Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
|---|---|---|
Find a real-valued vector solution to a system of differential equations Given:
$$\vec x'(t) = \begin{bmatrix}
4&-1\\
13&0
\end{bmatrix} \vec x(t) $$
Evaluating to find eigenvalues:
$$ (4-\lambda)(-\lambda)+13=0 $$$$ (\lambda-2)^2=-9$$ $$\lambda=2\pm3i$$
Finding the eigenvector for the eigenvalue $\lambda = 2+3i$:
$$ \left[
\begin{array}{cc|c}
2-3i&-1&0\\
13&-2-3i&0
\end{array}
\right] $$
Since, for a 2x2, the rows must be complex scalar multiples of each other (here: R2 = (2 + 3i)R1 ), then picking row 2:
\begin{align*} \begin{cases} v_1=\frac{2+3i}{13}v_2 \\ v_2=free \end{cases} \end{align*}
$$\vec v= v_2 \begin{bmatrix}
\frac{2+3i}{13}\\
1
\end{bmatrix} = v_2\begin{bmatrix}
2+3i\\
13
\end{bmatrix}$$
The solution takes the form of:
$$e^{\lambda_1t}\vec v_1 = e^{(2+3i)t}\begin{bmatrix}
2+3i\\
13
\end{bmatrix}
$$
However, this is not a real-valued vector solution, since the eigenvector has an imaginary part. How should I proceed?
| Your matrix is actually similar to one of the form $\begin{bmatrix}
2&-3\\
3&2
\end{bmatrix}$ with transition matrix $\begin{bmatrix}
2&3\\
13&0
\end{bmatrix}$ given respectively by the eigenvalues' real and imaginary parts and the transition is given (in columns) by real and imaginary parts of the first eigenvector.
The solution to the equation will then be given by $\vec x(t)=e^{2t}\begin{bmatrix}
cos(3t)& sin(3t)\\
-sin(3t)&cos(3t)
\end{bmatrix}.\vec x(0)$
To give a proof of this fact, apply Taylor expansion and definition of exponential of a matrix.
Regards
| {
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"url": "https://math.stackexchange.com/questions/2913610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
} |
Are these vectors linearly independent? 3 vectors. I have this vector set S = {(2,-1,2), (1,0,3), (3,-2,1)}. Are they linearly independent?
$$(0, 0, 0) = c_1(2,-1,2) + c_2(1,0,3) + c_3(3,-2,1)$$
Here are the corresponding equations:
$$2c_1 + c_2 + 3c_3= 0$$
$$-c_1 - 2c_3 = 0$$
$$2c_1 + + 3c_2 + c_3 = 0$$
->
$$\begin{bmatrix} 2 & 1 & 3 & 0 \\ -1 & 0 & -2 & 0 \\ 2 & 3 & 1 & 0 \end{bmatrix}$$
$$ -> \begin{bmatrix} 2 & 1 & 3 & 0 \\ 0 & 0.5 & 0.5 & 0 \\ 0 & 2 & -2 & 0 \end{bmatrix}$$
$$-> \begin{bmatrix} 1 & 0.5 & 1.5 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$
Let $c_3 = t, c_2 = t, c_1 = -2t$
So, there are infinite non-trivial solutions and as a result the vectors are linearly dependent. Is this right?
| Personally, I wouldn't use matrices for a simple set of equations like these. You have
$2c_1+ c_2+ 3c_3= 0$
$-c_1- 2c_3= 0$
$2c_1+ 3c_2+ c_3= 0$
From the second equation $c_1= -2c_3$. So the first equation is $-4c_3+ c_2+ 3c_3= c_2- c_3= 0$ and the second is
$-4c_3+ 3c_2+ c_3= 3c_2- 3c_3= 0$.
Both of those equations give $c_2= c_3$ so, yes, there are an infinite number of solutions and these vectors are dependent.
| {
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"url": "https://math.stackexchange.com/questions/2913717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integrating sum
*
*$\displaystyle \sum_{k=1}^n k=\dfrac{n(n+1)}{2}$
Which is equivalent of saying, $$1+2+3+4+5+\cdots+n=\dfrac{n(n+1)}{2}$$
Now how do I integrate the left side of the equality and how that would look like?
Integrating the right side gives $\int \dfrac{n(n+1)}{2} \, \mathrm dn =\dfrac{1}{2}(\dfrac{n^3}{3}+\dfrac{n^2}{2})$
Which is quite similar to the sum $\dfrac{1}{2}\displaystyle \sum_{k=1}^n k^2=\dfrac{1}{2}(\dfrac{n^3}{3}+\dfrac{n^2}{2}+\dfrac{n}{6})$
Can someone explain these?
| I'm choosing to interpret "taking the integral" as finding an anti-derivative, in the sense of indefinite integrals. So, we're looking for a function $F$ such that
$$F'(n) = 1 + 2 + \ldots + n.$$
In that sense, $\frac{n^3}{6} + \frac{n^2}{4}$ fits the bill; it's a function whose derivative agrees with the sum at integer points.
It is worth noting that we'll never get a function that agrees with the sum at non-integer points, as the sum doesn't make sense for non-integer points. Note that
$$1 + 2 + \ldots + 7.5$$
is a poorly-formed expression. The derivative of a function must be defined on an interval, including non-integer points (except if it's defined at a single point!). So, $\frac{n^3}{6} + \frac{n^2}{4} + C$ is about the best you're going to do.
Note also that, with the sum $\frac{1}{2} \sum_{k=1}^n k^2$, you are not integrating with respect to $n$. You are integrating the dummy variable $k$, which doesn't seem to have any conceptual connection with the original sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2913931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $T^9=T^8$, prove that $T^2=T$. Suppose $V$ is a complex inner product space and $T $ is a normal operator on $V$ such that $T^9=T^8$. Prove that $T^2=T$. Give an example of a non-normal $T$ such that $T^9=T^8$ but $T^2 \ne T$.
| The polynomial $x^9 - x^8 = x^8(x-1)$ annihilates $T$ so the minimal polynomial $m_T$ of $T$ divides $x^8(x-1)$.
On the other hand, $T$ is normal and hence diagonalizable so $m_T$ consists only of linear factors. Therefore $$m_T \in \{x, x-1, x(x-1)\}$$
and therefore $0 = T(T-I) = T^2 - T$.
As an example of a non-normal operator $T$ such that $T^9 = T^8$ but $T^2 \ne T$, we can take
$$T = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{pmatrix}$$
with $m_T(x) = x^2(x-1)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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What is this function related with continued fractions? Playing with continued fractions, I came with the idea of iterating the limit of the simplest one:
$$1 + \cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cdots}}}}}\ = \Phi$$
And then I thoght about iterating the result:
$$\Phi + \cfrac{1}{\Phi+\cfrac{1}{\Phi+\cfrac{1}{\Phi+\cfrac{1}{\Phi+\cfrac{1}{\Phi+\cdots}}}}}\ = ?$$
And after that keep iterating again and again. Essentially, what I am doing is solving this:
$$x = a + \frac{1}{x}$$
for different values of $a$. Whose solution is:
$$ x = \frac{a \pm \sqrt{a^2 + 4}}{2}$$
And I'm just staying with the positive solution. At the end, I'm just exploring this sequence defined recursively:
$$ f(1) = \Phi \\
f(n+1) = \frac{f(n) + \sqrt{f(n)^2 + 4}}{2}
$$
When you see the plot of the terms of the sequence, you get this picture:
I would like to know which is the function that generates that curve. It really seems to be a continuos function (notice that the picture is a set of points so close, that may seem to be continuous, but it's not).
So... What's the function underlying this curve and how can we find it?
Many thanks in advance!
| I think you can set up a differential equation and solve it.
\begin{eqnarray}
2f'(n) &= -f(n)+\sqrt{f(n)^2+4}\\
\text{d}n&=\frac{2\, \text{d}\! f}{-f+\sqrt{f^2+4}} \\
n &= \frac{1}{4} (f (f + \sqrt{4 + f^2} + 4 \sinh^{-1}\left(\frac{f}{2}\right)
\end{eqnarray}
Solving for $f$ should result in the function you wanted.
Wolfram plot
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "13",
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Parabola touching a circle
The parabola $\dfrac {x^2}{d^2}+\dfrac yh=1$ touches the circle $x^2+(y-R)^2=R^2$ at two points, $(\pm p, q)$. It can be easily shown geometrically that $p=\sqrt{2qR-q^2}$. Can it be shown geometrically that $q=d$?
See desmos implementation here. The purple dotted line forms a square.
(See also the solution to this other question on projectiles here).
| In parabola below, $AM=d$, $VM=h$ and $VF=VN=d^2/(4h)$, where $F$ is the focus and $NH$ the directrix of the parabola. We have $FP=PH$ and $\angle FPC=\angle KPC$, because radius $PC$ is normal to the parabola. Hence $CPHF$ is a parallelogram and
$$
CF=PH={h\over d^2} HN^2+VN=
{h\over d^2}(FH^2-FN^2)+{d^2\over4h}={h\over d^2}R^2.
$$
But $CF=h-R-d^2/(4h)$, which gives: $R=d-d^2/(2h)$.
And finally we can compute $q=PK$:
$$
PK=h-PH+VN=h-{h\over d^2}R^2+{d^2\over4h}=d.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Doubt in high school permutation question(seating arrangement)
$6$ boys and $6$ girls form a line with boys and girls alternating.Find the number of ways of making the line.
Answer $2(6!)^2$
I was trying to solve this question using Permutation and Combination
$$\square B \square B \square B \square B \square B \square B \square$$
Blank boxes shown above are for girls.Therefore girls can be chosen in $C(7,6)$. Now boys and girls can be arranged in $6!$ ways.
Therefore, it's answer must be $C(7,6) \cdot 6! \cdot 6!$.
Please point out my mistake.
Note- I got correct answer using Permutation and logical thinking but I want to solve by Combination too.
|
Therefore girls can be chosen in C(7,6)
Note that C(7,6) (out of seven, pick six to include) is the same as C(7,1) (out of seven, pick one to exclude), which is just 7. There are seven boxes, so it doesn't make sense to say that this is how many ways there are of choosing girls. It would make more sense to say that this is how many ways of choosing boxes for girls. But that still isn't correct, because every box between two boys must be chosen (otherwise, you would end up with two boys next to each other). So there are only two choices for which box to exclude: either the first box can be excluded, or the last.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is this set a basis for $R^3$. How to verify? Say I have $S = {(1,0,-1), (2,1,1), (-3,0,2)}$. Is my method correct for determining if it's a basis? How do I check if I'm right?
We need to determine if $S $ spans $R^3$ and if it's linearly independent.
First, check if $S$ spans $R^3$:
Let $u_1, u_2, u_3$ be a random vector.
$$(u_1, u_2, u_3) = c_1(1,0,-1) + c_2(2,1,1) + c_3(-3,0,-2)$$
$$ = (c_1 + 2c_2 - 3c_3, c_2, -c_1 + c_2 + 2c_3$$
the corresponding equations:
$$c_1 + 2c_2 - 3c_3 = u_1$$
$$c_2 = u_2$$
$$-c_1 + c_2 + 2c_3 = u_3$$
and the resulting matrix:
$$\begin{bmatrix} 1 & 2 & -3 \\ 0 & 1 & 0 \\ -1 & 1 & 2 \end{bmatrix}$$
Determine determinant using expansion by cofactors in the second row:
determinant = $0$ * (cofactor of 2nd row, 1st column) + 1 * (cofactor of 2nd row, 2nd column) + $0$ * (cofactor of 2nd row, third column) = 5
So the fact that the determinant is non-zero means that there is a unique solution for this system of equations for any random vector in $R^3$
Is it linearly dependent?
$$\begin{bmatrix} 1 & 2 & -3 \\ 0 & 1 & 0 \\ -1 & 1 & 2 \end{bmatrix}$$
$$\begin{bmatrix} 1 & 2 & -3 \\ 0 & 1 & 0 \\ 0 & 3 & -1 \end{bmatrix}$$
$$\begin{bmatrix} 1 & 2 & -3 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix}$$
Because the last row is $[0,0,1]$ this implies that the only solution is trivial so it is linearly dependent.
So $S $ is a basis for $R^3$. Is this logic right?
| Take a look at the zero-components. The second vector can’t be a linear combination of the other vectors. Since these other vectors aren’t parallel, the three vectors are linearly independent.
| {
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"timestamp": "2023-03-29T00:00:00",
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The ambigous definition of vacuous truth It is no doubt that the vacuous truth is related to material implication "$P\Rightarrow Q$". We say the material implication statement is true when $P$ is false. However, seems that this is not the definition for vacuously truth. Do we call it vacuously true only when $P$ can never be true? (Seems that all examples are in this way).
More clearly, suppose we have a statement "$P(x)\Rightarrow Q(x)$", and at some domains of $x$ (we may denote the domain by $T_x$), $P(x)$ is true; at some other domains of $x$ (denote the domain by $F_x$), $P(x)$ is false. Can we say the statement is vacuously true in the domain $F_x$? Is there any real example?
To make the problem more clear, we may check the statement "For all $x$, $P(x)\Rightarrow Q(x)$". Can we say this statement is vacuously true in the domain $F_x$ ($F_x$ is defined as above)?
| *
*I use this inclusive definition:
*
*A vacuous truth is an implication or universally-quantified implication whose antecedent is true or universally true, respectively.
As such, both $$P{\implies}Q$$ and $$\forall x\;[P(x){\implies}Q(x)]$$ are vacuously true. (Wikipedia gives both types of examples.)
*In particular, $$(P\land\lnot P){\implies}Q$$ is a vacuous tautology, while $$\forall x\;[x\ne x{\implies}Q(x)]$$ is a vacuous validity.
| {
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"timestamp": "2023-03-29T00:00:00",
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The limit of $\frac{n^3-3}{2n^2+n-1}$ I have to find the limit of the sequence above.
Firstly, I tried to multiply out $n^3$, as it has the largest exponent.
$$\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} =
\lim_{n\to\infty}\frac{n^3(1-\frac{3}{n^3})}{n^3(\frac{2}{n} + \frac{1}{n^2} - \frac{1}{n^3})} =
\lim_{n\to\infty}\frac{1-\frac{3}{n^3}}{\frac{2}{n} + \frac{1}{n^2} - \frac{1}{n^3}}$$
$$
\begin{align}
\lim_{n\to\infty}1-\frac{3}{n^3} = 1 \\[1ex]
\lim_{n\to\infty}\frac{2}{n} + \frac{1}{n^2} - \frac{1}{n^3} = 0 \\[1ex]
\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} = \frac{1}{0}
\end{align}
$$
Then, after realizing $\frac{1}{0}$ might not be a plausible limit, I tried to multiply out the variable with the largest exponent in both the dividend and the divisor.
$$\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} = \lim_{n\to\infty}\frac{n^3(1 - \frac{3}{n^3})}{n^2(2 + \frac{1}{n} - \frac{1}{n^2})} =
\lim_{n\to\infty}n\cdot\frac{1 - \frac{3}{n^3}}{2 + \frac{1}{n} - \frac{1}{n^2}}$$
$$
\begin{align}
\lim_{n\to\infty}1-\frac{3}{n^3} = 1 \\
\lim_{n\to\infty}2 + \frac{1}{n} - \frac{1}{n^2} = 2 \\
\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} = \frac{1}{2} \\
\lim_{n\to\infty}n = \infty
\end{align}
$$
So, my questions about this problem:
*
*Could $\frac{1}{0}$ be a valid limit?
*Does $\infty\cdot\frac{1}{2}$
equal to $\infty$?
*In conclusion, what is the limit of the sequence
above? $\infty?$
Thank you!
| You have done everything.I just want to represent it graphically.Then you will answer your own question.
Now,you tell me.what are you seeing here?
And ofcourse the answer of your your questions is simply "yes".
| {
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"timestamp": "2023-03-29T00:00:00",
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Showing that $ \lim_{x \to \infty} x(1/2)^x = 0$ Can someone explain to me why $$ \lim\limits_{x \to \infty} x\bigg(\frac{1}{2}\bigg)^x = 0$$
Is it because the $\big(\frac{1}{2} \big)^x$ goes towards zero as $ x $ approaches $\infty$, and anything multiplied by $0 $ included $\infty$ is $0$ ?
Or does this kind of question require using l'hopital's rule because it is in the form (0*$\infty$)?
I thought it could be solved this way. Please let me know if it is correct:
$$ \lim\limits_{x \to \infty} \bigg(\frac{x}{\frac{1}{2^{-x}}}\bigg)$$
L'hopitalize the above and get:
$$ \lim\limits_{x \to \infty} \bigg(\frac{1}{x(2^{x-1})}\bigg) = 0$$
| Let
$a_x = \dfrac{x}{2^x}$.
Then
$\dfrac{a_{x+1}}{a_x}
=\dfrac{\dfrac{x+1}{2^{x+1}}}{\dfrac{x}{2^x}}
=\dfrac{1+1/x}{2}
$.
Therefore,
for $x > 3$,
$\dfrac{a_{x+1}}{a_x}
=\dfrac{1+1/x}{2}
\lt\dfrac{1+1/3}{2}
=\dfrac46
=\dfrac23
$.
Therefore,
for $x > 3$,
$\begin{array}\\
\dfrac{a_x}{a_3}
&=\prod_{y=3}^{x-1}\dfrac{a_{y+1}}{a_y}\\
&\lt\prod_{y=3}^{x-1}\dfrac23\\
&=(2/3)^{x-3}\\
\text{so}\\
a_x
&\lt a_3(2/3)^3(2/3)^x\\
&= (3/8)(8/27)(2/3)^x\\
&= (1/9)(2/3)^x\\
\end{array}
$
To show
in an elementary way
that $c^x
\to 0$
for any $0 < c < 1$,
let
$c = 1/(1+d)$
where $d > 0$.
Note:
What follows is not original.
Then,
$d = 1/c-1$
so,
by Bernoulli's inequality,
$(1+d)^x \ge 1+xd > xd$
so
$c^x
=1/(1+d)^x
\lt 1/(xd)
=1/(x(1/c-1))
=c/(x(1-c))
\to 0$
as $x \to \infty$.
With $c = 2/3$,
$(2/3)^x
\lt (2/3)/(x(1/3))
=2/x
$
so that
$a_x \lt (1/9)(2/x)
=2/(9x)
$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Vector norm such that $\| (1,i) \| = 6$
Is there a norm $\| \cdot \|: \mathbb{C}^2 \rightarrow \mathbb{R}^+$ such that $\bigg\| \pmatrix{1\\i} \bigg\| = 6$?
I have come across this question and I don't really know what to look for here. Trying out random norms to see if one of them give $6$ doesn't look like a good idea, and I can't think of any properties of norms that the function in the example would not obey to prove such norm doesn't exist.
| Consider
$$\Vert x \Vert = 3\sqrt{x^H x}$$
where $x^H$ is the transpose - conjugation operation, i.e. Hermitian operator. This norm satisfies the three basic norm properties, i.e.
*
*$ \Vert x \Vert = 0 $ only if $x = 0$
*For $\alpha \in \mathbb{C}$,$ \Vert \alpha x \Vert = 3\sqrt{(\alpha x)^H \alpha x} = 3 \sqrt{ \vert \alpha \vert^2 x^H x } = \vert \alpha \vert 3\sqrt{ x^H x } = \vert \alpha \vert \Vert x \Vert$
*$ \Vert x + y \Vert \leq \Vert x \Vert + \Vert y \Vert $
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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"answer_id": 1
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Probability problem involving mappings of a finite set into itself. One mapping is selected at random from all the mappings of the set $\{1,2,\ldots,n\}$ into itself. What is the probability that
$(i)$ a specified element $i$ is transformed into another specified element $j$?
$(ii)$ the elements $i_1,i_2,\ldots,i_h$ are transformed into the elements $j_1,j_2,\ldots,j_h$ respectively?
| For (ii) suppose that the $i_k$ are all distinct, and the $j_k$ are all distinct.
There are $n(n-1)(n-2)\cdots(n-h+1)=n!/(n-h)!$ $h$-tuples of distinct elements
from $\{1,2,\ldots,n\}$. It is equally likely that your first $h$-tuple is taken
to any one of them, so the probability you seek is $(n-h)!/n!$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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The possible values of $x$, if $\tan^{-1} x>\cot^{-1}x$ What are the possible values of $x$, if $\tan^{-1}x >\cot^{-1}x$?
We have $\tan^{-1}x >\cot^{-1}x\implies \tan^{-1}x -\cot^{-1}x>0 \implies \tan^{-1} x-\tan^{-1}1/x>0\implies \tan^{-1}\dfrac{x-1/x}{1-x.1/x}>0$.
