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 |
|---|---|---|---|---|
1,728,662 | <p>Each set has bijective ordinal and cardinality is defined as the least one of such ordinals. I know that a set of ordinals is well-ordered by $\subseteq$ (inclusion) and thus has $\subseteq$-least element. However, I wonder which axiom guaranteed that the bijective ordinals above really construct to be a set. I worry about this since I noticed that all ordinals is not a set.
THX. my friends. </p>
| BrianO | 277,043 | <p>Given a set $X$, by the Axiom of Choice (AC) there's some ordinal $\alpha$ that's bijectable with $X$. </p>
<p>By AC, $\mathcal{P}(\alpha)$ (the powerset of $\alpha$) is bijectible with some ordinal $\beta$. Now, $\alpha$ can be injected into $\beta$, but by Cantor's theorem, $\beta$ can't be injected into $\alpha$. Thus $\alpha < \beta$: if not, then $\beta \le \alpha$, hence $\beta \subseteq \alpha$ and there <em>would</em> be an injection $\beta\to\alpha$.</p>
<p>Similarly, if $\gamma\ge\beta$, then $\gamma$ can't be injected into $\alpha$, otherwise $\beta$ could be.</p>
<p>So the class of all ordinals $\{\xi\in On\mid \text{$\xi$ bijectable with $X$}\}$ is contained in $\beta$, and therefore by the Comprehension Axiom it's a set.</p>
|
1,786,421 | <p>I have the following equality to prove. </p>
<p>Given $X \sim Bin(n, p)$ and $Y \sim Bin(n, 1 - p)$ prove that $P(X \leq k) = P(Y \geq n - k)$. I have been trying to come up with a solution but cannot find one. I am looking for suggestions and not a complete answer as this is a homework question.</p>
<p>What I did until now is the following:</p>
<p>$P(X \leq k) = \sum_{i = 0}^{k} \binom{n}{i} p^i (1 - p)^{n - i} = 1 - \sum_{i = k + 1}^{n} \binom{n}{i} p^i (1 - p)^{n - i} = \sum_{j = 0}^{n - k - 1} \binom{n}{j + k + 1} p^{j + k + 1} (1 - p)^{n - j - k - 1}$ </p>
<p>Here , $j = i - k - 1$. </p>
| André Nicolas | 6,312 | <p>Hint: Alicia flips a biased coin $n$ times. The coin has probability $p$ of landing heads. Let $X$ be the number of heads she gets, and $Y$ the number of tails.</p>
|
1,650,277 | <p>Does the category of partial orders have a subobject classifier? (Edit: No, see Eric's answer.)</p>
<p>If not, what is a category which is "close" to the category of partial orders (e.g. it should consists of special order-theoretic constructs) and has a subobject classifier? Bonus question: Is there also such an elementary topos? Notice that the category of partial orders has all limits, colimits and it is cartesian closed.</p>
| David | 297,532 | <p>For all $\alpha$, we have $X \cap X_{\alpha} = X_{\alpha} \in \tau_{\alpha}$.</p>
|
551,252 | <p>I'm sure I've made a trivial error but I cannot spot it.</p>
<p>Fix R>0
Consider the cube $C_R$ as the cube from (0,0,0) to (R,R,R) (save me from listing the 8 vertices) </p>
<p>Consider $S_R$ as the surface of $C_R$</p>
<p>Consider the vector field $v:\mathbb{R}^3\rightarrow\mathbb{R}^3$
given by $v(x,y,z) = (3x+z^2,2y,R-z)$</p>
<p><strong>Part 1</strong></p>
<p>Calculate $\nabla.v$</p>
<p>$\nabla.v=\frac{\partial{v_x}}{\partial{x}}+\frac{\partial{v_y}}{\partial{y}}+\frac{\partial{v_z}}{\partial{z}}=3+2-1=4$</p>
<p><strong>Part 2</strong></p>
<p>Calculate $\iiint_{C_R}\nabla.vdV$</p>
<p>$=\int_0^R\int_0^R\int_0^R4dxdydz$</p>
<p>This is trivial it is $=4R^3$</p>
<p><strong>Part 3</strong></p>
<p>Calculate the flux $\iint_{S_R}v.ndA$ where n is the unit normal to $S_R$ at the point.</p>
<p>Every keystroke is lagging now.</p>
<p>I did this by doing it over all 6 sides of the cube. The normals are trivial and the sides look like $R(1,s,t)$ for $s,t\in[0,1]$ (this is the right side) or $R(0,s,t)$ which is the left side.</p>
<p>I'm not even sure how you'd get an $R^3$ in there, I can provide more working if needed but it really isn't hard.</p>
<p>What have I done? Is this result right and perhaps I have misunderstood something?</p>
<p><strong>Addendum</strong></p>
<p>I think my error may come from my parameter ranges, I should be going from 0 to R not 0 to 1. This is essentially a substitution where I didn't multiply by the rate of that substitution with respect to the thing it replaces. </p>
| Robert Israel | 8,508 | <p>For example, the integral over the side $z=R$ is $0$, while the integral over
$z=0$ is $-R^3$ (integrating $-R$ over a square of side $R$).</p>
|
158,451 | <p>Suppose that the contents of an urn are $w$ red balls, $x$ yellow balls, $y$ green balls, and $z$ blue balls collectively, where $w \geq 3$, $x\geq 1$, $y\geq 1$, and $z\geq 1$. We draw balls randomly from this urn without replacement.</p>
<p>What is the probability of our having drawn at least 1 yellow ball by (and including) the seventh draw, at least 1 green ball by (and including) the eight draw, and at least 3 red balls and 1 blue ball by (and including) the ninth draw? </p>
<p>Note that this is one single event and not four separate events.</p>
| Community | -1 | <p>This follows from Cauchy–Schwarz inequality. The Cauchy–Schwarz inequality states that for any two vectors $a$ and $b$ in an inner product space, we have that
$$\lvert \langle a, b \rangle \rvert^2 \leq \lvert \langle a, a \rangle \rvert \lvert \langle b, b \rangle \rvert$$
In your case, the vector $a$ is taken as $a_i = (x_i-\bar{x})$ and the vector $b$ is taken as $b_i = (y_i-\bar{y})$ and the inner product of $a$ and $b$ is taken as $\displaystyle \langle a, b \rangle = \sum_{i=1}^n a_i b_i$. Hence, we get that
$$\displaystyle \langle a, b \rangle = \sum_{i=1}^n a_i b_i = \sum_{i=1}^n (x_i - \bar{x})(y_i - \bar{y})$$
$$\displaystyle \langle a, a \rangle = \sum_{i=1}^n a_i a_i = \sum_{i=1}^n (x_i - \bar{x})^2$$
$$\displaystyle \langle b, b \rangle = \sum_{i=1}^n b_i b_i = \sum_{i=1}^n (y_i - \bar{y})^2$$
Hence, by Cauchy–Schwarz inequality, we get that
$$\left(\sum_{i=1}^n (x_i - \bar{x})(y_i - \bar{y})\right)^2 \leq \left( \sum_{i=1}^n (x_i - \bar{x})^2 \right) \left( \sum_{i=1}^n (y_i - \bar{y})^2\right)$$
Taking the squareroot, we get that
$$\left|\sum_{i=1}^n (x_i - \bar{x})(y_i - \bar{y})\right| \leq \sqrt{\left( \sum_{i=1}^n (x_i - \bar{x})^2 \right) \left( \sum_{i=1}^n (y_i - \bar{y})^2\right)}$$
Hence, we can conclude that
$$|r_{xy}| = \dfrac{\left|\displaystyle \sum_{i=1}^n (x_i - \bar{x})(y_i - \bar{y})\right|}{\displaystyle \sqrt{\left( \sum_{i=1}^n (x_i - \bar{x})^2 \right) \left( \sum_{i=1}^n (y_i - \bar{y})^2\right)}} \leq 1$$</p>
<p><strong>EDIT</strong></p>
<p><strong>Proof of Cauchy Schwarz inequality:</strong></p>
<p>First note that if the vector $b$ is zero, then the inequality is trivially satisfied since both sides are zero. Hence, we can assume that $b \neq 0$. Now look at the component of $a$ orthogonal to $b$ i.e. $$c = a - \dfrac{\langle a, b \rangle}{\langle b, b \rangle} b$$
i.e.
$$a = c + \dfrac{\langle a, b \rangle}{\langle b, b \rangle} b$$
You can check that $c$ is orthogonal to $b$ by computing $$\langle c, b \rangle = \langle a,b \rangle - \dfrac{\langle a, b \rangle}{\langle b, b \rangle} \langle b, b \rangle = \langle a,b \rangle - \langle a,b \rangle = 0$$
You can also check that $\langle c, \alpha b \rangle = 0 = \langle \beta c, b \rangle$.</p>
<p>We now have that
\begin{align}
\langle a,a \rangle & = \left \langle c + \dfrac{\langle a, b \rangle}{\langle b, b \rangle} b, c + \dfrac{\langle a, b \rangle}{\langle b, b \rangle} b \right \rangle\\
& = \langle c,c \rangle + \left \langle c,\dfrac{\langle a, b \rangle}{\langle b, b \rangle} b \right \rangle + \left \langle \dfrac{\langle a, b \rangle}{\langle b, b \rangle} b, c \right \rangle + \left \langle \dfrac{\langle a, b \rangle}{\langle b, b \rangle} b, \dfrac{\langle a, b \rangle}{\langle b, b \rangle} b \right \rangle\\
& = \langle c,c \rangle + \left \lvert \dfrac{\langle a, b \rangle}{\langle b, b \rangle} \right \rvert^2 \langle b, b \rangle = \langle c,c \rangle + \dfrac{\left \lvert \langle a, b \rangle \right \rvert^2}{\langle b, b \rangle}
\end{align}
Now $\langle c,c \rangle \geq 0$. This gives that
$$\langle a,a \rangle \geq \dfrac{\left \lvert \langle a, b \rangle \right \rvert^2}{\langle b, b \rangle}$$
Rearranging, we get what we want, namely
$$\lvert \langle a, b \rangle \rvert^2 \leq \lvert \langle a, a \rangle \rvert \lvert \langle b, b \rangle \rvert$$</p>
|
27,951 | <p>Something I notice is when there's an advanced/specialized question, it often receives very few upvotes. Even if it is seemingly well written. I try to upvote advanced questions <strong>that I might not even understand</strong>, if they appear well written. </p>
<p>Is this good behaviour? Should we encourage upvoting seemingly well-written questions <strong>even if you don't understand it</strong>? </p>
| samerivertwice | 334,732 | <p>Not a perfect remedy for precisely this issue, but a nice feature that would help with this and bring other benefits for the community, would be to give a few users exceeding some reputation threshold or some maths skill threshold (e.g. holding certain badges) limited numbers of say $+5$ votes.</p>
<p>Not sure how well the sits with the idea of democracy.</p>
<p>To some degree I've seen users use the bounty functionality to this end which impacts reputation but doesn't have the same longstanding effect on question rating.</p>
|
2,861,443 | <p>Let $C_c(\mathbb{R})$ be the following:</p>
<p>$$C_c = \{ f \in C(\mathbb{R}) \mid \exists \text{ } T > 0 \text{ s.t. } f(t) = 0 \text{ for } |t| \geq T\}$$</p>
<p>Let $T_n \in L(C_c(\mathbb{R}))$ be a linear operator such that:</p>
<p>$$T_n u = \delta_n *u, \forall u \in C_c(\mathbb{R}),$$</p>
<p>where</p>
<p>$$
\delta_n(t)= \begin{cases}
n^2(t+1/n) & -1/n \leq t \leq 0 \\
-n^2(t-1/n) & 0 < t \leq 1/n \\
0 & \text{elsewhere}
\end{cases}
$$</p>
<p>I have to prove that with respect to the 2-norm, the operator $T_n$ has $\|T_n\| = 1$ and I have succeeded in proving that $\|T_nu\|_2 \leq \|u\|_2, \forall u \in C_c(\mathbb{R}).$</p>
<p>Now, I remained to prove that $\exists \text{ } u \in C_c(\mathbb{R}) \text{ s.t. } \|T_nu\|_2 = \|u\|_2$, but I really don't get how $u$ should be shaped in order to satisfy it.</p>
| Kelvin Lois | 322,139 | <p>I guess the theorem that you mention is Proposition 3.18 on Lee's book.</p>
<p>For any smooth chart $(U_{\alpha},\varphi_{\alpha})$ of $M$, we have a map $\widetilde{\varphi}_{\alpha} : \pi^{-1}(U_{\alpha}) \to \Bbb{R}^{2n}$ defined as
$$
\widetilde{\varphi}_{\alpha} (v_p) = (x^1(p),\dots,x^n(p),v^1,\dots,v^n) .
$$
This map is a bijection onto its image $\varphi_{\alpha}(U) \times \Bbb{R}^n$.
The topology defined on $TM$ is the one that generated by $\{\widetilde{\varphi}_{\alpha}^{-1}(V) : \forall \alpha \in A, V\subseteq \Bbb{R}^{2n} \text{ is open}\}$. You can check directly that the collection above indeed generate a topology.</p>
<p>To show the Hausdorff property, let $p \in U \subseteq M$, where $U$ is the domain of some smooth chart $(U,\varphi)$ of $M$. And $v_p,w_p \in \pi^{-1}(p) \subseteq \pi^{-1}(U)\subseteq TM$, be a pair of points that lie in the same fiber.</p>
<p>With this topology, the map $\widetilde{\varphi} : \pi^{-1}(U) \to \Bbb{R}^{2n}$ is a homeomorphism onto its image. We can write $\widetilde{\varphi}(v_p) = (\hat{p},v)$, with $\hat{p} = \varphi(p)$. So let $\widetilde{\varphi}(v_p)=(\hat{p},v)$ and $\widetilde{\varphi}(w_p) = (\hat{p},w)$ be their images and $v,w \in \Bbb{R}^n$ be two distinct vectors. To obtain disjoint neighbourhoods of $v_p$ and $w_p$, just take disjoint neighbourhoods $B_1,B_2 \subseteq \Bbb{R}^n$ of $v$ and $w$ resp. and maped back the disjoint open sets $\varphi(U)\times B_1$ and $\varphi(U)\times B_2$ to $\pi^{-1}(U)$. So $\widetilde{\varphi}^{-1}(\varphi(U) \times B_1)$ and $\widetilde{\varphi}^{-1}(\varphi(U) \times B_2)$ are the desired neighbourhoods for $v_p$ and $w_p$.</p>
|
1,494,167 | <p>Using only addition, subtraction, multiplication, division, and "remainder" (modulo), can the absolute value of any integer be calculated?</p>
<p>To be explicit, I am hoping to find a method that does not involve a piecewise function (i.e. branching, <code>if</code>, if you will.)</p>
| ASKASK | 136,368 | <p>I'm going to go with no.</p>
<p>My (not super rigorous) proof would be that any function of the variable $x$ using only addition, subtraction, multiplication, and division would have to be a differentiable function on its domain, but $|x|$ is not a differential function on all of its domain, and its domain is all of $\mathbb{R}$. </p>
<p>The introduction of an expression like $(x\mod 4)$ would introduce points of discontinuity (and hence not differential), but it would introduce an infinite amount of them (at all multiples of four), and the function $|x|$ only has one point of not being differentiable (at $x=0$).</p>
|
2,891,444 | <p>For the intersection of two line segments, how was it know to use the determinants shown <a href="http://mathworld.wolfram.com/Line-LineIntersection.html" rel="nofollow noreferrer">here</a>? </p>
<p>I'm trying to determine how it was shown that they could be used to compute the intersection point.</p>
| Deepesh Meena | 470,829 | <p>Suppose both proof readers missed the error the probability of this event is$$ = \frac{2}{100}\cdot \frac{2}{100}=\frac{4}{10000}=0.04 \% $$</p>
<p>thus the probability of both working independently and detecting the error is $100-0.04=99.96\%$</p>
|
524,073 | <p>Hey can some help me with this textbook question</p>
<p>Let $R^{2×2}$ denote the vector space of 2×2 matrices, and let</p>
<p>$S =\left\{
\left[\begin{matrix}
a \space b \\
b \space c \\
\end{matrix}\right]\mid a,b,c \in \mathbb{R}\right\}$</p>
<p>Find (with justication) a basis for $S$ and determine the dimension of $S$.</p>
| bradhd | 5,116 | <p>Observe that</p>
<p>$$S = \{\left[\begin{array}{cc}a&b\\b&c\end{array}\right]: a,b,c\in\mathbb{R}\}$$
$$ = \{\left[\begin{array}{cc}a&0\\0&0\end{array}\right]+\left[\begin{array}
{cc}0&b\\b&0\end{array}\right]+\left[\begin{array}{cc}0&0\\0&c\end{array}\right]: a,b,c\in\mathbb{R}\}$$
$$ = \{a\left[\begin{array}{cc}1&0\\0&0\end{array}\right]+b\left[\begin{array}
{cc}0&1\\1&0\end{array}\right]+c\left[\begin{array}{cc}0&0\\0&1\end{array}\right]: a,b,c\in\mathbb{R}\}$$
$$ = \operatorname{span}\left(\left[\begin{array}{cc}1&0\\0&0\end{array}\right],\left[\begin{array}{cc}0&1\\1&0\end{array}\right],\left[\begin{array}{cc}0&0\\0&1\end{array}\right]\right).$$</p>
<p>Moreover, the list $\left(\left[\begin{array}{cc}1&0\\0&0\end{array}\right],\left[\begin{array}{cc}0&1\\1&0\end{array}\right],\left[\begin{array}{cc}0&0\\0&1\end{array}\right]\right)$ is linearly independent (why?). Hence it is a basis for $S$.</p>
|
1,748,542 | <p>So a friend of mine has a little project going, and needs some help.</p>
<p>Basically, we want to create a function that takes two variables; One $X$, and one that we call $DC$ ("Difficulty Class, as this is for a pen-and-paper game).</p>
<p>The output should be $0$ if $X\leq (1/2)DC$, and it should be $100$ if $X\geq 2 DC$.</p>
<p>The rest of the curve should look roughly like an $X^3$ curve, centered around $DC$ (So it goes downwards below $DC$, and upwards over $DC$, with the derivative $F'(DC) = 0$ )</p>
<p>Thank you in advance!</p>
| Doug M | 317,162 | <p>How about:</p>
<p>$y = \begin{cases} 0,&x\leq \frac{DC}{2}\\\frac{100}{1.5^3}\times(\frac{X}{DC}-\frac{1}{2})^3,&\frac{DC}{2}<X\leq 2\times DC\\ 100,&X>2\times DC \end{cases}$</p>
|
392,020 | <p>It is well known that the minimum number of monochromatic triangles in a red/blue coloring of the edges of the complete graph <span class="math-container">$K_n$</span> is given by Goodman's formula
<span class="math-container">$$M(n)=\binom n3-\left\lfloor\frac n2\left\lfloor\left(\frac{n-1}2\right)^2\right\rfloor\right\rfloor;$$</span>
see OEIS sequence <a href="http://oeis.org/A014557" rel="nofollow noreferrer">A014557</a> or the original paper by A. W. Goodman, <a href="https://www.jstor.org/stable/2310464?seq=1#page_scan_tab_contents" rel="nofollow noreferrer">On sets of acquaintances and strangers at any party</a>, Amer. Math. Monthly 66 (1959), 778–783, or my answer to <a href="https://math.stackexchange.com/questions/2216943/counting-triangles-in-a-graph-or-its-complement">this math.stackexchange question</a>.</p>
<blockquote>
<p>Is there any literature on the more general question, what is the minimum number of monochromatic triangles in a red/blue coloring of the edges of the complete graph <span class="math-container">$K_n$</span> <strong>with a prescribed number of edges of each color?</strong></p>
</blockquote>
<p>...</p>
<blockquote>
<p>Is there any literature on the related question, given natural numbers <span class="math-container">$n$</span> and <span class="math-container">$k$</span>, what is the minimum value of the quantity <span class="math-container">$m+kb$</span> over all red/blue colorings of the edges of the complete graph <span class="math-container">$K_n$</span>, where <span class="math-container">$m$</span> is the number of monochromatic triangles (of either color) and <span class="math-container">$b$</span> is the number of blue edges?</p>
</blockquote>
| RobPratt | 141,766 | <p>Here are results, obtained via integer linear programming, for the first question for <span class="math-container">$n \le 10$</span> and <span class="math-container">$b$</span> blue edges, where <span class="math-container">$b \le \binom{n}{2}/2$</span>:
<span class="math-container">\begin{matrix}
n\backslash b & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 \\
\hline
2 & 0 \\
3 & 1 & 0 \\
4 & 4 & 2 & 0 & 0 \\
5 & 10 & 7 & 4 & 2 & 1 & 0 \\
6 & 20 & 16 & 12 & 8 & 6 & 4 & 2 & 2 \\
7 & 35 & 30 & 25 & 20 & 16 & 13 & 10 & 7 & 6 & 5 & 4 \\
8 & 56 & 50 & 44 & 38 & 32 & 28 & 24 & 20 & 16 & 14 & 12 & 10 & 8 & 8 & 8 \\
9 & 84 & 77 & 70 & 63 & 56 & 50 & 45 & 40 & 35 & 30 & 27 & 24 & 21 & 18 & 16 & 15 & 14 & 13 & 12 \\
10 & 120 & 112 & 104 & 96 & 88 & 80 & 74 & 68 & 62 & 56 & 50 & 46 & 42 & 38 & 34 & 30 & 28 & 26 & 24 & 22 & 20 & 20 & 20 \\
\end{matrix}</span></p>
<p>And here are results for the second question:
<span class="math-container">\begin{matrix}
n\backslash k & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline
2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
3 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
4 & 0 & 2 & 4 & 4 & 4 & 4 & 4 & 4 & 4 & 4 & 4 \\
5 & 0 & 5 & 8 & 10 & 10 & 10 & 10 & 10 & 10 & 10 & 10 \\
6 & 2 & 8 & 14 & 17 & 20 & 20 & 20 & 20 & 20 & 20 & 20 \\
7 & 4 & 14 & 21 & 28 & 32 & 35 & 35 & 35 & 35 & 35 & 35 \\
8 & 8 & 20 & 32 & 40 & 48 & 52 & 56 & 56 & 56 & 56 & 56 \\
9 & 12 & 30 & 44 & 57 & 66 & 75 & 80 & 84 & 84 & 84 & 84 \\
10 & 20 & 40 & 60 & 75 & 90 & 100 & 110 & 115 & 120 & 120 & 120 \\
\end{matrix}</span>
For <span class="math-container">$k=0$</span>, the minimum is <span class="math-container">$m+0=M(n)$</span>. For large enough <span class="math-container">$k$</span>, all edges are red, yielding <span class="math-container">$m=\binom{n}{3}$</span> and <span class="math-container">$b=0$</span>.</p>
|
3,857,698 | <p>Let <span class="math-container">$D_1, ..., D_n$</span> be arbitrary <span class="math-container">$n$</span> sets where <span class="math-container">$D_i \cap D_j \neq \emptyset$</span>. In the simplified case where <span class="math-container">$n = 2$</span>, we have that
<span class="math-container">$$
\begin{split}
| X \cap D_1 | + | X \cap D_2 | = &| X \cap (D_1 \setminus D_2) | + | X \cap (D_1 \cap D_2) | \\
&+ | X \cap (D_2 \setminus D_1) | + | X \cap (D_1 \cap D_2) | \\
= & |X| + | X \cap (D_1 \cap D_2) | \\
\leq & |X| + | D_1 \cap D_2 |.
