What is the conformal map of a rectangle into a band? What is the conformal map from the rectangle to the "band" (space between two parallel lines) in the complex plane? By the Riemann mapping theorem, such a map exists.

We can define a rectangle as $\{a+bi: a \in (-1,1),b\in(-w,w)\}$ for some $w \in \mathbb R_+$, and the "band" as $\{a+bi: a \in (-1,1), b \in (-\infty,\infty)\}$.

Note in particular that the function should map the rectangles vertical axis to the bands vertical axis (i.e., it should map $\{0\} \times (-w,w)$ to $\{0\} \times (-\infty,\infty)$).

(The reason I want such a map is that I want to transform the band model of Circle Limit III into a desktop wallpaper.)

EDIT: I found this article claiming (on page 4) that you can map from a rectangle to a disk of the form

$$w = \frac{1 + i\sqrt k sn(\frac 1 \alpha (z + ib))}{1 + \sqrt k sn(\frac 1 \alpha (z + ib))}$$

where $z$ is a point the rectangle $[-a,a] \times [0,2b]$ and $w$ is a point in the unit circle ($\alpha = \frac K a$, $K$ is a quarter of the real period of $sn$, and $k$ is parameter drawn from $[0,1]$). It is trivial then to map the disk into the band.

The problem is that I looked up the $sn$ function, and it has two arguments, but in the equation above, only one argument is being given.

EDIT: $sn$ does not preserve symmetry.

Here's just a sketch of an argument. A suitable elliptic function will map the interior of the rectangle conformally and bijectively to the upper half plane $H$. The principal logarithm will map $H$ conformally and bijectively to a strip $\{x+yi:0<y<\pi\}$.

OK, I am skimping on details of the elliptic function. Roughly speaking
you take a fundamental region built of four copies of the given rectangle.
Apply the Weierstrass $\wp$-function. Avoiding half-period points, and using symmetry should give you what you want. I don't have the time or
inclination to check this fully right now $\ddot\frown$.

I used the same formula
w=(1+ik√sn(1α(z+ib)))/(1+k√sn(1α(z+ib)))

sn function has two arguments, the other argument is k, the elliptic modulus. It was written vaguely in that paper that you have to choose the value of k, such that K'/K has to be 2*b/a.
K' is the elliptic integral of the first kind with parameter 1-m,
and K is the elliptic integral of the first kind with parameter m.
m is the elliptic parameter and related to k by the equation:
m = k^2
I have used Matlab function "ellipticK" to get the value of K and K' and then found a suitable value of k.

Hope, that helps.
Thank you.


Stuck on Differential Equation (Tried Substitution) $$\frac {dy}{dx}=\frac{x^2-xy-y^2}{x-y}$$

My work:

$$\frac{dy}{dx}=\frac{x^2-xy-y^2}{x-y}=\frac{(x+y)(x-y)-xy}{x-y}$$

$$\frac{dy}{dx}=x+y-\frac{xy}{x-y}=x+y\left[1-\frac{x}{x-y}\right]$$

$$\frac{dy}{dx}-y\left(1-\frac{x}{x-y}\right)=x $$

I tried to make it first order linear  differential equation and then I also tried to make it exact, but I still had trouble.

I tried substitution :  $Z=1/(x-y)$

Hint:

$\dfrac{dy}{dx}=\dfrac{x^2-xy-y^2}{x-y}$

$(y-x)\dfrac{dy}{dx}=y^2+xy-x^2$

This belongs to an Abel equation of the second kind.

Let $u=y-x$ ,

Then $y=u+x$

$\dfrac{dy}{dx}=\dfrac{du}{dx}+1$

$\therefore u\left(\dfrac{du}{dx}+1\right)=(u+x)^2+x(u+x)-x^2$

$u\dfrac{du}{dx}+u=u^2+2xu+x^2+xu+x^2-x^2$

$u\dfrac{du}{dx}=u^2+(3x-1)u+x^2$

Let $u=e^xv$ ,

Then $\dfrac{du}{dx}=e^x\dfrac{dv}{dx}+e^xv$

$\therefore e^xv\left(e^x\dfrac{dv}{dx}+e^xv\right)=e^{2x}v^2+(3x-1)e^x+x^2$

$e^{2x}v\dfrac{dv}{dx}+e^{2x}v^2=e^{2x}v^2+(3x-1)e^x+x^2$

$e^{2x}v\dfrac{dv}{dx}=(3x-1)e^x+x^2$

$v\dfrac{dv}{dx}=(3x-1)e^{-x}+x^2e^{-2x}$

It doesn't seem to have closed-form solutions.


Find number of elements of order p in group G where p is prime and G has m subgroups of order p Find the number of elements of order p in group G, where p is prime and G has m subgroups of order p.

Suppose G is a group, and p is prime. Then the number of elements of G of order p is multiple of p-1. is an equivalent problem to this, but it was closed (also I have a more specific question).

The problem

Suppose that a finite group $G$ has exactly $m$ subgroups of order $p$, where $p$ is a prime. Show that the number of elements of order $p$ in $G$ is $m(p-1)$.

My solution

Let $H_i, 1 \leq i \leq m$, be every subgroup with order $p$ in $G$.

Theorem:

  If $G$ is a group whose order is a prime $p$, then it is isomorphic to $C_p$
  (the cyclic group of order $p$).

Thus every $H_i$ is isomorphic to $C_p$.

Every cyclic group contains its generating element (of order $p$) exactly once, thus there must be at least $m$ elements in $G$ which has order $p$.

And that's as close as I can get to a solution.

Two subgroups of order $p$ are either equal or have trivial intersection.

Every nontrivial element in a subgroup of order $p$ has order $p$.

Each subgroup of order $p$ has $p-1$ elements of order $p$.

Every element of order $p$ belongs to a subgroup of order $p$.

Therefore, the number of elements of order $p$ is $m(p-1)$.

