$.</p>
<pre><code>\usepackage{mathabx}
\def\acts{\lefttorightarrow}
</code></pre>
<p>(3) <img src="https://lh6.googleusercontent.com/_7jyZyirE1is/TcM6JeflvjI/AAAAAAAAACQ/n_dfYCfeQEc/s800/A.png" alt="alt text"> </p>
<p>I saw this one in print, I guess it is used since there is no better symbol in "amssymb".</p>
<pre><code>\usepackage{amssymb}
\def\acts{\curvearrowright}
</code></pre>
| Pete L. Clark | 1,149 | <p>As it happens, I can still remember being confused the first few times I saw this notation: not to put too fine a point on it, but there is something syntactically new going on there beyond the usual function / arrow notation.</p>
<p>In my opinion this is not notation at all but rather <strong>shorthand</strong>. In other words, it is something that you can use in your own handwritten notes and something that you can write on the board <em>if</em> you are confident that your listeners will understand you. (I suppose if I start telling stories about all the talks I witnessed as a Harvard graduate student that sailed over my head whether the speaker intended them to or not, I will not get enough "Oh, you poor dear" responses to justify the effort. But it happened quite a lot!) It's nice to have an agreed upon shorthand. For instance, in the commutative algebra class I just taught, when things were hot and heavy I didn't want to keep writing out "$I$ is an ideal of $R$", so I used a shorthand for it and explained it the first 20 times I used it. (The students used it too when presenting solutions to problems.) But in my lecture notes it appears nowhere: if I have the time to tex up lecture notes at all, then I certainly have the time to write out "$I$ is an ideal of $R$". </p>
<p>Thus I would not recommend that anyone include this notation in their formal mathematical writings. Note that Dick Palais has said essentially the same thing above, so: don't listen to me, but listen to him!</p>
|
1,695,261 | <p>Is it true that for every $ε > 0$, there is $δ > 0$, such that $0 < |x−2| < δ ⇒ |(x^2 −x)−2| < ε$?</p>
<p>Now I know that $|(x^2 −x)−2|$ is same as $|(x-2)(x+1)|$, but I am not sure how to link that with the first bit of info given. In general epsilon-delta proofs confuse me. </p>
<p>So I start by saying that there is an epsilon s.t $|(x^2 −x)−2| < ε$. And if this is true then there is a delta s.t $0 < |x−2| < δ$. Or is it the other way around? </p>
<p>Now, if $|(x^2 −x)−2| < ε$ then $|(x-2)(x+1)| < ε$ and $|x-2||x+1| < ε$ and
$$|x-2|<\frac{ε}{|x+1|}$$ But since epsilon is always positive and so is $|x+1|$ then a delta always exists. </p>
<p>Is my proof correct or totally wrong? I feel as though all I have done is rearranged the equation, and not really proved anything. </p>
| Clarinetist | 81,560 | <p>I always like to refer people to my answer <a href="https://math.stackexchange.com/questions/418961/epsilon-delta-proof-that-lim-limits-x-to-1-frac1x-1/418991#418991">here</a> when it comes to simple polymonial $\delta$-$\epsilon$ proofs. Read this link so that you understand my methodology here.</p>
<p><strong>Scratch work</strong>:</p>
<p>$$|x^2-x-2| = |(x-2)(x+1)| = |x-2||x+1|\text{.} $$
Take $\delta = 1$. Then
$$|x-2| < 1 \Longleftrightarrow -1 < x-2 < 1 \Longleftrightarrow 1 < x < 3 \Longleftrightarrow 2 < x+1 < 4 \implies |x+1| < 4\text{.}$$