\end{split}
$$</span></p>
<p>My question is that, can we generalize the above upper bound to something like
<span class="math-container">$$
\sum_{i = 1}^n | X \cap D_i | \leq |X| + c,
$$</span>
where <span class="math-container">$c$</span> is dependent on <span class="math-container">$(D_1, D_2, ..., D_n)$</span>? It is self-evident that, if <span class="math-container">$D_1, ..., D_n$</span> is a disjoint partition of a universe, then we have <span class="math-container">$\sum_{i = 1}^n | X \cap D_i | = |X|$</span>. However, it seems difficult for me to bound <span class="math-container">$c$</span> when <span class="math-container">$D_1, ..., D_n$</span> are not disjoint.</p>
<p>It would be appreciated if you could give me any hint.</p>
| Thomas | 284,057 | <p>From the point of view of group theory there is a deep reason : the group of similitudes ( ratio-of-lengths preserving maps) of a (Euclidean) plane is isomorphic to the group of affine (or anti-affine) transformation of a complex line <span class="math-container">$(z\to az+b$</span> or <span class="math-container">$z\to a \bar z+b$</span>). This (exceptional) isomorphism enables us to do geometry by using complex numbers.</p>
<p>This is even more clear if we go to the projective line (the Riemann Sphere). The group of projective transformations of a projective line <span class="math-container">$PGL(2,C)$</span> is isomorphic to Möbius group of conformal maps of a sphere <span class="math-container">$PSO(3,1)$</span>.</p>
|
3,021,631 | <p>I've been strongly drawn recently to the matter of the fundamental definition of the exponential function, & how it connects with its properties such as the exponential of a sum being the product of the exponentials, and it's being the eigenfunction of simple differentiation, etc. I've seen various posts inwhich clarification or demonstration or proof of such matters as how such & such a property <em>proceeds</em> from its definition as <span class="math-container">$$e^z\equiv\lim_{k\rightarrow\infty}\left(1+\frac{z}{k}\right)^k .$$</span> I'm looking at how there is a <em>web</em> of interconnected theorems about this; and I am trying to <em>spin</em> the web <em>as a whole</em>. This is not necessary for proving <em>some particular item to be proven</em> - a path along some one thread or sequence of threads is sufficient for that; but I think the matter becomes 'unified', and by reason of that actually <em>simplified</em>, when the web is perceived as an entirety. This is why I <em>bother</em> with things like combinatorial demonstrations of how the terms in a binomial expansion <em>evolve towards</em> the terms in a Taylor series as some parameter tends to ∞, when the matter at hand is actually susceptible of a simpler proof by taking the logarithm of both sides ... & other seeming redundancies.</p>
<p>To this end, another 'thread' I am looking at is that of showing that the coefficients in the Taylor series for <span class="math-container">$\ln(1+z)$</span> actually are a <em>consequence</em> of the requirement that the logarithm of a product be the sum of the logarithms, and ... if not quite a combinatorial <em>derivation</em> of them <em>from</em> that requirement, at least a <em>reverse- (or sideways-) engineering <strong>equivalent</em></strong> of it - the combinatorially showing that if the coefficients <em>be plugged into</em> the Taylor series, then the property follows</p>
<p>Taking the approach that <span class="math-container">$$(1+x)(1+y) = 1+x+y+xy ,$$</span> plugging <span class="math-container">$z=x+y+xy$</span> into the series for <span class="math-container">$\ln(1+z)$</span> and hoping that all non-fully-homogeneous terms cancel out, leaving sum of the series for <span class="math-container">$\ln(1+x)$</span> & that for <span class="math-container">$\ln(1+y)$</span>, we are left with proving that</p>
<p><span class="math-container">$$\sum_{k=1}^\infty\left((-1)^k(k-1)!×\sum_{p\inℕ_0,q\inℕ_0,r\inℕ_0,p+q+r=k}\frac{x^{p+r}y^{q+r}}{p!q!r!}\right)$$</span><span class="math-container">$$=$$</span><span class="math-container">$$\sum_{k=1}^\infty(-1)^k\frac{x^k+y^k}{k} .$$</span></p>
<p>The <em>inner</em> sum of the LHS of this quite appalling-looking theorem is the trinomial expansion of <span class="math-container">$(x+y+xy)^k$</span> for arbitrary <span class="math-container">$k$</span>, & the outer sum is simply the logarithm Taylor series expansion (with its <span class="math-container">$k$</span> in the denominator 'absorbed' into the combinatorial <span class="math-container">$k!$</span> in the numerator of the inner sum) . This theorem can be quite easily verified by 'brute force' - simply <em>doing</em> the expansions at the first (very!) few terms; but the labour of it escalates <em>extremely</em> rapidly. An algebraic manipulation package would no-doubt verify it at a good few more terms; but what I am looking-for is a showing of the fully general case: but I do not myself have the combinatorial toolage for accomplishing this. </p>
<p>So I am asking whether anyone can show me an outline of what to do ... or even actually <em>do</em> it for me, although that would probably take up a <em>very</em> great deal of space and be an <em>extremely</em> laborious task for the person doing it ... so I'm content to ask for just an outline of doing it, or for some 'signposts' as to how to do it - maybe someone knows some text on this kind of thing that they would recommend.</p>
| AmbretteOrrisey | 613,228 | <p>Maybe it's not <em>too</em> bad actually. What what needs to be shown reduces to is that <span class="math-container">$$\forall m\in ℕ_1\&n\in ℕ_1 $$</span><span class="math-container">$$\sum_{p\in ℕ_0,q\in ℕ_0,r\in ℕ_0,p+r=m,q+r=n}\frac{(-1)^{p+q+r}(p+q+r-1)!}{p!q!r!} =0 ,$$</span> whence that</p>
<p><span class="math-container">$$\sum_{r=0}^{\min(m,n)}\frac{(-1)^{m+n-r}(m+n-r-1)!}{(m-r)!(n-r)!r!} =0 .$$</span></p>
<p>And this can be <em>essentialised</em> yet further: basically it's tantamount to saying that the sum of the alternating (in sign) sequence of binomial numbers of degree <span class="math-container">$n$</span>, which is <em>unweighted</em> well known to be zero, is also zero when weighted by <em>any sequence of consecutive Pochhammer numbers of degree</em> <span class="math-container">$n-1$</span>. If <em>this</em> be proven, then my combinatorial proof of the at-the-outset-mentioned property of logarithms is upheld. </p>
|
197,393 | <p>Playing around on wolframalpha shows $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi$. I know $\tan^{-1}(1)=\pi/4$, but how could you compute that $\tan^{-1}(2)+\tan^{-1}(3)=\frac{3}{4}\pi$ to get this result?</p>
| JSCB | 25,841 | <p><strong>Proof without word</strong></p>
<p>$\tan^{-1} 1+\tan^{-1} 2+\tan^{-1} 3 =\pi$.</p>
<p><img src="https://i.stack.imgur.com/Cvd3u.png" alt=""></p>
|
197,393 | <p>Playing around on wolframalpha shows $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi$. I know $\tan^{-1}(1)=\pi/4$, but how could you compute that $\tan^{-1}(2)+\tan^{-1}(3)=\frac{3}{4}\pi$ to get this result?</p>
| Tryst with Freedom | 688,539 | <p>Consider, <span class="math-container">$z_1= \frac{1+2i}{\sqrt{5}}$</span>, <span class="math-container">$z_2= \frac{1+3i}{\sqrt{10} }$</span>, and <span class="math-container">$z_3= \frac{1+i}{\sqrt{2} }$</span>, then:</p>
<p><span class="math-container">$$ z_1 z_2 z_3 = \frac{1}{10} (1+2i)(1+3i)(1+i)=-1 $$</span></p>
<p>Take arg of both sides and use property that <span class="math-container">$\arg(z_1 z_2 z_3) = \arg(z_1) + \arg(z_2) + \arg(z_3)$</span>:</p>
<p><span class="math-container">$$ \arg(z_1) + \arg(z_2) + \arg(z_3) = -1$$</span></p>
<p>The LHS we can write as:</p>
<p><span class="math-container">$$ \tan^{-1} ( \frac{2}{1}) +\tan^{-1} ( \frac{3}{1} ) + \tan^{-1} (1) = \pi$$</span></p>
<p>Tl;dr: Complex number multiplication corresponds to tangent angle addition</p>
|
88,880 | <p>In a short talk, I had to explain, to an audience with little knowledge in geometry or algebra, the three different ways one can define the tangent space $T_x M$ of a smooth manifold $M$ at a point $x \in M$ and more generally the tangent bundle $T M$:</p>
<ul>
<li>Using equivalent classes of smooth curves through $x$</li>
<li>Using derivations near $x$</li>
<li>Using cotangent vectors at $x$</li>
</ul>
<p>Just by looking at the definition, it is not at all clear why they should all define the same object. I went through the proof, but judging from their reaction, it was not very meaningful. I wonder if there is any way I can let them "see", with just intuition, that the three definitions are, in certain sense, the same.</p>
| Paul Siegel | 4,362 | <p>What all three definitions have in common is that they each try to capture the first order behavior of a smooth function on $M$.</p>
<ol>
<li><p>The derivative of a smooth function $f$ along a curve $\gamma$ with $\gamma(0) = p$ depends on $\gamma$ only insofar as it depends on $\gamma'(0)$, and indeed it recovers the directional derivative of $f$ at $p$ in the direction $\gamma'(0)$. The directional derivatives of $f$ determine the total derivative of $f$ which in turn determines the first order behavior of $f$ (more or less by the definition of the total derivative).</p></li>
<li><p>Since a derivation $D$ at $p$ sees only the values of a function $f$ and its derivatives at $p$ (not near $p$), we can replace $f$ by a polynomial by Taylor's theorem. By the Leibniz rule, $D(P)$ depends only on the linear part of a polynomial $P$ and hence $D(f)$ depends only on the first order part of $f$.</p></li>
<li><p>Recall that the cotangent bundle of $M$ at $p$ is the space $I/I^2$ where $I$ is the ideal in $C^\infty(M)$ consisting of functions $f$ such that $f(p) = 0$. If we imagine replacing $C^\infty(M)$ by a polynomial ring then $I$ represents the ideal of polynomials whose lowest order part has degree $1$ and $I^2$ is the ideal of polynomials whose lowest order part has degree $2$. In this case $I/I^2$ is naturally identified with the space of linear polynomials. Thus the cotangent bundle at $p$ is in a sense the space of "first order parts" of smooth functions on $M$.</p></li>
</ol>
<p>This intuition acutally allows us to be a little more explicit about how the relevant identifications are made.</p>
<p>It's very easy to go from 1 to 2: if $\gamma$ is a curve in $M$ with $\gamma(0) = p$ then $D(f) = (f \circ \gamma)'(0)$ is a point derivation at $p$ which depends only on the equivalence class of $\gamma$ in $T_p M$.</p>
<p>To go from 2 back to 1, let $f$ be a smooth function and let $f(x) \sim \sum_\alpha c_\alpha x^\alpha$ (multi-index notation) be its Taylor series in a coordinate system centered at $p$. Then for any derivation $D$ at $p$ we have $D(f) = c_1 D(x_1) + \ldots + c_n D(x_n)$ by the Leibniz rule, so $D$ corresponds to the tangent vector $(D(x_1), \ldots, D(x_n))$.</p>
<p>To go from 2 to 3, let $D$ be a derivation at $p$ and observe that $D(f) = 0$ for any $f \in I/I^2$ by Taylor's theorem and the Leibniz rule. Thus $D$ determines a linear functional in $(I/I^2)^*$.</p>
<p>Finally, to go from 3 back to 2, let $\ell \in (I/I^2)^*$ and define a point derivation by $D(f) = \ell(f - f(p) + I^2)$.</p>
|
2,576,466 | <p>One says a bounded $f$ is Riemann integrable on [a,b] if the Upper and lower Riemann integrals are equal. Another sufficient condition for Riemann integrablity is that the set of discontinuity of $f$ must be countable set. The following function is continuous only at $x=1/2$ and so the set of discontinuity of $f$ is uncountable set. This shows that $f$ is not Riemann integrable. But I want to justify this by computing the upper and lower Riemann integrals. Any one who can help me how to get the upper and lower Riemann integrals.
Let $f\colon[0,1]\rightarrow\mathbb{R}$ such that
$$f(x)=\left\{\begin{array}{c}
x, x\in\mathbb{Q}\\ 1-x, x\notin\mathbb{Q}
\end{array}\right.$$ Show that $f$ is not Riemann integrable on $[0,1].$</p>
| Przemysław Scherwentke | 72,361 | <p>HINT: The upper Riemann integral is the integral of $\max(x,1-x)$ on $[0,1]$ and the lower: $\min(x,1-x)$. (Why?)</p>
|
1,041,134 | <p>I need to show if $a$ is in $\mathbb{R}$ but not equal to $0$, and $a+\dfrac{1}{a}$ is integer, $a^t+\dfrac{1}{a^t}$ is also an integer for all $t\in\mathbb N$.
Can you provide me some hints please?</p>
| Steven Alexis Gregory | 75,410 | <p>Let $T_n = r^n + \dfrac{1}{r^n}$, $T_0 = 2$, and $T_1 = \alpha$</p>
<p>\begin{align}
r + \dfrac 1r &= \alpha \\
r &= \alpha - \dfrac 1r \\
\dfrac 1r &= \alpha - r\\
\hline
r^{n+2} &= \alpha r^{n+1} - r^n \\
\dfrac{1}{r^{n+2}} &= \dfrac{\alpha}{r^{n+1}} - \dfrac{1}{r^n} \\
r^{n+2} + \dfrac{1}{r^{n+2}}
&= \alpha\left( r^{n+1} + \dfrac{1}{r^{n+1}} \right)
- \left( r^n + \dfrac{1}{r^n} \right) \\
T_{n+2} &= \alpha T_{n+1} - T_n \\
\hline
\end{align}</p>
<p>Clearly, if $\alpha$ is an integer, then $T_n$ is an integer for all non negative integers $n$.</p>
|
139,135 | <p>Suppose $R = \mathbb{Q}[x_1, ..., x_n]/I$, and $J \subset R$ is a given height one ideal. Is there a quick algorithm one could write to determine if $J$ is a principal ideal or necessarily not principal? Is it not possible to do this with Groebner bases?</p>
| Karl Schwede | 3,521 | <h2>Locally principal</h2>
<p>At least in height-1 ideal case, in a normal domain (or at least G1+S2 domain), the following should let you know whether the ideal is <em>locally</em> principal. </p>
<p>Let $J$ be the ideal in question and let $J_2$ be another ideal isomorphic to $Hom_R(J, R)$ (which you can do in a number of ways, say by forming a colon or by embedding the module back into R). </p>
<p>Then $J\cdot J_2$ is an ideal. If $J$ is locally principal, it corresponds to a Cartier divisor $D$. Then $J_2$ corresponds to $-D$. It's a fact (first pointed out to me by Tommaso de Fernex) that $D$ is Cartier if and only if </p>
<p>$$O(D) \cdot O(-D) = J\cdot J_2 = \langle g \rangle$$</p>
<p>is principal. </p>
<h3>That doesn't help...</h3>
<p>But you say, this doesn't help us at all. All I've done is give you another ideal we need to check whether or not it is principal! </p>
<p>The point however, is that we know that the reflexification of $I\cdot J$ is principal! (Recall reflexification just means applying $Hom(\bullet, R)$ twice, Macaulay2 can do this for instance). Also remember that locally principal ideals are always reflexive.</p>
<p>On the other hand $J \cdot J_2$ agrees in codimension 1 with its reflexification (since reflexification won't change anything in codimension 1 for a normal domain). </p>
<p>It follows that $J \cdot J_2$ is principal if and only if $J\cdot J_2$ is reflexive. </p>
<p><strong>Proposition:</strong> <em>Hence $J$ is locally principal if and only if $J \cdot J_2$ is reflexive.</em></p>
<p>Perhaps its worth noting that you can also use this to identify the locus where $J$ is not locally principal. If $L$ is the reflexification of $J \cdot J_2$, then compute the
$$\text{Ann}_R(L/(J \cdot J_2)).$$
That is just the locus where $J$ is not locally principal.</p>
|
1,682,961 | <p>I was making a few exercises on set proofs but I met an exercise on which I don't know how to start:</p>
<blockquote>
<p>If $A \cap C = B \cap C $ and $ A-C=B-C $ then $A = B$</p>
</blockquote>
<p>Where should I start? Should I start from $ A \subseteq B $ or should I start from this $ ((A\cap C = B\cap C) \land (A-C = B-C)) \Rightarrow (A = B)$ ? </p>
| Akiva Weinberger | 166,353 | <p>\begin{align}
A\cap C&=B\cap C\\
A-C&=B-C
\end{align}
Take the union of the left- and right-hand sides:
\begin{align}
(A\cap C)\cup(A-C)&=(B\cap C)\cup(B-C)\\
(A\cap C)\cup(A\cap\overline C)&=(B\cap C)\cup(B\cap\overline C)\\
A\cup(C\cap\overline C)&=B\cup(C\cap\overline C)\\
A&=B
\end{align}</p>
|
293,234 | <p>I recently asked a question about <a href="https://math.stackexchange.com/questions/287116/proof-that-mutual-statistical-independence-implies-pairwise-independence">pairwise versus mutual independence</a> (also related to <a href="https://math.stackexchange.com/questions/281800/example-relations-pairwise-versus-mutual">this</a> and <a href="https://math.stackexchange.com/questions/283579/how-to-model-mutual-independence-in-bayesian-networks">this</a> q). </p>
<p>However, </p>
<p>(1) I inadvertently used incorrect terminology:</p>
<blockquote>
<p>three events, A, B, C are mutually independent when:</p>
<p>P[A,B]=P[A]P[B], P[B,C]=P[B]P[C], P[A,C]=P[A]P[C],
P[A,B,C]=P[A]P[B]P[C]</p>
</blockquote>
<p>Did and others pointed out that</p>
<blockquote>
<p>"Mutual independence means the four identities you copied, pairwise
independence means the first three of these identities." -- Did</p>
</blockquote>
<p>Note that the term <em>mutual</em> has varying definitions across math. For example, <a href="http://en.wikipedia.org/wiki/Mutual_information" rel="nofollow noreferrer">mutual information</a> is a pairwise relation. </p>
<p>(2) Going back to probability, GC Rota said the theory can be approached two ways: focusing on random variables (event algebra) or focusing on distributions. Here I am interested in distributions, where independence can be interpreted as factorization of the probability distribution function. The conditions are the same as above, where P is interpreted as the PDF function. </p>
<p>The following graphic based on a standard example from <a href="http://rads.stackoverflow.com/amzn/click/0412989018" rel="nofollow noreferrer"><em>Counterexamples in Probability and Statistics</em></a> of a 3-dimensional binomial PDF that factorizes pairwise (ie, each of the 3 pairs of random variables are independent and the 2-dim joint distributions can all be written as the product of the respective marginals) but not 3-way independent (the joint distribution cannot be written as the product of the individual marginal distributions)</p>
<p><img src="https://i.stack.imgur.com/IBCra.jpg" alt="enter image description here"></p>
<p>My question is whether the opposite can happen, ie if the 3-dim (or perhaps higher) joint distribution factorizes into the 1-dim marginals, does that imply the pairwise factorization of <em>all</em> 2-dim joint distributions into the marginals? </p>
| Community | -1 | <p>Let $K = \mathbb{Q}(\alpha^2, \beta^2, \gamma^2, \delta^2) = \mathbb{Q}(\sqrt[4]{2}, i)$ be the splitting field of $t^4 - 4 t^2 + 8t + 2$. Then $[K:\mathbb{Q}] = 8$ and $L/K$ is a Kummer extension formed by adjoining four square roots.</p>
<p>As per Kummer theory, we are thus interested in the subgroup of $K^*$ modulo squares that is generated by $\alpha^2, \beta^2, \gamma^2, \delta^2$. This is an abelian group of exponent 2 and order $1$, $2$, $4$, or $8$. (It can't be $16$ as we know their product is $2$, which is square)</p>
<p>We have $\alpha^2 = \sqrt[4]{2} (\sqrt[4]{2} + 1)$. The factor of $\sqrt[4]{2}$ shows that $\alpha^2$ is not a square, and so the group order cannot be $1$.</p>
<p>The group order cannot be 4 because of symmetry.</p>
<p>If the group order were $2$, it would mean $\alpha^2 \beta^2 = (\sqrt[4]{2})^2 (\sqrt{2} - 1)$ is a square. As it is in the real subfield, we can deduce that this in means $(\sqrt{2} - 1) = (\sqrt[4]{2}-1)(\sqrt[4]{2}+1)$ is a square in $\mathbb{Q}(\sqrt[4]{2})$.</p>
<p>Alas, these are units, which leaves me stuck. I want to believe $\sqrt[4]{2} \pm 1$ are the fundamental units and therefore this expression can't be square, but I don't know how to go about showing that. Mapping into a finite field might work, but the smallest prime that splits is $341$, I think, and that's beyond what I want to compute by hand at the moment.</p>
|
146,075 | <p>Within my limited experience, I have only known free groups to occur through two mechanisms: as fundamental groups of trees (graphs) and ping-pong. And sometimes only through one way: the fact that sufficiently high-powers of hyperbolic elements in a Gromov-hyperbolic group generate a free group arises via ping-pong. I know of no tree to represent the situation. </p>
<p>To the experts, the following question is surely either an obvious yes or no: </p>
<p>Can we explicitly describe all mechanisms by which finitely generated free subgroups of the Artin braid groups $B_n$ (for $n=2,3,\ldots$) arise ? </p>
<p>More specifically, seeing $B_n$ as the isotopy space of all $n$-pointed braids in the closed 2-disk, can we characterize all configurations generating the rank $k$ free group?</p>
<p>Much less specifically, do we know, say, that all finitely-generated free subgroups arise from ping-pong and can we describe all ping-pong games? </p>
| Andy Putman | 317 | <p>A theorem of Leininger-Margalit says that any two elements of the pure braid group either commute or generate a free group; see <a href="http://www.math.uiuc.edu/~clein/pbrels.pdf">here</a>. But I don't think there is any hope for classifying free subgroups that are generated by more than $2$ elements.</p>
|
2,835,290 | <p>I am given two variables $Y_1$ and $Y_2$ obeying an exponential distribution with mean $\beta= 1$</p>
<p>We are asked what the distribution of their average is and the solution must be found using moment generating functions.</p>
<p>The solution to this exercise says:</p>
<p><a href="https://i.stack.imgur.com/eOilN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eOilN.png" alt="enter image description here"></a></p>
<p>Well first, the solution already has a typo since it should be $E[e^{t((1/2)Y_1+(1/2)Y_2)}]$ but moreover, </p>
<p>I don't agree that $E[e^{t((1/2)Y_1+(1/2)Y_2)}]=M_{Y_1}(t)M_{Y_2}(t)$</p>
<p>Since $M_{Y_1}(t)M_{Y_2}(t)=E[e^{t(Y_1+Y_2)}]$, from my understanding.</p>
<p>So what is the actual solution to this problem?</p>
| xpaul | 66,420 | <p>Under $x\to\sin x$, one has
$$I=\int_{0}^{1}\ln\left(\frac{a-x^2}{a+x^2}\right)\cdot\frac{\mathrm dx}{x^2\sqrt{1-x^2}}=\int_{0}^{\pi/2}\ln\left(\frac{a-\sin^2x}{a+\sin^2x}\right)\cdot\frac{\mathrm dx}{\sin^2x}.$$
Let
$$I(k)=\int_{0}^{\pi/2}\ln\left(\frac{a-k\sin^2x}{a+k\sin^2x}\right)\cdot\frac{\mathrm dx}{\sin^2x}$$
and then
\begin{eqnarray}
I'(k)&=&\int_{0}^{\pi/2}\left(-\frac1{a-k\sin^2x}-\frac1{a+k\sin^2x}\right)\mathrm dx\\
&=&-\frac{\pi}{2}\left(\frac1{\sqrt{a(a+k)}}+\frac1{\sqrt{a(a-k)}}\right).
\end{eqnarray}
So
$$ I=I(1)=-\frac{\pi}{2\sqrt a}\int_0^1\left(\frac1{\sqrt{a+k}}-\frac1{\sqrt{a-k}}\right)\mathrm dk=-\frac{\pi}{\sqrt a}(\sqrt{a+1}-\sqrt{a-1}).$$</p>
|
1,270,584 | <p>I tried googling for simple proofs that some number is transcendental, sadly I couldn't find any I could understand.</p>
<p>Do any of you guys know a simple transcendentality (if that's a word) proof?</p>
<p>E: What I meant is that I wanted a rather simple proof that some particular number is transcendental ($e$ or $\pi$ would work), not a method to prove that any number is transcendental, sorry for the confusion.</p>
<p>Or even a proof about transcendental numbers being 'as common' as algebraic numbers?</p>
| marty cohen | 13,079 | <p>There is a BIG difference
between showing that
a particular number
that naturally occurs
(like $e$ or $\pi$)
is transcendental
and showing that
some number is.</p>
<p>The existence of transcendental
numbers was first shown in 1844
by Liouville. In 1851 he proved that
$\sum_{k=1}^{\infty} \frac1{10^{k!}}
$ is transcendental.</p>
<p>In 1873,
Hermite proved that
$e$ is transcendental.
The next year,
Cantor showed that the
algebraic numbers were countable,
so that almost all numbers are transcendental.
