Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
|---|---|---|
One to one correspondence between transcendental and uncomputable numbers I know that both sets are uncountable infinite but the transcendentals are not a subset of the uncomputables. I don’t know if there exist uncomputable numbers that are not transcendental. But my question is whether the two sets have the same cardinality.
| Here
https://en.wikipedia.org/wiki/Computable_number
under "properties" it is stated the the set of computable numbers is countable.
Hence, the set of uncountable numbers must be uncountable infinite and hence have the same cardinality as the set of transcendental numbers which is uncountable infinite as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3194389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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A game with an $n$-sided die
Suppose you play a game with a fair $n$ sided die (that, if being rolled yields us a discrete random variable uniformly distributed on $\{k \in \mathbb{N}| k \leq n\}$).
You play the following game:
You start by rolling this die. Every time you roll it you may choose either to quit the game or take a reroll. When you quit the game your score becomes equal to $(N_l - N_r)$, where $N_l$ is the result of your last roll and $N_r$ is the number of rerolls you have taken.
What is the optimal way to play to maximize your expected score?
What thoughts do I have on that matter:
If you have already more than $n - 1$ times, as it would guarantee your score to be non-positive (and thus definitely less than it could have been if you have stopped before)
If $n$ or $n - 1$ is the result of your last roll, then there is no need to reroll, as any score you will be able to get after retooling will definitely not exceed the score, that you will get, if you quit immediately.
However, those two thoughts are clearly not enough to construct the optimal strategy.
| Assuming for simplicity you are not allowed to play game infinitely (for example you are forced to quit if you roll $n$), then the game is "either take what you rolled or pay $1$ and play again". Lets say that we get maximum expected score using strategy $f$ "take roll if you rolled at least $k$, otherwise switch to strategy $g$".
$\mathbb E f = \frac{n - k + 1}{n} \cdot \frac{k + n}{2} + \frac{k - 1}{n} \left(\mathbb E g - 1\right)$
So, as $f$ gave as maximum score, we would be at least not worse if we switch back to $f$ instead of $g$. This gives us equation $x = \frac{n - k + 1}{n} \cdot \frac{k + n}{2} + \frac{k - 1}{n}\left(x - 1\right)$, so our score is $\frac{k^2 + k - n^2 - n - 2}{2(k - n - 1)}$. Maximizing it gets $k = n - \sqrt{2n} + 1$ (of course we need to round it up or down depending on $n$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3194536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Normal subgroup with index that divides n! Is this argument valid?
If $G$ is a finite group with $n$ Sylow $p$-subgroups (in particular $n = 1$ mod $p$ and $n$ divides $|G|$), then $G$ permutes them acting by conjugation. Therefore there is a homomorphism $\phi: G \to S_n$ (the symmetric group on $n$ elements). If we consider the kernel of this map, then it is a normal subgroup of $G$ with index $|G|/|\phi(G)|$, which in particular divides $|S_n| = n!$
| Hint:
What does the first isomorphism theorem tell you?
Note that the size of the kernel is $|G|/|\phi(G)|$, not the index of the kernel.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3194632",
"timestamp": "2023-03-29T00:00:00",
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How to solve this probability/combinatorics question about balls in bins Below is the problem that I need to solve
Suppose you blindly place five balls labeled A, B, C, D and E inside five bins labeled A, B, C, D and E. What are the chances that, in your selection, no ball has a label that matches its box?
I tried to break it down and think about the probability of one ball being placed blindly inside its matching bin as simply $\frac{1}{5}$ and the complementary event, the ball not being placed inside its matching bin, as $\frac{4}{5}$. But I know this is overly simplistic for the problem and any help on how to actually start approaching it would be appreciated.
| That sounds like a question of finding permutations without a fixed point.
The idea
Let's just assume we've got our balls sitting in a row, from A to E. Below, we've got our bins arranged from A to E. Once we've blindly done our permutation on the balls (more on that later), we'll put the leftmost ball in the row into the leftmost bin (bin A), the next ball into the next bin (bin B), and so on.
Now, all we have to do is think about how many permutations (rearrangements) of balls there are such that no ball is put into the same bin, i.e. such that no ball is sitting in the same position as it did before the permutation. We later divide the number of permutations that fulfill this condition by the total number of permutations of our balls to get the probability for this event.
The approach
Obviously, we've got an order of $5$ balls to permutate. For this, there are $5!$ possibilities in total ($5$ possibilities to choose the leftmost ball, multiplied by the remaining $4$ possibilities to choose the next ball, and so on).
For the amount of permutations without a fixed point (i.e., the amount of rearranged ball orders in which no ball is sitting in the same place as before), we need to construct the derangement number. Coming up with a way to do this is a little harder, but Wikipedia sketches a proof for it. This is the resulting formular:
$$ !n = n! \cdot \sum_{i=0}^{n} \frac{(-1)^i}{i!} $$
where $n$ is the number of objects to rearrange.
Then you just plug $n=5$ into this formular and you receive the value for $!5$. Your probability will be $\frac{!5}{5!}$.
Fun fact: $\frac{!n}{n!} \approx \frac{1}{e}$ for $n \geq 5$!
| {
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"timestamp": "2023-03-29T00:00:00",
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About an equality of fractional Laplacian on a bounded domain Let $0<s<1$. Let $\Omega\subset\mathbb{R}^n$ be a bounded domain.
We know that $$\|(-\Delta)^{s/2}u\|_{L^2(\mathbb{R}^n)}^2=\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}\frac{|u(x)-u(y)|^2}{|x-y|^{n+2s}}dxdy$$
See e.g. Hitchhiker's guide to the fractional Sobolev spaces, page 16.
I am wondering if the equality still holds when $\mathbb{R}^n$ is replaced by $\Omega$. Namely
$$\|(-\Delta)^{s/2}u\|_{L^2(\Omega)}^2=\int_{\Omega}\int_{\Omega}\frac{|u(x)-u(y)|^2}{|x-y|^{n+2s}}dxdy\ ?$$
We may assume that $u\in L^2(\mathbb{R}^n)$ and supp $u$ $\subset\Omega$.
Thanks!
| Nice question, with negative answer (unless $s=0$ or $2$, of course).
(I am unsure about this formula. The non-shaded part of this post is fine). The correct answer is $$\|(-\Delta)^{s/2} f\|_{L^2(\Omega)}^2=\int_\Omega \int_{\mathbb R^n} \frac{ |f(x)-f(y)|^2}{|x-y|^{n+2s}}\, dxdy; $$ note that only one of the integrals is on $\Omega$.
The point is that the fractional Laplacian is non-local; the value of $(-\Delta )^{s/2}f $ at a point depends on $f$ at all points. In particular, a norm of $(-\Delta)^{s/2} f$ must take into account the values of $f$ everywhere; you cannot arbitrarily choose to consider only the values in $\Omega$.
There are different versions of the fractional Laplacian adapted to domains. If you take a complete orthonormal system of eigenfunctions $\phi_0, \phi_1, \phi_2\ldots$ for the Dirichlet problem on $\Omega$, you can define the so-called Dirichlet Laplacian by
$$
(-\Delta)^{s}_{\mathrm{Dir}} f:=\sum_{\ell=0}^\infty \lambda_\ell^s \hat{f}(\ell) \phi_\ell, $$
where $-\Delta \phi_\ell=\lambda_\ell \phi_\ell$, and
$$
\hat{f}(\ell):=\int_{\Omega} f \phi_\ell\, dx.$$
This is a version of the fractional Laplacian that is localized on $\Omega$.
I don't know what is the precise relationship between the Dirichlet Laplacian and the free-space fractional Laplacian.
| {
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"timestamp": "2023-03-29T00:00:00",
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$H$ is a subgroup of $G$ with $\phi(H) = H$ $\forall \phi \in Aut(G) \Rightarrow $ $H$ is normal in $G$
$H$ is a subgroup of $G$ with $\phi(H) = H$ $\forall \phi \in Aut(G) \Rightarrow $ $H$ is normal in $G$
I'm not sure where to begin.
Perhaps we can use this corollary of the First Isomorphism Theorem
$N_G(H)/C_G(H) \cong Aut(H)$
But I'm not sure if this is helpful.
| Hint
$$h\mapsto ghg^{-1}$$
is an automorphism for all $g\in G$.
| {
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"timestamp": "2023-03-29T00:00:00",
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What method does a calculator use to calculate a linear regression line? Take three coordinates $(1,1)$, $(3,2)$ and $(4,3)$.
My calculator returns the linear regression line: $$y=0.6429x+0.2857$$ of the form $$y = ax +b$$ correct to four significant figures for constants $a$ and $b$.
How can I do this calculation by hand?
I've heard of least square fitting but I haven't learned how to do that and I'm not sure if it is the method or not.
Can someone point me in the right direction?
Also, please don't suggest I plot the points and draw a best fit line by eye and then get my line from the graph. I want to know what method calculators use to calculate the constants $a$ and $b$.
| (Taken from a previous writeup)
How to do
linear least squares fitting.
To fit
a linear sum of
$m$ functions
$f_k(x), k=1$ to $m$
to $n$ points
$(x_i, y_i), i=1$ to $n$,
we want to find the
$a_k, k=1$ to $m$
so that
$\sum_{k=1}^m a_kf_k(x)
$
best fits the data.
Let
$S
=\sum_{i=1}^n(y_i-\sum_{k=1}^m a_kf_k(x_i))^2$.
$\begin{array}\\
\dfrac{\partial S}{\partial a_j}
&=D_jS\\
&=D_j\sum_{i=1}^n(y_i-\sum_{k=1}^m a_kf_k(x_i))^2\\
&=\sum_{i=1}^nD_j(y_i-\sum_{k=1}^m a_kf_k(x_i))^2\\
&=\sum_{i=1}^n2(y_i-\sum_{k=1}^m a_kf_k(x_i))D_j(y_i-\sum_{k=1}^m a_kf_k(x_i))\\
&=\sum_{i=1}^n2(y_i-\sum_{k=1}^m a_kf_k(x_i))(-D_j a_jf_j(x_i))\\
&=\sum_{i=1}^n2(y_i-\sum_{k=1}^m a_kf_k(x_i))(- f_j(x_i))\\
&=-2\sum_{i=1}^nf_j(x_i)(y_i-\sum_{k=1}^m a_kf_k(x_i))\\
&=-2\left(\sum_{i=1}^ny_if_j(x_i)-\sum_{i=1}^nf_j(x_i)\sum_{k=1}^m a_kf_k(x_i)\right)\\
&=-2\left(\sum_{i=1}^ny_if_j(x_i)-\sum_{k=1}^m a_k\sum_{i=1}^nf_j(x_i)f_k(x_i)\right)\\
\end{array}
$
Therefore,
if $D_jS = 0$,
then
$\sum_{i=1}^ny_if_j(x_i)
=\sum_{k=1}^m a_k\sum_{i=1}^nf_j(x_i)f_k(x_i)
$.
Doing this
for $j=1$ to $m$
gives $m$ equations
in the $m$ unknowns
$a_1, ..., a_m$.
Example:
To fit a polynomial
of degree $m-1$,
let $f_j(x) = x^{j-1}$.
The equations are then
$\begin{array}\\
\sum_{i=1}^ny_ix_i^{j-1}
&=\sum_{k=1}^m a_k\sum_{i=1}^nx_i^{j-1}x_i^{k-1}\\
&=\sum_{k=1}^m a_k\sum_{i=1}^nx_i^{k+j-2}\\
\end{array}
$
For a line,
$m=2$
and the equations are,
for
$j = 1, 2$,
$\begin{array}\\
\sum_{i=1}^ny_ix_i^{j-1}
&=\sum_{k=1}^2 a_k\sum_{i=1}^nx_i^{k+j-2}\\
&= a_1\sum_{i=1}^nx_i^{j-1}+a_2\sum_{i=1}^nx_i^{j}\\
\end{array}
$
Explicitly these are
$j=1:\sum_{i=1}^ny_i
= a_1n+a_2\sum_{i=1}^nx_i\\
j=2:\sum_{i=1}^nx_iy_i
= a_1\sum_{i=1}^nx_i+a_2\sum_{i=1}^nx_i^{2}\\
$
These should look familiar.
For a quadratic,
$m=3$
and the equations are,
for
$j = 1, 2, 3$,
$\begin{array}\\
\sum_{i=1}^ny_ix_i^{j-1}
&=\sum_{k=1}^3 a_k\sum_{i=1}^nx_i^{k+j-2}\\
&= a_1\sum_{i=1}^nx_i^{j-1}+a_2\sum_{i=1}^nx_i^{j}+a_3\sum_{i=1}^nx_i^{j+1}\\
\end{array}
$
Example 2.
To fit a line
through the origin,
$y = ax$,
$m=1$ and
$f_1(x) = x$.
The equation is then
$\sum_{i=1}^ny_ix_i
=a_1\sum_{i=1}^nx_i^2
$
so the result is
$a
=\dfrac{\sum_{i=1}^nx_iy_i}{\sum_{i=1}^nx_i^2}
$.
| {
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"url": "https://math.stackexchange.com/questions/3195184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Totally ramified extension of $\mathbb{Q}_{p}$ which is not of a form $\mathbb{Q}_{p}(\sqrt[n]{pu})$ It is known that a finite extension $K/\mathbb{Q}_{p}$ is totally ramified if and only if $K = \mathbb{Q}_{p}(\alpha)$ where $\alpha$ is a root of Eisenstein polynomial. Is there any totally ramified extension that is not of the form $\mathbb{Q}_{p}(\sqrt[n]{pu})$ for some $u\in \mathbb{Z}_{p}^{\times}$? Every degree 2 totally ramified extensions has this form, but I don't know whether this is also true for degree 3 or higher. Thanks in advance.
| There is a general theorem that every tamely totally ramified extension of $\mathbf Q_p$ with degree $n$ has the form $\mathbf Q_p(\sqrt[n]{\pi})$ for some prime $\pi$ in $\mathbf Z_p$, so $\pi = pu$ for a unit $u$ in $\mathbf Z_p$. (There is a similar theorem over other local fields.) So if you want a totally ramified extension not of that form you need $n$ to be divisible by $p$.
Let's try $n=p$. Something we can say about extensions $\mathbf Q_p(\sqrt[p]{pu})$ for $p>2$ is that they are not Galois over $\mathbf Q_p$: a field $K$ containing a full set of roots of $x^p - pu$ must contain the nontrivial $p$th roots of unity, and those have degree $p-1$ over $\mathbf Q_p$ so $[K:\mathbf Q_p]$ is divisible by $p-1$. Therefore $[K:\mathbf Q_p] \not= p$ when $p>2$. Thus a Galois totally ramified extension of $\mathbf Q_p$ having degree $p$ can't have the form $\mathbf Q_p(\sqrt[p]{pu})$.
Every totally ramified abelian Galois extension of $\mathbf Q_p$ with degree divisible by $p$ contains a subextension with degree $p$ since the Galois group has a subgroup of index $p$: in an abelian group of order $n$ there is a subgroup of each order dividing $n$ and thus also a subgroup of each index dividing $n$ by using a subgroup of order equal to the complementary factor in $n$ of the desired index. Subextensions of totally ramified extensions are totally ramified and subextensions of abelian Galois extensions are abelian Galois extensions. Thus all we need to do now is find a totally ramified abelian Galois extension of $\mathbf Q_p$ with degree divisible by $p$ and inside of it there are extensions of degree $p$, all of which are examples of the kind being sought (not having the form $\mathbf Q_p(\sqrt[n]{pu})$).
The easiest choice is a cyclotomic extension: $\mathbf Q_p(\zeta_{p^2})$ where $\zeta_{p^2}$ is a root of unity of order $p^2$. This field has degree $p^2-p$ over $\mathbf Q_p$, with cyclic Galois group $(\mathbf Z/p^2\mathbf Z)^\times$, so the field contains a unique subextension with degree $p$ over $\mathbf Q_p$, namely the field fixed by the unique subgroup of the Galois group with order $(p^2-p)/p = p-1$. That subgroup is the solutions to $a^{p-1} \equiv 1 \bmod p^2$, and a generator of this extension over $\mathbf Q_p$ is $\sum_{a^{p-1} = 1} \zeta_{p^2}^a$ where the sum runs over all solutions of $a^{p-1} \equiv 1 \bmod p^2$.
Example when $p=3$: $a^2 \equiv 1 \bmod 9$ has solutions $\pm 1 \bmod 9$ and $\zeta_{9} + \zeta_9^{-1}$ has minimal polynomial $f(x) = x^3 - 3x + 1$. Then $f(x-1) = x^3 - 3x^2 + 3$ is Eisenstein at $3$; the polynomial $f(x+1)$ is not. I did my calculation of the minimal polynomial in $\mathbf C$, which is okay since a primitive $p$th-power root of unity has the same degree over $\mathbf Q_p$ as it does over $\mathbf Q$, so the structure of the intermediate fields in a $p$th-power cyclotomic extension over $\mathbf Q_p$ and over $\mathbf Q$ are the same.
Example when $p=5$: solutions to $a^4 \equiv 1 \bmod 25$ are $1, 7, 18$, and $24$, and $\zeta_{25} + \zeta_{25}^7 + \zeta_{25}^{18} + \zeta_{25}^{24}$ has minimal polynomial over $\mathbf Q_5$ equal to $g(x) = x^5 - 10x^3 + 5x^2 + 10x + 1$. (Note $g(x-1) = x^5 - 5x^4 + 25x^2 - 25x + 5$ is Eisenstein at 5; the polynomial $g(x+1)$ is not Eisenstein at $5$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3195349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve $f(x)=c \times f(\frac{x}{2})$ for $c$ Given:
*
*Function $f(x)$ is infinitely differentiable
*equation (1)
$f(x)=c \times f(\frac{x}{2})$
We have to find all $c$, for which the (1) has non-zero solutions
Any hints on theorems to apply here, I reckon it's somehow related to ODEs
| If
$ f(x)=c*f(x/2)
$
then
$\begin{array}\\
f(x)
&=cf(x/2)\\
&=c^2f(x/4)\\
&=c^3f(x/8)\\
&...\\
&=c^nf(x/2^n)\\
\end{array}
$
If $|c| < 1$ then
$f(x) \to 0$
so
$f(x) = 0$
for all $x$.
If $f(0) \ne 0$,
$\dfrac{f(x)}{c^n}
\to f(0)
$.
If $|c| > 1$,
$\dfrac{f(x)}{c^n}
\to 0
$
which contradicts
$f(0) \ne 0$.
If $f(0) = 0$,
then,
for small $x$,
$f(x) = xf'(0)+O(x*2)
$
so
$f(x/2^n)
=xf'(0)/2^n+O(x^2/4^n)
$
so
$f(x)
=c^n(xf'(0)/2^n+O(x^2/4^n))
=xf'(0)(c/2)^n+O(x^2(c/4)^n))
$.
This only works if
$c=2$;
it goes to zero if
$|c| < 2$
and to $\infty$ is
$|c| > 2$.
Therefore we must have
$c = 2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Probability of a false negative A population includes a fraction m of individuals carrying a disease exists and has the following characteristics:
P(positive test | individual with the disease) = p
P(positive test | individual without the disease) = r
What is the probability of a false negative : P(individual with the disease | negative test) ?
