Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
|---|---|---|
How about the converse of the property that a product of manifolds is a manifold? We know that the Cartesian product of two manifolds is a manifold, but is the converse true? Let us assume that we have $A$ and $B$ two second countable Hausdorff topological spaces, and $M = A \times B$. If we assume that $M$ is a $n$-manifold, with $n \geq 0$ finite, do we obtain that $A$ and $B$ are $k$- and $l$-manifolds with $k+l = n$?
| The dogbone space is not a manifold but its product with $\mathbb{R}$ is homeomorphic to $\mathbb{R}^4$. This was proved in this paper by Bing.
| {
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Logarithm power law discrepancy According to the power law:-
$$\log_a (x^k) =k\log_a x $$
So take the following example:-
$$\log_2 ((-2)^2) $$
On solving $\log_2 4=2$
However, if we use the power law, then on simplifying, $2\log_2 (-2)$ is not defined.
So how do I justify this?
| This dicrepancy is due to the fact that
*
*$\log x^2$ is defined for $x\neq 0$ but
*$2\log x$ is defined for $x>0$
then the two expression are equal $\iff x>0$.
What is true $\forall x\neq 0$ is that
*
*$\log x^2=2\log |x|$
where we have used that $\sqrt {x^2}=|x|$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2694848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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general equation to find cubic polynomial from two minimums? I tried researching and found that I can use a system of linear equations and solve by an inverse matrix to find the cubic equation given 4 points which satisfy the function f(x) of the general form $f(x)=ax^3+bx^2+cx+d$
I can also find a cubic of the form $ax^3 + d$ with no $x^2$ or $x$ term from 2 points, however I was wondering how one would go about finding a full general form cubic given only the minimum and maximum.
Example minimums could be (-1,4) and (2,3)
| Well, since the extrema are the roots of the derivate, it would be nice to have the minimum and the maximum (and an initial condition too.)
| {
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Find a 1-dimensional sufficient statistic for theta. $X_1, X_2,\ldots,X_n$ is a sample from random variable $X$.
$$f(x\mid\theta)=\frac 1 {\sqrt{2\pi}}\exp{(-\frac{1}{2}(x-\theta)^2)}$$
I have to find a 1-dimensional sufficient statistic for $\theta$ and I'm not sure how to go about doing this. Any help would be appreciated.
I've got that the log-likelihood function is
$$\ell(\theta)=\sum^n_{i=1}(-\frac{1}{2}(x-\theta)^2)$$
| Use the factorization theorem.
Hint #$1$:
$$
\sum(x_i - \theta)^2 =\sum x_i^2-2\theta n\bar{x}_n+n\theta^2,
$$
Hint #$2$:
The MLE is always a function of the MSS. What is the MLE of $\theta$?
| {
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How to decide the neighbourhood when proving a function is locally Lipschitz-continuous. I have come across a problem in one of my textbooks that I understand but could not replicate.
Show that the function $f : \mathbb{R}^+ → \mathbb{R}$ where $f(x) = {1\over x}$ is locally Lipschitz-continuous in $\mathbb{R}^+ = (0, \infty)$.
$$|f(y) − f(z)| = |{1\over y}−{1\over z}|={|y − z|\over yz}≤{|y − z|\over (x/2)^2}$$
where it chooses the neighbourhood of U $\subset$ $\mathbb{R}^+$ as U =$ ({x\over 2}, 2x)$ proving that it is locally Lipschitz-continuous in $\mathbb{R}^+ = (0, \infty)$ with $L = {4\over x^2}$
I am very confused on how they choose this value of U, if anyone could explain that would be amazing.
| Fact: If the derivative of a function is bounded over an interval, then the function is Lip on that interval with bound equal to the supremum of the derivatie there.
Derivative of $1/x$ is $1/x^2$ which is bounded from above as long as we're away from zero. So, in fact $1/x$ is Lipschitz over any $[\delta, +\infty)$, $\delta >0$.
| {
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Orthonormal basis. Linear Algebra. Hint.
Let $\lbrace u_{1}, ..., u_{n} \rbrace$ an orthonormal basis of $\mathbb{R}^{n}$, then $\displaystyle x = \sum_{i=1}^{n}\langle x, u_{i} \rangle u_{i}$ for all $x \in \mathbb{R}^{n}$.
I know that $\langle u_{i}, u_{i} \rangle = 1$ and $\langle u_{i}, u_{j} \rangle = 0$ if $i \neq j$, but I don't know how to use this information in this question. I didn't want the solution, just a hint.
| Hint: Write $x=x_1u_1+\dots+x_nu_n$ and take its inner product with $u_i$.
| {
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How to find the image under a Möbius transformation. Find the image of the set $D_1\cap D_2$, where $$D_1 = \{z : |z| < 1\}$$ and
$$D_2 = \{z : |z + 1/2| > 1/2\}$$ under the transformation
$$f(z) = \frac{z − i}{z + 1}$$
I have done the picture of $D_1\cap D_2$, but I don't know how to do it. If anybody could help me, please. Thanks!
| We have $f^{-1}(z) = \frac{z + i}{1 - z}$.
Let $w \in f(D_1)$. Then $f^{-1}(w) \in D_1$ and by definition we have $|f^{-1}(w)| = \left|\frac{w+i}{1-w}\right| < 1$, that is
$$|w+i| < |w-1|$$
So we can conclude that $f(D_1)$ contains all points $w$ that are closer to $-i$ than they are to $1$, i.e. the "half plane" $\{x+iy \mid y<-x\}$. Now you can use the same method to determine $f(D_2)$.
Then you only need to think about how $f(D_1)$, $f(D_2)$ and $f(D_1 \cap D_2)$ are related.
| {
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prove that $\{a_n\}_{n=1}^{\infty}$ is bounded and monotone increasing. I am trying to prove the sequence $\{a_n\}_{n=1}^{\infty}$, which is defined by $a_{n+1} = \frac {2(a_n+1)}{a_n+2}$, and $a_1=1$.
(1) prove that it is monotone increasing. ($a_{n+1} \ge a_n$)
proof by induction.
$P(1)$ : $a_2= \frac {2(1+1)}{2+1} = \frac 43 \ge a_1 = 1$
Suppose $P(k)$ holds true : $a_k \le a_{k+1}$.
$P(k+1)$ : $a_{k+2}= \frac {2(a_k+1)}{a_k+2}$. Then, how can I proceed from here??
(2) prove that it is bounded ($a^2_n<2$)
$P(1): a^2_1=1<2 $
Suppose $P(k)$ holds true: $a^2_k<2$
$P(k+1): a^2_{k+1}= (\frac {2(a_k+1)}{a_k+2})^2= \frac {2a_k^2+4a_k+2}{a^2_k+4a_k+4}$ I also don't know how to proceed from here.
Thank you in advance !!
| I'll play around
and see what happens.
$a_{n+1} = \frac {2(a_n+1)}{a_n+2},
a_1=1
$
$\begin{array}\\
a_{n+1}^2-2
&= \dfrac {4(a_n+1)^2}{(a_n+2)^2}-2\\
&= \dfrac {4a_n^2+8a_n+4-2(a_n^2+4a_n+4)}{(a_n+2)^2}\\
&= \dfrac {4a_n^2+8a_n+4-2a_n^2-8a_n-8}{(a_n+2)^2}\\
&= \dfrac {2a_n^2-4}{(a_n+2)^2}\\
&= 2\dfrac {a_n^2-2}{(a_n+2)^2}\\
\end{array}
$
So $a_{n+1}^2-2$
has the same sign as
$a_n^2-2$.
Since $a_1 = 1$,
all $a_n$ satisfy
$a_n^2 < 2$.
$\begin{array}\\
a_{n+1} -a_n
&= \dfrac {2(a_n+1)}{a_n+2}-a_n\\
&= \dfrac {2a_n+2-a_n^2-2a_n}{a_n+2}\\
&= \dfrac {2-a_n^2}{a_n+2}\\
&> 0
\qquad\text{since } a_n^2 < 2\\
\end{array}
$
Note that if
$a_1^2 > 2$
then
all $a_n^2 > 2$
and the $a_n$
are decreasing.
In either case,
the $a_n$ are bounded and
monotonic,
so they approach a limit.
If $L$ is this limit,
since
$a_{n+1} -a_n
= \dfrac {2-a_n^2}{a_n+2}
$,
$\dfrac {2-a_n^2}{a_n+2} \to 0$
so
$a_n^2 \to 2$.
| {
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Whats the probability you lose money? Suppose you are playing a game that costs $ $8 to play. You flip 10 coins and, for every head, you win $2. Whats the probability you lose money ?
$$
\begin{array}{c|cccccccccc}
x& 1 & 2 & 3 & 4 & 5 & 6 & 7 &8 &9&10 \\ \hline
p(x)
& 5/512 & 45/1024 & 15/128 & 63/256 & 105/512 &
\end{array}
$$
Since the loosing money means getting less than 4 heads I just did an addition of the $$P_X(1)+P_X(2)+P_X(3)=0.1708$$
With pmf of X being $P_X(x)=\binom{10}{x}(0.5)^x(0.5)^{10-x}$
And g(X)=Y=-8+2X the equation that define the net losses or winnings.
can this be a viable solution the way I presented it ?
If yes do you know any good shortcut for this exercise ?
| Don't forget about the possibility of a super bad luck event, that is you might get all tails.
probability of losing money is equal to
$$\frac{1}{2^{10}}\sum_{i=0}^3 \binom{10}{i}= \frac{1+10+45+120}{1024}\approx 0.1719$$
| {
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Name of the numerical method I came across one numerical method, however I have lost the link or is not able to find it on my history.
It used the following variables:
$$G=\frac{p'(x_k)}{p(x_k)} \ \& \ \ H=G^2-\frac{p''(x_k)}{p(x_k)}$$
Or something to that effect, not entirely sure. Thanks in advance!
| That looks like part of the Laguerre method for polynomial roots.
$$
x_{k+1}=x_k-\frac{n}{G\pm\sqrt{(n-1)(nH-G^2)}}.
$$
| {
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How to show that $f$ is monotone?
Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be defined by
$$f(x) = \int_{e^{x^3 +x}}^{1+e^{x^3+x}}e^{r^2} dr$$
for all $x\in\mathbb{R}$. Prove that $f$ is monotone.
I was thinking about Leibniz rule. Actually my main problem is that how to show that f is continuous...
I am struck at this problem as I am in fear to solve this problem.
Please help me and remove the fear from my mind.....
Thank you.
| For the first derivative of your integral we get $$e^{x^3+e^{2 \left(x^3+x\right)}+x} \left(e^{2
e^{x^3+x}+1}-1\right) \left(3 x^2+1\right)$$ this derivative is positive.
| {
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What is the probability of drawing a "conditionally" specific set of cards in a hand of 7? With the following specifics:
*
*Deck of 60 cards:
4 copies of $\mathcal A$,
3 copies of $\mathcal B$,
16 copies of $\mathcal C$ and
37 copies of $\mathcal Z$
*Hand of 7 cards drawn from the deck of 60, shuffled beforehand
*Order in the hand does not matter
1. Preliminary Problem
As a first step, I'm trying to express the probability of drawing a hand of $\mathcal A$, $\mathcal B$, $\mathcal C$, $\mathcal C$, $\mathcal Z$, $\mathcal Z$, $\mathcal Z$.
$$
P = \frac{\binom{4}{1}\binom{3}{1}\binom{16}{2}\binom{37}{3}}{\binom{60}{7}}
$$
If I was not to care about the 3 remaining cards (the 3 $\mathcal Z$) drawn in a way that they could be any combination of $\mathcal A$, $\mathcal B$, $\mathcal C$ and/or $\mathcal Z$, how should I fix my caclulation?
2. Follow-up Problem
Deck is now:
4 copies of $\mathcal A$,
3 copies of $\mathcal B$,
4 copies of $\mathcal C$,
12 copies of $\mathcal D$,
4 copies of $\mathcal E$,
15 copies of $\mathcal F$ and
18 copies of $\mathcal Z$
Now what if the valid cases I'm interested in are either:
*
*A hand of $\mathcal A$, $\mathcal B$, $\mathcal C$, $\mathcal C$, $\mathcal E$, $\mathcal ?$, $\mathcal ?$
*A hand of $\mathcal A$, $\mathcal B$, $\mathcal C$, $\mathcal C$, $\mathcal F
$, $\mathcal ?$, $\mathcal ?$
*A hand of $\mathcal A$, $\mathcal B$, $\mathcal C$, $\mathcal D$, $\mathcal E$, $\mathcal F$, $\mathcal ?$
Do I just sum these 3 probabilities together or is there a less verbose/slow known way of doing it? For context, I have small Python script that calculates probability of the hand seen in 1. and there are many more scenarii than just these 3.
| Sketch:
Let the $4-$ tuple $(a,b,c,z)$ represent the number of cards of each type in your hand. These are the patterns you wish to count. Of course we must have $a+b+c+z=7$ but there are other restrictions as well.
The good hands satisfy: $$4≥a≥1\quad 3≥b≥1\quad c≥2$$
We remark that we can drop the $4≥a$ restriction as it is implied by the others.
First step: Just count the number of patterns. Using the restrictions we see that we are trying to count the $4-$tuples $(a',b',c',z)$ of non-negative integers which add to $3$ and which satisfy $2≥b'$. By Stars and Bars that count, without the $2≥b'$ constraint is given by $$\binom {7-1}3=20$$ substracting the single case with $b'=3$ we see that there are $19$ patterns to consider. Not too bad. we just have to enumerate these and count the ways to populate each. Defining $$a=a'+1\quad b=b'+1\quad c=c'+2$$ we remark that the number of ways to populate the pattern $(a,b,c,z)$ is $$\binom 4a\times \binom 3b\times \binom {16}c\times \binom {37}z$$
I don't see a quick way to simplify the sum. Still, it's only $19$ terms. It could be done by hand.
Added: just did it by hand (mostly). I got $$\boxed {19771136}$$ but I did it hastily and there could easily be blunders.
Remark: might help an automated solution to note that defining $\binom nk=0$ when $k>n$ lets you ignore the cap on $b'$ (there would then be $0$ ways to populate that pattern). For a more complex configuration this might simplify the coding.
| {
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I don't understand how the diameters of $ \bar A$ and $A$ are equal I'm given the assignment to prove that $\operatorname{diam}\bar A = \operatorname{diam}A$, where $\operatorname{diam}A=\sup\{\rho(a,b): a,b \in A\}.$
How can they be equal if $\bar A= \partial A \cup A$:
Say $x\in \partial A$, then does it not follow that $x > a, \; x> b$, thus $ x \notin A$ ?
| We don't necessarily have any ordering among the points of the metric space (consider e.g. $\Bbb R^n$), so your statements $x>a$ and $x>b$ are meaningless.
To prove the claim, take any $a, b\in\bar A$ and consider sequences $a_n, b_n\in A$ with $a_n\to a, \ b_n\to b$, and note that $\rho(a_n, b_n) \to\rho(a, b) $.
| {
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Solving $2^x \cdot 5^y = 0,128$ $x,y \in \mathbb Z$
$2^x \cdot 5^y = 0,128$
$x+y = ?$
My attempt:
I know that
$$0,128 = \frac{128}{1000}$$
$$5^3 = 125$$
$$2^{-3} = \frac{1}{8}$$
EDIT:
$2^7 = 128$
Then we need to get
$0,128$
| You can write 5 in terms of 10 and 2... so you can write the problem as:
$2^x \cdot (\frac{10}{2})^y=.128$
$2^{x-y} \cdot 10^y=.128$
Find $y$ first and then $x $
| {
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A pattern appearing in a dynamical sequence of polynomials This question came up while I was looking at the dynamics of the map in this question
Let $f : \Bbb C[x,y,z] \to \Bbb C[x,y,z]$ given by $f(P)(x,y,z) = P(y(x+y),2z^2,z(x+y))$.
Consider the sequence of polynomials obtained by iterating $f$ on $z$ :
$P_0(x,y,z) = z$ and $P_{n+1} = f(P_n)$ for $n \ge 0$.
The first terms are $P_1 = z(x+y), P_2 = z(x+y)(y(x+y)+2z^2),$
$P_3 = 2z^3(x+y)(y(x+y)+2z^2)(x^2+3xy+2y^2+2z^2)$
Let $a_n$ be the numbers of $z$ factors in the prime factorisation of $P_n$.
Starting from $a_0$, the sequence is $1,1,1,3,5,9,19,37,73,\ldots$
let $A(t)$ be the power series $\sum_{n \ge 0} a_n t^n$.
Is it true that $A(t) = \frac {1-t-t^2}{(1-t^3)(1-2t)}$ ?
$f$ has a some kind of pseudo-inverse (after all it comes from an automorphism of $\Bbb P^2(\Bbb C)$).
Define $g$ by $g(P)(x,y,z) = P(2z^2-xy,xy,yz)$.
Then $g(f(P))(x,y,z) = f(P)(2z^2-xy,xy,yz) = P(2xyz^2,2y^2z^2,2yz^3)$.
And so if $P$ is homogeneous of degree $d$ then $g(f(P)) = (2yz^2)^d P$
Something similar happens for $f \circ g$, with a factor of $(2(x+y)z^2)^d$.
This should show that if $P$ is homogeneous and irreducible, then $f(P)$ (and $g(P)$) is the product of some irreducible polynomial with some extra small factors $(x+y)$ or $z$ ($y$ and $z$ in the case of $g$).
Then it is computationally easier to only keep that irreducible factor.
Define sequences $Q_n$ and $b_n$ with $Q_0 = z$ and $z^{b_{n+1}}Q_{n+1} = f(Q_n)$
(it does seem like, in this case, no new $(x+y)$ factor appears except in $Q_1$ so I am focusing only on $z$. A proof of this fact would also be welcome)
Then if we let $B(t) = \sum_{n \ge 1} b_nt^n$, $A$ and $B$ are related by the equation $A(t)B(t) = A(t)-1$, and so my question has the equivalent form :
Is it true that $B(t) = \frac{A(t)-1}{A(t)} = \frac{t-t^2+t^3-2t^4}{1-t-t^2}$ ?
| The results for this problem are typical. Define the homogeneous quadratic polynomial:
$\;L(x,y,z)=(y(x+y),2z^2,z(x+y)),\,$ and using initial value $L_0(x,y,z) = (x,y,z),\,$ define sequence $\,L_n\,$ by recursion $\,L_{n+1}=L(L_n).$
Now name the three components: $\;(r_n,q_n,p_n):=L_n(x,y,z).\;$ For example, $\,r_2=2z^2(xy+y^2+2z^2),\,q_2=2(x+y)^2z^2,\,p_2=(x+y)(xy+y^2+2z^2)z.$
Note that
$\;p_n = 2 p_{n-1}(p_{n-2}^3 + p_{n-1}p_{n-3}^2)/p_{n-2}\;$ for $n\ge 3,\;$ and $ r_{n+1}=(r_n+q_n)q_n,\;q_{n+1}=2 p_n^2.$
Define a sequence of irreducible factors of the $\,p\,$ polynomials: $\;f_0:=z,\;f_1:=x+y,\;f_2:=xy+y^2+2z^2,\;$ and then
$\;f_n := (U_n2^{u_n} f_{n-1} g_{n-4}^2 + V_n2^{v_n} g_{n-2}^2)/2^{w_n},\;$ with
$\;g_{-1}:=1,\;g_0:=f_0,\;g_1:=f_1,\;g_2:=f_2,\;$ $g_n:=f_n g_{n-3}\;$ if $n>2,\;$
with $\;U_2:=y,\;V_2:=4,\;U_3:=U_4=V_3=2,\;V_4:=1,\;U_n:=V_n=1,\;$ if $n>4,\;$
and $\;u_n\!:=\!(n-4)2/3\;$ if $\;n\equiv1{\pmod 3},$ else $0$,
$\;v_n\!:=\!1+(n-5)2/3\;$ if $\;n\equiv2{\pmod 3},$ else $0$,
$\;w_n\!:=\!1\;$ if $\;n\equiv0{\pmod 3},$ else $0$.