In 1882,
Lindemann showed that
$\pi$ is transcendental.</p>
<p>Look up more if you want.</p>
|
1,282,420 | <p>In his book of Differential Topology, Hirsch starts a little detour in his theory in order to present a way to see things in a general perspective. The precise fragment I'm referring to is the following:</p>
<p><img src="https://i.stack.imgur.com/rfzFr.png" alt="enter image description here"></p>
<p>I have never encountered a inverse limit before, and I don't have any intuition about what it is. I've searched the internet and it didn't help me either (for instance, the Wikipedia Page on <a href="http://en.wikipedia.org/wiki/Inverse_limit" rel="nofollow noreferrer">Inverse Limit</a>). He then proceeds to give an example:</p>
<p><img src="https://i.stack.imgur.com/5V3tB.png" alt="enter image description here"></p>
<p>It is not clear at all to me why it is continuous. My question is three-fold:</p>
<ol>
<li>Is there a (less general) definition of inverse limit adequate to this setting, or is the general definition adequate and acessible enough?</li>
<li>Is there an intuition?</li>
<li>Why is the structure functor above obviously continuous?</li>
</ol>
| oxeimon | 36,152 | <p>Perhaps this will help.</p>
<p>Basically an inverse limit is like this.</p>
<p>Fix a category (for example the category of sets, groups, rings, topological spaces, vector spaces). It's easy to define the inverse limit if your objects have an underlying "set", upon which possibly additional structure is imposed. (ie, groups are a set together with a group operation, topological spaces are a set together with a topology...etc)</p>
<p>Suppose you have a bunch of objects $X_i$ (possibly infinitely many) of that category. Suppose you have a bunch of morphisms $\phi_{ij} : X_i\rightarrow X_j$.</p>
<p>Then the inverse limit $\lim X_i$ over the objects $X_i$ and morphisms $\phi_{ij}$ is by definition the subset of the product $\prod_i X_i$ consisting of those tuples $(\ldots,x_i,\ldots,x_j,\ldots)$ such that for each morphism $\phi_{ij}$, $\phi_{ij}(x_i) = x_j$.</p>
<p>As a subset of the product, it comes with natural projection maps $pr_i : \lim X_i\rightarrow X_i$.</p>
<p>Then on this set there is usually a canonical "minimal" structure (group structure, topology,...etc) which makes the inverse limit an object in your category such that all the projection maps are morphisms in your category (ie, group/ring homomorphisms, or continuous maps...etc).</p>
<p>In other words, an inverse limit is a subset of the product consisting of tuples where the coordinates are all "compatible" with the morphisms $\phi_{ij}$. Note that if you take the inverse limit over a system $(\{X_i\}_i, \{\phi_{ij}\}_{ij})$, where the set of morphisms $\phi_{ij}$ is empty, then the inverse limit is precisely the product $\prod_i X_i$.</p>
<p>The $p$-adic numbers that Thomas Andrews describes is a good example. In that case, you're taking an inverse limit of infinitely many finite rings of the form $\mathbb{Z}/p^n\mathbb{Z}$ (each having positive characteristic), and ending up with an infinite ring of characteristic 0.</p>
<p>Another example is the classical cantor set. This can be seen as the 3-adic integers, where each $\mathbb{Z}/3^n\mathbb{Z}$ is seen as a finite set equipped with the discrete topology. Then the topology on the inverse limit is defined to be the smallest/weakest topology such that all the projection maps are continuous (it's a good exercise to work through the definitions to prove to yourself that this is homeomorphic to the usual definition of the cantor set)</p>
|
156,769 | <p>Let $\{a_{n}\}$ be a sequence of real numbers, where $0<a_{n}<1$, such that $\lim_{n\to \infty} a_{n}=0$, (then every subsequence will converges to zero). Is there any way to find a subsequence of $a_{n}$ which is decreasing to 0?</p>
| Arturo Magidin | 742 | <p>Yes: since the sequence converges to $0$, for every $\epsilon\gt 0$ there exists $N\in\mathbb{N}$ such that for all $n\geq N$, we have $0\lt a_n\lt\epsilon$.</p>
<p>So define the sequence recursively: take $a_1$. Then let $\epsilon = \frac{a_1}{2}$; we know there is an $n_1\gt 1$ such that $a_{n_1}\lt \frac{a_1}{2}$. Now take $\epsilon = \frac{a_{n_1}}{2}$; we know there exists $n_2\gt n_1$ such that $a_{n_2}\lt \frac{a_{n_1}}{2}$. </p>
<p>Lather, rinse, and repeat.</p>
|
2,885,453 | <p>Evaluate $$\lim _{x \to 0} \left[{\frac{x^2}{\sin x \tan x}} \right]$$ where $[\cdot]$ denotes the greatest integer function.</p>
<p>Can anyone give me a hint to proceed?</p>
<p>I know that $$\frac {\sin x}{x} < 1$$ for all $x \in (-\pi/2 ,\pi/2) \setminus \{0\}$ and $$\frac {\tan x}{x} > 1$$ for all $x \in (-\pi/2 ,\pi/2) \setminus \{0\}$. But will these two inequalities be helpful here?</p>
| jim | 289,829 | <p>Can you just use the known expansions for $\sin x, \tan x$ for small $x$? Then have $$\frac{x^2}{\sin x \tan x} = \frac{x^2}{(x - \frac{x^3}{6} + \cdots)(x + \frac{x^3}{3} + \cdots)}.$$ The <em>rhs</em> only including terms up to $x^2$ can be written $$\frac{1}{(1 - x^2/6)(1 + x^2/3)}$$ and the denominator to order $x^2$ is $1 + x^2/6$ so that finally have $$\frac{x^2}{\sin x \tan x} = \frac{1}{1 + \frac{x^2}{6} + \cdots} = 1 - x^2/6$$ and the integer part is 0.</p>
<p>Alternatively write the original expression as $$\frac{x^2 \cos x}{\sin^2 x}.$$</p>
|
2,309,613 | <p>Find a general solution to $x^2-2y^2=1$</p>
<p>I found that (3,2) is a solution.
Now what should I do?
I can not catch what the question really want.</p>
<p>It is about pell's equation. Would you give me a form of general solution? </p>
| fonfonx | 247,205 | <p><strong>Hints</strong></p>
<p>I would use the fact that $f$ is equal to its Taylor series on $G$ to write that $$f(z)=\sum_{n=0}^{\infty} a_n z^n$$</p>
<p>Then using the fact that $f(z) \in \mathbb R$ for $z \in \mathbb R$ you should be able to prove that the $a_n$ are real.</p>
<p>Using the fact that $f(z) \in i\mathbb R$ for $z \in i\mathbb R$ you can prove that for $n$ even $a_n$ is zero.</p>
|
125,862 | <p>I have been given that I am working with the space of all 2x2 matrices. The basis $B$ for this space is given as a set of four 2x2 matrices, each with an entry of 1 in a unique position and zeroes everywhere else (sorry about the description in words - I don't know how to format matrices for this site).</p>
<p>I have also been given the basis $B' = ({1, x, x^2})$ for the space of all polynomials of degree 2 or less and the basis $B'' = ({1})$ for $R$.</p>
<p>Then I am given a series of linear transformations and asked to find the matrices associated with them with respect to the bases above. I am completely lost as to how to do this! I would like help with how to achieve one of them so that I can then go and apply what I learn here to the other transformations.</p>
<p>The example I've chosen is the transformation T that maps 2x2 matrices to their transposes. I can't seem to construct a matrix that will bring the element in position '21' up to position '12'.</p>
<p>Can anyone give me some direction with this? Many thanks!!</p>
| geo909 | 27,882 | <p>You essentially have two ways of representing a Linear Tranformation (say, $T$ from now on):</p>
<ol>
<li>Using a "formula" or a kind of description (e.g. "the transpose")</li>
<li>Using a matrix (which depends on the basis that we choose; see below)</li>
</ol>
<p>In the second case, when you want to evaluate $T(u)$ where $u$ is an element of your vector space, you have to use the <em>vector representation</em> of $u$ with respect to the basis you chose! If for instance your vector space is a space of polynomials and $u$ is a polynomial, you cannot multiply a matrix with $u$; you <em>can</em> however multiply the vector representation of $u$ with the matrix. </p>
<p>In case you don't remember what a vector representation is: Let's <em>choose</em> the basis of the vector space of $2 \times 2$ matrices to be
$\left\{ B_1,B_2,B_3,B_4\right\}$, where
$$
B_1=
\begin{bmatrix}
1 & 0\\
0 & 0
\end{bmatrix},
B_2=
\begin{bmatrix}
0 & 1\\
0 & 0
\end{bmatrix},
B_3=
\begin{bmatrix}
0& 0\\
1& 0
\end{bmatrix},
B_4=
\begin{bmatrix}
0& 0\\
0& 1
\end{bmatrix}.
$$
Note that the order of the basis matters, and thus we call it an <em>ordered basis.</em></p>
<p>In our case, those $B_i$'s are the four $2\times 2$ basis matrices you mentioned above, in an order that we arbitrarily decided. For example, the vector representation of the matrix $\left(\begin{array}\11 & 2 \\ 3 & 4 \end{array}\right)$ would be $\left[1,2,3,4\right]$ because it can be written as $1\cdot B_1+2\cdot B_2 + 3\cdot B_3 + 4\cdot B_4$.</p>
<p>So now do the following: </p>
<ol>
<li><p>Plug in the basis elements in your T, that is, evaluate the $T(B_i)$'s for $i=1,2,3,4$. The result in each case is going to be of course a $2\times 2$ matrix (the transpose). </p></li>
<li><p>Find the vector representation of that matrix; it's going to be a vector of 4 coordinates, as in the example above. Say that you find the vectors $a_1, a_2, a_3, a_4$, respectively.</p></li>
<li>Put those vectors as columns in a matrix $A=\left( a_1| a_2 | a_3 | a_4 \right)$ (note that this is a $4\times 4$ matrix; your $T$ goes from a $4$-dimensional v.s. to itself). </li>
</ol>
<p>That $A$ is going to be the desired matrix. Note that you are <em>"asked to find the matrices associated with them with respect to the bases above"</em>. So, when you want to find the transpose of a matrix $B$ by using the matrix $A$ above, you will multiply the vector represenation of $B$ with $A$ and not try to multiply the <em>matrix</em> $B$ with $A$ (you can't anyway).</p>
<p>It is pretty much the same with any vector space: evaluate $T$ of each of the basis elements, write the results as vector representations, and put all those as columns in a matrix and you're done.</p>
<p>Note that the resulting matrix depends on the choice of your basis, as well as the order of the basis that you choose. For every vector space though, we have some "standard basis" that we use often. For example the one that I gave is often used for the vector space of $2\times 2$ matrices. But you may also see it in a different order, e.g.
$\left\{ B_1, B_3, B_2, B_4\right\}$, and that will give a different answer. This is fine, as long as you are clear about what basis you have chosen.</p>
<p>Edit: observe that if you follow the steps you will get the matrix that Rasmus gave you.</p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| algori | 2,349 | <p>There are $n$ balls rolling along a line in one direction and $k$ balls rolling along the same line in the opposite direction. The speeds of the balls in the first group and in the second group are equal. Initially the two groups of balls are separated from one another and at some point the balls start colliding. The collisions are assumed to be elastic. How many collisions will there be?</p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Qiaochu Yuan | 290 | <p>Adam Hesterberg told me this one ages ago. It apparently used to circulate around MOP. </p>
<p>Three spiders and a fly are placed on the edges of a regular tetrahedron, and travel only on those edges. The fly travels at the rate of $1$ edge/s, whereas the spiders travel at the rate of $1 + \epsilon$ edge/s for some $\epsilon > 0$. The spiders want to agree beforehand on a deterministic strategy for capturing the fly, whose location they do not know (but they do know each others' locations). The fly is invisible and omniscient; in particular, it is aware of the locations of the spiders and of their strategy at all times. (It also cannot fly.) </p>
<p>Can the spiders guarantee that they will catch the fly in finite time, regardless of the initial positions of the spiders and the fly? Does the answer depend on the value of $\epsilon$? </p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Roland Bacher | 4,556 | <p>Not a very difficult one but I like it since it is even suitable for non-mathematicians: </p>
<p>A small boat carrying a heavy stone is floating in a swimming pool. What happens to the level of water (up, down or remains equal) in the swimming pool if one removes the stone from the boat and throws it in the swimming pool?</p>
<p>The very easy solution suggests the following joke (illustrating the well-known ignorance of mathematicians of reality): Instead of sending scores of ships for saving passengers from the Titanic, one should have sunk all possible rescue-ships in order to lower the sea-level.</p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Gerald Edgar | 454 | <p>I had a good outcome with this one once. It probably helped that the other two mathematicians had a drink or two before dinner. (Otherwise they would have solved it in 5 seconds...) When salad was served, somebody had oil and vinegar in separate little pitchers...</p>
<p>Suppose you have two containers, one with oil, one with vinegar, equal volume. Take one teaspoon of the oil, put it into the vinegar, stir. Than take one teaspoon of the mixture, put it into the oil, stir. Now: is there more vinegar in the oil or more oil in the vinegar?</p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Alex R. | 934 | <p>Bob and Alice want to marry each other, so Bob decides to send Alice a ring. The problem is that they both live in different countries, and any valuables they send through the mail are sure to be stolen, unless they are sent in a locked box. The box can be locked by a padlock which can only be opened by the right key. Both Alice and Bob have an infinite supply of boxes and padlocks with corresponding keys. However, neither Alice nor Bob have keys to each other's padlocks, only for their own. Suppose you can put boxes inside each other. How can Bob send Alice the ring? Of course, the solution to the problem must end with Alice putting the ring on her finger. To reiterate, anything outside of a padlocked box is guaranteed to be stolen. </p>
<p>This problem has numerous solutions as well as interpretations which makes for a fun discussion. It can also be solved in your head without pencil or paper. </p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Douglas S. Stones | 2,264 | <p>What is four thousand and ninety-nine plus one?</p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Thierry Zell | 8,212 | <p>Simple puzzles; unfortunately, I do not know how to formulate them in a whimsical fashion suitable for a dinner, they very much sound like math puzzles.</p>
<ol>
<li><p>Take n labeled points $x_1, \dots, x_n$ in the plane. How do you construct a n-gon $a1, \dots, an$ such that for all i, $x_i$ is the midpoint of $[a_i, a_{i+1}]$ (with the convention $a_{n+1}=a_1$ of course).
I was surprised to come across this problem in the puzzle pages of <i>Le Monde.</i> I think non-mathematicians would have a hard time with it.</p></li>
<li><p>For mathematicians who don't already know it, the <a href="http://en.wikipedia.org/wiki/Sylvester%25E2%2580%2593Gallai_theorem" rel="nofollow">Sylvester Gallai Theorem</a> can offer stimulating after dinner discussions (or during those long proctoring sessions).</p></li>
<li><p>A napkin should be enough for this one (if even needed!). Consider a map f from the plane to the reals such that the sum of the values of f on the vertices of any square is zero. Find all such maps.</p></li>
</ol>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Jesus Martinez Garcia | 1,887 | <p>I think this has not been published yet. Apologies otherwise. I learnt it from Antonio Sánchez Calle in my first year of undergraduate and I had 3 non-mathematicians thinking about it for about 4 hours, so there is a guaranteed success if you tell around :)</p>
<p>5 people are shipwrecked in a deserted island. They find a monkey and lots of coconuts. They spend the whole day collecting coconuts that they keep together and since they are tired they go to sleep. The first person wakes up, attempts to divide the amount of coconuts in five parts, but one of them is spared, so he gives to the monkey. Then he eats one fifth of the coconuts and goes back to sleep.</p>
<p>The second person wakes up and follows the same procedure. He divides the coconuts in five, one is spared and he gives it to the monkey, eats his share and goes back to sleep.</p>
<p>The third, forth and fifth people do the same thing. How many coconuts were there at the beginning? (modulo something, of course).</p>
|
1,744,760 | <h2>Question</h2>
<p>What do we gain or lose, conceptually, if we consider <em>scalar multiplication</em> as a special form of <em>matrix multiplication</em>?</p>
<h2>Background</h2>
<p>The question bothers me since I have been reading about <em>dilations</em> and <em>scaling</em> of geometrical objects in Paul Lockhart's book "Measurement". Geometrically, dilation is a transformation that stretches an object in <strong>one</strong> dimension by a certain factor. Analogously, the linear transformation
$$
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & \lambda
\end{pmatrix}
\cdot
\begin{pmatrix}
x_1 \\
x_2 \\
x_3
\end{pmatrix}
$$
"stretches" the third component by the factor $\lambda$. Scaling is a geometric transformation that stretches an object in <strong>all</strong> dimensions by a certain factor. Analogously, the linear transformation
$$
\begin{pmatrix}
\lambda & 0 & 0 \\
0 & \lambda & 0 \\
0 & 0 & \lambda
\end{pmatrix}
\cdot
\begin{pmatrix}
x_1 \\
x_2 \\
x_3
\end{pmatrix}
$$
"stretches" all three components by the factor $\lambda$. This, however, can be written more succinctly using scalar multiplication:
$$
\lambda
\cdot
\begin{pmatrix}
x_1 \\
x_2 \\
x_3
\end{pmatrix}.
$$
In fact, every scalar multiplication can be expressed as a multiplication with a special matrix, and it turns out to be a mere shortcut. On the face of it, this observation is not very spectacular; however, it raises interesting philosophical and conceptual questions as to the foundations of linear algebra.</p>
<p>For example, if scalar multiplication is only a nice-to-have shortcut, then isn't it in fact superfluous conceptually? Currently, scalar multiplication is taught as if it was a distinct concept, independent of matrix multiplication. What would change if we got rid of this shortcut? What could alternative axioms of vector spaces and moduls look like? What about linear transformations? What is easier, what is harder$-$not to write down, but conceptually?</p>
<p>I know that this topic is very broad, but I would like to collect opinions, ideas, examples.</p>
| Seven | 135,088 | <p>I disagree that "scalar multiplication is only a nice-to-have shortcut", or that is "superfluous conceptually". In fact the very definition of a vector space $V$ requires there to be a scalar multiplication.</p>
<p>After that comes the concept of a linear transformation, which again requires the scalar multiplication to be defined. Matrix multiplication is a convenient way to represent these, and that too only in case of finite dimensional vector spaces (or certain infinite dimensional vector spaces). </p>
<p>So defining matrix multiplication and then saying that scalar multiplication is a special case is putting the cart in front of the horse in my opinion. </p>
|
191,373 | <p>I have a usual mathematical background in vector and tensor calculus. I was trying to use the differential operators of Mathematica, namely <code>Grad</code>, <code>Div</code> and <code>Curl</code>. According to my knowledge, the definitions of Mathematica for <code>Grad</code> and <code>Div</code> coincides with those usually employed in tensor calculus, that is to say</p>
<p><span class="math-container">\begin{align*}
\text{grad}\mathbf{T}&:=\sum_{k=1}^{3}\frac{\partial\mathbf{T}}{\partial x_k}\otimes \mathbf{e}_k\\
\text{div}\mathbf{T}&:=\sum_{k=1}^{3}\frac{\partial\mathbf{T}}{\partial x_k}\cdot\mathbf{e}_k \\
\tag{1}
\end{align*}</span></p>
<p>for any tensor <span class="math-container">$\mathbf{T}$</span> of rank <span class="math-container">$n\ge1$</span>. <span class="math-container">$x_k$</span>'s are Cartesian coordinates and <span class="math-container">$\mathbf{e}_i$</span>'s are the standard basis for <span class="math-container">$\mathbb{R}^3$</span>. <span class="math-container">$\otimes$</span> and <span class="math-container">$\cdot$</span> are the usual generalized outer and inner products which are also defined in Mathematica by <code>Outer</code> and <code>Inner</code>. The usual definition that I know from tensor calculus for the <code>Curl</code> is as follows
<span class="math-container">\begin{align*}
\text{curl}\mathbf{T}&:=\sum_{k=1}^{3}\mathbf{e}_k\times\frac{\partial\mathbf{T}}{\partial x_k}.
\tag{2}
\end{align*}</span>
However, it turns out that Mathematica's definition for curl is totally different. For example, it returns the <code>Curl</code> of a second order tensor as a scalar, while according to <span class="math-container">$(2)$</span> it should be a second order tensor.</p>
<blockquote>
<p>I couldn't find a precise definition of Mathematica for <code>Curl</code> in the documents. I am wondering what this definition is. What is the motivation for this? and How it can be related to the definition given in <span class="math-container">$(2)$</span>?</p>
</blockquote>
<p>Below is a simple piece of code for you to observe the outputs of Mathematica when we apply the <code>Grad</code>, <code>Div</code> and <code>Curl</code> operators to scalar, vector and second order tensor fields. I would like to draw your attention to some observations. <code>Curl</code> of a scalar is returned as a second order tensor, which I don't understand why! <code>Curl</code> of a vector coincides with the usual definition of <code>Curl</code> used in vector calculus. <code>Curl</code> of second order tensor is returned as a scalar, which I don't understand again.</p>
<pre><code>Var={Subscript[x, 1],Subscript[x, 2],Subscript[x, 3]};
Sca=\[Phi][Subscript[x, 1],Subscript[x, 2],Subscript[x, 3]];
Vec={Subscript[v, 1][Subscript[x, 1],Subscript[x, 2],Subscript[x, 3]],Subscript[v, 2][Subscript[x, 1],Subscript[x, 2],Subscript[x, 3]],Subscript[v, 3][Subscript[x, 1],Subscript[x, 2],Subscript[x, 3]]};
Ten=Table[Subscript[T, i,j][Subscript[x, 1],Subscript[x, 2],Subscript[x, 3]],{i,1,3},{j,1,3}];
MatrixForm[Grad[Sca, Var]]
MatrixForm[Grad[Vec, Var]]
MatrixForm[Grad[Ten, Var]]
MatrixForm[Div[Sca, Var]]
MatrixForm[Div[Vec, Var]]
MatrixForm[Div[Ten, Var]]
MatrixForm[Curl[Sca, Var]]
MatrixForm[Curl[Vec, Var]]
MatrixForm[Curl[Ten, Var]]
</code></pre>
<p>I will be happy if someone can reproduce the following result for the curl of a second order tensor with Mathematica's <code>Curl</code> function.</p>
<p><span class="math-container">\begin{align*}
\text{curl}\mathbf{T}&:=\sum_{k=1}^{3}\mathbf{e}_k\times\frac{\partial\mathbf{T}}{\partial x_k}=\sum_{k=1}^{3}\mathbf{e}_k\times\frac{\partial}{\partial x_k}\left(\sum_{i=1}^{3}\sum_{j=1}^{3}T_{ij}\mathbf{e}_i\otimes\mathbf{e}_j\right)\\
&=\sum_{k=1}^{3}\sum_{i=1}^{3}\sum_{j=1}^{3}\frac{\partial T_{ij}}{\partial x_k}(\mathbf{e}_k\times\mathbf{e}_i)\otimes\mathbf{e}_j\\
&=\sum_{k=1}^{3}\sum_{i=1}^{3}\sum_{j=1}^{3}\sum_{m=1}^{3}\epsilon_{kim}\frac{\partial T_{ij}}{\partial x_k}\mathbf{e}_m\otimes\mathbf{e}_j
\tag{3}
\end{align*}</span></p>
<p>where <span class="math-container">$\epsilon_{kim}$</span> is the <code>LeviCivitaTensor</code> for <span class="math-container">$3$</span> dimensions. Consequently, we get</p>
<p><span class="math-container">\begin{align*}
\left(\text{curl}\mathbf{T}\right)_{mj}=\sum_{k=1}^{3}\sum_{i=1}^{3}\epsilon_{kim}\frac{\partial T_{ij}}{\partial x_k}.
\tag{4}
\end{align*}</span></p>
<p>Implementing <span class="math-container">$(4)$</span> in Mathematica, we obtain</p>
<pre><code>CurlTen = Table[
Sum[
LeviCivitaTensor[3][[k, i, m]]
D[Subscript[T, i, j][Subscript[x, 1], Subscript[x, 2], Subscript[x, 3]], {Subscript[x, k]}], {k, 1, 3}, {i, 1, 3}],
{m, 1, 3}, {j, 1, 3}];
MatrixForm[CurlTen]
</code></pre>
| OA Fakinlede | 65,652 | <p>I am limiting myself to fields defined in the three dimensional Euclidean point space. The curl of a tensor can be found in these simple steps:</p>
<ol>
<li><p>Take the simple composition of the second-order tensor, <code>T</code>, with the <code>LeviCivitaTensor[3]</code>. This is effected by the command, </p>
<pre><code>Dot[T, LeviCivitaTensor[3]]
</code></pre></li>
<li><p>Take the transpose of the divergence of the above result using the command:</p>
<pre><code>curl[T_] := Transpose[Div[Dot[T, LeviCivitaTensor[3]], {x, y, z}]]
</code></pre></li>
</ol>
<p>The result you get here is the second-order tensor that is the curl of the tensor <code>T</code>.</p>
|
2,004,895 | <p>In my textbook there is a question like below:</p>
<p>If $$f:x \mapsto 2x-3,$$ then $$f^{-1}(7) = $$</p>
<p>As a multiple choice question, it allows for the answers: </p>
<p>A. $11$<br>
B. $5$<br>
C. $\frac{1}{11}$<br>
D. $9$</p>
<p>If what I think is correct and I read the equation as:</p>
<p>$$f(x)=2x-3$$
then,<br>
$$y=2x-3$$
$$x=2y-3$$
$$x+3=2y$$
$$\frac {x+3} {2} = y$$</p>
<p>therefore:</p>
<p>$$f^{-1}(7)=\frac {7+3}{2}$$
$$=5$$</p>
| Community | -1 | <p>(note: this answer was formulated in response to the original version of the question, which has since been edited by a third party)</p>
<p>You're parsing the expression wrong; it's not</p>
<p>$$ \color{red}{\mathbf{ f : x }} \mapsto 2x-3 $$</p>
<p>instead, it is </p>
<p>$$ f : \color{red}{\mathbf{ x \mapsto 2x-3 }}$$</p>
<p>That is, $x \mapsto 2x-3$ is one particular notation for "the function which, for all $x$, sends the value $x$ to the value $2x-3$". The colon notation is one particular way to attach a name to the mapping; to say "$f$ is the function which, for all $x$, ...". You might read it as "$f$ sends $x$ to $2x-3$".</p>
<p>The colon notation is more often used when you include a third term expressing the domain and codomain. e.g. if we are talking about a real-valued function on the reals, we would wrote</p>
<p>$$ f : \mathbb{R} \to \mathbb{R} : x \mapsto 2x-3 $$</p>
<p>Here:</p>
<ul>
<li>$\mathbb{R} \to \mathbb{R}$ denotes the <em>type</em> of mathematical object we call "real-valued functions of the reals"</li>
<li>$f : \mathbb{R} \to \mathbb{R}$ says that $f$ is a variable of that type — i.e. that $f$ denotes a real-valued function of the reals.</li>
<li>The whole expression additionally indicates <em>which</em> function $f$ specifically is.</li>
</ul>
|
915,414 | <p>I recently did some work to try to find $\int{\frac{dx}{Ax^3 - B}}$, but I'm always paranoid that my solution has some minor trivial error in the middle of the process that screwed up the end result entirely, so could someone please help check my solution?</p>
<p>The first step to my solution is to eliminate $A$ and $B$ from the integrand by pushing them out as constants and leaving $\frac{dx}{x^3 - 1}$. We start with
$$
\begin{align}
\int{\frac{dx}{Ax^3 - B}} & = \int{\frac{dx}{A(x^3 - B/A)}} \\
& = \frac{1}{A}\int{\frac{dx}{x^3 - B/A}}.