My solution is almost correct, however there is a fault in my reasoning:

A cyclic group of order $n$ doesn't have just 1 generator, it has $\phi(n)$ generators where $\phi(n)$ is Euler's phi function. Or in other words: The amount of numbers which are coprime to the order of the cyclic group is equal to the amount of generators for that cyclic group.

Since $p$ is prime we have $p-1$ such numbers, thus we have $m(p-1)$  such elements in $G$.


Why does this sequence converge? I have to deal with the following sequence :

$\lim \limits_{x \to \infty}\sqrt{x+\sqrt{x}} - \sqrt{x}$

If I factorize it to $\sqrt{x}(\sqrt{\sqrt{x}+1}-1)$, I would say it diverges since both factors diverge: $\lim \limits_{x \to \infty}\sqrt{x}= \infty  $ and $\lim \limits_{x \to \infty} \sqrt{\sqrt{x}+1} = \infty$

But if I type it in WolframAlpha, I get $\frac12$ as limit. Can you help me out?

Multiply with the conjugate $\sqrt{x+\sqrt{x}}+\sqrt{x}$ to obtain that $$\begin{align*}\sqrt{x+\sqrt{x}}-\sqrt{x}&=\frac{\left(\sqrt{x+\sqrt{x}}-\sqrt{x}\right)\left(\sqrt{x+\sqrt{x}}+\sqrt{x}\right)}{\sqrt{x+\sqrt{x}}+\sqrt{x}}\overset{(1)}=\frac{x+\sqrt{x}-x}{\sqrt{x+\sqrt{x}}+\sqrt{x}}=\\[0.2cm]&=\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{\frac{1}{\sqrt{x}}+1}+1\right)}=\frac{\sqrt{\not x}}{\sqrt{\not x}\left(\sqrt{\frac{1}{\sqrt{x}}+1}+1\right)}=\\[0.2cm]&=\frac{1}{\sqrt{\frac{1}{\sqrt{x}}+1}+1} \to \frac{1}{\sqrt{0+1}+1}=\frac12\end{align*}$$ as $x \to \infty$. In (1) we used the identity $$(a-b)(a+b)=a^2-b^2$$ with $a=\sqrt{x+\sqrt{x}}$ and $b=\sqrt{x}$.

A slightly more readable solution results if one sets $y=\sqrt{x}$. Then by general reasoning the limit becomes $\lim_{y\to\infty}\sqrt{y^2+y}-y$.  Then simplify further by setting $z=\frac{1}{y}$, obtaining $\lim_{z\to 0}\sqrt{1/z^2+1/z}-1/z$.
This can be evaluated as follows:
$$
\lim_{z\to 0}\sqrt{1/z^2+1/z}-1/z=\lim_{z\to0}\frac{1}{z}\left(\sqrt{1+z}-1\right)=\lim_{z\to0}\frac{\sqrt{1+z}-1}{z},
$$
which is recognizably the derivative of the square root function at $1$, and therefore the answer is 1/2.


Is there cases where interpolation is impossible? I just learned interpolation and have been looking through it.

It seems like for every data set, interpolation is possible -- is this true? Is there any cases it is not possible to use interpolation?

Any hints or help on this question will greatly help!!!

Consider the data set $x_0 = 0$, $y_0 = 0$, $x_1 = 0$, $y_1 = 1$. No function can serve as the interpolation here.

Do you know the tangent function?

You have $\tan 0 = 0$ and $\tan \pi = 0$. Dare to interpolate for $\frac\pi 2$...?


Why is the number of possible subsequences $2^n$? If anyone here is familiar with the Lowest Common Subsequence problem, they probably know that the number of posibble subsequences in a sequence is $2^n$; $n$ being the length of the sequence.

Another way of phrasing the statement is: given the sample space $\{a,b,c,d\}$, there are $2^4 = 16$ possible events (each event being defined as a subset of the whole sample space).

I know it sounds a fairly basic question, but can anybody please explain why is this so?

Either include A or exclude A.  2 possibilities.

Either include B or exclude B.  2 possibilities.

Either include C or exclude C.  2 possibilities.

Either include D or exclude D.  2 possibilities.

So the answer is $2\times2\times2\times2$.

Its not necessay that evey number has total subsequence is equal to 2^n.
For e.g: number 12345 has following subsequence equal to 25 only. 2^n are number of subsets...
--> 1 2 3 4 5
--> 12 13 14 15 23 24 25 34 35 45
--> 123 124 125 234 235 345
--> 1234 1235 1245 1345 2345
--> 12345


How should I solve this integral with changing parameters? I can't solve this. How should I proceed?

  $$\iint_De^{\large\frac{y-x}{y+x}}\mathrm dx\mathrm dy$$

$D$ is the triangle with these coordinates $(0,0), (0,2), (2,0)$ and I've changed the parameters this way $u=y-x$ and $v= y+x$ and the Jacobian is $-\frac{1}{2}$ but I have problem finding the range of $u$ and $v$ to calculate the integral.

Start by making a drawing of your domain. You can see that the $v$ is along the diagonal in the first quadrant, and $u$ is along the diagonal in the second quadrant. You can also see that the line between $(0,2)$ and $(2,0)$ is parallel to $u$, and intersects $v$ axis at $v=1$. So $v$ varies between $0$ and $1$ and $u$ varies between $-v$ and $v$.

Computer algebra gives (for the general case):

$$e \left(x^2 \text{Ei}\left(-\frac{2 x}{x+y}\right)-\frac{y^2 \text{Ei}\left(\frac{2
   y}{x+y}\right)}{e^2}\right)+\frac{1}{2} e^{1-\frac{2 x}{x+y}} (x+y)^2$$

over your specified region:

$$e-\frac{1}{e}$$


How to prove that $\lim_{k\to+\infty}\frac{\sin(kx)}{\pi x}=\delta(x)$? It is well-known that:
$$\lim_{k\to+\infty}\frac{\sin(kx)}{\pi x}=\delta(x).$$
This can also be written as
$$ 2\pi\delta(x)=\int^{+\infty}_{-\infty}e^{ikx}\,\mathrm dk.$$
However, I don't know how to prove this without using Fourier Transform. I have already searched google and looked for some books, but I just get nothing.

