Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
|---|---|---|
Find period of the following function Find period of :
$$f(x)=| \sin(x) + \cos(x) |$$
Attempt:
Let $$f(x)=|\sin x +\cos x|\tag{I}$$
Let $t$ be the required period,
So,
$f(x)=f(x+t)$
Putting $x=0$ we get,
$f(0)=f(t)$
f(0)=1 (from 'I')
Now we have the following,
$1=| \sin t + \cos t |$
If we put $t=\pi/2$ ,that would satisfy the equation. But when I put $t=\pi/2$ in $f(x+t)$,
I do not get $f(x)$ ,that means $\pi/2$ is not the period of the function. What I have done wrong?
| Let's assume you do not know the shape of the graphs but want to find the period. We will assume that you can evaluate the functions and know the sum and difference formulas for sine and cosine.
\begin{eqnarray}
|\sin x+\cos x|&=&|\sin(x+t)+\cos(x+t)|\\
&=&|\sin x\cos t+\cos x\sin t+\cos x\cos t-\sin x\sin t|\\
&=&|\sin x\cos t+\cos x\sin t+\cos x\cos t+\sin x\sin t-2\sin x\sin t|\\
&=&|(\sin x+\cos x)(\sin t+\cos t)-2\sin x\sin t|
\end{eqnarray}
Since $\sin\pi=0$ and $\cos\pi=-1$we see that when $t=\pi$ we get
$$|(\sin x+\cos x)(\sin \pi+\cos \pi)-2\sin x\sin \pi|=|-(\sin x+\cos x)|$$
But $|-(\sin x+\cos x)|=|\sin x+\cos x|=f(x)$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate integral for $\int \sin^2 (x+\frac{\pi}{6}) dx$ Can someone walk me through how to evaluate the integral $$\int \sin^2 (x+\frac{\pi}{6}) dx?$$
I get as far as
$$\int \frac {1 - \cos(x + \frac{\pi}{6})}2dx,$$
but I am not sure how to proceed.
| Note that in the particular case of $\sin^2(x)$ and $\cos^2(x)$ there is a trick you can use.
$\begin{cases}\cos^2(x)+\sin^2(x)=1\\\cos^2(x)-\sin^2(x) = \big(\sin(x)\cos(x)\big)' \end{cases}$
Thus if we call $C=\int\cos^2$ and $S=\int\sin^2$ we get the system below
$\begin{cases}C(x)+S(x)=x+cst\\C(x)-S(x)=\sin(x)\cos(x)+cst\end{cases}$
Which easily solves to
$\begin{cases}C(x)=\frac x2+\frac 12\sin(x)\cos(x)+cst\\S(x)=\frac x2-\frac 12\sin(x)\cos(x)+cst\end{cases}$
In your problem, you just translate by $\frac\pi6$ and get $I = \frac{x+\frac\pi6}2-\frac 12\sin(x+\frac\pi6)\cos(x+\frac\pi6)+cst$
I let convince yourself it is the same answer than Atmos's ($\frac{\pi}{12}$ is absorbed by the $cst$ term, and $\sin\cos$ is transformed using the doubled angle).
| {
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"timestamp": "2023-03-29T00:00:00",
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Eigenvalue problem corresponding to a univariate differential operator Let $Af=f''(x)-xf'(x)$ and $\mu$ be the Gaussian measure, i.e, $\,d\mu(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}$. We consider the following eigenvalue problem $$Af=\lambda f,$$ where $f\in C^\infty(\mathbb{R})\cap L^p(\mu)$ for all $p>1$. Show that any solution to the above problem must be a polynomial.
Note: I know of one way (which uses Orstein-Uhlenbeck process) where this result comes as a by-product. I would like to see if it can be done directly.
| The function $f$ will be a polynomial when $\lambda\in\mathbb N$. That can be seen from the fact that if we substitute a series $y=\sum_0^\infty a_nx^n$, we get the recursions
$$\tag1
a_{2n+2}=\frac{(2n-\lambda)(2n-2-\lambda)\cdots(2-\lambda)\lambda}{(2n+2)!}\,a_0,
$$
$$\tag2
a_{2n+1}=\frac{(2n-1-\lambda)(2n-3-\lambda)\cdots(1-\lambda)}{(2n+1)!}\,a_1,
$$
that can only stop when $a_0=0$ and $\lambda\in\mathbb N$ is odd, or when $a_1=0$ and $\lambda\in\mathbb N$ is even.
If, on the other hand, $\lambda\not\in \mathbb N$, either $\lambda<0$ or $\lambda\in(2k,2k+2)$ for some $k\in \mathbb N\cap \{0\}$. Consider separately the analytic solutions $\sum_na_{2n}x^{2n}$ and $\sum_na_{2n-1}x^{2n-1}$. Then, with $r_k=(2k+2-\lambda)\cdots(2-\lambda)$, and $x>0$,
$$
\left|\sum_{n\geq k}a_{2n}x^{2n}\right|
=|r_k|\,\sum_{n\geq k}\frac{(2n-2-\lambda)\cdots(2k+4-\lambda)\,x^{2n}}{(2n)!}
\geq|r_k|\,\sum_{n\geq k}\frac{x^{2n}}{(2n)!}.
$$
So the solution behaves like an exponential, and for $p$ big enough the power will overcome the $e^{-x^2}$ from the measure, and the solution cannot be in $L^2(\mu)$. A similar argument can be applied to the other solution.
So the only eigenvalues are the positive integers, and the eigenvectors are polynomials satisfying the recursions $(1)$ or $(2)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrating $\int_0^1 \frac {\log(1-x)\log^2(1+x)}x \mathrm{d}x$
Question: Integrate$$\int\limits_0^1dx\,\frac {\log(1-x)\log^2(1+x)}x=-\frac {\pi^4}{240}$$
I'm curious as to if there is a way to integrate this. I've tried using integration by parts to get$$I=-\frac {\pi^2}6\log^22+2\int\limits_0^1dx\,\frac {\operatorname{Li}_2(x)\log(1+x)}{1+x}$$However, I'm not sure how to continue even further. The polylog in the second integrand seems a bit intimidating and I don't see how the first term even helps.
| The integral is hard to tackle directly (without using Euler sums), but there is a nice trick (which is literally the same as posed above).
Let $$I = \int_0^1 {\frac{{\ln (1 - x){{\ln }^2}(1 + x)}}{x}dx} \qquad J = \int_0^1 {\frac{{{{\ln }^2}(1 - x)\ln (1 + x)}}{x}dx} $$
We have
$$\begin{aligned} 3I + 3J + \int_0^1 {\frac{{{{\ln }^3}(1 - x)}}{x}dx} + \int_0^1 {\frac{{{{\ln }^3}(1 + x)}}{x}dx} &= \int_0^1 {\frac{{{{\ln }^3}(1 - {x^2})}}{x}dx} \\ &= \frac{1}{2}\int_0^1 {\frac{{{{\ln }^3}(1 - u)}}{u}du} \end{aligned}$$
the substitution $x^2 = u$ is used.
Hence $$\tag{1}3I + 3J + \int_0^1 {\frac{{{{\ln }^3}(1 + x)}}{x}dx} = \frac{{{\pi ^4}}}{{30}}$$
On the other hand,
$$\begin{aligned}\int_0^1 {\frac{{{{\ln }^3}(1 - x)}}{x}dx} - 3J + 3I - \int_0^1 {\frac{{{{\ln }^3}(1 + x)}}{x}dx} &= \int_0^1 {\frac{{{{\ln }^3}(\frac{{1 - x}}{{1 + x}})}}{x}dx} \\ &= \int_0^1 {\frac{{2{{\ln }^3}u}}{{(1 - u)(1 + u)}}du} \\ &= \int_0^1 {\frac{{{{\ln }^3}u}}{{1 - u}}du} + \int_0^1 {\frac{{{{\ln }^3}u}}{{1 + u}}du}
\end{aligned}$$
the substitution $u=\frac{1-x}{1+x}$ is used.
Giving $$\tag{2} - 3J + 3I - \int_0^1 {\frac{{{{\ln }^3}(1 + x)}}{x}dx} = - \frac{{7{\pi ^4}}}{{120}} $$
Adding $(1)$ and $(2)$ together gives $I=-\frac{\pi^4}{240}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why can we not find n+ 2 vectors in $R^n$ so that the dot product of any two of them is negative? Take for example 3-space, where you can arrange 4 vectors so that the dot product of any two is negative, but when you add a fifth vector there is a least one non-negative dot-product.
I have seen examples of proofs using induction but they are too advanced for my background in linear algebra (very new).
Is there a way that this can be proved using angles for the example above, and also more generally.
| It might be better to just bite the bullet and try to understand the inductive proof. It's much simpler (and, once you understand it, more intuitive) than any proof you will get based on angles. In outline, in the inductive proof, you basically show that you can fix one of the vectors $x$ and sequentially modify another $n-1$ (by adding positive multiples of $x$ and of the vectors that have already been modified) to create an orthogonal basis such that these still have negative dot products with the other unmodified original vectors. If there are still two other vectors left, each having negative dot products with an orthogonal (and hence with the related orthonormal) basis, they each have all negative coordinates when written with respect to that basis and so much have a positive dot product with each other.
The induction is used to show that you can construct the basis. Without the inductive step, you'd have a hard time showing that the original set of vectors span the space.
| {
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"url": "https://math.stackexchange.com/questions/2788391",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Determinate $\lambda\in R$ so that the following equation has 2 real,distinct solutions. Determinate $\lambda\in R$ so that the following equation has 2 real,distinct solutions. $$2x+\ln x-\lambda(x-\ln x)=0$$
I think this should be solved using Rolle property for finding intervals with solutions.So i calculated $f^|(x)=\frac{2x-\lambda x+1+\lambda}{x}$.So $x=\frac{\lambda+1}{\lambda -2}$ so $f(x)=-\lambda-1+(1-\lambda )\ln \frac{2}{\lambda-2}-2$ Here I got stuck.Any help?
| Hint: defining $$f(x)=2x+2\ln(x)-\lambda(x-\ln(x))$$ then $$f'(x)=2+\frac{2}{x}-\lambda\left(1-\frac{1}{x}\right)$$ then $$f'(x)=0$$ if $$x_E=\frac{\lambda+2}{\lambda-2}$$ and $$f''(x_E)=-\frac{(-2+\lambda)^2}{\lambda+2}$$ then must hold
$$f''(x_E)<0$$ and $$f(x_E)=(\lambda+2)\left(\ln(x_E)-1\right)>0$$
| {
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Solution to Dirichlet boundary value problem on upper halve plane using Green's function Currently I am studying for an exam about partial differential equations. While looking through some of the exercises concerning Green's function on the plane, I came across a rather impenetrable-seeming integral connected to a Dirichlet BVP. As the textbook we are using (Partial Differential Equations, Peter Olver) does not provide clear examples on this topic, I figured that the internet would be a fine next step towards a solution. I also think that some experienced mathematicians out there might enjoy an exercise, and explaining it. Let me be clear: this exercise will not be examined. The problem is as follows:
Solve for $u$ where
\begin{align}
-\Delta u (x,y)= \frac{1}{1+y}\text{ on }\{(x,y):y>0\}\text{ and }u(x,0) = 0\text{ for all }x\in \mathbb{R}.
\end{align}
Now, we are asked to use the Green's function for the upper half plane to find a solution. Using the method of images, one finds such a function quite easily:
\begin{align}
G(x,y;\theta,\eta) = \frac{1}{4\pi}\log\left(\frac{(x-\theta)^2 + (y-\eta)^2}{(x-\theta)^2+(y+\eta)^2}\right).
\end{align}
By Green's representation formula, the solution is then given by
\begin{align}u(x,y) = \frac{1}{4\pi}\int_{-\infty}^{\infty}\int_0^{\infty}\frac{1}{1+\eta}\log\left(\frac{(x-\theta)^2 + (y-\eta)^2}{(x-\theta)^2+(y+\eta)^2}\right)\,\mathrm{d}\eta\,\mathrm{d\theta}.
\end{align}
I am wondering, did the author perhaps choose this example problem because the resulting integral would be solvable? If so, could someone shed light on the problem? And if not, is there a better way to solve it?
My thanks.
| I'll assume you made a typo in the equation, since the Green's function you have is for $\nabla$, not $-\nabla$
The inhomogeneous part only depends on $y$, so you can guess a solution of the form $u(x,y) = g(y)$, where
$$ -g''(y) = \frac{1}{1+y}, \ g(0) = 0 $$
Then integrating twice gives
$$ g(y) = (1+y)\ln(1+y) - y + cy $$
where $c$ is some arbitrary constant.
Change the order of integration to.
$$ \frac{1}{4\pi}\int_0^\infty \frac{1}{1+\eta}\int_{-\infty}^\infty \ln \left(\frac{(\theta-x)^2 + (\eta-y)^2}{(\theta-x)^2 + (\eta+y)^2}\right)\ d\theta\ d\eta $$
Substitute $u = \theta-x$ and rewrite $a = |\eta-y|$, $b=|\eta+y|$. The inner integral can be evaluated as
\begin{align}
\int_{-\infty}^\infty \ln\left(\frac{u^2+a^2}{u^2+b^2}\right) du &= u\ln\left(\frac{u^2+a^2}{u^2+b^2}\right)\Bigg|_{-\infty}^\infty - \int_{-\infty}^\infty \left(\frac{2u^2}{u^2+a^2} - \frac{2u^2}{u^2+b^2}\right)du \\
&= 2\int_{-\infty}^\infty \left(\frac{a^2}{u^2+a^2} - \frac{b^2}{u^2+b^2}\right) du \\
&= 2\left(a\arctan\frac{u}{a} - b\arctan\frac{u}{b}\right)\Bigg|_{\infty}^\infty \\
&= 2\pi (a-b)
\end{align}
So we're left with
\begin{align}
\frac12 \int_0^\infty \frac{1}{1+\eta}\big(|\eta-y|-|\eta+y|\big)
d\eta
&= \int_0^y \frac{1}{1+\eta}(-\eta)d\eta + \int_y^\infty \frac{1}{1+\eta}(-y)d\eta \\
&= -\int_0^y \left(1 - \frac{1}{1+\eta}\right)d\eta - y\int_y^\infty \frac{1}{1+\eta}d\eta \\
&= -y + \ln(1+y)+y\ln(1+y) + y\lim_{b\to\infty}\ln(1+b)
\end{align}
You may notice that the last term of the integral does not converge. This is because the solution is not unique. However, if you add a second boundary condition such that $u_y(x,b) = g'(b) = 0$, then the solution is indeed
$$ u(x,y) = (1+y)\ln(1+y) - y + y\ln(1+b), \ y \in (0,b) $$
for any $b$ arbitrarily large. This matches the solution obtained through integration as above.
| {
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Find $\lim \mathbb{E} \frac{X_{n}^{2}}{\log(1+X_{n}^{2})}$ Consider $Z_{n} = \frac{X_{n}^{2}}{\log(1+X_{n}^{2})}$, in which $X_n$ denotes a Gaussian random variable with zero expected value and variance equals $1/n$. Now we want find
$\lim_{n \to \infty} \mathbb{E} \frac{X_{n}^{2}}{\log(1+X_{n}^{2})}$.
I thought about find characteristic function of $Z_{n}$ and so it will be easy to find expectation. But I guess there should be easier approach. Any ideas ?
| I think it could be cracked by real analysis method. For convenience note $X_n$ by $f_n$. Then $EX_n^2=\int f_n^2dP=1/n$, and $f_n\overset{P}{\to}0$.
For $\delta>0$, set $$M(\delta)=\sup_{|x|\leq\delta}\frac{x^2}{\log(1+x^2)}$$ $$m(\delta)=\inf_{|x|\leq\delta}\frac{x^2}{\log(1+x^2)}$$
Note that $$\lim_{\delta\to0}M(\delta)=\lim_{\delta\to0}m(\delta)=1$$
We estimate this formula separately
\begin{eqnarray}
\int\frac{f_n^2}{\log(1+f_n^2)}dP&=&\int_{\{|f_n|\leq\delta\}}\frac{f_n^2}{\log(1+f_n^2)}dP+\int_{\{|f_n|>\delta\}}\frac{f_n^2}{\log(1+f_n^2)}dP
\end{eqnarray}
Fix $\delta$, as $n\to\infty$, for the upper bound
$$\leq P\{|f_n|\leq\delta\}M(\delta)+\frac{1}{\log(1+\delta^2)}\frac{1}{n}
\to M(\delta)$$
for the lower bound
$$\geq P\{|f_n|\leq\delta\}m(\delta)
\to m(\delta)$$
Let $\delta\to 0$, so we proved
$$\lim_{n\to\infty}\mathbb{E}\frac{X_n^2}{\log(1+X_n^2)}=1.$$
By the way, actually we don't need $X_n$ to be Gaussian r.v.
| {
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Solving a 2 variable integral with a delta function How can this integrals that include the dirac delta function be solved?
$$\int_{-\infty}^{+\infty}dq\int_{-\infty}^{+\infty}dp\cdot p^n\cdot \delta(p^2+q^2-E)$$
| Note that the the integral is zero by symmetry if $n$ is odd. Assume from now on that $n\geq 0$ is even.
One idea is to use polar coordinates $$(p,q)~=~(r\cos\theta,r\sin\theta).\tag{1}$$
Then
$$I~:=~ \iint_{\mathbb{R}^2}\! \mathrm{d}p~\mathrm{d}q~p^n~\delta(p^2+q^2-E)
~=~ \int_{\mathbb{R}_+}\! \mathrm{d}r~r^{n+1}\delta(r^2-E)\int_{[0,2\pi]}\! \mathrm{d}\theta~\cos^n\theta $$
$$~=~\ldots~=~\frac{1}{2}H(E) |E|^{n/2} 2\pi \frac{(n-1)!!}{n!!}.\tag{2}$$
| {
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Permutation Problem - Seating with Empty Chairs There are 3 men and 3 women to be seated in a row of 10 chairs. In how many different ways can they be seated if one man must be seated at each end of the row?
I began by calculating $_3P_2 = 6$ for the possible combinations for the end seats. The book gives the answer as 10,080, which I obtained by multiplying the first result by $_8P_4 = 1680$. I am unsure, however, about this second part of the problem, and I really only got there after trial-and-error, so I don't really understand why it works.
My understanding of the permutation formula $_nP_k$ is that it is for selecting $k$ objects from $n$ objects when order matters. So, if I were selecting $4$ people to fill $4$ remaining seats, it would make sense to me, but I don't understand where the $4$ empty seats come into play. Shouldn't there be many more possible permutations when accounting for the different positions of the empty chairs?
| Ignore "permutation" formulas and just do this directly via rule of product. (It is afterall from the rule of product that we get the permutation formulas in the first place).
*
*Choose which man sits at the far left end (three options)
*Choose which man sits at the far right end (two remaining options)
*Choose which seat from those left available the remaining man sits (eight remaining options)
*Choose which seat from those left available the youngest woman sits (seven remaining options)
*Choose which seat from those left available the youngest remaining woman sits (six remaining options)
*Choose which seat from those left available the youngest remaining woman sits (five remaining options)
Multiplying the number of options for each step together yields the total:
$3\cdot 2\cdot 8\cdot 7\cdot 6\cdot 5 = 10080$
| {
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"timestamp": "2023-03-29T00:00:00",
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$AA^*A=A$ with eigenvalues $1$ and $0$, prove that $A$ is unitarily diagonalizable.
Let $A$ be a $2$ by $2$ complex matrix such that $AA^*A=A$, and has eigenvalues of $1$ and $0$. Prove that $A$ is unitarily diagonalizable.
Well, if a matrix is Hermitian, then it is unitary diagonalizable. It also has real eigenvalues which is another property of Hermitian matrices. Hence, I want to prove Hermitian.
I want to prove it by showing that $\langle Az , w\rangle = \langle z, Aw\rangle$ but I am unable to find a clear way to showing this given the information. Is this the right approach?
| Here's a proof that works:
Every matrix is unitarily triangularizable. So, there exists a unitary $U$ such that $A = UTU^*$, where
$$
T = \pmatrix{1&t\\0&0}
$$
If we show that $t$ is necessarily zero, then we may conclude that $A$ is unitarily diagonalizable (since $T$ would then be diagonal).
With that goal in mind, we note that
$$
AA^*A = A \implies\\
UTU^*UT^*U^*UTU^* = UTU^* \implies\\
TT^*T = T
$$
Moreover, we compute
$$
TT^*T = T \pmatrix{1&t\\ \bar t & |t|^2} = \pmatrix{1 + |t|^2 & t(1 + |t|^2)\\0 & 0}
$$
We see that $TT^*T = T$ indeed implies that $t = 0$, which means that $A$ is unitarily diagonalizable.