So for $\epsilon > 0$,
$$|x-2||x+1| < 4|x-2| < 4\delta = 4\left(\dfrac{\epsilon}{4}\right)=\epsilon$$
if $\delta = \dfrac{\epsilon}{4}$, so we choose $\delta = \min\left(1, \dfrac{\epsilon}{4}\right)$.</p>
<p><strong>Proof</strong>:</p>
<p>Let $\epsilon > 0$ be given. Choose $\delta = \min\left(1, \dfrac{\epsilon}{4}\right)$. Then
$$|(x^2-x)-2| = |x^2-x-2| = |x-2||x+1| < 4|x-2|$$
(since if $|x-2| < 1$, $|x+1| < 4$), and
$$4|x-2| < 4\delta \leq 4\left(\dfrac{\epsilon}{4}\right) = \epsilon\text{.}$$</p>
|
2,755,143 | <p>Find Number of integers satisfying $$\left[\frac{x}{100}\left[\frac{x}{100}\right]\right]=5$$ where $[.]$ is Floor function.</p>
<p>I assumed $$x=100q+r$$ where $0 \le r \le 99$</p>
<p>Then we have </p>
<p>$$\left[\left(q+\frac{r}{100}\right)q\right]=5$$ $\implies$</p>
<p>$$q^2+\left[\frac{rq}{100}\right]=5$$</p>
<p>Since $rq$ is an integer we have $$rq=100p+r_1$$ where $0 \le r_1 \le 99$</p>
<p>Then we have</p>
<p>$$q^2+p+\left[\frac{r_1}{100}\right]=5$$ $\implies$</p>
<p>$$q^2+p=5$$ so the possible ordered pairs $(p,q)$ are</p>
<p>$(1,2)$, $(1,-2)$, $(-4, 3)$ i am getting infinite pairs.</p>
<p>How to proceed?</p>
| Jason | 130,776 | <p>A less-than-satisfying, but simple and fast computational approach -- this double-flooring operation is increasing in $x$. When $x=200$, you get 4. When $x=300$, to get 9. This just leaves 100 integers to check with a simple program. Here is one such program in Mathematica:</p>
<blockquote>
<p>Tally@Table[Floor[x/100*Floor[x/100]], {x, 200, 300}]</p>
</blockquote>
<p>Which yields</p>
<blockquote>
<p>{{4, 50}, {5, 50}, {9, 1}}</p>
</blockquote>
<p>[edit] A quick edit -- if you're also concerned with negative integers, a similar approach can be taken to see that we need only check between $x=-300$ and $x=-200$ to get all of those too.</p>
|
3,519,515 | <p>Here, I wonder what is a good way to use the epsilon delta definition or converging sequences to show that the set S containing quotients on [0,1] have/does not have volume 0, (i.e. whether there exist a <strong>finite</strong> number of intervals which union contain all of S such that the <strong>sum</strong> of length of all intervals is less than any <span class="math-container">$\epsilon > 0$</span> you fix). I believe it more likely does not have volume 0 from my intuition . I am lost on where to start the proof. </p>
<p>Does the idea of closure of S play a part in this proof? and how?</p>
<p>Also, is it possible to prove this using pigeonhole principle involving infinite rationals in one interval?</p>
| Henno Brandsma | 4,280 | <p>Suppose <span class="math-container">$\lambda(\Bbb Q \cap [0,1]) < 1$</span>, where by <span class="math-container">$\lambda$</span> I mean the volume of the set (as (sort of) defined in the comments to the question.</p>
<p>This would mean that we can cover <span class="math-container">$\Bbb Q \cap [0,1]$</span> by finitely many intervals <span class="math-container">$I_1,\ldots, I_n$</span> such that <span class="math-container">$\sum_{j=1}^n \lambda(I_j) < 1$</span> too, and we can assume these are closed intervals (as <span class="math-container">$\lambda([a,b])=\lambda((a,b))=b-a$</span> etc.) but then <span class="math-container">$[0,1]\setminus (\bigcup_{j=1}^n I_j)$</span> is a non-trivial open subset of <span class="math-container">$[0,1]$</span> that misses the dense set <span class="math-container">$\Bbb Q \cap [0,1]$</span>, which is a contradiction. So <span class="math-container">$\lambda(\Bbb Q \cap [0,1])=1$</span>. </p>