In short, I want to know the proof of this equation:
$$\lim_{k\to+\infty}\int^{+\infty}_{-\infty}\frac{\sin(kx)}{\pi x}f(x)dx=f(0).$$

I will prove that the limit is true in distribution sense: for all $\varphi \in C_{c}^{\infty}(\Bbb{R})$ we have

$$ \lim_{k\to\infty} \int_{\Bbb{R}} \frac{\sin (kx)}{\pi x} \varphi(x) \, \mathrm{d}x = \varphi(0). $$

The standard approximation-to-the-identity argument splits the integral into near-zero part and away-from-zero part, and estimate each part separately. In this case, however, this approach is a bit painful due to the fact that $\sin(kx)/\pi x$ is not absolutely integrable. So we adopt an indirect but easier approach.

Proof. Define $F(x) = \frac{1}{\pi}\int_{-\infty}^{x} \frac{\sin t}{t} \, \mathrm{d}t$. It is easy to check that $F$ is bounded, $F(-\infty) = 0$ and $F(+\infty) = 1$. Then performing integration by parts, we can write

$$ \int_{\Bbb{R}} \frac{\sin (kx)}{\pi x} \varphi(x) \, \mathrm{d}x = - \int_{\Bbb{R}} F(k x) \varphi'(x) \, \mathrm{d} x. $$

Now notice that the integrand of the RHS is uniformly bounded by $\| F \|_{\mathrm{sup}} |\varphi'| $, which is integrable. Thus taking $k \to \infty$, the dominated convergence theorem shows that

$$ \lim_{k\to\infty} \int_{\Bbb{R}} \frac{\sin (kx)}{\pi x} \varphi(x) \, \mathrm{d}x
= - \int_{\Bbb{R}} \lim_{k\to\infty} F(k x) \varphi'(x) \, \mathrm{d} x
= - \int_{\Bbb{R}} H(x) \varphi'(x) \, \mathrm{d} x, $$

where $H(x)$ is the Heaviside step function. Now manually computing the last integral gives

$$ - \int_{\Bbb{R}} H(x) \varphi'(x) \, \mathrm{d} x
= - \int_{0}^{\infty} \varphi'(x) \, \mathrm{d} x
= \varphi(0) $$

and therefore the claim follows.

Hint:You just need to prove that for any $f(x)$ which is continuous at $x=a$, the following result holds:

$\int_{-\infty}^{\infty} f(x)$.$\delta(x-a)dx=f(a)$.

Added-Substituting $\delta(x-a)$ using its definition, one gets

$\int_{a}^{a+\epsilon}$$f(x).\frac{1}{\epsilon}dx$$=\frac{1}{\epsilon}\int_{a}^{a+\epsilon}$$f(x)dx=\frac{1}{\epsilon}.(a+\epsilon-a).f(c)$ (where $a\le c\le a+\epsilon$)$=f(a)$ for $\epsilon→0$


Let $f:X\times [0,1] \to Y$ and let $X$ be compact. There exist points $0= t_0 < t_1 < · · · < t_k = 1$ such that $F(a, t_{i-1}) $,$F(a, t_{i})\in U$ Let $f:X\times [0,1] \rightarrow Y$ be a continuous function, let $X$ and $Y$ be compact Hausdporff, and let  $\mathcal{U}$ be an open cover of $Y$.  If $a\in X$, then there exist points  $0= t_0 < t_1 < · · · < t_k = 1$ and an open subset $U\in \mathcal{U} $ of $Y$  such that  $F(a, t_{i-1}) $,$F(a, t_{i})\in U$ for each $i$.

I need this result to prove a theorem about simplicial approximations of homotopic maps. Here it is:

I understand your problem to be:

compact X, f in C(Xx[0,1],Y), C open cover Y, a in X
implies some t0,.. t_k in [0,1], U in C with t0 = 0, t_k = 1
and for j = 1,.. k, f(a,t_(j-1)), f(a,t_j) in U.

Note that implies for j = 0,1,.. k, f(a,t_j) in U.
A counter example is X = {a}, Y = [0,1], C = { [0,1), (0,1] }

Could there be a quantification problem?
Is this a correct statement of your problem?

compact X, f in C(Xx[0,1],Y), C open cover Y, a in X
implies some t0,.. t_k in [0,1], with t0 = 0, t_k = 1 and
for j = 1,.. k, some U_j in C with f(a,t_(j-1)), f(a,t_j) in U.


Evaluate the integral along a contour containing 2 interior points by using Cauchy's Integral Formula Evaluate the integral $$\int_{C}\frac{z^2}{z^2+9}dz$$
where C is the circle $|z|=4$

I know that if f is analytic in simply connected domain $D$, $C$ a simple closed positively oriented contour that lies in D and $z_o$ lies interior to $C$, then
$$\int_{C}\frac{f(z)}{z-z_o}dz=2\pi i f(z_o)$$

But for this problem, the circle contains both interior points which is $3i$ and $-3i$. And I found that reducing the fraction into partial fraction seems to be useless in solving the problem. So what is the $f(z)$ here?

Hint:

Note that $z^2+9 = (z+3i)(z-3i)$.  So with partial fraction decomposition, we get:

$$\frac{z^2}{(z^2+9)} = \frac{iz^2}{6(z+3i)} - \frac{iz^2}{6(z-3i)}$$

Therefore:

$$\int_C \frac{z^2}{z^2 + 9} \ dz = \int_C \frac{iz^2}{6(z+3i)} \ dz - \int_C \frac{iz^2}{6(z-3i)} \ dz$$

And you can apply Cauchy's integral formula to each of the two pieces on the right hand side.

Alternative:

You can split the contour into two pieces $C_1$ and $C_2$, where $C_1$ is the upper half of the circle $|z| = 4$ together with the segment $[-2, 2]$, and likewise $C_2$ is the lower half of the circle $|z|=4$ together with the same segment.  If both are oriented counterclockwise, then $C = C_1 + C_2$, and so the integral splits accordingly, and you'll notice that each of $C_1$ and $C_2$ contains only one point where analyticity fails.  I won't write out all the details, but you can refer to my previous post here for more information.


