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 |
|---|---|---|---|---|
3,014,253 | <p>I am looking for examples of non-abelian groups of arbitrarily large size with the following properties </p>
<ol>
<li>Have order <span class="math-container">$p^a$</span>, where <span class="math-container">$a$</span> is a positive integer and <span class="math-container">$p$</span> is prime.</li>
<li>Contain an abelian subgroup of order <span class="math-container">$p^{a-2}$</span>.</li>
</ol>
<p>I know one example which is the quaternion group. I am looking for more examples of groups of arbitrarily large size.</p>
| user3482749 | 226,174 | <p>Dihedral groups of order <span class="math-container">$2^a$</span> have both properties (the subgroup being the cyclic one generated by the square of a highest-order element), but they also have a larger abelian subgroup of order <span class="math-container">$2^{a-1}$</span>, so might not be what you're after. </p>
|
3,014,253 | <p>I am looking for examples of non-abelian groups of arbitrarily large size with the following properties </p>
<ol>
<li>Have order <span class="math-container">$p^a$</span>, where <span class="math-container">$a$</span> is a positive integer and <span class="math-container">$p$</span> is prime.</li>
<li>Contain an abelian subgroup of order <span class="math-container">$p^{a-2}$</span>.</li>
</ol>
<p>I know one example which is the quaternion group. I am looking for more examples of groups of arbitrarily large size.</p>
| user10354138 | 592,552 | <p>Take, for example, the direct product of a nonabelian group of order <span class="math-container">$p^3$</span> with an abelian group of order <span class="math-container">$p^{a-3}$</span>.</p>
|
4,269,434 | <p>I want to compute the gradient of the vector function <span class="math-container">$f(\vec{x}) = \|\vec{x} - \vec{a}\|$</span>, I have a try, but the result is kind of strange to me.</p>
<p>so here is my steps
<span class="math-container">\begin{align*}
\nabla\|\vec{x} - \vec{a}\| & = \nabla\sqrt{\sum_{i = 1}^{n}(x_i - a_i)^2} \\
& = \frac{\nabla \sum_{i = 1}^{n} (x_i - a_i)}{2\sqrt{\sum_{i = 1}^{n}(x_i - a_i)^2}}\\
\end{align*}</span></p>
<p>since the gradient of <span class="math-container">$\sum_{i = 1}^{n} (x_i - a_i)$</span> will give back a vector, so I think this should be really the original vector scaling by 2, <span class="math-container">$2(\vec{x} - \vec{a})$</span>, and then
<span class="math-container">\begin{align*}
\nabla\|\vec{x} - \vec{a}\| & = \frac{\nabla \sum_{i = 1}^{n} (x_i - a_i)}{2\sqrt{\sum_{i = 1}^{n}(x_i - a_i)^2}}\\
& = \frac{2(\vec{x}-\vec{a})}{2\sqrt{\sum_{i = 1}^{n}(x_i - a_i)^2}}\\
& = \frac{\vec{x} - \vec{a}}{\| \vec{x} - \vec{a} \|_2}
\end{align*}</span>
so it's seems that the gradient of function <span class="math-container">$f(\vec{x}) = \|\vec{x} - \vec{a}\|$</span> is really doing the vector normalization? Am I making something mistake at here can someone check this with me?</p>
| Ted Shifrin | 71,348 | <p>This is perfectly correct. If you remember the conceptual interpretation of the gradient, it makes good sense: The norm increases most rapidly, and with rate 1, if you move from <span class="math-container">$x$</span> radially outward from <span class="math-container">$a$</span>.</p>
|
1,794,855 | <p>I need to prove that for every three integers $(a,b,c)$, the $\gcd(a-b,b-c) = \gcd(a-b,a-c)$. Assuming that a $a \ne b$.</p>
<p>Having:</p>
<p>$d_1 = \gcd(a-b,b-c)$</p>
<p>$d_2 = \gcd(a-b,a-c)$</p>
<p>How do i prove $d_1 = d_2$?</p>
| Maria | 308,011 | <p>The trick is the using the identity $k { n \choose k} = n {n-1 \choose k-1}$.
$$\begin{align*}
&\sum_{k=1}^n k { n \choose k } p^k (1-p)^{n-k}\\
&= \sum_{k=1}^n n { n-1 \choose k-1} p^k (1-p)^{n-k}\\
&=np \sum_{k=1}^n {n-1 \choose k-1} p^{k-1} (1-p)^{(n-1)-(k-1)}\\
&=np (p+(1-p))^n\\
&=np
\end{align*}$$</p>
|
4,238,241 | <blockquote>
<p>Let tangents <span class="math-container">$PA$</span> and <span class="math-container">$PB$</span> on hyperbola from any point <span class="math-container">$P$</span> on the Director Circle of hyperbola such that <span class="math-container">$d(P,AB).d(C,AB)=4d(S_1,PA).d(S_2,PA)$</span> and <span class="math-container">$|AS_1-AS_2|=4$</span> where <span class="math-container">$d(P,AB)$</span> denotes distance of point <span class="math-container">$P$</span> to the line <span class="math-container">$AB$</span> and <span class="math-container">$S_1$</span> and <span class="math-container">$S_2$</span> are the foci and <span class="math-container">$C$</span> is centre of the hyperbola. If asymptotes of hyperbola pass through the point <span class="math-container">$(1,2)$</span> and a line <span class="math-container">$y=4$</span> intersects the branch of hyperbola which lies entirely on the first quadrant only at a point then find the length of traverse axis of the hyperbola and its equation.</p>
</blockquote>
<p>This is originally a question i had encountered in a test I had given, and i do know however that the director circle is the circle from which a pair of perpendicular tangents can be drawn and from the question it becomes clear that <span class="math-container">$2a=4$</span> but I am unable to make any use of the equation <span class="math-container">$d(P,AB).d(C,AB)=4d(S_1,PA).d(S_2,PA)$</span> maybe it is some property of hyperbola I am unaware of, in the given 'solution' they had directly stated the value of <span class="math-container">$b$</span> (of hyperbola) but I have no idea how they got that and I would highly appreciate any help with this question as it had been bugging me for quite a while now, thanks!</p>
<p><a href="https://i.stack.imgur.com/1twDX.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1twDX.jpg" alt="enter image description here" /></a></p>
| user10354138 | 592,552 | <p><strong>Claim</strong>: Assuming a hyperbola has a (real) director circle and satisfies <span class="math-container">$d(P,AB)\cdot d(C,AB)=K\cdot d(S_1,PA)\cdot d(S_2,PA)$</span> for all <span class="math-container">$P$</span> on the director circle (<span class="math-container">$A,B,C,S_1,S_2$</span> same definition as in question, <span class="math-container">$K$</span> is a constant), then the semimajor axis <span class="math-container">$a$</span> and the semi-minor axis <span class="math-container">$b$</span> are related by
<span class="math-container">$$
\frac{b^2}{a^2}=\frac{K-1}{K}.
$$</span></p>
<p><em>Proof</em>: Choose Cartesian coordinates <span class="math-container">$(x',y')$</span> so the hyperbola is
<span class="math-container">$$
\frac{x'^2}{a^2}-\frac{y'^2}{b^2}=1,\quad a>b.
$$</span>
Its director circle is <span class="math-container">$x'^2+y'^2=a^2-b^2$</span>. Pick point <span class="math-container">$P$</span> to be <span class="math-container">$(x'_P,y'_P)=(\sqrt{a^2-b^2},0)$</span>. The tangents from <span class="math-container">$P$</span> to the hyperbola is <span class="math-container">$y'=\pm (x'-x'_P)$</span>. This gives the points <span class="math-container">$A,B$</span> are on <span class="math-container">$x'=\frac{a^2}{x'_P}$</span>. We also know the foci <span class="math-container">$S_1,S_2$</span> are <span class="math-container">$(\pm\sqrt{a^2+b^2},0)$</span>, so feeding that in <span class="math-container">$d(P,AB)d(C,AB)=Kd(S_1,PA)d(S_2,PA)$</span> we have
<span class="math-container">$$
\left\lvert x'_P-\frac{a^2}{x'_P}\right\rvert\cdot\frac{a^2}{x'_P}=K\frac{\lvert x'_P-\sqrt{a^2+b^2}\rvert}{\sqrt2}\cdot\frac{\lvert x'_P+\sqrt{a^2+b^2}\rvert}{\sqrt2}
$$</span>
yielding the stated relation upon simplification. QED.</p>
<p>Everything else should follow after noting the comments above.</p>
|
3,065,572 | <p>Function <span class="math-container">$f: (0, \infty) \to \mathbb{R}$</span> is continuous. For every positive <span class="math-container">$x$</span> we have <span class="math-container">$\lim\limits_{n\to\infty}f\left(\frac{x}{n}\right)=0$</span>. Prove that <span class="math-container">$\lim\limits_{x \to 0}f(x)=0$</span>. I have tried to deduce something from definition of continuity, but with no effect.</p>
| Yanko | 426,577 | <p>The only way to answer this question is to ask yourself at every given point "what does it mean?"</p>
<p>We're given that <span class="math-container">$A\subseteq B$</span>. <strong>What does it mean?</strong></p>
<p>It means that every element in <span class="math-container">$A$</span> belongs to <span class="math-container">$B$</span>, formally for all <span class="math-container">$x\in A$</span> we have that <span class="math-container">$x\in B$</span>.</p>
<p>We're also given that <span class="math-container">$B\subseteq C$</span>. <strong>What does it mean?</strong></p>
<p>It means that every element in <span class="math-container">$B$</span> belongs to <span class="math-container">$C$</span>, formally for all <span class="math-container">$x\in B$</span> we have that <span class="math-container">$x\in C$</span>.</p>
<p>We need to prove that <span class="math-container">$A\subseteq C$</span>. <strong>What does it mean?</strong></p>
<p>It means that given an <span class="math-container">$x\in A$</span> we need to show that <span class="math-container">$x\in C$</span>.</p>
<p>Can you finish now?</p>
|
2,091,589 | <p>The remainder when a polynomial $f(x)$ is divided by $(x-2)(x+3)$ is $ax+b$. When $f(x)$ is divided by $(x-2)$, then remainder is $5$. $(x+3)$ is a factor of $f(x)$. Find the values of $a$ and $b$. I am thinking of using the remainder and factor theorem to solve this however their quotients are different. Can anyone please show me how? Thanks</p>
| Andreas Caranti | 58,401 | <p>Your assumptions are that there are polynomials $q,s,t$ such that
$$
\begin{cases}
f(x) = (x-2)(x+3) q(x) &+ a x + b\\
f(x) = (x -2) s(x) &+ 5\\
f(x) = (x + 3) t(x)
\end{cases}
$$</p>
<p>Now calculate $f(2)$ and $f(-3)$.</p>
|
2,091,589 | <p>The remainder when a polynomial $f(x)$ is divided by $(x-2)(x+3)$ is $ax+b$. When $f(x)$ is divided by $(x-2)$, then remainder is $5$. $(x+3)$ is a factor of $f(x)$. Find the values of $a$ and $b$. I am thinking of using the remainder and factor theorem to solve this however their quotients are different. Can anyone please show me how? Thanks</p>
| Bernard | 202,857 | <p>Simple: first write $\;f(x)=(x+3)g(x)$, and perform the <em>Euclidean division</em> of $g(x)$ by $x-2$:
$$g(x)=(x-2)q(x)+g(2).$$
Now $f(2)=5g(2)$, so that
$$f(x)=(x-2)(x+3)q(x)+ g(2)(x+3)=(x-2)(x+3)q(x)+ \underbrace{\frac{f(2)}5(x+3)}_{\text{remainder}}. $$</p>
|
1,330,858 | <p>Suppose I'm at $(x=0,y=0)$ and I want to get to $(x=1,y=1)$. The shortest path is the diagonal and it has length $\sqrt{2}$. But what if I'm only allowed to make moves in coordinate directions---e.g., $1/2$ along $x$, $1/2$ along $y$, another $1/2$ along $x$, and a final $1/2$ along $y$. Then the length of my path is $2$. In fact, any coordinate-constrained path has length $2$. Let the path $p_n$ be $1/n$ along $x$, followed by $1/n$ along $y$, followed by $1/n$ along $x$, etc., until I get to $(1,1)$. Presumably, the limit of $p_n$ as $n\rightarrow\infty$ is the diagonal line. But the path length of each $p_n$ is $2$, while the path length of the limit is $\sqrt{2}$.</p>
<p>Weird, right? Is this just an example that shows that you can't exchange limit and path length?</p>
| Zach Stone | 38,565 | <p>It comes down to what kind of approximation you're working with. The sum of sides of a right triangle do NOT approximate it's hypotenuse. Even it the triangle is really small, the sides are still a bad approximation. To more specific, in an isocolese right triangle, with sides lengths $x$ and hypotenuse, $y$, we have $2x^2 = y^2$. So $y= \sqrt{2}x$. So
$$
\lim_{x\to 0} \frac{x}{y} = \lim_{x\to 0}\frac{x}{\sqrt{2}x} = 1/\sqrt{2}
$$</p>
<p>This tells you your approximation will always be bad. We want the limit to be 1.</p>
<p>To contrast this, let's look at an example of good approximation. On the unit circle, consider two points $A$ and $B$. We want to approximate the arclength from $A$ to $B$ by the chord from $A$ to $B$. Call the arclength $\theta$ (we're mathematicians, we use radians) and call the length of the chord $x$. Doing some geometry, it's not hard to show that </p>
<p>$$
x = 2\sin(\theta/2)
$$ </p>
<p>A standard result from calculus will tell us that
$$
\lim_{\theta\to 0} \frac{\theta}{2\sin{\theta/2}}=1
$$</p>
<p>So the chord is a good approximation of the arc length for small arclength!</p>
|
2,063,380 | <p>I am absolutely stuck, reading Bott and Tu isomorphism of de Rham cohomology. Please help. On page 92,</p>
<p><a href="http://www.maths.ed.ac.uk/~aar/papers/botttu.pdf" rel="nofollow noreferrer">http://www.maths.ed.ac.uk/~aar/papers/botttu.pdf</a></p>
<p>Step 2. $r^{*}$ is injective. </p>
<p>$$ r(\omega)=D\phi^{'}=d\phi^{'},
~~~\delta \phi^{'}=0 $$ </p>
<p>I think the claim that $ \delta \phi^{'}=0 $ is not correct and hence neither is the proof. Can someone check the proof. It starts from page 91. </p>
<p>Edit: Here is why I think the proof is not correct.
Let's follow the proof from page 91 to 92 for two columns only.
In the injective part, we argue that we can write $\phi$ as a sum: </p>
<p>$$\phi=\phi^{'}+D\phi^{''}$$</p>
<p>where $\phi^{'}$ has only top component. This is shown on the previous page as follow. Let $\phi=(\alpha_0, \alpha_1)$ then by exactness there exists $\beta$ such that $\alpha-D\beta$ has only top component. Let's write this explicitly.
We set $\beta=\phi^{''}=(\beta_0, 0)$ and $\delta\beta_{0}=\alpha_{1}$ to get the notation on page 92.</p>
<p>$$D\phi^{''}=D(\beta_{0}, 0) = (d\beta_0, \delta \beta_0)$$ and thus:</p>
<p>$$\phi^{'}=\phi-D\phi^{''}=(\alpha_0-d\beta_0, \alpha_1-\delta \beta_0)=(\alpha_0-d\beta_0, 0)$$</p>
<p>so $$\delta\phi^{'}= \delta(\alpha_0-d\beta_0)=\delta\alpha_{0}-d\alpha_1$$ and there is no reason why we should assume $\delta\alpha_0-d\alpha_1=0$.</p>
| Pedro | 23,350 | <p>The crucial step is that one can always replace a cochain in $K^{**}$ by one with one with last component zero. </p>
<p>To see that $r*$ is onto, take a cocycle in $K^{**}$. By the first remark, this is represented by a pair of forms $(\eta_1,\eta_2)$, and the cocycle condition means in the vertical direction, that each $\eta_i$ is closed, and second, that $\eta_1=\eta_2$ in $U\cap V$. Thus there is a global form $\omega$ that restricts to $(\eta_1,\eta_2)$.</p>
<p>To see that $r*$ is injective, take a global form $\omega$ and suppose the restrictions $(\omega_1,\omega_2)$ have trivial class in $\operatorname{Tot} K^{**}$, so this is of the form $D\varphi$ for some $\varphi$. By the first remark, one can assume $\varphi$ has zero last component. Now $r^*\omega$ <strong>also</strong> has zero last component -- it has $(q,0)$ component $r^*\omega$ and $(q-1,1)$ component $0$, and because $D\varphi=r^*\omega$, it follows that $\delta\varphi=0$, so $\varphi$ is a globally defined form. But we also have $d\varphi=r^*\omega$, so $\omega$ is exact. </p>
<hr>
<p>Their proof can be phrased as follows. There is a double complex $C^{**}$ obtained by looking at the exact sequence of complexes as such a double complex:</p>
<p>$$ \Omega^*(M)\stackrel{r}\longrightarrow \Omega^*(U)\oplus \Omega^*(V) \longrightarrow \Omega^*(U\cap V)$$ </p>
<p>This is a first quadrant cohomological double complex, and it has exact rows. Now there is induced a map $$r: \Omega^*(M)\to \operatorname{Tot}(K^{**})$$</p>
<p>as described by Bott and Tu. Now one checks the cone of $r$ is exactly the total complex of $C^{**}$, and because $C^{**}$ has exact rows -- by virtue of the exactness of the Mayer-Vietoris sequnce --- its total complex is acyclic. Because $\operatorname{cone}(r^*)$ is acyclic, it follows that $r$ is a quasi-isomorphism, as desired. </p>
<p>So, let us show that if a first quadrant cohomological complex $C^{**}$ has exact rows, then $\operatorname{Tot}(C^{**})$ is acyclic. Write $d'$ for the vertical differential and $d''$ for the horizontal one. </p>
<p>Consider a cocycle $c=(c_{0,q},c_{1,q-1},\ldots,c_{q,0})$. Then $d''c_{q,0}=0$. Because the rows are exact, there is $b_{q,0}$ such that $d''b_{q,0}=c_{q,0}$. Then $c'=c-Db$ where $b=(0,\ldots,0,b_{q,0})$ has last component $0$, and $Dc'=0$. Continue in this way: inductively assume you have killed the last $j$ components, so now $d''c_{q-j,j}=0$, so there is $b_{q-j,j}$ such that $d''b_{q-j,j}=c_{q-j,j}$. Inductively, we reach $c''$ with only the <em>top component</em> $(0,q)$ nonzero. But if $Dc''=0$, this means that $d''_{0,q}c''_{0,q}=0$, and since $d''$ is injective at $(0,*)$ because our complex has exact rows, it follows that $c''=0$. This is <strong>exactly</strong> what Bott and Tu are doing. </p>
|
2,063,380 | <p>I am absolutely stuck, reading Bott and Tu isomorphism of de Rham cohomology. Please help. On page 92,</p>
<p><a href="http://www.maths.ed.ac.uk/~aar/papers/botttu.pdf" rel="nofollow noreferrer">http://www.maths.ed.ac.uk/~aar/papers/botttu.pdf</a></p>
<p>Step 2. $r^{*}$ is injective. </p>
<p>$$ r(\omega)=D\phi^{'}=d\phi^{'},
~~~\delta \phi^{'}=0 $$ </p>
<p>I think the claim that $ \delta \phi^{'}=0 $ is not correct and hence neither is the proof. Can someone check the proof. It starts from page 91. </p>
<p>Edit: Here is why I think the proof is not correct.
Let's follow the proof from page 91 to 92 for two columns only.
In the injective part, we argue that we can write $\phi$ as a sum: </p>
<p>$$\phi=\phi^{'}+D\phi^{''}$$</p>
<p>where $\phi^{'}$ has only top component. This is shown on the previous page as follow. Let $\phi=(\alpha_0, \alpha_1)$ then by exactness there exists $\beta$ such that $\alpha-D\beta$ has only top component. Let's write this explicitly.
We set $\beta=\phi^{''}=(\beta_0, 0)$ and $\delta\beta_{0}=\alpha_{1}$ to get the notation on page 92.</p>
<p>$$D\phi^{''}=D(\beta_{0}, 0) = (d\beta_0, \delta \beta_0)$$ and thus:</p>
<p>$$\phi^{'}=\phi-D\phi^{''}=(\alpha_0-d\beta_0, \alpha_1-\delta \beta_0)=(\alpha_0-d\beta_0, 0)$$</p>
<p>so $$\delta\phi^{'}= \delta(\alpha_0-d\beta_0)=\delta\alpha_{0}-d\alpha_1$$ and there is no reason why we should assume $\delta\alpha_0-d\alpha_1=0$.</p>
| Pedro | 23,350 | <p><em>To address the edit</em>: you're misunderstanding what they are doing. A generic element in the total complex <span class="math-container">$C^*(\mathfrak U)$</span> of <span class="math-container">$K^{**}$</span> is of the form <span class="math-container">$(a,b)$</span> with <span class="math-container">$a\in \Omega^*(U)\oplus \Omega^*(V)$</span> and <span class="math-container">$b\in \Omega^*(U\cap V)$</span>.</p>
<p>The map <span class="math-container">$r^* : \Omega^*(M)\longrightarrow C^*(\mathfrak U)$</span> sends <span class="math-container">$\omega$</span> to <span class="math-container">$(a,0)$</span> where <span class="math-container">$a=(\omega\mid_U,\omega\mid_V)$</span>. Take a closed form <span class="math-container">$\omega$</span>, and asume that <span class="math-container">$r^*(\omega)=(a,0)$</span> is a boundary in <span class="math-container">$C^*(\mathfrak U)$</span>. Then <span class="math-container">$(a,0) = D\varphi$</span> for some <span class="math-container">$\varphi$</span>. Now <span class="math-container">$\varphi$</span> is an element of <span class="math-container">$C^*(\mathfrak U)$</span>, and by the argument in the book, there is <span class="math-container">$\varphi'$</span> of the form <span class="math-container">$(\eta',0)$</span> so that <span class="math-container">$\varphi-\varphi'=D\nu$</span> in <span class="math-container">$C^*(\mathfrak U)$</span>. Thus <span class="math-container">$(a,0) = D\varphi'$</span>. Now <span class="math-container">$D\varphi'=D(\eta',0)=(d\eta',\delta \eta')$</span>, and since this equals <span class="math-container">$(a,0)$</span>, then
<span class="math-container">$$d\eta'=a$$</span>
<span class="math-container">$$\delta \eta'=0$$</span></p>
<p>The first equation says that <span class="math-container">$\omega$</span> is the boundary of <span class="math-container">$\eta'$</span> on each <span class="math-container">$U,V$</span>, the second says that <span class="math-container">$\eta'=(\eta_1,\eta_2) \in \Omega^*(U)\oplus \Omega^*(V)$</span> is a globally defined form, so <span class="math-container">$\omega$</span> is a boundary, as desired.</p>
|
4,416,001 | <p>Given <span class="math-container">$f(xy) = f(x+y)$</span> and <span class="math-container">$f(11) = 11$</span>, what is <span class="math-container">$f(49)$</span>?</p>
| Adrian Anelta Kacaribu | 1,042,585 | <p>f(11) = 11</p>
<p>f(49)=f(49.1)=f(50)
f(50)=f(25.2)=f(27)
f(27)=f(27.1)=f(28)
f(28)=f(4.7)=f(11)</p>
<p>f(49)=f(11)</p>
<p>So, f(49)= 11</p>
|
2,709,273 | <blockquote>
<p>Determine the kernel of the following group homomorphism:
$$
\phi\colon\mathbb Z/270\mathbb Z\to\mathbb Z/270\mathbb Z\colon\overline x\mapsto\overline{6x}.
$$
Then find the solutions of the following system of equations in $\mathbb Z/270\mathbb Z$:
\begin{align}
6x=3\mod 27\\
6x=2\mod 10
\end{align}</p>
</blockquote>
<p>Since $6*45=270$, the kernel is $\{\overline{45}, \overline{90}, \overline{135}, \overline{180}, \overline{225}, \overline{270}\}$. Instead of working with $6x$, I solved the following equations:
\begin{align}
a=3\mod 27\\
a=2\mod 10,
\end{align}
using the Chinese remainder theorem. I found: $a=192=32*6$. So I would guess they are looking for this answer:
$$
\left\{\overline{32+k45}:k\in\{1,2,3,4,5,6\}\right\}.
$$
But I'm a bit confused by their phrasing, because we initially solved for $x\in\{0,1,\dots,269\}$, and not for $\overline a\in\mathbb Z/270\mathbb Z$... So I could only sort of guess what I had to do, but could someone clarify their wording? Why can it be interpreted as: find $\overline a\in\mathbb Z/270\mathbb Z$, such that for
$$
\phi'=\theta\circ\phi,
$$
where
$$
\theta\colon\mathbb Z/270\mathbb Z\to\mathbb Z/27\mathbb Z\times\mathbb Z/10\mathbb Z\colon a\mapsto (a,a),
$$
we have $\phi'(\overline a)=(3,2)$.</p>
<p><em>edit</em></p>
<p>I think I got it: to find solutions in $\mathbb Z/n\mathbb Z$ just means to find solutions modulo $n$.</p>
| Dr. Sonnhard Graubner | 175,066 | <p>we can write $$6x=3+27m$$ and $$6x=2+10n$$ where $m,n$ are integers, from here we get
$$1=10n-27m$$ solving this Diophantine equation we get $$m=7+10k,n=19+27k$$ and from here you will get $x$</p>
|
251,559 | <p>First of all, this is a very silly question, but I finished high school a long time ago and I really don't remember much about some basic stuff.</p>
<p>I have:</p>
<p>$$T(x) = -\frac1{10}x^2 + \frac{24}{10}x - \frac{44}{10}$$</p>
<p>$T$ = temperature</p>
<p>$x$ = hour</p>
<p>And they ask me to find the hour in which the temperature reached zero. So I know I must replace:</p>
<p>$$0 = -\frac1{10}x^2 + \frac{24}{10}x - \frac{44}{10}$$</p>
<p>It's very easy, I know, but I really don't remember how to clear/free the $x$ in this case.