What can I do now?
| HINT:
$$\cot^{-1}x=\frac\pi2-\tan^{-1}x\implies \tan^{-1}x-\cot^{-1}x=2\tan^{-1}x-\frac\pi2>0$$
| {
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Is Taylor Series changing from an uncountable basis to a countable basis? Say we've got an analytic function $f(x)$ from $\mathbb R$ to $\mathbb R$. It has an uncountable number of components in this basis, since there is one value of $f(x)$ for each $x$ and $x$ varies continuously.
When we do a Fourier transform and hence change the variable to $w$, there is still an uncountable number of components, because $w$ too varies continuously.
But, in case of a Taylor series, the function can be represented by a countable number of components: $[f(0)\,\, f'(0)\,\, f''(0)\,\, f'''(0)\, \cdots]$. In this case, the number of components changes from uncountable infinite to countably infinite, when we change the basis.
Am I interpreting this right? The number of components never changes when we change the basis of a finite-dimensional vector space. Why does the number of components change here?
| It is true that the domain ${\mathbb R}$ of a function $f:\>{\mathbb R}\to{\mathbb R}$ has uncountably many points. If you want to produce the function values by a random number generator independently for each point then you would have to call this routine an uncountable number of times. But already a continuous function, let alone an analytic function, is completely determined by its values on a dense subset of ${\mathbb R}$, e.g., the countable set of rationals: Given a counting ${\mathbb Q}=\bigl\{ r_n\bigm| n\geq1\bigr\}$ it is sufficient to store the list $(y_n)_{n\geq1}$ of function values $y_n:=f(r_n)$. In the case of an (entire) analytic function you can store the list of Taylor coefficients $a_n:={1\over n!} f^{(n)}(0)$ $(n\geq0)$ and have enough information to produce the exact value of $f(z)$ even for all $z\in{\mathbb C}$. Note that there are some very special functions where even a few dozen "tokens" suffice to define them at all points $x\in{\mathbb R}$, e.g., $\exp(x):=\sum_{k=0}^\infty{1\over k!} x^k$.
| {
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Sum of the series $\sum_{k=0}^n \frac{1}{k+1} \binom{n}{k}$ The task is to transform $\sum_{k=0}^n \frac{1}{k+1} \binom{n}{k}$ into a compact and not recursive formula.
I've done a bunch of similar (as I thought at first) series. I'll describe my method. I name $\sum_{k=0}^n f(k)=a_{n}$, and then describe $a_{n}=a_{n-1}+f(n)$. Then I solve recursive formula by, for example, generating functions. It works for a series like $\sum_{k=0}^n k^2(k+2)$, so a series which doesn't contain $n$ inside.
About the main question, I tried to transform it with some formulas I know, like $\frac{n-k}{k+1}\binom{n}{k}=\binom{n}{k+1}$ or $\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}$, but none of them worked. With the last one I was able to transform it to like $a_{n}=a_{n+1}+\sum_{0}^{n-1}\frac{1}{k+2}\binom{n-1}{k}$, but it doesn't do anything, because there is $\frac{1}{k+2}$ and not $\frac{1}{k+1}$.
I would appreciate any hints!
Also, I thought about using the series $\sum_{k\geq0} \binom{n}{k}x^{k}=(1+x)^{n}$, because it looks like $\sum_{k=0}^n \binom{n}{k} \frac{1}{k+1}=\sum_{k\geq 0}\binom{n}{k} \frac{1}{k+1}$, and it looks pretty familiar.
| Hint:
$$\frac1{k+1}\binom{n}{k}=\frac1{n+1}\binom{n+1}{k+1}$$
| {
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An Olympiad question about equality How many $(x, y)$ positive integer pairs have $y^2-x^2=2y+7x+4$ equality?
I can't solve this Olympiad question.
$\textbf{Solution:}$ If above equality is regulated,we obtain $(2x+2y-5)(2x-2y+9)=29$. And, $29$ is prime we can easily find solution.
*
*$\textbf{1-)}$ How can we regulate it to obtain $(2x+2y-5)(2x-2y+9)=29$?
*$\textbf{2-)}$ Are there the general method for this kind of questions?
Thank you for help...
| Let the factors $u:=y+x$ and $v:=y-x$ appear.
$$uv=\frac{9u-5v}2+4.$$
Then,
$$\left(2u+5\right)\left(2v-9\right)=61.$$
Update to please Misha Lavro:
The factorization of the LHS hints the introduction of these intermediate variables: as the RHS is linear, we know that we can transform to $uv=au+bv+c$, which factors as $(u-p)(u-q)=r$.
| {
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2 squres with 1 common corner The line HC is a median in triangle $\triangle BCE$
How would I prove that for two arbitrary sized squares any angle $\alpha$, $\angle DIC$ is 90 degrees?
I have tried playing around with this in geogebra but to no avail.
|
Choose point $J$ on line $BC$ such that $BC=CJ$. Let us focus on (green) triangle $DCG$ and (red) triangle $JCE$.
$$\angle DCG = \angle JCE = \angle JCG + 90^\circ$$
$$DC=JC,\space CG=CE$$
By SUS, triangles $DCG$ and $JCE$ are congruent and therefore $JE=DG$ and $\angle CDG = \angle CJE$. But $CD\bot CJ$ so it must be that:
$$JE\bot DG$$
(This is also obvious form the fact that triangle $JCE$ can be obtained by rotating triangle $DCG$ for $90^\circ$).
$C$ is the midpoint of $BJ$ and $H$ is the midpoint of $BE$. So lines $CH$ and $JE$ must be parallel. $JE$ is perpendicular to $DG$ and therefore $CH$ must be perpendicular too.
This also proves an interesting fact that $DG=2CH$
| {
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What is the meaning of the difference between the equations of two non-intersecting circles represent? The difference between the equations of two intersecting circles gives a linear equation which represents the common chord or the common tangent.
But what about two non-intersecting circles?
I experimented with a number of different circles of varying centre and radius but I can't find an answer. The resulting line is perpendicular to the line joining the centres but its position doesn't make much sense to me.
Any ideas?
| Difference of two non intersecting circles is a line perpendicular to the line joining the centre of the two circles.
The line always lies between the two circles(near to the smaller one) and touches neither of them(for non intersecting circles).
The distance of the line from mid-point of centres is equal to "difference of squares of radii, divided by twice the distance between centres". It can be written as :
$$
d=\frac{|a^2-b^2|}{2L}
$$
where
*
*$d$ the distance of line from mid of centres
*$a$ and $b$ are radii
*$L$ distance between centre of circles
| {
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How to tell if $\sum_{n=1}^\infty\frac{\ln(n)}{n^2}$ converges using Integral Test? So I have this problem:
Determine whether or not the following infinite series converge or diverge. State what test you used.
$$\sum_{n=1}^\infty\frac{\ln(n)}{n^2}$$
So I decided to do the Integral Test for $$\int^\infty_1\frac{ln(n)}{n^2}dx$$
I made $u = \ln(n)$ and $du = \frac{1}{n}dx$ and got:
$$\int^\infty_1\frac{u}{n}*du$$
which would become:
$$u[\ln(n)]|^\infty_1$$
This is incomplete, but does this look right so far? If not, how do I fix it?
| By integral test we should obtain
$$\int_1^\infty \frac{\ln x}{x^2} dx=\left[-\frac{1+\log x}{x}\right]_1^\infty$$
or by $\ln x=u \implies \frac1x dx=du$
$$\int_0^\infty \frac{u}{e^u} du=\left[-\frac{u+1}{e^u}\right]_0^\infty$$
If you are not forced to use integral test, as an effective alternative, we can use limit comparison test with
$$\sum_{n=1}^\infty\frac{1}{n^p}$$
with $p>1$ such that
$$\frac{\frac{ln(n)}{n^2}}{\frac{1}{n^p}}\to 0$$
| {
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Does differentiability on $(a, b)$ implies continuity on $[a, b]$? Provided that $f$ is defined on $[a, b]$, is it true that the differentiability of $f$ on $(a, b)$ implies that $f$ is continuous on $[a, b]$ ?
I'm asking this question because I'm reading Salas's calculus and, as he goes trough some theorems, it's not clear on which interval (closed or open) the function is defined.
For example, in one of the proofs the following is asserted: $f$ is differentiable on $(a,b)$; let $x_1$ and $x_2$ be any two numbers on $(a,b)$; then f is continuous on $[ x_1$ , $x_2]$. That would seem to be true if $x_1$ is different from $a$ and $x_2$ is different from $b$. And nothing is said explicitly about which values $f$ may take on $a$ and $b$.
| Take the function differentiable on $(0,1); f(x)=x$, $f(0)=f(1)=2$ it is not continuous on $[0,1]$.
| {
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What is the proof of finding the LCM of numbers by prime factorization method? Fir example on the internet I found this method to find the LCM :
Let's find the LCM of 30 and 45. One way to find the least common multiple of two numbers is to first list the prime factors of each number.
30 = 2 × 3 × 5
45 = 3 × 3 × 5
Then multiply each factor the greatest number of times it occurs in either number. If the same factor occurs more than once in both numbers, you multiply the factor the greatest number of times it occurs.
2: one occurrence
3: two occurrences
5: one occurrence
2 × 3 × 3 × 5 = 90 <— LCM
How would I go about proving that this method works for all numbers ? Is there a way ?
| By having all the prime factors of each, one has a common multiple.
By having no additional (prime) factors, one has the least such.
| {
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Elementary divisor rings (Smith normal form) Is the ring $R={\mathbb Z}/n{\mathbb Z}$ an elementary divisor ring? Can you provide an easy proof from well-known results, or a reference to this result?
(Recall that an elementary divisor ring $R$ is one for which every matrix over $R$ is equivalent to a diagonal matrix. The existence of Smith normal form shows that when $R$ is a principal ideal domain, it is an elementary divisor ring. Of course, when $n$ is composite then $R={\mathbb Z}/n{\mathbb Z}$ is not even a domain, although it is a principal ideal ring.)
| This follows from the fact that Smith normal forms over $\Bbb{Z}$ exist.
If $\overline{A}$ is matrix over $\Bbb{Z}_n$ then we can, arbitrarily, lift the entries to integers and form a matrix $A$ with integer entries. By the existence of Smith normal forms over $\Bbb{Z}$ there exists invertible (i.e. determinant $=\pm1$) integer matrices $P$ and $Q$ such that $PAQ=D$ is a diagonal matrix.
Reducing that matrix equation modulo $n$ gives
$$
\overline{D}=\overline{P}\overline{A}\overline{Q}.
$$
Here $\overline{D}$ is obviously a diagonal matrix, and because $\overline{P}$ and $\overline{Q}$ have determinants $=\pm\overline{1}$ they are invertible over $R$ (alternatively you can reduce their inverses in a matrix ring over $\Bbb{Z}$ modulo $n$). The claim follows from this.
You may lose the divisibility relations among the diagonal entries. I do suspect the invariant factors to survive in some form (think: finitely generated abelian groups), but divisibility arguments in the presence of zero divisors need a little bit of extra care.
| {
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Orbit of conjugation on subgroups of $D_8$ Let $X$ be the set of all subgroups of $D_8$ with order $2$. For fixed $g\in D_8$, and for all $x\in X$, conjugation by $g$ is defined by
$$x\mapsto gxg^{-1}$$
What is the orbit of this group action? I have that $X=\{\{e,r^2\},\{e,s\},\{e,sr\},\{e,r^2s\},\{e,rs\}\}$
| Orbit is that of an element of the set $X$ on which the group is acting. For example,
$$\text{Orb}(\{e,r^2\})=g\{e,r^2\}g^{-1}=\{e,gr^2g^{-1}\}.$$
The second element $gr^2g^{-1}$ will run over all conjugates of $r^2$, thus you need the conjugacy class of $r^2$, which is just....... Can you take it from here?
Added explanation:
Either $g=r^k$ or $g=sr^k$, with $k \in\{0,1,2,3\}$. Thus, if $g=r^k$, then the only conjugate we get is $r^kr^2r^{-k}=r^2$.
With $g=sr^k$, we get $(sr^k)r^2(sr^k)^{-1}=sr^{2}s=r^{-2}=r^2$.
So only conjugate of $r^2$ is $r^2$. This means
$$\text{Orb}(\{e,r^2\})=\{\{e,r^2\}\}.$$
| {
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Let $p$ be an odd prime and $P(x)$ a polynomial of degree $p-2$. If $p\mid P(n)+P(n+1)+...+P(n+p-1) \forall n$, must $P$ have integer coefficients? I am stuck on this question from Zeit's book The Art and Craft of Problem Solving.
Question
Let $p$ be an odd prime and $P(x)$ a polynomial of degree at most $p-2$.
If $P(n)+P(n+1)+...+P(n+p-1)$ is an integer divisible by $p$ for every integer $n$, must $P$ have integer coefficients?
Attempt
I think $P$ must have integer coefficients. Suppose $P(x)=\sum_{k=0}^{p-2} a_k \cdot x^k$. Define $H(n)=P(n)+P(n+1)+...+P(n+p-1)$. Then we can expand $H(n)$ into a polynomial $H(n)=\sum_{k=0}^{p-2} c_k \cdot n^k$, where each of the coefficients $c_k$ are some linear combination of $a_k$.
If $c_i$ is not an integer for some $i$, then I think by choosing some suitable $n_i$, we can cancel all denominators of $c_j$ where $j \neq i$, while leaving $c_i$ still a fraction. Then $H(n_i)$ will be non-integer. But I don't know how to prove this properly. The $c_i$ are already very complicated to express.
However, if $c_i$ is integer for all $i$, then I am stuck.
I also don't know where to use the fact that $p\mid H(n)$ for all $n$.
Thanks for the help!
| The answer is no. For example, $$P(x):=\frac{x\,(x+1)\,(x+5)}{6}$$
is not in $\mathbb{Z}[x]$. However, for $p=5$, we see that $P(x)$ satisfies the required conditions. To see this, we observe that
$$\sum_{r=0}^4\,P(n+r)=5\,\left(\frac{n(n-1)(n+1)}{6}+2n^2+8n+11\right)$$
for all $n\in\mathbb{Z}$.
| {
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Proposition 3.6 - " Functional Analysis , Sobolev Spaces and Partial Differential Equations ",Brezis "When $E$ is finite-dimensional, the weak topology $\sigma ( E,E') $ and the usual topology are the same."
We just need to show that every strongly open set ( I think it's the same of open set in the usual topology, that is the topology generated by a norm ) is weakly open. Let $x_{0} \in E$ and let $U$ be a neighborhood of $x_{0}$ in the strong topology. We have to find a neighborhood $V$ of $x_{0}$ in the weak topology $\sigma ( E,E') $ such that $V\subset U$. Let's assume we have already proved this thing, why would the topologies be the same?
I think I have some problem with the definition of basis of neighborhoods. Could anyone help me?
| A set $M$ is open if for every $x_0\in M$ it contains a neighborhood $U$ of $x_0$:
$$x_0\in U\subseteq M.$$
Brezis shows that if $E$ is finite dimensional and $U$ is a neighborhood in the strong topology it contains a neighborhood $V$ in the weak topology:
$$x_0\in V\subseteq U\subseteq M$$
thus showing that every strongly open set is also weakly open in the finite dimensional case. Since the weak topology is always coarser than the strong topology (also in the infinite dimensional case ) it follows that the set of weakly open sets and strongly open sets is the same , i.e. the topology is the same (in the finite dimensional case!).
| {
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Finding the limit by using the definition of derivative. Here is the problem.
Let $f$ be the function that has the value of $f(1)=1$ and $f'(1)=2$. Find the value of
$$ L = \lim_{x \to 1} {\frac{\arctan{\sqrt{f(x)}-\arctan{f(x)}}}{ \left (\arcsin{\sqrt{f(x)}}-\arcsin{f(x)}\right)^2}} $$
I have tried using
$$
L=\lim_{x\to 1}
\frac{1}{x-1}\frac{\frac{\arctan{\sqrt{f(x)}}-\arctan{\sqrt{f(1)}}}{x-1}-\frac{\arctan{{f(x)}}-\arctan{{f(1)}}}{x-1}}
{\left [\frac{\arcsin{\sqrt{f(x)}}-\arcsin{\sqrt{f(1)}}}{x-1}-\frac{\arcsin{{f(x)}}-\arcsin{{f(1)}}}{x-1} \right ]^2}
$$
and reduced that big chunks by using $f'(a)=\lim_{x \to a} \frac{f(x)-f(a)}{x-a}$ which I got
$$
\begin{split}
L&=\lim_{a\to 1}
\frac{1}{a-1}\frac{\Big [ \arctan\sqrt{f(x)} \Big ]'_{x=a} - \Big [ \arctan{f(x)} \Big ]'_{x=a}}
{\Big [ \arcsin\sqrt{f(x)} \Big ]'_{x=a} - \Big [ \arcsin{f(x)} \Big ]'_{x=a}}\\[2em]
&=\lim_{a\to 1}
\frac{1}{a-1} \frac{\frac{1}{1+f(a)}\frac{1}{2\sqrt{f(a)}}f'(a)-\frac{1}{1+(f(a))^2}f'(a)}
{\left [ \frac{1}{\sqrt{1-f(a)}}\frac{1}{2\sqrt{f(a)}}f'(a)-\frac{1}{\sqrt{1-(f(a))^2}}f'(a) \right ]^2}\\[2em]
&=\lim_{a\to 1}
\frac{1}{f'(a)\frac{a-1}{1-f(a)}} \frac{\frac{1}{1+f(a)}\frac{1}{2\sqrt{f(a)}}-\frac{1}{1+(f(a))^2}}
{\left [ \frac{1}{2\sqrt{f(a)}}-\frac{1}{\sqrt{1+f(a)}} \right ]^2}\\[2em]
&=\lim_{a\to 1}{-\frac{\frac{1}{1+f(a)}\frac{1}{2\sqrt{f(a)}}-\frac{1}{1+(f(a))^2}}
{\left [ \frac{1}{2\sqrt{f(a)}}-\frac{1}{\sqrt{1+f(a)}} \right ]^2}}\\[2em]
&=\boxed{(\sqrt{2}+1)^2}
\end{split}
$$
but the answer keys tell me that the answer of this problem is $L=\left( \frac{\sqrt{2}+1}{2}\right)^2$.
So, Can someone please explain to me what did I do wrong?
| We shall need several times the following principle: If $f(0)=0$, $\>f'(0)=1$, and $\lim_{x\to0} g(x)=0$, then we may write
$$f\bigl(g(x)\bigr)=g(x)\>h(x)\ ,\tag{1}$$ whereby the function $x\mapsto h(x)$ defined by $(1)$ satisfies $\lim_{x\to0}h(x)=1$.
We only need that $f(x)\nearrow1$ when $x\nearrow 1$ (note that your expression is undefined when $x>1$). There are functions $x\mapsto \alpha:=\alpha(x)$ and $x\mapsto\beta:=\beta(x)$ with $$0<\alpha<\beta\>,\qquad \lim_{x\to1-}\alpha(x)=\lim_{x\to1-}\beta(x)=0\ ,$$ such that $$\sqrt{f(x)}=\cos\alpha\>,\qquad f(x)=\cos\beta\ .$$
This ensures
$$\arcsin\sqrt{f(x)}-\arcsin f(x)=\beta-\alpha\ .$$
On the other hand
$$\arctan\sqrt{f(x)}-\arctan f(x)=\arctan{\cos\alpha-\cos\beta\over 1+\cos\alpha\cos\beta}={1\over2}(\cos\alpha-\cos\beta)\> g(x)\>,$$
whereby $\lim_{x\to1-}g(x)=1$. Now
$${1\over2}(\cos\alpha-\cos\beta)=\sin{\beta+\alpha\over2}\sin{\beta-\alpha\over2}={1\over4}(\beta+\alpha)(\beta-\alpha) h(x)\ ,$$
whereby $\lim_{x\to1-} h(x)=1$. It follows that
$$L={1\over4}\lim_{x\to1-}{\beta+\alpha\over\beta-\alpha }={1\over4}\lim_{x\to1-}{1+{\alpha\over\beta}\over1-{\alpha\over\beta}}.$$
As $\cos^2\alpha=\cos\beta$ for all $x<1$ we have
$\>\sin^2\alpha=2\sin^2{\beta\over2}$,
which then leads to $\lim_{x\to1-}{\alpha\over\beta}={1\over\sqrt{2}}$, so that we end up with
$$L={1\over4}\bigl(3+2\sqrt{2}\bigr)\ .$$
| {
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Roll a die probability question
An unbiased six-sided die is to be rolled five times. Suppose all these trials are independent. Let $E_1$ be the number of times the die shows a 1, 2 or 3. Let $E_2$ be the number of times the die shows a 4 or a 5. Find $P(E_1 = 2, E_2 = 1)$.
I have tried to solve this question this way:
Total $6P6 = 720$.