|
58,525 | <p>I am trying to make surface plots of squashed spheres. The spheres are defined by a list of points. For simplicity, consider the round sphere:</p>
<pre><code>pts = Flatten[
Table[{Sin[θ] Cos[ϕ], Sin[θ] Sin[ϕ],
Cos[θ]}, {θ, 0, π, π/14}, {ϕ, 0,
2 π, 2 π/14}], 1];
</code></pre>
<p>One way to plot this is:</p>
<pre><code>ListPlot3D[{pts,
-pts,}
BoundaryStyle -> None,
ColorFunction -> "Rainbow",
InterpolationOrder -> 2,
BoxRatios -> {1, 1, 1}]
</code></pre>
<p><code>ListPlot3D</code> has the nice feature that it interpolates the data to make a smooth surface. However, the northern and southern hemispheres are shaded differently, so there is a clear break at the equator:</p>
<p><img src="https://i.stack.imgur.com/A7jdx.png" alt="enter image description here"></p>
<p>An alternative is to do </p>
<pre><code>ListSurfacePlot3D[pts]
</code></pre>
<p>Now the shading is uniform (there is no break at the equator). However, the data is no longer interpolated (and interpolation is not an option for <code>ListSurfacePlot3D</code>), so the surface looks rough and lumpy:</p>
<p><img src="https://i.stack.imgur.com/xkfXr.png" alt="enter image description here"></p>
<p>I am trying to find a solution that combines the best of both world: the smooth surface of <code>ListPlot3D</code> with the uniform shading of <code>ListSurfacePlot3D</code>.</p>
| Dr. belisarius | 193 | <p>Just format your list such that <code>SphericalPlot3D[]</code> can handle it:</p>
<pre><code>pts = Flatten[Table[{Theta, Phi, Cos[Theta]}, {Theta, 0, Pi, Pi/14}, {Phi, 0, 2 Pi, 2 Pi/14}], 1];
f = Interpolation[pts];
SphericalPlot3D[f[Theta, Phi], {Theta, 0, Pi}, {Phi, 0, 2 Pi}, ColorFunction -> "Rainbow",
Mesh -> None]
</code></pre>
<p><img src="https://i.stack.imgur.com/7q4eo.png" alt="Mathematica graphics"></p>
|
1,981,360 | <blockquote>
<p>Given function $f:\mathbb{R}_0^+ \to \mathbb{R},~f(x) = x^2 + 4x + 4$ prove that it is injective.</p>
</blockquote>
<p>Using definition of injectivity $(\forall x_1, x_2 \in \mathbb{R}_0^+)(x_1 \neq x_2 \implies f(x_1) \neq f(x_2))$ I'm doing the following:</p>
<p>$$x_1^2 + 4x_1 + 4 = x_2^2 + 4x_2 + 4$$
$$x_1^2 - x_2^2 = -4(x_1 - x_2)$$
$$x_1 + x_2 = -4$$
$$x_1 = -4 - x_2.$$
Since domain is $\mathbb{R}_0^+$ it is apparent that $x_1 \neq -4 - x_2$ and hence function is not injective.</p>
<hr>
<p>Is my final argument correct? In cases like that, shall I use definition instead of counterpositive?</p>
| hamam_Abdallah | 369,188 | <p>$\exists (x_1,x_2) \in [0,+\infty)^2 : x_1^2+4x_1+4=x_2^2+4x_2+4 \; \implies$</p>
<p>$(x_1-x_2)(x_1+x_2+4)=0 \implies$</p>
<p>$x_1=x_2 $ since $x_1+x_2+4\geq4$.</p>
<p>thus $f$ is injective.</p>
|
191,548 | <p>Say I have a list:</p>
<pre><code>{{Line[{{-Sqrt[5/8 - Sqrt[5]/8], 1/4 (-1 - Sqrt[5])}, {0, 1}}],
Line[{{Sqrt[5/8 - Sqrt[5]/8], 1/4 (-1 - Sqrt[5])}, {0,1}}]},
{Line[{{-Sqrt[5/8 + Sqrt[5]/8],1/4 (-1 + Sqrt[5])}, {Sqrt[5/8 - Sqrt[5]/8], 1/4 (-1 - Sqrt[5])}}],