We have $\;p_n= 2^{c_n} \prod_{k=0}^n f_{n-k}^{a_k},\;$ for $\;n\ge0\;$ where $\;a_n,c_n\;$ are non-negative integer sequences. The generating function (g.f.) for $\;a_n\;$ is $\;A(t) = (1-t-t^2)/((1-t^3)(1-2t)),\;$ while the g.f. for $\;c_n\;$
is $\;t^3(1-t^4-t^5)/((1-t^3)^2(1-2t)).\;$
| {
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Evaluating the lambda expression $(.)(.)(.)$ In the book of Haskell Programming by C. Allen, at page 39, it is given the following lambda expression
$$(.)(.)(.)$$
According to me, this equals to by applying the left two expression as an input for the rightmost expression
$$(.)(.)(.) = (.)(.)y = (.)yz = yzy,$$
and if we read the main expression as
However, the book the result is (I did not understand the argument they provide)
$$.,$$
so which one of us is correct ?
Edit:
Now, after reviewing my calculation, I have
\begin{align}
(.)(.)(.) &= [(x(y.xxy))(a.ab)](c.cd) \\
&= (y.(a.ab)(a.ab)y)(c.cd) \\
&= (a.ab)(a.ab)(c.cd) \\
&= [(a.ab)(a.ab)](c.cd) \\
&= [(a.ab)b](c.cd) = bb(c.cd)
\end{align}
however, this result does not make any sense, so where is the mistake that I'm doing in here ?
Edit 2:
| It appears that the book first presents one example and then just moves to another one, as $(λxy.xxy)(λx.xy)(λx.xz)$ is not equivalent to $(λxyz.xz(yz))(λmn.m)(λp.p)$.
In order to not get confused with clashing variable names I prefer to use De Bruijn indices; the initial expression would then be:
$(λλ221)(λ12)(λ12)$
and it would reduce (with normal order) as follows:
$(λ(λ13)(λ13)1)(λ12)$
$(λ12)(λ12)(λ12)$
$(λ12)1(λ12)$
$11(λ12)$ - equivalent to your result, $bb(λc.cd)$
| {
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How do I verify if $\phi : \mathbb{Z}_{6} \rightarrow \mathbb{Z}_{36}$ is well-defined? Given $\phi : \mathbb{Z}_{6} \rightarrow \mathbb{Z}_{36}$ and $\phi ([a]_{6})=[a]_{36}$, verify that $\phi$ is a well-defined function.
My understanding is that well-defined is the converse of injective. So given a function $f:A\rightarrow B$, $f$ is well-defined when for $a = b$ in $A$, then $f(a) = f(b)$ in $B$.
Attempting to apply this definition, this is where I end up:
Say $[a]_6 = [b]_6$
$\Rightarrow 6$ | $(a - b)$
$\Rightarrow 36$ | $6(a - b)$
$\Rightarrow 36$ | $(6a - 6b)$
So $[6a]_{36} = [6b]_{36}$
Which is not $\phi([a]_{6}) = \phi([b]_{6})$
Does this show that $\phi$ is not a well-defined function? If not, where am I going wrong?
| To prove that a statement is false, you just need to find a counter example. You cannot say "this statement is false because I cannot prove it". For here, you just need to find some $[a]_{6}=[b]_{6}$ such that $\phi([a]_{6})\neq \phi([b]_{6})$.
| {
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Convex sets. Real Analysis.
Let $C \subset \mathbb{R^{n}}$ a convex set. Fixed $p \in \mathbb{R}^{n}$, let $\varphi: C \longrightarrow \mathbb{R}$ the function defined by $\varphi(x) = |x-p|=\sqrt{\langle x-p,x-p\rangle}$. Theres exists as most one point $a \in C$ such that $\varphi(a) = \inf \lbrace \varphi(x) | x \in C\rbrace$.
I don't know how to relate convexity to other hypothesis. I thought suppose $a \neq b$ such that $\varphi(a) = \varphi(b) = \inf\lbrace \varphi(x) | x \in C\rbrace$ and use $\varphi(a) \leq \varphi(b)$, $\varphi(b) \leq \varphi(a)$ so $\varphi(a) = \varphi(b)$. But I can't do anything with it anymore. I appreciate any hint.
| As mentioned, we need $C$ to be closed for this. Assume $a, b \in C$ are both distance minimizers to $p$. Let $\varphi(a) = \varphi(b) = d$ be the minimum value.
Using the paralellogram identity, we obtain:
\begin{align}
4d^2 &= 2\|a - p\|^2 + 2\|b-p\|^2 \\
&= \|(a-p)+(b-p)\|^2 + \|(a-p)-(b-p)\|^2 \\
&= 4\underbrace{\left\|\frac{a+b}2 - p\right\|^2}_{\ge d^2} + \|a - b\|^2\\
&\ge 4d^2 + \|a - b\|^2
\end{align}
Because $C$ is convex, we also have $\frac{a+b}2 \in C$ so $\left\|\frac{a+b}2 - p\right\| = \varphi\left(\frac{a+b}2\right) \ge d$.
We conlude $\|a-b\| = 0$ which implies $a = b$.
| {
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Inequality: $0<\alpha<1$, $(\sum_{i=1}^na_i)^{\alpha} \leq \sum_{i=1}^na_i^\alpha$ I would like to know if the following if true for $0<\alpha<1$, $a_i>0$ for all $i$.$$(\sum_{i=1}^na_i)^{\alpha} \leq \sum_{i=1}^na_i^\alpha$$
This looks like Jensen's inequality with counting measure. But $x^\alpha$ is
a concave function, which means the direction should be reversed . However, I checked $\sqrt{(1+1)}\leq \sqrt1+\sqrt1$ which admits the same inequality. Can anybody prove or disprove the inequality?
| For $n=2$, and $a>0$
\begin{align*}
(1+a)^{\alpha}-a^{\alpha}&=\int_{0}^{1}\dfrac{d}{dt}((1-t)a+t(a+1))^{\alpha}dt\\
&=\int_{0}^{1}\alpha((1-t)a+t(a+1))^{\alpha-1}dt\\
&=\int_{0}^{1}\alpha(a+t)^{\alpha-1}dt\\
&\leq\int_{0}^{1}\alpha t^{\alpha-1}dt\\
&=1,
\end{align*}
so $(1+a)^{\alpha}\leq 1+a^{\alpha}$.
So $(a_{1}+a_{2})^{\alpha}=a_{1}^{\alpha}\left(1+\dfrac{a_{2}}{a_{1}}\right)^{\alpha}\leq a_{1}^{\alpha}\left(1+\left(\dfrac{a_{2}}{a_{1}}\right)^{\alpha}\right)=(a_{1}^{\alpha}+a_{2}^{\alpha})$.
| {
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Prove linear transformation is one to one The linear operator T: R2→R2 defined by the equations
w1 = 4x1 - 6x2
w2 = -2x1 + 3x2
is not one-to-one. Using the methods in class, show why this is true. Once you have done this, provide a simple, specific, numerical example, where the output vector is not the zero vector, that illustrates why the transformation is not one-to-one.
Okay, so I was able to answer the first part (proving the transformation is not one to one).
But for the second part, I'm not exactly sure what to do. Do I just give any input vector that produces a nonzero output?
| You are suppose to find $x_1, x_2, y_1, y_2$ such that $(x_1, x_2) \neq (y_1, y_2), (w_1, w_2) \ne (0,0),$ and
$$4x_1-6x_2=w_1 = 4y_1-6y_2$$
$$-2x_1+3x_2=w_2 = -2y_1+3y_2$$
| {
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Showing that $(a,b)=1$ where $d=(m,n)$ and $m=ad$ and $n=bd$ If $d=(m,n)$, then $d|m$ and $d|n$ so there must exist integers $a$ and $b$ such that $m=ad$ and $n=bd$. Now $d=(m,n)=(ad,bd)=d(a,b)$ and so $(a,b)=1$. Is this correct?
| Yes, the proof is flawless.You have proved that$$ d=(m,n)=(ad,bd)=d(a,b)\implies (a,b)=1$$
| {
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Exponentials inequality: $|e^x + e^y| \leq |x - y|$ for $x,y<0$. I am trying to prove $|e^z - e^\omega| \leq |z - \omega|$ for $z,\omega\in\{z\in\mathbb{C}\;:\;\Re(z)<0\}$ and I get stuck in an inequality of the stated in the title : $|e^x + e^y| \leq |x - y|$ for $x,y<0$.
This is what I have tried:
\begin{equation}\left|e^z - e^\omega\right| = \left|e^{i\Im(z)}\right|\left|e^{\Re(z)} - e^{\Re(\omega)}e^{i(\Im(\omega) - \Im(z))}\right| \leq \left|e^{\Re(z)} + e^{\Re(\omega)}\right|
\end{equation}
Is it true that $|e^x + e^y| \leq |x - y|$ for $x,y<0$? If yes, any help will be appreciated.
Thank you!
| Um, take $x=y=-1$;
$$|e^{-1}+e^{-1}|>|x-y|=0$$
seems like your theorem isn't true.
| {
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For what non trivial values of $a$ and $k$, $a^k + 1$ will be prime?
For what non trivial values of $a$ and $k$, $a^k + 1$ will be prime?
If $a=1$, then $1^k+1=1+1=2$, where 2 is prime.
If $k=1$ then we have$a^k+1=a+1$, but if some prime $p$ is equal to $a+1$, then only we can say $a+1$ is prime.
But for the rest part I am clueless to proceed. Any suggestion is highly appreciated.
Is it similar to the problem $2^{2^n}+1$ is prime for even $n$?
| consider the case $k=3$ then we get $$a^3+1=(a+1)(1-a+a^2)$$
| {
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Showing that a linear map does not achieve its norm If we take the Banach spaces $(X, \|\cdot \|_X )$ and $(Y, \|\cdot \|_Y )$ and the bounded linear map $F: X \rightarrow Y$, then the norm of $F$ is $$\|F\| = \underset{\|x\|_X = 1}{\textrm{sup}} \|F(x)\|_Y.$$ My question is how we can determine, generally, whether or not the map $F$ actually achieves that supremum. This is the example I have in mind: $(X, \|\cdot \|_X ) = (C([0, 1], \mathbb{R}), \|\cdot \|_{\infty})$ and $(Y, \|\cdot \|_Y)= (\mathbb{R}, |\cdot |)$, and the map $F: X \rightarrow Y$ is given by $F(f) = \int_{0}^{1} xf(x)dx$. It can be shown in a standard way that $\|F\| = \frac{1}{\sqrt3}$ and that $F$ actually achieves this norm when $f(x) = x$.
But now consider the situation where instead of $X$ being all continuous functions from $[0,1]$ to $\mathbb{R}$, it's only those functions where $f(1)=0$. I believe it can be shown that $F$ doesn't achieve its norm in this case, although we can show that the norm of $F$ remains $\frac{1}{\sqrt3}$ by taking the limit of $|F(f_n )|$ where $$f_n (x) = \begin{cases}
x & 0\leq x\leq 1-\frac{1}{n} \\
(1-n)x + (n-1)& 1-\frac{1}{n}< x\leq 1
\end{cases}
$$
and this limit will be $\frac{1}{\sqrt3}$ because $f_n \rightarrow x$ on $[0,1]$. But since $f(x) = x$ isn't in our new space, we can't use it to prove that $F$ reaches its norm here. But we don't necessarily know that there's only one function that lets $F$ achieve its norm, do we? Maybe a theorem of Riesz can be used to show this? Otherwise, how can we show that $F$ won't achieve its norm on this new space? And how can we do that generally?
$\bf{\textrm{EDIT}}$: I miscalculated the norm of $F$ -- it is actually $\frac{1}{2}$.
| Firstly, when considering $F$ as a map from $C([0,1])$ we have $\|F\|=\frac{1}{2}$. One way to see this is that
$$\left|\int_0^1 xf(x)\ dx\right|\leq\int_0^1|x||f(x)|\ dx\leq\|f\|\int_0^1x\ dx=\frac{1}{2}\|f\|,$$
and the constant function $f(x)=1$ gives us this upper bound.
Now let's write $\tilde X=\{f\in C([0,1]):f(1)=0\}$ and talk about $F$ restricted to $\tilde X$. We still have $\|F\|=\frac{1}{2}$, since we can take approximations to the constant $1$ function while staying in $\tilde X$.
But to show that $F$ does not achieve this norm is a different matter. To answer your last question, there isn't a general way to do this, but your best bet is to go by contradiction since you at least get a function to play with. So assume $f\in\tilde X$ with $\|f\|=1$ and $F(f)=\frac{1}{2}$. Since $f(1)=0$, take any $\varepsilon>0$ and use continuity to show there is some $x_0\in (0,1)$ such that $|f(x)|<\varepsilon$ whenever $x_0<x\leq1$. Then we have
$$\frac{1}{2}=|F(f)|\leq\int_0^1|xf(x)|\ dx=\int_0^{x_0}|xf(x)|\ dx+\int_{x_0}^1|xf(x)|\ dx<\frac{1}{2}x_0+\varepsilon(1-x_0).$$
If $\varepsilon<\frac{1}{2}$ (which is OK, since it was arbitrarily positive), we obtain a contradiction.
| {
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Find all positives integers $n$ such that $n^3+1$ is a perfect square A solution as follows:
$n^3+1=x^2$
$n^3=x^2-1$
$n^3=(x-1)(x+1)$
$x-1=(x+1)^2~~or~~x+1=(x-1)^2$
$x^2+x+2=0~~or~~x^2-3x=0$
$x(x-3)=0$
$x=0~~or~~x=3~~\Longrightarrow~~n=2$
Does it cover all possible solutions? How to prove that 2 is the only which solves the problem.
| Hint: see that $m^2=n^3+1$ gives $(m-1)(m+1)=n^3$. What factors can $m-1$ and $m+1$ have in common? How can their product be a perfect cube?
| {
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Why does this integral not depend on the parameter? While working with stable distributions, we for $\alpha \in (1,2) $ have encountered the following integral:
$$\lim_{\varepsilon \to 0} \int_{(\varepsilon, \pi/2 - \varepsilon) \cup (\pi/2 + \varepsilon, \pi-\varepsilon)} \frac{\log \left(|\cos \theta|^\alpha + |\sin \theta |^\alpha + |\cos \theta + \sin \theta|^\alpha \right)}{\alpha \cos \theta \sin \theta} d\theta .$$
Mathematica suggests that it is independent of $ \alpha $, and we have other resons to believe that this is also the truth, but we do not know neither how to formally solve the integral now how to prove this independence. Is there any simple way to do either of these things which we (clearly) haven't thought of?
| A partial answer, designed for simplifying further attempts. Now an answer, with some detours not really needed but left here for documentation purposes.
$\cos\left(\theta+\frac{\pi}{2}\right)=-\sin\theta$ and $\sin\left(\theta+\frac{\pi}{2}\right)=\cos\theta$, hence the given integral can be written as
$$ \int_{0}^{\pi/2}\log\left(\frac{\sin^\alpha\theta+\cos^\alpha\theta+|\cos\theta+\sin\theta|^{\alpha}}{\sin^\alpha\theta+\cos^\alpha\theta+|\cos\theta-\sin\theta|^{\alpha}}\right)\frac{d\theta}{\alpha\sin\theta\cos\theta}$$
in the improper Riemann sense. By substituting $\theta=\arctan u$ we are left with
$$ \int_{0}^{+\infty}\log\left(\frac{1+u^\alpha+|1+u|^{\alpha}}{1+u^\alpha+|1-u|^\alpha}\right)\frac{du}{\alpha u}=2\int_{0}^{1}\log\left(\frac{1+u^\alpha+(1+u)^{\alpha}}{1+u^\alpha+(1-u)^\alpha}\right)\frac{du}{\alpha u}$$
or
$$\frac{2}{\alpha}\int_{0}^{+\infty}\log\left(\frac{2\cosh\frac{\alpha x}{2}+\left(2\cosh\frac{x}{2}\right)^\alpha}{2\cosh\frac{\alpha x}{2}+\left(2\sinh\frac{x}{2}\right)^{\alpha}}\right)\,dx=\frac{4}{\alpha}\int_{0}^{+\infty}\log\left(\frac{2\cosh\alpha x+\left(2\cosh x\right)^\alpha}{2\cosh\alpha x+\left(2\sinh x\right)^{\alpha}}\right)\,dx.$$
We already know some integrals that, due to the substitution $u\mapsto\frac{1}{u}$, do not really depend on their parameter $\beta>0$:
$$ \int_{0}^{+\infty}\frac{du}{(1+u^2)(1+u^\beta)}=\frac{\pi}{4},\qquad \int_{0}^{+\infty}\frac{\log(u)\,du}{(1-u^2)(1+u^\beta)}=-\frac{\pi^2}{8} $$
$$\int_{0}^{+\infty}\frac{du}{(1+u+u^2)(1+u^\beta)}=\frac{\pi}{3\sqrt{3}},\qquad \int_{0}^{+\infty}\frac{g\left(\frac{u-1}{u+1}\right)\,du}{(1-u^2)(1+u^\beta)}\;\text{ with }g\text{ odd}$$
hence it is not unlikely that by choosing a suitable $g$, applying $\int_{0}^{\alpha}(\ldots)d\beta$ and performing a substitution we can prove that
$$\frac{\partial}{\partial\alpha}\int_{0}^{1}\log\left(\frac{1+u^\alpha+(1+u)^{\alpha}}{1+u^\alpha+(1-u)^\alpha}\right)\frac{du}{u}=\text{ const.}$$
as wanted. Indeed
$$\begin{eqnarray*} \int_{0}^{1}\log\left(\frac{1+u^\alpha+(1-u)^\alpha}{2}\right)\frac{du}{u}&=&\int_{1}^{+\infty}\log\left(\frac{1+\frac{1}{u^\alpha}+\left(1-\frac{1}{u}\right)^\alpha}{2}\right)\frac{du}{u}\\&=&\int_{0}^{+\infty}\log\left(\frac{1+u^\alpha+(u+1)^\alpha}{2(u+1)^\alpha}\right)\frac{du}{1+u}\end{eqnarray*}$$
but the LHS, by symmetry, is also
$$\begin{eqnarray*} \int_{0}^{1}\log\left(\frac{1+u^\alpha+(1-u)^\alpha}{2}\right)\frac{du}{1-u}&=&\int_{1}^{+\infty}\log\left(\frac{1+\frac{1}{u^\alpha}+\left(1-\frac{1}{u}\right)^\alpha}{2}\right)\frac{du}{u(u-1)}\\&=&\int_{0}^{+\infty}\log\left(\frac{1+u^\alpha+(u+1)^\alpha}{2(u+1)^\alpha}\right)\frac{du}{u(1+u)}\end{eqnarray*}$$
while
$$\begin{eqnarray*} \int_{0}^{1}\log\left(\frac{1+u^\alpha+(1+u)^\alpha}{2}\right)\frac{du}{u}&=&\int_{1}^{+\infty}\log\left(\frac{1+\frac{1}{u^\alpha}+\left(1+\frac{1}{u}\right)^\alpha}{2}\right)\frac{du}{u}\\&=&\int_{1}^{+\infty}\log\left(\frac{1+u^\alpha+(u+1)^\alpha}{2 u^\alpha}\right)\frac{du}{u}\end{eqnarray*}.$$
Hence
$$
\begin{eqnarray*} && 2\int_{0}^{1}\log\left(\frac{1+u^\alpha+(1+u)^{\alpha}}{1+u^\alpha+(1-u)^\alpha}\right)\frac{du}{\alpha u}\\
&=&
\int_{0}^{1}\log\left(\frac{1+u^\alpha+(1+u)^\alpha}{2}\right)\frac{du}{\alpha u}
+
\int_{1}^{+\infty}\log\left(\frac{1+u^\alpha+(u+1)^\alpha}{2 u^\alpha}\right)\frac{du}{\alpha u}
\\&\qquad-&
\int_{0}^{+\infty}\log\left(\frac{1+u^\alpha+(u+1)^\alpha}{2(u+1)^\alpha}\right)\frac{du}{\alpha (1+u)}
-
\int_{0}^{+\infty}\log\left(\frac{1+u^\alpha+(u+1)^\alpha}{2(u+1)^\alpha}\right)\frac{du}{\alpha u(1+u)}
\\&=&
\int_{0}^{\infty}\log\left(\frac{1+u^\alpha+(1+u)^\alpha}{2\max(1,u)^\alpha}\right)\frac{du}{\alpha u}
-
\int_{0}^{+\infty}\log\left(\frac{1+u^\alpha+(u+1)^\alpha}{2(u+1)^\alpha}\right)\frac{du}{\alpha u}
\\&=&
\int_{0}^{\infty}\log\left(\frac{(u+1)^\alpha}{\max(1,u)^\alpha}\right)\frac{du}{\alpha u}
\\&=&
\int_{0}^{\infty}\log\left(\frac{u+1}{\max(1,u)}\right)\frac{du}{u}
\\&=&
2\int_{0}^{1} \frac{\log\left( u+1 \right)}{u} du.