How is $i^3$ = $\sqrt{-i}$. I new to complex numbers.So please don’t mind.
We know $i^2$ = -1.
Then $i^3$ is written as $= -i$
Then does it’s steps of writing follow  this:
$-1^2$= $i^3$= 1
So in terms i = It says in my book
= -  $\sqrt{-i}$.
How did we get this ?
I have done calculation mistake while writing this question.I have now answered this question.

$i^3$ =   $\sqrt{-1}$. *  $\sqrt{-1}$.*  $\sqrt{-1}$.
Therefore it equals = -1 *  $\sqrt{-1}$.
Or we can write it also using the property he listed in his question.
$i^2$ * i = -i.
Remember:When a and b both are negative and are multiplied.
Therefore , it equals - $\sqrt{ab}$.
For the OP.

$i = \sqrt{-1~}$ so $i^2 = -1$ and $i^3 = i^2 \cdot i = -1 \cdot \sqrt{-1~} = -\sqrt{-1~}$
Hope This helps!


sum sequence nonnegative How to show the following:

If we have a convergent sum of nonnegative numbers $a_i$, then show that there exists a sequence of nonnegative numbers strictly monotone increasing positive $b_i$ where $b_i$ converging to $\infty$ such that

$$\sum_{i=0}^{\infty}{a_i  b_i} < \infty\quad?$$

Thank you

Hint:
Since
$\sum_{k=1}^{\infty} a_k$
converges,
$A_n = \sum_{k=n}^{\infty} a_k \to 0$

Choose another sequence
$B_n$ which converges to zero
more slowly than $A_n$
and define $b_n$
in terms of $B_n$.

If the sum converges, then
$$
\lim_{i \to \infty} \frac{a_{i + 1}}{a_i} = c < 1
$$
Take $b_i = c^{-i/2}$, so $b_i \to \infty$, while
$$
\lim_{i \to \infty} \frac{a_{i + 1} b_{i + 1}}{a_i b_i} = c^{1/2} < 1
$$


Cross product in $\mathbb R^n$ (from Spivak's book) Spivak defines cross product in this way:

  $\quad$ We conclude this section with a construction which we will restrict to $\mathbf{R}^n$. If $v_1,\ldots,v_{n-1}\in\mathbf{R}^n$ and $\varphi$ is defined by $$\varphi(w)=\det\pmatrix{v_1 \\ \vdots \\ v_{n-1} \\ w},$$ then $\varphi\in\Lambda^1(\mathbf{R}^n)$; therefore there is a unique $z\in\mathbf{R}^n$ such that $$\langle w,z\rangle=\varphi(w)=\det\pmatrix{v_1 \\ \vdots \\ v_{n-1} \\ w}$$ This $z$ is denoted $v_1\times\cdots\times v_{n-1}$ and called the cross product of $v_1,\ldots,v_{n-1}$.

Why such a $z$ exists and why is it unique? When solving problems involving this notion, how do I find this $z$ explicitly (if it's possible)?

Also, what's the meaning of this cross product? Many sources say that the usual cross product in $\mathbb R^3$ can't be generalized to higher dimensions.

Just for kicks, here's a  proof of the Riesz representation theorem that Ivo talks about in the (much easier) finite dimension case.

Suppose $V$ is a real, finite dimensional inner product space.  Let $f:V\rightarrow \mathbb{R}$ be a linear functional.  We claim that there is a unique $z\in V$ for which $f(w) = \langle w,z\rangle$ for all $w\in V$.

First, if $f$ is identically $0$, then $z=0$ works, and only $z=0$ works because $0 =f(z) = \langle z,z\rangle = |z|^2$.  Thus, we may assume that $f$ is not identically $0$.  Let $v\in V$ with $f(v) \neq 0$.

Linearity implies that $f$ is surjective.  In more detail, if $r\in \mathbb{R}$ is any real number, then $f\left(\frac{r}{f(v)} v\right) = \frac{r}{f(v)} f(v) = r$.

By the rank-nullity theorem (which uses finite dimensionality of $V$), it follows that $\ker f\subseteq V$ is a codimension $1$ subspace.  Then $(\ker f)^\bot$ is $1$-dimensional.  Choose a non-zero $y\in (\ker f)^\bot$.  Note that $\langle y,v\rangle \neq 0$, for if it is $0$, then $v\in ((\ker f)^\bot)^\bot = \ker f$, contradicting the fact that $f(v) \neq 0$.

Finally, set $z = \frac{f(v)}{\langle y,v\rangle} y$.  I claim that this $z$ works.

For any $w\in V$, we break it up into $w_1 + \lambda v$ where $w_1\in \ker f$.

Then $f(w) = f(\lambda v) = \lambda f(v)$.  On the other hand, $$\langle w,z\rangle = \langle w_1 + \lambda v, z\rangle = \lambda \left\langle v,  \frac{f(v)}{\langle v,w\rangle}y\right\rangle = \lambda f(v).$$  So they match.

What about uniqueness?  If $z_1$ and $z_2$ both work, then for any $w\in V$, we have $\langle w,z_1\rangle = f(w) = \langle w,z_2\rangle$ which implies that $\langle w, z_1 - z_2\rangle = 0$ for all $w\in V$.  Choosing $w= z_1 - z_2$, we see that $|z_1 - z_2|^2 = 0$, which means $z_1 = z_2$.

For $n=3$, Spivak's definition is basically the way that 3Blue1Brown describes a cross product in https://youtu.be/BaM7OCEm3G0, and I like it a lot more than the usual parallelogram-and-right-hand-rule approach.

As for generalizing, there is no way to combine two vectors in $\mathbb{R}^n$ into a new vector that behaves the way $\times$ does in $\mathbb{R}^3$ but this linear functional approach is a perfectly good way to combine $n-1$ vectors in $\mathbb{R}^n$ in a way that behaves just like $\times$ does in $\mathbb{R}^3$.