Here is what I have so far:
Let W = with the disease, ~W = without the disease;
Let + = positive test, - = negative test
From the given, we have: P(W) = m, P(~W) = 1-m
We want P(W | -) = (P(- |W) P(W))/(P(-)) = (P(- | W)P(W)/(P(- | W)P(W) + P(- ~W)P(~W))
We have: P (+) = P(+|W)P(W) + P(+|~W)P(~W) = p * P(W) + r * P(~W)
However, I do not know how to proceed from here. Could anyone please help me? Thank you very much!
| You have it almost right. You write $P(W|-)=P(-|W)P(W)/P(-)$, and you know everything except $P(-)$: $P(-|W)=1-p$ and $P(W)=m$.
On the other hand, you write the correct expression for $P(+)$, but then you know $P(-)=1-P(+)$. That is, $P(-)=1-pm-r(1-m)$.
Here is an alternate solution, where I compute directly $P(-)$ instead of $P(+)$.
The probability that the test is negative is
$$P(-)=P(W)P(-|W)+P(\sim W)P(-|\sim W)=m(1-p)+(1-m)(1-r)$$
The probability that the test is negative and the person has the disease is
$$P(-\cap W)=P(W)P(-|W)=m(1-p)$$
Therefore, the probability that a person has the disease, given that the test is negative, is
$$P(W|-)=\frac{P(-\cap W)}{P(-)}=\frac{m(1-p)}{m(1-p)+(1-m)(1-r)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3195583",
"timestamp": "2023-03-29T00:00:00",
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Mapping $\mathbb{Z}_k$ into $\{d,d+1,\ldots,d+k-1\}$ preserving value $\bmod k$ I want to give a (simple) map $f:\mathbb{Z}_k \to \{d,d+1,\ldots,d+k-1\}$ for $d,k \in \mathbb{N}$ and such that $\forall i \in \mathbb{Z}_k. f(i) \bmod k = i$.
Is there a particularly simple way of doing this?
Goal
The goal is to formally prove that if $f$ is periodic with period $k$ then $\sum_{m}^{m+k-1} f \; l = \sum_{m+d}^{m+d+k-1} f \; l$. The above is the particular case $m = 0$.
Current solution (suggested by @kingW3)
$f(i) = (i-d) \; mod \; k + d$ and if $g(j) = j \; mod \; k$ then $g \circ f = Id$.
I'm interested in showing $f \circ g = Id$. How can I do it?
| There's only one way to map $\Bbb{Z}_k$ onto a set of consecutive $k$ integer in such a way
$$
f(\bar x)\equiv x\bmod k.
$$
That's because in any given set $K$ of $k$ consecutive integers there's only one $y\in D$ satisfying the displayed congruence for any given $\bar x$.
Thus, let $f(\bar x)=y$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Variance of x_i chosen from uniformly distributed hypersphere I'm looking for an expression of the variance of a single component of a point chosen from within a uniformly distributed n-ball with radius r for any n.
There are a few proofs showing that components of a point chosen this way become increasingly normally distributed as n tends to infinity. I can't seem to find expressions for any n.
| I didn't totally follow @kimchi lover's answer (particularly his assumption of normality), so here is a similar approach.
Suppose X is hyperspherically uniformly distributed with r=1, so each $X_i$ has $E[X_i] = 0$, and we want to compute $VAR[X_i]$. Since $E[X_i]=0$, by the variance-squares identity, we have $VAR[X_i] = E[X^{2}_i] - E[X_i]^2 = E[X^{2}_i]$, so we really just need to compute $E[X^{2}_i]$.
Notice that $r^2 = \Sigma_{i=1}^n X^{2}_i$, so we have $E[r^2] = nE[X^{2}_i]$. Since we know that r has pdf $p_n (r) = nr^{n-1}$, we can trivially compute $E[r^2] = \int_0^1 r^2 nr^{n-1}dr = \frac{n}{n+2}$, so then $E[X^{2}_i] = VAR[X_i] = \frac{E[r^2]}{n} = \frac{1}{n+2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\operatorname{Cl }A$ in topological space $\mathbb{R^2}$ with dictionary order topology. Let $ A = \left\{ (x,y) \in \mathbb{R^2} \mid y= \sin ( \frac{1}{x}) , \ 0 < x \leq 1 \right\}$ . Find $\operatorname{Cl} A$ in topological space $\mathbb{R^2}$ with dictionary order topology.
I guess $ \operatorname{Cl} A = A $?
| You're quite right. To prove it, I would take an arbitrary point not in $A,$ and find an open interval around it containing no points of $A.$ This shows that the complement of $A$ is open, so that $A$ is closed.
| {
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Probability of matching pair of 10-item baskets out of 100M people shopping 100 times in a year This is from Exercise 1.2.2 of MMDS MMDS Book
Suppose we have information about the supermarket purchases of 100
million people. Each person goes to the supermarket 100 times in a
year and buys 10 of the 1000 items that the supermarket sells. We
believe that a pair of [malicious people] will buy exactly the same set of 10
items (perhaps the ingredients for a [bad thing]?) at some time during the
year. If we search for pairs of people who have bought the same set of
items, would we expect that any such people found were [actually malicious]?
My solution was:
$$\frac{{1e8 \choose 2}(\frac{100}{365})^2}{1e3 \choose 10} = 0.000000001.$$
My logic to find the expected number of instances (assuming randomness) was to multiply the number of possible pairs ${1e8 \choose 2}$ by the chances that two people go shopping on the same day $(\frac{100}{365})^2$ and then multiplying by the chance that they pick the same 10 items.
My question is if my formula is correct and if I should square the denominator? I'm asking because a similar example in the book doesn't square the denominator. I figured it should be squared because each customer can independently choose which 10 items they want.
| There are ${1000 \choose 10} \approx 2.634 \cdot 10^{23}$ possible shopping baskets. This is a generalized birthday paradox problem. We must be intended to assume each customer takes a random independent sample of the possible baskets one each visit. There are $10^{10}$ samples taken. The number of samples needed to get a $50\%$ chance of a match is $\sqrt {2 \ln (2) 2.634\cdot 10^{23}}\approx 6.042\cdot 10^{11}$ so we decide it is rather unlikely that two baskets match by chance.
| {
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Finding other generalized inverses besides the pseudoinverse? I have a $16\times 4$ matrix $A$ of rank $4$. Besides its Moore-Penrose pseudoinverse $A^+$, I'm also interested in other generalized inverses $A^g$ that satisfy $A^gA=I_4$.
Is there a way to get all of them (presumably in some analytical form with free variables)?
Are there any special inverses? By "special," I mean $A^+$ gives the solution to $Ax=y$ with minimum $\ell_2$ norm, so is there one such $A^g$ that gives the solution with minimum $\ell_0$ norm, for example?
If the answers to the two questions above are both "No," how do I find any generalized inverse other than the pseudoinverse?
| All generalized inverses can be written as
$$
A^g = A^+ + U(I_{16}-AA^+),
$$
where $U\in {\mathbb C}^{4\times 16}$ is arbitrary.
Note: $I_{16}-AA^+$ is the orthogonal projection onto the null space of $A$.
| {
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degree extension over filed of $p$-adic numbers Let $K = \mathbb{Q}(\theta)$ be a numberfield and $[K:\mathbb{Q}]=n$. When $\mathbb{Q}_p$ is the field of $p$-adic numbers and $K_p=\mathbb{Q}_p(\theta)$, what about $[K_p : \mathbb{Q}_p]$?
| I’ll give the results, with no hint of a proof:
The general situation is that if $G(X)=\text{Irr}(\theta,\Bbb Q[X])$ splits as a product of $\Bbb Q_p$ irreducibles $G=g_ig_2\cdots g_m$, then there are essentially $m$ ways of embedding $\Bbb Q(\theta)$ into some finite extension of $\Bbb Q_p$. Each of these embeddings has, associated to it, numbers $f_i$ and $e_i$, the residue field extension degree and the ramification index, respectively, and $\sum_ie_if_i=[\Bbb Q(\theta):\Bbb Q]=\deg(G)$.
You may look at this from the standpoint of the factorization of ideals in the integer-ring $I$ of $\Bbb Q(\theta)$ this way: $(p)=pI=\prod_0^m\mathfrak p_i^{e_i}$, where $\mathfrak p_1,\cdots,\mathfrak p_m$ are the distinct prime ideals of $I$ containing $p$. They’re all maximal ideals of $I$, and each $I/\mathfrak p_i$ is a finite extension $\kappa_i$ of $\Bbb F_p=\Bbb Z/p\Bbb Z$, with $[\kappa_i:\Bbb F_p]=f_i$, the number mentioned above.
For an example, let $\theta$ be a root of the $\Bbb Q$-irreducible polynomial $G=X^4 +3X^3+3X^2+6$. Then Hensel tells us that $G\equiv (X^2+9X-1)(X^2-6X-6)\pmod{16}$. The first factor is congruent to $X^2+X+1$ modulo $2$, so generates $\Bbb F_4$, giving $f=2$, while the second factor is Eisenstein for $2$, giving a ramification degree of $2$. Both quadratic factors are $\Bbb Q_2$-irreducible. We get the result that $f_1=2,e_1=1$ while $f_2=1,e_2=2$, so that $\Bbb Q(\theta)$ has two very different embeddings into $\Bbb Q_2$.
| {
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Converse linear quadratic optimal control It is well known that for a linear time invariant system
$$
\dot{x} = A x + B u \tag{1}
$$
with $(A, B)$ controllable, there exists a static state feedback $u = -K x$ such that the cost function
$$
J = \int_0^{\infty} x^T Q x + u^T R u \, dt \tag{2}
$$
is minimized, assuming $Q \geq 0$ (positive semi-definite) and $R > 0$ (positive definite). The gain $K$ is the solution of the algebraic Riccati equation:
$$
\begin{align}
0 &= A^T P + P A - P B R^{-1} B^T P + Q \\
K &= R^{-1} B^T P \\
P &= P^T \geq 0
\end{align}
$$
known as linear quadratic regulator (LQR). However, I wonder whether the converse also holds?
That is, given a stabilizing $K_s$ (such that $A - B K_s$ is Hurwitz), do there exist matrices $Q \geq 0$ and $R > 0$ such that $u = -K_s x$ minimizes $(2)$ given $(1)$? Or put differently:
Question: Is every stabilizing linear state feedback optimal in some sense?
| See the paper: Kalman, R. E. (1964). When is a linear control system optimal?. Journal of Basic Engineering, 86(1), 51-60.
The answer is positive at least for a class of systems. As far as I remember, the answer is also positive for a general LTI system, but I cannot find a reference at the moment.
UPDATE: Every linear system with nondynamic feedback is optimal with respect to a quadratic performance index that includes a cross-product term between the state and control, see [R1].
If you do not allow for the cross-product term, then several sufficient and necessary conditions are known, see for example [R2] and the references there.
[R1] Kreindler, E., & Jameson, A. (1972). Optimality of linear control systems. IEEE Transactions on Automatic Control, 17(3), 349-351.
[R2] Priess, M. C., Conway, R., Choi, J., Popovich, J. M., & Radcliffe, C. (2015). Solutions to the inverse lqr problem with application to biological systems analysis. IEEE Transactions on control systems technology, 23(2), 770-777.
| {
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Distribution of prime numbers modulo $4$ Are primes equally likely to be equivalent to $1$ or $3$ modulo $4,$ or is there a skew in one direction?
That is my specific question, but I would be interested to know if there exists a trend more generally, say for modulo any even.
| In general, if $\gcd(a,b) = 1$, the number of primes which are of the form $b$ modulo $a$ is asymptotic to $\dfrac{\pi(x)}{\varphi(a)}$ where $\pi(x)$ is the number of primes $\le x$. As you see the asymptotic formula is independent of $b$ hence for a given modulo all residues occur equally in the long run.
| {
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Intersection of connected components with discrete subgroup I am currently studying Harmonic Analysis and didn't quite understand part of the proofs for the structure theorems of locally-compact abelia groups (LCA).
Let $A$ be an LCA group such that $\hat{A}/\hat{A}_0$ is compact, where $A_0$ denotes the connected component of the unit in $A$ and $\hat{A}$ is the dualization of $A$.
Now suppose that there is an infinite compact subgroup $E \leq A$. We can write this as an exact sequence
$$ 0 \to E \to A\to A/E\to 0$$
and dualizing it to the exact sequenz
$$ 0 \to B \to \hat{A} \to \hat{E}\to 0 .$$
Where $\hat{E}$ is an infinite discrete group and as such non-compact.
This is supposedly a contradiction to $\hat{A}/\hat{A}_0$ being compact, but I don't see why. By that, we would know that no such group can exist in $A$.
The map $\Phi$ from $\hat{A} \to \hat{E}$ is surjective and continous by construction, and we can restrict it to a map
$$\varphi: \hat{A}/\hat{A}_0 \to \hat{E}/(\hat{A}_0\cap\hat{E}).$$
If we knew that $\hat{A}_0\cap\hat{E}$ is finite (or even better $=\{e\}$) then we would be done, having a surjection from a compact to a non compact group.
Is this the right way to go or am I missing something important?
Any help would be greatly appreciated.
| If by contradiction $A$ has an infinite compact subgroup $E$, then, as you said, $\hat{A}$ has the infinite discrete quotient $\hat{E}$. Since $\hat{A}_0$ has a trivial image in $\hat{E}$, this is an infinite discrete quotient of $\hat{A}/\hat{A}_0$. But the latter is compact, so its discrete image should be finite. Contradiction.
| {
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Find the interior $\{(x,y): 0 < x^2 + y^2 < 1\}$.
Find the interior $A = \{(x,y): 0 < x^2 + y^2 < 1\}$. I assume the metric is the standard Euclidean metric.
I know that the int$(A) = A$, but I don't know how to prove it. I can gather that the radius for any $x \in A$ should be $r = \min(\sqrt{x^2_1 + x_2^2},1 -\sqrt{x^2_1 + x_2^2})$ such that $B(x;r) \subseteq A$ by looking at a graph, but how do I show that this is true?
| This works in any metric space. So you have a distance $d$ and the ball of radius $1$ around $x_0$.
Let $r=1-d(x_1,x_0)$. If $x$ is at distance less than $r$ from $x_1$, then by the triangle inequality
$$
d(x,x_0)\leq d(x,x_1)+d(x_1,x_0) < r + 1-r = 1.
$$
In your example $x_0=(0,0)$, $x_1= (x_1,x_2)$ and $d((a,b),(c,d))=\sqrt{(a-c)^2+(b-d)^2}$.
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Derivatives without limits Update: H/t David K for pointing out that my assumption that I can force $a^2+b^2=r^2$ is wrong. This led to analyzing a cubic equation which is now moot, but I think the bulk of the question remains sensible.
I think I'm making some mistakes in the below, but I'm not sure where. Even if there are mistakes, I'm thinking there might not be any practical point to it, but I'm curious.
I think it's possible to find the slope of a tangent line without using limits in some cases. Further it's possible to generalize the procedure to many common curves. I am having trouble with a proof of this and I'm not sure where I am going wrong.
Proof we can find the slope of a tangent line to a parabola without taking a limit:
*
*First we consider a circle. If we know the center of a circle and we know a point on that circle, we know the equation of a line connecting the center to that point. The slope of the line perpendicular to this line is the slope of the tangent line to the circle at that point. So we have already proven the thesis in the case of a circle.
*A circle is tangent to some other curve if it only intersects that curve at a single point.
*A circle can be constructed to be tangent to certain points on certain curves. Tangent circles share tangent lines with the curves to which they are tangent. So the procedure above can find tangents to those curves by constructing the correct tangent circle.
Parabola case:
Consider $y=x^2$.
We want a circle such that $(x-a)^2+(x^2-b)^2=r^2$
So:
$$x^4+(1-2b)x^2-2ax+(a^2+b^2-r^2)=0$$
We either have 2 complex roots or no real solutions. Assuming we have any real solutions, having a unique one requires that we have a repeated root.
So to find the tangent to the curve at $x_0$ we want to divide the quartic by $(x-x_0)^2$. We'd want to have the remainder terms be zero and the discriminant of the resulting quotient to be negative.
EDIT:
Turns out Descartes developed this method centuries ago bearing some fruit. It's algebraically more intense than Newton's and Leibniz's approach to finding slopes of tangent lines, so fell by the wayside. It's called Descartes' Method of Normals
| This is very creative, but I think it's overly complicated. You could do something very similar by using lines instead of circles.
Consider the point $P=(a,a^2)$ on the parabola $y=x^2$. Which lines through $P$ only intersect the parabola once? The line is determined by its slope, so it has equation $y=m(x-a)+a^2$. This intersects $y=x^2$ where
\begin{gather*}
x^2 = m(x-a)+a^2 \\
\iff x^2 -mx + (am-a^2) = 0
\end{gather*}
The last equation is quadratic in $x$ and has discriminant
$$
\Delta = m^2 - 4(1)(am-a^2) = m^2-4am + 4a^2 = (m-2a)^2
$$
So there is a single solution if and only if $m=2a$. Thus the line is tangent to the parabola when $m=2a$.
For a general polynomial curve $y=f(x)$, you'd look for slopes $m$ such that the equation $f(x) = m(x-a) + f(a)$ (an equation in $x$) has multiple roots. You can probably also extend it to algebraic curves in the plane, too. However, I'm not sure it provides any more insight than ordinary calculus.
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Show that a number $n$ is divisible by 6 if and only if it can be written as a sum of three distinct divisors. If $6|n$ then $n=6k=3k+2k+k$. And $3k|n$, $2k|n$ and $k|n$.
Now let $p,q$ and $r$ be three distinct divisors of $n$ so that :
$$n=p+q+r$$
Because $p|n $, $ q|n$ and $r|n$ I figured that $p|q+r $, $ q|p+r$ and $r|p+q$.
I tried to prove that n is even or a multiple of 3 but without much luck.
How can I prove the statement?
| Without, loss of generality, presume that $p > q > r$. Observe that $$p|(q+r) \implies (q+r) = k_1 p \implies 1 \le k_1 = \frac{q+r}{p} < 2 \implies k_1 =1.$$
As such, $p = q+r$, and $n = 2(q+r)$.
Since $q|n=2(q+r)$, we have $q|2r$. That is, $$2r = k_2 q \implies 1 \le k_2=\frac{2r}{q}<2 \implies k_2 = 1.$$
Therefore, $$q = 2r \implies p = q+r =3r \implies n = 6r.$$
| {
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Cartan homotopy formula and curl In Topological Methods in Hydrodynamics, V. I. Arnol'd writes that the following expression
$$curl(\mathbf a \times \mathbf b)=[\mathbf a, \mathbf b]+ \mathbf a \ div \ \mathbf b - \mathbf b \ div \ \mathbf a$$
could be obtained "repeatedly applying" the Cartan homotopy formula
$$L_v = i_vd+di_v$$
And (in another book) adds some hints:
*
*$i_{curl(\mathbf a \times \mathbf b)}\tau = di_{\mathbf a} i_{\mathbf
b}\tau$
*$div \ \mathbf a = di_{\mathbf a} \tau$
*$[\mathbf a, \mathbf b] = L_{\mathbf a} \mathbf b$
(where $\mathbf a, \mathbf b$ are two vectors in $R^3$, $i$ is the interior product, and $\tau$ the volume element).
I really could not figure how to proceed. Any suggestions?
| Apply (plug in by means of the operator $i$) both sides to a volume 3-form $\tau$.
The equality of vector fields on both sides is equivalent to the equality of the obtained 2-forms.
On the left we have
$i_{curl(a\times b)}\tau= di_b i_a \tau$ (note the order: vector field $a$ is plugged in first).