Could you guys help me?</p>
| Cameron Buie | 28,900 | <p>Let's multiply by $-10$ to clear the fractions and get $$0=x^2-24x+44.$$ It follows, then, that $$\begin{align}100 &= x^2-24x+144\\ &=x^2+2(-12)x+(-12)^2\\ &= (x-12)^2.\end{align}$$ The two numbers whose square is $100$ are $\pm 10$, so we have $x-12=\pm 10$, from which we have $x=2$ or $x=22$.</p>
|
234,477 | <p>In non-standard analysis, assuming the continuum hypothesis, the field of hyperreals $\mathbb{R}^*$ is a field extension of $\mathbb{R}$. What can you say about this field extension?</p>
<p>Is it algebraic? Probably not, right? Transcendental? Normal? Finitely generated? Separable?</p>
<p>For instance, I was thinking: Would it be enough to adjoin an infinitesimal to $\mathbb{R}$ and get $\mathbb{R}^*$? The axioms of hyperreals in Keisler's Foundations of Infinitesimal Calculus seem to suggest so, but I'm not sure. </p>
<p>I don't know how to approach this question, since infinitesimals and such things don't "result" from polynomials (like, say, complex numbers do). </p>
<p>Is $\mathbb{R}^*$ maybe isomorphic to the field of fractions of some polynomial ring? (This occured to me after thinking about $\mathbb{R}(\epsilon)$ and such things, where $\epsilon$ is an infinitesimal).</p>
| ShyPerson | 1,884 | <p>Briefly, one way to approach this question would be to try to construct alternative models of the hyperreals via the Compactness Theorem that either satisfy or fail to satisfy the properties you want. The Enderton text, <em>A Mathematical Introduction to Logic</em>, uses this kind of construction and has a very rigorous treatment of nonstandard analysis that you can use as a guide. </p>
<p>Since I only have 44 minutes left on this question, it could easily contain errors so I'm retagging your question to include logic and set theory so it can be properly checked.</p>
|
234,477 | <p>In non-standard analysis, assuming the continuum hypothesis, the field of hyperreals $\mathbb{R}^*$ is a field extension of $\mathbb{R}$. What can you say about this field extension?</p>
<p>Is it algebraic? Probably not, right? Transcendental? Normal? Finitely generated? Separable?</p>
<p>For instance, I was thinking: Would it be enough to adjoin an infinitesimal to $\mathbb{R}$ and get $\mathbb{R}^*$? The axioms of hyperreals in Keisler's Foundations of Infinitesimal Calculus seem to suggest so, but I'm not sure. </p>
<p>I don't know how to approach this question, since infinitesimals and such things don't "result" from polynomials (like, say, complex numbers do). </p>
<p>Is $\mathbb{R}^*$ maybe isomorphic to the field of fractions of some polynomial ring? (This occured to me after thinking about $\mathbb{R}(\epsilon)$ and such things, where $\epsilon$ is an infinitesimal).</p>
| Community | -1 | <p>There isn't any trouble constructing $\mathbb{R}(\epsilon)$ as a <em>formally real field</em> that is isomorphic (as a field) to $\mathbb{R}(x)$, and with $\epsilon$ a positive infinitesimal. (it may be easier to see the ordering by writing a rational function as a formal Laurent series in $\epsilon$ about 0)</p>
<p>It's easy to see that this is not a hyperreal field: it only has finite powers of $\epsilon$. In particular, if there were a positive transfinite positive integer $H$, then there would be an element $\epsilon^H$ that is smaller than every power of $\epsilon$ appearing in $\mathbb{R}(\epsilon)$.</p>
<p>($\mathbb{R}(\epsilon)$ also fails to have a square root of $\epsilon$. This can be fixed by passing to its <em>real closure</em>, but that would still fail the above property)</p>
<p>If you tried adjoining a second infinitesimal, a similar argument proves it can't be hyperreal.</p>
|
2,861,293 | <p>I found this statement with the proof:</p>
<p><a href="https://i.stack.imgur.com/bGRiZ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bGRiZ.jpg" alt="enter image description here" /></a></p>
<p>But I don't understand the proof. Where is the contradiction? We have a nonempty interval <span class="math-container">$J$</span> contained in the nonempty interval <span class="math-container">$(f(a),f(b))$</span>. Where is the problem?</p>
| Paul Frost | 349,785 | <p>There have been two answers, but the comments have shown that the arguments <em>at first glance</em> appeared a little obscure or cumbersome. I believe that everything is clarified now, either by enhancing the answers or by additional comments. But let us give another proof which is hopefully more transparent from the beginning.</p>
<p>Observe that in the theorem it is not specified whether $I$ is an open, closed or half-open interval and whether it is bounded or not. We shall prove the stronger result that $f$ is continuous on $I$.</p>
<p>Assume that $f$ is not continuous in a point $x_0 \in I$. Then $y_0^- < f(x_0)$ or $f(x_0) < y_0^+$. Note that only one of $y_0^\pm$ is defined if $x_0$ is a boundary point of $I$. We only consider the case $y_0^- < f(x_0)$, the other case is similar. The case under consideration can only occur when $I' = I \cap (-\infty,x_0) \ne \emptyset$. For $x < x_0$ we have $f(x) \le y_0^-$, for $x > x_0$ we have $f(x) \ge f(x_0)$. Therefore $(y_0^-,f(x_0) \cap f(I) = \emptyset$. For $\xi \in I'$ we have $f(\xi) \le y_0^-$ so that $(y_0^-,f(x_0) \subset [f(\xi),f(x_0)] \subset f(I)$ because $f(I)$ is an interval. This is the desired contradiction.</p>
|
4,198,496 | <p>If the value of <span class="math-container">$\int_{1}^2{e^{x^{2}}}dx$</span> is <span class="math-container">$\alpha$</span>, then what is <span class="math-container">$$\int_{e}^{e^{4}}{\sqrt{\log x}}dx$$</span></p>
<p>It just seems that substitution of any sort does not help. Is there some other way in which these two integrals are related?</p>
<p>The answer is <span class="math-container">$2e^4 - e - \alpha$</span></p>
| David | 950,170 | <p><span class="math-container">$$y=e^{x^2}\iff log y =x^2\iff \sqrt{\log y}=x$$</span>
The rectangle bounded by <span class="math-container">$(1,0), (2,0),(1,e^4), (2,e^4)$</span> has area <span class="math-container">$e^4.$</span>
The regions represented by the two integrals fill this rectangle with
an excess of a <span class="math-container">$1\times(e^4-e)$</span>-rectangle that touches the Y-axis.
(In the second integral, interchange <span class="math-container">$x$</span> and <span class="math-container">$y$</span> and use the equivalence displayed above.)
Therefore, <span class="math-container">$$\int_{e}^{e^{4}}{\sqrt{\log x}}\,dx=e^4 + (e^4-e)-a=2e^4 -e-a.$$</span></p>
|
1,225,359 | <p>Q: The sum of all the coefficients of the terms in the expansion of $(x+y+z+w)^{6}$ which contain $x$ but not $y$ is:</p>
<p>What I tried to do was make pairs of two terms and the expand it as a binomial expression and then again expand the binomial in the resulting series which gave me an expression with lot of unknowns and I got stuck. </p>
<p>Any help would be appreciated.
Thanks.</p>
| Jack D'Aurizio | 44,121 | <p><strong>Hint:</strong> What happens if you evaluate your expression in $x=1,y=0,z=1,w=1$?
And in $x=0,y=0,z=1,w=1$?</p>
|
2,410,648 | <p>$$\left(\begin{array}{ccc|c}
-1 & 2 & 1 & 3\\
3 & \alpha & -2 & \beta\\
-1 & 5 & 2 & 9
\end{array}\right)$$</p>
<p>I am struggling to solve this system $Ax=b$. I understand the basics of Gauss elimination but am not sure how to handle it with the alpha and beta. It needs to be solved giving conditions for alpha and beta for no solutions, one solution and infinitely many solutions. </p>
<p>I also need to compute the determinant of A and give a condition on $\alpha$ such that $A$ is invertible.</p>
| Martin Sleziak | 8,297 | <p>Let me start with some steps. Perhaps you'll be able to finish.</p>
<p>$$\left(\begin{array}{ccc|c}
-1 & 2 & 1 & 3\\
3 & \alpha & -2 & \beta\\
-1 & 5 & 2 & 9
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 &-2 &-1 &-3\\
-1 & 5 & 2 & 9\\
3 & \alpha & -2 & \beta
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 &-2 &-1 &-3\\
0 & 3 & 1 & 6\\
0 & \alpha+6 & 1 & \beta+9
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 &-2 &-1 &-3\\
0 & 3 & 1 & 6\\
0 & \alpha+3 & 0 & \beta+3
\end{array}\right)
$$</p>
<p><em>Side note:</em> Some comments to the strategy for this: It is good to avoid computations involving parameters if possible. That's the reason why I put the row with parameters into the last place. Similarly, that's the reason why in the last step I used third column rather than second. (It is much easier to subtract second row from the third, than to subtract $(\alpha+6)/3$ multiple of second row. At least when doing things by hand, it might help if we try to get simpler expressions.) </p>
<p>If you look at the above matrix, the last row corresponds to the equations $(\alpha+3)x_2=\beta+3$. What can you say about solutions of this equation depending on $\alpha$ and $\beta$. (What happens if $\alpha=-3$? What happens in $\alpha\ne-3$? Think about similar possibilities for $\beta$.) </p>
<p>We could as well introduce new parameters $c=\alpha+3$ and $d=\beta+3$ to simplify notation a bit. (In this way we have the equation $cx_2=d$ and we consider what happens depending on whether $c$, $d$ is zero or non-zero.)</p>
<hr>
<p>If $c\ne0$, then the system will have unique solution. Let us denote $t=d/c$ to simplify the notation:
$$\left(\begin{array}{ccc|c}
1 &-2 &-1 &-3\\
0 & 3 & 1 & 6\\
0 & c & 0 & d
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 &-2 &-1 &-3\\
0 & 3 & 1 & 6\\
0 & 1 & 0 & t
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 0 &-1 &-3+2t\\
0 & 1 & 0 & t\\
0 & 0 & 1 & 6-3t
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 0 & 0 & 3-t\\
0 & 1 & 0 & t\\
0 & 0 & 1 & 6-3t
\end{array}\right)
$$
where $t=\frac dc = \frac{\beta+3}{\alpha+3}$. (You can check by direct computation that this is indeed a solution.)</p>
<p>If $c=0$, then we have
$$\left(\begin{array}{ccc|c}
1 &-2 &-1 &-3\\
0 & 3 & 1 & 6\\
0 & 0 & 0 & d
\end{array}\right).$$</p>
<p>What happens if $d\ne0$? What happens if $d=0$?</p>
<hr>
<p>You mentioned that you also want to compute determinant of $A$. This can be done using basically the same row operations as above - you just need to know how <a href="https://proofwiki.org/wiki/Effect_of_Elementary_Row_Operations_on_Determinant" rel="nofollow noreferrer">row operations influence the determinant</a>.</p>
|
2,410,648 | <p>$$\left(\begin{array}{ccc|c}
-1 & 2 & 1 & 3\\
3 & \alpha & -2 & \beta\\
-1 & 5 & 2 & 9
\end{array}\right)$$</p>
<p>I am struggling to solve this system $Ax=b$. I understand the basics of Gauss elimination but am not sure how to handle it with the alpha and beta. It needs to be solved giving conditions for alpha and beta for no solutions, one solution and infinitely many solutions. </p>
<p>I also need to compute the determinant of A and give a condition on $\alpha$ such that $A$ is invertible.</p>
| Raffaele | 83,382 | <p>$$\det\left(
\begin{array}{rrr}
-1 & 2 & 1 \\
3 & a & -2 \\
-1 & 5 & 2 \\
\end{array}
\right)=-3-\alpha$$
if $-3-\alpha\ne 0$ that is $\alpha\ne -3$ there exists <strong>one and only one</strong> solution</p>
<p>$$\left\{\frac{3 a-b+6}{a+3},\frac{b+3}{a+3},\frac{3 (2 a-b+3)}{a+3}\right\}$$</p>
<p>if $\alpha=-3$ the determinant of the matrix $A$ of coefficients is zero $\text{rank}(A)=2$, so we must consider the augmented matrix $A|B$
$$\left(
\begin{array}{rrr|r}
-1 & 2 & 1 & 3 \\
3 & -3 & -2 & b \\
-1 & 5 & 2 & 9 \\
\end{array}
\right)$$</p>
<p>if $\text{rank}(A|B)\ne \text{rank}(A)$ the system has no solutions, therefore to have infinite solutions we must have $\text{rank}(A|B)=2$ so all $3-$rd order determinants extracted from $A|B$ must be zero</p>
<p>$$
\det \left(
\begin{array}{rrr}
2 & 1 & 3 \\
-3 & -2 & \beta \\
5 & 2 & 9 \\
\end{array}
\right) =\beta+3
$$</p>
<p>if $\beta\ne -3$ the system is impossible</p>
<p>if $\beta+3=0\to \beta=-3$ the determinant is zero the rank of the augmented matrix is equal to the rank of $A$ and we have infinite solutions</p>
<p>$\{t,\;t,\;3-t\}$</p>
|
162,520 | <p>This is a cross-post from <a href="https://math.stackexchange.com/questions/738094/good-book-on-analytic-continuation">MSE</a>.</p>
<p>For my Bachelor's thesis, I am investigating divergent series summation methods. One of those is analytic continuation. There are quite a few books on complex analysis that include a chapter or two on analytic continuation, but I would like to study this phenomenon in more detail. Could you please suggest a book on analytic continuation? Or a book on complex analysis with a heavy focus on analytic continuation?</p>
| Alexandre Eremenko | 25,510 | <p>English books are Hardy, Divergent series,
and P. Dienes, Taylor series:
an introduction to the theory of functions of a complex variable. Dover Publications, Inc., New York, 1957. (The title is somewhat misleading. This is a large book that indeed
contains an "introduction to complex variables" but it is also the most comprehensive book
in English on analytic continuation).
But the best book on my opinion is L. Bieberbach, Analytische Fortsetzung, Springer Berlin,
1955, which is available in German and Russian only. (Russian translation by Evgrafov
is better because many mistakes are corrected, and some proofs simplified).</p>
|
1,116,435 | <p>How do I get the value of </p>
<p>$$\lim_{x \rightarrow \frac{1}{4} \pi } \frac{\tan x-\cot x}{x-\frac{1}{4} \pi }?$$ </p>
<p>I need the steps without using L'hospital.</p>
| Dustan Levenstein | 18,966 | <p>idm's answer is good, but L´Hôpital's rule is often more directly an instance of a limit actually being the limit of a difference quotient defining a derivative, or, in this case, a pair of difference quotients defining a pair of derivatives:</p>
<p>$$\frac{\tan x-\cot x}{x-\frac{1}{4} \pi } = \frac{\tan x - 1}{x-\frac{1}{4} \pi } - \frac{\cot x -1}{x-\frac{1}{4} \pi }.$$</p>
<p>then we have</p>
<p>$$\lim_{x \rightarrow \frac{1}{4} \pi } \frac{\tan x-\cot x}{x-\frac{1}{4} \pi } = \lim_{x \rightarrow \frac{1}{4} \pi} \frac{\tan x - 1}{x-\frac{1}{4} \pi } - \frac{\cot x -1}{x-\frac{1}{4} \pi }$$
$$=\tan'\left(\frac{1}{4}\pi\right) - \cot'\left(\frac{1}{4}\pi\right)$$</p>
|
3,489,642 | <blockquote>
<p>Let <span class="math-container">$D_{2\cdot 8}$</span> be given by the group presentation <span class="math-container">$\langle x,y\mid xy = yx^{-1} , y^2 = e, x^8 = e\rangle$</span>. Let <span class="math-container">$G = F_{\{x,y\}}$</span> be the free group on two generators and <span class="math-container">$N = \langle\{xyx^{-1}y,y^2,x^8\}\rangle$</span>. Then</p>
<ol>
<li><span class="math-container">$N$</span> is a normal subgroup of <span class="math-container">$G$</span>.</li>
<li>The quotient group <span class="math-container">$G/N$</span> is isomorphic to <span class="math-container">$D_{2\cdot 8}$</span>.</li>
</ol>
</blockquote>
<p>For 1. I need to show that <span class="math-container">$gng^{-1}\in N$</span> for all <span class="math-container">$g\in G$</span>, <span class="math-container">$n\in N$</span>, i.e. that <span class="math-container">$N$</span> is closed under conjugation by elements of <span class="math-container">$G$</span>. But for example I don't see how we could write <span class="math-container">$yx^8y^{-1}$</span> as a product of elements of <span class="math-container">$\{xyx^{-1}y, y^2,x^8\}$</span> and their inverses.</p>
<p>For 2. I see that the quotient group must consist of <span class="math-container">$2\cdot8=16$</span> equivalence classes. But I don't see how we find these explicitly.</p>
<p>I believe it suffices to exhibit a group homomorphism <span class="math-container">$\varphi$</span> such that <span class="math-container">$\ker\varphi = \langle\{xyx^{-1}y,y^2,x^8\}\rangle$</span> and <span class="math-container">$\mathrm{Im}\varphi\cong G/N$</span>, by the first isomorphism theorem. But I am not sure how to define this homomorphism exactly.</p>
| Community | -1 | <p>If I remember correctly, <span class="math-container">$N$</span> should be <em>defined</em> as the smallest normal subgroup containing <span class="math-container">$\langle x^8,y^2, xyx^{-1}y\rangle $</span>. That is <span class="math-container">$N$</span> is its normalizer; hence normal.</p>
<p>The homomorphism <span class="math-container">$\varphi$</span> is just the canonical projection onto the quotient.</p>
<p>The order of the quotient can be established by using the commutativity relation <span class="math-container">$xy=y^{-1}x$</span> to write every word in the group in the form <span class="math-container">$x^ny^m$</span>.</p>
<p>It remains to be seen that the symmetries of a regular octagon satisfy the given relations.</p>
|
463,619 | <p>let us consider following problem:</p>
<p>Roger sold a watch at a profit of $10$%. If he had bought it at $10\%$ less and sold it for $13$ dollar less,then he would have made a profit of $15$%. What is the cost price of the watch?</p>
<p>suppose that price of watch is $x$ dollar, $profit=sell -cost $</p>
<p>so let us denote price of watch by $x$,then if he would buy it $10$ less,then cost of this would be $x-0.1*x=0.9*x$, and if it sold $13$ dollar less,then sold price would be $x-13$, profit is $15$%,then it would be $0.15*x$ right? or we have</p>
<p>$(x-13)-0.9*x=0.15*x$</p>
<p>but it makes negative equation like $13=-0.05*x$,so what is my mistake,how can i used information like Ronger sold watch at a profit of $10\%$? Thanks in advance</p>
| TZakrevskiy | 77,314 | <p>We have costs $c$ and sell price $p$. We know that he had $10\%$ profit, so $$p-c=0.1c$$
On the other hand, $$(p-13) - 0.9c = 0.15\times0.9c$$
Can you take it from here?</p>
<p><strong>Edit</strong></p>
<p>On the calulation of profit:</p>
<p>Profit is the difference between the selling price of the good and costs to create this good. In our case the costs are price of watch when it was bought. So, if the agent of the market buys a good for $p_{buy}$ and then resells it for $p_{sell}$, then the absolute profit is $p_{sell}-p_{buy}$. If we want to calculate our profit in percent, then we take them relatively to the costs (it's logical, because they represent our starting money). We have then the formula for relative profit $$\frac{p_{sell}-p_{buy}}{p_{buy}}\times 100\%.$$</p>
<p>As a numerical example, say you spend $100$USD to purchase a bottle of wine (quite a wine, I must say). You let it age for a couple of years and resell it for $120$USD. Your absolute profit is $$120USD-100USD = 20USD,$$
and your relative profit is $$\frac{120USD-100USD}{100USD}\times 100\%=20\%.$$</p>
|
4,341,297 | <p>Evaluate the given expression <span class="math-container">$$\sqrt[n]{\dfrac{20}{2^{2n+4}+2^{2n+2}}}$$</span> The given answer is <span class="math-container">$\dfrac{1}{4}$</span>. My attempt:
<span class="math-container">$$\sqrt[n]{\dfrac{20}{2^{2n+4}+2^{2n+2}}}=\sqrt[n]{\dfrac{20}{2^{2n}\cdot2^4+2^{2n}\cdot2}}\\=\sqrt[n]{\dfrac{20}{2^{2n}\cdot18}}=\sqrt[n]{\dfrac{10}{9\cdot2^{2n}}}$$</span> This is as far I as I am able to reach. Thank you!</p>
<p>PS I don't see how one can get <span class="math-container">$\dfrac14$</span>. For that we have to get something like <span class="math-container">$\sqrt[n]{A^n}$</span>.</p>
| CHAMSI | 758,100 | <p>You made a small mistake :<span class="math-container">\begin{aligned}\sqrt[n]{\dfrac{20}{2^{2n+4}+2^{2n+2}}}&=\sqrt[n]{\dfrac{20}{2^{2n}\cdot2^4+2^{2n}\cdot2^{\color{red}{2}}}}\\&=\sqrt[n]{\dfrac{20}{2^{2n}\cdot\color{red}{20}}}=\sqrt[n]{\dfrac{1}{2^{2n}}}=\frac{1}{4}\end{aligned}</span></p>
|
700,673 | <p>I saw this notation many times, but I don't understand why the $y$ variable is missing in the first term of the first equation below.</p>
<p>$$
\frac{\mathrm{d}y(x)}{\mathrm{d}x} = f(x,y)
$$</p>
<p>It just mean:</p>
<p>$$
\frac{\mathrm{d}y(x,y)}{\mathrm{d}x} = f(x,y)
$$</p>
<p>?</p>
<hr>
<h2>Tell me if I'm wrong</h2>
<p>In my understanding this is a linear ODE of first order in which $y$ is the unknown function and both of them (the derivate and the unknown one) have just a single variable.</p>
<p>$$
y'(x) = f(x)
$$</p>
<p><em>A simple example</em></p>
<p>$$
\begin{align}
y'(x) &= 2x\\
\int y'(x)\!\mathop{}\mathrm{d}x &= \int 2x\!\mathop{}\mathrm{d}x\\
y(x)&= x^2+\mathrm{C}\\
\end{align}
$$</p>
<p>Is this case:</p>
<p>$$
\dfrac{\mathrm dy}{\mathrm dx}(x)=f(x,y(x))
$$</p>
<p>would it be something like this?</p>
<p>$$
\begin{align}
y'(x,y) &= 2x + 2y\\
\iint y'(x,y)\!\mathop{}\mathrm{d}x\!\mathop{}\mathrm{d}y &= \iint 2x + 2y\!\mathop{}\mathrm{d}x\!\mathop{}\mathrm{d}y\\
y(x,y) &= x^2 + y^2+\mathrm{C}\\
\end{align}
$$</p>
<hr>
<p><em>For Christian Blatter</em></p>
<blockquote>
<p>This is what I would expected to be your exaplanation. Can you tell me
why my version is wrong?</p>
</blockquote>
<p>When dealing with ODEs for the first time we are given a function $f:\ (x,y)\mapsto f(x,y)$ defined in some region $\Omega$ of the $(x,y)$-plane. For each $(x,y)\in\Omega$ the value $f(x,y)$ is to be interpreted as a slope assigned to the point $(x,y)$. Therefore we are given a field of "line elements" of various slopes covering $\Omega$.</p>
<p>Given this "slope field" we are interested in curves $$\gamma:\ x\mapsto \gamma(x)=\bigl(x,\gamma(x)\bigr)\qquad(a<x<b)$$ lying in $\Omega$ that have for each of their points $\bigl(x,\gamma(x)\bigr)$ the slope $f\bigl(x,\gamma(x)\bigr)$ prescribed there. This means that we should have
$$\gamma'(x)\equiv f\bigl(x,\gamma(x)\bigr)\qquad(a<x<b)\ .\tag{1}$$</p>
| Christian Blatter | 1,303 | <p>Your difficulty stems from the use of the letter $y$ for two different purposes: (a) as <em>coordinate variable</em> in the $(x,y)$-plane, and (b) as <em>variable for (unknown) functions</em> $x\mapsto y(x)$ whose graphs are lying in the $(x,y)$-plane.</p>
<p>When dealing with ODEs for the first time we are given a function $f:\ (x,y)\mapsto f(x,y)$ defined in some region $\Omega$ of the $(x,y)$-plane. For each $(x,y)\in\Omega$ the value $f(x,y)$ is to be interpreted as a <em>slope</em> assigned to the point $(x,y)$. Therefore we are given a field of "line elements" of various slopes covering $\Omega$.</p>
<p>Given this "slope field" we are interested in curves $$\gamma:\ y=\phi(x)\qquad(a<x<b)$$ lying in $\Omega$ that have at each of their points $\bigl(x,\phi(x)\bigr)$ the slope $f\bigl(x,\phi(x))$ prescribed there. Note that these curves $\gamma$ are considered as graphs of unknown functions $\phi:\>x\mapsto y=\phi(x)$. The condition about the slopes means that we should have
$$\phi'(x)\equiv f\bigl(x,\phi(x)\bigr)\qquad(a<x<b)\ .\tag{1}$$
Now since centuries it is common to denote the unknown functions in such a problem not by $\phi$, or some other letter, but by the same letter as "the coordinate in which they take values", i.e., $y$ in our case. In view of $(1)$ this means that we are looking for functions $x\mapsto y(x)$ such that
$$y'(x)\equiv f\bigl(x,y(x)\bigr)\qquad(a<x<b)\ .\tag{2}$$
The condition $(2)$ is then further abbreviated to
$$y'=f(x,y)\qquad\bigl((x,y)\in\Omega\bigr)\ .\tag{3}$$
The <em>ODE</em> $(3)$ encodes the ideas described here and condenses them into four letters plus some extra tokens. </p>
<p><strong>Update</strong> concerning your "understanding": The unknown functions are functions of <em>one</em> variable $x$, whether one writes them as $\phi:\>x\mapsto y=\phi(x)$ or as $x\mapsto y(x)$. These functions have derivatives with respect to $x$ that are again functions of the one variable $x$. Therefore expressions of the form $y(x,y)$ or ${dy(x,y)\over dx}$ don't make sense. On the other hand the so-called <em>right side</em> of the ODE, a given function $f(x,y)$ defining the desired slope at each point $(x,y)\in\Omega$ is (in general) a function of the <em>two</em> variables $x$ and $y$. In some special cases, as with the super-simple ODE $y'=f(x)$, or with the ODE $y'=g(y)$ only one variable appears in $f$. The interpretation of this phenomenon is the following: In the first case the assigned slope is constant along vertical lines, and in the second case it is constant along horizontal lines.</p>
|
3,933,011 | <p>I know how to solve for the basic about this. But in this problem</p>
<p><span class="math-container">$\displaystyle{\frac{1}{2\pi i} \oint_{|z|=1} \frac{\cos \left(e^{-z}\right)}{z^2} dz}$</span>,</p>
<p>I don't know how to start. Can somebody help me or guide me about this? Or just give me a hint. Thanks. I would appreciate the help. Thanks.</p>
| Raghav | 164,237 | <p>First, recall the following form of Cauchy's integral formula. Let <span class="math-container">$f$</span> be a function holomorphic on some neighborhood of unit disk, then
<span class="math-container">$$f^{(n)}(w)=\frac{n!}{2i\pi}\oint_{|z|=1}\frac{f(\zeta)}{(\zeta-w)^{n+1}}d\zeta.$$</span></p>
<p>The above theorem is sometimes only mentioned for <span class="math-container">$n=0,$</span> but it is true more generally. Once you have the above formula, you observe that <span class="math-container">$f(z)=\cos(e^{-z})$</span> is holomorphic in the disk, and taking <span class="math-container">$w=0,$</span> you obtain
<span class="math-container">$$f'(0)=\frac{1}{2i\pi}\oint_{|z|=1}\frac{f(z)}{z^2}dz.$$</span></p>
<p>This is what was mentioned in the comment above. Therefore, you job reduces to finding the derivative <span class="math-container">$\frac{d}{dz}\vert_{z=0}\cos(e^{-z}),$</span> which is a straightforward computation using chain-rule.</p>
|
454,426 | <blockquote>
<p>In set theory and combinatorics, the cardinal number $n^m$ is the size of the set of functions from a set of size m into a set of size $n$.</p>
</blockquote>
<p>I read this from this <a href="http://en.wikipedia.org/wiki/Empty_product#0_raised_to_the_0th_power" rel="nofollow noreferrer">Wikipedia page</a>.</p>
<p>I don't understand, however, why this is true. I reason with this example in which $M$ is a set of size $5$, and $N$ is a set of size $3$. For each element in set $M$, there are three functions to map the element from the set of size $5$ to an element in the set of size $3$. </p>
<p>By my reasoning, that means the total number of functions is just $3*5$, i.e. $3$ functions for each of the $5$ elements in the set. Why is it actually $3^5$? I saw on <a href="https://math.stackexchange.com/questions/209361/size-of-the-set-of-functions-from-x-to-y">this thread</a> that the number of functions from a set of size $n$ to a set of size $m$ is equivalent to "How many $m$-digit numbers can I form using the digits $1,2,...,n$ and allowing repetition?" I know how to answer that question, but I don't know why it's the same thing as finding the number of functions from the size $n$ set to the size $m$ set. </p>
| user84413 | 84,413 | <p>You can think of constructing your function by choosing the images of each element in the domain, one at a time. For the first element of the set, you have n choices (any element in the target space); for the second element of the set, you again have n choices, and so on. Therefore the number of ways to construct a function from the first set to the second is just $n\cdots n =n^m$.</p>
|
63,348 | <p>This question arises from a discussion with my friends on a commonly encountered IQ test questions: "What's the next number in this series 2,6,12,20,...". Here a "number" usually means an integer. I was wondering whether there is a systematical way to solve such problems.Let us call a point on a plane integer point if all the components of it are integers. I want to know the following:</p>
<p>Give a finite set of integer points, Can we always find a corresponding polynomial that passes all these points and maps integers to integers? </p>
| Lubin | 11,417 | <p>It's a topic I liked to cover when I was still teaching junior-level Algebra, even if it didn't fit in well with the other topics. You start with a function defined on the set of integers from $0$ to $n$ inclusive, and end with a polynomial of degree $\le n$ that agrees at those $n+1$ points, and if the values you started with are integers, your function always sends integers to integers. You take successive differences, as indicated by Robert Israel above, and then you list $f(0)$, $\Delta f(0)$, $\Delta^2f(0)$, up to $\Delta^n$, and use these as coefficients, which you multiply to $C_0(x)=1$, $C_1(x)=x$, $C_2(x)=x(x-1)/2$, etc., the binomial polynomials. Your assignment is to try it out for a few examples, and then prove that the method works.</p>
|
2,094,243 | <p>Given the norm $||(x,y)|| = 2|x| +\frac{1}{3}|y|$.