Inverting a non-invertible matrix (Get all possible solutions) Currently I'm struggling with a problem.. I got a matrix which is in $GF2$. It has the size of $128 \times 128$. After constructing the matrix I need to invert it. The Problem is that the matrix can not be inverted right away, because it's rank is $120$ and therefore not invertible. Now I have been said that it is still somehow possible to invert it if I leave room for more solutions.

Let's say I got something like this: $Ax = y$. Now I calculated $A$ and want to invert it so that $A^{-1}y = x$. And i don't need a single solution for $x$, all possible solutions would be fine (Are there $2^8$ possible solutions?)

Does anyone know how to go from here? I actually know that this is possible but I don't know how to do it. Everything that points me in the right direction would be helpful.

Consider the example
$$
\begin{align}
%
  \mathbf{A} x &= b \\
%
\left[
\begin{array}{ccc}
 1 & 0 & 0 \\
 0 & 1 & 0 \\
 0 & 0 & 0 \\
\end{array}
\right]
%
\left[
\begin{array}{ccc}
 x_{1} \\
 x_{2} \\
 x_{3}
\end{array}
\right]
%
&=
%
\left[
\begin{array}{ccc}
 b_{1} \\
 b_{2} \\
 b_{3}
\end{array}
\right].
%
\end{align}
$$
There is no exact solution such that $\mathbf{A}x - b =0$ when $b_{3}\ne 0$. We turn to the method of least squares which will have the solution
$$
  x_{LS} =
\left[
\begin{array}{ccc}
 b_{1} \\
 b_{2} \\
 0
\end{array}
\right]
%
  +
%
\alpha
\left[
\begin{array}{ccc}
 0 \\
 0 \\
 1
\end{array}
\right], \qquad
\alpha \in \mathbb{C}.
$$

The general least squares solution is
$$
x_{LS} = \mathbf{A}^{\dagger} b + \left( \mathbf{I}_{n} - \mathbf{A}^{\dagger}\mathbf{A} \right) y, \quad y\in\mathbb{C}^{n}.
$$

If $Ax=y$, then $A(x+v)=y$ for all $v \in Null(A)$.


How many arrangements are there of the letters in word with restriction How many arrangements are there of the letters in REPETITIVENESS such that
the Ss are consecutive AND the first I comes (somewhere) before the first E ?

What i though is by viewing ss as one letter , it would solve the first restriction .

And i have no idea how to count the second restriction .

there are 13 spaces considering that ss is one space

there are 2 i and 4 e

still cant figure this out :(

First treat I and E as the same letter. Then, once you have decided where in the final arrangement they go, how many ways are there to distribute Is and Es to those positions, keeping in mind that the first of them must be an I?

Count letters: 14 total- 4 E's, 2 I's, 2 T's, 2 S's, 1 R, 1 P, 1 V, 1 N

You are on the right track with making the S's one letter, and making therefore 13 available spaces instead of 14.

1) Make SS one letter- 1 way

2) Pick 6 spots for I's and E's-    C(13,6)

3) Put I in the first spot of those you just picked- 1 way

4) Arrange remaining I and E's into the 5 other spots-  5!/4!

5) Arrange other letters among the 7 remaining spaces, remembering to include SS as ONE letter-   7!/2!

Now just multiply those numbers together:

1* C(13,6)* 1* 5!/4!* 7!/2!


Covering $\mathbb{R}^2$ with uncountably many disjoint non-degenerate line segments Is it possible to cover $\mathbb{R}^2$ with uncountably many disjoint non-degenerate line segments?

If a formal definition is necessary, let's define a line segment as a set $\{(x, mx+c): x \in [a, b]\}$ for some fixed constants $m, c, a, b \in \mathbb{R}$ with or a set $\{(u, x): x \in [a, b]\}$ for some fixed constants $u, a, b \in \mathbb{R}$. We say a line segment so defined is non-degenerate if $a \neq b$, i.e. the line segment is not a point.

This question was vaguely motivated by the observation that it's possible to cover $\mathbb{R}$ with uncountably many disjoint non-degenerate points. YuvalFilmus points out that the answer is negative in the one-dimensional case.

The answer is YES, you can cover the plane by non-degenerate line segments.

It is clear one can cover

any $1 \times 1$ closed squares (i.e those isometric to $[0,1] \times [0,1]$ ) by line segments of length $1$.
any $1 \times n$ quarter-open rectangles ( i.e. those isometric to $[0,1) \times [0,n]$ ) by line segments of length $n$.

You then proceed to cover $\mathbb{R}^2$ by

first placing a $1 \times 1$ closed square in the center.
surround the $1 \times 1$ closed square by four $1 \times 2$ quarter-open rectangles.
The "open" sides of the rectangles will be facing "inwards" and the
five quadrilaterals together form a $3\times 3$ closed square.
surround the $3 \times 3$ closed square by four $1 \times 4$ quarter-open rectangles.
The "open" sides again facing "inwards" and the nine quadrilaterals together form a
$5 \times 5$ closed square.
Just repeat this procedure. If you have a $(2k-1) \times (2k-1)$ closed square, you
surround it by four $1 \times 2k$ rectangles to form a $(2k+1) \times (2k+1)$ closed square.

Following is a picture illustrating the arrangement of the squares/rectangles
of line segments.

The line segments are represented by color bars.
The black lines indicate the the boundaries of the squares/rectangles.

Slice $\mathbb R^2$ into uncountably many copies of $\mathbb R^1$.


Evaluating a limit as $x \to -\infty$ of a power of a rational function Sorry for the weird title, I don't know how to put the equation on the title.

$$\lim_{x\to-\infty}\left(\frac{1-x^3}{x^2+7x}\right)^5$$

Ok I divided inside the parenthesis by $x^2$, but now I am stuck.