\end{align}
$$
To get rid of the $B/A$ term, we make the substitution $x = \sqrt[3]{B/A}u, \ dx = \sqrt[3]{B/A}du$.
$$
\begin{align}
\frac{1}{A}\int\frac{dx}{x^3 - B/A} & = \frac{1}{A}\int{\frac{\sqrt[3]{B/A}du}{(B/A)u^3 - B/A}} \\
& = \frac{1}{A}\cdot\frac{\sqrt[3]{B/A}}{B/A}\int{\frac{du}{u^3 - 1}} \\
& = \mathcal{C}\int{\frac{du}{u^3 - 1}}
\end{align}
$$
with $\mathcal{C} = (AB^2)^{-1/3}$. Now we can just worry about solving $\int{\frac{du}{u^3 - 1}}$.</p>
<p>We can decompose $\frac{1}{u^3 - 1}$ using the fact that $u^3 - 1 = (u - 1)(u^2 + u + 1)$. So</p>
<p>$$
\begin{align}
\frac{1}{u^3 - 1} & = \frac{1}{(u - 1)(u^2 + u + 1)} \\
& = \frac{P}{u - 1} + \frac{Qu + R}{u^2 + u + 1}
\end{align}
$$
Here $P = 1/3$, but $Q$ and $R$ aren't as trivial to find. $u^2 + u + 1$ has no real roots, so I chose to sub in $u = -\sqrt[3]{-1}$ (it was the first root that I found). We then have to solve the equation
$$
(-\sqrt[3]{-1}-1)^{-1} = Q(-\sqrt[3]{-1}) + R
$$
which we can solve by setting $Q = R = [(-1)^{2/3} - 1]^{-1}$ (which will also help when integrating $\frac{Qu + R}{u^2 + u + 1}$ since we can factor $Q$ and $R$ out as a common constant from the numerator).</p>
<p>So
\begin{align}
\mathcal{C}\int{\frac{du}{u^3 - 1}} & = \mathcal{C}\left[\frac{1}{3}\int{\frac{du}{u - 1}} + Q\int{\frac{u + 1}{u^2 + u + 1}du}\right] \\
& = \mathcal{C}\left\{\frac{1}{3}\ln{|u - 1|} + Q\left[\frac{1}{2}\ln{|u^2 + u + 1|} + \frac{1}{\sqrt{3}}\arctan{\left(\frac{2}{\sqrt{3}}u + \frac{1}{\sqrt{3}}\right)} \right] \right\} \\
& = (AB^2)^{-1/3}\left\{\frac{1}{3}\ln{\left|\frac{x}{\sqrt[3]{B/A}} - 1\right|} + \frac{1}{(-1)^{2/3} - 1}\left[\frac{1}{2}\ln{\left|\left(\frac{x}{\sqrt[3]{B/A}}\right)^2 + \frac{x}{\sqrt[3]{B/A}} + 1\right|} \right. \right. \\
& \hspace{15mm} \left. \left. + \frac{1}{\sqrt{3}}\arctan{\left(\frac{2x}{\sqrt{3}\sqrt[3]{B/A}} + \frac{1}{\sqrt{3}}\right)}\right]\right\} + \text{constant}
\end{align} </p>
<p>(I know what $\int{\frac{x + 1}{x^2 + x + 1}dx}$ is from previous problems)</p>
<p>How does it look? Did I do anything severely wrong (I don't feel entirely confident about the partial fractions part)? Any suggestions for how I could get to an answer faster or more efficiently?</p>
| rogerl | 27,542 | <p>Everything looks fine until your partial fraction decomposition. Indeed $P=\frac{1}{3}$, but to find $Q$ and $R$, start with
$$1 = \frac{1}{3}(u^2+u+1) + (Qu+R)(u-1)$$
and choose $u=0$ to get
$$1 = \frac{1}{3} - R,$$
so that $R = -\frac{2}{3}$. You then get $Q = -\frac{1}{3}$. So the decomposition is
$$\frac{1}{u^3-1} = \frac{1}{3(u-1)} - \frac{u+2}{3(u^2+u+1)}.$$
(Your solution involves complex numbers --- $(-1)^{2/3}$ is ambiguous, but must be complex since otherwise the denominator vanishes --- I doubt that was what you were looking for.)</p>
|
150,295 | <p>I will appreciate any enlightenment on the following which must be an exercise in a certain textbook. (I don't recognize where it comes from.)
I understand that the going down property does not hold since $R$ is not integrally closed (in fact, it is not a UFD), but I have no idea how to show that $q$ is such a counterexample.</p>
<blockquote>
<p>Let $k$ be a field, $A = k[X, Y]$ be a polynomial ring, $R = \lbrace f \in A \colon f(0, 0) = f (1, 1) \rbrace \subset A$ be a subring.
Define $q = (X)\cap R$, $p = (X - 1, Y - 1) \cap R$, $P = (X - 1, Y - 1)$.
Show that there is no $Q \in \operatorname{Spec} A$, $Q\subset P$ that goes down to $q$. </p>
</blockquote>
| Georges Elencwajg | 3,217 | <p>I'll show that the existence of $Q\in Spec(A)$ satisfying $Q\subsetneq P$ and $q= Q\cap R$ leads to a contradiction. </p>
<p>We have $X\cdot (X-1)\in q $ , so $X\cdot (X-1)\in Q$.<br>
Hence we have $(X-1)\in Q$ (since $X\notin Q$ because $X\notin P$).<br>
But this forces $Q=(X-1)A$, since $(X-1)A\subset Q\subsetneq P=(X-1,Y)$.<br>
But then $(X-1)Y\in Q\cap R \setminus q$ : contradiction . </p>
<p><strong>Edit</strong><br>
As wxu remarks in his comment, $q$ as defined by eltonjohn is not included in $p$.
The above answer remains correct (I am sure of that, because else wxu would have noticed!), but it is not a counterexample to Going Down.<br>
I advise users to read wxu's post: he modified eltonjohn's question precisely in order to give such a counterexample.</p>
|
915,016 | <p>Let $S$ be a non - empty set and $F$ be a field. Let $C(S,F)$ denote the set of all functions $f\in \mathcal F(S,F)$ such that $f(s)=0$ for all but a finite number of elements of $S$. Prove that $C(S,F)$ is a subspace of $\mathcal F(S,F)$</p>
<p>My progress:
I have shown that if $f,g\in C(S,F)$ and $a\in F$ then $f+g\in C(S,F)$ and $af\in C(S,F)$.</p>
<p>I need to show that the zero element is in $C(S,F)$. What is the zero element here and how do I show it's in $C(S,F)$</p>
| mvw | 86,776 | <p>Just checking if your solution candidate fullfills the PDE $(*)$ and boundary condition $(**)$. </p>
<p>The condition $(**)$ holds.</p>
<p>For the PDE we need to calculate $u_x$ and $u_t$. For calculating $u_t$ we use the <a href="http://en.wikipedia.org/wiki/Leibniz_integral_rule#General_form_with_variable_limits" rel="nofollow">Leibniz rule</a>.
We get
$$
\begin{align}
u_x(x,t)
&=
\cos(x-tb) +
\frac{\partial}{\partial x}
\int\limits_0^t \left[x-(t-\sigma)b+\sigma\right]
\cos\left(\left[x-(t-\sigma)b\right]\sigma\right) \, d\sigma \\
&=
\cos(x-tb) +
\int\limits_0^t
\frac{\partial}{\partial x}
\left[x-(t-\sigma)b+\sigma\right]
\cos\left(\left[x-(t-\sigma)b\right]\sigma\right) \, d\sigma \\
&=
\cos(x-tb) + \\
&\int\limits_0^t
\cos\left(\left[x-(t-\sigma)b\right]\sigma\right) -
\left[x-(t-\sigma)b+\sigma\right]
\sin\left(\left[x-(t-\sigma)b\right]\sigma\right) \sigma
\, d\sigma
\end{align}
$$
and
$$
\begin{align}
u_t(x,t)
&=
-b \cos(x-tb) +
\frac{\partial}{\partial t}
\int\limits_0^t \left[x-(t-\sigma)b+\sigma\right]
\cos\left(\left[x-(t-\sigma)b\right]\sigma\right) \, d\sigma \\
&=
-b \cos(x-tb) +
\int\limits_0^t
\frac{\partial}{\partial t}
\left[x-(t-\sigma)b+\sigma\right]
\cos\left(\left[x-(t-\sigma)b\right]\sigma\right) \, d\sigma + \\
& \left[x-(t-t)b+t\right]
\cos\left(\left[x-(t-t)b\right]t\right) \frac{dt}{dt} \\
&=
-b\cos(x-tb) + \\
&(-b) \int\limits_0^t
\cos\left(\left[x-(t-\sigma)b\right]\sigma\right) -
\left[x-(t-\sigma)b+\sigma\right]
\sin\left(\left[x-(t-\sigma)b\right]\sigma\right) \sigma \, d\sigma + \\
& (x+t) \cos(xt)
\\
&=
-b \, u_x(x,t) + (x+t) \cos(xt)
\end{align}
$$
So, assuming I made no mistake, your candidate solves
$$
u_t(x,t) + b \, u_x(x,t) = (x+t) \cos(xt)
$$
which is your $(*)$ if $b = 1$.</p>
|
915,016 | <p>Let $S$ be a non - empty set and $F$ be a field. Let $C(S,F)$ denote the set of all functions $f\in \mathcal F(S,F)$ such that $f(s)=0$ for all but a finite number of elements of $S$. Prove that $C(S,F)$ is a subspace of $\mathcal F(S,F)$</p>
<p>My progress:
I have shown that if $f,g\in C(S,F)$ and $a\in F$ then $f+g\in C(S,F)$ and $af\in C(S,F)$.</p>
<p>I need to show that the zero element is in $C(S,F)$. What is the zero element here and how do I show it's in $C(S,F)$</p>
| doraemonpaul | 30,938 | <p>Follow the method in <a href="http://en.wikipedia.org/wiki/Method_of_characteristics#Example" rel="nofollow">http://en.wikipedia.org/wiki/Method_of_characteristics#Example</a>:</p>
<p>$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$</p>
<p>$\dfrac{dx}{ds}=1$ , letting $x(0)=x_0$ , we have $x=s+x_0=t+x_0$</p>
<p>$\dfrac{du}{ds}=(x+t)\cos xt=(2s+x_0)\cos((s+x_0)s)$ , letting $u(0)=f(x_0)$ , we have $u(x,t)=f(x_0)+\int_0^s(2\sigma+x_0)\cos((\sigma+x_0)\sigma)~d\sigma=f(x-t)+\int_0^t(2\sigma+x-t)\cos((\sigma+x-t)\sigma)~d\sigma$</p>
<p>$u(x,0)=\sin x$ :</p>
<p>$f(x)=\sin x$</p>
<p>$\therefore u(x,t)=\sin(x-t)+\int_0^t(2\sigma+x-t)\cos((\sigma+x-t)\sigma)~d\sigma$</p>
|
1,534,246 | <p>I'm trying to simplify this boolean expression:</p>
<p>$$(AB)+(A'C)+(BC)$$</p>
<p>I'm told by every calculator online that this would be logically equivalent:</p>
<p>$(AB)+(A'C)$</p>
<p>But so far, following the rules of boolean algebra, the best that I could get to was this: </p>
<p>$(B+A')(B+C)(A+C)$</p>
<p>All of the above are logically equivalent (I've made a truth table for each) but I don't understand what steps am I missing trying to simplify the expression.</p>
<p>I also couldn't find an "expression simplifier" tool online that could show me the steps that I'm missing.</p>
<p>Help / directions to go to would be much appreciated, thanks in advance.</p>
| Lucian | 93,448 | <p><em>I would like to add the following explanation to the above answers:</em></p>
<blockquote>
<p>The first two terms translate as “If <em>A</em>, then <em>B</em>, else <em>C</em> ”. Notice, therefore, that <em>B</em> and <em>C</em> cannot <br> simultaneously coexist, meaning that the third term can be safely ignored or omitted, since it <br> is superfluous or redundant.</p>
</blockquote>
|
1,049,933 | <p>If $M$ is the transition matrix of a discrete Markov chain, and $M$ is both irreducible, symmetric and positiv-definite, is the resulting Markov chain necessarily aperiodic? </p>
<p>In my intuition, periodicity would correspond to an $-1$-eigenvalue of $M$, but I don't know if that is true or how to formalize it.</p>
| frogfanitw | 297,978 | <p>I believe there is a recursive formula for this:</p>
<p>Let L(k) = LCM({$p_1 - 1$, $p_2 - 1$, $p_3 - 1$, ..., $p_k$}</p>
<p>Then L(k+1) = L(k) ($p_{k+1}-1$) / GCD(L(k),$p_{k+1}-1$)</p>
<p>where GCD = greatest common divisor</p>
|
4,242,093 | <p><em><strong>Question:</strong></em></p>
<blockquote>
<p>Let <span class="math-container">$G=(V_n,E_n)$</span> such that:</p>
<ul>
<li>G's vertices are words over <span class="math-container">$\sigma=\{a,b,c,d\}$</span> with length of <span class="math-container">$n$</span>, such that there aren't two adjacent equal chars.</li>
<li>An edge is defined to be between two vertices that are different by only one char.</li>
</ul>
<p>A. Does the graph contain an Euler cycle?</p>
<ul>
<li>Find a pattern.</li>
</ul>
<p>B. Does the graph contain a Hamiltonian cycle</p>
<ul>
<li>This can be proven by induction.</li>
</ul>
</blockquote>
<hr />
<p><span class="math-container">$Solution.A.$</span></p>
<p>Now, when <span class="math-container">$n=1$</span>, we have 4 vertices: <span class="math-container">$$v_1= \ 'a'$$</span> <span class="math-container">$$v_2= \ 'b'$$</span> <span class="math-container">$$v_3= \ 'c'$$</span> <span class="math-container">$$v_4= \ 'd'$$</span></p>
<p>Therefore, for each <span class="math-container">$v\in \{v_1,v_2,v_3,v_4\}$</span>, <span class="math-container">$N(v)=\{v_1,v_2,v_3,v_4\}/ \{v\}$</span> so we get that their degree is 3, so by a theorem we get that there isn't an Euler cycle.</p>
<p>In addition, when <span class="math-container">$n=4$</span>, considering the string <span class="math-container">$"abad"$</span> we have 2 options to replace the edges of the string. In order to replace the second char we have 2 options, replacing it by <span class="math-container">$'c'$</span> and <span class="math-container">$'d'$</span>. For the third char, we can replace it only by <span class="math-container">$'c'$</span>. In total, we got 7 edges with this vertex, so by a theorem, we get that there isn't an Euler cycle.</p>
<p>I cannot find here a pattern, because if we take a look at <span class="math-container">$n=2$</span> we get an Euler cycle.</p>
<p><span class="math-container">$Solution.B.$</span></p>
<p>First, we examine whether each vertex has at least <span class="math-container">$\frac{n}{2}$</span> neighbors. Hence, we should take the vertex to have the least number of neighbors. This vertex should be the string with disjoint chars. i.e. the string "abcd" when <span class="math-container">$n=4$</span>. The first and last chars has always 2 neighbors, so we get that the least degree is: <span class="math-container">$$2+2+\binom{n-2}{n-3} \cdotp 1=4+n-2=n+2\geq \frac{n}{2}$$</span></p>
<p>Thus, we get that the graph always has a Hamiltonian cycle.</p>
<hr />
<p>I don't get why I didn't get a pattern in <span class="math-container">$A$</span>, and how <span class="math-container">$B$</span> can be proven by induction. In addition, is my answer correct?</p>
| 1Rock | 208,645 | <p>Your solution B doesn't work. The graph <span class="math-container">$G_n$</span> has <span class="math-container">$4\cdot3^n$</span> vertices rather than <span class="math-container">$n$</span> vertices, so you'd need to show each vertex had <span class="math-container">$4\cdot 3^n/2$</span> vertices to apply Dirac's theorem.</p>
|
2,390,036 | <blockquote>
<p><strong>Theorem</strong>. Let $G = (V,E)$ be a simple graph with $n$ vertices, $m$ edges and $\chi (G) = k$. Then, $$m \geqslant {k \choose 2}$$</p>
</blockquote>
<p>I tried proving myself but made little to no progress. I am aware of the inequality $$n/ \alpha (G) \leqslant \chi (G) \leqslant \Delta(G) + 1,$$ where $\Delta (G)$ is the maximum degree in $G$ and $\alpha (G)$ is the size of the maximum independent set of vertices in the graph. Does anybody have any tips? (<strong>tips</strong> are appreciated more than complete answers!)</p>
| Kuifje | 273,220 | <p>Color classes are necessarily pairwise connected by at least one edge (otherwise you could merge them), and since there are $k$ color classes ...</p>
|
1,930,933 | <blockquote>
<p>Does there exist an $n \in \mathbb{N}$ greater than $1$ such that $\sqrt[n]{n!}$ is an integer?</p>
</blockquote>
<p>The expression seems to be increasing, so I was wondering if it is ever an integer. How could we prove that or what is the smallest value where it is an integer?</p>
| Ege Erdil | 326,053 | <p>This is impossible due to Bertrand's postulate, since there will always be a prime $ p $ in $ n! $ occuring with multiplicity $ 1 $ as long as $ n \geq 2 $. This actually implies that $ n! $ is never a perfect power for $ n \geq 2 $.</p>
|
786,301 | <p>Is there a systematic way to express the sum of two complex numbers of different magnitude (given in the exponential form), i.e find its magnitude and its argument expressed in terms of those of the initial numbers?</p>
| Alex G. | 130,309 | <p>Not really. One has to express the complex numbers as the sums of their real and imaginary parts, and then add componentwise, like usual. The best we can really do is make use of the triangle inequality:</p>
<p>$| |z_1| - |z_2| | \leq |z_1 + z_2| \leq |z_1| + |z_2|$</p>
|
84,312 | <p>In the topological sense, I understand that the unit circle $S^1$ is a retract of $\mathbb{R}^2 \backslash \{\mathbb{0}\}$ where $\mathbb{0}$ is the origin. This is because a continuous map defined by $r(x)= x/|x|$ is a retraction of the punctured plane $\mathbb{R}^2 \backslash \{\mathbb{0}\}$ onto the unit circle $S^1 \subset \mathbb{R}^2 \backslash \{\mathbb{0}\}$. Does this mean that $S^1$ is not a retract of $\mathbb{R}^2$? I would appreciate some clarification here.</p>
| Mariano Suárez-Álvarez | 274 | <p>No, you cannot conclude that $S^1$ is not a retract of $\mathbb R^2$ that way. To prove that something is not a retract usually requires more machinery, and algebraic topology is more or less designed to be helpful for this. I'll explain an argument using the fundamental group $\pi_1$, but one could use other functors for the same purpose (homology being the most obvious alternative)</p>
<p>If a subspace $Y\subseteq X$ is a retract of $X$, there is a retraction $r:X\to Y$ such that the composition $r\circ i$ with the inclusion $i:Y\to X$ is the identity of $Y$. If we pick a base point $x_0\in Y$, then the inclusion map $i$ induces an homomorphism $\pi_1(i):\pi_1(Y,x_0)\to\pi_1(X,x_0)$ such that $$\pi_1(r)\circ\pi_1(i)=\pi_1(r\circ i)=\pi_1(\mathrm{id}_Y)=\mathrm{id}_{\pi_1(Y)}.$$ In particular, the map $\pi_1(i)$ is injective.</p>
<p>But for any choice of $x_0\in S^1$, the map $\pi_1(i):\pi_1(S^1,x_0)\to\pi_1(\mathbb R^2,x_0)$ is <em>not</em> injective. Indeed, $\pi_1(S^1,x_0)$ is a non-trivial group while $\pi_1(\mathbb R^2,x_0)$ is trivial.</p>
|
1,003,020 | <p>Without recourse to Dirichlet's theorem, of course.