It is interesting to see how this proof generalizes to the $n \times n$ case:
If $A$ has eigenvalues $0$ and $1$, then it is unitarily similar to the block matrix
$$
T = \pmatrix{I & Q\\0&N}
$$
where $N$ is strictly upper triangular (it is indeed true that the upper-left entry must be $I$; we note that the block associated with the eigenvalue $1$ is an invertible matrix satisfying $M = MM^*M$). As before, we may conclude that
$$
AA^*A = A \implies TT^*T = T
$$
and finally, we compute (with block-matrix multiplication)
$$
TT^*T = \pmatrix{I & Q\\0&N} \pmatrix{I & Q\\Q^* & Q^*Q + N^*N} =
\pmatrix{I + QQ^* & (I + QQ^*)Q + QN^*N\\NQ^*&N(Q^*Q + N^*N)}
$$
noting that $QQ^* = 0 \iff \operatorname{tr}(QQ^*) = 0 \iff Q = 0$, we see that $TT^*T = T$ implies that $Q = 0$.
However, we have failed to prove that $N = 0$; we merely know that $N$ is strictly upper triangular with $N = NN^*N$. This does not imply that $N$ is zero, as seen with the counterexample
$$
N = \pmatrix{0&1\\0&0}
$$
| {
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Stochastically dominating variables with same expectation Consider the random variables $X, Y$ with distributions $F_X, F_Y$ which satisfy $F_X(t) \leq F_Y(t), \; \forall t$. Additionally, we know $X, Y \geq 0$. The aim is to show that $\mathbb{E}(X) = \mathbb{E}(Y) \Rightarrow F_X(t) = F_Y(t), \forall t$.
My attempt:
Since $X, Y$ are positive, we may write $\mathbb{E}(X) = \int_{\mathbb{R}_+}
\mathbb{P}(X \geq t) \text{d} t$, likewise for $Y$, so we obtain
$$
\mathbb{E}(X) - \mathbb{E}(Y) = \int_{\mathbb{R}_+} F_Y(t) - F_X(t) \text{d} t
$$
Now, I consider the following partition of $\mathbb{R}_+$, defining
$D = \left\{ t : F_Y(t) > F_X(t) \right\}$ and have
$$
0 = \mathbb{E}(X) - \mathbb{E}(Y) = \int_D F_Y(t) - F_X(t) \text{d}t +
\int_{D^c} F_Y(t) - F_X(t) \text{d} t =
\int_D \underbrace{F_Y(t) - F_X(t)}_{> 0} \text{d} t
$$
Clearly, from the above we must have $F_Y(t) > F_X(t)$ on a set of measure $0$, otherwise $\int_D F_Y(t) - F_X(t) dt > 0$. However, the problem asks to conclude that $F_X(t) = F_Y(t)$ everywhere. How does one go from almost everywhere to everywhere in this situation?
| $g(t)=F_Y(t)-F_X(t)$ is a non-negative measurable function whose integral over $\mathbb R^{+}$ is zero. This implies that $g(t)=0$ almost everywhere. In turn this implies that it is zero on a dense subset of $\mathbb R^{+}$. Since $g$ is right continuous it follows that $g(t)=0$ for all $t$. [Details: if $A$ has Lebesgue measure 0 then $A^{c}$ is dense because no open interval can be contained in $ A$. Given any $t$ we can choose a sequence from this dense set which decreases to $t$ and right continuity can be applied to complete the proof].
| {
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How do I solve $\int_1^{2}\frac{(\ln{x})^2}{x^3} dx$? $\int_1^{2}\frac{(\ln{x})^2}{x^3} dx$
This seemed like an integration by parts problem, so I used it and got:
$\int_1^{2}\frac{(\ln{x})^2}{x^3} dx = [-\frac{1}{2}x^{-2}(\ln{x})^2 | _{1}^{2}] - \int_1^{2}\frac{2\ln{x}}{x^3}dx$
But I can't get anywhere with the $\int_1^{2}\frac{2\ln{x}}{x^3}dx$ part, making me think I didn't choose the right approach. Am I missing something simple?
| Hint :
Put $\ln x=t$ and hence $dx=e^t dt$
The integral then changes to $$\int_{0}^{\ln 2} t^2e^{-2t}dt$$
Now apply repeated integration by parts with $u=t^2$ and $dv=e^{-2t}dt$
| {
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How to solve this nonlinear least square optimization? Problem description
Given data at many time instance $t$,
$$\min _{\alpha, \xi, \beta} \lVert y(t) - \alpha e^{\Lambda t} \beta \rVert_F$$
with $$ \lVert \alpha \rVert_F^2 = 1 $$
where $y(t) \in \mathbb{R}^{n \times M}$, $\Lambda \in \mathbb{R}^{r\times r}$ is a diagonal matrix, $\alpha \in \mathbb{R}^{n \times r}$, $\beta \in \mathbb{R}^{r \times M}$
Special case
if $\alpha = I$, the problem becomes a classical problem in variable projection for nonlinear least square, which has been studied for decades.
The benefit of $\alpha = I$ is that, the nonlinear and linear part becomes separable through pseudo-inverse.
What I currently do
I can simply (blindly) use an optimization package to generally solve this problem by do naive gradient descent, it looks like a neural network training process, without leveraging the structure of the problem.
It works for toy problem, like when n,r less than 3.
The downside is that, since it doesn't assume any structure to leverage, the computational time is costly.
What I want to know
I want to know if any one know which category does this problem belongs to, and if there is better algorithm existing to solve this.
| First, note that the problem is non-convex, and I am not aware of a convex reformulation. That is, we resort to methods which might get stuck in a non-optimal point instead of converging to the optimum, including the naive gradient descent.
The first thing I would try is alternating minimization. Note that the Frobenius norm is the Euclidean norm of the vector of stacked matrix columns. Thus, everything can be expressed in terms of vectors.
When $\alpha$ is kept constant - we get a linear least squares problem, which is easy. When $\beta$ is kept constant, we get a GTRS (Generalized Trust-Region Subproblem) with interval inequality, which also has known global solution algorithms. Thus is one way to exploit problem structure.
Next thing I would try would be a proximal variant of the above.
Good luck!
| {
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A quick way, say in a minute, to deduce whether $1037$ is a prime number So with $1037 = 17 \cdot 61$, is there a fast method to deduce that it's not a prime number?
Say $1037 = 10^3+6^2+1$. Does $a^3 + b^2 + 1$ factorize in some way?
As part of their interviews, a company is asking whether a number is prime. I have never studied number theory, and I am not aware of a strategy for this apart from polynomial factorization.
I am guessing that for a number they give you, it would have to not be prime, as the only way to see that a number is a prime by hand is to test all primes below $\sqrt{N}$.
| $u\mid v$ is read as "$u$ divides $v$" and $u\nmid v$ is read as "$u$ does not divide $v$."
Obviously $2\nmid 1037$ since it has an odd last digit.
By the "Divisibility by $3$ Rule," it follows that $3\nmid 1037$ since $3\nmid 1+0+3+7=11$.
Obviously $5\nmid 1037$ since the last digit is not $0$ nor $5$.
If $7\mid 1037$ then $7\mid 1037-7=1030$ but $1030=103\times 10$, and $103$ is prime and $7\nmid 10$.
If $11\mid 1037$ then $11\mid 1037-77=960$ but $960=96\times 10$ and $11\mid 99$ so $11\nmid 96$ since it is off by a remainder of $3$ (and obviously $11\nmid 10$).
Try for $13$ this time, and then for $17$. You will see that it works for $17$, which indicates that $17\mid 1037$ and $1037$ is not prime.
You can try this method for all numbers, and you only need to try the first prime numbers less than the square root of the number you are testing with; for example, say you don't know if $61$ is a prime number. Then, apply this method.
Now, $64=8^2$ so $\sqrt{61}$ is pretty close to $8$. And since $7^2=49<61$, then all you need to do is see if $2,3,5$ or $7$ divide $61$. If they don't, then $61$ is prime (which it is).
This method does not have an official name, so it might as well be called a "trial divisibility check" by primes. Thanks to users that commented below who corrected me as I thought this method was the following.
A well known method is the "Sieve of Eratosthenes."
Tip:
A good number of primes to remember off the top of your head is all of the primes up to $127$. This has helped me, and $127$ is a pretty special number. There are many good properties about $127$; for example, $127$ is a Mersenne prime (a prime of the form $2^p-1$ for prime $p$), and it is the $31$st prime number, $31$ also being a Mersenne prime.
If you want to remember more prime numbers after $127$, you can. (I did all the way up to $383$, so it is possible.)
Apologies if this answer is primarily opinion-based.
| {
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Discrete Maths: Ways of sitting 3 exams during a 13 week semester. I had a discrete maths test today and this question was the last one and it really threw me off.
During a 13 week semester ( no breaks in between ), a student must sit 3 exams >for a particular course.
However, he must sit them in order, i.e 1st -> 2nd -> 3rd. In other words, you must first sit the 1st exam, then the 2nd then the 3rd. so can't sit the second exam without having sat the 1st one, etc..
If he can only sit one exam per week, in how many ways can the student sit the exams.
I tried during inclusion/exclusion principle but it got messy real quick. I also thought of doing the lines and dots method but figured it's wrong since the dots are not identical.
I'm just curious how to do this.
the second part of the question was
Suppose they are not allowed to sit two exams in consecutive weeks, now in how >many ways can the student sit the exams.
I didn't even attempt the second part as I did not have time.
Any help would be greatly appreciated.
| (Edited the answer to fit all criterias.)
Choose the 3 weeks out of 13 on which the student writes the exam. On the first chosen week the first exam, the second chosen week the second exam, third week the third.
There is a known mathematical formula for choosing $k$ instances from $n$, which is calculated by
$${n\choose k} = \frac{n!}{k!(n-k)!}$$
Note that this already fits the criteria of writing the exams in order. It just selects $3$ weeks out of $13$, and on the $n^{th}$ selected week ($n \in \{1,2,3\}$) you write the $n^{th}$ exam.
In this case:
$${13\choose 3} = \frac{13!}{3!(13-3)!} = 268$$
However, if you need to have at least $1$ week off before you write the next exam, you really only have $11$ objects to order (instead of $13$):
*
*(First exam's week + a week off)
*(Second exam's week + a week off)
*(Third exam's week)
*(Other exam free weeks)$\times 8$
You need to choose $3$ places out of $11$ to fit these weeks (or double weeks) into the semester, so the answer is:
$${11\choose 3} = \frac{11!}{3!(11-3)!} = 165.$$
| {
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Given that $u$ is harmonic. Prove $\Delta v \geq 0$ where $v = |\nabla u |^2$ Let $\Omega \subset \Re^2$ be open, and let $B_r(x) \subset \Omega$ be any open disc in $\Omega$.
Assume that $u \in C^2(\Omega)$ is harmonic in $\Omega$
Let $v = |\nabla u |^2$. Prove $\Delta v \geq 0$ in $\Omega$
My Attempt:
So what I have tried is that
By integrating $v = |\nabla u |^2$ over $\Omega$ I get that
$$\int_\Omega v \, dx = \int_\Omega |\nabla u|^2 \, dx \geq \frac{1}{C}\int_\Omega |u|^2 \, dx$$
By using the Poincare Inequality.
So What I need to do is that I need to show $\int_\Omega |u|^2 \, dx$ is equal to $0$, but I don't really know how to do about this. Can I just say that this must be greater than or equal to zero?
Edit I misread the question so now I am going to attempt this again
$$\int_\Omega \Delta v \, dx = \int_\Omega div(\nabla v) \, dx = \int_{\delta\Omega} \nabla v \cdot n \, ds $$
Now what I am saying is that $\nabla v = \nabla(|\nabla u|^2) = 2\nabla u \Delta u $ but I don't think this is correct?
Assuming this is correct I get that
$$ = \int_{\delta\Omega} 2\nabla u \Delta u \, ds$$
But once again I am stuck here
| Useful general formula: If $u,v\in C^2,$ then
$$\Delta (uv) = u\Delta v + v\Delta u + 2 \langle \nabla u, \nabla v\rangle.$$
Thus if $u$ is harmonic, then $\Delta (u^2) = 2 |\nabla u|^2.$ In your problem, apply this to each $(D_k u)^2,$ and recall that if $u$ is harmonic, then so is $D_ku.$
| {
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Area in $uv$ plane corresponding to two areas in $xy$? The inequalities $1 \leq x^2 - y^2 \leq 4$ and $3 \leq xy \leq 5$ describe the area $D$. See picture below.
The excerice is to calculate the integral: $\iint_D 2(x^4 - y^4) \,dx\,dy$
I would do a vaiable substitution where $u = x^2 - y^2$ and $v = xy$ where $1 \leq u \leq 4, 3 \leq v \leq 5$, with the corresponding Jacobian $\frac{\partial(u, v)}{\partial(x, y)} = 2(x^2 + y^2) = 2u \Leftrightarrow \frac{\partial(x, y)}{\partial(u, v)} = \frac{1}{2(x^2 + y^2)}$.
Doing the integral, we get
$\int_{1}^{4} \int_{3}^{5} 2 (x^4 - y^4) \, |\frac{\partial(x, y)}{\partial(u, v)}| \, du \, dv = \int_{1}^{4} \int_{3}^{5} 2 (x^2 + y^2) (x^2 - y^2) \frac{1}{2(x^2 + y^2)} \,du\,dv = \int_{1}^{4} \int_{3}^{5} u \,du \,dv = 15$.
As one clearly sees, the inequalities corresponds to two areas in $xy$; one in the first quadrant and one in the third quadrant.
How do I know if which area I've integrated over when I do the variable substitution? Is it the one in the first or third quadrant or both? Or is it indifferent?
To me it looks like the $uv$ inequalities correspond to both areas in $xy$. Is that correct?
| To integrate we need also to express the intagrand function $2(x^4 - y^4)$ in terms of $u$ and $v$
We can proceed as follow
*
*$x^4 - y^4=(x^2-y^2)(x^2+y^2)=u\sqrt{u^2+4v^2}$
| {
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Without-loss-of-generality question Going through solutions of IMO'09. Bumped into a without-loss-of-generality assumption that I can't comprehend.
Here's the statement of the problem:
Let $a,b,c$ be positive real numbers such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=a+b+c$. Prove that
$$
\frac{1}{(2a+b+c)^2}+\frac{1}{(2b+c+a)^2}+\frac{1}{(2c+a+b)^2}\leq\frac{3}{16}.
$$
Here's how they start in the solution:
We prove the homogenized inequality
$$
\frac{(a+b+c)^2}{(2a+b+c)^2}+\frac{(a+b+c)^2}{(2b+c+a)^2}+\frac{(a+b+c)^2}{(2c+a+b)^2}\leq\frac{3}{16}(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)
$$
for all positive real numbers $a,b,c$. Without loss of generality we choose $a+b+c=1$. Thus, the problem is equivalent to prove for all $a,b,c>0$, fulfilling this condition, the inequality
$$
\frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}\leq\frac{3}{16}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right).
$$
And so on... If someone's interested in the rest, it's problem A2, solution 2 in this pdf: https://www.imo-official.org/problems/IMO2009SL.pdf
Why is there no loss of generality in such a choice of $a,b,c$?
| There is no loss of generality because the inequality
$$
\frac{(a+b+c)^2}{(2a+b+c)^2}+\frac{(a+b+c)^2}{(2b+c+a)^2}+\frac{(a+b+c)^2}{(2c+a+b)^2}\leq\frac{3}{16}(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)
$$
is homogenized. This means: if you multiply $a,b$ and $c$ with the same constant $r$, the inequality does not change. Now give yourself some $a,b$ and $c$ where you have free choice of selecting these variables. You will have the sum $s = a + b+ c$. Now multiply $a,b$ and $c$ with the same constant $r=1/s$, then in the new (multiplied) variables you have $1 = a + b+ c$ which means you can always take that choice.
| {
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Having problems with the actual proof step of epsilon-N sequence convergence proofs I'm having problems actually completing epsilon-N proofs for the convergence of sequences, namely the part where, once I've found an N, I have to work backward to derive the original statement. I almost always up ending in a situation where I simply can't reverse one of the steps, logically speaking.
For instance, what I assume is a very easy example, that $\dfrac{2n-1}{4n^2}$ converges to $0$ and thus $\left|\dfrac{2n-1}{4n^2}\right| < \varepsilon $ for all $\varepsilon > 0$.
Finding an $N$ is easy enough:
$\left|\dfrac{2n-1}{4n^2}\right| < \varepsilon $ -- Assuming $n>0$ we can drop the absolute value:
$\dfrac{2n-1}{4n^2} < \varepsilon $ -- Assuming $n>1$ we can create a smaller fraction. This was a method gone over in class:
$\dfrac{1}{4n^2} < \dfrac{2n-1}{4n^2} < \varepsilon $ -- inverting and simplifying:
$n > \dfrac{1}{2\sqrt{\varepsilon}}$
So then I choose $N = \dfrac{1}{2\sqrt{\varepsilon}}$ and reverse the steps, which is all doable and straightforward up to $\dfrac{1}{4n^2} < \varepsilon $.
But at this point I'm stuck, because creating a smaller fraction only works one way; even if $\dfrac{1}{4n^2} < \varepsilon $, that doesn't mean I can make the left side larger and have it still be true. I don't see any other way to get back to the original statement, though. It's possible/likely I made a mistake deriving some part of this, but it happens with virtually every question I try where I have to use the "create a smaller quantity" method.
Thanks in advance.
| $a_n = \frac {2n-1}{4n^2}$
If $n>1$ then $0<a_n< \frac {1}{2n}$
For any epsilon, if $N > \frac {1}{2\epsilon}$ then $n>N \implies |a_n| < \epsilon$
| {
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Does $|a_n-a_{n+1}|\to 0$ imply $(a_n)$ is Cauchy? My textbook has this problem as a kind of "concept check", where one is supposed to find a counterexample to the following statement:
A sequence of real numbers is cauchy iff.
$$ \forall \epsilon>0, \, \exists N \in \mathbb{N}, \, \forall n \geq N: |a_n-a_{n+1}|< \epsilon $$
I know that a sequence of real numbers is cauchy if
$$ \forall \epsilon>0, \, \exists N \in \mathbb{N}, \, \forall n,m \geq N: |a_n-a_m|< \epsilon $$
Finding such a counterexample is probably trivial, but I haven't been able to think of one.
| $a_n = \ln n$ will do the job.
| {
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Which manifold is formed by the set of resolutions of the identity operator into orthogonal projectors? A resolution of the identity operator $I$ on $\mathbb{R}^n$ or $\mathbb{C}^n$ is a decomposition
$$I = \sum_{i=1}^n P_i,$$
where the $\{P_i\}$ are a set of orthogonal rank-one projection operators. What is the manifold (or Lie group) structure of the set of all resolutions of the identity, or equivalently sets $P_i$?
A naive guess is to express each $P_i$ as an outer product $|\psi_i\rangle \langle \psi_i|$ and identify every resolution of the identity with an orthonormal basis $\{|\psi_i\rangle\}$ - the set of which is diffeomorphic to $\mathrm{U}(n)$ (since you can rotate any orthonormal basis into another one via an appropriate unitary operator). But I don't think is quite right, because you could multiply any basis vector by a phase factor $e^{i \theta}$ (or in the real case, $-1$) without changing the corresponding projector $P_i$, so the orthonormal basis contains redundant information/degrees of freedom. (Permuting basis elements also leaves the decomposition unchanged, but we can ignore this possibility because it can't be done continuously.) Is the answer just $\mathrm{U}(n) / \mathrm{U}(1)^{\times n}$ to remove the $n$ redundant phase factors, or something more complicated? If so, is there a simpler expression for this quotient group?
| Yes, it's just $U(n)/U(1)^n$. This is not a quotient group because $U(1)^n$ is not a normal subgroup. This manifold is known as the (complete) flag variety of $\mathbb{C}^n$, and much is known about it.
| {
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Kolmogorov like Maximal inequality with exponential expected value Let $X_i$ a countable collection of independent random variables with symmetric distribution i.e. $P(X_i\in A)=P(X_i \in -A)$ for all $i\geq 1$. If $\lambda\in\mathbb{R}$ is such that $E(e^{\lambda X_i})$ is finite for all $i$ I want to prove the following Kolmogorov like inequality
$$P\left(\max_{1\leq k\leq n} \left|\sum_{i=1}^{k}X_{k}\right|>t\right)\leq e^{-\lambda t}\prod_{i=1}^{n}E(e^{\lambda X_i })$$
Have you some hint for solution?
| Using extrapolation I am able to prove your statement but only for $t$ large relative to $\lambda^{-1}$ and with the stronger condition of $E[e^{\lambda |X_i|}] < \infty$ (maybe you were missing an absolute value). Let us denote by $S$ the random variable
$$
S = \sum_{j = 1}^\infty X_j
$$
and by $\Sigma_n$ the $\sigma$-algebra generated by the random variables $X_1,...X_n$. Using the fact that each $X_j$ is of mean $0$, the conditional expectation $E_n: L^1(\Omega) \to L^1(\Omega; \Sigma_n)$ satisfies that
$$
E_n(S) = \sum_{j = 1}^n X_j := S_n.
$$
By Doob's Theorem and Marcinkiewicz interpolation we have that
\begin{align}
P \Big( \max_{1 \leq k \leq n} |E_k(S_n)| > t \Big) & \lesssim \frac{\|S_n\|_1}{t}\\
P \Big( \max_{1 \leq k \leq n} |E_k(S_n)| > t \Big) & \lesssim \Big( \frac{p}{p-1}\Big) \frac{\|S_n\|^p_p}{t^p}.