Sketch the open ball at the on the origin $(0,0)$, and radius $1$.</p>
<p>I understand that the sketch of an open ball withina set looks like the image attached, <a href="https://i.stack.imgur.com/j9HN4.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/j9HN4.jpg" alt="enter image description here"></a> in a general case, but have no idea how to sketch one applying the above norm to the situation. </p>
<p>i understand that in the case of $B_{r}(a)=\{x \in X | d(x,a) < r\}$ in this case, $a = (0,0)$, and $r = 1$. Could someone please help as to how to sketch it?</p>
<p>Thanks</p>
| pancini | 252,495 | <p>The example you drew is not a general case at all. That is what an open ball looks like in $\Bbb R^2$ under the euclidian metric.</p>
<p>We want to cover any point in the plane $(x,y)$ such that $2|x|+\frac{1}{3}|y|<1$.</p>
<p>In the first quadrant, $x$ and $y$ are positive, so we have $2x+y/3<1$, or the area under $y<3-6x$. In the second quadrant, $x$ is negative and $y$ is positive, so use $-2x+y/3<1$, or the line $y<3+2x$.</p>
<p>Et cetera. It should be fairly easy to cover all four cases and then make sure the boundary points are correct.</p>
|
1,577,174 | <p>If $a, b \in \mathbb{C}$, then we have the standard triangle inequality for the difference:</p>
<p>$$||a| - |b|| \le |a - b|.$$</p>
<p>I am wondering if this inequality generalizes to exponents greater that one.</p>
<blockquote>
<p>My question is, for $1 < p < \infty$, does there exists a constant $C_p$ such that $||a|^p - |b|^p| \le C_p|a - b|^p$ for all $a, b \in \mathbb{C}$?</p>
</blockquote>
<p>I am aware of the "standard" triangle inequality for $p > 1$:</p>
<p>$$|a +b|^p \le 2^{p}(|a|^p + |b|^p),$$</p>
<p>and that $2^p$ may not be sharpest constant possible. If I try to use this estimate to prove, say, that $|a|^p - |b|^p \le C_p |a - b|^p$, I get stuck with an extra term that I'm not sure what to do with.</p>
<p>$$|a|^p - |b|^p \le 2^p|a -b|^p + (2^p -1)|b|^p.$$</p>
<p>A resolution on this matter is greatly appreciated.</p>
| PhoemueX | 151,552 | <p>This is false for <span class="math-container">$p>1$</span>. In fact, by the mean value theorem, there is for <span class="math-container">$n\in \Bbb {N} $</span> some <span class="math-container">$\xi_n \in (n,n+1) $</span> with
<span class="math-container">$$
(n+1)^p - n^p = p \cdot \xi_n^{p-1}\to \infty
$$</span>
as <span class="math-container">$n\to \infty $</span>. But if your claim was true, the left-hand side would be bounded by <span class="math-container">$C_p $</span>.</p>
<hr>
<p>EDIT: For <span class="math-container">$p < 1$</span>, the inequality is true.</p>
<p>To see this, first note that <span class="math-container">$f(x) := (1 + x)^p \leq 1 + x^p =: g(x)$</span> for all <span class="math-container">$x \in [0,\infty)$</span>.
Indeed, <span class="math-container">$f,g : [0,\infty) \to [1,\infty)$</span> are continuous with
<span class="math-container">$f(0) = 1 = g(0)$</span> and that <span class="math-container">$f'(x) = p \cdot (1 + x)^{p-1} \leq p \cdot x^{p-1}$</span> since <span class="math-container">$p < 1$</span>.</p>
<p>Next, note that if <span class="math-container">$a,b > 0$</span> and <span class="math-container">$x := a/b$</span>, then multiplying the inequality <span class="math-container">$f(x) \leq g(x)$</span>
by <span class="math-container">$b^p$</span> leads to <span class="math-container">$(b + a)^p \leq b^p + a^p$</span> for all <span class="math-container">$a,b > 0$</span>.
For <span class="math-container">$b = 0$</span> or <span class="math-container">$a = 0$</span>, this trivially continues to hold.</p>
<p>Finally, for <span class="math-container">$a,b \in \Bbb{C}$</span>, we thus see
<span class="math-container">$$
|a|^p
= |(a - b) + b|^p
\leq (|a-b| + |b|)^p
\leq |a-b|^p + |b|^p
$$</span>
and hence <span class="math-container">$|a|^p - |b|^p \leq |a-b|^p$</span>.
By symmetry (swap <span class="math-container">$a,b$</span>), this implies <span class="math-container">$\big| |a|^p - |b|^p \big| \leq |a-b|^p$</span> as desired.</p>
|
677,859 | <p>$f(x)= f(x+1)+3$ and $f(2)= 5$, determine the value of $f(8)$.</p>
<p>I don't understand how $f(x)$ can equal $f(x+1)+3$</p>
| MPW | 113,214 | <p>This just means that if the point $(x,y)$ is on the graph of $f$, then so is the point $(x+1, y-3)$. So if you have a starting point, you can move to some specific other points on the graph by stepping $1$ unit to the right and $3$ units down. Then, since you are at a new point now, you can repeat this procedure: right $1$, down $3$; right $1$, down $3$; etc.</p>
<p>So, since you are told a starting point $(2,5)$, just step along until you reach a point with $x=8$. The sequence of points is $(\boxed{2},5),(\boxed{3},2),(\boxed{4},-1),(\boxed{5},-4),(\boxed{6},-7),(\boxed{7},-10),(\boxed{8},-13)$. So $f(8)=-13$.</p>
<p>(Remember, the points on the graph are pairs $\left( x,f(x)\right)$ for various values of $x$.)</p>
|
3,489,280 | <p>If <span class="math-container">$f_n → f$</span> and <span class="math-container">$g_n → g$</span>, does <span class="math-container">$f_n g_n → fg$</span> in the space <span class="math-container">$C[0, 1]$</span> for the norms <span class="math-container">$||.||_1$</span> and <span class="math-container">$||.||_∞$</span></p>
<p>Give a proof or counterexample for each.</p>
<p>I know that <span class="math-container">$||.||_1$</span> is the sum of the magnitudes of the values, and <span class="math-container">$||.||_∞$</span> is the biggest magnitude of the values. From this I assume <span class="math-container">$||.||_1$</span> DOES work whilst <span class="math-container">$||.||_∞$</span> doesn't, but I am unsure on if this is true, and if so, how to show it.</p>
<p>Thanks in advance</p>
| Wlod AA | 490,755 | <blockquote>
<p>In my earlier answer, I've addressed only the uniform norm; as, @zhw. noted, I sloppily missed the <span class="math-container">$\|.\|_1$</span>-norm (shame on me) which zhw. did solve. And still, to make up for my misdeed, let me present another example which shows that <span class="math-container">$\|.\|_1$</span>-norm is not continuously multiplicative.</p>
</blockquote>
<p>Let <span class="math-container">$\,a>0\,$</span> and <span class="math-container">$\,0<b\le 1\,$</span> be two otherwise arbitrary reals. Define:</p>
<p><span class="math-container">$$ \forall_{x\in[0;1]}\quad F_{a\,b}(x)
\ :=\ \max(a\!\cdot\!(b-x)\ \ 0) $$</span></p>
<p>hence
<span class="math-container">$$ \forall_{x\in[0;1]}\quad F^2_{a\,b}(x)\ =
\ \min(a^2\!\cdot\!(b-x)^2\ \ 0) $$</span></p>
<p>so that</p>
<p><span class="math-container">$$ \|F_{a\ b}\|_{_1}\ =\ \frac 12\cdot a\cdot b^2 $$</span></p>
<p>and</p>
<p><span class="math-container">$$ \|F^2_{a\ b}\|_{_1}\ =\ \frac 13\cdot a^2\cdot b^3 $$</span></p>
<p>Define</p>
<p><span class="math-container">$$ \forall_{n\in\Bbb N}\quad f_n := F_{_{n^5\,\ n^{-3}}} $$</span></p>
<p>Then</p>
<p><span class="math-container">$$ \|f_n\|_{_1} = \frac 1{2\cdot n}\quad\to\quad 0 $$</span>
while
<span class="math-container">$$ \|f^2_n\|_{_1} = \frac n3\quad\to\quad\infty $$</span></p>
<p>This shows the mentioned lack of continuity. <em><strong>Great!</strong></em></p>
|
1,914,752 | <p>dividing by a whole number i can describe by simply saying split this "cookie" into two pieces, then you now have half a cookie. </p>
<p>does anyone have an easy way to describe dividing by a fraction? 1/2 divided by 1/2 is 1</p>
| ZirconCode | 278,375 | <p>Dividing by a fraction is the same as multiplying by the reciprocal of the fraction. I think this is the most intuitive approach when trying to teach this concept.</p>
<p>For multiplication we first split the cookie up into denominator-number of pieces, and then take numerator-number of pieces of those pieces.</p>
|
162,611 | <p>I am working with the square-roots of square symmetric matrices. The answers are to be binary symmetric matrices.</p>
<p>If we take the matrix $$M = \begin{pmatrix}1&1&1&0&0&0&0\\1&0&0&1&1&0&0\\1&0&0&0&0&1&1\\0&1&0&0&1&0&1\\0&1&0&1&0&1&0\\0&0&1&0&1&1&0\\0&0&1&1&0&0&1 \end{pmatrix},$$ then $M^2$ will yield a $7 \times 7$ matrix with $3$ down the main diagonal and $1$ elsewhere.Once we have the matrix $M^2$ then we want to take the square root of it in order to get M.Now this is the simplest of all examples that I can give. M is the incidence matrix of a symmetric balanced incomplete block design. This design has 7 varieties, 7 blocks,each variety occurs in 3 blocks,each block contains 3 varieties, each pair of varieties occurs in exactly one block. If you are familiar with block designs then the parameters are (v,b,r,k,l)=(7,7,3,3,1) known as the fano plane.</p>
| Will Jagy | 10,400 | <p>Had to screw around with the field of four elements. Not that bad. The pattern is different, there are the expected five null vectors, but the vector $(1,1,1)$ is orthogonal to all five, so I put that one first. The result is a symmetric matrix $P,$ such that $P^2$ is a 21 by 21 matrix with 5's on the main diagonal and 1's elsewhere.</p>
<p>$$ P \; = \;
\left( \begin{array}{c|cccc|cccc|cccc|cccc|cccc}
0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ \hline
1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 \\ \hline
1 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 \\ \hline
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 \\ \hline
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \hline
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0
\end{array}
\right).
$$</p>
|
28,456 | <p>I built a <code>Graph</code> based on the permutations of city's connections from :</p>
<pre><code>largUSCities =
Select[CityData[{All, "USA"}], CityData[#, "Population"] > 600000 &];
uScityCoords = CityData[#, "Coordinates"] & /@ largUSCities;
Graph[#[[1]] -> #[[2]] & /@ Permutations[largUSCities, {2}] ,
VertexCoordinates -> Reverse[uScityCoords, 2], VertexStyle -> Red,
Prolog -> {LightBrown, CountryData["USA", "FullPolygon"]},ImageSize -> 650]
</code></pre>
<p>It looks like this:
<img src="https://i.stack.imgur.com/1ea6c.png" alt="graph"></p>
<p>My question, is there any way to have the Graph like this?
<img src="https://i.stack.imgur.com/ajwOT.png" alt="enter image description here"></p>
| bobthechemist | 7,167 | <p>Revised:</p>
<p>First, let's get some map data. This is an 8MB file and you might want to save it locally.</p>
<pre><code>map = First@
Import["http://www2.census.gov/geo/tiger/TIGER2009/tl_2009_us_\
state.zip", "Data"];
</code></pre>
<p>Then, look at the answers that are better than yours (nod nod <a href="https://mathematica.stackexchange.com/users/61/cormullion">cormullion</a>), and steal from them. I'm taking the <code>stateConnections</code> and <code>stateData</code> he defines. I won't reprint them here.</p>
<pre><code>shape = "Geometry" /. map;
names = "STUSPS" /. "LabeledData" /. map[[4]] // Quiet;
cshape = shape[[DeleteCases[Range[56],
Alternatives @@ {1, 5, 7, 19, 27, 28, 33, 38}]]];
cnames = names[[DeleteCases[Range[56],
Alternatives @@ {1, 5, 7, 19, 27, 28, 33, 38}]]]; shape =
"Geometry" /. map;
names = "STUSPS" /. "LabeledData" /. map[[4]] // Quiet;
themap = Show[
MapThread[
Graphics@{EdgeForm[{White, Thick}], FaceForm[LightGray],
Tooltip[#1, #2]} &, {cshape, cnames}]];
(* Make map points with Cormullion's data *)
cpoint2 = Reverse /@ Flatten[ToExpression[
Cases[
Map[StringSplit[#, ","] &,
stateData], {#, x__, y__} -> {x, y}]] & /@ cnames, 1];
(* Make adjacency matrix with Cormullion's data *)
Table[Map[{i, #} &,
Flatten@Position[cnames,
Alternatives @@
Cases[stateConnections, x_ /; x[[1]] == cnames[[i]]][[1,
2 ;;]]]], {i, 1, 48}];
sam2 = SparseArray[Flatten[%, 1] -> 1];
(* Make adjacency graph *)
graph = AdjacencyGraph[Range[Length[cpoint2]], sam2,
VertexCoordinates -> cpoint2];
Show[themap, graph, Background -> LightBlue]
</code></pre>
<p>Here's what we get:</p>
<p><img src="https://i.stack.imgur.com/4Q4EK.png" alt="connection map"></p>
|
2,300,855 | <p>Do there exist two non-equivalent knots which are indistinguishable by <a href="https://en.wikipedia.org/wiki/Fox_n-coloring" rel="nofollow noreferrer">Fox $n$-colouring</a> for every positive integer $n$?</p>
<p>That is, do there exist non-oriented knots $K_0$ and $K_1$ which are different (here I would like to exclude the case that one is the mirror image of the other) and for every $n\in\mathbb{Z}^+$, $\mathrm{col}_n(K_0)=\mathrm{col}_n(K_1)$, where $\mathrm{col}_n(K)$ denotes the number of distinct Fox $n$-colourings of a knot $K$? </p>
<p>How about the case of links?</p>
| Zuriel | 23,173 | <p>I just realised that the knot <a href="http://www.indiana.edu/~knotinfo/diagram_display/diagram_display_10_124.html" rel="nofollow noreferrer">$10_{124}$</a> and <a href="http://www.indiana.edu/~knotinfo/diagram_display/diagram_display_10_153.html" rel="nofollow noreferrer">$10_{153}$</a> both have <a href="http://www.indiana.edu/~knotinfo/descriptions/determinant.html" rel="nofollow noreferrer">determinant</a> $1$ and therefore, indistinguishable by Fox $n-$colouring for every positive integer $n$ since both cannot be coloured modulo $n$ for any $n$.</p>
|
2,300,855 | <p>Do there exist two non-equivalent knots which are indistinguishable by <a href="https://en.wikipedia.org/wiki/Fox_n-coloring" rel="nofollow noreferrer">Fox $n$-colouring</a> for every positive integer $n$?</p>
<p>That is, do there exist non-oriented knots $K_0$ and $K_1$ which are different (here I would like to exclude the case that one is the mirror image of the other) and for every $n\in\mathbb{Z}^+$, $\mathrm{col}_n(K_0)=\mathrm{col}_n(K_1)$, where $\mathrm{col}_n(K)$ denotes the number of distinct Fox $n$-colourings of a knot $K$? </p>
<p>How about the case of links?</p>
| N. Owad | 85,898 | <p>These are not prime knots, but the square knot and the granny knot I believe are indistinguishable by coloring. That is, $3_1\#\bar{3}_1$ and $3_1\#3_1$. You need peripheral subgroups (or at least that was the way it was done first). And they can be colored, if that helps your work.</p>
|
253,746 | <p>I am currently aware of the following two versions of the global Cauchy Theorem. Which one is stronger?</p>
<p>1.)If the region $U$ is simply connected, then for every closed curve contained therein, the integral of the holomorphic function $f$ defined on $U$ over the curve is zero.</p>
<p>2.) If the domain $D$ is an arbitrary open set, then for every closed curve contained therein with the property that the index is zero for points outside $D$, the integral of the holomorphic function $f$ defined on $D$ is zero.