1,2 and 3 can be placed in $2$ locations $= 3^2$
4 and 5 can be placed in 1 location $= 2^1$
and 6 can be placed in two locations $= 2^2$
The probability is $= 72/720 = 0.1$
Is this correct?
| The probability equals:$$\binom5{2,1,2}\left(\frac36\right)^2\left(\frac26\right)^1\left(\frac16\right)^2=\frac5{72}$$(where $\binom5{2,1,2}:=\frac{5!}{2!1!2!}$)
You can look at this as a case of trinomial distribution.
There are $5$ independent experiments and there are not $2$ possible and mutually exclusive outcomes (success and failure, as is the case by binomial distribution) but there are $3$ possible and mutually exclusive outcomes: the die shows a face in $\{1,2,3\}$, or in $\{4,5\}$ or in $\{6\}$. This with probabilities $\frac36$, $\frac26$ and $\frac16$ that sum up to $1$.
| {
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If $f : [a,b]\to\Bbb R$ is continuous, are there $x_1,x_2\in (a,b)$ such that $\tfrac{f(b)-f(a)}{b-a} = \tfrac{f(x_1)-f(x_2)}{x_1-x_2}$? I just thought about the mean value theorem and wondered whether the following statement is true:
If $f : [a,b]\to\Bbb R$ is continuous, then there are $x_1,x_2\in (a,b)$
such that $\tfrac{f(b)-f(a)}{b-a} = \tfrac{f(x_1)-f(x_2)}{x_1-x_2}$.
One way to look at it is to consider the function $F : \{(x,y) : x,y\in[a,b],\,y>x\}\to\Bbb R$, defined by $F(x,y) = \tfrac{f(x)-f(y)}{x-y}$. If there don't exist such $x_1,x_2$, then $(a,b)$ is a maximum of a minimum of $F$. But I don't know what to conclude from that. Does anybody have an idea?
| First, without loss of generality assume that $f(a) = 0$(why can we do this?).
Next, set $h(x) = \frac{f(b)(x-a)}{b-a}$ on the interval $[a,b]$. This function is continuous. Note that $h(a) = 0$ and $h(b) = f(b)$.
Therefore, so is $g(x) = h(x) - f(x)$ on $[a,b]$. Note that $g(a) = 0$ and $g(b) = 0$.
Suppose there existed $x,y$ such that $g(x) = g(y)$ on $[a,b]$. Then, $h(x) - h(y) = f(x) - f(y)$, resulting in $\frac{f(b) - f(a)}{b-a} =\frac{f(x) - f(y)}{x-y}$, while noting that $f(a) = 0$.
Therefore, all we are reduced to proving is that there exist $x,y \in (a,b)$ such that $g(x) = g(y)$, for any continuous function $g$ on $[a,b]$ satisfying $g(a) =g(b) = 0$.
First of all, note that if $g$ is constant then of course we are done.
Without loss of generality, assume that $\max_{[a,b]} g(x) > 0$. (Otherwise, take $-g$). Now, the maximum is attained at a point $z$, where $z \in (a,b)$ (because $g$ is zero at the endpoints). If it is attained in two different points, then we are done by taking those two points. So we assume only one maximum exists.
Let $\delta > 0$ be such that $|y-z| < \delta \implies f(z)> f(y) > 0$. Let $y$ satisfy $|y-z| < \delta$, and $b > y > z$. Then define $l : [a,y] \to \mathbb R$ by $l(x) = g(x) - g(y)$. Note that $l$ is continuous, $l(a) < 0$ and $l(z) > 0$. So there exists $b \in (a,z)$ such that $l(b) = 0$ or that $g(b) = g(y)$. Hence, we are done.
EDIT : Here is the strengthening.
Let $g$ be a non-constant continuous function on $[a,b]$ such that $g(a) = g(b) = 0$(WLOG , $g(y) > 0$ for some $y$) . Let $z$ be any point at which $g(z) = \max_{[a,b]} g$(such a point exists, definitely). Let $d_z = \min\{z-a,b-z\}$. Then for all $v < d_z$, there exist points $x_1,x_2 \in (a,b)$ such that $x_1 - x_2 = v$ and $g(x_1) = g(x_2)$.
Proof : let $J(x)$ be defined on $(z-d_z,z+d_z)$ and be given by $J(x) = g(x) - g(x+v)$. Then, $J(z-v) \leq 0$ and $J(z) \geq 0$, so there is some point $l \in (z-v,z)$ such that $J(l) = 0$, or that $g(l) = g(l+v)$.
Note that if there are many maximums, then choosing a very central maximum(thus maximizing $d_z$) allows us to expand the range of values that $x_1 -x_2$ can attain.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Help with Mellin-Barnes Integral (product of two Hypergeometrics) I am trying to prove that
$$\int_0^1 \frac{dz}{z^2} z^{h}\cdot {}_{2}F_{1}(h,h;2h;z) \cdot {}_{2}F_{1}\left(\frac{1+2a}{2},\frac{1-2a}{2};1;\frac{z-1}{z}\right) = -\frac{\Gamma(2h)}{\Gamma{(h)}^2} \cdot \left(\frac{1}{a^2-(h-1/2)^2} \right).$$
To begin I first use Mellin-Barnes representation of Hypergeometric representation:
$$ {}_{2}F_{1}(a,b;c;z)=\frac{\Gamma(c)}{\Gamma(a)\Gamma(b)}\int_{-i\infty}^{i\infty}ds\frac{\Gamma(-s)\Gamma(a+s)\Gamma(b+s)}{\Gamma(c+s)}(-z)^s$$
The integral over the z variable is (done in Mathematica):
$$\int_0^1 \frac{z^h(-z)^s(1-z)^t}{z^2z^t}=(-1)^s\frac{\Gamma(-1+h+s-t)\Gamma(1+t)}{\Gamma(h+s)} $$
Plugging this back in I get,
$$\frac{\Gamma(2h)}{\Gamma(h)^2\Gamma(1/2+a)\Gamma(1/2-a)}\int\int ds dt \frac{\Gamma(-1+h+s-t)\Gamma(-s)\Gamma(h+s)\Gamma(-t)\Gamma(1/2+a+t)\Gamma(1/2-a+t)}{\Gamma(2h+s)}(-1)^s$$
where s and t are the Mellin-Barnes variables.
Then I use Barnes lemma to integrate over the t-variable to obtain:
$$\frac{\Gamma(2h)}{\Gamma(h)^2}\int_{-i\infty}^{i\infty}\frac{\Gamma(-s)\Gamma(-1/2+h+s+a)\Gamma(-1/2+h+s-a)}{\Gamma(2h+s)} \, ds$$
I dont see how I will get to the final result from here.
Any help appreciated.
Thanks
| The hypergeometric function in contour form is
$${}_{2}F_{1}(a, b; c; x) = \frac{1}{2 \pi i} \, \int_{-i \infty}^{i \infty} \frac{\Gamma(-s) \Gamma(a+s) \Gamma(b+s)}{\Gamma(c + s)} \, ds.$$
The integral in question becomes:
\begin{align}
I &= \int_{0}^{1} {}_{2}F_{1}(p,p;2p;t) \cdot {}_{2}F_{1}\left(\frac{1+2a}{2},\frac{1-2a}{2};1;\frac{t-1}{t}\right) \, t^{p-2} \, dt \\
&= \left(\frac{1}{2 \pi i}\right)^2 \, \int \int \frac{\Gamma(-u) \Gamma^2(p+u)}{\Gamma(2p + u)} \cdot \frac{\Gamma(-s) \Gamma(\frac{1}{2} + a + s) \Gamma(\frac{1}{2} - a + s)}{\Gamma(s + 1)} \, J \, du \, ds,
\end{align}
where
\begin{align}
J &=\int_{0}^{1} t^{p-2} \, (-t)^{u} \, \left(-1 + \frac{1}{t}\right)^{s} \, dt = (-i)^{u} \, B(s+1, p+u-s-1) \\
&= (-1)^{u} \, \frac{\Gamma(s+1) \Gamma(p+u-s-1)}{\Gamma(p+u)}.
\end{align}
Now, by use of Barnes' Lemma (first) given by
$$\frac{1}{2 \pi i} \, \int_{-i \infty}^{i \infty} \Gamma(a+s) \Gamma(b+s) \Gamma(c -s) \Gamma(d -s) \, ds = \frac{\Gamma(a+c) \Gamma(a+d) \Gamma(b+c) \Gamma(b+d)}{\Gamma(a+b+c+d)},$$
with $d = 0$, $c = p+u-1$, $a = 1/2 +a$ and $b = 1/2 -a$, then
$$ \frac{1}{2 \pi i} \, \int_{- i \infty}^{i \infty} \Gamma(-s) \Gamma(\frac{1}{2} + a + s) \Gamma(\frac{1}{2} - a + s) \Gamma(p+u-s-1) \, ds = \frac{\Gamma(a + \frac{1}{2}) \Gamma(p+a-\frac{1}{2} + u) \Gamma(p - a - \frac{1}{2} +u)}{\Gamma(p+u)} $$
and
\begin{align}
I &= \left(\frac{1}{2 \pi i}\right) \, \int \frac{(-1)^u \Gamma(-u) \Gamma(p+u)}{\Gamma(2p + u)} \, du \\
& \hspace{5mm} \cdot \frac{1}{2 \pi i} \, \int_{- i \infty}^{i \infty} \Gamma(-s) \Gamma(\frac{1}{2} + a + s) \Gamma(\frac{1}{2} - a + s) \Gamma(p+u-s-1) \, ds \\
&= \frac{\Gamma(a + \frac{1}{2})}{2 \pi i} \, \int_{- i \infty}^{i \infty} \frac{\Gamma(-u) \Gamma(p-a-\frac{1}{2} + u) \Gamma(p +a-\frac{1}{2} +u)}{\Gamma(2 p + u)} \, (-1)^{u} \, du \\
&= \Gamma\left(a + \frac{1}{2}\right) \, {}_{2}F_{1}\left(p+a-\frac{1}{2}, p-a-\frac{1}{2}; 2p ; 1\right) \\
&= \frac{\Gamma(2p) \Gamma\left(a+\frac{1}{2}\right)}{\Gamma\left(p+a+\frac{1}{2}\right) \Gamma\left(p-a+\frac{1}{2}\right)}.
\end{align}
| {
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"question_score": "2",
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Hilbert space and orthonormal basis. Let $H$ be a Hilbert space and let ${e_n} ,\ n=1,2,3,\ldots$ be an orthonormal basis of $H$. Suppose $T$ is a bounded linear oprator on $H$. Then which of the following can not be true?
$$(a)\quad T(e_n)=e_1, n=1,2,3,\ldots$$
$$(b)\quad T(e_n)=e_{n+1}, n=1,2,3,\ldots$$
$$(c)\quad T(e_n)=e_{n-1} , n=2,3,4,\ldots , \,\,T(e_1)=0$$
I think $(a)$ is not true because $e_1$ can not span the range space. I really don't know how to approach to this problem. Could you please give me some hints? Thank you very much.
| You're questions been answered, but if you're curious about examples of bounded linear operators that satisfy the last two resp. Look no further than $l^2(\mathbb{N})$ and consider the left and right shift operators on the standard basis. That is
$$T_L( (a_1, a_2, ... a_n , ... )) = (a_2, a_3, ... a_n,\ ... ) \\
T_R( (a_1, a_2, ... a_n , ... )) = (0, a_1,\ ... \ a_n,\ ... ) $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find particular solution of $y''+4y=12$ if the point $(0,5)$ has horizontal tangent line Find the particular solution of $$y''+4y=12$$ if the point $(0,5)$ has horizontal tangent line (parallel to the $x$-axis).
I know the general solution of $y''+4y=12$ is $$y=C_1\cos{(2x)}+C_2\sin{(2x)}+3\quad\text{for some}~C_1,C_2\in\mathbb R.$$ Now we have to find the particular solution but I don't know if the initial conditions are correct: $$\begin{cases}y''+4y=12\\\color{red}{y(0)=5}\\\color{red}{y'(0)=k,\quad\text{for some}~k\in\mathbb R}.\end{cases}$$ Are the $\color{red}{\text{initial conditions}}$ correct?
Thanks!
| The equation of the tangent line at $(0,5)$ is $$y(x) = y'(0)(x-0) + y(0) = y'(0)x + 5$$
This is parallel to the $x$-axis if and only if $y'(0) = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2917181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Onto and one-to-one linear transformations My book seems a bit unclear about onto and one to one linear transformations. Here is the excerpt:
and this is what I understand to be the rank of T:
rank(T) = dimension of column space of the original matrix.
Is the rank of T not the same thing as the dimension of W? Are these not the same thing? Can someone give me an example?
| Yes of course for any matrix the dimension of $\operatorname{Col}(A)$ is equal the dimension of $\operatorname{Row}(A)$ and that number is by definition $\operatorname{rank}(A)$.
Therefore if $\operatorname{rank}(A)$ is equal to the dimension of $W$ it means that the set "columns of $A$" contains a basis for $W$ and therefore the transformaion is onto.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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It is the set $V(y-\sin(x))\subset k^2$ a variety? It is the set $V(y-\sin(x))\subset k^2$ a variety?
I know that $V(y-\sin(x))=\{...,-4\pi,-3\pi,-2\pi,-\pi,0,\pi,2\pi,3\pi,4\pi,...\}$. Could it be a set of points a variety? My intuition tells me what not. How can I do a formal test, what property should I use? Thank you very much.
| Suppose $X = V(y-\sin(x))\subset \Bbb A^2$ were a variety. Then $X$ also admits a description as $V(f_1,\cdots,f_n)$ for some finite list of nonzero polynomials $f_i(x,y)$, each of which vanishes on $V(y-\sin(x))$.
By examining $f_i(x,0)$, each single-variable polynomial $f_i(x,0)$ should vanish on $V(y-\sin(x))\cap V(y)$. On the other hand, as you (intended to) identify in your post, $V(y-\sin(x))\cap V(y)$ is infinite, which is a problem (Why? Try to identify for yourself before mousing over the following spoiler).
Any nonzero polynomial in one variable may have at most finitely many solutions. As $f_i(x,0)$ has infinitely many solutions, it must be the zero polynomial, which implies that $y|f_i$ for each $f_i$, or that $V(y-\sin(x))$ contains the $x$-axis, which it does not by your identification of $V(y-\sin(x))\cap V(y)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $2x^3+2x^2-10x+6$ is positive if $x>1$
Show that $2x^3+2x^2-10x+6$ is positive if $x>1$
I have tried to solve it using mean value theorem but it doesn't work. Please anyone help me to solve this.
| Alt. hint: let $\,x-1 = y \gt 0\,$, then:
$$
2x^3+2x^2-10x+6 = 2(y+1)^3+2(y+1)^2-10(y+1)+6=2 y^3 + 8y^2 \gt 0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2917546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
RSA: Prove/disprove that $M_1^e\equiv M_2^e\ (\textrm{mod}\ n),$ given $M_1\not=M_2$ Given that $M_1\not=M_2, n=pq$ where $p,q$ are large primes and $\gcd(e,\varphi(n))=1$, how to prove/disprove that
$$M_1^e\equiv M_2^e\ (\textrm{mod}\ n)$$
is impossible? That is: I'm thinking about the case that two distinct messages $M_1,M_2$ has the same encrypted-text $C=M_1^e\mod n=M_2^e\mod n$.
I've tried a proof that using CRT and lead to a contradiction $M_1=M_2$ but it is a little bit complicated so I want to know is there any possible (simpler) alternatives.
| I posted a proof here that RSA is a bijective map from $\mathbb{Z}_n$ to itself, provided that $(\phi(N),e) =1$, and where the message space is the set of classes modulo $N$, the modulus.
So if we assume (as we must) that we are working in that message set, the RSA mapping has an inverse (using the exponent $d$ as described) and is in particular injective.
I also use the CRT, but why not? It's a basic tool in number theory. It's not "complicated".
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Definite integral of $x\sin^n x$ from $0$ to $\pi/2$ How to find
\begin{equation*}
\int_0^{\pi/2} x\sin^n x dx
\end{equation*}
where $n$ is a positive integer? I tried $y=x-\pi/4$ and that gives
\begin{equation*}
\frac{1}{2^{n/2}}\frac{\pi}{4}\int_{-\pi/4}^{\pi/4} (\sin y+\cos y)^n dy+\frac{1}{2^{n/2}}\int_{-\pi/4}^{\pi/4} y(\sin y+\cos y)^n dy.
\end{equation*}
Although some terms may be canceled, the summation of the others seems terrible.
Another method is integration by parts. Here I found the recurrence formula for $\sin^n x$:
Integrating $\int \sin^n{x} \ dx$
So the first step is to have
\begin{equation*}
\left.\left(-\frac{1}{n}x\cos x\sin^{n-1} x+\frac{n-1}{n}x\int \sin^{n-2} xdx\right)\right|_0^{\pi/2}-\int_0^{\pi/2}\left(-\frac{1}{n}\cos x\sin^{n-1} x+\frac{n-1}{n}\int_0^x \sin^{n-2} t dt\right)dx
\end{equation*}
the evaluations at $0$ and $\pi/2$ and the integral of $\cos x\sin^{n-1}x$ seem OK, but what about the second term? Any better method for this problem?
| Do the same thing: let the integral be $I_n$,
$$
\begin{align*}
&\phantom{=}\int_0^{\pi/2} x \sin(x)^n \mathrm dx = -\int_0^{\pi/2} x \sin(x)^{n-1}\mathrm d(\cos(x)) \\
&=\left. x \sin(x)^{n-1} \cos(x) \right|_{\pi/2}^0 - \int_{\pi/2}^0 \cos(x) (\sin(x)^{n-1} + (n-1)x \sin(x)^{n-2} \cos(x))\mathrm d x \\
&= 0 + \int_0^{\pi/2} \sin(x)^{n-1}\mathrm d (\sin(x)) +(n-1) \int_0^{\pi/2} x\sin(x)^{n-2}(1 - \sin(x)^2)\mathrm dx\\
&= \frac 1n + (n-1) I_{n-2} - (n-1)I_n ,
\end{align*}
$$
i.e.
$$
\fbox{$nI_n =\dfrac 1n + (n-1)I_{n-2}, $}
$$
with
$$
I_1 = 1,\quad I_2 = \frac 14 + \frac {\pi^2} {16},
$$
since
\begin{align*}
I_2&=\int_0^{\pi/2} x\sin(x)^2\mathrm dx \\
&= \left. x \sin(x) \cos(x)\right|_{\pi/2}^0 + \int_0^{\pi/2} \cos(x) (\sin(x)+ x\cos(x)) \mathrm dx \\
&= 0 + \frac 12 + \int_0^{\pi/2} x(1 - \sin(x)^2) \mathrm dx\\
&= \frac 12 + \frac {\pi^2} 8 - I_2\\
&\implies I_2 = \frac 14 + \frac {\pi^2}{16}.
\end{align*}
To go further, we are going to solve the recursive relation, but seems gruesome. So I let you take it from here.
UPDATE
According to wiki, we have a systematic method to solve this. First consider the terms with even index,
Let $J_{n} = I_{2n}$, then
$$
J_n = \frac 1{4n^2} + \frac {2n-1}{2n}J_{n-1},
$$
then divide both side of the equation by $(2n-1)!!/(2n)!!$:
$$
\frac {J_n}{\dfrac {(2n-1)!!}{(2n)!!}} = \frac 1 {4n^2} \frac {(2n)!!}{(2n-1)!!}+ \frac {J_{n-1}}{\dfrac {(2n-3)!!}{(2n-2)!!}},
$$
now let $P_n = J_n (2n)!!/(2n-1)!!$, then
$$
P_n = \frac 1{2n} \frac {(2n-2)!!}{(2n-1)!!} + P_{n-1},
$$
then
$$
P_n = \sum_2^n \frac 1{2k}\frac {(2k-2)!!}{(2k-1)!!} + P_1,
$$
which is pretty much for deriving a concrete number of $I_n$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to evaluate $ \int\left[\left(\frac{x}{e}\right)^x+\left(\frac{e}{x}\right)^x\right]\ln(x)\,dx$? $$ \int \left[ \left( \frac {x}{e}\right) ^ x + \left( \frac {e}{x}\right) ^ x \right] \ln(x)\, dx$$
Here I have no clue from where to start with.
I already tried some
*
*substitutions
*taking out ${e} ^ {x}$
But none seems to work please help
| We have
$\frac{d}{dx}\left(x\ (\ln x - 1)\right) = (\ln x - 1) + 1 = \ln x$, and
$\left(\frac xe\right)^x = e^{x(\ln x - 1)}$
| {
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How to show that $P(A\cup B) = P(A) + P(B) - P(A\cap B)$
Say we have a venn diagram with A on the left, intersection C in the middle and B on the right.