\end{align}
Therefore
\begin{align}
(e^{t \lambda} -1) \, P \Big( \max_{1 \leq k \leq n} |E_k(S_n)| > t \Big)
& \lesssim \sum_{p=1}^\infty \frac{t^p \lambda^p}{p!} P \Big( \max_{1 \leq k \leq n} |E_k(S_n)| > t \Big)\\
& \lesssim \sum_{p=1}^\infty \frac{t^p \lambda^p}{p!} \frac{E [|S_n|^p ]}{t^p}\\
& = E \Big[ \sum_{p=1}^\infty \frac{\lambda^p |S_n|^p }{p!} \Big]\\
& = E \Big[ \mathrm{exp}\big(\lambda |S_n|\big) - 1 \Big]
\leq E \Big[ \mathrm{exp}\big(\lambda |S_n|\big) \Big].
\end{align}
But now, using independence, we obtain that
$$
E \Big[ \mathrm{exp}\big(\lambda |S_n|\big) \Big] \leq \prod_{k = 1}^n E \big[e^{\lambda |X_k|}\big].
$$
In sum, we have
$$
P \Big( \sup_{1 \leq k \leq n} \Big| \sum_{k=1}^n X_k \Big| > t \Big) \lesssim \frac{1}{e^{t \lambda} - 1} \prod_{k = 1}^n E \big[e^{\lambda |X_k|}\big].
$$
| {
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Value of Solving Unsolved / Edge-case Mathematical Problems I just came across Unsolved Problems in Group Theory, of which there are 100's of very specific, detailed problems, such as these:
...
15.68. Does there exist an infinite finitely generated 2-group (of finite exponent) all of whose proper subgroups are locally finite?
16.78. Do there exist linear non-abelian simple groups without involutions?
17.9. Is there a group containing a left Engel element whose inverse is not a left Engel element?
...
In learning about the basics of group theory, it is an interesting subject. It has applications in many fields. But the applications as far as I can tell rely on the common body of knowledge in group theory (or in a mathematical field). Basically stuff you can find in textbooks or Wikipedia.
So I was wondering if one could explain the value in answering these 100's of questions. I can see answering some questions like a very relevant theorem perhaps, but the detail of these questions is very deep and there are so many. It seems like many more questions could be proposed as well, lots of edge-cases etc.. I know I am missing a lot of context so I hope this comes across okay. I am hoping to find it interesting, right now it feels overwhelming :)
| Each of these problems comes with the name of the person who sent it. If one of them strikes you, it could be worthwhile to have a look at the kind of research this person does, either via their published papers or their personal webpage, to see what kind of mathematics they are doing and specifically what problems they are trying to solve from a broader perspective.
This can help give some context to a specific problem, but of course becomes horrifyingly time-consuming if you don't focus one one or two problems.
I think that this kind of list is made for researchers who need a break in their own work, to seek new ideas or see whether there is a question there that they happen to have the right tools to solve.
| {
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Sine of angle between direct common tangents of two circles? This is a question I found: Given two circles intersecting orthogonally having the length of common chord 24/5 units the radius of one of the circles is 3 units then what is the sine of the angle between the direct common tangents? The answer given is 4√6/25. How do I approach this problem? I know that direct common tangents meet at a point that externally divides line joining centres of circles in ratio of their radii, but I am getting negative values when I try to solve this. How do I do this?
| To make things simpler. There's this formula you can memorise or derive as you may like by similarity and Pythagoras.
Alpha{angle between DCT}= 2sin^-1[|r2-r1|/ d]
Where R1 and R2 are radii and d is distance between centres.
Here, sin a/2 comes out to be 1/5.
Thus sin a = 2 sin a/2 cos a/2
= 4✓6/25
| {
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Which of the following statements is FALSE?
Which of the following statements is FALSE? There exists an integer $x$ such that
$1.$ $x \equiv 23$ mod $1000$ and $x \equiv 45$ mod $6789$.
$2.$ $x \equiv 23$ mod $1000$ and $x \equiv 54$ mod $6789$.
$3.$ $x \equiv 32$ mod $1000$ and $x \equiv 54$ mod $9876$.
$4.$ $x \equiv 32$ mod $1000$ and $x \equiv 44$ mod $9876$.
I observe that the first and second options are correct. Because for the first option to hold we need to find integers $X$ and $Y$ for which he equation
$$6789X-1000Y=22$$ holds. By the virtue of the fact that $6789$ and $1000$ are coprime we can say by Euclid's algorithm that there exist $m$ and $n$ such that $$6789m-1000n=1.$$ Then clearly $X=22m$ and $Y=22n$ suit our purpose.
By similar reasoning it can said that option $(2)$ is also correct. I have found trouble when I was trying to determine which of the options between $(3)$ and $(4)$ is correct. Then the rest is false and we are through. But in those cases we can't use coprime argument which has been used to determine that the first two options are indeed correct.
So how do I proceed for the remaining two cases? Please help me in doing this.
Thank you very much.
EDIT $:$
I think option $(4)$ is also correct. Because $(250,2469)=1$. so there exist $m$ and $n$ such that $$250m-2469n=1$$ Multiplying both sides by $4$ we have $$1000m-9876y=4$$ Then clearly if we take $X=3m$ and $Y=3n$ then $$1000X-9876Y=12$$ which in turn implies that there exist $X$ and $Y$ such that $$1000X+32=9876Y + 44$$ holds. So there exists some $x$ such that
$$x \equiv 32\ \mathrm {mod}\ 1000\ \mathrm {and}\ x \equiv 44\ \mathrm {mod}\ 9876.$$ Which proves that $(4)$ is also a correct option.
| The first and second case clearly work out, since $6789$ and $1000$ are co prime, so one can solve the given equations simultaneously.
If $x \equiv 32 \mod 1000$ then $x$ is a multiple of $8$, since $x = 1000k + 32$ so it is a sum of multiples of $8$.
However, $9876 \equiv 4 \mod 8$, so for any $x = 9876k + 54$ , it leaves a remainder of either $2$ or $6$ when divided by $8$, so it is never a multiple of $8$, so the third case does not work out.
On the other hand, the fourth case does work out, since if $x = 1000k+32 = 9876l+44$ then one may transpose and divide by four to get $250k - 2469l = 12$, and this can be solved since $250$ and $2469$ are coprime.
Hence, $1,2,4$ are solvable while $3$ is not.
A short note
More precisely, suppose you are trying to solve two simultaneous congruences, say $x \equiv a \mod b$ and $x \equiv c \mod d$. Then, write as the conclusion, $x = a+bk = c + ld$.
Therefore, rewrite this to get $ld - bk = a-c$. That is, we want to find integers $k,l$ such that $ld - bk = a-c$.
In conclusion, the question is the following: can $a-c$ be expressed as a linear combination of $b$ and $d$? For this question, Bezout's lemma gives a very clear answer : it can, if and only if $a-c$ is a multiple of the greatest common divisor of $b$ and $d$.
Therefore, all you need to do, is to check that $a-c$ is a multiple of the greatest common divisor of $b$ and $d$!
For example, in the first two parts, $(1000,6789) = 1$, and $a-c$ is a multiple of $1$ always, so the answer is yes.
In the latter two parts, we have $(1000,9876) = 4$, so it is enough to see that $a-c$ is a multiple of $4$. In the first case, $54 - 32 = 22$ isn't a multiple of $4$, while in the second case $44-32 = 12$ is.
This short note provides a complete answer to the kind of question you have been asked. It can also be extended to three or more congruences.
| {
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How many spheres are needed to shield a point source of light?
How many spheres are needed to shield a point source of light?
I read this from a mathematical puzzle book. And book says the answer is six without explanation.
From the geometric point of view, I'm thinking of a tetrahedron where the point source is located at the center. Then we can decompose 4 sectors, thus it seems to me 4 spheres is enough to shield the light. So I'm confused with the answer given in the book. If someone understands this problem, can you explain it to me?
| The minimal number of spheres required to shield the origin is $4$.
4 spheres is sufficient.
Consider following $4$ points on unit sphere $S^2$,
$$
v_0 = \frac{1}{\sqrt{3}}( -1,-1, -1),
v_1 = \frac{1}{\sqrt{3}}( -1, 1, 1),
v_2 = \frac{1}{\sqrt{3}}( 1,-1, 1),
v_3 = \frac{1}{\sqrt{3}}( 1, 1, -1)
$$
they are forming the vertices of a regular tetrahedron. It is easy to check every point on $S^2$ is at an angular distance no more than $\cos^{-1}\frac13$ from one of these vertices.
For each $k = 0,1,2,3$, place a sphere of radius $( \frac{\sqrt{8}}{3} + \epsilon)\rho_k$ at $\rho_k v_k$. These 4 spheres will block every ray start at origin. In order for them not to overlap, a sufficient condition is $$\max\left(\frac{\rho_i}{\rho_j},\frac{\rho_j}{\rho_i}\right) > 5+2\sqrt{6} \approx 9.899$$ for every $i \ne j$. By setting $(\rho_0,\rho_1,\rho_2,\rho_3)$ to $(1,10,100,1000)$, we obtain $4$ non-overlapping spheres which completely shield the origin.
4 spheres is necessary.
Given any $3$ spheres $S_1, S_2, S_3$. Let $c_1, c_2, c_3$ be unit vectors pointing towards their centers. Given any ray pointing at direction $n$, if sphere $S_i$ block it, we have $n \cdot c_i > 0$. Given $c_1, c_2$, it is easy to find a unit vector $u$ such that $u \cdot c_1 = u \cdot c_2 = 0$, this means neither $S_1$ nor $S_2$ is blocking the two rays pointing at direction $\pm u$. Since $S_3$ can block at most one of these rays, $3$ spheres is not enough to shield the origin.
| {
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Complement of a projective two sided ideal is two sided again? If a finite dimensional algebra $A$ over a field $\mathbb{k}$ is semisimple then any two sided ideal of $A$ is generated (as a left module) by a central idempotent, so its (unique) complement is a two sided ideal again. In general, this observation is wrong as e.g., in the $\mathbb{k}$-algebra of upper triangular ($2 \times 2$)-matrices the (non-central) matrix unit $e_{2,2}$ generates a two sided ideal $Ae_{2,2}$ having a complement $Ae_{1,1}$ which is not two sided.
However, I am not aware of such an example if $A$ is a group algebra.
Question: Let $\mathbb{k}$ be a field, and let $G$ be a finite group such that the algebra $\mathbb{k} G$ is not semisimple. Let $e \in \mathbb{k} G$ be an idempotent generating a two sided ideal $\mathbb{k} G e$. Does this imply that $e$ is central?
Note that since group algebras are self-injective, any left ideal $L$ isomorphic to a two sided ideal $I$ must be equal to $I$. So if $\mathbb{k} G e$ has any two sided complement then it has to be unique, and $e$ must be central.
| If $A$ is any finite dimensional algebra and $e$ any idempotent, then $A=Ae\oplus A(1-e)$ as a direct sum of left ideals. The endomorphism ring of $A$ as a left module is $A^{op}$ acting by right multiplication, so the condition that $Ae$ is a two sided ideal is equivalent to $\text{Hom}_A\left(Ae,A(1-e)\right)=0$.
But for a finite group algebra this is equivalent to $\text{Hom}_A\left(A(1-e),Ae\right)=0$, and hence to $A(1-e)$ also being a two sided ideal, since finite group algebras are examples of symmetric algebras, for which $\text{Hom}_A(P,Q)$ is naturally dual to $\text{Hom}_A(Q,P)$ for finitely generated projective modules $P$ and $Q$.
| {
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Evaluate $\lim_{n\rightarrow\infty}\sum_{k=1}^n\arcsin(\frac k{n^2})$
Compute $$\lim_{n\to\infty}\sum_{k=1}^n\arcsin\left(\frac k{n^2}\right)$$
Hello, I'm deeply sorry but I don't know how to approach any infinite sum that involves $\arcsin$, so I couldn't do anything to this question. Any hints/tips would be appreciated. I know I have to make it somehow telescopic but I don't know how to use formulas like
$$\arcsin x-\arcsin y=\arcsin\left(x\sqrt{1-y^2}+y\sqrt{1-x^2}\right)$$
My knowledge level is 12th grade.
I tried to put it in between $$\arcsin\frac 1{n^2}< \sum_{k=1}^n\arcsin\frac k{n^2} <\arcsin\frac n{n^2}$$ so then $L=0$, but it's wrong.
| Another way to look at this is to observe that
$$\arcsin{\left ( \frac{k}{n^2} \right )} = \frac{k}{n^2} \int_0^1 \frac{du}{\sqrt{1-\frac{k^2}{n^4} u^2}} $$
Then you can reverse order of summation and integration and get that the sum equals
$$\int_0^1 du \, \frac1{n} \sum_{k=1}^n \frac{(k/n)}{\sqrt{1-\frac{k^2}{n^4} u^2}} $$
We almost have a Riemann sum, but not quite. The good news is that we can convert this to a Riemann sum by subbing $u=n v$ in the integral. The result is
$$n \int_0^{1/n} dv \, \frac1{n} \sum_{k=1}^n \frac{(k/n)}{\sqrt{1-\frac{k^2}{n^2} v^2}} $$
Now we have a Riemann sum, and as $n \to \infty$ it becomes the integral
$$\int_0^1 dx \, \frac{x}{\sqrt{1-v^2 x^2}} = \frac{1-\sqrt{1-v^2}}{v^2} $$
The limit we seek is then
$$\lim_{n \to \infty} \left (n \int_0^{1/n} dv \, \frac{1-\sqrt{1-v^2}}{v^2} \right ) = \lim_{n \to \infty} \left (n \int_0^{\arcsin{1/n}} d\theta \, \frac{\cos{\theta}}{1+\cos{\theta}} \right )$$
which is $1/2$.
| {
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Is a small linear invertible perturbation of a linear isomorphism also an isomorphism? Suppose we have a linear isomorphism $T_{0}: \mathbb{R}^{n} \mapsto \mathbb{R}^{n}$, and $T_{\epsilon}$ is a small linear perturbation of it, i.e.
$$T_{\epsilon} = T_{0} + \epsilon T_{1} + O(\epsilon^{2}),$$
where $T_{1}$ is also linear isomorphism and $0< | \epsilon | <<1$. Does it follow that $T_{\epsilon}$ is also an isomorphism? If so, I am wondering is it possible to use an implicit function theorem for this, or can some result from perturbation theory of linear operators be used? (Note that this question comes from an earlier one: How do I set up Implicit Function Theorem to verify this function is a $C^{r}$ diffeomorphism?)
Thanks!
| The answer is yes and it does not depend on $T_1$ being an isomorphism.
The reason is that the map $$\mathcal{L}(\Bbb R^n\to\Bbb R^n)\to\Bbb R: T\mapsto \det(T)$$ is continuous, and the set of isomorphisms is the preimage of $\Bbb R\setminus \{0\}$ which is an open set, hence the set of isomorphisms is open itself.
| {
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Leading terms in asymptotic expansion of modified bessel function of the first kind Show that leading and next-to-leading terms in an asymptotic expansion for large $x>0$ of the modified Bessel functions of the first kind $I_0(x)$ and $I_1(x)$ are:
$$I_0(x) \sim \frac{e^x}{\sqrt{2\pi x}}\left(1+\frac{1}{8x}\right) \ \ \ \text{ and } \ \ \ I_1(x) \sim \frac{e^x}{\sqrt{2\pi x}}\left(1-\frac{3}{8x}\right).\tag{1}$$
I am fairly sure I should use the method of steepest descent to evaluate the integral, but not sure how to proceed. Any help would be greatly appreciated.
Modified Bessel function of the first kind ($n$ is an integer):
$$I_n(x) = \frac{1}{\pi}\int_0^\pi e^{x \cos(\theta)} \cos(n\theta) d\theta.\tag{2}$$
Some helpful links:
http://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html
https://dlmf.nist.gov/10.40
| From
$$ I_n(x)
~=~\frac{1}{\pi}\int_0^{\pi} \! \mathrm{d}\theta ~\exp\left(x\cos\theta\right)\cos n\theta, \qquad n\in~\mathbb{N}_0,\tag{A} $$
we calculate
$$\begin{align}
\sqrt{x}\pi e^{-x}I_n(x)
&~~=~\sqrt{x} \int_0^{\pi} \! \mathrm{d}\theta ~\exp\left(- x(1-\cos\theta)\right)\cos n\theta \cr
&\stackrel{t=\sqrt{x}\theta}{=}~
\int_0^{\pi\sqrt{x}} \! \mathrm{d}t ~\exp\left(- x(1-\cos\frac{t}{\sqrt{x}})\right)\cos \frac{nt}{\sqrt{x}} \cr
&~~=~ \int_0^{\infty} \! \mathrm{d}t ~\exp\left(- \frac{t^2}{2} + \frac{t^4}{24 x} + O(x^{-2})\right) \left(1- \frac{(nt)^2}{2x} + O(x^{-2})\right) \cr
&~~=~ \int_0^{\infty} \! \mathrm{d}t ~\exp\left(- \frac{t^2}{2}\right) \left(1- \frac{(nt)^2}{2x}+ \frac{t^4}{24 x} + O(x^{-2})\right) \cr
&\stackrel{u=t^2/2}{=}~ \int_0^{\infty} \! \frac{\mathrm{d}u}{\sqrt{2u}} ~\exp\left(- u\right) \left(1- \frac{n^2u}{x}+ \frac{u^2}{6x} + O(x^{-2})\right) \cr
&~~=~\frac{1}{\sqrt{2}}\left( \Gamma(\frac{1}{2}) - \frac{n^2}{x}\Gamma(\frac{3}{2}) +\frac{1}{6x}\Gamma(\frac{5}{2}) + O(x^{-2})\right)\cr
&~~=~\sqrt{\frac{\pi}{2}}\left( 1+ \frac{1-4n^2}{8x} + O(x^{-2})\right), \end{align}\tag{B}$$
which agrees with OP's sought-for formulas (1).
| {
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Find all continuous functions $ f : \mathbb{R} \to \mathbb{R} $ such that $(f(x)g(x))' = f'(x)g'(x) $
Find all continuous functions $ f : \mathbb{R} \to \mathbb{R} $ such that $(f(x)g(x))' = f'(x)g'(x), f,g \neq const $
My solution: $ f'(x)g(x)+g'(x)f(x) = f'(x)g'(x) \\
g(x)+g'(x)f(x)/f'(x) = g'(x) \\
f(x)/f'(x) = (g'(x)-g(x))/g'(x) $
But how to proceed further to reduce to something concrete, I do not know
| $$f'(x)g(x)+f(x)g'(x)=f'(x)g'(x)$$
$$\frac{f(x)}{f'(x)}+\frac{g(x)}{g'(x)}=1$$
$$\frac{1}{[\ln f(x)]'}+\frac{1}{[\ln g(x)]'}=1$$
call $\ln f(x)=p(x)$ and $\ln g(x)=q(x)$
$$p'(x)=\frac{q'(x)}{q'(x)-1}=1+\frac{1}{q'(x)-1}$$
$$p(x)=x+\int\frac{1}{q'(t)-1}dt+c$$
| {
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Prove that $Z-(Y-X)=X\cup(Z-Y)$ if $X\subset Y\subset Z$
Let $X\subset Y\subset Z$. Prove that $Z-(Y-X)=X\cup(Z-Y)$.
Here, $A-B$ is the complement of $B$ in $A$.
To establish equality, I need to show that $Z-(Y-X)\subset X\cup(Z-Y)$ and $X\cup(Z-Y)\subset Z-(Y-X)$. Here's how I start the proof:
Let $\alpha\in Z-(Y-X)$. Then $\alpha\in Z$ and $\alpha\not\in Y-X$, which means $\color{red}{\alpha\not\in Y}$ and $\alpha\in X$. The fact that $\alpha\in X$ is sufficient to establish that $\alpha\in X\cup(Z-Y)$, and so $Z-(Y-X)\subset X\cup(Z-Y)$.
But $X\subset Y$, which would mean $\alpha\in Y$. This seems to contradict the highlighted portion above. What am I missing here?
I can see that equality holds by looking at the Venn diagram below. $X$ is green/innermost region, $Y$ is blue/middle, and $Z$ is red/outermost. Then $Y-X$ is the blue region alone, and so $Z-(Y-X)$ is made up of both the green and red regions. On the other hand, $Z-Y$ is the red region alone, so $X\cup(Z-Y)$ is that plus the green region. So it's (pictorially) true that $Z-(Y-X)=X\cup(Z-Y)$.
Based on this picture, it seems to be the case that $\alpha$ cannot belong to $X$. Then is the "contradiction" I point to above actually of no consequence to the proof?
| You can also use directly the De Morgan laws.