Thanks in advance.</p>
| Applied mathematician | 42,887 | <p>The way I learned it is:</p>
<blockquote>
<p><span class="math-container">$$\int_{\Gamma}f(z)dz=0 $$</span></p>
<p>Iff</p>
<p>1.) <span class="math-container">$f$</span> is analytic in a simply connected domain D</p>
<p>2.) <span class="math-container">$\Gamma$</span> is any closed contour in D,</p>
</blockquote>
<p>Where a domain D is defined by "an open connected set", D <span class="math-container">$\subseteq \mathbb{C}$</span>.</p>
<p>A function <span class="math-container">$f=u(x,y) +iv(x,y)$</span>, defined in some open set G, containing the point <span class="math-container">$z_0$</span> is differentiable at <span class="math-container">$z_0$</span> if the first partial derivatives of <span class="math-container">$u$</span> and <span class="math-container">$v$</span> exists in G, are continuous at <span class="math-container">$z_0$</span> and satisfiy the Cauchy-Riemann Equations.</p>
<p>Consequently, if the first partial derivatives are continuous and satisfy the Cauchy-Riemann Equations at all points of G, then <span class="math-container">$f$</span> is analytic in G</p>
|
984,852 | <p>Let $p$ be a univariate polynomial over a field $F$, and let $K$ be an extension of $F$. </p>
<p>If $p(x) = 0$ for all $x \in F$, does this imply that $p(x) = 0$ for all $x \in K$? How about if $p$ is multivariate?</p>
<p>For context, I'm trying to understand if doing Schwartz-Zippel-style arithmetic circuit identity-testing over a large enough extension field gives the right answer when the degree of the expression may be high.</p>
| orangeskid | 168,051 | <p>All you know is this: a nonzero polynomial of degree $d$ cannot have more than $d$ roots in some (extension) field. </p>
|
497,546 | <p>Let $A$ be an infinite set.</p>
<p>Then, we can construct an injective function $f:\omega \rightarrow A$. </p>
<p>But how do i construct this via orginal statement of $AC_\omega$? (i.e. $\forall countable X, [\emptyset \notin X \Rightarrow \exists f:X\rightarrow \bigcup X \forall A\in X, f(A)\in A$)</p>
<p>So my question is:</p>
<p>(1) how do i prove that every infinite set has a countable subset?</p>
<p>(2) how do i write down a situation such as: "after choosing $x_1,...,x_k$, choosing $x_k$ satisfying a given condition. Continue this and form a sequence"</p>
| user88595 | 88,595 | <p>I do not know any quick proofs from scratch to show this formula but a general formula to find generating functions is the following: $$G(x,y) = \frac{\omega(z)}{\omega(x)}\frac1{1-y\cdot p'(z)}$$
Where $z = x + y\cdot p(z)$ and $p(z)$ is from the Legendre ODE $\frac{d}{dx}[p(x)y_n'(x)] + q(x)y_n'(x) + \lambda_ny_n(x) = 0$ in its standard form.</p>
<p>So here you have $p(z) = 1-z^2$ for the Legendre equation so you find that $p'(z) = -2z$ and solve $z = x+y + 1 - z^2$ to get $z$ as a function of $x$ and $y$. (Nasty but that's where you get the square root from.)</p>
<p>Sub back everything in the formula and you get your answer.</p>
<p>PS : Generating functions are not unique.</p>
|
497,546 | <p>Let $A$ be an infinite set.</p>
<p>Then, we can construct an injective function $f:\omega \rightarrow A$. </p>
<p>But how do i construct this via orginal statement of $AC_\omega$? (i.e. $\forall countable X, [\emptyset \notin X \Rightarrow \exists f:X\rightarrow \bigcup X \forall A\in X, f(A)\in A$)</p>
<p>So my question is:</p>
<p>(1) how do i prove that every infinite set has a countable subset?</p>
<p>(2) how do i write down a situation such as: "after choosing $x_1,...,x_k$, choosing $x_k$ satisfying a given condition. Continue this and form a sequence"</p>
| Felix Marin | 85,343 | <p>$\newcommand{\+}{^{\dagger}}%
\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
\newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
\newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
\newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}%
\newcommand{\dd}{{\rm d}}%
\newcommand{\ds}[1]{\displaystyle{#1}}%
\newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
\newcommand{\fermi}{\,{\rm f}}%
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
\newcommand{\half}{{1 \over 2}}%
\newcommand{\ic}{{\rm i}}%
\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}%
\newcommand{\isdiv}{\,\left.\right\vert\,}%
\newcommand{\ket}[1]{\left\vert #1\right\rangle}%
\newcommand{\ol}[1]{\overline{#1}}%
\newcommand{\pars}[1]{\left( #1 \right)}%
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}%
\newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
\newcommand{\sech}{\,{\rm sech}}%
\newcommand{\sgn}{\,{\rm sgn}}%
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}%
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
This is a very popular proof in Electromagnetism ( see, for example, <a href="http://rads.stackoverflow.com/amzn/click/047130932X" rel="nofollow">Jackson book</a> ):
$\verts{\vec{r} - \vec{r}'}^{-1}$ satisfies the Laplace Equation when
$\vec{r} \not= \vec{r}'$. It means $\verts{\vec{r} - \vec{r}'}^{-1}$ can be expanded in a serie of Legendre Polynomials:
$$
\left.\nabla^{2}\pars{{1 \over \verts{\vec{r} - \vec{r}'}}}
\right\vert_{\vec{r}\ \not=\ \vec{r}'} = 0
\quad\imp\quad{1 \over \verts{\vec{r} - \vec{r}'}} = \sum_{\ell = 0}^{\infty}A_{\ell}{\rm P}_{\ell}\pars{\cos\pars{\theta}}
$$
where $\theta \equiv \angle\pars{\vec{r},\vec{r}'}$:
$$
{1 \over \verts{\vec{r} - \vec{r}'}}
=
{1 \over r_{>}}\,{1 \over \root{1 - 2\pars{r_{<}/r_{>}}\cos\pars{\theta} + \pars{r_{<}/r_{>}}^{2}}}\,,\qquad r_{< \atop >} = {\min \atop \max}\braces{r,r'}
$$
Set $r_{>} = 1$, $h \equiv r_{<}/r_{>}$ and $x = \cos\pars{\theta}$. We get
$$
{1 \over \root{1 -2xh + h^{2}}} = \sum_{\ell = 0}^{\infty}A_{\ell}{\rm P}_{\ell}\pars{x}
$$
With $x = 1$, $\pars{~{\rm P}_{\ell}\pars{1} = 1\,,\ \forall\ \ell = 0,1,2,\ldots~}$:
$$
\sum_{\ell = 0}^{\infty}A_{\ell} = {1 \over 1 - h} = \sum_{\ell = 0}^{\infty}h^{\ell}
$$</p>
|
3,540,045 | <p>The definite integral, <span class="math-container">$$\int_0^{\pi}3\sin^2t\cos^4t\:dt$$</span></p>
<p><strong>My question</strong>: for the trigonometric integral above the answer is <span class="math-container">$\frac{3\pi}{16}$</span>. What I want to know is how can I compute these integrals easily. Is there more than one way to solve it? If so, is the key to solving these integrals, just recognizing some trig identities and using u-sub until it looks like a simpler integral?</p>
<p><strong>Here's what I tried (Why doesn't it work!):</strong></p>
<p>I rewrote the integrand as: <span class="math-container">$3(1-\cos^2t)\cos^4t\:dt$</span> then foiled it in,</p>
<p><span class="math-container">$3\cos^4t-3\cos^6t dt$</span>
, then I used the power rule and multipied through by chain rule and then did </p>
<p><span class="math-container">$F(\pi)-F(0)$</span> and got the answer: <span class="math-container">$\frac{6}{35}$</span></p>
<p>Why does this not work?!?</p>
| user5713492 | 316,404 | <p>I see <span class="math-container">$2$</span> quick methods. Firstly, one could use <span class="math-container">$\sin t\cos t=\frac12\sin2t$</span> and <span class="math-container">$\cos^2t=\frac12(1+\cos2t)$</span> to get
<span class="math-container">$$\int_0^{\pi}3\sin^2t\cos^4t\,dt=\frac38\int_0^{\pi}\left(\sin^22t+\sin^22t\cos2t\right)dt$$</span>
The period of <span class="math-container">$\sin2t$</span> and <span class="math-container">$\cos2t$</span> is <span class="math-container">$\pi$</span> so the integral is the average value of the integrand times the length of the interval. Now <span class="math-container">$\langle\sin^22t\rangle=\langle\cos^22t\rangle=\frac12\langle\sin^22t+\cos^22t\rangle=\frac12\langle1\rangle=\frac12$</span> and <span class="math-container">$\langle\sin^22t\cos2t\rangle=0$</span> because it has on odd power of <span class="math-container">$\cos2t$</span> so
<span class="math-container">$$\int_0^{\pi}3\sin^2t\cos^4t\,dt=\frac38\pi\left(\frac12+0\right)=\frac3{16}$$</span>
The other easy approach is via the Beta and Gamma functions, but it's a little higher level math:
<span class="math-container">$$\begin{align}\int_0^{\pi}3\sin^2t\cos^4t\,dt&=3\times2\int_0^{\pi/2}\sin^2t\cos^4t\,dt=3\operatorname{B}\left(\frac32,\frac52\right)\\
&=3\frac{\operatorname{\Gamma}\left(\frac32\right)\operatorname{\Gamma}\left(\frac52\right)}{\operatorname{\Gamma}\left(4\right)}=3\frac{\sqrt{\pi}\left(\frac12\right)\sqrt{\pi}\left(\frac12\right)\left(\frac32\right)}{3!}=\frac{3\pi}{16}\end{align}$$</span></p>
|
3,540,045 | <p>The definite integral, <span class="math-container">$$\int_0^{\pi}3\sin^2t\cos^4t\:dt$$</span></p>
<p><strong>My question</strong>: for the trigonometric integral above the answer is <span class="math-container">$\frac{3\pi}{16}$</span>. What I want to know is how can I compute these integrals easily. Is there more than one way to solve it? If so, is the key to solving these integrals, just recognizing some trig identities and using u-sub until it looks like a simpler integral?</p>
<p><strong>Here's what I tried (Why doesn't it work!):</strong></p>
<p>I rewrote the integrand as: <span class="math-container">$3(1-\cos^2t)\cos^4t\:dt$</span> then foiled it in,</p>
<p><span class="math-container">$3\cos^4t-3\cos^6t dt$</span>
, then I used the power rule and multipied through by chain rule and then did </p>
<p><span class="math-container">$F(\pi)-F(0)$</span> and got the answer: <span class="math-container">$\frac{6}{35}$</span></p>
<p>Why does this not work?!?</p>
| Deepak | 151,732 | <p>Here's a method that can tackle more general problems of this nature.</p>
<p>First, let's consider the integral <span class="math-container">$I = \int \sin^2 x \cos^4 x dx$</span> (I just omitted the constant multiplier).</p>
<p>We can write <span class="math-container">$I = \int \sin x \cdot \sin x \cos^4 x dx$</span> and use integration by parts with <span class="math-container">$u = \sin x, dv = \int \sin^2 x \cos^4 x dx$</span> to get <span class="math-container">$I = \sin x \cdot (-\frac 15)\cos^5 x - \int\cos x \cdot (-\frac 15)\cos^5 xdx = \sin x \cdot (-\frac 15)\cos^5 x +\frac 15 \int\cos^6 xdx$</span></p>
<p>Applying the bounds, we get <span class="math-container">$I_{0}^{\pi} = \frac 15 \int_{0}^{\pi}\cos^6 xdx$</span></p>
<p>Now we're left with the problem of working out <span class="math-container">$J = \int \cos^6 xdx$</span></p>
<p>From Euler's identity and De Moivre's theorem (complex number theory), we can write <span class="math-container">$\cos x = \frac 12(e^{ix} + e^{-ix})$</span>. We also know that <span class="math-container">$\cos nx = \frac 12(e^{nix} + e^{-nix})$</span></p>
<p>Applying binomial theorem and the above, we can write <span class="math-container">$\cos^6x = (\frac 12(e^{ix} + e^{-ix}))^6= \frac 1{64}(e^{6ix} + 6e^{4ix} + 15e^{2ix} + 20 + 15e^{-2ix} + 6e^{-4ix} + e^{-6ix}) = \frac 1{64}(2\cos 6x + 12\cos 4x + 30\cos 2x + 20)$</span>, and you can then compute <span class="math-container">$J = \frac 1{64}(\frac 13 \sin 6x + 3\sin 4x + 15\sin 2x + 20x) + c$</span>, and after applying the bounds, you get <span class="math-container">$J_{0}^{\pi} = \frac {20\pi}{64} = \frac{5\pi}{16}$</span>.</p>
<p>You can now compute <span class="math-container">$I_{0}^{\pi} = \frac 15 J_{0}^{\pi} = \frac{\pi}{16}$</span> and the original integral is <span class="math-container">$3I_{0}^{\pi} =\frac{3\pi}{16}$</span>.</p>
|
893,822 | <p>If $p(x)$ has integer coefficients and $p(100)$ equals $100$ what is the maximum number of integer solutions $k$ to the equation $p(k)=k^3$.</p>
<p>I have tried hard to solve this problem but I could not figure it out. I tried some particular cases but got nowhere, could someone please show me how to get the answer.</p>
| Jack D'Aurizio | 44,121 | <p>As pointed out by r9m in the comments, by setting $x=\cosh t$ the inequality is equivalent to:
$$ \cosh(n t)\leq \left(\cosh t+(n-1)(\cosh t-1)\right)^n \tag{1}$$
or to:
$$ \cosh(n t)\leq \left(1+2n\sinh^2\frac{t}{2}\right)^n \tag{2}$$
or to:
$$ 1+2\sinh^2(nz)\leq \left(1+2n\sinh^2 z\right)^n \tag{3}$$
or to:
$$ \frac{1}{n}\log\left(1+2\sinh^2(nz)\right)\leq \log\left(1+2n\sinh^2 z\right) \tag{4}.$$
Equality is achieved in $z=0$; by considering the difference of the derivatives it is sufficient to prove that:
$$ (n-1)\sinh(nu)\geq n\sinh((n-1)u).\tag{5}$$
Equality is achieved in $u=0$, by differentiating again we are left to prove that:
$$ \cosh(nu)\geq \cosh((n-1)u) \tag{6}$$
that is trivial.</p>
|
4,363,327 | <p>Consider the one-layer neural network <span class="math-container">$y=\mathbf{w}^T\mathbf{x} +b$</span> and the optimization objective <span class="math-container">$J(\mathbf{w}) = \mathbb{E}\left[ \frac12 (1-y\cdot t) \right]$</span> where <span class="math-container">$t\in\{-1,1\}$</span> is the label of our data point. I am asked to compute the Hessian of <span class="math-container">$J$</span> at the current location <span class="math-container">$\mathbf{w}$</span> in the parameter space. I know that the correct solution is <span class="math-container">$H=\frac{\partial^2J}{\partial \mathbf{w}^2} = \mathbb{E}\left[ \mathbf{x}\mathbf{x}^T \right]$</span>.</p>
<p>I am having issues to arrive at this exact formulation because of how derivation of row/column vectors work. My solution goes as follows:
We first determine the first derivative.</p>
<p><span class="math-container">\begin{align*}
&\frac{\partial \mathbb{E}\left[ \frac12 \left( 1 - yt \right)^2 \right]}{\partial \mathbf{w}}\\
&= \mathbb{E}\left[ \frac{\partial \frac12 \left( 1 - yt \right)^2 }{\partial \mathbf{w}} \right]\\
&= \mathbb{E}\left[ \frac{\partial \frac12 \left( 1 - yt \right)^2 }{\partial y} \frac{\partial y}{\partial \mathbf{w}} \right]\\
&= \mathbb{E}\left[ -t\cdot(1-yt) \frac{\partial \mathbf{w}^T\mathbf{x}+b}{\partial \mathbf{w}} \right]\\
&= \mathbb{E}\left[ -t\cdot(1-yt) \mathbf{x} \right]\\
\end{align*}</span></p>
<p>Note that I think <span class="math-container">$\mathbf{x}\in\mathbb{R}^d$</span> is considered a <em>column</em> vector, and according to the matrix cookbook, <span class="math-container">$ \frac{\partial \mathbf{w}^T\mathbf{x}+b}{\partial \mathbf{w}} = \mathbf{x}$</span>, not <span class="math-container">$\mathbf{x}^T$</span> (I have found sources saying otherwise...)</p>
<p>We now differentiate this again, in order to derive the Hessian.
<span class="math-container">\begin{align*}
&\frac{\partial \mathbb{E}\left[ -t\cdot(1-yt) \mathbf{x} \right]}{\partial \mathbf{w}}\\
&= \mathbb{E}\left[ \frac{\partial -t(1-yt)\mathbf{x}}{\partial y} \frac{\partial \mathbf{w}^T\mathbf{x}+b}{\partial \mathbf{w}} \right]\\
&= \mathbb{E}\left[ \frac{\partial -t(1-yt)\mathbf{x}}{\partial y} \mathbf{x} \right]\\
&= \mathbb{E}\left[ \frac{\partial (-t+yt^2)\mathbf{x}}{\partial y} \mathbf{x} \right]\\
&= \mathbb{E}\left[ \underbrace{t^2}_{= 1} \mathbf{x} \mathbf{x} \right]\\
&= \mathbb{E}\left[ \mathbf{x} \mathbf{x} \right]\\
&\neq \mathbb{E}\left[ \mathbf{x}\mathbf{x}^T \right]
\end{align*}</span></p>
<p>So here, we have column vector times column vector which is not really defined. I do not know where to get the transpose from, though. I tried deriving the whole thing again with the assumption that <span class="math-container">$\frac{\partial \mathbf{w}^T\mathbf{x}+b}{\partial \mathbf{w}} = \mathbf{x}^T$</span>, instead of <span class="math-container">$\mathbf{x}$</span>. Then we either get <span class="math-container">$\mathbb{E}\left[ \mathbf{x}^T\mathbf{x}^T \right]$</span> similar to before, or, <em>if</em> we assume that the derivative of a row vector w.r.t. scalar is a column vector, we get <span class="math-container">$\mathbb{E}\left[ \mathbf{x}\mathbf{x}^T \right]$</span>, which is what we want. However,this would be a very weird assumption, to me, because why would the derivative of a row vector w.r.t. to a scalar be a column vector? And furthermore, it contradicts the matrix cookbook, which says that <span class="math-container">$\frac{\partial \mathbf{w}^T\mathbf{x}+b}{\partial \mathbf{w}} = \mathbf{x}$</span>.</p>
<p>I would be very glad for help here. Where did I go wrong? Which assumptions of row/column vectors are correct? Thank you so much for your help!</p>
<p>Last, I found an alternative way of solving it where you don't have that issue of transposing or not, by just inserting the definition of the prediction <span class="math-container">$y$</span>, but I still would like to know where the issue in my solution above lies.</p>
<p><span class="math-container">\begin{align*}
&\frac{\partial \mathbb{E}\left[ -t\cdot(1-yt) \mathbf{x} \right]}{\partial \mathbf{w}}\\
&= \frac{\partial \mathbb{E}\left[ -t\mathbf{x} + t^2y\mathbf{x} \right]}{\partial \mathbf{w}}\\
&= \frac{\partial \mathbb{E}\left[ -t\mathbf{x} + t^2(w^T\mathbf{x} + b)\mathbf{x} \right]}{\partial \mathbf{w}}\\
&= \frac{\partial \mathbb{E}\left[ -t\mathbf{x} + t^2(w^T\mathbf{x})\mathbf{x} + b\mathbf{x} \right]}{\partial \mathbf{w}}\\
&= \frac{\partial \mathbb{E}\left[ -t\mathbf{x} + t^2\mathbf{x}^Tw\mathbf{x} + b\mathbf{x} \right]}{\partial \mathbf{w}}\\
&= \mathbb{E}\left[ \underbrace{t^2}_{= 1} \mathbf{x}\mathbf{x}^T \right]\\
&= \mathbb{E}\left[ \mathbf{x}\mathbf{x}^T \right]\\
\end{align*}</span></p>
| Steph | 993,428 | <p>Forgetting the expectation operator (for a while),
the differential of the cost function writes
<span class="math-container">$$
d\phi
= t(t\cdot y-1)dy
= (y-t) \mathbf{x}^Td\mathbf{w}
$$</span>
since <span class="math-container">$dy=\mathbf{x}^Td\mathbf{w}$</span>.</p>
<p>The gradient is <span class="math-container">$\mathbf{g}
=
(y-t) \mathbf{x}
$</span></p>
<p>It follows that
<span class="math-container">$d\mathbf{g}
=
dy\cdot \mathbf{x}
=
(\mathbf{x} \mathbf{x}^T) d\mathbf{w}
$</span>
which gives the Hessian <span class="math-container">$\mathbf{H}=\mathbf{x} \mathbf{x}^T$</span>.</p>
|
467 | <p>Moderators have started incorporating the old faq material in the new <a href="https://mathoverflow.net/help">help system</a>. It wasn't a perfect fit, a lot of stuff is no longer relevant, redundant, missing or broken. You can help by going through the <a href="https://mathoverflow.net/help">help center</a> and post anything that needs fixing here.</p>
<p>In case substantial editing is needed it would help to add a proposal; we may not use it verbatim but it will nevertheless save a lot of effort on our end.</p>
<p>Note that this is not the right place for policy discussion. If you think we should be wearing shirts instead of wearing pants, start a new question with your proposal...</p>
| Community | -1 | <p>Some quick observations on in <a href="https://mathoverflow.net/help/on-topic">https://mathoverflow.net/help/on-topic</a> </p>
<ol>
<li><p>The part Where's the rule that says I have to wear pants? seems not good there anymore it should be merged into <a href="https://mathoverflow.net/help/behavior">https://mathoverflow.net/help/behavior</a> (or just ommitted but there are some things in it that some care about a lot namely "real names" so it should likely be merged). Also, you might or might not want to change 'bahavior' (to be clear, <em>I</em> do not need or even want a change, but since some where so concerned about not having greetings and related you might want to remove mention that they are not allowed)</p></li>
<li><p>The part What kind of questions should I not ask here? likely should replace or rather augment (some things seem good some not fitting but could perhaps be modified) <a href="https://mathoverflow.net/help/dont-ask">https://mathoverflow.net/help/dont-ask</a> </p></li>
<li><p><strike>The info on accounts seems deleteable it is under My account in the help center.
</strike></p></li>
<li><p><strike>The point-privileges list seems deleteable, it is both outdated and redundant with other things.</strike></p></li>
<li><p><strike>Community Wiki. It is now <em>wrong</em> one can make a question CW as a user. The documenetatio still says so. There is also a <em>broken</em> link to the old meta. I would suggest to remove this link rather then to update it. This discussion is ancient.
But I could imagine to have a general discussion on CW could be good before finalizing the description. But this is not for this thread.</strike> </p></li>
<li><p>There is a broken link to the old meta in 'MO is not a discussion forum' Again I think the link is better removed than updated.</p></li>
<li><p><strike>Description of "bounty" is wrong. Can be deleted as explained elsewhere.</strike></p></li>
<li><p>Open problems also contains a broken link. But there is a suggestion for a complete rewrite (without this link) in another answer.</p></li>
</ol>
<p>Added: TriG mentioned broken links in 'How can I become an MO ninja' the tips and tricks one seems still broken. The copy is in this new meta thread <a href="https://meta.mathoverflow.net/questions/367/where-is-the-tips-and-tricks-list-from-mo-1-0">where is the Tips and Tricks list from MO 1.0?</a> but also I started an attempt to redo/augment it (many of the old tips are outdated and it felt better to start over than to edit this one answer) see <a href="https://meta.mathoverflow.net/questions/389/what-are-tips-and-tricks-to-use-mo-more-effectively">What are Tips and Tricks to use MO more effectively?</a> This stalled a bit lately, but not completely and I intend at least to get more or less everything from the old list in the new one over time. So one might use this instead though it is not quite there yet. (But certainly I am also happy if somebody else does this.)</p>
|
467 | <p>Moderators have started incorporating the old faq material in the new <a href="https://mathoverflow.net/help">help system</a>. It wasn't a perfect fit, a lot of stuff is no longer relevant, redundant, missing or broken. You can help by going through the <a href="https://mathoverflow.net/help">help center</a> and post anything that needs fixing here.</p>
<p>In case substantial editing is needed it would help to add a proposal; we may not use it verbatim but it will nevertheless save a lot of effort on our end.</p>
<p>Note that this is not the right place for policy discussion. If you think we should be wearing shirts instead of wearing pants, start a new question with your proposal...</p>
| TRiG | 6,395 | <p>Under <a href="https://mathoverflow.net/help/on-topic">On Topic</a>:</p>
<blockquote>
<p>If you're just really interested in how the underlying Stack Exchange software works, consider visiting meta.stackoverflow or meta.stackexchange.</p>
</blockquote>
<p>This may change in <em>ahem</em> six to eight weeks, but for now MSE merely redirects to MSO, so listing the two separately seems wrong.</p>
<blockquote>
<p><strong>How do I get/use an OpenID?</strong></p>
<p>If you already have a Google or Yahoo account, then you already have an OpenID. Just click the login link at the top of the page and click the Google (resp. Yahoo) button. You'll be asked for your Google (resp. Yahoo) username and password (this information is never sent to MathOverflow) then you'll be returned to MathOverflow.</p>
<p>If you don't have a Google or Yahoo account (or don't want to use it), you can sign up for an OpenID with myOpenID. After you sign up, you can enter your OpenID at the login screen (your OpenID will look something like <a href="http://username.myopenid.com" rel="nofollow noreferrer">http://username.myopenid.com</a>)</p>
</blockquote>
<p>Three problems with this:</p>
<ol>
<li>I've never before seen this use of <em>resp.</em>, and am not sure what it means. <em>Respective</em>?</li>
<li>This is out-of-date, as it doesn't mention that you can now sign in with Stack Exchange directly, not using a third-party login.</li>
<li>This information is duplicated on the <a href="https://mathoverflow.net/help/creating-accounts">Create Account</a> page.</li>
</ol>
<blockquote>
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<p>This formatting is completely broken. Also, I'm pretty sure this information is duplicated elsewhere.</p>
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<p>When you make a post, you have the option of making it "community wiki" by checking a box at the bottom right of the input field.</p>
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<p>Do you? I thought that existed only for answers these days.</p>
<blockquote>
<p>Slice off a bit of your own hard-earned reputation -- anywhere from 50 to 500 -- and attach it to the question as a bounty.</p>
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<p>Other places in the help centre use dashes. Here, you use double hyphens instead.</p>
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<p>What's that comma doing there? Get rid of it. Also, the MathJax example following looks a bit broken to my eyes, but I could be wrong.</p>
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<p>Just be really awesome at asking and answering math questions. Supposing you've done that, you can have a look at the Tips and Tricks page.</p>
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<p>This line contains two broken links.</p>
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<p>Is the term <em>hosted service</em> still accurate?</p>
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<p>I thought MO had special agreements about advertising, which might mean this boilerplate text should be modified.</p>
|
467 | <p>Moderators have started incorporating the old faq material in the new <a href="https://mathoverflow.net/help">help system</a>. It wasn't a perfect fit, a lot of stuff is no longer relevant, redundant, missing or broken. You can help by going through the <a href="https://mathoverflow.net/help">help center</a> and post anything that needs fixing here.</p>
<p>In case substantial editing is needed it would help to add a proposal; we may not use it verbatim but it will nevertheless save a lot of effort on our end.</p>
<p>Note that this is not the right place for policy discussion. If you think we should be wearing shirts instead of wearing pants, start a new question with your proposal...</p>
| Kaveh | 7,507 | <p>List of moderators on help/on-topic page needs to be updated.
