Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
|---|---|---|
Line intersecting spheroid I have two planes $(A): u_{1}x + v_{1}y + w_{1}z = d_{1}$ and $(B): u_{2}x + v_{2}y + w_{2}z = d_{2}$.
They intersect together, then they yield a line $(L)$ that has a direction vector $M (x_{M},y_{M},z_{M})$
$M$ is the cross product of the normal vectors of $A$ and $B$
$M = (u_{1},v_{1},w_{1})×(u_{2},v_{2},w_{2})$
The parametric equations of the line $L$ of a parameter $t$ are:
$x = x_{P} + x_{M}.t$
$y = y_{P} + y_{M}.t$
$z = t$
I intersect this line $L$ with an spheroid $E$ of semi-axis major $a$ in the equatorial plane $xoy-Plane$ along $x-axis$ and $y-axis$ and semi-axis minor $b$ along $z-axis$
My problem is how to find the coordinates of the two points of intersection?
| $(x_p, y_p, z_p)$ is any point on the line of intersection.
For example, suppose the two planes are x+ 2y+ z= 1 and 2x- y+ z= 3. Subtracting the first equation from the second gives x- 3y= 2 so x= 3y+ 2. Taking, arbitrarily, y= 0, x= 2. Then both 2+ 0+ z= 1 and 4- 0+ z= 3 give z= -1. (2, 0, -1) is one of the infinitely many points on the line of intersection.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is it possible to determine the largest number $\tau$ such that the spectral radius $\rho(A\pm \tau ee^T) < 1$ Let $A \in M_n(\mathbb R)$ with no particular structure assumed and the spectral radius $\rho(A) < 1$. Let us denote the all $1$ vector by $e = (1, \dots, 1)^T$. I would like to determine a number $\tau \in \mathbb R_+$ such that $\rho(A \pm \tau e e^T) < 1$. Since we know $\{X \in \mathbb R: \rho(X) < 1\}$ is open, we could always have some positive number $\tau$. I would be happy if we can find a number $\tau = \tau(A)$ dependent on $A$ such that is sufficient to guarantee $\rho(A\pm \tau ee^T)<1$.
| In this post, we assume that $A$ is symmetric (otherwise I do not see what to show).
Let $spectrum(A)=\{\lambda_1\geq \cdots\geq\lambda_n\}$. We may assume that $\rho(A)=\lambda_1\geq 0$.
$\textbf{Proposition 1}.$ If $\tau< \dfrac{1-\rho(A)}{n}$, then $\rho(A\pm \tau ee^T)<1$.
$\textbf{Proof}$.
Let $U=ee^T$ (symmetric $\geq 0$). Note that $\rho(U)=n$ and $\rho(A+\tau U)\leq \lambda_1+\tau n<1$.$\square$
We can do better when $\rho(A)$ is closed to $1$, that we assume in the sequel.
$\textbf{proposition 2}$. Let $\theta =\sup\{|u_1+\cdots+u_n|; Au=\rho(A) u,||u||=1\}$.
then $\rho(A\pm \tau ee^T)<1$ when $\tau<\tau_0$ where $\tau_0\approx \dfrac{1-\rho(A)}{\theta^2}$.
$\textbf{Proof}$. Let $A(\tau)=A+\tau U$; it's an analytic function. Then the eigenvalues and a basis of (unit length) eigenvectors of $A(\tau)$ are globally analytically parametrizable (even if the eigenvalues (for example $\rho(A)$) present some mutiplicities.
Let $u(\tau)$ be a unitary vector function ($u^Tu'=0$) s.t. $A(\tau)u(\tau)=\lambda(\tau)u(\tau)$. Then $A'u+Au'=\lambda'u+\lambda u'$ and consequently $\lambda'=u^TA'u=u^TUu=(u_1+\cdots+u_n)^2$.
Finally $\lambda(\tau)\approx \rho(A)+\tau \theta^2$. Note that $\theta^2\leq n$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to cut a square on $5$ squares? We can cut any square on $n$ squares if $n>5$ and $n=4$.
The proof is easy by induction. Base cases $n=6,7,8$ are easy to find and then since we can cut a square on $4$ squares we get $3$ new squares, so we go from $n\to n+3$ and we are done.
But I can not find a proof that we can't cut it on $5$ squares. I suppose we should search for some contradiction, but...?
| If you had glue, @greedoid, it would be too easy:
| {
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How do I find the equation of a tangent to a hyperbola whose centre is (h,k)? Given that $\frac{(x-3)^2}{9} - \frac{(y-2)^2}{4} = 1$ is equation a hyperbola,
I have to find its tangent at the point $\left(-2,\frac{14}{3} \right)$.
I know about the equations $c^2=(am)^2-b^2$ and $\frac{xx1}{a^2} - \frac{yy1}{b^2} = 1$ but cant figure how to apply those here since the centre is not at origin.
| Your hyperbola is $\frac{(x- 3)^2}{9}- \frac{(y- 2)^2}{4}= 1$ and you want to find the tangent line to it "at (-2, 14/3)". The first thing I would do is check to make sure that point is on the hyperbola. With x= -2, $(x- 3)^2= 25$ and $\frac{(x- 3)^2}{9}= \frac{25}{9}$. With $y= \frac{14}{3}$, $(y- 2)^2= \frac{64}{9}$ and $\frac{y- 14/3}{4}= \frac{16}{9}$. Yes, a bit to my surprise, $\frac{(x- 3)^2}{9}- \frac{(y- 2)^2}{4}= \frac{25}{9}- \frac{16}{9}= \frac{9}{9}= 1$ and the point is on the hyperbola!
Now, do you know what a "tangent line" is and how to find the slope of a tangent line without simply plugging into formulas? Using "implicit differentiation" to differentiate both sides with respect to x, we have $\frac{2}{9}(x- 3)- \frac{1}{2}(y- 2)y'= 0$. In particular, at x= -2, y= 14/3, $\frac{2}{9}(-5)- \frac{1}{2}\frac{8}{3}y'= -\frac{10}{9}-\frac{4}{3}y'= 0$ so $y'= -\left(\frac{3}{4}\right)\left(\frac{10}{9}\right)= -\frac{5}{6}$. The tangent line is $y= -\frac{5}{6}\left(x+ 2\right)+ \frac{14}{3}$
| {
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What is the degree of the following map?
Let $a \notin S^1$, $f:S^1 \rightarrow S^1$, be given by $z \mapsto \frac{z-a}{|z-a|}$. What is the degree of $f$?
Where by degree we mean under the usual definition here
or general case here .
My current thoughts are usuing the definition that it is the value of the unique lift, when $f$ is idenitfied with $\hat{f}:I \rightarrow S^1$, $\hat{f}(1)=\deg f$. But this requires me to explicitly construct a lift of the map
$$ \hat{f}:I \rightarrow S^1, s \mapsto \frac{e^{2 \pi i s} - a }{ |e^{2 \pi i s} - a | }.$$
which seems plausible but messy. I would be interested for an elegant proof.
| This is the answer proposed by Steve D.
By the homotopy axiom, the map $H(f):H_1(S^1) \rightarrow H_1(S^1)$ is the same for homotopic maps $g \simeq f$.
When $|a|<1$, we give the homotopy
$$ H_t(z):= \frac{z-ta}{|z-ta|}.$$
So the degree of $f$ is equivalent to that of $g(z)=z$, which is $1$.
When $|a|>1$, we give the homotopy
$$H_t(z):= \frac{tz-a}{|tz-a|}$$
so the degree of $f$ is equivalent to that of the constant map, which has degree $0$.
| {
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Compute $\lim_{x\to0} \ln^x(x)$
$$\lim_{x\to0} \ln^x(x)$$
I know I can use L'Hospital and get an answer equal to one.
But how can he ask a question like this even if the function is not continuous from $[0, 1]$. It has only discrete solutions not a graph
| The given function doesn't have discrete values if you plot it on its natural domain.
Speaking about the natural domain of the function, since you're using a real exponent, which is typically defined as $a^x=e^{a\ln x}$, the natural domain of your function is $[1,+\infty)$. Hence you can't talk about a limit around $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2926703",
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Find a Jordan form of a non-diagonazable matrix I am trying to find Jordan's form of this matrix:
\begin{pmatrix}
0 & 1 & 0\\
-4& 4& 0\\
-2 & 1& 2
\end{pmatrix}
The only eigenvalue $r$ is 2 and therefore the simplest eigenvector $v_{1}$ is (0, 0, 1)
To get the other two independent vectors (generalized eigenvectors) for the P matrix we need to compute $(A-rI)v_{2}=v_{1}$ and $(A-rI)v_{3}=v_{2}$ .
However, in the calculation for 2 I get the following system which has no solution:
$$ \left[
\begin{array}{ccc|c}
-2 & 0& 0&0 \\
0& 1& 0&0 \\
0& 0& 0 & 1
\end{array}
\right] $$
Any Idea on what I am doing wrong? If the eigenvalue is 2, when we substract it the lowest row will always contain all zeros...
| I like this method for hand calculations: first, calling your matrix $A,$ let
$$ B = A - 2 I $$
$$
B =
\left(
\begin{array}{ccc}
-2&1&0 \\
-4&2&0 \\
-2&1&0
\end{array}
\right)
$$
A basis for the genuine eigenvectors is given by the convenient
$$
E =
\left(
\begin{array}{cc}
1&0 \\
2&0 \\
0&1
\end{array}
\right)
$$
We may or may not be using these eigenvectors in the form shown. We know that the characteristic equation for $A$ is just showing $B^3 = 0.$ However, the minimal polynomial for $A$ is $B^2 = 0,$ which you can check easily enough.
We are going to make a matrix $R$ with columns $u,v,w;$ on the far right, we take $w$ as any vector with $B^2 w = 0$ (automatic) but $Bw \neq 0.$ Then $v = Bw$ will be a genuine eigenvector. Finally, we will choose an independent eigenvalue $u.$
I like ones and zeros, I choose
$$
w =
\left(
\begin{array}{c}
0 \\
1 \\
0
\end{array}
\right)
$$
Then from $v = B w$ we get
$$
v =
\left(
\begin{array}{c}
1 \\
2 \\
1
\end{array}
\right)
$$
This $v$ is a genuine eigenvector, it is the sum of the two columns of my $E.$
At last, we get to choose some $u$ eigenvector that is not a multiple of $v,$ I choose
$$
u =
\left(
\begin{array}{c}
0 \\
0 \\
1
\end{array}
\right)
$$
$$
R =
\left(
\begin{array}{ccc}
0&1&0 \\
0&2&1 \\
1&1&0
\end{array}
\right)
$$
Next we find $R^{-1} $ and $J = R^{-1}A R,$ which will be the Jordan form if we did it correctly. A piece of luck , because of choosing ones and zeros, the determinant of $R$ is small, actually $1,$ and we calculate
$$
R^{-1} =
\left(
\begin{array}{ccc}
-1&0&1 \\
1&0&0 \\
-2&1&0
\end{array}
\right)
$$
and
$$
J = R^{-1} A R =
\left(
\begin{array}{ccc}
2&0&0 \\
0&2&1 \\
0&0&2
\end{array}
\right)
$$
$$ $$
$$ R^{-1} A R = J $$
$$
\left(
\begin{array}{ccc}
-1&0&1 \\
1&0&0 \\
-2&1&0
\end{array}
\right)
\left(
\begin{array}{ccc}
0&1&0 \\
-4&4&0 \\
-2&1&2
\end{array}
\right)
\left(
\begin{array}{ccc}
0&1&0 \\
0&2&1 \\
1&1&0
\end{array}
\right) =
\left(
\begin{array}{ccc}
2&0&0 \\
0&2&1 \\
0&0&2
\end{array}
\right)
$$
$$ $$
$$ R J R^{-1} = A $$
$$
\left(
\begin{array}{ccc}
0&1&0 \\
0&2&1 \\
1&1&0
\end{array}
\right)
\left(
\begin{array}{ccc}
2&0&0 \\
0&2&1 \\
0&0&2
\end{array}
\right)
\left(
\begin{array}{ccc}
-1&0&1 \\
1&0&0 \\
-2&1&0
\end{array}
\right) =
\left(
\begin{array}{ccc}
0&1&0 \\
-4&4&0 \\
-2&1&2
\end{array}
\right)
$$
COMMENT: the line above "as any vector with $B^2 w = 0$ (automatic)" may appear silly. If, however, you were given a 5 by 5 matrix $A$ with characteristic polynomial $(x - 5)^3 (x-7)^2$ and minimal polynomial
$(x - 5)^2 (x-7),$ the demand for a vector $w$ with $(A - 5I)^2 w = 0$ but $(A - 5I) w \neq 0$ would make some sense, as the eigenvalue $7$ is ignored in this condition.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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How would one prove that a linear combination of convex functions is also convex? As above, how would one mathematically prove that a linear combination of convex functions is also convex?
We know a function defined on a convex set $S$ is convex if:
$$f(tx_1+(1-t)x_2)\leq tf(x_1)+(1-t)f(x_2)$$
where $t$ is from $0$ to $1$
We must prove that $\sum_{i=1}^n a_i f_i(x)$ is also convex given a bunch of functions $f_1, f_2$ etc.
How do i approach this problem? I could say the following:
$tf(x_1)+(1-t)f(x_2)+tg(x_1)+(1-t)g(x_2)=t(f(x_1)+g(x_1))+(1-t)(f(x_2)+g(x_2))$
$f(tx_1+(1-t)x_2)+g(tx_1+(1-t)x_2)\leq t(f(x_1)+g(x_1))+(1-t)(f(x_2)+g(x_2))$
Is this how we show it?
Secondly, how would we prove the same thing for a concave function? Isn't it just adding a - sign? How would i mathematically prove it?
| hint
You should assume the coefficients positive.
Your sum is finite, so you just need to prove that
$$a_1f_1+a_2f_2$$ is convexe.
$$a_i>0.$$
| {
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Solution to this Differential Equation $f''(x)=f(x)f'(x)$ needed I came up with this differential equation and I don't know how to solve it.
$$f''(x)=f(x)f'(x)$$
I attempted to solve it several times, but they were all fruitless. Wolfram Alpha says that the solution is
$$f(x)=\sqrt{2a} \tan\left({\frac{\sqrt{2a}}{2} \cdot (x+b)}\right),$$ where $a$ and $b$ are constants.
How does one get this solution?
| Let us consider your differential equation:
$$f''(x)=f(x)\cdot f'(x)$$
Integrate with respect to $x$ on both sides. Recognize that $df'(x)=f''(x)\ dx$ and $df(x)=f'(x)\ dx$:
$$\int f''(x)\ dx=\int f(x)\cdot f'(x)\ dx\rightarrow \int df'(x)=\int f(x)\ df(x).$$
It follows that
$$f'(x)=\frac{(f(x))^2}{2}+a=\frac{(f(x))^2+2\cdot a}{2}.$$
Divide by $(f(x))^2+2\cdot a$ on both sides:
$$\frac{f'(x)}{(f(x))^2+2\cdot a}=\frac{1}{2}.$$
Integrate with respect to $x$ on both sides. Recognize that $df(x)=f'(x)\ dx$:
$$\int \frac{f'(x)\ dx}{(f(x))^2+2\cdot a}=\int \frac{dx}{2}\rightarrow \int \frac{df(x)}{(f(x))^2+2\cdot a}=\frac{x}{2}+b=\frac{x+2\cdot b}{2}.$$
Redefine $2\cdot b$ as $b$, since it is a constant:
$$\int \frac{df(x)}{(f(x))^2+2\cdot a}=\frac{x+b}{2}.$$
Let $f(x)=\sqrt{2\cdot a}\cdot \tan(s)$ such that $df(x)=\sqrt{2\cdot a}\cdot (\tan^2(s)+1)\ ds$:
$$\int \frac{ds}{\sqrt{2\cdot a}}=\frac{x+b}{2}\rightarrow \frac{s}{\sqrt{2\cdot a}}=\frac{x+b}{2}.$$
Isolate $s$ and let $s=\arctan \left (\frac{f(x)}{\sqrt{2\cdot a}}\right)$:
$$s=\frac{\sqrt{2\cdot a}}{2}\cdot (x+b)\rightarrow \arctan \left (\frac{f(x)}{\sqrt{2\cdot a}}\right)=\frac{\sqrt{2\cdot a}}{2}\cdot (x+b).$$
Therefore, an expression for your function $f(x)$ can be written as
$$f(x)=\sqrt{2\cdot a}\cdot \tan \left (\frac{\sqrt{2\cdot a}}{2}\cdot (x+b)\right).$$
The derived expression is equivalent to what you found with Wolfram Alpha.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that if the functional sequence $xf(n) + \sum_{k=1}^n f(k)$ is convergent then the sequence $f(n)$ is convergent Let $f:D_f\rightarrow\mathbb{R}$ be a function such that $\mathbb{N}\subseteq D_f$. My question is proving or disprving the following\
"If the functional sequence $xf(n) + \sum_{k=1}^n f(k)$ is convergent as $n\to{\infty}$ then the sequence $f(n)$ is convergent." ($x\in\mathbb{R}$)\
Note that, if the series $\sum_{n=1}^{\infty} f(n)$ is convergent then getting the result is clear. So, we should find the other cases.
| If $g_n(x):=xf(n)+\sum_{k=1}^{n} f(n)$ then:
$$2g_n(x) - g_n(2x) = \sum_{k=1}^{n} f(k)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to easily create a polynomial function that gives a desired output? I am looking for an easy way (formula or algorithm) to create a polynomial function that gives the arbitrary preset output for the first values of x. For instance, the desired output can be $y = 1, 2, 3, 4, 5, 6, 100$ for $x = 1, 2, 3, 4, 5, 6, 7$. I can create one of the many functions by hand as:
$y = x + (x-1)(x-2)(x-3)(x-4)(x-5)(x-6)\cdot \frac{93}{720}.$
But for other desired outputs it is really painstaking. For example, I am working now with the desired output $y = 2, 10, 12, 16, 17, 18, 19, 200, 300$ when $x = 1, 2, 3, 4, 5, 6, 7, 8, 9$.
Notes:
$x$ is always a natural positive number ($x > 0$). It doesn't matter what happens for other values of $x$ (greater than the ones given).
- I am looking for a easy way, as I don't know / I don't have / I have no experience with math related software.
- Also for a polynomial (or similar simple) function, as my mathematical background is quite limited, so I don't know any gamma, delta or other functions, series, etc.
- This quest is simply for recreation, to show friends how there are different solutions to the problems of the type: "Find the next term in this succession"
| As Gabriel Romon commented: Lagrange interpolation does the trick.
The polynomials form a vector space (which works very similar to the 3D coordinate space) where the coefficients work as coordinates and the monomials including $x^0=1$ as orthogonal standard base. The Lagrange polynomials form another one. They are constructed from your given sampling points ($x$-values). The nice thing is that each Lagrange polynomial is zero in all but one sampling points. And in this sampling point it has the value $1$. So you can multiply each of the Lagrange polynomials with the function value at the sampling point where it has a nonzero value and sum over all these polynomials.
This looks as easy as
$$
y = \sum_{i=1}^n y_i \cdot \ell_i(x).
$$
As you can see there is no $x_i$ in this formula any more. This makes it possible to precompute the Lagrange polynomials $\ell_i$ and solve all your problems together in a secound step, as long as all functions are defined using the same sampling points.
| {
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"timestamp": "2023-03-29T00:00:00",
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Minimum of $\left(\frac{1+\sin^2x}{\sin^2x}\right)^n+\left(\frac{1+\cos^2x}{\cos^2x}\right)^n$
I would like to find the minimum of
$$f(x)=\left(\frac{1+\sin^2x}{\sin^2x}\right)^n+\left(\frac{1+\cos^2x}{\cos^2x}\right)^n,$$
where $n$ is a natural number.