$\begin{align}
Z-(Y-X)&=Z\cap(Y-X)^\complement & \text{subtraction formula : } A-B=A\cap B^\complement\\
&=Z\cap(Y\cap X^\complement)^\complement & \text{subtraction formula} \\
&=Z\cap(Y^\complement\cup X) & \text{inversion : } (A\cap B)^\complement=(A^\complement\cup B^\complement)\\
&=(Z\cap X)\cup(Z\cap Y^\complement) & \text{distribution : }A\cap(B\cup C)=(A\cap B)\cup(A\cap C)\\
&=(Z\cap X)\cup(Z-Y) & \text {subtraction formula} \\
&=X\cup(Z-Y) & X\subset Z\implies Z\cap X=X\end{align}$
Note: we didn't use the particular condition on $Y$ being inside $Z$ and containing $X$, so the formula is a bit more general.
| {
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To what number does $ n^{-2} \times \sum_{m=1}^{n-1} n \bmod m$ converge, as $n$ gets large? I evaluated the expression $$ n^{-2} \times \sum_{m=1}^{n-1} n \bmod m$$ for "large" $n$ values ($10^3$, $10^4$, $10^5$, $...$) and it seems to converge to the number approximately $0.17753188$. I tried to search for this number on the internet, found nothing and I also tried to analyze the expression, but my mathematical knowledge seems to be too small for this problem. Does anybody have an idea what this number could be (if it has a closed form)?
| Write $n\bmod m = n -m\cdot\left\lfloor\frac{n}m\right\rfloor$ and hence your partial sums are
\begin{align}
s_n
&=
\sum_{m= 1}^{n-1}\,\frac1n - \frac{m}{n^2}\cdot\left\lfloor\frac{n}m\right\rfloor
\\&=
1 - \frac{1}{n} -\frac1{n^2} \left( \sum_{m= 1}^{n-1}\,m \cdot \left\lfloor\frac{n}m\right\rfloor \right).
\end{align}
Now, write
\begin{align}
c_n = \frac1{n^2} \left( \sum_{m= 1}^{n-1}\,m \cdot \left\lfloor\frac{n}m\right\rfloor \right)
&=
\sum_{m= 1}^{n-1}\,\left(\frac{m}n \cdot \left\lfloor\frac1{m/n}\right\rfloor \right)\cdot\frac1n
\end{align}
so that as $n\to\infty$ we have
$$\lim_{n\to\infty} c_n =\int_0^1\,x\left\lfloor 1/x\right\rfloor\,dx.$$
It follows that $\lim_{n\to\infty}s_n = 1 - \int_0^1\,x\left\lfloor 1/x\right\rfloor\,dx$, so an expression for the limit of your sum hinges on our ability to provide a closed form expression for this integral.
The change of variable $x=1/t$ yields
\begin{align}
\int_0^1\,x\left\lfloor 1/x\right\rfloor\,dx
&=
\int_1^\infty\,\frac1{t^3}\left\lfloor t\right\rfloor\,dt
\\&=
\sum_{n=1}^{\infty}\,n\,\int_n^{n+1}\,t^{-3}\,dt
\\&=
\sum_{n=1}^{\infty}\,n\,{\left[-\frac12t^{-2}\right]}_n^{n+1}
\\&=
\sum_{n=1}^{\infty}\,n\,\left(\frac1{2n^2}-\frac1{2(n+1)^2}\right)
\\&=
\frac12\,\left(\sum_{n=1}^{\infty}\,\frac1n-\frac{n}{(n+1)^2}\right).
\end{align}
Observe that
$$\frac{n}{(n+1)^2} = \frac1{n+1}-\frac1{(n+1)^2}$$
to obtain that
$$\int_0^1\,x\left\lfloor 1/x\right\rfloor\,dx
=
\frac12\,\left(\sum_{n=1}^{\infty}\,\frac1n-\frac1{n+1}+\frac1{(n+1)^2}\right).
$$
The first bit we can recognize as a telescoping series and the second bit as the infamous Basel problem, and hence
$$\int_0^1\,x\left\lfloor 1/x\right\rfloor\,dx = \frac12\,\frac{\pi^2}{6}.$$
It follows that our limit sum equals
$$1-\frac{\pi^2}{12} \simeq 0.17753296657588678.$$
| {
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Homotopy equivalence of pairs Let $ I = [0,1] \subseteq \mathbb{R}$. I want to prove that the pair $(I^n,\partial I^n)$ is homotopy equivalent to $(\mathbb{R}^n,\mathbb{R}^n\setminus\{ 0,0,...,0 \})$, but I have a problem with the definition itself.
Can someone please state the definition of homotopy equivalence of pairs? Apparently, I couldn't find it anywhere.
| Let $A\subset X$ and $B\subset Y$ be CW-pairs (or any pairs of topological spaces such that the inclusions are cofibrations, see at the end).
The map $f:X\to Y$ is a homotopy equivalence between the pairs $(X,A)$ and $(Y,B)$ if there is a map $g:Y\to X$ such that:
*
*$f(A)\subset B$
*$g(B)\subset A$
*There is a homotopy $H:X\times[0,1]\to X$ such that
*
*$H(x,0)=x$ for all $x\in X$
*$H(x,1)=g\circ f(x)$ for all $x\in X$
*$H(x,t)\in A$ for all $x\in A, t\in [0,1]$
*There is a homotopy $K:Y\times[0,1]\to Y$ such that
*
*$K(y,0)=y$ for all $y\in Y$
*$K(y,1)=f\circ g(y)$ for all $y\in Y$
*$K(y,t)\in B$ for all $y\in B, t\in [0,1]$
You can find a broader definition, where the inclusion maps $i:A\to X$ and $j:B\to Y$ are not inclusions, but only cofibrations, here as part of Proposition 1.15.
Note that not all inclusions are cofibrations, but it is the case for CW-pairs.
| {
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Show that if $p$ is a prime such that $p|(2^{64}+1)$ then $p \equiv 1 $ (mod 128) I'm not sure if I'm on the right track with this problem. So far I've said: $2^{64} = (2^{32})^2 \equiv -1$ (mod p). Then by Fermat's two square theorem $p = 2$ or $p \equiv 1$ (mod 4). We know $p \not = 2$ because $p|(2^{64}+1)$. Then $p \equiv 1$ (mod 4). From here I'm unsure on how to proceed.
| Edit: I misread and thought you wrote $p|2^{2^{64}}+1$. The result happens to hold as well, and it is actually the number Euler was trying to factor when he proved the theorem below.
Hint: You are on the right track since what you need is a generalization of Fermat's two-square theorem, due to Euler who used it to prove that not all Fermat's numbers are primes.
Theorem (Euler). For any coprime integers $a$ and $b$ with $a$ even, and any integer $n$, the prime factors of the sum $$a^{2^n}+b^{2^n}$$ will all be
congruent to $1\bmod2^{n+1}$.
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Show that there does not exist $g\in C_{\Bbb R}([-1,1])$ such that $f(0)=\langle f,g\rangle$ for every $f\in C_{\Bbb R}([-1,1])$ . Let $C_{\Bbb R}([-1,1])$ be the vector space of continuous real valued functions on the interval $[-1,1]$ with inner product given by $\langle f,g\rangle=\int _{-1}^1f(x)g(x)\,dx$
for $f,g\in C_{\Bbb R}([-1,1])$ .
Show that there does not exist $g\in C_{\Bbb R}([-1,1])$ such that $f(0)=\langle f,g\rangle$ for every $f\in C_{\Bbb R}([-1,1])$ .
Assume the contrary. Then $\exists g\in C_{\Bbb R}([-1,1])$ such that $f(0)=\langle f,g\rangle$ for every $f\in C_{\Bbb R}([-1,1])$ .
I took $f(x)=$
\begin{cases} 0 & -1\le x\le 0\\mx& 0\le x\le \frac{1}{m}\\1& \frac{1}{m} \le x\le 1\end{cases}
Then $0=f(0)=\int _{-1}^1 fg=\int _0^1 fg=\int _0^{\frac{1}{m}}fg+\int _{\frac{1}{m}}^1fg$
$\implies m|\int _0^{\frac{1}{m}}xg(x)|=|\int _{\frac{1}{m}}^1g(x)|$
Also $g$ is uniformly continuous on $[-1,1]$ .
I am not able to proceed further.
Should I use any other example for $f$?Also how to use the infinite dimensionality of $C_{\Bbb R}([-1,1])$?
NOTE: I took the above since for easy examples of $f$ like polynomials etc. I did not get anything.
| Take $f(x)=x^2g(x)$, you get $$\int_0^1(xg(x))^2dx=0.$$
The continuity and nonnegativity of the integrand imply then that $xg(x)=0$ for all $x\in [-1,1]$. Thus $g(x)=0$ for all $x\in [-1,1]\setminus\{0\}$. But, since $g$ is contiuous we conclude that $g\equiv0$.
Now, testing with the constant function $f\equiv1$ we get the contradiction $1=\int_{-1}^1g(x)dx=0$. So, no such $g$ exist.
| {
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Is it true that every compact subset of $\mathbb{R}$ (with a Euclidean metric) is homeomorphic to a closed and bounded interval? Could anyone please give me a hint about how to start to prove that or a counter-example if it's false? It sounds true, since both sets are compact, but it is hard to construct a bijection between two abstract sets.
Thank you in advance.
| Every interval $[a,b]$ for $a< b$ is homeomorphic to $[0,1]$ via $x\mapsto (x-a)/(b-a)$. And therefore any two closed, bounded and non-singleton intervals are homeomorphic.
Now not every compact subset of $\mathbb{R}$ is homeomorphic to a closed and bounded interval. The simpliest example is a subset of two points, e.g. $\{0, 1\}$.
But this is true for connected subsets, i.e. every connected and compact subset of $\mathbb{R}$ is exactly $[a,b]$ for some (possibly equal) $a,b\in\mathbb{R}$. To see that first you need to prove that connected subsets of $\mathbb{R}$ are intervals (meaning convex subsets of $\mathbb{R}$). This can be easily proved by considering non-intervals and a point "inside" that disconnects the subset. Then you note that for an interval to be compact it needs to be of the form $[a,b]$. Which again is not difficult, since it has to be bounded and there are $4$ possibilites for a bounded interval: $(a,b), [a,b), (a,b], [a,b]$.
| {
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Sum of $2008$ consecutive positive integers
The sum of $2008$ consecutive positive integers is a perfect square. What is the
minimum value of the largest of these integers?
I understand this means that I need the sum of numbers, where $n=2008$ and I believe that the nearest perfect square might be $1936$ or $2025$. Am I correct? Then, I can equate the sum of $n$ numbers formula to the sum and find the first number in the series, which should be the minimum value, I believe!
| The sum is:
$$\sum_{k=0}^{2007} (n+k)=x^2 \Rightarrow 2008n+1004\cdot 2007=x^2.$$
Note that $x=2k$. Then:
$$502n+251\cdot 2007=k^2 \Rightarrow 251\cdot (2n+2007)=k^2.$$
Since $251$ is a prime number, then $k=251m$. Then:
$$2n+2007=251m^2.$$
The smallest $m=3$ and $n=126$. Hence, the minimum of the largest of them is $126+2007=2133$.
| {
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Can we view a vector space as a field extention? We can view field extentions as vector spaces over a field (an idea that in my experience has not really been explained but I get it more or less).
But can any vector space over a field also be seen as an extention of that field? I think no, since vector spaces are not fields. And frankly I don't see how any $r$ dimension vector space $V$ over a field $K$ of the form $\{\alpha_1v_1+...+\alpha_rv_r\ :\alpha_i\in K, v_i\in B\}$ and $B$ is some $K$-basis of $V$ can be a field...
| A vector space over a field can be regarded as an extension of that field as long as you can equip the vector space with an appropriate concept of product that satisfies the field axioms.
For example, equipping $\mathbb{R}^2$ with the product $(a,b) \times (c,d) = (ac-bd, ad+bc)$ creates a field which is an extension of $\mathbb{R}$. Since $(0,1) \times (0,1) = (-1,0)$ we realise that we can identify $(0,1)$ with $i=\sqrt{-1}$ and we have created the field extension $\mathbb{R}(i)$ which is the field of complex numbers.
Or we can equip $\mathbb{Q}^2$ with the product $(a,b) \times (c,d) = (ac+2bd, ad+bc)$. Now $(0,1) \times (0,1) = (2,0)$ so we can identify $(0,1)$ with $\sqrt{2}$ and we have created the field extension $\mathbb{Q}(\sqrt{2})$.
Defining a concept of product that satisfies the field axioms is not always possible. For $n>2$ there is no product rule that makes $\mathbb{R}^n$ a field. It is possible to create interesting mathematical structures in this way, but they are not fields - for example, the quaternions can be defined by equipping $\mathbb{R}^4$ with a product, but this product is not commutative.
| {
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Finding the smallest n satisfying $S_n > 10$ Let $S_n = 1 + \frac 12 + \frac 13 + \cdots + \frac 1n$, where $n \in \{ 1,2,3,\cdots\}$ Find the smallest $n$ satisfying $S_n > 10$.
Sorry, it's my first time asking and I don't know how to format this thing. I still don't see anything even after staring at this for really long. Any clues?
| Should be near
$$ \lfloor{ \frac{e^9}{2} }+1 $$
I use
$$ \int_k^{k+1} \frac{1}{t} dt \leq \frac{1}{k} \leq \int_{k-1}^{k} \frac{1}{t} dt $$
You sum and integrate which give you log then you solve for upper and lower bound and verify.
| {
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Properties of a polynomial with zero discriminant In Wikipedia it says[1] that
"The discriminant of a polynomial over an integral domain is zero if and only if the polynomial and its derivative have a non-constant common divisor."
How does one prove this fact?
[1] https://en.wikipedia.org/wiki/Discriminant#Zero_discriminant
| If $P(x)$ is a polynomial with degree $n$ with coefficients in an integer domain $D$, if $K$ is the algebraic closure of the ring of fractions of $D$ and if $r_1,\ldots,r_n\in K$ are the roots of $P(x)$ (there may be repeated roots, of course), then the discriminant $\Delta$ of $P(x)$ is $\left(\prod_{k=1}^n(r_i-r_j)\right)^2$. Therefore\begin{align}\Delta=0&\iff\text{there are repeated roots}\\&\iff\text{there are roots with degree >1}\\&\iff P(x)\text{ and }P'(x)\text{ have a common root.}\end{align}
| {
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Find all positive integers $x,y$ such that $\frac{x^2-1}{xy+1}$ is a non-negative integer. QUESTION: Find all positive integers $x,y$ such that $\frac{x^2-1}{xy+1}$ is a non-negative integer.
ATTEMPT (AND SOME SOLUTIONS): So, for a positive integer $x$, $x^2-1\ge0$. In the case $x^2-1=0$ i.e $x=1$ (since it's a positive integer), we get
$\frac{x^2-1}{xy+1}=0$ which is a non-negative integer, regardless of $y$. So, one solution is $(x,y)=(1,k)$ where $k$ is a positive integer.
Now, let's say $y=1$. Then $\frac{x^2-1}{xy+1}=\frac{x^2-1}{x+1}=x-1$ which is always a non-negative integer so $(x,y)=(k,1)$ is also a solution. However, I don't know how to find the other or prove that those are the only ones.
| $x^2-1=(x-1)(x+1)$
$\frac{x^2-1}{xy+1}$ can be an integer if: | {
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A compact in a measurable set The problem is as follows:
Given a Lebesgue-measurable set $M\subset\mathbb{R^n}$ find a compact $C_\epsilon$ and an open $A_\epsilon$, $C_\epsilon\subset M\subset A_\epsilon$, such that $\lambda(A_\epsilon)-\lambda(C_\epsilon)<\epsilon$.
$\lambda$ is the Lebesgue-measure, in particular, if $M$ is not bounded, $k\in\mathbb{N}$,
\begin{equation}
\lambda(M)=\lim_{k\to+\infty}\lambda(M\cap[-k,k]^n).
\end{equation}
Now, if $M$ is bounded I can find $A_\epsilon$ and $C_\epsilon$ with the required properties.
if $M$ is not bounded but $\lambda(M)<\infty$, how can I find the open and the compact?
Furthermore, if $\lambda(M)=\infty$,
I think we can not found a compact such that the difference between its measure and the measure of $M$ is small, is it correct?
| If $M$ is unbounded and $\lambda(M)<\infty$, start by taking a $k>0$ such $\lambda(M)-\lambda\bigl(M\cap[-k,k]\bigr)<\frac\varepsilon2$.
If $\lambda(M)=\infty$, then you are right: there is no such compact set.
| {
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What's the intuition behind using Law of Excluded Middle in Natural Deduction? I've recently started learning First-Order Logic and I have been doing some Natural Deduction exercises. I understand the principles behind most of the Inference Rules but when it comes to applying Classical Rules such as the law of excluded middle, I struggle to reason why it has been used.
For example:
In the proof for:
(φ → ∃x. ψ) ⊢ ∃x. (φ → ψ)
*
*φ → ∃xψ (hypothesis)
*φ ∨ ¬φ (law of excluded middle)
...
∃x. (φ → ψ)
The solution proceeds by using law of excluded middle for φ so that you can use the ∃ elimination rule to reach the conclusion. I understand the solution but I cannot understand why someone has thought to use law of excluded middle to proceed. Is there any intuition behind this, or is it just a 'trick'?
| As per @Sudix's answer above, the intuition behind the use of the Law of Excluded Middle in the proof of :
$(φ → ∃x ψ) ⊢ ∃x (φ → ψ)$
is to apply a "case analysis".
(i) Assume that $φ$ does not hold, i.e. assume $¬φ$.
This means (by the truth-table for the conditional) that $φ → ψ$ is TRUE, and thus also $∃x (φ → ψ)$ is TRUE.
(ii) Now assume that $φ$ holds, i.e. assume $φ$.
We know that the premise $(φ → ∃x ψ)$ holds, and this means (again by the truth-table for the conditional) that also $∃x ψ$ is TRUE, i.e. that $ψ$ is TRUE for some $x$.
Thus, $φ → ψ$ is TRUE for some $x$, i.e. $∃x (φ → ψ)$ is TRUE.
| {
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Finding $n$-th power of transition matrix Is there any shortcut to find $P^{n}=\begin{pmatrix} 1-p & p \\ q & 1-q \end{pmatrix}^n$ quickly and elegantly?
This type of matrix often comes up while dealing with Markov chains. Diagonalization takes way too much time. I need to find eigenvalues, eigenvectors and all that...
| Let $S$ be the matrix
$$
S=
\begin{bmatrix}
1&p\\1&-q
\end{bmatrix}
\ ,
$$
which has on the columns the eigenvectors to the eigenvalues $1, 1-p-q$. Then
$$
\begin{aligned}
S^{-1}PS &=
\begin{bmatrix}
1&0\\0&1-p-q
\end{bmatrix}
\ ,
\\
P &=
S
\begin{bmatrix}
1&0\\0&1-p-q
\end{bmatrix}
S^{-1}\ ,
\\[2mm]
&\qquad\text{ so }
\\[2mm]
P^n &=
S
\begin{bmatrix}
1^n&0\\0&(1-p-q)^n
\end{bmatrix}
S^{-1}\ .
\end{aligned}
$$
| {
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Group homology Künneth Formula I am trying to understand Weibel's proof of Künneth Formula for group homology (Prop. 6.1.13 from An Introduction to Homological Algebra), and I am struggling with following statement:
Let $G,H$ be groups. Let $P\rightarrow\mathbb{Z}$ and $Q\rightarrow\mathbb{Z}$ be free resolutions of $\mathbb{Z}$ respectively over $\mathbb{Z}G$ and $\mathbb{Z}H$. Then, by the Kunneth Formula for chain complexes, we get that the homology of $P\otimes_\mathbb{Z}Q$ (seen as total complex of tensor product double complex) is $0$ for non-zero degree.
How one can derive this statement from Künneth Formula for chain complexes?
| We have, by Künneth formula,
$$H_n(P_{\cdot} \otimes_{\Bbb Z} Q_{\cdot}) \cong
\left(
\bigoplus_{p+q=n} H_p(P_{\cdot}) \otimes_{\Bbb Z} H_q(Q_{\cdot})
\right) \oplus
\bigoplus_{p+q=n-1} \mathrm{Tor}_1^{\Bbb Z}(H_p(P_{\cdot}), H_q(Q_{\cdot}))$$
When $r>0$, we know that $H_r(P_{\cdot}) = H_r(Q_{\cdot}) = 0$, since these complexes are free resolutions.
When $n>0$ and $p+q=n$, we must have $p>0$ or $q>0$, thus the first direct summand is $0$.
When $n>1$ and $p+q=n-1$, we must have $p>0$ or $q>0$, thus the second direct summand is $0$. If $n=1$, we are left with
$\mathrm{Tor}_1^{\Bbb Z}(P_0, Q_0) = 0$, since both modules are free over $\Bbb Z[G]$, hence over $\Bbb Z$ (hence flat over $\Bbb Z$). In all cases, when $n>0$, the second direct summand also vanishes.