I want to show that $P(A\cup B) = P(A) + P(B) - P(A\cap B)$
I know what it can be written as A+B+C, but a lot of the proofs go from:
$P(A\cup B) = P(A) + P(B) + P(C)$ to
$P(A) + P(B) = P(A\cup B) + P(C) $
and then rearrange to get the proof and get $P(A\cup B) =P(A) + P(B)- P(C) $ . I just don't get how the left side of the inequality get go from $P(A\cup B)$ to $P(A) + P(B)$. Doesn't that only happen if we assume that they are disjoint. In this case, do we assume that they are disjoint? I know how to write the proof but just having a hard time actually understanding it.
| One of the reasons that Venn Diagrams are effective in visualizing probability problems as it poses a rather difficult concept to grasp (probability) in terms of a more intuitive concept (area). So, why don't we use area as a tool to help us prove this concept informally.
Let $P(A)$ be the area enclosed by circle $A$, and $P(B)$ the area enclosed by circle $B$. We seek to find the total area covered by both circles, i.e., $P(A\cup B)$. As you noted, $P(A\cup B)\neq P(A)+P(B)$ since the circles overlap, so we double count the area contained by both circles. How much area is double counted? The area contained by both circles, otherwise denoted by $P(A\cap B)$. This should help you develop intuition for why $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$Now, this is by no means a proper proof, but it should develop your intuition on the issue. And actually, this concept of representing probability as area is not as crazy as it may seem. As you'll learn if you continue studying probability, probability is closely related to a concept of measure, which is a representation of how much "space" a set contains.
Another way to see the proof is the following. Denote the area contained by both circles as $P(C)$. So, our equation becomes $$P(A\cup B) = P(A-C)+P(B-C)+P(C)$$Now since $C\subset A$ and $C\subset B$, $$P(A-C)=P(A)-P(C)\quad P(B-C)=P(B)-P(C)$$So, when we substitute into our equation, we get $$P(A\cup B) = P(A)+P(B)-P(C)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Min, max and saddle points of a function I want to find all local minima, maxima and saddle points of the function $f(x,y)=(x-y)(1-xy)$
Therefore I wanted to set the partial derivatives to zero to get all possible points.
First I calculated the derivatives $f_x(x,y) = y^2-2xy+1$ and $f_y(x, y) = -x^2+2xy-1$
From $f_x$ I got $x=\frac{y^2+1}{2y}$ and I used it in $f_y$ to be able to determine the values of $y$.
$-\frac{y^2+1}{2y}^2+2y\frac{y^2+1}{2y}-1=0$
Then I got $y^4-\frac{10}{3}y^2-\frac{1}{3}=0$
I substituted $z = y^2$ and got $z^2-\frac{10}{3}z-\frac{1}{3}=0$
When I wanted to solve this, it leads to $z_1= \frac{5+\sqrt(28)}{3}$ and $z_2= \frac{5-\sqrt(28)}{3}$. As I am not allowed to use a calculator, I think that I've done something wrong as it's for example not very easy for me to determine the result of $\sqrt(28)$ without using one, but I'm not able to find what I may have done wrong.
| You are ok expressing it as a square root...no more is needed. By the way, using taylor series is a great way of approximating functions. Or using differentials (easier):
Call $f(x)=\sqrt{x}, \space x=9 ,\space dx=1 \space:$
$$f(x+dx)\approx f(x)+f'(x)dx$$
$$\sqrt{10}\approx\sqrt{9}+\frac {1}{2\sqrt{9}}*1$$
$$\sqrt{10} \approx \frac {19}{6} \approx 3,16$$
If you use the binomial expansion (taylor) you can get as close as you want. But in math problems, it is ok to leave answers expressed in square roots, exponentials, or whatever function you can plug in a calculator later.
| {
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"timestamp": "2023-03-29T00:00:00",
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Functional equation $f(a)·f(b)=\Bigl\lbrace\frac{1}{ab}\Bigr\rbrace$ I am trying to get all possible solutions of the following functional equation:
$$f(a)·f(b)= \Bigl\lbrace\frac{1}{ab}\Bigr\rbrace$$
Where {} mean fractional part function.
Solutions only need to be valid inside open interval $a,b \in (0,1)$.
I am not really fluent in functional equations, any help would be useful.
| We have
$$\Biggl(f\left(\frac{1}{2}\right)\Biggr)^2=\left\{\frac{1}{\left(\frac{1}{2}\right)^2}\right\}=\{4\}=0\,,$$
so
$$f\left(\frac{1}{2}\right)=0\,.$$
Consequently, for $a\in(0,1)$, we get
$$0=f(a)\,f\left(\frac{1}{2}\right)=\left\{\frac{1}{a\cdot\left(\frac12\right)}\right\}=\left\{\frac{2}{a}\right\}\,.$$
However, this is not true, for example, when $a=\frac{3}{4}$. Hence, there is no such function $f:(0,1)\to\mathbb{R}$.
| {
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Basics of Quadratic Sieve algorithm I'm trying to understand Quadratic Sieve algorithm for integer factorization, I follow the description in the book "Prime Numbers" by Crandall and Pomerance, specifically the Algorithm 6.1.1. (Even though the question below apply to any description of QS, as far as I can see.)
Please, I have a few very basic questions about the method:
*
*in the initialization phase one finds the square roots ($a_i$) on the number being factored ($N$) modulo some prime ($p_i$). But in the description the roots are never mentioned again, what are they good for?
*sieve phase, this is possibly just me not understanding the text fully, that I should "sieve for $B$-smooth values". I think it is possible to read the "smooth" word that I have to try all the primes up to the bound, but I think what is meant here that I only should try the primes having Legendre symbol equal to one, from the previous phase. (I try both and there was no difference..?)
*why do I have to have $(K+1) \times K$ matrix in the linear algebra step ($K$ is the number of primes found during the initialization)? I mean, if I got lucky I could end up having just three exponent vectors summing to zero vector so I assume it is because the algorithm uses linear algebra method to find the right subset, correct?
*in the final step, what to do if the GCD does not reveal non-trivial factor? Simply compute more exponent vectors for the linear algebra step? Remove some of the already computed?
| $(1)$ The purpose of the roots is to accelerate the search for numbers $x$ with the property that $\ \ x^2\mod N\ \ $ has small factors. The algorithm would also work for random numbers, but the quadratic residues are a significant improvement concerning speed.
$(2)$ A number is $B$-smooth , if it has no prime factor larger than $B$. The goal of the number field sieve is to find enough congruences $x^2\equiv y\mod N$ with $B$-smooth $y$
$(3)$ In fact, we could have a zero vector with less than $K$ rows, but the chance will not be very large , if the matrix is large. Moreover, we may need many congruences of the form $\ \ x^2\equiv y^2\mod N\ \ $. Theoretically, there is a chance of $\frac{1}{2}$ that such a congruence gives a non-trivial factor, but the congruences constructed are not random.
$(4)$ It is as you say. In fact, sometimes , no factor is found, some congruences are deleted and new congruences are calculated. I do not know however the details.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that exists $u_m$ such that $\phi(u_m) = \max\{\phi(u),u\in E\}$
Let $E = \{u:[0,1]\to \mathbb R | u(0)=0, lip(u)\leq 1\}$. Define $\phi(u) = \int_0^1 (u^2(x)-u(x))dx$
Show that exists $u_m\in E$ such that $\phi(u_m) = \max\{\phi(u),u\in E\}$
I intended to show that $\{\phi(u),u\in E\}$ is compact, in order to find that $\sup \{\phi(u),u \in E\} = \max\{\phi(u), u \in E\}$.
I can think of two ways of proving this, showing that $\phi$ is continuous and $E$ is a compact set, or showing that $\{\phi(u), u \in E\}$ is closed and limited.
I'm trying to use the second Idea:
Let $u \in E \implies |u(x)-u(y)|\leq |x-y|\; \forall x,y \in [0,1]$. As $u(0) = 0$, $|u(x)|\leq |x| = x\; \forall x \in [0,1]$.
So $\int_0^1((u^2(x)-u(x))dx \leq \int_0^1(|x|^2-|x|)dx = -\frac{1}{6}$, therefore $\{\phi(u),u\in E\}$ is limited.
Now, how can I show it is closed? I don't believe that showing it's complement is open is a good idea and I can't think of a function to which this set would be the pre-image of something closed.
Any tips on how to continue OR alternative ways of showing that this $u_m$ exists?
| This is an easy consequence of Arzela -Ascoli Theorem. The set $E$ is equi-continuous and bounded at $0$ which implies it is uniformly bounded. Hence any sequence $\{u_n\} \subset E$ has a subsequence which converges uniformly. Can you now prove that $\{\phi (u):u\in E\}$ is closed?
| {
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Upper bound of sub-gaussian norm of bounded random variable? I am reading the High-Dimensional Probability by Dr.Roman Vershynin , where I stuck on some statement at page 28. where state as below:
Any bounded random variable $X$ is sub-gaussian with: $$\newcommand\norm[1]{\left\lVert#1\right\rVert}
\norm{X}_{\psi_2}\leq \frac{\norm{X}_{\infty}}{\sqrt{\log2}}
$$
where $\newcommand\norm[1]{\left\lVert#1\right\rVert}
\norm{X}_{\psi_2}$ is the sub-gaussian norm define as:
$$\newcommand\norm[1]{\left\lVert#1\right\rVert}
\norm{X}_{\psi_2} =\inf \left\{ t>0 : \mathbb{E} \left[\exp{\left(\frac{X^2}{t^2}\right)} \right] \leq 2 \right\}
$$
where $\newcommand\norm[1]{\left\lVert#1\right\rVert}
\norm{X}_{\infty} :=( \mathbb{E} |X|^p)^{1/p}$ as $p \to \infty$
I can see how why the bounded random variable is sub-gaussian (hoeffing lemma ),but How could I see this upper bound of sub-gaussian norm?
| If $t=(\sqrt {\log 2})^{-1} \|X\|_{\infty}$ then $Ee^{\frac {X^{2}} {t^{2}}} \leq e^{\log 2}= 2$ and hence $\|X\|_{\psi_2} \leq t$. I have used the fact that $\|X\|_{\infty}$ is nothing but the essential supremum of $X$.
| {
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Summation With Binomial Coefficient I am wondering how to estimate the following summation.
For $p \ge 0$,
$$
\sum_{i=0}^n \binom{2n}{i} (-1)^i (n-i)^p.
$$
When $p$ is a fixed integer this seems easy to do. But what if $p$ is a general real number?
I performed some numerical simulation and it seems this summation equals zero iff p is even. Can we give some lower bound to this summation, in general?
Update: the lower bound of the summation should be i=0, sorry for the confusion.
| Working with the question asked about $p$ an even integer we evaluate
$$S_{n,p} = \sum_{q=0}^{n} {2n\choose q} (-1)^q (n-q)^p.$$
Note that
$$\sum_{q=n}^{2n} {2n\choose q} (-1)^q (n-q)^p
= \sum_{q=0}^{n} {2n\choose n+q} (-1)^{n+q} (-q)^p
\\ = \sum_{q=0}^{n} {2n\choose 2n-q} (-1)^q (q-n)^p.$$
With $p$ even this is
$$\sum_{q=0}^{n} {2n\choose q} (-1)^q (n-q)^p.$$
Since the duplicate term for $q=n$ from the lower and upper sum is
zero we thus have
$$2S_{n,p} = \sum_{q=0}^{2n} {2n\choose q} (-1)^q (n-q)^p
\\ = p! [z^p] \sum_{q=0}^{2n} {2n\choose q} (-1)^q
\exp((n-q)z)
\\ = p! [z^p] \exp(nz) \sum_{q=0}^{2n} {2n\choose q} (-1)^q
\exp(-qz)
\\ = p! [z^p] \exp(nz) (1-\exp(-z))^{2n}.$$
Since $1-\exp(-z) = z + \cdots$ and hence $(1-\exp(z))^{2n} = z^{2n}
+\cdots$ this is zero when $p\lt 2n.$ In other words, $S_{n,p} = 0$
when $p$ is even and $2n\gt p.$ We also find that for $p=2n$ the value
is
$$\frac{1}{2} (2n)! [z^{2n}] \exp(nz) (1-\exp(-z))^{2n}
= \frac{1}{2} (2n)! [z^0] \exp(nz) = \frac{1}{2} (2n)!.$$
| {
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Finding $f$ satisfies $\limsup_{n\to\infty}\frac{\tau(n)}{f(n)}=1$ Is there a known result on finding the function $f$ satisfies
$$\limsup_{n\to\infty}\frac{\tau(n)}{f(n)}=1?$$
where $\tau(n)$ is the number of factor(s) of $n$.
Related:Prove that $d(n)\leq 2\sqrt{n}$
It shows that $f(n)=O(\sqrt n)$.
Some Ideas
Also, letting $n=2^m,m\in\mathbb Z$ gives $f(n)=\Omega(m)$, or $f(n)=\Omega(\ln n)$.
Factorize $n$ into $\prod_ip_i^{\alpha_i}$ gives $\tau(n)=\prod_i(\alpha_i+1)$. So probably the next step to do is to estimate $p_i$.
Wikipedia gives $p_n\le n\ln n+n\ln\ln n$ when $n\ge 6$. Can we use this result to give the better bounds for $f$?
| The best result in this direction is due to S. Wigert (1907):$$\limsup_{n\to\infty}\frac{\log\tau (n) \log\log(n)}{\log(2)\log(n)}=1.$$
It comes as a corollary to certain estimates of $\max_{n\le x} \tau(n)$, the strongest of which is due to S. Ramanujan (1915).
| {
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There are infinity many numbers when added to Three distinct integers making them pair-wise relatively prime Given $0<a<b<c$ three distinct integers, prove that there exists infinity many numbers $n$ such that $a+n,b+n,c+n$ are relatively prime to each other.
For the case when $(a,b)=(b,c)=(a,c)=1$ which means they are already relatively prime the answer is easy, i found that $n= (c-b)(c-a)(b-a)k $ for $k \in \mathbb{N} $ works, for instance $gcd((c-b)(c-a)(b-a)k + a, (c-b)(c-a)(b-a)k +b) = (b-a,(c-b)(c-a)(b-a)k +a)= (b-a,a) = (b,a) = 1$
And so for the rest.
But what to do when one or more are not relatively prime ?
How to solve it in more general way?
| Note: this does not work with $4$ numbers. For example, if you start with $(2,3,4,5)$ then whatever you add to each term, two of the set will be even. Whatever argument one constructs must fail if you try to extend it to $4$ terms.
Your remark shows that it would suffice to find a single such $n$.
Without loss of generality, suppose that $a>b>c$.
Note: if a prime $p$ divides $\gcd (a+n,b+n)$ then $p\,|\,a-b$.
Use the Chinese Remainder Theorem to solve the system of congruences $n\equiv 1-a\pmod {p_i}$ for each of the primes $p_i$ which divide $a-b$ or $a-c$, and $n\equiv 1-c\pmod {q_j}$ for each of the primes $q_j$ which divide $b-c$. Note that if $p_i=q_j$ for some pair $(i,j)$ then the congruences are consistent. (N.B. This is the bit that fails if you extend to $4$ terms).
We claim that $n$ works.
Indeed, If a prime $\psi$ divides $\gcd(a+n,b+n)$ then $\psi\,|\,a-b$ but in that case $a+n\equiv 1 \pmod {\psi}$, a contradiction.
| {
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Non trivial “multiplications” on $\mathbb{R}^2$ What are all the continuous and associative binary maps $M: \mathbb{R}^2 \times \mathbb{R}^2 \rightarrow \mathbb{R}^2$ that satisfy $M(x,y+z) = M(x,y) + M(x,z)$?
Besides the 0 constant function, and the identity multiplicative map $M((x_0,x_1),(y_0,y_1)) = (x_0y_0, x_1y_1)$ The theory of the complex numbers gives our first non trivial specimen: $M((x_0,x_1),(y_0, y_1)) = (x_0y_0-x_1y_1, x_0y_1+x_1y_0)$
How to find the rest? If there are any? Posing this as a system of two functional equations (associativity, distributivity over addition) seems like a natural direction but i lack machinery to turn that into a general solution.
Along the functional equation approach: we are looking for solutions to the following system of equations
$$M(x,M(y,z)) = M(M(x,y),z)$$
$$M(x,y+z) = M(x,y)+M(x,z)$$
| Let $z=a+bi$ with $b\ne0$. Then $\mathbb C=\mathbb R[z]$. In particular, $1,z$ is a basis for $\mathbb C$ over $\mathbb R$. In this basis, multiplication is given by
$$
\begin{align}
(x_1,y_1)(x_2,y_2)
&=(x_1+y_1 z)(x_2+y_2 z)
\\&=x_1 x_2 + y_1 y_2 z^2 +(x_1 y_2 +x_2 y_1)z
\\&=(x_1 x_2 - y_1 y_2(a^2+b^2))+(x_1 y_2 +x_2 y_1+2a)z
\\&=(x_1 x_2 - y_1 y_2(a^2+b^2),x_1 y_2 +x_2 y_1+2a)
\end{align}
$$
| {
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prove a function is lower semicontinuous $E = \ell^p$, with $1\le p<\infty$. For $x\in\ell^p$, $x = (x_1,x_2,\dots,x_n,\dots)$, check function
$$\varphi(x) = \begin{cases}\sum_{k=1}^{\infty}k|x_k|^2 &\text{ if } \sum_{k=1}^{\infty}k|x_k|^2<\infty \\
+\infty &\text{ otherwise}\end{cases}$$
is convex, l.s.c.
I have shown that $\varphi$ is convex, but I'm stuck in how to show it is l.s.c.. I attempt to use the definition to check that $\forall\lambda\in\mathbb{R}$, $A=\{\varphi\le\lambda\}$ is closed. Take a sequence $\{x^{(n)}\}$ in A s.t. $x^{(n)}\to x$ for some $x\in \ell^p$. I'm wondering how to use sum over power p to get the sum over squares. Or is there some other ways to prove it?
| The function $\phi$ is the supremum of the functions $\phi_n(x) = \sum_{k=1}^n k |x_k|^2$. Each $\phi_n$ is lower semicontinuous, and a supremum of l.s.c. functions is l.s.c.
| {
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how many numbers in (not-strictly) descending order of string length k Given K numbers $n_1, \, \ldots \,, n_k$ s.t. $0 \leq n_i \leq b$
How many ways are there to write the numbers $n_1$ to $n_k$ such that they are descending. (Descending here is used in the analysis sense)
Given $k=4$ and $b=4$ we could say that:
$0000$ is descending
$3321$ is descending
yet $2321$ is not.
I know this is a combinatorics problem and I have encountered solutions for strictly descending, but not descending.
Thank you in advance for any help.
| Notice that such a string is completely determined by how many times each digit appears in the string. For instance, if we have a descending string of length $6$ consisting of three $3$s, two $2$s, and one $1$, then it must be $333221$ since there is only one way to arrange the digits in descending order.
Let's consider the concrete example of the number of strings of length $12$ that can be formed using the decimal digits. Let $x_i$, $0 \leq i \leq 9$, be the number of times the digit $i$ appears in the string of descending decimal digits. Since there are a total of $k$ digits in the string,
$$x_0 + x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 + x_9 = 12$$
which is an equation in the nonnegative integers. A particular solution of the equation corresponds to the placement of nine addition signs in a row of twelve ones. For instance,
$$1 1 + 1 + + + 1 1 1 + 1 + + 1 1 1 1 + 1 +$$
corresponds to the solution $x_0 = 2$, $x_1 = 1$, $x_2 = 0$, $x_3 = 0$, $x_4 = 3$, $x_5 = 1$, $x_6 = 0$, $x_7 = 4$, $x_8 = 1$, $x_9 = 0$ and the string $877775444100$. The number of such solutions is the number of ways we can place nine addition signs in a row of twelve ones, which is
$$\binom{12 + 9}{9} = \binom{21}{9}$$
since we must select which $9$ of the $21$ positions required for twelve ones and nine addition signs will be filled with addition signs.
We wish to count the number of descending strings of length $k$ that can be formed with digits no larger than $b$. Let $x_i$, $0 \leq i \leq b$, be the number of times the digit $i$ appears in the string. The number of such strings that can be formed is the number solutions of the equation
$$x_0 + x_1 + x_2 + \cdots + x_b = k$$
in the nonnegative integers. A particular solution of this equation requires placing $b$ addition signs in a row of $k$ ones. Therefore, the number of such solutions is
$$\binom{k + b}{b}$$
since we must choose which $b$ of the $k + b$ positions required for $k$ ones and $b$ addition signs will be filled with addition signs.
| {
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Is it known that $\sum_{i=1}^\infty \frac{if\;(i \pmod n=0)\;then\;(1-n)\;else\;(1)}{i}=log(n)$? I found this general infinite sum:
$\sum_{i=1}^\infty \frac{ \mathtt i \mathtt f \; (i \pmod n = 0) \; \mathtt t \mathtt h \mathtt e \mathtt n \;(1-n) \; \mathtt e \mathtt l \mathtt s \mathtt e \; (1)}{i} = log(n)$
Sample with n = 2:
log(2) = 1 - $\frac{1}{2}$ + $\frac{1}{3}$ - $\frac{1}{4}$ + $\frac{1}{5}$ - $\frac{1}{6}$ + $\frac{1}{7}$ - ...