If we divide numerator and denominator in the parentheses by $x^3$ we get

$$\frac{1-x^3}{x^2+7x}=\frac{ 1/x^3 - 1}{1/x + 7/x^2}$$

Now note that as $x\to -\infty$ the numerator approaches $-1$ and the denominator approaches $0$. So overall the fraction diverges.
This already shows that the original limit diverges since the exponential function is continuous.

Since $X^5$ is continuous, it suffice to calculate $$\lim_{x\to-\infty}\left(\frac{1-x^3}{x^2+7x}\right)$$. Then by using L'Hospital's rule twice, you get the answer $\infty$.

Without using LHospital's rule, divide $x^2$ on the numerator and denominator.
$$\lim_{x\to-\infty}\left(\frac{\frac{1}{x^2}-x}{1+\frac{7}{x}}\right)$$

As $x\to -\infty$, $\frac{1}{x^2}\to 0, \frac{1}{x}\to 0$, hence numerator $\to \infty$, denominator $\to 1$, so the limit is $\infty$


Proving that $e^{\pi}-{\pi}^e\lt 1$ without using a calculator Prove that $e^{\pi}-{\pi}^e\lt 1$ without using a calculator.

I did in the following way. Are there other ways?

Proof : Let $f(x)=e\pi\frac{\ln x}{x}$. Then,
$$e^{\pi}-{\pi}^e=e^{f(e)}-{e}^{f(\pi)}\tag1$$
Now,
$$f'(x)=\frac{e\pi(1-\ln x)}{x^2},\quad f''(x)=\frac{e\pi (2\ln x-3)}{x^3},\quad f'''(x)=\frac{e\pi (11-6\ln x)}{x^4}.$$
Since $f'(x)\lt 0$ for $e\lt x\lt\pi$, one has $f(e)\gt f(\pi)$. By Taylor's theorem, there exists a point $c$ in $(e,\pi)$ such that
$$\begin{align}f(\pi)&=f(e)+(\pi-e)f'(e)+\frac{(\pi-e)^2}{2}f''(c)\\&\gt f(e)+(\pi-e)\cdot 0+\frac{(\pi-e)^2}{2}\cdot \frac{e\pi(2\ln e-3)}{e^3}\\&=f(e)-\frac{\pi(\pi-e)^2}{2e^2}\tag2\end{align}$$
because $f'''(x)\gt 0\ (e\lt x\lt \pi)$ implies $f''(c)\gt f''(e)$.

By the mean value theorem and $(2)$,
$$e^{f(e)}-e^{f(\pi)}\lt (f(e)-f(\pi))e^{f(e)}\lt \frac{\pi (\pi-e)^2}{2e^2}\cdot e^{\pi}=\frac{e\pi(\pi-e)^2}{2}\cdot e^{\pi-3}\tag3$$

Since $e^x\lt \frac{1}{1-x}\ (0\lt x\lt 1)$ and $0\lt \pi-3\lt 1$,
$$e^{\pi-3}\lt\frac{1}{4-\pi}\tag4$$

From $(1)(3)(4)$,
$$e^{\pi}-{\pi}^e\lt \frac{e\pi(\pi-e)^2}{2}\cdot\frac{1}{4-\pi}\lt\frac{3\times\frac{22}{7}\left(\frac{22}{7}-2.718\right)^2}{2}\cdot\frac{1}{4-\frac{22}{7}}\lt 0.993\lt 1$$

I looked at a couple of ways of examining this using series expansions.  A difficulty that I noted in a comment to mathlove's answer is that this requires rather careful selection of estimators, since the permissible "error budget" is very limited if we are to show that the difference is less than 1 , when its actual value is about 0.682.  Another issue is that we would like to use rational approximations in a way that makes these computations look credibly like ones that can be managed without a calculator.  I use some "number facts" here that are either generally familiar or can be found readily enough (while these can be obtained using a calculator, that can be avoided if one is acquainted with approximate values); it proves to be very important to use values that closely approximate the needed numbers.

We can attempt to estimate an upper limit the difference $ \ e^{\pi} \ - \ \pi^e \ $ by setting up an inquality in which $ \ e^{\pi} \ $ is overestimated and $ \ \pi^e \ $ is underestimated.  In this first one, we will start from

$$ \left( \frac{68}{25} \right)^{22/7} \ > \ e^{\pi} \ > \ \pi^{e} \ > \ \left( \frac{157}{50} \right)^{19/7} \ \ $$

$$ \Rightarrow \ \ \left( 3 \ - \ \frac{7}{25} \right)^{3 + \frac{1}{7}} \ > \ e^{\pi} \ > \ \pi^{e} \ > \ \left( 3 \ + \ \frac{7}{50}  \right)^{3 - \frac{2}{7}} \ \ $$

$$ \Rightarrow \ \ 3^3 \ \cdot \ 3^{1/7} \ \cdot \left( 1 \ - \ \frac{7}{75} \right)^3 \ \left( 1 \ - \ \frac{7}{75} \right)^{\frac{1}{7}} \ > \ e^{\pi} \ > \ \pi^{e} \quad > \ 3^3 \ \cdot \ 3^{-2/7} \ \cdot \left( 1 \ + \ \frac{7}{150}  \right)^3 \ \left( 1 \ + \ \frac{7}{150}  \right)^{-\frac{2}{7}} \ \ .  $$