In other words, we've got : $H_n(P_{\cdot} \otimes_{\Bbb Z} Q_{\cdot}) = 0$ when $n>0$.
| {
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Prove that $f(x) = \frac{(x+1)\sin x}{x}$ is bounded at $(0,1)$. Is it a valid proof? Proof:
We assume that $f(x)$ is unbounded at $(0, 1)$. So for any $M\in \mathbb{R}$
there exists an $x\in (0,1)$ such that:
$|f(x)| = |\frac{(x+1)\sin x}{x}| > M \Rightarrow^{(1)} \frac{(x+1)\sin x}{x}>M
\Rightarrow x\sin x+\sin x > Mx \Rightarrow^{(2)} x(\sin x + 1)>Mx \Rightarrow
\sin x+1 > M \Rightarrow \sin x>M-1.$
(1).$\sin x > 0$ for any $x\in(0,1)$
(2).$x>\sin x$ for $x\in(0,1)$
We can choose $M=3$, so:
$\sin x>2.$
And this is a contradiction because $\sin x \leq 1$ for any $x\in(0,1)$. $\Box$
| Here's a direct proof, similar to yours, but perhaps a little simpler . . .
On the interval $(0,1)$, we have $0 < \sin x < x$, hence
$$0 < \frac{(x+1)\sin x}{x} < \frac{(x+1)\,x}{x} = x+1 < 1 + 1 = 2$$
so on the interval $(0,1)$, we have $0 < f(x) < 2$.
| {
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Does the sequence of r.v. given by the law of large numbers converge almost surely to the mean in an oscillatory fashion? I state the very famous strong law of large numbers in it's simplest form:
Given an IID sequence of random variables $\{ X_n\}_{n \in N}$ then
$$ \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{i= 1}^n X_i = E[X_0] \qquad a.s.$$
I am wondering if the sequence given by $Y_n = \frac{1}{n} \sum_{i= 1}^n X_i $ converges almost surely oscillating to the mean (for every $n_1 \in N$ s.t. $E[Y_{n_1}] > E[Y_0]$ there exists a $n_2> n_1$ s.t. $E[Y_{n_2}] <E[Y_0]$ , and vice-versa).
In particular how would one go about proving such a statement?
| We consider the case $E(X_1^2)<\infty$. We reformulate the problem, by defining $U_i=X_i-E(X_i)$. So, we wish, to show that $\sum_{i}^nU_i$ oscillates from zero.
We know that $\limsup_{n}\frac{U_n}{\sqrt{n}}=\infty$, almost surely. One way to see this is by using the CLT theorem and Kolmogorov's zero-one law.
By symmetry, we also obtain that $\liminf_{n}\frac{U_n}{\sqrt{n}}=-\infty$
More specifically, we have that $\limsup_{n} U_n=\infty$ and $\liminf_{n} U_n=-\infty$ almot surely. $\square$
| {
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$\lim_{n\rightarrow\infty}\frac{a_{n}}{n}=1$ implies that $\lim_{n\rightarrow\infty}\sup_{0\leq k\leq n}\frac{|a_{k}-k|}{n}=0$ I'm trying to prove the following:
If $$\displaystyle\lim_{n\rightarrow\infty}\frac{a_{n}}{n}=1$$ then $$\displaystyle\lim_{n\rightarrow\infty}\sup_{0\leq k\leq n}\frac{|a_{k}-k|}{n}=0.$$
So, the hypotesis $\displaystyle\lim_{n\rightarrow\infty}\frac{a_{n}}{n}=1$ implies that $\forall\epsilon>0,\exists N:=N(\epsilon)\in\mathbb{N}$ such that $\forall n\geq N$ implies $\frac{|a_{n}-n|}{n}<\epsilon.$
Fixed $\epsilon>0,$ we have $\displaystyle\sup_{n\geq N}\frac{|a_{n}-n|}{n}<\epsilon.$
Now, for such $N$ I'd like to prove that $\sup_{0\leq k\leq n}\frac{|a_{k}-k|}{n}<\epsilon$ $\forall n\geq N,$ but I'm stuck in this.
I think that if $n\geq N$ then $$\sup_{0\leq k\leq n}\frac{|a_{k}-k|}{n}\leq\sup_{0\leq k\leq n}\frac{|a_{k}-k|}{N}.$$
Then,for example, if $\sup_{0\leq k\leq n}\frac{|a_{k}-k|}{N}=\frac{|a_{N}-N|}{N}$ it's done because of the hypotesis. If $\sup_{0\leq k\leq n}|a_{k}-k|$ is achieved in $0\leq k<n$ then $\sup_{0\leq k\leq n}\frac{|a_{k}-k|}{N}<\epsilon$ only for large enough $N.$
Because of the above I think such limit have sense and it's equal to $0$ but I don't get how to prove it exactly.
Any kind of help is thanked in advanced.
| Let $\varepsilon>0$ be arbitary. Then there exists $N_{1}\in\mathbb{N}$
such that $\left|\frac{a_{n}-n}{n}\right|<\varepsilon$ whenever $n\geq N_{1}$.
Let $M=\max_{0\leq k<N_{1}}|a_{k}-k|$. Choose $N_{2}\in\mathbb{N}$
such that $\frac{M}{N_{2}}<\varepsilon$. Let $N=\max(N_{1},N_{2})$.
Let $n\geq N$ be arbitrary. Let $0\leq k\leq n$. If $0\leq k<N_{1}$,
we have that $\left|\frac{a_{k}-k}{n}\right|\leq\frac{M}{N}<\varepsilon$.
If $N_{1}\leq k\leq n$, then $\left|\frac{a_{k}-k}{n}\right|\leq\left|\frac{a_{k}-k}{k}\right|<\varepsilon$
because $k\geq N_{1}$. It follows that $\sup_{0\leq k\leq n}\left|\frac{a_{k}-k}{n}\right|<\varepsilon$.
Hence $\lim_{n}\sup_{0\leq k\leq n}\left|\frac{a_{k}-k}{n}\right|=0$.
| {
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Clarification needed: Smallest subfield of a field
Q: Given a field does there exist a smallest subfield ? | Once you prove that the intersection of all subfields of a field is itself a field, you see that every field has a single smallest subfield. The function that Lubin suggests gives you a way to understand the structure of that smallest field. Lubin is defining the function $\varphi: \mathbb{Z} \rightarrow F$ (where $F$ is your original field) by $$\varphi(n) = \underbrace{1 + 1 + \dots + 1}_{\text{$n$ times}}.$$
Equivalently, you can also define this function as the group homomorphism $\varphi: (\mathbb{Z}, +) \rightarrow (F, +)$ with the property $\varphi(1) = 1_F$. (Since $1$ generates $\mathbb{Z}$ as a group, knowing where $\varphi$ sends $1$ is enough to define the whole function.)
At this point, you have to prove that $\varphi$ is also a ring homomorphism (which I'll leave to you).
Consider the kernel of $\varphi$. Since $\mathbb{Z}$ is a principal ideal domain, we have that $\ker \varphi = p\mathbb{Z}$ for some nonnegative integer $p$. This $p$ will, in fact, be the characteristic of the field $F$, meaning $p = 0$ or is a prime.
Now, suppose $p = 0$. Then consider the subset $$S = \left\{\varphi(a) \varphi(b)^{-1}: a, b \in \mathbb{Z}, b \not = 0\right\}$$ of $F$. You can show that $S$ is isomorphic to $\mathbb{Q}$. Since $\mathbb{Q}$ has no proper subfields, $S$ must be the smallest subfield of $F$.
On the other hand, suppose $p$ is a prime. Then $\varphi(\mathbb{Z}) \cong \mathbb{Z}/p\mathbb{Z}$. Since $\mathbb{Z}/p\mathbb{Z}$ has no proper subfields, $\varphi(\mathbb{Z})$ must be the smallest subfield of $F$.
Now, how do we know that a field $F$ with $p^n$ elements (where $p$ is prime) has characteristic $p$ (and hence a smallest subfield isomorphic to $\mathbb{Z}/p\mathbb{Z}$)? Since $F$ is a finite field, it must have some non-zero prime characteristic, $q$. Thus, $F$ has a subfield isomorphic to $\mathbb{Z}/q\mathbb{Z}$, and hence $F$ is a vector space over $\mathbb{Z}/q\mathbb{Z}$. Let $m = \dim F$ as a $\mathbb{Z}/q\mathbb{Z}$-vector space. Because $F$ is finite, $m$ must be finite as well. That means $p^n = q^m$. Hence $p = q$ and $m = n$.
| {
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multiplication in $\mathbb{Z}_p$ I'm not even sure what would be a good title for my question, so feel free to edit it (if you can).
I want to prove (or dispute) the correctness of the following lemma:
Let $p$ be a prime number, and let $i,j,c \in \mathbb{Z}_p \setminus
\{0\}$ such that $i \neq j$. Is it true that $ic \neq jc \pmod p$?
In other words, what I want to show is that given any $c \in \mathbb{Z}_p \setminus \{0\}$, the results of $c, 2c, \cdots , (p-1)c$ are pairwise different.
First - is this even correct? a short python program I wrote for small prime numbers gives me the feeling it is.
Second - if it's correct -how do I prove it? I tried assuming that $ic = jc \pmod p$ but got stuck with it.
Thanks
| Recall that an integral domain is a commutative ring such that $ab=0$ implies $a=0$ or $b=0$.
The essential property of primes I'm going to use is the fact that if $p$ is prime and $p|ab$ then $p|a$ or $p|b$.
Lemma. $\mathbb{Z}_n$ is an integral domain if and only if $n$ is prime.
Proof. "$\Rightarrow$" assume $n$ is not prime and take $p|n$ a prime divisor. Then both $p, n/p\in\mathbb{Z}_n$ are non-zero but
$$p\cdot n/p=n=0\text{ (mod n)}$$
Contradiction.
"$\Leftarrow$" Let $ab=0 \text{ (mod p)}$ for prime $p$, i.e. $p|ab$. Since $p$ is prime then this implies that either $p|a$ or $p|b$, i.e. $a=0$ or $b=0$ in $\mathbb{Z}_p$. $\Box$
With that it is easy to prove the statement: assume that $i,j\in\mathbb{Z}_p$ and $c\in\mathbb{Z}_p\backslash\{0\}$ are such that
$$ic=jc\text{ (mod p)}$$
Then
$$ic-jc=0\text{ (mod p)}$$
$$(i-j)c=0\text{ (mod p)}$$
and since $\mathbb{Z}_p$ is an integral domain then $i-j=0\text{ (mod p})$ because $c\neq 0$. Therefore $i=j\text{ (mod p)}$ which completes the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2793951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Identity Theorem: Extending from $\mathbb{R}$ to $\mathbb{C}$ Suppose we have $f_1, f_2: \mathbb{C} \rightarrow \mathbb{C}$ holomorphic, and $f_1 = f_2$ on $\mathbb{R}$. Can we then say $f_1 = f_2$ identically on $\mathbb{C}$?
This appears to be true by the uniqueness of analytic continuations, but the identity theorem requires that the functions must agree on an open set, and clearly $\mathbb{R}$ is not open with respect to $\mathbb{C}$.
Assuming the above is true, why can this not be shown using the identity theorem?
| The identity theorem does not require that the functions must agree on an open set !
Let $A:=\{z \in \mathbb C:f_1(z)=f_2(z)\}$ .
If $A$ has an accumulation point in $ \mathbb C$, then , by the identitiy theorem, $f_1=f_2$ on $ \mathbb C$.
In your case we have $A = \mathbb R$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2794024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Finding the sum of $\cos\frac{\pi}{7}$, $\cos\frac{3\pi}{7}$, $\cos\frac{5π}{7}$ by first finding a polynomial with those roots
Without using tables, find the value of $$\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7}$$
This is a very common high school trigonometric problem, and the usual way to solve this is by repeated application of trigonometric identities. But I thought of a bit different approach.
Somehow, if we can find a polynomial whose roots are the three terms of the above expression, then we can apply Vieta's formula to find the value.
So please help me with it. (Any hint will be appreciated.)
| $\text{Using Complex Number}$
Let $z=e^{\frac{\pi i}{7}},$ so $z^7=-1.$ Let $Q$ be the desired quantity. Then
$$2Q=z+\frac{1}{z}+z^3+\frac{1}{z^3}+z^5+\frac{1}{z^5} = \frac{z^{10}+z^8+z^6+z^4+z^2+1}{z^5} = \frac{z^{12}-1}{z^5(z^2-1)}$$
$$=\frac{-z^5-1}{z^7-z^5} = \frac{-z^5-1}{-1-z^5} = 1$$ $$\therefore Q=\frac{1}{2}\ \ \square$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2794176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Identity for triple integrals I found an identity but I haven't been able to prove it.
This is the identity
$\int _{ 0 }^{ x }{ \int _{ 0 }^{ y }{ \int _{ 0 }^{ z }{ f(t)dtdzdy\quad =\quad \frac { 1 }{ 2 } \int _{ 0 }^{ x }{ (x-t)^{ 2 }f(t)dt } } } } $
A teacher told me that the first step to achieve the result is to change $x$ for $t$ and then draw the solid. I have no idea how to draw the solid and what to do with that. Any ideas?
| The region $z\in[0,y]$ and $t\in[0,z]$ is a triangular region in the $t$-$z$ plane bounded by $t=0$, $z=y$, and $z=t$.
By changing the order of integration we can write
$$\int_0^y\int_0^zf(t)\,dt\,dz=\int_0^y \int_t^yf(t)\,dy\,dt=\int_0^y(y-t)f(t)\,dt$$
Can you proceed by evaluating the integral $\int_0^x \int_0^y(y-t)f(t)\,dt\,dy$ by changing the order of integration in an analogous way?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2794308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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reference for special relativity I got a student finishing its first year of Math and she would like to study a bit of special relativity from a mathematics point of view. I know the subject quite well but I don't know any basic references. what I look for her his some basic on Minkowsky? and if possible the Maxwell equation from the differential form point of view? Of course she don't know differential form yet, but she is motivated to learn. any idea willbe welcome. Thx!
| There is a really lovely book by David Bressoud called Second Year Calculus: From Celestial Mechanics to Special Relativity. He is an excellent writer and the book is a joy to read. It gives the gentlest introduction to differential forms and special relativity that you could hope for.
https://www.springer.com/us/book/9780387976068
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Evaluate $\int_0^1x(\tan^{-1}x)^2~\textrm{d}x$
Evaluate $\int\limits_0^1x(\tan^{-1}x)^2~\textrm{d}x$
My Attempt
Let, $\tan^{-1}x=y\implies x=\tan y\implies dx=\sec^2y.dy=(1+\tan^2y)dy$
$$
\begin{align}
&\int\limits_0^1x(\tan^{-1}x)^2dx=\int\limits_0^{\pi/4}\tan y.y^2.(1+\tan^2y)dy\\
&=\int\limits_0^{\pi/4}\tan y.y^2dy+\int\limits_0^{\pi/4}\tan^3y.y^2dy\\
&=\bigg[y^2.\log|\sec y|-\int2y.\log|\sec y|dy\bigg]+\int_0^{\pi/4}\tan^3y.y^2dy\\
&=\bigg[y^2.\log|\sec y|-y^2.\log|\sec y|+\int\frac{\tan y\sec y}{\sec y}y^2dy\bigg]+\int_0^{\pi/4}\tan^3y.y^2dy\\
\end{align}
$$
How do I proceed further and solve the integration?
| $$\int_ {0}^{1}x(arctan(x))^2dx$$
First solve the integral without boundaries
$$\int x(arctan(x))^2dx$$
Apply Integration By Parts, where $u=\arctan^2(x),v^{\prime}=x$
$$=\arctan^2(x)\frac{x^2}{2}-\int \frac{2\arctan(x)}{1+x^2} (\frac{x^2}{2})dx$$
$$=\frac12 x^2\arctan^2(x)-\int \frac{x^2\arctan(x)}{x^2+1}dx$$
Note that $\int \frac{x^2\arctan(x)}{x^2+1}dx= x\arctan(x)-\arctan^2(x)+\ln|\frac{1}{\sqrt{1+x^2}}|+\frac12\arctan^2(x)$
So,
$$=\frac12 x^2\arctan^2(x)-(x\arctan(x)-\arctan^2(x)+\ln|\frac{1}{\sqrt{1+x^2}}|+\frac12\arctan^2(x))$$
After simplifying we get,
$$=\frac12 x^2\arctan^2(x)-x\arctan(x)+\frac12\arctan^2(x)-\ln|\frac{1}{\sqrt{1+x^2}}|$$
Now compute the boundaries and we get
$$=\frac{\ln(2)}{2}+\frac{\pi^2}{16}-\frac{\pi}{4}$$
$$\int_ {0}^{1}x(arctan(x))^2dx=\frac{\ln(2)}{2}+\frac{\pi^2}{16}-\frac{\pi}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2794560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Is $\mathbb{F}_{p^r}/\mathbb{F}_{p}$ a Galois extension? Let $r>0$. Is $\mathbb{F}_{p^r}/\mathbb{F}_{p}$ a Galois extension? If so, why?
I know that it is a finite extension, with $[\mathbb{F}_{p^r}:\mathbb{F}_{p}]=r$. To show that it is a Galois extension, it suffices to show that $|Aut(\mathbb{F}_{p^r}/\mathbb{F}_{p})|=r$.
The notaion $Aut(K/F)$ indicates the group of field automorphisms $K\to K$ such that the automorphism fixes every element of $F$.
But, how do I show $|Aut(\mathbb{F}_{p^r}/\mathbb{F}_{p})|=r$? A simple, easy to grasp proof without holes in it would be ideal.
| Every field $K$ of characteristic $p$ has the Frobenius endomorphism
$F:x\mapsto x^p$. This is a homomorphism of fields, and so is injective.
If $K$ is finite, then $F$ must be bijective, so an automorphism.
On $\Bbb F_p$, $F$ acts trivially.
The fixed points of $F^t$ are the solutions of $x^{p^t}-x$.
Every element of $\Bbb F_{p^r}$ is a solution of $x^{p^r}-x$,
if $t<r$ then not all elements of $\Bbb F_{p^r}$ is a solution of
$x^{p^t}-x$ since that polynomial has fewer than $p^r$ zeros.
Thus $F$ has order $r$ on $\Bbb F_{p^r}$. As $|\Bbb F_{p^r}:\Bbb F_p|
=r$, then the Galois group must consist of the powers of $F$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Probability: How to get to a percentage? Exercice: We have a class of $16$ students with $2$ of them not having phones. Every class in the school is like that ($14$ with phones, $2$ without). How many students in the school do I have to pick for the probability of having at least one student without a phone to be at $99,9\%$ ?
Answer: $52$
I have no Idea how to solve this, please help.
| The question asks for at least one student without a phone, it could be 1 student, 2 student ,... So for solving this, we can say if the probability of having at least one student without a phone is 99.9% (or higher) then the probability of ALL having phone must be 0.1% or less. If you choose one student it has $\frac{14}{16}$ probability that he has phone. So if we choose $n$ students, then the probability of all having phone is $(\frac{14}{16})^n$
so $(\frac{14}{16})^n <0.001$ by using logarithm, we get the answer $n>51.73$ so $n\geq52$
Solving the inequality by using base 10 logarithm:
$(\frac{14}{16})^n <0.001 \Rightarrow n\times\log(\frac{14}{16}) < -3 \Rightarrow -0.05799 \times n < -3 \Rightarrow n> 51.73$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Optimizing singular Rayleigh quotient subject to linear constraint I want to numerically solve
$$\min_x \frac{x^TAx}{x^TBx} \quad \mathrm{s.t.}\quad Cx=0,$$
where $A$ and $B$ are large sparse matrices, $A$ is positive semi-definite, $B$ is positive-definite, and $C$ is sparse.
One approach would be to find a basis $N$ for the nullspace of $C$, write $x=Ny$, and then reduce the above to an unconstrained eigenvector problem. But if I try to compute $N$ (using e.g. a QR decomposition of $C$) it ends up intractably dense.
So instead I want to incorporate constraint projection into the power iteration for finding $x$: more specifically, I start with a random vector $x^0$, then iterate:
*
*Solve $A\tilde x^{i+1} = Bx^i$ for $\tilde x^{i+1}$.
*Solve $CC^T\lambda = C\tilde{x}^{i+1}$ for $\lambda$.
*Compute $x^{i+1} = \frac{\tilde{x}^{i+1}-C^T\lambda}{\|\tilde{x}^{i+1}-C^T\lambda\|_B}.$
However I'm running into a couple of problems:
A) The matrix $A$ is singular, so the linear system in step (1) does not have a solution;
B) The matrix $CC^T$ may be singular.
Concern B is less serious than concern A since the linear system in step (2) is guaranteed to always have at least one solution. So I can solve the linear system in step (2) using a robust method for underdetermined least squares, e.g. QR decomposition. What should I do to fix step (1), though? Of course, there is a strong temptation to "fix" $A$ by adding a small multiple of the identity; is this fix mathematically sound?
| Answering my own question as I've found a rather satisfactory solution to this problem.