On the right first use the Cartan calculus formula
$$
i_{[a,b]}\tau=[i_b, L_a]\tau = i_b L_a \tau - L_a i_b\tau
= i_b d i_a\tau - di_a i_b\tau -i_a d i_b\tau
$$
(note that $d\tau = 0$).
Since $L_b\tau = ({div}\,b)\tau$, we have
$i_{a\, {div}\, b}\tau = i_a(L_b\tau)= i_a d i_b\tau$
and similarly for $i_{b\, {div}\, a}\tau = i_b d i_a\tau$.
Hence the inner multiplication of $[a,b]+ a\, {div}\, b - b\, {div}\, a$ to the 3-form $\tau$ is
$$
i_b d i_a\tau - di_a i_b\tau -i_a d i_b\tau + i_a d i_b\tau - i_b d i_a\tau = - di_a i_b\tau,
$$
which proves the formula.
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If $f(x,y)=9-x^2-y^2$ if $x^2+y^2\leq9$ and $f(x,y)=0$ if $x^2+y^2>9$ study what happens at $(3,0)$
If$$f(x,y)=\begin{cases}9-x^2-y^2&\text{if }x^2+y^2\leq9\\0&\text{if }x^2+y^2>9\end{cases}$$study the continuity and existence of partial derivative with respect to $y$ at point $(3,0)$.
The graph of the domain of $f$ is:
Continuity study at $(3,0)$:
$f(3,0)=9-3^2-0=0$, but I do not know how to find $$\lim_{(x,y)\to(3,0)}f(x,y).$$ I tried the following: $$\lim_{(x,y)\to(3,0)}f(x,y)=\left\{\begin{array}{l}\displaystyle\underset{C_1\colon x^2+y^2=9}{\lim_{(x,y)\to(3,0)}}f(x,y)=\underset{C_1\colon x^2+y^2=9}{\lim_{(x,y)\to(3,0)}}(9-(x^2+y^2))=9-9=0\\\displaystyle\underset{C_2\colon x^2+y^2\neq9}{\lim_{(x,y)\to(3,0)}}f(x,y)=\underset{C_2\colon x^2+y^2\neq9}{\lim_{(x,y)\to(3,0)}}\text{??}=\text{??????}\end{array}\right.$$ but then I realized that the "curve" $C_2$ is actually NOT a curve but a set of infinite points, as shown in the previous image.
Existence of partial derivative with respect to $y$ at $(3,0)$:
I know that I need to study whether $$\frac\partial{\partial y}f(3,0)=f'((3,0);(0,1))=f_y(3,0)=\lim_{h\to0}\frac{f((3,0)+h(0,1))-f(3,0)}{h}=\lim_{h\to0}\frac{f(3,h)}{h}$$ exists or not, but I am not able to even find that limit.
Any help? Thanks!!
| Continuty at (3,0) : When $|(x,y)-(3,0)|=\sqrt{(x-3)^2+y^2}<\delta<1$ for some $\delta>0$ then $|f(x,y)|\leq 9-x^2-y^2=(3-x)(3+x)\leq 6 (3-x)\leq 6\sqrt{(x-3)^2+y^2}<6\delta.$ This implies that $f$ is continuous at $(3,0).$
Partial differential wrt $y$ : $f(3,y)=0$ for any $y$ and hence $f$ has a partial derivative at $(3,0)$ wrt $y$ and the partial derivative is $0.$
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Does $x^n$ converges uniformly on $ [0,1)?$ Doing the limit we can see that in the open interval it converges pointwise to the constant function $f(x) = 0$. In the closed interval it doesn't converge uniformly because in $x=1$ $f(x) =1$ and when $0<x<1$ then $f(x)=0$. It isn't a continuous function although $f_n(x)$ is continuous for all $n$. So it can't be uniform.
However,let $0<ε<1$, we know the function increases as $x$ increases. So,
$|f_n(x) - f(x)| \leq ε^{n}$
And ε^(n) goes to 0 as n tends to infinity. Doesn't it mean that the convergence is uniform?
| No, $x^n$ does not converge to zero uniformly on $[0,1)$, because
$$\lim_{n\to\infty}\sup_{0\leq x<1}|x^n|=1\neq 0$$
| {
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Intuition behind determinant of a matrix with $2$ equal rows In my linear algebra course, we have just proved that if a matrix $A$ contains $2$ equal rows, then $\det(A)=0$.
I understand how the proof works, but could somebody offer a more intuitive explanation of why this is the case?
| If 2 rows are equal then they are linearly dependant so you'll have a zero eigen value and so the determinant (which is the product of the eigen values) will be zero.
| {
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For which $\alpha$ will take the cake ever be again with chocolate on the bottom and cream on the top Question: A bored kid left alone at home decides to take a chocolate cream cake (chocolate on the bottom, cream on top) and his protractor and spend the day as follows: He cuts a slice of angle $\alpha$ put it back up-side-down (i.e. cream on the bottom) rotate the cake clockwise by $\alpha$ and repeat the same procedure again and again. For which $\alpha$ will take the cake ever be again with chocolate on the bottom and cream on the top?
My approach: Well, intuitively $\alpha$ could be 180 degrees, and 360 degrees, my approach is that $\alpha$ could be anything that divides 360 degrees without a remainder. Not quite sure how to show/prove that mathematically, maybe there could be some trick here, any help is appreciated.
| EDIT: this answer assumes model bad model of piece inverting (that putting piece up-side-down is equal to reversing colors). In real pie we not only inverse colors but also mirror them. See @user2323232's link for correct answer.
You need $\alpha$ to be rational (I assume slice is half-open segment, otherwise we will never get the entire cake right).
Lets reformulate a bit: we take a red line, start at point $0$, then color segment $[0, \alpha)$ in green, then color segment $[\alpha, 2\alpha)$ and so on. Point at $x$ degrees is cream-top after $n$ steps iff even number of points $x, x + 360, x + 720, \ldots$ is green (it means that we made an even number of flips of this point)
If $\alpha = \frac{n}{m}$ is rational, then after $720m$ steps, we will have segment $[0, 720n)$ green and the rest is red. Then for any $x$ there will be exactly $2n$ green points: $x, x + 360, \ldots, x + 360 \cdot (2n - 1)$. So after $720m$ steps our cake will be cream-top again.
Lets assume that after $k$ steps our cake is cream-top. We will have green segment $[0, k\alpha)$. If $k\alpha = 720n$ for some integer $n$, then $\alpha$ is rational. Otherwise choose $n$ such that $720n < k\alpha < 720(n + 1)$. Take some $x^\prime \in (720n, k\alpha)$ such that $x^\prime + 360 > k\alpha$ and let $x = x^\prime - 720n$. Then point at $x$ degrees is cream-bottom, as $x, x + 360, \ldots, x + 720n$ are green, but $x + 720n + 360$ isn't. So $k$ steps didn't left us with cream-top cake.
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$11x + 13 \equiv 4$ (mod 37) $11x + 13 \equiv 4$ (mod 37)
My "solution" to the problem
$11x + 13 \equiv 4$ (mod 37) $\rightarrow$
$11x + 13 = 4 + 37y $
$11x - 37y = - 9$
Euclid's algorithm.
$37 = 11*3 + 4$
$11 = 4*2 + 3$
$4 = 3*1 + 1 \rightarrow GCD(37,11) = 1$
$3 = 1*3 + 0$
Write as linear equation
$1 = 4 - 1*3$
$3 = 11 - 2*4$
$4 = 37-3*11$
$1 = 4 -1*3 = 4-1(11-2*4) = 3*4-1*11 = 3(37-3*11)-1*11 = 3*37 -10*11$
$1 = 3*37 -10*11$
$11(10) - 37(3) = -1$
$11(90) - 37(27) = -9$
x = 90
That is my answer. But the correct answer should be:
x = 16
| Note: $90\mod37 =16$, so the answer you found was correct, just not fully reduced.
| {
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Polynomial Roots with no complex roots Let $(a_1,b_1),$ $(a_2,b_2),$ $\dots,$ $(a_n,b_n)$ be the ordered pairs $(a,b)$ of real numbers such that the polynomial
$$p(x) = (x^2 + ax + b)^2 +a(x^2 + ax + b) - b$$has exactly one real root and no nonreal complex roots. Find $a_1 + b_1 + a_2 + b_2 + \dots + a_n + b_n.$
I have no idea how to do this. Can someone help please?
| We expand the OP's polynomial and write it as
$\tag 1 p(x) = (a b + b^2 - b) + (a^2 + 2 a b) x + (a^2 + a + 2 b) x^2 + 2 a x^3 + x^4 $
We also have (see Robert Israel's answer)
$\tag 2 (x-r)^4 = r^4 - 4 r^3 x + 6 r^2 x^2 - 4 r x^3 + x^4$
Fortunately, we can find an easy 'coefficient hook', and so $2a = -4r$, or
$$\tag 3 r = -\frac{a}{2}$$
Plugging this back into $\text{(2)}$, we get
$$\tag 4 (x+a/2)^4 = \frac{a^4}{16} + \frac{a^3}{2} x + \frac{3 a^2}{2} x^2 + 2 a x^3 + x^4$$
Once again, there is an easy way to proceed, and we find
$$\tag 5 b = \frac{a^2 - 2a}{4}$$
Once again, there is a (relatively) easy way to proceed, and we find that
$$\tag 6 \frac{a^4}{16} - \frac{a^2}{2} + \frac{a}{2} = \frac{a^4}{16}$$
must be true.
So there are at most two possible solutions:
$(a_1, b_1) = (0, 0)$
and
$(a_2, b_2) = (1, -\frac{1}{4})$
You will find that by plugging into $p(x)$ that they both work - the polynomial will only have one real root with multiplicity $4$.
I have no idea why we are interested in the sum of these coordinates. Perhaps I need to review algebra-precalculus.
| {
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Combinatorial proof for $\sum_{k=0}^p (-1)^k {n \choose k} = (-1)^p {n-1 \choose p}$ I am trying to give a combinatorial proof for: $$\sum_{k=0}^p (-1)^k {n \choose k} = (-1)^p {n-1 \choose p}$$
Where $p$ and $n$ are natural numbers.
We could easily see that if $p=n$ this reduces to the fact that a set has as many subsets of even cardinality, as those with odd cardinality. However, this formula suggests a relation between the even and odd cardinalities less that $p$.
Note: I'm not interested in an algebraic proof, as Pascal's identity gives a telescopic sum on the RHS.
Thank you in advance.
| The LHS counts subsets of $\{1,2,\dots,n\}$ whose size is at most $p$, with even subsets counted positively and odd subsets counted negatively.
We define a sign reversing involution of such subsets; if $1$ is in the subset, remove it, otherwise, add it in. This involution partitions almost all of the subsets of size at most $p$ into pairs which cancel each other in the sum.
The only unpaired subsets are those with size exactly $p$ which do not contain $1$, as adding $1$ would make a subset of size $p+1$. The number of such subsets is $\binom{n-1}p$, and they are counted with sign $(-1)^p$.
| {
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Proof of a Corollary to Hall's Marriage Theorem I'm trying to prove Corollary 3.3 from this paper (http://www.sfu.ca/~mdevos/notes/graph/345_matchings.pdf) and am lost at the step that says: "Similarly, every vertex in N(X) has degree k, so t is less than or equal to k|N(X)|. It follows that |X| is at most |N(X)|."
How do we know that t is less than or equal to k|N(X)|?
Thanks so much!
| Every vertex in $X$ has at least $k$-neighbors (that are all in $Y$ because the graph is bipartite
So if $|X| \le k$ we are done since $|N(X)| \ge k$
If $|X|=n>k$ then there is at least $n$ vertices in $N(X)$ since if there is less than $n$ vertices then you'll find a vertex in $N(X)$ of degree $> k$, a contradiction.
Thus, $|X| \le |N(X)|$
| {
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Vanishing of rotation of timelike unit vector in numerical relativity I'm a bit confused. I don't think it is that difficult, but still don't manage :-(
So the question is why the rotation of the timelike unit-vector in numerical relativity
$$n_\mu=-\alpha \nabla_\mu t$$
vanishes. $\alpha=\left(-g^{\mu\nu} \nabla_\mu \nabla_\nu t\right)^{-\frac{1}{2}}$ is a normalization constant and $t$ the coordinate time.
The rotation then is $$\nabla_\mu n_\nu - \nabla_\nu n_\mu$$ as far as I expect, which becomes
$$-\alpha \nabla_\mu \nabla_\nu t - (\nabla_\mu \alpha)(\nabla_\nu t) + \alpha \nabla_\nu \nabla_\mu t + (\nabla_\nu \alpha)(\nabla_\mu t) = (\nabla_\nu \alpha)(\nabla_\mu t) - (\nabla_\mu \alpha)(\nabla_\nu t)$$
because $$\nabla_{[\mu}\nabla_{\nu]}t=0 \, .$$ But I think $\alpha$ can depend on the coordinates so it doesn't seem clear why the entire expression should vanish, right?
edit: So to add further context this is basically a question from "Shapiro - Numerical Relativity" (p.27, Eq.2.23) or see also exercise 2.10 where it is to show that the twist $$\omega_{ab} = \gamma_a^c \gamma_b^d \nabla_{[c}n_{d]}$$ vanishes ($\gamma_a^b = \delta_a^b + n_a n^b$ is the projector onto the hypersurface). Shapiro also defines
$$ \omega_a = \alpha \nabla_a t = -n_a$$ and claims $$\omega_{[a}\nabla_b \omega_{c]}=0$$
so it doesn't really matter if I write $n$ or $\omega$. What matters is that either vector is normal to the $t=$const hypersurfaces. I also think the above bracketing must contain a typo, because $[abc]$ doesn't make sense. For the specific case of the SS-metric in isotropic coordinates we have
$$\alpha=\frac{1-M/2r}{1+M/2r}$$
so it does depend on the coordinates.
| Hint:
$$n_{[i} \nabla_j n_{k]} = n_i \nabla_j n_k + n_j \nabla_k n_i + n_k \nabla_i n_j - n_i \nabla_k n_j - n_j \nabla_i n_k - n_k \nabla_j n_i $$
| {
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Find distribution of $Z=\frac{X+Y}{2}$ given $f_{X,Y}(x,y)=e^{-(x+y)}$ Excercise
Let $X, Y$ be random variables such that their joint density function is defined by:
$f_{X,Y}(x,y)=e^{-(x+y)}, \enspace x,y>0$.
Find the distribution of Z defined as:
$Z=\frac{X+Y}{2}$.
Attempt of solution
I am using the fact that:
$F_Z(z)=P\{\frac{X+Y}{2}<z\}$, which implies that:
$F_Z(z)=\int_{0}^{\infty}\int_{0}^{2z-y}e^{-(x+y)}dxdy$.
My problem is that this leads to:
$F_Z(z)=\int_{0}^{\infty}e^{-y}-e^{-2z}dy$,
and this integral does not converge.
What am I getting wrong here? Thanks in advance.
| Note that marginal densities of $X$ and $Y$ are $f_X(x)=e^{-x}$ and $f_Y(y)=e^{-y}$.
Then, $Z=X+Y$ has density according to the convolution:
$$f_Z(z)=\int_0^\infty f_X(x)f_Y(z-x)dx = \int_0^z e^{-x}e^{-(z-x)}dx=ze^{-z}.$$
Density of $W=Z/2$ is
$$f_W(w)=f_Z(2w)|dz/dw| = 4we^{-2w}.$$
| {
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Showing that for $x \ge 7$, $x\# \ge x^2+x$ Let $x\#$ be the primorial of $x$.
I am trying to show that if $x \ge 7$:
$$x\# \ge x^2+x$$
Is there a straight forward argument?
Here's what I came up with:
(1) From Bertrand's Postulate, for any $x$, there exists a prime $p$ such that $x < p < 2x$
(2) Base Case: $7\# = 210 \ge (14)^2 + (14) = 210$
(3) Assume for a prime $p \ge 7$, $p\# \ge (2p)^2 + (2p)$
(4) Let $p+c$ be the lowest prime greater than $p$ so that from step 1 above, $2 \le c < p$
(5) $(2p+2c)^2 + 2(p+c) = (4p^2+2p+2c) + 8pc + 4c^2 < (4p^2 + 4p) + (8p^2 + 8p) + (4p^2 + 4p) < 2p\# + 4p\#+ 2p\# = (2+4+2)p\# < (p+c)p\#$
| The idea is pretty good, but I feel the solution was difficult to read. I will just rephrase your ideas more clearly.
1) Claim: It is enough to prove that $p\#\geq (2p)^2+2p$ for $p$ prime.
Indeed, let $x$ any natural number, then we can find $p_{n}\leq x<p_{n+1}$. By Bertrand postulate, we have $ p_{n+1}<2p_n$. Hence,
$$x\#=p_{n}\#\geq (2p_n)^2+(2p_n)\geq x^2+x.$$
2) We are going to prove the claim by induction:
Case p=7 is easy to check.
Assume hypothesis for $p_n$, let's prove it for $p_{n+1}$:
Recall $p_{n+1}\leq 2p_n$. Then, we have$$ (2p_{n+1})^2+(2p_{n+1})\leq (4p_n)^2+4p_n\leq 4((2p_n)^2+(2p_n))\leq 4p_n\#\leq p_{n+1}\#$$
Which is what we wanted to prove.
| {
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Number of possible chess pairs where order and position matter
Given 11 chess players and 5 distinct tables, in how many ways can we pair them to play (color does matter)?
My problem is that I have found two approaches, both of which give different numbers, and I am not sure what is missing in one or double-counted in the other.
The first approach is to just take any permutation of the $11$ players, as the tables are distinct there are 11 unique spots (one of them is not playing), so we can just place the players according to the permutation. This gives $11!=39916800$ possible games.
The second approach is to first choose pairs, the first player has 10 choices, the next has 8, ... and then we multiply by $2^5$ to account for the colors of each pair, and finally multiply by the number of ways to seat them at tables (so multiply by $5!$), giving
$$10\cdot8\cdot6\cdot4\cdot2\cdot2^5\cdot120=14745600$$
My intuition is that the first approach is correct, but then I am not sure which pairings are missing in the second one..
| The first approach perfectly makes sense.
In the second approach, when we say first player $p$ has $10$ options, we explicitly choose a player and choose it's opponents from $10$ options. But then we skip the case where $p$ does not play at all. This shows there are some cases that are not counted but still does not cover all the cases that are not counted. Missing cases probably come from other assumptions like second chosen player having $8$ options, third chosen player having $6$ options, etc.
| {
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Why we never use the product between vectors like between elements of direct groups product? We can say, that any field $\mathbb{K}$ -- $1$-dim vector space on itself: $\mathbb{K}_{\mathbb{K}}$. So any vector of one another finite-dimensional vector space $V_{\mathbb{K}}$, after choosing the some basis can be represented as the element of isomorphic space $\mathbb{K}_{\mathbb{K}}^{n} = \prod_{i=1}^n\mathbb{K}_i$, where $n$ -- dimension of $V_{\mathbb{K}}$. But we can determine operations on elements of $\mathbb{K}_{\mathbb{K}}^{n}$ like on the direct group product: $(x_1,\dots,x_n) + (x'_1, \dots, x'_n) = (x_1 + x'_1, \dots ,x_n + x'_n)$ and similar for second field operation: $(x_1,\dots,x_n) \times (x'_1, \dots, x'_n) = (x_1 \times x'_1, \dots ,x_n \times x'_n)$.
But usually we doing it only with one field operation $+$. Why?
| On the vector space, there is an addition that does not depend on the choice of basis you make, sometimes called a "natural" addition.