I know there is possible by derivate, but
$$f'(x)=n \left(\left(\cos ^2(x)+1\right) \sec ^2(x)\right)^{n-1} \left(2 \left(\cos ^2(x)+1\right)\tan (x) \sec ^2(x)-2 \tan (x)\right)+n \left(\left(\sin ^2(x)+1\right) \csc^2(x)\right)^{n-1} \left(2 \cot (x)-2 \left(\sin ^2(x)+1\right) \cot (x) \csc^2(x)\right).$$
I think this is not the best way.
| Using A.M $\geq G.M.$ on the first and second inequalities yields
$$
\begin{aligned}
f(x) &=\left(\csc ^{2} x+1\right)^{n}+\left(\sec ^{2} x+1\right)^{n} \\
& \geqslant 2 \sqrt{\left[\left(\csc ^{2} x+1\right)\left(\sec ^{2} x+1\right)\right]^{n}} \\
&=2\left[\left(2+\cot ^{2} x\right)\left(2+\tan ^{2} x\right)\right]^{\frac{n}{2}} \\
&=2\left(4+2 \cot ^{2} x+2 \tan ^{2} x+1\right)^{\frac{n}{2}} \\
& \geqslant 2\left(4+2 \cdot 2 \sqrt{\cot ^{2} x \tan ^{2} x}+1\right)^{\frac{n}{2}} \\
&=2\cdot 9^{\frac{n}{2}} \\
&=2\cdot 3^{n}
\end{aligned}
$$
$\therefore f(x)$ attains its minimum value $2 \cdot 3^{n}$ when $$\cot ^{2} x=\tan ^{2} x \textrm{ and } \csc^2 x=sec^2 x\Leftrightarrow x=\pm \frac{\pi}{4}.$$
| {
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prove that $X^2 \equiv 35 \pmod{100}$ has no solutions This problem is from 'A Survey of Modern Algebra' - Garret Birkoff, and Saunders Mac Lane in Section 1.9.
I'm an autodidact and there are no answers in the back so I need you guys to look at my proofs every once in a while to verify them. As you might be able to tell from my previous posts my mathematical maturity is not too high so I appreciate your patience. I give you permission to laugh at
this post from 6 months ago but I'm getting better.
The truth of the congruence
$X^2 \equiv 35 \pmod{100}$ can be inferred from looking at $100|(x^2-35)$ which has the equivalent of saying $100k = x^2 -35$ for some $k \in \Bbb{Z}$
by basic algebra $ x^2 =100k + 35 $ which I will now write as a function of k and attempt an induction.
Let
$P(k) = 100k + 35$
$P(0) = 35$ which is not a square.
$P(k+1) = 100(k + 1) + 35 = 100k + 100 +35 = (100k + 35) + 100$
by substitution from the induction hypothesis
$ P(0) + 100 = 35 + 100 = 135 $
which is also not a square. Q.E.D.
| $X^2=100k+35=4(25k)+35=4(25k+8)\color{red}{+3}$.
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Extensions of amenable groups Let $1 \to N \to G \to Q \to 1$ be an extension of (discrete) groups, where $N$ and $Q$ are amenable. Using the fixed-point theorem, I know how to show that $G$ is amenable. However, I was wondering if there is a proof which only uses the Følner property, namely $G$ is amenable if and only if: for every finite $S \subset G$ and for every $\epsilon > 0$ there exists some finite $A \subset G$ such that $|SA \Delta A| < \epsilon |A|$. I am asking this since in some texts this is used as the actual definition of an amenable group.
| Here's a direct proof.
Let $G$ be a group with $N$ amenable normal subgroup, such that $G/N$ is amenable; write $p:G\to G/N$. Let $S$ be a finite subset of $G$ and $\varepsilon>0$.
There exists a nonempty finite subset $F$ of $G$ such that $p(F)\equiv_\varepsilon p(sF)$ for all $s\in S$, and $p|_F$ is injective. Here $A\equiv_t B$, for finite nonempty subsets $A,B$ means $|A\Delta B|\le t|A|$ (beware that this is a symmetric relation only when $|A|=|B|$).
For $s\in S$ and $g\in F$ such that $p(sg)\in p(F)$, write $sg=f(s,g)\nu(s,g)$, with $\nu(s,g)\in N$ and $f(s,g)\in F$. Let $T\subset N$ be the (finite) set of $\nu(s,g)$ when $(s,g)$ range over such pairs
Let $M$ be a nonempty finite subset of $N$ such that $M\equiv_\varepsilon tM$ for all $t\in T$.
Then, for $s\in S$ and $g\in F$ such that $p(sg)\in p(F)$, we have $sgM=f(s,g)\nu(s,g)M\equiv_\varepsilon f(s,g)M$. Hence, if $F_s\subset F$ is the union of such $g$ and $F'_s\subset F$ is the union of such $f(s,g)$, we have $sF_sM\equiv_\varepsilon F'_sM$.
Also, we have $sFM\equiv_\varepsilon sF_sM$ and $FM\equiv_\varepsilon F'_sM$.
So
$$|sFM\smallsetminus FM|\le |sFM\smallsetminus F'_sM|\le |sFM\smallsetminus sF_sM|+|sF_sM\smallsetminus F'_sM|\le \varepsilon|FM|+\varepsilon|F_sM|\le 2\varepsilon|FM|$$ and similarly $|FM\smallsetminus sFM|\le 2\varepsilon|FM|$. Therefore $|FM\Delta sFM|\le 4\varepsilon$.
So there's a $(S,4\varepsilon)$-invariant subset. Hence $G$ is amenable.
| {
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An expression for the sum $\sum\limits _{k=1}^{n-1} k \, (n-k)^2$ I really don't know how to find the sum of the series: $$\sum\limits _{k=1}^{n-1} k \, (n-k)^2 = 1(n-1)^2+2(n-2)^2+3(n-3)^2+\dots+(n-1)1^2.$$
My attempt:
I tried to approach the old school approach of how we find the sum of arithmetic-co geometric progression but unable to do so.
Any help is appreciated!
| For the series
$$\sum_{k=1}^{n-1} k \, (n-k)^{2}$$
consider
$$\sum_{k=1}^{n-1} k \, (n-k)^{2} = n^{2} \, \sum_{k=1}^{n-1} k - 2n \, \sum_{k=1}^{n-1} k^{2} + \sum_{k=1}^{n-1} k^{3}$$
and use
\begin{align}
\sum_{k=1}^{n} k &= \frac{n(n+1)}{2} \\
\sum_{k=1}^{n} k^{2} &= \frac{n(n+1)(2n+1)}{6} \\
\sum_{k=1}^{n} k^{3} &= \frac{n^2 (n+1)^2}{4}
\end{align}
to obtain
$$\sum_{k=1}^{n-1} k \, (n-k)^{2} = \frac{n^2 \, (n^2 -1)}{12}.$$
| {
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Why isn't this approach in solving $x^2+x+1=0$ valid? There is this question in which the real roots of the quadratic equation have to be found:
$x^2 + x + 1 = 0$
To approach this problem, one can see that $x \neq 0$ because:
$(0)^2 + (0) + 1 = 0$
$1 \neq 0$
Therefore, it is legal to divide each term by $x$:
$x + 1 + \frac{1}{x} = 0$
$x = -1 - \frac{1}{x}$
Now, substitute $x$ into the original equation and solve:
$x^2 + (-1-\frac{1}{x}) + 1 = 0$
$x^2-\frac{1}{x} = 0$
$x^3 = 1$
$x = 1$
to get $x = 1$. Clearly this isn't the right answer. But why? Thanks.
| Transformations you apply to an equation may introduce alien solutions.
Taking an extreme example,
$$x=0\implies 0=0$$ which is satisfied by all $x$ !
So you may apply transformations, but validate the solutions using the original equation.
In your example, you establish
$$x^2+x+1=0\implies x^3-1=0.$$
But as $$x^3-1=0=(x-1)(x^2+x+1),$$ nothing is wrong if you ignore the first factor.
| {
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Compactness of $\ A= \{f: f$ is power series with infinite radius of convergence and is bounded by $1\}$ Consider the set $\ K=C[0,1]$, the set of continuous functions on $\ [0,1]$, with the supremum norm. Let
$\ A= \{f: f$ is power series with infinite radius of convergence and is bounded by $1\}$.
I am asked to prove or disprove whether A is compact or not.
I tried noting a counterexample to show that the set isn't closed.
$$ \sum^{\infty}_{k=0}\frac{(-1)^{k}x^{2k+1}}{(2k+1)!}=\sin(x) $$
This is a power series which converges to $\sin(x)$ which isn't in the set. Does this work?
| Let $f_n(x) = x^n, n=1,2,\dots $ Then each $f_n\in A.$ If $A$ were compact, then some subsequence $f_{n_k}$ would converge uniformly to some $f\in A.$ This would imply $f_{n_k}\to f$ pointwise on $[0,1].$ But the full sequence $f_n,$ hence the subsequence $f_{n_k},$ converges pointwise to the function that equals $0$ on $[0,1)$ and equals $1$ at $1.$ That function isn't even in $K,$ much less $A,$ contradiction. Thus $A$ is not compact.
| {
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Does every two variable equation are separable? I am a first year undergraduate student(Not undergone group theory yet) and i describe my situation as follows:
I was trying to trace curve of equation say :
$$x(y-x)^2=ay^2$$
So as we are taught in school by simple steps to work like :
(1) Symmetry
(2) Origin & Tangent at origin
(3) Point of Intersection with axes.
(4) Asymptotes
(5) Region of absence.
(6) Some optimization techniques etc.
I found the (5) as very helpful if you can express y in terms of then you have a very firm idea about its nature.
Say i have an equation :
$$2x^2-2xy+y^2-4x+2y+1=0$$
So at first look it seems to not expressed in terms of $y=f(x)$
But it is actually ;
$$y=(x-1)\pm(2x-x^2)^\frac{1}{2}$$
QUESTION
Is it possible to express every equation in the form of $y=f(x)$ say of degree 2,3,4 and 5 ?
Can you express these ?
$$x(y-x)^2=ay^2$$
$$a^5y^2+2a^3x^3y+x^7=0$$
I guess for these type of equation say more than 5 i must use polar co-ordinates.
Please let me know what i can do in such situation to trace its graph ?
Thanks
| In general no, because to put an equation in the form y=f(x) you need for there to only be one y value for each x value.
You can cheat a little on this, as you have in one example, by using ±, but this won't work in general. For example, it has been proven that there is no closed form solution to 5th order (and beyond) polynomials, so most equations involving $y^5$ and beyond can't be expressed this way at all (in the sense that such an expression doesn't even exist, not just that you can't find it reliably), even if you did allow multi-value operators in your equation.
That said, your two examples are quadratic in y, and so they can be solved. Hint: express them in the form $ay^2+by+c=0$, where 'a', 'b' and 'c' may each be functions of x, then use the quadratic formula to get:
$y = \frac{-b±\sqrt{b^2-4ac}}{2a}$
If you want to plot functions you can't write as y = f(x), you could try writing x = f(y), but that will only work sometimes. Any other tricks I can imagine would be specific to particular equations. Personally, at some point I would just stop trying to do it by hand and use a suitable software tool.
| {
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Why can the set of all natural numbers and omega be put in one-to-one correspondence with natural numbers? If $\omega$ comes literally after we've run out of all natural numbers, then why can the set of all natural numbers and omega be put in one-to-one correspondence with natural numbers? I feel the existence of $\omega$ is somewhat contradictory for this reason. Please explain.
| There are two distinct notions that are relevant here:
*
*Sets
*Ordered sets
When you talk about "$\omega$ coming after the natural numbers", you are talking about ordered sets — specifically, the ordered set $\omega + 1$. (the underlying set of $\omega + 1$ is is $\mathbb{N} \cup \{ \omega \}$)
There does not exist an order-preserving bijection between $\mathbb{N}$ and $\omega+1$.
When you talk about "$\omega+1$ can be put into one-to-one correspondence with natural numbers", you are talking about sets.
There does exist a bijection between $\mathbb{N}$ and (the underlying set of) $\omega + 1$. But by the above remarks, no such bijection can be order-preserving. As mentioned in the comments, an easy-to-consider bijection is the following correspondence:
$$ \begin{matrix}
0 & 1 & 2 & 3 & \ldots
\\ \updownarrow &\updownarrow &\updownarrow &\updownarrow &
\\ \omega & 0 & 1 & 2 & \ldots
\end{matrix} $$
See how it doesn't preserve order: we've corresponded $0 \leftrightarrow \omega$ and $1 \leftrightarrow 0$, but as to the ordering on these two sets we have $0<1$ and $\omega > 0$.
| {
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Higher differentiability of weak solutions to 2nd order elliptic PDEs with mixed boundary conditions I am interested in regularity results for 2nd order elliptic PDEs with mixed boundary conditions like
$$\left\{\begin{array}{rl}-\text{div}(a\nabla u) =& f &\text{in }\Omega, \\
u=&\varphi &\text{on }\Gamma_D, \\
\frac{\partial u}{\partial\nu}=& g & \text{on }\Gamma_N,
\end{array}\right.$$
where $\Omega\subset\mathbb{R}^N$ denotes a bounded domain and $\Gamma_D$ with positive surface measure and $\Gamma_N:=\partial\Omega/\Gamma_D$ the Dirichlet and Neumann part of the boundary $\partial\Omega$, respectively. So my question is the following:
What assumptions regarding regularity and compatibility do I have to make to ensure $u\in H^s(\Omega)$ holds for some given $s>1$?
I am aware that there are such results when dealing with a purely Dirichlet or Neumann boundary value problem. However, there are simple examples in a mixed boundary value setting, where smooth data and smooth boundary are not enough to ensure higher regularity.
| Unfortunately, there is no easy answer to your question. Mixed Dirichlet-Neumann problems have singular solutions even when the boundary conditions are regular. Take $f=\varphi=g=0$ then the function $u(r,\theta)=r^{1/2}\sin\frac\theta2$ is harmonic in the half-space $y>0$ and satisfies the Dirichlet-Neumann boundary conditions. This paper Costabel-Dauge has a few references you might want to look at. I don't know how to kill the singular part.
| {
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Find a number which satisfies the equation $|x^6-3x^3+5x| \leq Q$ whenever $|x|$ $\leq$ $2$? NOTE (EDIT): I have noticed that no one noticed that a TRIANGLE INEQUALITY is being used. Hence, part of the confusion in my comments of the solution.
TRIANGLE INEQUALITY:
$|x+y| \leq |x|+|y|$, important part to note that there is $\leq$ symbol.
So there this question in which a number has to be found such that it satisfys the following equation (whenever $|x|$ $\leq$ $2$):
Also: this is strictly a proofs question and no calculus is allowed. Thanks.
$|x^6-3x^3+5x| \leq Q$ whenever $|x|$ $\leq$ $2$
$(1)$ $\leq |x^6+(-3x^3)+5x|$
$(2)$ $\leq |x^6|+|-3x^3|+|5x|$
$(3)$ $\leq |x|^6+3*|x|^3+5*|x|$
$(4)$ $\leq 2^6 + 3*2^3 + 5*2 = 98$
The question is: what is the problem asking for? Does it say to find a number in which it satisfies an expression between the interval of $x$ in $[-2, 2]$? Why do you substitute $|x|$ with $2$ specifically and not any other numbers? And, apparently, this solution can be done in another way to end up with another answer... because any number higher than 98 is also the answer? Thanks.
| Yes we have that for $|x|\le 2$
$$0-24-10\le x^6-3x^3+5x\le 64+24+10 \implies |x^6-3x^3+5x|\le 98$$
Note that the problem is not asking for the "minimum value $Q$ such that..." but simply for any value $Q$ which satisfy the inequality. Therefore the solution $Q=98$, even if it is not the minimal possible answer, is correct as any other $Q\ge 98$.
| {
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A cofinal function into a limit ordinal Let $\alpha$ be a limit ordinal. Let $f\colon\beta\to\alpha$ be a cofinal function, that is, a function so that $f(\beta)$ is unbounded in $\alpha$. That is, we have $(\forall \gamma < \alpha)(\exists \eta < \beta)(\gamma \leq f(\eta))$.
It is true that $\bigcup_{\eta < \beta} f(\eta) = \alpha$? If yes, why?
| Apparently, yes. It is clear that $\bigcup_{\eta < \beta} f(\eta) \subseteq \alpha$.
Let $\gamma < \alpha$. Since $f$ is cofinal, let $\eta < \beta$ so that $\gamma \leq f(\eta)$. Then $\gamma < f(\eta) + 1$. Note that since $f(\eta) < \alpha$, $f(\eta) + 1 \leq \alpha$. Since $\alpha$ is a limit ordinal, $f(\eta) + 1 \neq \alpha$, hence $f(\eta) + 1 < \alpha$. Again, since $f$ is cofinal, let $\xi < \beta$ so that $f(\eta) + 1 \leq f(\xi)$. Then
$$\gamma < f(\eta) + 1 \leq f(\xi).$$
| {
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How to find the equivalence classes on a relation? Suppose we define the relation $∼$ by $v∼w$ (where $v$ and $w$ are arbitrary elements in $R^n$) if there exists a matrix $$A∈ GL_n(R)$$ such that $v=Aw$. What are the equivalence classes for $∼$ in this case? NOTE:$GL_n(R)$ is a set that contains all the $n×n$ matrices with $det≠0$
| Given any unit vectors, there is a nonsingular rotation matrix that takes the first vector to the other. Given any non-zero, non-unit vector, there is some non-singular scaling matrix that takes the vector to a unit vector. So given two arbitrary non-zero vectors $v$ and $w$, there are $S_v$ and $S_w$ such that $S_vv$ and $S_ww$ are unit vectors, and $R$ such that $R(S_vv)=S_ww$. Then $w=(S_w^{-1}RS_v)v$. So taking $A=S_w^{-1}RS_v$, we have that $v=Aw$, so given any non-zero vectors $v$, $w$, $v$~$w$; the set of non-zero vectors is an equivalence class.
| {
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Function transformation: shrink horizontally Write the formula for $f(x)$, if the graph of $f$ can be obtained from the graph of $y = g(x)$ by shrink horizontally by a factor of $5$ then shift left $3$ units
The equation should be
$f(x) = g(5(x+3))$ or $g\left(\frac{1}{5}(x+3)\right)$?
I prefer the second answer but my teacher said the correct is the first one? Can anyone explain for me why it is $5$ instead of $\frac{1}{5}$ while we are dealing with horizontal shrinking?
Thanks a lot
| To shrink a function means to make the graph of the function seems narrower.
For example, consider the function
$$f(x)=x^2$$
If you want to make the function shrink horizontally by a factor of 2 you would want the function
$$f(2x) = (2x)^2 = 4x^2$$
On the other hand, you would argue that
$$f\left(\frac{1}{2}x\right) = \left(\frac{1}{2}x\right)^2 = \frac{1}{4}x^2$$
is correct.
If you graph the functions, you would get
Obviously, the function $f(2x) = (2x)^2 = 4x^2$ seems narrower. Similarly, your question is asking you to shrink the function by a factor of five, so it should be $f(5x)$ instead of $f\left(\frac{1}{5}x\right)$.
| {
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How many integers from $1$ through $1000$ are not divisible by any one of $4, 5$, and $6$? I hate to use this resource as an argument settler but my friend and I have come across this question and we cannot agree on an answer.