From the KKT conditions of the original problem one sees that the minimizer is the $x$-part of the eigenvector with smallest eigenvalue in the generalized eigenvalue problem
$$\begin{bmatrix} A & C^T \\ C & 0\end{bmatrix}\begin{bmatrix}x\\\mu\end{bmatrix} = \lambda \begin{bmatrix} B & 0\\0 & 0\end{bmatrix} \begin{bmatrix} x \\ \mu \end{bmatrix}.$$
It follows that the following power iteration method will find $x$:
*
*Initialize $x^0$ to a random vector.
*Until converge, repeat:
*
*
*Compute $y^i = Bx^i$
*
*
*Solve $$\begin{bmatrix} A & C^T \\ C & 0\end{bmatrix}\begin{bmatrix}z^i\\\mu^i\end{bmatrix} = \begin{bmatrix}y^i\\0\end{bmatrix}.$$
In the case that $A$ and $CA^{-1}C^T$ are invertible, the solution has expression
$$z^i = A^{-1}y^i - A^{-1}C^T(CA^{-1}C^T)^{-1}CA^{-1}y^i.$$
*
*
*Set $x^{i+1} = z^i/\sqrt{(z^i)^TBz^i}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2794900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Why is there a pattern to the last digits of square numbers? I was programming and I realized that the last digit of all the integer numbers squared end in $ 0, 1, 4, 5, 6,$ or $ 9 $.
And in addition, the numbers that end in $ 1, 4, 9, 6 $ are repeated twice as many times as the numbers that end in $ 0, 5$
I checked the numbers from $1$ to $1000$, and the results are:
$1.$ The numbers on the left are the last digit of each digit squared.
$2.$ The numbers on the right are the number of times that the last digit is repeated.
$$
\begin{array}{cc}
0: &100, \\
1: &200, \\
4: &200, \\
5: &100, \\
6: &200, \\
9: &200
\end{array}
$$
So, why does this happen? What is the property that all integers have?
| Consider:
$$(x+k)^2=(x+k)(x+k)=x^2+2xk+k^2$$
In your case, $x=10z, z\in \Bbb Z$, and $0\le k\le9, k\in \Bbb Z$. Thus it becomes:
$$(x+k)^2=100z^2+20zk+k^2$$ for which the only possible unit is the unit from $k^2$, and so the facts that:
$$1^2,9^2\space\text{end in}\space 1$$
$$2^2,8^2\space\text{end in}\space 4$$
$$3^2,7^2\space\text{end in}\space 9$$
$$4^2,6^2\space\text{end in}\space 6$$
$$5^2\space\text{ends in}\space 5$$
$$0^2\space\text{ends in}\space 0$$
means that those are your only possibilities.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2795029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
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equidimensional equation For $x^2y''+axy'+by=0$ where $a,b\in\mathbb{R}$, find conditions on $a,b$ such that all solutions are bounded as $x\to0$.
First write $y''+\frac{a}{x}y'+\frac{b}{x^2}y=0$ and guess $y=x^r$. Plug it in and simplify I got $r^2+(a-1)r+b=0$, thus $r=\frac{1-a}{2}\pm\sqrt{\left(\frac{1-a}{2}\right)^2-b}$. Since the solution to this equation looks like $y=c_1x^{r_1}+c_2x^{r_2}$, I think if $r_1,r_2\geq 0$, all solutions are bounded as $x\to0$. Similarly, if one of $r_1,r_2$ is positive, then the solution is partially bounded as $x\to0$.
My question is, how can I find the condition on $a,b$ such that $r_1,r_2\geq 0$? I figured if $a=1,b=0$, $r_1=r_2=0$ and all the solutions are bounded. But what's next?
| I would assume that your solution is required to be real.
Does your solution need to be well defined at $x = 0$? Nothing was stated about the range of validity.
Your statement that if one is positive then it is partially bounded is suspect. If one of the roots is less than zero then one term will be complex for values of $x < 0$ so the limit from the left hand side may not be well defined in $\Bbb R$ (in the complex plane maybe).
So let's say the two conditions are positivity and real valued.
The discriminant needs to be positive or zero
$$ \implies \frac{(1 - a)^2}{4} \ge b $$
if b is negative then the negative root will be less than zero, only one possible solution with the positive root (thus you'd need to search for a second solution by some other means).
If $b$ is positive then for both roots to be positive we need $a \le 1$
For $b = 0$, $a = 1$, as you pointed out.
For now let's look at the positive solutions. The conditions are:
$$a \ge 1$$ and $$\frac{(1 - a)^2}{4} \ge b > 0 $$
So there is a whole family of solutions.
I am not sure that there is a next step. We used the desire to have $\lim_\limits{x\to 0}y(x)$ be finite and to impose these conditions. You should check that the limits are the same form both sides (but we used that to constrain a and b).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Minimal sufficient statistic for normal distribution with known variance Let $X_1, ..., X_n$ be a random sample from the $N(\theta,1)$ distribution. Find a minimal sufficient statistic for $\theta$.
Now, I can find a sufficient statistic using the factorisation theorem ($\sum X_i$), but I don't think that this statistic is in fact minimal sufficient.
The question seems to imply that there exists a minimal sufficient statistic, but I'm not even sure that there is one.
MY QUESTION: How would I go about proving that there is no minimal sufficient statistic, or if there is one, what is it!?
Any hints greatly appreciated!
| By the factorization criterion
$$
\mathcal{L}(\theta)=\frac{1}{(2\pi)^{n/2}}\exp\{-\sum_{i=1}^nX_i^2/2 +\bar{X}_n \theta -n\theta^2/2\}
$$
$$
\qquad = \exp\{\bar{X}_n\theta-n\theta^2/2\}\times(2\pi)^{-n/2}\exp\{-\sum X_i^2/2\}.
$$
So $\bar{X}_n$ is sufficient statistic.
| {
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"timestamp": "2023-03-29T00:00:00",
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Determining the points at constant distance $d$ from parabola $y = x^2$ My friends recently learned about Locus and how parabola are the points equdistant from a fixed point (focus) and a straight line (directrix). However, now we're trying to find an equation or set of equations for the points at a constant distance from the parabola of function y = x^2 and cannot be closer/further from any point on the curve than any arbitrary distance (just use variable d).
As they have more experience with derivatives, they've gone with using the equation of the normal to find these equations. I've tried to use the distance formula, but with no success.
It would be awesome if we got some help on this!
| We have $f(x) = x^2$. A tangent vector at each point $P(x)= (x, f(x))$ is $$
t(x)=(1, f'(x)).
$$
The normal vectors are orthogonal to the tangent vectors which gives
$$
n_i(x) = (\mp f'(x), \pm 1)
$$
and satisfies
$$
t(x) \cdot n_i(x)
= 1 \cdot (\mp f'(x)) + f'(x) \cdot (\pm 1)
= 0.
$$
As we want to specify a distance along a normal, it is easier to use normal vectors with length $1$ (unit vectors):
$$
n_i(x) = \frac{(\mp f'(x), \pm 1)}{\sqrt{1+f'(x)^2}}.
$$
The two cases of sign stand for the normal vector pointing inwards and outwards. Then a point $Q(x)$ with distance $d$ along a normal to $P(x)$ is
$$
Q(x)
= P(x) + d \, n_i(x)
= (x, f(x)) + d \, \frac{(\mp f'(x), \pm 1)}{\sqrt{1+f'(x)^2}}.
$$
In this case
$$
Q(x) = (x, x^2) + d \frac{(\mp 2x, \pm 1)}{\sqrt{1+4x^2}}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2795313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Probability that $z$ is EVEN satisfying the equation $x + y + z = 10$ is Question
Three randomly chosen non negative integers $x, y \text{ and } z$ are found to satisfy the equation $x + y + z = 10$. Then the probability that $z$ is even, is"?
My Approach
Calculating Sample space -:
Number of possible solution for $x + y + z = 10$
$$=\binom{10+3-1}{10}=12 \times 3=66$$
Possible outcome for $z$ to be even =$6(0,2,4,6,8,10)$
Hence the required probability$$=\frac{6}{66}=\frac{1}{11}$$
But the answer is $\frac{6}{11}$
Am I missing something?
| Using computer power:
sage: allCases = len( [ S for S in cartesian_product( [ [0..10], [0..10], [0..10] ] ) if sum(S) == 10 ] )
sage: goodCases = len( [ S for S in cartesian_product( [ [0..10], [0..10], [0..10] ] ) if sum(S) == 10 and S[2] % 2 == 0 ] )
sage: allCases
66
sage: goodCases
36
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $\frac{\cos(\alpha -3\theta)}{\cos^3 \theta}=\frac{\sin(\alpha -3\theta)}{\sin^3 \theta}=m$ prove that $\cos\alpha=\frac{2-m^2}{m}$
If $$\frac{\cos(\alpha -3\theta)}{\cos^3 \theta}=\frac{\sin(\alpha -3\theta)}{\sin^3 \theta}=m$$
prove that
$$\cos\alpha=\frac{2-m^2}{m}$$
My approach:
$$\cos^2(\alpha-3\theta)+\sin^2(\alpha-3\theta)=m^2(\sin^6\theta+\cos^6\theta)$$
$$\Rightarrow \frac{1}{m^2}=\sin^6\theta+cos^6\theta=1-\frac{3}{4}\sin^22\theta$$
$$\Rightarrow \sin^22\theta=\frac{4}{3}᛫\frac{m^2-1}{m}$$
I can not proceed further, please help.
| Hint for your last equation:
$$\sin(2\theta)=2sin(\theta)\cos(\theta)$$
so $$\sin(2\theta)^2=4\sin^2(\theta)\cos^2(\theta)$$
and you can eliminate $\theta$
So you get $$4\left(1-\cos(\theta)^2\right)\cos(\theta)^2=\frac{4}{3}\frac{m^2-1}{m}$$
Solve this for $\cos(\theta)$
I know this, when you get $\theta$ then you can compute $\alpha$ with the equations above!
Expanding the term
$$\cos(\alpha-3\theta)\sin(\theta)^3-\sin(\alpha-3\theta)\cos(\theta)^3$$
we get
$$4\, \left( \sin \left( \theta \right) \right) ^{3}\cos \left( \alpha
\right) \left( \cos \left( \theta \right) \right) ^{3}-3\, \left(
\sin \left( \theta \right) \right) ^{3}\cos \left( \alpha \right)
\cos \left( \theta \right) +4\, \left( \sin \left( \theta \right)
\right) ^{4}\sin \left( \alpha \right) \left( \cos \left( \theta
\right) \right) ^{2}- \left( \sin \left( \theta \right) \right) ^{4
}\sin \left( \alpha \right) -4\, \left( \cos \left( \theta \right)
\right) ^{6}\sin \left( \alpha \right) +3\, \left( \cos \left( \theta
\right) \right) ^{4}\sin \left( \alpha \right) +4\, \left( \cos
\left( \theta \right) \right) ^{5}\cos \left( \alpha \right) \sin
\left( \theta \right) - \left( \cos \left( \theta \right) \right) ^{
3}\cos \left( \alpha \right) \sin \left( \theta \right)=0
$$
This term can we solve for $$\alpha$$
$$\alpha=-\arctan \left( 3\,{\frac {\cos \left( \theta \right) \sin \left(
\theta \right) \left( 2\, \left( \sin \left( \theta \right) \right)
^{2}-1 \right) }{6\, \left( \sin \left( \theta \right) \right) ^{4}-6
\, \left( \sin \left( \theta \right) \right) ^{2}+1}} \right)
$$ and now you can use your $$\theta$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Divisibility of Sum of Equally Spaced Binomial Coefficients According to a numerical calculation I did for small values of $k$, it appears that the following is true.
$$4|\left[\sum_{j=1}^{n-1}\binom{3n}{3j}\right]$$ or $$\sum_{j=1}^{n-1}\binom{3n}{3j}=4p, p\in\mathbb{Z}$$
Ex.
If $n=2, \binom{6}{3}=20=4\cdot 5$
If $n=3, \binom{9}{3}+\binom{9}{6}=2\cdot 84=168=4\cdot 42$
If $n=4, \binom{12}{3}+\binom{12}{6}+\binom{12}{9}=2\cdot 220+924=1364=4\cdot341$
If $n=5, \binom{15}{3}+\binom{15}{6}+\binom{15}{9}+\binom{15}{12}=2\cdot 455+2\cdot 5005=10920=4\cdot 2730$
Is there a way to prove this? Using induction, as above I've shown the base case is true. Then if we assume that
$$S_m=\sum_{j=1}^{m-1}\binom{3m}{3j}=4q, q\in\mathbb{Z}$$
Then
$$S_{m+1}=\sum_{j=1}^{m}\binom{3m+3}{3j}=?$$
And I have no idea how to go forward. Perhaps its not true? Is there a counterexample?
| Let $$S=\sum_{j=1}^{n-1}\binom {3n}{3j}$$
Adding terms for $j=0$ and $j=n$ gives
$$\begin{align}
S+2
&=\sum_{j=0}^n \binom {3n}{3j}\\
&=\sum_{r=0}^{3n}\binom {3n}r-\left[\sum_{j=0}^{n-1}\binom {3n}{3j+1}+\binom {3n}{3j+2}\right]\\
&=2^{3n}+2\Re\left[\sum_{j=0}^{n-1}\binom {3n}{3j+1}e^{i2\pi/3}+\binom {3n}{3j+2}e^{i4\pi/3}\right]\\
&=8^n+2\Re\left[\sum_{j=0}^{n-1}\binom {3n}{3j+1}e^{i2\pi(3j+1)/3}+\binom {3n}{3j+2}e^{i2\pi(3j+2)/3}\right]\\
&=8^n+2\Re\left[\underbrace{\binom {3n}{3n}+\sum_{j=0}^{n-1}\sum_{k=1}^3\binom {3n}{3j+k}e^{i2\pi(3j+k)/3}}-\sum_{j=0}^{n}\binom {3n}{3j}\right]\\
&=8^n+2\Re\left[\qquad\ \ \quad\overbrace{\sum_{r=0}^{3n}\binom {3n}r e^{i2\pi r/3}}\qquad \qquad -\left(S+2\right)\right]\\
&=8^n+2\Re\big[\big(1+e^{i2\pi /3}\big)^{3n}-\left(S+2\right)\big]\\
&=8^n+2\Re\big[\big(e^{i\pi /3}\big)^{3n}\big]-2\big(S+2\big)\\
&=8^n+2\Re \big[e^{i\pi n}\big]-2\big(S+2\big)\\
&=8^n+2(-1)^n-2\big(S+2\big)\\
3\big(S+2\big)
&=2(-1)^n+8^n\\
3S&=2(-1)^n+8^n-6\\
&\equiv 2(-1)^n+0+2\mod 4\\
&\equiv 2\pm 2\mod 4\\
&\equiv 0\mod 4\\
S=\sum_{j=1}^{n-1}\binom {3n}{3j}&\equiv 0\mod 4\\
\Longrightarrow 4\bigg|&\sum_{j=1}^{n-1}\binom {3n}{3j}\; \blacksquare\end{align}$$
| {
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inverse Fourier transform of product of two functions What is the inverse Fourier transform of $i\omega f(\omega)g(\omega)$?
is it just $\frac{d}{dt}(f(t)\cdot g(t))$ or will I end up with some kind of convolution?
| It will be a convolution by the convolution theorem for inverse Fourier transforms:
$\mathcal{F}^{-1}\left(i\omega \hat{f}(\omega)\hat{g}(\omega)\right)(t)=\frac{d}{dt}\mathcal{F}^{-1}\left(\hat{f}(\omega)\hat{g}(\omega)\right)(t)=\frac{d}{dt}\mathcal{F}^{-1}\left(\widehat{(f*g)}(\omega)\right)(t)=\frac{d}{dt}(f*g)(t)$
Just remember (and never forget) that convolution turns into multiplication after applying the Fourier transform and vice versa.
| {
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Subspace $\ell^2$ is of first category
Let $\ell^2$ be the Hilbert space of square summable sequences, and $\mathcal{H}$ be the subspace consisting of sequences $\{x_n\}$ with $\sum_{n=1}^\infty n^2|x_n|^2<\infty$
Show that $\mathcal{H}$ is of the first category.
I unsure how to start this problem.
| Hint: Consider $E_M=\{(x_n)\in l^2: \sum n^2x_n^2\le M\}.$ Show $E_M$ is closed in $l^2.$ (Fatou's lemma). Then show $E_M$ has no interior.
| {
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How does GIMPS work and what are these iterations? I downloaded GIMPS today just out of curiosity and have been running it. On my machine it is checking $M_{52898149}=2^{52898149}-1$.
From what I could find on Wikipedia I suppose that GIMPS uses Lucas-Lehmer primality test which means it will pick $s_{52898147}$ term in Lucas Lehmer sequence and will see whether $M_{52898149}=2^{52898149}-1$ divides it or not?
Now my question is, what does the iterations that my software is doing, means? Here is a pic :
What does iteration $160000/52898149$ means here? I suppose, by the pattern, I must be getting iterations upto 5290 such iterations, and each one of them will take around a couple of weeks, which may increase with higher sequences. Can someone explain what this software is doing to check the primality of this Mersenne prime.
Also, why was I assigned this number, is everyone randomly assigned a prime number for which $M_p$ has not been verified/tested yet as in case of this number it's primality has been tested before as the main page of the site says that all exponents below 78 421 769 have been tested at least once, so I must be just verifying it, meaning most likely it won't come out as a prime, otherwise they must have had it tested on different types of software by now.
Also, can I choose my own prime exponent larger than $78421769$ to check the primality?
P.S.- If this question doesn't fit on this site, please direct me to appropriate place. Also, let me know if I should create a GIMPS tag?
| Each one of those iterations is just squaring the number then put it in modulo of whatever your prime exponent is. Iteration 160000/52898149 simply means that your computer system has completed 160000 such iterations. ms/iter counts how many milliseconds are needed on average for the last 10000 iterations, and ETA estimates the time needed to finish the calculations using the current ms/iter.
A prime number exponent is randomly assigned to users using the software for the first time according to the assignment rules. Since M52898149 is checked once using LL in 2012-12-17 and not checked twice, double-checking is commenced to make sure that the results(residue) of the first test is correct.
If you want to create custom assignments, such as to check the primality of larger prime exponents, you can use the Assignments tab under the Manual Testing column, then choose your number of workers available, number of jobs for a worker, your preferred work type and the (optional) exponent range which you get your exponents.
| {
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change in unit radial vector and unit polar angle due to change in polar angle In cylindrical coordinates, the set B of basis vectors is
$B=\left \{ \vec{e}_{r},\vec{e}_{\theta} \right \}$.
It is clear, geometrically, that
$\frac{\partial e_{r}}{\partial r}=0=\frac{\partial e_{\theta}}{\partial r}$
However, I am unable to convince myself that
$\frac{\partial e_{r}}{\partial \theta}=e_{\theta}$ and $\frac{\partial e_{\theta}}{\partial \theta}=-e_{r}$
Any help to shed light on this is appreciated.
Thanks in advance
| This is for polar coordinates, cylindrical would have a third basis vector $e_z$, but regardless, as the angle $\Delta\theta$ goes to $0$ in the ccw direction, the differential angle makes an appearance resulting in a differential change in the direction of the radial vector
So wherever $e_r$ was pointing, it now has to rotate over an amount of $d\theta$ (ccw) to point in the new correct direction, and the direction of where it should go is already accounted for by $e_\theta$
Likewise, $e_\theta$ is perpendicular to $e_r$, so when going counterclockwise, the new $e_\theta$ is more flattened horizontally than the original $e_\theta$, so the direction $e_\theta$ must go when rotated by $d\theta$ is in the direction of $-e_r$
This is to say
$$\underline{e_r}=\cos\theta\underline{i}+\sin\theta\underline{j}$$
By definition, they are orthogonal
$$\underline{e_r}\cdot\underline{e_\theta}=0$$
And we want $e_\theta$ to have positive orientation sense, so
$$\underline{e_\theta}=-\sin\theta\underline{i}+\cos\theta\underline{j}$$
And we can now clearly see
$$\frac{d\underline{e_r}}{d\theta}=-\sin\theta\underline{i}+\cos\theta\underline{j}=\underline{e_\theta}$$
$$\frac{d\underline{e_\theta}}{d\theta}=-\cos\theta\underline{i}-\sin\theta\underline{j}=-\underline{e_r}$$
| {
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Finitely-generated abelian groups in a long exact sequence I learned here that if the outer groups in a short exact sequence are finitely-generated, then the middle group is, too.
Question. Is there a generalization of this to long-exact sequences?
Julian Kuelshammer mentioned the horseshoe lemma but I'm not really sure how to apply it.
| Well, if you have an exact sequence $$A\stackrel{f}\to B\stackrel{g}\to C$$ where $A$ and $C$ are finitely generated, then so is $B$, This follows from the short exact sequence $$0\to \ker(g) \to B\to \operatorname{im}(g)\to 0$$ where $\ker(g)=\operatorname{im}(f)$ and $\operatorname{im}(g)$ are finitely generated because $A$ and $C$ are. So you don't need to start with a short exact sequence: in an exact sequence of any length, any term trapped between two finitely generated terms is finitely generated.