It might be better to remove the list from there and
replace it with a link to
<a href="https://mathoverflow.net/users?tab=moderators">https://mathoverflow.net/users?tab=moderators</a></p>
|
2,506,182 | <p>The question speaks for itself, since the question comes from a contest environment, one where the use of calculators is obviously not allowed, can anyone perhaps supply me with an easy way to calculate the first of last digit of such situations?</p>
<p>My intuition said that I can look at the cases of $3$ with an index ending in a $1$, so I looked at $3^1=3$, $3^{11}=177,147$ and $3^{21}= 10,460,353,203$. So there is a slight pattern, but I'm not sure if it holds, and even if it does I will have shown it holds just for indices with base $3$, so I was wondering whether there is an easier way of knowing.</p>
<p>Any help is appreciated, thank you.</p>
| Guy Fsone | 385,707 | <p>the last digit correspond to
$$3^{2011}\equiv x\mod 10$$
We have,
$$3^4\equiv 1\mod 10$$</p>
<p>But, $$2011 = 4 * 502+3$$</p>
<p>Thus,
$$3^{2011}=3^{4*502+3} \equiv 3^3\mod 10~~~\text{ that is }~~
3^{2011} \equiv 7\mod 10 $$</p>
<p>Hence $x=7 $ is the last digit</p>
|
104,592 | <p>Hello everyone,</p>
<p>Suppose that I am defining a function which embeds a surface (manifold) in $\mathbb{R}^3$.</p>
<p>Is there a standard symbol or letter that is used for this function?</p>
<p>Additionally, is there any other classical or standard notation (such as the hooked arrow for inclusion maps) of which I ought to be aware?</p>
<p>Regards,</p>
<p>Christopher</p>
| Wolfgang Loehr | 15,327 | <p>I usually use $\iota$ (\iota) for all kinds of embeddings. </p>
|
9,010 | <p>I hear people use these words relatively interchangeably. I'd believe that any differentiable manifold can also be made into a variety (which data, if I understand correctly, implicitly includes an ambient space?), but it's unclear to me whether the only non-varietable manifolds should be those that don't admit smooth structures. I'd hope there's more to it than that.</p>
<p>I've heard too that affine schemes are to schemes local coordinates are to manifolds, so maybe my question should be about schemes instead -- I don't even know enough to know...</p>
| Community | -1 | <p>If you are French, a variety <em>is</em> a manifold.</p>
<p>If not, the connection is a bit more subtle. The correct analogue of a manifold is a scheme of finite type over a field. That is, it has a compatible covering by ring spectra (which is similar to the requirement that a manifold have a covering by compatible copies of euclidean space), and it lives over a field (the base space is an honest "point"). Here, I am being intentionally vague about "compatibility", since this would lead me into a discussion of <em>sheaves</em>, which I would rather not discuss unless you feel it's necessary. </p>
<p>The classical notion of a variety much more resembles the classical notion of a manifold (see, for instance, Milnor's book "Topology from a differentiable viewpoint"). The classical notion of a manifold is simply a locally closed subset of $\mathbf{P}^n(\mathbf{R})$ (or just $\mathbf{R}^n$, although looking at locally-closed subsets of projective space makes the analogy that much stronger), and the notion of a variety (quasi-projective, I guess) comes from considering locally closed subsets of a projective space.</p>
<p>Lastly, there is a theorem of Serre called GAGA (Géométrie algébrique et géométrie analytique), which says that in nice cases, we can "replace" complex varieties with complex-analytic manifolds to "do" cohomology and get a better picture of the real underlying space. (Here, by <em>complex variety</em>, we simply mean finite-type schemes over $\mathbf{C}$)</p>
|
3,784,484 | <p>Is there a bounded linear operator <span class="math-container">$T \in l^2(\mathbb{N})$</span> have for the essential spectrum the unit disk of <span class="math-container">$\mathbb{C}$</span>; i.e, such that <span class="math-container">$\sigma_{e}(T)=\textbf{D}(0, 1)$</span>; where <span class="math-container">$\textbf{D}(0, 1):=\{\lambda \in \mathbb{C} : |\lambda|\leq 1\}.$</span></p>
| Stephen Montgomery-Smith | 22,016 | <p>Consider <span class="math-container">$T$</span> on <span class="math-container">$\ell^2(\mathbb N \times \mathbb N)$</span>, with <span class="math-container">$T(a)_{m,n} = a_{m,n+1}$</span>. Then for <span class="math-container">$|\lambda| < 1$</span>, elements of the form <span class="math-container">$(b_m \lambda ^n)_{m,n}$</span> are in the kernel of <span class="math-container">$T - \lambda I$</span>. So the kernel has infinite dimension.</p>
<p>Also <span class="math-container">$T$</span> on <span class="math-container">$L^2(D(0,1))$</span> (which is isomorphic to <span class="math-container">$\ell^2(\mathbb N)$</span>) given by <span class="math-container">$Tf(z) = z f(z)$</span>. Then the cokernel of <span class="math-container">$T - \lambda I$</span> is infinite dimensional for all <span class="math-container">$\lambda \in D(0,1)$</span>.</p>
|
1,725,945 | <p>I'm reading the proof of "Fundamental Theorem of Finite Abelian Groups" in Herstein Abstract Algebra, and I've found this statement in the proof that I don't see very clear.</p>
<p>Let $A$ be a normal subgroup of $G$. And suppose $b\in G$ and the order of $b$ is prime $p$, and $b$ is not in $A$. Then $A \cap (b)=(e)$.</p>
| Martín Vacas Vignolo | 297,060 | <p>Suppose that exists $0<k<p$ such that $b^k\in A\to b^{km}\in A \forall m\in\mathbb{Z} *$.</p>
<p>Now, $p$ is prime, $(k,p)=1\to ak+cp=1$ for any $a,c$. </p>
<p>If we put $m=a$ in $*$, obtain that $b^{ka}=b^{1-bp}=b.(b^p)^{-b}=b\in A$, an absurd.</p>
|
50,547 | <p>let $X$ be a $n$-manifold. let $A=\{(x,y,z) \, |\,x=y\}$. I want to see if $A$ is a submanifold of $X^3$.</p>
<p>Consider the map $\Delta\times 1:X\times X\rightarrow X \times X\times X;\, (x,y)\mapsto (x,x,y)$. </p>
<p>If $U_x$ and $U_y$ are neighborhoods of $x$ and $y$ in $X$ then $\Delta\times 1 (U_x\times U_y)=\Delta(U_x)\times U_y$ is a neighborhood of
$\Delta(x,y)=(x,x,y)$ in $X^3$, and we can see that $\Delta(U_x)\times U_y$ is in $A$. So this is a neighborhood of $(x,x,y)$ in $A$. </p>
<p>Now $U_x\cong \mathbb R^n$ and $\Delta(U_x)\cong U_x\cong \mathbb R^n$ and so $A$ is an $2n$-manifold. </p>
<p>Is my argument correct?and does it matter that the diagonal $\Delta(U_x)$ is a closed set, I mean don't we always want a neighborhood to be open? </p>
| Listing | 3,123 | <p>Those integrals were discussed by us in detail already.</p>
<p>To be more explicit you are asking for the special case of <a href="https://math.stackexchange.com/questions/44928/interesting-integral-formula">Interesting integral formula</a> for $m=2$, $n=4$ and $a=1$. Just directly plugging in those values in the proof you get what you want (they are not really used anyways).</p>
|
2,494,232 | <p>I'm trying to find the out if $\sum_{n=1}^\infty {{1\over \sqrt{n}}-{1\over{\sqrt{n+1}}}}$ is divergent or convergent.</p>
<p>Here are some rules my book gives that I will try to follow:</p>
<p><a href="https://i.stack.imgur.com/TNV6x.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TNV6x.png" alt="enter image description here"></a></p>
<p>Looking at those I can see that it isn't #1, not a p-series form; It isn't geometric so that rules out #2; #3 looks like a good fit, anyways I don't think the next 5 steps will apply.</p>
<p>Using #3, I'll need to split it up into $a_n$ and $b_n$ so I'll go ahead and combine them into one term: $${ \sqrt{n+1} - \sqrt{n} } \over {\sqrt{n}\sqrt{n+1}}$$ And it is at this point that I'm stuck. I need to figure out the highest powers of n in the numerator and denominator.</p>
<p>If I'm on the wrong path I would appreciate some help, otherwise, I just need to figure how to get $a_n$ and $b_n$. Thank you.</p>
| operatorerror | 210,391 | <p>Telescoping series, while not mentioned in your list, are important and I hope were taught to you.</p>
<p>You can write the $N$th partial sum as
$$
\bigg(\frac{1}{1}-\frac{1}{\sqrt{2}}\bigg)+\bigg(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\bigg)+\cdots+\bigg(\frac{1}{\sqrt{N}}-\frac{1}{\sqrt{N+1}}\bigg)
$$
where it isn't too hard to see all the terms but the first and last cancel, leaving you with
$$
1-\frac{1}{\sqrt{N+1}}\to 1
$$
as $N\to \infty$.</p>
|
2,494,232 | <p>I'm trying to find the out if $\sum_{n=1}^\infty {{1\over \sqrt{n}}-{1\over{\sqrt{n+1}}}}$ is divergent or convergent.</p>
<p>Here are some rules my book gives that I will try to follow:</p>
<p><a href="https://i.stack.imgur.com/TNV6x.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TNV6x.png" alt="enter image description here"></a></p>
<p>Looking at those I can see that it isn't #1, not a p-series form; It isn't geometric so that rules out #2; #3 looks like a good fit, anyways I don't think the next 5 steps will apply.</p>
<p>Using #3, I'll need to split it up into $a_n$ and $b_n$ so I'll go ahead and combine them into one term: $${ \sqrt{n+1} - \sqrt{n} } \over {\sqrt{n}\sqrt{n+1}}$$ And it is at this point that I'm stuck. I need to figure out the highest powers of n in the numerator and denominator.</p>
<p>If I'm on the wrong path I would appreciate some help, otherwise, I just need to figure how to get $a_n$ and $b_n$. Thank you.</p>
| Guy Fsone | 385,707 | <p>The series obviously converges to $1$ by telescoping the partial sum: </p>
<p>$$\sum_{n=1}^k \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}
=1-\frac{1}{\sqrt{k+1}}$$</p>
|
2,991,719 | <p>How to prove/disprove</p>
<blockquote>
<p>If <span class="math-container">$f : (0, \infty) \to \mathbb{R}^n$</span> is continuous on <span class="math-container">$[a, \infty) , \forall a>0$</span> then <span class="math-container">$f$</span> is continuous on <span class="math-container">$(0, \infty)$</span></p>
</blockquote>
<hr>
<p><strong>My try</strong></p>
<p>Assume it is not continuous on <span class="math-container">$(0, \infty)$</span>.</p>
<p>Let's denote a discontinuity point <span class="math-container">$x_0 \in (0,\infty)$</span>. <span class="math-container">$\exists \epsilon >0$</span> s.t. <span class="math-container">$\forall \delta >0$</span>, <span class="math-container">$|x - x_0| < \delta$</span> and <span class="math-container">$|f(x) - f(x_0)| \ge \epsilon $</span>. </p>
<p>But since <span class="math-container">$f$</span> is continuous on <span class="math-container">$[x_0/2, \infty)$</span>, it is contradiction.</p>
<p>This proof seems doubtful for me, but I cannot specify the doubts.</p>
<p>Is there anyone to clarify this?</p>
| NoChance | 15,180 | <p>Sometimes you can't tell if matrix <span class="math-container">$B$</span> can be generated from matrix <span class="math-container">$B$</span> or not. In such a case, one may expand each determinant and compare the result of each. However, this approach requires careful attention to signs!</p>
<p>Let <span class="math-container">$X$</span> be the determinant of the first matrix and <span class="math-container">$Y$</span> be the determinant of the 2nd matrix, then we have:</p>
<p><span class="math-container">$ X = a(ei - fh) - b(di - fg) + c(dh - eg) $</span> </p>
<p><span class="math-container">$ X = aei - afh - bdi + bfg + cdh - ceg $</span></p>
<p><span class="math-container">$ Y = 3afh - 3eai + 3ecg + 3bdi - 3bfg - 3cdh $</span></p>
<p><span class="math-container">$ \frac{Y}{-3} = -afh + eai -ecg -bdi + bfg + cdh $</span></p>
<p>You could re-arrange the terms of any of the equations to see that:</p>
<p><span class="math-container">$X=\frac{Y}{-3}$</span></p>
<p>Since X=4, </p>
<p><span class="math-container">$Y=-12$</span></p>
|
738,122 | <p>I have been looking at Kolmogorov's book "Introductory Real Analysis" and have become stuck at the problem 4a) on page 301.</p>
<p>In this problem we are given $f$ a nonnegative and integrable function on $A$, a set of finite measure. We are asked to show that</p>
<p>$\int_A f(x) d \mu \ge 0$</p>
<p>However, how would we prove this? In this text the Lebesgue integral of a function $f$ is defined to be (when the limit exists);</p>
<p>$\int_A f(x) d\mu = \lim\limits_{n \rightarrow \infty} \int_{A} f_n(x) d\mu$</p>
<p>where $f_n$ is a sequence of integrable simple functions that converges uniformly to $f$.</p>
<p>I have attempted to show this by showing that the sequence of simple functions converging to a nonnegative function, is a sequence of nonnegative functions. However this isn't true, take $f_n = -1/n$ and $f = 0$. How else could I go about this? </p>
<p>Thanks.</p>
| Ellya | 135,305 | <p>Simply state that you are restricting to a sequence simple functions that are non negative.</p>
<p>Or use the fact that $f\ge g\Rightarrow \int f\ge \int g $</p>
<p>Then $\int f = \sup\{\int s|s, simple\}=\sup\{\int s|s, simple,positive\}$</p>
|
3,647,772 | <p>In positional representations, there are always some rational numbers which have multiple representations. For example, in base 10, 1 can be written as 1 or as <span class="math-container">$0.\overline{9}$</span>. Do there exist any numerical representations in which all rationals have exactly one representation?</p>
| Andrew Chin | 693,161 | <p>The fundamental theorem of arithmetic states that every natural number greater than <span class="math-container">$1$</span> can be written as a unique product of powers of prime numbers. Extending this to the rationals simply requires us to allow the negative powers of prime factors. For example, <span class="math-container">$$60=2^2\times3\times5$$</span> and <span class="math-container">$$\frac{64}{375}=2^6\times 3^{-1}\times5^{-3}.$$</span></p>
<p>We know that these are unique due to the monotone nature of exponentials.</p>
|
3,647,772 | <p>In positional representations, there are always some rational numbers which have multiple representations. For example, in base 10, 1 can be written as 1 or as <span class="math-container">$0.\overline{9}$</span>. Do there exist any numerical representations in which all rationals have exactly one representation?</p>
| user780256 | 780,256 | <p>You can express rational numbers uniquely by terminating continued fractions. Every rational number can be expressed as a finite sequence of integers <span class="math-container">$[a_0, a_1, \ldots, a_n]$</span> containing at least one term, where <span class="math-container">$a_i \in \Bbb{Z}$</span> for <span class="math-container">$i \ge 0$</span>, and <span class="math-container">$a_i > 0$</span> for <span class="math-container">$i > 0$</span>. We associate <span class="math-container">$[a_0, a_1, \ldots, a_n]$</span> with</p>
<p><span class="math-container">$$\huge{ a_0 + \frac{1}{a_1 + \frac{1}{a_2 + \frac{1}{\ddots_{a_{n-1}+\frac{1}{a_n}}}}}}.$$</span></p>
<p>The way you convert a rational number to its unique continued fraction is you let <span class="math-container">$a_0$</span> be the floor function of the given rational. Subtract <span class="math-container">$a_0$</span> from this number, leaving you the fractional part, a number in <span class="math-container">$[0, 1)$</span>. If the number is <span class="math-container">$0$</span>, then stop. If the number is positive, then compute its reciprocal to obtain a number greater than <span class="math-container">$1$</span>. You then start over, letting <span class="math-container">$a_1$</span> be the floor of this new number, etc.</p>
<p>As an example, consider <span class="math-container">$\frac{67}{21}$</span>. The floor of this number is <span class="math-container">$3$</span>, so we subtract <span class="math-container">$\frac{67}{21} - 3 = \frac{4}{21}$</span>. We reciprocate this fractional part, which produces <span class="math-container">$\frac{21}{4}$</span>. We let <span class="math-container">$a_1$</span> be the integer part of this number, i.e. <span class="math-container">$5$</span>, leaving us a fractional part <span class="math-container">$\frac{1}{4}$</span>. Reciprocating this gives us <span class="math-container">$4$</span>, which has an integer part of <span class="math-container">$4$</span> (so <span class="math-container">$a_2 = 4$</span>), and a fractional part of <span class="math-container">$0$</span>, telling us to stop. So, <span class="math-container">$\frac{67}{21}$</span> is represented by <span class="math-container">$[3, 5, 4]$</span>, i.e.
<span class="math-container">$$\frac{67}{21} = 3 + \frac{1}{5 + \frac{1}{4}}.$$</span>
Note that the denominator of the fraction under consideration is always strictly decreasing, since the fractional part is always less than <span class="math-container">$1$</span>, so the process will terminate for any rational number.</p>
|
744,787 | <p>I need the equation of the line passing through a given point $A(2,3,1)$ perpendicular to the given line
$$
\frac{x+1}{2}= \frac{y}{-1} = \frac{z-2}{3}.
$$</p>
<p>I think there must bee some kind of rule to do this, but I can't find it anyway.</p>
| Alan | 92,834 | <p><img src="https://i.stack.imgur.com/h1zw8.jpg" alt="enter image description here"></p>
<p>The system looks over determined by the line's equation. However, one chooses the line between (2,3,1) and (-1,0,2). (I minimized the distance and got a result in agreement.) I thought it a bit of a trick question. </p>
|
285,719 | <p>Why is $(-3)^4 =81$ and $-3^4 =-81 $?This might be the most stupidest question that you might have encountered,but unfortunately i'am unable to understand this.</p>
| A.D | 37,459 | <p>Note that $(-3)^4=((-1)\cdot3)^4=(-1)^4\cdot3^4=81$</p>
<p>Again note that $-3^4=(-1)\cdot3^4=-81$</p>
|
285,719 | <p>Why is $(-3)^4 =81$ and $-3^4 =-81 $?This might be the most stupidest question that you might have encountered,but unfortunately i'am unable to understand this.</p>
| guest196883 | 43,798 | <p>Because multiplication doesn't distribute into exponentiation. That is, $a(b^c) = (ab)^c$ doesn't always hold. And there's no reason why it should, unless $c = 1$. So $(-3)^4 = (-1 \cdot3)^4 = (-1)^4 \cdot (3)^4 = 81$, and not $-81$, as the multiple of $-1$ doesn't factor out. </p>
|
3,456,351 | <pre><code>Apples: 1
Apple Value: 2500
Pears: lowest = 1, highest = 10
</code></pre>
<p>If I have 1 apple and my apple is worth <strong>2500</strong> if I have 1 pear, how can I calculate the value of my apple if I have X pears, at a maximum of 10 pears and a minimum of 1 pear, where 1 pear represents 100% value and 10 pears represents an increase of that value by 150%?</p>
<pre><code>Apple Worth = 2500 * (1+((pears*15)/100))
</code></pre>
<p>This formula sort of works. The problem here is that if I have 1 pear, the value becomes <code>2500 * 1.15 = 2875</code> which is incorrect, as the value of my apple always is 100% of it's original value if I have at least 1 pear (pear count can never go below 1).</p>
<p>What am I missing here?</p>
<p>Edit:</p>
<p>Some clarification.</p>
<pre><code>2500 = 1p = 100% (do not add 0.15 until p > 1)
2875 = 2p = 115%
3250 = 3p = 130%
3625 = 4p = 145%
4000 = 5p = 160%
4375 = 6p = 175%
4750 = 7p = 190%
5125 = 8p = 205%
5500 = 9p = 220%
5875 = 10p = 235%
6250 = 11p = 250% (this is what 10p should be)
</code></pre>
<p>At 1 pear, my apple is worth 100% of it's original value, so this means that unless I have more than 1 pear, I can't add 0.15% per pear. The first pear is, in other words, not worth 0.15%, but rather 0%. This also means that the 10th pear should be 250% and not 235%.</p>
| Joseph Desaulniers | 1,111,320 | <p>Here is an (maybe) easier approach.</p>
<p>To prove <span class="math-container">$\lfloor\sqrt n+\sqrt {n+1}\rfloor=\lfloor\sqrt{4n+2}\rfloor$</span>, and it's easy to notice that <span class="math-container">$\lfloor\sqrt{4n+2}\rfloor>\lfloor\sqrt n+\sqrt {n+1}\rfloor$</span> and that <span class="math-container">$\lfloor\sqrt{4n+2}\rfloor-\lfloor\sqrt n+\sqrt {n+1}\rfloor<1$</span>, it suffices to prove that there's no integer between <span class="math-container">$\sqrt n+\sqrt {n+1}$</span> and <span class="math-container">$\sqrt{4n+2}$</span>.<br />
<br />
We suppose that there is an integer <span class="math-container">$k$</span> between <span class="math-container">$\sqrt n+\sqrt {n+1}$</span> and <span class="math-container">$\sqrt{4n+2}$</span>. Then<br />
<span class="math-container">\begin{equation*}
\sqrt n+\sqrt {n+1}\leq k\leq\sqrt{4n+2}\\
n+n+1+2\sqrt{n(n+1)}\leq k^2\leq 4n+2
\end{equation*}</span>
Since <span class="math-container">$\sqrt{n(n+1)}>n$</span>, we have<br />
<span class="math-container">\begin{equation*}
4n+1<k^2\leq4n+2
\end{equation*}</span>
So the only possibility is <span class="math-container">$k^2=4n+2$</span>. So <span class="math-container">$k=\sqrt{4n+2}=2\sqrt{n+1/2}=2\sqrt{(2n+1)/2}$</span>. So
<span class="math-container">\begin{equation}
\sqrt{\frac{2n+1}{2}}=x\quad (1),\quad or\\
\sqrt{\frac{2n+1}{2}}=y+\frac{1}{2}\quad(2)
\end{equation}</span>
for integers <span class="math-container">$x$</span> and <span class="math-container">$y$</span>. Since <span class="math-container">$2n+1$</span> is an odd number, <span class="math-container">$(2n+1)/2$</span> cannot be an integer, so <span class="math-container">$(1)$</span> cannot be the case. For <span class="math-container">$(2)$</span>, we notice that <span class="math-container">$(y+1/2)^2=(2y+1)^2/4$</span>, which is an odd number over <span class="math-container">$4$</span>. But <span class="math-container">$(2n+1)/2$</span> is an odd number over <span class="math-container">$2$</span>, namely an even number over <span class="math-container">$4$</span>, thus they can not be equal. We have the desired contradiction.</p>
|
2,696,579 | <p>So I understand that in order to have a particular solution you have to have a non homogenous second order differential equation. However I have a slightly difficult time comprehending how to pick the particular solution given $g(t)$. </p>
<p>The reason I ask is our teacher skimmed over it and hardly covered it in class and I would like some more information if possible. The question in short is given the forms and the function $g(t)$ in a second order non homogenous differential equation what is the best way to pick your potential particular solution?</p>
| José Carlos Santos | 446,262 | <p>It is easy to prove that, for all $x\in\mathbb Q$, $f(x)=ax$, with $a=f(1)$. It follows from this and from the fact that $f$ is monotonic that, for each $y\in\mathbb R$,$$\lim_{x\to y}f(x)=f(y).$$Putting all this together, one gets that $(\forall x\in\mathbb{R}):f(x)=ax$.</p>
|
165,328 | <p>What is the difference between $\cap$ and $\setminus$ symbols for operations on sets?</p>
| JMP | 210,189 | <p>We have:</p>
<p>$$A\cap B = \{x|x\in A \land x\in B\}$$
$$A\setminus B = \{x|x\in A \land x\not\in B\}$$</p>
<p>from here we can say:</p>
<p>$$A\cap B=A\setminus B^c=B\setminus A^c$$
$$A\cap B^c=A\setminus B=B^c\setminus A^c$$</p>
<p>where:</p>
<p>$$B^c=\{x|x\not\in B\}$$</p>
|
361,740 | <p>Spivak's <em>Calculus on Manifolds</em> asks the reader to prove this (problem 1-8, pp.4-5):</p>
<blockquote>
<p>If there is a basis $x_1, x_2, ..., x_n$ of $\mathbb{R}^n$ and numbers $\lambda_1, \lambda_2, ..., \lambda_n$ such that $T(x_i) = \lambda_i x_i$, $1 \leq i \leq n$, prove that $T$ is angle-preserving iff $\left| \lambda_i \right| = c, 1 \leq i \leq n$.</p>
</blockquote>
<p>Here "angle-preserving" means that the linear map $T$ satisfies $$\frac{ \langle x, y \rangle}{\|x\| \|y\|} = \frac{ \langle T(x), T(y) \rangle}{\|T(x)\| \|T(y)\|},$$
and that $T$ is injective.</p>
<p>My first problem with this question is that the claim is false. Taking $n = 2$, $x_1 = (1, 0)$, $x_2 = (1,1)$, $T(x_1) = -x_1$, $T(x_2) = x_2$, and setting $x = x_1$, $y = x_1 + x_2$, the expression in the RHS above evaluates to $0$, while the expression in the LHS evaluates to $\frac{1}{\sqrt{2}}$.</p>
<p>My second, bigger problem is that I'm not really understanding what's going on. An earlier part of the problem had me show that norm-preserving matrices are angle-preserving; this I'm not sure I get. Thus, I'm not sure what true "version" of this statement the author had in mind (was he trying to get a converse?) and I don't know what to do.</p>
<hr>
<p>Here's my guess:</p>
<p>Looking at some transformations in $\mathbb{R}^2$ (just drawing them), it looks like some of them could be "flip the sign of a basis vector" <strong>IF</strong> it's an orthogonal basis. However, I can't seem to recover the usual "rotation through an angle $\theta$" transformation this way, so I'm not sure that requiring the basis be orthogonal makes the statement true. </p>
<p>Also, I'm not sure how to take the inner product of vectors which aren't in standard coordinates. Or am I missing something here? </p>
| Alexander Jones | 86,853 | <p>I noticed this too. If you ignore the absolute value on lambda, it's a true statement which can be easily proven.</p>
|
2,071,332 | <p>I am working on implementation of a machine learning method that in part of the algorithm I need to calculate the value of $\Gamma (\alpha) / \Gamma (\beta) $. $\alpha$ and $\beta$ are quite large numbers (i.e. bigger than 200) and it causes the python $gamma$ function to overflow. However, as the difference of $\alpha$ and $\beta$ is relatively small (e.g. $|\alpha-\beta|<5$), the final result is not such a big number and can be used for later purposes. So, I am trying to calculate (or approximate) the value of $\Gamma (\alpha) / \Gamma (\beta) $ without going through the calculation of $\Gamma (\alpha)$ and $\Gamma (\beta)$ directly. If $\alpha$ and $\beta$ were integers, the result would be simple equal to $\alpha . \alpha+2. \alpha+3... \beta-1$, But I can not imagine how this formula will be changed if we let $\alpha$ and $\beta$ to be real numbers.</p>
| Claude Leibovici | 82,404 | <p>I think that a good solution would be Stirling approximation that is to say $$\log(\Gamma(x))=x (\log (x)-1)+\frac{1}{2} \left(-\log \left({x}\right)+\log (2 \pi
)\right)+\frac{1}{12 x}+O\left(\frac{1}{x^3}\right)$$ Now, consider $$y=\frac{\Gamma(\alpha)}{\Gamma(\beta)}\implies \log(y)=\log(\Gamma(\alpha))-\log(\Gamma(\beta))$$ Apply the formula (even with more terms) and use later $y=e^{\log(y)}$.</p>
<p>You are then able to control overflows and underflows if required.</p>
<p>By the way, why not to use in Python function <strong>lgamma(x)</strong> ?</p>
|
290,527 | <p>What would be a good metric on $C^k(0,1)$, space of $k$ times continuously differentiable real valued functions on $(0,1)$ and $C^\infty(0,1)$, space of infinitely differentiable real valued functions on $(0,1)$? </p>
<p>It is of course open to interpretation what good would mean, I want it to bring a good notion of convergence and unitize the openness of the interval as well as the $k$ times/infinite continuously differentiable property. This is a question that I want to think more about to understand metric spaces better. Thank you.</p>
<p>EDIT: And what changes if it is on $[0,1]$?</p>
| Davide Giraudo | 9,849 | <p>We can define the metrics
$$d(f,g):=\sum_{n=1}^{+\infty}2^{-n}\min\left\{1,\max_{0\leqslant k\leqslant n}\max_{n^{-1}\leqslant x\leqslant 1-n^{-1}}|f^{(k)}(x)-g^{(k)}(x)|\right\}\quad\mbox{on } C^\infty(0,1)$$
$$d_N(f,g):=\sum_{n=1}^{+\infty}2^{-n}\min\left\{1,\max_{0\leqslant k\leqslant N}\max_{n^{-1}\leqslant x\leqslant 1-n^{-1}}|f^{(k)}(x)-g^{(k)}(x)|\right\}\quad\mbox{on } C^N(0,1).$$
This gives information about the behavior of the functions and their derivatives on compact intervals. Summing this help us to know the behavior on $(0,1)$. Furthermore, one can check that $C^\infty(0,1)$ and $C^N(0,1)$ endowed with the corresponding metrics are complete. </p>
<p>Note that I just used the fact that the natural topologies on $C^k(0,1)$ and $C^\infty(0,1)$ are generated by a countable family of semi-norms. </p>
<p>We can give a generalization when we replace $(0,1)$ by any open subset of $\Bbb R^d$, $d\geqslant 1$.</p>
|
290,527 | <p>What would be a good metric on $C^k(0,1)$, space of $k$ times continuously differentiable real valued functions on $(0,1)$ and $C^\infty(0,1)$, space of infinitely differentiable real valued functions on $(0,1)$? </p>
<p>It is of course open to interpretation what good would mean, I want it to bring a good notion of convergence and unitize the openness of the interval as well as the $k$ times/infinite continuously differentiable property. This is a question that I want to think more about to understand metric spaces better. Thank you.</p>
<p>EDIT: And what changes if it is on $[0,1]$?</p>
| mdg | 64,184 | <p>Ask yourself if the spaces $C^k(0,1)$ are of interest. </p>
<p>EDIT: I also suggest thinking more about and therefore formalising the notion of a 'good' metric on a space. You might find that a good metric is one that induces a topology on the space that has desirable properties for doing analysis, like local convexity, being complete, the Heine-Borel property, an interesting dual space, etc.</p>
<p>Since $(0,1)$ is the union of countably many compact sets $K_n$ such that $K_n\subset\text{int}K_{n+1}$, then we define $C(0,1)$ as the space of all continuous real-valued functions on $(0,1)$ with topology induced by the family of separating semi-norms:
$$
p_n(f):=\sup_{x\in K_n}|f(x)|.
$$
Then the sets
$$
V_n:=\{f\in C(0,1)\;|\;p_n(f)<1/n\},\qquad n=1,2,...
$$
form a local convex base for $C(0,1)$. This topology is in fact induced by the metric
$$
d(f,g):=\max_n\frac{1}{2^n}\frac{p_n(f-g)}{1+p_n(f-g)}
$$
This metric is complete, so $C(0,1)$ becomes a Frechet space.</p>
<p>The space $C^\infty(0,1)$ is defined to be the space of all real-valued functions $f$ on $(0,1)$ with the property that $D^kf\in C(0,1)$ for all $k\in\mathbb{N}_0$. Choosing the same compact sets $K_n$ as above, the following family of semi-norms undices a metrizable, locally-convex topology on $C^\infty(0,1)$:
$$
p_n(f):=\max_{x\in K_n,\;k\leq n}|D^kf(x)|.
