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Cesaro limit of analytic functions Let $f_n$ be a uniformly bounded sequence of analytic functions on $\Omega\subset\mathbb C$.
If $f_n(z)\to f(z)$ forall $z\in\Omega$, then by the Montel's theorem I know that the convergence is uniform on compact sets, hence $f$ is an analytic function.
Assume instead only that
$$ \frac{1}{n}\sum_{k=1}^nf_k(z) \to f(z) $$
for all $z\in\Omega$. May I say that $f$ is analytic?
|
Yes, The sequence $T_n (z) =\frac{1}{n} \sum_{j=1}^n f_j (z)$ is also uniformly bounded sequence of analytic functions.
|
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|
Help understanding Weyl's proof of Heisenberg's Inequality https://www.math.unl.edu/~scohn1/8423/heisenberg.pdf
Can someone help me understand how they go from line to line in this proof? I'm confused by it. The proof is on the last page. Specifically how do they go from $$\left(\int \left\vert xf^*(x)f'(x)\right\vert\, dx\right)^2 \geq \frac14 \left(\int x\frac{d}{dx}\left\vert f(x)\right\vert^2\,dx\right)^2 = \frac14\left(\int \left\vert f(x)\right\vert^2 \, dx\right)^2$$ Also why were they able make the following substitution? $$\int \left\vert\widehat{f'}(k)\right\vert^2\, dk = \int\left\vert f'(x)\right\vert^2\,dk$$
|
For the first question, note that $\frac{1}{2}\frac{d}{dx}|f|^2 = ff'$. For the second question, use the Plancherel Theorem.
|
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|
is there a relation between the fact that the derivative of x^2 is 2x and that the difference between 1,4,9,16, ... is 3, 5, 7, 9, ...? Is there a relation between the fact that the derivative of x^2 is 2x and that the difference between 1,4,9,16, ... is 3, 5, 7, 9, ...?
And why is the difference always 2?
I think there is a relationship, but I can't get how and why...
|
Yes. The derivative of $x^2$ tells you how fast the function $x^2$ increases. The second derivative of $x^2$ tells you how fast the derivative of $x^2$ increases. The second derivative is 2, meaning that the speed at which the rate of growth of $x^2$ increases is fixed. Therefore, the difference of the difference of consecutive squares is constant.
For your second question, let $d_n = (n+1)^2 - n^2 = 2n+1$. Then
$$
d_{n+1}-d_n = 2(n+1)+1 - (2n+1) = 2n+2+1-2n-1 = 2.
$$
|
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|
Power series expansion of $x\ln(\sqrt{4+x^2}-x)$
Find $a_n $ where $x \ln(\sqrt{4+x^2}-x) =\sum_{n=0}^{\infty} a_nx^n$.
I know that I must find power series expansion of $\ln(\sqrt{4+x^2})$ but it doesn't help. Can anyone give me a hint? many thanks
|
The factor $x$ can be momentarily disregarded; consider $f(x)=\ln(\sqrt{4+x^2}-x)$ and note that
$$
f'(x)=\frac{\dfrac{x}{\sqrt{4+x^2}}-1}{\sqrt{4+x^2}-x}=-\frac{1}{\sqrt{4+x^2}}=-\frac{1}{2}(1+(x/2)^2)^{-1/2}
$$
The Taylor development of the derivative can be written down. Integrate and multiply by $x$.
$$
f'(x)=-\frac{1}{2}\sum_{n\ge0}\binom{-1/2}{n}\frac{x^{2n}}{2^{2n}}
$$
So
$$
f(x)=\ln2-\frac{1}{2}\sum_{n\ge0}\binom{-1/2}{n}\frac{x^{2n+1}}{2^{2n}(2n+1)}
$$
I leave to you determining $a_n$.
|
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|
Mean of overcooking time This question came up this week when I had to put my rice in the microwave for a third time.
Suppose the perfect cooking time for a meal is given by a random variable $X$ with values in seconds. Now suppose a quick check allows to determine if the food is :
*
*Uncooked,
*Perfectly cooked,
*Overcooked.
What is the estimated overcooking time in seconds if one uses the following technique :
Start by cooking for $T$ seconds, then
a) Check food state.
b) If food if perfectly cooked or overcooked, stop.
c) If food is uncooked, double the last $T$ used.
An answer could also hint for a better technique or optimize the choice of $T$.
EDIT : As suggested bellow, let us assume that $X\sim N(\mu;\sigma^2)$.
|
This is not a complete answer, but it is too long to fit in the comments. (No reference here to a similar sentence by a chap named Fermat.)
Let $\{ t_n \}_{n=1}^N$ be the sequence of times at which you plan to stop and check, where $N$ may be finite or infinite. In your question, for instance, this is defined recursively as $a_1 = T$ and $a_n = a_{n-1} + T$.
(Incidentally, for well-behaved distributions, I'd expect that it is better to have $a_n - a_{n-1}$ decreasing, because the longer you have cooked, the higher the risk of overcooking.)
Let $X$ the "perfect time". If $X \le a_1$, you stop cooking at $a_1$: this occurs with probability $P(X \le a_1)$ and gives you an expected overcooking time $E(a_1-X|X\le a_1)$. If If $a_{n-1} < X \le a_n$, you stop cooking at $a_n$: this occurs with probability $P(a_{n-1} < X \le a_n)$ and gives you an expected overcooking time $E(a_n-X|a_{n-1}<X \le a_n)$.
Define $a_0=0$ for convenience. Adding up across all cases, the formula for the expected overcooking is
$$\sum_{n=1}^N P(a_{n-1} < X \le a_n) E(a_n-X|a_{n-1}<X \le a_n)$$
For instance, if $X$ is uniformly distributed between 0 and 100 and we take $T=1$ in your checking schedule and thus $N=100$, we find
$$\sum_{n=1}^N P(a_{n-1} < X \le a_n) E(a_n-X|a_{n-1}<X \le a_n) = \\ \sum_{n=1}^{100} \frac{\left(n - (n-1) \right)}{100} E(n-X|n-1<X \le n) = \\ \sum_{n=1}^{100} \frac{1}{100} E(n-X|n-1<X \le n) = \\ \sum_{n=1}^{100} \frac{1}{100} \frac{1}{2} = \frac{1}{2}$$
|
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Parametrization of a cylinder that is parallel to x axis The answer is no it does not matter.
The surface is $y^2+z^2=4$, I parametrized it so:
$\mathbf r=x \mathbf i +2\cos\theta \mathbf j + 2\sin\theta \mathbf k$
But Pauls Outline works through the problem with the j and k components switched. Does it matter which way I go about parametrizing this surface?
|
It really doesn't matter. Thinking of a similar problem in two dimensions, both
$$\cos\theta\,\mathbf{i} + \sin\theta\,\mathbf{j}\text{ and }
\sin\theta\,\mathbf{i} + \cos\theta\,\mathbf{j}$$
parametrize the unit circle, but in different ways.
|
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Indefinite Integral $\int\frac{ \sin^3 2x-x \cos^3 x}{\cos^2 x}~dx $ How can I calculate $\displaystyle \int\frac{ \sin^3 2x-x \cos^3 x}{\cos^2 x} dx$? Help me I'm quite blur and how should I start. A little show on working are gladly appreciated.
|
Hint:
Expand and arrive at (which should prove easier)
$$ \int 8 \sin^3(x) \cos(x)~dx-\int x \cos(x) ~dx$$
|
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|
Normed Vector Space Let $V$ be a real normed vector space.
Suppose that $A$ is an open set from $V$. Show that the set $\frac12 A = \left\{ \frac12 x \, : \, x \in A \right\}$ is also open.
Let $V$ be a complex vector space. A norm on $V$ is a function $|| \cdot || : V \to R$ that satisfies the following three conditions:
*
*$||v|| \geq 0$, $\forall v \in V$, and $||v|| = 0 \Leftrightarrow v = 0$
*$||αv|| = |α| \cdot ||v||$, $\forall v \in V, α \in C$
*$||v + w|| \leq ||v|| + ||w||$, $\forall v, w \in V$
I'm not sure how to approach this problem, any pointers are appreciated!
|
Let $x \in \dfrac12A$, then $2x \in A$. Since $A$ is open, $\exists\epsilon>0$ such that the open neighbourhood $B(2x,\epsilon)\subseteq A \iff B(x,\dfrac\epsilon 2) \subseteq \dfrac12A$, so $\dfrac12A$ is open, where the open ball $B(x,a)$ denotes the set $\{y \in V:\lVert y-x\rVert<a\}$ for any $a\in \Bbb R$
|
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|
Sketching $y=a^x-a^{2x}, a>0$ When I need to sketch: $y=a^x-a^{2x}, a>0$, do I need to graph two functions?
One when $0<a<1$ and the other when $a>1$?
I didn't get any difference in the ascending descending intervals.
Because $y'=lna(a^x-2a^{2x})$ so y'=0
when $x=log_a0.5$ and $y''(log_a0.5)<0$
So it has a max point.
Thank you.
|
Yes, the behavior of the function differs in the cases $0<a<1$ and $a>1$. To see why, consider the function
$$
f_a(x)=a^x-a^{2x}
$$
and note that
$$
f_{1/a}(x)=f_a(-x)
$$
For instance, the graph you get for $a=1/2$ is symmetric with respect to the $y$-axis to the graph for $a=2$.
However, after noting this, you can avoid separating the study in the two cases, because you can appeal to this symmetry.
For instance, since
$$
\lim_{x\to\infty}f_a(x)=-\infty
$$
for $a>1$, you already know that
$$
\lim_{x\to-\infty}f_a(x)=-\infty
$$
for $0<a<1$.
Here's a picture for $a=3$:
and the corresponding picture for $a=1/3$:
As you correctly point out, the maximum of $f_a$ is at $-\log_a2$, which is positive for $0<a<1$ and negative for $a>1$. The symmetry above explains it.
|
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maximum Difference between two zeros What is the maximum difference between the two consecutive zeros of the solutions of $y''+(1+x)y=0$ on $0\leq x<+\infty$? I have applied the Strum's comparison theorem (by comparison with $y''+y=0$), and I obtained this maximum as $2\pi$ but I think it may be $\pi$. Is it necessarily true? (According to the Strum comparison theorem we know that the solution of the above equation have many infinite zeros on $R$).
|
Equation $$y''+y=0$$
has a general solution
$$y=a\cos t + b\sin t$$
Which obtain zero at
$$t=\mathrm{atan2}\left(\pm\frac{a}{\sqrt{a^2+b^2}},\mp\frac{b}{\sqrt{a^2+b^2}}\right)=\begin{cases}
\arctan\frac{b}{\pm a}&\pm a>0\\
-\pi+\arctan\frac{b}{a}&\pm a<0,\mp b<0\\
\pi+\arctan\frac{b}{a}&\pm a<0,\mp b>0\\
\end{cases}$$
where $\mathrm{atan2}$ is a special version of $\arctan$ that work in all quadrants.
For example $a=1=b$ you have difference between zeros $=\pi$.
The situation for $y''+(1+x)=0$ is much more complicate if $y=y(x)$
$$y=c_1 \text{Ai}\left(-\sqrt[3]{-1} (-x-1)\right)+c_2 \text{Bi}\left(-\sqrt[3]{-1} (-x-1)\right)$$
See
http://mathworld.wolfram.com/AiryFunctionZeros.html
http://onlinelibrary.wiley.com/doi/10.1002/zamm.19960760714/abstract
|
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|
Obtaining the Fundamental Polygon of $\mathbb{R}P^2$ On this page, Wikipedia shows, under the "Examples" heading, the fundamental polygons of the Sphere and the Real Projective Plane. Can we obtain the latter diagram from the former? I thought that this might be possible since $\mathbb{R}P^2$ is defined as the quotient of the sphere under the equivalence relation that identifies antipodal points, but I couldn't find a way to make it work. Any insight is appreciated!
|
It suffices to convince yourself the diagram below represents the antipodal map, with the shaded triangle as fundamental domain.
On the subject, there's a nice way to visualize antipodal identification of a cuboctahedron (a polyhedral model of a sphere) yielding a tetrahemihexahedron (a polyhedral model of a real projective plane).
|
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|
Multiple eigenvectors for an eigenvalue and how to know We have the following matrix:
$$\begin{pmatrix}
0 & -1 & 0 \\
4 & 4 & 0 \\
2 & 1 & 2
\end{pmatrix} $$
This matrix has characteristic equation $- \lambda ^3 + 6 \lambda ^2 - 12 \lambda + 8$, which gives us the eigenvalue $\lambda = 2$. However, this eigenvalue has two eigenvectors, namely $(1,-2,0)$ and $(0,0,1)$.
How would you mathematically be able to know how many eigenvectors are associated with a single eigenvector? Is there a way to find out?
|
Yes let $v=(x,y,z)^T$ an eigenvector for the given matrix $A$ associated to $2$ then we have
$$Av=2v$$
and if we solve the system of equations we find that $v$ is a linear combination of the two vectors given in your question.
|
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|
Linear Algebra - seemingly incorrect result when looking for a basis
For the following matrix $$ A = \begin{pmatrix}
3 & 1 & 0 & 0 \\
-2 & 0 & 0 & 0 \\
-2 & -2 & 1 & 0 \\
-9 & -9 & 0 & -3
\end{pmatrix} $$
Find a basis for the eigenspace $E_{\lambda}(A)$ of each eigen value.
First step is to find the characteristic polynomial to find the eigenvalues. $$\det(A- I\lambda) = 0 \implies \lambda ^4 - \lambda^3 - 7 \lambda^2 + 13 \lambda -6 = 0 $$ $$\lambda = 1 \lor \lambda = 2 $$ Now that we have the eigen values, to find a basis for $E_{\lambda}(A)$ for the eigen values $\lambda$, we find vectors $\mathbf{v}$ that satisfy $(A-\lambda I)\mathbf{v} = 0$:
$$(A - 2I)\mathbf{v} = \begin{pmatrix}
1 & 1 & 0 & 0 \\
-2 & -2 & 0 & 0 \\
-2 & -2 & -1 & 0 \\
-9 & -9 & 0 & -5
\end{pmatrix} \begin{pmatrix}
v_1 \\
v_2 \\
v_3 \\
v_4
\end{pmatrix} = \begin{pmatrix}
0 \\
0 \\
0 \\
0
\end{pmatrix} \implies
\begin{cases}
v_1 + v_2 = 0\\
-2v_1 -2v_2 = 0 \\
-2v_1 - 2v_2 -v_3 = 0 \\
-9v_1 -9v_2 - 5v_4 = 0
\end{cases}$$ $\implies v_2 = -v_1 \land v_3 = 0 \land v_4 = 0$. So $(v_1, v_2, v_3, v_4) = (v_1, -v_1, 0, 0) = v_1(1, -1, 0, 0)$ so $(1, -1, 0, 0)$ is the basis.
However, this is apparently incorrect. What have I done wrong here?
Also, for the eigenvalue $\lambda = 1$ I do get the correct answer, so only the part above gives me problems.
|
You're missing an eigenvalue! I think you've made a typo in your characteristic polynomial, writing $-\lambda^4$ instead of $\lambda^4$. The characteristic polynomial factorizes to $(x-2)(x+3)(x-1)^2$, hence the eigenvalues are $-3$, $2$, and $1$. The vector you have found is indeed a basis for the 2-eigenspace, and you can easily to compute that the basis vectors of the eigenspaces of $-3$ and $1$ are
$\left( \begin{array}{ccc}
0 \\
0 \\
0 \\
1 \end{array} \right)$ and $\left( \begin{array}{ccc}
0 \\
0 \\
1 \\
0 \end{array} \right)$ respectively.
|
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|
Adding and multiplying piecewise functions How do I add and multiply two piecewise functions?
$$
f(x)=
\begin{cases}
x+3 &\text{if }x<2\\
\dfrac{x+13}{3} &\text{if }x>2
\end{cases}
$$
$$
g(x)=
\begin{cases}
x-3 &\text{if }x<3\\
x-5 &\text{if }x>3
\end{cases}
$$
|
It helps to think about the fact that you can only add and multiply stuff that actually exists (i.e. is defined):
We can only add and multiply these functions in places where they both exist at the same time, namely:
$x<2\\
2<x<3\\
x>3$
The final step is to check what the functions equal at each of these segments, then put it all together:
$$
f(x)\times g(x)=
\begin{cases}
(x+3)\times(x-3) &\mbox{if } \quad x<2\\
(\frac{x+13}{3})\times(x-3) &\mbox{if } \quad 2<x<3\\
(\frac{x+13}{3})\times(x-5) &\mbox{if } \quad x>3
\end{cases}
$$
$$
f(x)+g(x)=
\begin{cases}
(x+3)+(x-3) &\mbox{if } \quad x<2\\
(\frac{x+13}{3})+(x-3) &\mbox{if } \quad 2<x<3\\
(\frac{x+13}{3})+(x-5) &\mbox{if } \quad x>3
\end{cases}
$$
BONUS ROUND: What about when functions have different arguments, like $f(x)$ and g(y)? Since there is no $x$ in $g()$ and there is no $y$ in $f()$, we don't have to worry about one function being defined when the other one isn't. They live in different worlds (i.e. different domains) so we just have to add or multiply each case of one with each case of the other. For example:
$$
f(x)=
\begin{cases}
x+3 &\mbox{if } \quad x<2\\
\frac{x+13}{3} &\mbox{if } \quad x>2
\end{cases}
$$
$$
g(y)=
\begin{cases}
y-3 &\mbox{if } \quad y<3\\
y-5 &\mbox{if } \quad y>3
\end{cases}
$$
$$
f(x)\times g(y)=
\begin{cases}
(x+3)\times(y-3) &\mbox{if } \quad x<2,y<3\\
(x+3)\times(y-5) &\mbox{if } \quad x<2,y>3\\
(\frac{x+13}{3})\times(y-3) &\mbox{if } \quad x>2,y<3\\
(\frac{x+13}{3})\times(y-5) &\mbox{if } \quad x>2,y>3
\end{cases}
$$
|
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|
Easy way of memorizing or quickly deriving summation formulas My math professor recently told us that she wants us to be familiar with summation notation. She says we have to have it mastered because we are starting integration next week. She gave us a bunch of formulas to memorize. I know I can simply memorize the list, but I am wondering if there is a quick intuitive way of deriving them on the fly. It has to be a really quick derivation because all of her test are timed.