We're going to go over the problems in class but I'd prefer to know the answer today.</p>
<p>Let $S = \{3n+2 \in \mathbb P: n \in \mathbb N_{\ge 1}\}$</p>
<p>edit:</p>
<p>The original question is "the set of all primes of the form $3n + 2$, but I was only considering odd primes because of a reason I don't remember anymore.</p>
<p>How can I show $S$ is infinite? I start by assuming it's finite. </p>
<p>I've tried: </p>
<p>Assuming that the product $(3n_1+2)(3n_2+2)...(3n_m+2)$ is of the form "something" so I can show the product contains a prime factor not in the finite list of primes, which would be a contradiction, but came up with no useful "something".</p>
<p>I tried showing that the product would not be square-free, but couldn't show that.</p>
<p>I tried showing the product or sum was both even and odd, but couldn't show that.</p>
<p>What else should I try? Or was one of the above methods correct?</p>
| user2345215 | 131,872 | <p>If $n$ in your question is even, the number can't be a prime. So it suffices to prove there are infinitely many primes of the form $6n-1$.</p>
<p>Assume there are only finitely many of them let $n_1\ldots,n_k$ be their representants. Then
$$6\bigl((6n_1-1)(6n_2-1)\ldots(6n_k-1)\bigr)-1$$
is a number of the form $6n-1$ which is not divisible by $2,3,$ and any of the primes of the form $6n-1$.</p>
<p>But all primes are either $2$, $3$, or of the form $6n\pm1$, so the number must be a product of primes of the form $6n+1$, but that's impossible because product of such numbers has again the form $6n+1$.</p>
|
1,895,721 | <p>How to find $3\times3$ matrices that satisfy the matrix equation $A^2=I_3$? Can anyone please show me steps to do this question?</p>
| Learnmore | 294,365 | <p>Note that $x^2+1=0$ is a annihilating polynomial of $A$. Now minimal polynomial of a matrix divides a annihilating polynomial.</p>
<p>So the possible choices are $x=-1,x=1,x^2+1$.</p>
<p>Hence the matrices in first two cases will be $A=I,A=-I$.</p>
<p>Consider the third case.Since the minimal polynomial and characteristic polynomial have the same roots so characteristic polynomial will be $(x-1)^2(x+1),(x+1)^2(x-1)$.</p>
<p>Hence the possible choices for $A$ will be :</p>
<p>\begin{bmatrix} 1&0 &0\\0&1&0\\0&0&-1\end{bmatrix}</p>
<p>or \begin{bmatrix} -1&0 &0\\0&-1&0\\0&0&1\end{bmatrix}</p>
|
1,103,478 | <p>$ r = 2\cos(\theta)$ has the graph<img src="https://i.stack.imgur.com/yvLb1.png" alt="enter image description here"></p>
<p>I want to know why the following integral to find area does not work $$\int_0^{2 \pi } \frac{1}{2} (2 \cos (\theta ))^2 \, d\theta$$</p>
<p>whereas this one does:</p>
<p>$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2} } \frac{1}{2} (2 \cos (\theta ))^2 \, d\theta$$</p>
<p>Why do the limits of integration have to go from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$? Doesn't going from $0$ to $2\pi$ also sweep out the full circle?</p>
| rlartiga | 93,314 | <p>Start with $\theta=0$ you get $r=2$ then move to $\theta=\frac{\pi}{2}$ you will get $r=0$ (then you have from here the upper half of the circle). From $\frac{\pi}{2}$ to $\pi$ you have a negative $r$ which don't have really much sense. Then you have to consider it from $\frac{3\pi}{2}$ to $2\pi$ to complete the lower half.</p>
<p>$$\int_{0}^{\frac{\pi}{2} } \frac{1}{2} (2 \cos (\theta ))^2 \, d\theta+\int_{\frac{3\pi}{2}}^{2\pi} \frac{1}{2} (2 \cos (\theta ))^2 \, d\theta=\int_{0}^{\frac{\pi}{2} } \frac{1}{2} (2 \cos (\theta ))^2 \, d\theta+\int_{-\frac{\pi}{2}}^{0} \frac{1}{2} (2 \cos (\theta ))^2 \, d\theta$$
$$=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{2} (2 \cos (\theta ))^2 \, d\theta$$</p>
|
3,532,173 | <p>I have seen this problem somewhere on the internet but I could not prove it.</p>
<p>Let <span class="math-container">$$I_{0}=\int^{\infty}_{0}\frac{\sin x}{x}dx$$</span> and then define
<span class="math-container">$$I_{n+1}=\int^{I_{n}}_{0}\frac{\sin x}{x}dx.$$</span></p>
<p>Show that
<span class="math-container">$$\lim_{n\rightarrow\infty}\sqrt{n}\ I_{n}=3.$$</span></p>
| approxolotl | 747,548 | <p>Denote the integral by <span class="math-container">$S(I_n)$</span></p>
<p><span class="math-container">$$
I_{n+1}+I_n=S_{n+1}+S_n\\
I_{n+1}-I_n=S_{n+1}-S_n \\
\rightarrow
2 I_{n}= 2 S_{n}
$$</span>
writing (Taylor expansion)
<span class="math-container">$$
S_{n}=I_{n-1}-\frac1{18}I_{n-1}^3+O(I_{n-1}^5)
$$</span></p>
<p>in the limit of big <span class="math-container">$n$</span> this leads to a "continuum approximation" (justification follows)</p>
<p><span class="math-container">$$
18\,\partial_nI_n\sim-I^3_n+O(I_{n}^5) \quad (\star)
$$</span></p>
<p>or (the constant of Integration can be neglected in the limit <span class="math-container">$n\rightarrow\infty$</span>)</p>
<blockquote>
<p><span class="math-container">$$
I_n \sim\frac{3}{\sqrt{n}} \quad (\star \star)
$$</span></p>
</blockquote>
<hr>
<p>Justification of <span class="math-container">$(\star)$</span>:</p>
<p>We assume that <span class="math-container">$I_n$</span> is in the form of <span class="math-container">$(\star \star)$</span>:</p>
<p><span class="math-container">$$
I^{(k)}_n\ll I^{(m)}_n\quad (1)
$$</span>
for <span class="math-container">$m<k$</span>, so we can approximate the finite difference by a continious derivative.</p>
<p><span class="math-container">$$
I_n-I_{n-1}\ll 1/\sqrt{n}\quad (2)
$$</span>
so we can replace <span class="math-container">$I_{n-1}$</span> by <span class="math-container">$I_n$</span>.</p>
<p><span class="math-container">$$
I^{k}_n\ll I^{m}_n\quad (3)
$$</span>
for <span class="math-container">$m<k$</span>, so we can neglect higher order Terms in the Taylor approximation to <span class="math-container">$S(I_n)$</span>.</p>
<p>Combining <span class="math-container">$(1)$</span> & <span class="math-container">$(2)$</span> & <span class="math-container">$(3)$</span> shows that our approximation <span class="math-container">$(\star)$</span> is consistent. </p>
|
156,013 | <p>I would like to <code>FoldList</code> a simple function, with desired output:</p>
<pre><code>f[a,b,1]
f[b,c,1]
f[c,a,1]
f[f[a,b,1],f[b,c,1],2]
f[f[b,c,1],f[c,a,1],2]
f[f[c,a,1],f[a,b,1],2]
f[f[f[a,b,1],f[b,c,1],2],f[f[b,c,1],f[c,a,1],2],3]
f[f[f[b,c,1],f[c,a,1],2],f[f[c,a,1],f[a,b,1],2],3]
f[f[f[c,a,1],f[a,b,1],2],f[f[a,b,1],f[b,c,1],2],3]
</code></pre>
<p>which I can hack with</p>
<pre><code>t2[li_] := Take[#, 2] &@RotateLeft[li, #] & /@ Range[0, Length@li - 1];
sf[n_] := "" <> ToString[# - 1] <> "], " <> ToString[# - 2] <> "]" ->
"" <> ToString[# - 1] <> "], " <> ToString[#] <> "]" &@n;
nl = ToExpression@With[{string = ToString /@
Rest[FoldList[f[#[[1]], #[[2]]] & /@ t2@# &, {a, b, c},
Range@#]]}, Fold[StringReplace, string,
Flatten[{{"]" -> ", 1]", "1], 1]" -> "1], 2]"},
sf /@ Range[3, #]}, 1]]] &[5];
Column[Column /@ Take[nl, 3]]
</code></pre>
<p>with huge inefficiencies for large <code>n</code> (there is no problem with the <code>t2</code> part of the function). All I want to do is add in indexing into the last position of <code>f[_,_,_]</code>.</p>
<p>I'd really like to do it with something simple like</p>
<pre><code>FoldList[f[#[[1]], #[[2]],#2] & /@ t2@# &, {a, b, c}, Range@3]
</code></pre>
<p>but I can't use <code>#2</code> since <code>MapThread</code> gets in the way. I know I am over-complicating things - what is a more sensible way to achieve the above?</p>
| Alucard | 18,859 | <p>here is another option, though it is not as good as the code provided by Carl </p>
<pre><code>t3[j_List, l_, m_] := Append[ Take[RotateLeft[j, l], 2], m]
h[p_List, i_ ] := f @@@ MapIndexed[ t3[p, #, i] & , Range[0, 2]]
FoldList[h, {a, b, c}, {1, 2, 3 }]
</code></pre>
|
3,443,226 | <blockquote>
<p>Prove that
<span class="math-container">$$
\int_{0}^{2\pi}\frac{d\theta}{(a+\cos\theta)^2}=\frac{2\pi a}{(a^2-1)^{\frac{3}{2}}}.
$$</span></p>
</blockquote>
<p>This is an exercise in Stein's <em>Complex Analysis</em>.</p>
<p>By letting <span class="math-container">$z=e^{i\theta}$</span>, we have
<span class="math-container">$$
\int_{0}^{2\pi}\frac{d\theta}{(a+\cos\theta)^2}=4\int_C\frac{dz}{iz(z+\bar z+2a)^2},
$$</span>
where <span class="math-container">$C$</span> is the unit circle.</p>
<p>I find that <span class="math-container">$z+\bar z+2a$</span> has no zero in the closure of the unit disc. So by the residual formula, I get
<span class="math-container">$$
\int_C\frac{dz}{iz(z+\bar z+2a)^2}=\frac{2\pi}{4a^{2}},
$$</span>
which is false.
Note that <span class="math-container">$\bar z$</span> is not a holomorphic function, so how to use the residual formula?</p>
| José Carlos Santos | 446,262 | <p>Let <span class="math-container">$R(x)=\frac1{(x+a)^2}$</span>. You want to compute <span class="math-container">$\int_0^{2\pi}R(x)\,\mathrm dx$</span>. Now, let<span class="math-container">$$g(z)=\frac1zR\left(\frac{z+z^{-1}}2\right)=\frac{4z}{\left(2az+z^2+1\right)^2}.$$</span>Then<span class="math-container">$$\int_0^{2\pi}R(x)\,\mathrm dx=i\int_{\lvert z\rvert=1}g(z)\,\mathrm dz$$</span>and, by the residue theorem, this is integral is the sum of the residues of <span class="math-container">$g$</span> in the open unit disk times <span class="math-container">$2\pi$</span>. Can you take it from here?</p>
|
166,666 | <p>For which values of the coefficient $c$ does the quantity
$$
\cos\alpha\cos\beta- c\sin\alpha\sin\beta
$$
depend on $\alpha$ and $\beta$ only through their sum?</p>
<p>(I'll post a quick answer below. This will be the first time I've posted a question with intent to immediately post an answer.)</p>
| celtschk | 34,930 | <p>Well, let's define the new quantities $s=\frac{\alpha+\beta}{2}$ and $d=\frac{\alpha-\beta}{2}$. With those quantities the expression reads $$E:=\cos(s+d)\cos(s-d)-c\sin(s+d)\sin(s-d).$$ The question now is: For which values of $c$ is this expression independent from $d$?</p>
<p>Let's apply the standard addition theorems to get
$$E =(\sin s\cos d+\cos s\sin d)(\sin s\cos d-\cos s\sin d)
-c(\cos s\cos d-\sin s\sin d)(\cos s\cos d+\sin s\sin d).$$</p>
<p>Using $(a+b)(a-b) = a^2-b^2$, we therefore get $$E=\sin^2 s\,\cos^2 d-\cos^2s\, \sin^2d - c(\cos^2 s\,\cos^2 d-\sin^2 s\,\sin^2 d).$$</p>
<p>Now we collect the functions of $s$ to get $$E = \sin^2 s(\cos^2 d+c\,\sin^2 d) - \cos^2s(\sin^2 d+c\,\cos^2 d).$$</p>
<p>Now it is easy to see that the only possibility that $E$ is independent from $d$ (and therefore in the original form depends only on the sum) is $c=1$, where $\sin^2 d+\cos^2d=1$</p>
|
2,767,392 | <p>I have the following curve:</p>
<p>$x^4=a^2(x^2-y^2)$</p>
<p>Prove that the area of its loop is $\frac{2a^2}{3}$.</p>
<p><strong>My approach</strong></p>
<p>This curve has four loops. So the required area should be:</p>
<p>$4\int_{0}^{a}\frac{x}{a}\sqrt{a^2-x^2} dx$</p>
<p>But, After solving, the area turned out to be $\frac{4a^2}{3}$. What am I doing wrong here?</p>
<p>Thanks</p>
| epi163sqrt | 132,007 | <p>Here is answer where we follow the advice in the solution and use $(1-x)^n$ and $\frac{1}{1-x}$, but <em>inside out</em>. It is convenient to use the <em>coefficient of</em> operator $[x^k]$ to denote the coefficient of $x^k$ in a series. This way we can write for instance
\begin{align*}
[x^k](1+x)^n=\binom{n}{k}\tag{1}
\end{align*}</p>
<blockquote>
<p>We obtain for non-negative integers $n,k$:
\begin{align*}
\color{blue}{\sum_{j=0}^k\binom{n}{j}(-1)^j}&=\sum_{j=0}^k[x^j](1-x)^n\tag{2}\\
&=\sum_{j=0}^{k}[x^{k-j}](1-x)^n\tag{3}\\
&=[x^k](1-x)^n\sum_{j=0}^kx^j\tag{4}\\
&=[x^k](1-x)^n\frac{1-x^{k+1}}{1-x}\tag{5}\\
&=[x^k](1-x)^{n-1}\tag{6}\\
&\,\,\color{blue}{=(-1)^k\binom{n-1}{k}}\tag{7}
\end{align*}
and the claim follows.</p>
</blockquote>
<p><em>Comment:</em></p>
<ul>
<li><p>In (2) we use the <em>coefficient of</em> operator according to (1).</p></li>
<li><p>In (3) we change the order of summation for convenience by letting $j\to k-j$.</p></li>
<li><p>In (4) we use the linearity of the <em>coefficient of</em> operator and apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$.</p></li>
<li><p>In (5) we apply the <em><a href="https://en.wikipedia.org/wiki/Geometric_series#Formula" rel="nofollow noreferrer">finite geometric series formula</a></em>.</p></li>
<li><p>In (6) we note that the term $x^{k+1}$ do not contribute to $[x^k]$ and we can simplify accordingly.</p></li>
<li><p>In (7) we select the coefficient of $x^k$.</p></li>
</ul>
<p><em>Hint:</em> Note that $\frac{1}{1-x}=1+x+x^2+\cdots$.</p>
|
811,535 | <p>I want to show that every prime power $p^k$ that divides $\binom{2m}{m}$ is smaller than or equal to $2m$.</p>
<p>As a first step, I looked at
$$\binom{2m}{m}
= \frac{(2m)!}{(m!)^2}
= \frac{2m(2m-1) \ldots (m+2)(m+1)}{m!} \, .$$
Here I'm essentially stuck. I can apply the prime factorization to numerator and denominator, then I can cancel and I know that $p^k$ is left over in the numerator. But I cannot conclude $p^k \leq 2m$.</p>
<p>I feel that some vital ingredient is missing here, but I don't know what it is.</p>
<p>(Post edited with respect to the helpful comment.)</p>
| Shaun | 104,041 | <p><strong>Hint:</strong> Let $\alpha_{n, p}$ be the largest natural number such that $p^{\alpha_{n, p}}\mid n!$ for $n\in\mathbb{N}$ and $p$ prime. Then $$\alpha_{n, p}=\sum_{k=1}^{\infty}{\left[\frac{n}{p^k}\right]},$$ where $[x]$ is the integer part of $x$.</p>
|
866,144 | <p>Prove that
$$1-2^{-x}\geq \frac{\sqrt 2}{2}\sin\left(\frac{\pi}{4} x\right)$$
for $x\in[0,1]$.
Any suggestions please?</p>
| AsdrubalBeltran | 62,547 | <p>If $$F(x)=\frac{\sqrt{2}}{2}\sin\left(\frac{\pi}{4} x\right)+2^{-x}$$</p>
<p>$F(x)>0$.
How $F(0)=F(1)=1$, then by value mean theorem there is a $x_0\in [0,1]$ such that $F'(x_0)=0$
$$F'(x)=\frac{\pi\sqrt{2}}{8}\cos\left(\frac{\pi}{4} x\right)-2^{-x}\ln(2)$$
Note that $F'(0)<0$ and $F'(1)>0$, also $F'''(x)<0, \forall x\in [0,1]$ then $F'(0)$ is concave up, then there is a only $x_0$ such that $F'(x_0)=0$. Then $F(x_0)$ is a maximun or is a minimun in $[0,1]$, how $F(\frac{2}{3})<1$ then $F(x_0)$ is a minimun, then if $x\in[0,1]$:</p>
<p>$$F(x)\leq1$$
$$\frac{\sqrt{2}}{2}\sin\left(\frac{\pi}{4} x\right)+2^{-x}\leq1$$
$$\frac{\sqrt{2}}{2}\sin\left(\frac{\pi}{4} x\right)\leq1-2^{-x}$$</p>
|
131,435 | <p>Wikipedia is a widely used resource for mathematics. For example, there are hundreds of mathematics articles that average over 1000 page views per day. <a href="http://en.wikipedia.org/wiki/Wikipedia:WikiProject_Mathematics/Popular_pages" rel="noreferrer">Here is a list of the 500 most popular math articles</a>. The number of regular Wikipedia readers is increasing, while the number of editors is decreasing (<a href="http://stats.wikimedia.org/EN/SummaryEN.htm" rel="noreferrer">graphs</a>), which is causing growing concern within the Wikipedia community. </p>
<p><a href="http://en.wikipedia.org/wiki/Wikipedia_talk:WikiProject_Mathematics" rel="noreferrer">WikiProject Mathematics</a> is relatively active (compared to other WikiProjects, but not compared to MathOverflow!), and there is always the need for more experts who really understand the material. An editor continually faces the tension between (1) providing a lot of advanced material and (2) explaining things well, which generates many productive discussions about how mathematics articles should be written, and which topics should have their own article. </p>
<p>Regardless of the long term concerns raised about whether Wikipedia is capable of being a resource for advanced mathematics (see <a href="https://mathoverflow.net/questions/19631/can-wikipedia-be-a-reliable-and-sustainable-resource-for-advanced-mathematics">this closed question</a>), the fact is, people are attempting to learn from Wikipedia's mathematics articles right now. So improvements made to articles today will benefit the readers of tomorrow. </p>
<hr>
<p>Wikipedia is a very satisfying venue for summarizing topics you know well, and explaining things to other people, due to its large readership. Based on the number of mathematicians at MathOverflow, who are willing to spend time (for free!) answering questions and clarifying subjects for other people, it seems like there is a lot of untapped volunteer potential here. So in the interests of exposing the possible obstacles to joining Wikipedia, I would like to know:</p>
<blockquote>
<p>Why don't mathematicians spend more time improving Wikipedia articles?</p>
</blockquote>
<p>Recent efforts intended to attract new participants and keep existing ones include the friendly atmosphere of the <a href="http://en.wikipedia.org/wiki/Wikipedia:Teahouse" rel="noreferrer">Teahouse</a>, as well as WikiProject Editor Retention. </p>
<p>In case anyone is interested, my Wikipedia username is User:Mark L MacDonald (which is my real name). </p>
| Alexander Chervov | 10,446 | <p>+1 for question. Hope it will stimulate us to contribute more, it would be great. I feel great respect to those who contribute to Wikipedia. My experience about Wikipedia and its quality is highly positive, probably one of the reasons, is that many many articles are written by one of the Fields Medalists. </p>
<p>Personally I have contributed some material to Wikipedia, the main reason that my contribution is not so big - is lack of time. I feel somewhat ashamed about it, and some excuse which I see for myself - is that I am currently in industry and cannot enjoy academic freedom.</p>
<p>I would also be second on Jérémy Blanc's words: "One difference between MO and wikipedia is that here you get very quickly answers or remarks on what you have written, and it is then very much more attractive..."
Wikipedia pages have view-counters, which somehow indicates how useful is what you have done,
but it is not clear what exactly is counted ? I guess some views come from robots or other automatic engines.
The page about <a href="http://en.wikipedia.org/wiki/Capelli's_identity" rel="nofollow noreferrer">Capelli identity</a> (which is mostly written by me) is visited about 5 times per day according to <a href="http://stats.grok.se/en/latest/Capelli%27s_identity" rel="nofollow noreferrer">http://stats.grok.se/en/latest/Capelli%27s_identity</a>
Are there really so many people who know what is it ? :)))
Although, it is not very important for me, but just having more feedback would be nicer.</p>
<p>Let me also mention that when I was in academia, I did not see big use of Wikipedia for me - when you are deep in one field Wikipedia cannot help you.
But when I switched into industry and I often got a new task about things which I know nothing about - then Wikipedia became extremely useful.
Moreover I see that for hundreds of engineers Wikipedia is main window to mathematical knowledge. </p>
<p>I am probably not the only person who finds that many math textbooks are quite boring
<a href="https://mathoverflow.net/questions/13089/why-do-so-many-textbooks-have-so-much-technical-detail-and-so-little-enlightenmen">Why do so many textbooks have so much technical detail and so little enlightenment?</a>.
In my experience Wikipedia is an exception.