$$
The metric on this space is of the same form as above. It can be shown that this space is also a Frechet space and additionally has the Heine-Borel property, so it follows that it is not normable (the metric is not induced by a norm). On the space $C^\infty(0,1)$ is constructed the space of distributions with compact support, which are linear functionals that are continuous with respect to the above defined topology.</p>
<p>To generalise the above, you should replace $(0,1)$ with an arbitrary open subset of $\Omega\subset\mathbb{R}^m$, and allow $k$ to be a multi-index.</p>
<p>Other continuous function spaces that you might be interested in are: </p>
<p>(i) the space $\mathcal{D}_K$ which is the space of all $f\in C^\infty(\mathbb{R}^m)$ such that $\text{supp}f\subset K$ where $K\subset\Omega$ is compact. It can be shown that $\mathcal{D}_K$ is a closed subspace of $C^\infty(\Omega)$.</p>
<p>(i) the Schwartz space $\mathcal{S}(\mathbb{R}^m)$, which is important for Fourier analysis. On this space is consructed the space of tempered distributions, which is composed of linear functionals that are continuous with respect to the appropriate topology defined on the Schwartz space.</p>
<p>(ii) the space of test functions $\mathcal{D}(\Omega)=C^\infty_0(\Omega)$, which is important for Sobolev space and PDE theory. The space of distributions is constructed on this space, and consists of all linear functionals that are continuous with respect to the appropriate topology on $\mathcal{D}(\Omega)$. This topology is a challenge to define and understand, but is easily characterised in terms of convergence.</p>
<p>By far the best reference for all of this stuff is Rudin's book on functional analysis. Also see Yosida's book on functional analysis, DiBenedetto's book on real analysis and Knapp's book on advanced analysis. </p>
|
2,725,839 | <p>The question is below.<a href="https://i.stack.imgur.com/k3UMf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/k3UMf.png" alt="enter image description here"></a></p>
<p>I was able to solve part (a) because the $x$-coordinate would just be the circumference of the circle, which is $2\pi$. Therefore, $P = (2\pi, 0)$. </p>
<p>I am confused with parts (b) and after, but I know that the $y$-coordinate will remain to be $1$ because that's the radius of the circle and how far it is off from the $x$-axis. Any help with how to find the $x$-coordinate will help for part (b). Thank you in advance. </p>
| Marin | 549,731 | <p>The centre point travels 2π<em>r</em> for every complete turn along <em>x</em>. Similarly it travels (1/4)2π<em>r</em> for a quarter turn.</p>
<p>Then turn the figure vertically with the <em>x</em> axis pointing down. Imagine the angle of reference as the second hand of a clock turning clockwise, starting at 9 o’clock. This means the <em>y</em> coordinate of the point is (the absolute value of) the sine of the angle, scaled by the radius (take another look at the definition of sine in the unit circle). In this case, after a quarter turn the <em>y</em> coordinate is <em>r</em>⋅sin((1/4)2π)</p>
|
3,090,879 | <p>Is there prime number of the form <span class="math-container">$1+11+111+1111+11111+...$</span>. I've checked it up to first 2000 repunits, but i found none. If <span class="math-container">$R_1=1$</span>, <span class="math-container">$R_2=1+11$</span>, <span class="math-container">$R_3=1+11+111$</span>, <span class="math-container">$R_n=1+11+111+...+$</span>nth repunit. What is the smallest n such that <span class="math-container">$R_n$</span> is prime ?</p>
| David G. Stork | 210,401 | <p>Here's the smallest such prime, found using this code:</p>
<pre><code>Select[Accumulate[Table[
Sum[10^i, {i, 0, n}],
{n, 0, 10000}]], PrimeQ]
</code></pre>
<p><span class="math-container">$$1234567901234567901234567901234567901234567901234567901234567901234567
9012345679012345679012345679012345679012345679012345679012345679012345
6790123456790123456790123456790123456790123456790123456790123456790123
4567901234567901234567901234567901234567901234567901234567901234567901
2345679012345679012345679012345679012345679012345679012345679012345679
0123456790123456790123456790123456790123456790123456790123456790123456
7901234567901234567901234567901234567901234567901234567901234567901234
5679012345679012345679012345679012345679012345679012345679012345679012
3456790123456790123456790123456790123456790123456790123456790123456790
1234567901234567901234567901234567901234567901234567901234567901234567
9012345679012345679012345679012345679012345679012345679012345679012345
6790123456790123456790123456790123456790123456790123456790123456790123
4567901234567901234567901234567901234567901234567901234567901234567901
2345679012345679012345679012345679012345679012345679012345679012345679
0123456790123456790123456790123456790123456790123456790123456790123456
7901234567901234567901234567901234567901234567901234567901234567901234
5679012345679012345679012345679012345679012345679012345679012345679012
3456790123456790123456790123456790123456790123456790123456790123456790
1234567901234567901234567901234567901234567901234567901234567901234567
9012345679012345679012345679012345679012345679012345679012345679012345
6790123456790123456790123456790123456790123456790123456790123456790123
4567901234567901234567901234567901234567901234567901234567901234567901
2345679012345679012345679012345679012345679012345679012345679012345679
0123456790123456790123456790123456790123456790123456790123456790123456
7901234567901234567901234567901234567901234567901234567901234567901234
5679012345679012345679012345679012345679012345679012345679012345679012
3456790123456790123456790123456790123456790123456790123456790123456790
1234567901234567901234567901234567901234567901234567901234567901234567
9012345679012345679012345679012345679012345679012345679012345679012345
6790123456790123456790123456790123456790123456790123456790123456790123
4567901234567901234567901234567901234567901234567901234567901234567901
2345679012345679012345679012345679012345679012345679012345679012345679
0123456790123456790123456790123456790123456790123456790123456790123456
7901234567901234567901234567901234567901234567901234567901234567901234
5679012345679012345679012345679012345679012345679012345679012345679012
34567901234567901234567901234567901234567900957$$</span></p>
|
3,148,094 | <p>In class, my professor computed the density of a gamma random variable by taking the derivative of its CDF, but he skipped many steps. I am trying to go through the derivation carefully but cannot reproduce his final result.</p>
<p>Let <span class="math-container">$k$</span> be the shape and <span class="math-container">$\mu$</span> be the scale. Then the CDF for a gamma random variable <span class="math-container">$T$</span> is</p>
<p><span class="math-container">$$
F(t) = 1 - \sum_{i=0}^{k-1} \frac{e^{-\mu t} (\mu t)^{i}}{i!}
$$</span></p>
<p>Using the product rule, I get</p>
<p><span class="math-container">$$
\begin{align}
f(t)
&= \frac{\partial}{\partial t} F(t)
\\
&= \frac{\partial}{\partial t} \Big( 1 - \sum_{i=0}^{k-1} \frac{e^{-\mu t} (\mu t)^{i}}{i!} \Big)
\\
&= - \sum_{i=0}^{k-1} \frac{\partial}{\partial t} \Big( \frac{e^{-\mu t} (\mu t)^{i}}{i!} \Big)
\\
&= - \sum_{i=0}^{k-1} \frac{1}{i!} \frac{\partial}{\partial t} \Big( e^{-\mu t} (\mu t)^{i} \Big)
\end{align}
$$</span></p>
<p>where</p>
<p><span class="math-container">$$
\frac{\partial}{\partial t} \Big( e^{-\mu t} (\mu t)^{i} \Big) = e^{-\mu t} (-\mu)(\mu t)^i + e^{-\mu t} i (\mu t)^{i-1}
$$</span></p>
<p>Putting everything together, we get</p>
<p><span class="math-container">$$
f(t)
=
\sum_{i=0}^{k-1} \frac{\mu e^{-\mu t} (\mu t)^i}{i!}
-
\sum_{i=0}^{k-1} \frac{e^{-\mu t} (\mu t)^{i-1}}{(i-1)!}
$$</span></p>
<p>But here I am stuck. I know that</p>
<p><span class="math-container">$$
f(t) = \frac{\mu e^{-\mu t} (\mu t)^{k-1}}{(k-1)!}
$$</span></p>
<p>but am not sure how to get there.</p>
| MarianD | 393,259 | <p><a href="https://i.stack.imgur.com/lIbHV.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lIbHV.jpg" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/d6JoF.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/d6JoF.jpg" alt="enter image description here"></a></p>
<p><span class="math-container">$$\left .
\begin{array}{l}
x \in \color{green}B \implies 2 <x <=3 \implies x > 2 \implies x^2 > 4 \implies x^2-4>0\\
x \in \color{green}B \implies 2 <x <=3 \implies x \le 3 \implies x^2 \le 9 \implies x^2 -9 \le 0
\end{array}
\right\}
\implies x \in \color{red}A$$</span></p>
<p>So <span class="math-container">$\color{green}B \subset \color{red}A$</span>.</p>
<p>But the reverse inclusion <span class="math-container">$\left (\color{red}A \subset \color{green}B\, \right)$</span> is not true, because for <span class="math-container">$x=-3:$</span> </p>
<p><span class="math-container">$$x \in \color{red}A,\quad x \notin \color{green}B$$</span></p>
<p>So <span class="math-container">$\color{green}B$</span> is a <em>proper</em> subset of <span class="math-container">$\color{red}A$</span>.</p>
|
1,557,039 | <h2>Background</h2>
<p>I am a software engineer and I have been picking up combinatorics as I go along. I am going through a combinatorics book for self study and this chapter is absolutely destroying me. Sadly, I confess it makes little sense to me. I don't care if I look stupid, I want to understand how to solve these problems. </p>
<p>I am studying counting with repetition. That is, generalized binomial coefficients and generating functions. </p>
<h2>Problem</h2>
<p>Suppose that an unlimited amount of jelly beans is available in each of the five different colors: red, green, yellow, white, and black. </p>
<ul>
<li>How many ways are there to select from twenty jelly beans? </li>
<li>How many ways are there to select twenty jelly beans if we must select at least two jelly beans of each color? </li>
</ul>
<h2>Attempted Solution</h2>
<ul>
<li><p>How many ways are there to select from twenty jelly beans?</p>
<p>$5^{20}$</p></li>
<li><p>How many ways are there to select twenty jelly beans if we must select at least two jelly beans of each color? </p></li>
</ul>
<p>I keep thinking of applying the hypergeometric distribution here, but I think this is dead wrong. The entire chapter is on series, so I am confused as to what these series are and why they are being applied to solve these problems? The above solutions (some I am too embarrassed to share) didn't pass the smell test at all :( </p>
| Peter | 82,961 | <p>Base case : $9\equiv 1\ (\ mod\ 4\ )$ is true.</p>
<p>Suppose $9^n\equiv 1\ (\ mod\ 4\ )$</p>
<p>Then $9^{n+1}=9\times 9^n\equiv 9\equiv 1\ (\ mod\ 4\ )$</p>
|
1,278,442 | <p>I was wondering if the singular homology functor preserve injectivity and surjectivity? I've been trying to figure out a proof or counterexample for ages now and I just can't.</p>
<p>This came up when I was looking at the reduced homology $H_p(S^{n},S^{n-1})$. To calculate it, I have looked at the canonical injection $$\iota: S^{n-1} \longrightarrow S^{n}.$$ I'm viewing $S^{n-1}$ as the equator of $S^n$, in this case does the functor preserve injectivity? That is, I want to say that $\iota_*$ is injective. Is this true? Thanksarinos</p>
| Qiaochu Yuan | 232 | <p>I don't understand what you mean by relative homology preserving injectivity and surjectivity, so consider instead the following general point. </p>
<p>No interesting homotopy-invariant functor on spaces can preserve injections or surjections. The reason is that every map is homotopy equivalent to an injection and also to a surjection! There are very nice explicit constructions accomplishing this called the <a href="http://en.wikipedia.org/wiki/Mapping_cylinder" rel="nofollow">mapping cylinder</a> and <a href="http://ncatlab.org/nlab/show/cocylinder" rel="nofollow">mapping cocylinder</a> respectively. </p>
<p>In other words, "surjection" and "injection" aren't homotopy-theoretically meaningful concepts. </p>
|
1,569,476 | <p>We throw fair dice until $6$ will appear. Let $X$ denote total number of throws and $Y$ - number of $5$ we received.</p>
<ol>
<li>Find distribution $(X,Y)$</li>
<li>Are variables $X$ and $Y$ independent?</li>
</ol>
<p>I have to say that I have utterly no idea how to proceed with this question, detailed explanation appreciated.</p>
| BGM | 297,308 | <p>For the first part you need the following:
$$X \sim \text{Geometric}\left(\frac {1} {6}\right) $$ and
$$Y|X = x \sim \text{Binomial}\left(x, \frac {1} {6}\right)$$ </p>
<p>and the joint pmf is just the product of these pmf.</p>
|
3,797,724 | <blockquote>
<p>Find all real continuous functions that verifies :
<span class="math-container">$$f(x+1)=f(x)+f\left(\frac{1}{x}\right) \ \ \ \ \ \ (x\neq 0) $$</span></p>
</blockquote>
<p>I found this result <span class="math-container">$\forall x\neq 1 \ \ f(x)=f\left(\frac{x}{x-1} \right)$</span> and I tried to study the behaviour of the function <span class="math-container">$g$</span> defined as <span class="math-container">$g(x)=\frac{x}{x-1}$</span> and compare it with <span class="math-container">$x$</span> in order to use fixed point theorem but it won't work.</p>
<p>I need a hint and thanks.</p>
| nonuser | 463,553 | <p>Some partial results:</p>
<ul>
<li>Letting <span class="math-container">$x+1= {1\over x} $</span> we get <span class="math-container">$$x_{1,2}= {-1\pm \sqrt{5}\over 2}$$</span> are zeroes for <span class="math-container">$f$</span>.</li>
<li>If <span class="math-container">$x\to {1\over x}$</span> we get <span class="math-container">$$f(1+{1\over x}) = f({1\over x})+f(x)=f(1+x)$$</span> so <span class="math-container">$$\boxed{f(1+{1\over x}) = f(1+x);\;\;\;\forall x}$$</span></li>
</ul>
<p>So taking <span class="math-container">$x\to 0$</span> in this equation we get <span class="math-container">$$f(1) = \lim _{x\to
\pm \infty}f(x)$$</span> and on the other hand we have, from starting equation <span class="math-container">$$f(1)-f(0) =\lim _{x\to
\pm \infty}f(x)$$</span>
so <span class="math-container">$f(0)=0$</span> and we have <span class="math-container">$x_3=0$</span>.</p>
<ul>
<li>From boxed equation we get <span class="math-container">$x\to x-1$</span> we get, for <span class="math-container">$\forall x\ne 1$</span>: <span class="math-container">$$f(x) = f\Big({x\over x-1}\Big)$$</span></li>
</ul>
<blockquote class="spoiler">
<p> Since last formula is not defined for <span class="math-container">$x=1$</span> it is tempting to check whether <span class="math-container">$$\boxed{g(x) = {x^2+x-1\over x-1}}$$</span> satisfies the condition, which does, but unfortunately, it is no defined on whole <span class="math-container">$\mathbb{R}$</span>.</p>
</blockquote>
<ul>
<li>Finally, if we set <span class="math-container">$V$</span> as a set of all solution to this equation, we see that <span class="math-container">$V$</span> is a vector space over <span class="math-container">$\mathbb{R}$</span>.</li>
</ul>
|
3,237,242 | <p>I have the following problem:</p>
<p>I need to prove that given the following integral</p>
<p><span class="math-container">$\int_{0}^{1}{c(k,l)x^k(1-x)^l}dx = 1$</span>,</p>
<p>we the constant <span class="math-container">$c(k,l) = (k+l+1) {{k+l}\choose{k}} = \frac{(k+l+1)!}{k!l!}$</span>,</p>
<p>with the use of two dimensional mathematical induction on <span class="math-container">$min(k,l)$</span>.
Here <span class="math-container">$k$</span> and <span class="math-container">$l$</span> are two nonnegative integers.</p>
<p>(THUS: I need to proof that <span class="math-container">$c(k,l)$</span> is equal to <span class="math-container">$(k+l+1) {{k+l}\choose{k}}$</span>)</p>
<p>For the base step I have proved that <span class="math-container">$c(k, 0) = c(0, k) = k + 1$</span> for all <span class="math-container">$k$</span>.</p>
<p>I am given a hint that for the induction step I could try using integration by parts to show <span class="math-container">$c(k,l) = \frac{k+1}{l} c(k+1,l−1)$</span>.</p>
<p>By integrating the following by parts I indeed managed to show the latter:</p>
<p><span class="math-container">$\int_{0}^{1}{c(k+1,l-1)x^{k+1}(1-x)^{l-1}dx}=1$</span>.</p>
<p>However, I don't really see how this helps me to complete my proof, since I don't really get the idea of two dimensional induction.</p>
<p>Can someone maybe clarify this a bit for me, and help me further with my proof?</p>
| G Cab | 317,234 | <p>Let me give you an intuitive hint on how to deal with 2D induction</p>
<p><a href="https://i.stack.imgur.com/l3CTh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/l3CTh.png" alt="2D_Induction_1"></a> </p>
<p>You have demonstrated that the hypothesis is true on the axis <span class="math-container">$k=0$</span>, for whichever <span class="math-container">$l$</span> in the domain of interest (<span class="math-container">$[0,\infty )$</span> in this case).</p>
<p>You have demonstrated that if the hypothesis is true for <span class="math-container">$(l,k)$</span> then it is true for <span class="math-container">$(l-1,k+1)$</span>.<br>
Therefore it is true for the whole diagonal line shown in the sketch, going from <span class="math-container">$(l,0)$</span> to <span class="math-container">$(0,l)$</span>.</p>
<p>Thus, it is true for any such diagonal line, starting from whichever point <span class="math-container">$(l,0)$</span>.</p>
<p>These lines cover all the points in <span class="math-container">$[0, \infty)^2$</span> and thus the hypothesis is true over all such a domain.</p>
<p>If the recursion had been <span class="math-container">$(l,k)\, \to \, (l+1,k+1)$</span> then it is clear that you should have started from demonstrating that the hypothesis is true on <span class="math-container">$(l,0)$</span> and <span class="math-container">$(0,k)$</span>
to be able to cover all the quadrant.</p>
|
4,000,089 | <p>Let <span class="math-container">$x$</span> be an element of a Banach Algebra. Let <span class="math-container">$\lambda \in \rho(x)$</span>, where <span class="math-container">$\rho(x)$</span> is the resolvent of <span class="math-container">$x$</span>.</p>
<p>Let <span class="math-container">${d(\lambda, \sigma(x))}$</span> be the distance between <span class="math-container">$\lambda$</span> and <span class="math-container">$\sigma(x)$</span>.</p>
<p>Show that <span class="math-container">$\| (\lambda-x)^{-1} \| \geq \frac{1}{d(\lambda, \sigma(x))}$</span>.</p>
<p>Information I know:</p>
<p><span class="math-container">$\rho(x)$</span> is an open set.</p>
<p><span class="math-container">$\sigma(x)$</span> is compact.</p>
<p>I've seen many places using this inequality, but I cannot find a proof of it. And I'm not sure how to prove this by breaking down the definitions I know.</p>
<p>Thanks in advance!</p>
| Ng Chung Tak | 299,599 | <p><strong>Equation of tangent</strong></p>
<p><span class="math-container">$$0=T(x,y)\equiv \frac{x\cos t}{a}+\frac{y\sin t}{b}-1$$</span></p>
<p><strong>Equation of normal</strong></p>
<p><span class="math-container">$$0=N(x,y) \equiv \frac{ax}{\cos t}-\frac{by}{\sin t}-(a^2-b^2)$$</span></p>
<p>Note that <span class="math-container">$N(u,0)=0$</span> and</p>
<p><span class="math-container">\begin{align}
u(t) &= \frac{(a^2-b^2)\cos t}{a} \\
d(t) &= \frac{|T(u,0)|}{\sqrt{\dfrac{\cos^2 t}{a^2}+\dfrac{\sin^2 t}{b^2}}} \\
&= \frac{\left| -\sin^2 t-\dfrac{b^2\cos^2 t}{a^2} \right|}
{\sqrt{\dfrac{\cos^2 t}{a^2}+\dfrac{\sin^2 t}{b^2}}} \\
&= b^2\sqrt{\frac{\cos^2 t}{a^2}+\frac{\sin^2 t}{b^2}} \\
\kappa &= \frac{1}{a^2 b^2
\left(
\dfrac{\cos^2 t}{a^2}+\dfrac{\sin^2 t}{b^2}
\right)^{3/2}} \\
&= \frac{b^4}{a^2 d^3}
\end{align}</span></p>
|
2,074,276 | <p>How would I go about computing $\displaystyle\int_{10}^{16}\sin(\cosh^{-1}(x)+7)\mathrm dx$?</p>
<p>I haven't attempted anything yet, because I don't even know how to integrate the inverse hyperbolic cosine.</p>
| Fawad | 369,983 | <p><a href="https://i.stack.imgur.com/A052G.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/A052G.png" alt=""></a></p>
<p>Let's assume you have a number,for example 5. When you are asked 5 is in which circle you will say it is in B. Then by this statement one can say your number is also in A.</p>
<p>But lets assume you have number 3. And you are asked in which circle number is? You day it is in circle A. So one can't say the number is in B!</p>
<p>This case is same , assume circle B as Paris and fields A as France and see your statement.</p>
|
2,074,276 | <p>How would I go about computing $\displaystyle\int_{10}^{16}\sin(\cosh^{-1}(x)+7)\mathrm dx$?</p>
<p>I haven't attempted anything yet, because I don't even know how to integrate the inverse hyperbolic cosine.</p>
| Chas Brown | 167,790 | <p>Let's agree that 'if you are in Paris, then you are in France' ($A \implies B$). (We could get picky, and say maybe you're in Paris, Texas; but let's not!).</p>
<p>But then the <strong>converse</strong> ($B \implies A$) is not automatically true: for example, we can't then deduce from 'if you are in Paris, then you are in France' that 'if you are in France, then you are in Paris'. In that sense you're right that 'it's not logical' to say the converse is always true. </p>
<p>Now, bear in mind that the converse <em>might</em> be true, or it <em>might not</em>. It must be proven either way based on other information. And it turns out that we know enough other things, about European geography in this case, that we can also <em>prove</em> that the converse is not true.</p>
<p>On the other hand, what we can <em>always</em> deduce is called the <strong>contrapositive</strong>: once we accept the truth of 'if you are in Paris, then you are in France', then we always automatically can say 'if you are <em>not</em> in France, then you are <em>not</em> in Paris' ($\neg B \implies \neg A$). That will always be true (at least, in the world of mathematical language).</p>
|
4,568,221 | <p><span class="math-container">$$\iint\limits_{D} \left(x^2+y^2\right)\mathrm{d}x \mathrm{d}y$$</span>
where <span class="math-container">$D$</span> is given each time by
<span class="math-container">$D=x^2-y^2=1,\hspace{0.5cm} x^2-y^2=9,\hspace{0.5cm} xy=2,\hspace{0.5cm} xy=4$</span></p>
<p>I try to use Polar coordinate transformation
<span class="math-container">\begin{cases}
x &= \rho \cos \theta \\
y &= \rho \sin \theta
\end{cases}</span>
but I don’t know how to find the range of <span class="math-container">$\rho$</span> and <span class="math-container">$\theta$</span></p>
| user170231 | 170,231 | <p><span class="math-container">$D$</span> is made of two disconnected but symmetric regions. <span class="math-container">$xy$</span> is positive, so either both <span class="math-container">$x,y$</span> are positive or they are both negative. Let <span class="math-container">$D_+$</span> be the part of <span class="math-container">$D$</span> in the first quadrant. The other part is mirrored in the fourth quadrant. The integrand <span class="math-container">$x^2+y^2$</span> is symmetric about the origin, so its integral over <span class="math-container">$D$</span> is exactly twice the integral over <span class="math-container">$D_+$</span>.</p>
<p>Using polar coordinates:</p>
<p>First rewrite the boundaries of <span class="math-container">$D_+$</span>.</p>
<p><span class="math-container">$$x^2 - y^2 = a \implies r^2 \left(\cos^2(\theta) - \sin^2(\theta)\right) = a \implies r = \sqrt a \sqrt{\sec(2\theta)} \\
xy = b \implies r^2 \cos(\theta) \sin(\theta) = b \implies r = \sqrt{2b} \sqrt{\csc(2\theta)}$$</span></p>
<p>Split up <span class="math-container">$D_+$</span> into three regions:</p>
<p><span class="math-container">$$D_+ = D_1 \cup D_2 \cup D_3 \\
D_1 = \left\{(r,\theta) \mid 2 \sqrt{\csc(2\theta)} \le r \le 3\sqrt{\sec(2\theta)} \text{ and } \theta_1 \le \theta \le \theta_2\right\} \\
D_2 = \left\{(r,\theta) \mid 2 \sqrt{\csc(2\theta)} \le r \le 2\sqrt2 \sqrt{\csc(2\theta)} \text{ and } \theta_2 \le \theta \le \theta_3\right\} \\
D_3 = \left\{(r,\theta) \mid \sqrt{\sec(2\theta)} \le r \le 2\sqrt2 \sqrt{\csc(2\theta)} \text{ and } \theta_3 \le \theta \le \theta_4\right\}$$</span></p>
<p>where <span class="math-container">$\theta_1,\theta_2,\theta_3,\theta_4$</span> are the values of <span class="math-container">$\theta$</span> corresponding to the vertices of <span class="math-container">$D_+$</span>, represented by the rays in the plot in counter-clockwise order:</p>
<p><a href="https://i.stack.imgur.com/ah9rPm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ah9rPm.png" alt="enter image description here" /></a></p>
<p>The specific coordinates are given in the table below.</p>
<p><span class="math-container">$$\begin{array}{c|cr}
(x,y) & (r,\theta) \\
\hline
\left(\sqrt{\frac{9+\sqrt{97}}2}, \frac{\sqrt{97}-9}4 \sqrt{\frac{9+\sqrt{97}}2}\right) & \left(\sqrt[4]{97}, \tan^{-1}\left(\frac{\sqrt{97}-9}4\right)\right) & (\theta_1) \\
\left(\sqrt{\frac{9+\sqrt{145}}2}, \frac{\sqrt{145}-9}8 \sqrt{\frac{9+\sqrt{145}}2}\right) & \left(\sqrt[4]{145}, \tan^{-1}\left(\frac{\sqrt{145}-9}4\right)\right) & (\theta_2) \\
\left(\sqrt{\frac{1+\sqrt{17}}2}, \frac{\sqrt{17}-1}8 \sqrt{\frac{1+\sqrt{17}}2}\right) & \left(\sqrt[4]{17}, \tan^{-1}\left(\frac{\sqrt{17}-1}4\right)\right) & (\theta_3) \\
\left(\sqrt{\frac{1+\sqrt{65}}2}, \frac{\sqrt{65}-1}8 \sqrt{\frac{1+\sqrt{65}}2}\right) & \left(\sqrt[4]{65}, \tan^{-1}\left(\frac{\sqrt{65}-1}8\right)\right) & (\theta_4)
\end{array}$$</span></p>
<p>Then the integral is</p>
<p><span class="math-container">$$\begin{align*}
\iint_D (x^2+y^2) \,dx\,dy &= 2 \iint_{D_+} r^3 \, dr \, d\theta \\[1ex]
&= 2 \left\{\int_{\theta=\theta_1}^{\theta_2} \int_{r=2\sqrt{\csc{2\theta}}}^{3\sqrt{\sec(2\theta)}} + \int_{\theta=\theta_2}^{\theta_3} \int_{r=2\sqrt{\csc{2\theta}}}^{2\sqrt2\sqrt{\csc(2\theta)}} + \int_{\theta=\theta_3}^{\theta_4} \int_{r=\sqrt{\sec{2\theta}}}^{2\sqrt2\sqrt{\csc(2\theta)}}\right\} r^3 \, dr \, d\theta
\end{align*}$$</span></p>
<p>Doable, yes, but you're much better off using Cartesian coordinates or making the substitution suggested in comments and the other answer.</p>
|
2,289,777 | <p>I have a function $f(x)=(8x^2+7)^3(x^3-7)^4$</p>
<p>I have differentiated it using the chain rule and arrived at:</p>
<p>$3(8x^2+7)^2 \cdot 16x \cdot 4(x^3-7)^3 \cdot 3x^2$ And apparently this is wrong?</p>
<p>What am I missing here? </p>
| Claude Leibovici | 82,404 | <p>When you face problem like this one which only contains products, ratios and powers, your best friend is logarithmic differentiation $$f=(8x^2+7)^3(x^3-7)^4\implies \log(f)=3\log(8x^2+7)+4\log(x^3-7)$$ Differentiate both sides $$\frac{f'}f=3 \frac{16x}{8x^2+7}+4\frac{3x^2}{x^3-7}=\frac{12 x \left(12 x^3+7 x-28\right) }{(8x^2+7)(x^3-7)}$$ Now $$f'=f\times \frac{f'}f=(8x^2+7)^3(x^3-7)^4\frac{12 x \left(12 x^3+7 x-28\right) }{(8x^2+7)(x^3-7)}$$ $$f'=12 x \left(8 x^2+7\right)^2 \left(x^3-7\right)^3 \left(12 x^3+7 x-28\right)$$</p>
|
449,296 | <p>I am trying to deduce how mathematicians decide on what axioms to use and why not other axioms, I mean surely there is an infinite amount of available axioms. What I am trying to get at is surely if mathematicians choose the axioms then we are inventing our own maths, surely that is what it is but as it has been practiced and built on for so long then it is too late to change all this so we have had to adapt and create new rules?? I'm not sure how clear what I am asking is or whether it is even understandable but would appreciate any answers or comments, thanks.</p>
| Greebo | 64,519 | <p>There's one important thing that any choice of axioms should at least do, which is to be consistent, i.e. you cannot prove both a theorem and it's inverse, because if you could, you could prove anything at all, which would not give you any interesting theory.</p>