Otherwise is there an easy way, you guys remember these formulas.
$\begin{align}
\displaystyle
&\sum_{k=1}^n k=\frac{n(n+1)}{2} \\
&\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6} \\
&\sum_{k=1}^n k^3=\frac{n^2(n+1)^2}{4} \\
&\sum_{k=1}^n k(k+1)=\frac{n(n+1)(n+2)}{3} \\
&\sum_{k=1}^n \frac{1}{k(k+1)}=\frac{n}{n+1} \\
&\sum_{k=1}^n k(k+1)(k+2)=\frac{n(n+1)(n+2)(n+3)}{4} \\
&\sum_{k=1}^n \frac{1}{k(k+1)(k+2)}=\frac{n(n+3)}{4(n+1)(n+2)} \\
&\sum_{k=1}^n (2k-1)=n^2
\end{align}$
Note: Sorry if there is an easy and obvious answer to this question. Most of the students in my class already know these formulas from high school, but I only went up to Algebra 2 and Trig when I was in high school. PS: This is for a calculus class in college.
|
I would like to share the way I ended up remembering these formulas. Most of them are geometric ways of remembering these summation formulas. I still like Raymond Manzoni answer, so I will leave that as my accepted answer! He really helped me on my test.
$\sum \:_{n=a}^b\left(C\right)=C\cdot \:\left(b-a+1\right)$: This is one where it is quite easy to remember by just understanding what summation definition means. It is basically saying keep on adding C, so $C+C+C+C+C+...+C$. You adding c, $b-a+1$ times. So, $C\cdot \left(b-a+1\right)$.
$\sum _{n=1}^k\left(n\right)=\frac{\left(1+k\right)\cdot \:k}{2}$ You can make a dot plot in the form of a triangle. This first column will have 1 dot, the second 2, the third 3, the fourth 4, and you will keep on doing it k times. The total amount of dots in one triangle is what we want to find out. If we make to identical triangles of this format and flip one over and overlap it to make a rectangle. We get a rectangle with height k and with k+1. But this give us the total form two triangles. We only care about 1, so we divide it by two.
$\sum _{n=1}^k\left(2n-1\right)=k^2$: For this, I like the picture Raymond gave. Look at his answer for the explanation of this.
$\sum _{n=1}^k\left(n^2\right)=\frac{\left(\left(2k+1\right)\cdot \:\frac{k\left(k+1\right)}{2}\right)}{3}=\left(\frac{\left(2k+1\right)\cdot k\left(k+1\right)}{6}\right)$: Create rectangles. First 1x1,2x2,3x3,...until you get up to k rectangles.Now, take three copies of each of you rectangles and arrange them as the picture. Take two copies and arrange them in decreasing order upward. Now cut the third copy of each rectangle and arrange them in the middle. They are color coded so you can see that all these dots appear from one of these rectangles. Now you form with a new rectangle with base 2*k+1 and width of $\sum _{n=1}^k\left(n\right)$. You already know $\sum _{n=1}^k\left(n\right)$ is equal to $\:\frac{k\left(k+1\right)}{2}$ from the triangle picture above. So now multiply length and width to find area(total dots), and you get $\left(2k+1\right)\cdot \:\frac{k\left(k+1\right)}{2}$, but this is three copies of each. So we divide by 3. Leaving us with $\frac{\left(\left(2k+1\right)\cdot \:\frac{k\left(k+1\right)}{2}\right)}{3}$
$\sum \:_{n=1}^k\left(n^3\right)=\left(\sum \:\:\:_{n=1}^k\left(n\right)\right)^2=\left(\frac{\left(1+k\right)\cdot \:\:\:k}{2}\right)^2$:Draw a cube with $1^3$ dots,$2^3$ dots,$3^3$ dots,$4^3$ dots,...till you get to k cubes. Now rearrange them,to squares and rectangles as shown in the picture. Every other square is cut in half. while others are stacked by planes of the next layer. Odd Layers=Stack. Even Layers=Cut. You get a resulting square of width $\sum _{n=1}^k\left(n\right)=\frac{\left(1+k\right)\cdot \:k}{2}$ and height $\sum _{n=1}^k\left(n\right)=\frac{\left(1+k\right)\cdot \:k}{2}$.
|
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What would this homotopy look like? I found a very weird definition of null-homotopy.
Definition.
Let $M$ be a $C^k$ surface which is a subset of $\mathbb{R}^3$. Let $\mathscr{R}$ be an open connected subset of $M$
Let $\gamma$ be a closed curve whose image is the boundary of $\mathscr{R}$
Let $\sigma$ be any closed curve of period $L$ which is either $\gamma$ or lies in $\mathscr{R}$. Let $\sigma(0)=x_0$. Then $\sigma$ is null-homotopic iff there exists a continuous function $\Gamma:[0,L]\times [0,1]\rightarrow M$ such that $\Gamma(t,0)=\sigma(t)$ and $\Gamma(0,s)=\Gamma(t,1)=x_0$, and for all $0<s\leq 1$ and $t\in (0,L)$, $\Gamma(t,s)$ lies in $\mathscr{R}$.
This is an extremely wierd definition. Anyway, how does this homotopy look like? It deforms $\sigma$ to $x_0$ continuously with only one side fixed.
That is,
What happens at the right side?
Is it possible that a closed curve $\sigma$ becomes a non-closed curve at some time while deforming to $x_0$?
|
This is almost certainly a typo. Indeed, as written, every $\sigma$ is "null-homotopic", via the map $\Gamma(t,s)=\sigma(t(1-s))$. It should additionally say that $\Gamma(L,s)=x_0$ for all $s$.
|
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|
Prove that $S=((x,y) \in \mathbb{R}^2 : e^x+e^y\le 100$ and $x+y\geq 0)$ is compact I want to use the fact that a subset of $R^2$ is compact if it is exactly closed and bounded. For the closed part, I'm defining the continuous functions $f(x,y)=e^x+e^y$ and $g(x,y)=x+y$ and I'm saying
$S=A\cap B$, where
$A=((x,y)\in \mathbb{R}^2: 0\leq f(x,y) \leq 100$) and
$B=((x,y)\in \mathbb{R}^2: 0\leq g(x,y) \leq b$), with $b\in \mathbb{R}: B \supset S$
with these functions and the fact that the target set is closed and they are continuous, then A and B are also closed and the intersection of closed sets is closed.
For the bounded part I don't know how to prove it. Any help with this guys? am I righ when proving S is closed?
|
You have that $e^x\leq e^x+e^y\leq 100$ so $x\leq\ln(100)$, and $y\leq\ln(100)$ too. As $x\geq -y\geq -\ln(100)$ and $y\geq -x\geq -\ln(100)$ you get $S\subset[-\ln(100),\ln(100)]^2$ which is bounded so is $S$.
|
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|
Solve for $f(x)=f'(x)$ without previous knowledge.
Solve for $f(x)=f'(x)$ without previous knowledge.
I know it is obviously $f(x)=e^x$, but could you prove this without knowing $\frac d {dx}e^x=e^x$?
And does there exist a $g(x)=g'(x)$ but $g(x)\ne f(x)$?
|
$$
f = \frac{df}{dx}
$$
$$
dx = \frac{df} f
$$
$$ x + \text{constant} = \log_e |f|
$$
$$
e^x \cdot\text{positive constant} = |f|
$$
$$
e^x \cdot\text{constant} = f.
$$
(One must check separately that $f=0$ is a solution.)
Suppose there were some other solution $g$ (maybe equal to $0$ at some points and not at others? or whatever $\ldots$). Then
$$
\underbrace{\frac d {dx}\ \frac{g(x)}{e^x} = \frac{e^x g'(x) - g(x)e^x}{e^{2x}}}_\text{quotient rule}\ \ \underbrace{ = \frac{e^xg(x) - e^x g(x)}{e^{2x}}}_{\text{since }g\,'\,=\,g} = 0,
$$
so $\dfrac{g(x)}{e^x} = \text{constant}$, and we can rule out other solutions.
|
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|
Prove $a^2+b^2\geq \frac{c^2}{2}$ and friends if $a+b\geq c\geq0$ Sorry for my inequality spam, but I got to prepare for my exams today :( Here's another:
Problem:
Prove$$a^2+b^2\geq \frac{c^2}{2}$$$$a^4+b^4\geq \frac{c^4}{8}$$$$a^8+b^8\geq \frac{c^8}{128}$$ if $a+b\geq c\geq0$
Attempt:
Working backwards:
$$a^2+b^2\geq \frac{c^2}{2}$$
$$\implies 2a^2+2b^2\geq c^2$$
I am stuck on the first one, let alone the others. I know by AM-GM, $\frac{a^2+b^2}{2}\geq ab$ but how can I use it here?
|
Using the information you supplied, you could also just square both sides of the original inequality to obtain $a^2+2ab+b^2 \geq c^2$. Divide the inequality by 2 and subtract $ab$ to the other side. Given the AM-GM, we know that $\frac{a^2+b^2}{2} \geq ab $. Adding this inequality to the previous one we obtain $2\frac {a^2+b^2}{2} \geq \frac {c^2}{2} +ab-ab $. Simplifying gives us our desired result of $a^2+b^2\geq \frac {c^2}{2}$.
As macavity said, you obtain the rest of the inequalities by substituting $a^2$ for $a $, $b^2$ for $b $, and so on.
|
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|
If $\sigma$ takes squares to squares, conclude $a < b$ implies $\sigma a < \sigma b$. Prove that any $\sigma \in $ Aut$(\mathbb{R/\mathbb{Q}})$ takes squares to squares and takes positive reals to positive reals. Conclude that $a < b$ implies $\sigma a < \sigma b$ ,$\forall a,b
\in R$.
proof: Suppose $\sigma: a → b$, such that $a = c^2$ where $b$ is a square.
Then $\sigma(a) = \sigma(c^2) = \sigma(c *c) = \sigma (c) \sigma(c) = (\sigma(c))^2$ which is a square. So it takes positive reals to positive reals.
Then suppose that if $a < b$ , where $a, b\in R$.
Then there is a $q \in \mathbb{Q}$ such that $a < q < b$. So $q = \sigma q = \sigma ( q + a - a) = \sigma (q - a) + \sigma a > \sigma a.$
So $q > \sigma(a)$.
And $q = \sigma (q) = \sigma (q + b - b) = \sigma( q + b ) - \sigma (-b)$
can someone please help me with $q < \sigma(b)$? and please verify I am on the right track. Anything could help thank you !
|
Instead of splitting $\sigma(q+b-b)$ as $\sigma(q+b)-\sigma(b)$, split it as $\sigma(q-b)+\sigma(b)=-\sigma(b-q)+\sigma(b)$. Since $b>q$, $\sigma(b-q)>0$, so you get $\sigma(q)<\sigma(b)$.
(Actually, you can do this without introducing $q$ at all! Just notice that $\sigma(b)=\sigma(a+b-a)=\sigma(a)+\sigma(b-a)$, and $\sigma(b-a)>0$.)
|
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|
Area bounded between the curve $y=x^2 - 4x$ and $y= 2x/(x-3)$ I've determined the intersects to be $x = 0, 2, 5$ and that $\frac{2x}{x-3}$, denoted as $f(x)$, is above $x(x-4)$, denoted as $g(x)$, so to find the area, I'll need to find the integral from $0$ to $2$ of $f(x) - g(x)$. But I've been stuck for a while playing around with this question.
|
Notice, the area bounded by the curves from $x=0$ to $x=2$, is given as $$\int_{0}^{2}\left(\frac{2x}{x-3}-(x^2-4x)\right)\ dx$$
$$=\int_{0}^{2}\left(\frac{2(x-3)+6}{x-3}-x^2+4x\right)\ dx$$
$$=\int_{0}^{2}\left(2+\frac{6}{x-3}-x^2+4x\right)\ dx$$
$$=\left(2x+6\ln|x-3|-\frac{x^3}{3}+2x^2\right)_{0}^{2}$$
$$=4-\frac{8}{3}+8-6\ln (3)$$$$=\color{red}{\frac{28}{3}-6\ln (3)}$$
|
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|
If $|G|=30$ and $|Z(G)|=5$, what is the structure of $ G/Z(G)$? The question is:
If $|G|=30$ and $|Z(G)|=5$, what is the structure of $G/Z(G)$?
I don't know what do we mean by 'structure' asked in the question. Please help.
|
$|G|=30$, $|Z(G)|=5$. $|G/Z(G)|=6$
We know that all groups of order 6 are isomorphic to $S_3$ or $\mathbb{Z} /6\mathbb{Z}$.
well known result:
If $G/Z(G)$ is cyclic, then $G$ is abelian.
If $G/Z(G) \cong \mathbb{Z} /6\mathbb{Z} $ ,then $G$ is abelian which is contradiction to $|Z(G)|=5$ . Therefore $G/Z(G) \cong S_3$
|
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|
Image of a circle under conformal map $1/z$ The image of a circle under conformal map $1/z$ should be a circle, but how to prove it (or how to find the relationship between the two circles)?
$z = x + iy = d + a\exp(i\theta)$, where $a$ is the radius of the circle and $d$ is a real number, represents a circle with radius $a$ displaced along the real axis.
Now $w = u + iv = 1/z$; how to find the center position and radius of the transformed circle?
I just write $w = 1/(d + a\exp(i\theta))$ and after some algebra I get $$
u = \frac{d + a\cos(\theta)}{d^2 + a^2 + 2d\cos(\theta)}\quad\text{and}\quad
v = -\frac{a\cos(\theta)}{d^2 + a^2 + 2d\cos(\theta)},
$$
but then I don't know how to simplify it and get it to the form that looks like a circle.
|
The real points $d \pm a$ lie on a diameter of your circle. The image points $1/(d \pm a)$ lie on a diameter of the image circle, so the image circle has center
$$
\frac{1}{2}\left[\frac{1}{d - a} + \frac{1}{d + a}\right]
= \frac{d}{d^{2} - a^{2}}
$$
and radius
$$
\frac{1}{2}\left|\frac{1}{d - a} - \frac{1}{d + a}\right|
= \frac{a}{|d^{2} - a^{2}|}.
$$
Alternatively, starting from
$$
(x - d)^{2} + y^{2} = a^{2}
$$
and substituting
$$
x = \frac{u}{u^{2} + v^{2}},\qquad
y = -\frac{v}{u^{2} + v^{2}}
$$
works.
|
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|
Prove that $( 1 + n^{-2}) ^n \to 1$. I need to prove that $\left(1 + \frac 1 {n^2} \right)^n \to 1$.
I tried to use Bernoulli's inequality, but that is not very useful since in the original sequence there is a plus sign. I then tried to use the Sandwich Theorem by finding two sequences which would make bounds for the original one. The lower bound is obvious, the upper bound not so much. I tried using the sequence $\left( \frac 1 {n+1} \right) ^{\frac 1 n}$, but I could not show that this sequence is bigger than the original one for all $n$. Could anyone help me with this?
|
Use $e^x \geq 1 + x$ for all $x \in \mathbb{R}$. There are many ways to see this. The easiest, in my opinion, is to note that $y = x + 1$ is tangent to $y = e^x$ and $x \mapsto e^x$ is convex, then use definition of convexity.
It follows that
$$
1 \leq
\left(1 + \frac{1}{n^2}\right)^n \leq
\left(\exp\frac{1}{n^2}\right)^n =
\exp\frac{1}{n} \rightarrow
1
$$
as $n \rightarrow \infty$.
|
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|
How to maximize $\frac 1 {PC}+\frac 1 {PB}$? Assume $P$ is a point inside the angle $\hat A$. How to draw a line that intersects with lines $a$ and $b$ AND maximizes $\frac 1 {PC}+\frac 1 {PB}$ ?
Here is the picture:
EDIT:
The point $P$ is given. the only thing changeable is the angle of the line passing the point $P$.
|
I claim that $AP \perp BC$.
We set $X \in AB$ such that $AC \parallel PX$, and $Y \in AP$ such that $XY \parallel BC$.
Now, we have $$\triangle AYX \sim \triangle APB$$ $$\triangle PYX \sim \triangle APC$$
This gives us $$\frac{XY}{PB}=\frac{AY}{AP}$$ $$\frac{XY}{PC}=\frac{YP}{AP}$$
Summing them, we have $$\frac{1}{PB}+\frac{1}{PC}=\frac{1}{XY}$$
We need to minimize $XY$. Since $X$ is fixed and $Y$ is on $AP$, we need $XY \perp AP$, i.e. $AP \perp BC$.
|
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|
A question on proof of the Zermelo's theorem "Every set is well-orderable." There exists some proof for the theorem. Some of them use Transfinite Recursion.
Some of them use same argument with the following.
Proof:(copied from proofwiki)
"Let $S$ be a set.
Let $\mathcal{P}(S)$ be the power set of $S$.
By the Axiom of Choice, there is a choice function $c$ defined on $\mathcal{P}(S)-\{∅\}$.
We will use $c$ and the Principle of Transfinite Induction to define a bijection between $S$ and some ordinal.
Intuitively, we start by pairing $c(S)$ with $0$, and then keep extending the bijection by pairing $c(S∖X)$ with $α$, where $X$ is the set of elements already dealt with.
Base case
$α=0$
Let $s_0=c(S)$.
Inductive step
Suppose $s_β$ has been defined for all β<α.
If $S−\{s_β:β<α\}$ is empty, we stop.
Otherwise, define:
$s_α:=c(S−\{sβ:β<α\})$.
The process eventually stops, else we have defined bijections between subsets of S and arbitrarily large ordinals."
My question is that why "The process eventually stops".