I got $500$ through the use of inclusion and exclusion and he got $466$ through the use of gathering LCM's and such.
Who is correct?
| Inclusion Exclusion makes more sense to me than "the use of gathering LCM's and such"
But you do need LCM to do inclusion exclusion.
There are $1000$ integers. $250$ divisible by $4$ and $200$ by $5$ and $166$ by $6$. There are $1000/20 = 50$ there are divisible by both $4$ and $5$ (divisible by $4$ and $5$ means divisible by $20$). But to be divisible by both $4$ and $6$ means to be divisible by $12$, not $24$ so there are $83$ divisible by both $4$ and $6$. And $33$ by both $6$ and $5$. And to be divisible by all $3$ is to be divisible by $60$ and there are $16$ of those.
So the answer is $1000 - 250 - 200 - 166 + 50 + 83 + 33 - 16 = 534$.
| {
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What is the probability that a randomly chosen $10$-card hand has exactly three three- of-a-kinds (and no four-of-a-kinds)? This is my attempt:
For the first three-of-a-kind:
There are ${13}\choose{1}$ options for the three cards alike and ${4}\choose{3}$ for the suits
For the second three-of-a-kind:
There are ${12}\choose{1}$ options for the three cards alike and ${4}\choose{3}$ for the suits
Similarly for the third: ${11}\choose{3}$ $\times$ ${4}\choose{3}$
Finally, for the remaining card, there are ${43}\choose{1}$ options for the remaining card and ${4}\choose{1}$ for the suit.
So there are $\displaystyle\frac{C(13,1)\times C(4,3)\times C(12,1)\times C(4,3) \times C(11,1)\times C(4,3)\times C(43,1)\times C(4,1)}{C(52,10)}$
Can someone please check and verify? If this is wrong, can someone help me through this problem.
| We'll divide the number of successful hands by the total number of hands.
The total number of hands is $$\binom{52}{10}$$
The number of successful hands is the number of ways to choose $3$ card values that will be repeated $3$ times, then $1$ card value that will occur once, then the excluded suit for each of the sets of $3$, and finally the suit of the singleton card:
$$\binom{13}{3}\binom{10}{1}\binom{4}{1}^3\binom{4}{1}$$
Our final probability is
$$\displaystyle\frac{\displaystyle\binom{13}{3} \cdot 10 \cdot 4^4\,\,}{\displaystyle\binom{52}{10}}$$
| {
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Find all functions $f:\mathbb{N}^+\to\mathbb{N}^+$ such that $f\big(f(n)\big)+f(n)=2n$ for every $n\in\mathbb{N}^+$.
Find all functions $f:\mathbb{N}^+\to\mathbb{N}^+$ such that $$f\big(f(n)\big)+f(n)=2n$$ for every $n\in\mathbb{N}^+$.
I think the answer is $f(n)=n$,We prove this by induction. (at last step I can't induction it）
It is true for $n=1$ because
Let $f(1)=x\ge 1,$and let $n=1$,we have
$$f(x)+x=2\Longrightarrow x=1$$
(2) Suppose it is true for
$f(n)=n(n\le k)$,
then for $n=k+1$,let $f(n+1)=y$,so we have
$$f(y)+y=2(k+1)\Longrightarrow f(y)=2(k+1)-y$$
it is easy to have $k+1\le y\le 2k+1$,and
$$f(f(y))+f(y)=2y\Longrightarrow f(2k+2-y)=2y-f(y)=3y-2(k+1)$$
since $2k+2-y\in [1,k+1]$,then I can't it ,can you help? Thanks
(if $2k+2-y\in [1,k]$,I have done it! bacuase $f(2k+2-y)=2k+2-y$,so $y=k+1$)
| At first, we will show that $f$ is injective.
Suppose, for some $m,n$
$f(m)=f(n)\implies f(f(m))=f(f(n))$ therefore, by the given condition we have, $m=n$
Now, we claim, $f(n+1)\ge n+1$
Otherwise,
suppose $f(n+1)=n+1-k (1\ge k\le n)=f(n+1-k)$ (by induction)
Therefore, $n+1=n+1-k$ (as $f$ is injective)
So, a contradiction arise.
Hence, $f(n+1)\ge n+1$
In a similar way we can show that $f(f(n+1)\ge n+1$
Hence, $2(n+1)=f(n+1)+f(f(n+1))\ge 2(n+1)$
Therefore, we must have, $f(n+1)=n+1$ and hence we are done!
| {
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Prove that there exists a number $x$ such that $x^3 = 6$ I want to show that there exists a real number $x$ such that $x^3 = 6$. Here is what I have so far. $\\$
Let $S = \{x \mid x \in \mathbb{R}, x \geq 0, x^3 < 6\}$. By this definition, $S$ is nonempty since $0 \in S$, and also $S$ is bounded above since $2^3 = 8 > 6$. Thus, by the Completeness Axiom, $S$ has a least upper bound; call it $b$. We will show that $b^3 = 6$ (and hence, there exists a real number such that $x^3 = 6$) by showing that we cannot have $b^3 > 6$ or $b^3 < 6$.
First, for the sake of contradiction, suppose we had $b^3 > 6$. Then, we will show that we can choose a suitably small positive number $\epsilon$ such that $b - \epsilon$ is also an upper bound for $S$, which contradicts $b$ being the least upper bound. But, I'm not sure about how to find $\epsilon$. I tried expanding:
$$(b - \epsilon)^3 = b^3 - 3b^2\epsilon + 3b\epsilon^2 - \epsilon^3,$$
and from here, I think I'm supposed to use greater-than equalities to try and come up with $\epsilon$, but I'm not really sure how to do that. Any help is appreciated.
| Note that$$b^3 - 3b^2\varepsilon + 3b\varepsilon^2 - \varepsilon^3>6\iff b^3-6>3b^2\varepsilon-3b\varepsilon^2+\varepsilon^3.$$Now, take $\varepsilon\in\left(0,1\right)$ such that$$\varepsilon<\dfrac{b^3-6}{6b^2\varepsilon}\tag1$$and that$$\varepsilon<\dfrac{b^3-6}2.\tag2$$Then\begin{align}3b^2\varepsilon+\varepsilon^3&<3b^2\varepsilon+\varepsilon\text{ (because $\varepsilon<1$)}\\&<\frac{b^3-6}2+\frac{b^3-6}2\text{ (by $(1)$ and $(2)$)}\\&=b^3-6.\end{align}Therefore$$3b^2\varepsilon-3b\varepsilon^2+\varepsilon^3<3b^2\varepsilon+\varepsilon^3<b^3-6.$$
| {
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Prove $\frac{d}{dx} x^n=nx^{n-1} : \forall n\in \mathbb{Z}_{+}$ by induction Problem
Prove $$\frac{d}{dx} x^n=nx^{n-1} : \forall n\in \mathbb{Z}_{+}$$ by mathematical induction.
Attempt to solve
Base case
when $n=1$
$$ \frac{d}{dx} x^1 = 1 \cdot x ^{0}=1 $$
which is true
Induction step
$$\frac{d}{dx}x^{n+1}=(n+1)x^{n+1-1}= (n+1)x^{n}$$
At this point not quite sure how to prove this with induction without proving operator $\frac{d}{dx}$ with
$$ \frac{d}{dx}f(x)=\lim_{h \rightarrow 0}\frac{f(a+h)-f(a)}{h} $$
and then proving existence of such limit with:
$$ 0<|x-a|< \delta \implies|\frac{f(a+h)-f(a)}{h}-\frac{d}{dx}f(a)| < \epsilon
$$
and then we can arrive at implication that $$\frac{d}{dx}x^n=nx^{n-1} \implies \frac{d}{dx}x^{n+1}=(n+1)x^n$$
Most likely there is easier by induction which is capable of showing that $$ \frac{d}{dx}x^n=nx^{n-1} $$ is applicable $\forall n \in \mathbb{Z}_+$?
| Assume that we have
$(*) \quad \frac{d}{dx} x^n=nx^{n-1}$
for some $ n\in \mathbb{Z}_{+}$.
For the induction step use $(*)$ and the product rule !
| {
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What does it mean if the fundamental group is Abelian? On proving the problem that $\pi_1(X,x_0)$ is Abelian iff for every pair $\alpha$ and $\beta$ if paths form $x_0$to$x_1$, we have $\hat \alpha=\hat\beta$ where $X$ is path-connected space
,
I'm curious of meaning for what fundamental group is Abelian. If possible, could you explain the case of product of given two distinct loop would not be commutable on the $R^2$?.
| Consider the wedge sum of two circles (that is, two circles that are connected by one point, let's call it the base point). By Van Kampen's theorem, you can see that the fundamental group of this space is the free product $\mathbb{Z} * \mathbb{Z}$.
Obviously this is not abelian. In fact, if you consider the loop $\alpha$ which does one turn of the first circle, from the base point, and the loop $\beta$ which does one turn on the second cicrle from the base point, the images of these two loops in the fundamental group are not commuting.
| {
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Orthogonal Block Matrix Given $A \in \mathbb{R^{\text{nxn}}}$, $B \in \mathbb{R^{n\text{x}m}}$ and $C \in \mathbb{R^{m\text{x}m}}$ such that
$$ M = \begin{bmatrix}
A & B \\
0 & C
\end{bmatrix} \in \mathbb{R^{(n+m)\text{x}(n+m)}} $$
a block Matrix.
Prove that: $M$ is orthogonal $\iff$ $A$ and $C$ are orthogonal and $B = 0$
My solution idea:
$\Longleftarrow$
Since $A$ is orthogonal, its columns form a linearly independent orthonormal set of vectors. It follows that the first $n$ columns of $M$ are orthonormal, because a vector $v_{i}$ of the form $(a_{1,i},a_{2,i},...a_{n,i},0,..,0)$ for $i = 1,..n$ still has Norm equal to $1$ and $\langle\ v_{i},v_{j} \rangle = 0$ for $j = 1,..,n$ given $i \neq j$
Same logic applies to columns vectors of $M$ from $n+1$ to $m$ since $C$ is orthogonal. So, taken together $\{v_{1},...,v_{n},v_{n+1},...,v_{m}\}$ they build an orthonormal Basis for $\mathbb{R^{(n+m)}}$ and it follows that $M$ is orthogonal.
$\Longrightarrow$
Reversing the argument:
Since $M$ is an orthogonal Matrix, its columns vectors build an orthonormal Basis for $\mathbb{R^{(n+m)}}$
Given that $a_{i,j} = 0$ for $i = m,..,n+m$ and $j =1,..,n$ It follows that the columns vectors of $A$ build an orthonormal Basis for $\mathbb{R^{n}}$ and $A$ is orthogonal.
Now to prove $B = 0$ confuses me. Because the columns vectors of $M$ are orthonormal, it has to be that the inner product of each column vector of $A$ with the columns vectors of $B$ is zero. Is this enough to follow that $B= 0$?
If $B = 0$ then it's easy to follow that $C$ is orthogonal.
Any help/comments would be great!
| The $n+i$-th column of $M$ must be orthogonal to all the first $n$ columns of $M$. If you write the dot product, it's clear that the dot product of the $n+i$-th column of $M$ and the $k$-th column of $M$ (for $k\leq n$) is equal to the dot product of the $i$-th column of $B$ and the $k$-th column of $A$.
Therefore, the above demand is equivalent to demanding that the $i$-th column of $B$ must be orthogonal to all the $n$ columns of $A$.
Since the columns of $A$ span $\mathbb R^n$ (orthogonal nonzero vectors are always linearly independent), the only vector orthogonal to all the columns of $A$ is the zero vector.
| {
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General formula for nth element of the sequence 0, 1, 0, 1, ... The sequence is $f = 0, 1, 0, 1, \ldots$
I want to find a general formula for the $n$th element. The sequence starts at $n = 0$ (the $0$ here is not the first element $0$ but rather denotes the $0$th position).
One easy and obvious solution is: $n$th $f = n \bmod 2$. This works because even positions have $0$ and odd positions have $1$.
However, this question is part of a homework and modulus has not been discussed (or part of the syllabus or even a prerequisite). And so I am hesitant to use it.
Is there another way to solve this problem using only basic arithmetic operations (one that a beginning high schooler knows of)?
| Is the sequence just an eternal alternation of 0 and 1? Then all you need to do is put in a couple dozen alternations in the OEIS, find https://oeis.org/A000035, then just sit back and read until you find a formula you like.
Least significant bit of $n$, lsb(n).
This works even if $n$ is negative, but it gets a little bit into computer science.
Also decimal expansion of $\frac{1}{99}$.
Yeah... if you ignore the integer 0 in 0.01010101010101010101010101010101...
Also the binary expansion of $\frac{1}{3}$.
Though I'm not sure this applies under the IEEE 754 floating point format specification.
There's a whole bunch more to look at. I personally like the recurrence relation $a(n) = 1 - a(n - 1)$, which is of course initialized with $a(0) = 0$.
| {
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Find a closed form of $x_{n+2} = x_{n+1}+2x_n +2$ where $x_1 = x_2 = 1$, $n \in \mathbb N$
Find a closed form $x_n$ for the following recurrence relation:
$$
x_{n+2} = x_{n+1} + 2x_n + 2 \\
x_1 = x_2 = 1,\;\;n\in \mathbb N
$$
I'm trying to understand why I get different results for different guesses of a solution for particular part of the recurrence relation.
I've started by splitting the solution into two parts: homogenous and particular ones. It's known that $$x_n = x_n^{(h)} + x_n^{(p)}$$
Having the above in mind lets solve for homogenous. I've done this with the help of characteristic polynomial:
$$
\lambda^2 - \lambda - 2 = (\lambda + 1)(\lambda - 2) = 0
$$
By this we obtain the form of the homogenous solution:
$$
x_n^{(h)} = C_1\cdot(-1)^n + C_2 \cdot2^n
$$
Here is where things get vague for me. Let's try to guess the form of the non-homogenous solution. I've started with $B\cdot n$, then
$$
B\cdot(n+2) = B\cdot(n+1) + 2Bn + 2 \\
B = \frac{2}{1-2n}
$$
Therefore:
$$
x_n = C_1 \cdot (-1)^n + C_2\cdot2^n + \frac{2}{1-2n}
$$
Using the initial conditions one may find $C_1 = -{13 \over 9}$ and $C_2 = {7\over 9}$. Obtain the final form:
$$
x_n = -{13 \over 9} \cdot (-1)^n + {7 \over 9} \cdot 2^n + \frac{2}{1-2n}
$$
And that result doesn't match the answer in the book. However if I assume that the solution for non-homogenous part is in the form $B$, then $C_1 = -{2\over 3}$ and $C_2 = {2 \over 3}$, which by the steps above results in:
$$
x_n = {1\over 3} \cdot (2^{n+1} - 2\cdot(-1)^n) - 1
$$
being a match with the answer from the book.
I'm trying to understand where it got wrong and why the answers are different. I could have assumed non-homogenous solution to be in various forms which I believe would still lead to some form of solution. Have I made a mistake in my first assumption?
| If you put $x_n = y_n+c$ so that $y_n$ is a solution to homogenus equation we get $c=-1$.
Since $y_1=y_2= 2$ and $$y_n = a(-1)^n+b2^n$$ we get $$y_n = {2\over 3}(2^n-(-1)^n)$$
or $$x_n = {2\over 3}(2^n-(-1)^n)-1$$
| {
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Understanding the field of fractions of $F[[x]]$ (the ring of formal power series in the indeterminate x with coefficents in F) Let $F[[x]]$ be the ring of formal power series in the indeterminate x with coefficients in F. Show that the field of fractions of $F[[x]]$ is the ring $F((x))$ of formal Laurent series.
I've been going in circles on this one for what seems like ever now! If anyone could lend me some insight I would be exceedingly grateful! Thanks!
| $\sum_{k=-n}^\infty a_kx^k=\frac{\sum_{k=0}^\infty a_{k-n}x^k}{x^n}$, so every Laurent series can be written as a quotient of power series.
Conversely, given a fraction $\frac{F(x)}{G(x)}$, write $G(x)=a_nx^n(1+xH(x)),$ where $H(x)$ is a power series. Then
$$
\frac1{G(x)}=\frac1{a_nx^n}\cdot \frac{1}{1+xH(x)}=\frac1{a_nx^n}\sum_{k=0}^\infty(-1)^kx^kH(x)^k
$$
When $H(x)^k$ is expanded into powers of $x$, and like terms are collected, the above writes $\frac1{G(x)}$ as a Laurent series. Since Laurant series are closed under multiplication, $F(x)\cdot \frac1{G(x)}$ is also a Laurent series.
| {
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Proving that $\tau = \{N,\varnothing\}\cup \{S_n:n\in\mathbf{N}\}$ is a topology on $\mathbf{N}$. I am required to show that $\tau = \{N,\varnothing\}\cup \{S_n:n\in\mathbf{N}\}$ where $S_n := \{1,\dots,n\},\forall n\in\mathbf{N}$ is a topology on $\mathbf{N}$, the part where i am experiencing some dfficulty is in proving that the union of a countably infinite subset of $\{S_n:n\in\mathbf{N}\}$ belongs to $\tau$.
Here is my attempt so far, Is it correct?
Let $\mathcal{A} =\{S_{\lambda_1},S_{\lambda_2},S_{\lambda_3},\dots\}$ be an infinite subset of $\{S_n:n\in\mathbf{N}\}$, where we assume without loss of generality that $\lambda_j<\lambda_k$ whenever $j<k$. We demonstrate that
$$\bigcup_{r=1}^{\infty}S_{\lambda_r} = \mathbf{N}$$
Proof. Prior to addressing the main claim, we first prove that result that $\lambda_{n}\ge n,\forall n\in\mathbf{N}$, by recourse to Mathematical Induction. The base case is trivial, so we assume that $\lambda_k\ge k$ for some arbitrary $k\in\mathbf{N}$, thus $\lambda_k+1\ge k+1$, but $\lambda_{k+1}>\lambda_k$, implying $\lambda_{k+1}\ge \lambda_k+1$, so $\lambda_{k+1}\ge k+1$, completing the induction. Now let $n\in\mathbf{N}$, obviously $n\in S_n$, but since $n\leq\lambda_n$, it follows that $S_n\subseteq S_{\lambda_n}$, consequently $n\in S_{\lambda_n}$ and thus $\mathbf{N}\subseteq\bigcup_{r=1}^{\infty}S_{\lambda_r}$, the converse of which is trivial.
$\blacksquare$
| The idea is OK, but the proof should be set up for arbitary unions, and then the union is not always $\mathbb{N}$, though it often is:
Let $O_i$, $ i \in I$ be an arbitary union of open sets. We can WLOG omit any $O_i = \emptyset$, because they do not contribute to the union, and if any $O_i = \mathbb{N}$, so is the union and we are done. So all $O_i$ are then of the form $O_i = \{1, \ldots, n(i)\}$ for some $n(i) \in \mathbb{N}$.
Now there are two cases when considering the set $N = \{n(i): i \in I\} \subseteq \mathbb{N}$.
a) $N$ is bounded above and thus $\max N$ exists. If this is the case, $\bigcup_i O_i = \{1,\ldots, \max N\}$, as all $O_i$ are subsets of that set (as $n(i) \le \max N$) and at least one $O_i$ actually equal to that right hand set.
b) $N$ is not bounded above and then $\bigcup_i O_i = \mathbb{N}$: let $n \in \mathbb{N}$, then for some $i$, $n(i) > n$ by unboundedness, and then $n \in O_i = \{1,\ldots, n(i) \} \subseteq \bigcup_i O_i$.