<p>That said, thanks to Gödel, we know that ZFC cannot be proved to be consistent using ZFC (unless it is inconsistent), so we have little way of truly knowing whether the axioms we have chosen 'work', so to say, but intuitively they don't seem to be inconsistent.</p>
<p>I would say that when one creates axioms, one creates them with a specific goal in mind, so while we have an infinite amount of axioms to choose from, we're only interested in the subset that describes some object which behaves in a way which fits with our intuition. </p>
<p>Take the Peano axioms as an example, we want to describe the natural numbers, and we have some intuition on how these should behave. We want a number 0, we want a way to count upwards, we don't want to end up back at 0 if we count for long enough, etc. The axioms is just a way to use logic to explain what we already 'knew'. If I created a different set of axioms for the natural numbers, but where, once I got to 17, I started back at 3, I think we'd all agree that while the mathematical object I've created might be perfectly valid, it is not the natural numbers I've described.</p>
<p>Any other set of axioms is chosen in much the same way. We have a mathematical object we want to describe, we try to create a set of axioms which does the job, and see where the axioms lead us. When we see where the axioms go, we may decide from 'outside' the logic whether this set of axioms really was what we wanted, or if they should be changed to better suit what our intuition tells us should happen.</p>
<p>Hope this answered at least parts of your questions :)</p>
|
449,296 | <p>I am trying to deduce how mathematicians decide on what axioms to use and why not other axioms, I mean surely there is an infinite amount of available axioms. What I am trying to get at is surely if mathematicians choose the axioms then we are inventing our own maths, surely that is what it is but as it has been practiced and built on for so long then it is too late to change all this so we have had to adapt and create new rules?? I'm not sure how clear what I am asking is or whether it is even understandable but would appreciate any answers or comments, thanks.</p>
| Doug Spoonwood | 11,300 | <p>I will speak boldly and assert that the very purpose of an axiomatic system lies in developing a theory deductively. This implies that the axiomatic system will basically help you or your machine generate theorems which require little to no insight on your part (though of course, any insight you have, if correct, will help). Now what properties will axioms prefeably have if this holds true? I suggest the following:</p>
<ol>
<li>Axioms preferably will be few in number. The more axioms you have, the more you take for granted. The point of an axiomatic system lies in the deductions made and having the ability to tell what deductions can get made. This becomes clearer when you have fewer axioms.</li>
<li>The axioms preferably will be <strong>independent</strong> for system S in the sense that using the same rules of inference, for any axiom A will not come as derivable from a system (S-A) with all the other axioms of S, but not having A as an axiom (some authors do NOT seem to realize this as important). In other words, using the rules of inference you've taken for granted, you can't logically deduce any axiom from the other axioms. If you can, then you could eliminate one of the axioms, and you could have deduced more, in some sense, if you had worked with the smaller set of axioms instead of the full set of axioms.</li>
<li>The rules of inference for the theory will get kept to a minimum also. In effect, if you have a lot of rules of inference, you take a lot for granted and consequently in some sense deduce less than you could have if you had fewer rules of inference. (this part may come as the most contentious.)</li>
<li>The rules of inference will come as sufficiently powerful to deduce a lot of theorems.</li>
<li>The axioms preferably will be consistent in the sense you can't derive a proposition and its negation in the theory. If you do have inconsistent axioms, then you may not have deduced anything.</li>
<li>The axioms preferably will be at least relatively easy to use, so that you can deduce more. This in part will depend on who or what does the deducing, so this does imply different axiom sets as potentially feasible, since not everyone thinks the same way, and not all machines behave the same way. For this reason short axioms oftentimes will come as preferable to long axioms, though again NOT ALWAYS, because everyone thinks differently to some degree, and every machine differs from other machines.</li>
</ol>
|
4,122,419 | <blockquote>
<p>From the triangle <span class="math-container">$\triangle ABC$</span> we have <span class="math-container">$AB=3$</span>, <span class="math-container">$BC=5$</span>, <span class="math-container">$AC=7$</span>. If
the point <span class="math-container">$O$</span> placed inside the triangle <span class="math-container">$\triangle ABC$</span> so that
<span class="math-container">$\vec{OA}+2\vec{OB}+3\vec{OC}=0$</span> , then what is the ratio of the area
of <span class="math-container">$\triangle ABC$</span> to the area of <span class="math-container">$\triangle AOC$</span> ?</p>
<p><span class="math-container">$1)\frac32\qquad\qquad2)\frac53\qquad\qquad3)2\qquad\qquad4)3\qquad\qquad5)\frac72$</span></p>
</blockquote>
<p>By knowing the length of the sides of <span class="math-container">$\triangle ABC$</span> I concluded it is an obtuse triangle (Because <span class="math-container">$3^2+5^2<7^2 $</span> ). I'm not sure how to use <span class="math-container">$\vec{OA}+2\vec{OB}+3\vec{OC}=0$</span> to solve the problem, but from the forth choices I realized it happens when the point <span class="math-container">$O$</span> be the centroid of <span class="math-container">$\triangle ABC$</span> so this might be the answer.</p>
| JMP | 210,189 | <p>The starting position for <span class="math-container">$B$</span> for a draw with <span class="math-container">$A$</span> can be found by comparing the ratios of the race lengths for <span class="math-container">$A$</span> and <span class="math-container">$B$</span>:</p>
<p><span class="math-container">$$\frac{\text{distanceA}}{\text{distanceB}}=\frac{\text{new distanceA}}{\text{new distanceB}}$$</span>
<span class="math-container">$$\frac{L}{L-10}=\frac{L+10}{L+k}$$</span></p>
<p><span class="math-container">$$L^2+kL=L^2-100$$</span></p>
<p><span class="math-container">$$k=\frac{-100}{L}$$</span></p>
<p>As <span class="math-container">$k$</span> is always negative, <span class="math-container">$B$</span> must always start in front of the starting line in order to draw.</p>
<p>Similarly for <span class="math-container">$A$</span> and <span class="math-container">$C$</span>.</p>
|
908,083 | <p>I'd like to know what methods can I apply to simplify the fraction $\frac{4x + 2}{12 x ^2}$ </p>
<p>Is it valid to divide above and below by 2? (I didn't know it but Geogebra's Simplify aparantly does this)</p>
<p>Thanks in advance</p>
| Cookie | 111,793 | <p>Factor out the $2$ from both the numerator and denominator. That is,
$$\require{cancel}\frac{4x+2}{12x^2}=\frac{\cancel{2}(2x+1)}{\cancel{2}\cdot6x^2}=\frac{2x+1}{6x^2}.$$</p>
|
4,013,796 | <p>If we make a regular polygon with n vertices (n edges) and triangulate on the inside with n-3 edges, then triangulate on the outside with (n-3) edges (or draw dotted lines inside again), a Maximal Planar Graph is formed. Edges shouldn't be repeated and there's no loops or directions.</p>
<p>How many distinct graphs of this type are there?</p>
<p>It's connected to an earlier question where it was asked 'How many Distinct Maximal Planar Graphs are there? <a href="https://math.stackexchange.com/questions/4009628/how-many-distinct-maximal-planar-graphs-exist-with-n-vertices">How many distinct Maximal Planar Graphs exist with $n$ vertices?</a></p>
<p>Will Orrick gave the OEIS numbers A000109. It was then wondered if there was a known formula for those numbers or bounds on them. It's conjectured that graphs of the type in this question might constitute most Maximal Planar Graphs, so a formula for the 'polygon' types might be an approximate formula or a lower bound for all types of Maximal Planar Graphs.</p>
<p>Dividing the polygon on the inside can be done in C(n-2) ways, where C(n) are the Catalan numbers A000108, so a connection was looked for - and A000109 seem to be close to C(n-2)*2^(n-13) at least for n=9 to 23.</p>
<p>So, the answer to this question would be of interest and any thoughts on connecting the A000108 and A000109 numbers. Lots of ways have been tried so far, e.g. since the next Catalan number can be formed by adding the product of ones before, e.g. C(4) = C(3)*C(1) + C(2)*C(2) + C(1)*C(3), perhaps something similar happens for A000109, or by incorporating numbers from both sequences. Lots of coincidences (probably) have been found including the 233 number in A000109 say X(10) is C(9-2) - the sum of the Catalan numbers before it.</p>
<p>Think I'm going to go crazy looking for patterns any longer! Any suggestions please!</p>
| John Hunter | 721,154 | <p>Here is a partial answer.</p>
<p>In what follows the number of distinct Hamiltonian Maximal Planar Graphs with n vertices is <span class="math-container">$X(n)$</span>. It should follow A000109 <a href="https://oeis.org/A000109/list" rel="nofollow noreferrer">https://oeis.org/A000109/list</a> up to <span class="math-container">$n=11$</span> when the first non Hamiltonian occurs.</p>
<p>Also used are the number of distinct ways to divide a polygon <a href="https://oeis.org/A000207" rel="nofollow noreferrer">https://oeis.org/A000207</a> . <span class="math-container">$A(n)$</span> , offset by 2 from the list, will be a number from this list corresponding to the number of distinct ways to triangulate a polygon with <span class="math-container">$n$</span> sides There is a formula for these, not written here.</p>
<p>Since a Maximal Planar Hamiltonian graph can be represented by a regular polygon triangulated by 'inside' edges and also triangulated again with 'outside' edges, without repeating edges, the following is suggested.</p>
<p>For example the hexagon. Let's start with the maximum number of added edges leaving a vertex, it's <span class="math-container">$3$</span>, making a fan <span class="math-container">$F$</span>. Let the hexagon be <span class="math-container">$ABCDEF$</span>. If the <span class="math-container">$3$</span> inside (say) edges leave node <span class="math-container">$B$</span>. The outside edges can't repeat any of these, one of the outside edges must also join <span class="math-container">$A$</span> to <span class="math-container">$C$</span>, to ensure it's triangulated and not repeat edges. This leaves a pentagon to triangulate by the outside edges, but using one less edge (one was used for <span class="math-container">$AC$</span>) - the number of distinct ways for that is <span class="math-container">$A(5)$</span></p>
<p>So we would expect <span class="math-container">$X(6)$</span> to be <span class="math-container">$A(5) + e(2)$</span> where the <span class="math-container">$e(2)$</span> is the number of graphs where the maximum number of edges leaving any node is 2. For the hexagon <span class="math-container">$e(2)$</span> is 1.
After <span class="math-container">$n=6$</span> the minimum maximum number of edges leaving any node is 3.</p>
<p><span class="math-container">$X(7) = A(6) + e(3)$</span><br />
<span class="math-container">$X(8) = A(7) + e(4) + e(3)$</span>,</p>
<p>From <span class="math-container">$n=13$</span> onwards the minimum maximum number of edges leaving any node is 4</p>
<p>Here is a table of <span class="math-container">$A(n)$</span> and <span class="math-container">$X(n)$</span>.
<span class="math-container">\begin{array}{c|cc}
n & A(n) & X(n)\\
\hline
3 & 1 & 1\\
4 & 1 & 1\\
5 & 1 & 1\\
6 & 3 & 2\\
7 & 4 & 5\\
8 & 12 & 14\\
9 & 27 & 50\\
10 & 82 & 233\\
11 & 228 & 1249\\
\end{array}</span></p>
<p>The pattern for the <span class="math-container">$e(i)$</span> isn't clear but some have been found using a computer program.
<span class="math-container">$$ X(6)=1+1 $$</span>
<span class="math-container">$$ X(7)=3+2 $$</span>
<span class="math-container">$$ X(8)=4+8+2 $$</span>
<span class="math-container">$$ X(9)=12+23+14+1 $$</span>
<span class="math-container">$$ X(10)=27+84+92+29+1 $$</span>
<span class="math-container">$$ X(11)=82+281+508+333+44+0 $$</span></p>
<p>the <span class="math-container">$X(11)$</span> total is one short of the A000109 numbers so the program confirms that there is at most one non-Hamiltonian graph for n=11. One graph is [was], missing for n=10 which the computer couldn’t find for Perhaps there's an undiscovered non-Hamiltonian graph with 10 nodes! [edit 14/02/2021, thanks to Will Orrick's suggestion of improving isomorphism testing, it's been found]</p>
<p>Now for the cases where one of the 'fan' is missing, e.g the <span class="math-container">$e(3)$</span> for the heptagon. The inner edges are now 3 for the fan, from, say vertex <span class="math-container">$B$</span>, and one somewhere else. The outer edges cannot include the missing fan element (that would cause the maximum number leaving a vertex to be bigger than 3), so must include <span class="math-container">$AC$</span>, leaving a shape with 6 sides to triangulate (with 3 more edges) which we would expect is again to do with <span class="math-container">$A(6)$</span>, but this time with extra restrictions. The remaining outer edges can't triangulate this remaining hexagon in a way that repeats the one inner edge that's inside it, or causes more than 3 edges from a vertex.</p>
<p>Even if many edges were missing from a fan of inner edges coming from vertex <span class="math-container">$B$</span>, none of them can be used by the outer edges without causing too many edges to leave a vertex, so again <span class="math-container">$AC$</span> is needed as one of the outer edges. If for example we were trying to find <span class="math-container">$e(16)$</span> for <span class="math-container">$X(30)$</span>, then it would still seem to involve <span class="math-container">$A(29)$</span>, but with many extras and restrictions. Extras meaning we might have to reintroduce reflections and rotations (of distinct triangulations of the remaining 29 a-gon), restrictions due to keeping the maximum number of edges leaving a vertex to 16 (other cases would be covered by <span class="math-container">$e(17)$</span> etc...) and also avoiding the 11 inside edges which can be inside the 29 a-gon.</p>
<p>Alternative system:</p>
<p>Now also found are a set of <span class="math-container">$e(i)$</span>, where the maximum must come from at least one set of triangulations, inner or outer (but not both combined). i.e. one that was previously <span class="math-container">$e(5)$</span> in the above, having a vertex of maximum degree 7, might count as <span class="math-container">$e(4)$</span> below, if its connections are 2 in the polygon, 4 inner connections and one outer connection.</p>
<p>For this system,</p>
<p><span class="math-container">$$ X(6)=1+1 $$</span>
<span class="math-container">$$ X(7)=3+2 $$</span>
<span class="math-container">$$ X(8)=4+7+3 $$</span>
<span class="math-container">$$ X(9)=12+22+15+1 $$</span>
<span class="math-container">$$ X(10)=27+77+96+32+1 $$</span>
<span class="math-container">$$ X(11)=82+261+498+361+46+0 $$</span></p>
<p><span class="math-container">$X(6)$</span> and <span class="math-container">$X(7)$</span> are the same as before, <span class="math-container">$X(8)$</span> is the first different one. It is the graph with the lowest number of vertices that can't be drawn in polygon form, where the edges from a vertex of maximum degree can all be inside edges. It has an interesting structure (below) with lots of symmetry. The <span class="math-container">$X(11)$</span> numbers were from a run of 600,000 graphs, it's possible there could be some 'shuffling left' by one or two (for rows above n=10), but since the numbers didn't change during the second half of the run it's likely that they are exact.</p>
<p>The <span class="math-container">$X(8)$</span> first different one, can be drawn as a regular octagon (vertices 1-8) with these inside edges (1,3), (1,4), (1,5), (5,7), (5,8) and these outside edges, which are best dawn inside again with dotted lines to appreciate the symmetry (3,5), (3,6), (3,7), (7,1), (7,2).</p>
<p>On the computer program and the missing graph...</p>
<p>Despite repeated runs, one of the n=10 Hamiltonian graphs is missing. The program finds 232 instead of 233. Perhaps there is an undiscovered n=10 non-Hamiltonian graph! To help others check this and in other ways, here are the details. [Edit: the list is complete now, thanks Will].</p>
<p>To check for isomorphism the program makes an ID set and used two 'tests' initially. Test1 was the sum of the squares of the degrees of all nodes, typically about 250 for n=10. If two graphs are different for Test1, they are not isomorphic.</p>
<p>The second test assigned a value test2(i) to each node, made up of summing, for each connecting node, j, (excluding itself) deg(j)*[11+deg(j)-deg(i)]. Where deg(j) is the degree of node j. Test2 is the sum of all these values. If two graphs are different for Test2, again,they are not isomorphic. the 11 is random to increase the size and make it less likely that the same number can be got by summing a different set of numbers the -deg(i) is included for a similar reason.</p>
<p>So for two graphs to be isomporphic they should have identical Test1 and Test2.</p>
<p>This system didn't distinguish and find all 50 for n=9, it found 48. So another test, Test3, was introduced.</p>
<p>The third test assigned a value test3(i) to each node, made up of summing, for each connecting node, j, (excluding itself) test2(j)*[test2(j)+2-deg(i)]. Test3 is the sum of all these values.</p>
<p>Now all 50 were found for n=9, but 232 out of 233 for n=10. So Test4 was introduced, which is like Test3, test4(i) is by summing test3(j)*[test3(j)+2-deg(i)] and Test4 is the sum of these. [Edit, after Will's 'edge ID' was added, all 233 have been found].</p>
|
34,775 | <p><strong>What is the goal of MSE? Is it to get a repository of interesting questions and well-written answers. Or are we instead an online math tutoring site where we help anyone as long as they seem to be trying. These two goals are often in contradiction with each other!</strong></p>
<p>I am afraid that we are headed in the direction of being an online tutoring site, at least for a couple months in the spring and fall when school is in session. What I have noticed this past spring was that MSE was inundated with "newbie" users coming on here and asking on average a high volume of problem-set questions each--oftentimes 10 questions/week per person. Now, in all fairness, the users were demonstrating some effort in their questions. But it was clear that they were struggling with the basics, so their questions were hardly what you would consider to be "good" questions. And yet, these users still received a lot of help on their questions from the more established posters on MSE nonetheless. And so it continued on through March and April. It is like MSE was filling the role of Teaching Assistant or whatever for these students.</p>
<p>[It seems to have quieted down now that the most recent semester is about to end, but it will pick up again. Just wait until the fall! Or maybe even later this summer. If not even sooner than that.]</p>
<p>If the desire is to move back away from being a homework-tutoring site, it is probably going to be hard for the site to stop this without making changes on the admin level. [A possibility would be lowering the number of votes to close from 5 to 3. Another possibility would be to make a tag or section of MSE dedicated for someone learning the basics.] <strong>Meanwhile, I'm not seeing how the EoQS currently implemented, is changing this.</strong> This site is nonetheless being clogged with many boring or poorly-written questions, which are still getting rewarded with a long string of comments doing their best to tutor the student, and at least one of those comments [are comments under the purview of EoQS<span class="math-container">$^1$</span>] has the answer to the student's questions. These questions may get a couple votes to close and maybe a downvote too, but then they also get a pity upvote. And so we get many more such questions, because users are being rewarded for asking them--whether there is an "Answer" or not. There does seem to be a critical mass of users on MSE who do feel that this should be a site where struggling students can come for help with their basic homework even if their questions don't meet the MSE Guidelines, as long as they are demonstrating some effort.</p>
<p>If you cannot already tell, my vote would be MSE moving towards a repository of high-quality questions and answers, and away from being a homework-tutoring site.</p>
<p><strong>ETA: In any event though, I do think EoQS would work better if the way it were administered were shifted.</strong> What if the following changes were implemented:</p>
<p>(a) Reduce the number of votes needed to <em>close</em> [<strong>NOT</strong> delete!] a question from 5 down to 4 or 3. I think a reason why EoQS came to be in the first place was the proliferation of too many really bad questions that get too much oxygen.</p>
<p>(b) Enforce <em>comments</em> as much as answers. In particular, no more rewarding bad questions by answering in the comments. If we don't want a bad question answered in the answer box, then we don't want a bad question answered in the comments either. Likewise, if a question is worth keeping around, then it is worth being answered, <em>in the answer box</em>, as answering in the comments really helps no one.</p>
<p>I'm not necessarily for more enforcement, I am for smarter enforcement. The net result of what we are doing now w EoQS are question after question of debatable quality, with a long string of comments--in place of a well-written answer written where it is supposed to be--the answer box. The worst of both worlds--still no quality control but now messy flow. Should those questions be allowed to stay? Maybe. I get from the comments and whatnot that it is a debate. But if so, then at the very least, the formatting should be right.</p>
<p>Please advise.</p>
<p><strong>ETA 5/20/2022 18:30 EDT: Reading the other posts and comments here, I think the biggest problem with EoQS as I see it, is in unclear and contradictory objectives of here, and so what gets enforced as bad content is often absurd.</strong> I understand that really confused students are going to end up asking questions that are really duplicates [even with context]. For example, every semester we see a bunch of question such as:</p>
<p><em>Is <span class="math-container">$\{(x_1,x_2) \in \mathbb{R}^2; 2x_1+x_2=5\}$</span> a vector space?</em></p>
<p>We will also get a bunch of questions about the probability of drawing <span class="math-container">$2$</span> red cards or a certain hand from a deck of <span class="math-container">$52$</span> cards, and so on. Just as we did last semester and the semester before that. The consensus on here, going by what I'm reading in the comments anyway, is that those questions should get respect on here if the student is showing effort. Alright, fine and great. If this is what the board decides then let's give those questions respect.
<strong>But then if these questions are fine and allowed, then what is the point of EoQS again? What is the point of shutting down more interesting questions again then? Sometimes an answer to a duplicate gives a different take that may be useful to the next person. And just as much, why are the ones who <em>answer</em> a lot of hard questions getting put into the corner then. They are the ones contributing to the knowledge base here! And they were never really contributing to the problem EoQS was supposedly about fixing.</strong></p>
<p>It often just all seems to arbitrary and capricious....</p>
<p><span class="math-container">$^1$</span> <strong>EoQS = Enforcement of Quality Standards</strong></p>
| Tryst with Freedom | 688,539 | <p>There are three main categories of questions which I feel are worth talking about:</p>
<ol>
<li><p>Conceptually deep questions which can't be answered from standard books</p>
</li>
<li><p>Conceptually deep questions which can be answered from standard books</p>
</li>
<li><p>Standard problems which come as exercises</p>
</li>
</ol>
<p>In my personal experience, I find it that 90% of mse users can only handle 2. and 3. . There are only handful I met on this site who I really felt" gets it" beyond the book and can answer the type in 1. .</p>
<p>And depending on the subtopic of math, the number of people that maybe capable to answer in this category here is non existent.</p>
<p>Now, for the direction of mse , everything I said in the above paragraph is mostly irrelevant because firstly these questions are in minority, and, I think the people who are capable of writing deep answers won't be attracted more to the site more than they are right now by policy shifts. Infact, I have actually found that the people who answer in 1. often do not engage much in meta.</p>
<p>For 2. and 3., I'll just say let them be to be honest. Students need help and it is certainly valuable to provide a free reliable resource like mse to them. I think we should aim for inclusivty rather than exclusivity at this point.</p>
<p>To begin with mse is already very exclusive because it is not at all simple to use it. Firstly you need an Internet connection, then some command of English, and finally a willingness to learn some mathjax.</p>
<p>Most people don't speak English, and also may not have an Internet connection.</p>
<p>To achieve inclusivity, the website should be made more accessible in some way. I see no future for it if the philosophy is to keep it like some elitist mathematics circle for the English speaking privileged.</p>
<hr />
<p>Now where is it actually going towards? That I believe is unanswerable. Only the future can say.</p>
|
4,394,676 | <p>Solve</p>
<p><span class="math-container">$$\frac{dy}{dx}=\cos(x-y)$$</span></p>
<p>So I know I need to make the substitution <span class="math-container">$u=x-y$</span> but then what's <span class="math-container">$du$</span>, is it <span class="math-container">$du=dx-dy$</span>?</p>
<p>Or do I rewrite <span class="math-container">$$\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$$</span></p>
<p>Really stuck on this one.</p>
| PierreCarre | 639,238 | <p>You just need to note that when <span class="math-container">$u = x - y$</span> (<span class="math-container">$x$</span> being the independent variable) you have that <span class="math-container">$u' = 1-y'$</span>, i.e. <span class="math-container">$y' = 1-u'$</span>. The equation then becomes <span class="math-container">$u' = 1 - \cos u$</span>. This is an equation with separable variables that is given implicitly by
<span class="math-container">$$
-\cot \frac u2 = x + C
$$</span></p>
<p>You can also have constant solutions <span class="math-container">$u = u_0$</span> whenever <span class="math-container">$\cos u_0 =1$</span>. Below you can find a stream plot for this equation (<span class="math-container">$u$</span>), and for the original equation (<span class="math-container">$y$</span>).