My attempt: By assuming the process doesn't stop. Then we have a one to one function from $Ord$ to $S$. If you say this is contradict because of $|S|\in Ord$. The idea is not true, since cardinal numbers defines after the theorem and using it. I want to find a contradiction that by using ZFC axioms.
($Ord:$the class of all ordinals)
|
The process has to stop because of Hartogs theorem:
For every set $X$ there exists a well-ordered set $Y$ such that there is no injection from $Y$ into $X$.
Or, if you have the axiom of replacement, you can replace $Y$ by an ordinal and obtain the following:
For every set $X$ there exists a least ordinal $\alpha$ such that there is no injection from $\alpha$ into $X$. This $\alpha$ is called the Hartogs number of $X$ and is sometimes denoted by $\aleph(X)$.
|
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|
Use the change of variables to evaluate the double-integral $\iint_R4(x+y)e^{x-y}\;dA$
Use the change of variables indicated to evaluate the double-integral $\iint_R4(x+y)e^{x-y}\;dA$ where $R$ is the interior of the triangle whose vertices are $(-1,1)$, $(0,0)$, and $(1,1)$; $x=\frac{1}{2}(u+v)$ and $y=\frac{1}{2}(u-v)$
I know I have to find the following components for the formula:
$$\iint_S f(g(u,v),h(u,v)) \left|\frac{\partial(x,y)}{\partial(u,v,)}\right| \ du \; dv$$
The Jacobian is $\frac{1}{2}$
Next I replaced $x$ and $y$ in my integral with the change of variable to give me the following double integral:
$$\iint 4\left[\frac{u}{2}+\frac{v}{2}+\frac{u}{2}-\frac{v}{2}\right]
e^{\frac{u}{2}+\frac{v}{2}-\frac{u}{2}+\frac{v}{2}}\;\left(\frac{1}{2}\right)\;du\;dv$$
After simplifying the above I got:
$$2\iint ue^v\;du\;dv$$
I'm completely confused with how to map $(x,y)$ to $(u,v)$ so that I can identify my limits of integration.
|
If I'm not mistaken:
Notice how
$$u = x +y \qquad \text{and} \qquad v = x-y.$$
Then notice how $v = \alpha$ implies parallell lines to the first bisector of the plane. This means $v = 0$ results in the first bisector.
Furthermore $u =0$ results in the second bisector.
All that is remaining is the straight line $y = 1$. You can write this using the variables $u$ and $v$ as $1 = \dfrac{1}{2}(u-v)$ or $2 = u-v$.
This all implies the following:
Let $$v: 0\to u-2 \qquad \text{and} \qquad u = 0 \to 2.$$
Added picture:
By using the change of variables you integrate along the red arrows.
These red arrows start at $v = 0$, but and stop at $v = y-2$. Integration goes as follow:
$$\int_0^2 \operatorname du \int_0^{u-2} \operatorname dv$$
For example: At $u=0$ (the first red arrow) integration goes from $v = 0$ to $v = 0-2 = -2$ (or $x-y = -2$ or $y = x+2$ the blue line).
At $u =1$ (the second red arrow) integrations goes from $v=0$ to $v = 1-2 = -1$ (or $x-y = -1$ or $y = x+1$ the pink line).
When $u \approx 2$ then integration goes from $v=0$ to $v \approx 2-2 =0$ or the red arrow is getting tiny small.
|
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|
Linear Algebra and Quadratic Equations I'm just wondering if Linear Algebra is concerned only with Linear equations? Can quadratic equations(or any higher power) also be considered under Linear Algebra? What does the term Linear stand for?
|
Linear algebra very much included Quadratic equations. in fact. Polynomial equations of any degree. For example $ax^2+bx+c=0$ may be written as product of a row vector $(a, b, c)$ with a column vector $(x^2,x,1)^\top$.
|
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|
Monoids where $\operatorname{Hom}(M,M) \cong M$ What are some examples of monoids where $\operatorname{End}(M) \cong M$? Is there a nice characterization of such monoids? E.g., they will necessarily have a zero element, since $\forall x (x \mapsto 1_M)$ is a zero of $M^M$ (so no groups satisfy $G^G \cong G$).
Examples:
*
*$\{1\}$
*$(\mathbb{F}_2, \cdot)$
Non-examples:
*
*$(\mathbb N, +) \mapsto (\mathbb N, \cdot)$
*$(\mathbb F_2, +_2) \mapsto (\mathbb F_2, \cdot)$
|
Partial answer. If $M$ is a finite monoid such that $\operatorname{End}(M) \cong M$, then either $M = \{1\}$ or $M = \{0, 1\}$ with the usual multiplication of integers.
Proof. Let $G$ be the group of invertible elements of $M$. Since $M$ is finite, $M - G$ is an ideal of $M$. Let $E(M)$ be the set of idempotents of $M$. For each $e \in E(M)$, let $\bar e$ be the endomorphism of $M$ defined by
$$
\bar e(x) = \begin{cases}
1 & \text{if $x \in G$}\\
e & \text{otherwise}
\end{cases}
$$
Then $\bar e \in E(\operatorname{End}(M))$ and moreover $\bar e \bar f = \bar e$ for all $e, f \in E(M)$. Thus the map $e \to \bar e$ is an injection from $E(M)$ to $E(\operatorname{End}(M))$. Since $M$ and $\operatorname{End}(M)$ are isomorphic, this injection is a bijection. In particular, there exists an idempotent $e$ in $M$ such that $\bar e$ is the identity map on $M$. Since the range of $\bar e$ is $\{1, e\}$, this means that $M = \{1, e\}$, which gives the two possible solutions.
|
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Proof limit $b^n n^{\alpha} \to 0$, when $n \to \infty$ and $\forall |b| < 1 $. $$\lim_{n \to \infty} b^n n ^{\alpha} = 0, \forall |b| < 1, \alpha > 0$$
I have proved it for cases when $\alpha \le 0$, but I am not sure where to go further. I have tried D'Alembert's test, but I believe that it does not work for $\alpha > 0$.
Can you give me any direction where to go ? At least, some nominal ideas.
|
Because the limit of $b^n$ is $0$, since $|b| < 1$ and $n$ approaches infinity
$$\lim_{n \to \infty} (b^n) (n^a) = \lim_{n \to \infty} (b^n) \cdot \lim_{n \to \infty}(n^a)$$
since limit of $b^n$ is $0$, the limit of the previous one is $0$.
|
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Expectation of normal distributions Let $X$ have a normal distribution with mean $µ$ and variance $σ^2$.
Find $E[X^3]$ (in terms of $µ$ and $σ^2$).
the pdf of this function is $$\frac{1}{σ\sqrt{2\pi}} e^{\frac{-(x-µ)^2}{2σ^2}}$$
so $$E[X^3] = \int_{-\infty}^\infty x^3\frac{1}{σ\sqrt{2\pi}} e^{\frac{-(x-µ)^2}{2σ^2}} dx$$
How do I compute this integral?
|
Using the symmetry of $\frac{X-\mu}{\sigma}$, one can conclude that
$$
E\left(\frac{X-\mu}{\sigma}\right)^{2n+1}=0, \text{for}\quad n=0,1,2,...
$$
Hence, using $Var(X)=\sigma^2$ and $EX=\mu$, you can show that $E\left(\frac{X-\mu}{\sigma}\right)^{3}=\frac{EX^3-3\mu\sigma^2-\mu^3}{\sigma^3}=0$, thus
$$
EX^3 = 3\mu\sigma^2 + \mu^3.
$$
|
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|
Group Theory - Cartesian Product Given $G = C_3 \times C_2 \times C_4 \times C_5$ where $C_2 = \langle\{0,1\}, +2\rangle$; $C_3 = \langle\{0, 1, 2\}, +3\rangle$; and so on. Consider the group $(G, *)$, here the group operator $*$ is defined by the constituent groups in the Cartesian product.
i) Write an expression for $t \in G$ as a vector and show where the components of vector $t$ are drawn from?
ii) For the vectors $t, u \in G$, Compute $t*u = w \in G$ by showing the components of $w$.
iii) What is the group identity $e_G$ ? Choose any $t \in G$, $t \neq e_G$ and compute $t^{-1}$
iv) List all the cosets of $G$ relative to the subgroup $C_3\times C_2 \times C_4$
Here's a picture for better formatting: http://imgur.com/ebgiHta
I've been attempting to gather my thoughts on this one but I'm struggling.
I've been thinking:
$t$ in $G$ is of the form $(T\bmod{3}, T \bmod 2, T \bmod 4, T \bmod 5)$ where $T$ is an integer.
$t*u = (T+U \bmod 3, T+U \bmod 2, T+U \bmod 4, T+U \bmod 5)$ where $t = (T \bmod 3, T \bmod 2, T \bmod 4, T \bmod 5)$ for some integer $T$, and similarly for $u$ and $U$.
$(0,0,0,0)$ is the identity
$t-1$ = $(3-T \bmod 3, 2-T \bmod 2, 4-T \bmod 4, 5-T \bmod 5)$ where $t = (T \bmod 3, T \bmod 2, T \bmod 4, T \bmod 5)$ for some integer $T$.
Any help would be awesome!
|
A cartesian product $X \times Y$ consists of all ordered pairs $(x,y)$, where $x$ is in $X$ and $y \in Y$. So $x$ and $y$ don't have to be the same in any way. Similarly, $G = C_3 \times C_2 \times C_4 \times C_5$ should consist of all ordered quadruples $(t_1, t_2, t_3, t_4)$, where $t_1 \in C_3, t_2 \in C_2, t_3 \in C_4$, and $t_4 \in C_5$. Your answer seemed to suggest that an element should have all the components be the same integer, which need not be the case.
Yes, if $t = (t_1, t_2, t_3,t_4)$, and $u = (u_1, u_2, u_3, u_4)$, then $t \ast u$ should be $(t_1 + u_1, t_2 + u_2, t_3 + u_3, t_4 + u_4)$, and $(0,0,0,0)$ is the identity of $G$.
The problem statement is committing a slight abuse of notation, because $C_3 \times C_2 \times C_4$ is not really a subgroup of $G$ (it is literally not even a subset). What they really mean is the subgroup $H:= C_3 \times C_2 \times C_4 \times \{0\}$, where by $\{0\}$ I mean the subset of $C_5$ whose only element is the zero element of $C_5$ (that is, $0 \pmod{5}$).
A trick with cosets: since $H$ is a group with $3 \cdot 2 \cdot 4 = 24$ elements, and $G$ is a group with $3 \cdot 2 \cdot 4 \cdot 5 = 120$ elements, a complete set of coset representatives for $H$ in $G$ should have exactly $\frac{|G|}{|H|} = \frac{120}{24} = 5$ elements. So what you are looking for here are $5$ distinct elements $g_1, ... , g_5$ of $G$ such that $g_i \ast g_j^{-1}$ is not a member of $H$ for any $i \neq j$. There is more than correct answer for a complete set of coset representatives, but there is a somewhat obvious choice you can make.
Let me know if you have any questions about anything I said.
|
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|
If the difference of the function only dependent on the difference of input, can we say it's linear without assuming it's continuous? As state in the title, if for a function $$f(t+\tau)-f(t)=R(\tau), \forall t,\tau$$
$R(\tau)$ is just a general function showing the difference has nothing to do with $t$.
then I can prove $f(t)=at+b$ where $a,b$ is some certain constant, if assuming $f$ continuous, I prove it by first proving it's true over rational number set.
However, is this also true even if we don't specify the function to be continuous but only measurable?
|
Yes, measurability implies that $f$ is affine. Proof: Let $g(t) = f(t) - f(0)$. Then $g$ is additive, because $$ g(a + b) = f(a + b) - f(0) = f(a) + R(b) - f(0),$$ and $R(b) = f(0 + b) - f(0)$, so $g(a + b) = f(a) + f(b) - 2f(0) = g(a) + g(b)$. Furthermore, $g$ is measurable. Every additive, measurable function is linear (see e.g. Theorem 5.5 in Horst Herrlich's "Axiom of Choice".) Therefore, there is some $a$ so that $g(t) = at$ for all $t$, and hence $f(t) = at + f(0)$ for all $t$.
|
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|
To evaluate limit $\lim_{n \to \infty} (n+1)\int_0^1x^{n}f(x)dx$
Let $f:\Bbb{R} \rightarrow \Bbb{R}$ be a differentiable function whose derivative is continuous, then
$$\lim_{n \to \infty} \left((n+1)\int_0^1x^{n}f(x)dx\right).$$
I think I have to use L'Hop rule, but I don't see how.
|
We can prove that if $f$ is continuous function ($f$ being differentiable function is not needed)
$$
\lim_{n \to \infty} (n+1)\int_0^1x^{n}f(x)dx=f(1)
$$
Since $f$ is continuous, $f$ is bounded on $[0,1]$. Also
$$
\forall \epsilon>0, \exists \delta>0, \forall x\in(1-\delta,1],\text{there is } |f(x)-f(1)|<\epsilon
$$
\begin{align}
\left|(n+1)\int_{0}^{1}x^nf(x)dx-f(1)\right|&=
\left|\int_{0}^{1-\delta} (n+1)x^nf(x)dx+\int_{1-\delta}^{1} (n+1)x^nf(x)dx-f(1)\right|
\\
&=\left|f(t_1) \int_{0}^{1-\delta} (n+1)x^ndx+f(t_2)\int_{1-\delta}^{1} (n+1)x^ndx-f(1)\right| \hspace{5 mm}
\\ &\hspace{10 mm}(t_1\in(0,1-\delta),t_2\in(1-\delta,1) \text{ and by IMVT})
\\
&=\left|f(t_1)(1-\delta)^{n+1}+f(t_2)(1-(1-\delta)^{n+1})-f(1)\right|
\\
&\leqslant 2M(1-\delta)^{n+1}+|f(t_2)-f(1)|
\\&<2M(1-\delta)^{n+1}+\epsilon
\end{align}
So
$$
\varlimsup\limits_{n\to\infty}\left|(n+1)\int_{0}^{1}x^nf(x)dx-f(1)\right|\leqslant \varlimsup\limits_{n\to\infty}2M(1-\delta)^{n+1}+\epsilon=\epsilon
$$
Since $\epsilon$ is arbitrary small
$$
\varlimsup\limits_{n\to\infty}\left|(n+1)\int_{0}^{1}x^nf(x)dx-f(1)\right|=0 \hspace{5 mm} \text{and}\hspace{5 mm} \varliminf\limits_{n\to\infty}\left|(n+1)\int_{0}^{1}x^nf(x)dx-f(1)\right|=0
$$
So
$$
\lim\limits_{n\to\infty}\left|(n+1)\int_{0}^{1}x^nf(x)dx-f(1)\right|=0\hspace{5 mm} \text{or} \hspace{5 mm}\lim\limits_{n\to\infty}(n+1)\int_{0}^{1}x^nf(x)dx=f(1)
$$
|
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|
If $f''+f'=f$ then $f\equiv 0$ Let , $f:\mathbb R\to \mathbb R$ be a two times continuously differentiable function that satisfies the differential equation $f''+f'=f$ for all $x\in [0,1]$. If $f(0)=f(1)=0$ then , which is correct ?
(A) There exists $(a,b)\subset [0,1]$such that $f(x)>0$ in $(a,b)$.
(B) There exists $(a,b)\subset [0,1]$such that $f(x)<0$ in $(a,b)$.
(C) $f\equiv 0$ in $[0,1]$.
(D) such function does not exists.
If $f(x)=0$ for all in $x\in[0,1]$ then $f$ satisfies the given conditions. So option (D) is false. I think option (C) is correct , but I can't prove this.
|
The function, which is continuous on $[0,1]$, must have a maximum point $x$. In the maximum point (if $x\in(0,1)$) you have: $f'(x)=0$ and $f''(x)\le 0$. So the equation says $f(x) = f''(x) \le 0$. With a similar reasoning you find that on the minimum you have $f(x)\ge 0$. Hence the function is identically zero.
|
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|
Does showing term-by-term differentiation of a series via uniform convergence imply that the series defines an infinitely differentiable function? If I show that
$$\sum 2^{n\alpha} \sin\left(\frac{x}{\alpha}\right)$$
converges uniformly by the M-test, for all x in $R$, then we may differentiate term-by-term everywhere.
Can I conclude from here that the power series must define a $C^{\infty}$ function?
Or must I show that the differentiated series is also uniformly convergent and hence a continuous function, and then use an induction argument?
|
1) This is not a power series.
2) It is not enough to show that the series converges uniformly with each summand being differentiable.
3) It is enough to show that the series of derivatives converges uniformly to show that the limit function is differentiable.
More precisely, if each $f_n : I \to \Bbb{R}$ is $C^1$, if $\sum_n f_n(x)$ converges for at least one $x\in I$ and if $\sum_n f_n'$ converges (locally) uniformly, then the series $\sum_n f_n$ converges (locally) uniformly on $I$ and defines a $C^1$ function.
4) You can then in principle use an induction argument to show that the limit is $C^\infty $.
5) In your case, the series is very strange, since the sine term apparently does not depend on $n $ and can thus be pulled out of the series.
|
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|
Finding the extremal point of a function I have to find and classify all extremal points of a few problems and I am finding it difficult as I don't know how to even start. I can't find any examples of what to do. So if someone could help with one question then I can use that to figure out the rest. One of the questions is
$$f(x, y) = 5xy − 7x^2 + 3x − 6y + 2.$$
Thanks in advance.
|
In general, when having real functions $ f: \mathbb{R}^m \rightarrow \mathbb{R}^n $, which are continuous and twice differentiable, to find their extreme values, you may follow the following strategy:
*
*Find the gradient of $f$, and study its critical points (i.e. points where $\nabla f =0$). Note, it may happen that $f$ has no critical points, and hence no extreme points in the open domain it is defined on.