Line[{{Sqrt[5/8 - Sqrt[5]/8], 1/4 (-1 - Sqrt[5])}, {0, 1}}]}}
</code></pre>
<p>So that each sublist of that list consist of, in this case, two lines. All the points that tell us about the position of the line appear in another list, call this list <code>points</code>. Now, I want to extract the position of all the points from the above list in that <code>points</code> list. I'm aware of <code>Position</code> function but I'm not sure how to effectively apply it to my big list above in order to get the list of positions. AI'd very much appreciate some help. </p>
| Sjoerd Smit | 43,522 | <p>You can use <code>Cases</code> for this:</p>
<pre><code>lines = {{Line[{{-Sqrt[5/8 - Sqrt[5]/8],
1/4 (-1 - Sqrt[5])}, {0, 1}}],
Line[{{Sqrt[5/8 - Sqrt[5]/8], 1/4 (-1 - Sqrt[5])}, {0,
1}}]}, {Line[{{-Sqrt[5/8 + Sqrt[5]/8],
1/4 (-1 + Sqrt[5])}, {Sqrt[5/8 - Sqrt[5]/8],
1/4 (-1 - Sqrt[5])}}],
Line[{{Sqrt[5/8 - Sqrt[5]/8], 1/4 (-1 - Sqrt[5])}, {0, 1}}]}};
Catenate @ Cases[lines, Line[pts : {{_, _} ..}] :> pts, Infinity]
</code></pre>
<blockquote>
<blockquote>
<p>Out[5]= {{-Sqrt[5/8 - Sqrt[5]/8], 1/4 (-1 - Sqrt[5])}, {0, 1}, {Sqrt[
5/8 - Sqrt[5]/8], 1/4 (-1 - Sqrt[5])}, {0,
1}, {-Sqrt[5/8 + Sqrt[5]/8], 1/4 (-1 + Sqrt[5])}, {Sqrt[
5/8 - Sqrt[5]/8], 1/4 (-1 - Sqrt[5])}, {Sqrt[5/8 - Sqrt[5]/8],
1/4 (-1 - Sqrt[5])}, {0, 1}}</p>
</blockquote>
</blockquote>
|
191,548 | <p>Say I have a list:</p>
<pre><code>{{Line[{{-Sqrt[5/8 - Sqrt[5]/8], 1/4 (-1 - Sqrt[5])}, {0, 1}}],
Line[{{Sqrt[5/8 - Sqrt[5]/8], 1/4 (-1 - Sqrt[5])}, {0,1}}]},
{Line[{{-Sqrt[5/8 + Sqrt[5]/8],1/4 (-1 + Sqrt[5])}, {Sqrt[5/8 - Sqrt[5]/8], 1/4 (-1 - Sqrt[5])}}],
Line[{{Sqrt[5/8 - Sqrt[5]/8], 1/4 (-1 - Sqrt[5])}, {0, 1}}]}}
</code></pre>
<p>So that each sublist of that list consist of, in this case, two lines. All the points that tell us about the position of the line appear in another list, call this list <code>points</code>. Now, I want to extract the position of all the points from the above list in that <code>points</code> list. I'm aware of <code>Position</code> function but I'm not sure how to effectively apply it to my big list above in order to get the list of positions. AI'd very much appreciate some help. </p>
| gwr | 764 | <h2>Answer to what you want</h2>
<pre><code>lines = {
{ Line[{{-Sqrt[5/8 - Sqrt[5]/8], 1/4 (-1 - Sqrt[5])}, {0,1}}]
, Line[{{Sqrt[5/8 - Sqrt[5]/8], 1/4 (-1 - Sqrt[5])}, {0,1}}]
}
, { Line[{{-Sqrt[5/8 + Sqrt[5]/8], 1/4 (-1 + Sqrt[5])}, {Sqrt[5/8 - Sqrt[5]/8], 1/4 (-1 - Sqrt[5])}}]
, Line[{{Sqrt[5/8 - Sqrt[5]/8], 1/4 (-1 - Sqrt[5])}, {0, 1}}]
}
};
lines /. Line[ l : { {_, _} ..} ] :> l
(* or as WReach has pointed out: lines /. Line -> Sequence *)
</code></pre>
<p>Simple and expressive - no remembering whether its <code>Flatten[ ... , 1 ]</code> or <code>Flatten[ ... , {1} ]</code>, or <code>Flatten[ ... , -1 ]</code> and what have you.</p>
<h2>Different Approach Making Use of Datatypes</h2>
<p>While having lists is very compact we can easily get lost in different levels. Why not simply give each single point the head <code>Point</code> (<a href="https://reference.wolfram.com/language/ref/Point.html" rel="nofollow noreferrer"><code>Point</code></a> can also contain a list of <code>{x,y}</code> tuples like <code>Line</code>, but giving each point a head of its own simplifies counting and identification of duplicates imo). You can then use pattern matching quite easily, count points, and immediately visualize the data:</p>