So currently my feeling is that
If I need to understand some new topic or an idea - I will look to Wikipedia, if it is described there - then it is possible to learn the subj. <strong>If something is missing on Wikipedia - no chance to understand an idea in a reasonable amount of time.</strong></p>
|
597,949 | <p>$\begin{cases}x\equiv 1 \pmod{3}\\
x\equiv 2 \pmod{5}\\
x\equiv 3 \pmod{7}\\
x\equiv 4 \pmod{9}\\
x\equiv 5 \pmod{11}\end{cases}$ </p>
<p>I am supposed to solve the system using the Chinese remainder theorem but $(3,5,7,9,11)\neq 1$
How can I transform the system so that I will be able to use the theorem?</p>
| lab bhattacharjee | 33,337 | <p>Observe that $\displaystyle x\equiv4\pmod 9\implies x\equiv4\pmod3\equiv1$</p>
<p>Now, we can safely apply <a href="http://www.cut-the-knot.org/blue/chinese.shtml" rel="nofollow">C.R.T</a> on $$\begin{cases}
x\equiv 2 \pmod5\\
x\equiv 3 \pmod7\\
x\equiv 4 \pmod9\\
x\equiv 5 \pmod{11}\end{cases}$$ as $5,7,9,11$ are pairwise relatively prime</p>
|
2,156,606 | <p>I am stuck in this exercise of calculus about solving this indefinite integral, so I would like some help from your part: </p>
<blockquote>
<p>$$\int \frac{dx}{(1+x^{2})^{\frac{3}{2}}}$$</p>
</blockquote>
| R.W | 253,359 | <p><strong>Hint 1:</strong> $\cosh^2(x)-\sinh^2(x)=1$ and use change of variables.</p>
<p><strong>Hint 2:</strong> </p>
<blockquote>
<p>$$\frac{\mathrm{d}}{\mathrm{d}x}\left(\tanh(x)\right) =\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\sinh(x)}{\cosh(x)}\right) = \frac{1}{\cosh^2(x)}\left(\cosh^2(x)-\sinh^2(x)\right) = \text{sech}^2(x)$$</p>
</blockquote>
|
354,885 | <p>Let <span class="math-container">$X_1,...,X_n$</span> be iid normal random variables. </p>
<p>I am looking for a strategy to establish the following limit for fraction of expectation values</p>
<p><span class="math-container">$$\lim_{N \rightarrow \infty} \frac{E(\prod_{1\le i < j\le n} \vert X_i-X_j \vert^{1/n})}{E(\prod_{1\le i < j\le n-1} \vert X_i-X_j \vert^{1/n})}=1.$$</span></p>
<p>Does anybody have any ideas what to use for this limit?</p>
| Iosif Pinelis | 36,721 | <p>The <a href="https://en.wikipedia.org/wiki/Selberg_integral#Mehta's_integral" rel="nofollow noreferrer">Mehta integral</a> is
<span class="math-container">$$M_n(\gamma):=E\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma}
=\prod_{j=1}^n\frac{\Gamma(1+j\gamma)}{\Gamma(1+\gamma)}.$$</span>
So, your fraction under the limit sign is
<span class="math-container">$$\frac{M_n(1/(2n))}{M_{n-1}(1/(2n)}=\frac{\Gamma(3/2)}{\Gamma(1+1/(2n))}\to\Gamma(3/2)\approx0.886227.$$</span></p>
|
566 | <h3>We all love a good puzzle</h3>
<p>To a certain extent, any piece of mathematics is a puzzle in some sense: whether we are classifying the homological intersection forms of four manifolds or calculating the optimum dimensions of a cylinder, it is an element of investigation and inherently puzzlish intrigue that drives us. Indeed most puzzles (cryptic crosswords aside) are somewhat mathematical (the mathematics of Sudoku for example is hidden in <a href="http://en.wikipedia.org/wiki/Latin_squares" rel="nofollow noreferrer">latin squares</a>). Mathematicians and puzzles get on, it seems, rather well.</p>
<h3>But what is a good puzzle?</h3>
<p>Okay, so in order to make this question worthwhile (and not a ten-page wade-athon through 57 varieties of the men with red and blue hats puzzle), we are going to have to impose some limitations. Not every puzzle-based answer that pops into your head will qualify as a good puzzle—to do so it must</p>
<ul>
<li><strong>Not be widely known:</strong> If you have a terribly interesting puzzle that motivates something in cryptography; well done you, but chances are we've seen it. If you saw that <em>hilarious</em> scene in the film 21, where Kevin Spacey explains the Monty hall paradox badly and want to share, don't do so here. Anyone found posting the liar/truth teller riddle will be immediately disembowelled.</li>
<li><strong>Be mathematical (as much as possible):</strong> It's true that logic <em>is</em> mathematics, but puzzles beginning '<em>There is a street where everyone has a different coloured house …</em>' are tedious as hell. Note: there is a happy medium between this and trig substitutions.</li>
<li><strong>Not be too hard:</strong> Any level of difficulty is cool, but if coming up with an answer requires more than two sublemmas, you are misreading your audience.</li>
<li><strong>Actually have an answer:</strong> Crank questions will not be appreciated! You can post the answers/hints in <a href="http://en.wikipedia.org/wiki/Rot_13" rel="nofollow noreferrer">Rot-13</a> underneath as comments as on MO if you fancy.</li>
<li><strong>Have that indefinable spark that makes a puzzle awesome:</strong> Like a situation that seems familiar, requiring unfamiliar thought …</li>
</ul>
<p>Ideally include where you found the puzzle so we can find more cool stuff like it. For ease of voting, one puzzle per post is best.</p>
<hr />
<h1>Some examples to set the ball rolling</h1>
<blockquote>
<p>Simplify <span class="math-container">$\sqrt{2+\sqrt{3}}$</span></p>
</blockquote>
<p><strong>From:</strong> problem solving magazine</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> Try a two term solution</p>
</blockquote>
<hr />
<blockquote>
<p>Can one make an equilateral triangle with all vertices at integer coordinates?</p>
</blockquote>
<p><strong>From:</strong> Durham distance maths challenge 2010</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> This is equivalent to the rational case</p>
</blockquote>
<hr />
<blockquote>
<p>The collection of <span class="math-container">$n \times n$</span> <a href="http://en.wikipedia.org/wiki/Magic_squares" rel="nofollow noreferrer">Magic squares</a> form a vector space over <span class="math-container">$\mathbb{R}$</span> prove this, and by way of a linear transformation, derive the dimension of this vector space.</p>
</blockquote>
<p><strong>From:</strong> Me, I made this up (you can tell, can't you!)</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> Apply the rank nullity theorem</p>
</blockquote>
| BlueRaja - Danny Pflughoeft | 136 | <p><strong>Frk n th rd 1</strong></p>
<p>Y'r n pth n n slnd, cme t frk n th rd. Bth pths ld t vllgs f ntvs; th ntr vllg thr lwys tlls th trth r lwys ls <em>(bth villgs cld b trth-tllng r lyng vllgs, r n f ch)</em>. Thr r tw ntvs t th frk - thy cld bth b frm th sm vllg, r frm dffrnt vllgs <em>(s bth cld b trth-tllrs, both lrs, r ne f ch)</em>.</p>
<p>n pth lds t sfty, th thr t dm. Y'r llwd t sk nly n qstn t ch ntv t fgr t whch pth s whch.</p>
<p>Wht d y sk?</p>
|
566 | <h3>We all love a good puzzle</h3>
<p>To a certain extent, any piece of mathematics is a puzzle in some sense: whether we are classifying the homological intersection forms of four manifolds or calculating the optimum dimensions of a cylinder, it is an element of investigation and inherently puzzlish intrigue that drives us. Indeed most puzzles (cryptic crosswords aside) are somewhat mathematical (the mathematics of Sudoku for example is hidden in <a href="http://en.wikipedia.org/wiki/Latin_squares" rel="nofollow noreferrer">latin squares</a>). Mathematicians and puzzles get on, it seems, rather well.</p>
<h3>But what is a good puzzle?</h3>
<p>Okay, so in order to make this question worthwhile (and not a ten-page wade-athon through 57 varieties of the men with red and blue hats puzzle), we are going to have to impose some limitations. Not every puzzle-based answer that pops into your head will qualify as a good puzzle—to do so it must</p>
<ul>
<li><strong>Not be widely known:</strong> If you have a terribly interesting puzzle that motivates something in cryptography; well done you, but chances are we've seen it. If you saw that <em>hilarious</em> scene in the film 21, where Kevin Spacey explains the Monty hall paradox badly and want to share, don't do so here. Anyone found posting the liar/truth teller riddle will be immediately disembowelled.</li>
<li><strong>Be mathematical (as much as possible):</strong> It's true that logic <em>is</em> mathematics, but puzzles beginning '<em>There is a street where everyone has a different coloured house …</em>' are tedious as hell. Note: there is a happy medium between this and trig substitutions.</li>
<li><strong>Not be too hard:</strong> Any level of difficulty is cool, but if coming up with an answer requires more than two sublemmas, you are misreading your audience.</li>
<li><strong>Actually have an answer:</strong> Crank questions will not be appreciated! You can post the answers/hints in <a href="http://en.wikipedia.org/wiki/Rot_13" rel="nofollow noreferrer">Rot-13</a> underneath as comments as on MO if you fancy.</li>
<li><strong>Have that indefinable spark that makes a puzzle awesome:</strong> Like a situation that seems familiar, requiring unfamiliar thought …</li>
</ul>
<p>Ideally include where you found the puzzle so we can find more cool stuff like it. For ease of voting, one puzzle per post is best.</p>
<hr />
<h1>Some examples to set the ball rolling</h1>
<blockquote>
<p>Simplify <span class="math-container">$\sqrt{2+\sqrt{3}}$</span></p>
</blockquote>
<p><strong>From:</strong> problem solving magazine</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> Try a two term solution</p>
</blockquote>
<hr />
<blockquote>
<p>Can one make an equilateral triangle with all vertices at integer coordinates?</p>
</blockquote>
<p><strong>From:</strong> Durham distance maths challenge 2010</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> This is equivalent to the rational case</p>
</blockquote>
<hr />
<blockquote>
<p>The collection of <span class="math-container">$n \times n$</span> <a href="http://en.wikipedia.org/wiki/Magic_squares" rel="nofollow noreferrer">Magic squares</a> form a vector space over <span class="math-container">$\mathbb{R}$</span> prove this, and by way of a linear transformation, derive the dimension of this vector space.</p>
</blockquote>
<p><strong>From:</strong> Me, I made this up (you can tell, can't you!)</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> Apply the rank nullity theorem</p>
</blockquote>
| BlueRaja - Danny Pflughoeft | 136 | <p><strong>Fork in the road 2</strong></p>
<p>You're once again at a fork in the road, and again, one path leads to safety, the other to doom.</p>
<p>There are three natives at the fork. One is from a village of truth-tellers, one from a village of liars, one from a village of random answerers. Of course you don't know which is which.</p>
<p>Moreover, the natives answer "pish" and "posh" for yes and no, but you don't know which means "yes" and which means "no."</p>
<p>You're allowed to ask only two yes-or-no questions, each question being directed at one native.</p>
<p>What do you ask?</p>
|
566 | <h3>We all love a good puzzle</h3>
<p>To a certain extent, any piece of mathematics is a puzzle in some sense: whether we are classifying the homological intersection forms of four manifolds or calculating the optimum dimensions of a cylinder, it is an element of investigation and inherently puzzlish intrigue that drives us. Indeed most puzzles (cryptic crosswords aside) are somewhat mathematical (the mathematics of Sudoku for example is hidden in <a href="http://en.wikipedia.org/wiki/Latin_squares" rel="nofollow noreferrer">latin squares</a>). Mathematicians and puzzles get on, it seems, rather well.</p>
<h3>But what is a good puzzle?</h3>
<p>Okay, so in order to make this question worthwhile (and not a ten-page wade-athon through 57 varieties of the men with red and blue hats puzzle), we are going to have to impose some limitations. Not every puzzle-based answer that pops into your head will qualify as a good puzzle—to do so it must</p>
<ul>
<li><strong>Not be widely known:</strong> If you have a terribly interesting puzzle that motivates something in cryptography; well done you, but chances are we've seen it. If you saw that <em>hilarious</em> scene in the film 21, where Kevin Spacey explains the Monty hall paradox badly and want to share, don't do so here. Anyone found posting the liar/truth teller riddle will be immediately disembowelled.</li>
<li><strong>Be mathematical (as much as possible):</strong> It's true that logic <em>is</em> mathematics, but puzzles beginning '<em>There is a street where everyone has a different coloured house …</em>' are tedious as hell. Note: there is a happy medium between this and trig substitutions.</li>
<li><strong>Not be too hard:</strong> Any level of difficulty is cool, but if coming up with an answer requires more than two sublemmas, you are misreading your audience.</li>
<li><strong>Actually have an answer:</strong> Crank questions will not be appreciated! You can post the answers/hints in <a href="http://en.wikipedia.org/wiki/Rot_13" rel="nofollow noreferrer">Rot-13</a> underneath as comments as on MO if you fancy.</li>
<li><strong>Have that indefinable spark that makes a puzzle awesome:</strong> Like a situation that seems familiar, requiring unfamiliar thought …</li>
</ul>
<p>Ideally include where you found the puzzle so we can find more cool stuff like it. For ease of voting, one puzzle per post is best.</p>
<hr />
<h1>Some examples to set the ball rolling</h1>
<blockquote>
<p>Simplify <span class="math-container">$\sqrt{2+\sqrt{3}}$</span></p>
</blockquote>
<p><strong>From:</strong> problem solving magazine</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> Try a two term solution</p>
</blockquote>
<hr />
<blockquote>
<p>Can one make an equilateral triangle with all vertices at integer coordinates?</p>
</blockquote>
<p><strong>From:</strong> Durham distance maths challenge 2010</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> This is equivalent to the rational case</p>
</blockquote>
<hr />
<blockquote>
<p>The collection of <span class="math-container">$n \times n$</span> <a href="http://en.wikipedia.org/wiki/Magic_squares" rel="nofollow noreferrer">Magic squares</a> form a vector space over <span class="math-container">$\mathbb{R}$</span> prove this, and by way of a linear transformation, derive the dimension of this vector space.</p>
</blockquote>
<p><strong>From:</strong> Me, I made this up (you can tell, can't you!)</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> Apply the rank nullity theorem</p>
</blockquote>
| Ross Millikan | 1,827 | <p>From New Scientist some years ago: 20 teams play a round robin tournament, each gets 1 point for a win, 0 for a loss, and there are no ties. Each team's score is a square number. How many upsets occurred? An upset is defined as team A defeating team B where B scored more total points than A.</p>
|
566 | <h3>We all love a good puzzle</h3>
<p>To a certain extent, any piece of mathematics is a puzzle in some sense: whether we are classifying the homological intersection forms of four manifolds or calculating the optimum dimensions of a cylinder, it is an element of investigation and inherently puzzlish intrigue that drives us. Indeed most puzzles (cryptic crosswords aside) are somewhat mathematical (the mathematics of Sudoku for example is hidden in <a href="http://en.wikipedia.org/wiki/Latin_squares" rel="nofollow noreferrer">latin squares</a>). Mathematicians and puzzles get on, it seems, rather well.</p>
<h3>But what is a good puzzle?</h3>
<p>Okay, so in order to make this question worthwhile (and not a ten-page wade-athon through 57 varieties of the men with red and blue hats puzzle), we are going to have to impose some limitations. Not every puzzle-based answer that pops into your head will qualify as a good puzzle—to do so it must</p>
<ul>
<li><strong>Not be widely known:</strong> If you have a terribly interesting puzzle that motivates something in cryptography; well done you, but chances are we've seen it. If you saw that <em>hilarious</em> scene in the film 21, where Kevin Spacey explains the Monty hall paradox badly and want to share, don't do so here. Anyone found posting the liar/truth teller riddle will be immediately disembowelled.</li>
<li><strong>Be mathematical (as much as possible):</strong> It's true that logic <em>is</em> mathematics, but puzzles beginning '<em>There is a street where everyone has a different coloured house …</em>' are tedious as hell. Note: there is a happy medium between this and trig substitutions.</li>
<li><strong>Not be too hard:</strong> Any level of difficulty is cool, but if coming up with an answer requires more than two sublemmas, you are misreading your audience.</li>
<li><strong>Actually have an answer:</strong> Crank questions will not be appreciated! You can post the answers/hints in <a href="http://en.wikipedia.org/wiki/Rot_13" rel="nofollow noreferrer">Rot-13</a> underneath as comments as on MO if you fancy.</li>
<li><strong>Have that indefinable spark that makes a puzzle awesome:</strong> Like a situation that seems familiar, requiring unfamiliar thought …</li>
</ul>
<p>Ideally include where you found the puzzle so we can find more cool stuff like it. For ease of voting, one puzzle per post is best.</p>
<hr />
<h1>Some examples to set the ball rolling</h1>
<blockquote>
<p>Simplify <span class="math-container">$\sqrt{2+\sqrt{3}}$</span></p>
</blockquote>
<p><strong>From:</strong> problem solving magazine</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> Try a two term solution</p>
</blockquote>
<hr />
<blockquote>
<p>Can one make an equilateral triangle with all vertices at integer coordinates?</p>
</blockquote>
<p><strong>From:</strong> Durham distance maths challenge 2010</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> This is equivalent to the rational case</p>
</blockquote>
<hr />
<blockquote>
<p>The collection of <span class="math-container">$n \times n$</span> <a href="http://en.wikipedia.org/wiki/Magic_squares" rel="nofollow noreferrer">Magic squares</a> form a vector space over <span class="math-container">$\mathbb{R}$</span> prove this, and by way of a linear transformation, derive the dimension of this vector space.</p>
</blockquote>
<p><strong>From:</strong> Me, I made this up (you can tell, can't you!)</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> Apply the rank nullity theorem</p>
</blockquote>
| Henry | 6,460 | <p>You are on the surface of a cube, starting at the midpoint of one of the edges. Which point(s) on the cube is furthest away from you if you are constrained to travel on the surface of the cube? </p>
|
566 | <h3>We all love a good puzzle</h3>
<p>To a certain extent, any piece of mathematics is a puzzle in some sense: whether we are classifying the homological intersection forms of four manifolds or calculating the optimum dimensions of a cylinder, it is an element of investigation and inherently puzzlish intrigue that drives us. Indeed most puzzles (cryptic crosswords aside) are somewhat mathematical (the mathematics of Sudoku for example is hidden in <a href="http://en.wikipedia.org/wiki/Latin_squares" rel="nofollow noreferrer">latin squares</a>). Mathematicians and puzzles get on, it seems, rather well.</p>
<h3>But what is a good puzzle?</h3>
<p>Okay, so in order to make this question worthwhile (and not a ten-page wade-athon through 57 varieties of the men with red and blue hats puzzle), we are going to have to impose some limitations. Not every puzzle-based answer that pops into your head will qualify as a good puzzle—to do so it must</p>
<ul>
<li><strong>Not be widely known:</strong> If you have a terribly interesting puzzle that motivates something in cryptography; well done you, but chances are we've seen it. If you saw that <em>hilarious</em> scene in the film 21, where Kevin Spacey explains the Monty hall paradox badly and want to share, don't do so here. Anyone found posting the liar/truth teller riddle will be immediately disembowelled.</li>
<li><strong>Be mathematical (as much as possible):</strong> It's true that logic <em>is</em> mathematics, but puzzles beginning '<em>There is a street where everyone has a different coloured house …</em>' are tedious as hell. Note: there is a happy medium between this and trig substitutions.</li>
<li><strong>Not be too hard:</strong> Any level of difficulty is cool, but if coming up with an answer requires more than two sublemmas, you are misreading your audience.</li>
<li><strong>Actually have an answer:</strong> Crank questions will not be appreciated! You can post the answers/hints in <a href="http://en.wikipedia.org/wiki/Rot_13" rel="nofollow noreferrer">Rot-13</a> underneath as comments as on MO if you fancy.</li>
<li><strong>Have that indefinable spark that makes a puzzle awesome:</strong> Like a situation that seems familiar, requiring unfamiliar thought …</li>
</ul>
<p>Ideally include where you found the puzzle so we can find more cool stuff like it. For ease of voting, one puzzle per post is best.</p>
<hr />
<h1>Some examples to set the ball rolling</h1>
<blockquote>
<p>Simplify <span class="math-container">$\sqrt{2+\sqrt{3}}$</span></p>
</blockquote>
<p><strong>From:</strong> problem solving magazine</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> Try a two term solution</p>
</blockquote>
<hr />
<blockquote>
<p>Can one make an equilateral triangle with all vertices at integer coordinates?</p>
</blockquote>
<p><strong>From:</strong> Durham distance maths challenge 2010</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> This is equivalent to the rational case</p>
</blockquote>
<hr />
<blockquote>
<p>The collection of <span class="math-container">$n \times n$</span> <a href="http://en.wikipedia.org/wiki/Magic_squares" rel="nofollow noreferrer">Magic squares</a> form a vector space over <span class="math-container">$\mathbb{R}$</span> prove this, and by way of a linear transformation, derive the dimension of this vector space.</p>
</blockquote>
<p><strong>From:</strong> Me, I made this up (you can tell, can't you!)</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> Apply the rank nullity theorem</p>
</blockquote>
| MJD | 25,554 | <p>There is a square table with a pocket at each corner; in each pocket is a drinking glass, which you cannot see. Each glass might be right-side up ("up") or upside-down ("down").</p>
<p>You and an adversary will play the following game. You select exactly two of the pockets, withdraw the two glasses, thus learning their orientations. You then replace the glasses in their pockets, altering their orientations in any way you desire.</p>
<p>If at this point, all four glasses are oriented the same way, you win.</p>
<p>Otherwise, you look away, and the adversary rotates the table. All the pockets are indistinguishable, so you cannot tell what multiple of $\frac\pi2$ the table has been rotated.</p>
<p>Provide a strategy that is guaranteed to win in bounded time.</p>
|
566 | <h3>We all love a good puzzle</h3>
<p>To a certain extent, any piece of mathematics is a puzzle in some sense: whether we are classifying the homological intersection forms of four manifolds or calculating the optimum dimensions of a cylinder, it is an element of investigation and inherently puzzlish intrigue that drives us. Indeed most puzzles (cryptic crosswords aside) are somewhat mathematical (the mathematics of Sudoku for example is hidden in <a href="http://en.wikipedia.org/wiki/Latin_squares" rel="nofollow noreferrer">latin squares</a>). Mathematicians and puzzles get on, it seems, rather well.</p>
<h3>But what is a good puzzle?</h3>
<p>Okay, so in order to make this question worthwhile (and not a ten-page wade-athon through 57 varieties of the men with red and blue hats puzzle), we are going to have to impose some limitations. Not every puzzle-based answer that pops into your head will qualify as a good puzzle—to do so it must</p>
<ul>
<li><strong>Not be widely known:</strong> If you have a terribly interesting puzzle that motivates something in cryptography; well done you, but chances are we've seen it. If you saw that <em>hilarious</em> scene in the film 21, where Kevin Spacey explains the Monty hall paradox badly and want to share, don't do so here. Anyone found posting the liar/truth teller riddle will be immediately disembowelled.</li>
<li><strong>Be mathematical (as much as possible):</strong> It's true that logic <em>is</em> mathematics, but puzzles beginning '<em>There is a street where everyone has a different coloured house …</em>' are tedious as hell. Note: there is a happy medium between this and trig substitutions.</li>
<li><strong>Not be too hard:</strong> Any level of difficulty is cool, but if coming up with an answer requires more than two sublemmas, you are misreading your audience.</li>
<li><strong>Actually have an answer:</strong> Crank questions will not be appreciated! You can post the answers/hints in <a href="http://en.wikipedia.org/wiki/Rot_13" rel="nofollow noreferrer">Rot-13</a> underneath as comments as on MO if you fancy.</li>
<li><strong>Have that indefinable spark that makes a puzzle awesome:</strong> Like a situation that seems familiar, requiring unfamiliar thought …</li>
</ul>
<p>Ideally include where you found the puzzle so we can find more cool stuff like it. For ease of voting, one puzzle per post is best.</p>
<hr />
<h1>Some examples to set the ball rolling</h1>
<blockquote>
<p>Simplify <span class="math-container">$\sqrt{2+\sqrt{3}}$</span></p>
</blockquote>
<p><strong>From:</strong> problem solving magazine</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> Try a two term solution</p>
</blockquote>
<hr />
<blockquote>
<p>Can one make an equilateral triangle with all vertices at integer coordinates?</p>
</blockquote>
<p><strong>From:</strong> Durham distance maths challenge 2010</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> This is equivalent to the rational case</p>
</blockquote>
<hr />
<blockquote>
<p>The collection of <span class="math-container">$n \times n$</span> <a href="http://en.wikipedia.org/wiki/Magic_squares" rel="nofollow noreferrer">Magic squares</a> form a vector space over <span class="math-container">$\mathbb{R}$</span> prove this, and by way of a linear transformation, derive the dimension of this vector space.</p>
</blockquote>
<p><strong>From:</strong> Me, I made this up (you can tell, can't you!)</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> Apply the rank nullity theorem</p>
</blockquote>
| Asaf Karagila | 622 | <p><em>Not a mathematical puzzle in the classical sense, but it is an interesting variation on the men with hats problem. It has a much "quicker" solution which to some degree is a bit surprising. If you are willing to indulge me, I will also add a rich background story.</em></p>
<p>After Hilbert passed away in 1943 his grand hotel stood empty for some years until it was bought by a mad version of Stalin and was turned into a crazy prison for the infinitely many enemies of the state.</p>
<p>Several days later the prison is full, and Stalin being less-mathematically inclined than Hilbert decides that instead of moving all the prisoners he will execute them. Since mathematicians <em>can</em> be useful to the Soviet Union he decides to play a game.</p>
<p>He announces that in the morning of the next day every prison will be given a hat either black or white (but not both), and the prisoners will be standing in line by their room number, each seeing all those whose room number is larger. Without talking to each other they will have to guess whether or not they wear a white hat or a black hat. If someone guesses the correct color they go free, otherwise they have a meeting with the executioner.</p>
<p>The prisoners all meet at the dining room later that day and devise a strategy in which no infinite number of prisoners will die. What is it?</p>
<blockquote class="spoiler">
<p> The answer is as follows:<br>
<br>
The prisoners consider all the infinite black-white (binary) sequences, they define an equivalence class that two sequences are equivalent if and only if they differ at finitely many coordinates.<br>
<br>
Using the axiom of choice they choose a representative from each equivalence class. The next morning each of the prisoners see a tail of the sequence of the hats and each prisoner knows its index number, therefore they all know a specific sequence which is equivalent to the sequence of hats, and each prisoner can say the color appearing in the coordinate of their room.<br>
<br>
Since the representative is only different in finitely many places than Stalin's choice, almost all prisoners gets to live.<br>
<br>
It is interesting to remark that the prisoners <strong>have</strong> to use the axiom of choice, as in models of ZF+AD such choice is impossible.</p>
</blockquote>
|
2,674,938 | <blockquote>
<p>Let $L:C^2(I)\rightarrow C(I), L(y)=x^2y''-3xy'+3y.$ Find the kernel of the linear transformation $L$. Can the solution of $L(y)=6$ be expressed in the form $y_H$+$y_L$, where $y_H$ is an arbitrary linear combination of the elements of ker L.</p>
</blockquote>
<p><strong>What I have tried:</strong></p>
<p>Since ker L is subspace of $C^2(I)$ and dim $C^2(I)=2$, dim ker $L\leq 2$.</p>
<p>Let $y(x)=x^r$. Then substituting gives $L(x^r)=x^rr^2-3rx^r+3x^r$, hence $L(x^r)=0$ <strong>iff</strong> $x^rr^2-3rx^r+3x^r=0.$ $r$ can be solved using the quadratic formula.</p>
<p>$$r=\frac{3+i\sqrt3}{2} \vee r=\frac{3-i\sqrt3}{2}$$ </p>
<p>$$y_1(x)=x^\frac{3+i\sqrt3}{2}, y_1(x)=x^\frac{3-i\sqrt3}{2} $$which are linearly independent(?).</p>
<p>How would one show that $y_1$ and $y_2$ are <strong>LI</strong>? </p>
| user577215664 | 475,762 | <p>Another way</p>
<p>$$x^2y''-3xy'+3y=0$$
For $x \neq 0$
$$y''-3 \left (\frac {xy'-y}{x^2} \right )=0$$
$$y''-3\left (\frac {y}{x}\right )'=0$$
$$y'-3 \left (\frac {y}{x} \right )=K_1$$
$$x^3y'-3{y}{x^2}=K_1x^3$$
$$\frac {x^3y'-3{y}{x^2}}{x^6}=\frac {K_1}{x^3}$$
$$\left (\frac {y}{x^3} \right )'=\frac {K_1}{x^3}$$
$$\frac {y}{x^3}=K_1\int\frac {dx}{x^3}=\frac {K_1}{x^2}+K_2$$
$$\boxed {y=K_1x+K_2x^3}$$
Both functions $(x,x^3)$ are linearly independant.</p>
<p>Show that $ \forall x \,\, c_1x+c_2x^3=0 \implies c_1=c_2=0$</p>
|
724,302 | <p>I have read the page about category theory in wikipedia carefully, but i don't really get what this theory is.</p>
<p>Is category theory a content in ZFC-set theory? (Just like measure theory, group theory etc.) If not, is it just another formal logic system independent from the standard ZFC-set theory?</p>
<p>Following wikipedia, i think it is the latter one.</p>
<p>If category theory is another formal logic system, then is there any theorem that relates category theory and ZFC-set theory? For example, there is a theorem states that "every theorem about sets in NBG is provable in ZFC, assuming consistency of ZFC" even though NBG and ZFC are completely different set theory. Moreover, is there a theorem which tells the relation between consistencies of ZFC and category theory? Also, is the usual independence proof in ZFC no more available in category theory?</p>
<p>Moreover, how does one write a <em>formula</em> in ZFC-set theory in category theory and vise versa? Since there is no <strong>undefined notion: set</strong> in category theory, i think this is not possible..</p>
| mathematician | 98,943 | <p>I know only very basic things about category theory, but from what I understand category theory is an axiomatic approach to proving things that various mathematical objects have in common. Like instead of proving a theorem about group homomorphisms and ring homomorphisms and maps between sets, you prove all those different flavors of what is essentially the same theorem at once in a general setting.</p>
<p>Apparently there is also a way to start from scratch using category theory instead of set theory.</p>
|
594,811 | <p>This is my first post, sorry for my naiveness..</p>
<p>I know a basic equation that relates Gram-schmidt matrix and Euclidean distance matrix:</p>
<p>$XX'=-0.5*(I-J/n)*D*(I-J/n)'$</p>
<p>Where $X$ is centered data (is $d \times n$), $I$ is identity matrix, $J$ is a matrix filled with ones (1), $n$ is the number of columns in $X$, and $D$ is the distance matrix (with dimensions $n \times n$).</p>
<p>My question is:</p>
<p>How can I derive Euclidean distance matrix $D$ from this equation? I would like something like:</p>
<p>$D=$ (something ¿?)</p>
<p>For example, I can see that:</p>
<p>$D=-2*inv((I-J/n))*XX'*inv((I-J/n)'$</p>
<p>But $(I-J/n)$ is a singular matrix. I am still interested in some approximation.</p>
<p>Thanks a lot!!</p>
<p>Mark</p>
| Aydin | 68,715 | <p>There are simple linear relationships between an Euclidean distance matrix and a Gramian matrix in both directions.