However, you usually can't define a natural multiplication in your way, it would always depend on a basis and different bases will give different multiplications. Therefore, it is mostly not used when dealing with vector spaces.
| {
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Show that the order of an element g is well-defined Suppose $G$ is a group and let $g∈G$, explain why the order of $g$ is well-defined, while the definition of the order is the following:
The smallest positive r such that $g^r=e$, if no such r is found then we say g has infinite order.
My Question:
What strategy should I adopt to check the well-definedness? I know we are essentially checking if the output is unique or not.
My Attempt: Suppose the order of g is finite then consider the set $\{r>0:g^r=e\}$. We know this set is non-empty since $g^k=e$ for some $k$. Then by Well-Ordering Principle there exists smallest such $r$ and hence the order is well-defined.
| Your approach is good, but not correct. You cannt say “Suppose the order of $g$ is finite”, since this already assumes that the order exists. You can consider two cases:
*
*$(\forall k\in\mathbb N):g^k\neq e$: then, by definition, $\operatorname{ord}f=\infty$.
*$g^k=e$ for some natural $k$. Then, by the Well-Ordering Principle, there is a smallest such $k$, in which case $\operatorname{ord}g=k$.
| {
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Alpha-beta quadratic equation Question: The equation $$3x^2-6x-4=0$$ has roots α and β. Find the value of 1/α + 1/β.
I'd just like confirmation on my answer, as I've already found the answer but am not confident in it.
since αβ=c/a and
α+β= -b/a
1/α +1/β= (α+β)/αβ= (-b/a)/(c/a)=2/(-4/3)=-3/2
| Yes, it is correct.
You could do also like this. You have to find $a+b$ if $a,b$ are solution to $$3{1\over x^2}-6{1\over x}-4=0$$that is $$3-6x-4x^2=0$$
so $$a+b =-{-6\over -4}=-{3\over 2}$$
| {
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Prove that there is a differentiable function $f$ such that $[f(x)]^5 + f(x) + x = 0$. Spivak Ch. 12, Q. 14. Prove that there is a differentiable function $f$ such that $[f(x)]^5 + f(x) + x = 0$. (My textbook offers the following hint: Show that $f$ can be expressed as an inverse function.)
My Progress
$f(x) = -[f(x)]^5 - x$
$x = -[f \ (f^{-1} (x) )]^5 - f^{-1}(x)$
$-x^5 - x = f^{-1}(x)$
However, I do not know where to go next. The only thing that comes to mind is that $f$ is one-one since $f^{-1}$ is certainly a function.
Any hints please?
| I think the answer of Alex R. is on track, but I want to make things a bit clearer.
Let $g(y) = y^5 + y.$ Argue from one variable results that $g$ is a bijection of $\mathbb R$ onto $\mathbb R,$ with a differentiable inverse $g^{-1}.$ Define $f(x) = g^{-1}(-x).$ Then $f$ is differentiable on $\mathbb R,$ and we have $g(f(x)) = -x.$ This is the desired result.
| {
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Is the Lagrange multiplier $\lambda$ always positive? In the Lagrange function
$${\mathcal {L}}(x,y,\lambda )=f(x,y)-\lambda g(x,y),$$
is the Lagrange multiplier $\lambda$ term supposed to always be positive or can't it take negative values?
| Geometrically, the Lagrange multiplier method has the following interpretation:
*
*We want a point on the level curve $g(x,y)=0$. This comes from differentiating with respect to $\lambda$ and setting to zero.
*We want that, at the point $p$ under consideration, the level curve of $f(x,y)$ is tangent to the level curve $g(x,y)=0$. This is equivalent to the normal vectors to the level curves, given by the gradients, are parallel. In other words, we require the existence of a scalar $\lambda$ (necessarily different from zero if we assume a minimum of regularity for the level curves) such that
$$\nabla_{p} f = \lambda\nabla_{p} g$$
From condition $2$, it is clear that there is no restriction telling us that $\lambda$ must e positive or negative. On the contrary, with this intuition it is easy to construct examples for both cases.
| {
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Can a nonabelian group $G$ have a normal abelian subgroup $H$ with $[G:H]=3$? If $H$ is a cyclic group, for instance, and $[G:H］=2$, then $G$ is a dihedral group with elements equal to the possible states of a regular polygon upon which rotations and reflections are applied. In this case, the quotient group $G/H$ of order $2$ can be interpreted as the group whose elements are the states "not reflected" and "reflected."
I'm having lots of trouble imagining what it would mean if $H$ were a cyclic normal subgroup of $G$ and $[G:H]=3$, or any odd number, for that matter. For even numbers, however, I can imagine the group whose elements are states of a regular polygon upon which rotations and reflections are applied, and whose vertices change color whenever two reflections are applied. In this case, if $H$ is the subgroup of rotations, then $[G:H］$ could be made to be any even numbers. However, I can't imagine any such thing for odd numbers.
Any hints or advice would be greatly appreciated. For context, I'm doing research about Cayley graphs, and my group theory knowledge isn't as good as it should be. I tried looking through textbooks to find the answer to this question, but I haven't been successful.
| Yes. The Klein subgroup $\{1, (1~2)(3~4), (1~3)(2~4), (1~4)(2~3)\} \triangleleft A_4$. This subgroup is abelian, as the title of the question requests, but it's not cyclic.
| {
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Characterizing a module of Kahler differentials
Consider the $\mathbb C$-algebra $R=\mathbb C[x,y,z]/(z(y^2-x^3)-1)$. How to prove that the module of Kahler differentials $\Omega_{R/\mathbb C}$ of $R$ over $\mathbb C$ is a free $R$-module of rank 2?
This $R$-module is generated by $\{d(f):f\in R\}$ modulo the relations $$d(bb')=bd(b')-d(b)b'\\d(ab+a'b')=ad(b)+a'd(b')$$ for all $a,a'\in \mathbb C,b,b'\in R$ where $$d:R\to \Omega_{R/\mathbb C}$$ is a derivation (a group homomorphism such that $d(fg)=fd(g)+d(f)g$ for all $f,g\in R)$.
| There are two observations which make this calculation go much more smoothly.
*
*First, note that $$R\cong\Bbb{C}\left[x,y,\frac{1}{y^2 - x^3}\right]\cong S^{-1}\Bbb{C}[x,y],$$
where $S = \{1, y^2 - x^3, (y^2 - x^3)^2,\dots\}.$
*Secondly, K\"ahler differentials are compatible with localization: namely, if $B$ is an $A$-algebra and $S\subseteq B$ is a multiplicative subset, then
$$
S^{-1}\Omega^1_{B/A}\cong\Omega^1_{S^{-1}B/A}.
$$
(For a proof, see here.)
Now we can show the claim. Recall that $\Omega^1_{\Bbb{C}[x,y]/\Bbb{C}}\cong\Bbb{C}[x,y]dx\oplus\Bbb{C}[x,y]dy.$ Then if $S = \{1, y^2 - x^3, (y^2 - x^3)^2,\dots\}$ as above,
\begin{align*}
\Omega^1_{R/\Bbb{C}}&\cong \Omega^1_{S^{-1}\Bbb{C}[x,y]/\Bbb{C}}\\
&\cong S^{-1}\Omega^1_{\Bbb{C}[x,y]/\Bbb{C}}\\
&\cong S^{-1}\left(\Bbb{C}[x,y]dx\oplus\Bbb{C}[x,y]dy\right)\\
&\cong \left(S^{-1}\Bbb{C}[x,y]dx\right)\oplus\left(S^{-1}\Bbb{C}[x,y]dy\right)\\
&\cong Rdx\oplus Rdy\\
&\cong R^2.
\end{align*}
Thus, $\Omega^1_{R/\Bbb{C}}$ is a free $R$-module of rank $2.$
| {
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How do I prove this combinatorial identity Show that
$${2n \choose n} + 3{2n-1 \choose n} + 3^2{2n-2 \choose n} + \cdots + 3^n{n \choose n} \\ = {2n+1 \choose n+1} + 2{2n+1 \choose n+2} + 2^2{2n+1 \choose n+3} + \cdots + 2^n{2n+1 \choose 2n+1}$$
One way that I did it was to use the idea of generating functions.
For the left hand side expression, I can find 2 functions. Consider;
$$f_1 (x) = \frac{1}{(1-3x)} \\ = 1 + 3^1x + 3^2x^2 + 3^3x^3 + \cdots + 3^nx^n + \cdots \\ f_2(x) = \frac{1}{(1-x)^{n+1}} \\ = {n \choose n} + {n+1 \choose n}x + {n+2 \choose n}x^2 + \cdots + {2n-1 \choose n}x^{n-1} + {2n \choose n}x^n + \cdots + $$
Consider the coefficient of $x^n$ in the expansion of $f_1 (x) . f_2 (x)$. Then the coefficient will be the expression on the left hand side.
Now we further consider 2 functions for the right-hand side expression.
Consider;
$$f_3 (x) = \frac {1}{(1-2x)} \\ = 1 + 2^1x + 2^2x^2 + \cdots + 2^{n-1}x^{n-1} + 2^nx^n + \cdots \\ f_4 (x) = (1+x)^{2n+1} \\= 1 + {2n+1 \choose 1}x + \cdots + {2n+1 \choose n-1}x^{n-1} + {2n+1 \choose n}x^n +\cdots + {2n+1 \choose 0}x^{2n +1} \\ = {2n+1 \choose 2n+1} + {2n+1 \choose 2n}x + {2n+1 \choose 2n-1}x^2 + \cdots + {2n+1 \choose n+2}x^{n-1} + {2n+1 \choose n+1}x^{n} + \\
+ {2n+1 \choose n}x^{n+1} +\cdots + {2n+1 \choose 0}x^{2n +1}$$
Hence the coefficient of $x^n$ is the coefficient of $x^n$ in the expansion of $f_3(x) . f_4(x)$
This is what I managed to do so far. I'm not sure if $f_1(x) .f_2(x) = f_3(x).f_4(x)$. If the two functions are indeed equal, then I can conclude that their coefficient of $x^n$ must be equal, which will immediately answer the question. If they are equal, how do I show that they are?
If the two functions are not equal? How do I proceed to show this question?
Edit: It might not be true that the product of the two functions are equal. I tried substituting $x=0.1, n=1$. Seems like the two values are not equal. How do I proceed with this question?
| We seek to show that
$$\sum_{q=0}^n {2n-q\choose n} 3^q
= \sum_{q=0}^n {2n+1\choose n+1+q} 2^q.$$
We have for the LHS
$$\sum_{q=0}^n {2n-q\choose n-q} 3^q
= \sum_{q=0}^n 3^q [z^{n-q}] (1+z)^{2n-q}
\\ = [z^n] (1+z)^{2n} \sum_{q=0}^n 3^q z^q (1+z)^{-q}.$$
The coefficient extractor controls the range and we obtain
$$[z^n] (1+z)^{2n} \sum_{q\ge 0} 3^q z^q (1+z)^{-q}
= [z^n] (1+z)^{2n} \frac{1}{1-3z/(1+z)}
\\ = [z^n] (1+z)^{2n+1} \frac{1}{1-2z}.$$
We could conclude at this point by inspection. Continuing anyway we
get for the RHS
$$\sum_{q=0}^n {2n+1\choose n-q} 2^q
= \sum_{q=0}^n 2^q [z^{n-q}] (1+z)^{2n+1}
\\ = [z^n] (1+z)^{2n+1} \sum_{q=0}^n 2^q z^q.$$
The coefficient extractor once more controls the range and we obtain
$$[z^n] (1+z)^{2n+1} \sum_{q\ge 0} 2^q z^q
= [z^n] (1+z)^{2n+1} \frac{1}{1-2z}.$$
The two generating functions are the same and we have equality for LHS
and RHS.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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Matrix-valued function: understanding
$A(x)=(a_{i,j}(x))_{i,j}: \mathbb{R}^n\to \mathbb{R}^{m\times k}$ is a matrix-valued function.
From my understanding it is a function that maps a vector to a matrix. How does this happen? Is the 'function' a matrix of which the elements are functions? If so, are these just $\mathbb{R}^n-\mathbb{R}$-functions?
I don't really understand the form of a matrix-valued function and how it 'takes' the required input. Maybe an example could be helpful?
Thanks a lot.
| As an example: Imagine $x=[x_1,x_2]^T\in \mathbb{R}^2$ Then the function
$$A(x)=
\begin{bmatrix}
x_1x_2 & x_1^2\sin x_2 & x_2 \\
x_1x_2^3 & x_2\exp(x_1) & 2x_2x_1\sin(x_1) \\
\end{bmatrix}$$
is a map from $\mathbb{R}^2\to \mathbb{R}^{2\times3}$. The entries of the matrix are multivariate scalar functions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3201625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Asymptotic distribution of the MLE of $\theta$ such that $\log X_i$ is distributed as $N(\theta, \theta)$ Let $X_1, . . . , X_n$ be a random sample such that $\log X_i$ is distributed as $N(θ, θ),$ $θ > 0$ is unknown.
I've calculated the MLE and I got $$\hat\theta = \frac{-1 + \sqrt{1 + 4n^{-1} \sum_{i=1}^{n} Y_i^2}}{2} $$ where $Y_i = \log X_i$ for $i = 1, ..., n$.
I was wondering what the asymptotic distribution for this MLE would be as $n \rightarrow \infty$ . I've double-checked this MLE and I'm almost certain it is correct but I don't recognise this distribution nor what the sample mean looks like. I've tried the central limit theorem but I'm unsure whether it'd apply here. Thoughts?
| Under certain regularity conditions (like the ones mentioned here on page 1), maximum likelihood estimators have an asymptotic normal distribution. In particular, distributions which are members of the regular exponential family satisfy these conditions.
For $Y_i=\log X_i$, joint density of $Y_1,\ldots,Y_n$ is
\begin{align}
f_{\theta}(y_1,\ldots,y_n)&=\frac{1}{(\sqrt{2\theta\pi})^n}\exp\left[-\frac{1}{2\theta}\sum_{i=1}^n (y_i-\theta)^2\right]
\\&=\frac{1}{(\sqrt{2\theta\pi})^n}\exp\left[-\frac{1}{2\theta}\sum_{i=1}^n y_i^2+\sum_{i=1}^n y_i-\frac{n\theta}{2}\right]\quad,\small (y_1,\ldots,y_n)\in\mathbb R^n,\,\theta>0
\end{align}
This shows that $f_{\theta}$ is a member of a regular one-parameter exponential family. So we can say that the MLE $\hat\theta$ of $\theta$ has an asymptotic normal distribution, given by
$$\sqrt n(\hat\theta-\theta)\stackrel{L}\longrightarrow N\left(0,\frac{1}{I_{Y_1}(\theta)}\right)\,,$$
where $I_{Y_1}(\theta)=-E_{\theta}\left[\frac{\partial^2}{\partial\theta^2}\ln f_{\theta}(Y_1)\right]$ is the information contained in $Y_1$.
A routine calculation gives $I_{Y_1}(\theta)=\frac{2\theta+1}{2\theta^2}$, so that the limiting distribution is eventually
$$\sqrt n(\hat\theta-\theta)\stackrel{L}\longrightarrow N\left(0,\frac{2\theta^2}{2\theta+1}\right)$$
| {
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Finding minimum value of x such that GCD(A+x,B+x) = C where A , B ,C are given I need to add Minimum non-negative Integer such that I can get the desired GCD(a+x,b+x)
Let say A=12 & B=26
For GCD(12+x,26+x) = 1 , x should be 1
For GCD(12+x,26+x) = 2 , x should be 0
For GCD(12+x,26+x) = 7 , x should be 9
For GCD(12+x,26+x) = 14 , x should be 2
| Your question title says you want to find the minimum value of a non-negative integer $x$ such that $\gcd(A + x, B + x) = C$, where $A, B, C$ are given. First, note there's no such integer in certain cases. In particular, since $C$ divides $A + x$ and $B + x$, it must also divide their difference, i.e., $C \mid (A + x) - (B + x) = A - B$. Thus, a minimum requirement is that
$$A \equiv B \pmod C \tag{1}\label{eq1}$$
Next, consider the special case of $A = B$. If $A \le -C$, then $x = -C - A$ works. Otherwise, if $A \le C$, then $x = C - A$ works. Finally, if $A \gt C$, then the gcd would always be $A + x \gt C$, so there are no solutions.
If \eqref{eq1} is satisfied and $A \neq B$, next note $A$ and $B$ can be rewritten as
$$A = mC + r, B = nC + r, \; \text{ where } \; m,n,r \in \mathbb{Z}, \; m \neq n, \; 0 \le r \lt C \tag{2}\label{eq2}$$
Thus, $A + x = mC + (r + x)$, so $C \mid A + x$ requires $C \mid r + x$. In general, this requires
$$x = kC - r, \; k \ge 0 \tag{3}\label{eq3}$$
The smallest non-negative value of $x$ which might work is $0$ if $r = 0$, else it's $C - r$. For $B + x = nC + (r + x)$, it's also true that $C \mid B + x$ in any general case.
The one thing you need to be careful of is that there are no other common factors so the GCD would then be a multiple of $C$. In particular, \eqref{eq3}, gives $A + x = C(m + k)$ and $B + x = C(n + k)$. Thus, you want to find the smallest non-negative integer $k$ such that $x$ from \eqref{eq3} is non-negative and $\gcd(m + k, n + k) = 1$. Note there'll always be such a $k$. To see this, the gcd of $m + k$ and $n + k$ must divide the difference, i.e., $m - n$. Since $m \neq n$, then $\left|m - n\right|$ is either $1$, in which case the gcd is always $1$, or it's a product of $1$ or more primes, i.e., $\left|m - n\right| = \prod_{i = 1}^j p_i^{e_i}$ for some $j \ge 1$, primes $p_i$ and integers $e_i \ge 1$. In the latter case, for each $i$, note that $m \equiv n \equiv r_i \pmod p_i$ for some $0 \le r_i \le p_i - 1$. Since all primes are $\ge 2$, there are always at least $2$ possible values for each $r_i$. For each one, choose any value such its sum with $r_i$ is not congruent to $0$. No $p_i$ would divides $n + k$ or $m + k$. Thus, any prime factors of $\gcd(n + k, m + k)$ must be different than any $p_i$. However, since any prime factors of the gcd must divide the difference, this means there are no prime factors of the gcd, i.e., $\gcd(n + k, m + k) = 1$. For each possible combination of these congruences, the method described to handle more than $2$ numbers section in Extended Euclidean algorithm shows you can always find non-negative integers $k$ satisfying those congruences. Thus, choose the smallest non-negative integer $k$ among all those values so that $x$ given in \eqref{eq3} is non-negative.
To demonstrate how this works, here's how to apply this to your examples with $A = 12$ and $B = 26$. First, as $26 - 12 = 14 = 2 \times 7$, the only possible values of $C$ are $1, 2, 7, 14$, as you've shown. For $C = 1$, since $\gcd(12, 26) = 2$, as $12 = 12 \times 1 + 0$ and $26 = 26 \times 1 + 0$, but $\gcd(12 + 1, 26 + 1) = 1$, you need to use $x = 1$. For $C = 2$, just $x = 0$ works. For $C = 7$, since $12 = 1 \times 7 + 5$, $26 = 3 \times 7 + 5$, and $\gcd(1 + 1, 3 + 1) = 2$, you can't use $x = 7 - 5 = 2$. Instead, since $\gcd(1 + 2, 3 + 2) = 1$, you need to use $x = 14 - 5 = 9$ instead. Finally, for $C = 14$, since $12 = 0 \times 14 + 12$, $26 = 1 \times 14 + 12$ and $\gcd(0 + 1, 1 + 1) = 1$, you can just use $x = 14 - 12 = 2$.
| {
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Logic: "the cubic root of a rational number is also a rational number" I was attempting an online logical and mathematical statements self-test from the University of Toronto and came across the following statement in question 1:
The cubic root of a rational number is also a rational number.