<p>Here I turn a list of lines into a list of points, thus not adding another level for each line:</p>
<pre><code>points = lines /. Line[ l : { { _, _ }.. } ] :> ( Sequence @@ Point /@ l )
(* if you do not want this you can simply do: lines /. Line -> Point *)
</code></pre>
<blockquote>
<p>{{Point[{-Sqrt[5/8-Sqrt[5]/8],1/4 (-1-Sqrt[5])}],Point[{0,1}],Point[{Sqrt[5/8-Sqrt[5]/8],1/4 (-1-Sqrt[5])}],Point[{0,1}]},{Point[{-Sqrt[5/8+Sqrt[5]/8],1/4 (-1+Sqrt[5])}],Point[{Sqrt[5/8-Sqrt[5]/8],1/4 (-1-Sqrt[5])}],Point[{Sqrt[5/8-Sqrt[5]/8],1/4 (-1-Sqrt[5])}],Point[{0,1}]}}</p>
</blockquote>
<p>Now it is easy to count points:</p>
<pre><code>Count[ points, Point[ _ ], Infinity ]
</code></pre>
<blockquote>
<p>8</p>
</blockquote>
<p>It is easy to discover that there are only four unique points:</p>
<pre><code>Flatten @ points // Union
</code></pre>
<blockquote>
<p>{Point[{0, 1}], Point[{-Sqrt[5/8 - Sqrt[5]/8], 1/4 (-1 - Sqrt[5])}],
Point[{Sqrt[5/8 - Sqrt[5]/8], 1/4 (-1 - Sqrt[5])}],
Point[{-Sqrt[5/8 + Sqrt[5]/8], 1/4 (-1 + Sqrt[5])}]}</p>
</blockquote>
<p>And we can immediately visualize the results:</p>
<pre><code>Graphics[ {Red, PointSize -> Large, points} ]
</code></pre>
<p><a href="https://i.stack.imgur.com/N6P6z.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/N6P6z.png" alt="Plot of Points"></a></p>
<p>Don't care for <code>Point</code> anymore? Get back to lists:</p>
<pre><code>points /. Point[ pos : { _, _ } ] :> pos
</code></pre>
|
2,256,973 | <p>I'm writing a computer algorithm to do binomial expansion in C#. You can view the code <a href="https://gist.github.com/jamesqo/01015428601641347e436129c1ae0079#file-multinomial-cs-L29-L39" rel="nofollow noreferrer">here</a>; I am using the following identity to do the computation:</p>
<p>$$
\dbinom n k = \frac n k \dbinom {n - 1} {k - 1}
$$</p>
<p>Currently, the algorithm computes $\dbinom {n - 1} {k - 1}$, multiplies it by $n$, then divides by $k$. Since signed integers are capped at $2^{32} - 1$ on a computer, which can be a very small limit for large $n$, I want to change the algorithm to divide by $k$ before mixing in $n$. However, computer integer division floors the result if the dividend isn't a multiple of the divisor; for example, $11 / 5 = 2$. So I want to come up with a criteria for whether $k$ divides $\dbinom {n - 1} {k - 1}$ that can be cheaply checked.</p>
<p>Here's my work so far: assuming $n > k$, then $n - 1 \geq k$. Then $(n - 1)!$ will have at least 1 factor of $k$, and $k \mid \dbinom {n - 1} {k - 1}$ if the $(k - 1)!(n - k)!$ factor does not cancel those factors out. If $k$ is prime, then $(k - 1)!$ will have no factors of $k$. $(n - k)!$ will often have less factors of $k$ than $(n - 1)!$, but that is not the case if $n \equiv -1 \mod k$. For example, $5 \not\mid \dbinom 9 4$, because both the numerator and denominator have exactly one factor of $5$.</p>
<p>So I figured out that $n \not\equiv -1 \mod k$ must be true for primes, but I've no idea how to deal with composites. I played around a bit with them, though, and $6$ is another value of $k$ for which $6 \not\mid \dbinom 11 5$, so clearly $k$ doesn't have to be prime. Any tips?</p>