*Find the Hessian matrix of $f$ at each of the critical points. If the Hessian matrix is positive definite, then the corresponding critical point is a local minimum, and if it is negative definite then the corresponding critical point is a local maximum. If the Hessian matrix associated to a critical point is neither positive nor negative definite then your point is neither a local min nor a local max.
Some Special cases:
*
*If your function is convex (resp. concave) then your function has a unique global min (resp. max).
*If your function is defined on a closed domain, the you may study optimization first in the interior of your closed domain (thus considered as optimization on an open domain), then study optimization of $f$ on the boundary, using for example Lagrange multipliers.
Hope this helps you, try for the above defined function in your question, and I am ready for any clarification.
|
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|
Finding the sum of n terms $S_n$ starting from sigma $k=0$ $$\sum_{k=0}^{n} ((4k-3)\cdot 2^k)+4=(2^{n+3}+4)n-7\cdot2^{n+1}+15$$
How? I've tried everything but i don't see it. Any equivalent solutions are also welcome, thanks.
|
You can exploit a telescoping sum:
$$
\sum_{k=0}^n\bigl((k+1)2^{k+1}-k2^k\bigr)=(n+1)2^{n+1},
$$
because you also have:
$$
\sum_{k=0}^n\bigl((k+1)2^{k+1}-k2^k\bigr)=
\sum_{k=0}^n\bigl(k2^k-2^{k+1}\bigr)
$$
so that
$$
\sum_{k=0}^n k2^k=
\sum_{k=0}^n\bigl((k+1)2^{k+1}-k2^k\bigr)+\sum_{k=0}^n2^{k+1}
=(n+1)2^{n+1}+2(2^{n+1}-1)
$$
|
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|
Series' convergence - making my ideas formal
Find the collection of all $x \in \mathbb{R}$ for which the series $\displaystyle \sum_{n=1}^\infty (3^n + n)\cdot x^n$ converges.
My first step was the use the ratio test:
$$ \lim_{n \to \infty} \dfrac{(3^{n+1}+n+1) \cdot |x|^{n+1}}{(3n+n) \cdot |x|^n} = \lim_{n \to \infty} \dfrac{3^{n+1}|x|+n|x|+|x| }{3^n+n} = \lim_{n \to \infty} ( \dfrac{3^{n+1}|x|}{3^n+n} + \dfrac{n|x|}{3^n+n} + \dfrac{|x|}{3^n+n}) $$
$$ = \lim_{n \to \infty} (\dfrac{3|x|}{1+\dfrac{n}{3^n}} + \dfrac{|x|}{\dfrac{3^n}{n} + 1} + \dfrac{|x|}{3^n + n}) = 3|x| + 0 + 0 = 3|x| $$
So we want $3|x| < 1$, i.e. $|x| < \dfrac{1}{3}$. But we have to also check the points $x=\dfrac{1}{3}, -\dfrac{1}{3}$, where the limit equals $1$.
For $x=\dfrac{1}{3}$ we have $\displaystyle \sum_{n=1}^\infty (3^n + n)\cdot (\dfrac{1}{3})^n = \displaystyle \sum_{n=1}^\infty (1 + n \cdot \dfrac{1}{3}^n) $ which obviously diverges.
For $x=-\dfrac{1}{3}$ we have $\displaystyle \sum_{n=1}^\infty (3^n + n)\cdot (-\dfrac{1}{3})^n = \displaystyle \sum_{n=1}^\infty ((-1)^{n} + n \cdot (-\dfrac{1}{3})^n)$. Now I know this diverges because $(-1)^n$ and $(-\dfrac{1}{3})^n$ have alternating coefficients. But is there a theorem that I can use here?
So we conclude that $|x| < \dfrac{1}{3}$, but is the above work enough to show it conclusively?
|
The series $\displaystyle \sum_{n=1}^\infty (3^n + n)\cdot x^n$ converges if $|x|<\frac{1}{3}$
Indeed $$(3^n + n)\cdot x^n\sim_{\infty} 3^n \cdot x^n$$ but this is a geometric series that converges only if $|x|<\frac {1}{3}$
|
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|
Exercise 1.2.19 of Hatcher's Algebraic Topology I've been trying to prove exercise $1.2.19$ of Hatcher's algebraic topology:
Show that the subspace of $\mathbb{R}^3$ that is the union of the spheres of radius $\frac{1}{n}$ and center $(\frac{1}{n}, 0, 0)$ for $n = 1, 2, ···$ is simply-connected.
I was thinking of applying Seifert-Van Kampen inductively somehow, but I can't conclude using just that argument. I saw this proof but I would like to see a proof of this exercise without using $1-$skeletons.
How can I prove this using just Seifert-Van Kampen?
|
Let $S_n$ be the sphere of radius $1/n$, and let $U_n$ be an open neighborhood of the origin $(0,0,0)$ within $S_n$. Then we can cover the space $X$ by $V_n=(\bigcup_{n=1}^\infty U_n)\cup S_n$ for $n=1,2,\cdots$. Since $V_n$ are path-connected and have trivial fundamental groups, the intersection of any two or more $V_n$'s is $\bigcup_{n=1}^\infty U_n$, which is also path-connected and has trivial fundamental group, $\pi_1(X)=*_n\pi_1(V_n)=0$.
|
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|
There are 31 houses on north street numbered from 1 to 57. Show at least two of them have consecutive numbers. I thought to use the pigeon hole principle but besides that not sure how to solve.
|
There is a unique subset with the maximum number of non-adjacent houses: {1,3,5,...,57}. The number of houses (29) in this set is less than 30.
This observation solves the problem, but it also indicates that with 30 houses there are at least 2 adjacencies. With only 1 adjacency $(i-1,i)$ you could add $+1$ to all the house numbers $\geq i$ and get a set of house numbers between 1 and 58 with no adjacencies, but there again the maximum size of set is 29, not 30. To have 30 nonadjacent numbers we need to choose them from an interval of size at least 59.
Continuing the argument to the general case, to have $H$ houses with distinct numbers between $1$ and $n$, and with $\leq k$ adjacencies, one needs $(n+k) \leq 2H - 1.$ This is optimal; the equality can hold, and in a unique way.
|
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|
Find $\lim_{x\to0}\frac{\sin\left(1-\frac{\sin(x)}{x}\right)}{x^2}$. Is my approach correct? Find:
$$
L = \lim_{x\to0}\frac{\sin\left(1-\frac{\sin(x)}{x}\right)}{x^2}
$$
My approach:
Because of the fact that the above limit is evaluated as $\frac{0}{0}$, we might want to try the De L' Hospital rule, but that would lead to a more complex limit which is also of the form $\frac{0}{0}$.
What I tried is:
$$
L = \lim_{x\to0}\frac{\sin\left(1-\frac{\sin(x)}{x}\right)}{1-\frac{\sin(x)}{x}}\frac{1}{x^2}\left(1-\frac{\sin(x)}{x}\right)
$$
Then, if the limits
$$
L_1 = \lim_{x\to0}\frac{\sin\left(1-\frac{\sin(x)}{x}\right)}{1-\frac{\sin(x)}{x}},
$$
$$
L_2 = \lim_{x\to0}\frac{1}{x^2}\left(1-\frac{\sin(x)}{x}\right)
$$
exist, then $L=L_1L_2$.
For the first one, by making the substitution $u=1-\frac{\sin(x)}{x}$, we have
$$
L_1 = \lim_{u\to u_0}\frac{\sin(u)}{u},
$$
where
$$
u_0 = \lim_{x\to0}\left(1-\frac{\sin(x)}{x}\right)=0.
$$
Consequently,
$$
L_1 = \lim_{u\to0}\frac{\sin(u)}{u}=1.
$$
Moreover, for the second limit, we apply the De L' Hospital rule twice and we find $L_2=\frac{1}{6}$.
Finally, $L=1\frac{1}{6}=\frac{1}{6}$.
Is this correct?
|
By L' Hospital anyway:
$$\frac{\sin\left(1-\frac{\sin(x)}{x}\right)}{x^2}$$ yields
$$\cos\left(1-\frac{\sin(x)}x\right)\frac{\sin(x)-x\cos(x)}{2x^3}.$$
The first factor has limit $1$ and can be ignored.
Then with L'Hospital again:
$$\frac{x\sin(x)}{6x^2},$$
which clearly tends to $\dfrac16$.
|
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|
Cauchy sequence in $L^1$ Space I am learning about the $L^1$ space (the complete Riemann integrable functions) and I am not used to using $\epsilon, \delta$ in these type of problems yet. Here is my attempt. Below I want to that $f_n$ is a Cauchy sequence.
Let $\epsilon > 0$ be given.
$$
\int |f_n - f_m| = \int_{1/n}^{1/m} x^{-1/2} dx = 2 \cdot \left[\left(\frac{1}{m}\right)^{1/2} - \left(\frac{1}{n}\right)^{1/2}\right] < \epsilon $$
I do not know how to find $N$ such that $n, m \geq N$ the above happens.
Also if I just say as $n, m$ approaches infinity, then we can easily see that $\int |f_n - f_m|$ approaches zero. Is this okay enough so that we don't really have to use $\epsilon, \delta$ here?
|
Let $f_n = x^{-1/2}\chi_{[1/n, 1]}$.
Let $\epsilon >0$. Choose $N>\frac{4}{\epsilon^2}$ such that when $n,m > N$, then
$$\begin{align}\int |f_n - f_m| &= \int |x^{-1/2}\chi_{[1/n, 1]}-x^{-1/2}\chi_{[1/m, 1]}| \\
&= \int_{1/n}^{1/m} x^{-1/2} dx \\
&= 2 \cdot \left[\left(\frac{1}{m}\right)^{1/2} - \left(\frac{1}{n}\right)^{1/2}\right] \\
&< 2 \cdot \frac{1}{m}^{1/2} < 2 \frac{1}{N^{1/2}} < \epsilon
\end{align}$$
|
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|
Math induction problem with large numbers I am trying to figure out how to prove $17^{200} - 1$ is a multiple of $10$. I am talking simple algebra stuff once everything is set in place.
I have to use mathematical induction.
I figure I need to split $17^{200}$ into something like $(17^{40})^5 - 1$ and have it as $n = 17^{40}$ and $n^5 - 1$.
I just don't know if that's a good way to start.
|
Since it seems a bit strange to use induction to solve for particular case ($17^{200} - 1$), and it seems from your question that you want to see this solved via an inductive proof, let's use induction to solve a somewhat more general problem, and recover this particular example as a special case.
So, let's make the conjecture that
$$4 \mid n \implies 10 \mid (17^n - 1).$$
This means we want to show 10 divides $17^{4k} - 1$ for all integers $k$. Now, the base case is $k = 1$ and we have
$$ 17^{4k} - 1 = 17^4 - 1 = 83520 = 10 * 8352.$$
Indeed, the hypothesis holds in the base case.
Now, assume the statement is true for some integer $m$. We then have
\begin{align*}
&&17^{4m} - 1 &= 10 \cdot a &\text{for some } a \in \mathbb{Z} \\
\implies && 17^4 (17^{4m} - 1) &= 10 \cdot a \\
\implies && 17^{4m+1} - 83521 &= 10 \cdot a \\
\implies && 17^{4m+1} - 1 &= 10\cdot a + 83520 \\
\implies && 17^{4m+1} - 1 &= 10\cdot (a + 8352)
\end{align*}
Thus, 10 divides $17^{4m+1}$ if it divides $17^{4m}$; hence, we have shown that 10 divides $17^{4k}$ for every integer $k$.
Since 200 is a multiple of 4, the problem at hand is then solved as a special case of this theorem.
|
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If $\lambda$ is an eigenvalue of $A$, determine eigenvalues of $A^2$ and $A^3$ Also what is an eigenvalue of $A^n$?
I know you can square the eigenvalues, etc... but how do I prove this for a general case?
|
If $p(x)$ is a rational function defined on the spectrum of $A$ then the spectrum $$\sigma(p(A))=p(\sigma(A))$$
The spectrum being the set with all the eigenvalues or in a more precise way $\sigma(A)$={ $\lambda I-A$ is not invertible}.
Such definition is because this theorem comes from operator theory where in principle $A$ doesn't have to be a matrix, but in you case you can take the ordinary definition of spectrum being all the eigen values of $A$. So in your special case: $\sigma(A^m)=(\sigma(A))^m$ where the last is an abuse of notation to indicate $\sigma(A)$= {$\lambda^n$ : $\lambda$ eigenvalue of $A$ }
If you'd like some demonstration, it does need a little bit of effort but essentially you can work it out starting from the "Jordan Canonical Form" noticing that the matrix $J$ which represent the canonical form of $A$ is the result of $J=D+N$ in other word is the result of a $D$ diagonal matrix and a $N$ nihilpotent matrix. Then you apply the function and you will work out more or less the theorem.
|
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the first player to win three games in a row or a total of four games wins. In a competition between players X and Y, the first player to win three games in a row or a total
of four games wins.
a. How many ways can the competition be played in total?
b. How many ways can the competition be played if X wins the first game and Y wins the
second and third games?
|
Suppose four games are played and one player wins them all. WWWW. There are 2 ways that can happen. (Either one player wins all four or the other player does.
Suppose five games are played and one player winning 4 and losing one. LWWW or WLWW or WWLW. The winner can't lose the last game (because then the winner would have already won four games and they'd never play the fifth game). So there are three choices for the winner to lose. That's ${3 \choose 1}$. And then there are 2 possible winners so $2{3 \choose 1}$
Suppose six games are played, one player wins 4 and loses two. There are ${4 \choose 2}$ to do this and two players. $2{4 \choose 2}$.
Suppose seven games... winner wins 4 loses 3. $2{5 \choose 3}$.
Total; $2 + 2{3 \choose 1}+ 2{4 \choose 2} + 2{5 \choose 3}$.
|
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Correlation Coefficient I am trying to understand the following equation for Correlation Coefficient:
$r = \frac{\sum_{i=1}^{n}(x_i-\bar x)(y_i-\bar y)}{\sqrt(\sum_{i=1}^{n}(x_i - \bar x)^2\sum_{i=1}^{n}(y_i-y)^2)}$
Can someone dissect this equation and provide reasoning as to why this equation does what it does, producing $-1 \le r \le 1$ and showing the relationship between two variables? Or how this was derived?
Thank you as always.
|
Let $\mathbf{z}$ be a vector in $n$-dimensional space (think in two-dimensions, $n=2$, if that is easier). In terms of a given coordinate system with unit vectors $(\mathbf{e}_1, \mathbf{e}_2, \dots, \mathbf{e}_n$), the vector can be expressed as the sum of its components
$$
\mathbf{z} = z_1\mathbf{e}_1 + z_2\mathbf{e}_2 + \dots + z_n\mathbf{e}_n \,.
$$
The magnitude (length) of the vector is
$$
\lVert \mathbf{z} \rVert = \sqrt{\mathbf{z}\cdot\mathbf{z}} = \sqrt{z_1^2 + z_2^2 + \dots + z_n^2} \,.
$$
Therefore, the unit length vector in the direction of $\mathbf{z}$ is
$$
\hat{\mathbf{z}} = \frac{\mathbf{z}}{\lVert \mathbf{z} \rVert}
$$
Now consider two such vectors $\mathbf{v}$ and $\mathbf{w}$. The unit vectors in the directions of these two vectors are
$$
\hat{\mathbf{v}} = \frac{\mathbf{v}}{\lVert \mathbf{v} \rVert} \quad \text{and} \quad \hat{\mathbf{w}} = \frac{\mathbf{w}}{\lVert \mathbf{w} \rVert}
$$
The inner product (dot product) of these unit vectors is
$$
\hat{\mathbf{v}}\cdot\hat{\mathbf{w}} = \frac{\mathbf{v} \cdot \mathbf{w}}{\lVert \mathbf{v} \rVert \lVert \mathbf{w} \rVert} = \frac{v_1w_1 +v_2 w_2 + \dots + v_n w_n}{\sqrt{v_1^2+v_2^2+\dots+v_n^2}\sqrt{w_1^2+w_2^2+\dots+w_n^2}} = \cos\theta
$$
where $\theta$ is the angle between these vectors. Also $\cos\theta$, by definition, must lie between $-1$ and $+1$.
In your case,
$$
\mathbf{v} = \mathbf{x} - \bar{x}\mathbf{1} \quad \text{and} \quad
\mathbf{w} = \mathbf{x} - \bar{y}\mathbf{1}
$$
where $\mathbf{1} = \mathbf{e}_1 + \mathbf{e}_2 + \dots \mathbf{e}_n$. Therefore,
$$
\hat{\mathbf{v}}\cdot\hat{\mathbf{w}} = \frac{(\mathbf{x} - \bar{x}\mathbf{1})\cdot (\mathbf{y}- \bar{y}\mathbf{1})}{\lVert (\mathbf{x} - \bar{x}\mathbf{1})\rVert \lVert \mathbf{y}- \bar{y}\mathbf{1} \rVert} := r
$$
which implies that $r$ must lie between $-1$ and $1$. The shift centers the data makes sure that none of the components of the two vectors $\mathbf{v}$ and $\mathbf{w}$ is to far from the mean.
So all the calculation does is find the projection of the $x$ values on the $y$ values in a $n$-dimensional space.
|
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$H$ is a finite abelian of order $k$, also $q \in Z^{+}$ with $(k,q)=1$. $\phi:H \rightarrow H$ by $\phi(h)=h^q,\forall h \in H$ belongs to $Aut(G)$. $H$ is a finite abelian group of order $k$, also $q \in Z^{+}$ with $(k,q)=1$. $\phi:H \rightarrow H$ by $\phi(h)=h^q,\forall h \in H$ belongs to $Aut(G)$.
What I did so far:
$\phi(h_{1}h_{2})=(h_{1}h_{2})^q=h_{1}^qh_{2}^q=\phi(h_{1})\phi(h_{2})$
If $\phi(h_{1})=\phi(h_{2})$, then $h_{1}^q=h_{2}^q$,therefore $h_{1}=h_{2}$(one to one)
$\forall h\in H$, we have $\phi(h)=h^q$ (onto)
I think I did something wrong here, since there are two conditions that I do not use:1. abelian group; 2. order $k$ and $(k,q)=1$
Does anyone could help me? Thanks!
|
The condition that $H$ is abelian is used in proving that $H$ is a homomorphism.