\end{eqnarray*}
$$
The last expression is clearly independent of $\alpha$, and in particular, for the original integral we get
$$
\int_{0}^\pi \frac{\log(|\cos \theta|^\alpha+|\sin\theta|^\alpha+|\cos \theta + \sin \theta|^\alpha)}{\alpha \cos \theta \sin \theta} d\theta
=
\int_{0}^{1} \frac{\log\left( u+1 \right)}{u} du
= \frac{\pi^2}{12}.
$$
The case $\alpha=3$ is not instantly recognized by (my version of) Mathematica. It turns out to be a sort of Ahmed-like integral.
| {
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Maximal ideals in a polynomial ring I am having a lot of trouble figuring out which ideals are maximal in rings that aren't fields. I read that maximal ideals of $\mathbb{Z}[x]$ are of the form $(p,f(x))$ where $p$ is a prime number and $f(x)$ is a polynomial in $\mathbb{Z}[x]$ which is irreducible modulo $p$.
$\\$I am trying to apply this to figuring out if the ideals generated by $x+1$ and $x^2+x+1$ are maximal in $\mathbb{Z}$. I've been told to consider the ideals $(2,x+1)$ and $(2, x^2+x+1)$ respectively, but I don't get where the two comes from. Can you pick any prime number to check this condition? Any explanation of how to proceed would be appreciated. Thanks
| We show that $(x+1)$ is not a maximal ideal in $\mathbb{Z}[x]$. Since $(x+1)\subset (2,x+1)\subset \mathbb{Z}[x]$, we just need to show that $(x+1)\neq (2,x+1)$ and $(2,x+1)\neq \mathbb{Z}[x]$. The first statement holds because $2\notin (x+1)$. The second statement is true by the result you said in the first paragraph.
| {
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Inequality $\frac{a^3+b^3+c^3}{3}\geq\sqrt{\frac{a^4+b^4+c^4}{3}}$ Let $a$, $b$ and $c$ be non-negative numbers such that $(a+b)(a+c)(b+c)=8$. Prove that:
$$\frac{a^3+b^3+c^3}{3}\geq\sqrt{\frac{a^4+b^4+c^4}{3}}$$
Some attempts:
*
*From the condition follows $a^3+b^3+c^3 = (a+b+c)^3 -24$
*It is known (see here)
$$\frac{a+b+c}{3}\geq\sqrt[27]{\frac{a^3+b^3+c^3}{3}}$$
*Setting $2x=a+b$, $2y = b+c$, $2z = a+c$, we can express $a =x+z-y$ etc. The condition then becomes $xyz = 1$ which can be parametrized with free variables $0\leq q \leq 2 \pi /3 $ and arbitrary $r$ by
$$
x = \exp(r \cos q) \; ; \; y = \exp(r \cos (q + 2 \pi /3)) \; ; \; z = \exp(r \cos (q - 2 \pi /3))
$$
Using that, the condition can be removed and then calculus may be used.
*The question may be interpreted geometrically. Expressions such as $a^3+b^3+c^3 = $const. and $a^4+b^4+c^4 =$ const. can be interpreted as hypersurfaces of what has been called an N(3)-dimensional ball in p-norm, see here. A nice visualization is given in here. Then properties such as extrema, convexity etc. of these surfaces can be used.
I couldn't put the pieces together.
| I repeat the above hints:
Setting $2x=a+b$, $2y = b+c$, $2z = a+c$, we can express $a =x+z-y$ etc. Further, we have $a^3+b^3+c^3 = (a+b+c)^3 -24 = (x+y+z)^3 -24$.
The condition then becomes $xyz = 1$ which can be parametrized with free variables $0\leq q \leq 2 \pi /3 $ and arbitrary $r$ by
$$
x = \exp(r \cos q) \; ; \; y = \exp(r \cos (q + 2 \pi /3)) \; ; \; z = \exp(r \cos (q - 2 \pi /3))
$$
Using that, the condition can be removed and then calculus may be used.
So we have to show
$$f = ((x+z+y)^3 -24)^2- 3 \sum_{cyc}(x+z-y)^4 \geq 0$$
From the parametrization, we see that $f$ is periodic in $q$ with period length $2 \pi/3$. The maxima are at $q_+ = n 2 \pi/3$ and the minima are at $q_- = \pi/3 + n 2 \pi/3$. Hence, it is enough to investigate $f$ at $q = \pi/3$. This gives
$$f(r,q) \ge f(r,q = \pi/3) = \\
(\exp(-3r)(2\exp(3r/2) + 1)^3 - 24)^2 - 6\exp(-4r) - 3\exp(-4r)(2\exp(3r/2) - 1)^4$$
It is now a matter of calculus to show that the minimum of $f(r,q = \pi/3)$ occurs at $r=0$, giving $f(r,q) \ge f(r=0,q = \pi/3) = 0 \geq 0$ which proves the claim. $\qquad \Box$
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Showing $P(S_mLet $X_1, X_2,\ldots$ be a sequence of iid random variables such that for each $i$, $X_i$ takes value as nonnegative integer and is in $L^1$. Let $ S_n = \sum_{i=1}^n X_i$. How to show that
\begin{equation}
P(S_m<m, \forall\ 1\leq m\leq n | S_n)=\max\{0, 1-S_n/n\} ?
\end{equation}
I think that there is something to do with martingales, but I am not really sure where to start.
Thanks!
| Let $E_n = \bigcap_{i=1}^n \{S_i<i\}$. We will prove $P(E_n|S_n=k)=(1-k/n)^+$ by induction on $n$, where $x^+=\max(0,x)$.
Given a list of numbers $X = (X_1,\dots,X_n)$, let $Y = (Y_1,\dots,Y_n)$ be the same list rearranged in weakly increasing order. We will prove the stronger fact that for any deterministic $y=(y_1,\dots,y_n)$, where $y_1\le y_2\le \dots\le y_n$, that
$$
P(E_n|Y=y)=(1-k/n)^+, \text{ where }k=y_1+\dots+y_n
$$
In other words, we are conditioning on everything except the order of the $X_i$, but the resulting probability only depends on the sum of the values, so it is still true when you only condition on the sum.
The result is obvious when $k\ge n$ since in that case, $S_n=k\ge n$, so assume $k<n$. Given that the sorted list of $(X_1,\dots,X_n)$ equals $(y_1,\dots,y_n)$, $X_n$ is equally likely to be any of the $y_i$. Therefore,
$$
P(E_n|Y=y) = \sum_{i=1}^nP(E_n|Y=y,X_n=y_i)\cdot\tfrac1{n}
$$
Now, when we are given that $X_n=y_i$, the remaining numbers $(X_1,\dots,X_{n-1})$ are equally likely to be any rearrangement of the list $y$ with $y_i$ removed. Letting $\hat y_i$ be this list, and letting $\hat Y$ be the weakling increasing rearrangement of $(X_1,\dots,X_{n-1})$,
$$
P(E_n|Y=y) = \sum_{i=1}^nP(E_{n-1}|\hat{Y}=\hat y_i)\cdot \tfrac1n=\sum_{i=1}^n\left(1-\frac{k-y_i}{n-1}\right)^+\cdot \frac1n
$$
where the last equality follows by the induction hypothesis. Since we assumed $k<n$, it follows each $1-\frac{k-y_i}{n-1}\ge 0$, so we can remove the $^+$ from the above:
$$
P(E_n|Y=y) =\sum_{i=1}^n\left(1-\frac{k-y_i}{n-1}\right)\cdot \frac1n
$$
After some algebra, and recalling that $\sum_{i=1}^n y_i=k$, the above simplifies to $1-k/n$.
| {
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How does WolframAlpha get an exact answer here? ${}{}$ I had a simple thing to compute with a calculator:
$$\sin\left(2\cos^{-1}\left(\frac{15}{17}\right)\right)$$
I got the decimal answer of about $0.83044983$, but when I typed it in WolframAlpha, it also gave an exact answer of $\frac{240}{289}$. How in the world would one get an exact answer here?
| $ (8,15,17)$ are lengths of a Pythagorean triple right triangle. A narrow right triangle of these side lengths can be drawn if needed.
$$\sin(2\cos^{-1}\frac{15}{17}) = \sin(2\sin^{-1}\frac{8}{17}) = 2 \cdot \frac{8}{17}\cdot \frac{15}{17} =\frac{240}{289}.$$
| {
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What groups have the property: All nontrivial isomorphic subgroups have a nontrivial intersection? This is a generalization of my previous question: What are all finite groups such that all isomorphic subgroups are identical?
Specifically, what finite groups $G$ have the following property: For any two subgroups $1 < H, H' < G$ that are isomorphic, then $H \cap H'$ is non-trivial.
Some quick results: This property is equivalent to all prime order cyclic subgroups being unique. Also, each of these subgroups are normal.
If $G$ is an abelian group, it turns out $G$ is cyclic. (Proof: $G=G_1 \times \cdots \times G_n$, each cyclic, with $|G_i|$ dividing $|G_{i+1}|$. Two factors share a common prime $p$, hence two disjoint copies of $C_p$, so no two factors exist.)
However, there are non-abelian groups that satisfies this condition (if I've thought through it right). Take the Quaternion group, which from the subgroup lattice is easily seen to have the desired property.
Can the (non-abelian) groups with this property be easily characterized?
| The question has an answer here.
[In such a group] every Sylow $p$-subgroup is either cyclic or generalized quaternion (because these are the only $p$-groups that have a unique subgroup of
order $p$). Apparently, Zassenhaus (in the 1930s) classified all
solvable groups where all the Sylow $p$-subgroups for odd primes are
cyclic, and where the Sylow 2-subgroup contains an index $2$ cyclic
subgroup (and Michio Suzuki, in the paper ''On finite groups with
cyclic Sylow subgroups for all odd primes'' from Amer. J. of Math in
1955 classified the non-solvable groups with the same properties).
This gives a complete classification of the groups that meet your
condition.
| {
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Minkowski's triangle inequality for $0I have a trouble in the following theorem introduced "Minkowski's inequality for $0<p<1$":
Let $0<p<1$ and let $x,y\ge0$. Then, $|x^{p}-y^{p}|\le|x-y|^{p}$.
I have proved the case for $p=\frac{1}{2}$.
How do I prove it in the general case ?
Give some advice! Thank you!
| A concave function $f:[0,\infty)\to\Bbb R$ such that $f(0)\ge0$ must be subadditive: if by any chance $f(a+b)>f(a)+f(b)$, then $\frac{f(a+b)-f(a)}{b}>\frac{f(b)-f(0)}{b}$, against the hypothesis of concavity.
So, for $p\in(0,1)$ and $x,y\ge 0$, we have that $$(\max\{x,y\}-\min\{x,y\})^p+(\min\{x,y\})^p\ge (\max\{x,y\})^p\\(\max\{x,y\}-\min\{x,y\})^p\ge(\max\{x,y\})^p-(\min\{x,y\})^p\\ \lvert x-y\rvert^p\ge(\max\{x,y\})^p-(\min\{x,y\})^p$$
And, since $x^p$ is an increasing function, the RHS is $\lvert x^p-y^p\rvert$.
| {
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In general, what techniques can be used to show that 2 groups are not isomorphic? Say I have 2 groups $G$ and $H$, what techniques can be used to show that they are not isomorphic?
A simple one I can think of is proving that their order is different, thus showing there cannot be a bijection in between the 2. However I am interested in other approaches as well.
| Look for any difference in the groups such as
*
*Order of the groups.
*One group has an element of order $n$, and the other does not have an element of order $n$.
*One group has a subgroup of order $n$, and the other does not.
*The orders of the centers of the groups are different.
*One group is Abelian, the other is not.
*One group is cyclic, the other is not.
| {
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Minimizing $f(x,y,z)=y$ In the problem of minimizing $f(x,y,z)=y$ over the constraint set $z=y^3 - x^2$ and $z=x^2$, I have managed to solve the problem directly and obtain that the minimum occurs at $x=0, y=0, z=0$, yielding a value of $f(0,0,0)=0$, but when I write the Lagrangian and try to solve it, I realize that it has no solutions. I am trying to understand why the Lagrange multipliers didn't work here and any insights on this would be helpful.
Edit: $\nabla f+λ_1∇g_1+λ_2∇g_2=0$ yields: $x:−2xλ_1+2xλ_2=0$ $y:1+3y^2λ_1=0$ $z:−λ_1−λ_2=0$ Which gives: $λ_1=−λ_2$ and $λ_2x=0$ but since $λ_2$ cannot be $0$, $x=0$, which gives $y^3−z=0$ and $−z=0$, giving $y=0$ and $z=0$, but $1+3y^2λ_1=0$ is not satisfied.
Thank you for your help!
| Let $g(x,y,z):=(z-y^3+x^2, z-x^2)$, $h(x,y,z, \lambda, \mu):=f(x,y,z)+\lambda(z-y^3+x^2)+ \mu(z-x^2)$ and $\phi:= \nabla h$.
Lagrange says: if(!) $f$ has in $(x_0,y_0,z_0) $ a local extremum under the constraints $g(x,y,z)=(0,0)$ and if(!) $rank g'(x_0,y_0,z_0)=2$, then there are $\lambda_0, \mu_0$ such that $\phi(x_0,y_0,z_0, \lambda_0, \mu_0)=0$.
In your case we have $x_0=y_0=z_0=0$, but $rank g'(0,0,0)=1 \ne 2$.
| {
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Prove $\cos(x)$ is continuous I'm stuck at a particular step and could use some help.
By definition, a function is continuous at $x=a$ iff $\lim_{x \to a} f(x) = f(a)$.
So I assume to prove $\cos(x)$ is continuous we must use the definition of a limit to show that:
$$\lim_{x \to a} \cos(x) = \cos(a) \iff \forall \epsilon > 0, \exists \delta>0 : 0 < |x - a| < \delta \implies |\cos(x)-\cos(a)| < \epsilon$$
From some triangle proofs it can be shown that:
$\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha) \sin(\beta)$
$\cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha) \sin(\beta)$
Subtract the two equations:
$$\cos(\alpha + \beta) - \cos(\alpha - \beta) = -2\sin(\alpha) \sin(\beta)$$
Let $x = \alpha + \beta$ and $a = \alpha - \beta$.
Then $\alpha = \frac{x+a}{2}$ and $\beta = \frac{x-a}{2}$, implying:
$$\cos(x) - \cos(a) = -2\sin\left(\frac{x+a}{2}\right)\sin\left(\frac{x-a}{2}\right)$$
So
$$\left|\cos(x)-\cos(a)\right| = 2\left|\sin\left(\frac{x+a}{2}\right)\right| \cdot \left|\sin\left(\frac{x-a}{2}\right)\right|$$
I feel like I am close because I was able to change it so at least we have an $x-a$ term but now I'm not sure where to go from here.
| Use that $|\sin u|\leq\min\{|u|,1\}$, then $2\left|\sin\left(\dfrac{x+a}{2}\right)\right|\cdot\left|\sin\left(\dfrac{x-a}{2}\right)\right|\leq 2\left|\dfrac{x-a}{2}\right|=|x-a|<\epsilon$ if we take $\delta=\epsilon$.
| {
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Can I combine two functions (express them as a single function) where they meet a certain criteria at different times. Say I have a function $F(x) = 7x$. That meets my criteria when $0 \leq x \leq 10$.
And another function $G(x) = x^2$. That meets my criteria when $x > 10$.
Is there a way to combine these two into one function?
| One way to do this is to define a piecewise function like N. F. Taussig said in the comments above. However I have my own alternative way of defining such funcion, using the properties of floor and ceiling functions.
First notice one special function that I found:
$$y=\Bigg\lceil\frac{\lfloor x\rfloor}{x}\Bigg\rceil$$
What's special about it is that it outputs 0 for all numbers between zero and one, and one for the rest of numbers.
We can invert the behavior of this function by negating it and adding one. This will output one for the range of 0 to 1, and zero for other numbers.
$$y=-\Bigg\lceil\frac{\lfloor x\rfloor}{x}\Bigg\rceil+1$$
Now we can stretch this by ten dividing x by 10 everywhere:
$$y=-\Bigg\lceil\frac{10\lfloor {x\over10}\rfloor}{x}\Bigg\rceil+1$$
Now we simply multiply this by 7x, our function between 0 and 10.
$$y=-7x\Bigg\lceil\frac{10\lfloor {x\over10}\rfloor}{x}\Bigg\rceil+7x$$
This is one part of our function, now we do the similar thing with the second part, except this time we want our special floor-ceiling function to be 1 everywhere except in the range 0 to 10, so we do:
$$y=\Bigg\lceil\frac{10\lfloor {x\over10}\rfloor}{x}\Bigg\rceil$$
We simply multiply this part by our second function, $x^2$:
$$y=x^2\Bigg\lceil\frac{10\lfloor {x\over10}\rfloor}{x}\Bigg\rceil$$
Now, when we got two parts of the final function. We simply add them together to get:
$$f(x)=x^2\Bigg\lceil\frac{10\lfloor {x\over10}\rfloor}{x}\Bigg\rceil-7x\Bigg\lceil\frac{10\lfloor {x\over10}\rfloor}{x}\Bigg\rceil+7x$$
There it is, an alternative definition of piecewise function. Now there are some tweaks we could add to make this better, for example make this defined at 0, because now it isnt, and make negatives undefined. However this will nee some more playing with floor, ceiling, square root, and modulo functions.