(the minus are given at (i MOD 2)=0 --> (1 minus 2) = -1)
Sample with n = 3:
log(3) = 1 + $\frac{1}{2}$ - $\frac{2}{3}$ + $\frac{1}{4}$ + $\frac{1}{5}$ - $\frac{2}{6}$ + $\frac{1}{7}$ + $\frac{1}{8}$ - $\frac{2}{9}$ + ...
(the minus are given at (i MOD 3)=0 --> (1 minus 3) = -2)
and so on.
Is this formula already known?
Can that be proven?
Thanks in advance.
| You have to start with partial sums. Choose some $M$ to get:
$$
S_M = \sum_{i=1}^{nM} \frac{if\;(i \pmod n=0)\;then\;(1-n)\;else\;(1)}{i}= \sum_{i=1}^{nM} \frac1i - n \sum_{i=1}^M \frac1{n i} = \sum_{i=1}^{nM} \frac1i - \sum_{i=1}^M \frac1{i}
$$
Now observe the well-known limit (see here) $\lim_{k \to \infty}\sum_{i=1}^k \frac1{i} = \ln(k) + \gamma$, where $\gamma$ is the Euler–Mascheroni constant.
This gives
$$
S = \lim_{M \to \infty} S_M = \ln(nM) + \gamma - (\ln(M) + \gamma) = \ln(n)
$$
| {
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Going from Fourier sum to Fourier integral - confusion on intermediate step How does it solve the $A_0$ term?
Can one prove this with an example? I don't know how to prove it to myself with an example
In my lecture notes, my lecturer is trying to justify $C_n$ as weighting the sine and cosine terms in the Fourier integral differently despite how it looks explicitly in the Fourier infinite sum, that is, in:
$$\sum_{n=-\infty}^{\infty} C_n \ e^{ik_nx} \ $$
I have two main confusions with this:
1) How does this rewriting with $a_n$ and $b_n$ resolve the $A_0$ term?
2) I don't see how the substitution of $A_n$ and $B_n$ for $a_n$ and $b_n$ is valid, even given the changed summation ranges. I also don't know how I'd prove it with an example. How can one do this?
| The "trick" is in the sum from $-\infty$ to $\infty$ where $|n|$ is counted once for $|n|=0$ , and twice for $|n| \ge 1$.
| {
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Prime and rational numbers I came across the following question while studying that is stumping me. Can anyone please help me solve it?
Let "$a$" be a prime number greater than $10,000$ and let $x=\sqrt{a}$. Which of the following expressions represents a rational number?
F) $x/2$
G) $\sqrt{x}$
H) $2x$
J) $x^2$
K) $x+2 $
| Assume $x$ is rational. Then we can express $x$ as:
$$x=\frac{p}{q}$$
where $p,q$ are co-prime integers. $x=\sqrt a$ can be written as:
$$x^2=a$$
$$\frac{p^2}{q^2}=a$$
$$p^2=aq^2$$
As the prime $a$ divides the RHS, it divides the LHS too. $p=ka$ for some integer $k$. Substituting:
$$k^2a^2=aq^2$$
$$k^2a=q^2$$
Hence, $a$ also divides $q$, which implies $p$ and $q$ are not co-prime. Contradiction.
Thus, $x$ is irrational.
We can go ahead proving by contradiction that F,G,H,K are all irrational.
J is the answer as $x^2=a$ is prime and thus integer and rational.
| {
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How many ways are there for 50 people to divide them into three groups, A, B and C such that each consists of 20, 18, and 12, respectively? I have tried to solve this problem but I can not figure out where to start. Any help would be appreciated.
Thanks,
EDIT: After another attempt I am leaning towards the answer $\frac{50!}{20!18!12!}$
EDIT2: I know that the answer that I arrived at is correct but I still do not understand why it is correct.
| As indicated in the comments, your answer is correct.
We can select $20$ of the $50$ people for group $A$ in $\binom{50}{20}$ ways, $18$ of the remaining $30$ people for group $B$ in $\binom{30}{18}$, and all $12$ of the remaining $12$ people for group $C$ in $\binom{12}{12}$ ways. Hence, the number of ways of selecting $20$ people for group $A$, $18$ people for group $B$, and $12$ people for group $C$ is
$$\binom{50}{20}\binom{30}{18}\binom{12}{12} = \frac{50!}{20!30!} \cdot \frac{30!}{18!12!} \cdot \frac{12!}{12!0!} = \frac{50!}{20!18!12!0!} = \frac{50!}{20!18!12!}$$
as you found.
| {
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Evaluating $\sum_{n=1}^{\infty} \frac{\phi(n)}{7^n + 1}$, where $\phi(n)$ is Euler's totient function
Evaluate $$\sum_{n=1}^{\infty} \frac{\phi(n)}{7^n + 1}$$
where $\phi(n)$ is Euler's totient function.
I found this problem on a Discord server I just joined. The first time I saw this sum, I was daunted. After gathering courage, I started my work on this problem.
I tried various things, like using the definition
$$\phi(n) = n\left(1 - \frac{1}{p_1}\right)\cdots\left(1-\frac{1}{p_m}\right)$$ (where the $p_i$ are the prime factors of $n$) and what not, but it lead to nothing.
Then I started thinking about the $7$ in the denominator; why only $7$? What is so special about $7$? What is the difference between
$$\sum_{n=1}^{\infty}\frac{\phi(n)}{6^n + 1}$$ and
$$\sum_{n=1}^{\infty}\frac{\phi(n)}{7^n + 1}$$.
To answer that question, I took the literal difference of the two sums, but the denominator blew up so I started to think again. Because of that, I defined the two sequences and their sum:
$$a_n = \frac{\phi(n)}{7^n + 1} \textrm{ and }b_n= \frac{\phi(n)}{7^n - 1}$$
$$S_1 = \sum_{n=1}^{\infty} a_n\textrm{ and }S_2 = \sum_{n=1}^{\infty} b_n$$
and took THEIR difference, which lead to
$$\sum_{n=1}^{\infty}(b_n - a_n) = 2\sum_{n=1}^{\infty}\frac{a_nb_n}{\phi(n)}$$
Now I can say that I'm genuinely stuck. Can someone provide a hint as to what to do next?
| Another approach is to use Lambert Series generating function:
$$
\begin{align}
\frac{q}{(1-q)^2} &=\sum_{n=1}^{\infty}\frac{\phi(n)q^n}{1-q^n}=\sum_{n=1}^{\infty}\frac{\phi(n)}{q^{-n}-1} \space\colon\space |q|\lt1\implies \\
\frac{q^2}{(1-q^2)^2} &=\sum_{n=1}^{\infty}\frac{\phi(n)}{q^{-2n}-1} =\frac12\sum_{n=1}^{\infty}\left[\frac{\phi(n)}{q^{-n}-1}-\frac{\phi(n)}{q^{-n}+1}\right]
\end{align}
$$
Hence:
$$
\begin{align}
\color{red}{\sum_{n=1}^{\infty}\frac{\phi(n)}{q^{-n}+1}}&=\sum_{n=1}^{\infty}\frac{\phi(n)}{q^{-n}-1}-2\sum_{n=1}^{\infty}\frac{\phi(n)}{q^{-2n}-1} \\
&=\frac{q}{(1-q)^2}-\frac{2\,q^2}{(1-q^2)^2}=\color{red}{q\,\frac{1+q^2}{(1-q^2)^2}}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2919929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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$K = \mathbb{Z} / 2 \mathbb{Z}$. How many subspaces does the $K$-vector space $K^2$ have?
Let $K = \mathbb{Z} / 2 \mathbb{Z}$, How many subspace does the $K$-vector space $K^2$ have?
I (hope to) already know the following
*
*When diving a whole number by 2, we can only obtain 0 or 1, so $K = \{0, 1\}$ and $K^2 = \{0, 1\} \times \{0, 1\}$. Therefore $K^2$ has only four elements, $(0,0)$, $(0,1)$, $(1,0)$ and $(1,1)$.
*Every vector space has the two trivial subspaces: itself and the zero vector space.
*A subspace has to be not empty and closed under addition and multiplication with scalars from $K$.
My Question
Is $V = \{0\} \times \{0,1\} \not= \emptyset$ a $K$-vector subspace? Using the addition from $K$, we obtain $(0,0) + (0,1) = (0,1) \in V$ And $\alpha \cdot (0,0) = (0,0)$ for all scalars $\alpha \in K$ and we obtain $0 \cdot (0,1) = (0,0) \in V$ as well as $1 \cdot (0,1) = (0,1) \in V$ etc.
Short Summary of answers given
Besides the two trivial subspaces mentioned above there are three other subspaces, the reasoning being analogous to the one for $V = \{0\} \times \{0,1\}$ above. They are
*
*$V_1 = \{0\} \times \{0,1\}$
*$V_2 = \{0, 1\} \times \{0\}$
*$V_3 = \{ (0,0), (1,1) \} $
| Yes, $\{0\}\times\{0,1\}$ is indeed a $K$-vector subspace. You almost completed your proof, but it has some flaws you should still fix. Remember, to prove that $U\subset V$ is a vector subspace (over some field $F$), you need to prove two things:
*
*For every pair $u_1, u_2\in U$, the element $u_1+u_2$ is also in $U$. In other words, $$\forall u_1, u_2\in U: u_1+u_2\in U$$
*For every $u\in U$, and every $\alpha\in F$, the element $\alpha u$ is also in $U$. In other words, $$\forall u\in U, \alpha\in F: \alpha u\in U$$
So far, you have proven that $(0,0) + (0,1)$ is indeed an element of $V$, but you still have to prove that all pairs of elements from $V$ add up to elements of $V$. This includes pairs of identical elements!
Also, I don't like the sentence
$\alpha \cdot (0,1)$ will be either $(0,0)$ or $(0,1)$ because of the way addition in $K$ works.
which I think you should rewrite. First of all, it's not because of the way addition works, it's about how multiplication works (sure, the two are related, but it's better form). Second, it's better if you just write down all the possibilities, since there are only two options for $\alpha$ anyway.
| {
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Finding the sum of squares of roots of a quartic polynomial. What is the sum of the squares of the roots of $ x^4 - 8x^3 + 16x^2 - 11x + 5 $ ?
This question is from the 2nd qualifying round of last year's Who Wants to be a Mathematician high school competition which can be seen here:
I know the answer (32) because that is also given in the link, and I have checked by brute force that the given answer is correct.
However, I have made no progress at all in figuring out how to calculate the sum of squares of the roots - either before or after knowing the answer! I was expecting there to be a nice "trick" analagous to the situation if they had given a quadratic and asked the same question -- in that case I know how to get the sum and product of the roots directly from the coefficients, and then a simple bit of algebraic manipulation to arrive at the sum of squares of the roots.
In this case (the quartic) I have no idea how to approach it, and I have not spotted any way to simplify the problem (e.g. I cannot see an obvious factorisation, which might have helped me).
I've looked on the web at various articles which dicuss the relationships between the coefficients of polynomials and their roots and - simply put - I found nothing which gave me inspiration for this puzzle.
Given the audience for this test, it should be susceptible to elementary methods ... I would appreciate any hints and/or solutions!
Thank you.
| Hint:
\begin{align}
\sum_{i=1}^4a_i^2 &= \left(\sum_{i=1}^4a_i \right)^2-2\sum_{i< j}a_ia_j
\end{align}
Also Vieta's formula might help.
| {
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How to calculate the derivative of $x^x$? I'm trying to follow an example in my textbook.
$$y=x^x$$
$$\ln (y)=\ln (x) \cdot x$$
We want to calculate the derivative with respect to x
The book makes quite a leap here and states that:
$$\frac{y'}{y}=\frac{1}{x}\cdot x+1\cdot \ln(x)$$
Since $y=x^x$ this means that:
$$y'=x^x(1+\ln(x))$$
Is this correct?
If I start from the beginning then:
$$y=x^x$$
$$\ln (y)=\ln (x) \cdot x$$
Only if we want to take the derivative this expression isn't useful, we'll have to use the full expression:
$$e^{\ln (y)}=e^{\ln (x) \cdot x}$$
Now, if we take the derivative of this with respect to x we find that:
$$\frac {1}{y}\cdot e^{\ln (y)}=e^{\ln (x) \cdot x}\cdot (1+ln(x))$$
The left hand expression would simplify to $\frac {y'}{y}$, since $e^{\ln(y)}=y$ and the derivative of $y=y'$
so:
$$\frac {y'}{y}=e^{\ln (x) \cdot x}\cdot (1+\ln(x))$$
Which can be written as:
$$\frac {y'}{x^x}=x^x\cdot (1+\ln(x))$$
Which simplifies to:
$$y'=x^{2x}\cdot (1+\ln(x))$$
| You missed out $y'$ in this line:
$$\frac {1}{y}e^{\ln y}=e^{x\ln x}(1+\ln x)$$
as $$\left(e^{\ln y}\right)'=e^{\ln y}\cdot\frac1y\cdot y'=y'$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simplicial homology and homeomorphisms In Hatcher's book, in the introduction page of singular homology, he mentions that "it is obvious that homeomorphic spaces have the same singular homology, in contrast to simplicial homology". However I thought that this was also true for simplicial homology (and looking at the construction I don't see why this would not be true for simplicial homology).
What does he mean by this statement? Does he point at the fact that not all spaces are triangulable and hence do not admit the construction of simplicial homology?
| Neal's answer explains the problem of different triangulations. Nevertheless it turns out that the simplicial homology $H_*(\mathcal{T})$, where $\mathcal{T}$ is a simplicial complex triangulating $X$, is a topological invariant of $X$. That is, if $X_1, X_2$ are homeomorphic and $\mathcal{T}_i$ are triangulations of $X_i$, then $H_*(\mathcal{T}_1) \approx H_*(\mathcal{T}_2)$. The standard proof relies on "identifying" the simplicial homology of a triangulation of $X$ with the singular homology of $X$. This is a genuine topological proof.
Historically, the need for a topological proof was not so obviuos. In the "early days" mathematicians conjectured that a combinatorial proof was possible. It is a simple observation that if $\mathcal{T}$ is a triangulation of $X$ and $\mathcal{T}'$ is a subdivision of $\mathcal{T}$, then $H_*(\mathcal{T}) \approx H_*(\mathcal{T}')$. Now the so-called Hauptvermutung said that any two triangulations of a triangulable space have a common subdivision. This would obviuosly prove that $H_*(\mathcal{T})$ is a topological invariant of $X$.
Unfortunately the Hauptvermutung fails as was shown by John Milnor in 1961. Therefore there is no combinatorial proof.
See for example https://en.wikipedia.org/wiki/Hauptvermutung.
| {
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"timestamp": "2023-03-29T00:00:00",
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In how many ways can ten people be seated in a row so that a certain two of them are not next to each other? What I'm thinking: Find total ways that the ten people can be seated, which is 10!.
Then I take that number and subtract the ways the these two people would be seated next to each other. I do this by treating these two people as a single space, which leaves the eight other students plus that space consisting of the two. This would mean 9!
Then, 10! - 9! = 3265920 ways for the ten people to be seated so that a certain to are not next to each other.
| Call the two special people $A$ and $B$. There are nine seats where the "left" one can sit and the right next to him. There are two such cases $AB$ and $BA$. For each of these conditions there are 8 remaining slots that can be filled by the remaining $8$ folk arbitrarily.
Thus: $9 \cdot 2 \cdot 8! = 725,760$
The total number of ways the chosen two people do not sit next to each other is $10! - 9 \cdot 2 \cdot 8! = 2,903,040$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2920517",
"timestamp": "2023-03-29T00:00:00",
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Construct an order on a field $\mathbb{Q}^2$ Let $\mathbb{Q}$ be the field of rational numbers. Let $\mathbb{Q}^2=\{(a,b):a,b\in\mathbb{Q}\}$ and define addition and multiplication as follows:
$$
(a,b)+(c,d)=(a+c,b+d)\\
(a,b)\cdot(c,d)=(ac+2bd, ad+bc)
$$
Then $(\mathbb{Q}^2, +, \cdot)$ is a field, and $(0,0)$ is its zero element while $(1,0)$ is its unit element. The inverse of $(a,b)$ is $(\frac{a}{a^2-2b^2},-\frac{b}{a^2-2b^2})$. I'm asked to construct an order on it such that it becomes an ordered field, namely invariant under addition and multiplication with a positive element.
What I know is $(1,0)>(0,0)$, so $(n,0)>(0,0)$ for any natural number $n$. I don't quite know how to proceed. Is there any rules to follow to construct an order on a field?
Thank you for any help!
| The quick and easy answer is to notice that this field is isomorphic to $\mathbb Q[\sqrt{2}]$ by sending $(a,b)$ to $a+b\sqrt 2$, so $(a,b)<(c,d)$ if $a+b\sqrt 2<c+d\sqrt 2$.
| {
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Why does an indeterminate cause a hole in a rational function and not a vertical asymptote? My intermediate Algebra textbook gave the following example and a graph of this function:
$$f(x)=\frac{2x+1}{2x^2-x-1}$$
The factored form of this is:
$$\frac{2x+1}{(2x+1)(x-1)}$$
I know that a vertical asymptote is caused when a number is divided by $0$, since division by $0$ is undefined, and the vertical asymptote is located at the value of $x$ that causes a $0$ in the denominator. Therefore we set the factors of the denominator to $0$ and solve for $x$.
$x-1=0$
$x=1$
And
$2x + 1 = 0$
$x=-\frac{1}{2}$
I can see why there's a vertical asymptote at $x = 1$, since $x =1$ will result in division by $0$, thus $x$ can approach $1$ but never be $1$ and as a result $y$ goes to either $\infty$ or $-\infty$ depending on which side $x$ approaches from.
However, taking $x=-\frac{1}{2}$ and plugging it back it will cause an indeterminate, but an indeterminate is still division by $0$, so shouldn't there be a vertical asymptote at $x = -\frac{1}{2}$ and shouldn't $y$ also be going to either $\infty$ or $-\infty$ as $x$ approaches $-\frac{1}{2}$? Why is it just a hole instead?
| Simply,which value of x makes the numerator zero of any function is called zero point and which x value makes denominator zero means for what values of x the function becomes undefined is called poles.Here,we first have to check whether the numerator is zero or not.In this case,if the numerator would not be zero means at $x=\dfrac{1}{2}$ if
$$f(x)=\dfrac{something}{0}~~~[something~~\ne~~0]$$
then we would surely call it a pole means there would be a vertical asymptote at $x=\dfrac{1}{2}$.But,it has also maked the numerator zero. So we have got $$f(x)=\dfrac{0}{0}$$ this is not possible...so it is undefined or there is no value for $f(x)$ there because this is not a valid expression.so,the function has a discontinuity there.
| {
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String powers. Lemma 1.3.5 in word processing in groups. Epstein.
I can't understand this demonstration.
Why if ${w'}_1$ is different from ´${w'}_2$ then we have that $f(u)$ and $v'$ are powers of some string $z$?
| The way induction is used is not described in much detail. I try to understand and explain it. The base case where $|f(u)|+|f(v)| = 0$ is not even mentioned.
For the induction step, instead of the alphabet $\{u,v\}$ the alphabet $\{u,v'\}$ is used, supposedly with the map $f'(u) = f(u); f'(v') = v'$. Since $f(u)v' = f(v)$ by the definition of $v'$, $|f'(u)| + |f'(v')| = |f(v)|$. Unless $f(u)$ is empty, $|f'(u)| + |f'(v')| < |f(u)|+|f(v)|$ and therefore the induction hypothesis applies to $f(u)$ and $v'$. If $f(u)$ is empty it is a common power of any string.
| {
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No analytic function $f$ has modulus $|f(z)|=1/\cosh(\Re z)$ An analytic function $f(z) = f(x+iy)$ in $\mathbb{C}$ cannot have modulus $\frac{A}{\cosh(x)}$ for some constant $A \neq 0$.
Can we do so simply using the Cauchy-Riemann Equations?
I tried working by contradiction:
Say there is such a $f(z)$. Given $f(z) = u(x,y) + i v(x,y)$ is analytic, it satisfies: $u_x = v_y$ and $u_y = -v_x$
We also see that: $$|f(z)|^2 = u(x,y)^2 + v(x,y)^2 = (\frac{A}{\cosh x})^2$$
I tried reaching a contradiction, but seem to be getting lost in a mess of reformulations.