Since $(k,q) = 1$, $\exists x,y \in \Bbb Z$ such that $qx + ky = 1$.
If $\phi(h_{1})=\phi(h_{2})$, then $h_{1}^q=h_{2}^q$.
Since $|H| = k$, $h_1^k = h_2^k = e$.
$$h_1 = h_1^{qx+ky}=h_1^{qx}=h_2^{qx} = h_2^{qx+ky}=h_2.$$
Similarly, surjectivity can be proved.
$$\phi(h^x) = h^{qx} = h^{qx + ky} = h$$
Hence $\phi:H\to H$ is an automorphism on $H$.
|
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Multiplicity of 0 eigenvalue of directed graph Laplacian matrix I am looking for a result (if it exists) for directed graphs relating the multiplicity of 0 eigenvalues of the directed Laplacian matrix.
Consider a directed graph $\mathcal{G}=(\mathcal{V},\mathcal{E})$ and define the in-degree Laplacian as $L_{in}(\mathcal{G}) = \Delta_{in}(\mathcal{G}) - A_{in}(\mathcal{G})$, where $\Delta_{in}(\mathcal{G})$ is the diagonal in-degree matrix, and $A_{in}(\mathcal{G})$ is the adjacency matrix.
A well-known result states that $L_{in}(\mathcal{G})$ has rank $|\mathcal{V}|-1$ if and only if $\mathcal{G}$ contains a rooted out-tree.
For undirected graphs, is is also known that the rank of $L(\mathcal{G})$ is $|\mathcal{V}|-c$ where $c$ is the number of connected components.
Is there a similar result for directed graphs? It is straight-forward to show, for example, that $L_{in}(\mathcal{G})$ loses rank for every node with in-degree 0 that is, if there are $p$ nodes with in-degree 0, then $\mbox{rk}[L_{in}(\mathcal{G})] \leq n-p$. However, I am not sure if this is in fact an equality.
I hope this is clear. Thanks!
|
I'm a few years late but I think tst conjecture is correct, according to https://arxiv.org/pdf/2002.02605.pdf (which also cites other references).
They use the notion of Reach :
Definition 2.3 :
i) Let $i\in V$. The reachable set $R(i)$ consists of all $j\in V$ such that there exists a path from $i$ to $j$
ii) A reach $R$ is a maximal reachable set
so basically the number of reach is the number of (oriented) trees needed to cover the graph (notice that the number of reaches might not be the number of strongly connected components).
The paper then claim
Theorem 4.6:
Given a digraph $G$. The algebraic and geometric multiplicity of the eigenvalue $0$ of $L$ equals the number of reaches.
The paper does contain a proof (which is actually quite nice) and detailled examples
|
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Sum of all values of $b$ if the difference between the largest and smallest values of the function $f(x)=x^2-2bx+1$ in the segment $[0,1]$ is $4$ Find sum of all possible values of the parameter $b$ if the difference between the largest and smallest values of the function $f(x)=x^2-2bx+1$ in the segment $[0,1]$ is $4$.
I found that the smallest value of $f(x)=x^2-2bx+1$ is $1-b^2$
But i do not know what will be the largest value of the quadratic expression,whether it is at $x=0$ or $x=1$.
Please help me.Thanks.
|
Since $f(x) = (x-b)^2 + (1 - b^2)$, the vertex will be at $(b, 1-b^2)$.
Also $f(1) = 2-2b$ and $f(0) = 1$.
When the vertex is at or to the left of $x=0$, then $f(x)$ is increasing on $[0, 1]$. So
$$ \max_{x \in [0,1]}f(x) - \min_{x \in [0,1]}f(x) = f(1) - f(0) = 1-2b$$
and $1-2b=4$ when $b = -\frac 32$
When the vertex is at or to the right of $x=1$, then $f(x)$ is decreasing on $[0, 1]$. So
$$ \max_{x \in [0,1]}f(x) - \min_{x \in [0,1]}f(x) = f(0) - f(1) = 2b-1$$
and $2b-1=4$ when $b = \frac 52$
When the x-coordinate of the vertex is in the interval $[0,\frac 12]$ then
$\begin{align}
\max_{x \in [0,1]}f(x) - \min_{x \in [0,1]}f(x)
&= \max\{f(0),f(1)\} - (1-b^2)\\
&= f(1) - (1-b^2)\\
&= (b-1)^2
\end{align}$
This does not equal $4$ when the x-coordinate of the vertex, b, is in the interval $[0,\frac 12]$
Finally, when the x-coordinate of the vertex is in the interval $(\frac 12, 1]$ then
$\begin{align}
\max_{x \in [0,1]}f(x) - \min_{x \in [0,1]}f(x)
&= \max\{f(0),f(1)\} - (1-b^2)\\
&= f(0) - (1-b^2)\\
&= b^2
\end{align}$
This does not equal $4$ when the x-coordinate of the vertex, b, is in the interval $(\frac 12, 0]$.
So the requested sum is -$\frac 32 + \frac 52 = 1$
|
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|
Group of order $255$ is cyclic
Let $G$ a group and its order is $255$.
Prove that $G$ is cyclic.
I easily demonstrated that the group has only one $17$-Sylow subgroup $P$ that is normal in $G$ and it's cyclic since it is of a prime order. Then $G/P$ is also cyclic since a group of order $15$ is cyclic. Then $G$ can be seen as $G=P(G/P)$ since the orders are coprime and then $G$ is cyclic. Is it correct?
|
Yes I think it is correct. Because of the orders of all elements must divide to the order of the group and gcd(15,17)=1 you can prove that all the group is generated by only one element (by the same way you prove that a group of order 15 is cyclic)
|
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Series of functions convergence: $a_n \sin\left(\frac 1{tn}\right),\, t\geq 2/\pi$ Problem: If the partial sums of $\sum_{n=1}^{\infty} a_n$ are bounded, then prove that $\sum_{n=1}^{\infty} a_n \sin\left(\frac{1}{tn}\right)$ converges for $t\geq\frac{2}{\pi}$.
I'm trying to use Dirichlet's Test because the partial sums of $\sum_{n=1}^{\infty} a_n$ are bounded and $\lim \sin(\frac{1}{tn})=0$, but I am having trouble showing that $\sin\left(\frac{1}{tn}\right)$ is decreasing for $t\geq\frac{2}{\pi}$. Any help would be much appreciated.
|
Hint: Show that if $\pi/2 \ge x_1 \ge x_2 \ge \cdots \to 0,$ then $\sin x_n$ decreases to $0.$ Apply this to $x_n = 1/(tn).$
|
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Prove the uniqueness of the Markov distribution Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space. I want to prove the uniqueness of a probability distribution under which a process $(X_n)_{n\geq0}$ is a Markov chain with transition probabilities $P$ and initial distribution $\mu_0$.
As an hint I have that I have to use Dynkin's lemma. In order to do this I write the statement below:
Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space. Let $\alpha$ be the statement which I want to prove, and which should hold for some set in $\Omega$. Let $\Sigma$ a $\sigma$-algebra with a generator stable under intersections for which $\alpha$ holds. Then check that $\mathcal{D}:=\{A\in \Sigma\mid A \text{ satisfies } \alpha\}$ is a Dynkin's system.
Let now in our situation $P_{\mu_0}^1,P_{\mu_0}^2$ be two probability distributions under which the process $(X_n)_{n\geq 0}$ is a Markov chain with transition probabilities $P$ and initial distribution $\mu_0$.
We already know that $\forall x_0,...,x_n\in E$, $E$ is the state space of the chain, $P_{\mu_0}^i=\mu_0(x_0)p_{x_0x_1}\dots p_{x_{n-1}x_n}$ for $i=1,2$. I took $\Sigma$ to be the sigma algebra of the probability space where my Markov chain is defined. I have troubles to find my generating set of the sigma-algebra $\Sigma=\mathcal{F}$.
thanks for any help.
|
The phrase "the sigma algebra of the probability space where my Markov chain is defined" is pretty vague. Specifically, you need to use the $\sigma$-algebra generated by the process, i.e., ${\cal F}=\sigma(X_0,X_1,\dots).$ A generating set $\cal D$ for $\cal F$ is the collection of all sets of the form $$\bigcap_{j=0}^n\,\{\omega\in \Omega: X_j(\omega)=x_j\}.$$
As you know, both measures $P^1_{\mu_0}$ and $P^2_{\mu_0}$ assign the same probability to such sets, so you are done!
|
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Galois Field to bits, implementation is fine, but the mathematics is not. I am working on a hardware implementation of the SIMON cipher and the key expansion is based on GF(2). The original paper is here, https://eprint.iacr.org/2013/404 I have successfully created the hardware, but the fact that I cannot grasp the mathematics is bothersome.
The LFSR in the paper lists the initial condition as 00001 going right where 1 is the MSB.
$W = \left[ \begin{array}{ccccc}
0&1&0&0&0\\
0&0&1&0&0\\
1&0&0&1&0\\
0&0&0&0&1\\
1&0&0&0&0\end{array} \right]$
The matrix produces this bitstream w= 10000100101100111....
I can successfully create this bitstream in hardware, which I will detail below; however, my abstract algebra is not really good enough apparently. I even read a book, but now that I'm up on GF(2), I'm still not able to grasp how this bistream is mathematically created.
The LFSR in my implementation is initially loaded with 10000 in hardware, where 1 is the MSB. As far as the hardware, the state of my LFSR is follows where the MSB is the "bit" in the output stream follows.
count 5-bit
00 ---10000
01 ---00001
02 ---00010
03 ---00100
04 ---01001
05 ---10010
06 ---00101
07 ---01011
08 ---10110
09 ---01100
(down to 31)
The point is, I have successfully created the bitstream because I realized that the stream was just a 5-bit shift register with an XOR.
If anyone mathematician can bridge my gap between theory and reality, I would greatly appreciate it.
|
Of course, after writing this up, I came up with a resolution. Via Matlab:
W=[0 1 0 0 0; 0 0 1 0 0; 1 0 0 1 0; 0 0 0 0 1; 1 0 0 0 0 ]
A_o=[0;0;0;0;1];
for i=1:31
A_o=W*A_o;
A_o=mod(A_o,2)
end
The result is that the MSB matches the bitstream.
|
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If $a_{n+1}=\frac {a_n^2+5} {a_{n-1}}$ then $a_{n+1}=Sa_n+Ta_{n-1}$ for some $S,T\in \Bbb Z$. Question
Let $$a_{n+1}:=\frac {a_n^2+5} {a_{n-1}},\, a_0=2,a_1=3$$
Prove that there exists integers $S,T$ such that $a_{n+1}=Sa_n+Ta_{n-1}$.
Attempt
I calculated the first few values of $a_n$: $a_2=7,a_3=18, a_4=47$ so I'd have the system of diophantine equations:
$$
7=3S+2T\\
18=7S+3T\\
47=18S+7T
$$
Now: it seems that all of the $a_i$ are pairwise coprime, so these equations should always have solutions, but how could I check that the intersection of all the solutions is not $\emptyset$?
|
The result is $a_{n + 1} = 3 a_n - a_{n - 1}$.
HINT: Let $a_{k + 1} = a_k + b_k$. Then, by the original equation, we get $$a_{n - 1} + b_{n - 1} + b_n = \frac {(a_{n - 1} + b_{n - 1})^2 + 5} {a_{n - 1}} = a_{n - 1} + 2 b_{n - 1} + \frac {b_{n - 1}^2 + 5} {a_{n - 1}}.$$ Hence, $$b_n = b_{n - 1} + \frac {b_{n - 1}^2 + 5} {a_{n - 1}}.$$
|
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Combinatorics - how many possible solutions are there for: $|x_1| + x_2+x_3 = 16$ How many possible solutions are there for this equation:
$|x_1| + x_2+x_3 = 16$ ; $x_1 \in Z$ $x_2,x_3 \in N$
I know it's a simple combinatorics question but I'm still having trouble figuring it out.
Since $x_1$ is an integer, there could be infinite solutions for this equation (I know this is probably not true but that is how I see it).
I know that in the case that $x_1 = 0$ there could be ${16+2-1 \choose 2-1}$ possible solutions.
The other case is when $x_1$ is non zero, when it is positive, the total # of combinations is:
${16+3-1 \choose 3-1}$ and another case where $x_1$ is negative.
Hence: ${16+2-1 \choose 2-1} + 2 \cdot {16+3-1 \choose 3-1}$
However, I checked and in my book it says that the solution is:
${16+2-1 \choose 2-1} + 2 \cdot {15+3-1 \choose 3-1}$
I'm having trouble understanding the ${15+3-1 \choose 3-1}$ part, can anyone please explain this to me?
I also noticed that this same question was posted in this site but wasn't explained appropriately, so I posted again.
Many thanks.
|
If $x_1 = 0$ then there are $15$ pairs for $x_2, x_3$.
If $x_1 = \pm 1$ then there are $14$ pairs.
If $x_1 = \pm 2$ then there are $13$ pairs.
if $x_1 = \pm 13$ then there are $2$ pairs.
if $x_1 = \pm 14$ then there is $1$ pair.
So it's a stars and bars problem for the second part. Each bin contains at least one of the $15$ object (because you're already handled the zero case) and there are three bins. This is expressed (verbatim!) as
$${15 + 3 - 1 \choose 3-1},$$
and you get the factor of $2$ from the absolute value.
|
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$A_4$ is not the direct product of its sylow subgroups? This is an exercise in Hungerford. I've tried proving this as follows (forgetting for the moment that in $A_4$, $n_2=1$ and $n_3=4$):
By Sylow III, $n_2=1$ or $3$.
If $n_2=3$, the number of elements of order $2$ in $A_4$ is $n_2 (2-1)=3(2-1)=3.$ Then the number of elements of order not $3$ is $12-3=9$. Let $P$ be a Sylow $3$-subgroup. $|P|=3$ Number of elements we have to fill up this subgroup with exceeds the number of elements of this subgroup, hence the Sylow $3$-subgroup is not normal and $A_4$ not the direct product of its sylow subgroups.
Suppose next $n_2=1$. Again by Sylow III, $n_3=1$ or $4$. If $n_3=1$, then there are 6 elements in $A_4$, which is a contradiction. Hence $n_3 = 4$ then the Sylow $3$-subgroup is not normal and again $A_4$ is not the direct product of its Sylow subgroups.
Any corrections and/or alternative proofs are welcome.
|
There are many incorrect statements in your proof.
Anyway, this is beside the point as you can prove this statement quickly without counting Sylow subgroups, or doing any other hard analysis:
Assume $A_4$ is a direct product of its Sylow subgroups. The Sylow 2-subgroups of $A_4$ have order 4, hence are abelian. Similarly, the Sylow 3-subgroups are abelian. If $A_4$ is a direct sum of Sylow subgroups, then $A_4$ must be abelian, a contradiction.
OR
If $A_4$ is a direct product of its Sylow subgroups, then each Sylow subgroup is normal (i.e. $n_p=1$ for $p=2,3$). However, $(123)$ and $(124)$ generate distinct Sylow 3-subgroups. Hence, $A_4$ is not a direct sum of its Sylow subgroups.
|
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Few group theory questions I am trying to solve the following;
First, given $G$ is a group and $H$ a subgroup of $G$, what can we say about the relation $a \cong b$ if $b^{-1}a \in H$
I can show that it is reflexive as the identity is always in the subgroup.
if $a \cong b$ then $b^{-1}a \in H$ and so $(b^{-1}a)^{-1}=a^{-1}b \in H$ so $b \cong c$.
Now I must determine if $G$ being abelian is required for this to be transitive.
My thought would be that if $ a \cong b $ then $b^{-1}a \in H $ and if $b \cong c$ then $c^{-1}b \in H$ and by properties of subgroups so to is $c^{-1}bb^{-1}a=c^{-1}a$ which would seem to imply $a \cong c$, but I don't know if I am somehow doing something that couldn't be done if $G$ was not abelian. What do you guys think?
And second part asks to discuss the possible homomorphism (group) from $\mathbb{Z}/n\mathbb{Z}$ to $\mathbb{Z}$ and it says for $n \ge 1$, is every homomorphism the zero one. Either yes, no, or it depends on $n$.
My thoughts are that only the zero homomorphism is possible, since if $\phi$ was a homomorphism then $\phi(ab)=\phi(a)\phi(b)$ and this would quickly result in having $ab=n=0$ but $\phi(a)$ and $\phi(b)$ not being $0$.
If $n$ was prime, then I am not so sure, but wouldn't this still end up mapping many things to the identity etc?
Thank you all
|
*
*Correct, commutativity is not needed and not used in your proof.
*The group operation is rather addition.
Hint: Where can the equivalence class $[1]_{n\Bbb Z}$ be mapped by a homomorphism to $\Bbb Z$?
|
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$\lim\limits_{n\to\infty}$ $(1+ {2\over n})^n$ = $e^2$ If I set N=$2\over n$, the equation becomes $(1+ N)^{2\over N}$, which I can take the natural log of and work down until I reach $e^2$. However, my teacher wants me to use subsequences, starting with $x_n$ = $(1+ {1\over n})^n = e$, to prove this and I am stuck.
|
You could rewrite this as
$$\lim_{n \to \infty}\left(1+\frac{1}{n/2}\right)^n = \lim_{n \to \infty}\left(\left(1+\frac{1}{n/2}\right)^{n/2}\right)^2 = \lim_{n\to\infty} (x_n)^2$$
where $x_n=\left(1+\frac{1}{n/2}\right)^{n/2}$. Now as $n \to \infty$, $\frac{n}{2} \to \infty$ as well. Hence $\lim_{n\to\infty} x_n = e$. Thus, bringing the limit inside the square (which we can do since squaring is continuous):
$$\lim_{n \to \infty}\left(1+\frac{1}{n/2}\right)^n = \lim_{n \to \infty}(x_n)^2= \left(\lim_{n\to\infty} x_n \right)^2=e^2.$$
This doesn't "use subsequences," but it makes use of the sequence whose limit you already know, if that was your teacher's intention.
|
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Fake proof that there don't exist complicated numbers So there's this false proof going around that I can't seem to find now that says that complicated numbers don't exist. So let me explain what it's about (I've added some technical details of my own, but the idea is the same).