The best way is to simply define a piecewise function, but alternative I presented is also an interesting one.
Edit: here's a graph of the function: http://www.wolframalpha.com/input/?i=y%3Dx%5E2%5Clceil%5Cfrac%7B10%5Clfloor+%7Bx%5Cover10%7D%5Crfloor%7D%7Bx%7D%5Crceil-7x%5Clceil%5Cfrac%7B10%5Clfloor+%7Bx%5Cover10%7D%5Crfloor%7D%7Bx%7D%5Crceil%2B7x+from+0+to+20
| {
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Help me understand how we define closed sets in real analysis I'm currently reading the second version of Understanding Analysis by Stephen Abbott and on page $90$ theorem $3.2.8.$ states that "A set $F \subseteq \mathbb{R}$ is closed if and only if every Cauchy sequence contained in $F$ has a limit that is also an element of $F$". According to definition $3.2.7.$ on the same page "A set $F \subseteq \mathbb{R}$ is closed if it contains its limit points".
The proof was left as an exercise. I think I pretty much proved it but I need help understanding some of the logic involved here.
If $F$ only contains a single point then there are no sequences in $F$ that do not contain their limit so there are no limit points. Is $F$ still closed? I assume that some sort of logical mambo jambo can be applied here to show that this is the case. To say that $F$ contains all its $0$ limit points does not sit very well with me since we're treating "nothing" as a "something". We're essentially saying that $F$ contains this "nothing". It makes no sense to me. Personally I decided to just embellish the definition with "If $F$ only contains one point then it is closed, otherwise blablabla...". Is this equivalent to the previous definition?
| You are correct when you say that a singleton set does not contain any limit points.
However, think of the proposition in the following fashion : $F$ is closed, if the set of limit points of $F$ is a subset of $F$.
Then, the set of limit points of a singleton set, is the empty set, and is therefore contained in $F$. This notion clarifies the doubt of how $F$ can contain nothing, to a better notion of set containment.
| {
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unexplained modulo, rs ≡ 1 (mod m) where r = 3, s= 59 and m = 176 I'm trying to work through the math behind public key encryption, I'm a computer programmer, but not a mathematician. I came across this wonderful example, but I'm confused about the use of (mod x) my understanding of modulus is that it is the remainder after devision. e.g. 5 % 2 = 1. but this is odd:
rs ≡ 1 (mod m) where r = 3, s = 59 and m = 176
how can s = 59 my understanding is that 1 mod 176 would be 1.
| You should know that $a \equiv b (\text{mod $m$)}$ means that $a-b$ is divisible by $m$.
For this modulo, you should note that $0$ is divisible by all positive natural numbers (result would be $0$), so $1$ when divided by any positive natural numbers (except $1$) will have $1$ as a remainder $\Rightarrow 1 \equiv 1 (\text{mod $m$)}$.
For this problem, $177$ when divided by $176$ will have a remainder of $1$, or $177-1$ is divisible by $176$ $\Rightarrow 3 \times 59 \equiv 1 (\text{mod $176$)}$.
| {
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How to evaluate $\int_a^b(b-x)(x-a)dx$ in a smart way? It holds that $\int_a^b(b-x)(x-a)dx=1/6(b-a)^3$. I was wondering if there is a smart way of seeing this, instead of brute-force solving the intergral (it's not that tedious, but if there is a quick way about it, I would like to know it)
If it's not possible to do it smartly, then my other question would be how we go about factorizing the following expression:
$$
1/6(b^3-a^3+3a^2b-3ab^2).
$$
I mean, I know now that it equals $1/6(b-a)^3$, but what are the steps for factorizing this? Should we think of it as a polynomial in $a$ and $b$? Whenever I think of a factorization of a polynomial, I think of zeroes, but I'm not sure how that would work here.
| You may use the fact that for quadratic polynomials the Simpson $1-4-1$ rule is an exact quadrature formula. In particular
$$ \int_{a}^{b}(b-x)(x-a)\,dx = \frac{b-a}{6}\cdot 4\left(\frac{a-b}{2}\right)^2 =\frac{1}{6}(b-a)^3.$$
As an alternative, you may substitute $x=a+\lambda(b-a)$ and exploit
$$ \int_{0}^{1}\lambda(1-\lambda)\,d\lambda = B(2,2) = \frac{\Gamma(2)^2}{\Gamma(4)}=\frac{1}{6}.$$
In a elementary geometric fashion, the identity above is a consequence of the Archimedean quadrature of the parabolic segment.
| {
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Counting 3-cycles in $S_6$ I've bolded the incorrect part.
Find the number of order $3$ elements in $S_6$
My "Solution":
There are two ways to obtain these, either a 3-cycle or product of 2 disjoint 3-cycles.
Counting 3-cycles:
Choose 3 of 6 elements, this will determine the first 3 elements. We don't worry about the order of the rest. To get the distinct cycles (noting $(1 2 3) = (2 3 1)$), we fix the first element and permute the remaining elements. Thus:
$\frac{6!}{3!3!} (3-1)!=40$
Counting Disjoint 3-cycles:
Choose 3 of 6 elements, this will determine the first 3 elements. Using the above computation we obtain $40$. We need to account for of the rest. Noting again that some cycles are similar, we fix the first element and permute the remaining to get distinct cycles, thus: $40\times (3-1)! = 80$
Thus: $120$ elements of order $3$.
The bolded reasoning is incorrect. The answer key says it should be 40. How am I over counting?
| Two elements of $S_{n}$ are conjugate if and only if they have the same cycle structure. There are two possible cycle structures for an element of order $3$ in $S_{6}$, as you noted yourself. Hence we need to count the conjugates of $x = (123)$ and of $y = (123)(456)$ in $S_{6}$ ( and add these numbers to get the number of elements of order $3$ in $S_{6}$). Now $C_{S_{6}}(x)$ has order $18$ (it's isomorphic to $\langle x \rangle \times S_{3}$) so $x$ has $6!/18 = 40$ conjugates, and $C_{S_{6}}(y)$ also has order $18$ (it's isomorphic to $C_{3} \wr C_{2},$ which is in fact isomorphic to $C_{3} \times S_{3}$), so $y$ also has $40$ conjugates. The key point to note is that the element $(14)(25)(36)$ commutes with $(123)(456) = y,$ which gives a slightly less obvious element of $C_{S_{6}}(y)$ and in one sense explains why the reasoning in bold in the question statement is incorrect (since there you get twice as many elements there as you should).
Hence $S_{6}$ contains $80$ elements of order $3$ in total, as already noted above by a different method.
| {
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The right adjoint of a functor is unique up to a natural isomorphism I'm having trouble showing that the right adjoint of a functor $F \colon \mathcal{C} \to \mathcal{D}$ is unique up to isomorphism. (My definition of adjoint functors is the one using units and counits.)
Let $G_1$ and $G_2$ be two right adjoints of $F$, with units $e_1,e_2$ and counits $\epsilon_1,\epsilon_2$ respectively. We can construct natural transformations
$$f \colon G_1 \overset{e_2G_1}\longrightarrow G_2FG_1 \overset{G_2\epsilon_1}\longrightarrow G_2$$
and
$$g \colon G_2 \overset{e_1G_2}\longrightarrow G_1FG_2 \overset{G_1\epsilon_2}\longrightarrow G_1.$$
I'm not quite sure how to show that these are inverses of each other. For example, consider $g \circ f \colon:G_1\to G_1,$ which is the map
$$G_1\epsilon_2 \circ e_1G_2 \circ G_2\epsilon_1 \circ e_2G_1.$$ How do I show that this is the identity? I know that $G_i\epsilon_i \circ e_iG_i$ is the identity natural transformation on $G_i$, but how do I use that information here, since the subscripts in the composite are alternating?
| Hint 1: Find the arrows in the following diagram which make its squares commutative, using naturality.
$$\matrix{G_1 & \to & G_2FG_1 & \to & G_2\\
\downarrow && \downarrow && \downarrow \\
G_1FG_1 & \to & G_1FG_2FG_1 & \to & G_1FG_2 \\
&&\downarrow && \downarrow \\
&&G_1FG_1 & \to & G_1}$$
Hint 2: Alternatively, we can draw these natural transformations as diagonal morphisms $\swarrow$ in a square of functors, expressing $\alpha:UF\to GV$ as
$$\matrix{&\overset F\longrightarrow&\\ {}_V\downarrow&\alpha&\downarrow_{\,U}\\
&\underset G\longrightarrow}$$
Then draw the units $\eta_i:=e_i$ as squares for $1\ \!1\to G_iF$ and the counits as squares for $FG_i\to 1\ \!1$, and consider the following arrangement:
$$\pmatrix{ \varepsilon_1 &\eta_2 \\ &\varepsilon_2&\eta_1} $$
Determine the functors on the borders, and show that, in general, pasting the squares horizontally commutes with pasting them vertically.
| {
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Find the equation of the normal at the point $(1,2)$ to the curve $y=x+\frac{1}{x}$. The answer in the textbook is listed as $x=1$, however I don't understand why this is.
My working:
$$y=x+\frac{1}{x}$$
$$f'(x)=1-x^{-2}$$
$$f'(x)=1-\frac{1}{x^{2}}$$
$$f'(1)=1-\frac{1}{1}=0$$
Therefore the gradient of the tangent to the curve at $x=1$ is $0$.
The gradient of the normal is given by:
$$m_1m_2=-1$$
$$m_2=\frac{-1}{0}$$
As anything divided by $0$ is undefined, how do I then use this to find the the normal to the tangent, to the curve, at $x=1$? Is there a way of analytically working out tangents/normals when they are undefined, or is an alternative approach required?
EDIT:
My alternative approach would entail ascertaining the tangent at $(1,2)$, which turns out to be $y=2$ - a horizontal line; which means that the normal would be perpendicular to this and is thus given by $x$-value at this point, i.e. $x=1$.
Is a visual way the only method of producing an answer? Is there a better way of demonstrating this?
| Perhaps a picture will clarify the situation.
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Can this statement about supremum be proven directly? Theorem (I.32, pg 27, Apostol's Calculus Vol.1): Let $\varepsilon \in \mathbf{R}^+$ and let $S \subset \mathbf{R}$. If $\sup S$ exists, then there exists $x\in S$ such that $$x > \sup S -\varepsilon.$$
(Reminder: The real number $B = \sup S$ is the least upper bound of $S$; that is, $B$ is an upper bound for $S$ and no number less than $B$ is an upper bound for $S$.)
Now this is proven very elegantly in 1 line, via contradiction. For if $x\leq \sup S - \varepsilon$ for all $x\in S$ then $\sup S-\varepsilon$ would be an upper bound for $S$ smaller than its least upper bound. I am wondering if this statement could be proven directly, with out arguing by contradiction.
The reason I am interested in a direct proof is because I think I get a better, more permanent understanding of things when I hear their direct proof.
| For a direct prove:
Let $\epsilon >0$ be given.
Note that $ (\sup S - \epsilon) < \sup S $
Therefore $ (\sup S - \epsilon)$ is not an upper bound of S, because $\sup S$ is the least upper bound of S.
Thus there is an element of S, say $x$ such that $x> \sup S - \epsilon$
| {
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Finding value of $\int\frac{\ln(x)}{1+x^2}dx$
Finding value of $\displaystyle \int\frac{\ln(x)}{1+x^2}dx$
Try: let $$I=\int\frac{\ln(x)}{1+x^2}dx=\frac{1}{2}\int\bigg[\frac{\ln x}{1-ix}+\frac{\ln x}{1+ix}\bigg]dx$$
$$I=\frac{1}{2}\int\frac{\ln x}{1-ix}dx+\frac{1}{2}\int\frac{\ln x}{1+ix}dx$$
Put $1+ix =t$ Then $dx=-idt$ and $1-ix=u$ Then $dx=idu$
So $$I=\frac{i}{2}\int \frac{\ln(1-u)-\ln(i)}{u}du-\frac{-i}{2}\int \frac{\ln(1-t)-\ln(i)}{t}dt$$
Could some help me to solve it , Thanks
| Note that
$$\ln(x)+C=\int\frac1x{\rm~d}x$$
and
$${\rm Li}_2(x)+C=\int\frac{\ln(1-x)}x{\rm~d}x$$
where ${\rm Li}$ is the polylogarithm, since
$${\rm Li}_1(x)=\ln(1-x)$$
and
$${\rm Li}_{s+1}(x)=\int_0^x\frac{{\rm Li}_s(t)}t{\rm~d}t$$
| {
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Is Cartesian product of Graph and Composition of Graph the same thing? As you must have seen from my previous questions, I am mostly dealing with probabilistic maths so Graph is completely new to me. Still, I am trying to understand a paper that utilizes Graph theory.
I understand the Cartesian product of Graphs, my question is...Is the "Cartesian Product" and "Composition " of the graph the same thing?
| No, the composition of two graphs $G_1$ and $G_2$ is their Cartesian product plus an edge between every pair of vertices $(u_1,v_1)$ and $(u_2,v_2)$ where $u_1$ and $u_2$ are adjacent in $G_1$ and $v_1\neq v_2$.
| {
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The equality of two conditional probability Suppose that a company has had $70\%$ female applicants and $30\%$ male applicants since it was founded. We also assume that $70\%$ of all staff are female.
Let $F$ denote female and $M$ denote male and $S$ denote the event that an applicant is successful.
The question is: Is $P(S|F)=P(S|M)$?
Here is my working but perhaps it is not correct.
By the conditional probability we have
$$
\begin{align*}
P(S|F)&=\frac{P(S\cap F)}{P(F)}=\frac{0.7}{0.7}=1\\
P(S|M)&=\frac{P(S\cap M)}{P(M)}=\frac{0.3}{0.3}=1
\end{align*}
$$
So $P(S|F)=P(S|M)$.
| You're saying that given an applicant is female, they are guaranteed to get the job, and similarly for males. That is probably not true. $P(S\cap F)$ does not necessarily equal $0.7$. If there were $70$ female applicants of $100$ total applicants and $7$ females got the job then $P(S\cap F)=0.7\cdot0.1=0.07.$
We have
$$
\begin{align*}
P(S\mid F)&=\frac{P(F\mid S)\cdot P(S)}{P(F)}=\frac{0.7\cdot P(S)}{0.7}\\
P(S\mid M)&=\frac{P(M\mid S)\cdot P(S)}{P(M)}=\frac{0.3\cdot P(S)}{0.3}
\end{align*}
$$
From here the $P(S)$'s cancel and you can deduce the desired result.
| {
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Gradient of a function with respect to a matrix How can I compute the gradient of the following function with respect to $X$,
$$g(X) = \frac{1}{2}\|y-AX\|^2$$
where $X\in\mathbb{R}^{n\times n}$, $y\in\mathbb{R}^m$, and $A:\mathbb{R}^{n\times n}\to \mathbb{R}^m$ is linear. We can assume that $A$ is of the form,
$$A = \begin{pmatrix}\langle X| A_1\rangle\\\vdots\\\langle X|A_m\rangle\end{pmatrix}$$
where $A_1,\ldots,A_m$ are $n\times n$ real matrices and the inner product is the Frobenius inner product.
Edit: my attempt at finding the gradient,
$$g(X+H) = \frac{1}{2}\langle y-A(X+H), y-A(X+H)\rangle,\\
= \frac{1}{2} \langle y-AX-AH, y-AX-AH\rangle,\\
=\frac{1}{2} \left(\langle y-AX, y-AX\rangle -\langle y-AX,AH\rangle -\langle AH, y-AX\rangle +o(\|H\|)\right),\\
=g(X) - \langle y-AX, AH\rangle,\\
=g(X)-\langle A^*\left(y-AX\right),H\rangle,\\
\implies \nabla g(X) = -A^*\left(y-AX\right)$$
Now I must compute the adjoint operator $A^*$ of $A$.
To find $A^*$ we do the following,
$$\langle y, AX\rangle = \sum\limits_{i=1}^m y_i\langle X, A_i\rangle=\sum\limits_{i=1}^m \langle X, y_iA_i\rangle = \langle X, \sum\limits_{i=1}^my_iA_i\rangle$$
to see that $A^*y = \sum\limits_{i=1}^m y_iA_i$. Applying this to the expression we found above gives,
$$\nabla_Xg(X) = -A^*(y-AX) = -\sum\limits_{i=1}^m\left(y_i-\mbox{tr}(X^TA_i)\right)A_i.$$
| Write the cost function as
$\phi
= \frac{1}{2} \sum_i z_i^2
$
with $z_i=
\mathrm{tr}
\left( \mathbf{A}_i^T \mathbf{X} \right)
-y_i
$.
Then taking a differential approach,
we obtain
$$
d\phi
= \sum_i z_i
\mathrm{tr}
\left( \mathbf{A}_i^T d\mathbf{X} \right)
=
\mathrm{tr}
\left(
\sum_i z_i
\mathbf{A}_i^T d\mathbf{X} \right)
$$
The gradient is thus
$\sum_i z_i \mathbf{A}_i$.
| {
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Analyzing a function with abstract algebra For an $n$-tuple $S$ of decreasing positive integers, we can define $f(S)$ as subtracting $1$ from every element of $S$, prepending $n$, and then removing $0$s and re-ordering in decreasing order if neccecary. For example, $f((4,2,1))=(3,3,1)$
We are just learning group theory, and our teacher asked us if we can use it to prove that all tuples will eventually reach a cycle when repeatedly applying this function.
My idea is using $f$ as our sort of group action acting on the set of finite decreasing tuples of positive integers, and then the proof would just be proving all elements have finite order, but I've hit a speedbump - a group action has to be a composition of two elements, i.e $x\cdot y$. This is a function that's just applied to one. I thought of maybe symmetric groups because I heard those have composition as a group action but their elements are functions so I don't think that's helpful here.
So how can this function be analyzed from a group-theoretic perspective?
EDIT: Accidentally deleted this, I just un-did it. Hope I didn't mess up anything.
| You just have to prove that if $x=(x_1,\dots,x_n)$ is your tuple then you only have a finite number of elements in the sequence $$x,f(x),f(f(x)),\dots $$
Hint: look at the sum of the elements constituting the tuple in $x$ and in $f(x)$.
If you do this then you will necessarily reach twice the same element in the sequence and thus it will lead to a cycle.
Remark: I wouldn't say it is about group theory (maybe about the monoidal action of $\{Id,f,f^2,f^3,\dots\}$). However, I can see why this kind of reasoning is used to show basic group theoretic statements such as: in a finite group $G$, for any $g\in G$, there exists $n_g\in\mathbb{N}$ such that $g^{n_g}=g^{-1}$.
| {
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Given a straight line with some distance, a fixed rate of acceleration/deceleration, and a time, find the velocity needed I have a path which is a straight line with some distance d.