However, you cannot separate the finitely generated terms by more than one term in between. For instance, for any $A$, there is an exact sequence $$0\to A\to A\to 0,$$ where $0$ is finitely generated by $A$ need not be.
| {
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argument and atan/arctan As I have to works with sine voltages and currents, I often have to use complex numbers.
I know that $\tan\arg(Z)=\left(\frac{\Re(Z)}{\Im(z)}\right)^{-1}$
but, how to prove that we need to add $ \pi$ in case $\Re(Z)\leq0$ ?
I know that $\arctan(x) + \arctan(\frac{1}{x})=\pm \frac\pi2, \forall x \in \mathbb{R}^*$ but I cant go any further.
| That is because, when you want to determine the argument of a complex number $z=x+iy$, really you have to solve, not a single trigonometric equation, but a system of trigonometric equations:
$$\begin{cases}\cos\theta=\dfrac x{\sqrt{x^2+y^2}}\\[1ex] \sin\theta=\dfrac y{\sqrt{x^2+y^2}}\end{cases} $$
This system implies the equation $\;\tan\theta=\dfrac yx$, but the converse is not true, and the latter equation determines $\theta$ modulo $\pi$, whereas the system determines it modulo $2\pi$.
To know if you have to add (or subtract a $\pi$), you can use the signs of $y$ and $y$:
*
*if $\dfrac y x >0$, either $x$ and $y$ are both positive, or they're both negative.
*if $\dfrac y x <0$, a single one of them is positive.
Thus considering the signs of $x$ and $y$ lets you know whether you have to add $\pi$ to the solution found from the tangent, or not.
| {
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Why can't we write $\sin x$ as $\prod_{n=0}^{\infty}\left(x^2-n^2\pi^2\right)$? Why can't we write $\sin x$ as $\prod_{n=0}^{\infty}\left(x^2-n^2\pi^2\right)$?
Since $ n\pi$ where $n \in \mathbb{N}$ are all the roots of $\sin{x}$, then by the fundamental theorem of arithmetic it may be written as
$$\sin x=\prod_{n=0}^{\infty}\left(x^2-n^2\pi^2\right)$$
But this does not actually represent $\sin{x}$ as shown by graph below
Then what modifications have to be made to make both graphs look similar, Please Explain?
| The answers so far (which essentially say: "sine is not a polynomial") are ok. However, they fail on the heuristic side. Thinking like this Euler would have never solved the Basel problem.
Here is my attempt:
There are two main problems with your answer:
1) sin only has a single root at $x=0$ whereas your expression has a double root
2) one can expand a polynomial in terms of factors however there is an overall factor that has to be adjusted
To 2): we have that $4x^2-1$ has the solutions $x=\pm 1/2$. However, it is of course not true that $4x^2-1 = (x-1/2)(x+1/2)$ but rather $4x^2-1 =\mathbf{4} (x-1/2)(x+1/2)$
So let us try to fix the problems and propose
$$\sin x = \lim_{N\to \infty} c x \prod_{n=1}^N (x^2-n^2\pi^2) \tag{1}$$
with $c$ a constant that has to be still determined. (It will turn out that the constant $c$ has to depend on $N$ and that $c_N \to \infty$ for $N\to\infty$)
Now, how to set the overall constant $c$: The simplest way is to remember that $\sin x \sim x$ for $x\to0$. Looking at small $x$, we immediately recognize that
$$c = c_N = \prod_{n=1}^N ({-n^2\pi^2})^{-1}$$
is the proper normalization factor. With that, we obtain
$$\sin x = x \lim_{N\to\infty} \prod_{n=1}^N\left(\frac{x^2-n^2 \pi^2}{-n^2 \pi^2}\right) = x\prod_{n=1}^\infty \left(1- \frac{x^2}{n^2 \pi^2}\right). $$
| {
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Does $\sum _{n=1}^{\infty }\left(-1\right)^{n+1}\left(1-\cos\left(\frac{1}{\sqrt{n}}\right)\right)$ converges conditionally? I'm trying to understand whether the following series converges absolutely, conditionally or diverges.
$$
\sum_{n=1}^{\infty}\left(-1\right)^{n+1}\left(1-\cos\left(\frac{1}{\sqrt{n}}\right)\right)
$$
I think that it converges conditionally. But it's getting complicated for me to show that it does not converge absolutely. I've already tried the limit comparison test with $b_n = \frac{1}{\sqrt{n}}$ but the limit results in $0$. I've also tried the integral test, as the conditions for the test hold, but the integral became very complicated.
Is there an easier way?
| Note that
$$1-\cos\left(\frac{1}{\sqrt{n}}\right)= \frac1{2n}+O\left(\frac1{n\sqrt n}\right)$$
then the given series doesn’t converge absolutely by limit comparison test with $\sum \frac1{n}$ and then
$$
\sum _{n=1}^{\infty }\left(-1\right)^{n+1}\left(1-\cos\left(\frac{1}{\sqrt{n}}\right)\right)=\sum _{n=1}^{\infty }\left(-1\right)^{n+1} \frac1{2n}+\sum _{n=1}^{\infty }\left(-1\right)^{n+1} O\left(\frac1{n\sqrt n}\right)
$$
which converges since
*
*$\sum _{n=1}^{\infty }\left(-1\right)^{n+1} \frac{1}{2n}$ converges by alternating series test
*$\sum _{n=1}^{\infty }\left(-1\right)^{n+1} O\left(\frac1{n\sqrt n}\right)$ converges absolutely by limit comparison test with $\sum \frac1{n\sqrt n}$
| {
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image of homomorphism contains free group, then so does domain This question is a question which I had on my abstract algebra exam, I could not solve it:
True or false: suppose $f:G \to H$ is a group homomomorphism and assume that $\operatorname{Im}(f)$ contains a free group of rank $2$, then $G$ contains a free group of rank $2$.
I tried proving this statement, as I thought it is correct (at least intuitively this seems correct to me...). I did not manage, but still would like to find the answer.
I attempted to prove this statement (actually, I tried proving a more general statement, as I thought there would not be much difference between rank $2$ or any rank at least $2$.)
This is my attempt: by the first isomorphism theorem, we have that
$$F_n \leq \operatorname{Im}(f) \cong G/\ker f$$
and therefore, there is a subgroup $G_1$ of $G$ with $\ker f \leq G_1 \leq G$ and $F_n \cong G_1/\ker f$. In particular, $G_1/\ker f$ is finitely generated, with generating elements $g_1\ker f, \ldots, g_n \ker f$.
At this point, I was stuck: I tried doing something with the $g_i$: I tried proving that $\langle g_1, \ldots, g_n \rangle$ was a free group of rank $n$, but this did not work out...
any hints on this question would be appreciated, as I still would like to know the answer (counterexamples
| Your proof is most of the way there.
Let $F = \langle x_1, \dots, x_n\rangle$ be a copy of the free group of rank $n$, and write $H = \langle g_1, \dots, g_n\rangle$ for your unknown group. Then the map $x_i \mapsto g_i$ is a surjective group homomorphism $F\to H$.
What's its kernel? Well, whatever it is, it must be contained in the kernel of the composite map $F\to H\to G/\ker f$ sending each $x_i$ to $g_i\ker f$. But the kernel of this map is trivial, because the $g_i \ker f$ generate a free group by assumption. Hence the map $F\to H$ is an isomorphism.
| {
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Dividing concentric super ellipses to equal area slices How the concentric super ellipses as shown in the figure can be divided into parts containing equal area such that the total area of the superellipse A = A1 + A2 + ... An where n = 60 in the shown figure given its semi-major axis a and semi-minor axis b. The origin of the sectors lies at the center of the superellipse. Any help is appreciated.
| Solve the problem for concentric circles in the unit circle centered at the origin. Then stretch the $x$-axis by $a$ and the $y$-axis by $b$ to turn the circles into ellipses. All the areas will be scaled by the same stretch factor.
This is essentially the same question you asked here: Dividing an ellipse into equal area
| {
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Manipulating a factorial equation How do I convert $\dfrac{1\cdot2\cdot3\cdots2n}{n!}$ to $(n+1)\cdot(n+2)\cdot\cdots\cdot(2n)$?
I don't understand how the $n + x$ terms have appeared in this equation.
| You can write :
$$\begin{align}
1\cdot 2\cdot 3\cdots (2n) &= (1\cdot 3\cdot 5\cdots (2n-1))\cdot(2\cdot 4\cdot 6\cdots (2n))\\ &= (1\cdot 3\cdot 5\cdots (2n-1))\cdot 2^n\cdot(1\cdot 2\cdot 3\cdots n)\\ &= (1\cdot 3\cdot 5\cdots (2n-1))\cdot 2^n\cdot n!
\end{align}$$
Note that this goes beyond the question, as it shows that you can rewrite the product in a certain ways which suits your question situation. The "other" answer better suits the need of the problem.
| {
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Limit of $n(1-x)^n$ as $n\to\infty$ when $0Wolfram alpha gives that this is $0$, but I'm not sure how to show it.
I Tried writing as $\frac{(1-x)^n}{1/n}$ and using L'Hopital's rule, but a new $n$ term shows up every time I take the derivative.
I also try to set $h=\frac1n$ and then write as
$$
\lim_{n\to\infty} n(1-x)^n = \lim_{h\to 0} \frac{(1-x)^\frac1h}{h}
$$
but same problem, numerator goes to $0$, denominator to $0$, and L'Hôpital doesn't seem to work because the derivative of the numerator gives $\frac1h (1-x)^{\frac1h -1}$, which has the initial trouble
| You can solve this by taking the logarithm of each side. Let $y = \lim_{n \rightarrow \infty} n(1-x)^n$, and then we see that $$\ln(y) = \ln(\lim_{n \rightarrow \infty} n(1 -x)^n) = \lim_{n\rightarrow \infty } \ln(n (1-x)^n)$$ by using the continuity of the natural log function. Now, we use log rules and see $$\ln(y) = \lim_{n \rightarrow \infty} \ln(n) + n\ln(1-x).$$ Since $1 - x$ is less than 1, $\ln(1-x) < 0$. The $\ln(n)$ term grows logarithmically in $n$ and the $n\ln(1-x)$ term grows linearly in $n$. So $\ln(y)$ tends to $- \infty$, hence $y = e^{\ln(y)}$ tends to 0.
| {
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Number of ways to flip a coin 10 times with no consecutive heads The problem statement is as follows: A fair coin is to be tossed $10_{}^{}$ times. Let $i/j^{}_{}$, in lowest terms, be the probability that heads never occur on consecutive tosses. Find $i+j_{}^{}$.
My solution was to consider the sequence of flips as a string of either [Head then Tail] or [Tail]. Let $x$ represent the number of [Head then Tail] and $y$ represent the number of [Tail]. Then $2x$ + $y$ = $10$.
Then I did casework for each value of $x$:
When $x = 0$ it is bijective to the number of arrangements of $AAAAAAAAAA$, which is $1$.
Then, when $x = 1$ it is bijective to the number of arrangements of $AAAAAAAAB$, which is 9 and so on...
The sum of these values turns out to be $89$ and the number of ways to flip is $1024$, but that is wrong. What is wrong with my solution? Thanks!
| Your $89$ is presumably $1+9+28+35+15+1$
It should be $1+10+36+56+35+6$
Since your $89$ would be correct for the numerator with nine coin tosses, you have presumably missed all those starting with heads, or all those finishing with heads
It is not a coincidence that $89$ and $144$ are consecutive Fibonacci numbers, and an alternative approach would use a simple recurrence
| {
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Suppose we are dealt five cards from an ordinary 52-card deck. What is the probability that Suppose we are dealt five cards from an ordinary 52-card deck. What is the
probability that
(c) we get no pairs (i.e., all five cards are different values)?
(d) we get a full house (i.e., three cards of a kind, plus a different pair)?
Solution:
(c) The number of ways this can happen is equal to $\frac{52 \cdot 48 \cdot44 \cdot 40 \cdot 36}{5!}$. Therefore $1317888/{52\choose5}$.
(d) The number of ways this can happen is equal to $(13)(12) {4 \choose 3}{4 \choose 2} = 3744$. Therefore the probability is $3744/{52\choose 5}$
I don't understand the solutions visually.
For (c) I don't understand why they do $52 \cdot 48 \cdot 44 \cdot 40 \cdot 36$. I suppose they divide by $5!$ because there are five categories of cards?
For (d) I don't understand how they got $(13)(12) {4 \choose 3}{4 \choose 2} = 3744$.
|
Suppose we are dealt five cards from an ordinary $52$-card deck. What is the probability that we get no pairs?
We must select cards from five of the thirteen ranks. For each selected rank, we must select one of the four suits. Hence, the number of favorable cases is
$$\binom{13}{5}4^5$$
Since there are $\binom{52}{5}$ ways to select five of the $52$ cards in the deck,
$$\Pr(\text{five cards of different ranks}) = \frac{\dbinom{13}{5}4^5}{\dbinom{52}{5}}$$
As for the given solution: The first card that is selected can be any of the $52$ cards in the deck. Since the second card that is selected must be of a different rank, it can be selected in $48$ ways. Since the third card that is selected must be of a different rank than each of the first two cards, it can be selected in $44$ ways. Continuing in this way, we get $52 \cdot 48 \cdot 44 \cdot 40 \cdot 36$ ordered selections of five cards of different ranks. However, the order of selection does not matter, so we must divide by the $5!$ orders in which the same five cards could be selected, so the number of favorable cases is
$$\frac{52 \cdot 48 \cdot 44 \cdot 44 \cdot 36}{5!}$$
Dividing by $\binom{52}{5}$ gives the probability that each card is of a different rank.
You should check that
$$\binom{13}{5}4^5 = \frac{52 \cdot 48 \cdot 44 \cdot 44 \cdot 36}{5!}$$
Suppose we are dealt five cards from an ordinary $52$-card deck. What is the probability that we get a full house?
There are $13$ ways to select the rank from which three cards are selected and $\binom{4}{3}$ ways to select three of the four cards of that rank. There are $12$ ways to select the rank from which two cards are selected and $\binom{4}{2}$ ways to select two cards of that rank. Hence, the number of favorable cases is
$$\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{2}$$
Since there are $\binom{52}{5}$ ways to select five cards from the deck,
$$\Pr(\text{full house}) = \frac{\dbinom{13}{1}\dbinom{4}{3}\dbinom{12}{1}\dbinom{4}{2}}{\dbinom{52}{5}}$$
| {
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Upper bound in an integral with exponential I'm looking to find an upper bound on the following integral
$$\int_0^\infty K(u)S(t-u)du\,, $$
where
$$ K(u) = e^{-u}(u-u^2/2), $$
and $ S(t) < C$ for some constant $C$.
Could someone help?
| Observe the fact that since $S(t)<C$, we can use the estimate that $$\int_{0}^\infty K(u)S(t-u)du<|C|\,|\int_0^\infty e^{-u}(u-u^2/2)du|\leq|C|\int_0^\infty |K(u)|du$$
Now to finish your proof, you have to show that $\int_{0}^\infty |e^{-u}(u-u^2/2)|du$ exists and is finite.
Edit: Consider the derivative of $\frac{1}{2}e^{-u}u^2$ and the limit for $u\to\infty$.
| {
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Limit $ \sum_{k=0}^∞ \left( \sum_{j=0}^k \binom{k}{j} \left(-\frac{1}{3}\right)^j \right) $ I have to find the limit of the following series:
$$ \sum_{k=0}^∞ \left( \sum_{j=0}^k \binom{k}{j} \left(-\frac{1}{3}\right)^j \right) $$
I don't even know how to approach this... Any help would be very appreciated
| Since,
by the binomial theoram,
$ \sum_{j=0}^k \binom{k}{j}x^j
=(1+x)^k
$,
$\begin{array}\\
\sum_{k=0}^∞ \left( \sum_{j=0}^k \binom{k}{j}x^j \right)
&=\sum_{k=0}^∞ (1+x)^k\\
&=\dfrac{1}{1-(1+x)}
\qquad\text{geometric series with ratio }1+x\\
&=\dfrac{-1}{x}\\
\end{array}
$
Putting $x=-\frac13$,
this gives
$\dfrac{-1}{-\frac13}
=3$.
Note that the sum
does not converge
if $x > 0$,
or else you would
get the nonsensical result
(if, say,
$x=\frac12$),
$\sum_{k=0}^∞ \left( \sum_{j=0}^k \binom{k}{j}(1/2)^j \right)
=-2$.
| {
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Let $\varphi(x) = \frac{1 - e^{-ax}}{1 + e^{-ax}}$, proof that $\varphi'(x) = \frac{a}{2}(1-\varphi^2(x))$ I am trying to find the required steps to reach that derivative, but I am not finding the right way for that. My current development has the following steps:
$\varphi(x) = \dfrac{1 - e^{-ax}}{1 + e^{-ax}}$, then
$\varphi'(x) = \dfrac{(0-(-ae^{-ax}))(1 + e^{-ax})-(1-e^{-ax})(0-ae^{-ax})}{(1+e^{-ax})^2}$
$\varphi'(x) = \dfrac{ae^{-ax}(1 + e^{-ax})+(1-e^{-ax})ae^{-ax}}{(1+e^{-ax})^2}$
$\varphi'(x) = \dfrac{2ae^{-ax}}{(1+e^{-ax})^2}$.
Now, I am trying to find some way assuming that
$$\dfrac{1 - e^{-ax}}{1 + e^{-ax}} = \tanh\left(\frac{ax}{2}\right)$$
but without success yet.
| You can approach the problem through basic differential equation theory by going "backwards". Suppose that you have the separable ODE
$$
\tag{$\star$}
\frac{\varphi'}{1-\varphi^2}=\frac{a}{2}
$$
with boundary condition $\varphi(0)=0$ given by the form of $\varphi$ (just calculate it in $0$).
By integrating on both sides you have
$$
\int_0^{\varphi(x)} \frac{\mathrm dy}{1-y^2} = \frac{a}{2}x
$$
so calculating the integral on the left yields
$$
\text{arctanh}(\varphi(x))=\frac{a}{2}x,
$$
which by using the definition of $\tanh$
$$
\tanh(x)=\frac{1-e^{-2x}}{1+e^{-2x}},
$$
becomes
$$
\varphi(x)=\frac{1-e^{-ax}}{1+e^{-ax}}
$$
This shows that if we take $\varphi$ as given by the problem, then $\varphi$ satisfies $(\star)$.
| {
"language": "en",
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} |
Rademacher theorem for 2nd order derivative The (simplest form of the) Rademacher theorem reads as follows:
Any Lipschitz continuous function $f: \mathbb{R} \to \mathbb{R}$ is Lebesgue-almost everywhere differentiable.
In other words: If the finite difference $\Delta_h^1[f](x) := f(x+h)-f(x)$ satisfies $$|\Delta_h^1[f](x)| \leq C |h|, \qquad x,h \in \mathbb{R} \tag{1}$$ for some absolute constant $C>0$, then $f$ is almost everywhere differentiable.
Question: Is there a generalization for derivatives of second order?
More precisely, if we replace the finite difference $\Delta_h^1$ by a second-order difference, for instance
$$\Delta_h^2[f](x) := f(x+h)-2f(x)+f(x-h), \tag{2}$$
then does
$$|\Delta_h^1[f](x)| \leq C_1|h| \qquad \quad |\Delta_h^2[f](x)| \leq C_2 |h|^2, \qquad x,h \in \mathbb{R} \tag{3}$$
imply that $f$ is almost everywhere twice differentiable? If not, then what additional information on the regularity gives the estimate $(3)$ compared to the almost everyhwere differentiability which follows from the weaker assumption $(1)$?
The obvious idea would be to try to apply the Rademacher theorem twice (first to $f$ and then to its derivative $f'$), but unfortunately the estimate
$$|\Delta_h^1[f'](x)| \leq C |h|$$
will, in general, only hold for Lebesgue almost every $x,h$, and therefore it is not possible to apply the Rademacher theorem directly to $f'$.
| I believe the answer is yes. Writing $D$ for the derivative in the sense of distributions, and $f'$ for the pointwise derivative:
It's easy to see that $\frac1{h^2}\Delta_h^2 f\to D^2f$ in the sense of distributions (or $2D^2f$ or $\frac 12D^2f$ or whatever it is). So the hypothesis implies $$D^2f\in L^\infty.$$This implies that $Df$ is a Lipshitz function. In particular, $Df$ is continuous, which implies that $f$ is differentiable everywhere and $f'=Df$. Since $Df$ is Lipschitz this implies that $f'$ is differentiable almost everywhere.
| {
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"source": "stackexchange",
"question_score": "2",
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} |
Exponential equation with double radical I'm trying ti solve this exponential equation:
$(\sqrt{2+\sqrt{3}})^x+(\sqrt{2-\sqrt{3}})^x=2^x$.
Here my try:
$\sqrt{2+\sqrt{3}}=\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}$ and
$\sqrt{2-\sqrt{3}}=\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}$
So i get this relation:
$\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}}$
Using the substitution $t=\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}$ the equation can be written as
$(t)^x+(\frac{1}{t})^x=2^x$.
And from this i wrote:
$t^{2x}-(2t)^x-1=0$.