Definition A number $a \in \mathbb{N}$ is said to be complicated if it cannot be expressed with less than 40 ASCII characters.
Statement There aren't any complicated numbers.
Proof Suppose there are. Then there is one which is the first complicated number. But we can call that one "The first complicated number", which has less than 40 ASCII characters. We get a contradiction, and this proves the statement.
So, I know this is false because I can proof the opposite in a way that seems (to me) more robust, plus intuition tells me there are complicated numbers. Here's a proof that there are complicated numbers.
Proof We assume there are 255 ASCII characters. Assume you can represent all natural numbers with a string of length 40. In particular, you have a unique representation for the $255^{40} + 1$ first natural numbers. By the pigeonhole principle, since there are $255^{40}$ possible strings of length 40, there is two numbers that have the same representation, which contradicts the statement.
Can someone point out the fallacy in the first (or in the second, if there happens to be one) proof? I just can't see why it's wrong.
|
This paradox is called the Berry paradox. The problem is that you haven't been specific enough about what it means to express a number using ASCII characters. Once you fix the meaning of "express," the first proof becomes a proof by contradiction which shows that "the first complicated number" is not a well-defined expression.
This is a version of the diagonal argument powering Cantor's theorem, Russell's paradox, the unsolvability of the halting problem, the incompleteness theorem, etc.
|
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|
Prove the set $\{x \in \mathbb{R} | f(x) > 0\}$ is open Let $f: \mathbb{R} \to \mathbb{R}$ be continuous. I need to use the definition to prove that the set where $f(x) > 0$ is open. I think it's true that a set defined by $f(x)$ is closed if it contains its limit points. I'm not sure what do with the proof however.
|
Approach 1: pick a point $x_0$ in $\{x \mid f(x)>0\}$. By the definition of continuity, some small neighborhood around $x_0$, say, $(x_0-\delta,x_0+\delta)$ is mapped by $f$ to a $(f(x)/2, 3f(x)/2) \subset (0,\infty)$. Thus, $(x_0-\delta,x_0+\delta) \subset \{x \mid f(x)>0\}$.
Approach 2: equivalently we can show the complement is closed. Let $x^* \in \mathbb{R}$ and suppose $(x_n)$ is a sequence in $\{x \mid f(x) \le 0\}$ such that $x_n \to x$. By continuity, $f(x_n) \to f(x^*)$, so $f(x^*) \le 0$ and thus $x^* \in \{x \mid f(x) \le 0\}$, so the set is closed.
Approach 2, without sequential characterization: Let $x^*$ be a limit point of $\{x \mid f(x) \le 0\}$. Suppose for sake of contradiction that $f(x^*) > 0$. Then some neighborhood of $x^*$ maps to $(f(x^*)/2,3f(x^*)/2) \subset (0,\infty)$, which is a contradiction because any neighborhood of $x^*$ contains a point of $\{x \mid f(x) \le 0\}$ which could not possibly map to $(f(x^*)/2,3f(x^*)/2)$.
|
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|
Martingale related question Let $S_n = Z_1 + · · · + Z_n$ where ${Z_k}$ are i.i.d. $N (0, 1)$ variables. Find constants $c_n$ such that $M_n = e^{S_n+c_n}$ is a martingale. How can I show using martingale theory that there exists a finite random variable $M_{\infty}$ such that $M_n → M_{\infty}$ a.s.
|
$M_n$ is nonnegative. Hence $\lim M_n$ exists a.s.
From Probability with Martingales:
|
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|
Double integral over annulus
Evaluate $$\iint_D\ e^{-x^2 - y^2}\ dA,$$
where $D$ is annulus $a \le x^2 + y^2 \le b$
My understanding is it involves polar coordinates but I don't understand how to convert it.
|
Since $r^2 = x^2 + y^2$ (do you see why?), we're considering the annulus $\sqrt{a} \le r \le \sqrt{b}$. Then the integral is
$$\iint_D\ e^{-r^2}\ dA,$$
The area element is $dA = r\ dr\ d\theta$ in polar coordinates. Now what do you make the limits of integration?
|
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|
Show $f$ is constant zero function on $[a, b]$ if $\int_a^x f(t) dt = \int_x^b f(t) dt \forall x \in [a, b].$ I've started by trying to use the Fundamental Theorem of Calculus and have that
$$\int_a^x f(t) dt - \int_x^b f(t) dt = 0$$
$$= F(x) - F(a) - (F(b) - F(x))$$
$$0 = 2F(x) - F(a) - F(b)$$
but am not sure how to proceed to show that $f$ is the constant zero function.
I saw this very similar question: Suppose $f:[0,1] \Rightarrow \mathbb{R}$ is continuous and $\int_0^x f(x)dx = \int_x^1 f(x)dx$. Prove that $f(x) = 0$ for all $x$, but am not sure how to form the "one line proof" using the Fundamental Theorem of Calculus mentioned in the comments.
Thanks for any tips!
|
Just differentiate both sides to get:
$$f(x)=-f(x)$$
for all $x$, then $f(x)=0$ for all $x$.
Recall that the fundamental theorem of calculus tells you that for $f(x)$ continuous:
$$\frac{d}{dx}\int_a^x f(t) dt=f(x)$$
Then note that:
$$\int_x^b f(t) dt=-\int_b^x f(t) dt$$
|
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Describe $\text{Jac}(n\mathbb Z)$ in terms of the prime factorization of $n$ Let $I$ be an ideal of the commutative ring $R$ and define
$\mathrm{Jac}(I)$ to be the intersection of all maximal ideals of $R$ that contain $I$.
Let $n > 1$ be an integer. Describe $\text{Jac}(n\mathbb Z)$ in terms of the prime factorization of $n$. (in ring $\mathbb Z$)
I have no idea how to start it.
|
Hints :
*
*In $\mathbb Z$, an ideal $I\ne ${$0$} is a prime ideal if and only if it is a maximal ideal, if and only if $I=p\mathbb Z$ for some prime number $p$.
*An ideal $I=m\mathbb Z$ in $\mathbb Z$ contains $n\mathbb Z$, if and only if $m|n$
*Jac $n\mathbb Z=m\mathbb Z$, where $m$ is the product of the prime factors of $n$.
|
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How to take an integral using half angle trigonometric substitution. So i have this question which is asking to take the integral using a predefined trigonometric substitution which is $$u=\tan\frac{x}{2}$$
and the integral equation is $$\int\frac{\sin x\ dx}{(6\cos x-2)(3-2\sin x)}$$ How would i go on about this problem? Because to begin with i do not know how i would even use the given substitution method. Any help is appreciated thank you.
|
We have $$x=2arctan(u)$$, that gives $$dx=\frac{2}{u^2+1}du$$
Furthermore, we have $$cos(x)=\frac{1-u^2}{1+u^2}$$
and $$sin(x)=\frac{2u}{u^2+1}$$
Inserting those terms, you get a rational function of $u$. Try it from here.
|
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Why is weak topology considered weak while others are strong. I know for a banach space $\mathbf{X}$ that the norm $||\cdot||$ produces a topology for the space. I also know this is considered "strong". I also understand that the dual, $\mathbf{X}^\ast$, is the set of all continuous linear functionals. I also understand that the set of functions in $\mathbf{X}^\ast$ can be used to create a topology on $\mathbf{X}$ such that they are all continuous. This one being called "Weak topology", I don't understand this, why is it called weak? Why is it viewed as being "weaker" than the norm based one? What is the motivation behind this?
|
We speak of weak/strong topologies and also of coarse/fine topologies. Either metaphor relates to the same relationship.
A topology on a set $X$ is a subset $\tau$ of $\mathfrak{P}X$ - the power set of $X$. To qualify as a topology the collection $\tau$ must satisfy the usual axioms.
If $\sigma$ is another topology on $X$ then to say $\sigma$ is weaker than $\tau$ means exactly that, as sets
$$
\sigma \subseteq \tau.
$$
You may find it a useful exercise to explore how this relation induces a partially-ordered set structure on the topologies on $X$, and whether you can find suitable definitions of $\sigma \land \tau$ and $\sigma \lor \tau$ which make this poset into a lattice.
|
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Optimization of Rectangle A landscape architect wished to enclose a rectangular garden on one side by a brick wall costing \$50 per foot and on the other three sides by a metal fence costing \$10 per foot. If the area is 24 square feet, what are the dimensions of the garden that minimize the cost? (Let x= side with bricks; y= adjacent side)
I tried finding the length of one side as 24/x and the other as 24/y, but it
hasn't gotten me far. I also thought about finding the variables through the perimeter, but I realized all they gave was the area.
Thank you!
|
Let $C(x)$ be the the cost of covering the garden of a side $x$. Since $y=24/x$ (by area), $C(x)=50x+10x+2(10\frac{24}{x})=60x+\frac{480}{x}$. Since $0<x<240$, our restriction will be $x\in(0,240)$. To find for critical number/s,
$$C'(x)=60-\frac{480}{x^2}=0\ (or\ undefined)$$
$$x^2=\frac{480}{60}=8$$
$$\therefore x=2\sqrt{2}\ or\ -2\sqrt{2}$$but $-2\sqrt{2}\notin(0,240)$, so $x=2\sqrt{2}$. To test if it is a relative extremum:
$$C''(x)=\frac{960}{x^3}$$
$$C''(2\sqrt{2})=\frac{960}{(2\sqrt{2})^3}>0\ (rel. min.)$$
Since, there is only one relative extremum in $(0,240)$, therefore it will the absolute extremum, too.
Therefore, the $x$ (side with bricks) is $2\sqrt{2}ft.$, and the other side is $6\sqrt{2}ft.$ Sorry for a not so good solution, but hope it helps. Sorry my previous answer :(
|
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Why is this Conditional Density Function correct? This answered question shows how to solve the problem but I still do not understand how to get the conditional density function, i.e.
${"}$Let $Z=X+Y$, then the density $f_{X,Z}$ of $(X,Z)$ is defined by $f_{X,Z}(x,z)=f_X(x)f_Y(z-x)$ because $X$ and $Y$ are independent hence the conditional distribution of $X$ conditionally on $Z=z$ is proportional to $f_X(x)f_Y(z-x)$, that is,
$$
f_{X\mid Z}(x\mid z)=\frac1{c(z)}f_X(x)f_Y(z-x),\qquad
c(z)=\displaystyle\int f_X(t)f_Y(z-t)\mathrm dt."
$$
I keep getting that
$$
f_{X\mid Z}(x\mid z)=\frac1{c(z)}f_X(x)f_Z(z)=\frac1{c(z)}f_X(x)\int f_X(x)f_Y(z-x),\qquad
c(z)=\displaystyle\int f_X(x)f_Y(z-x)\mathrm dx.
$$
which just ends up being
$$
f_{X\mid Z}(x\mid z)=f_X(x).
$$
I know that I'm making a fundamental error, could someone please explain in more detail what it is?
|
The general formula for conditional density is
$$
f_{X\mid Z}(x\mid z) = {f_{X,Z}(x,z)\over f_Z(x)}.\tag1
$$
Your error is in replacing the numerator in (1) with
$$f_{X,Z}(x,z)=f_X(x)f_Z(z),
$$
which is true only if $X$ and $Z$ are independent.
What you should do is replace the numerator in (1) with
$$
f_{X,Z}(x,z)=f_X(x)f_Y(z-x)
$$
and the denominator in (1) with
$$
f_Z(z)=\int f_{X,Z}(x,z)\,\mathrm dx = \int f_X(x)f_Y(z-x)\,\mathrm dx=:c(z),
$$
and you'll get the desired conditional density function.
|
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Prove that $e^\frac{x+y}{2} < \frac{1}{2}(e^x+e^y)$ for $x\neq y$ Should I use the taylor series expansion of the exponential function?
|
In general if $f''>0$ and $x>y$ then there exist $a\in ([x+y]/2,x)$ and $b\in (y,[x+y]/2)$ and $c\in (a.b)$ such that $$[f(x)+f(y)]/2 -f([x+y]/2)=$$ $$[f(x)-f([x+y]/2)]/2-[f([x+y]/2)-f(y)]/2=$$ $$=[(x-y)f'(a)]/4-[(x-y)f'(b)]/4=(x-y)^2f''(c)/4>0.$$
|
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Finding the value of Inverse Trigonometric functions beyond their Real Domain I wanted to ask how can we calculate the values of the inverse of trigonometric functions beyond their domain of definition, for example $\arcsin{2}$ beyond its domain of $-\frac{\pi}{2}<x<\frac{\pi}{2}$.
I tried to use the Euler form but did not get much. Thanks.
|
@Neelesh Vij, he makes an interesting point. I will do my best to explain.
We have as follows:$$\sin(z)=x, z=\arcsin(x)$$$$e^{iz}=\cos(z)+i\sin(z)$$$$e^{-iz} =\cos(-z)+i\sin(-z)=\cos(z)-i\sin(z)$$And we are trying to solve for an unknown z, given x.
Made clear, the last line has the equation $e^{-iz}=\cos(z)-i\sin(z)$. This was found with trigonometric identities.
Now subtract the last two equations to get the following:$$e^{iz}-e^{-iz}=2i\sin(z)$$$$\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$$Recall from the very top that $\sin(z)=x$.$$x=\frac{e^{iz}-e^{-iz}}{2i}=\frac1{2i}e^{iz}-\frac1{2i}e^{-iz}$$Multiply both sides by $e^{iz}$ to get a solvable quadratic.$$xe^{iz}=\frac1{2i}e^{2iz}-\frac1{2i}\to0=\frac1{2i}e^{2iz}-xe^{iz}-\frac1{2i}$$We can now use quadratic formula to solve for $e^{iz}$. If you are confused, try substituting $e^{iz}=y$.$$e^{iz}=\frac{x\pm\sqrt{x^2-1}}{\frac1i}=i(x\pm\sqrt{x^2-1})$$Now, solve for z.$$iz=\ln[i(x\pm\sqrt{x^2-1})]=\ln(i)+\ln[(x\pm\sqrt{x^2-1})]=\pi(\frac12+2in)+\ln[x\pm\sqrt{x^2-1}],n=\pm(0,1,2,3,4,\ldots)$$
We get a grand and wondrous solution:$$z=\frac{\pi(\frac12+2in)+\ln[x\pm\sqrt{x^2-1}]}i,n=\pm(0,1,2,3,4,\ldots)$$So for $x=2$, we can find $z=\arcsin(2)$.$$\arcsin(2)=\frac{\pi(\frac12+2in)+\ln[2\pm\sqrt{2^2-1}]} i,n=\pm(0,1,2,3,4,\ldots)$$
Lastly, take note that $\arcsin(x)=\arccos(x)-\frac{\pi}2\pm2\pi m, m=0,1,2,3,4,\ldots$ so that we have:$$\arccos(x)=\frac{\pi(\frac12+2in)+\ln[x\pm\sqrt{x^2-1}]}i+\frac{\pi}2\pm2\pi m, m=0,1,2,3,4,\ldots,n=\pm(0,1,2,3,4,\ldots)$$
|
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Trig : With point $(X,Y)$ and a end point $(x_2,y_2)$, work out the 2 angles So imagine your arm, you have a shoulder point be $X,Y (0,0)$ you have a angles off this, then you have a elbow point $(x_1,y_1)$ and another angles which ends at your hand.
If you know the arm length (4 and 4), you know your starting point $(0,0)$ and you want to get the hand to say $(6,2)$ is there some maths to work out what the angles at the elbow and shoulder should be?
|
Draw a line from the origin to the hand, called the distance line, then you have (based on your image, I assumed $y = -2$):
$$
\begin{eqnarray}
l &=& 4 & \textrm{length of an arm segment} \\
d &=& \sqrt{x^2 + y^2} = 2\sqrt{10} & \textrm{distance from the origin to the hand} \\
\tan \theta_1 &=& \frac{y}{x} = -\frac{1}{3} & \textrm{angle between x-axis and distance line} \\
\cos \theta_2 &=& \frac{d/2}{l} = \frac{\sqrt{10}}{4} & \textrm{angle between distance line and arm} \\
\end{eqnarray}
$$
Express $\cos \theta_2$ in terms of $\tan$ and then apply the tangent sum identity to get the angle at the shoulder:
$$
\begin{eqnarray}
\tan \theta_2 &=& \frac{-\sqrt{1 - \cos^2 \theta_2}}{\cos \theta_2} &=& -\frac{\sqrt{15}}{5} & \textrm{negative in fourth quadrant} \\
\tan (\theta_1 + \theta_2) &=& \frac{\tan \theta_1 + \tan \theta_2}{1 - \tan \theta_1 \tan \theta_2} &=& -\frac{12 + 5\sqrt{15}}{21} & \textrm{tangent sum identity} \\
\theta_1 + \theta_2 &=& \arctan -\frac{12 + 5\sqrt{15}}{21} &\approx& -0.98 \, \textrm{rad}
\end{eqnarray}
$$
Apply the double angle formula to get the angles with the distance line at the shoulder and hand, then remains the angle at the elbow:
$$
\begin{eqnarray}
\cos 2\theta_2 &=& 2 \cos^2 \theta_2 - 1 &=& \frac{1}{4} & \textrm{double angle formula} \\
\theta_3 &=& \pi - \arccos \frac{1}{4} &\approx& 1.82 \, \textrm{rad} \\
\end{eqnarray}
$$
|
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Is it possible to simplify $3\sin(z)\cos^2(z)-\sin^3(z)$? I have to write $\sin(3z)$ only with $\sin$'s and $\cos$'s. By using the additions-theorem I obtained $3\sin(z)\cos^2(z)-\sin^3(z)$ and I don't know If I can continue simplifying it.
|
It's not completely clear what "simplify" means in this context -- very often, two equivalent trigonometric formulas can be equally "simple" -- but one thing you can do is replace $\cos^2(z)$ with $1-\sin^2(z)$. Then you'll have
$$3\sin(z)(1-\sin^2(z))-\sin^3(z)$$
$$3\sin(z)-3\sin^3(z)-\sin^3(z)$$
$$3\sin(z)-4\sin^3(z)$$
|
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Triangle of numbers sum What is the total of all these numbers?