My start point is:
$$u_x = v_y \text{ and } u_y = -v_x$$
along with
$$u(x,y)u_x + v(x,y)v_x = -A^2 \frac{sinh x}{(cosh x )^3}$$
and
$$u(x,y)u_y + v(x,y)v_y = 0$$
I would appreciate some hints!
| So
$$
\begin{align*}
uu_x+vv_x&=-A^2\frac{\sinh x}{\cosh^3 x}\\
-uv_x+vu_x&=0\\
\end{align*}
$$
So
$$
u_x=-u\frac{\sinh x}{\cosh x}
$$
(remember $u^2+v^2=\dfrac{A^2}{\cosh^2 x}$) and similarly $v_x=-v\dfrac{\sinh x}{\cosh x}$. So
$$
u=\frac{1}{\cosh x}\cdot f(y)\text{ and }v=\frac{1}{\cosh x}\cdot g(y)
$$
and there are no $g$ that gives $u_x=v_y$.
| {
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definite integral $\int_{0}^{\frac{\pi}{4}} \frac{\sin^2x\cos^2x}{\sin^3x+\cos^3x}dx$
$$\int_{0}^{\frac{\pi}{4}} \frac{\sin^2x\cos^2x}{\sin^3x+\cos^3x}dx$$
I tried $2$ or $3$ trigonometric transformations but it did not work. One of them is as follows
$$\frac{\sin^2x\cos^2x}{(\sin x+\cos x)(1-\sin x\cos x)}$$ after that I am not able to figure out what to do. If I use double angle formula then
$$\frac{\frac{\sin2x}{4}}{(\sin x+\cos x)\left(1-\frac{\sin2x}{2}\right)}$$ again i am clueless
| Using identities
$$\sin x+\cos x=\sqrt{2}\cos(x-\dfrac{\pi}{4})$$
$$\sin x\cos x=\cos^2(x-\dfrac{\pi}{4})-\dfrac12$$
and then substitution $\dfrac{\pi}{4}-x=u$ gives
\begin{align}
I
&= \int_{0}^{\frac{\pi}{4}} \frac{\sin^2x\cos^2x}{\sin^3x+\cos^3x}dx \\
&= \int_{0}^{\frac{\pi}{4}}\frac{\sin^2x\cos^2x}{(\sin x+\cos x)(1-\sin x\cos x)}dx \\
&= \dfrac{1}{2\sqrt{2}}\int_{0}^{\frac{\pi}{4}}\frac{(2\cos^2u-1)^2}{\cos u(3-2\cos^2u)}\ du \\
&= \dfrac{1}{2\sqrt{2}}\int_{0}^{\frac{\pi}{4}}\frac{(2\cos^2u-1)^2}{\cos^2u(3-2\cos^2u)}\cos u\ du
\end{align}
now let $\sin u=w$ therefore
\begin{align}
I
&= \dfrac{1}{2\sqrt{2}}\int_{0}^{\frac{\sqrt{2}}{2}} \frac{(2w^2-1)^2}{(1-w^2)(2w^2+1)}\ dw \\
&= \dfrac{1}{2\sqrt{2}}\int_{0}^{\frac{\sqrt{2}}{2}} -2+\frac13\frac{1}{1-w^2}+\frac83\frac{1}{2w^2+1}\ dw \\
&= \color{blue}{-\dfrac12+\dfrac{\pi}{6}+\dfrac{\sqrt{2}}{12}\operatorname{arccoth}\sqrt{2}}
\end{align}
| {
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Two-player game in $19$ rounds
Aashna and Radhika see the integers $1$ to $211$ written on a blackboard. They alternate turns and in every step each of them wipes out any $11$ numbers until only $2$ numbers are left on the blackboard. If the difference of these $2$ numbers (by subtracting the smaller from the larger) is $\geq 111$, the first player wins, otherwise the second. If you were Aashna, would you chose to play 1st or 2nd and why?
By intuition only – I don’t see how I can prove it:
Since there are 19 turns ($19\times 11=209$ numbers + 2 on the blackboard), I would choose to play 1st.
In my 1st move I would wipe out numbers 101 to 111 since these are the only ones that do not have a pair to meet the rule.
Then for whichever numbers Radhika would remove, I would respond by removing the numbers that would have a difference of 111, for example for 92 I wipe out 203 etc. If Radhika chose to remove pairs with difference equal to 111, there would still be a single number, for which I would remove its pair and then I would also remove pairs with difference 111 or 112 and so on.
Does this method guarantee a win?
| If you chose to play first you would have 10 turns. -> 10 x 11 = 110
If you chose to play second you would have 9 turns. -> 9 x 11 = 99
If you are second and remove every high number starting with 211 while the first removes the low numbers you would stop at 211 - 99 = 112.
The last left numbers would be 112 and one number (between 1 and 112).
The first would then only win if that number is 1, because 112 - 1 = 111 (which is >= 111).
If it was the opposite and you would play first you could remove 110 numbers and you would win if the difference is >= 111. There are only 211-111 = 100 numbers with a difference >= 111. Remove all of them and you win.
So your chance of winning by playing first is 100%.
| {
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"timestamp": "2023-03-29T00:00:00",
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Rank of a Hermitian matrix in terms of Eigen values? I have a complex Hermitian matrix, say W, which is obtained by solving a convex optimization problem. In order for this matrix to be the result of my original problem, W must satisfy the following condition rank ( W ) = 1. When I checked this condition, MATLAB gives me an answer 3 for 3x3 W matrix. However, eigenvalues of this matrix are [-2.04e-11,-1.92e-12,2.81]. Now, my question is should I consider W as a rank-1 matrix based on the eigenvalues result or should I consider it as a rank 3 matrix as provided by MATLAB rank condition? (I have learnt that the rank of a symmetric matrix is the number of non-zero eigenvalues. If this is the case, can I also extend this def to Hermitian matrix? and If this def is also true for hermitian matrix, then what should be the rank of W (1 or 3))?
| If the matrix has three non-zero eigenvalues, then its rank is $3$. The rank of a $3 \times 3$ matrix is $3$ minus the dimension of the eigenspace corresponding to $0$ (see here for more detail).
However, if you look at the three eigenvalues produced, you'll notice that two of them are very small. I'd wager that they are non-zero because of numerical error; whatever method Matlab has used to arrive at the solution, its approximation is close to, but not, a matrix of rank $1$.
As for how you should proceed, we don't really have the context to answer this. It depends for what purpose you plan to use this matrix.
| {
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Is this Riemann sum formula for definite integral using of prime numbers true? While answering another question in MSE, I had used the following result which I thought was a trivial consequence of the prime number theorem and equidistribution. However, I realized from the comments that many people thought that this was not either true or counter intuitive. Hence I am posting this as a question looking for a proof or disproof.
Let $p_k$ be the $k$-th prime and $f$ be a continuous function
Riemann integrable in $(0,1)$ such that
$$\lim_{n \to \infty}\frac{1}{n}\sum_{r = 1}^{n}f\Big(\frac{r}{n}\Big)
= \int_{0}^{1}f(x)dx. $$
Then, $$ \lim_{n \to \infty}\frac{1}{n}\sum_{r =
1}^{n}f\Big(\frac{p_r}{p_n}\Big) = \int_{0}^{1}f(x)dx. $$
My proof was based on showing that as $n \to \infty$, the ratios $p_r/p_n$ approached equidistribution in $(0,1)$ hence the integral follows as a trivial property of equidistributed sequence.
Motivation: There are several identities, limits etc on prime numbers which can be easily proven using this simple formula, including all answers to all three questions on the arithmetic, geometric and harmonic means of primes mentioned in the above link.
| Too long for a comment
Divide $[0,1]$ into $$[0,p_1/p_n],[p_1/p_n,p_2/p_n],\cdots,[p_{n-1}/p_n,1]$$
Then, using Riemann sum, we have
$$I:=\int^1_0f(x)dx=\lim_{n\to\infty}\sum^n_{k=1}f\left(\frac{p_k}{p_n}\right)\frac{p_{k+1}-p_k}{p_n}$$
If we assume that $p_j=j\ln j$,
$$I=\lim_{n\to\infty}\sum^n_{k=1}f\left(\frac{p_k}{p_n}\right)h(k,n)+\lim_{n\to\infty}\frac1n \sum^n_{k=1}f\left(\frac{p_k}{p_n}\right) \qquad{(1)}$$
where
$$h(k,n)=\frac{(k+1)\ln(k+1)-k\ln k}{n\ln n}-\frac1n$$
It can be shown that $$h(k,n)\le h(n,n)=O(\frac1{n\ln n})$$
Therefore, the absolute value of the first term in $(1)$ is upper bounded by
$$h(n,n)\cdot nM\to 0$$ where $M$ is a positive constant. This leads us to our desired result.
I am not sure if this argument can be made rigorous. I will review it when I have leisure time.
| {
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How to prove that $\text{null} \;T_1 \subset \text{null} \;T_2$ implies $\text{dim(range}\; T_1) \geq \text{dim(range}\ T_2)$?
If $W$ is finite dimensional, $T_1, T_2 \in L(V,W)$ and $\text{null}\; T_1 \subset \text{null}\; T_2$ then $$\text{dim}( \text{range}\; T_1) \geq \text{dim}(\text{range}\; T_2)$$
I'm solving a problem in Linear Algebra Done Right and I'm still struggling with a problem. And in my approach I current facing this problem. I can prove it easily when $V$ is finite dimensional but things get ugly when I attempt to prove for more general case. Any help will be appreciated.
| Since you can prove it for $V$ finite-dimensional, the easiest way out is to look at the induced linear maps on the quotient $V/\ker T_1\to W$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality for the maximum of the absolute value of two normal distributed random-variables I would like to show following statement:
For $M\geq 2,\ X_1,\dots,X_M\sim^{iid}\mathcal{N}(0,1)$ independent, it holds $P(\max_{i=1,\dots,M}\lvert X_i\rvert\geq y)\leq Me^{-y^2/2}$.
I think it is possible to show it via induction, but I have got problems at the start:
So, I want to show $P\left(\max(\lvert X_1\rvert,\lvert X_2\rvert)\geq y \right)\leq 2 e^{-y^2/2}$. One can show $P\left(\max(\lvert X_1\rvert,\lvert X_2\rvert )\geq y\right)=4\Phi(y)(1-\Phi(y))$ where $\Phi$ denotes the distribution function. I know that $1-\Phi(y)\leq e^{-y^2/2}$, so this inequality seems to be a little sharper to me. Maybe someone knows how to prove this or an alternative prooving strategy.
| Direct proof: Let $G(y)=1-4\Phi(y)(1-\Phi(y))=1+4\Phi(y)^2-4\Phi(y)$, the distribution function for $max(|X_1|,|X_2|)$. The density function $g(y)=8\Phi(y)\phi(y)-4\phi(y)$. Now $\Phi(y)=\frac{1}{2}+\frac{1}{\sqrt{2\pi}}\int_0^y e^{-\frac{u^3}{2}}du$, so $g(y)=\frac{4}{\pi}e^{-\frac{y^2}{2}}\int_0^y e^{-\frac{u^2}{2}}du$. Therefore $G(y)=\frac{4}{\pi}\int_0^y\int_0^ve^{-\frac{u^2+v^2}{2}}dudv$. Switching to polar coordinates (note), where $0\le \theta \le \frac{\pi}{4}$, we get $G(y)\gt \int_0^yre^{-\frac{r^2}{2}}dr=1-e^{-\frac{y^2}{2}}$, or $P(max|X_1|,|X_2|)\ge y)\le e^{-\frac{y^2}{2}}$
note: The domain of the double integral is a right triangle in the $(u,v)$ plane with the hypotenuse along the line $u=v$, from $(0,0)$ to $(y,y)$ and base along the $v$ axis. To get the inequality, the integral is over the sector of radius $y$ within the triangle.
| {
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How can I find the residue of this removable singularity? We have the following function : $$f(z)=\frac{z^2}{1-\cos z}$$ where $z_0=0$ is a removable singularity since the limit as $z$ goes to $0$ is $2$.
In such cases, in order to find the residue I proceed by trying to find the Laurent series around the singularity and checking the coefficient of the $z^{-1}$ term. However, the cosine is in the denominator and I can't properly find the first negative power of $z$. Will I have to resort to a polynomial division to get my negative $z$ powers?
| Mark's answer is, of course, the best to approach this. Supposing it were not a removable singularity, then what I'll show will still sometimes work. In specific, you'll get the Laurent Series.
Recall Geometric Series.
$$ \frac{x^2}{1-\cos{x}} = x^2 (1+\cos(x)+\cos(x)^2+\dots) = x^2 \sum_{n=0}^\infty \left( \sum_{k=0}^\infty (-1)^k \frac{x^{2k}}{(2k)!} \right)^n $$ That's your Laurent Series.
| {
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Expectation of product of more than two independent random variables I am trying to determine whether independence of random variables changes when multiplied with other, potentially dependent variables. The question isn't in a measure-theoretic context.
In particular, I have 3 random variables, which we can call A, B, C
A is independent of both B and C, but B and C are not independent. Is it true then that:
$$\mathbb{E}[ABC] = \mathbb{E}[A]\mathbb{E}[BC]?$$
If it's true, I would appreciate if someone could either show me a proof (preferably one for someone without a strong measure-theoretic probability background) or point me to one.
Thanks
| Let's assume you have discrete random variables so that we do not deal with integrals. First use the independence assumption:
$$
\mathbb E(ABC)=\sum_{a,b,c} abc P(A=a,B=b,C=c)=\sum_{a,b,c} abc P(A=a)P(B=b,C=c).
$$
And then use the following identity to get the result:
$$
\sum_{a,b,c} abc P(A=a)P(B=b,C=c)=\sum_{a} a P(A=a)\sum_{b,c} bc P(B=b,C=c)=\mathbb E(A)\mathbb E(BC).
$$
| {
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Conditional Probability Problem (3 Cards, 1 red-red, 1 red-black, 1 black-black) Our class got into a heated debate over a question that even my professor couldn't answer with certainty. So here's the problem:
"Of three cards, one is painted red on both sides; one is painted black on both sides; and one is painted red on one side and black on the other. A card is randomly chosen and placed on a table. If the side facing up is red, what is the probability that the other side is also red?"
Our class is split into the "1/2" and the "2/3" camp. I lean more towards 1/2. My argument is that when you uncover the red card, the black-black card is no longer considered in the sample space. So there are only 2 cards remaining, and since only one of those cards will have red on its other side, the final probability is 1/2.
My professor leans more towards 2/3. He didn't have time to formulate his argument fully, however. I've been trying to think about an alternative explanation which would lead to 2/3 as an answer, but it always leads back to the same argument I used for 1/2. Which one would be correct, and why?
|
My argument is that when you uncover the red card, the black-black card is no longer considered in the sample space.
It eliminates more than that.
The card is selected (1 from 3), and a side from that card is shown (1 from 2), without bias. So this experiment may be represented using a sample space consisting of six unbiased outcomes. Of these, witnessing a red side eliminates not just the two outcomes for the double black card, but also the outcome of "the two-colour card shows black".
$$\{R_1(R_2), R_2(R_1), R_3(B_3),\require{cancel} \cancel{B_3(R_3), B_2(B_1), B_1(B_2)}\}$$
So there are three red sides you could be looking at with equal probability. Two of them have another red side on the flipside. Thus giving the answer.
$$\dfrac 23$$
| {
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Sum of series $\frac{ 4}{10}+\frac{4\cdot 7}{10\cdot 20}+\frac{4\cdot 7 \cdot 10}{10\cdot 20 \cdot 30}+\cdots \cdots$
Sum of the series
$$\frac {4}{10}+\frac {4\cdot7}{10\cdot20}+ \frac {4\cdot7\cdot10}{10\cdot20\cdot30}+\cdots $$
Sum of series $\frac {4}{10}+\frac {4\cdot7}{10\cdot20}+ \frac {4\cdot7\cdot10}{10\cdot20\cdot30}+\cdots$
But i want to solve it using Beta ganmma function.
My Process follows
Let $$S = \frac{1}{10}\bigg[\frac{ 4}{1}+\frac{4\cdot 7}{1\cdot 2}+\frac{4\cdot 7 \cdot 10}{1\cdot 2 \cdot 3}+\cdots \cdots \bigg]$$
Let $\displaystyle a_{n} = \prod^{n}_{k=1}(3k+1)=3^n \prod^{n}_{k=1}\bigg(k+\frac{1}{3}\bigg)$
$\displaystyle =3^n\Gamma\bigg(n+\frac{4}{3}\bigg)\cdot \frac{1}{\Gamma \bigg(\frac{4}{3}\bigg)}$
And Let $\displaystyle b_{n} = \prod^{n}_{k=1}k = \Gamma (n+1)$
So $\displaystyle S =\frac{1}{10} \sum^{\infty}_{n=1}3^n \frac{\Gamma \bigg(n+\frac{4}{3}\bigg)\cdot \Gamma \bigg(\frac{2}{3}\bigg)}{\Gamma \bigg(\frac{2}{3}\bigg)\Gamma \bigg(\frac{4}{3}\bigg)\cdot \Gamma(n+1)}$
using $\displaystyle \Gamma(x)\cdot \Gamma(1-x) = \frac{\pi}{\sin (\pi x)}$ and
$\displaystyle \frac{\Gamma (a) \Gamma(b)}{\Gamma(a+b)} =B(a,b)= \int^{1}_{0}x^{a-1}(1-x)^{b-1}dx$
So $\displaystyle S = \frac{1}{10}\sum^{\infty}_{n=1}3^n\int^{1}_{0}x^{n+\frac{1}{3}}(1-x)^{-\frac{1}{3}}dx$
$\displaystyle =\frac{1}{10}\int^{1}_{0}\frac{3x}{1-3x}\cdot \bigg(\frac{x}{1-x}\bigg)^{\frac{1}{3}}dx$
Now put $\displaystyle \frac{x}{1-x} = t\Rightarrow x = \frac{t}{1+t} = 1-\frac{1}{1+t}$
So $\displaystyle S = \frac{3}{10}\int^{\infty}_{0}\frac{t^{\frac{4}{3}}}{(2t-1)(1+t)^2}dt$
Could some Help me to solve it, Thanks
although it has solved Here
| $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
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$\ds{{4 \over 10} + {4 \cdot 7 \over 10 \cdot 20} +
{4 \cdot 7 \cdot 10 \over 10 \cdot 20 \cdot 30} + \cdots =
\sum_{n = 1}^{\infty}{\prod_{k = 1}^{n}\pars{3k + 1} \over
\prod_{k = 1}^{n}10k}:\ {\LARGE ?}}$.
\begin{align}
&\bbox[10px,#ffd]{\ds{\sum_{n = 1}^{\infty}
{\prod_{k = 1}^{n}\pars{3k + 1} \over \prod_{k = 1}^{n}10k}}} =
\sum_{n = 1}^{\infty}
{3^{n}\prod_{k = 1}^{n}\pars{k + 1/3} \over 10^{n}n!} =
\sum_{n = 1}^{\infty}
{\pars{3/10}^{n} \over n!}\,\pars{4 \over 3}^{\large\overline{n}}
\\[5mm] = &\
\sum_{n = 1}^{\infty}
{\pars{3/10}^{n} \over n!}\,{\Gamma\pars{4/3 + n} \over \Gamma\pars{4/3}} =
\sum_{n = 1}^{\infty}
\pars{3 \over 10}^{n}\,{\pars{n + 1/3}! \over n!\pars{1/3}!}
\\[5mm] = &\
\sum_{n = 1}^{\infty}\pars{3 \over 10}^{n}\,{n + 1/3 \choose n} =
\sum_{n = 1}^{\infty}\pars{3 \over 10}^{n}
\bracks{{-4/3 \choose n}\pars{-1}^{n}}
\\[5mm] = &\
\sum_{n = 1}^{\infty}
{-4/3 \choose n}\pars{-\,{3 \over 10}}^{n} =
\bracks{1 + \pars{-\,{3 \over 10}}}^{-4/3} - 1
\\[5mm] = &\
\bbx{\pars{10 \over 7}^{4/3} - 1} \approx 0.6089
\end{align}
| {
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Continuous group actions and transitivity Let $G$ be a topological group and $M$ a topological space. Suppose that $$\circ: M\times G\longrightarrow M$$ is a continuous group action of $G$ on $M$. If $\circ$ is transitive, then we can take an arbitrary $m\in M$ and the set $\mathfrak{H}$ of all cosets of $C_G(m)$ in $G$ is in a natural bijection $$\varphi:\mathfrak{H}\longrightarrow M$$ with $M$. I've been said that, in this case, $M$ is even homeomorphic to $\mathfrak{H}$ with the natural topology inherited by $G$ (an open set here is such if and only if the union of the cosets is open in $G$). I can prove that $\varphi$ is continuous composing the maps $$G\longrightarrow M\times G\longrightarrow M$$ but I cannot prove that $\varphi$ is also open. Any ideas?
| This is not true unless you impose further assumptions: both spaces should be locally compact and Hausdorff and $G$ should be second countable. You can find a proof of this fact in the online notes by Paul Garret about 2-Solenoids. See also this related question (where I found the link to the previous notes when I had a similar doubt myself).