We are asked to select an equivalent statement from the provided list. The answer turned out to be:
If $x$ is a rational number, then $\sqrt[3]{x}$ is a rational number.
Why is that?
The given statement seems to be of the form "A is also a B".
A statement fitting that form would be "a feline is also a mammal". But this wouldn't be equivalent to "if $x$ is a mammal, then $x$ is a feline".
I actually translated the given statement as follows, but it wasn't one of the available options:
$\forall x \in \mathbb{Q}, x = \sqrt[3]{y} \Longrightarrow x \in \mathbb{Q}$
Can someone help me identify the flaws in my reasoning, and why the given answer is the correct one?
| Let's firstly put aside the completely irrelevant question herein of whether the statements are true; obviously, it's not true in general that a rational number's cube root is rational. Let's look into how we rewrite statements with algebra.
While "a feline is also a mammal" becomes "if $x$ is a feline, then $x$ is a mammal", "the cubic root of a rational number is also rational" means "if $y$ is the cube root of a rational $x$, then $y$ is a rational number". Since this time two objects are named in algebra, it's more natural to rearrange the second example as "if $x$ is a rational number and $y=\sqrt[3]{x}$, then $y$ is a rational number". And it's even more natural to only have one named variable, viz. "if $x$ is a rational number, then $\sqrt[3]{x}$ is a rational number".
| {
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"timestamp": "2023-03-29T00:00:00",
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Complex numbers $z=-3-4i$ polar form Let $z=-3-4i$ . In polar form this becomes $[5, 233° ]$. The question then asks for $z^2$, so the polar form becomes $[25,466]$ However in the solution they did $466° -360° $ and I am unsure why they did this. Is it a rule that needs to be applied.
| It's because$$\cos(x-360^\circ)=\cos(x)\text{ and }\sin(x-360^\circ)=\sin(x).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Convergence for power series $\sum_{n=0}^{\infty}z^n(5+e^{i\pi n})^n$? Consider the power series
$$\sum_{n=0}^{\infty}z^n(5+e^{i\pi n})^n \tag{1}.$$
Identifiying $e^{i\pi n}$ as $(-1)^n$ for all integers $n$, highlights the problem with using the ratio test or the root test for convergence of series, since $(-1)^n$ does not converge, right?
How can the convergence be determined for this power series? Does it exist?
Does it exist for other than $z=0$? Can it be determined by setting $(-1)^n=-1$ for all $n$ and therefore calculate the "smallest" radius or something like that?
| Let's set $a_n=(5+(-1)^n)z^n$.
We have $|4z|^n\le|a_n|\le |6z|^n$ so the series converges for sure whenever $|z|<\frac 16$ and diverges for sure whenever $|z|>\frac 14$
What about values in the ring $\frac 16\le |z|\le\frac 14$ ?
To show that $r=\frac 16$ is the actual radius of convergence, we just need to exhibit one value on the border of the convergence disk that makes the series divergent.
Note that it doesn't prevent the series to converge for other values of $z$ in the concerned ring, but the radius of convergence is defined as the larger $r$ such that the series for $|z|<r$ is guaranteed to converge.
So we try with $z=\frac 16$ and the since the series is with all positive terms we can split it to $\displaystyle \sum\limits_{k=0}^{\infty} \left(\frac 46\right)^{2k+1}+\sum\limits_{k=0}^{\infty} \left(\frac 66\right)^{2k}$
The first series is convergence since it is a geometric series with $\left(\frac 46\right)^2<1$ and the second one is trivially divergent since its terms are all equal to $1\not\to0$.
Thus the series for $z=\frac 16$ is divergent, and it fixes the radius of convergence to its lower bound.
| {
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Using knowledge of derivatives, what is $f(x)$ if $f'(x) = 4x + e^x$ with initial condition $f(0)=2$? How to solve this with knowledge of derivatives? Is it asking to solve this without using anti-derivatives?
| Integrate both sides to get $f(x)$ and the equation becomes
$$f(x) = 2x^2 + e^x + c.$$
Put $x=0$ and $f(x=0)=2$ to get $c=1$. Substitute the value of $c$ and get
$$f(x)=2x^2+e^x+1.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Any comprehensive books on global smooth optimization? Can you share any information about books that review most of the existing numerical methods for global minimization of a multivariable objective function? The objective and constraints are assumed to be smooth, I'm not into integer programming of any type at the moment.
I'm aware of many local optimization methods, such as steepest descent, Newton and quasi-Newton, conjugate gradients, trust-region methods, etc. But currently I would like to learn some practical ones used specifically for global minimization. Equality and inequality constraints are assumed. Most information I find comes in form of narrow-focused academic papers that are disconnected from each other.
Thanks!
| Try "Metaheuristics". Simulated Annealing, Genetic Algorithm, Tabu search, Ant colony algorithm, swarm intelligence and so on. There will be many books, but I do not know which book is best.
| {
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Stuck on candy bowl fraction I am really stuck on this problem because I'm not even sure where to start.
Larissa has a bowl of candies. On the first day, she eats 1/2 of the candies plus one more. On the second day, she eats 1/3 of the remaining candies plus one more. On the third day, she eats 1/5 of the remaining candies plus one more. On the fourth day she eats the three remaining candies. How many candies did she have at the start?
| Hints
Let $N$ be the number of candies.
Day 1: she eats $\frac{1}{2}N + 1$: remainder $N_{1}$
Day 2: she eats $\frac{1}{3}N_{1} + 1$: remainder $N_{2}$
Day 3: she eats $\frac{1}{5}N_{2} + 1$: remainder $N_{3}$
Day 4: she eats $3$: $N_{3} = 3$
Can you go from here? (answer $=20$)
Step 1: English $\to$ algebra
Create a table which defined the key variables of $e_{k}$, the amount of candies eaten on day $k$, and $a_{k}$, the amount available at the end of day $k$.
Step 2: Solution criteria
We need to find $a_{0}$, the initial amount of candies, such that $a_{3}=3$.
Step 3: Solution strategy
We all agree that the solution involves expressing the intermediate variables $a_{k}$, and $e_{k}$, $k=1,2,3$ in terms of $a_{0}$.
The are a few ways to do this. We chose a top down approach.
Day 1
$$a_{1} = a_{0} - e_{1} = \tfrac{1}{2}a_{0}-1$$
Day 2
$$e_{2} = \tfrac{1}{3}a_{1} + 1 = \tfrac{1}{6}a_{0}+\tfrac{2}{3}$$
$$a_{2} = a_{1} - e_{2} = \tfrac{1}{3}a_{0}-\tfrac{5}{3}$$
Day 3
$$e_{3} = \tfrac{1}{5}a_{2} + 1 = \tfrac{1}{15}a_{0}+\tfrac{2}{3}$$
$$a_{3} = a_{2} - e_{3} = \tfrac{4}{15}a_{0}-\tfrac{7}{3}$$
Step 4: Solve
$$
a_{3} = 3
$$
implies
$$
\tfrac{4}{15}a_{0}-\tfrac{7}{3} = 3 \quad \implies \quad a_{0} = 20
$$
Hopefully, this is reflected in your class notes.
Check the solution
| {
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limit of the sequence $(x_{n+1})=a^{x_n}$ with $x_1=a$ We are to find the limit of the sequence $(x_{n+1})=a^{x_n}$, where $x_1=a$ for $a>0$.
Attempt:
Case (1): $0<a<1$
We have $\frac{1}{a}=(1+ |r_1|)>1$ for some real $r$.
Hence, $\frac{1}{a^a}=(1+ |r_1|)^a=1+a|r_1|+\frac{a(a-1)}{2}|r_1|^2+...>1$
Evidently, $1+a|r_1|> \frac{1}{a^a}>1$ [$f(x) =1+ax-(1+x)^a \implies f'(x)<0]$. Repeating, $ 1+a^n|r_1|> \displaystyle\frac{1}{a^{a^{a^{{{.}^{{.}^{.}}}}}}}>1$. But, $\lim|a|^n =1$.
$\therefore$ by Sandwich Theorem we have the result $\lim x_{n+1}=1$
Case (2): $a=1$
Proof is evident.
Case (3): $a>1$
$1> (1/a) \implies 1>(1/a)>(1/a)^a>0$. Clearly, $1/x_n>1/x_{n+1}>0$.
But I want to show that $0$ is indeed an infimum or in some other way, that $(x_n)$ diverges to $\infty$ for $a>1$.
Also, consider checking the other cases.
| If the limit $L$ does exist, it is given by the solution of
$$L=a^L\implies L=-\frac{W(-\log (a))}{\log (a)}$$ where appears Lambert function.
However, in the real domain, $W(t)$ requires $t \geq -\frac 1e$ that is to say $a \leq e^{\frac{1}{e}}$
| {
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Does the order of the subgroup generated by two elements divide the product of the element orders? Let $(G, \cdot)$ be a group, $a,b\in G$ such that $\DeclareMathOperator{\ord}{ord}\ord(a),\ord(b)<\infty$.
Do we then have that $\left|\left<a,b\right>\right|$ divides $\ord(a)\ord(b)$? (Where $\left< a,b \right>$ denotes the subgroup generated by $a$ and $b$)
I managed to show this for the case that $a$ and $b$ commute, however I would be interested if this holds even if they don't.
| $S_3=\langle (12), (23)\rangle$.
| {
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Derivation of a well-known property of the standard normal distribution I have found the following property of the standard normal distribution:
$ \int_r^\infty xf(x) dx = f(r) $
where $f(.)$ is the pmf of the standard normal distribution.
$f(r)$ and $F(r)$ are defined as follows:
$f(r) = \cfrac{1}{\sqrt {2π}} e^{-r^2/2} $ ; $ \int_r^\infty f(x) dx $
$ F(.) $ is the cdf of the standard normal distribution
Does anybody knows how this property is derived or where I can find this derivation?
Steven
| It is obtained by just noting that $-xe^{-x^{2}/2}$ is the derivative of $e^{-x^{2}/2}$, so $\int_r^{\infty} xf(x)dx =f(r)$.
| {
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If $r+r’\leq d(a,a’)$ then $B(a,r)\cap B(a’,r’)= \emptyset$. I have to prove this assertion :
Suppose that $r+r’\leq d(a,a’)$
Show that $B(a,r)\cap B(a’,r’)= \emptyset$.
What I have to prove is that if I have $p \in B(a,r)\cap B(a’,r’) $ then $p=0$.
If $p \in B(a,r)\cap B(a’,r’)$ then :
$$d(a,p)<r $$
$$d(a’,p)<r’$$
Then :
$$0\leq d(a,p)+d(a’,p)\leq r+r’ \leq d(a,a’) \leq d(a,p)+d(p,a’)$$
Know I’m stuck, can I Say that $d(a,a’)=d(a,p)+d(a’,p)$ according to the last inequality ? If yes, I don’t know I to prove that $p=0$...
| I assume the balls are open, otherwise the statement is not true.
Let $p\in B(a,r)$, i.e. $d(a,p)\lt r$. Then the triangle inequality states that
$$d(a,p)+d(p,a')\ge d(a,a')\ge r+r'$$
Hence
$$d(p,a')\ge (r-d(a,p))+r'>r'$$
Hence $p\notin B(a',r')$.
Therefore, $B(a,r)\cap B(a',r')=\emptyset$.
Note that it's absolutely not true that you have to show that if $p \in B(a,r)\cap B(a’,r’)$ then $p=0$. If $p=0$ were in $B(a,r)\cap B(a’,r’)$, then obviously the intersection would not be empty.
| {
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A consequence of Riemann-Lebesgue lemma Riemann Lebesgue lemma states that: For a function $f\in L_1([-\pi,\pi])$ we have
\begin{align}
\lim_{|n|\to \infty} \int_{[-\pi,\pi]} f(t) e^{-int} \text{ d}t =0.
\end{align}
The integral here is Lebesuge integral. As consequences of this lemma,
\begin{align}
\lim_{|n|\to \infty} \int_{[-\pi,\pi]} f(t) \cos(nt) \text{ d}t =0 ,
\end{align}
and
\begin{align}
\lim_{|n|\to \infty} \int_{[-\pi,\pi]} f(t) \sin(nt) \text{ d}t = 0.
\end{align}
How to prove these two corollaries. I need any help.
Thanks in advance.
| Assuming $\lim_{|n|\rightarrow\infty}\int f e^{inx}dx = 0$ gives
$$
\lim_{n\rightarrow\infty}\int f e^{inx}dx = 0 \\
\lim_{n\rightarrow\infty}\int f e^{-inx}dx = 0
$$
Adding these results in
$$
\lim_{n\rightarrow\infty}\int f \cos(nx)dx = 0
$$
I'll let you tackle the other combinations.
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Geometric interpretation of an elliptic point on a Riemann surface / hyperbolic surface Let $\Gamma$ be a Fuchsian group of signature $[g;m_1,\dots,m_r;s]$. When we quotient $\mathbb{H}^2$ by $\Gamma$ we obtain a genus $g$ surface with $s$ cusps and $r$ elliptic points with orders $m_1,\dots,m_r$.
Now, the concepts of genus and cusp have a geometric description i.e handles/holes and well, cusps. What is the equivalent description of an elliptic point and what does it look like?
| They are sometimes called cone points, because they look like taking a piece of paper with a corner of angle $2\pi/m_i$ and folding it over to identify opposite sides of the vertex. This produces something which looks like a cone at the singular point. They are also sometimes called pillowcase points as they look like the corners of a pillow.
So, for example, if $m_i=2$ then take a piece of paper, pick a point on one side, and identify the sides opposite the point to make a cone. If $m_i=3$ take a piece with a $120^\circ$ angle and identify the opposite sides. The cone gets "sharper" as the angle decreases, i.e. as $m_i$ increases.
| {
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Finding a pattern, I'm stuck I have a friend who knows how much I love math, (I imagine new problems just to do the math behind them and to see if I can expand my understanding) and so he brings me the stuff that stumps him. Usually I end up finding the answer for him and then explaining how to him, but he gave me one 2 weeks ago and I cant figure it out. It's part of some math for calculating the area (or maybe perimeter?) of an elipse, the part he wanted me to figure out is: there is a part that goes π * (a + b) * (1 + 1/4 * h + 1/64 * h^2 + 1/256 * h^3 + 1/16384 * h^4 + 1/ 65536 * h^5 + 1/1048576 * h^6 ...) we only have those six, and he wanted me to find the pattern to calculate the rest of the denominators to extend it. But everything I have tried adds extra values within the first six or skips values within the first six. Does anyone know what the pattern is and what formula can be used to calculate the rest?
| If I could give some advice, these kinds of questions generally benefit from a look at the Online Encyclopedia of Integer Sequences.
Considering it has to do with ellipses, the sequence of denominators should be A056982.
| {
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evaluating $\int_{-\infty}^{\infty}{\frac{x\sin(2x)}{x^2+4}}\mathrm dx$ I want to calculate $\int_{-\infty}^{\infty}{\frac{x\sin(2x)}{x^2+4}}\mathrm dx$. Now I think the answer is $\pi/e^4$. Now the poles are cleary $2i, -2i$. When I try to then use the residue theorem I get that the integral is $0$, which I think is wrong.
What am I doing wrong exactly, can I not use the residue theorem here?
| The trick to apply contours in this case is to write $\sin 2x = \frac1{2i} e^{2ix} + \frac1{2i} e^{-2ix} $ and do two separate integrals.
For the former, we have a contour going from $-\infty$ to $\infty$ along the $x$ axis, then returning n a large semicircle in the upper half plane. The semicircle contributes zero because until the imaginary part is much bigger than the real part, the $\frac1 x$ behavior of $\frac{x}{x^2+4}$ will make the integral arbitrarily small, while after the imaginary part is much bigger than the real part, the exponential fall-off of $e^{2iz}$ as the imaginary part of $z$ grows positive will dominate over the linear growth of the circumference length. This contour encloses just the pole at $z=2i$ so this integral is
$$
\left. 2\pi i \frac{z\frac1{2i} e^{2iz}}{z+2i} \right|_{z = 2i} = 2\pi i\frac{(2i)\frac1{2i} e^{2i(2i)}}{(2i)+2i} = \frac{2\pi i e^{-4}}{4i}=\frac12\pi e^{-4}
$$
For the $-\frac12 e^{-2ix} $ term, similar arguments have the contour returning in the lower half plane. However, if we go from $-\infty$ to $+\infty$ along the real axis, we will have a clockwise path; so we will have to take the negative of the residue. This term then gives
$$
\left. -2\pi i \frac{z\frac{-1}{2i} e^{-2iz}}{z-2i} \right|_{z = -2i} = -2\pi i\frac{(-2i)\frac{-1}{2i} e^{-2i(-2i)}}{(-2i)-2i} = -\frac{2\pi i e^{-4}}{-4i}=\frac12\pi e^{-4}
$$
and we can then add those two terms giving $\pi e^{-4}$.
Your mistake may have been not to noice that in the lower half plane, if you go from $-\infty$ to $+\infty$ along the real axis, the direction of integration is not counterclockwise.
| {
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Classification of surfaces The Classification Theorem for surfaces says that a compact connected surface $M$ is homeomorphic to $$S^2\# (\#_{g}T^2)\# (\#_{b} D^2)\# (\#_{c} \mathbb{R}P^2),$$ so $g$ is the genus of the surface, $b$ the number of boundary components and $c$ the number of projective planes.
From there, it is easy to compute $\chi(M)=2-2g-b-c$.
Nevertheless, I have read another statement of The Classification Theorem that states that a compact connected surface is determined by its orientability (yes/no), the number of boundary components and its Euler characteristic.
I do not understand how is it possible to know the decomposition of $M$ as a connected sum by knowing that. By knowing $b$, there are still two variables, $c$ and $g$ which have to be known from $\chi(M)$, and orientability only tells us if $c=0$ or $c\geq 1$. Can someone help me, please?
| If I read correctly you are trying to determine $c$ and $g$, given a compact connected surface $M$ for which you know $b$ the number of boundary components, $\chi(M)$ the Euler characteristic the and whether or not $M$ is orientable.
This can't be done in a unique manner, as the connected sum of a tori and a projective plan is homeomorphic to the connected sum of three projective planes.
Please correct me if I misunderstood your question.
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Solving polynominals equations (relationship of roots)
The roots of $x^3-4x^2+x+6$ are $\alpha$, $\beta$, and $\omega$.
Find (evaluate):
$$\frac{\alpha+\beta}{\omega}+\frac{\alpha+\omega}{\beta}+\frac{\beta+\omega}{\alpha}$$
So far I have found:
$$\alpha+\beta+\omega=\frac{-b}{a} = 4 \\
\alpha\beta+\beta\omega+\alpha\omega=\frac{c}{a} = 1 \\
\alpha×\beta×\omega=\frac{-d}{a} = -6$$
And evaluated the above fractions creating
$$\frac{\alpha^2\beta+\alpha\beta^2+\alpha^2\omega+\alpha\omega^2+\beta^2\omega+\beta\omega^2}{\alpha\beta\omega}$$
I don't know how to continue evaluating the question.