| yberman | 92,108 | <p>You might be better off memoizing the traditional formula? If you have a 32-bit roof there isn't much that can take that long.</p>
<pre><code> #include <stdio.h>
#define N 50
int cache[N][N];
int f(int n, int k) {
if (k == 0 || k == n) {
return 1;
}
if (cache[n][k] != 0) {
return cache[n][k];
}
cache[n][k] = f(n-1, k-1) + f(n-1, k);
return cache[n][k];
}
int main() {
printf("%d\n", f(2*15, 15));
}
</code></pre>
<p>Edit: as per comment, got rid of bad idea for dealing with leafs of recursion (sorry wrote the original when sleep deprived)</p>
|
2,781,017 | <p>I known that $\sum a_i b_i \leq \sum a_i \sum b_i$ for $a_i$, $b_i > 0$. It seems this inequality will also hold true when $a_i$, $b_i \in (0,1)$. However, I am unable to find out if</p>
<p>$\sum \frac{a_i}{b_i} \leq \frac{\sum a_i}{\sum b_i}$ </p>
<p>holds true for $a_i$, $b_i \in (0,1)$.</p>
| Ingix | 393,096 | <p>$a_1=0.1, a_2=0.2, b_1=0.3, b_2=0.4$ lead to the incorrect statement</p>
<p>$$\frac13 + \frac24 \le \frac37$$</p>
<p>In reality, the opposite inequality is true. You can see that if you rename $a,b$ to $x,y$ and rewrite it as</p>
<p>${\sum y_i} \sum \frac{x_i}{y_i} \geq \sum x_i$. This is the first inequality you said you knew, with $a_i:=y_i$ and $b_i:=\frac{x_i}{y_i}$,</p>
|
2,948,045 | <p>In Eric Gourgoulhon's "Special Relativity in General Frames", it is claimed that the two dimensional sphere is not an affine space. Where an affine space of dimension <em>n</em> on <span class="math-container">$\mathbb R$</span> is defined to be a non-empty set E such that there exists a vector space V of dimension <em>n</em> on <span class="math-container">$\mathbb R$</span> and a mapping </p>
<p><span class="math-container">$\phi:E \times E \rightarrow V,\space\space\space
(A,B) \mapsto \phi(A,B)=:\vec {AB}$</span></p>
<p>that obeys the following properties:</p>
<p>(i) For any point O <span class="math-container">$\in E$</span>, the function </p>
<p><span class="math-container">$\phi_O: E \rightarrow V,\space\space\space
M \mapsto \vec {OM}$</span></p>
<p>is bijective. </p>
<p>(ii) For any triplet (A,B,C) of elements of E, the following relation holds:</p>
<p><span class="math-container">$\vec {AB} + \vec {BC} = \vec {AC}.$</span></p>
<p>I would like to show that the sphere is not an affine space using this definition. My approach has been to assume that such a <span class="math-container">$\phi$</span> exists and then seek a contradiction. I can construct specific <span class="math-container">$\phi_O$</span>'s that are bijective and I can show that a contradiction arises if I use the same construction centered at a new point A, wtih <span class="math-container">$\phi_A$</span>, but this only invalidates the specific construction I made. I am having trouble generalizing this to any <span class="math-container">$\phi$</span>. </p>
| J. Darné | 611,408 | <p>The thing is, you might want to get some topology in the picture. In fact, if you do not, you can choose any bijection between the sphere and a <span class="math-container">$\mathbb R$</span>-vector space, and you end up with a structure of vector space on your "sphere" (by transporting the structure). My point is, there exist such <span class="math-container">$\varphi$</span>, but what you really want is not for <span class="math-container">$ \varphi_O$</span> to be only bijective : if your space already has a shape, you want it to be a homeomorphism.</p>