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
. . . . . .
1 2 ........N
The answer should be a single binomial coefficient. It looks pretty easy, but I cannot come to a solution.
|
The sum of the $k$-th row is $\sum_{h=1}^k h=\frac{k(k+1)}{2}$.
Recalling that $\sum\limits_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6}$, we have that the sum of the numbers in the triangle is $$\sum_{k=1}^n \left(\frac{k^2}{2}+\frac{k}{2}\right)=\frac{n(n+1)(2n+1)}{12}+\frac{n(n+1)}{4}=\\=\frac{n(n+1)(2n+4)}{12}=\frac{n(n+1)(n+2)}{6}=\\=\binom{n+2}{3}$$
|
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the function of a complex variable $\int\limits_{0}^{\pi}\frac{cos^2{\varphi}}{2-sin^2{\varphi}}d\varphi$
I think
$e^{i\varphi}=z$ $\to d\varphi=\frac{dz}{iz}$
$cos\varphi=\frac{z^2+1}{2z}$
$sin\varphi=\frac{z^2-1}{2iz}$
$\oint\limits_{|z|=1}^{}\frac{(z^2+1)^2}{iz(z^4+6z^2+1)}dz$
and then get 4 roots that are not good
and it is not clear how things
maybe I made the wrong conversion
|
You see, upper limit is only $\pi$. Do substitution after doubling of the argument:
$$\int\limits_0^{\pi}\dfrac{\cos^2 \phi}{2-\sin^2\phi}d\phi=\int\limits_0^{\pi}\dfrac{2\cos^2 \phi}{2+2\cos^2\phi}d\phi=\int\limits_0^{\pi}\dfrac{1+\cos 2\phi}{3+\cos2\phi}d\phi,$$
$$\quad e^{2i\phi}=z,\quad d\phi=\dfrac{dz}{2iz},\quad \cos 2\phi = \dfrac{z^2+1}{2z},\dots $$
|
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|
Finding a vector field Give a formula $F=M(x,y)i + N(x,y)j$ for the vector field in the plane that has the property that F points toward the origin with magnitude inversely proportional to the square of the distance from (x,y) to the origin. (The field is not defined at (0,0)).
I first found the norm of the vector $|F|=k/(x^2+y^2)$ with k>0.
Then I set $F=|F|n$ with n being the direction of the vector F. I put $n=(-x)i+(-y)j$ but the solution manual set $n=(-x/\sqrt{x^2+y^2})i+(-y/\sqrt{x^2+y^2})j$ and I don't understand why.
|
$n$ should be so that $||n||= 1$
|
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|
What is a formal proof for why there are exactly $\frac{p-1}{2}$ elements that are not quadratic residues? If $p$ is prime, I was trying to understand why $Z^*_N$ has $\frac{p-1}{2}$ non-quadratic elements exactly.
The current proof that I have is sort of more of an intuitive argument (or at least it doesn't feel rigorous enough to me) but I wanted to make it more formal.
Currently this is what I have:
The square root function maps 1 element to 2 elements. i.e. given $a = x^2 \in Z^*_N $ it maps it to either $x$ or $-x$. Therefore, if each element has 2 square roots, eventually, we only need have the number of elements to have an element in the co-domain assigned (since the mapping is 1 to 2 to a finite set and both the domain and co-domaine are finite and the same size $p-1$). Since the mapping covers all its target set, then we can argue its a surjection (i.e. all elements of the co-domain are mapped) and (p-1)/2 elements are covered.
This argument feels correct, except I am not sure if there might be some subtle number theoretic thing or math detail that I am missing. Is there such a thing missing?
|
Your argument, as well as I can follow it, would seem to extend to a proof that $\mathbb{Z}_N^*$ has $\phi(N)/2$ non-residues for any $N$, not just when $N$ is prime. But that's not true in general: If $N=15$, for example, there are $2$ quadratic residues and $6$ non-residues, not $4$ of each. So what seems to be missing is some essential use of the fact that you're working with primes (beyond the fact that $\phi(p)=p-1$).
|
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|
Connecting the formula for the surface area of a sphere to differential forms I have a formula for computing the area of the surface of any object in $\Bbb R^3$, namely given some parametrization $\Phi(s,t)$, I take the cross product of the partial derivative with respect to each variable and norm it.
This looks suspiciously like a wedge product, especially when you think of the cross product as the determinant of the $3\times 3$ with the top row as the $e_i$ vectors.
How do I connect the two concepts? Concretely, I tried to integrate the surface area of a sphere using the differential 2-form $dx\wedge dy$ on the obvious parametrization and kept getting stuck, finding the area was zero.
|
Here's the calculus method:
Begin with parametrization of a sphere with radius R.
$$S=\big(R\sin(\phi)\cos(\theta),R\sin(\phi)\sin(\theta), R\cos(\phi)\big)$$
For $\theta\in[0, 2\pi], \phi\in[0, \pi]$.
The equation for the surface area $A$ of a parametric surface $S$ is:
$$A=\iint_D \big|\,S_\phi \times S_\theta\big|\,dA$$
Order doesn't matter because you are taking the magnitude.
$\big|\,S_\phi\times S_\theta\big|=R^2\sin(\phi).$ - This part sucks, but just use $\sin^2(x)+\cos^2(x)=1$ a few times and you'll get there.
$$A=R^2\int_0^\pi\int_0^{2\pi}\sin(\phi)\,d\theta\,d\phi=2\pi R^2\int_0^\pi \sin(\phi)\,d\phi=4\pi R^2.$$
|
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|
Express the following complex numbers in standard form $$\left(\frac{\sqrt 3}{2}+\frac{i}{2}\right)^{25}$$
I know that you have to put it in the form $\cos\theta+i\sin \theta$ but I'm not sure how to go about it.
|
When you want to raise a number to an integer power, the polar form is very handy. When in doubt, draw a picture-plot $\frac {\sqrt 3}2+\frac i2$. You should notice that the radius is about $1$. Do you know how to compute the radius and find it is exactly $1$? Then you should know that $(re^{i\theta})^n=r^ne^{in\theta}$ You should recognize the angle.
|
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|
Quantifier: "For all sets" I've seen the following statement a few times:
"Let $A$ be a set, then $\emptyset\subseteq A$".
Or, written 'more formally':
$$
\forall A\,\, \emptyset\subseteq A
$$
My doubt is: I've always seen the quantifier $\forall x$ to mean "for all $x$ elements of some set $S$". However, when talking about all the sets, how do we define this quantification?
|
$\forall A$ means "for all $A$" or more completely "for all objects $A$ that the theory we are working in is talking about". If the theory is set theory (say, we started off by writing down the axioms of ZFC), then our $A$ is a set; when working in other theories it may instead mean that $A$ is a natural number or that $A$ is a real number. However, this doesn't not happen by magic or even by convention, it happens by the very fact that we accept the appropriate axioms to hold for our thingies.
So by the very fact that we work with $\forall A\forall B\exists C\colon ((C\in A\leftrightarrow C\in B)\leftrightarrow A=B)$ and $\forall A\forall B\exists C\forall D\colon(D\in C\leftrightarrow (D=A\lor D=B))$ and so on as axioms, it happens that our thingies behave exactly as sets are supposed to behave.
|
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|
Find the limit of $\dfrac{e^{x^2}}{xe^x}$ Please help me find this limit. I have already tried to use L'Hopitals rule and logarithmic differentiation but it turns into a quadratic.
|
I will assume you mean to evaluate the limit as $x\to\infty$.
Note, the expression is actually $\dfrac{e^{x^2-x}}{x}$. Now, we know for large $y$, the inequality $e^y\geq 1+y$ is true.
So noting that our expression is always non-negative, and putting $y=x^2-x$ in the above inequality, $\dfrac{e^{x^2-x}}{x}\geq \dfrac{1+x^2-x}{x}=\dfrac{1}{x}+x-1$.
As $x\to\infty$, the right side goes to $\infty$, so the limit is indeed $\infty$.
|
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|
Let $f(x),g(x),h(x)$ are three differentiable functions satisfying:
Let $f(x),g(x),h(x)$ are three differentiable functions satisfying:
$\int (f(x)+g(x))dx=\frac{x^3}3+C_1, \int(f(x)-g(x))dx=x^2-\frac{x^3}{3}+C_2$, $\int\frac{f(x)}{h(x)}dx=-\frac1x+C_3$.
$(1)$ The value of $\int (f(x)+g(x)+h(x))dx$ is equal to-
$(2)$ Number of points of non differentiability of function $\phi(x)=\min\{f(x),f(x)+g(x),h(x)\}$ is equal to-
$(3)$ If number of distinct terms in the expansion of $((1+f(x))+(\frac{f(x)+g(x)}{h(x)}))^{\sum n}$, ($n\in N$) is $31$, then the value of $n$ is-
We get by differentiating first and second equations, $f(x)+g(x)=x^2\tag1$$f(x)-g(x)=2x-x^2\tag2$
So $f(x)=x$ and $g(x)=x^2-x$. And by a similar argument, $h(x)=x^3$. So answer to the first question is $\frac{x^3}{3}+\frac{x^4}{4}+C$.
For the second one $\phi(x)=\min\{x,x^2,x^3\}$ and all of these are differentiable at every point, and thus, answer to the second question is $0$.
For the third one, the expression becomes $(1+x+\frac1x)^{\sum n}$, but here is where I got stuck, so how to solve this. And are my previous answers correct?
|
For the second part, $x=-1,0,1$ are not differentiable as the limit from left and right are not equal.
For the third part, you have $x^0,x^{\pm1},x^{\pm2},...,x^{\pm15}$ so $\Sigma n=15\implies n=5$.
|
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|
group actions on spheres Let $\mathbb{Z}/2$ act on the $m$-sphere $S^m$ freely and properly discontinuously. If the action is not trivial, can we conclude that the action is homotopy equivalent to the antipodal action? That is, the following diagram
commutes up to homotopy?
Can we generalize this to $S^1$-actions on $S^{2m+1}$ and $S^3$-actions on $S^{4m+3}$? Is the uniqueness true?
|
If $\Bbb Z/2$ acts freely on $S^n$, then acting by the nontrivial element of $\Bbb Z/2$ one gets a map $f:S^n \to S^n$ which has no fixed point.
Thus, $f(x) \neq x$ for all $x \in S^n$. The homotopy $$F(x, t) = \frac{(1-t)x - tf(x)}{|(1-t)x - tf(x)|}$$
between $f$ and the antipodal map $-\text{id}$ is well-defined. Thus the map induced by that action is indeed homotopic to the antipodal map.
|
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|
Write $\frac {1}{1+z^2}$ as a power series centered at $z_0=1$ I'm trying to solve a question where I need to write $\frac {1}{1+z^2} $ as a power series centered at $z_0=1$
I'm not allowed to use taylor expansion. So my first thought was to rewrite the function in a form where I can apply the basic identity:
$$ \frac{1}{1-x} = \sum_{n=0}^\infty x^n $$
So let's rewrite the original function $f(z) = \frac {1}{1+z^2} $ and use above identity:
$$\frac {1}{1+z^2} = \frac {1}{1-(-z^2)} $$
But observe that I need to center it at $z_0=1$ so I rewrote it to:
$$\frac {1}{1-(-z^2)}=\frac {1}{2z-(-(z-1)^2)} $$
I can factor out $\frac {1}{2z}$ and use the basic identity to get :
$$\frac {1}{2z}\frac {1}{1-\frac{(z-1)^2}{2z}} = \frac {1}{2z}\sum_{n=0}^\infty \left(\frac{-(z-1)^2}{2z}\right)^n =\frac {1}{2z}\sum_{n=0}^\infty \frac{(-1)^n(z-1)^{2n}}{(2z)^n} = \sum_{n=0}^\infty \frac{(-1)^n(z-1)^{2n}}{(2z)^{n+1}}$$
The problem is if I ask wolfram alpha: power series$ \frac{1}{1+z^2} $centered at z = 1 it will give me a different answer that doesn't match the sum I found.
Can anyone help me figure this one out? What am I doing wrong?
|
$$
\begin{align}
\frac1{1+z^2}
&=\frac1{2+2w+w^2}\tag1\\[9pt]
&=\frac1{2i}\frac1{1-i+w}-\frac1{2i}\frac1{1+i+w}\tag2\\[6pt]
&=\frac{1-i}4\frac1{1+\frac{1+i}2w}+\frac{1+i}4\frac1{1+\frac{1-i}2w}\tag3\\
&=\sum_{k=0}^\infty\frac{(-1)^k}4\left[\left(\frac{1+i}2\right)^{k-1}+\left(\frac{1-i}2\right)^{k-1}\right](z-1)^k\tag4\\
&=\frac1{\sqrt2}\sum_{k=0}^\infty\left(-\frac1{\sqrt2}\right)^k\cos\left(\frac{\pi(k-1)}4\right)(z-1)^k\tag5
\end{align}
$$
Explanation:
$(1)$: substitute $z=w+1$
$(2)$: partial fractions
$(3)$: use the forms $\frac1{1+aw}$ for easier series
$(4)$: write $\frac1{1+aw}$ as geometric series and undo $(1)$
$(5)$: convert to cosines
|
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|
Kernel in Modern Algebra What is a Kernel and how can it be describe in the real world and how can it be defined well and precisely. Tried asking my professor and he just tell us it is just abstract idea.
|
The kernel of a linear map ($T$) between two vector spaces $U$ and $V$ are those elements ($v$) of $V$ such that $T(v)=0$. It is defined analogously for other structures.
You can see it as a measure of how far is the map from being one to one.
|
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|
Probability :Knock Out Tournament Of Ranked Players
Thirty-two players ranked 1 to 32 are playing in a knockout
tournament. Assume that in every match between any two players, the
better-ranked player wins, the probability that ranked 1 and ranked 2
players are winner and runner up respectively, is ?
I dont understand. Should'nt the player ranked 1 always be the winner as the better ranked guy always wins?
|
If you picture the usual tree diagram for a tournament bracket, you'll see that the 1 and 2 seeds will meet in the final if and only if they start on opposite sides of the bracket. By symmetry, it doesn't matter where you position the 1 seed (so you may as well place her say at the top left). This leaves $31$ possible starting positions for the 2 seed, $16$ of which are on the opposite side from the 1 seed. So the probability they'll meet in the final (where the 1 seed will prevail) is
$${16\over31}\approx.516129$$
|
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|
Does the sequence $f_n(x)={1+x^n\over 2+x^n}$ converge uniformly on $[0,1)$? Does the sequence of functions $$f_n(x)={1+x^n\over 2+x^n}$$
converge uniformly to ${1\over 2}$ on the interval $0\le x\lt1$?
I don't think so because $\sup\left|f_n(x)-{1\over2}\right|=1$. If the interval is $[0,a]$ where $0\lt a\lt1$, then I think this is true.
|
You're right, it is not uniformly convergent. And given $\varepsilon>0$, you can even find how badly $N(x)$ degenerates as $x$ approaches 1, where $N(x)\in\mathbb{N}$ is such that $|f_n(x)-0.5|<\varepsilon,\forall n>N(x)$. Notice that
$$
f_n(x) = 1 - \frac{1}{2+x^n}
$$
so
$$
\left|f_n(x)-\frac{1}{2}\right| = \frac{x^n}{4+2x^n}<\varepsilon
$$
If you solve for $x$ you obtain $x^n<\frac{4\varepsilon}{1-2\varepsilon}$, which gives (recall that $x<1$, so $\log_x(b)$ is decreasing)
$$
n>\log_x\frac{4\varepsilon}{1-2\varepsilon}=\frac{\ln\frac{4\varepsilon}{1-2\varepsilon}}{\ln x}
$$
So $N(x)$ blows up roughly as $1/(x-1)$ as $x$ approaches 1.
Edit: yes, if you restrict your attention to $[0,a]$ with $0<a<1$, then $N(x)$ as a maximum, so the sequence uniformly converges.
|
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|
(Perhaps) An Easy Combinatorics Problem: (Perhaps not so easy...)
Still have some difficulty with problems like this.
Suppose I have Box with various buttons inside. For the sake of example, let the box contain: $$[ 4 \textrm{ Red}, 5 \textrm{ Blue}, 7 \textrm{ Green} ]$$ How many distinct tuples of, say, 5 buttons can I draw from the box? In the coordinates $(r,b,g)$ I can enumerate them:
$$\{(4,0,1),(4,1,0),(3,2,0),(3,1,1),(3,0,2),(2,3,0),(2,2,1),(2,1,2),(2,0,3),(1,4,0),(1,3,1),(1,2,2),(1,1,3),(1,0,4),(0,5,0),(0,4,1),(0,3,2),(0,2,3),(0,1,4),(0,0,5)\}$$
And I count twenty.
How do I do this in general?
Given a box (multiset) of buttons $B$,
$$B = \{B_1,B_2,\cdots,B_n\}$$
composed of $n$ different styles with respective sizes $|B_j|$ how many distinct $n$-tuples can I get by drawing $k$ buttons from such a box?
From the Selected Answer:
In the general case with my notation, we want the coefficient of $x^k$ in the expansion
$$P(x) = \Pi_{j=1}^n \frac{1-x^{|B_j|}}{1-x} $$
for the number of distinct $k$-tuples selected from the box.
In the cases that I'm dealing with, I think I'll have a big $n$ but maybe a smallish $k$. It appears that writing out the whole inclusion-exclusion form is a bit of a bear. But I get the general way to solve this now.