Note:
The answer I have been given is $-\dfrac{11}{3}$
| Hint: We can write $$\frac{4-w}{w}+\frac{4-\beta}{\beta}+\frac{4-\alpha}{\alpha}$$ and this is $$4\left(\frac{\alpha\beta+\alpha w+w\beta}{\alpha \beta w}\right)-3$$ and this is $$-\frac{2}{3}\left(1-\beta w-\alpha w+\alpha w+\beta w\right)$$
This simplifies to $$-\frac{2}{3}-3=-\frac{11}{3}$$
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Number of ways to pick a team of 4 with at least 1 girl and 1 boy from 4 girls and 4 boys. To find the number of ways to pick a team of $4$ with at least $1$ girl and $1$ boy from $4$ girls and $4$ boys, I thought to manually assign the team with $1$ girl and $1$ boy to begin with.
$$G, \quad B, \quad G \textrm{ or } B, \quad G \textrm{ or } B$$
The number of ways to pick $1$ girl and $1$ boy from $4$ girls and $4$ boys is
$${4 \choose 1} \times {4 \choose 1}$$
The number of ways to pick the $2$ remaining team members from $6$ people is
$${6 \choose 2}$$
Using this logic the number of teams with at least $1$ girl and $1$ boy is $4 \times 4 \times 15 = 240$, but the total number of teams with no restrictions is ${8 \choose 4} = 70$, so there is definitely something wrong with my thinking.
How do I arrive to the correct answer with this line of reasoning. I already know you can use the fact that it is $70 - (\textrm{number of teams with all boys or all girls}) = 70 - 2 = 68$, but I would like to understand how to do it the way I thought of.
| A team of four with at least one boy and at least one girl will have either three boys and one girl, two boys and two girls, or one boy and three girls. Hence, the number of admissible teams is
$$\binom{4}{3}\binom{4}{1} + \binom{4}{2}\binom{4}{2} + \binom{4}{1}\binom{4}{3} = 68$$
which agrees with the answer you obtained by subtracting those teams composed only of boys or only of girls from the total number of teams that could be formed.
Why was your attempt wrong?
In designating a particular boy and a particular girl as the boy and girl who must be on the team, you counted each team with more than one boy or more than one girl multiple times. In particular, you counted each team with three boys and one girl three times, once for each way of designating one of the boys as the boy on the team.
\begin{array}{c c}
\text{designated boy} & \text{designated girl} & \text{additional boys}\\ \hline
\text{Alexander} & \text{Barbara} & \text{Clifford, David}\\
\text{Clifford} & \text{Barbara} & \text{Alexander, David}\\
\text{David} & \text{Barbara} & \text{Alexander, Clifford}
\end{array}
By symmetry, you also counted each team with one boy and three girls three times, once for each way you could have designated one of the girls as the girl on the team.
You counted each team with two boys and two girls four times, once for each of the two ways you could have designated one of the boys as the boy on the team and once for each of the two ways you could have designated one of the girls as the girl on the team.
\begin{array}{c c}
\text{designated boy} & \text{designated girl} & \text{additional children}\\ \hline
\text{Alexander} & \text{Barbara} & \text{Claire, David}\\
\text{David} & \text{Barbara} & \text{Alexander, Claire}\\
\text{Alexander} & \text{Claire} & \text{Barbara, David}\\
\text{David} & \text{Claire} & \text{Alexander, Barbara}
\end{array}
Notice that
$$\color{red}{\binom{3}{1}}\binom{4}{3}\binom{4}{1} + \color{red}{\binom{2}{1}\binom{2}{1}}\binom{4}{2}\binom{4}{2} + \color{red}{\binom{3}{1}}{\binom{4}{1}\binom{4}{3}} = \color{red}{240}$$
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Why isn't the definition of absolute value applied when squaring a radical containing a variable? I recently learned about the following definition of absolute value:
$|a| = \sqrt{a^2}$
Then I came across a solution to a problem that had the following step:
$5 \geq \sqrt{5 - x}$
In order to proceed, we had to square both sides:
$5^2 \geq (\sqrt{5 - x})^2$
With the aforementioned definition of absolute value in mind, I wrote:
$25 \geq |5 - x|$
But the actual solution turned out to be:
$25 \geq 5 - x$
I don't understand why the absolute value definition wasn't applied here. Can anyone tell me why?
| From the fact that you can take $\sqrt {5-x}$ you know that $5-x \ge 0$ so you don't need the absolute value signs.
| {
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What is the probability there is no prime between $n$ and $n+\ln(n)$? Consequences of the Prime Number Theorem tell us the probability of $n$ being prime is $1/\ln(n)$. This also means that the number of expected primes between $n$ and $n+\ln(n)$ is close to $1$, but not always. What is the probability there is no prime between $n$ and $n+\ln(n)$?
| This is currently a question we cannot answer unconditionally; however, there is a heuristic answer, which we can prove if we assume another result.
The naive probabilistic model is that each integer $k$ between $n$ and $n+\log n$ independently has a probability $1/\log k \sim 1/\log n$ of being prime. According to this model, the probability that none of these integers are prime (as alluded to in Sil's comment) is
$$
\prod_{n<k<n+\log n} \bigg( 1-\frac1{\log n} \bigg) \sim \bigg( 1-\frac1{\log n} \bigg)^{\log n} \sim e^{-1}.
$$
But this model actually gives something more: for any nonnegative integer $m$, the probability that there are exactly $m$ primes between $n$ and $n+\log n$ is $e^{-1}/m!$. In other words, intervals of this length should give rise to a Poisson distribution. (And this can be generalized to the primes between $n$ and $n+C\log n$ for any constant $C>0$.)
In 1976, Gallagher proved that this is indeed the case if you assume a suitably strong version of the Hardy–Littlewood prime $k$-tuples conjecture. This paper of Goldston and Ledoan describes the result more precisely on its first page and gives the exact reference.
One can criticize the naive model by noting that adjacent integers can never both be prime, nor can three consecutive integers, etc. But a more advanced probabilistic model that takes divisibility by small primes into account eventually leads to the same heuristic answer. And Gallagher's result shows that the naive heuristic is correct on average on this scale (I doubt it would be on significantly smaller scales, for instance).
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Big $O$ small $o$ notation True or False I'm stuck on a few big-$O$ and small $o$ notation True or False questions, any insight would be appreciated.
$4n^3+6n+17=O(n^4)$ True. The degree of complexity is directly related to n, but it is not represented there. The function grows much faster (left side is like n^2, since it grows faster).
$4n^3+6n+17=O(n^2)$ I think this is false, but no idea why
$n^{17}=o(2^n)$ True. if n^17 increase by 1, an increase is no more n^18, whereas if 2^n increase by 1, an incease is double: 2^n+1. Somewhere 2^n begans to grow faster, because small o notion includes funkction, that grows only slower like in this case.
$5^n=o(3^n)$
False. if lim n->inf (5^n / 3^n) = 0. Since(5^n / 3^n) = (5/3)^n and since for any 0 <= x < 1, lim n->inf (x^n) = 0, is false 5^n = o(3^n).
Is here anything correct?
| [When using this notation, it is important to specify which limit you are thinking about. I assume from your attempted solutions that you mean $n\to\infty$.]
You have got the correct answers. For the second one, you're right that it is false. How can you prove it? Suppose it's true. $$4n^3+6n+17=O(n^2)$$ Then $\exists M$ such that for sufficiently large $n$, $$4n^3+6n+17\le Mn^2\\4n+\frac6n+\frac{17}{n^2}\le M$$ Clearly LHS grows unboundedly, so this is not true.
| {
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Reference Request: Convergence of Series over Integer Lattice I am writing a school paper on Modular forms, and I want to use the following result without including a proof, since it is not directly related to my main topic.
For $s>0$, $d \in \mathbb{N}$, the series $\sum\limits_{v\in\mathbb{Z}^d\setminus\{0\}} \frac{1}{{\left \| v \right \|}^s}$ converges iff $s>d$.
However I am struggling to find a good reference for this to include in my paper. Can anyone point me to a book/paper that proves this?
| It is enough to prove the convergence of the sum over tuples of nonzero integers.
$$
\sum_{v\in (\mathbb{Z}-\{0\})^d} \frac1{\|v\|^s} \ \ \ \tag{1}
$$
Apply the AM-GM inequality, we have for $v=(v_1,v_2,\ldots, v_d)$,
$$
\frac1{\|v\|^2} \leq \frac d{(v_1^2v_2^2\cdots v_d^2)^{1/d} }
$$
The $s/2$-th powers of these satisfy
$$
\frac1{\|v\|^s}\leq \frac{d^{s/2}}{|v_1v_2\cdots v_d|^{s/d}}
$$
If $s>d$, then the sum $(1)$ converges by comparison test.
| {
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Convergence of series of quotients of gamma functions I am interested in determining if the following series converges:
$$\sum_{k=0}^{\infty} \frac{\Gamma(k+2\delta)}{\Gamma(k+1)} \;,$$
where $\delta > 0$ is a real number and the gamma function is defined by $$\Gamma(\alpha) = \int_0^{\infty} x^{\alpha-1} e^{-x} \, dx \;,$$ for $\alpha > 0$. Any thoughts on this would be much appreciated.
| For $\delta>0$ and $k\geq1$ we have $\Gamma(k+2\delta)>\Gamma(k)>0$ and $\Gamma(k+1)>0$, so
$$\frac{\Gamma(k+2\delta)}{\Gamma(k+1)}>\frac{\Gamma(k)}{\Gamma(k+1)}=\frac1k.$$
It follows that the series diverges because the harmonic series diverges.
| {
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$(b_n) \rightarrow b$ then $(1/b_n) \rightarrow 1/b$ proof. Let $(b_n)$ be a sequence that converges to $b$. Show that $(1/b_n)$ converges to $1/b$ when $b_n \neq 0$ for all $n$ and $b \neq 0$.
First see that
$$
|\frac{1}{b_n} - \frac{1}{b}| = |\frac{b - b_n}{b_nb}| = \frac{|b_n - b|}{|b_n||b|}
$$
Now from convergence of $(b_n)$ it follows that $\forall \epsilon \geq 0$ there is $N \in \mathbb{N}^{+}$ such that $|b_n - b| \leq \epsilon$ when $n \geq N$.
That implies that $\forall \epsilon \geq 0$ when $n \geq N$, it is true that
$$
\frac{|b_n - b|}{|b_n||b|} \leq \frac{\epsilon}{|b_n||b|} = \epsilon'
$$
Clearly given any $\epsilon$ we have a positive $\epsilon'$, therefore for any
$\epsilon' \geq 0$ it follows that there is $N \in \mathbb{N}^{+}$ such that
$$
|\frac{1}{b_n} - \frac{1}{b}| \leq \epsilon'
$$
Hence $(1/b_n) \rightarrow 1/b$.
Correction based on yousef magableh answer:
I didn't notice that my $\epsilon'$ was defined in terms of $n$. To avoid that suppose that $b > 0$. Let's pick an $\epsilon = 2b$. Hence that will exists $N_1 \in \mathbb{N}^{+}$ that guarantees that:
$$
|b_n-b| \leq 2b \rightarrow -3b \leq -b \leq b_n \leq 3b \rightarrow |b_n| \leq 3b
$$
Therefore, for any positive $\epsilon$, there is a positive $\epsilon'$ such that:
$$
\frac{|b_n - b|}{|b_n||b|} \leq \frac{\epsilon}{|b_n||b|} = \frac{\epsilon}{3b|b|} = \epsilon'
$$
holds for $n \geq N_1$.
If $b < 0$ just pick $\epsilon = -2b$.
Is it correct? If so, what would you do in order to improve it? Thanks.
| I would recommend not assuming anything is positive. It's fine to do that, but is very important to be able to work with absolute values carefully. And absolute value inequalities specifically.
You need to be sure to know when you want a lower vs upper bound. Often the game is bounding something by a series of upper bounds: $<...<...<...$ etc. However to bound something in a denominator, we will need a lower bound on it in order to keep the overall flow of upper bounds. Also we will need such a lower bound (on the thing on the denominator) to be positive, so we stay away from zero.
$$0<a<x<b ~ \text{ implies } ~ 0<\frac 1b < \frac 1x <\frac 1a$$
So a lower bound on $x$ is transformed into an upper bound on $\frac1x$.
Start with $\epsilon=|b|/2$ as already suggested. This will allow us to keep things away from zero.
Then $|b_n-b|<|b|/2$ implies $b-|b|/2<b_n<b+|b|/2$ so $|b_n|>m=\min(|b\pm|b|/2|)$ which we can be certain is positive by the construction. This may seem a bit hard to follow. If so, plug in numbers and try a few examples with specific sequence formula like $1-\frac1n=b_n$ or $-1+\frac1n=b_n$.
Then for any arbitrary $\epsilon$, choose $N$ large enough so that $n>N$ implies both $|b_n-b|<\epsilon m |b|$ and $|b_n-b|< |b|/2$.
Then we have
$$
\frac{|b_n-b|}{|b_n||b|}<\frac{\epsilon m |b|}{|b_n||b|}=
\frac{\epsilon m}{|b_n|}<\frac{\epsilon m}{m}=\epsilon.$$
Of course there are many ways to do this kind of problem, and this is only one possibility.
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What is the largest square that can fit in a dodecagon and not rotate? I am an engineer in the oil field and I am trying to find a socket that can fit around a square drive pin. I am trying to prove this for fun before I just draw it on CAD and measure it. I believe sockets have six or twelve sides. I have had some success calculating the largest hexagonal wrench that can fit on a square drive, but it came out very large and I remember that it was not this large of a tool when I worked in the field. The problem is that when you add this extra point in a hexagon making it a decagon, it allows the square to rotate. The square drive is d=.750 and I came out with x=(sqrt(5)/(3-sqrt(3))d, where x is the width across two sides of a hexagon. The proof of the hexagon starts with the fact that the largest square inside of a hexagon shares the same center. Its a strange proof that I dont understand, but it can be found here: http://www.drking.org.uk/hexagons/misc/deriv3.html.
This web page gives us the result that (1) d=(3-sqrt(3))a. Then I derived (2) a=x/sqrt(5) using the Pythagorean therom. From there, I substituted (2) into (1). and got the result x~1.764d, x~1.323 when d=.750.
I know this cant be right. I could calculate the diagonals of the square drive and set that equal to the diagonals of the dodecagon. From there I could probably find the distance between two sides of the dodecagon and that would be a starting point, but I'd rather find a relation and proof for largest square inside a dodecagon that cannot rotate. I know aligning the square diagonals with the dodecagon diagonals would not be correct because I do not think this is how you'd want to start this proof.
| Wouldn't it be easier with a regular dodecagon, since 4 divides evenly into 12?
The largest square which will not rotate will fit snugly into four corners of the dodecagon.
Note that in $\triangle AOM$, side $OM$ must be the largest side. Since $\angle MOA<30^\circ$ then $\angle MAO>75^\circ$.
| {
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Minimum value of $4$-digit number divided by sum of its digits? If x was a positive 4 digit number and you divide it by the sum of its digits to get the smallest value possible, what is the value of x? For example (1234 = 1234/10)
I got 1099 as my answer however I don't know if this is right or how to prove it.
| Let $x$ be the 4-digit number ending in digit not equal to 9 with the sum of digits equal to $s$. If you increase the last digit by one, the number becomes $x+1$ and the sum of digits becomes $s+1$.
You can easily prove that:
$$\frac xs>\frac{x+1}{s+1}$$
This is true because $x>s$. So to reduce the ratio as much as possible, the last digit has to be as big as possible, which is $9$.
Now consider number $x$ with the third digit not eqaul to 9. Increase the third digit by one and the number becomes $x+10$ and the sum increases by 1. Again, you can show that:
$$\frac xs>\frac{x+10}{s+1}$$
This is true because $x>10s$. So the third digit also has to be as big as possibe. So the last two digits must be equal to 9.
Consider now the first digit and decrease it by one. The number becomes $x-1000$ and the sum becomes $s-1$.
$$\frac xs>\frac{x-1000}{s-1}$$
This is true because it is equivalent to:
$$x<1000s$$
...and you know that $s$ is certainly greater than 18 (the last two digits must be 9). So the first digit has to be as small as possible i.e. 1 and the number is of the form:
$$1a99$$
The ratio you want to minimize now becomes:
$$\frac{100a+1099}{a+19}=100-\frac{801}{a+19}$$
Minimum value is reached for minumum value of $a$ which is 0. So the solution is: 1099.
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"source": "stackexchange",
"question_score": "9",
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Can a function be of class C1 even if its partial derivatives are not continuous? I know that if all partial derivatives of a function f exist and are continuous then the function is said to be of class C1 (continuously differentiable).
However, I was not able to find whether this is a necessary or a sufficient condition. What I mean is: Could a function be of class C1 even though its partial derivatives are not continuous?
PS. Can a function be continuous/differentiable even if its partial derivatives are not continuous?
Thank you!
| No. A function is of class $C^{1}$ if and only if it has continuous partial derivatives. You can refer to Rudin's book for a proof.
| {
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Show that the limit $\lim_{z \to i}{\frac{1-|z|}{i-z}}$ does not exist. Show that the limit $\lim_{z \to i}{\frac{1-|z|}{i-z}}$ does not exist.
I tried by putting $z =ix $, where $x \to 1$ and got that the limit is $\frac{1}{i}$, but can't think of another example.
| Hint:
$$\lim_{z \to i}{\frac{1-|z|}{i-z}}= \lim_{s \to 0}{\frac{|s+i|-1}{s}}.$$
Choose $s$ real number then the limit will be 0.
| {
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$\lim_{n \to \infty }n\int_{0}^{\pi}\left \{ x \right \}^{n}dx$ Calculate
$$\lim_{n \to \infty }n\int_{0}^{\pi}\left \{ x \right \}^{n}dx$$
My try: $x-1<[x]<x$
$-x<-[x]<1-x$
$0<x-[x]<1$ then I applied an integral but I get the result pi which is the wrong answer.
| Break the integral into four parts; from $0$ to $1$, from $1$ to $2$, from $2$ to $3$ and from $3$ to $\pi$. Then $\{x\}=x,\ \{x\}=x-1,\ \{x\}=x-2,\ \{x\}=x-3$ in those intervals respectively. Then, you can easily calculate the three integrals to find that
$$n\int_{0}^{\pi}\{x\}^ndx=\frac{3n}{n+1}+\frac{n(\pi-1)^{n+1}}{n+1}.$$
Now you can take it from here. Note that $0<\pi-3<1,$ so that $(\pi-1)^{n+1}\rightarrow 0.$
With time delay. I apologize for that, because all the answers here are the same.
| {
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Calculating tangent line at $t=2$
Calculate the tangent line at $t=2$ of the curve
$$x(t)=t^2+2t+4,\ y(t)=1+te^t$$
How would you go about determining this? So far I have
$$x'(t)=2t + 2\\
y'(t) = te^t+e^t$$
Where do I go from here to get the tangent line?
| The following expression is how you can find the slope of a parametric curve:
$$
\frac{dy}{dx}=\frac{\left(\frac{dy}{dt}\right)}{\left(\frac{dx}{dt}\right)},\ \frac{dx}{dt}\ne0.
$$
When you have found the slope of the curve at $t=2$, use the equation of a straight line to form the actual line:
$$
y-y_0=m(x-x_0).
$$
Use your original parametric equation to find the point $(x_0,y_0)$ at $t=2$.
| {
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How can $\pi$ be defined as a surreal number. I want to express $\pi$ in the Surreal Number notation $\{L|R\}$. What is the most natural or intuitive way of doing so, seeing as there are many (possibly infinite) ways of expressing the same surreal number.
| If you want to have a more explicit expression, start with the largest integer below $\pi$, that is $a_0=3$, and then iterate the following algorithm:
*
*If $a_n<\pi$, it goes into the left set, and $a_{n+1} = a_n + 1/2^{n+1}$.