I have a particle which is at position 0 in the line, with a resting velocity of 0 m/s. The particle has a fixed rate of acceleration at 2 m/s^2 (this rate is set to 0 once a desired velocity has been reached). The particle has a deceleration rate of -2 ms/s^2 as well.
The time taken for the particle to straight moving, maintain some velocity, and come back to a velocity of 0 after d meters have been traversed is fixed.
I am trying to solve for the velocity that the particle must accelerate to, along with the points in time where the particle needs to stop accelerating, and the point where the particle needs to begin decelerating.
When approaching this problem I can divide distance by time, and get the velocity that would be maintained if the point could instantaneously accelerate to a given velocity, and instantaneously stop at the end.
I then am trying to adjust this by accounting for the time needed to reach the desired velocity, and the time needed to return to 0 from that desired velocity. Am I on the right track here in solving this problem?
| You can draw a line AB, then draw points AC and DB, where AC = DB, which is the distance between the point where the particle starts accelerating to the point where it reaches max speed (same thing for deceleration). So, CD would be the distance it travels at maximum speed.
AC and DB would be:
$t(t-1)$ Since the distance that the particle travels every second is a series of arithmetic progression (0 + 2 + 4 + 6 + 8 +...).
$ S = \frac{t(t-1)*2}{2}$
So if you want to reach to the desired velocity and decelerate at the same rate, you would nee a distance of at least $2*(t^2 - t)$, and the distance it remains top speed at is $x - 2*(t^2 - t)$
| {
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Topological space , containing a point whose closure is the whole space, is contractible? Let $X$ be a topological space containing a point whose closure is the whole space $X$. Then is $X$ contractible ?
I feel it is, but I am unable to come up with a proof. Please help.
| If $\eta$ is the dense point and $I=[0,1]$ the map $f:X\times I\to X$ defined by $$f(x,0)=x \quad \operatorname {and} \quad f(x,t)=\eta \quad \operatorname {for} t\gt0$$ is the required contraction of $X$ to $\eta$ .
| {
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How to find two monic polynomials with the same number of roots mod p for each prime p Given an arbitrary monic polynomial $f_1$, is there an elegant way of identifying another $f_2$ such that both have the same number of roots (counting repeated roots as distinct) of $f_i \equiv 0 \bmod p$ for each prime $p$, excluding the trivial transformations $f_2(x)=f_1(ax+b)$?
A sufficient condition is that their real roots can be related by rationals, but identifying a suitable $f_2$ from a given $f_1$ seems to require trial and error and much messy algebra. For example, $f_1=x^3+756x+1302$ and $f_2=x^3-126x+714$ have discriminants $-193^2 \times 2^2 3^5 7^2$ and $-11^2 \times 2^2 3^5 7^2$ respectively, and their real roots $\alpha,\beta$ are related by $\beta=\dfrac{-5\alpha^2+29\alpha-2520}{193}$, $\alpha=\dfrac{5\beta^2+37\beta-420}{11}$. The polynomial $f_3=x^3+42x-182$ has discriminant $-5^2 \times 2^2 3^5 7^2$ but its real root doesn't appear to have any similar relation to $\alpha$ or $\beta$. As an example of correspondence of number of roots, $f_1 \equiv 0 \bmod 19$ has three roots: 9,14,15, $f_2 \equiv 0 \bmod 19$ also has three roots: 1,5,13, but $f_3 \equiv 0 \bmod 19$ has no solutions.
If $f_1(r) \equiv 0 \bmod p$, then $(-5 \times r^2+29 \times r-2520)193^{-1} \bmod p$ is a root of $f_2 \equiv 0 \bmod p$, so it's clear why $f_1$ and $f_2$ have the same number of roots.
| Note that the better question here is why $f_1$ and $f_2$ have the same number of roots over $\mathbb{F}_p$ for every prime $p$, although $f_1$ and $f_2$ are not related by a transformation $x\mapsto x+a$. Indeed,
$$
f_1(x+a)=x^3 + 3x^2a + 3x(a^2 + 252) + a^3 + 756a + 1302,
$$
so that $f_2=f_1(x+a)$ would imply $a=0$, a contradiction.
| {
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Prox Operator of a First Order Perturbation (Adding Linear Term to the Function) Given a function $f$ we can describe its proximal operator as,
$$\mbox{prox}_{\frac{1}{\rho}f}(x) = \arg\min\limits_{u} f(u) + \frac{\rho}{2}\|x-u\|^2$$
How does this change if we introduce a linear perturbation, i.e. we replace $f$ with $f(x) + \langle\mu,x\rangle$? Assuming $f$ is simple, i.e. the prox of $f$ has a closed form, does the perturbation also have a simple prox?
I am curious how easy it is to calculate,
$$\arg\min\limits_{u}f(u)+\langle \mu, u\rangle +\frac{\rho}{2}\|x-u\|^2$$
in particular when $x$ is an $n\times n$ matrix and $f(x) = \lambda \|\mbox{vec}(x)\|_{1}$
| [Corrected via comments] Sure. The optimality condition for your original prox function is
$$0 \in \partial f(u) - \rho ( x - u)$$
For the perturbation, it is
$$0 \in \partial f(u) + \mu - \rho ( x - u ) = \partial f(u) - \rho ( x - \rho^{-1} \mu - u)$$
So basically, your perturbation is solved by $$\textstyle\mathop{\textrm{prox}}_{\tfrac{1}{\rho}f}(x-\rho^{-1}\mu)$$
| {
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Show probability bound. Let $Y$ be the number of successes in $n$ trials of a Bernoulli experiment with success probability $p$. Show that:
$$ Pr(|\frac{Y}{n} - p |<e) \geq 1 - \frac{1}{4ne^2}$$
I tried starting with Chebychev and work backwards but I got stuck:
I used $np(1-p)$ for variance since binomial.
$$
\begin{align}
Pr(|Y-\mu|\geq a) &\leq \frac{np(1-p)}{a^2}\tag1\\
Pr(|\frac{Y}{n}-\frac{\mu}{n}|\geq \frac{a}{n}) &\leq \frac{np(1-p)}{a^2}\tag2\\
\end{align}
$$
Since for binomial $\mu/n=p$
$$
\begin{align}
Pr(|\frac{Y}{n}-p|\geq \frac{a}{n}) &\leq \frac{np(1-p)}{a^2}\tag3\\
1-Pr(|\frac{Y}{n}-p|\geq \frac{a}{n}) &\geq 1-\frac{np(1-p)}{a^2}\tag4\\
Pr(|\frac{Y}{n}-p|< \frac{a}{n}) &\geq 1-\frac{np(1-p)}{a^2}\tag5\\
\end{align}
$$
I then set $e=a/n$
$$
\begin{align}
Pr(|\frac{Y}{n}-p|< e) &\geq 1-\frac{np(1-p)}{(en)^2}\tag6\\
Pr(|\frac{Y}{n}-p|< e) &\geq 1-\frac{p(1-p)}{e^2n}\tag7\\
Pr(|\frac{Y}{n}-p|< e) &\geq \frac{e^2n-p(1-p)}{e^2n}\tag8\\
\end{align}
$$
I gave up here because I still need to flip the greater than in the prob and get the RHS to be exactly the same (how??) and I still don't really know if this is the right approach to start with anyway.
| Check out Hoeffding's inequality on Wikipedia. It won't give you exactly the answer that you're looking for: instead, it will give you a far stronger one (asymptotically, as $n$ gets large)! Regarding the precise formlation that you have, you're almost there: all you need to do is note that
$$ p(1-p) \le \tfrac14 \quad \text{for all} \quad p \in [0,1]. $$
I'd definitely advise looking at the Wiki page though. There you'll find a proof of a slightly more general result -- it generalised from Binomial, ie sum of independent Bernoullis, to a sum of independent bounded random variables.
I leave the proof to your reading of that page, and just look here at how to apply it. It says that
$$ P( |Bin(n,p) - np| \ge \epsilon n ) \le 2 \exp(-2 \epsilon^2 n). $$
Note that, $|Y/n - p| < e$ if and only if $|Y - pn| < en$. Hence, applying this, we find that
$$ P( |Y/n - p| \ge e ) = P( |Y - pn| \ge en ) \le 2 \exp(- 2 e^2 n). $$
Taking the compliment gives the probability you desire.
This actually shows a much better bound that you've been asked to show for large $n$. Note that Hoeffding is valid for all $n$.
| {
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$G$ is a group of order $pq$ and $P_q$ and $P_p$ are Sylow subgroups... If $(G,*)$ is a group of order $pq$, then it is clear that there are Sylow subgroups $P_q$ and $P_p$ of order $q$ and $p$ in $G$. If $q>p$ then $P_q$ is normal.
I want to find a decomposition for all $g \in G$, there exists $(q',p') \in P_q \times P_p$ such that $g = q'*p'$, but I have been stuck for about 8 hours. How can I find such decomposition?
| Since $p$ and $q$ are different primes, Bézout's Lemma guarantees the existence of $m,n \in \mathbb{Z}$, with $1=mp+nq$. Hence $g=(g^p)^m \cdot (g^q)^n$. Now prove that $g^p \in P_q$ and $g^q \in P_p$. Hence powers of these elements are in the same respective subgroups.
| {
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Solving following System of ordinary differential equations. How to solve system of ODEs which contains independent variable like t in equation as in this particular case :
$$\frac{dx}{dt}= -x + ty $$ $$ \frac{dy}{dt} = tx-y$$ can we solve them or additional information may be needed (may be about t) ?
| Hint...you can subtract the equations and write $u=y-x$ to obtain a seperable variable DE in $u$ and $t$
| {
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How to show that the normalizer is the largest subgroup of a group in which a set is normal? Assume we have $H \trianglelefteq K\subseteq G$
We want to prove the $K \subseteq N_G(H)$
However all I have so far is:
Assume $K \subseteq G$ such that $H$ is normal in $K$. Then it must be that $kHk^{-1} \subseteq H$
If $kHk^{-1} = H$ Then trivially $k \in N_G(H)$, but I do not know what to do if that is not the case.
| As $H$ is a normal subgroup of $K$, for $k \in K$, you have that $kHk^{-1} \subseteq H$ and $k^{-1}Hk \subseteq H$. So
$$H = (kk^{-1})H(kk^{-1}) = k(k^{-1}Hk)k^{-1} \subseteq kHk^{-1} \subseteq H.$$
This gives you that $kHk^{-1} = H$.
| {
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Orthogonal vector in a plane I have two vectors $\boldsymbol{a}$ and $\boldsymbol{b}$. Now I want to find a vector $\boldsymbol{v}$ orthogonal to $\boldsymbol{a}$ that is in the plane spanned by $\boldsymbol{a}$ and $\boldsymbol{b}$. Is it ok if I do $\boldsymbol{v} = \boldsymbol{a} \times (\boldsymbol{a} \times \boldsymbol{b})$?
| Yes. $\boldsymbol a\times (\boldsymbol a\times \boldsymbol b)$ is perpendicular to both $\boldsymbol a$ and $\boldsymbol a\times \boldsymbol b$. Being perpendicular to $\boldsymbol a\times \boldsymbol b$ means being on the plane generated by $\boldsymbol a$ and $\boldsymbol b$.
See the other answer for a less expensive computation that outputs such a vector.
| {
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How to find the general solution of $x^2y'' - x(x+2)y' + (x+2)y = 0$ $$x^2y'' - x(x+2)y' + (x+2)y = 0$$
where a particular solution is $y_1(x) = x$
So, can I start off by subbing in $x$ for $y$ since I have a particular solution?
Which would give me:
$\frac{x}{x+2}y'' - y' + y = 0$
If this is a valid approach, then solving this DE would not be too difficult.
| Here $y_1(x)=x$ is your known integral. For finding complete solution in terms of known integral you put: $y=vx$, where $v$ is also function of $x$; then the above ODE becomes $\dfrac{d^2v}{dx^2}-\dfrac{dv}{dx}=0$, which can be easily solve for $v$. Then substitute $v$ in $y=vx$ you got general solution.
| {
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When every minimal subgroup is contained in the center
Let $G$ be a finite group and $(*)$ be the property:
$(*)$: Every minimal normal subgroup is contained in the center.
$(a)$ Let $N$ and $M$ be normal subgroups of $G$, both of which satisfy $(*)$. Then prove:
$NM$ satisfies $(*)$.
$(b)$ If $G$ satisfies $(*)$, then prove: every normal subgroup of $G$ satisfies $(*)$.
I would also agree that it may be a possible duplicate of this post. However, unfortunately, I could find that post a little complicated to understand and not well developed, of which the logic, notation, and language might not have been so polished.
It would be greatly appreciated, if you could throw light on this question and be kind enough to give an elegant proof. Thanks a lot! $\ddot\smile$
| I have an argument for (a) in the case $N \cap M = 1$, which implies $NM = N \times M$. Let $K$ be some minimal normal subgroup of $NM$. Then by considering the quotient $KN / N \unlhd NM / N \cong M$ either $KN \le N$ or $KN / N$ is minimal normal in $MN / N$, in the last case $[x,y] \in N$ for each $x \in K, y \in NM$ by assumption. Similar by looking at $NM / M$ we find $K \le M$ or $[x,y] \in M$ for each $x \in K, y \in NM$. As $[N, G] \le N$ and $[M, G] \le M$ by normality, in all cases we have $[x,y] \le M \cap N = 1$ for $x \in K, y \in NM$, which gives the claim.
If $N \cap M \ne 1$ maybe (b) can help in the sense that we can deduce that property $(*)$ then holds for $M \cap N$.
EDIT (5/11/19): I found a proof for the general case which works quite different. Let $K$ be some minimal normal subgroup in $NM$. Then as $N$ is normal we have that $[K, N] \le N \cap K$ is normal in $NM$, which implies $[K,N] = 1$ or $[K,N] = K$. Assume $[K,N] = K$. Let $L \le K$ be some minimal normal subgroup of $N$ in $[K,N]$ and consider $H = \langle L^g \mid g \in NM \rangle$, which is normal in $NM$. Hence $H = [K,N] = K$.
The groups $L^g$ are also minimal normal in $N$ by normality of $N$ and the fact that non-trivial homomorphic images are minimal normal in its image, hence $H$ is a direct product of some subset of these conjugages. By assumption the factors are all in the center of $N$, hence $K$ is in the center of $N$, i.e. $[K,N] = 1$ contradicting the assumption. Simlar we find $[K,M] = 1$. So $[K,MN] = [K,M][K,N] = 1$, i.e. $K$ is in the center of $NM$. $\square$
| {
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Simplex Solver - is it possible to enforce maximum number of non-zero weights? I have a problem which I am trying to solve using a simplex solver (but I am happy to use any approach that works).
$$
a_1 x_1 + a_2 x_2 + ... + a_n x_n= TotalCost
$$
Where $a_i$ are constant inputs and $x_i$ are the weights that I am optimizing.
I have a constraint that the sum of weights needs to add up to one
$$
\sum^n_{i=1} x_i = 1
$$
I wish to have the largest possible number of non-zero weights. However, some weights will be zero (because they have a very high cost). There are solutions to this (such as this Q). However, this doesn't seem to work with my constraint that the weights always sum to one.
How to impose sparsity (or the opposite) in this case?
--- Added later ---
Take for example $n=4$
$$ a = (0.2, 0.25, 0.25, 1.0)$$
The lowest cost solution is $x = (1, 0, 0, 0)$ but I would like to biase the solution such that $x = (0.8, 0.1, 0.1, 0)$ becomes optimal.
My first instinct would be to add a term of the form
$$
\lambda \sum^n_{i=1} x_i^2
$$
where $\lambda$ is a constant (i.e. something aking to a $L_2$ regularization) but that is no longer linear.
One can add an $ \alpha \ z$ term, where $ z \ge x_i \ \forall \ i$ but such a constraint seems to either do nothing (when $\alpha$ is too small) or force all $x_i$ that are non-zero to be equal.
| You could add extra binary variables $y_i$ that take value $0$ when $x_i=0$ and penalize $1-y_i$ in the objective function:
Add the following constraint, where $\varepsilon$ is a small constant $<1$:
$$
x_i\ge \varepsilon \;y_i\\
y_i \in \{0,1\}
$$
Then minimize the following term in the cost function, where $p$ is another constant (the penalty):
$$
p\sum_{i}(1- y_i)$$
When $x_i>\varepsilon$, $y_i=1$ and no penalty is accounted for in the objective function. When $x_i=0$, $y_i=0$ and the cost function is penalized. The difficulty may be in finding the best value for $\varepsilon$.
| {
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Discontinuity set of an increasing right continuous function Let $f:[0,\infty) \rightarrow [0,\infty] $ be both right-continuous and increasing. Is it possible that $f$ is discontinuous at dense subset of $\mathbb R$? I already know that if we drop right-continuity, then $f$ can be discontinuous at every rational.
Any hint would be appreciated. Thanks and regards.
| This should work:
$f(x)=\sum_{q_n \leq x} 2^{-n},$ where $\{q_n\}$ is an enumeration of the rationals $\mathbb{Q} \cap [0,+\infty)$.
Tell me if you have some doubts.
| {
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Shouldn't $2^x \sin\frac{180}{2^x}$ approach $\pi$ as $x$ gets large? Take a circle with diameter $1$. Obviously its circumference is $\pi$.
Draw a square inside this circle (biggest such). Since the circle's diameter is 1, we can work out that each side of the square would be $\frac{1}{\sqrt{2}}$. Its perimeter, therefore, is $2\sqrt{2}$.
Imagine a triangle being drawn over each side of the square, so as to create a regular octagon. Using the law of sines, we can work out the measure of each side of the octagon. The base of each triangle would be $\frac{1}{\sqrt{2}}$, and the angle opposite to it $135^o$. Each of the other angles would be $\frac{180-135}{2} = 22.5^o$.
So,
$$
\frac{1}{\sin135(\sqrt2)} = \frac{x}{\sin22.5}
$$
$$
1 = \frac{x}{\sin22.5}
$$
$$
x=\sin22.5
$$
Therefore the perimeter of this octagon is $8\sin22.5$.
We can continue drawing triangles on this octagon, and more triangles on that, and so on. Once the number of sides reaches infinity (which, of course, is not possible in the physical world), the perimeter of the said figure would be $\pi$.
I am trying to write a similar function $f(x)$ such that
$$
\lim_{x \to 0}f(x) = \pi
$$
Note that $f(x)$ is a function of levels. Level 1 is for 4 sides, level 2 is for 8 sides, then 16 sides, 32 sides and so on.
Insofar:
$$f(1) = 2\sqrt2 = 4 \sin 45$$
$$f(2) = 8 \sin22.5$$
$$...$$
I work out that
$$f(x) = 2^x \sin\frac{180}{2^x}$$
When graphing that function:
Given this, it doesn't seem to limit on the graph. Say, if it does limit to $\pi$, would it approximate $\pi$ for a relatively large $x$?
Please note. There might be a lot of mistakes here. I'm not studying math at the college-level and is a careless person.
| The limit should be $180$ if you are taking the sine in radians. Yes, it should be $\pi$ if you are taking the sine in degrees. You should be graphing for larger values of $x$ as even $\frac {180}{2^8} \gt \frac 12$ is not so small. Below is a plot from Alpha that shows nice convergence to $180$ using radians
| {
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how many fish are needed to have probability 2/3 of at least two fish in a bucket?
Lets say we have $f$ fish and $b = 10$ buckets. You toss the fish independetly and uniformly into the buckets.