So the topology is closed under all unions, as we needed to show.
| {
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Show that $\vec a \cdot \vec b = {1 \over 4} |\vec a + \vec b|^2 - {1 \over 4} |\vec a - \vec b|^2$
Show that
$$\vec a \cdot \vec b = {1 \over 4} |\vec a + \vec b|^2 - {1 \over 4} |\vec a - \vec b|^2$$
I have tried:
$$\vec a \cdot \vec b = {1 \over 4} |\vec a + \vec b|^2 - {1 \over 4} |\vec a - \vec b|^2
\\ \vec a \cdot \vec b = {1\over 4}(|\vec a|^2+2|\vec a||\vec b|+|\vec b|^2) - {1\over 4}(|\vec a|^2 -2|\vec a||\vec b|+|\vec b|^2)
\\ \vec a \cdot \vec b = |\vec a||\vec b|$$
However it seems that the equation only holds true when $\vec a$ and $\vec b$ are collinear.
Is that true? Or did something go wrong?
Thanks
| HINT
Such formula is known as the Polarization Identity. Due to the inner product properties, we have:
\begin{align*}
\lVert\textbf{a}+\textbf{b}\rVert^{2} = \langle\textbf{a}+\textbf{b},\textbf{a}+\textbf{b}\rangle = \langle\textbf{a},\textbf{a}\rangle + 2\langle\textbf{a},\textbf{b}\rangle + \langle\textbf{b},\textbf{b}\rangle = \lVert\textbf{a}\rVert^{2} + 2\langle\textbf{a},\textbf{b}\rangle + \lVert\textbf{b}\rVert^{2}
\end{align*}
Analogously, we have
\begin{align*}
\lVert\textbf{a}-\textbf{b}\rVert^{2} = \langle\textbf{a}-\textbf{b},\textbf{a}-\textbf{b}\rangle = \langle\textbf{a},\textbf{a}\rangle - 2\langle\textbf{a},\textbf{b}\rangle + \langle\textbf{b},\textbf{b}\rangle = \lVert\textbf{a}\rVert^{2} - 2\langle\textbf{a},\textbf{b}\rangle + \lVert\textbf{b}\rVert^{2}
\end{align*}
Can you proceed from here?
| {
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Simple Cubic Polynomial not Yielding Expected Results So I've got this beautiful little piece of math, $n$ in this formula can be substituted with any non-negative number between 1 and 100.
$$
\frac65 n^3 - 15n^2+100n-140
$$
The expected results are kind of as follows...
*
*$n = 1 \to 9$
*$n = 2 \to 48$
*$n = 3 \to 39$
*$n = 4 \to 39$
The results I got after plugging in the respective values are such.
*
*$n = 1 \to -54$
*$n = 2 \to 10$
*$n = 3 \to 57$
*$n = 4 \to 97$
What am I not doing right? How do I figure out where the error is?
Included now are graphs of the expected results according to the person who first found this formula within a game. The formula I am posting about is represented in these graphs by the line with the color purple.
Graph of expected results
Graph of expected results, ratio of n to "level" cubed
| If you want to define a cubic polynomial
$$
p(n) = an^3+bn^2+cn+d
$$
satisfying $p(1)=9,p(2)=49,p(3)=p(4)=39$, there is a unique polynomial like that. You can construct the coefficients by solving the Vandermonde system
$$
\begin{pmatrix}
1 & 1 & 1 & 1\\
1 & 2 & 4 & 8\\
1 & 3 & 9 & 27 \\
1 & 4 & 16 & 64
\end{pmatrix}
\times
\begin{pmatrix} d \\ c \\ b \\ a \end{pmatrix}
= \begin{pmatrix} 9 \\ 48 \\ 39 \\ 39 \end{pmatrix}
$$
which yields the solution
$$
\begin{pmatrix} d \\ c \\ b \\ a \end{pmatrix}
= \begin{pmatrix} -135 \\ 215.5 \\ -81 \\ 9.5 \end{pmatrix}
$$
and the corresponding polynomial
$$
p(n) = \frac{19}{2} n^3-81n^2+\frac{431}{2}n-135
$$
but this seems to have nothing to do with your polynomial.
| {
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Shortest distance between two lines and find points on each line I'm given the following two lines:
$L_1$: $P_1=(−13, 3, 14)$ with direction vector $d_1=(2, −1, −2)$
$L_2$: $P_2=(5, 4, 4)$ with direction vector $d_2=(−2, 1, 0)$
I'm then asked to find the shortest distance $d$ between these two lines, and then find a point, $Q_1$, on $L_1$, and a point, $Q_2$, on $L_2$ so that $d(Q_1,Q_2) = d$.
So far, I've determined that the shortest distance between these two points can be solved with a projection of the vector $\vec{P_1P_2}$ onto the direction vector found by the cross product of $d_1$ and $d_2$. In this case, it is $(4,8,0)$, or a magnitude of $4\sqrt{5}$.
I'm not really sure how to determine the two points now that I've gotten the distance, any tips or explanations would be appreciated.
| HINT
Let consider the plane orthogonal to a line and containing the other one.
| {
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Poker combinatorics: chances opponent has a full house given that you have 2 pairs I've read the answers to some very similar problems to this, but I can't manage to apply the knowledge to this specific one, sorry for this redundance. Here's the question:
What is the probability that your opponent is dealt a full house, given that you have two pairs?
My attempt: | You have 2 pairs
B: The opponent has a full house
$P(B|A) = \frac{P(BA)}{P(A)}$
P(A) is easy enough to calculate, but it's P(BA) that's causing me problems. I'm interpreting it as the probability of the dealer selecting 10 cards that will provide the opponent with a full house and you with 2 pairs. Right now I have:
$P(BA) = \frac{ {13\choose 2}{13\choose 1}{4\choose 2}^{3}{10\choose 3}{4\choose 3}{??\choose??} }{52\choose 10}$
The ${??\choose ??}$ is supposed to represent the final card in the 2 pair hand, but there seem to be so many different cases that I don't know how to calculate it.
I don't even know if I'm close. Thanks for any help!
| {
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Why is the dot product of two vectors a scalar value? I'm having some trouble seeing why dot products are said to give scalar values. As a far as I can see, it just gives another vector that is projected onto one of the 2 original vectors. How, then, is the result a scalar quantity. Can someone please explain this to me? Thank you.
| $$(1,2)\cdot (3,4) = 1 (3) + 2(4) = 11$$
is a scalar.
I think you are confusing dot product with projection.
Suppose $u$ is a unit vector, we can project $v$ onto $u$ and its length would be $|u\cdot v|$ while the projection would be $(u\cdot v) u$.
| {
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Solve $\sin^{3}x+\cos^{3}x=1$
Solve for $x\\ \sin^{3}x+\cos^{3}x=1$
$\sin^{3}x+\cos^{3}x=1\\(\sin x+\cos x)(\sin^{2}x-\sin x\cdot\cos x+\cos^{2}x)=1\\(\sin x+\cos x)(1-\sin x\cdot\cos x)=1$
What should I do next?
| Hint $$\sin^3(x)+\cos^3(x)=\sin^2(x)+\cos^2(x).$$
| {
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Solve $7^x+x^4+47=y^2$
Solve $$7^x+x^4+47=y^2$$ where $x, y \in \mathbb{N}^*$
If $x$ is odd then the left term is congruent with $3$ mod $4$ so it couldn't be a perfect square, so we deduce that $x=2a$ and the relation becomes $$49^a+16a^4+47=y^2$$ and it is easy to see that the left term is divisible by $16$ so we obtain that $y=4b$, so we have to find $a$ and $b$ such that $$49^a+16a^4+47=16b^2$$From this point I was completely stuck. I think that there are no solutions but how can I prove it?
| When $x$ is odd then $7^x+x^4+47\equiv 3(\mod 4)$. So, it is not possible $y^2\equiv 3(\mod 4)$.
Let $x=2k$, then for $k\geq 4$
$$(7^k)^2<7^{2k}+(2k)^4+47<(7^k+1)^2$$
This mean we have $k\leq 3$.
if try to $k=1,2,3$ only $x=4$ is a solution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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- Optimization - Standard Grid Search I'm struck into an portfolio opt. problem and the paper I'm replicating (or, better, trying to) is using a "Standard Grid Search".
Since I never encountered it before, I would like to ask you about: what's the intuition behind this numerical method? What's it used for? How could it be helpful for optimization problem?
Please, share with me some insights!
If you have any link (no wikipedia, already tried) or material, it's very welcome!
Thank you in advance!!
p.s. I understand the purpose of this numerical method is to avoid numerical convergence problems and local optima issues. But I would like to have a longer explanation (or at least less cryptic).
| The grid search is, in essence, systematically search through all possible (hyper)parameters to find the best one.
So, for example, if your portfolio depends on two hyperparameters $A,B$ (say taking values in $[0,1]$), then you might search through the $1001^2=1\,002\,001$ possible pairs
$$
(A,B)=(0,0), (0,0.001), (0,0.002),\dots,(0,1),(0.001,0),(0,001,0.001),\dots,(1,1)
$$
to find the optimum $A,B$ to 3 decimal places.
This method is easy to implement, but is very inefficient when the number of hyperparameters grows or you want higher precision. However, this is the only method that will guarantee you the absolute optimum.
| {
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Coding Theory Set Problem Supose that $\mathbb{A}$ is a finite set and take $\overline{u},\overline{v} \in \mathbb{A}^n$. Let:
$$X=\{\overline{x} \in \mathbb{A}^n\mid d(\overline{u},\overline{x})<d(\overline{v},\overline{x})\}$$
$$Y=\{\overline{y} \in \mathbb{A}^n\mid d(\overline{u},\overline{y})>d(\overline{v},\overline{y})\}$$
Prove that $Card(X)=Card(Y)$
I am really stuck on this problem. I thought about creation some bijection between both sets. But I don't really know how to do that, what makes me feel it's not the good path. Maybe I have to use some propiety about distances.
| It's not necessary that the underlying set is a field (or abelian group). Just show that the cardinality of the set $\{x\in A^n\mid d(u,x)=i\}$ with $i\geq 0$ fixed is independent of the choice of $u\in A^n$.
Then we can write $d_i = |\{x\in A^n\mid d(u,x)=i\}|$ for $i\geq 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How do I solve for $g$ in $12g = 12 \left(\frac{2}{3g} - 1\right) + 11$? For this specific problem, I somehow keep coming up with the wrong answer. Can someone help me?
For the problem, I need to solve for $g$.
$$12g = 12 \left(\frac{2}{3g} - 1\right) + 11$$
Here is how I am trying to solve it:
$$12 \cdot \frac{2}{3g} = 8g$$
$$12 \cdot -1 = -12$$
$$12g = 8g - 12 + 11$$
(Then I subtract $8g$ from $12g$ and $8g$)
$4g = -12 + 11$
(Then I add $12$ to $-12$ and $11$)
$$4g = 23$$
(Then I divide $4$ from $4g$ and $23$)
$$g = 5.75$$
So, shouldn't $g$ equal $5.75$? But when I plug it in the equation to check my answer, both answers do not match. Where am I going wrong?
| So you have the equation $12g = 12 (\frac{2}{3g} - 1) + 11$. You want to find $g$. $$12g=\frac{24}{3g}-12+11$$ $$\implies 12g=\frac{8}{g}-1$$ $$\implies 12g^2=8-g$$ $$12g^2+g-8=0$$ $$g=\boxed{\frac{-1\pm\sqrt{385}}{24}}$$
However, if you mean $12g=12(\frac{2}{3g-1})+11$, then we have $$12g=\frac{24}{3g-1}+11$$ $$\implies 12g(3g-1)=24+33g-11$$ $$\implies 36g^2-12g=33g+13$$ $$\implies 36g^2-45g-13=0$$ This is a quadratic, so use the quadratic formula getting $$g=\boxed{\frac{15\pm\sqrt{433}}{24}}$$
I’m sure there is some limit of $g$ to eliminate on of the solutions, such as “no negative numbers”, so I think you can take it from there... given that I’m right :P.
| {
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Relation between determinant of a Jacobi matrix and its minor Let
$$
A = \begin{bmatrix}
a_1 & b_1 \\
b_1 & a_2 & b_2 \\
& b_2 & \ddots & \ddots \\
& & \ddots & \ddots & b_{n-1} \\
& & & b_{n-1} & a_n
\end{bmatrix},\ B = \begin{bmatrix}
a_2 & b_2 \\
b_2& a_3 & \ddots & \\
&\ddots & \ddots & b_{n-1} \\
& & b_{n-1} & a_n
\end{bmatrix},
$$
where $b_i \neq 0$.
Can we prove that for any real value of $\lambda$, $\det(\lambda I_n-A) \neq \det(\lambda I_{n-1}-B)$ where $I_k$ is identity matrix of order $k$? It has been a few days I am trying to prove it but I failed. I would be appreciated for any help.
I could only prove that no two immediate consecutive minors of a Jacobi matrix share eigenvalues. I mean there is no $\lambda \in \mathbb{R}$ such that $\det(\lambda I_n-A) = 0$ and $\det(\lambda I_{n-1}-B) = 0$. I proved it by contradiction which will leads to some $b_i = 0$, while $b_i \neq 0$.
If $P_{n}(\lambda) = \det(\lambda I_n-A)$ then $P_{n}(\lambda) = (\lambda-a_n)P_{n-1}(\lambda)-b_{n-1}^2P_{n-2}(\lambda)$. By this equation I proved the contradiction.
Thanks in advance.
| This proposition is not necessarily true in general.
Lemma: If $B$ and $C$ are square matrices and $a \in \mathbb{R}$, then there exists $b \in \mathbb{R}^*$ such that the equation$$
(λ - a) |λI - B| = b^2 |λI - C|
$$
has a real solution.
Proof: It suffices to prove that there exist $λ \in \mathbb{R}$ and $b \in \mathbb{R}^*$ satisfying the equation. Note that $|λI - B|$ is a polynomial of $λ$ with a positive leading coefficient. If it has real roots, then the largest one $λ_0 \leqslant ρ(B)$ and $|λI - B|$ is positive for $λ > λ_0$. Otherwise $|λI - B| > 0$ for all $λ \in \mathbb{R}$. Thus $|λI - B|$ is always positive for $λ > ρ(B)$.
Now take an arbitrary $λ > \max(a, ρ(B), ρ(C))$, then $λ - a > 0$, $|λI - B| > 0$, $|λI - C| > 0$, and take $b = \sqrt{(λ - a)\dfrac{|λI - B|}{|λI - C|}} > 0$. Thus $(λ - a) |λI - B| = b^2 |λI - C|$.
Now back to the question. Denote$$
C = \begin{pmatrix}
a_3 & b_3 &&\\
b_3 & \ddots & \ddots &\\
& \ddots & \ddots & b_{n - 1}\\
&& b_{n - 1} & a_n
\end{pmatrix},
$$
then$$
|λI - A| = (λ - a_1) |λI - B| - b_1^2 |λI - C|.
$$
Thus$$
|λI - A| = |λI - B| \Longleftrightarrow (λ - a_1 - 1) |λI - B| = b_1^2 |λI - C|.
$$
The lemma implies for any $a_1, \cdots, a_n$ and $b_2, \cdots, b_{n - 1}$, there exists $b_1 \in \mathbb{R}^*$ such that $|λI - A| = |λI - B|$ has a real solution. Therefore, $|λI - A| ≠ |λI - B|\ (\forall λ\in \mathbb{R})$ is not necessarily true.
| {
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Commutativity in Mapping Class Groups I am trying to understand the behavior of finite order mapping classes for surfaces of genus g>=2.
After fiddling for a while I started to think that no finite order mapping class commutes with any Dehn twist for g>=2. Is there a simple proof or disproof? Otherwise I'd appreciate a pointer to a reference!
| The thing to know is that, if $D_\alpha$ denotes the Dehn twist along a simple loop $\alpha$, then for an orientation-preserving homeomorphism $f$ of the surface $S$ we have $f D_\alpha f^{-1}= D_{f(\alpha)}$ (up to isotopy, of course). Thus, the issue reduces to finding a reducible mapping class of finite order $\ge 2$, i.e. a finite order homeomorphism which preserves an essential simple loop on the surface. One, quite famous, example is the hyperelliptic involution $\theta$ of $S$. For instance, if $S$ is closed and oriented of genus $\le 2$ then $\theta$ commutes with every element of the mapping class group. If you want higher order examples, take (similarly what Mike Miller suggested) an order $n$ diffeomorphism $h$ of a surface $F$ which has at least two distinct fixed points $x, y\in F$. Next, remove small $h$-invariant disks $D_x, D_y$ around these fixed points and identify the boundary components of $F'=F- (D_x\cup D_y)$ by a suitable orientation-reversing diffeomorphism of the boundary curves ("suitable" simply means that it commutes with $h$ restricted to the boundary of $F'$). The result is a surface $S$ of genus equal to 1+ genus($F$). Then $h$ yields a reducible homeomorphism $f: S\to S$ of order $n$.
| {
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Confusion regarding the sample space of a experiment
First, of all, I need to realize what sample space is and this is causing me some trouble. Perhaps I am misunderstading the problem? But this is my reasoning:
Since player 1 has two choices at each stage and there is only 4 games he can have provided he wins to player 2,3 and 4, then the size of the sample space is $2^4=16$ And for example,
$$ P(X=0) = \frac{1}{16} $$
since there is only a way player 1 loses that is if he loses agains player 2 at the beginning. However, in my notes it says the sample space size is $5!$ and $P(X=0) =1/2$. How come this is true? I just don't see it. Am I overthinkin this problem?
| There are $5$ discreet numbers which can be randomly arranged in $5! = 120$ ways which is your sample space.
I take it that $P(x=i)$ means $i =$ exactly $0,1,2,3,4$.
In the $120$ different sequences of $5$ numbers, we have $60$ sequences where the second number is larger than the first. ($1,2$ etc)
$P(0) = \frac{60}{120} = \frac{1}{2}$
Then we have $20$ sequences where the first number is larger than the second but smaller than the third. ($2,1,3$ etc)
$P(1) = \frac{20}{120} = \frac{1}{6}$
Then we have $10$ sequences where the first number is larger than the second and third but smaller than the fourth. ($3,1,2,4$ etc)
$P(2) = \frac{10}{120} = \frac{1}{12}$
$6$ sequences where the first number is larger than the second, third and fourth but smaller than the fifth. ($4,1,2,3,5$ etc)
$P(3) = \frac{6}{120} = \frac{1}{20}$
And finally, $24$ sequences (all start with $5$) where the first number is larger than the other $4$. ($5,1,2,3,4$ etc)
$P(4) = \frac{24}{120} = \frac{1}{5}$
Notice the sequences total $120$ and so the probabilities sum to $1$.
| {
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Using trigonometry to prove $\frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2}=\frac{4abc}{\left(1-a^2\right)\left(1-b^2\right)\left(1-c^2\right)}.$
For the numbers $a,b,c$ with $ab+ac+bc=1$, prove that
$$\frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2}=\frac{4abc}{\left(1-a^2\right)\left(1-b^2\right)\left(1-c^2\right)}.$$
This is a question from the trigonometric section of my textbook. But is it possible to use trigonometry here?