*If $a_n>\pi$, it goes to the right set, and $a_{n+1} = a_n - 1/2^{n+1}$.
This gives
$$\pi = \{3, 3.125, 3.140625, \ldots |\, 3.5, 3.25, 3.1875, 3.15625, \ldots\}$$
| {
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Show that if $\alpha \vDash \beta$ and $\beta \vDash \alpha$, then $\vDash (\alpha \leftrightarrow \beta)$ I would like to show that if $\alpha \vDash \beta$ and $\beta \vDash \alpha$, then $\vDash (\alpha \leftrightarrow \beta)$, and I'm thinking that the argument below fails.
Since $\alpha \vDash \beta$ and $\beta \vDash \alpha$, we have for every truth assignment $v$ that $\overline{v}(\alpha) = \overline{v}(\beta) = T$. Therefore, $\overline{v}(\alpha \leftrightarrow \beta) = T$.
However, I think it would be incorrect to conclude that $\vDash(\alpha \leftrightarrow \beta)$, because I have only shown that $v$ satisfies $\alpha \leftrightarrow \beta$ if $v$ also satisfies $\alpha$ and $\beta$.
Thoughts?
| In your attempt, the problem is with the line
Since $\alpha \vDash \beta$ and $\beta \vDash \alpha$, we have for every truth assignment $v$ that $\overline{v}(\alpha) = \overline{v}(\beta) = T$.
We do not have $v(\alpha) = v(\beta) = T$ for every $v$. We only know If $v(\alpha) = T$ then $v(\beta) = T$ and vice versa. As you notice, this excludes the case where an assignment does not satisfy $\alpha$ or $\beta$.
Instead, we need to consider two cases:
Consider an arbitrary assignment $v$.
Case 1: $v(\alpha) = T$. Then by $\alpha \vDash \beta$, we also have $v(\beta) = T$, since for all $v$, if $v(\alpha) = T$ then $v(\beta) = T$.
Case 2: $v(\alpha) = F$. Then by $\beta \vDash \alpha$, $v(\beta) = F$, since otherwise there would be an assignment $v$ s.t. $v(\beta) = T$ but $v(\alpha) = F$ which contradicts the premise that if for all $v$, if $v(\beta) = T$ then $v(\alpha) = T$.
So in any of the cases, $v(\alpha) = v(\beta)$, thus by definition of $\leftrightarrow$, $v(\alpha \leftrightarrow \beta) = T$. Since $v$ arbitrary, the above holds for all assignments $v$, therefore $\vDash \alpha \leftrightarrow \beta$.
(The proof could be shortened a bit; i.m.o. the "since"-parts in the two cases are rather self-evident.)
Another possiblity would be to use the import-export theorem (sometimes referred to as the deduction theorem) which states that
$$A \vDash B\ \Longleftrightarrow\ \vDash A \to B$$
With that, we have
$$\alpha \vDash \beta \text{ and } \beta \vDash \alpha$$
$$\Longrightarrow\ \vDash \alpha \to \beta \text{ and } \vDash \beta \to \alpha$$
i.e.
$$\text{"For all valuations $v$, $v(\alpha \to \beta) = T$ and $v(\beta \to \alpha) = T$"}$$
from which
$$\vDash \alpha \leftrightarrow \beta$$
follows more or less immediately by definition of $\to$ and $\leftrightarrow$.
| {
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Prove that if $f$ is not surjective, then $f$ is homotopic to a constant via a homotopy that fixes a point I'm trying to prove that if $f:S^1 \rightarrow S^1$ is not surjective, then $f$ is homotopic to a constant function via a homotopy that fixes a point $\theta \in S^1$. Showing that it is homotopic to a constant function is simple, but showing that there exists a homotopy that fixes a point is proving to be a bit tricky... Is showing that $f$ must have a fixed point enough?
I can extend $f$ to a map on the disk $g:D^2 \rightarrow S^1$. If $i:S^1 \rightarrow D^2$ is the inclusion mapping, then $i \circ g: D^2 \rightarrow D^2$ is a map on the disk that must have a fixed point, so that $g(\theta) = \theta$ for some $\theta \in D^2$. This was my idea at a proof, but I fail to see how it has a connection, if any, to a homotopy...
| More generally consider $f:S^n\to S^n$ and let $P\in S^n$ be such that $P\not\in f(S^n)$. For any $\theta\in S^n$ we have a homotopy
$$H:I\times S^n\to S^n$$
$$H(t, s)=\pi^{-1}\big(t\cdot \pi(f(s))+(1-t)\cdot \pi(f(\theta))\big)$$
where $\pi:S^n\backslash\{P\}\to\mathbb{R}^n$ is the stereographic projection which is a homeomorphism. Note that $H$ is well defined, continuous and we have
$$H(0, s)=f(\theta)$$
$$H(1, s)=f(s)$$
$$H(t,\theta)=f(\theta)$$
Finally since $\theta$ was arbitrary then all you need now is a fixed point $f(\theta)=\theta$. And the existance of such point follows from the Brouwer's fixed point theorem as you've mentioned yourself.
| {
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Use derivative to find maxima, minima of the function : $n^\frac 1n: n \in \mathbb{N}$ This post is in ref. to my earlier post's row #10.
*$\{n^\frac 1n: n \in \mathbb{N} \}$: Domain of values is in the set of naturals.
I am unable to find minimum, maximum (in range), and list a few values below:
$$\begin{array}{c|c|} & \text{$n\in \mathbb{N}$}& \text{$n^{\frac1n}$}\\ \hline
a & 1& 1\\ \hline
b & 2& \sqrt{2}\\ \hline
c & 3& 3^{\frac13}\\ \hline
d & 4& 4^{\frac14}\\ \hline
\end{array}$$
The max. /min. value in range of function is unknown by me, hence it needs finding derivative (both first & second).
The eqn. would be :
$y = n^\frac 1n: n \in \mathbb{N}$, with steps to solve being in-complete:$ \implies \ln y = \frac 1n \ln n \implies \frac 1y y' = \frac 1{n^2}(1-\ln n ) \implies y' = n^\frac 1n\frac 1{n^2}(1-\ln n )$
Need to double differentiate the above, but don't know how to pursue further.
Edit:
Based on the responses (comments, answers) have modified my attempt, that is still incomplete. Request vetting the contents also.:
As the function is exponential, so continuous one; but consider restricted domain of natural numbers, as given:
$y = n^\frac 1n: n \in \mathbb{N}$
As $\log$ is a monotonic function, so $\log y$ will be too.
$ \implies \ln y = \frac 1n \ln n \implies \frac 1y y' = \frac 1{n^2}(1-\ln n ) \implies y' = n^\frac 1n\frac 1{n^2}(1-\ln n )$
In $3$ product terms of $y' = n^\frac 1n\frac 1{n^2}(1-\ln n )$, only last term $(1-\ln n)$ can reduce to $0$ for finite values, i.e. at $x=e$, as $\ln e = 1$.
First approach is to confirm that at $e$ if there is a maxima / minima, & need find by 2nd derivative.
Second approach (as shown in the selected answer) is to take value of fn. at integers surrounding $e$ at $x=2,3$, i.e. $3^{\frac13}, 2^{\frac12}$; it shows max. value at $x=e$.
Coming back to the first approach: if $y'$ max at $x=e$, then $y''$ is negative there, & vice versa.
$ \implies \ln y = \frac 1n \ln n \implies \frac 1y y' = \frac 1{n^2}(1-\ln n ) \implies y' = n^\frac 1n\frac 1{n^2}(1-\ln n )$
Need to double differentiate the above.
$y' = n^\frac 1n\frac 1{n^2}(1-\ln n) \implies \ln y' = \frac 1n \ln n\frac 1{n^2}(1-\ln n)$
Differentiating w.r.t. $n$ again:
$y'' = \frac{d}{dn}(y'.\frac 1n \ln n\frac 1{n^2}(1-\ln n))\implies \frac{d}{dn}(n^\frac 1n\frac 1{n^2}(1-\ln n).\frac 1n \ln n\frac 1{n^2}(1-\ln n))$
Need help in completing finding the second derivative.
| Hint: Check the derivatives at n=2 and 3
(specifically at $2.718281828459045235360\cdots$)
| {
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Proving $\int_{S^{n-1}}x_1^2dS =\int_{S^{n-1}}x_k^2dS$ Denote $x = (x_1,...,x_n)$. I'm trying to prove the following:
$$\int_{S^{n-1}}x_1^2dS =\int_{S^{n-1}}x_k^2dS \; , \; 2\leq k\leq n $$
Intuitively this equality is due to the symmetry of the sphere, but I'm looking for a formal explanation. I thought about using the definition with a parametrization, but I'm not sure how to find a good paramterization of the sphere for that purpose.
| A formal explanation uses the invariance of the measure $dS$ under orthogonal transformations. That is, given a continuous function $f:\mathbb{S}^{n-1}\to\mathbb{R}$ and an orthogonal matrix $U$ (that is, $U^tU=I$), one has
$$\int_{\mathbb{S}^{n-1}}f(Ux)\,dS(x)=\int_{\mathbb{S}^{n-1}}f(x)\,dS(x)\quad (*)$$
In your case, take $f(x)=\langle x,e_1\rangle^2$, where $\langle\cdot\rangle$ denotes the standard inner product. Then $f(x)$ returns the square of the first coordinate of $x$. Pick any $1\leq k\leq n$. Take an orthogonal matrix $U$ such that $U^te_1=e_k$. (That is, rotate the vector $e_k$ to the vector $e_1$). Then, by invariance, you have
$$
\begin{aligned}
\int_{\mathbb{S}^{n-1}}x_1^2\,dS&=\int_{\mathbb{S}^{n-1}}\langle x,e_1\rangle^2\,dS\\
&=\int_{\mathbb{S}^{n-1}}\langle Ux,e_1\rangle^2\,dS
=\int_{\mathbb{S}^{n-1}}\langle x,U^te_1\rangle^2\,dS\\
&=\int_{\mathbb{S}^{n-1}}x_k^2\,dS
\end{aligned}
$$
| {
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Express a trig equation in terms of $k$ The acute angle $x$ radians is such that $\tan x$ = $k$, where $k$ is a positive integer. Express in terms of $k$.
i) $\tan (\pi - x)$
ii) $\tan (\frac{1}{2}\pi - x )$
iii) $\sin x$
I don't understand what the question is stating or asking. I got the answer $\tan \pi - k$ for the first one but that is wrong :(
Edit: Correct answer for the first one is $\tan(\pi - x) = -k$, I don't know how to get there.
| Hint: Use $\tan (\alpha +\beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$.
| {
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How to find the following summation How to find the following sum $$\sum\limits_{n=1}^{\infty}\dfrac{\left(\dfrac{3-\sqrt{5}}{2}\right)^n}{n^3}$$
I have tried by rationalizing. But after that I got stuck.
| By definition, for $|r|<1$,
$$ \sum_{n=1}^\infty r^n/n^3 = \text{polylog}(3,r)$$
This is not an elementary function.
| {
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Show that $T \mathbb{R}^n$ is diffeomorphic to $\mathbb{R}^{2n}$ $T \mathbb{R}^n$ is the tangent bundle to $\mathbb{R}^n$. How do I show that is is diffeomorphic to $\mathbb{R}^{2n}$? I know that there is an isomorphism (and therefore a diffeomorphism) between $\mathbb{R}^n$ and $T_p \mathbb{R}^n$, can I conclude from here, using the fact that $TM$ is the disjoint union of all $T_pM$s?
Alternatively: I know that, if $M$ is an $n$-dimensional differentiable manifold, then $TM$ is a $2n$-dimensional differentiable manifold. I have seen the construction given in Warner, Foundations of Differentiable Manifolds and Lie Groups, page 19, complete with proof. It really shouldn't be hard to explicitely construct the diffeomorphism, but I'm struggling for ideas.
| Going by Lee's definition, the tangent bundle is, by definition, disjoint union of all $T_pM $s.
Let $(U,\phi)$ be in atlas of M. Define map $f: \pi^{-1}(U) \rightarrow R^{2n} $ as $$f\Bigg(v^i|\frac{\partial}{\partial x^i}\Bigg|_p \Bigg)=(p,v) $$
Now write down the inverse of $f$ and observe that it is a bijection. Now write down the map $ id\circ f\circ\phi^{-1}$. This should in fact turn out to be identity map.
| {
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$e(3,6,17)\geq 40$, minimum number of edges possible in an $(3, 6)$-graph on $17$ points the article is Some Graph Theoretic Results
Associated with Ramsey's Theorem for JACK E. GRAVER AND JAMES YACKEL pp: 144-145: https://core.ac.uk/download/pdf/82034211.pdf
I'm studying graph theory and the author makes a statement that I can not understand
$G$ is an $(x, y)$-graph if $x > C(G)$ and $y > I(G)$, where $I(G)$ is the
maximum number of points of $G$ that can be chosen so that no two are
joined by an edge and $C(G)$ is the maximum number of points in any complete subgraph of G. (in other word, $G$ is a graph that does not have cliques of size $x$ neither independent sets of size $y$)
$e(x, y, n)$ is the minimum number of edges possible in an $(x, y)$-graph on $n$ points.
Theorem: $e(3,6,17)\geq 40$
previously it is proved that $e(3,6,17)\geq 38$ using the simplex method in python.
the author proves that is not possible that there are points with valence $3$ in a $(3,6)$-graph $G$ on $17$ points and $39$ edges.
Now the author affirms that there must be a $ 0- $ point in $ H_2 $ to then extend the graph to $ 39 $ edges joining $p$ and the $0$-point. Let $G$ be a $(3, 6)$-graph on $17$ points with $38$ edges. It is easy to
show that if $p$ is a preferred point of valence $4$ then $H_2$ must contain
a $0$-point.
Notation:
where: a point p of $H_2$ which has exactly $j$ edges to $H_1$
will be called a $j$-point.
Let $G$ be a graph and $p$ a point of $G$. By $H_1$ we will mean
the graph spanned by all points of $G$ which are joined to $p$ by an edge.
$H_2$ will denote the graph spanned by all points different from $p$ which are
not joined to $p$ by an edge. In the proof to Lemma $2$ we proved that if
$G$ is an $(x, y)$-graph then $H_1$ is an $(x -1, y)$-graph and $H_2$ is an $(x, y-1)$-graph. We will indicate this disection of $G$ by stating that $p$ is the preferred point.
for more than I do accounts I can not find why there must necessarily be a $0$ -point in $ H_2 $, if i suppose there is not $ 0 $ -point,
I do not come to any contradiction.
| I don't know if my argument is the same as the authors' "easy" argument, but here is a proof.
(Note that in modern terminology, "points" are called "vertices", we say "degree" rather than "valence", and probably other terminology in the paper is similarly outdated. I will stick to the paper's terminology.)
We assume for the sake of contradiction that $H_2$ has no $0$-points.
First, we show that there are many edges between $H_1$ and $H_2$.
If a point in $H_1$ is adjacent to three or more $1$-points in $H_2$, then we find an independent set of size $6$: those three $1$-points, together with the other three points of $H_1$. This is impossible. So a point in $H_1$ can be adjacent to at most two $1$-points in $H_1$, and the total number of $1$-points is at most $8$. If there are no $0$-points, then the remaining $4$ points of $H_2$ are at least $2$-points, giving at least $1 \cdot 8 + 2 \cdot 4 = 16$ edges between $H_1$ and $H_2$.
Second, we show that there cannot be many edges between $H_1$ and $H_2$.
The graph $H_2$ is a $(3,5)$-graph, and it is mentioned earlier in the paper that a $(3,5)$-graph must have at least $20$ edges. There are $4$ more edges between $p$ and $H_1$. If there are $38$ edges total in the graph, this leaves at most $38 - 20 - 4 = 14$ edges between $H_1$ and $H_2$.
There cannot simultaneously be at least $16$ and at most $14$ edges between $H_1$ and $H_2$, contradiction.
| {
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If your category has pullbacks and equalizers, do you get products? I've proven that products + pullback gives equalizers, and products + equalizer gives pullback. So, can we get products out of pullback + equalizer?
| Short answer: no. There is actually a very trivial counterexample. Take the category $\mathbf 2$ with just two objects (say $A$ and $B$) and only the identity arrows. This category has all pullbacks and equalizers, but there is no product $A \times B$. This also indicates the general problem: a product is a limit of a disconnected diagram, while equalizers and pullbacks are limits of connected diagrams.
You may be interested in the following though. The following are equivalent for any category $\mathcal{C}$:
*
*$\mathcal{C}$ has all finite limits;
*$\mathcal{C}$ has finite products (including the empty product, i.e. terminal object) and equalizers;
*$\mathcal{C}$ has pullbacks and a terminal object.
See for example, this nLab page. So this means that we can either use finite products + equalizers to build any finite limit, or we can use pullbacks + a terminal object.
Exercise: suppose we have a category $\mathcal{C}$ that has pullbacks and a terminal object, how can you use this to form the product of two objects $A$ and $B$?
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Locally connected and connected Let X be locally connected, $A \subset X$ arbitrary. Let $S \subset A$ be connected and open in A. Show $S=U\cap A$ where $U$ is connected and open.
I think that I have solved already but I never used that X was locally connected so if you could help me a little bit....
Like $S$ is open in $A$ there exists $K\subset X$ open such that $S=K\cap A$. Let suppose K is disconnected in $X$. Thus there exist $J,L\subset X$ open and $K\cap J\neq \emptyset$ $\neq$ $K\cap L$ and $J\cap L \cap K= \emptyset$ such that $K=(J\cap K)\cup (L\cap K)$. Then we have $S=((J\cap K)\cup (L\cap K))\cap A=(J\cap K \cap A)\cup (L\cap K \cap A)$. So $S$ would be disconnected in $A$, hence $K$ is connected in $X$.
I rushed in the conclusion, let see....
If both $(J\cap K \cap A) =\emptyset= (L\cap K \cap A)$, then $U=\emptyset$ would do the job. So lets suppose $(J\cap K \cap A)\neq \emptyset$, like $J,K$ are open in X then $J\cap K$ is open in X, so $U=J\cap K$ its what we need. Not is not what we need, I do not know if $U=J\cap K$ is connected, so here I must use the X is locally connected....
| As far as what you've done goes, I'd remark that
... since $J\cap K\cap A$ happens to be empty because I said so, there is no absurd here.
Rather, I think you should observe that connected components of open sets in a locally connected topological space are open, and then you should consider the connected component of $K$ that contains $S$.
| {
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scheme without global sections Let $R$ be an arbitrary non zero ring and $X$ a $R$-scheme. Is it possible that the ring of global sections $H^0(X, \mathcal{O}_X)$ might be a zero ring? The cruical point is that despite of the $R$-scheme structure for $X$ providing the existence of ring map $R \to H^0(X, \mathcal{O}_X)$ I see no obstruction for this map to be a zero map.
| Yes. The empty scheme has zero global sections, because for any scheme $X$, $\mathcal{O}_X(\emptyset)$ is the zero ring.
Conversely, any scheme with zero global sections is empty.