<a href="https://i.stack.imgur.com/0Rdej.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0Rdej.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/sPblr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sPblr.png" alt="enter image description here" /></a></p>
|
2,832,614 | <blockquote>
<p>Prove by induction that
$$\lim_{x \to a} \frac{x^n-a^n}{x-a}=na^{n-1}.$$</p>
</blockquote>
<p>I did a strange proof using two initial results: We know that result is true for $n=1$ and $n=2$. Assuming the result is true for $n=k-1$ and $n=k$, I can prove the result for $n=k+1$. For this I used my assumption for the case of $n=k$, and used a random multiplier to get the desired result:
$$\lim_{x \to a} \frac{x^{k}-a^{k}}{x-a}·\frac{x+a}{x+a}.$$
During the simplification I had to use my $n=k-1$ case result as well.</p>
<p>Is this proof OK in terms of induction principle? I can show my whole working if needed. </p>
<p>Thank you.</p>
<p><strong>Edit</strong></p>
<p>whole proof:</p>
<p>$\lim_{x \to a} \frac{x^{k}-a^{k}}{x-a}·\frac{x+a}{x+a}=k.a^{k-1}$ (I'll omit limit notation for clarity)</p>
<p>$$\frac{x^{k}-a^{k}}{x-a}·\frac{x+a}{x+a}=k.a^{k-1}$$</p>
<p>$$\frac{1}{2a}\frac{x^{k+1}+ax^k-a^kx-a^{k+1}}{x-a}=k.a^{k-1}$$</p>
<p>$$\frac{1}{2a}(\frac{x^{k+1}-a^{k+1}}{x-a}+\frac{ax(x^{k-1}-a^{k-1})}{x-a})=k.a^{k-1}$$</p>
<p>$$\frac{1}{2a}[\frac{x^{k+1}-a^{k+1}}{x-a}+{a^2(k-1).a^{k-2}}]=k.a^{k-1}$$</p>
<p>Hence, $$\lim_{x \to a} \frac{x^{k+1}-a^{k+1}}{x-a}=(k+1)a^{k}.$$</p>
| Bumblebee | 156,886 | <p>$$x^{n+1}-a^{n+1}=(x-a)(x^n+a^n)+ax(x^{n-1}-a^{n-1})$$</p>
|
77,136 | <p>I have a multidimensional-variable list, suppose for example {{x1,y1},{x2,y2}..}. I have duplicate values for the 'x' coordinates and I need to find for the duplicate 'x' elements the corresponding minimum 'y'. A sample of this list is the following:</p>
<pre><code>l1={{1, 1.43E-46}, {21, 2.79E-48}, {41, 3.22E-45}, {41,1.74E-46}, {81, 2.77E-46}, {121,9.97E-48}, {161, 1.24E-45}, {181,1.19E-45}}
</code></pre>
<p>I need to rewrite the list with only those elements from duplicate 'x' coordinates with a minimum value of 'y', for the former list this would be:</p>
<pre><code>l1={{1, 1.43E-46}, {21, 2.79E-48}, {41,1.74E-46}, {81, 2.77E-46}, {121,9.97E-48}, {161, 1.24E-45}, {181,1.19E-45}}
</code></pre>
<p>Since from the duplicate coordinates 'x', the minimum 'y' is:</p>
<pre><code> {41,1.74E-46}
</code></pre>
| Jens | 245 | <p>This works:</p>
<pre><code>l1 = {{1, 1.43 10^-46}, {21, 2.79 10^-48}, {41, 3.22 10^-45}, {41,
1.74 10^-46}, {81, 2.77 10^-46}, {121, 9.97 10^-48}, {161,
1.24 10^-45}, {181, 1.19 10^-45}};
Map[First[SortBy[#, Last]] &, SplitBy[l1, First]]
(*
==> {{1, 1.43*10^-46}, {21, 2.79*10^-48}, {41, 1.74*10^-46}, {81,
2.77*10^-46}, {121, 9.97*10^-48}, {161, 1.24*10^-45}, {181,
1.19*10^-45}}
*)
</code></pre>
<p>The first step is to use <a href="http://reference.wolfram.com/mathematica/ref/SplitBy.html" rel="nofollow"><code>SplitBy</code></a> with the first element as the criterion. Then the size of the last element is tested in <a href="http://reference.wolfram.com/mathematica/ref/SortBy.html" rel="nofollow"><code>SortBy</code></a>.</p>
|
563,927 | <p>Show that $\mathbb{R}$
is not a simple extension of $\mathbb{Q}$
as follow:</p>
<p>a. $\mathbb{Q}$
is countable.</p>
<p>b. Any simple extension of a countable field is countable.</p>
<p>c. $\mathbb{R}$
is not countable.</p>
<p>I 've done a. and c. Can anyone help me a hint to prove b.?</p>
| Johannes Kloos | 26,325 | <p>Here is a <strong>hint</strong>: Let $F$ be a countable field, $F(a)$ a simple extension. Then every element in $F(a)$ can be written as a quotient $\frac{p(a)}{q(a)}$, where $p$ and $q$ are polynomials. Now, combine known results on countability.</p>
|
2,329,751 | <p>I have this definition: $f:R^n → R^m$ is differentiable at $a∈R^n$, if there exists a linear transformation $μ:R^n→R^m$ such that</p>
<p>$\lim_{h \to 0} \frac{|f(a+h)-f(a)-\mu(h)|}{|h|} = 0$.</p>
<p>My questions are what's the linear transformation $μ(h)$ for? What does it mean and where does it come from? Why is it necessary?</p>
<p>Can anyone explain the definition to me a bit better? Thanks</p>
| Peter | 409,941 | <p>I'll try to give you an equivalent definition, which I think is a bit clearer, I will leave the equivalence to you in first instance. I might add more explanation later.</p>
<p>First let's consider the case that $n=2$ and $m=1$, (actually what I will say works equally well for arbitrary $n$, but $n=2$ is the simplest case that is new to us).</p>
<p>Let me try to be concrete, suppose we have a function $f: \mathbb{R}^{2} \rightarrow \mathbb{R}$, and we want to compute the derivative of $f$, say at $0 \in \mathbb{R}^{2}$. That is, we want to figure out how the output of $f$ changes if we change the input. The idea is to use the fact that we know how to differentiate functions from $\mathbb{R}$ to $\mathbb{R}$. Suppose we have any $v \in \mathbb{R}^{2}$, then we can get a function from $\mathbb{R}$ to $\mathbb{R}$ as follows
\begin{align}
f_{v}: \mathbb{R} &\rightarrow \mathbb{R}, \\
t &\mapsto f(tv).
\end{align}
This function is (a reparametrization of) the <em>restriction</em> of $f$ to the line spanned by $v$.
We know how to differentiate such a function with respect to $t$ (it might not be differentiable, in which case $f: \mathbb{R}^{2} \rightarrow \mathbb{R}$ is not differentiable). We thus get a <em>number</em>
\begin{equation}
D_{v} f := \frac{\text{d}}{\text{d}t}\bigg|_{t=0} f(tv),
\end{equation}
which tells us how quickly the output of the function $f$ changes if we vary the input along the line spanned by $v \in \mathbb{R}^{2}$.</p>
<p>What was described above makes sense for <em>any</em> vector $v \in \mathbb{R}^{2}$, so we have a map from $\mathbb{R}^{2}$ to $\mathbb{R}$,
\begin{align}
\mu: \mathbb{R}^{2} &\rightarrow \mathbb{R}, \\
v &\mapsto D_{v}f = \frac{\text{d}}{\text{d}t}\bigg|_{t=0} f(tv).
\end{align}
Now, I claim that the map $\mu$ is linear, and furthermore satisfies the equation that you wrote down (with $a=0$):
\begin{equation}
\lim_{h \rightarrow 0} \frac{\|f(h) - f(0) - \mu(h)\|}{\|h\|} = 0.
\end{equation}
Note that $h \in \mathbb{R}^{2}$. It's up to you to show that $\mu$ satisfies this equation, (and that any $\mu$ that satisfies this equation is the derivative). (I <em>might</em> add some steps to show this later).</p>
<hr>
<p>Now I have claimed that $\mu: \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}$ is the derivative of $f$ at $0$ (or $a$), I should maybe tell you a bit about what it would look like in the case that $m=n=1$. In this case we see (exercise!) that for $s \in \mathbb{R}$
\begin{equation}
\mu(s) = s \frac{\text{d}}{\text{d}t}\bigg|_{t=0} f(t).
\end{equation}</p>
<hr>
<p>Like remarked above, I have not really <em>used</em> the fact that $n=2$ and the entire story holds equally well for arbitrary $n$. For higher $m$, we should view a function $f: \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}$ as $m$ functions $f_{i}: \mathbb{R}^{n} \rightarrow \mathbb{R}$.</p>
|
1,748,914 | <blockquote>
<p>Describe the structure of the Galois group of the splitting field $L$ of the polynomial $X^4-7$ over $\mathbb{Q}$</p>
</blockquote>
<p>I believe that $L=\mathbb{Q}(\sqrt[4]{7}, i)$ is the splitting field of the polynomial</p>
<p>Now I need to describe the structure of $Gal(L, \mathbb{Q})$</p>
<hr>
<p>$H=Gal(L, \mathbb{Q}(i))$ is a cyclic group of order $2$ generated by $\tau_1$ such that:</p>
<p>$\tau_1(\sqrt[4]{7})=i\sqrt[4]{7}$</p>
<p>$Gal(L, \mathbb{Q}(\sqrt[4]{7}))$ is a cyclic group generated by $\tau_2$ such that:</p>
<p>$\tau_2(\sqrt[4]{7})=\sqrt[4]{7}$ and $\tau_2(i)=-i$</p>
<hr>
<p>I believe that $Gal(L, \mathbb{Q})$ consists of the identity element $e$ and some multiplied combinations of $\tau_1$ and $\tau_2$ but could do with some help filling in the gaps and improving my understanding</p>
| Noble Mushtak | 307,483 | <p>We know that the minimal polynomial of $\sqrt[4] 7$ has degree $4$ and the minimal polynomial of $i$ has degree $2$, so $[\Bbb{Q}(\sqrt[4] 7, i) : \Bbb{Q}]=[\Bbb{Q}(\sqrt[4] 7, i) : \Bbb{Q}(i)][\Bbb{Q}(i) : \Bbb{Q}]=4*2=8$. Thus, the Galois group has order $8$ because of the Fundamental Theorem of Galois Theory.</p>
<p>Now, this is the splitting field of $x^4-7$, meaning all of the roots of $x^4-7$ must permute with each other. Thus, the Galois group is a subgroup of $S_4$. However, according to <a href="http://groupprops.subwiki.org/wiki/Subgroup_structure_of_symmetric_group:S4" rel="nofollow">this really helpful table</a>, the only subgroup of $S_4$ of order $8$ is $D_8$, so this is the structure of the Galois group.</p>
<p>This can also be shown with your work: $\tau_1$ represents a rotation of 90 degrees in the complex plane while $\tau_2$ represents a reflection over the real number line in the complex plane. You are rotating and reflecting $4$ elements (the roots of $x^4-7$) without just permuting them at free will, which means the group has a structure of $D_8$.</p>
|
1,748,914 | <blockquote>
<p>Describe the structure of the Galois group of the splitting field $L$ of the polynomial $X^4-7$ over $\mathbb{Q}$</p>
</blockquote>
<p>I believe that $L=\mathbb{Q}(\sqrt[4]{7}, i)$ is the splitting field of the polynomial</p>
<p>Now I need to describe the structure of $Gal(L, \mathbb{Q})$</p>
<hr>
<p>$H=Gal(L, \mathbb{Q}(i))$ is a cyclic group of order $2$ generated by $\tau_1$ such that:</p>
<p>$\tau_1(\sqrt[4]{7})=i\sqrt[4]{7}$</p>
<p>$Gal(L, \mathbb{Q}(\sqrt[4]{7}))$ is a cyclic group generated by $\tau_2$ such that:</p>
<p>$\tau_2(\sqrt[4]{7})=\sqrt[4]{7}$ and $\tau_2(i)=-i$</p>
<hr>
<p>I believe that $Gal(L, \mathbb{Q})$ consists of the identity element $e$ and some multiplied combinations of $\tau_1$ and $\tau_2$ but could do with some help filling in the gaps and improving my understanding</p>
| Pedro | 23,350 | <p>There's a general tool to deal with this type of extensions. Suppose you have a diamond of extensions $F \subseteq L,L' \subseteq E$ (make the drawing) where $L\cap L'=F,LL'=E$. </p>
<p>Suppose moreover that $L/F$ is normal, and $E/F$ is Galois. Let $G={\rm Gal}(E/F)$, $H={\rm Gal}(E/L)$ and $K={\rm Gal}(E/L')$. Then show that $HK=G, H\cap K=1$ and $K\lhd G$. All this means that $G$ is the semidirect product of $H$ and $K$, in particular every element $\sigma$ of $G$ is written uniquely as $\tau\rho$ with $\tau\in H,\rho\in K$. In your case one gets $G$ is a semidirect product $C_2\rtimes C_4$ which is the dihedral group of order four. But one gets <em>more</em> than this, the above gives an explicit "decomposition" of the Galois group in terms of the subextensions $L,L'$.</p>
<p>To be more precise, take $L=\Bbb Q(i),L'=\Bbb Q(7^{1/4}),F=\Bbb Q$ and $E=LL'$ (your extension). Then $L/\Bbb Q$ is normal and $L\cap L'=\Bbb Q$ (this is not hard to prove, since $L$ is obtained by adjoining $i$, and its a degree two extension). One can see that $E/L$ is of degree four and $E/L'$ is of degree two. </p>
<p>The technique above is very useful when one adjoins more than two elements, since one can iterate the decomposition. For example, one can explicitly calculate the Galois group of $(X^4-3)(X^3-5)$ over $\Bbb Q$ as a semidirect product $(C_3\rtimes C_4)\rtimes C_2$ explicitly: it is generated by $\omega,\eta,\psi$ such that $\omega^3=\eta^4=\psi^2=1$</p>
<ol>
<li>$\eta(3^{1/4})=3^{1/4}i$ and $\eta$ fixes $i,5^{1/3}$</li>
<li>$\psi(i)=-i$ and $\psi$ fixes $3^{1/4},5^{1/4}$,</li>
<li>$\omega(5^{1/3})=5^{1/3}\xi$ and $\omega$ fixes $i,3^{1/4}$.</li>
</ol>
<p>Here $\xi$ a primitive cuberoot of $1$, and every element of $G$ is written uniquely as $\omega^i\eta^j\psi^k$ with $i=0,1,2,j=0,1,2,3,k=0,1$. Moreover $\psi\omega\psi =\omega^{-1},\eta\omega\eta^{-1}=\omega^{-1}$. A bit more calculations can give you a presentation for $G$, which has order $24$. </p>
|
109,922 | <p>Let $B_n$ denote the group of signed permutations on $n$ letters. Is there a good explanation or understandable way to see why
$$
\sum_{w\in B_n}q^{\text{inv}(w)}=(2n)_q(2n-2)_q\cdots(2)_q?
$$</p>
<p>I've been thinking about it on and off while reading through Taylor's <em>Geometry of the Classical Groups</em>, but don't understand why this identity holds. I appreciate any explanation. Thanks!</p>
| hoyland | 14,722 | <p>I can only offer a rough idea (and hope that I have the same definition of inv as you do). The proof in type A is on page 36 of the PDF version of EC 1 available on Stanley's website (<a href="http://www-math.mit.edu/~rstan/ec/ec1/" rel="nofollow">here</a>). Basically, any permutation can be encoded via its inversion table, a sequence $(a_1,\ldots,a_n)$, where $0\leq a_i\leq n-1$, and $\mathrm{inv}(w)=a_1+\cdots+a_n$, so the sum $\sum_{w \in S_n}q^{\mathrm{inv}(w)}$ can be converted to a sum over inversion tables.</p>
<p>One should be able to define the inversion table of a signed permutation similarly and push a similar proof through, but I can't get the right definition of inversion table. (The identity suggests one needs only $n$ entries in the table, which makes perfect sense, and that they can range between 0 and $2n-1$, which also makes sense, but I can't put the pieces together, nor find a reference.)</p>
|
4,057,255 | <p>Suppose we have 10 items that we will randomly place into 6 bins, each with equal probability. I want you to determine the probability that we will do this in such a way that no bin is empty. For the analytical solution, you might find it easiest to think of the problem in terms of six events Ai, i = 1, . . . , 6 where Ai represents the event that bin i is empty, and calculate P(AC1 ∩···∩AC6)using DeMorgan’s law.</p>
<p>Analytical= 1- P(bin one empty) + P(bin 2 empty) + P(bin 3 empty) + P(bin 4 empty) + P(bin 5 empty) + P(bin 6 empty) – P(1 and 2 empty) – P(1 and 3 empty) – P(1 and 4 empty) – P(1 and 5 empty) – P(1 and 6 empty) – P(2 and 3 empty) – P(2 and 4 empty) – P(2 and 5 empty) – P(2 and 6 empty) – P(3 and 4 empty) – P(3 and 5 empty) – P(3 and 6 empty) – P(4 and 5 empty) – P(4 and 6 empty) – P(5 and 6 empty) + P( 1 2 3 empty) + P(1 2 4 empty) + P (1 2 5) empty + P (1 2 6 empty) + P( 1 3 4 empty) + P(1 3 5 empty) + P( 1 3 6 empty) + P(1 4 5 empty) + P(1 4 6 empty) + P(1 5 6 empty) – P(1 2 3 4 empty) – P(1235 empty) – P(1236 empty) – P(2345 empty) – P(2346 empty) – P(3456 empty) + P(12345 empty) + P(23456 empty) + P(34561 empty) + P(45612 empty) – P(123456 empty)</p>
<p>1-(6<em>P(1 specific empty bin) – 15</em>P(2 specific empty bins) + 20<em>P(3 specific empty bins) – 15</em>P(4 specific empty bins) +6*P(5 specific empty bins)</p>
<p>This is what I have so far but I don't know how to determine the individual probabilities above.</p>
| user2661923 | 464,411 | <p>Addendum added to examine the danger of overcounting from a different perspective.</p>
<hr />
<p>This is a reaction to the answers of Math Lover and true blue anil.</p>
<p>I definitely regard Inclusion-Exclusion as the preferred approach. Math Lover's answer completely covers that.</p>
<p>If you wish to consider the direct approach, then the Stirling Numbers of the Second Kind are dead on point. This is exactly what true blue anil's answer details.</p>
<p>My answer is a manual exploration of what Stirling Numbers of the Second Kind represent,
how they are computed, and what their relevance is to this problem.</p>
<p><a href="https://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind" rel="nofollow noreferrer">This Wikipedia Article</a>
indicates that :</p>
<ul>
<li><p><span class="math-container">$\{ {n\atop k} \} = S(n,k) = $</span> the number of ways to partition a set of n objects into k non-empty subsets.</p>
</li>
<li><p><span class="math-container">$\displaystyle S(n,k) = \frac{1}{k!}\sum_{i=0}^k (-1)^k \binom{k}{i} (k-i)^n.$</span></p>
</li>
</ul>
<p>Suppose that you wanted to use the direct approach to the query's problem. Further suppose that you wanted to manually derive each of the numbers, rather than relying on the given formulas. This answer will illustrate the procedure, using the OP's constraints of <span class="math-container">$(10)$</span> balls going into <span class="math-container">$(6)$</span> rooms, and calculating the probability that each room gets at least one ball.</p>
<p>The probability will be expressed as</p>
<p><span class="math-container">$$\frac{N\text{(umerator)}}{D\text{(enominator)}}
~~\text{where}~~ D = 6^{(10)} ~~\text{and}~~ N = N_1 + N_2 + \cdots + N_5.$$</span></p>
<p><span class="math-container">$D = 6^{(10)}$</span>, which was chosen for convenience, represents the total number of ways that each of the <span class="math-container">$10$</span> balls might be randomly distributed in the <span class="math-container">$6$</span> rooms. Because of this choice, <span class="math-container">$N$</span> must be computed in a manner <strong>consistent</strong> with how <span class="math-container">$D$</span> is computed. This means (for example) that [(ball-1 : room-1), (ball-2 : room-2)] must be considered distinct from
[(ball-1 : room-2), (ball-2 : room-1)].</p>
<p>There are 5 distinct ways that the 10 balls can be partitioned into 6 non-empty subsets of balls and then distributed into the various rooms. These various ways are enumerated below.</p>
<p><span class="math-container">$N_1 : ~~$</span> 5-1-1-1-1-1 <br>
This represents sending <span class="math-container">$5$</span> balls into one of the rooms, and <span class="math-container">$1$</span> ball into each of the remaining rooms. <br>
There are <span class="math-container">$\binom{10}{5}$</span> ways of selecting <span class="math-container">$5$</span> balls into 1 <em>unit</em>. <br>
Once this is done, you then have <span class="math-container">$6$</span> <em>units</em> that must be distributed into the 6 rooms. <br>
There are <span class="math-container">$6!$</span> ways of doing this.</p>
<p>Therefore <br>
<span class="math-container">$N_1 = \binom{10}{5} \times 6!.$</span></p>
<p><span class="math-container">$N_2 : ~~$</span> 4-2-1-1-1-1 <br>
This represents sending <span class="math-container">$4$</span> balls into one of the rooms, <span class="math-container">$2$</span> balls into one of the other rooms, and <span class="math-container">$1$</span> ball into each of the remaining rooms. <br>
There are <span class="math-container">$\binom{10}{4} \times \binom{6}{2}$</span> ways of <br>
selecting <span class="math-container">$4$</span> balls into 1 <em>unit</em> and then selecting <span class="math-container">$2$</span> balls into another <em>unit</em>. <br>
As before, there are <span class="math-container">$6!$</span> ways of distributing the <em>units</em> among the rooms.</p>
<p>Therefore <br>
<span class="math-container">$N_2 = \binom{10}{4} \times \binom{6}{2} \times 6!.$</span></p>
<p><span class="math-container">$N_3 : ~~$</span> 3-3-1-1-1-1 <br>
<strong>Warning</strong> - overcounting danger.<br>
Superficially, one would suppose that the enumeration for <span class="math-container">$N_3$</span> would be:
<span class="math-container">$\binom{10}{3} \times \binom{7}{3} \times 6!.$</span></p>
<p>The best way to demonstrate why this is wrong, is by considering a specific example.<br>
Consider sending the <em>units</em> <span class="math-container">$\{1,2,3\},\{4,5,6\}$</span> into Room-1 and Room-2, respectively.
In one instance, <span class="math-container">$\{1,2,3\}$</span> will be the first <em>unit</em> created, <span class="math-container">$\{4,5,6\}$</span> will be the second <em>unit</em> created, and the first <em>unit</em> will be sent to Room-1, while the second <em>unit</em> will be sent to Room-2.</p>
<p>In a different instance, <span class="math-container">$\{4,5,6\}$</span> will be the first <em>unit</em> created, <span class="math-container">$\{1,2,3\}$</span> will be the second <em>unit</em> created, and the first <em>unit</em> will be sent to Room-2, while the second <em>unit</em> will be sent to Room-1.</p>
<p>The <strong>overcounting</strong> danger is the price you pay for the convenience of using <span class="math-container">$(6!)$</span> to enumerate the number of ways of distributing the <em>units</em> into the rooms. As long as you are aware of the danger, there is an <strong>easy fix</strong>.</p>
<p>By symmetry, because two of the <em>larger</em> units (more than 1 ball) have exactly the same size, you have to apply the scaling factor of <span class="math-container">$\frac{1}{2!}.$</span></p>
<p>Therefore <br>
<span class="math-container">$N_3 = \binom{10}{3} \times \binom{7}{3} \times 6! \times \frac{1}{2!}.$</span></p>
<p><span class="math-container">$N_4 : ~~$</span> 3-2-2-1-1-1 <br>
Again, the scaling factor of <span class="math-container">$\frac{1}{2!}$</span> will be needed, <br>
because two of the <em>larger</em> units (more than 1 ball) have exactly the same size.</p>
<p>Therefore <br>
<span class="math-container">$N_4 = \binom{10}{3} \times \binom{7}{2} \times \binom{5}{2}
\times 6! \times \frac{1}{2!}.$</span></p>
<p><span class="math-container">$N_5 : ~~$</span> 2-2-2-2-1-1 <br>
Here, the scaling factor of <span class="math-container">$\frac{1}{4!}$</span> will be needed, <br>
because <strong>four</strong> of the <em>larger</em> units (more than 1 ball) have exactly the same size.</p>
<p>Therefore <br>
<span class="math-container">$N_5 = \binom{10}{8} \times \binom{8}{2} \times \binom{6}{2} \times \binom{4}{2}
\times 6! \times \frac{1}{4!}.$</span></p>
<hr />
<p>Putting this all together:</p>
<p><span class="math-container">$$N ~=~ N_1 + N_2 + N_3 + N_4 + N_5$$</span></p>
<p><span class="math-container">$$=~ 6! \times \left\{\left[\binom{10}{5}\right] ~+~
\left[\binom{10}{4} \binom{6}{2}\right] ~+~
\left[\frac{1}{2!}\binom{10}{3} \binom{7}{3}\right] ~+~
\left[\frac{1}{2!}\binom{10}{3} \binom{7}{2} \binom{5}{2}\right] ~+~
\left[\frac{1}{4!}\binom{10}{2} \binom{8}{2} \binom{6}{2} \binom{4}{2}\right]\right\}$$</span></p>
<p><span class="math-container">$$=~ 6! \times \left[(252) + (3150) + (2100) + (12600) + (4725)\right]
~=~ 6! \times (22827) ~=~ (16435440).$$</span></p>
<hr />
<p><strong>Addendum</strong><br>
Examine the danger of overcounting from a different perspective.</p>
<p>Consider the subsets that are structured 4-2-1-1-1-1 :<br>
against the subsets structured 3-3-1-1-1-1.</p>
<p>In order for the algorithm to work, each possible subset of the given structure must be counted exactly once.</p>
<p>With the 4-2-1-1-1-1, and the counting algorithm of
<span class="math-container">$\binom{10}{4} \binom{6}{2}$</span>, <br>
the specific collection of subsets <br>
<span class="math-container">$\{1,2,3,4\}, \{5,6\}, \{7\}, \{8\}, \{9\}, \{10\}$</span>, <br>
will be counted exactly one time.</p>
<p>With the 3-3-1-1-1-1, and the counting algorithm of
<span class="math-container">$\binom{10}{3} \binom{7}{3}$</span>, <br>
the specific collection of subsets <br>
<span class="math-container">$\{1,2,3\}, \{4,5,6\}, \{7\}, \{8\}, \{9\}, \{10\}$</span>, <br>
will be counted <strong>twice</strong>.</p>
<p>The first time that it will be counted is when <span class="math-container">$\{1,2,3\}$</span> is the first subset formed and <span class="math-container">$\{4,5,6\}$</span> is the second subset formed.</p>
<p>The second time that it will be counted is when <span class="math-container">$\{4,5,6\}$</span> is the first subset formed and <span class="math-container">$\{1,2,3\}$</span> is the second subset formed.</p>
<p>This explains the scaling factor of <span class="math-container">$\frac{1}{2!}.$</span></p>
<p>Examining the subsets structured as 2-2-2-2-1-1, <br>
and the counting algorithm of
<span class="math-container">$\binom{10}{2} \binom{8}{2} \binom{6}{2} \binom{4}{2}$</span><br>
the specific collection of subsets <br>
<span class="math-container">$\{1,2\}, \{3,4,\}, \{5,6\}, \{7,8\}, \{9\}, \{10\}$</span>, <br>
will be counted (4!) times. <br></p>
<p>This is because with respect to the counting algorithm, any one of the four 2-element subsets might be created, first. Then, any one of the three other 2-element subsets might be created second, and so forth.</p>
<p>This explains the scaling factor of <span class="math-container">$\frac{1}{4!}.$</span></p>
<hr />
<p>One more example: <br>
suppose that instead of forming 10 balls into 6 subsets, <br>
you were forming 10 balls into only <strong>4</strong> subsets.</p>
<p>Consider the subsets structured as 3-3-2-2, <br>
and the counting algorithm of
<span class="math-container">$\binom{10}{3} \binom{7}{3} \binom{4}{2}.$</span><br>
How many different ways would the following collection of subsets be counted:<br>
<span class="math-container">$\{1,2,3\}, \{4,5,6\}, \{7,8\}, \{9,10\}$</span>?</p>
<p>The ordering of the creation of the two 3-element subsets could occur in <span class="math-container">$(2!)$</span> different ways.<br>
Similarly, the ordering of the creation of the two 2-element subsets could also occur in <span class="math-container">$(2!)$</span> different ways.</p>
<p>This means that the correct enumeration for the structure of 3-3-2-2 would be <br>
<span class="math-container">$\binom{10}{3} \binom{7}{3} \binom{4}{2} \times \frac{1}{2!} \times \frac{1}{2!}.$</span></p>
|
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