Edit:
Using the Dinesh Shankar's hint with $a=\tan\frac{x}{2}$, $b=\tan\frac{y}{2}$ and $c=\tan\frac{z}{2}$, the equation becomes
$$\tan x+\tan y+\tan z=\tan x\cdot\tan y\cdot\tan z.$$
But $z=\pi-(x+y)$, then
$$\tan x + \tan y + \tan \left(\pi-(x+y)\right)=\tan x\cdot\tan y\cdot\tan\left(\pi-(x+y)\right),$$
which implies
$$\tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\cdot\tan y}.$$
Therefore, the original equation is also valid.
Please feel free to give another solution.
| Hint:
You probably know that if $x+y+z=\pi$, then
$$\tan\frac{x}{2}\cdot\tan\frac{y}{2}+\tan\frac{x}{2}\cdot\tan\frac{z}{2}+\tan\frac{y}{2}\cdot\tan\frac{z}{2}=1.$$
Now, since $ab+ac+bc=1$, use $a=\tan\frac{x}{2}$....
I hope this help you.
| {
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Calculating the value of the limit $$\lim_{(x,y)\to(1,2)}(\sin(y)-\sin(x))$$
My try:
I got as $$\sin(2)-\sin(1)$$
But I cannot calculate the exact value of the given limit. Can anyone please explain this.
| Yes your result is correct, indeed since the function is continuous at the point we have
$$\lim_{(x,y)\to(1,2)}(f(x,y))=f(1,2)$$
that is
$$\lim_{(x,y)\to(1,2)}(\sin(y)-\sin(x))=\sin 2 -\sin 1$$
where $2$ and $1$ are expressed in radians, a numerical evaluation leads to $\approx 0.068$.
| {
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In $ \mathbb{Z}_p $ where $ p $ is prime, what is the number of roots of $ x^2+1\equiv 0 \mod p $?
In $ \mathbb{Z}_p $ where $ p $ is prime, what is the number of roots of $ x^2+1\equiv 0 \mod p $?
Since $ x^2\equiv -1 \mod p $, then $ x^4\equiv 1 \mod p $ and we have $ 4 | \phi(p) $. If $ p=5 $, then $ 2,3 $ are two roots of $ x^2+1\equiv 0 \mod 5 $, but what are the situations for other prime $ p $ like $ p=13 $? we can't check that all by hand right? Or does it just depend without a generous law?
| Explicitly, when $p \equiv 1 \bmod 4$, the solutions of $x^2 \equiv 1 \bmod p$ are $\pm \left(\frac{p-1}{2}\right)!$. See here.
| {
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Simplify $x^\frac{1}{(\log_a x)}$ Simplify $x^\frac{1}{(\log_a x)}$
The solution in my textbook is the following:
Since $\log_a (x^\frac{1}{(\log_a x)}) = \frac{1}{\log_a x}$ $\log_a x
= 1$,
therefore $x^\frac{1}{(\log_a x)} = a^1 = a.$
I understand the law used in the first line is $\log_a (x^y) = y \log_a x$. I also understand that $\log_a x$ multiplied by it's reciprocal equals to 1. However, I do not understand why it makes sense to take the $\log_a$ of our expression in the first place and neither do I understand how the conclusion in the first line helps us arrive at the conclusion in the second line that $x^\frac{1}{(\log_a x)} = a^1 = a.$
| First, the expression has sense only for $x > 0$.
So for $x > 0$, the number $y= x^{\frac{1}{\log_a(x)}}$ exists and is $> 0$, so you can compute its $\log_a$. The computation shows that $\log_a(y)=1$.
Now, do you know a lot of numbers whose $\log_a$ are equal to $1$ ?
| {
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Is the standard model for the language of number theory elementarily equivalent to one with a nonstandard element? On page 89 in A Friendly Introduction to Mathematical Logic, the author writes that the standard model $\mathfrak{N}$ for $\mathcal{L}_{NT}$ is elementarily equivalent to a model $\mathfrak{A}$ that has an element of the universe $c$ that is larger than all other numbers.
I'm new to mathematical logic, but I understand that elementarily equivalent means the two structures have the same set of true sentences. However, it seems to me that the following sentence is true in $\mathfrak{A}$ but not in $\mathfrak{N}$. What am I missing?
$\exists x\ \forall y\ (x=y \vee y<x)$
| In the notes, I don't see the claim that $c$ is larger than all other numbers of $\mathfrak{A}$. The number $c$ in $\mathfrak{A}$ is larger than $0$, $S(0)$, $S(S(0))$, etc., - so $c$ is greater than every element of $\mathfrak{N}$. But there will be other elements of $\mathfrak{A}$ that are larger than $c$. Not every element of $\mathfrak{A}$ is of the form $S^n(0)$ for some $n \in \mathfrak{N}$.
| {
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How is $\sin^2(x)+\cos^2(x)=1$ where $x$ is an obtuse angle? Ok, so I know that $\sin^2(x)+\cos^2(x)=1$ for all angles. If $x$ is an acute angle in a right angled triangle it's a straightforward proof.
But what about if $x$ is obtuse ? How do I mathematically prove it plus get a visual analysis of the same so you can use it in $2D$ or $3D$ geometry using the co-ordinate axes.
Edit: I know the proofs in which they show a unit circle. But I thought the obtuse angles were only designed to incorporate the sign.
For example say a point on the unit circle is (-0.5,0.866). This shows up for x=120. However, we do the calculations for the acute angle from the negative x axis and then just put sign for sin(x) or cos(x) if x is any angle so that when we resubstitue x=r*sin(x) we get the polarity of x because sin(x) can be positive or negative but r is always considered postive.
r is the distance from the origin.
But what if we have a triangle in space which has coordinates say (1,1),(1,5) and (-2,8). How do we use trigonometry in those cases ? Since the angle made with the x axis for the above points doesn't matter since the angle between the linea joining those points are something totally different
| I can't think of a geometric interpretation right now, but I would argue with the angle reduction formulas.
Let $x \in [\frac{\pi}{2}; \pi]$ be an obtuse angle. This means $x$ can be rewritten in terms of an acute angle $y \in [0; \frac{\pi}{2}]$ plus $\frac{\pi}{2}$, meaning $x=y+\frac{\pi}{2}$. Then, with the reduction formulas it holds that $\sin(x)=\sin(y+\frac{\pi}{2})=\cos(y)$ and $\cos(x)=\cos(y+\frac{\pi}{2})=-\sin(y)$.
Plugging this in yields
$\sin^2(x)+\cos^2(x)=(\sin(x))^2+(\cos(x))^2=(\cos(y))^2+(\sin(y))^2=\cos^2(y)+\sin^2(y)=1$.
The last equality being true, because of $y$ being an acute angle.
| {
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Does $K[\alpha]$ has an inverse of $\alpha$ if this element is algebraic? I was asking myself a question :
Let $K$ be a field and $K[\alpha]$ the smallest ring containing $K$ and $\alpha$. If $\alpha$ is algebraic, can we always find $\alpha^{-1}$ in $K[\alpha]$.
My thoughts were as following. I $p$ is the minimal polynomial of $\alpha$, than if we suppose moreover (do we have to???)
$gcd(x,p) = 1$
the gcd = 1 can be rewritten as :
$x f(x) + p(x) g(x) = 1$
$\alpha f(\alpha) + p(\alpha) g(\alpha) = 1$
so because $p(\alpha) = 0$
$f(\alpha) = \alpha^{-1}$
But can we not assume the gcd with $x$ is 1?
| Now let's see . . .
First of all, the method proposed by our OP roi_saumon appears to be correct in its essentials, though lacking in a proof that $\gcd(x, p(x)) = 1$; how may we show this is the case? Well, if
$d(x) \mid x, \; d(x) \mid p(x) \; \text{with} \; x, \; d(x), \; p(x) \in K[x], \tag 1$
there is some $q(x) \in K[x]$ with
$q(x)d(x) = x; \tag 2$
it then follows that
$\deg q(x) + \deg d(x) = \deg x = 1, \tag 3$
and thus either
$\deg q(x) = 0, \; \deg d(x) = 1 \; \text{or} \; \deg q(x) = 1, \; \deg d(x) = 0; \tag 4$
we may rule out the former case since, by (2), we then necessarily have
$d(x) = ax, \; q(x) = a^{-1}, \; 0 \ne a \in K; \tag 5$
but then $d(x) \mid p(x)$ implies the existence of
$t(x) = \displaystyle \sum_0^{\deg p - 1} t_i x^i \in K[x] \tag 6$
with
$p(x) = ax t(x) = \displaystyle \sum_0^{\deg p - 1} a t_i x^{i + 1}, \tag 7$
according to this equation,
$a \alpha t(\alpha) = p(\alpha) = 0, \tag 8$
whence if $\alpha \ne 0$ we have
$t(\alpha) = 0; \tag 9$
but
$\deg t(x) \le \deg p(x) - 1 < \deg p(x), \tag{10}$
and this contradicts the minimality of $p(x)$ with respect to $\alpha$; therefore, we may rule out the case $\deg q(x) = 0$, $\deg d(x) = 1$; on the other hand, with $\deg q(x) = 1$, $\deg d(x) = 0$ for every common divisor $d(x)$ of both $x$ and $p(x)$, it is certainly true that we may take
$\gcd(x, p(x)) = 1, \tag{11}$
and from this point our OP's approach to the problem is seen to be both rigorous and fruitful.
Another way to approach this problem, one which in fact develops a explicit formula for $\alpha^{-1}$, is based upon the observation that the minimal polynomial $p(x) \in K[x]$ of $\alpha$ must have a non-vanishing constant term. Otherwise, we may write
$p(x) = \displaystyle \sum_1^{\deg p} p_i x^i = x \sum_1^{\deg p} p_i x^{i - 1}, \tag{12}$
which implies
$\displaystyle \alpha \sum_1^{\deg p} p_i \alpha^{i - 1} = p(\alpha) = 0; \tag{13}$
under the assumption $\alpha \ne 0$, (13) additionally implies
$\displaystyle \sum_1^{\deg p} p_i\alpha^{i - 1} = 0, \tag{14}$
which asserts that $\alpha$ satisfies a polynomial of degree less that $\deg p(x)$, contradicting the minimality of $p(x)$; therefore
$p_0 \ne 0 \tag{15}$
and we have
$\displaystyle \sum_0^{\deg p} p_i \alpha^i = 0, \tag{16}$
which may be written
$\alpha \displaystyle \sum_1^{\deg p} p_i \alpha^{i - 1} = \sum_1^{\deg p} p_i \alpha^i = -p_0, \tag{17}$
or
$\alpha \displaystyle \sum_1^{\deg p} -\dfrac{p_i}{p_0} \alpha^{i - 1} = 1, \tag{18}$
which shows both that $\alpha^{-1}$ exists and is given by the formula
$\alpha^{-1} = -\displaystyle \sum_1^{\deg p} \dfrac{p_i}{p_0} \alpha^{i - 1} \in K[\alpha]. \tag{19}$
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Closed form of $\int_0^\infty \left(\frac{\arctan x}{x}\right)^ndx$ I know that for $n=1$ the integral is divergent and that for $n=2$ the integral has a closed form. However, I wonder if the general expression has a closed form.
My attempt: $$\int_0^\infty \left(\frac{\arctan(x)}{x}\right)^ndx=\frac{n}{1-n}\int_0^\infty\frac{\arctan^{n-1}(x)}{x^{n-1}(x^2+1)}dx=\frac{n}{1-n}\int_0^{(\frac{\pi}{2})^{n-1}} u^{n-1}\cot^{n-1}\left(u^{1/(n-1)}\right)du$$
I don't know if I'm on the right track here or not but I do not know through what methods to evaluate the last integral. Any help is appreciated.
| It wasn't requested, but instead of exact representations the OP might want an asymptotic expression for $n \to \infty.$ This one works well with the technique of Depoissonization. Make an exponential power series and analyze it asymptotically:
$$ \sum_{n=0}^\infty \frac{y^n}{n!} C_n = \int_0^\infty \exp{\big(\frac{y}{x} \text{arctan}(x) \big)} dx , \quad C_n=\int_0^\infty \Big(\frac{\text{arctan}(x)}{x}\Big)^n dx$$
Now $\text{arctan}(x)/x = 1-x^2/3+x^4/5+...$ and with $y$ large and keeping on the first term in the asymptotic expansion
$$ e^{-y} \sum_{n=0}^\infty \frac{y^n}{n!} C_n \sim \int_0^\infty \exp{\big(-y\,\frac{x^2}{3}\big)} dx = \frac{1}{2} \sqrt{\frac{3\pi}{y}}.$$
By Depoissonization we can conclude that
$$ C_n \sim \frac{1}{2} \sqrt{\frac{3\pi}{n}} .$$
For $n=50$ the asymptotic expression is within 2% of the value from a numerical integration.
| {
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When given a language, does the algebra follow when raising a symbol in the alphabet to 0 or a negative number? For example, I'm given the language {$a^j b a^k | j < k + 4$}. Do I have to worry about cases like $a^0$? Does it come out to 1 or $\epsilon$? What about things like $a^{-1}$?
| $a^0$ means the empty word $\epsilon$, though it's not sure whether they allowed $j=0$ or not.
Negative powers are not defined in this setting, the words only form a semigroup (actually, monoid with $\epsilon$), not a group, so that we don't have inverses.
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$A\cap G=A\cap H=1$ implies $A=1$? Suppose $G\times H$ is an abelian group and $A\le G\times H$.
Is it true that $A\cap G=A\cap H=1$ implies $A=1$? ($1$ is the trivial subgroup)
| In general, no. Take for example $\langle (1,1) \rangle \leq \mathbb{Z}_2 \times \mathbb{Z}_2$.
| {
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Evaluate $\lim\big(\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2n}\big)$ using sequential methods
Evaluate $$\lim\Big(\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2n}\Big)$$ using sequential methods.
Of course: $$\frac{1}{n}\bigg(\frac{1}{1+1/n}+\frac{1}{1+2/n}+\ldots+\frac{1}{1+n/n}\bigg) \rightarrow \int_{0}^{1}\frac{dx}{1+x}=\ln 2$$
but by using only sequences, I don't know where to start from. I thought of the squeeze theorem, but i am not sure how to get $\ln 2$ on the way.
Thanks in advance.
| Using the fact that $\ln(1 + x) \leq x$, we have
$$
\ln(k + 1) - \ln k \leq \frac{1}{k} \leq \ln{k} - \ln(k - 1),
$$
hence
$$
\ln \frac{2n + 1}{n + 1} \leq \sum_{k = {n + 1}}^{2n} \frac{1}{k} \leq \ln \frac{2n}{n} = \ln2,
$$
and the squeeze theorem will give the result.
| {
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About dense subspaces in Banach spaces I have the following problem:
Let $E$ a Banach space and $X_1,X_2$ dense subspaces. Is $X_1\cap X_2$ dense in $X$? What is the answer if $X_1,X_2$ has codimension 1?
I don't know how to start. If anyone can give me a hint it will be appreciated.
Thanks in advance!
| If $X_1$ and $X_2$ are dense, $X_1 \cap X_2$ need not be dense, in fact it might be just $\{0\}$. Consider any infinite-dimensional separable Banach space $E$. Start with a dense countable set $\{w_1, w_2, \ldots\}$. Inductively choose two sequences $x_j$ and $y_j$ such that $\|x_j - w_j\| < 1/j$ and $\|y_j - w_j\| < 1/j$ and $\{x_1, y_1, \ldots, x_j, y_j\}$ is linearly independent. Let $X_1$ be the linear span of the $x_j$ and $X_2$ the linear span of the $y_j$. Then $X_1$ and $X_2$ are dense in $E$ and $X_1 \cap X_2 = \{0\}$.
| {
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Isomorphisms in a reflective subcategory Let $S$ be a small family of arrows in a locally presentable category $\mathcal{K}$.
It is known that the category $\mathcal{K}[S^{-1}]$ is reflective in $\mathcal{K}$ and correspond to the solution of the orthogonality problem associated to $S$.
Can I infer that a map is an iso in $\mathcal{K}[S^{-1}]$ if and only if it comes from a map in $S$?
This question is strongly related to this other one. In that case, the localization might not be reflective, and this difference might be quite relevant in the answer.
| No, you cannot infer this. The class of morphisms inverted by any functor must satisfy the two-for-three, even the two-for-six, property. But $S$ is completely arbitrary. Consider, for a dramatic example, $\mathrm{Set}[(\emptyset \to \{*\})^{-1}]$, where inverting a single arrow is equivalent to inverting all arrows. Furthermore, there are many more conditions than just two-for-$n$ in this situation. Indeed, the class of morphisms inverted by any left adjoint is closed under colimits in the arrow category.
| {
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$18a$ and $25a$ both integers, then so is $a$ Let $a\in \mathbb{Q}$ such that $18a$ and $25a$ are integers, then we wish to prove that $a$ must be an integer itself. What that means is that $a=\frac{p}{1}$ where $p \in \mathbb{Z}$. What we do know is that we can express the $\gcd(18,25)$ as:
$$ \gcd(18,25)=18x +25y$$ Now if $x=y=a$, we are done, since:
$$ \gcd(18,25)=18a +25a=43a$$ as the $\gcd$ is always an integer and so is 43, so $a$ is also an integer.
But, how would I generalise this?
| Another way to look at this: $18a$ and $25a$ are integers. Therefore, so is $25a-18a = 7a$.
Therefore so is $18a-2(7a) = 4a.$
Therefore so is $7a-4a = 3a.$
Therefore so is $4a-3a = a.$
| {
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Finding the parametric and vector forms of the line is perpendicular to two lines Finding the parametric and vector forms of the line is perpendicular to lines $(4t,1+2t,3t)$ and $(−1+s,−7+2s,−12+3s)$
And passes through the point of the intersection of two lines
A vector perpendicular to these lines is $$v = (4, 2, 3) \times (1, 2, 3) = [0, -9, 6]$$
How would I write the vector / parametric form?
| HINT
We have found the direction vector $\vec v$ for the perpendicular line, now we need the intersection point $P_0$ to determine the parametric equation
$$P(t)=P_0+t\vec v$$
| {
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How to evaluate contour integral I have a homework problem to evaluate the integral
$$
\oint_{\gamma}\frac{\cos z}{(z+i)^3}dz
$$
along the curve $\gamma(t)=-i+e^{it}, t\in[0,2\pi]$. I proceeded to plug the given information into the definition of a contour integral and got to the expression
$$
\oint_{\gamma}=i\int_{0}^{2\pi}\cos(-i+e^{it})e^{-2it}dt
$$
which seems hardly helpful. I don't know how to evaluate this or manipulate it any further and I suppose there is some trick earlier on to make the integration more manageable. I just can't figure it out so I'd be grateful for any help.
| Hint: As Jose hints in the comments, we want to use the fact that
$$f^{(k)}(z_0)=\frac{k!}{2\pi i}\oint_{\partial B_r(z_0)}\frac{f(z)}{(z-z_0)^{k+1}}dz$$
with the correct holomorphic function $f(z)$
| {
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How to solve $y=\tan(θ)$ for $θ$? If $\arctan(\tan(\theta))$ is not necessarily equal to $\theta$, how come if we are given $y=\tan(θ)$ the solution in terms of $\theta$ is $\theta=\arctan(y)$?
I'm trying to intergrate $1/(1+y^2)$ using trig substitution and I am trying to get my solution, $\theta$, in terms of $y$, $y=\tan(\theta)$
| Hint
$$\tan x=\tan (x+k\pi), k\in \mathbb{Z}$$
$$\int \frac{dx}{1+x^2}=\arctan x + C.$$
$C$ plays the role of not having $\arctan(\tan \theta)=\theta$ depending on the range.
| {
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Projectile Envelope
Consider a projectile launched from the origin at velocity $v$ and angle $\theta$. From this other question we see several approaches to arrive at the equation for the envelope of different trajectories with varying $\theta$.