Proof: Suppose $X$ has zero global sections. Let $U\subseteq X$ be an affine open set, isomorphic to $\text{Spec}(A)$. Then there is a restriction map $\rho^X_U\colon \mathcal{O}_X(X)\to \mathcal{O}_X(U)$, so $A = \mathcal{O}_X(U)$ is the zero ring (because $1_A = \rho^X_U(1) = \rho^X_U(0) = 0_A$). Since the zero ring has no prime ideals, $U$ is empty. Now $X$ admits a cover by affine open sets, all of which must be empty, so $X$ is empty.
| {
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"url": "https://math.stackexchange.com/questions/3207629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the general solution of differential equation $y\frac{d^{2}y}{dx^2} - (\frac{dy}{dx})^2 = y^2 log(y)$ What is the general solution of differential equation $y\frac{d^{2}y}{dx^2} - (\frac{dy}{dx})^2 = y^2 log(y)$.
The answer to this DE is $log(y) = c_1 e^x + c_2 e^{-x}$
I don't know the method to solve differential equation with degree more than 1. Please tell me how to solve these types of equations.
| Hint. The equation can be written as
$$\frac{d}{dx}\Bigl(\frac{\frac{dy}{dx}}{y}\Bigr)=\log x\ .$$
There is in general no specific procedure for solving this kind of thing and you have to rely on "spotting" something like the above.
BTW I don't think the answer you have given is correct. Probably the $\log x$ should be $\log y$ as suggested by LutzL.
| {
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"url": "https://math.stackexchange.com/questions/3207767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral of $\frac{1}{100-x}$ I don't understand how the integral below:
\begin{align}
\int{\frac{1}{100-x}dx} = \ -\log|x-100| +c
\end{align}
When I integrate I get the answer:
\begin{align}
\ -\log|100-x| +c
\end{align}
I understand that it probably has something to do with the absolute value function, would someone kindly explain it to me? Thanks!
| The absolute value function disregards sign, so |4| = |-4| for example.
That also means that $|x-100| = |100-x|$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3207934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
estimating $\prod_{i=1}^k a_i\leq \max_i a_i^{p_i}$, where $\sum_i 1/p_i=1$ Let $1<p_1,\dots,p_k<\infty$ such that $\sum_{i=1}^k\frac{1}{p_i}=1$.
Moreover, let $a_1,\dots,a_k\geq 0$.
I have to show that
$$\prod_{i=1}^k a_i\leq\max_i a_i^{p_i}$$
I want to use induction over $k$, but I am struggling with it.
For $k=2$ I only got to here:
Since $p_1$ and $p_2$ are Hölder conjugates, we have
$\frac{1}{p_1}+\frac{1}{p_2}=1\Leftrightarrow p_1=\frac{p_2}{p_2-1}$.
I have been experimenting with the case when $a_1 a_2>a_2^{p_2}$, but I didn't get anywhere
\begin{align*}
1<a_2^{p_2-1}a_1^{-1}
\Leftrightarrow
1<a_2 a_1^{-1/(p_2-1)}
\Leftrightarrow
1<a_2\frac{a_1^{p_1}}{a_1}
\end{align*}
...
I thought maybe case distinction between the four cases $a_i\leq 1$ and $a_i>1$ could work, but I don't see how this works, since we also have to take care of the $p_i$...
| Assuming $a_i > 0$ for $i=1,\ldots , k$ (otherwise the inequality is trivial) you may set
*
*$a_i = x_i^{\frac{1}{p_i}}$
So, it is enough to show for $x_i > 0$
$$ \prod_{i=1}^k x_i^{\frac{1}{p_i}}\leq\max_i x_i$$
But this follows immediately by the concavity and monotonicity of $\log x$:
$$\sum_{i=1}^k\frac{1}{p_i}\log x_i \leq \log \left(\sum_{i=1}^k\frac{1}{p_i}x_i \right)\leq \log \left(\max_i x_i \cdot \sum_{i=1}^k\frac{1}{p_i} \right) = \log\max_i x_i $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3208063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can't use Euler's Theorem? I want to use Euler's theorem to calculate the following:
$14^{(2019^{2019})} \mod 60$
$a^{\phi(n)} \equiv 1 \mod n \iff \gcd(a,n) = 1$
Now I start with the outer one first:
$14^{(2019)}\mod 60 \iff \gcd(14,60) = 2$
Now I don't know how can I solve it ...
Can anyone suggest me any hints/further calculations?
Thank you in advance,
Kind Regards.
| Like How to find last two digits of $2^{2016}$,
Get the last two digits of $16^{100}$ and $17^{100}$
what are the last two digits of $2016^{2017}$?
last two digits of $14^{5532}$?,
As $(14^n,60)=2^2$ for $n\ge2$
and as $14\equiv-1\pmod{15}\implies14^m\equiv(-1)^m$
$\implies F_{m+2}=14^{m+2}=14^2\cdot14^m\equiv14^2(-1)^m\pmod{15\cdot14^2}$
$F_{m+2}\equiv14^2(-1)^m\pmod{15\cdot4}$ as $15\cdot4$ divides $15\cdot14^2$
$F_{m+2}\equiv16(-1)^m\pmod{60}$
If $m$ is odd like here $2019^{2019}-2,$ $$F_{m+2}\equiv16(-1)\pmod{60}\equiv-16+60$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3208183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Universality of uniform: plugging a random variable into its CDF? Consider the universality theorem of the uniform distribution. One way to formulate it is the following: Let $F:\mathbb{R}\rightarrow [0,1]$ be a right continuous, increasing function. Then, if $X\sim F$ (ie. $X$ is a random variable that has CDF $F$) then $F(X)\sim Uniform(0,1)$.
While I can prove it, I am confused about the concept of plugging $X$ (a random variable) into its CDF.
$X$ is a random variable, that is, X is a mapping from the sample space $S$ to $\mathbb{R}$ (considering 1-dimensional scenario). $X\sim F$ means $F(t)=P(X\le t)$ where "$X\le t$" is an event, "$X\le t$"$=\{s\in S: X(s)\le t\}$. The domain of $F$ is $\mathbb{R}$. How could we even imagine the concept of $F(X)$? $X$ is a function, coming from the space of functions, not the domain of $F$. And what would $F(X)$ even be, $P(X\le X)$? What event is "$X\le X$"? This is so weird and confusing.
| $X$ is a function from $S$ to $\mathbb R$, $F$ is a function from $\mathbb R$ to $\mathbb R$, so $F(X)$ is just a composition of functions $F(X(s))$.
For example, let $X$ be an exponential random variable with unit mean. Then $F(t)=1-e^{-t}$ for $t\geq 0$. And $F(X)=1-e^{-X}$.
More concrete, for each $s\in S$, $X(s)\in\mathbb R$ is a real number, so
$$
F(X(s)) = \mathbb P\{s'\in S~:~X(s')\leq X(s)\}
$$
The event $\{X\leq X\}=S$ is not appeared here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3208277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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} |
Calculating Launch Angle Given Distance with Added Height (which itself depends on launch angle) In my physics class we have learned to calculate a desired launch angle to allow a projectile to hit a target given the target’s distance away and the initial velocity.
Now in this case the initial velocity of the projectile occurs at the axis of rotation of the so called “cannon” that is launching the projectile. But what if the initial velocity occurs at the tip of the cannon? When the launch angle changes, so does the launch height and the distance to the target.
With this added information I was NOT able to solve for the desired launch angle using the kinematic equations of motion. Is this problem possible to solve?
Please see the attached pdf for a better visualization and the equations
As requested by a comment, here is the equation I couldn’t solve for $\theta$:
$$d - r \cos(\theta) = \frac{ v \cos(\theta) }{-g} \cdot \left( -v \sin(\theta)\pm \sqrt{\bigl(v \sin(\theta)\bigr)^2 + 2g \bigl( h + r \sin(\theta) \bigr) \vphantom{\Big|}}\right)$$
This question has been asked before here but not answered.
| In the simple case
$x(\theta, t) = v (\cos \theta) t\\
y(\theta, t) = v (\sin \theta)t - \frac 12 g t^2$
Where $v$ is your launch velocity. $\theta$ is your launch angle, $t$ is time, and $g$ is your gravitational constant.
Based on the picture, you have the cannon at some initial altitude, $y_0$.
$x(\theta, t) = v (\cos \theta) t\\
y(\theta, t) = v (\sin \theta)t - \frac 12 g t^2+y_0$
Then end of your cannon is just one more translation.
$x(\theta, t) = v (\cos \theta) t + d\\
y(\theta, t) = v (\sin \theta)t - \frac 12 g t^2+y_0 + h$
But it might help if you see that $\frac {h}{d} = \tan \theta$
and the length of the barrel $l = \sqrt {h^2+ d^2}$
$x(\theta, t) = v (\cos \theta) t + l\cos\theta\\
y(\theta, t) = v (\sin \theta)t - \frac 12 g t^2+y_0 + l\sin\theta$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3208443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Another combinatorial identity involving product of binomial coefficients While answering a question I came in a side calculation across the following identity valid by numerical evidence for integer $m$ and $n$:
$$
\sum_{k=m}^{\left\lfloor\frac n2\right\rfloor} \binom km\binom n{2k}=\frac{n(n-m-1)!}{m!(n-2m)!}2^{n-2m-1}.
$$
I would appreciate any hints and suggestions on combinatorial as well as algebraic proofs of the identity.
| Here is an algebraic proof. With $n\ge 2m$ we claim that
$$\sum_{k=m}^{\lfloor n/2 \rfloor}
{n\choose 2k} {k\choose m} =
\frac{n}{m} {n-m-1\choose m-1} 2^{n-2m-1}.$$
The LHS is
$$\sum_{k=m}^{\lfloor n/2 \rfloor}
{k\choose m} {n\choose n-2k}
= [z^n] (1+z)^n \sum_{k=m}^{\lfloor n/2 \rfloor}
{k\choose m} z^{2k}.$$
The coefficient extractor enforces the upper limit and we get
$$[z^n] (1+z)^n \sum_{k\ge m}
{k\choose m} z^{2k}
= [z^n] (1+z)^n z^{2m} \sum_{k\ge 0}
{k+m\choose m} z^{2k}
\\ = [z^{n-2m}] (1+z)^n \frac{1}{(1-z^2)^{m+1}}
= [z^{n-2m}] (1+z)^{n-m-1} \frac{1}{(1-z)^{m+1}}
\\ = \sum_{q=0}^{n-2m} {n-m-1\choose q}
{n-m-q\choose m}.$$
Now we have
$${n-m-1\choose q} {n-m-q\choose m}
= \frac{n-m-q}{m} {n-m-1\choose q} {n-m-1-q\choose m-1}
\\ = \frac{n-m-q}{m}
\frac{(n-m-1)!}{q! \times (m-1)! \times (n-2m-q)!}
\\ = \frac{n-m-q}{m}
{n-m-1\choose m-1} {n-2m\choose q}.$$
We get for our sum
$$\frac{1}{m} {n-m-1\choose m-1}
\sum_{q=0}^{n-2m}
(n-m-q) {n-2m\choose q}
\\ = \frac{1}{m} {n-m-1\choose m-1}
\sum_{q=0}^{n-2m}
(m+q) {n-2m\choose q}.$$
For the sum without the scalar we obtain
$$m 2^{n-2m} + \sum_{q=1}^{n-2m}
q {n-2m\choose q}
= m 2^{n-2m} + (n-2m) \sum_{q=1}^{n-2m}
{n-2m-1\choose q-1}
\\ = m 2^{n-2m} + (n-2m) 2^{n-2m-1}
= n 2^{n-2m-1}.$$
Collecting everything we find
$$\bbox[5px,border:2px solid #00A000]{
\frac{n}{m} {n-m-1\choose m-1} 2^{n-2m-1}.}$$
as claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3208566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How is the Tonnetz grid isomorphic to a torus? In Western music, there are $12$ notes up to transposition. In other words, music notes can be represented as integers modulo $12$ : $\big\{[0],[1],\dots,[11]\big\}$.
In Tonnetz grid, moving one step to the right gives the note "perfect fifth" above. That is, given a note $[a]$, the note on the right is $[a+7]$. Obviously, moving $12$ steps to the right gives the same note.
Moreover, given a note $[a]$, the note on the top-right is a "major third" above, that is $[a+4]$.
In the bottom right, the note "minor third" above lies, that is $[a+3]$.
Clearly, moving top-right $3$ times, or bottom-right $4$ times will give the same note.
Now, I am wondering, how this graph looks like if we glue the same notes together. It is stated that it is isomorphic to a torus, but I fail to see that result because there seems to be three independent cycles.
And if it is isomorphic to a torus, how should I place these $12$ points on the torus? I know, it does not really matter (because topologically they are the same) but I want to know if there is a "reasonable" way to place these notes on the torus?
For example, all of them on the top of the torus as a cycle. Or maybe $6$ of them on the top, $6$ of them on the bottom of the torus perhaps?
| There are actually only two independent directions. If you can go $\nearrow$ and $\searrow$, then you create a step in the rightward direction. So there are two independent circles; the circle of three steps in the $\nearrow$ direction, and the circle of four steps in the $\searrow$ direction. This forms a torus:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3208700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Lagrange's theorem in Elementary Number Theory Can someone please explain it to me like you would explain to an idiot?
tried to read about it in Burton's, watch videos and read answer's from here on questions about the subject and I don't get it yet.
Here's a document presenting the proof I found on the internet:
https://people.maths.bris.ac.uk/~mazag/nt/lecture6.pdf
we try to prove by induction that any function $f(x)$ in some degree n has n solutions or less mod p, when p is a prime. its easy to prove it for the 1st degree.
then we assume solution a for the polynomial $f(x)$ and we define $f(x)-f(a)=(x-a)g(x)$ , where starts the part that really confuses me. how is $-ag(x)=-f(a)\,$? shouldn't it be $g(a)$ instead?...
the reasoning behind our choice to proceed through this idea is really unintuitive for me.
Thanks a lot!
| A few things:
*
*not all functions are polynomials, it applies to polynomials.
*by division by $x-a$ we have $g(x)=\frac{f(x)-f(a)}{x-a}$
*$f(x)=a_nx^
n + a_{n−1}x
^{n−1} +\cdots + a_0$ which evaluated at a is $f(a)=a_na^
n + a_{n−1}a
^{n−1} +\cdots + a_0$
*This makes their difference, the coefficients times a difference of relevant powers termwise.
*those power differences have $x-a$ as a factor. factoring it out, leaves a polynomial with smaller degree.
*We've assumed that all lower degrees work.
*Therefore, our next higher degree is proven.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Why irreducible manifolds are prime? In wikipedia there is a proof for 3-manifolds that I don't understand. It says that if $M$ is an irreducible manifold and we express $M=N_1\sharp N_2$, then $M$ is obtained by removing a ball each from $N_1$ and $N_2$ and then gluing the resulting spheres together. These united spheres form a 2-sphere in $M$, and the fact that $M$ is irreducible means that this sphere must bound a ball. Undoing the gluing operation (how?), either $N_1$ ir $N_2$ is obtained by gluing that ball to the previously removed ball on their borders.
I understand until here. Now, it says:
This operation though simply gives a 3-sphere. That means that one of the two factors was in fact a trivial 3-sphere.
But why gluing a 2-sphere to one of the factors gives a 3-sphere? I can't understand this part.
Thanks for the help!
| This is not about "gluing a 2-sphere to one of the two factors". Instead its about reconstructing one of the two factors $N_1$ or $N_2$ by gluing two 3-balls together. Here's a bit more detail.
Suppose that $B \subset M$ is the 3-ball bounded by the "united sphere".
Also, for $k \in \{1,2\}$ let $B_k \subset N_k$ be the 3-ball that was removed from $N_k$.
As you say, from irreducibility of $M$ it follows that for either $k=1$ or $k=2$, the manifold $N_k$ is obtained, or is reconstructed, by gluing $B$ and $B_k$. That gluing identifies the 2-sphere boundary of $B$ and the 2-sphere boundary of $B_k$. Any manifold that is obtained by gluing two 3-balls, identifying their 2-sphere boundaries, is homeomorphic to $S^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3208971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Using Euler's Formula to prove $e^{i\theta}e^{i\alpha}=e^{i(\theta+ \alpha)}$ I have a homework question thats been puzzling me. It says:
Using Euler's Formula, prove the product property of the complex exponential:
$$e^{i\theta}e^{i\alpha}=e^{i(\theta+ \alpha)}$$
Besides knowing Euler's formula, I have no idea where to start so any help is appreciated. :)
| Fix $y$ and let $f(x) = e^{ix} e^{iy}- e^{i(x+y)}$.
Note that $f(0)=0$ and
$f'(x) = (\sin x - i \cos x) (\cos y + i \sin y) - \sin (x+y) +i \cos (x+y) = 0$.
Hence $f(x) = 0$ for all $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3209084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Prove that $\overline{ \lim} ~u_n\overline{\lim}~v_n\geq \overline{\lim}~u_nv_n$ Problem: Let $\{u_n\}_n$ and $\{v_n\}_n$ be two bounded sequences and $u_n>0, v_n>0$ for all $n\in \mathbb{N}$. Prove that $$\overline{\lim }~u_n\cdot \overline{\lim}~v_n\geq \overline{\lim}~u_n v_n.$$
Progress:
Since $\{u_n\}_n$ and $\{v_n\}_n$ are bounded , there exis $K_1,K_1>0$ such that $|u_n|\leq K_1$ and $|v_n|\leq K_2$ for all $n\in \mathbb{N}$. What can I do next to prove that result?
| For $n \geq m$ we have $u_ny_n \leq (\sup \{u_k: k \geq m\}) (\sup \{v_k: k \geq m\})$. Taking sup over $n$ we get $(\sup \{u_kv_k: k \geq m\}\leq (\sup \{u_k: k \geq m\}) (\sup \{v_k: k \geq m\})$. Now take limit as $m \to \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3209182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Minimizing $\left ( \sin^2(x) + \frac{1}{\sin^2(x)} \right )^2 + \left ( \cos^2(x) + \frac{1}{\cos^2(x)} \right )^2$ While solving a problem I came across this task, minimizing
\begin{align}
\left ( \sin^2(x) + \frac{1}{\sin^2(x)} \right )^2 + \left ( \cos^2(x) + \frac{1}{\cos^2(x)} \right )^2.
\end{align}
One can easily do it with calculus to show that the minimum value is $12.5$.
I tried to do it using trigonometric identities and fundamental inequalities (like AM-GM, Cauchy-Schwarz, etc.) but failed. Can someone help me to do it using trig identities and inequalities?
| Another method:
$$\left ( \sin^2(x) + \frac{1}{\sin^2(x)} \right )^2 + \left ( \cos^2(x) + \frac{1}{\cos^2(x)} \right )^2=\\
\sin ^4x+\cos ^4x+4+\frac1{\sin^4 x}+\frac1{\cos^4 x}=\\
(\sin ^2x+\cos ^2 x)^2-2\sin^2x\cos^2x+4+\frac{(\sin ^2x+\cos ^2 x)^2-2\sin^2x\cos^2x}{\sin^4x\cos^4x}=\\
1-\frac12\sin^2 2x+4+\frac{1-\frac12\sin 2x}{\frac1{16}\sin^4 2x}=\\
\left(1-\frac12\sin^2 2x\right)\left(1+\frac{16}{\sin^4 2x}\right)+4\ge \\
\left(1-\frac12\cdot 1\right)\left(1+\frac{16}{1^2}\right)+4=12.5.
$$
Note: $1-\frac12\sin ^2 2x>0$, so $\sin^2 2x$ should be maximized to $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3209521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
} |
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