<p>And there is no homeomorphism between the sphere and a <span class="math-container">$\mathbb R$</span>-vector space (for example because a vector space is contractible - you can shrink it continuously into a point - whereas the sphere is not ; you can look that up in any basic course of algebraic topology)</p>
|
312,878 | <p>Why is $\mathbb{Z} [\sqrt{24}] \ne \mathbb{Z} [\sqrt{6}]$, while $\mathbb{Q} (\sqrt{24}) = \mathbb{Q} (\sqrt{6})$ ?</p>
<p>(Just guessing, is there some implicit division operation taking $2 = \sqrt{4}$ out from under the $\sqrt{}$ which you can't do in the ring?)</p>
<p>Thanks. (I feel like I should apologize for such a simple question.) </p>
| Zev Chonoles | 264 | <p>No need to apologize; and your instinct is correct. Note that $\sqrt{6}\notin\mathbb{Z}[\sqrt{24}]$; indeed,
$$\mathbb{Z}[\sqrt{24}]=\{a+b\sqrt{24}\mid a,b\in\mathbb{Z}\}=\{a+2c\sqrt{6}\mid a,c\in \mathbb{Z}\}.$$
Thus, $\mathbb{Z}[\sqrt{24}]$ consists of the elements of $\mathbb{Z}[\sqrt{6}]$ for which the number of times $\sqrt{6}$ occurs is even. However, when thinking about $\mathbb{Q}$, we now have $\frac{1}{2}$ available to us, and
$$\mathbb{Q}(\sqrt{24})=\{a+b\sqrt{24}\mid a,b\in\mathbb{Q}\}=\{a+2c\sqrt{6}\mid a,c\in \mathbb{Q}\}=\{a+d\sqrt{6}\mid a,d\in\mathbb{Q}\}=\mathbb{Q}(\sqrt{6})$$
because we can take $c=\frac{d}{2}$.</p>
|
1,061,311 | <p>Suppose $\sum_{n=0}^\infty a_n$ and $\sum_{m=0}^\infty b_m$ converge absolutely. I have to show that $$\left(\sum_{n=0}^\infty a_n\right) \cdot \left(\sum_{m = 0}^\infty b_m\right) = \sum_{m, n}^\infty a_nb_m.$$ But I do not understand what the sum on the right-hand side means (i.e. what limit this represents). Could anyone help explain it?</p>
| Etienne | 80,469 | <p>You are perfectly right in non-understanding what this "double sum" means.</p>
<p>Here is one possible interpretation of what you have to prove. This relies on the notion of <em>summability</em> for a family of real numbers. Let $(c_i)_{i\in I}$ be a family of real numbers indexed by some set $I$. Then, $(c_i)_{i\in I}$ is said to be <em>summable</em> if there exists a real number $S$ such that the following holds : for any $\varepsilon>0$, one can find a finite set $F\subset I$ such that $\vert \sum_{i\in F'} c_i-S\vert<\varepsilon$ for any finite set $F'\subset I$ containing $F$. In this case, $S$ is uniquely determined, and denoted by $S:=\sum_{i\in I} c_i$. In your situation, the index set is $\mathbb N\times \mathbb N$ and $c_i=a_nb_m$ for $i=(n,m)\in\mathbb N\times \mathbb N$. So, what you have to prove could read as follows: show that the family $(a_nb_m)_{(n,m)\in\mathbb N\times \mathbb N}$ is summable with sum the product of the two sums on the left-hand side.</p>