We know from standard projectile formulas that
(i) at $\theta=\frac \pi 4$, the projectile reaches maximum horizontal range of $\frac {v^2}g$, and
(ii) at $\theta=\frac \pi2$, the projectile is launched vertically and reaches a
maximum height of $\frac {v^2}{2g}$.
Hence the $x$-axis intercept of the envelope is $\dfrac {v^2}g$ and the $y$-axis intercept is $\dfrac{v^2}{2g}$.
Can it be concluded, without first working out its equation, that the envelope is a parabola symmetrical about the $y$-axis?
If so, then, together with the axes intercepts deduced above, the formula for the envelope can be derived directly.
NB - if we know that the envelope is a parabola symmetrical about the $y$-axis, with $x$- and $y$- intercepts $\dfrac {v^2}g$ and $\dfrac {v^2}{2g}$ respectively, then the formula for the envelope is
$$\frac {x^2}{\left(\dfrac {v^2}{g}\right)^2}+\frac y{\left(\dfrac {v^2}{2g}\right)}=1$$
which reduces to the standard result for the envelope
$$y=\frac {v^2}{2g}-\frac {g}{2v^2}x^2$$
Note: Just found a usefeul reference here, with an alternative approach.
| I put together an argument that I think relies only on true facts and avoids actually computing either the envelope's equation or the equation of any trajectory. I suspect that the proofs of the facts used in this argument require (at least in some cases) more sophistication than the calculation using the discriminant (mentioned in the linked question), and altogether this argument may take more effort than simply working out the equation. But here it is:
Select a line through the origin at an arbitrary angle $\alpha$ counterclockwise from the positive $x$ axis, with $0 < \alpha < \frac\pi2.$
The farthest point on that line that can be hit by the projectile launched to the right of vertical is achieved when the initial velocity angle of the projectile is
$\frac12\left(\frac\pi2 - \alpha\right)$ clockwise from vertical.
That is, in order to maximize the distance reached on a sloped line,
we bisect the angle between the sloped line and the vertical line through the launch point.
Moreover, if we take the velocity with which the projectile arrives at $P$ and consider all possible trajectories outward from $P$ at that velocity with an initial angle to the left of vertical, the farthest point on the line that we can reach is the origin, reached by retracing the original projectile's path from the origin to $P.$ Therefore the tangent to the projectile's parabolic path at $P$ (which is also the tangent to the envelope at $P$) bisects the angle between the sloped line and the vertical line through $P.$
The tangent line is therefore sloped downward to the right at an angle
$\frac12\left(\frac\pi2 - \alpha\right)$ from horizontal.
The reflection of the ray from the origin to $P$ off the tangent line at $P$ is a vertical line.
This holds for every sloped line through the origin: wherever that line meets the envelope, the line is reflected by the tangent to the envelope into a vertical line.
The only continuous curves that have this property (all lines outward from the origin are reflected by the tangent of the curve into vertical lines) are parabolas with focus at the origin, symmetric across the $y$ axis. You only need to examine one point on the envelope (such as the highest point reached by shooting straight up) in order to determine which parabola is the envelope.
| {
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The problem of dividing 3kg flour to three 1kg parts using a balance Suppose you have 3kg flour, and you are asked to divide it to three 1kg parts using a balance scale.
It seems to me that it's impossible to do with a finite number of weighing, but I can't see how to prove it. Is this a known problem? Any hint is appreciated.
Please comment if the question is vague or needs further clarification.
Edit: I'd appreciate it if you formulate this problem with reasonable assumptions into a pure math problem.
| Theoretically, assuming it is possible to use the balance scale to divide any flour quantity to two equal parts. Make a continuous sets of two half dividing rounds. At end of each set, sum one side and use the other side to continue the two dividing rounds sets. This shloud give you the sum:
$$ \color{red}{S} =\frac{3}{4}+\frac{3}{16}+\frac{3}{64}+\cdots =\sum_{n=1}^{\infty}\frac{3}{4^n} =\color{red}{1} $$
| {
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$A^TA=B^TB$. Is $A=QB$ for some orthogonal $Q$? Suppose that $A$ and $B$ are two real square matrices and $A^TA=B^TB$. Can we say that $A=QB$ for some orthogonal matrix $Q$?
If they are vectors we have $\|a\|^2=a^Ta=b^Tb=\|b\|^2$, so intuitively clear, since we just have to rotate. But it is hard to picture the matrix case but I have not been able to show.
| There is also a nice geometric way to see this.
The geometric interpretation of the singular value decomposition says that a $(n\times n)$-matrix $A$ maps the unit $(n-1)$-sphere in $\mathbb{R}^n$ to a hyperellipsoid. The lengths of the axes of this ellipsoid are the roots of the eigenvalues of $A^T A$.
So if $A^T A = B^T B$, the matrices $A$ and $B$ map the unit $(n-1)$-sphere to congruent hyperellipsoids. Hence $A$ and $B$ must be the same up to an orthogonal matrix, which represents the isometry that maps one hyperellipsoid into the other.
| {
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Solving green function for first quadrant with boundary conditions (dirichlet problem?) Let $\Omega = \{x_1,x_2\in\mathbb{R}^2: x_1>0, x_2>0\}$. Solve the problem $$\Delta u = 0 \mbox{ in $\Omega$ }, u\in C^2(\Omega)\cap C(\overline\Omega) \mbox{ bounded}\\u(x_1,0)=u_0(x_1),x_1\ge 0\\u(0,x_2) = 0, x_2\ge 0$$
Where $u_0$ is continuous, bounded from $0$ to $\infty$ and $u_0(0) = 0$
I opened a bounty in Green function of the first quadrant because I thought it would help be solve this problem which is a slighty different one but I don't think it helps.
I truly have no idea how to modify the answer there to obtain these boundary conditions.
UPDATE:
The only thing I could think of is to solve a system of dirichlet problems:
$$\Delta u_1 = 0 \mbox{ in } \mathbb{R}_+^n\\ u_1 = -u_2 + u_0 \mbox{ in } \partial \mathbb{R}_+^n$$
$$\Delta u_2 = 0 \mbox{ in } \mathbb{R}_{++}^n\\ u_2 = -u_1\mbox{ in } \partial \mathbb{R}_{++}^n$$
Where $\mathbb{R}_+^n$ is the upper half plane, and $\mathbb{R}_{++}^n$ is the 'upper right plane'. That is $\{(x_1,x_2)| x_1>0\}$
Then if we sum these solutions, we have
$u_3 = u_1 + u_2 \implies \Delta u_3 = 0$ in $\{(x_1,x_2), x_1>0, x_2>0\}$
and at the first border, which is $\{(x_1,0), x_1>0\}$, we have
$u_3 = u_1 + u_2 = -u_2 + u_0 + u_2 = u_0$
and at the second border, which is $\{(0,x_2), x_2>0\}$, we have
$u_3 = u_1 + u_2 = u_1 + -u_1 = 0$
So we should find the solutions of $u_1$ and $u_2$ explictly. I found them to be
$$u_1(x) = \frac{x_2}{\pi}\int \frac{1}{|y-x|^2}(-u_2(x)+u_0(x))\ dy$$
$$u_2(x) = \frac{x_1}{\pi}\int \frac{1}{|y-x|^2}(-u_1(x))\ dy$$
so the solution would be
$$u_1 + u_2 = \\ \frac{x_2}{\pi}\int \frac{1}{|y-x|^2}\left(-\frac{x_1}{\pi}\int \frac{1}{|y-x|^2}(-u_1(x))\ dy + u_0(x)\right)\ dy + \frac{x_1}{\pi}\int \frac{1}{|y-x|^2}(-u_1(x))\ dy$$
Is it right?
| It is well known that the solution of the Dirichlet problem in a half plain
$$\Delta u=0,\ x_2>0,\quad u(x_1,0)=\psi(x_1),$$
is given by the integral
$$
u(x_1, x_2) = \frac1\pi
\int_{-∞}^∞
\frac{x_2\psi(y)}{(x_1-y)^2+x_2^2}\,dy.
$$
Now extending function $u_0$ to $\tilde u_0$ on $\mathbb R$ as odd:
$\tilde u_0(x_1)=u_0(x_1)$, $x\ge0$,
$\tilde u_0(x_1)=-u_0(-x_1)$, $x<0$,
will give the required solution:
$$
u(x_1, x_2) = \frac1\pi
\int_{-∞}^∞
\frac{x_2\tilde u_0(y)}{(x_1-y)^2+x_2^2}\,dy.
$$
| {
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Prove that there is no $\{3,5\}$-Hall subgroup in $A_{5}$ My approach.
A $\{3, 5\}$-Hall subgroup $K$ of $A_{5}$ has order $3\cdot 5$ and index $2^{2} = 4$. Note that $A_{5}$ acts in cosets of $K$, with $4$ distincts cosets, this way:
$$A_{5}/K = \{a_{1}K, a_{2}K, a_{3}K, a_{4}K\}.$$
Thus, we have a homomorphism $\varphi: A_{5} \to S_{4}$. Since $A_{5}$ is a simple group, $\ker \varphi = \{e\}$ or $\ker \varphi = A_{5}$. Then
Case 1: $\ker \varphi = \{e\}$.
So, $\ker \varphi = \{e\}$ implies $\varphi$ injective, an absurd because $|A_{5}| = 60 > 24 = |S_{4}|$.
Case 2: $\ker \varphi = A_{5}$.
So, for any $b \in A_{5}$ we have $b(a_{i}K) = a_{i}K$ iff $b \in a_{i}K$, where $1 \leq i \leq 4$, an absurd because $\bigcap(a_{i}K) = \emptyset$.
Therefore, there is no $\{3, 5\}$-Hall subgroup of $A_{5}$.
Is there an error? Or is it possible to improve this proof without using overkill results?
| Even easier is to note that all groups of order 15 are cyclic, and $A_5$ has no element of order 15. Thus $A_5$ has no subgroup of order 15.
| {
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Let $\mathcal F$ be the set of mappings $f:\Bbb N \to \Bbb N$ for which $f(m) \ge f(n)$ for $m \le n$. Show that $\mathcal F$ is countable
Let $\mathcal F$ be the set of mappings $f:\Bbb N \to \Bbb N$ for which $f(m) \ge f(n)$ for $m \le n$. Show that $\mathcal F$ is countable.
My attempt:
For all $f\in \mathcal F$, $f$ will be eventually constant, i.e. there exists $N\in \Bbb N$ such that $f(n)=f(N)$ for all $n>N$. Let $N_f$ be the least element of such $N$ for each $f\in \mathcal F$.
Let $\operatorname{Seq}(\Bbb N)$ be the set of all finite sequences from $\Bbb N$. We define a mapping $G:\mathcal F \to \operatorname{Seq}(\Bbb N)$ by $G(f)=f_{\restriction \{0,\cdots,N_f\}}$. Then $G$ is clearly injective. Hence $|\mathcal F| \le |\operatorname{Seq}(\Bbb N)|$. We already know that $\operatorname{Seq}(\Bbb N)$ is countable. It follows that $\mathcal F$ is countable.
My questions:
*
*Does my proof look fine or contain gaps?
*I feel that this proof depends on the theorem If $A$ is countable, then the set of finite sequences from $A$ is countable, which in turn requires several heavy lemmas. I would like to ask for a simpler proof.
| An alternative approach written for fun and/or clarity:
This produces a sequence with maximum $f(1)$ that eventually becomes constant (as pointed out in the OP). Let us just consider the cases where $f(1) = 3$ for concreteness. This means all sequences that are of one of the following forms:
*
*all $3$s
*finitely many $3$s, all $2$s
*finitely many $3$s, finitely many $2$s, all $1$s
Respectively, the number of these sequences is:
*
*one
*countably infinite: one for each possible place at which it switches from $3$ to $2$, which has size $|\mathbb{N}|$
*countably infinite: one for each possible place at which it switches from $3$ to $2$, and another for each possible place at which it switches from $2$ to $1$, which has size $|\mathbb{N} \times \mathbb{N}| = |\mathbb{N}|$
Altogether, we conclude that the number of sequences with $f(1) = 3$ constitutes a countably infinite collection: it corresponds to a a finite union of countable sets. But, a similar line of reasoning works for $f(1) = n$ for each $n \in \mathbb{N}$. Since the countable union of countable sets is countable, the assertion follows.
| {
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Finding a bijection from $\{1,2,...,nm\}$ to $X \times Y$ I'm trying to prove that for two finite sets $X,Y$, where $|X|=n$, $|Y|=m$, $|X||Y|=|X \times Y|$. I know that there exists bijections $f:\{1,2,...,n\} \rightarrow X$ and $g:\{1,2,...,m\} \rightarrow Y$ and I'm trying to find a bijection $h:\{1,2,...,nm\} \rightarrow X \times Y$. I know that this is equivalent to showing there exists a bijection $k: X \times Y \rightarrow \{1,2,...,nm\}$. One such bijection that seemingly works is $k(x_i,y_j) = (i-1)m + j$, where $1 \leq i \leq n$ and $1 \leq j \leq m$. I've tried to prove the injectivity and surjectivity of this function but to no avail:
Injecivity: Suppose $k(x_a,y_b) = k(x_c,y_d)$ for $x_a, x_c \in X$ and $y_b, y_d \in Y$. This gives $(a-1)m + b = (c-1)m + d$ but I've not been able to show that $a = c, b = d$ from here.
Surjectivity: Suppose $z \in \{1,2,...,nm\}$ then $\exists i,j$ s.t. $ z = (i-1)m + j$ for some $i,j$ satisfying $1 \leq i \leq n$ and $1 \leq j \leq m$. Now, since $X, Y$ can be written as $X = \{x_1,x_2,...,x_n\}$, $Y = \{y_1,y_2,...,y_m\}$ and $X \times Y = \{(x,y) | x \in X \wedge y \in Y\}$ then we can say $\exists(x_i, y_j) \in X \times Y$ and by the definition of $k$ we have $k((x_i, y_j)) = (i-1)m + j$. But I'm not sure if this all watertight since I've assumed that $z$ can be written in the desired form.
| First let me do a little change: instead of $\{1,2,...,n\},\{1,2,...,m\},\{1,2,...,nm\}$ I will use $\{0,1,...,n-1\},\{0,1,...,m-1\},\{0,1,...,nm-1\}$.
Now, to prove $k$ is bijective I will find the inverse of $k$: $k(x_i,y_j)=(i-1)m + j$, to isolate $i$ I will use the fact that $j<m$, first let's divide by $m$: $\frac{(i-1)m + j}m=i-1 + \frac jm$, now notice that $i-1$ is an integer and $\frac jm$ is a fraction, so if I use the floor function on this expression I will get rid from $j$:$\lfloor i-1 + \frac jm\rfloor=i-1$, now adding to this $1$ we get $$\left\lfloor\frac{k(x_i,y_j)}{m}\right\rfloor+1=i\implies f^{-1}\left(\left\lfloor\frac{k(x_i,y_j)}{m}\right\rfloor+1\right)=x_i$$
Now we can easily isolate the $j$: $k(x_i,y_j)=(i-1)m + j=\left(\left\lfloor\frac{k(x_i,y_j)}{m}\right\rfloor+1-1\right)m+j=\left\lfloor\frac{k(x_i,y_j)}{m}\right\rfloor m+j\\\implies j=k(x_i,y_j)-\left\lfloor\frac{k(x_i,y_j)}{m}\right\rfloor m$
With this we only left to do is to use the inverse of $g$:$$j=k(x_i,y_j)-\left\lfloor\frac{k(x_i,y_j)}{m}\right\rfloor m\implies g^{-1}\left(k(x_i,y_j)-\left\lfloor\frac{k(x_i,y_j)}{m}\right\rfloor m\right)=y_j$$
Combining those 2 we get $$k^{-1}(z)=\left(f^{-1}\left(\left\lfloor\frac{z}{m}\right\rfloor+1\right),g^{-1}\left(z-\left\lfloor\frac{z}{m}\right\rfloor m\right)\right)$$
Thus we can conclude that $k$ is bijective
Here is a different proof that use induction.
It is trivial that $X\times \{a\}$ is the same size as $\{0,1,...,n-1\}$
Assuming that $X\times Y$ is the same size $\{0,1,...,nm-1\}$ I'll show that $X\times (Y\cup \{a\})$ where $a\notin Y$ is the same size of $\{0,1,...,n(m+1)-1\}$:
Let's call $h$ the bijective from $X\times Y$ to $\{0,1,...,nm-1\}$, then set $k(x_i,y_j)=h(x_i,y_j)$ if $(x_i,y_j)\in X\times Y$ and $k(x_i,y_j)=nm-1+i$ elsewhere, try to show that $k$ is bijective from $X\times (Y\cup \{a\})$ to $\{0,1,...,n(m+1)-1\}$ and then you are done with the induction
!
| {
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Whittaker-Shannon-Kotel’nikov theorem in higher dimensions The fundamental result in sampling theory states that if a signal $f(t)$
contains no frequencies higher than $\omega$ cycles per second, then $f(t)$ is completely
determined by its values $f(\frac{k}{2\omega})$ at a discrete sequence of sample
points with spacing $\frac{1}{2\omega}$ and can be reconstructed from these values by
the formula$\dagger$
$$\label{1}\tag{1} f(t)=\sum_{k=-\infty}^{\infty}f(\frac{k}{2\omega})\frac{\sin(\pi(2\omega t-k))}{\pi(2\omega t -k)}$$
Is something like $(\ref{1})$ still valid if instead of the signal
$f(t)\colon t \in \mathbb{R} \to f(t) \in \mathbb{R}$ we have signal defined in higher dimension? For example the signal
$f(x,y)\colon x,y \in \mathbb{R} \to f(x,y) \in \mathbb{R}$
or
$f(x,y,z)\colon x,y,z \in \mathbb{R} \to f(x,y,z) \in \mathbb{R}$
$\dagger$, Literally taken from the incipit of:
CIAURRI, Óscar; VARONA, Juan. A Whittaker-Shannon-Kotel’nikov sampling theorem related to the Dunkl transform. Proceedings of the American Mathematical Society, 2007, 135.9: 2939-2947.
| After a bit of search I have found the answer:
THEOREM 1.6$\dagger$
Let $f(t_1 , t_2 ,\ldots , t_n )$ be a function of $n$ real variables, whose
$n$-dimensional Fourier integral, $g$, exists and is identically zero outside an $n$-dimensional rectangle and is symmetrical about the
origin, that is,
$g(y_1 , y_2 , \ldots , y_n )=0$, $\mathopen|y_k\mathclose|\gt\mathopen|\omega_k\mathclose|$, $k=1,2,\ldots,n$
Then
$$f(t_1 , t_2 ,\ldots , t_n )=f(t)=\sum_{m_1=-\infty}^{\infty}\ldots \sum_{m_n=-\infty}^{\infty}f(\frac{\pi m_1}{\omega_1},\ldots,\frac{\pi m_n}{\omega_n})\times \frac{\sin(\omega_1 t_1-m_1\pi)}{\omega_1 t_1-m_1\pi}\times\ldots\times\frac{\sin(\omega_n t_n-m_n\pi)}{\omega_n t_n-m_n\pi}$$
$\dagger$
POULARIKAS, Alexander D. Transforms and applications handbook. CRC press, 2010.
| {
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How can I prove that if $A$ is a singular square matrix, then $\det A = 0$ without using Binet's Theorem? Well, I need to prove that if $A$ is a $n \times n$ matrix and it is singular, then $\det A = 0$, in order to show Binet's Theorem $\det(AB) = \det(A)\cdot\det(B)$ in the case when both $A$ and $B$ are singular square matrices.