We can construct Taylor polynomials for the powers of $ \ 3 \ $ based on the function $ \ 3^x \ $ [using $ \ \ln 3 \ \approx \ 1.1 \ $ ], and can apply the (generalized) binomial series for the other factors:

$$  3^3 \ \cdot \ 3^{1/7} \ \cdot \left( 1 \ - \ \frac{7}{75} \right)^3 \ \left( 1 \ - \ \frac{7}{75} \right)^{\frac{1}{7}} $$ $$ \approx \ 27 \   \left( 1 \ + \ \ln 3 \cdot \frac{1}{7} \ + \ \frac{(\ln 3)^2}{2} \cdot \frac{1}{7^2} \right) \   \left( 1 \ - \ 3 \cdot \frac{7}{75}\ + \ 3 \cdot \frac{7^2}{75^2} \right)  $$ $$ \cdot   \left( 1 \ - \ \frac{1}{7} \cdot \frac{7}{75} \ + \ \frac{\frac{1}{7} \cdot  \frac{-6}{7}}{2} \cdot \frac{7^2}{75^2} \right)  \ $$

$$ \approx \ 27 \   \left( 1 \ + \ \frac{11}{10} \cdot \frac{1}{7} \ + \  \frac{6}{10} \cdot \frac{1}{7^2} \right) \   \left( 1 \ - \  \frac{7}{25}\ + \ 3 \cdot \frac{7^2}{75^2} \right)  \  \left( 1 \ - \  \frac{1}{75} \ - \  \frac{3}{75^2} \right)  \ \ , $$

and similarly,

$$  3^3 \ \cdot \ 3^{-2/7} \ \cdot \left( 1 \ + \ \frac{7}{150} \right)^3 \ \left( 1 \ + \ \frac{7}{150} \right)^{-\frac{2}{7}} $$ $$ \approx \ 27 \   \left( 1 \ - \ \ln 3 \cdot \frac{2}{7} \ + \ \frac{(\ln 3)^2}{2} \cdot \frac{4}{7^2} \ - \ \frac{(\ln 3)^3}{6} \cdot \frac{8}{7^3} \right) \   \left( 1 \ + \ 3 \cdot \frac{7}{150}\ + \ 3 \cdot \frac{7^2}{150^2} \right)  $$ $$ \cdot   \left( 1 \ - \ \frac{2}{7} \cdot \frac{7}{150} \ + \ \frac{\frac{-2}{7} \cdot  \frac{-9}{7}}{2} \cdot \frac{7^2}{150^2} \right)  \ $$

$$ \approx \ 27 \   \left( 1 \ - \ \frac{11}{10} \cdot \frac{2}{7} \ + \  \frac{6}{10} \cdot \frac{4}{7^2} \ - \ \frac{4 \cdot 4}{3 \cdot 3} \cdot \ \frac{1}{7^3}  \right) \   \left( 1 \ + \  \frac{7}{50}\ + \ 3 \cdot \frac{7^2}{150^2} \right) $$ $$ \cdot \ \left( 1 \ - \  \frac{1}{75} \ + \  \frac{9}{150^2} \right)  \ \ . $$

In order for the series for $  \ 3^{-2/7} \ $ to remain an underestimate, it is necessary to extend the Taylor polynomial to third-degree.  Simple linearizations give far too "coarse" approximations to resolve the inequality we desire clearly enough; we are also unable to obtain sufficient precision by only keeping the "linear" terms in the product.  There isn't much for it other than to arrange each factor to have a single denominator, which gives us

$$ 27 \ \left[ \ \left(  \frac{573}{490}  \right) \   \left( \frac{4197}{5625} \right)  \  \left(  \frac{5547}{5625} \right) \  -  \  \left(  \frac{11260}{15435}  \right) \ \left( \frac{25797}{22500} \right)  \  \left(  \frac{22209}{22500} \right) \right] \ > \ e^{\pi} \ - \ \pi^e \ \ . $$

From this point, it becomes a question of how much calculation one has the patience to do without the aid of a device; it would likely be easiest to divide out each ratio, multiply the three factors in each term, take the difference between terms, and multiply the result by $ \ 27 \ $ .  In the pre-calculator era, this would not have been considered unreasonable: after all, the Ludolphine number $ ( \pi ) $ was computed to 35 decimal places over 400   years ago using polygon-approximations over the course of some years. If we avail ourselves of a calculator at this point, we see that we have

$$ 27 \ \left( \ 1.16939 \cdot   0.74613  \cdot  0.98613 \  -  \  0.72951 \cdot 1.14653  \cdot  0.98707 \right) \ \approx \ 0.9403 \ > \ e^{\pi} \ - \ \pi^e \ \ . $$

This may be compared with the first inequality we wrote, for which our result is an intended overestimate,

$$ \left( \frac{68}{25} \right)^{22/7} \ - \ \left( \frac{157}{50} \right)^{19/7} \ \approx \ 0.8901  > \ e^{\pi} \ - \ \pi^{e} \  \ . $$

$$ \ \ $$

We might well wish for something more tractable, which can be obtained by making the estimate a bit differently.  If we're permitted the "number facts" $ \ e^3 \ \approx \ 20.1 \ \ , \ \ \pi^3 \ \approx \ 31 \ \ , \ \pi \ < \ \frac{22}{7} \ $ and $ \ e \ > \ \frac{19}{7}  \ $ , we can proceed from

$$ \frac{201}{10} \ \cdot \ e^{1/7} \ > \  e^{3 \ + \ \frac{1}{7}} \ > \ e^{\pi} \ > \ \pi^{e} \ > \ \pi^{3 \ - \ \frac{7}{25}} \ > \ 31 \ \cdot \ \pi^{-7/25} \ > \ 31 \ \cdot \ \left( \frac{\pi}{e} \right)^{-7/25} \ \cdot \  e^{-7/25} \ \ . $$

We would like the factor $ \ \left( \frac{\pi}{e} \right)^{-7/25} \ $ to be underestimated, so we will use the overestimate

$$ \ \left( \frac{\pi}{e} \right) \ < \ \frac{22/7}{19/7} \ = \ \frac{22}{19} \ \ . $$