The linear transformation $G\rightarrow D$ is given by
$$ D = \delta(G)\mathbf{1}^\mathrm{T} + \mathbf{1}\delta(G)^\mathrm{T} - 2G,$$
where $G=XX^\mathrm{T}$, and $\delta(G)$ denotes the diagonal elements of $G$.</p>
<p>Your formula for $G\rightarrow D$ may also be rewritten as
$$ G = -\frac{1}{2}VDV,$$
where $V=I_n-\frac{1}{n}\mathbf{1}\mathbf{1}^\mathrm{T}$ is a geometric centering matrix.</p>
|
4,561,863 | <p>This limit is proposed to be solved without using the L'Hopital's rule or Taylor series:
<span class="math-container">$$
\lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}},
$$</span>
where <span class="math-container">$a>0$</span>, <span class="math-container">$b>0$</span> are some constants.
I know how to calculate this limit using the L'Hopital's rule:
<span class="math-container">$$
\lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}=
e^{\lim\limits_{x\to 0} \ln\left(\left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}\right)};
$$</span>
<span class="math-container">$$
\lim\limits_{x\to 0} \ln\left(\left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}\right)=
\lim\limits_{x\to 0} \frac{\ln\left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)}{x}$$</span>
<span class="math-container">$$
=
\lim\limits_{x\to 0} \frac{2}{a^{\sin x}+b^{\sin x}}\cdot\frac12\cdot
\left( a^{\sin x}\cos x \ln a+b^{\sin x}\cos x \ln b \right)=
\frac12\left( \ln a+ \ln b \right)
$$</span>
<span class="math-container">$$
\Rightarrow
\lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}=
e^{\frac12\left( \ln a+ \ln b \right)}=\sqrt{ab}.
$$</span>
I'm allowed to use the limits <span class="math-container">$\lim_{x\to0}\frac{\sin x}{x}=1$</span>,
<span class="math-container">$\lim_{x\to0}\frac{a^x-1}{x}=\ln a$</span>,
<span class="math-container">$\lim_{x\to0}\frac{\log_a(1+x)}{x}=\log_a e$</span> and <span class="math-container">$\lim_{x\to0} (1+x)^{1/x}=e$</span>.</p>
| insipidintegrator | 1,062,486 | <p><a href="https://math.stackexchange.com/questions/1987215/1-to-the-power-of-infinity-formula">This post</a> has multiple answers that explain why, if <span class="math-container">$\displaystyle\lim_{x\to a} f(x)=1$</span> and <span class="math-container">$\displaystyle\lim_{x\to a}g(x)=\infty$</span> then <span class="math-container">$$\lim_{x\to a} (f(x))^{g(x)}=e^{\lim\limits_{x\to a} (f(x)-1)g(x)}$$</span> Using this formula, calling the limit as L, we have <span class="math-container">$$L=e^{\lim\limits_{x\to 0}\left(\frac{a^{\sin x}+b^{\sin x}-2}{2x}\right)}$$</span> so <span class="math-container">$$\ln L=\lim_{x\to 0}\left(\dfrac{a^{\sin x}-1}{2x}+ \dfrac{b^{\sin x}-1}{2x}\right)$$</span><span class="math-container">$$\ln L=\lim_{x\to 0}\left(\dfrac{a^{\sin x}-1}{2\sin x}\cdot\frac{\sin x}{x}+ \dfrac{b^{\sin x}-1}{2\sin x} \frac{\sin x}{x}\right)$$</span><span class="math-container">$$=\frac{\ln a}{2}+\frac{\ln b}{2}=\frac{\ln ab}{2}=\ln\sqrt {ab}$$</span> whence <span class="math-container">$L=\sqrt{ab}$</span></p>
|
4,561,863 | <p>This limit is proposed to be solved without using the L'Hopital's rule or Taylor series:
<span class="math-container">$$
\lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}},
$$</span>
where <span class="math-container">$a>0$</span>, <span class="math-container">$b>0$</span> are some constants.
I know how to calculate this limit using the L'Hopital's rule:
<span class="math-container">$$
\lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}=
e^{\lim\limits_{x\to 0} \ln\left(\left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}\right)};
$$</span>
<span class="math-container">$$
\lim\limits_{x\to 0} \ln\left(\left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}\right)=
\lim\limits_{x\to 0} \frac{\ln\left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)}{x}$$</span>
<span class="math-container">$$
=
\lim\limits_{x\to 0} \frac{2}{a^{\sin x}+b^{\sin x}}\cdot\frac12\cdot
\left( a^{\sin x}\cos x \ln a+b^{\sin x}\cos x \ln b \right)=
\frac12\left( \ln a+ \ln b \right)
$$</span>
<span class="math-container">$$
\Rightarrow
\lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}=
e^{\frac12\left( \ln a+ \ln b \right)}=\sqrt{ab}.
$$</span>
I'm allowed to use the limits <span class="math-container">$\lim_{x\to0}\frac{\sin x}{x}=1$</span>,
<span class="math-container">$\lim_{x\to0}\frac{a^x-1}{x}=\ln a$</span>,
<span class="math-container">$\lim_{x\to0}\frac{\log_a(1+x)}{x}=\log_a e$</span> and <span class="math-container">$\lim_{x\to0} (1+x)^{1/x}=e$</span>.</p>
| ZGperx | 708,955 | <p>Let's prove this expression:</p>
<p>Given positive <span class="math-container">$a$</span> and <span class="math-container">$b$</span>:
<span class="math-container">$$\lim_{n\rightarrow\infty}\bigg(\frac{\sqrt[n]{a}+\sqrt[n]{b}}{2}\bigg)^n=\sqrt{ab}$$</span>.</p>
<p>In order to show this we are going to use the following:</p>
<p><span class="math-container">$$\lim_{n\rightarrow\infty}n(\sqrt[n]{a}-1)=\ln a$$</span> thus by AM-GM</p>
<p><span class="math-container">$$\frac{1}{n}\ln\sqrt{ab}\leq\ln\frac{1}{2}(\sqrt[n]{a}+\sqrt[n]{b})=\ln\bigg(\frac{1}{2}(\sqrt[n]{a}-1)+\frac{1}{2}(\sqrt[n]{b}-1)+1\bigg)<\frac{1}{2}\bigg((\sqrt[n]{a}-1)+(\sqrt[n]{b}-1)\bigg)$$</span></p>
<p>now multiply both sides by <span class="math-container">$n$</span>, we get</p>
<p><span class="math-container">$$\ln\sqrt{ab}\leq n\ln\frac{1}{2}(\sqrt[n]{a}+\sqrt[n]{b})\leq\frac{n}{2}\bigg((\sqrt[n]{a}-1)+(\sqrt[n]{b}-1)\bigg)$$</span>
Taking <span class="math-container">$n\rightarrow\infty$</span> and using the squeeze theorem:
We get
<span class="math-container">$$\lim_{n\rightarrow\infty}\bigg(\frac{\sqrt[n]{a}+\sqrt[n]{b}}{2}\bigg)^n=\sqrt{ab}.$$</span></p>
<p>Let <span class="math-container">$n=\frac1x$</span> and the rest is done.</p>
|
1,790,222 | <p>I know that $[0,1]$ and a unit circle $\mathbb{S}^1$ are one-point compactifications of $\mathbb{R}$ under some suitable homeomorphism. But how does one construct the Stone–Čech compactification? </p>
| Henno Brandsma | 4,280 | <p>You cannot really construct it, as such. You can define it, and prove its existence (using the Axiom of Choice) but you cannot give a concrete, definable example of a point in the remainder $\beta\mathbb{R}\setminus\mathbb{R}$. It is common to define the half-line $\mathbb{H} = [0,\infty)$ and consider $\beta\mathbb{H} \setminus \mathbb{H}$, because one can show that $\beta \mathbb{R} \setminus \mathbb{R}$ is a disjoint union of two copies of $\beta\mathbb{H} \setminus \mathbb{H}$ (one on the right, and one on the left). This sort of makes sense, because the reals have no holes to fill internally (it's already locally compact and complete), it's compactified at the boundaries, so to say. Adding a single point at infinity yields the (essentially unique) one-point compactification which is homeomorphic to $\mathbb{S}^1$, and because it is ordered we can add two points (at both ends) to get an orderable compactification which is homeomorphic to $[0,1]$.</p>
<p>The Cech-Stone compactification of $X$ is characterised by the function extension property (if we assume for simplicity that $X \subseteq \beta X$): every continuous function $f: X \rightarrow Y$, where $Y$ is any compact Hausdorff space, can be extended to $\beta f: \beta X \rightarrow Y$. This extension $\beta f$ is automatically unique because the co-domain $Y$ is Hausdorff and $X$ is dense in $\beta X$. In fact we could suffice to check this for $Y = [0,1]$ only, it turns out, to have it for all compact Hausdorff $Y$.</p>
<p>This gives unicity: if $\gamma X$ has the same property (where again we assume for simplicity that $X \subseteq \gamma X$), we extend $1_{X,\beta}: X \rightarrow \beta Y$ with $1_X(x) = x$ to $\gamma 1_{X,\beta} : \gamma X \rightarrow \beta X$, and in the same way we have a continuous $\beta 1_{X,\gamma} : \beta X \rightarrow \gamma X$. As these maps are each other's inverses on the dense set $X$ (which we have in both), because the maps are just the identity there, they are each other's inverses on $\gamma X$ and $\beta X$ as well, making these spaces homeomorphic. </p>
<p>This extension property means that we must extend $f(x) = \sin(x)$ from the reals to $[-1,1]$ to all of $\beta \mathbb{R}$, and it's clear we cannot extend it just using one or two points, so that the well-known compactifications are certainly not the Cech-Stone one. E.g. if $f(\infty)$ should equal the limit of $f(2n\pi)$ as well as $f(2n\pi + \frac{1}{2})$, and these are already different. A function can only be extended to the one-point compactification of the reals, if the values outside of $f[-n,n]$ vary less and less for larger $n$, roughly speaking. So this is just a limit class. We need to be able to extend <em>all</em> bounded continuous functions. </p>
<p>Constructions (sketches of them) can be found <a href="https://en.wikipedia.org/wiki/Stone%E2%80%93%C4%8Cech_compactification" rel="noreferrer">on Wikipedia</a> or good books like Engelking's "General Topology", or Gillman and Jerrison's "Rings of continuous functions". For the reals (which is normal) the new points of $\beta \mathbb{R}$ correspond to ultrafilters of closed subsets of $\mathbb{R}$ (free ones, i.e. their intersection is empty). But these cannot even be proved to exist without some form of the Axiom of Choice, and no explicit examples of these can be given. </p>
<p>We can e.g. define a filter base $\mathscr{F}$ of closed sets by taking the complements of all open intervals of the form $(a,b)$, where $a < b, a, b \in \mathbb{R}$. These have empty intersection in the reals (as $x \notin \mathbb{R}\setminus (x-1,x+1)$) but any two or finitely many of them intersect (as can easily be checked). Then Zorn's lemma (or some other such principle) tells us there exists some maximal filter of closed sets that contains $\mathscr{F}$, in fact plenty of them, and each of those gives us a point of $\beta \mathbb{R} \setminus \mathbb{R}$. Note that in a compact space any family of closed sets with the finite intersection property has non-empty intersection (so $\mathscr{F}$ witnesses the non-compactness of the reals), so the closure of the sets in $\mathscr{F}$ in $\beta \mathbb{R}$ <em>should</em> have non-empty intersection, and in fact those ultrafilters (considered as points) will be in that intersection in $\beta \mathbb{R}$. </p>
<p>In fact there are $2^\mathfrak{c}$ many new points in $\beta \mathbb{R} \setminus \mathbb{R}$ (as many as points added to $\mathbb{N}$, and as many as there are subsets of $\mathbb{R}$). It's a very large, very non-metrisable space, that still has the reals (and thus $\mathbb{Q} \subseteq \mathbb{R}$ as well) as a dense subspace.</p>
<p>One of the few spaces that we have that the Cech-Stone equals the one-point compactification is $\omega_1$ (the first uncountable ordinal in the order topology) which has $\omega_1+1$ as its unique compactification. So here it is concrete, but this situation is quite rare.</p>
|
42,258 | <p>I have a Table of values e.g. </p>
<pre><code>{{x,y,z},{x,y,z},{x,y,z}…}
</code></pre>
<p>How do I replace the the "z" column with a List of values?</p>
| halirutan | 187 | <p>With the notation package something like this is easy. I would never use this by myself, because IMO such <em>sugar</em> can easily introduce bugs and undesired behavior if one is not cautious. I will paste a screenshot so that you see how I used the <code>Notation`</code> package, but first of all you have to load it:</p>
<pre><code><< Notation`
</code></pre>
<p>then you can use </p>
<blockquote>
<p><img src="https://i.stack.imgur.com/IkJ4e.png" alt="Mathematica graphics"></p>
</blockquote>
<p>Testing it</p>
<blockquote>
<p><img src="https://i.stack.imgur.com/JpMmv.png" alt="Mathematica graphics"></p>
</blockquote>
<p>or </p>
<blockquote>
<p><img src="https://i.stack.imgur.com/iTgfA.png" alt="Mathematica graphics"></p>
</blockquote>
|
4,353,891 | <p>How to prove that <span class="math-container">$A=\{(x,y)\in \mathbb{R}^2: |x|+|y|^{1/2}<1\}$</span> is convex? I tried using the definition but couldn’t go far, since the second component involves square root(tried squaring, that made it complicated).</p>
<p>I plotted it in mathematica and the graph comes out to be convex, but how to prove it mathematically?</p>
<p>I started with two points (x,y) and (a,b) in A, and tried to prove (ta+(1-t)x, tb +(1-t)y) is in A, for which I need to show, <span class="math-container">$| ta+(1-t)x|+|t b +(1-t)y|^{1/2}<1$</span>, but since the square root term is there, I couldn’t simplify this.</p>
| Essaidi | 708,306 | <p>It's not convex :<br>
<a href="https://i.stack.imgur.com/lqRPa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lqRPa.png" alt="Surface <span class="math-container">$\{|x| + \sqrt{|y|} < 1\}$</span>" /></a></p>
|
76,378 | <p>I am trying to simplify the following expression I have encountered in a book</p>
<p>$\sum_{k=0}^{K-1}\left(\begin{array}{c}
K\\
k+1
\end{array}\right)x^{k+1}(1-x)^{K-1-k}$</p>
<p>and according to the book, it can be simplified to this:</p>
<p>$1-(1-x)^{K}$</p>
<p>I wonder how is it done? I've tried to use Mathematica (to which I am new) to verify, by using</p>
<p>$\text{Simplify}\left[\sum _{k=0}^{K-1} \left(\left(
\begin{array}{c}
K \\
k+1
\end{array}
\right)*x{}^{\wedge}(k+1)*(1-x){}^{\wedge}(K-1-k)\right)\right]$</p>
<p>and Mathematica returns</p>
<p>$\left\{\left\{-\frac{K q \left((1-q)^K-q^K\right)}{-1+2 q}\right\},\left\{-\frac{q \left(-(1-q)^K+(1-q)^K q+(1+K) q^K-(1+2 K) q^{1+K}\right)}{(1-2 q)^2}\right\}\right\}$</p>
<p>which I cannot quite make sense of it.</p>
<p>To sum up, my question is two-part:</p>
<ol>
<li><p>how is the first expression equivalent to the second?</p></li>
<li><p>how should I interpret the result returned by Mathematica, presuming I'm doing the right thing to simplify the original formula?</p></li>
</ol>
<p>Thanks a lot!</p>
| J. M. ain't a mathematician | 498 | <p><code>Simplify[PowerExpand[Simplify[Sum[Binomial[K, k + 1]*x^(k + 1)*(1 - x)^(K - k - 1), {k, 0, K - 1}], K > 0]]]</code> works nicely. The key is in the use of the second argument of <code>Simplify[]</code> to add assumptions about a variable. and using <code>PowerExpand[]</code> to distribute powers.</p>
|
76,378 | <p>I am trying to simplify the following expression I have encountered in a book</p>
<p>$\sum_{k=0}^{K-1}\left(\begin{array}{c}
K\\
k+1
\end{array}\right)x^{k+1}(1-x)^{K-1-k}$</p>
<p>and according to the book, it can be simplified to this:</p>
<p>$1-(1-x)^{K}$</p>
<p>I wonder how is it done? I've tried to use Mathematica (to which I am new) to verify, by using</p>
<p>$\text{Simplify}\left[\sum _{k=0}^{K-1} \left(\left(
\begin{array}{c}
K \\
k+1
\end{array}
\right)*x{}^{\wedge}(k+1)*(1-x){}^{\wedge}(K-1-k)\right)\right]$</p>
<p>and Mathematica returns</p>
<p>$\left\{\left\{-\frac{K q \left((1-q)^K-q^K\right)}{-1+2 q}\right\},\left\{-\frac{q \left(-(1-q)^K+(1-q)^K q+(1+K) q^K-(1+2 K) q^{1+K}\right)}{(1-2 q)^2}\right\}\right\}$</p>
<p>which I cannot quite make sense of it.</p>
<p>To sum up, my question is two-part:</p>
<ol>
<li><p>how is the first expression equivalent to the second?</p></li>
<li><p>how should I interpret the result returned by Mathematica, presuming I'm doing the right thing to simplify the original formula?</p></li>
</ol>
<p>Thanks a lot!</p>
| André Nicolas | 6,312 | <p>The following answer makes sense only with some background in probability.
Suppose first that $0 \le x\le 1$.</p>
<p>A possibly biased coin has probability $x$ of landing "heads." Toss the coin $K$ times. We compute the probability $P$ of <strong>one or more</strong> heads in two different ways. </p>
<p>The probability we get exactly $k+1$ heads is
$$\binom{K}{k+1}x^{k+1}(1-x)^{K-1-k}.$$
Add up, from $k=0$ to $k=K-1$. We get the probability of $1$ head, plus the probability of $2$ heads, plus the probability of $3$ heads, and so on, up to the probability of $K$ heads. Thus
$$P=\sum_{k=0}^{K-1}\binom{K}{k+1}x^{k+1}(1-x)^{K-1-k}.$$</p>
<p>The probability of $0$ heads (all tails) is $(1-x)^K$, so the probability of one or more heads is $1-(1-x)^K$, and therefore
$$P=1-(1-x)^K.$$</p>
<p>The argument so far only proves the desired result for $0\le x\le 1$. But note that for any $K$, each of the expressions we are trying to prove equal is a <em>polynomial</em> of degree $K$. But if two polynomials $A(x)$ and $B(x)$ with real coefficients, each of degree $\le K$, are equal at $K+1$ values of $x$, then they are identically equal.</p>
|
2,976,613 | <p>I want to show, that <span class="math-container">$a:=\sum \limits_{n=0}^{\infty} \left(\dfrac{2n+n^3}{3-4n}\right)^n$</span> is not converging, because <span class="math-container">$\lim \limits_{n \to \infty}(a)\neq 0 \; (*)$</span>. Therefore, the series can't be absolute converge too.</p>
<p>Firstly, I try to simplify the term. After that I want to find the limit.</p>
<p>Unfortunately, I can't seem to find any good equation with that I can clearly show <span class="math-container">$(*)$</span>.
<span class="math-container">\begin{align}
\sum \limits_{n=0}^{\infty} \left(\dfrac{2n+n^3}{3-4n}\right)^n&=\left (\dfrac{\not{n}\cdot (2+n^2)}{\not n \cdot (\frac{3}{n}-4)}\right)^n\\
&=\left(\dfrac{2+n^2}{\frac 3n-4}\right)^n\\
&= \cdots
\end{align}</span></p>
<p>How to go on?</p>
| OgvRubin | 468,471 | <p>I'm a bit confused by your choice of notation for it seems like you write that a series diverges if its limit is not <span class="math-container">$0$</span> what is not true. So i think you mean that we want to show that </p>
<p><span class="math-container">$$\frac{(2n+n^3)^n}{(3-4n)^n}\not\rightarrow 0$$</span></p>
<p>as <span class="math-container">$n\rightarrow \infty$</span> for this implies that the series does not converge. Note also that</p>
<p><span class="math-container">$$\left(\frac{2+n^2}{3/n-4}\right)^n\neq \frac{2^n+n^{2n}}{3^n\cdot (1/n)^n-4^n}$$</span></p>
<p>see <a href="https://en.wikipedia.org/wiki/Freshman%27s_dream" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Freshman%27s_dream</a> </p>
<p>Instead what can you say about</p>
<p><span class="math-container">$$\frac{2n+n^3}{3-4n}\rightarrow ?$$</span></p>
<p>as <span class="math-container">$n\rightarrow \infty$</span>?</p>
|
2,237,963 | <p>One-point compactification of $S_{\Omega}$ is homeomorphic with $\bar S_{\Omega}$.</p>
<p>Let $X$ be a topological space. Then the One-point compactification of $X$ is a certain compact space $X^*$ together with an open embedding $c : X \to X^*$ such that the complement of $X$ in $X^*$ consists of a single point, typically denoted $\infty$. </p>
<p>Let $X$ be a well-ordered set. Given $\alpha \in X$, let $S_{\alpha}$ denote the set $S_{\alpha} = \{x \mid x \in X \text{ and }x < \alpha \}$.