I'll thank any help with this issue!
| I would use two proof
*
*Permutation change the sign of the determinant
*Multiplying a row by a constant multiply the determinant by this constant
Then, you can just say if you have a singular matrix S, 2 row $r_i$ and $r_j$ are linearly dependent. You can multiply $r_i$ by a constant C to get a new matrix S' where $r_i$ = $r_j$.
We know
\begin{equation}
det(S') = C det(S)
\end{equation}
Then, you can swap $r_i$ and $r_j$ but you have the same matrix has before so
\begin{equation}
det(S') = -det(S') = -C det(S) = C det(S)
\end{equation}
The equation det(S) = -det(S) means det(S) is 0.
| {
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Removing factors to make the number ends with'2225' We have the integers 1, 2, ..., 50000.If we multiply them together, the last four digits will be '0000'. At least how many integers must be removed so that the product of the remaining integers ends with '2225'?
I know that if an integer ends with '2225', it's factors don't contain any '2', therefore at least 25000 integers need to be remove. Also, '2225' is divisible by 25 but not divisible by 125. Therefore, 5000−2=4998 multiples of 5 have to be removed. Thus, at least 25000+4998=29998 was cancelled out. However, how can I prove that the multiple of the remaining integers ends with '2225'(I can choose to reserve which multiples of 5), or there is any other integers to be removed? Thanks in advance.
| Suppose the product of the $20000$ with no $2$ or $5$ is $A$, and we choose $5B$ and $5C$. We want to choose $B$ and $C$ so that $ABC=400D+89$. Let $B=1$.
• Let $C_1$ be whatever digit makes $AC_1$ end in $9$.
• Let $C_2$ be whatever digit makes $A(C_1+10C_2)$ end in 89.
• Let $C_3$ be whichever of $1,2,3,4$ makes $AC=89\pmod{400}$.
| {
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Name and derivation of the approximation $\frac{1+x}{1+y} - 1 \approx x-y$? I am wondering if there is a name and way to derive the following approximation:
$$\frac{1+x}{1+y} - 1 \approx x-y$$
I'm essentially interested in how to refer to this.
| I would call this a Taylor approximation. When $|y|\lt1$,
$$
\begin{align}
\frac{1+x}{1+y}-1
&=-1+(1+x)\left(1-y+y^2-y^3+\dots\right)\\
&=x-y-xy+y^2+xy^2-y^3-xy^3+\dots\\[6pt]
&=x-y+O\!\left(\max(|x|,|y|)^2\right)
\end{align}
$$
| {
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I can't seem to undersatnd how did we compute $\frac{}{}$ from the function $\mathbf T=\frac{v }{1+uvw}$ $$ \mathbf T = \frac{v}{1+uvw}$$
Solution:
$$d\mathbf T =
\frac{\partial\mathbf T} {\partial u}\,du +\frac{\partial\mathbf T}{\partial v} \, dv +\frac{\partial\mathbf T}{\partial w} \, dw $$
$$d\mathbf T =
\frac{−v^2w }{(1+uvw)^2}\,du +
\frac{1}{(1+uvw)^2} \, dv +\frac{−uv^2}{(1+uvw)^2} \, dw$$
I get this instead of the above solution in computing:
$$\frac{\partial\mathbf T}{\partial v} = \frac{-vuw}{(1+uvw)^2}+\frac{1}{(1+uvw)}$$
| Quotient rule:
\begin{align}
& \frac\partial {\partial v} \, \frac v {1+uvw} = \frac{(1+uvw) \dfrac\partial{\partial v} v - v \dfrac\partial {\partial v} (1+uvw)}{(1+uvw)^2} \\[12pt]
= {} & \frac{(1+uvw) - v(uw)}{(1+uvw)^2} = \frac 1 {(1+uvw)^2}.
\end{align}
| {
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Find coefficients to a quadratic equation knowing roots and a point... Given a standard quadratic equation:
$$p(z) = az^2 + bz + c$$
We know that $-10$ and $10-i$ are roots.
We know that $p(i)=-10$
What are $a$, $b$ and $c$?
| Try $p(z)=a(z-i)(z-10+i)$
Multiply and find coefficients using the information at $z=i$
| {
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Does $3-4+1$ equal $0$ or $-2$.. or maybe $2$? Does $3 - 4 + 1 = 0$ or $3 - 4 + 1 = -2$?
Makes sense that $(3 - 4) + 1 = 0$ and $3 - (4 + 1) = -2$, but what if there are no parenthesis?
Also, if I have $4$ apples and I add $1$ more apple, then I have $5$ apples, but if I eat $3$ apples, then I have $2$ left.
Any ideas?
| Add positives together, and you will have $3+1=4$
The only negative that you have is $-4$
The sum is then $4-4=0$
If you do not have parenthesis you add positives and keep the result.
Then you add negatives and keep the result. Then find the total result by the algebraic sum of those two.
For example $$1-5-3+12-23=?$$
We have $1+12=13$ positives. We also have $-5-3-23=-31$ negatives.
Thus the result is $13-31 = -18$
| {
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A box has three coins. One has two heads, another two tails and the last is a fair coin. I am stuck on this question:
A box has three coins. One has two heads, one has two tails, and the other is a fair coin with one head and one tail. A coin is chosen at random, and comes up head.
a) What is the probability that the coin chosen is the two headed coin
b) What is the probability that if it is thrown another time it will come up heads
c) What is the probability that the coin chosen is the two headed coin, supposing that the coin is thrown a second time and comes up heads again
I have solved part a, and the answer is $\frac 23$.
However I am stuck on parts b and c. This is my thought process:
part b: P(Heads on the second throw) = P($H_2$|$H_1$) $\cdot$ P(fair coin) + P($H_2$|$H_1$) $\cdot$ P(coin with two heads) = $\frac 12$ $\cdot$ $\frac 13$ + 1 $\cdot$ $\frac 13$ = $\frac 12$, and I'm not sure how to even start with part c.
The answer to part b should be $\frac 56$, but can anyone explain why? Thanks.
| Your answer to part (a) is incorrect. If you pull out a random coin and flip, you have six scenarios:
*
*Head 1 of 2-headed coin
*Head 2 of 2-headed coin
*Head of fair coin
*Tail of fair coin
*Tail 1 of 2-tailed coin
*Tail 2 of 2-tailed coin
3 of those are "heads", and 2 of those 3 correspond to the 2 headed coin. Thus, the answer to part (a) is $\frac 23$
Part (b): Probability of getting double headed * getting head from that double headed + Probability of getting fair * getting head from that fair coin
$$\frac 23 \cdot 1+\frac 13 \cdot \frac 12=\frac 56$$
Part (c): Now you have $3\cdot 2\cdot 2=12$ possiblities; check all of them and see which of the ones involving two heads involve the double headed coin.
You can also look at Bayes's Theorem; it covers problems like this one.
| {
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Understand a dynamical figure: stable manifold, node, domain of attraction Could someone please point out where the node, saddle and stable manifold are in this figure (as indicated in the caption)? I can see there is a focus, but I am not sure about the others. Thank you in advance.
Source: http://www.staff.science.uu.nl/~kouzn101/NBA/LAB2.pdf
| The focus is at the center where the flow is spiraling in.
The stable manifold is the slant line coming down from the focus.
The saddle is the intersection of the slant line and the middle trajectory.
The node is intersection of the slant with the lower trajectory.
| {
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Solving cos(x)=x Algebraically I think a graphical approach is the best way, but can algebra be used? With IVT, from x:[0,1] a solution must exist.
| As said in comments and answers, you need a numerical method such as Newton.
Consider the function $$f(x)=x-\cos(x)\qquad f'x)=1+\sin(x)$$ The iterates will be given by
$$x_{n+1}=\frac{x_n \sin (x_n)+\cos (x_n)}{1+\sin (x_n)}$$ You can have a very good starting point building the $[2,2]$ Padé approximant of the function at $x=0$; this would give
$$f(x)\approx \frac{-1+\frac{7 }{6}x+\frac{1}{4}x^2 } { 1-\frac{1}{6}x+\frac{1}{12}x^2}$$ and then $x_0=\frac{1}{3} \left(\sqrt{85}-7\right)$.
Now, the iterates would be
$$\left(
\begin{array}{cc}
n & x_n \\
0 & 0.73984815243096243667 \\
1 & 0.73908526166480706885 \\
2 & 0.73908513321516428479 \\
3 & 0.73908513321516064166
\end{array}
\right)$$ which is the solution for twenty significant figures.
Edit
For the fun of it, let us build the sama Pade approximant around $x=\frac \pi 4$. This will give as solution
$$4 \left(48+34 \sqrt{2}+7 \pi \right)x_0=$$ $$-720-408 \sqrt{2}+\pi \left(36+46 \sqrt{2}+7 \pi \right)+4 \sqrt{65976+45744
\sqrt{2}-3 \pi \left(1328+940 \sqrt{2}+\left(131+62 \sqrt{2}\right) \pi
\right)}$$ which is $\approx 0.73908513347523013264$. Quite close, isn't it ?
| {
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Subcategories of T1, Hausdorff and TD spaces are reflective subcategories of Category *Top*. How do I construct a T1, Hausdorff and TD reflections to show that subcategories of T1, Hausdorff and TD spaces are reflective subcategories of Category Top?
The T0 reflection is just the T0 quotient of a space X. What about the others?
| This question and its answers talk in detail about the Hausdorff reflection.
The $T_1$ reflection of $X$ is the quotient under the intersection of all equivalence relations on $X$ that have closed equivalence classes, see the post here with an attempted proof (the previous one has the statement).
My details: Let $X$ be any space and define $$\mathcal{C} = \{S \subseteq X \times X: S \text{ is an equivalence relation and } \forall x \in X: [x]_S \text{ closed in } X\}$$ where $[x]_S$ is the class of $x \in X$ under $S$.
This is a non-empty set of equivalence relations as the trivial relation $R= X \times X$ is in $\mathcal{C}$. As the intersection of any family of equivalence relations is again an equivalence relation $R= \bigcap\{S \in \mathcal{C}\}$ is well- defined. As for any $x$ we have $$[x]_R = \bigcap \{[x]_S: S \in \mathcal{C}\}$$ and intersections of closed sets are closed, we have that $R$ also has the property that all classes are closed and so $X/R$ is a well-defined $T_1$ space in the quotient topology (with quotient map $q: X \to X/R$ defined as usual by $q(x)= [x]_R$). This $(X/R,q)$ is the $T_1$-reflection of $X$: let $Y$ be any $T_1$ space and $f: X \to Y$ continuous. Then $R_f = \{(x,x') \in X \times X: f(x)=f(x')\}$ is an equivalence relation and as $[x]_{R_f} = f^{-1}[\{f(x)\}]$, $R_f \in \mathcal{C}$, and so $R \subseteq R_f$, which implies that if $[x]_R =[x']_R$ we have that $f(x) = f(x')$ too, so that $\tilde{f}([x]_R) = f(x)$ is well-defined and continuous as $q^{-1}[\tilde{f}^{-1}[O]] = f^{-1}[O]$ for all $O$ and $X/R$ has the quotient topology w.r.t. $q$. It's clear that $\tilde{f}$ is unique.
That same linked thread claims that $T_D$ is not reflective because $S_2^{\aleph_0}$, where $S_2$ is the Sierpiński two-space, is not $T_D$. More is explained in the introduction of this paper.
| {
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Prove that $b_k \to 1/a$ if $a_k \to a$ where $b_0 = 0$ and $b_k = 1/a_k$ for $k>0$
Let $a_n$ where $n \in \mathbb {N}$ be a sequence of rational numbers converging to $a$. Suppose $a \neq 0$, for $k = 1, 2, ...$ let
$$b_k=\begin{cases} 0 & \text{if}\;a_k=0\\\\\frac{1}{a_k} &\text{if}\;a_k \neq 0\end{cases}$$
Prove that $b_n$ converges to $\frac{1}{a}$.
I was studying real analysis and got stuck on this problem.
Can you help me solve this problem or give me some hints?
Thanks
edit: is it possible to solve this in terms of Cauchy Sequence?
| Hint
Because $a\not=0$ (let's say $a>0$), show that $\exists n_0\in\mathbb{N}: \forall n\geq n_0\quad a_n>0$
| {
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7 fishermen caught exactly 100 fish and no two had caught the same number of fish. Then there are three who have together captured at least 50 fish.
$7$ fishermen caught exactly $100$ fish and no two had caught the same number of fish. Prove that there are three fishermen who have captured together at least $50$ fish.
Try: Suppose $k$th fisher caught $r_k$ fishes and that we have
$$r_1<r_2<r_3<r_4<r_5<r_6<r_7$$
and let $r(ijk) := r_i+r_j+r_k$.
Now suppose $r(ijk)<49$ for all triples $\{i,j,k\}$.
Then we have $$r(123)<r(124)<r(125)<r(345)<r(367)<r(467)<r(567)\leq 49$$
so $$300\leq 3(r_1+\cdots+r_7)\leq 49+48+47+46+45+44+43= 322$$
and no contradiction. Any idea how to resolve this?
Edit: Actually we have from $r(5,6,7)\leq 49$ that $r(4,6,7)\leq 48$ and $r(3,6,7)\leq 47$ and then $r(3,4,5)\leq r(3,6,7) - 4 \leq 43$ and $r(1,2,5)\leq r(3,4,5)-4\leq 39$ and $r(1,2,4)\leq 38$ and $r(1,2,3)\leq 37$ so we have:
$$300\leq 49+48+47+43+39+38+37= 301$$
but again no contradiction.
| If the maximum number of fish caught is $m$, then the total number of fish caught is no more than $m+(m-1)+...+(m-6)$. So there is one fisherman that caught at least 18 fish. Repeat this process for the second and third highest number of fish caught and you should be good.
I should add that this is a common proof technique in combinatorics and graph theory. To show that something with a certain property exists, choose the "extremal" such something, and prove that property holds for the extremal object. For instance, to show in a graph where each vertex has degree at least $d$ there is a path of length at least $d$, and one proof starts by simply showing a maximal path has length at least $d$.
| {
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Prove this problem using Mean - Value theorem let $m$ and $n$ are in $\mathbb{N}$ with $m>n$ . Use the Mean Value Theorem, to prove that $(1+x)^\frac{m}{n}\ge 1+\frac{m}{n}x, x \ge -1$
since $(1+x)^\frac{m}{n}$ is continuous on $[-1,x]$ and differentiable on $(-1,x)$ so by mean value theorem $\frac{f(x)-f(-1)}{x+1}=f'(c)$ for $-1\le c\le x$
then $\frac{(1+x)^{\frac{m}{n}}}{x+1}=\frac{m}{n}(1+c)^{\frac{m}{n}-1}$
form here how to we processed
| Let $f(x) = (1+x)^{\frac mn}$. Then
$$f'(x) = \frac mn (1+x)^{\frac mn -1} =\frac mn (1+x)^{\frac {m-n}n} $$
$f'(x)$ is increasing for $x\ge -1$, and in particular, $f'(0) = \frac mn$.
Case 1: $x > 0$. Consider $f(0)$ and $f(x)$. By mean value theorem, there exists a $c\in (0, x)$ that satisfies
$$\begin{align*}
f'(c) &= \frac{f(x)-f(0)}{x-0}\\
&= \frac{(1+x)^{\frac mn}-1}{x}\\
\end{align*}$$
Since $f'(x)$ is increasing for $x > -1$,
$$\begin{align*}
f'(c) &\ge f'(0)\\
\frac{(1+x)^{\frac mn}-1}{x} &\ge \frac mn\\
(1+x)^{\frac mn} &\ge 1 + \frac mn x
\end{align*}$$
Case 2: $-1 \le x < 0$. Consider $f(x)$ and $f(0)$, and there exists a $c\in(x, 0)$ that satisfies
$$\begin{align*}
f'(c) &= \frac{f(0)-f(x)}{0-x}\\
&= \frac{1-(1+x)^{\frac mn}}{-x}\\
\end{align*}$$
Since $f'(x)$ is increasing for $x > -1$,
$$\begin{align*}
f'(c) &\le f'(0)\\
\frac{1-(1+x)^{\frac mn}}{-x} &\le \frac mn\\
1-(1+x)^{\frac mn} &\le -\frac mn x\\
(1+x)^{\frac mn} &\ge 1 + \frac mn x
\end{align*}$$
Case 3: $x=0$ is trivial.
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Evaluating $\lim_{n \to \infty}\left({^n\mathrm{C}_0}{^n\mathrm{C}_1}\dots{^n\mathrm{C}_n}\right)^{\frac{1}{n(n+1)}}$
$\lim_{n \to \infty}\left({^n\mathrm{C}_0}{^n\mathrm{C}_1}\dots{^n\mathrm{C}_n}\right)^{\frac{1}{n(n+1)}}$ is equal to:
a) $e$
b) $2e$
c) $\sqrt e$
d) $e^2$
Though it looks really innocent at first sight, it certainly isn't.
Attempt: It's $\infty^{\infty}$ form.
I had tried taking the product raised to the power $\frac{1}{n(n+1)}$ as function $f(n)$. Then I took logarithm of both sides to see if things simplifying. Even after factoring out the extra factorials it wasn't easy.
Note that ${^n\mathrm{C}_x} = \binom{n}{x}$.
| $$\prod_{k=0}^{n}\binom{n}{k}=\frac{n!^{n}}{\prod_{k=0}^{n}k!^2}=\frac{n!^n}{\left[\prod_{k=1}^{n}k^{n+1-k}\right]^2}=\frac{n!^n}{n!^{2n+2}}\prod_{k=1}^{n}k^{2k} \tag{1}$$
hence the outcome depends on the asymptotic behaviour of the hyperfactorial.Since by Riemann sums
$$ \lim_{n\to +\infty} \frac{1}{n}\sum_{k=1}^{n}\log\frac{k}{n}=\int_{0}^{1}\log(x)\,dx = -1, $$
$$ \lim_{n\to +\infty} \frac{1}{n}\sum_{k=1}^{n}\frac{k}{n}\log\frac{k}{n}=\int_{0}^{1}x\log(x)\,dx = -\frac{1}{4}\tag{2} $$
we have
$$ \frac{1}{n^2}\log\prod_{k=0}^{n}\binom{n}{k} = \frac{1}{n^2}\left[2\sum_{k=1}^{n}k\log k-(n+2)\log n!\right]\to \frac{1}{2}\tag{3} $$
and the correct option is c).
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Are there uncountably many injective functions from $\mathbb N$ to $\mathbb N$? I think there are, but I haven't been able to prove this. I tried to make two injections, but I get stuck on trying to map all functions f from $\mathbb N$ to $\mathbb N$ onto the injective ones. How do you make sure f becomes injective, while making sure that the bigger injection stays injective? Or am I wrong and there are not even uncountably many functions?
| Each injection from $\mathbb{N} \to \mathbb{N}$ can be thought of as an infinite sequence. Now use Cantor's diagonal argument to show uncountably many functions.
| {
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Limit of a sum using complex analysis. I'm trying to find the limit of this sum: $$S_n =\frac{1}{n}\left(\frac{1}{2}+\sum_{k=1}^{n}\cos(kx)\right)$$
I tried to find a formula for the inner sum first and I ended up getting zero as an answer.