Now i don't how to proceed. Any suggestions?
Thanks!
| Hint:
Write $s= \sqrt{2+\sqrt{3}}$, then $\sqrt{2-\sqrt{3}} = {1\over s}$
Then $$s^x+({1\over s})^x\geq 2 \implies x\geq 1$$
| {
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Compatibility of a Kähler form On a complex manifold $M$ a Kähler form is a symplectic form $\omega$ which is compatible with the canonical almost complex structure $J$ in the following sense
$$\omega({}\cdot{},J{}\cdot{})$$
is a Riemannian metric tensor, i.e. symmetric and positive definite.
From $J^*\omega=\omega$ we have $\omega\in\Omega^{1,1}$ then I find the following weird
$$\omega(\partial_{z_1},J \partial_{z_1})= i\omega(\partial_{z_1}, \partial_{z_1}) =0$$
because $\omega\in\Omega^{1,1}$ but it should be positive .
Can someone tell me where I am wrong?
| You are not wrong. You can double check this as follows:
$$g(\partial_z,\partial_z) = \frac{1}{4}g(\partial_x-i\partial_y,\partial_x-i\partial_y) = \frac{1}{4}(1-1)=0.$$
The metric $g$ I use here is the Riemannian metric on your manifold extended by $\mathbb{C}$-linearity on the complexified tangent bundle. This suggests you that this metric is actually of type $(1,1)$.
| {
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What is the base of $\log x$? I've seen "$\log x$" being used in some papers (and by Wolfram|Alpha), and I was confused because so far I have only ever seen the $\log$ used with a base ( so e.g. $\log_y x$).
Am I correct that $\log x = \log_e x = \ln x$?
*
*If so, why is $\log x$ used over $\ln x$? Isn't the letter more expressive
and less confusing?
*If not, what is the base of $\log x$?
| On a standard scientific calculator, the log button denotes the "common logarithm", i.e. $\log_{10}$. This is consistent with the common usage in engineering and the natural sciences; for example, the pH scale used for measuring acidity, the Richter scale used for measuring earthquake intensity, and the decibel scale used for measuring sound intensity are all defined using a base-10 logarithm. Scientific calculators use the ln button to indicate the "natural logarithm", i.e. $\log_e$.
In contrast, mathematicians tend to use the symbol $\log$ to refer to $\log_e$. That's because (from the point of view of pure mathematics) there is nothing special about the number $10$, and no real reason to define a logarithm to a a single, arbitrary privileged base. From a pure mathematics standpoint, the only logarithm that really matters is $\log_e$, so this is what the generic symbol $\log$ refers to.
For some reason many mathematicians tend to be oblivious to the fact that outside of our own tribe nearly everyone uses the symbols log and ln to refer to different things, and rather obstinate in insisting that $\log = \log_e$, as if notation were not merely a convention but rather somehow a law of nature or a logical necessity. Notation is always conventional, and all conventions are local.
| {
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Prove that a strongly convex function imples $2c(F(w)-F_*) \leq ||\nabla F(w)||_2^2$
The proof is given as follows:
My question is why is the unique minimizer $\bar{w}_* = w - \frac{1}{c} \nabla F(w)$?
| There is an easier argument for quadratic functionals: completing the square. The following identity is just as easy to prove as the 1D version you know from highschool:
$$ \alpha ‖x‖^2 + \beta · x = \alpha\left\|x+\frac1{2\alpha}{\beta} \right\|^2 - \frac{‖\beta‖^2}{4\alpha}$$
In your case, this yields (remember, $w$ is fixed, $\bar w$ is varying) with $$x = w - \bar w, \quad \alpha = \frac{c}{2},\quad \beta = \nabla F(w)$$
$$⟹ q(\bar w) = C(w) + \frac c{2}\left\|\bar w - \left(w-\frac1c∇ F(w)\right)\right\|^2 $$
where $C(w) = F(w)-\frac{‖∇ F(w)‖^2}{2c}$. Its now obvious what the minimiser and minimum is.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2798749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Hypothesis testing - Critical region and confidence level Exercise :
For the estimation of the unknown rate of votes $p$ that a political group $A$ will gather in the following elections, suppose we selected a random sample of $n=15$ voters. Suppose that you want to check the null hypothesis $H_0 : p = 0.5$ against the alternative $H_1 : p< 0.5$. Suppose that the critical region of this specific hypothesis test is $K=\{y\leq 2\}$, where $y$ is the observed number of voters who voted for the political group $A$.
i) Calculate the confidence level $a$ of the above hypothesis test.
ii) If the political group $A$ eventually gathers a rate of $30\%$ in the elections, calculate the probability of the type II error for the above hypothesis test.
Question - Request : For our upcoming exams, we are supposed to be able to handle such problems but we weren't really introduced to any of them due to shortage of time. I kindly request some tips or a thorough explanation such as I will be able to get a grip on such problems. (This is a past exams exercise)
| So $\alpha$ represents the probability of making a Type I error, that is, rejecting $H_0$ when $H_0$ is true, that is why we say "we are 95% confident that $H_0$ is true", because there is a $1-95\%=\alpha$ probability of this conclusion not being true. Therefore:
\begin{align}
\alpha&=Pr(\text{Reject } H_0 | H_0) \\ \\
&=Pr(y \leq 2 | p=0.5) \\ \\
&= \sum_{i=0}^2 {15\choose{i}}0.5^i(1-0.5)^{15-i} \\ \\
&=0.5^{15}\sum_{i=0}^2 {15\choose{i}} \\
\end{align}
Now $\beta$ is the probability of doing a Type II error given $p=0.3$:
\begin{align}
\beta&=Pr(\text{Accept }H_0|H_1) \\ \\
&=Pr(y > 2 | p=0.3) \\ \\
&=1-Pr(y \leq 2 | p=0.3) \\ \\
&=1-\sum_{i=0}^2{15\choose{i}}0.3^i(1-0.3)^{15-i}
\end{align}
| {
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Find a Basis for $W=\{p(x)\in V: p(1)=p'(1)=0\}$
Let $V=\mathbb{P_4}$ and $W=\{p(x)\in V: p(1)=p'(1)=0\}$. Assuming that $W$ is a subspace of $V$, find a basis for $W$ and thereby determine the dimension of $W$.
I think that $\dim(W)=3$ as there are two restrictions enforced upon $W$ $($ $p(1)=1$ and $p'(1)=0$$)$ and $\dim(\mathbb{P_4})=5$
However, I'm unsure of how to find a basis for $W$. I have tried to create components for my basis that satisfy the conditions of $W$, but is there a more concrete/routine way of finding my basis? I could not find solutions on this problem, so it is difficult to work backwards from a solution.
EDIT:
My solution is as follows.
Let $p(x)\in\mathbb{P_4}$, then $p(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4$. Now \begin{align*}
p(1)&=a_0+a_1+a_2+a_3+a_4\\
p'(1)&=a_1+2a_2+3a_3+4a_4\\
\end{align*}
Hence $a_0=a_2+2a_3+3a_4$. Substituting this back into $p(x)$, we find $p(x)=a_1x+a_2(1+x^2)+a_3(2+x^3)+a_4(3+x^4)$.
Do the coefficients form my basis? If so, why?
| Consider the linear map $F\colon\mathbb{P}_4\to\mathbb{R}^2$ defined by
$$
F(p)=\begin{bmatrix} p(1) \\ p'(1) \end{bmatrix}
$$
Then $W=\ker F$. The matrix of $F$ with respect to the standard basis $\{1,x,x^2,x^3,x^4\}$ is
$$
\begin{bmatrix}
1 & 1 & 1 & 1 & 1 \\
0 & 1 & 2 & 3 & 4
\end{bmatrix}
$$
which has the RREF
$$
\begin{bmatrix}
1 & 0 & -1 & -2 & -3 \\
0 & 1 & 2 & 3 & 4
\end{bmatrix}
$$
A basis for the null space of this matrix consists of three vectors, for instance
$$
\begin{bmatrix}
1 \\
-2 \\
1 \\
0 \\
0
\end{bmatrix}
\qquad
\begin{bmatrix}
2 \\
-3 \\
0 \\
1 \\
0
\end{bmatrix}
\qquad
\begin{bmatrix}
3 \\
-4 \\
0 \\
0 \\
1
\end{bmatrix}
$$
which correspond to the polynomials
$$
1-2x+x^2
\qquad
2-3x+x^3
\qquad
3-4x+x^4
$$
of which the vectors above are the coordinate vectors with respect to the standard basis, so they form a basis for $W$.
| {
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Evaluate the limit with exponents using L'Hôpital's rule or series expansion Evaluate the limit$$\lim_{x\to 0}\dfrac{\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}}
-\sqrt{ab}}{x}$$ It is known that $a>0,b>0$
My Attempt:
I could only fathom that $$\lim_{x\to 0}\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}}=\sqrt{ab}$$
| just use the l-H hospital for it,a tip for that is you can write the $((a^x+b^x)/2)^{1/x}$ to this inform: $e^{[\ln((a^x+b^x)/2)]/x}$,so you can use chain rule to differeitiate it,here you are,the answer leaves for you
| {
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What is the integral of the indicator function of VITALI? For a positive function $f$, the Lebesgue integral is the supremum of the integral of all simple functions below $f$. So if $f$ is the indicator function for the VITALI set, the lebesgue integral for it must exist. But in general, the integral of an indicator function of a set is the measure of that set. But VITALI is unmeasurable!?
| Any measurable subset $U$ of Vitali's set $V$ has measure zero. Therefore, if you take the supremum of the integrals of simple functions not larger than the indicator of vitali's set, their integrals are zero, and so is the supremum.
If $U\subset V$ is measurable, then for an enumeration $q_i$ of the rationals we have $U+q_i\pmod{1}$ disjoint for different $i$'s, measurable and of the same measure. Therefore $\sum_{i=0}^{\infty}\nu(U)=\sum_{i=0}^{\infty}\nu(U+q_i\pmod{1})=\nu(\bigcup_{i=0}^{\infty}(U+q_i\pmod{1}))\leq\nu([0,1])=1$. Therefore $\nu(U)=0$.
| {
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Can a quotient of polynomials be simplified before been analyzed? Having the function $f(x) =\frac{X^3+3X^2}{2X^2+4x}$, why it is not the same to analyze $\frac{X^2+3X}{2X+4}$, if it verifies $\frac{X^3+3X^2}{2X^2+4x}=\frac{X^2+3X}{2X+4}$ ? In this case, the first one has only one root, while the second one has another one in $0$.
In general, can a function be simplified before been analyzed? (I mean, find roots, continuity, maxima and minima...)
| Hardy is fairly robust on this issue (and while possibly slightly at variance with current set-theoretic dogma, nevertheless quite consistent with the standard calculus interpretations):
The function $\frac{x^2-1}{x-1}$ has no value for $x=1$; for $x=1$, $\frac{x^2-1}{x-1}$ is strictly and absolutely meaningless. The fact that its limit for $x=1$ is $2$ is entirely irrelevant. The functions $\frac{x^2-1}{x-1}$ and $x+1$ are different functions. They are equal when $x$ is not equal to $1$. Similarly the function $y=\frac{x}{x}$ is $=1$ when $x \neq 0$ and undefined for $x=0$. To calculate $f(x)$ for $x=0$ we must put $x=0$ in the expression of $f(x)$ and perform the arithmetical operations which the form of the function prescribes, and this we cannot do in this case.
(Mathematical Gazette 1907, 4 pp. 13–14)
| {
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Is it a contraction map? I have that map
$$
f:(\mathbb{R}^2,d_1)\to(\mathbb{R}^2,d_1)\\
(x,y)\mapsto \left(y-\frac13 \tanh(x)+\frac14 Argsh(y),\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 4x-\tanh(y)+\frac43 Argsh(x)\right)
$$
where $d_1((x,y),(x',y'))=|x-x'|+|y+y'|$
I calcultate $$ d_1(f(x,y),f(x',y'))\leq \frac{43}{12}|y-y'|+\frac{13}{3}|x-x'|$$ like this it is not contraction
Can someone tel me if what i do is correct ?
$$
d_1(f(x,y),f(x',y'))\leq |y-y'|+\frac13|\tanh(x)-\tanh(x')|+\frac14 |Argsh(y)-Argsh(y')|+4|x-x'|+|\tanh(y')-\tanh(y)|+\frac43 |Argsh(x)-Argsh(x')|\\ \leq |y-y'|+\frac13|x-x'|+\frac14|y-y'|+4|x-x'|+|y'-y|+\frac43|x-x'|$$
Thank you
| You work with the metric $d_1((x,y),(x',y') = \lvert x - x' \rvert + \lvert y - y' \rvert $ (typo in your question!). Your inequality
$$d_1(f(x,y),f(x',y')) \le 43/12\lvert y - y' \rvert + 13/3\lvert x - x' \rvert$$
does not help because you cannot be sure that it best possible.
It suffices to look at special values. Let $(x,y) = (a,0)$ with $a >0$ and $(x',y') = (0,0)$. We have $f(a,0) = (-1/3\tanh(a),4a + 4/3Argsh(a))$ and $f(0,0) = (0,0)$. The distance between these two points is $1/3 \tanh(a) + 4a + 4/3 Argsh(a) > 4a$. Therefore $f$ is no contraction.
One more remark: $Argsh$ seems to be used only in French literature. It is the inverse of the hyperbolic sine $\sinh$ and is most frequently denoted by $arsinh$.
| {
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Why does divergence represent expansion or contraction? Why does $\mathrm{div}\ V$ represent how much $V$ is expanding or contracting? By its definition I know that diverging means deviating from its original path, but what about $\mathrm{div}\ V$ makes it so $V$ expands or contracts, is there a $\mathrm{div}$ formula that explains it?
| Let $D$ be a small spherical region centered at point $P$.
By the divergence theorem,
$$(\nabla\cdot{\bf V})_P \approx \frac{1}{\mathrm{vol}(D)} \iint_S {\bf V}\cdot{\bf n}\, dS.$$
But $\iint_S {\bf V}\cdot{\bf n}\, dS$ is the net flux of ${\bf V}$ through the surface $S$ of $D$.
Thus, $\nabla\cdot{\bf V}$ is a measure of source of the field ${\bf V}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2799582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Proving $\lim\limits_{n\to\infty}\int_0^{\pi/4} \tan^n{x}\,dx=0$ How would you prove that $\displaystyle\lim\limits_{n\to\infty}\int_0^{\pi/4} \tan^n{x}\,dx=0$.
It is obvious if you see the graph of $\tan^n{x}$ on $(0, \pi/4)$ as $n$ increases but i'm looking for a more algebraic way.
This result is for connecting the power reduction formula for $\displaystyle\int_0^{\pi/4} \tan^n{x}\,dx$ to Leibniz formula for $\pi$.
| Squeezing is straightforward:
$$ 0\leq \int_{0}^{\pi/4}\tan^n(x)\,dx \stackrel{x\mapsto\arctan u}{=}\int_{0}^{1}\frac{u^n}{1+u^2}\,du \leq \int_{0}^{1}u^n\,du = \frac{1}{n+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2799740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Inner circle of torus of revolution is calibrated I'm working on the following problem from Lee's "Introduction to Smooth Manifolds":
Let $D \subseteq \mathbb R^3$ be the surface obtained by revolving the circle $(r-2)^2 + z^2 = 1$ around the z-axis, with the induced Riemannian metric from $\mathbb R^3$, and let $C \subseteq D$ be the “inner circle” defined by $C = \{(x,y,z) : z=0, \, x^2 + y^2 = 1\}$. Show that $C$ is calibrated, and therefore is the shortest curve in its homology class.
In this case, a calibration of a Riemannian manifold $M$ is a closed $p$-form $\omega$ on $M$ so that $\omega(v_1, \ldots, v_p) \leq 1$ for every orthonormal set $\{v_1, \ldots, v_p\}$, and a Riemannian submanifold $S \subseteq M$ is calibrated if there is a calibration $\omega$ so that $\iota_S^*\omega$ is the induced Riemannian volume form on $S$.
So I need to find a 1-form $\omega \in \Omega^1(D)$ for which $\omega(v) \leq 1$ for every unit tangent vector $v\in TD$, and $\iota_C^* \omega$ is the induced Riemannian volume form on $C$. Let $F(\theta, t) = \big((2t+\cos t)\cos\theta, (2+\cos t)\sin\theta, \sin t\big)$; then $F : [0,2\pi]^2 \to \mathbb R^2$ parametrizes $D$. My original thought was to let $\omega$ be the 1-form $(\overline F_\theta)^\flat$, i.e. $(\overline F_\theta)^\flat(v) = g_D (v, \overline F_\theta)$ for every $v \in TD$, where $g_D$ is the induced Riemannian metric on $D$ and $\overline F_\theta = F_\theta / |F_\theta|$ is the normalization of the tangent vector $F_\theta$. It seems clear to me in this case that $\iota^*_C \omega$ is the induced volume form on $C$, and that $\omega(v) \leq 1$ for unit tangent vectors $v$. However, I have two concerns:
*
*I'm not sure how to show $\omega = (\overline F_\theta)^\flat$ is closed, and
*if I could show the above, why would a similar argument not apply to the outer circle of $D$, which is clearly not minimizing in its homology class?
I would appreciate any thoughts or clarifications.
| Assume that $\omega (x)= g(x,\frac{\partial_\theta}{|\partial_\theta|^2})$.
Then $d\omega (\partial_t ,\partial_\theta )=0$ so that it is a closed
form.
If $c$ represents $C$ and $ c\sim \alpha$ and $\alpha$ has a unit
speed, then $$ {\rm length\ of}\ c=\int_0^{2\pi}\ \omega ( c' )=
\int_0^{2\pi}\ \omega(\alpha')\ dt \leq {\rm length\ of}\ \alpha $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2799872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Use $ \ \nabla f(3,2) \ $ to find a vector normal to the curve at $ \ (3,2)\ $ View the curve $ \ (y-x)^2+2=xy-3 \ $ as a contour of $ \ f(x,y) \ $
Use $ \ \nabla f(3,2) \ $ to find a vector normal to the curve at $ \ (3,2)\ $
Answer:
Let $ \ f(x,y)=(y-x)^2-xy+5=0 \ $
Then,
$ \nabla f(x,y)=\left\langle f_x,f_y \right\rangle \ = \left\langle 2(x-y)-y,2(y-x)-x \right\rangle $
Therefore,
$ \nabla f(3,2)=\left\langle0,-5 \right\rangle \ $ , which is normal t the level curves but not on the curve $ \ f(x,y) \ $
How to find find the vector normal to the curve $ \ f(x,y) \ $ using $ \ \nabla f(3,2) \ $ ?
Help me doing this.
| Here is a graph of your curve, plotted by Maple.
It shows clearly that the curve is horizontal at $(3,2)$, so the normal is vertical, so your answer is correct and the software marking it is wrong.
The only suggestion I could make is that any vertical vector is normal to the curve at this point, that is, any vector $(0,b)$ with $b\ne0$. Try a unit vector, $(0,1)$ or $(0,-1)$ and see if that gets marked correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2799962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Given a set $U=\{1,2,3,...n\}$. How to partition elements in $U$ into sets $A$ and $B$ such that the sum of the elements in $A$ and $B$ is in minimum. I came across a particular problem wherein I think the answer would necessitate me to partition $U=\{1,2,3,...n\}$ into subsets $A$ and $B$ such that the sum of the elements in $A,B$ is at minimum.
To show that it is in minimum, the following should hold:
Let:
$sumA$ be the sum of the elements in $A$
$sumB$ be the sum of the elements in $B$
Then $sumA,sumB$ is at $minimum$ when $|sumA-sumB|$ is at $minimum$.
Possible Cases:
We are actually looking for all subsets of $U$. The number of subsets of $U$ is $2^n$.
Example: Let $U$={1,2,}. Then the number of subsets are $2^2=4$:
$\{1,2\}$,$\{1\}$,$\{2\}$,$\{\}$
We will drop $\{1,2\}$,$\{\}$ since they irrelevant to our task. So there should be $2^n-2$ possible subsets to be paired, in this case $\{1\}$,$\{2\}$.
Let $p$ be the possible subsets, then:
$$p=2^n-2$$
The number of ways to arrange these subsets into $2$ sets is $p\cdot\ (p-1)=2^n-2 \cdot\ (2^n-3)$.
Problem: My problem is to figure out what will be the exact combination of subsets $A,B$ such that $sumA$ and $sumB$ is at the minimum. Any help would be very helpful.
| Split them into consecutive groups of $4$ starting with the largest ( so the first block is $\{n-3,n-2,n-1,n\}$ ) and in each group $\{a,a+1,a+2,a+3\}$ split them so $a$ and $a+3$ are in $A$ and $a+1$ and $a+2$ are in $B$.
If the number of elements is a multiple of $4$ we are done.
If the remainder is $1$ put $1$ in $A$.
If the remainder is $2$ put $1$ in $A$ and $2$ in $B$.
If the remainder is $3$ put $1$ and $2$ in $A$ and $3$ in $B$.
It is not hard to see this is optimal.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2800037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
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