Define $A_{f} =$ "There exists a bucket that has at least two fish"
Let $p_{f} = Pr(A_{f})$
What is the lowest value of $f$ such that $p_{f} \ge 2/3$ ?
Here's my intuition:
$f = 1:$
When we are just about to throw the first fish, there are no fish in any buckets, so the probability that it will land in a bucket with another fish is $0$.
$f = 2:$
When we are just about to throw the second fish, the probability that it will land in a bucket with another fish is $1/10$ because there is one bucket with one fish. Similairly, there are $9/10$ other buckets with no fish in them, so $9/10$ is the probability that it will land in an empty box.
$f = 3:$
Proability of at least two fish is $2/10$, probability of landing in empty bucket is $8/10$.
$...$
I think there is a pattern here: $p_{f} = {\frac{f-1}{10}}$
So for the probability of $2/3$, thats:
$2/3 = \frac{f-1}{10}$
$20/3 = f-1$
$\lfloor23/3\rfloor = f$
$\lfloor23/3\rfloor = 7$ fish
It cannot be $\ge 10$ because if it were, then the probability would be $1$ (pigeonhole principle). So there must be a lower bound.
Which makes sense to me. Now my question is how can generalize this and turn it into a proper solution, assuming I am correct? If I am not, can anyone explain?
| This is sometimes called the birthday problem.
In general if there are $f$ fish and $b$ buckets then the probability that no bucket has two fish is:
$(1-\frac{1}{b})(1-\frac{2}{b})\dots (1-\frac{f-1}{b})$
So the probability that there is at least one bucket is:
$1-(1-\frac{1}{b})(1-\frac{2}{b})\dots (1-\frac{f-1}{b})$.
So we want the smallest $f$ such that:
$(1-\frac{1}{10})(1-\frac{2}{10})\dots (1-\frac{f-1}{10})\leq \frac{1}{3}$
Notice:
$9>3.\overline 3$
$9*8=72>33.\overline 3$
$9*8*7=504>333.\overline 3$
$9*8*7*6=3024\leq3333.\overline 3$
So the answer is $5$ fish
| {
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Question on projection operator and direct sum Here is question 10,section 6.6,Hoffman and Kunze:
Let $F$ be a field of characteristic 0.Let $V$ be a finite dimensional vector space over $F$.Suppose that $E_1,..,E_k$ are projections of V such that $E_1+..+E_k=I$.Prove that $E_i E_j$=0.
My attempt:
Let $W_i=E_i(V)$.
Then we get $V=W_1+...+W_k$.
Again from matrix representation on both sides of $E_1+..+E_k=I$ we get $dim W_1+..dim W_k=dim V$(since trace of matrix rep.of $E_i$ is dim $W_i$).Hence we have $V$ is the direct sum of $W_1,..W_k$.
How do i proceed after this? $V$ being the direct sum of $W_i$,we will obtain k projections,but they may not equal $E_i$,right?
| Hint:
Assume that for some $\alpha\in V$, $\ E_iE_j\alpha\neq 0$, and then conclude that $E_j\alpha\in W_i\cap W_j$ which contradicts $V$ being the direct sum of $W_1,..., W_k$.
| {
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Intuition behind the Banach fixed-point theorem The theorem appeared as an exercise in my real analysis book and only considered functions in $\mathbb{R}$ but the proof of the general theorem seems to be almost identical after looking up the wikipedia article. Is there a somewhat intuitive way of thinking about this theorem? I understand the proof and what it entails but I don't see why the result ought to hold given the necessary conditions. Is there a way of convincing someone that the theorem ought to be true without actually proving it? Some vague geometric intuition would be nice to have a picture in my head of what's going on.
| Generally, a function between metric spaces can be very wild with respect to the distances. But if we demand that the function actually decreases the distance between any two points, then it becomes difficult to actually construct such functions. Looking at a few examples of such functions one sees that one easy way to obtain such a function from a space to itself is to choose a point and treat it as a sort of magnet, where the function describes how points move toward it, as if the point exerts a gravitaional field, thus shrinking distances. If the gravitational pull is strong enough, the shrinking will be at least by a factor smaller than 1. The Banach fixed point theorem says that any endo-function shrinking distances sufficiently fast must be the result of such a point pulling things to it. If the space is not complete then that point may not be in the space but rather in its completion, but it’s still ‘there’. So, if globally the function behaves like the result of a pulling point with sufficient force, then that point is really there, since (intuitively?) if there is no point pulling everything to it, how can all the distances be tapidly shrinking no matter where you are?
| {
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Linear mapping of a binary vector to a decimal-based binary vector Given a binary vector $\mathbf{v}$, where $\mathbf{v} \in \{0,1\}^N$ and the binary-to-decimal conversion of $\mathbf{v}$ is equal to $j$, is there a way to linearly map the vector $\mathbf{v}$ to a binary vector $\mathbf{e}_j$ with $\mathbf{e} \in \{0,1\}^{2^N}$, and its $(j+1)$-th element as $1$ (starting the index from zero)?
For example if $\mathbf{v} = [1 \quad 0]$ (that represents 2 in decimal) how $\mathbf{v}$ can be linearly mapped to $\mathbf{e}_3$, that is, $\mathbf{e}^\mathrm{T} = [0\quad 0 \quad 1 \quad 0]$.
The use case is in the objective function of an ILP.
| If by "linearly map" you mean a real matrix $H$ such that $\mathbf{e} = H \mathbf{v}$ yields what you want, the answer is no. If $$\left(\begin{array}{c} 0\\ 0\\ 1\\ 0 \end{array}\right)=H \left( \begin{array}{c} 0\\ 1 \end{array}\right)$$ (encoding of 2) and $$\left(\begin{array}{c} 0\\ 1\\ 0\\ 0 \end{array}\right)=H \left( \begin{array}{c} 1\\ 0 \end{array}\right) $$(encoding of 1), then what you get for the encoding of 3 is$$\left(\begin{array}{c} 0\\ 1\\ 1\\ 0 \end{array}\right)=H \left( \begin{array}{c} 1\\ 1 \end{array}\right) = H \left( \begin{array}{c} 1\\ 0 \end{array}\right) + H \left( \begin{array}{c} 0\\ 1 \end{array}\right),$$whereas you want$$\left( \begin{array}{c} 0\\ 0\\ 0\\ 1 \end{array}\right).$$
That said, the conversion from $\mathbf{v}$ to $\mathbf{e}$ can be handled by linear constraints. Treating the components of both vectors as binary variables, you just need$$\sum_{i=1}^{2^N}e_i \le 1$$and $$\sum_{i=1}^{2^N} i \cdot e_i = \sum_{j=1}^N 2^{j-1} \cdot v_j.$$
| {
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Find the equilibrium position of an object which satisfies the equation of motion Find the equilibrium position of an object which satisfies the equation of motion. I know how that there is a formula $$ d^{2}x/dt^{2} = - \omega^{2}x $$ but I can't see the usage of it when there are so many "things" on the right side of the equation.
$$ 4d^{2}x / dt^{2} = −x^{3} + x^{2} − x + 1 $$
| Recall Newton's law
$$
\frac{{\rm d}^2 x}{{\rm d}t^2} = -\frac{{\rm d}V}{{\rm d}x}
$$
In you case, it is pretty straightforward to find the potential $V$
$$
V(x) = \frac{x^4}{16} - \frac{x^3}{12} + \frac{x^2}{8} - \frac{x}{4}
$$
You see that the dominant term is $\sim x^4$, this means that the potential goes to infinite, therefore, it must have an equilibrium point somewhere, to find it just calculate the locations where
$$
\frac{{\rm d}V}{{\rm d}x} = 0
$$
If you do the math, you'll find the only real root is $x=1$ (red point in the graph)
| {
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Find the value of $\frac{\tan A}{\tan B}$, given $\frac{\sin A}{\sin B}=5$
If $\displaystyle \frac{\sin A}{\sin B}=5$, then find the value of $\displaystyle \frac{\tan A}{\tan B}$
Try using the Componendo and Dividendo formula:
$$\frac{\sin A+\sin B}{\sin A-\sin B}=\frac{3}{2}$$
$$\frac{\tan(A+B)/2}{\tan(A-B)/2}=\frac{3}{2}$$
Can someone help me find: $$\frac{\tan A}{\tan B}$$
Thanks
| suppose $\dfrac{SinA}{SinB}=\dfrac{x}{y}=5$
so from right tringle we know $cosA=\sqrt{1-x^{2}}$ and so $$tanA=\dfrac{x}{\sqrt{1-x^{2}}}.$$ the same for y so :
$$\dfrac{tanA}{tanB}=\dfrac{\frac{x}{\sqrt{1-x^{2}}}}{\frac{y}{\sqrt{1-y^{2}}}}$$
and we have $x=5y$ so $x^{2}=25y^{2}$ , $1-x^{2}=1-25y^{2}$
$$\dfrac{tanA}{tanB}=\dfrac{\frac{5y}{\sqrt{1-25y^{2}}}}{\frac{y}{\sqrt{1-y^{2}}}}=\dfrac{5y\sqrt{1-y^{2}}}{y\sqrt{1-25y^{2}}}$$
now you need to know what is $y$!
| {
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An urn has 4 balls of 4 different colours Red,Blue,Green,Yellow.
An urn has $4$ balls of $4$ different colours; red, blue, green, and
yellow. I pick one ball at random at first and if it is red, I paint
it blue and return it to the urn. If it is blue, I paint it green. If
it is green, I paint it yellow. If it is yellow, I paint it red. What
is the expected number of trials to get all $4$ balls of the same
colour?
Reminder:
$$\color{red}{red}\to \color{blue}{blue}$$
$$\color{blue}{blue}\to \color{green}{green}$$
$$\color{green}{green}\to \color{yellow}{yellow}$$
$$\color{yellow}{yellow}\to \color{red}{red}$$
I am really stuck with this problem. Help!
| This problem is equivalent to one in which there are four people in four rooms that are joined cyclically by corridors. Initially, each room has one of the four people, and at each turn, one person (not one room) is chosen at random, and this person moves counterclockwise. How long before they end up in the same room?
One can define a Markov chain that records the relative position of the four people. It will be easier to explain this first for the case of three people. There are only four possible states:
*
*One person in each of the three rooms, which we denote $111$.
*Two people in one room, and one person in the next room, which we denote $21$.
*One person in one room, and two people in the next room, which we denote $12$.
*All three people in one room, which we denote $3$.
The dynamics of the chain are also fairly simple:
*
*From $111$, we can only move to $21$. Because we only care about the relative positioning of the people, all three possible resulting arrangements are identical (up to rotation).
*From $21$, we can only move to $12$.
*From $12$, we move to $3$ with probability $1/3$, and to $111$ with probability $2/3$.
For any state $k$, let $t_k$ denote the expected time to reach state $3$. Then $t_3 = 0$, and
$$
t_{111} = 1+t_{21} \\
t_{21} = 1+t_{12} \\
t_{12} = 1+\frac{2t_{111}}{3}
$$
This yields $t_{12} = 7$, $t_{21} = 8$, and in particular, $t_{111} = 9$.
With four people, we obtain the following equations:
$$
t_{1111} = 1+t_{211} \\
t_{211} = 1+\frac{3t_{121}}{4}+\frac{t_{202}}{4} \\
t_{121} = 1+\frac{3t_{112}}{4}+\frac{t_{31}}{4} \\
t_{31} = 1+\frac{3t_{22}}{4}+\frac{t_{103}}{4} \\
t_{202} = 1+t_{112} \\
t_{112} = 1+\frac{t_{1111}}{2}+\frac{t_{22}}{4}+\frac{t_{103}}{4} \\
t_{22} = 1+\frac{t_{211}}{2}+\frac{t_{13}}{2} \\
t_{103} = 1+\frac{3t_{211}}{4}+\frac{t_{13}}{4} \\
t_{13} = 1+\frac{3t_{121}}{4}
$$
When one solves this stack of equations, one obtains $t_{1111} = \frac{1042}{15} = 69\frac{7}{15}$, well in line with Remy's simulation value.
| {
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Is $\frac{cos^4(x)}{4}-\frac{cos^2(x)}{2}+C$ a correct evalution of $\int sin^3(x)cos(x)\ dx$ The answer on Khan Academy states that the integral evaluates to $1 \over 4$$\sin^4(x)+C$
However, I performed a u-substitution that I cannot find a mistake in (maybe I am blind).
Here's the working:
$I=\int \sin^3(x)\cos(x)dx$
let $\ u=\cos(x)$
$du=-\sin(x)dx$
$dx= \frac{-1}{\sin(x)}\ du$
$I=\int \sin^3(x)u\cdot-\frac{1}{\sin(x)}dx$
$I=\int -\sin^2(x)\cdot u\ du$
$I=\int -(1-\cos^2(x))\ u\ du$
$I=\int (\cos^2(x)-1)\cdot u\ du$
$I=\int u(u^2-1)\ du$
$I=\int u^3-u\ du$
$I=\frac{u^4}{4}-\frac{u^2}{2}$
$I=\frac{\cos^4(x)}{4}-\frac{\cos^2(x)}{2}+C$
| Your solution is correct (with minor error at the end) and it is equivalent to the one given in Khan Academy:
$$\begin{align}I&=\frac{\cos^4(x)}{4}\overbrace{\require{cancel}\cancel{+}}^{-}\frac{\cos^2(x)}{2}+C=I= \\
&=\frac{(1-\sin^2 x)^2}{4}-\frac{\cos^2(x)}{2}+C=\\
&=\frac14-\frac{\sin^2 x}{2}+\frac{\sin^4 x}{4}-\frac{\cos^2 x}{2}+C= \\
&=\frac{\sin^4 x}{4}+C+\frac14-\frac12=\frac{\sin^4 x}{4}+A.\end{align}$$
| {
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What is wrong in $(\frac{1}{8})^{-\frac{1}{3}}$? I tried this:
$$\biggl (\frac{1}{8}\biggr)^{-\frac{1}{3}}= \frac{1^{-\frac{1}{3}}}{8^{-{\frac{1}{3}}}}=\frac{-\sqrt[3]{1}}{-\sqrt[3]{8}}=\frac{-1}{-2}$$
Also, is it possible to see what I don't understand here or are there maybe several things..
| $$(1/8)^{-1/3}=((1/8)^{1/3})^{-1}=(1/2)^{-1}=2$$
| {
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Prove $4^n-1$ is divisible by $3$, for all $n\in\Bbb N$? Prove $4^n-1$ is divisible by $3$, for all $n\in\Bbb N$?
I started by assuming there exists some $k\in\Bbb N$ s.t. $4^n-1=3k\iff \dfrac{4^n}3-\dfrac 13=k$, so for $k$ to be a natural number, $4^n\equiv 1\mod 3$ must be true, but this tells us no new information, and I don't know how to follow from there. Any help would be appreciated.
| With only middle school tools:
$$4^n-1=4^n-1^n=(\underbrace{4-1}_{\textstyle 3})(4^{n-1}+4^{n-2}+\dots +4+1).$$
| {
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Solve $\sqrt {x^2-3}=x-3$ in $\mathbb R$
Solve $\sqrt {x^2-3}=x-3$ in $\mathbb R$
My attempt:
$|x^2-3|=(x-3)^2$
So $-(x^2-3)=(x-3)^2$ or $(x^2-3)=(x-3)^2$
If $-(x^2-3)=(x-3)^2=x^2+9-6x$
So no solutions in $\mathbb R$
And if $(x^2-3)=(x-3)^2$
So $x^2-3=x^2+9-6x$
Now, can I delete $x^2$ with $x^2$ ? Like this
$x^2-x^2-3-9+6x=0$
$6x=12$
$x=2$
But $f(2)$ isn’t equal to $0$?
| I was taught to always find the domain of possible solutions first. We have
\begin{cases}
x-3 \ge 0, \\
x^2-3 \ge0
\end{cases} or \begin{cases}
x \ge 3, \\
-\sqrt{3} \le x \le \sqrt{3}
\end{cases}
which has no solutions
| {
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Find $\oint\limits_{|z-\frac{1}{3}|=3} z \text{Im}(z)\text{d}z$ In my test on complex analysis I encountered following problem:
Find $\oint\limits_{|z-\frac{1}{3}|=3} z \text{Im}(z)\text{d}z$
So first I observed that function $z\text{Im}(z)$ is not holomorphic at least on real axis. Therefore we have to intgrate using parametrization.
First, let's change variable $w = z - \frac{1}{3}$. So we got $\oint\limits_{|w|=3} (w+\frac{1}{3}) \text{Im}(w+\frac{1}{3})\text{d}w = \oint\limits_{|w|=3} (w+\frac{1}{3}) \text{Im}(w)\text{d}w = \frac{1}{2i}\oint\limits_{|w|=3} (w+\frac{1}{3}) (w-\bar w)\text{d}w$.
Then by letting $w=3e^{i \phi}$ we transform integral to the form
$\frac{1}{2}\int\limits_{0}^{2\pi}(3e^{i \phi}+\frac{1}{3})(3e^{i \phi}-3e^{-i \phi})ie^{i \phi}\text{d}\phi = -\frac{1}{2}\int\limits_{0}^{2\pi}\text{d}\phi=-\pi$.
Is my reasoning correct? I don't quite sure about change of variable I made since function is not holomorphic at real axis. Is there any other way how this integral can be evaluated? Thanks!
| Note that since $\text{Im}(z)=\frac1{2i}(z-\bar z)$, that
$$z\text{Im}(z)=\frac1{2i}(z^2-|z|^2)$$
Since $z^2$ is analytic, we have
$$\begin{align}
\oint_{|z-\frac13 |=3}z\text{Im}(z)\,dz&=\frac i2\oint_{|z-\frac13 |=3}|z|^2\,dz\\\\
&=-\frac {3}2 \int_0^{2\pi} \left|\frac13 +3e^{i\phi}\right|^2 e^{i\phi}\,d\phi\\\\
&=-\frac {3}2 \int_0^{2\pi} \left(\frac{10}9 +e^{i\phi}+e^{-i\phi}\right)e^{i\phi}\,d\phi\\\\
&=-3\pi
\end{align}$$
| {
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perturbation of differential equation I am struggling with the following problem, which on first site looks easy, but I can't see it.
Given the DE:
$$\frac {d^2y}{dx^2} +y =\frac{\cos 2x}{a+ \epsilon y}$$
with initial conditions: $y(-\pi/4) =y(\pi/4) = 0$, $a>0$ and $|\epsilon| \ll1$
By using the scaling: $y=\alpha z$ this may written in the form: $$\frac {d^2z}{dx^2} +z =\frac{\cos 2x}{1+ \delta z}$$
How is this done? And can $\alpha$ be expressed in terms of $a$? and $\delta$ in terms of $a$ and $\epsilon$?