The sum is supposed to converge to $\cot(x\over 2)$ and that appears in my last expression but it goes to zero.
Here is what I got with a rather long and clumsy reasoning:
$$S_n = \frac{1}{2n}\left(\cot\left(\frac{x}{2}\right)\sin(nx)+\cos(nx)\right)$$
Thanks in advance.
| Just write $\cos (kx)$ as the real part of $e^{ikx}$.
| {
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What is the formal definition of a Mackey Topology? I believe the question is self-explanatory. After browsing in the web the following definitions stand out:
Assume we have a dual pair $(X,X')$
*
*The topology of uniform convergence on the convex balanced subsets of $X'$ that are compact in the weak topology $\sigma(X,X')$. (from Encyclopedia of Mathematics)
*Is a polar topology defined on $X$ by using the set of all absolutely convex and weakly compact sets in $X'$. (from Wikipedia)
However, neither can be considered ''formal'' definition and for an amateur in functional analysis the definitions are very obscure.
Wikipedia's definition seem to be more suitable for a layperson, but it does not specify how to ''use'' the absolutely compact sets and the weakly compact sets.
Furthermore, I would greatly appreciate some intuition regarding the relevance of the concept and some solid references to learn about the subject. I came across this concept while researching social choice [Shinotsuka, Tomoichi. "Equity, continuity, and myopia: a generalization of Diamond’s impossibility theorem." Social Choice and Welfare 15.1 (1997): 21-30].
| Both definitions are perfectly `formal', provided you know what all the terms mean. The wikipedia definition links to polar topology where you find an explanation how the topology is generated.
Both the wikipedia page and the Encyclopedia page point to textbooks on topological vector spaces where you can learn more.
Probably Schaefer's book is a good place to start.
| {
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Proving $x$ between $\alpha - \epsilon$, if $\alpha$ is supremum Let $S\subseteq\Bbb R$ and $\alpha \in \Bbb R$. If $\alpha = \sup(S)$, then show that for any $\epsilon > 0$, there is some $x \in S$ such that $\alpha - \epsilon < x$.
What I have done :
Since $\alpha$ is the supremum, $x<\alpha$ and $\alpha - \epsilon < \alpha$ I want to show $x$ is in between $\alpha - \epsilon$ and $\alpha$ , but can't and not even sure if it is possible.
Its been a long I posted a question, so sorry if this seems a total homework question.
| We want to show the existence of such $x$.
Suppose it doesn't exist, then we have $\alpha - \epsilon$ being an upper bound of $S$ which contradicts that $\alpha$ is the least upper bound.
| {
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If Rolle's Theorem is assumed to be true, doesn't that prove the MVT? If we assume that Rolle's Theorem is true is it practical to say that it also proves the MVT?
My reasoning is that even though Rolle's Theorem is the special case for when $f(a)=f(b)$ and the secant line between $(a,f(a))$ and $(b,f(b))$ is horizontal, doesn't that mean you can 'rotate' and 'stretch' the function $f$ and thus 'stretch' and 'rotate' the straight line between $(a,f(a))$ and $(b,f(b))$ as well as the line tangent to $(c,f(c))$ to maintain its symmetry to the line secant between $(a,f(a))$ and $(b,f(b))$ to thus prove that for the straight line intersecting points $(a,f(a))$ and $(b,f(b))$, there must be a point $c$ on $[a,b]$ for which $f'(c)$ is parallel to the aforementioned line between $(a,f(a))$ and $(b, f(b))$?
I feel that this could be better explained with something like linear algebra. I do not know it well enough to use it.
EDIT: My question was just as unclear as I thought it was. I am going to try and explain it differently:
For a function $f$, if it is differentiable/continuous over $[a,b] $/$(a,b)$ etc. then there is a straight line between $(a, f(a))$ and $(b,f(b))$. You can rotate $f$ around the origin until this straight line is parallel to the x-axis. By Rolle's Theorem, there must exist a $c$ on the transformed $(a,b)$ such that $f'(c)=0$ and thus there is at least one line tangent to $f$ on $(a,b)$ that is parallel to the secant of $(a,f(a))$ and $(b,f(b))$. Then, you can 'undo' the original rotation of $f$ and know that the lines proven to exist by Rolle's Theorem exist with the same symmetries with respect to $f$, meaning the MVT must hold even when the secant line is not horizontal.
| A proof of MVT used Rolle's theorem explicitly.
Consider a continuous function $f:[a,b]\to\Bbb R$ that is differentiable on $(a,b)$. We are to prove MVT, i.e. to prove there exists a point $c\in(a,b)$ such that $f'(c)=\frac{f(b)-f(a)}{b-a}$, assuming that Rolle's theorem is true.
Define a new function $g:[a,b]\to\Bbb R$ by $g(x)=f(x)-\frac{f(b)-f(a)}{b-a}x$. Verify yourself that $g$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Verify also that $g(a)=g(b)$. So we can apply Rolle's theorem to this function $g$, to obtain a point $c\in(a,b)$ such that $g'(c)=0$. Note that the derivative of $g$ is $g'(x)=f'(x)-\frac{f(b)-f(a)}{b-a}$. The point $c$ also satisfies $f'(c)-\frac{f(b)-f(a)}{b-a}=0$.
| {
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Solution of a system of congruence equations Consider the following system out of which I want to find all possible values of $\lambda$.
\begin{eqnarray}
1287\lambda\equiv0 (\mathrm{mod}\ 6)\label{eq1}\\
165\lambda\equiv0 (\mathrm{mod}\ 4)\label{eq2}\\
9\lambda\equiv0 (\mathrm{mod}\ 2)\label{eq3}
\end{eqnarray}
Since I am not comfortable with the Chinese Remainder Theorem, I have tried to proceed with direct substitution.
\begin{eqnarray}
1287\lambda=6k\Rightarrow\lambda=\frac{2k}{429}\label{eq4}\\
165\frac{2k}{429}=4r\Rightarrow k=\frac{858r}{165}\label{eq5}\\
\lambda=\frac{4r}{165}\label{eq6}\\
9\lambda=2t\Rightarrow r=\frac{55t}{6}\label{eq7}\\
\lambda=\frac{2t}{9}\\
\end{eqnarray}
i.e. we seek $t$ such that $2t\equiv0 (\mathrm{mod}\ 9)\Rightarrow t\equiv0 (\mathrm{mod}\ 9)$ i.e. $t=9,18,\dots$. Thus, the possible values of $\lambda$ are $\lambda=2,4,\dots$
I have definitely done something wrong here since e.g. $\lambda=2$ cannot work, but I cannot spot the mistake. Alternative ways are also welcome.
| You are correct that $t$ must be a multiple of $9$ and $\lambda$ must be even, but this actually only uses the third equation. You also have the condition that $r=\frac{55t}6$ is an integer, and this means $t$ also has to be a multiple of $6$, so $t$ is a multiple of $\operatorname{lcm}(9,6)=18$. (You also need the first equation to be satisfied, so $k=\frac{858r}{165}=\frac{143t}3$ has to be an integer, but that just means $t$ has to be a multiple of $3$, which you already know.)
Thus $t=18,36,...$ giving $\lambda=4,8,...$ (together with $0$ or negative values if they're allowed).
| {
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Counterexamples Of The Infinite Dimensional Linear Space Let $V$ and $W$ be two finite dimensional linear spaces over the field $\mathbb{F}$ and $\mathscr{A} :V\rightarrow W$ be a linear map between $V$ and $W$. Then we have
*
*$\mathscr{A}$ is an injective linear map if and only if there exists a linear map $\mathscr{B}:W\rightarrow V$ such that $\mathscr{BA}=\mathit{Id}_{\mathbf{V}}$. ($\mathit{Id}_{\mathbf{V}}$ is the identity map on $V$)
*$\mathscr{A}$ is a surjective
linear map if and only if there exists a linear map $\mathscr{C}:W\rightarrow V$ such that $\mathscr{AC}=\mathit{Id}_{\mathbf{W}}$.
I think that 1 and 2 above don't hold if $V$ and $W$ are infinite dimensional linear spaces over $\mathbb{F}$.
I need some counterexamples to verify my idea. How can I find them?
| Exercise (1):
$\Leftarrow$: Assume $A$ was not injective, i.e. we had $v, v'$ with $Av = Av'$. Then $BAv = v = v' = BAv'$, which is a contradiction.
$\Rightarrow$: Define the linear map $B: Im(A) \to V$ by
$$B(w) \in A^{-1}(w)$$
for every $w \in Im(A)$. This definition is unique and well-defined since $A$ is injective.
$B$ is as the (effective) inverse of a linear map also linear.
Extend $B$ from the subspace $Im(A)$ to $W$ linearly. Then obviously $BAv = B(Av) = v$ by definition.
You can indeed extend $B$ linearly by noting that within vector spaces every subspace has a complement (e.g. see here): There is a subspace $M$, so that $W = Im(A) \oplus M$.
Now define $B': W \to V$ by $B'(w) = B'(x + y) := B(x) + 0$ with $x \in Im(A)$ and $y \in M$.
Alternative: Let $\{v_i | i \in I\}$ be a basis of $V$. Then $N := \{Av_i | i \in I\}$ is a set of linearly independent vectors as well:
$$0 = \sum_k \alpha_k Av_k \Rightarrow A^{-1}0 = A^{-1} \sum_k \alpha_k Av_k \Rightarrow 0 = \sum_k \alpha_k v_k \Rightarrow \forall k. \alpha_k = 0$$
Extend $N$ to a basis of $W$, namely $N \cup \{w_j | j \in J\}$ and define $B: W \to V$ using Ennar's suggestion
by its image on the basis elements: $B(Av_i) = v_i$ and $B(w_j) = 0$ (arbitrary).
Then $BAv = BA(\sum_k \alpha_k v_k) = \sum_k \alpha_k B(Av_k) = \sum_k \alpha_k v_k = v$ as required.
Exercise (2)
$\Leftarrow$: $AC = Id_W$ directly implies that $A$ is surjective.
$\Rightarrow$: Again, using Ennar's suggestion
, Define the linear map $C: W \to V$ by $$C(w_i) \in A^{-1}(w_i).$$
Then $ACw = AC(\sum \beta_j w_j) = \sum \beta_j ACw_j = \sum \beta_j w_j = w.$
| {
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Calculate limit of $\frac{1}{n}\cdot (1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n})$ How to prove that
$$\lim_{n\to\infty}(\frac{1}{n}\cdot(1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n})) = +\infty$$
using only basic limit operations and theorems?
| Creative telescoping does the job nicely. We have
$$(n+1)\sqrt{n+1}-n\sqrt{n} = \sqrt{n+1}+\frac{n}{\sqrt{n}+\sqrt{n+1}} \leq \frac{3}{2}\sqrt{n+1}$$
hence
$$\sum_{k=1}^{n}\sqrt{k}\geq \frac{2}{3}\sum_{k=1}^{n}\left[n\sqrt{n}-(n-1)\sqrt{n-1}\right] = \frac{2}{3}n\sqrt{n}.$$
| {
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Prove that upper triangular matrices are closed under inversion Prove that the group of upper triangle matrices $G\subset GL_n(F)$ is also a subgroup of $GL_n(F)$
Contains identity: $I_n\in G$ since $I_{ij}=0$ when $i>j$
Closed under multi: $C=AB, c_{ij}=\sum_{k=1}^na_{ik}b_{kj},i=1,2..,n,j=1,2..,n$ for all $c_{ij},i>j$ either $i\ge k>j$ or $i>k\ge j$. One factor in each addend is $0$: either $a_{ik}=0$, or $b_{kj}=0$ thus $c_{ij}=0$ if $i>j\rightarrow C\in G$
Closed under inversion: We know that a unique inverse exist since G is atleast a subset of $GL_n(F)$. Any tips on how do I proceed form here on? I thought about using my proof above to prove that $A^{-1}\in G$ How do I know there is not a fringe case where $AA^{-1}=I, A^{-1}\not\in G$
| Conceptually, the algebra of upper triangular matrices $UT_n(F)$ is the set of transformations which fixes a complete flag of subspaces of $F^n$.
That is, for all upper triangular matrices, there is a chain of subspaces of $F^n$ written as $0\subseteq V_1\subseteq V_2\subseteq\ldots\subseteq V_n$, where $\dim_F(V_i)=i$, such that for every upper triangular matrix $A$, and each index $i$, $A(V_i)\subseteq V_i$. Conversely given any such flag, you can choose a basis such that the flag-preserving transformations are upper triangular. You should have no problem describing such a flag of subspaces of $F^n$ such that the transformations are the upper triangular matrices.
If $A$ is invertible, then it preserves dimension, so $A(V_i)=V_i$ for all $i$. The inverse map, if it exists, would map $V_i$ right back into $V_i$, so the inverse is also an upper triangular matrix.
| {
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Is $\mathbb{Z}_p [x]$ finite? Seems to me like it is. There are only finitely many distinct powers of $x$ modulo $p$, by Fermat's Little Theorem (they are $\{1, x, x^2, ..., x^{p-2}\}$), and the coefficient that I choose for each of these powers can only be taken from $\{0,1,2,..., p-1\}$. So essentially I'm choosing amongst $p$ things $p-1$ many times, resulting in at most $p^{p-1}$ distinct polynomials.
Yet an assignment claims that $\mathbb{Z}_p[x]$ is infinite.
| $\Bbb Z_p[x]$ is indeed infinite, since it contains polynomials, with coefficients in $\Bbb Z_p$, of arbitrary high degree, e.g. $x^n \in \Bbb Z_p[x]$, where $n \in \Bbb N$; also, $x^n \ne x^m$ if $m \ne n$.
Fermat's Little Theorem, that $a^p = a$ for $a \in \Bbb Z_p$, does not apply to the indeterminate $x \in \Bbb Z_p[x]$. $x \in \Bbb Z_p[x]$ and $a \in \Bbb Z_p$ are definitely "birds of a different feather" in this regard.
So though $\Bbb Z_p$ is finite, $\Bbb Z_p[x]$ is not.
| {
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Partial sum of divergent series I am trying to find the nth partial sum of this series:
$S(n) = 2(n+1)^2$
I found the answer on WolframAlpha:
$\sum_{n=0}^m (1+2n)^2 =\frac{1}{3}(m+1)(2m+1)(2m+3)$
How can I calculate that sum, without any software?
| $\sum_\limits{i=0}^n 2(i + 1)^2 = 2\sum_\limits{i=1}^{n+1} i^2$
Which gets to the meat of the question, what is $\sum_\limits{i=1}^n i^2$?
There are a few ways to do this. I think that this one is intuitive.
In the first triangle, the sum of $i^{th}$ row equals $i^2$
The next two triangles are identical to the first but rotated 120 degrees in each direction.
Adding corresponding entries we get a triangle with $2n+1$ in every entry. What is the $n^{th}$ triangular number?
$3\sum_\limits{i=1}^n i^2 = (2n+1)\frac {n(n+1)}{2}\\
\sum_\limits{i=1}^n i^2 = \frac {n(n+1)(2n+1)}{6}$
To find: $\sum_\limits{i=1}^{n+1} i^2 $, sub $n+1$ in for $n$ in the formula above.
$\sum_\limits{i=0}^n 2(i + 1)^2 = \frac {(n+1)(n+2)(2n+3)}{3}$
Another approach is to assume that $S_n$ can be expressed as a degree $3$ polynomial. This should seem plausible
$S(n) = a_0 + a_1 n + a_2 n^2 + a_3n^3\\
S(n+1) = S(n) + 2(n+2)^2\\
S(n+1) - S_n = 2(n+2)^2\\
S(n+1) = a_0 + a_1 (n+1) + a_2 (n+1)^2 + a_3(n+1)^3\\
a_0 + a_1 n+a_1 + a_2 n^2 + 2a_2n+a_21 + a_3n^3 + 3a_3n^2 + 3a_3n + 1\\
S(n+1) - S(n) = (a_1 + a_2 + a_3) + (2a_2 + 3a_3) n + 3a_3 n^2 = 2n^2 + 4n + 2$
giving a system of equations:
$a_1 + a_2 + a_3 = 2\\
2a_2 + 3a_3 = 4\\
3a_3 = 1\\
a_0 = S(0)$
| {
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Sum involving Hermite polynomials I am wondering if there is a simpler form of the following summation involving the (physicists') Hermite polynomials:
$$\sum_{k=0}^{n}\frac{H_k(x)}{(2i)^k},$$
where $i=\sqrt{-1}$ is the imaginary unit. I would love to find a "closed form" solution for this summation, perhaps in terms of a product of one or more Hermite polynomials (or another polynomial that is easily computable in MATLAB). I've checked this website as well as my go-to sources for these kinds of problems (Gradshtein and Ryzhik, and DLMF) without any luck.
| An operator form, or symbolic for computation, can be seen as
$$\sum_{k=0}^{n} \frac{H_{k}(x)}{(2 \, i)^{n}} = (-i)^{n} \, e^{- D^{2}/4} \, \left( \frac{x^{n+1} - i^{n+1}}{x - i} \right).$$
This is obtained by using
$$H_{n}(x) = e^{- D^{2}/4} \, (2 x)^{n}$$ and then summing the resulting series.
| {
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"timestamp": "2023-03-29T00:00:00",
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Bit strings of length $n$ in which any two consecutive $1$'s are separated by an even number of $0$’s I want to find the number of bit strings of length $n$ in which any two consecutive $1$'s are separated by an even number of $0$’s
My attempt:
Let $a_n$ be the required number
Take two cases:
*
*The string ends with $1$. In this case, the string must end with $001$. Number of such strings is $a_{n-3}$
*The string ends with $0$. Number of such strings is $a_{n-1}$
Mistake which I found but can't fix in my approach:
For $n=5$, $10$ is a valid length 2 string, appending $001$ doesn't give a valid length $5$ string.
| Form a graph with $6$ nodes: $S, A, B, C, D, E$.
*
*$S$ the starting state (we're here when the string is empty)
*$A$ a state for beginning zeros
*$B$ a state where you have just gotten a one (and the string isn't already rejected)
*$C$ counting zeros after a one, have odd amount
*$D$ counting zeros after a one, have even amount
*$E$ the rejecting state
And the transition matrix, when indexed by this order of states (check each transition and justify them yourself)
$$ P =
\begin{bmatrix}
0 & 1 & 1 & 0 & 0 & 0 \\
0 & 1 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 1 \\
0 & 0 & 0 & 0 & 1 & 1 \\
0 & 0 & 1 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 2 \\
\end{bmatrix}
$$
Now $a_n = 2^n-P^n_{S, E}$. That is, we are looking for the top-right element of $P^n$, it gives the number of walks of length $n$ on the graph that end up in the rejecting state and start from the state $S$.
We can diagonilize $P$ to solve for a formula. Eigenvalues are
$\lambda_1 = 2$
$\lambda_2 \approx 1.32472$
$\lambda_3 = 1$
$\lambda_4 \approx 0.662359 + 0.56228 i$
$\lambda_5 = \bar{\lambda_4}$
$\lambda_6 = 0$
Here $\lambda_2, \lambda_4$ and $\lambda_5$ are the roots of $x^3-x-1$ which we get by dividing the exact roots away from the characteristic polynomial: $\frac{x^6 - 3x^5 + x^4 + 2x^3 + x^2 - 2x}{x(x-1)(x-2)}$.
But then you would need to solve the eigenvectors and the inverse of that matrix and for me it looks pretty messy. Anyway, I think the sequence $a_n$ is this OEIS sequence and there a recurrence is given and from that it is easier to solve for the formula (for example via generating functions).
| {
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