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Any rectangular shape on a calculator numpad when divided by 11 gives an integer. Why? I have come across this fact a very long time ago, but I still can't tell why is this happening.
Given the standard calculators numpad:
7 8 9
4 5 6
1 2 3
if you dial any rectangular shape, going only in right angles and each shape consisting of 4 points, then the dialed number is always divisible by 11 without a remainder.
Examples of the shapes: 1254, 3179, 2893, 8569, 2871, and so on. It is not allowed to use zero, only digits 1..9.
UPDATE: The part of the question below proved to be an error on my part because I did not double-check what the Programmers calculator was showing, and it turned out that it was rounding the results. !!! See the accepted answer for an interesting follow-up to this, which actually expands the usecase to a working hexadecimal "keypad", and the other answers for even more various interesting approaches, discovering different aspects of this problem !!!
The same rule also works even for the Programmer calculator layout on MS Windows 10 which looks like this:
A B 7 8 9
C D 4 5 6
E F 1 2 3
All valid rectangular shapes, for example, A85C, E25C, 39BF, and so on, being divided by 11 still give an integer result!
Initially I was thinking that it's just somehow tied to picking digits from the triplets and being just another peculiarity of the decimal base and number 11 and started looking this way, but discovering that it works for the hexidecimal base and even with the hex part of the keyboard layout not obeying exactly the pattern of the decimal part layout, I'm lost.
What law is this fun rule based on?
| Start at the degenerate rectangle 1111, a multiple of 11.
Each time you move a side of your rectangle by one step in one of the 4 directions (leaving the other side in place), you add or remove one of these numbers (leading zeroes added for clarity):
*
*0011 (horizontally) or 0033 (vertically)
*0110 (horz.) or 0330 (vert.)
*1100 or 3300
*1001 or 3003
which are all multiples of 11 (obviously for the first three lines; 1001 = 11 * 91).
Adding such a number preserves being a multiple of 11.
This way from the degenerate rectangle 1111 you can reach any other rectangle (aligned to the numpad orientation), which prove they're all multiple of 11.
| {
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Prove formula $\operatorname{arctanh} x = \frac12\,\ln \left(\frac{1+x}{1-x}\right)$ Problem
Prove formula $\operatorname{arctanh} x = \frac{1}{2} \ln \left(\frac{1+x}{1-x}\right)$
Attempt to solve
To start off with definition of functions $\sinh(x)$ and $\cosh(x)$
$$ \cosh(x)=\frac{e^x+e^{-x}}{2} $$
$$ \sinh(x) = \frac{e^x-e^{-x}}{2} $$
Hyperbolic tangent is defined as:
$$ \tanh(x)=\frac{\sinh(x)}{\cosh(x)}=\frac{e^x-e^{-x}}{e^{x}+e^{-x}} $$
Notation $\text{arctanh}(x)$ means area tangent which is inverse of hyperbolic tangent.
$$ \operatorname{arctanh}(x)=\tanh^{-1}(x) $$
Trying to invert the $\tanh(x)$ we get:
$$\begin{align}
\frac{e^x-e^{-x}}{e^x+e^{-x}} &= y &\implies\\
e^x-e^{-x}&=y(e^x+e^{-x}) &\implies\\
e^x-e^{-x}&=ye^{x}+ye^{-x} &\implies\\
e^x(1-y)&=e^{-x}(1+y) &\implies\\
\ln(e^x(1-y)) &= \ln(e^{-x}(1+y)) &\implies\\
\ln(e^x)+\ln(1-y) &= \ln(e^{-x})+\ln(1+y) &\implies\\
x + \ln(1-y) &= -x + \ln(1+y) &\implies\\
2x &= \ln(1+y)-\ln(1-y)&\implies\\
x &= \frac{1}{2} \ln \frac{1+y}{1-y}
\end{align}
$$
By switching variables we get:
$$ \implies \operatorname{arctanh}(y) = \frac{1}{2} \ln \left(\frac{1+y}{1-y}\right) $$
| Just to add to the responses already here. You can also show this by solving this integral two different ways, one with trigonometric substitutions and the other with partial fraction decomposition.
$$ \int \frac {dx}{x^2+1} = \int \frac {dx}{x^2+1}$$
$\qquad x \mapsto \tan u$
$$ \int\frac{dx}{(1+ix)(1-ix)}= \int \frac {\sec^2 u}{1+ \tan^2u}du$$
$$ \int \Big(\frac{\frac 12}{1+ix}+\frac{\frac 12}{1-ix}\Big)dx\ = \int du$$
$$\frac 1{2i}\big(\ln(1+ix)- \ln(1-ix)\big)=u+C$$
$$\arctan x=\frac i2\ln \Big( \frac{1-ix}{1+ix} \Big)$$
Plug in $x=ix.$
$$\arctan(ix)=\frac i2 \ln\Big(\frac{1+x}{1-x} \Big)$$
Now use the identity$ \space\operatorname {arctanh}(x)=-i\arctan(ix).$
| {
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Show that $\phi : H\to HN/N$ defined by $\phi(h)=hN$ is injective. I am trying to show the following:
Show that $\phi : H\to HN/N$ defined by $\phi(h)=hN$ is injective.
Note that $H\leq G$ and $N$ is normal in $G$.
My attempt so far: Let $\phi(h_{1})=\phi(h_{2})$. Then $h_{1}N=h_{2}N$ or there exists an $n\in N$ such that $h_{1}=h_{2}n$.
This is where I am stuck. I apologize for my elementary knowledge in cosets and quotient groups. Can anyone point me to the right direction, such as a property for cosets, to help me?
| It's not true. $\phi$ is a homomorphism and you have $h\in Ker(\phi)$ $\iff$ $hN=N$ $\iff$ $h\in H\cap N$. So the kernel is trivial only when $H\cap N=\{e\}$.
| {
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Using Theta Notation $\Theta$ how can I prove that $(13n + 3)(9n + 1)(\log(4n^2 + 100))$ is an element of $\Theta(n^2 \log n)$ I'm having a hard time figuring out what systematic approach I need to follow to solve questions like these :/ Do I expand and use the Max() principle to show that it reduces to the RHS?
Please help :(
| So you would like to show
$$f(n) = \Theta(n^2 \log n)$$
This means that we have to find $c_1,c_2,n_0$ for every $n \geq n_0$, we have
\begin{equation}
c_1 n^2 \log n \leq f(n) \leq c_2 n^2 \log n
\end{equation}
where
\begin{equation}
f(n) = (13n + 3)(9n + 1)(\log(4n^2 + 100))
\end{equation}
Notice that you can write
\begin{equation}
f(n) =(117n^2 + 40n + 3)\log(4n^2 + 100)
\end{equation}
Now it is clear that for $c_1 = 1$, we have
\begin{equation}
f(n) \geq n^2 \log n
\end{equation}
This leave us to find $c_2$ and $n_0$. Assuming the $\log$ is of base $2$ (even though it could be generalized for any base), we have
\begin{equation}
\begin{split}
f(n) &=(117n^2 + 40n + 3)\log(4n^2 + 100) \\
&\leq (117n^2 + 117n^2 + 117n^2)\log(4n^2 + 100) \\
&= 3(117n^2) \log(4n^2 + 100)\\
&\leq 3(117n^2) \log(4n^2 + 4n^2)\\
&= 351n^2 \log (8n^2)\\
&= 351n^2 \log (8) + 351n^2\log(n^2)\\
&= 3(351n^2) + 702n^2\log(n) \\
&= 1053n^2 + 702n^2\log(n) \\
&\leq 1053n^2 + 1053n^2\log(n) \\
&\leq 1053n^2\log(n) + 1053n^2\log(n) \\
&= 2016n^2\log(n) \\
&= c_2 n^2\log(n)
\end{split}
\end{equation}
So choosing $c_2 = 2016$ for $n_0 = 2$, we get that $f(n) \leq c_2 n^2\log(n)$
| {
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uniform convergence of $f_n = n \chi_{[1/n, 2/n]}$
$f_n = n \chi_{[1/n, 2/n]}$. Show that $f_n$ converge uniformly to $0$.
$f_n = n$ if $1/n \le x \le 2/n$, and $0$ otherwise.
But, I am a little bit confused here.
When $n$ goes to infinity,
$f_n \to \infty$ if $0\le x\le 0 \implies x=0$?
How can we deal with this case?
Another question is that for $\delta > 0$, $f_n$ is uniformly convergent on the complement of $[0, \delta]$. However, there does not exists a set of measure $0$ on the complement of which $f_n$ is uniformly convergent.
I think that in this case, when $n \to \infty$,
since $x=0$ is not considered, we can conclude that $f_n$ converge uniformly to $0$. But, since the measure of $[0, \delta]$ is $\delta$, such set does not exists. Am I understanding correct?
| The given sequence of functions converges pointwise and almost uniformly but NOT uniformly to $f \equiv 0$ on $\mathbb{R}$.
Notice that if $x \in \mathbb{R} \setminus (0,2]$ then $f_n(x)=0$ for every $n \in \mathbb N$.
Suppose $x \in (0,2]$. By the Archimedean Property there is a positive integer $N$ so that $Nx>2$. Hence $f_n(x)=0$ whenever $n \geq N$. Therefore, $f_n(x) \to 0$ pointwise.
Select $\hat{\varepsilon}=1$, and let $N \in \mathbb N$. Select $x=\frac{3}{2(N+1)}$. So $N+1>N$ but $|\,f_{N+1}(x)-0|=N+1\geq1$. Therefore, the sequence of functions $\{\, f_n\}_{n=1}^\infty$ does not converge uniformly to $0$.
I will leave the almost uniform convergence to the question asker. Hint: Given $\delta>0$, find $N \in \mathbb N$ so that $N\delta>2$.
| {
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Let $f(x) = 2x + 3$ and $g(x) = ax + b$. If $f(g(x)) = x$ for all $x$, determine $a$ and $b$ This is what I know:
$f(x) = 2x + 3$
$g(x) = ax+b$
$f(g(x)) = 2(ax+b) + 3 = x$
I am just not too sure what to do next to find $a$ and $b$.
| As mentioned in the comments, we have $f(g(x)) = 2ax + 2b + 3 = x$. This is only true if $2a = 1$ and $2b + 3 = 0$, otherwise the equality wouldn't hold. Solving those, we get that $a = \dfrac{1}{2}$ and $b = -\dfrac{3}{2}$.
| {
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Generating Function Sum and Combinotorics Problem:
The sum $\sum_{n=2}^{\infty} \frac{\binom n2}{4^n} ~~=~~ \frac{\binom 22}{16}+\frac{\binom 32}{64}+\frac{\binom 42}{256}+\cdots$ has a finite value. Determine that value.
I am quite stuck on how to do this. Can somebody give me <only> a hint or hints to get going?
Thanks!
| Consider
$$S=\sum_{n=2}^\infty \binom{n}{2}x^n =\frac 12 \sum_{n=2}^\infty n(n+1) x^n=\frac 12 \sum_{n=2}^\infty[n(n-1)+2n] x^n$$
$$S=\frac 12 \sum_{n=2}^\infty n(n-1) x^n+ \sum_{n=2}^\infty n x^n=\frac {x^2}2 \sum_{n=2}^\infty n(n-1) x^{n-2}+x\sum_{n=2}^\infty n x^{n-1}$$
$$S=\frac {x^2}2 \left(\sum_{n=2}^\infty x^n \right)''+x\left(\sum_{n=2}^\infty x^n \right)'$$ When finished, let $x=\frac 14$.
| {
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How to show example where convergence in $L^1$ norm does not hold for continuous functions of the random variables? Suppose a sequence of random variables $X_n$ converges to a random variable $X$ in $L^1$ norm, and that $g: \mathbb{R}\rightarrow\mathbb{R}$ is a continuous function.
It is not necessarily true that $g(X_n)$ converges to $g(X)$ in $L^1$ norm. I'm trying to figure out a counterexample that shows this.
| Let $x_n = n 1_{[0,{1 \over n^2}]}$ and $g(t) = t^2$.
Then $\|x_n-0\|_1 \to 0$ but $\|g\circ x_n - g \circ 0 \|_1 = 1$ for all $n$.
| {
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Find all the positive integers k for which $7 \times 2^k+1$ is a perfect square Find all the positive integers $k$ for which $7 \times 2^k+1$ is a perfect square.
The only value of $k$ I can find is $5$. I am not sure how to find every single one or the proof, I simply used trial and error.
| \begin{align}
a^2 &= 7*2^k + 1 \\
\Rightarrow (a^2 - 1) = (a - 1)(a+1) &= 7*2^k \\
\text{For } k > 2: (\frac{a-1 }{2}) (\frac{a+1}{2}) &= 7*2^{k-2} \\
\text{Let } d = (\frac{a-1 }{2}) &\Rightarrow d + 1 = (\frac{a+1}{2}).\\
\Rightarrow d(d+1) &= 7*2^n (n = k - 2\in \mathbb{N})\\
gcd(d,d+1) = 1 &\Rightarrow (d,d+1) = (7,2^n) \text{ or } (2^n,7) \\
\text{ Only answer possible is } n = 3 &\Leftrightarrow k = 5 \text{ if } k > 2.\\
k = 1,2 \text{ do not satisfy, check individually. }
\end{align}
| {
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Is the substitution of variables allowed in proofs? EDIT: I later realized no one saw that substituting $\sqrt{x} = x$ is not valid because they are not equal, you would have to assume $x$ as a completely different variable (i.e. $a$) and not the same one because they have no relation to each other. However, the solution to my proof problem is given.
So for example I have to prove:
$2y \leq \frac{y^2}{x^2} + x^2$ (for any real numbers $x$ and $y$ and $x$ $\neq$ $0$),
and I do some rough work at start with this and end up with the arithmetic-geometric mean inequality:
$\sqrt{xy} \le \frac{x+y}{2}$ (using substitution of variables).
Now to prove it, I start with the arithmetic-geometric mean inequality, then do the following steps:
$\sqrt{xy} \leq \frac{x+y}{2}$
$\sqrt{x}\sqrt{y} \leq \frac{x+y}{2}$
Now, substitute $\sqrt{x}$ with $x$ and so:
$xy \leq x^2+\frac{y^2}{2}$
Then throughout, I do one more substitution:
$\sqrt{x} = x$
to end up with the first (1) equation. Would this be allowed or is there another way of proving it? Such as squaring the right side of arithmetic-geometric mean inequality? I am not too sure. Thanks.
| HINT
We have that
$$2y\le \frac{y^2}{x^2}+x^2 \iff2yx^2\le y^2+x^4 \iff x^4-2yx^2+y^2\ge 0$$
| {
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Rank of Matrix Determination $X=(I+ab^T)A(I+ba^T)$;
$A$ is symmetric and positive definite matrix of $n \times n$. $I$ is the Identity matrix of $n \times n$.
$a$ and $b$ are vectors of $n \times 1$.
$a.b \neq -1$ and $a.b \neq 0$
$a$ is not parallel to $Ab$
How do we show that $X-A$ is a rank $2$ matrix ?
Efforts:
$$X-A= ab^T A + Aba^T+ ab^T A ba^T$$
Hence each of the terms are having rank $1$. so the sum of all the terms can have rank $\leq 3$
But I am not getting how it can be exactly of rank $2$..
| $(X-A)$ is spanned by two vectors only
We can solve this using projector matrices, i.e. matrices of the form
\begin{equation}
P = Z(Z^TZ)^{-1}Z^T
\end{equation}
The number of columns of $X$ will determine the rank of the matrix $X -A $. If we stack in $Z$,
\begin{equation}
Z =
\begin{bmatrix}
a & Ab
\end{bmatrix}
\end{equation}
We are sure that $Z$ is full column rank because $a$ is not parallel to $Ab$, Hence it is easy to see that
\begin{equation}
Pa = a \tag{1}
\end{equation}
and
\begin{equation}
PAb = Ab \tag{2}
\end{equation}
Hence
$$P(X-A) = P(ab^T A + Aba^T+ ab^T A ba^T)$$
which is
$$P(X-A) = Pab^T A + PAba^T+ Pab^T A ba^T$$
Using equations $(1,2)$, we get
$$P(X-A) = ab^T A + Aba^T+ ab^T A ba^T = X-A$$
Hence $X-A$ is spanned by two vectors, i.e. rank $2$.
Therefore $P$ spans a two dimensional space, which is the span of the column space of $X -A$. On the other hand, we can also show that
\begin{equation}
(I-P).(X-A) = 0
\end{equation}
where $I-P$ is of rank $n-2$.
| {
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Union of an increasing sequence on connected sets Let $(A_i)$ be a family of strictly increasing connected sets, then I have to prove that $A=\bigcup_{i\in I}A_i$ is connected.
By contradiction, I assume that $A$ is not connected. Then there exist two open, non empty and disjoint sets $U$ and $V$ such that $ A= U\cup V,$
as $A_i$ is connected, we have $$\forall i\in I, [A_i\subset U~\text{or}~A_i\subset V]$$
How to prove that
$$\forall i\in I, A_i\subset U~\text{or}~\forall i\in I, A_i\subset V$$
to get a contradiction?
Thank you.
| @Ashish's comment I think makes it easier to prove. If you have $$ \forall i \in I, [A_i \subset U \text{ or } A_i \subset V]$$ and we know that $\exists i , A_i \subset U$ and $\exists j, A_j \subset V$ (because otherwise $U$ or $V$ would be empty), then your sets can't be increasing $A_i \not\subset A_j$ and $A_j \not\subset A_i$.
P.S. I assume that by increasing you mean that $\forall i, A_i \subset A_{i+1}$.
| {
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Trouble understanding how the Transfer Principle is applied for the Extreme Value theorem. I am reading Keisler's Elementary Calculus (which can be downloaded here). I am having trouble understanding his proof sketch of Extreme Value Theorem and how he is applying the Transfer Principle.
For reference, he defines the "Transfer Principle" as:
Every real statement that holds for one or more particular functions holds for the hyperreal natural extension of these functions.
On page 164 (using left corner numbering) of the book he provides the following "sketch":
I understand the counter examples and I am able to understand the issues with them using standard tools. I don't understand, however, how one can immediately utilize the Transfer Principle. It is not immediately obvious to me that "there is a partition point $a + K\delta$ at which $f(a + K\delta)$ has the largest value."
To elaborate, the proof seems circular. In trying to to "expand" the sketch to be more precise. I ended up writing instead of:
By the Transfer Principle, there is a partition point $a + K\delta$ at which $f(a + K\delta)$ has the largest value.
To:
Applying the Transfer Principle to the Extreme Value Theorem we see that the Extreme Value holds for hyperreals as well. Hence, there is a partition point $a + K\delta$ at which $f(a + K\delta)$ has the largest value.
But this relies on a proof of the Extreme Value Theorem for reals.
Hopefully what I am saying makes sense, please ask for any clarification.
| Let me propose a different interpretation of transfer (in the formulation suggested by m_t_) for the Intermediate Value Theorem.
Consider the following "standard" argument about ordinary real numbers. For all $n \in \mathbb{N}$, it is possible to partition $[a,b]$ into $a, a+\frac{b-a}{n}, \ldots, a+n\frac{b-a}{n}=b$.
Since there is a finite number of partition points (they are $n+1$), it is well-defined the maximum among the values $f(a), f\left(a+\frac{b-a}{n}\right), \ldots, f(b)$.
Recall that, by transfer, finite and $^\ast$finite sets satisfy the same properties. As a consequence, you can always pick the biggest element of a $^\ast$finite set.
Hence, transfer entails also that for all $n \in\ \! ^\ast\mathbb{N}$ the number
$$\max\left\{ f(a), f\left(a+\frac{b-a}{n}\right) , \ldots, f\left(a+n\frac{b-a}{n}\right)=f(b) \right\}$$
is well-defined for every partition of $^\ast[a,b]$ into $n$ equal parts.
In other words, you are proving the Extreme Value Theorem by applying transfer to the statement "every finite set has a maximum element", and not to the Extreme Value Theorem for real numbers.
Let now $n \in\ \! ^\ast\mathbb{N}$ be infinite, and let $$f\left(a+K\frac{b-a}{n}\right) = \max\left\{ f(a), f\left(a+\frac{b-a}{n}\right) , \ldots, f\left(a+n\frac{b-a}{n}\right)=f(b) \right\}.$$ (What I call $\frac{b-a}{n}$ is the infinitesimal $\delta$ in the original proof by Keisler).