Any help appreciated..
| Let's substitute $y=\alpha z$ into your equation:
$$\alpha\left(\frac {d^2z}{dx^2} +z\right) =\frac{\cos 2x}{a+ \alpha\epsilon z}$$
$$\frac {d^2z}{dx^2} +z =\frac{\cos 2x}{\alpha a+ \alpha^2\epsilon z}$$
Let $\alpha = 1/a$. Then
$$\frac {d^2z}{dx^2} +z =\frac{\cos 2x}{1+ \frac{\epsilon}{a^2} z}$$
Introduce $\delta = \epsilon / a^2$. Finally:
$$\frac {d^2z}{dx^2} +z =\frac{\cos 2x}{1+ \delta z}$$
| {
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Evaluating $\int\limits_0^\infty{\frac{1}{1+x^2+x^\alpha}dx}$ I'm trying to evaluate$$f(\alpha)=\int\limits_0^\infty{\frac{1}{1+x^2+x^\alpha}dx}$$
I proved:
$f(\alpha)$ converges when $\alpha\in\mathbb{R}$
$f(2-\alpha)=f(\alpha)$
$f(0)=f(2)=\frac{\pi}{2\sqrt{2}}$
$f(1)=\frac{2\pi}{3\sqrt{3}}$
$f(-\infty)=f(\infty)=\frac{\pi}{4}$
Similar question:$$\int\limits_0^\infty{\frac{1}{1+x^\alpha}dx}=\frac{\pi}{\alpha}\csc\frac{\pi}{\alpha}$$
I tried all of the techniques can be used in evaluating this integral, but I still cannot get the answer.
When I was using complex analysis, I found that the poles of $\frac{1}{1+x^2+x^\alpha}$ is hard to be found.
| Not a full answer, some considerations.
First, we make a substitution:
$$x=\tan t$$
The integral becomes (I use $a$ instead of $\alpha$ for convenience):
$$f(a)=\int_0^{\pi/2} \frac{dt}{1+\cos^{2-a} t~ \sin^a t}$$
Which by the way, makes the functional equation from the OP very clear.
For the values of $a$ satisfying $|\cos^{2-a} t~ \sin^a t|\leq 1$, we can expand the integrated function as a geometric series:
$$f(a)=\sum_{n=0}^\infty (-1)^n \int_0^{\pi/2} \cos^{2n-an} t~ \sin^{an} t ~dt$$
However, the range of $a$ allowing this representation is quite restrictive: $0 \leq a \leq 2$.
The integral in the general terms is easy to recognize as Beta function:
$$\int_0^{\pi/2} \cos^{2n-an} t~ \sin^{an} t ~dt=\frac{1}{2} B \left(n-\frac{an-1}{2},\frac{an+1}{2} \right)$$
Now the original integral becomes:
$$f(a)=\frac{1}{2} \sum_{n=0}^\infty (-1)^n B \left(n-\frac{an-1}{2},\frac{an+1}{2} \right)$$
Let's use the Gamma function representation:
$$f(a)=\frac{1}{2} \sum_{n=0}^\infty (-1)^n \frac{\Gamma \left(n-\frac{an-1}{2} \right) \Gamma \left(\frac{an+1}{2} \right)}{\Gamma (n+1)} $$
For some $a$ this might be a Hypergeometric function. To find its form, let us consider the ratio of the terms:
$$\frac{c_{n+1}}{c_n}=\frac{\Gamma \left(n+1-\frac{an}{2}-\frac{a}{2}+\frac{1}{2} \right) \Gamma \left(\frac{an}{2}+\frac{a}{2}+\frac{1}{2} \right)}{\Gamma \left(n-\frac{an}{2}+\frac{1}{2} \right) \Gamma \left(\frac{an}{2}+\frac{1}{2} \right)} \frac{-1}{n+1}$$
If $a$ is even, the ratio will be a polynomial, and we will have a generalized Hypergeometric function.
But because of the condition on $a$ for which we derived the series, the only allowed even values are $a=0$ and $a=2$, for which cases the OP already provides a closed form.
| {
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How to prove divergent sequences? For this question, I know that the sequence diverges to infinity, but I'm not sure if I am doing it right. Here is what I have so far. Can anyone please help me out?
Determine whether the following sequence is convergent or divergent
$a_n = \{8n^3 + n^2 -2\}$
$\lim_{n \to \infty} a_n = \infty$
Wts for any $M>0$, there exists some $N>0$, st if $n>N$, then $a_n>M$.
$n^3 > N^3 > M$
$n^3(8 + \frac{1}{n} - \frac{2}{n^3}) > N^3(8 + \frac{1}{N} - \frac{2}{N^3}) > M$
$n^3(8 + \frac{1}{n} - \frac{2}{n^3}) > N > (\frac{M}{8+\frac{1}{N} -\frac{2}{N^3}})^\frac{1}{3}$
| We fix $M>2$ and we look for $N$ such that $\forall n>N$
$$a_n = 8n^3 + n^2 -2>M$$
then choose $n=M$ and check that
$$8M^3 + M^2 -2>M\iff 8M^3 + M^2>M+2$$
which is true.
Then it suffice to choose $N\ge M$.
| {
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Finding the locus of $z=at+\frac bt$ I have to find the locus of$$
z=at+\frac{b}{t},
$$
where $a, b \in \mathbb{C}$ are constants. I took $a=a_1+ia_2$ and $b=b_1+ib_2$, but could not get the solution.
| In a real setting, the curve given by parametric equations
$$\tag{1}x=t, \ \ \ y=\dfrac{1}{t},$$
i.e., with cartesian equation $y=\dfrac{1}{x}$, is a (equilateral) hyperbola.
Of course, (1) can be written
$$\tag{2}x+iy=t\color{red}{1}+\dfrac{1}{t}\color{red}{i}$$
Thus $z=t\color{red}{a}+\dfrac{1}{t}\color{red}{b}$ (see the analogy with (2)) is a hyperbola with respect to oblique axes defined by $\vec{OA}$ and $\vec{OB}$ ($A,B$ are points associated with complex numbers $a,b$). Thus it is also a hyperbola with respect to standard axes.
| {
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A Problem on Theory Of Equations Let $f(x) = x^2 + x$, for all real $x$. There exist positive integers $m$ and $n$, and distinct nonzero real
numbers $y$ and $z$, such that $f(y) = f(z) = m + \sqrt{n}$ and $f(1/y) + f(1/z) = 1/10$ . Compute $100m + n$.
| Hint. Let $a=m+\sqrt{n}>0$. Note that $y$ and $z$ are the two solutions of the quadratic equation $x^2+x-a=0$. Therefore $y+z=-1$ and $yz=-a$. Hence
$$\frac{1}{y}+\frac{1}{z}=\frac{y+z}{yz}=\frac{-1}{-a}=\frac{1}{a}.$$
Moreover
$$\frac{1}{y^2}+\frac{1}{z^2}=\frac{y^2+z^2}{y^2z^2}=\frac{(y+z)^2-2yz}{(yz)^2}=\frac{(-1)^2-2(-a)}{(-a)^2}=\frac{1}{a^2}+\frac{2}{a}.$$
Can you take it from here?
| {
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Expressing $\tan 20°$ in terms of $\tan 35°$ If $\tan 35^\circ = a$, we are required to express $\left(\frac{\tan 145^\circ - \tan 125^\circ}{1 + \tan 145^\circ\tan 125^\circ}\right)$ in terms of $a$.
Here's one way to solve this: $$\frac{\tan 145^\circ - \tan 125^\circ}{1 + \tan 145^\circ\tan 125^\circ} = \tan (145^\circ - 125^\circ) = \tan 20^\circ = \tan (90^\circ - 70^\circ) = \cot 70^\circ = \frac{1}{\tan 70^\circ} = \frac{1}{\tan (2 \times 35^\circ)} = \frac{1 - \tan^2 35^\circ}{2\tan 35^\circ} = \frac{1 - a^2}{2a}$$
However, I tried to solve it using another method as described below, and faced a problem:
$$\frac{\tan 145^\circ - \tan 125^\circ}{1 + \tan 145^\circ\tan 125^\circ} = \tan 20^\circ = \tan (35^\circ - 15^\circ) = \frac{\tan 35^\circ - \tan15^\circ}{1 + \tan 35^\circ\tan 15^\circ} = \frac{a - (2 - \sqrt3)}{1 + a(2 - \sqrt3)} = \frac{a - 2 + \sqrt3}{1 + 2a - \sqrt3a}$$
I tried to simplify it to get $\frac{1 - a^2}{2a}$, but I couldn't. So my question is, is there any way to show that $\frac{a - 2 + \sqrt3}{1 + 2a - \sqrt3a}$ is equal to $\frac{1 - a^2}{2a}$? If not, why are we getting two different answers?
| $\tan145^\circ=\tan(180^\circ-35^\circ) =-\tan35^\circ $ and $\tan125^\circ=\tan(90^\circ+35^\circ) =-\frac{1}{\tan35^\circ} $
For your two answers, have you find their values with a calculator?
Actually, you have proven an equality in $a$
| {
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Using differentiation for computing $\int_{0}^{1}\frac{\arctan(ax)}{x\sqrt{1-x^2}}\,dx$ How can I prove that I can differentiate the integral:
$$ \int_{0}^{1} \frac{\arctan(ax)}{x\sqrt{1-x^2}}\,dx $$
First I have to prove this integral converges.
Next I have to prove the integral $$ \int_{0}^{1} f_{a}'(x, a)\,dx $$ converges uniformly.
Then I can differentiate integral with Leibniz's rule.
I tried Weierstrass and Dirichlet's tests. Nothing succeeded.
Thank you for help in advance.
| The given hints contain pretty much everything. $\frac{d}{da}\arctan(ax)=\frac{x}{1+a^2 x^2}$ and
$$ \int_{0}^{1}\frac{dx}{(1+a^2 x^2)\sqrt{1-x^2}} \stackrel{x\mapsto\sin\theta}{=} \int_{0}^{\pi/2}\frac{d\theta}{1+a^2\sin^2\theta}=\int_{0}^{\pi/2}\frac{d\theta}{1+a^2\cos^2\theta}\\\stackrel{\theta\mapsto\arctan u}{=}\int_{0}^{+\infty}\frac{du}{(1+a^2)+u^2}=\frac{\pi}{2\sqrt{1+a^2}} $$
so by the dominated/monotone convergence theorem
$$ \int_{0}^{1}\frac{\arctan(ax)}{x\sqrt{1-x^2}}\,dx = \frac{\pi}{2}\int_{0}^{a}\frac{dv}{\sqrt{1+v^2}}=\color{blue}{\frac{\pi}{2}\text{arcsinh}(a)}.$$
Differentiation under the integral sign is also known as Feynman's trick: it is pretty efficient in dealing with integrals involving $\log$ or $\arctan$ or $\text{arctanh}$.
| {
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Graph of $\log(3-x)$ So if $y=\log(3-x) = \log(-x+3)$ then you reflect $\log(x)$ in the $y$ axis to get $\log(-x)$.
Then because it is $+3$ inside brackets you then shift to the left by $3$ giving an asymptote of $x=-3$ and the graph crossing the $x$ axis at $(-4,0)$.
However this does not work. The answer shows the $+3$ in the bracket shifting the curve to the right by $3$ giving an asymptote of $x=3$ and the curve crossing the $x$ axis at $(2,0)$.
Why does it do this? Can anyone please explain?
| Start with $y=\log(x)$. To shift this left three units, replace "$x$" with "$x+3$". Now you have $y=\log(x+3)$.
Now reflect over the $y$-axis. To do this, replace "$x$" with "$-x$". Now you have $y=\log(-x+3)$.
The order that the horizontal graph transformations happen is opposite from what you might think by the order of operations. If you first do "$x\mapsto-x$ and then do $x\mapsto x+3$, you get $\log(x)\mapsto\log(-x)\mapsto\log(-(x+3))$ which is not what you set out with.
| {
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Why is $M_n(A)$ a von Neumann algebra I'm trying to verify that for each von Neumann algebra $A$ the algebra of matrices with entries in $A$ is again von Neumann.
I do already know, that those kind of matrix algebras are again C-*-algebras.
I'd prefer a basic argument involving only weak or strong closedness over others.
Thanks!
| Strong convergence of a net in $M_n(A)$ is equivalent to entrywise strong convergence.
Assume $x_\lambda$ is a net in $M_n(A)$ converging strongly to $x \in B(H^n)$. Let $h,k \in H$ and $1 \leq i,j \leq n$. Let $\xi \in H^n$ be the vector which has $h$ in the $i$-th component and $0$ elesewhere, and let $\eta \in H^n$ which has $k$ in the $j$-th component and $0$ else. Then, in particular $\langle x_\lambda \xi, \eta \rangle \to \langle x \xi, \eta \rangle$. But this just means $\langle (x_\lambda)_{ij} h ,k \rangle \to \langle x_{ij} h, k \rangle$.
It follows that $x_{ij} \in A$.
| {
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What values of $a$ make matrix $A$ diagonalisable? I have the following question in an assignment paper.
Let $$A=\begin{bmatrix} 0 & a & 0\\ 1 & 0 & a\\ a & 1 & 0\end{bmatrix}$$ For what values of $a$ is $A$ diagonalisable?
Simply put, I don't know how to do it. In the $2 \times 2$ case we were asked, I completed the square of the characteristic polynomial and found that in all but $1$ choice of the unknown entry you got distinct eigenvalues and, therefore, distinct eigenvectors. At which point I just had to consider the one case for which I had eigenvalue of algebraic multiplicity $2$ and show that the geometric multiplicity of the eigenvector associated with it was $1$, I was done.
Any tips would be hugely appreciated, I've said it an assignment so reservation on full solution I understand but some hints would be amazing. Thank you.
| Taking $a^3 = \frac{32}{27} \; , $ this includes complex $a:$
$$
\frac{1}{864a}
\left(
\begin{array}{rrr}
40 & 36 a^2 & 24 a \\
-100 & -90 a^2 & 156 a \\
-9 a^2 & -24 a & 32
\end{array}
\right)
\left(
\begin{array}{rrr}
0 & a & 0 \\
1 & 0 & a \\
a & 1 & 0
\end{array}
\right)
\left(
\begin{array}{rrr}
6 a & -12 a & 54 a^2 \\
9 a^2 & 9 a^2 & -60 \\
10 & 4 & 0
\end{array}
\right) =
\left(
\begin{array}{ccc}
\frac{3}{2} a^2 & 0 & 0 \\
0 & - \frac{3}{4} a^2 & 1 \\
0 & 0 & - \frac{3}{4} a^2
\end{array}
\right)
$$
| {
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Spivak's Calculus: Chapter 1, Problem 18b (Quadratic determinant less than zero) The problem in question is as follows:
18b Suppose that $b^2 -4c \lt 0$. Show that there are no numbers $x$ that satisfy $x^2 + bx + c = 0$; in fact, $x^2 + bx + c \gt 0$ for all $x$. Hint: complete the square.
Trying to apply the hint, I began by constructing $b^2 - 4c < 0 \therefore (b-\frac{2c}{b})^2 - \frac{4c^2}{b^2} \lt 0$, but manipulating this ultimately just leads you to $b^2 \lt 4c$ which you didn't need to complete the square to get anyway.
The only other idea I had was that one could construct the quadratic equation beginning from the assumption that $x^2 + bx + c = 0$ and then go for proof by contradiction e.g.
$x^2 + bx + c =0$
$x^2 + bx = -c$
$x^2 + bx + (\frac{b}{2})^2 = -c + (\frac{b}{2})^2$
$(x + \frac{b}{2})^2 = \frac{b^2 - 4c}{4}$
$\therefore$ Given that for all real values of $x$ and $b$, $(x + \frac{b}{2})^2 \gt 0$, by transitivity of equality, $\frac{b^2 - 4c}{4} \gt 0$
$\therefore 4(\frac{b^2 - 4c}{4}) \gt 4(0)$
$\therefore b^2 - 4c \gt 0$ for all x such that $x^2 + bx + c = 0$
But that still leaves the statement "in fact, $x^2 + bx + c \gt 0$ for all $x$" unproven, unless it's supposed to obviously follow, in which case I'm not seeing how.
| What you are missing is that$$x^2+bx+c=\left(x+\frac b2\right)^2-\frac{b^2-4c}4.$$
| {
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probability - distinguishable vs indistinguishable objects/persons I am solving one simple problem just for fun.
The problem is as follows.
$n$ people have to be seated randomly in a cinema hall which has $(n+k)$ seats.
What is the probability that in this process, some fixed $m$ places ($m \leq n$) get occupied/taken?
The book gives this answer:
${{n} \choose {m}} \cdot {{n-k-m}\choose{n-m}} / {{n+k} \choose {n}}$
There are 2 things I don't like here:
1) I think that ${{n-k-m}\choose{n-m}}$ is obviously a typo and should read ${{n + k-m}\choose{n-m}}$
Why? Well to say the least $n-k-m \leq n-m$ so this binomial coefficient does not make much sense here.
2) This answer seems to treat the $n$ people as indistinguishable which is strange. Since these are $n$ people/persons these are $n$ different people (we can never have identical people, let alone $n$ of them, even twins are different persons). Thus it makes more sense to treat the people as distinguishable. So the order in which the people take the places also matters, right?
But in that case I am getting this answer:
${{n} \choose {m}} \cdot m! \cdot {{n + k-m}\choose{n-m}} \cdot (n-m)! / ( {{n+k} \choose {n}} \cdot n! ) $
What is your opinion of these two items?
Am I correct in both or at least in one of them?
| The term $\binom{n}{m}$ counts the number of ways $m$ of the $n$ people can occupy the specified $m$ seats. That leaves $n + k - m$ seats available. The remaining people can occupy $n - m$ of them. Hence, the number of favorable cases is
$$\binom{n}{m}\binom{n + k - m}{n - m}$$
as you suspected. The denominator $\binom{n + k}{n}$ represents the number of ways the $n$ people can select $n$ of the $n + k$ seats in the theater.
The people are distinguishable in the sense that we are selecting which $m$ of the $n$ people sit in the designated seats. What does not matter is which particular person sits in which particular seat, just which seats are occupied.
| {
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is $(1+\pi)/(1-\pi)$ a transcendental number? I know it's a open problem to show if $\frac{\pi}{e}$ is a trans. number or not. But what about quotient between numbers in function only of $\pi$, which is trans, such as $\frac{1+\pi}{1-\pi}$ or $\frac{1+i\pi}{1-i\pi}$. Wolfram says all theses numbers are trans.
$\frac{1+\pi}{1-\pi} = -1- \frac{2}{\pi-1}$
I'm summing a not trans number with a trans. That implies the result is trans? Also, how to be sure that $\frac{2}{\pi-1}$ is trans? Is a division between "equal" trans numbers also trans? by "equal" I'm mean when both numbers in the division are written in function of $\pi$. (except cases like $\pi/\pi=1$)
$\frac{1+i\pi}{1-i\pi} = -1+ \frac{2i}{\pi-i}$.
Same problem.
The same question can be asked about $e$ instead of $\pi$.
| $1-{2\over{\pi-1}}$ is algebraic implies that ${1\over{\pi-1}}=a$ where $a$ is algebraic, this implies that $\pi-1=1/a$ and $\pi=1+1/a$ contradiction
| {
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Riemann surface $y=(1-x^3)^{1/3}$ Let $X\subset \mathbb C^2$ be the Riemann surface given by （the multivalued function) $y=(1-x^3)^{1/3}$, and let $\phi:X \to \mathbb C $ be the induced map. Let $X'\subset P(\mathbb C^2)$ be the complex curve $x^3+y^3=z^3$ (in homogeneous coordinates). I want to define a complex manifold structure on $X'$ and a holomorphic $\phi':X'\to \mathbb C$ that extends $\phi$.