<p>As it turns out, if the index set $I$ is countable, then a family $(c_i)_{i\in I}$ is summable with sum $S$ if and only if, for <em>every</em> bijection $\phi:\mathbb N\to I$, the series $\sum c_{\phi(n)}$ is convergent with $\sum_1^\infty c_{\phi (n)}=S$. In other words, if you enumerate (in a 1-1 way) the set $I$ as $\{ i_n;\; n\in\mathbb N\}$ then the series $\sum_n c_{i_n}$ should be convergent with $\sum_0^\infty c_{i_n}=S$, independently of the enumeration you have chosen. This is also equivalent to <em>absolute convergence</em> of $\sum c_{i_n}$ for some enumeration of $I$ (and then this holds for every enumeration). So, what you are asked to do may be the following: show that for any enumeration of the set $\mathbb N\times \mathbb N$ by the integers, i.e. whenever you write $\mathbb N\times \mathbb N$ in a 1-1 fashion as $(n_k,m_k)_{k\in\mathbb N}$, the series $\sum a_{n_k}b_{m_k}$ is convergent with sum the product of the two sums in the left-hand side.</p>
<p>Another possibility is that you are "just" asked to prove the standard theorem on "product series"; namely, if you set $c_k=\sum_{n=0}^k a_nb_{k-n}$ then the series $\sum c_k$ is convergent with sum the product of the two sums in the left-hand side. This corresponds to the "usual" enumeration of $\mathbb N\times \mathbb N$ obtained by going through each "diagonal" $\Delta_k=\{ n+m=k\}$, $k\in\mathbb N$.</p>
|
1,821,437 | <p>I'm solving past exam questions in preparation for an Applied Mathematics course. I came to the following exercise, which poses some difficulty. <em>If it's any indication of difficulty, the exercise is only Part 3-A of the sheet, graded for 10%</em></p>
<blockquote>
<p>Solve the equation $z^5=-32$ and draw its solutions in complex space, then describe their characteristic geometrical property.</p>
</blockquote>
<p>Is it asking to convert z to polar form, then use DeMoivres theorem as I've seen in solutions around the net? If that is the case, how can I work out the $\theta$ angle to be used?</p>
<p>Additionally, what does it refer to as its characteristic geometrical property?</p>
<p>Any answers would be extremely appreciated as they'd help to get me out of the ditch. I'm completely stalled.</p>
| JasonM | 343,478 | <p>The solutions should just be $2e^{\frac{2\pi ki}{10}}$, for odd integers $0 \leq k <10$. They are the divisions of the circle of radius 2 centered at the origin, divided into 5 pieces, rotated about the origin $\frac{2 \pi}{10}$ in the complex plane. </p>
<hr>
<p>Here's more details:</p>
<p>$z^5=-32 \implies 32((\frac{z}{2})^5+1)=0$. Setting $y=z/2$, and noting that $y^{10}-1=0 \implies (y^5-1)(y^5+1)=0$ indicates that the values of $y$ are the $10$th roots of unity for which $y^5=-1$. So $y=e^{\frac{2\pi ki}{10}}$ for odd $k$ (even values of $k$ would be roots of the first factor instead). </p>
|
98,088 | <p>I can't understand a sentence in a textbook: if $x$ is a transitive set, then $\bigcup x^+=x$? Could someone help me to understand?</p>
<p><strong>added:</strong> $x^+=x\cup\{x\}$</p>
| Michael Greinecker | 21,674 | <p>The set $x$ is <em>transitive</em> if $z\in y \in x$ implies $z\in x$. Moreover, $x^+=x\cup\{x\}$. So let $x$ be a transitive set.</p>
<p>We have $x\subseteq\bigcup x^+$ since $x\in\{x\}$. Now let $z\in \bigcup x^+$. Then $z\in\bigcup x\cup x$. If $z\in\bigcup x$, then there exists $y\in x$ with $z\in y$. But since $x$ is transitive, $y\subseteq x$ and hence $z\in x$.</p>
|
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