It is called the section of $X$ by $\alpha$. </p>
<p>I am finding difficulty to do the problem!</p>
| Travis Willse | 155,629 | <p>Drafting behind Michael Rozenberg's clever answer, appealing to the concavity of <span class="math-container">$\sin$</span> on <span class="math-container">$[0, \pi]$</span> quickly reduces the problem to showing the inequality
<span class="math-container">$$2 \sin 1 > \frac{8}{5} .$$</span>
From <a href="https://en.wikipedia.org/wiki/Proof_that_22/7_exceeds_%CF%80" rel="noreferrer"><span class="math-container">$\pi < \frac{22}{7}$</span></a> we deduce <span class="math-container">$\frac{3 \pi}{10} < \frac{66}{70} < 1$</span>, and so
<span class="math-container">$$2 \sin 1 > 2 \sin \frac{3 \pi}{10} = 2 \cdot \frac{1}{4}(1 + \sqrt{5}) > \frac{8}{5}.$$</span> The last inequality (which itself is a reasonably tight bound on the Golden Ratio <span class="math-container">$\phi$</span>) follows from rearranging and squaring.</p>
|
3,176,482 | <p>In my Econometrics class yesterday, our teacher discussed a sample dataset that measured the amount of money spent per patient on doctor's visits in a year. This excluded hospital visits and the cost of drugs. The context was a discussion of generalized linear models, and for the purposes of Stata, he found that a gamma distribution worked best. Good enough.</p>
<p>But I started thinking about the dataset itself, and the pdf curve "total dollars per patient" might follow. I'm thinking of it as two variables: </p>
<p>1) The number of visits per person (k). Let's use a Poisson distribution with λ=1, so p(0)=0.368, p(1)=0.368, p(2)=0.184, etc.</p>
<p>2) The cost of each visit (x). Let's use a normal curve with a mean of 50 dollars and a SD of 5 dollars. A person's total expenditure would be the sum of normally distributed variables, or x = N(k50, k25) for k=0 to infinity visits. So a person with no visits would spend nothing, ~.95 of people with one visit would spend between 40 and 60, ~.95 of those with two visits would spend between ~85.86 and ~114.14, three visits would spend ~132.68 and ~167.32, etc.</p>
<p>I think a pdf curve would be some combination of these two functions. My guess is that it would look like a vertical line a 0 and a series of decreasing, ever flattening humps at 50, 100, 150 dollars, etc, where the integral of x=(0, inf.) equals 0.632 (because zero visits is p=.0368)</p>
<p>Can anyone help me put the pieces together? What would that function be? I've taken through Calc III (multivariate/vector calculus), FYI. Thanks.</p>
| BruceET | 221,800 | <p>I agree with @Michael that you might be thinking about a <strong>random sum of random variables</strong>. The number of visits <span class="math-container">$N \sim \mathsf{Pois}(\lambda)$</span> and the amount spent on the <span class="math-container">$i$</span>th visit is <span class="math-container">$X_i \sim \mathsf{Norm}(\mu, \sigma).$</span></p>
<p>There are formulas for <span class="math-container">$E(T)$</span> and <span class="math-container">$Var(T),$</span> where
<span class="math-container">$T = \sum_{i=1}^N X_i.$</span> For the mean it's <span class="math-container">$E(T) = E(N)E(X) = \lambda\mu.$</span> There is a more complicated
two-term formula for <span class="math-container">$Var(T),$</span> sometimes called the 'total variation'. I'll let you look for that in your text, notes, or online.</p>
<p>Simulating (in R statistical software) for a million hypothetical people, with <span class="math-container">$\lambda =5$</span>, <span class="math-container">$\mu = 100,$</span> <span class="math-container">$\sigma=20,$</span> we have the following results. (With <span class="math-container">$m = 10^6$</span> iterations, you can expect 2 or 3 "significant" digits of accuracy.)</p>
<pre><code>set.seed(2019) # for reproducibility
m = 10^6; lam = 5; mu = 100; sg = 20
t = numeric(m)
for(i in 1:m) {
n = rpois(1, lam); t[i] = sum(rnorm(n, mu, sg))
}
mean(t); var(t); sd(t); mean(t > 700)
[1] 500.0381 # aprx E(T) = 500
[1] 51976.34 # aprx V(T)
[1] 227.9832 # aprx SD(T)
[1] 0.185734 # aprs P(T > 700)
hist(t, prob=T, col="skyblue2", main = "Simulated Dist'n of T")
</code></pre>
<p><a href="https://i.stack.imgur.com/9IZqT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9IZqT.png" alt="enter image description here"></a></p>
<p>A normal approximation might not the best way
to evaluate probabilities such as <span class="math-container">$P(T > 700).$</span></p>
<p><strong>Addendum:</strong> Revised histogram as per Comment by @r.e.s.:</p>
<p><a href="https://i.stack.imgur.com/CQOjy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CQOjy.png" alt="enter image description here"></a></p>
<p>Also R's default kernel density estimator added to the earlier
histogram from <span class="math-container">$m = 10^6$</span> iterations, gives a much clearer picture then just the histogram (although, naturally enough, it misses
the spike at <span class="math-container">$0.)$</span> Additional code following the <code>hist</code> line is shown below:</p>
<pre><code>lines(density(t), col="red", lwd=2)
</code></pre>
<p><a href="https://i.stack.imgur.com/GNlJl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GNlJl.png" alt="enter image description here"></a></p>
|
4,320,849 | <p>I had this problem in an exam I recently appeared for:</p>
<blockquote>
<p>Find the range of
<span class="math-container">$$y =\frac{x^2+2x+4}{2x^2+4x+9}$$</span></p>
</blockquote>
<p>By randomly assuming the value of <span class="math-container">$x$</span>, I got the lower range of this expression as <span class="math-container">$3/7$</span>. But for upper limit, I ran short of time to compute the value of it and hence couldn't solve this question.</p>
<p>Now, I do know that one way to solve this expression to get its range is to assume the whole expression as equals to K, get a quadratic in K, and find the maximum/minimum value of K which will in turn be the range of that expression. I was short on time so avoided this long winded method.</p>
<p>Another guy I met outside the exam center, told me he used an approach of <span class="math-container">$x$</span> tending to infinity in both cases and got the maximum value of this expression as <span class="math-container">$1/2$</span>. But before I could ask him to explain more on this method, he had to leave for his work.</p>
<p>So, will someone please throw some light on this method of <span class="math-container">$x$</span> tending to infinity to get range, and how it works. And if there exists any other efficient, and quicker method to find range of a function defined in the form of a ( quadratic / quadratic ).</p>
| soupless | 888,233 | <p>In general, if <span class="math-container">$\deg f = 0$</span> where <span class="math-container">$$f(x) = \frac{a_nx^n + a_{n - 1}x^{n - 1} + \cdots + a_1x + a_0}{b_nx^n + b_{n - 1}x^{n - 1} + \cdots + b_1x + b_0},$$</span> the limit of <span class="math-container">$f$</span> as <span class="math-container">$x$</span> increases/decreases without bound is <span class="math-container">$a_n/b_n$</span>.</p>
<p>In your case, <span class="math-container">$a_2 = 1$</span> and <span class="math-container">$b_2 = 2$</span>. Hence, <span class="math-container">$a_2/b_2 = 1/2$</span>.</p>
<hr />
<p>We'll factor <span class="math-container">$f$</span> as <span class="math-container">$$\frac{x^2+2x+4}{2x^2+4x+9} = \frac{(x + 1)^2 + 3}{2(x + 1)^2 + 7}.$$</span></p>
<p>Notice that for all <span class="math-container">$x \in \mathbb{R}$</span>, <span class="math-container">$f > 0$</span>. Also, we can see that <span class="math-container">$(x+1)^2 + 2 < 2(x + 1)^2 + 7$</span>. This means that the range should be a part of <span class="math-container">$(0,1/2)$</span>. Since both numerator and denominator have <span class="math-container">$(x + 1)^2$</span> without any remaining <span class="math-container">$x$</span>'s, we can see that this will be at its minimum when <span class="math-container">$x = -1$</span>. Then, <span class="math-container">$$f(-1) = \frac{(-1 + 1)^2 + 3}{2(-1 + 1)^2 + 7} \\ = \frac{(0)^2 + 3}{2(0)^2 + 7} \\ \frac{3}{7}$$</span></p>
<p>Therefore, the range is <span class="math-container">$[3/7, 1/2)$</span>.</p>
|
794,301 | <p>I am trying to find out the sum (I just derived these from 2 + 0.5 + 0.125 + 0.03125 + ...):</p>
<p>$$\sum_{n=0}^{\infty} \frac{5^{2n-1}}{10^{2n -1}}$$</p>
<p>It's confusing me because it doesn't match $${ar}^{n-1}$$ the power by which $r$ is raised. </p>
| Henno Brandsma | 4,280 | <p>$\frac{a^k}{b^k} = (\frac{a}{b})^k$. Also $a^{2k} = (a^2)^k$, and $a^{2k-1} = \frac{1}{a} \cdot a^{2k}$. This should allow you to simplify it.</p>
|
125,399 | <p>How can I solve the following integral?
$$\int_0^\pi{\frac{\cos{nx}}{5 + 4\cos{x}}}dx, n \in \mathbb{N}$$</p>
| PAD | 27,304 | <p>Of course it can be solved easily with complex residues. Since $\cos x$ is even you replace it with 1/2 the integral from $0$ to $ 2 \pi$. Then you make the substitution $z=e^{i\theta}$. You end-up with an integral over the unit circle. You end-up with a function which has singularities only at $0$ and $-\frac{1}{2}$. Then find the residues.</p>
|
988,566 | <p>For x ∈ ℝ, define by:
⌊x⌋ ∈ ℤ ∧ ⌊x⌋ ≤ x ∧ (∀z ∈ ℤ, z ≤ x ⇒ z ≤ ⌊x⌋).</p>
<p>Use this definition to prove or disprove the following with a structured proof technique:
∀x ∈ ℝ, ∀y ∈ ℝ, x > y ⇒ ⌊x⌋ ≥ ⌊y⌋.</p>
<p>I understand I need to start with assuming the domain to be true aswell as the antecedent, then equating a value for x as something and continuing it onwards all the way towards the concluding statement, but I'm fairly new to this and could use some help in proving the thinking part of this</p>
| kolonel | 104,564 | <p>Using Cauchy-Shwartz's inequality we have:
$$\begin{align*}
\mathbf{x}^\intercal H \mathbf{x} &= (1-\rho)\sum_{i=1}^n x_i^2 + \rho\left(\sum_{i=1}^n x_i\right)^2\\
&\geq \frac{(1-\rho)}{n}\left(\sum_{i=1}^n x_i\right)^2 + \rho\left(\sum_{i=1}^n x_i\right)^2\\
&=\frac{1+(n-1)\rho}{n}\left( \sum^n_{i=1} x_i^2\right)^2\\
&\geq 0
\end{align*}$$
and you are done (you have to assume that $n>1$ otherwise your $\rho$ could evaluate to $1/0$).</p>
|
265,377 | <p>Let $k$ be a finite field, and let $G$ be the absolute Galois group of $k$, which is isomorphic to $\widehat{\mathbb{Z}}$. Let $\mathcal{C}$ be the category of $G$-modules. Then, we have the following:</p>
<p>For a finite $G$-module $N$, we have
$$
Ext^r_{\mathcal{C}}(N, \mathbb{Z}) \simeq H^{r-1}(G,~N^D),
$$
where $N^D:=\operatorname{Hom}_\mathbb{Z}(N, \mathbb{Q}/\mathbb{Z})$ is the Pontrjagin dual of $N$, with the natural $G$-action by $(g\phi)(x):=g(\phi(g^{-1}(x)))=\phi(g^{-1}(x))$ for $g\in G$, $x \in N$ and $\phi \in N^D$.</p>
<p>Why is this true?</p>
| nfdc23 | 81,332 | <p>As noted in R. van Dobben de Bruyn's answer, since $N$ is killed by some positive integer we can reformulate the problem as that of constructing natural isomorphisms ${\rm{Ext}}^i_G(N, \mathbf{Q}/\mathbf{Z}) \simeq {\rm{H}}^i(G, N^D)$ for any $i \ge 0$ and any finite discrete $G$-module $N$ (with $G := {\rm{Gal}}(k_s/k)$ for a finite field $k$). We will do this for any field $k$; finiteness is irrelevant.</p>
<p>It is convenient to work in the framework of abelian etale sheaves on ${\rm{Spec}}(k)$. From this viewpoint, a discrete $G$-module $M$ corresponds to an abelian etale sheaf $\mathscr{F}_M$. The local-to-global Ext spectral sequence (which uses that sheaf-Ext into an injective abelian sheaf is flasque and hence acyclic for etale cohomology) is
$${\rm{H}}^i(G, \mathscr{E}xt^j_k(\mathscr{F}_N, \mathscr{F}_M)) \Rightarrow {\rm{Ext}}^{i+j}_k(\mathscr{F}_N, \mathscr{F}_M) = {\rm{Ext}}^{i+j}_G(N,M)$$
for any discrete $G$-modules $N$ and $M$.</p>
<p>Now assume $N$ is finite, so $\mathscr{F}_N$ becomes constant over an etale cover of ${\rm{Spec}}(k)$. It is then not hard to show by an erasable $\delta$-functor argument that $\mathscr{E}xt^{\bullet}_k(\mathscr{F}_N,\mathscr{F}_M) = {\rm{Ext}}^{\bullet}_{\rm{Ab}}(N,M)$ as discrete $G$-modules. Hence, the spectral sequence takes the form
$${\rm{H}}^i(G, {\rm{Ext}}^j_{\rm{Ab}}(N,M)) \Rightarrow {\rm{Ext}}^{i+j}_G(N,M)$$
for any discrete $G$-module $M$ and finite discrete $G$-module $N$.</p>
<p>Now set $M$ to be the constant $G$-module $\mathbf{Q}/\mathbf{Z}$. Then ${\rm{Hom}}_{\rm{Ab}}(\cdot, M)$ is exact, so ${\rm{Ext}}^{\bullet}_{\rm{Ab}}(\cdot,M)$ vanishes in positive degrees. The spectral sequence with this $M$ therefore degenerates to give isomorphisms
$${\rm{H}}^i(G, N^D) \simeq {\rm{Ext}}^i_G(N, \mathbf{Q}/\mathbf{Z})$$
for all $i > 0$, as desired. (To be really useful one should address $\delta$-functoriality in $N$, but this was not requested in the question.)</p>
|
1,728,910 | <p>Whenever I get this question, I have a hard time with it. </p>
<p>An example of a problem:</p>
<p>In the fall, the weather in the evening is <em>dry</em> on 40% of the days, <em>rainy</em>
on 58% of days and <em>snowy</em> 2% of the days. </p>
<p>At noon you notice clouds in the sky. </p>
<p>Clouds appear
at noon on 10% of the days that have <strong>dry evenings</strong>, </p>
<p>25% of the days that have <strong>rainy evenings</strong>, </p>
<p>and
35% of the days that have <strong>snowy evenings</strong>.</p>
<p>From that I get:</p>
<p>$P(S)$ = "prob. snows" = $.02$</p>
<p>$P(R)$ = "prob. rains" = $.58$</p>
<p>$P(D)$ = "prob. is dry" = $.40$</p>
<p>$P(C|S)$ = "prob. that it is cloudy given that it snows" = $.35$</p>
<p>$P(C|R)$ = "prob. that it is cloudy given that it rains" = $.25$</p>
<p>$P(C|D)$ = "prob. that it is cloudy given that it is dry" = $.10$</p>
<p>However, the question given is:</p>
<p>Find the probability of snow in the evening given that you saw clouds at noon.</p>
<p>I interpret this as:</p>
<p>$$P(S|C)$$</p>
<p>I can't get it right. Is there a formula to it without having P(C)? </p>
| Feras | 215,405 | <p>$P\left(S \mid C \right) = \dfrac{P\left(C \mid S \right)P\left(S\right)}{P\left(C\right)} = \dfrac{P\left(C \mid S \right)P\left(S\right)}{P(C|S)P(S)+P(C|D)P(D)+P(C|R)P(R)}$.</p>
|
13,432 | <p>I want to turn a sum like this</p>
<pre><code>sum =a-b+c+d
</code></pre>
<p>Into a List like this: </p>
<pre><code>sumToList[sum]={a,-b,c,d}
</code></pre>
<p>How can I achieve this?</p>
| kglr | 125 | <pre><code>List @@ sum
</code></pre>
<blockquote>
<p>{a, -b, c, d}</p>
</blockquote>
<p>From the docs on <a href="http://reference.wolfram.com/mathematica/ref/Apply.html" rel="noreferrer">Apply (@@)</a>:</p>
<blockquote>
<p>f@@<em>expr</em> replaces the head of <em>expr</em> by f.</p>
</blockquote>
<p>So <code>List@@sum</code> replaces <code>Head[sum]</code> (that is, <code>Plus</code>) with <code>List</code>.</p>
<p>You can also get the same result by changing <code>0</code>th <code>Part</code> of <code>sum</code> (which is its <code>Head</code>) to <code>List</code>:</p>
<pre><code>sum[[0]] = List; sum
</code></pre>
<blockquote>
<p>{a, -b, c, d} </p>
</blockquote>
|
1,348,046 | <p>I ran into this sum $$\sum_{n=3}^{\infty} \frac{3n-4}{n(n-1)(n-2)}$$
I tried to derive it from a standard sequence using integration and derivatives, but couldn't find a proper function to describe it.
Any ideas?</p>
| mathlove | 78,967 | <p>Setting </p>
<p>$$\frac{3n-4}{n(n-1)(n-2)}=\frac{A(n-1)-B}{(n-2)(n-1)}-\frac{An-B}{(n-1)n}$$
gives you $A=3,B=2$, i.e.
$$\frac{3n-4}{n(n-1)(n-2)}=\frac{3(n-1)-2}{(n-2)(n-1)}-\frac{3n-2}{(n-1)n}.$$
Hence, we have
$$\begin{align}\sum_{n=3}^{\infty}\frac{3n-4}{n(n-1)(n-2)}&=\lim_{m\to\infty}\sum_{n=3}^{m}\left(\frac{3(n-1)-2}{(n-2)(n-1)}-\frac{3n-2}{(n-1)n}\right)\\&=\lim_{m\to\infty}\left(\frac{3\cdot 2-2}{1\cdot 2}-\frac{3m-2}{(m-1)m}\right)\\&=2\end{align}$$</p>
|
1,206,528 | <p>Find the matrix $A^{50}$ given</p>
<p>$$A = \begin{bmatrix} 2 & -1 \\ 0 & 1 \end{bmatrix}$$ as well as for $$A=\begin{bmatrix} 2 & 0 \\ 2 & 1\end{bmatrix}$$</p>
<p>I was practicing some questions for my exam and I found questions of this form in a previous year's paper.</p>
<p>I don't know how to do such questions.</p>
<p>Please assist over this question.</p>
<p>Thank You</p>
| paw88789 | 147,810 | <p>Look at the first few powers:
$$\left(\begin{matrix}2&-1\\0&1 \end{matrix} \right)$$
$$A^2=\left(\begin{matrix}2&-1\\0&1 \end{matrix} \right)\left(\begin{matrix}2&-1\\0&1 \end{matrix} \right)=\left(\begin{matrix}4&-3\\0&1 \end{matrix} \right)$$
$$A^3=\left(\begin{matrix}8&-7\\0&1 \end{matrix} \right)$$
This suggests that
$$A^n=\left(\begin{matrix}2^n&1-2^n\\0&1 \end{matrix} \right)$$
Which can be proved by induction.</p>
|
1,687,500 | <p>Prove that, any group of order $15$ is abelian (without help of Sylow's theorem or its application).</p>
<p>What I have done so far is, </p>
<p>by class equation we know that $|G|=|Z|+\sum\frac{|G|}{C(a_i)}$. Now if I can show that $|G|=|Z|$ then the theorem is proved. Now order of $|Z|$ can not be $3$ or $5$, because if it is then $(G/Z)$ will be cyclic,which can not be since $G$ non-abelian.So, only possibilities are $|Z|=1$ or $15$. Now haw can I conclude that $|Z|\not=1$?</p>
| Kuifje | 273,220 | <p>Assume we are dealing with a minimization problem. Any variable $x_r$ with negative ($\le 0$) reduced cost can enter the basis. Once this is done, another variable has to leave the basis. You can choose any variable $x_i$ such that
$$
\frac{b_i'}{a_{ir}'} \ge 0,
$$
i.e. any variable such that if $x_r$ replaces $x_i$, the solution remains feasible. Typically we choose the variable that minimizes $\frac{b_i'}{a_{ir}'}|a_{ir}'>0$, because it gives you the best possible value for the entering variable, before the solution becomes infeasible.</p>
<p>So in this particular case, $x_q$ will leave the basis if and only if </p>
<ol>
<li>There is another variable with a negative reduced cost (or positive if we are maximizing)</li>
<li>$\min_i\{\frac{b_i'}{a_{ir}'}|a_{ir}'>0\}=\frac{b_q}{a_{qr}'}=b_q$</li>
</ol>
|
1,358,927 | <p>I have to solve this problem using integration by parts. I am new to integration by parts and was hoping someone can help me.</p>
<p>$$\int\frac{x^3}{(x^2+2)^2} dx$$</p>
<p>Here is what I have so far:</p>
<p>$$\int udv = uv-\int vdu $$</p>
<p>$$u=x^2+2$$ Therefore, $$xdx=\frac{du}{2}$$
$$dv=x^3$$
Therefor, $$v=3x^2$$</p>
| Nathan Janos | 350,573 | <p>I believe what you are looking for is this:</p>
<p><span class="math-container">$$ x \bmod M = [\frac{1}{2} + \frac{i}{2\pi}\ln(-e^{-i2\pi x/M})]\times M $$</span></p>
<p>You can check it out graphed <a href="https://www.wolframalpha.com/input/?i=plot%20(1%2F2%20%2B%20(i%2F(2*pi))*log(-1*exp(-i*2*pi*x%2FM)))*M%20where%20M%20%3D%2010" rel="nofollow noreferrer">here:</a></p>
<p>I stumbled upon this post because I was playing around with the function <span class="math-container">$(-1)^x$</span> which has a real and imaginary part that are periodic and out of phase. Using Euler's formula I pulled out the part of the function that was cyclical and graphed it and saw that it was a saw wave. I generalized it by scaling it in the x and y direction by M.</p>
<p>Here is the Euler formula transform of the same function:</p>
<p><span class="math-container">$$ x \bmod M = [\pi + \frac{\ln(\cos(\frac{2\pi (x+M/2)}{M}+M+\ln(2))+i*\sin(\frac{2\pi (x+M/2)}{M}+M+\log(2)))}{i}]\times \frac{M}{2\pi} $$</span></p>
<p>You can also derive this by taking the closed form solution that drops out of the <a href="http://mathworld.wolfram.com/FourierSeriesSawtoothWave.html" rel="nofollow noreferrer">infinite Fourier series</a>:</p>
<p>Note that the Gibbs phenomenon that you get with a finite number of terms when approximating the saw wave disappears with the closed form solution.</p>
|
1,102,310 | <p>For the quadratic function $$-ax^2 + 1$$ an upside down parabola with $y(0) = 1,$ is there a way to compute <em>a</em> such that the definite integral of $y$ between the roots ($x_1, x_2: f(x_1) \land f(x_2)= 0$) equals $1?$</p>
| mickep | 97,236 | <p>Hint: $y=-ax^2+1$ is zero when $x=\pm1/\sqrt{a}$, so it might be so that you want to calculate the integral
$$
\int_{-1/\sqrt{a}}^{1/\sqrt{a}}1-ax^2\,dx.
$$</p>
|
2,579,137 | <p>According to the definition of harmonic number $H_n = \sum\limits_{k=1}^n\frac{1}{k}$.</p>
<p>How we can define $H_{n+1}$ and $H_{n+\frac{1}{2}}$? </p>
| Jack D'Aurizio | 44,121 | <p>Well, the question is more or less the same as <em>how do we define $n!$ for $n\not\in\mathbb{N}?$</em><br>
Intro: there are many functions agreeing with $n!$ over $\mathbb{N}$, but it we impose that </p>
<ol>
<li>$f(1)=1$ and $f(x+1)=x\,f(x)$ for any $x\in\mathbb{R}^+$</li>
<li>$\log f$ is convex over $\mathbb{R}^+$</li>
</ol>
<p>then we may prove that the $\Gamma$ function, defined through
$$ \Gamma(n+1)=\int_{0}^{+\infty} x^n e^{-x}\,dx $$
is <em>the only</em> natural extension of the factorial function. This is the <a href="https://en.wikipedia.org/wiki/Bohr%E2%80%93Mollerup_theorem" rel="nofollow noreferrer">Bohr-Mollerup theorem</a>.<br>
(Have a look at the section about <em>special functions</em> in <a href="https://drive.google.com/file/d/0BxKdOVsjsuEwdjBEM1dpRkhMa2s/view" rel="nofollow noreferrer">my notes</a>, too). The integral representation and the property $\Gamma(x+1)=x\,\Gamma(x)$ provide an analytic continuation to the complex plane: $\Gamma(x)$ turns out to be a meromorphic function with simple poles at $0,-1,-2,-3,\ldots$ and residues $\frac{1}{0!},-\frac{1}{1!},\frac{1}{2!},-\frac{1}{3!},\ldots$ Weierstrass factorization theorem then gives
$$\Gamma(z+1) = e^{-\gamma z}\prod_{n\geq 1}\left(1+\frac{z}{n}\right)^{-1}e^{z/n} $$
where $\gamma=\lim_{n\to +\infty}\left(H_n-\log n\right)$. Here we go: by applying $\frac{d}{dz}\log(\cdot)$ to both sides of the previous identity, and by defining $\psi(z)$ as $\frac{\Gamma'(z)}{\Gamma(z)}$, we have
$$ \psi(z+1) = -\gamma+\sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{n+z}\right) $$
so the most natural extension of $H_n$ is given by
$$ H_n = \gamma+\psi(n+1)=\gamma+\int_{0}^{+\infty}\sum_{m\geq 1}\left(e^{-mu}-e^{-(m+n)u}\right)\,du =\gamma+\int_{0}^{+\infty}\frac{1-e^{-nu}}{e^u-1}\,du.$$</p>
<p>The <a href="https://en.wikipedia.org/wiki/Gamma_function#Properties_2" rel="nofollow noreferrer">duplication and reflection formulas</a> for the $\Gamma$ function give duplication and reflection formulas for the $\psi$ function, too, such that</p>
<p>$$ H_{n+1/2} = 2H_{2n+1}-H_n-2\log 2.$$
That can be proved from the integral representation only, if needed. The ultimate tool is <a href="http://mathworld.wolfram.com/GausssDigammaTheorem.html" rel="nofollow noreferrer">Gauss' Digamma Theorem</a>, essentially stating that $\psi(x)$ (so $H_{x-1}$ too) has a not-so-involved closed form for any $x\in\mathbb{Q}^+$.</p>
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301,318 | <p>As the title stated , what is the meaning of infinitely many ? When we say a set contains infinitely many elements, does this mean we cannot finish counting all the elements in the set ? Does infinitely many same as $\forall$ ? </p>
| Emily | 31,475 | <p>$\forall$ is just a shorthand way of saying "for all". It can be used for infinite or finite sets.</p>
<p>For example:</p>
<p>$$\forall x \in \{ 1, 2, 3\},\ x > 0.$$</p>
<p>"Infinitely many" means that there are not finitely many. In other words, "infinitely many" means that there does not exist some real integer $n$ such that you can describe the objects with a set of cardinality (size) $n$.</p>
<p>There are different types of infinitely many: countably many, which means the infinite set can be counted with a bijection to the integers, and uncountably many, which means the infinite set cannot be counted with a bijection to the integers.</p>
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162,324 | <p>Let $x_n$ be a sequence in a Hilbert space such that
$\left\Vert x_n \right\Vert=1$ and $ \langle x_n,\ x_m \rangle =0 $, for all $n \neq m$.</p>
<p>Let $ K= \{ x_n/ n : n \in \mathbb{N} \} \cup \{0\} $.</p>
<p>I need to show that $K$ is compact, $\operatorname{co}(K)$ is bounded, but not closed and finally find all the extreme points of $ \overline{\operatorname{co}(K)} $ .</p>
| Michael R. Chernick | 30,995 | <p>Setting the partial derivative of L with respect to lambda f to 0 forces g(x,y)=0. Requiring partial of L with respect x and y to 0 will lead to a local extreme point subject to g(x,y) = 0. Because of the form of L this could be a minimum.</p>
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