Thanks!
|
If Im not wrong I see 5 different values for red (from 0 to 4), 6 different values for blue and 7 different values for green.
You want that they sum exactly 5 (or starting from 1 instead of 0, 8). This is the coefficient of $x^5$ of this generating function:
$$f(x)=(1+x+x^2+x^3+x^4)(1+x+x^2+x^3+x^4+x^5)(1+x+x^2+x^3+x^4+x^5+x^6)$$
Hacking a bit trough wolfram we have that $[x^5]f(x)=20$.
But it is a lot faster, for this example of RGB, just evaluate the number of weak compositions for $5$ in $3$ parts and subtract the impossible ones due to the constraint.
Because the unique real constraint is the limit of $4$ on Red you only must subtract one case, the $(5,0,0)$ tuple.
Then it is $\binom{7}{2}-1=20$.
|
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|
the space of lipschitz function is complete with respect to some norm Let $V$ be the space of real valued lipschitz functions over $[a,b]$,we define:
$M_f=sup_{x\neq y}\frac{|f(x)-f(y)|}{|x-y|}$
and lipschitz norm:
$||f||_{Lip}=|f(a)|+M_f$
prove that $V$ with lipschitz norm is a complete normed vector space.
it is easy to see $V$ is a vector space,but what about the completeness?
is there any hint?
thank you very much.
|
You should start from a hypothetical Cauchy sequence of functions (Cauchy in the sense of the newly defined norm). Prove that the values of that function at the left end $a$ converge as real numbers, and then prove that the functions converge pointwise at any $x\in[a,b].$ This allows you to write $f(x)$ for the pointwise limit. Now establish that the sequence actually converges to $f$ in the sense of the new norm.
Explicitly, let $\epsilon>0$ be given. Pick $N$ such that for $m,n\geq N,$ $M_{f_m-f_n}<\epsilon/4.$
For every $x,y\in[a,b]$ choose $m_{x,y}\geq N$ such that $|f_{m_{x,y}}(x)-f(x)|<\epsilon|x-y|/4$ and $|f_{m_{x,y}}(y)-f(y)|<\epsilon|x-y|/4.$ Then
$$\eqalign{
\forall x,y\in[a,b]:&|(f(x)-f_n(x))-(f(y)-f_n(y))|\\
&\leq |f(x)-f_{m_{x,y}}(x)|+|f_{m_{x,y}}(x)-f_n(x)+f_n(y)-f_{m_{x,y}}(y)|
+|f_{m_{x,y}}(y)-f(y)| \\
&\leq \epsilon|x-y|.
}$$
|
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|
Determine whether a matrix is othrogonal I need to determine if the following matrix is orthogonal
$A = \frac{1}{\sqrt{2}} \begin{pmatrix}
1 & 1 \\
0 & 0 \\
1 & -1
\end{pmatrix}$
Here is what I did:
$u \cdot v = (\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}) + (\frac{1}{\sqrt{2}} \cdot -\frac{1}{\sqrt{2}}) = 0$
$||u|| = {\sqrt{(\frac{1}{\sqrt{2}})^2 + 0 + (\frac{1}{\sqrt{2}}}})^2 = 1$
$||v|| = {\sqrt{(\frac{1}{\sqrt{2}})^2 + 0 + (-\frac{1}{\sqrt{2}}}})^2 = 1$
This should indicate that the matrix is orthogonal, however, the answer in the book says it is not orthogonal and I can't see where I went wrong
|
Your matrix satisfies $A^TA=I_2$, that is the columns are orthonormal. Such matrices are useful when you want to define the polar decomposition of a $m\times n$ matrix $M$ with $m\geq n$ and $rank(M)=n$. The decomposition is $M=US$ where $U$ is a $m\times n$ matrix with orthonormal columns and $S$ is a $n\times n$ SDP matrix. More precisely $S=\sqrt{M^TM}$ and $U=MS^{-1}$.
|
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|
Problem with $SL(2)$ isometric action on a compact homogeneous space Let $G=SL(2,\mathbb{R})$, fix any left-invariant Riemannian metric $g$ on $G$. Let $\Gamma$ be a cocompact discrete subgroup of $G$ and $X=G/\Gamma$. Because $\Gamma$ acts by isometries $g$ descends to $X$ (call it $g'$) and the action of $G$ on $X$ must preserve $g'$, since being an isometry (in the Riemannian sense) is a local fact and $X$ is locally just like $G$.
Hence, we get an isometric $G$-action on a compact Riemannian manifold $X$ which also preserves the Riemannian volume $\mu_{g'}$ on $X$. But this shouldn't be possible. The isometry group of a compact $X$ is compact, so the unitary representation of $G$ on $L^2(X,\mu_{g'})$ given by the aforementioned action factors through a compact group unitary representation and must then split into finite-dimensional components. This is a contradiction with the well-known fact that $G$ has no finite-dimensional unitary representations.
Where's the error in my reasoning?
EDIT: Just noticed my argument assumes $g$ is both left and right invariant.
|
Suppose that $G$ is endowed with a bi invariant metric, it induces on its Lie algebra $sl(2)$ a scalar product invariant by the adjoint representation. This implies that the image of $SL(2)$ by the adjoint representation is conjugated to a subgroup of $SO(3)$ and this is not true since the adjoint representation has unbounded orbits.
|
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|
How many ways to roll a die seven times How many ways to roll a die seven times and obtain a sequence of outcomes with three 1s, two 5s, and two 6s?
Ans: When I was thinking of a way to decompose the problem I first thought that if a die is getting rolled seven times then each roll is independent of the other thus: $6^7$ for all possible outcomes... then I was having trouble with how to take care of the constraint. The solution given is: $$\frac{7!}{3!2!2!}$$
The section I am covering right now is about repetition, but I do not see at al how this could be the result. I thought I am supposed to find the outcomes WITH three 1s, two 5s, and two 6s? Isn't dividing those sequemces out doing the exact opposite and now we are finding all the arrangements WITHOUT three 1s, two 5s, and two 6s?
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You want to know in how many ways you can get $1155666$ in any order.
But there are $7!$ possible permutations of these numbers, except that some of the numbers are equal, and you need to compensate for that.
Since there are two $1$'s, two $5$'s, and three $6$'s, I've counted the same permutation $3!2!2!$ times, which is why the answer is
$$\frac{7!}{3!2!2!}.$$
For instance, the following two permutations are really the same: $\color{red}{1}\color{blue}{1}\color{red}{5}\color{blue}{5}\color{red}{6}\color{blue}{6}\color{green}{6}$ and $\color{blue}{1}\color{red}{1}\color{red}{5}\color{blue}{5}\color{red}{6}\color{green}{6}\color{blue}{6}$, but both are contained in the $7!$.
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|
Why does $\sqrt{x^2}$ seem to equal $x$ and not $|x|$ when you multiply the exponents? I understand that $\sqrt{x^2} = |x|$ because the principal square root is positive.
But since $\sqrt x = x^{\frac{1}{2}}$ shouldn't $\sqrt{x^2} = (x^2)^{\frac{1}{2}} = x^{\frac{2}{2}} = x$ because of the exponents multiplying together?
Also, doesn't $(\sqrt{x})^2$ preserve the sign of $x$? But shouldn't $(\sqrt{x})^2 = (\sqrt{x})(\sqrt{x}) = \sqrt{x^2}$?
How do I reconcile all this? What rules am I not aware of?
Edit: Since someone voted to close my question, I should probably explain the difference between my question and Proving square root of a square is the same as absolute value, regardless of how much I think the difference should be obvious to anyone who reads the questions. Cole Johnson was asking if there's any way to prove that $\sqrt{x^2} = |x|$. I am not asking that; I already accept the equation as fact. I'm asking how to resolve some apparent contradictions that arise when considering square roots of squares, and how I should approach these types of problems. (Cameron, please read.)
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First of all, $\sqrt{a}\sqrt{b} = \sqrt{ab}$ only when $a$ and $b$ are non-negative. $\sqrt{a}$ and $a^{\frac{1}{2}}$ are not the same function, as Cameron Williams points out in the comments above. Secondly, we can break your square of square root up. For positive values, $\sqrt{x^2} = x$, as one would expect. However, since $\sqrt{(-x)^2} = \sqrt{x^2}$, we notice that we get a positive answer from a negative value, getting $x$ from $-x$. We can represent this as flipping the function over the $x$-axis, or in other-words throwing a negative sign out front (For example, $y=-(x^2+5x)$ flips $y=x^2+5x$ across the $x$-axis). Thus, we get
$$\sqrt{\gamma^2}=\begin{cases}\gamma & \gamma\geq 0\\-\gamma & \gamma<0,\end{cases}$$
And this is the definition of the absolute value function
|
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|
Finding the geometric series of the fraction I am confused as to how to turn a fraction into a sum using geometric series.
I have $\frac{z+2}{(z-1)(z-4)}=\frac{2}{z-4}+\frac{-1}{z-1}$
I do not know how I turn the last 2 fractions into geometric series and write them as sum. Can someone please help me?
|
You may have previously seen that
$$\sum_{n=0}^\infty z^n =\frac{1}{1-z}, \;\;|z|<1$$
Then you may use $\frac{2}{z-4}=-\frac{2}{4}\cdot\frac{1}{1-z/4}$ and $\frac{-1}{z-1}=\frac{1}{1-z}$.
|
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|
Statistics: Deriving a Joint Probability Function From a Definition of Other PDF's Here's a particular question I'm trying to understand from the lecture notes.
It says:
Assume that $Y$ denotes the number of bacteria per cubic
centimeter in a particular liquid and that $Y$ has a Poisson
distribution with parameter $x$. Further assume that $x$ varies from
location to location and has an exponential distribution with
parameter $β = 1$.
-Find $f_{X,Y}(x,y)$, the joint probability function of $X$ and $Y$.
In the lecture slides, it says:
$$f_{X,Y}(x,y) = f_{Y\mid X}(y\mid x)f_X(x)$$
Where $f_{Y\mid X}(y\mid x)$ is the PDF for the Poisson distribution, and $f_X(x)$ is the PDF for the exponential with $\beta = 1$.
I'm not sure how $f_{X,Y}(x,y) = f_{Y\mid X}(y\mid x)f_X(x)$ came to be. Is there anything for me to look for in the question that would hint to using this form?
Any help would be appreciated.
Thanks.
|
This follows from definition of conditional probability. If there are two events $A,B$, then the probability of their joint occurrence is $P(A\cap B)$, conditional probability of occurrence of $B$ given that event $A$ has taken place is written as $P(B\lvert A)$, which is defined as
\begin{equation}
P(B\lvert A)=\frac{P(A\cap B)}{P(A)}
\end{equation}
Now if you map the elements of event space to real numbers, this mapping is called random variables. Then we can define distribution of these random variables. From the distribution we can define density of random variables. Let $f_X(x)$ be density of random variable $X$, $f_Y(y)$ of r.v. $Y$ then their joint density is written as $f_{X,Y}(x,y)$. Now following the definition of conditional probability, we can write conditional distribution of $Y\lvert X$ as
\begin{equation}
f_{Y\lvert X}(y\lvert x)=\frac{f_{X,Y}(x,y)}{f_X(x)}
\end{equation}
Hope this clears the doubt. This definition is quite general, hence holds for the scenario you have mentions. In nut shell, in the lecture series the author is just using definition of conditional distribution which follows from conditional probability definition.
|
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|
$p$-adic logarithm, $|\log_p(1 + x)|_p = |x|_p$? Define the $p$-adic logarithm$$\log_p(1 + x) = \sum_{i =1}^\infty (-1)^{i-1}x^i/i.$$How do I see that if $p > 2$ and $|x|_p < 1$, then $|\log_p(1 + x)|_p = |x|_p$?
|
Here is an outline of a proof. First recall that the disc of convergence of $\log_p(1+x)$ is $D(0,1) = p\mathbb{Z}_p$. Let $v_p$ denote the $p$-adic valuation. Given $x \in p\mathbb{Z}_p$, then $v_p(x) \geq 1$.
($1$) Show that $v_p\left(\frac{x^n}{n}\right) > v_p(x)$ for all $n \geq 2$. Intuitively, this should hold because $x$ has at least $1$ factor of $p$, so $x^n$ has at least $n$ factors of $p$, while the number of factors of $p$ in $n$ grows at most logarithmically. More formally, write $n = up^k$ for some $k \in \mathbb{Z}_{\geq 0}$ and unit $u \in \mathbb{Z}_p^\times$. Then
\begin{align*}
v_p\left(\frac{x^n}{n}\right) = n v_p(x) - v_p(n) = up^k v_p(x) - k \, .
\end{align*}
($2$) Recall that $v_p(a \pm b) \geq \min\{v_p(a),v_p(b)\}$. Show if $a,b \in \mathbb{Q}_p$ and $v_p(a) \neq v_p(b)$, then $v_p(a \pm b) = \min\{v_p(a),v_p(b)\}$.
$(3)$ Consider a partial sum
$$
s_N = \sum_{n = 1}^N (-1)^{n-1} \frac{x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \pm \frac{x^N}{N} \, .
$$
Since $v_p\left(\frac{x^n}{n}\right) > v_p(x)$ for all $n \geq 2$ by $(1)$, then
$$
v_p(s_N) = \min\left\{v_p(x), v_p\left(\frac{x^2}{2}\right), \ldots, v_p\left(\frac{x^N}{N}\right)\right\} = v_p(x)
$$
by $(2)$, so $|s_N|_p = |x|_p$ for all $N$.
$(4)$ Since the absolute value $|\cdot|_p$ is continuous, then
$$
|x|_p = \lim_{N \to \infty} |s_N|_p = |\lim_{N \to \infty} s_N|_p = |\log_p(1+x)|_p
$$
as desired.
|
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|
Find the length of common chord of two intersecting circles Let us consider two intersection circles
$x^2+y^2+2g_1x+2f_1y+c_1=0$
and $x^2+y^2+2g_2x+2f_2y+c=0$.
Then equation of common chord of the above two circles is
\begin{equation}2(g_1-g_2)x+2(f_1-f_2)y+c_1-c_2=0............(1)
\end{equation}
I want to find the length of the chord (1).
Any one help me to find out the length of chord.
Thank you.
|
Equation is unnecessary. The length is given by $2\sqrt{(C_1P)^2-(C_1M)^2}$ where $C_1P$ is radius of circle 1 and $C_1M$ is length of perpendicular from $C_1$ to common chord $PQ$ can you take it from there.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Separation of variables to derive the solution $u(x, t)$ How do I solve this question, what are the steps ?
Use the method of separation of variables to derive the solution $u(x, t)$ to the equation for a vibrating string
$$\frac{\partial ^2u}{\partial \:t^2}=9\:\frac{\partial ^2u}{\partial x^2} \hspace{4em} (0 < x < 4, t > 0)$$
with fixed endpoints $u(0, t) = u(4, t) = 0$. The initial conditions are given by
$u(x, 0) =\frac{1}{2}\left(\frac{\pi \:x}{4}\right)$
$$\frac{\partial u}{\partial t}\:\left(x,\:0\right)\:=\:−\:\sin\left(\frac{\pi x}{2}\right)\:\hspace{4em} (0 ≤ x ≤ 4)$$
|
I advise you to go through the procedure outlined here (first hit on Google when you search for 'separation of variables'), which shows the method for the equation
\begin{equation}
\frac{\partial u}{\partial t} = \alpha \frac{\partial^2 u}{\partial x^2}.
\end{equation}
Adopting it to your problem shouldn't be too hard.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $(\mu \times \mu)(G) = 0$ where $\mu$ is the Lebesgue measure using Fubini
Suppose $f: [0,1] \rightarrow [0,1]$ is a Borel measurable function. Suppose $G = \{(x,y)\mid f(x) = y\} \subset [0,1] \times [0,1]$. Use Fubini to prove that
$$
(\mu \times \mu)(G) = 0.
$$
What I've tried:
$$(\mu \times \mu)(G) = \iint_{G}\ d\mu d\mu = \iint 1_G(x,y)\ d\mu d\mu.$$
I kind of get stuck here, how am i supposed to go further, is it for example smart to take the rieman integral, or can i say the following thing:
$$G \subset \{ 0\leq x \leq1, f(x)\leq y \leq f(x) \}$$ so we see that
$$\iint 1_G(x,y) d\mu d\mu \leq \iint 1_{\{ 0\leq x \leq1, f(x)\leq y \leq f(x) \}} = \int_{[0,1]} \int_{\{f(x)\}} 1_G(x,y) d\mu d\mu = 0.$$
Since $\{f(x)\}$ is a singleton...?
Kees
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Let $G_{x}=\lbrace y\in[0,1]:(x,y)\in G\rbrace$. Clearly, $\mu(G_{x})=0$ for all $x$ (even a.e. would be sufficient ). Now use the fact that
$$(\mu\times\mu)(G)=\int_{[0.1]}\mu(G_{x})d\mu(x)=0.$$
|
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|
What is the greatest $4$-digit integer for which the sum of its proper divisors is $n-1$
What is the greatest $4$-digit integer $n$ for which the sum of its proper divisors is $n-1$?
I can't think of a way to solve this without guessing and checking.
Edit: It is easy to show that all powers of $2$ are such numbers. But are they the only numbers?
|
I get $8192$. This is also $2^{13}$, which I'm sure is significant. I was lazy and programmed it. Its divisors are 1,2,4,8,16,32,64,128,256,512,1024,2048,4096.
using System;
class Program
{
static void Main()
{
int result = 9999;
for (int i = 9999; i >= 1000; i--)
{
if (SumOfDivisors(i) == i - 1)
{
result = i;
break;
}
}
Console.WriteLine(result);
Console.ReadKey();
}
//Sums the proper divisors of n
static int SumOfDivisors(int n)
{
int sum = 0;
for (int i = 1; i <= n / 2; i++)
{
if (n % i == 0) sum += i;
}
return sum;
}
}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1570953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
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