Define $c=\ \!^\circ\left(a+K\frac{b-a}{n}\right)$. By continuity of $f$ you have that $^\circ\left(^\ast f\left(a+K\frac{b-a}{n}\right) \right) = f(c)$. From here onward, it is possible to follow the original proof by Keisler.
| {
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Introductory recursion Proof. Suppose $f(n)$ is the number of strings of length $n$ with symbols from the
alphabet $\{a,b,c,d\}$ with an even number of $a$’s.
(b) Show that $f$ satisfies the recurrence
$f(n+1)=2f(n)+4^n$
My approach was to first to assume the case holds for $n=k$ and then consider the case for $n=k+1$(It is easy to show that it holds for the base case). I first considered the case for where strings have $a$ NOT as the last element. Then the strings with the previous sequence of elements (same strings from strings with length $k+1$ but without the last element ) have even $a$'s which means that there are $f(k)$ strings with length $k$. The last element can be $b,c,d$, so there are $3f(k)$ strings with length $k+1$ with $a$ NOT as the last element.
Now, the problem is that I am not sure if this method works for the case where the last element IS an a. also, are there any other ways to prove this? It seems to me that my approach lacks rigor.
| Your approach is quite right, and I believe its level of rigor fits the question.
As you correctly said, considering all possible strings of length $n+1$ with that property yields a partition in two sets: chopping off the last letter, we can either end up with a string of length $n$ which already has an even number of $a$, or we do not and the string has an odd number of $a$ - so we must have chopped off an $a$. So the question boils down to whether or not the string ends with an $a$.
The number of strings of the first kind, as you noted, is $3f(n)$ since the last letter can be $b, c, d$. The number of strings of the second kind is the exact same as the number of strings of length $n$ that do not contain an even number of $a$, since we know we chopped off one. And we can find this number by considering all possible strings of length $n$ and then subtracting the number of those that do have an even number of $a$: $4^n - f(n)$. This works because a string can either contain an even or an odd number of $a$, no in-between.
So putting it all together yields:
$$f(n+1)= 3f(n) + (4^n - f(n)) = 2f(n) + 4^n$$
| {
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Differentiating the sum of increasing nonnegative functions Let $f_n:[0,1] \to \mathbb R_{+}$ be a sequence of nondecreasing functions such that $f_n(0) = 0$ and $\sum_{n=1}^{\infty} f_n(1) \leq \infty$. Show that the sum can be differentiated term by term almost everywhere, i.e. $f' = \sum^{\infty}_{n=1} f'$.
It appears that the set up of the problem is hinting Dominated Convergence Theorem. So I want to view the summation as the integration against the counting measure. In order for the theorem to apply, I need a bound on the derivative of $f_n$. Is it possible to derive a bound from the information given?
| I have myself a different solution to the same problem.
Since each $F_n$ is increasing, $F$ is increasing (on $[0, 1]$). Furthermore, this means $F'$ exists on $(0, 1)$. Since we're only asked for a derivative formula to hold a.e. I'm going to discard the endpoints $0$ and $1$ now to avoid technicalities with one-sided derivatives/limits.
Now, fix $x \in (0, 1)$. By our given formula for $F$, we have the following formula for the symmetric difference quotient where $h > 0$ is sufficiently small so that $x + h, x - h \in (0, 1)$. Such $h$ exists since $(0, 1)$ is an open set. Then
$$\frac{F\left(x + \frac{h}{2}\right) - F\left(x - \frac{h}{2}\right)}{h} = \int_{\mathbb{N}}\frac{F_n\left(x + \frac{h}{2}\right) - F_n\left(x - \frac{h}{2}\right)}{h}d\mu(n),$$
where $d\mu(n)$ denotes the counting measure on $\mathbb{N}$. Now, by applying the Fundamental Theorem of Calculus on the right-hand side,
$$\frac{F\left(x + \frac{h}{2}\right) - F\left(x - \frac{h}{2}\right)}{h} = \int_{\mathbb{N}}\int^{x + \frac{h}{2}}_{x - \frac{h}{2}}\frac{F_n'(y)}{h}dyd\mu(n)$$
Since the functions $F_n$ are all increasing, the integrand on the right-hand side is nonnegative, and measurable with respect to the product measure (which I'll denote $d(y \times n)$). Then by the Fubini Theorem,
$$\frac{F\left(x + \frac{h}{2}\right) - F\left(x - \frac{h}{2}\right)}{h} = \int^{x + \frac{h}{2}}_{x - \frac{h}{2}}\frac{1}{h}\int_{\mathbb{N}}F_n'(y)dyd\mu(n)$$
The limit on the left-hand side as $h \rightarrow 0$ exists since $F$ is differentiable at $x$, so
$$F'(x) = \lim_{h \rightarrow 0}\int^{x + \frac{h}{2}}_{x - \frac{h}{2}}\frac{1}{h}\int_{\mathbb{N}}F_n'(y)dyd\mu(n)$$
The limit on the right-hand side actually exists as well. The inner integral is independent of $h$, so we only concern with the outer integral. The length of the interval is $h$, and we are dividing by $h$, so this is a mean-value integral in one dimension. Thus, as $h \rightarrow 0$, the integrand approaches the value at the center of the interval. Thus
$$F'(x) = \int_{\mathbb{N}}F_n'(x)d\mu(n),$$
completing the proof. $\square$
| {
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Denseness of a preimage. Continous and surjective function Let $F:\mathbb{R^2} \to \mathbb{R}$ surjective and continuous on $\mathbb{R^2} $.
Let $A\subset \mathbb{R}$, where $A$ is a dense set on $\mathbb{R}$.
Is $F^{-1}(A)$ a dense set on $\mathbb{R^2}$ ?
| Let $F(x,y)=xy$ if $x>0$ and $F(x,y)=0$ if $x \leq 0$. Let $A$ be any dense set in $\mathbb R$ which does not contain $0$, say non-zero rationals. Then $F^{-1}(A)$ does not intersect $\{(x,y): x \leq 0\}$ so it is not dense.
| {
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Prove that if in a tetrahedron if two pairs of opposite edges are perpendicular then the third pair is also perpendicular.
Prove that if in a tetrahedron if two pairs of opposite edges are
perpendicular then the third pair is also perpendicular.
Method:
$\vec a+ \vec b + \vec c + \vec d +\vec e + \vec f = 0 \tag0$ where the vectors represent the sides.
also, $(\vec a + \vec b). (\vec c+\vec d)= 0 \tag{1}$
$(\vec e+\vec f) .(\vec c+\vec d)= 0 \tag{2}$
From 1 and 2,
$\vec e + \vec f = k(\vec a+ \vec b)$
Substituting in $0$,
$\vec a + \vec b = -\dfrac{\vec c + \vec d}{k+1}$
Now, $(\vec e+ \vec f). (\vec a+\vec b) = -\dfrac 1{k+1}((\vec a + \vec b). (\vec c+\vec d))= 0$
Hence, the third pair too is perpendicular.
QED.
Is my method to solve it correct?
| Let the tetrahedron $T$ be spanned from $0$ by the three vectors $a$, $b$, $c$. The three equations
$$\eqalign{v+w&=a\cr u\!\qquad+w&=b\cr u+v\qquad &=c\cr}\tag{1}$$
determine three linearly independent vectors $u$, $v$, $w$. From $(1)$ it then follows that the vertices of $T$ are vertices of the parallelepiped $P$ spanned by $u$, $v$, $w$, and opposite edges of $T$ are diagonals of opposite faces of $P$. Two such edges of $T$ are orthogonal iff the corresponding faces of $P$ are rhombi. If we have two pairs of orthogonal opposite edges of $T$ we can therefore conclude that $|u|=|v|=|w|$, which then implies that the third pair of opposite edges of $T$ is orthogonal as well.
| {
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connected set of sum of upper semi continuous function Let $C(X)$:space of continuous functions on a compact space.
Consider $f$ and $g :C(X)\rightarrow \mathbb{R}$ are upper semi continuous.
suppose for every $T\in C(X)$ set of $f(T)$ and set $(f+g)(T)$ are closed interval(connected set).Can we say set of $g(T)$ is closed interval(connected set),as well?
If Not,under which condition we have it.
| The answer is no. First ask this question for functions from $\mathbb{R}$ to $\mathbb{R}$. There are easy counterexamples: take $f(t) = t$ and let $g$ be the characteristic function of the interval $(-\infty,0)$. Then the ranges of $f$ and $f + g$ are both $\mathbb{R}$ but the range of $g$ is not connected.
You can turn that into a counterexample on $C(X)$ by fixing a point $x \in X$ and composing the $f$ and $g$ described above with the evaluation functional $\phi \mapsto \phi(x)$ from $C(X)$ to $\mathbb{R}$ (or, really, any nonzero linear functional on $C(X)$).
I can't think of any good condition that would guarantee this. Just knowing that the range of a function is connected tells you very little about the function.
| {
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Find $n$ elements of $\mathbb{R_{\mathbb{Q}(\sqrt[3]{2})}}$ linearly independent We know that $\mathbb{R}$ is a linear space over $\mathbb{Q}$, denoted by $\mathbb{R_{\mathbb{Q}}}$,where the vector addition is the real numbers and scalar multiplication is the number in $\mathbb{Q}$ times the number in $\mathbb{R}$.
Then $ \forall n\in\mathbb{N^{+}}$,$\{1,\sqrt[n]{2},\sqrt[n]{2^2},\cdots,\sqrt[n]{2^{n-1}}\}$ is a linearly independent set in $\mathbb{R_{\mathbb{Q}}}.$
Further,we can get $\mathbb{R_{\mathbb{Q}}}$ is an infinite dimensional linear space.
Let $\mathbb{Q}(\sqrt[3]{2}):=\{a+b\sqrt[3]{2}+c\sqrt[3]{2^2} \Big|a,b,c\in \mathbb{Q}\}.$
Similarly，$\mathbb{R}$ is a linear space over $\mathbb{Q}(\sqrt[3]{2})$, denoted by $\mathbb{R_{\mathbb{Q}(\sqrt[3]{2})}}$, where the vector addition is the real numbers and scalar multiplication is the number in $\mathbb{Q}(\sqrt[3]{2})$ times the number in $\mathbb{R}$. We can proof that $\mathbb{R_{\mathbb{Q}(\sqrt[3]{2})}}$ also is an infinite dimensional linear space. In fact, If we suppose $\dim(\mathbb{R_{\mathbb{Q}(\sqrt[3]{2})}})\in \mathbb{N^{+}},$ then $\dim(\mathbb{R_{\mathbb{Q}}})\in\mathbb{N^{+}}$ holds. But forall $n\in\mathbb{N^{+}},$ how can I find $n$ elements of $\mathbb{R_{\mathbb{Q}(\sqrt[3]{2})}}$ such that these are linearly independent in $\mathbb{R_{\mathbb{Q}(\sqrt[3]{2})}}$?
| Here’s a way of looking at your problem that makes use of ramification theory. It may be too advanced for your taste, but at least it does provide an answer to your question.
You’re looking for many real quantities, preferably algebraic, that will be linearly independent over the real field $\Bbb Q\left(2^{1/3}\right)$. As an extension of $\Bbb Q$, this field is ramified over only $2$ and $3$. So if you get lots of real fields that are unramified over $2$ and $3$, you have a solution to your problem.
A quadratic extension $\Bbb Q(\sqrt n\,)$ of $\Bbb Q$ (with $n$ squarefree) is unramified above $2$ if $n\equiv1\pmod4$. And unramified above $3$ if $n$ is prime to $3$. So take positive values of $n$ satisfying these conditions, such as $5,13,17,41,77,\cdots$ As you see with $77$, they don’t need to be prime.
Anyway, adjoin the square roots of $m$ of these numbers, and you should get a field of dimension $2^m$ over $\Bbb Q\left(2^{1/3}\right)$.
| {
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If a fair die is rolled 3 times, what are the odds of getting an even number on each of the first 2 rolls, and an odd number on the third roll? If a fair die is rolled 3 times, what are the odds of getting an even number on each of the first 2 rolls, and an odd number on the third roll?
I think the permutations formula is needed i.e. $n!/(n-r)!$ because order matters but I'm not sure if n is 3 or 6 and what would r be?
Any help would be much appreciated!
| Assuming the die is just marked even and odd rather than with numbers, there are eight orderings. They range from all odd to all even: OOO, OOE, OEO, EOO, OEE, EOE, EEO, EEE. In your formula:
$$\frac{n!}{(n-r)!}$$
You are missing that you don't care about the orders of the duplicates. So you need
$$\frac{n!}{(n-r)!r!}$$
The $n$ is the total number of rolls, the $r$ is the number of either evens or odds (it's symmetric). But to get the total number of orderings, you need to add these:
$$\sum\limits_{r=0}^n\frac{n!}{(n-r)!r!}$$
Now substitute 3 for $n$.
$$\sum\limits_{r=0}^3\frac{3!}{(3-r)!r!}$$
Unrolling that, we get
$$\frac{3!}{3!0!} + \frac{3!}{2!1!} + \frac{3!}{1!2!} + \frac{3!}{0!3!}$$
or
$$1 + 3 + 3 + 1$$
So we have one all odds, three with one even, three with two evens, and one all even. That's eight total.
Another way of thinking of this is that there is only one ordering of all odd numbers or all even numbers while there are three places where the lone odd or even number can be.
Of those eight, how many fit your parameters? Exactly one, OOE. So one in eight or $\frac{1}{8}$.
As others have already noted, you could get that much more easily by simply figuring that you have a one in two chance of getting the result you need for each roll. There's three rolls, so $(\frac{1}{2})^3 = \frac{1}{8}$.
If you want to treat 1, 3, and 5 as different values and 2, 4, and 6 as different values, you can. But it is much easier to think of them as just odd or even. Because you don't want to try write this out for $6^3 = 216$ orderings. And in the end, you will get the same basic result. You will have twenty-seven OOE orderings, which is again one eighth of the total. This is because there are three possible values for each, 1, 3, and 5 for the two odds and 2, 4, and 6 for the even. And $\frac{27}{216} = \frac{1}{8}$.
Permutations leads you down a harder path. It's easier to think just in terms of probability or even ordering.
| {
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"url": "https://math.stackexchange.com/questions/2925298",
"timestamp": "2023-03-29T00:00:00",
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Inheriting complex structure from a covering space (Griffiths and Harris) On page 16 of Griffiths and Harris' Principles of algebraic geometry, they write
In general, if $\pi: M\to N$ is a topological covering space and $N$ is a complex manifold, then $\pi$ gives $M$ the structure of a complex manifold as well; if $M$ is a complex manifold and the deck transformations of $M$ are holomorphic, then $N$ inherits the structure of a complex manifold from $M$.
My question is for the last statement does one need more assumptions on the covering $\pi: M\to N$? Or could anyone please explain to me how to construct the complex structure on $N$?
| Let $X$ be a compact connected Riemann surface such that $\chi(X)$ is negative and divisible by $3$. Then $X$ admits an irregular topological covering $\pi: X\to Y$, where $Y$ is another Riemann surface. On the other hand, by the dimension count, a generic Riemann surface of hyperbolic type does not admit a nontrivial holomorphic covering over any surface. This is a counter example to the claim.
Edit. Regarding the dimension count. Let $X, Y$ be two compact Riemann surfaces of genera $g_X, g_Y$. Then every topolical covering map $p: X\to Y$ induces a holomorphic immersion $p^*: {\mathcal M}_{g_Y}\to {\mathcal M}_{g_X}$, where ${\mathcal M}$ denotes the moduli space. Here, $p^*$ sends a Riemann surface $Y\in {\mathcal M}_{g_Y}$ to the Riemann surface $X$ equipped with the pull-back of the complex structure on $Y$ via the covering $p$.
The (complex) dimensions of the respective moduli spaces are $3g_X -3 > 3g_Y-3$. Since there are only finitely many inequivalent covering maps $X\to Y$, it follows that a generic Riemann surface in ${\mathcal M}_{g_X}$ does not cover holomorphically any other Riemann surface.
A similar argument works for surfaces of finite type.
Next, it is a nice exercise in algebraic topology to prove the following: Every compact connected oriented $X$ surface of Euler characteristic which is not a power of $2$, and hence, divisible by some prime number $q\ge 3$, admits a nontrivial degree $q$ topological covering $p: X\to Y$ whose group of covering transformations is trivial.
Now, if you equip $X$ with a randomly chosen complex structure, then holomorphicity of the group of covering transformations of such $p: X\to Y$ is an empty condition, while $X$ does not cover any surface of smaller genus holomorphically. Hence, the claim made in Griffiths and Harris is false. They simply forgot to require the covering to be regular.
| {
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"timestamp": "2023-03-29T00:00:00",
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A Probability Problem About Seating Arrangements I'm having trouble with a probability question. It goes, "a class room has three rows of 4 seats. These 12 seats are randomly assigned to 4 male students and 5 female students. a)What is the probability that no female student sits in the front row?"
What I've tried to do is divide 8 choose 5 by 12 choose 9. But that isn't the answer. Apparently the answer is 7/99 and I've got no idea how the book got there.
Can someone please explain to me how to do it? Can you not only show me the solution, but also explain to me, how you got there? Thanks.
| The correct answer is:$$\frac{\binom85\binom40}{\binom{12}5}=\frac{\binom85}{\binom{12}5}$$
You only have to focus on females and this is the probability that $5$ chairs are selected for them that do not belong to the front row.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2925479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Lagrange inversion formula example unclear The following example is from De Bruijn's Asymptotic methods in analysis (page 24). The considered equation is
$x^t = e^{-x}$
The author wants to transform the equation into the form: $w=z/f(z)$, in order to use the Lagrange inversion formula. So he sets $x=1+z$ and $t^{-1}=w$. And then obtains the equation for $f(z)$ as: $f(z)=-z(1+z)/(\log{(1+z}))$. And then says:
The function $f(z)$ is analytic at $z=0$: $f(z)=-1+c_1z+...$.
It follows that $x=1-t^{-1}-c_1t^{-2}+...$
My question is: This function $f(z)$ at $z=0$ will be $0/0$. i.e. undefined! So how come he says the function is analytic at $z=0$?
| It's a removable singularity; we commonly identify functions with removable singularities with the function which agrees with them on the original domain and with the singularities removed. It's a removable singularity because the numerator and denominator are analytic in a neighborhood of $z=0$ and both scale as $O(z)$ as $z \to 0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Difference between $\operatorname{Var}(Y)$ and $\operatorname{Var}(Y\mid X)$? What is the difference between $\mathrm{var}(Y)$ and $\mathrm{var}(Y\mid X)$? If $Y = c + \beta X$ and $\operatorname{var}(X)=\sigma^2$, won't both come out to be the same, i.e., $\beta^2\sigma^2$?
| Note that we always have $E[f(X)|X] = f(X)$. Loosely speaking, given $X$ there is no randomness in $f(X)$, so we expect the conditional variance to be zero.
Let $f(x) = c+ \beta x$.
\begin{eqnarray}
\operatorname{var} (f(X)|X) &=& E [ (f(X)-E[f(X)|X])^2 | X] \\
&=& E [ (f(X)-f(X))^2 | X] \\
&=& E[0 | X] \\
&=& 0
\end{eqnarray}
On the other hand (note that in this case we have $E[f(X)] = f(EX)$).
\begin{eqnarray}
\operatorname{var} (f(X)) &=& E [ (f(X)-E[f(X)])^2 ] \\
&=& E [ (f(X)-f(EX))^2] \\
&=& E[(\beta(X-EX))^2] \\
&=& \beta ^2 E[(X-EX)^2] \\
&=& \beta^2 \operatorname{var} X \\
&=& \beta^2 \sigma^2
\end{eqnarray}
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the minimum or maximum of a bivariate function when $f_{xx}\times f_{yy}-f_{xy}^2=0$. I see that when $f_{xx}\times f_{yy}-f_{xy}^2<0$ then it is a saddle point.
Also when $f_{xx}\times f_{yy}-f_{xy}^2>0$ then it is a minima or maxima.
What is exactly happening when $f_{xx}\times f_{yy}-f_{xy}^2=0$?
| For a regular $f(x,y)$ we have around a point $p_0 = (x_0,y_0)$ with $p = (x,y)$
$$
f(x,y) = f(x_0,y_0) + f_x(p_0)(x-x_0)+f_y(p_0)(y-y_0) + \frac 12(p-p_0)^{\top}J(p_0)(p-p_0)+O(|p-p_0|^3)
$$
If at $p_0$ we have a relative minimum/maximum then $f_x(p_0) = f_y(p_0) = 0$ so the characterization is done with the help of
$$
J(p_0) = \left(\begin{array}{cc}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{array}\right)_{p_0}
$$
if $\det\left(J(p_0)\right) = 0$ then we have that the quadratic associated form is the product of two lines as for example $(x+y)^2$ characterizing a parabolic point. If $J(p_0) = 0$ the qualification should be done with the first non null high order differential included in $O(|p-p_0|^3)$
| {
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"timestamp": "2023-03-29T00:00:00",
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