So far I know the following,
(i) $(x,y) \mapsto [x:y:1]$ gives a homeomorphism $\psi$ of $\mathbb C^2$ onto its image.
(ii) $\psi(X)\bigcup \{[1:-\zeta_3:0], [1:-\zeta_3^2:0], [1:-1:0]\}=X'$.
So I want to first define a manifold structure for $X$ and push it to $X'$ by $\psi$ and use another chart to cover $\{[1:-\zeta_3:0], [1:-\zeta_3^2:0], [1:-1:0]\}$. I think $x$ is a coordinate chart around $x_0$ if we fix a branch (fix a $y_0$). So we need three charts to cover $X$ because we have three branches. Is this correct?
Also, how do I construct a compatible chart to cover the three points at infinity?
| If I got well, $X$ is the zero locus $x^3+y^3=1$ in $\mathbb{C}^2$ and your $\phi $ is the projection on one of the coordinate (I will assume is the $y$ just to be clear).
Now you take $X'$ that is ,as you can easily prove, the projective closure of $X$ in $\mathbb{P}^2(\mathbb{C})$.
As you correctly said , you need three charts because intuitively there are three branches. A standard way to proceed in the projective spaces is to use the standard charts $x_i \neq 0$ that are (in the projective plane) biholomorphic to $\mathbb{C}^2$.
If you restrict $X'$ to this charts , you will get three varieties: the first one (in the chart $z \neq0$) is your $X$ (and so you get your immersion) and the other two are very closely related to $X$ (so you can easily find complex structures on them in the same way you did with $X$).
| {
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$(p \wedge q) \wedge p$ convert to CNF I was doing some exercises to convert formulas to CNF by means of the axioms and I imagined the following exercise $(p \wedge q) \wedge p$, but for this exercise, I do not find an axiom that can take me to CNF. However, I know that if I use the truth table I could solve it. I would like to know if there is an axiom that I may not know, by which you can work the exercise.
| $\begin{align}(p\wedge q)\wedge p &= p\wedge(p\wedge q) &&\text{commutivity}\\ &= (p\wedge p)\wedge q &&\text{associativity}\\ &=p\wedge q && \text{idempotence}\end{align}$
| {
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What is wrong with my proof of Law of Large Numbers? Need help in deriving variance for LLN, but am getting something wrong here, can you help point out where I'm going wrong?
Thanks a ton!
For i.i.d random variables $X_1,X_2,\ldots,X_n$ with mean $\mu$ and variance $\sigma^2$.
Let $X=\frac{1}{n}\displaystyle\sum_{i=1}^nX_i$ and it's mean be $\mu$(from LLN) then
\begin{align}
\operatorname{var}(X)& =E[(X-\mu)^2]\\[10pt]
& =E[X^2]-\mu^2\\[10pt]
& =E\left[\left(\frac{1}{n}\sum_{i=1}^nX_i\right)^2\right]-\mu^2\\[10pt]
& =E\left[\frac{1}{n^2} \left( \sum_{i=1}^nX_i\right)^2\right]-\mu^2\\[10pt]
& =\frac{1}{n^2}E\left[\left(\sum_{i=1}^nX_i\right)^2\right]-\mu^2\\[10pt]
& =\frac{1}{n^2}E\left[\sum_{i=1}^nX_i^2+\sum_{i,j=1,i\neq j}^{i,j=n} X_iX_j\right]-\mu^2\\[10pt]
& =\frac{1}{n^2}\left(\sum_{i=1}^nE[X_i^2]+\sum_{i,j=1,i\neq j}^{i,j=n}E[X_iX_j]\right)-\mu^2\\[10pt]
& =\frac{1}{n^2}(n\sigma^2+(n^2-n)\mu^2-n^2\mu^2)\\[10pt]
& =\frac{1}{n^2}(n\sigma^2-n\mu^2)\\[10pt]
& =\frac{\sigma^2-\mu^2}{n}
\end{align}
| You say "from LLN", but you are not using LLN anywhere.
Moreover, it's much easier to do this:
$$
\text{Var}\left(\frac{1}{n}\sum X_{i}\right)=\frac{1}{n^{2}}\sum\text{Var}\left(X_{i}\right)=\frac{n\sigma^{2}}{n^{2}}=\frac{\sigma^{2}}{n}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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On which topological spaces, can we give a group structure to make it a topological group? Let $X$ be a non-empty set. It is known that we can give a group structure on $X$. Now let $X$ be a non-empty topological space. Then can we give a group structure on $X$ so that it becomes a topological group w.r.t. its original topology ?
| No, you cannot do that for all spaces $X$.
If $X$ has the structure of a topological group, it implies a lot of extra facts about it, and those give necessary conditions that $X$ should fulfill.
Some examples of such properties:
*
*If $X$ is $T_0$ it must also be $T_{3\frac{1}{2}}$ (Tychonoff). (it's uniformisable)
*$X$ is homogenous: for every $x, y \in X$ there is a homeomorphism $h:X \to X$ such that $h(x) = y$.
*$X$ does not have the fixed point property (any non-unit multiplication shows this)
*If $X$ is compact it is dyadic and thus ccc.
*If $X$ is first countable and $T_0$ it is metrisable. (Birkhoff metrisation theorem).
So e.g. $X= [0,1]^n$ cannot be made into a topological group, because of both 2 and 3. The Sorgenfrey line fails 5. The infinite cofinite topology fails 1.
So many spaces cannot have a structure of a topological group.
@orangeskid mentioned an algebraic topology reason of possible failure: $\pi_1(X)$ is Abelian when $X$ is a topological group. This makes the wedge sum of circles $S^1 \vee S^1$ another example, I believe.
| {
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"timestamp": "2023-03-29T00:00:00",
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Create Approximately Follows Distribution Symbol Using MathJax. I want to use MathJax to create an "approximately follows distribution" symbol. It is a tilde with a dot above and a dot below. The closest I get is $\overset{\cdot}{\underset{\cdot}{\sim}}$ but the distance from the bottom dot to the tilde is greater than the distance from the top dot to the tilde.
Thanks,
Jack
| Using the lower and raise command should give you a desirable result like this:
$\overset{\lower{0.5ex}{\cdot}}{\underset{\raise{1ex}{\cdot}}{\sim}}$.
Hope you find this useful, thanks!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2707198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Distributing $2$ identical balls to $8$ distinct boxes Find the number of ways to distribute $2$ identical balls to $8$ distinct boxes.
This is how I reasoned it:
As the boxes are distinct, the only way by which we could get different configurations would be by considering the number of ways in which $2$ distinct pairs of boxes can be chosen. This can be done in $\binom {8}{2} $ ways.
Do you think that I have reasoned it correctly?
| It depends on what you mean by "distribute":
a) if you mean "(randomly) throw the balls into the boxes" ,
then it means that you consider equi-probable that at each launch you can
choose one of the $8$ boxes, so a total of $8^2=64$ ways to do that;
b) if instead you mean "(randomly) pour the balls into the boxes" ,
meaning that you consider equi-probable any "occupation histogram"
such as $(2,0, \cdots,0), \cdots,(1,1,0,\cdots,0), \cdots$, then that is equivalent to the
number of weak compositions
of $2$ into exactly $8$ parts, which is $\binom{2+8-1}{8-1} = 36$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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A problem about the order of element in a group Assume $a \in G$, where $G$ is a group and element $a$'s order is $mn$ with $(m,n) = 1$, the problem is to prove there exist elements $b$ and $c$ in the group where $a = bc = cb$ , $b$'s order is $m$ and $c$'s order is $n$ and $b,c$ are unique.
I have proven that the existence of b and c using $ms+nt = 1$, but still don't know how to prove the uniqueness.
| Hint: $\mathbb Z_{mn}\cong \mathbb Z_m\times \mathbb Z_n$ where $(m,n)=1$. The uniqnes comes from the propery of "direct product".
| {
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Cancellation law for multiplication of natural numbers I'm trying to prove the following cancellation law for multiplication of natural numbers: if $xz=yz$ for natural numbers $x,y$ and $z$, where $z$ is non-zero, then $x=y$.
I'm working with the peano-axioms and I've already proven elementary properties of multiplication such as commutativity, left and right-distributivity etc.
I'm thinking that the proof could maybe be done by using induction on one of the variables while the other two are fixed, but I'm not sure where to start.
Thanks
| 1) Prove If $x \ne y$ then eithere there exists a $c \ne 0$ so that $x + c =y$ or that $x = y + c$.
2) Prove that if $c \ne 0$ then $z(c + x) \ne z(0 + x)$
=====
Actually You could use $1$ to define subtraction. If $a \le b$ then, by definition there is a $c$ so that $a + c = b$. If $a \le b$ then define $b - a= c$.
Prove that if $b-a = c$ then $b = a$ if and only if $c = 0$.
Then prove that for all $a,b$ either $a \le b$ or $b \le a$.
And prove subtraction distributes
If $xz = yz$ then $xz \le yz$ and so $0= xz - yz = z(x - y)$. And as $z$ is not $0$ then if $x - y\ne 0$ then $z(x-y) = 0$ we'd get a contradiction. So $x - y = 0$ and $x = y$.
| {
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What is the maximum value of $a + b + c$, given $ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{5}$ What is the maximum value of $a + b + c$, where $a, b, c\in \mathbb{Z}$, and
$$
\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{5}
$$
Note: I could solve the question if the question asks "minimum" instead of "maximum". The answer would be calculated as 45 with arithmetic mean - harmonic mean inequality, where all $a, b, c$ are equal to 15, and that would be the minimum value of $a+b+c$. However, the question asks for the maximum value. I could find some other valid solutions, such as $a=6$, $b=31$, $c=930$, giving the sum equal to 967. I cannot prove whether any larger integer solutions exist or not.
| Lacking any insight, what follows is a purely mechanical approach. We'll show that there are only finitely many possibilities for $a,b,c$. We will not assume that they are all positive.
Taking any solution, sort it so that $|a|≤|b|≤|c|$. We remark that $$\frac 15=\big \vert \frac 1a+\frac 1b+\frac 1c\big \vert≤ \frac 1{|a|}+\frac 1{|b|}+\frac 1{|c|}≤\frac 3{|a|}\implies |a|≤15$$
Thus there are only finitely many possible values for $a$.
Fix a choice of $a$. Now we have $\frac 1b+\frac 1c=\frac 15-\frac 1a$ and a similar argument shows that there are only finitely many choices for $b$. As $a,b$ determine $c$ we are done.
Note: I did the search via computer and it appears that the OP has the optimal solution in $(6,31,930)$. However I strongly advise checking this more carefully than I have done.
| {
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"url": "https://math.stackexchange.com/questions/2707707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate angle from accelerometer on a moving axis For the purpose of this question, let's say that I have an accelerometer that is on a foot of a human. ( red square on by the heel in the below photo )
I'd calculate the angle the following way, as the leg goes from step 1 to step 2.
$\theta = tan^{-1}(\frac{x}{y})$
At step 1, the angle would be $0$, as $y = -1$ and $x = 0$. As the leg moves towards step 2, it would come closer and closer to $90^{\circ}$. If the leg could be raised enough to be parallel to the waist, it would be a full $90^{\circ}$, given the assumption that if $y == 0$, the result equates to $90^{\circ}$.
Now that I have illustrated what I am doing, I am curious to monitor what happens on the z axis. The base assumption thus far has been that the person will go from Step 1, to Step 2, and not swing their leg back and forth. So what I want to do is, as the person is moving their leg, I want to measure the angular change on the Z axis. i.e Is the person moving their leg back and forth while going from Step 1 to Step 2.
This has me a bit stumped at the moment, because I do not know which of the other two axis I should use to determine the change in angle on the Z axis. I'd expect to see the following:
As the person moves the leg back over time, but continues from step 1 to step 2, that the angle goes from $0^{\circ}$ to $-90^{\circ}$. As the person moves the leg forwards, but continues from step 1 to step 2, that the angle on the z axis goes from $0^{\circ}$ to $90^{\circ}$.
| In the plane $y-z$ the rotation would be
$$\alpha = \tan^{-1}\left(\frac{z}{y}\right)$$
with positive values for $\alpha$ when moving back.
A general expression for the absolute angle between the leg and y axis is
$$\phi = \tan^{-1}\left(\frac{\sqrt{x^2+z^2}}{|y|}\right)$$
For a general description you could use spherical coordinates system using the standard axis system
$$\begin{align} r&=\sqrt{x^2 + y^2 + z^2} \\ \theta &= \arccos\frac{z}{\sqrt{x^2 + y^2 + z^2}} = \arccos\frac{z}{r} \\ \varphi &= \arctan \frac{y}{x} \end{align}$$
Note that the inverse tangent must be suitably defined, taking into account the correct quadrant.
Setting, with your reference system
*
*$z=-Y$
*$y= Z$
*$x=-X$
| {
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"timestamp": "2023-03-29T00:00:00",
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What is κ(G) and κ′(G) and δ(G) for graph G?
Am I correct to say that G is 4-connected?
So κ(G) = 4 but then κ′(G)=2 but that cannot happen since κ(G)<=κ′(G)<=δ(G)
I know δ(G)=4 so wouldn't κ′(G)=4 then? However, I don't see how that would happen.
Could someone please explain this?
| I hope it can help you
Vertex Connectivity: $\kappa(G)$ is the minimum size of a vertex set S s.t. G\S is disconnected.
Edge Connectivity: $\lambda(G) $ or $\kappa'(G)$ is the minimum size of edge set F s.t. G\F has more than one component.
in your graph $\kappa(G)=4$: for example $S=\{f,l,i,c\}$ and
$\lambda(G)=4 $ for example $F=\{\{f,e\},\{l,k\},\{i,g\},\{c,d\}\}$
$\kappa(G)\le\lambda(G)\le\delta(G)$
(Whitney 1932, Harary 1994):
for your graph $\kappa(G)=\lambda(G)=\delta(G)=4$
| {
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Generating function of a sequence with a sum I am new to generating functions and I know some of the basics how to create them. But now I am not sure how to create a generating function of this sequence:
$$a_0 = 0$$
$$ a_n = n + c (\sum_{k=0}^{n-1} a_k)$$ for n >=1
where c is a real non-zero constant.
I guess I need to add $ \frac {1}{(1-x)^2}$ (which is generating function of the n (sequence 1, 2, 3, 4, ...)) to the rest. But how do I evaluate the rest?
Thank you for any tips.
| Note that if
$$
A(x)=\sum_{n=0}^\infty a_n x^n;\quad
B(x)=\sum_{n=0}^\infty b_n x^n
$$
are two formal power series then (by definition)
$$
A(x)B(x)=\sum_{n=0}^\infty
\left(
\sum_{k=0}^na_kb_{n-k}
\right)
x^n.\tag{1}
$$
In particular taking $B(x)=(1-x)^{-1}$, we have that
$$
\frac{A(x)}{1-x}=\sum_{n=0}^\infty
\left(
\sum_{k=0}^na_k
\right)
x^n.\tag{2}
$$
by (1)
Further taking $A(x)=B(x)=(1-x)^{-1}$. we see that
$$
(1-x)^{-2}=\sum_{n=0}^\infty(n+1)x^n\tag{3}
$$
by (1). Finally observe that
$$
\sum_{n=0}^\infty a_{n+1}x^n=\frac{A(x)-a_0}{x}\tag{4}
$$
Now we have enough tools to tackle the problem. First rewrite the recurrence as
$$
a_{n+1} = n+1 + c \left(\sum_{k=0}^{n} a_k\right);\quad (n\geq0)\tag{5}
$$
where $a_0=0$. Let $A(x)$ be the generating function corresponding to the sequence $a_n$. Multiply both sides of (5) by $x^n$ and sum on $n\geq 0$ to obtain that
$$
\frac{A(x)}{x}=\frac{1}{(1-x)^2}+c\frac{A(x)}{1-x}
$$
where we have used (2), (3) and (4). We now proceed to solve for $A(x)$. Indeed,
$$
A(x)(1-x)^2=x+cA(x)x(1-x)\implies A(x)=\frac{x}{(1-x)(1-(c+1)x)}
$$
as desired.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Probability question regarding coin flips. There are 20 coins in a jar. Of these, 8 are quarters, 5 are dimes, 3 are nickels, and 4 are pennies. 8 coins are drawn at random, without replacement from the jar.
What is the chance that the fourth coin is a quarter and the eighth coin is a dime?
For this question I used combination method and I got:
$\frac{\binom{8}{4}\binom{5}{1}}{\binom{20}{8}}$
which is equal to 0.0028
The correct answer is 0.105
I don't know what I did wrong here.
What is the chance that the last two coins are of the same denomination?
for this question I got:
P(last 2 coins are quarters)+P(last 2 coins are dimes)+P(last two coins are nickels)+P(last two coins are pennies)
$\frac{\binom{12}{6}\binom{8}{2}+\binom{15}{6}\binom{5}{2}+\binom{17}{6}\binom{3}{2}+\binom{16}{6}\binom{4}{2}}{\binom{20}{8}}$
is this correct?
Thank you!
| What is the chance that the fourth coin is a quarter and the eighth coin is a dime?
There is actually a simpler way but let’s push forward with your combinatorics method. For the first 4, the cases are that there may be 1, 2, 3, or 4 quarters by the time the fourth coin is picked so we need to take care of that. We don't have to care about quarters from the 5th draw onwards.
Similarly, amongst the first 8 (except 4th) we also have to take note that there may be anything from 1 to 7 dimes in picked by the eighth draw.
Now let's look at some cases
Case A | No quarter, no dime excluding on 4th and 8th respectively.
$$p_A=\frac{8}{17}\cdot\frac{5}{13}\cdot\frac{7\cdot6\cdot5\cdot4\cdot3\cdot2}{20\cdot19\cdot18\cdot16\cdot15\cdot14}$$
Case B | 1 quarter, no dime excluding on 4th and 8th respectively.
$$p_B=\frac{8-1}{17}\cdot\frac{5}{13}\cdot\frac{\binom31(8)(6\cdot5\cdot4\cdot3\cdot2)}{20\cdot19\cdot18\cdot16\cdot15\cdot14}$$
Case C | 2 quarters, 3 dimes excluding on 4th and 8th respectively.
$$p_C=\frac{8-2}{17}\cdot\frac{5-3}{13}\cdot\frac{\binom32(8\cdot7)\binom{6-2}3(5\cdot4\cdot3)(7)}{20\cdot19\cdot18\cdot16\cdot15\cdot14}$$
We can then generalize this to the case with $n$ quarters and $m$ dimes (outside of 4th and 8th places) whose probability is
$$p_{n,m}=\frac{8-n}{17}\cdot\frac{5-m}{13}\cdot\frac{\binom3n\frac{8!}{(8-n)!}\binom{6-n}{m}\frac{5!}{(5-m)!}\frac{7!}{(7-(n+m))!}}{20\cdot19\cdot18\cdot16\cdot15\cdot14}$$
From here onwards, it's tedious work, or you can use the summation:
$$\text{Probability required}=\sum_{n=0}^{3}{\sum_{m=0}^{6-n}{p_{n,m}}}$$ to help you out!
What is the chance that the last two coins are of the same denomination?
You were almost correct but your approach would require you to take care of the permutation (i.e. different orders the coins can be) and the identical nature of the coins with the same denominations. An easier approach would be to do independent choosing for each placing! I'll let you try out yourself first and will put up the full approach in a while.
| {
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