As above, we will use Taylor polynomials for the powers of $ \ e \ $ and the generalized binomial series to produce

$$ \frac{201}{10} \ \cdot \ \left( 1 \ + \ \frac{1}{7} \ + \ \frac{1}{2 \cdot 7^2} \right) \ >  \ e^{\pi} \ > \ \pi^{e} \ > \  31 \ \cdot \ \left( 1 \ + \ \frac{3}{19} \right)^{-7/25} \ \cdot \  \left( 1 \ - \ \frac{7}{25} \ + \ \frac{7^2}{2 \cdot 25^2} \right)  $$

$$ \Rightarrow \ \ \frac{201}{10} \ \cdot \ \left( 1 \ + \ \frac{1}{7} \ + \ \frac{1}{98} \right) \ >  \ e^{\pi} \ > \ \pi^{e} \ > \  31 \ \cdot \ \left( 1 \ - \ \frac{7}{25} \cdot \frac{3}{19} \right) \ \cdot \  \left( 1 \ - \ \frac{7}{25} \ + \ \frac{49}{1250} \right)  $$

$$ \Rightarrow \ \ \frac{201}{10} \ \cdot \  \frac{113}{98}  \ >  \ e^{\pi} \ > \ \pi^{e} \ > \  31 \ \cdot \ \frac{454}{475}   \ \cdot \   \frac{949}{1250}  \ \ , $$

these ratios being obtainable by reasonably simple hand-calculation.  We can make the work we face in the right-hand term of this inequality a bit less onerous by using

$$ 31 \ \cdot \ \frac{454}{475}   \ \cdot \   \frac{949}{1250}  \ > \ 31 \ \cdot \ \frac{450}{475}   \ \cdot \   \frac{945}{1250}  \ = \ 31 \ \cdot \ \frac{18}{19}   \ \cdot \   \frac{189}{250} \ \ . $$

Re-arranging the inequality gives us

$$  \frac{201}{10} \ \cdot \  \frac{113}{98} \ - \ 31 \ \cdot \ \frac{18}{19}   \ \cdot \   \frac{189}{250} \ = \ \frac{201 \cdot  113 \cdot  19 \cdot 25 \ - \ 31 \cdot 18 \cdot 189 \cdot 98}{980 \cdot 19 \cdot 25}  \ >  \ e^{\pi} \ - \ \pi^{e}   \ \ . $$

While this looks a bit daunting, it is considerably easier to compute by hand than the earlier estimate:  a little work with binomials simplifies some of the multiplicaton and the rest is not too bad to work out "long-hand".  Hence,

$$ \frac{10,788,675 \ - \ 10,335,276}{465,500} \ = \ \frac{453,399}{465,500}  \ >  \ e^{\pi} \ - \ \pi^{e}   \ \ . $$

So we again establish the desired inequality, this time with more handiwork.  We can make use of the calculator (or divide long-hand if we really want to remain "pure") to find that this ratio is about 0.9740 .

I will admit to a bit of guidance from a calculator to make adequate selections for the rational approximations in the interests of saving some time.  It would take rather longer than  to refine the numbers used in this investigation if one had to work entirely without "machine assistance".

Take the roots against $\pi e$, to get $e^{1/e}$ v $\pi^{1/\pi}$.  These are instances if $x^{1/x}$ for $x=e$ and $x=\pi$.

For integers, we see that 9>8 and 2^4=4^2=16 imply that $\sqrt[3]{3} > \sqrt[4]{4} = \sqrt{2}$, from the sixth and eighth roots respectively.

Put $y=\sqrt[x]{x}$, whence $x = y^x$. We note that for values greater than 1, there are two solutions, the second is $z=y^z$.  We now suppose z>x, and consider the case w>z.  Where w>z, then $w<y^w$, which means that the value of y for w is less than that for z.

In the region where 1 < w < x  we note that $y^x > y^w$, still, and that $\sqrt[w]{w} < \sqrt[x]{x} < L$

If now we note that $\pi > 3 > e$, and if we find a corresponding value for 3, then the range from 3 to 3' will contain either pi or e.

We now seek a value r for which $r^3 = 3^r$.  Put r=2.5, gives r^3 = 125/8 = 15.625, and $9\sqrt{3}=1.732*9$, at 15.588.  This tells us that r<2.5, and we get $r=3'< 2.5 < e < 3 < \pi$.

Since this is a convergence of $x$ and $z$ converge on some limit $L$, we can use brackets of the form of ( and ) to show this convergance, viz

$ 3' (2.5  ( e )) 3 ) \pi$, and thus from $3' ( e ) 3 ) \pi$, $e^\pi > \pi ^e$.


if $X$ and $Y$ are Normal random variables with same parameters. is $(aX+bX)=(aX+bY)$? if $X$ and $Y$ are random variables and both are $\sim N(\mu,\sigma^2)$, ie share the same average and variance. Then we defined a new random variable as the linear combination of them as: $Z=(aX+bY)$ is that the same as saying $Z=(aX+bX)$? Is $Z=(aX+bX)$ the same as saying $Z=((a+b)X)$?

You have not said how $X$ and $Y$ are jointly distributed, but only how they are marginally distributed.

If they are both marginally distributed as $N(\mu,\sigma^2)$ then:

At one extreme you have $X=Y,$ in which case $X+Y= 2X \sim N(2\mu,4\sigma^2).$
At an opposite extremem you have $Y = 2\mu-X,$ in which case $X+Y\sim N(2\mu,0),$ i.e. the sum is constant.
Somewhere in between, $X$ and $Y$ are independent. In that case $X+Y\sim N(2\mu, 2\sigma^2).$
And there are cases where there is a nonzero correlation between $X$ and $Y$ but it's not $+1$ or $-1.$
There are also cases in which $X$ and $Y$ are not jointly normally distributed (but still have that same marginal distribution). One of those is where $Y = \pm X$ with plus and minus having equal probabilities and being chosen independently of $X$. In that case $X+Y$ is not normally distributed, since $\Pr(X+Y=0) = 1/2.$

I'm pretty sure the answer is no. An easy way to think about this is physically. AX+BX corresponds to the same event happening twice, whereas AX+BY corresponds to two separate, INDEPENDENT events occuring. With AX+BX, you pull the same point out of the distribution (causing an amplification), whereas with AX+BY, you pull different points (probably causing an increase in variance).

However, I'm not an expert, and there might be some weird mathematical symmetry I don't know about, but I would put it at 94.9% confidence that it is different.


Can math be subjective? Often times in math, ever since Kindergarten and before, math has been defined by the fact that there are only one answer for problems. For example: $1+1=2$ and $\frac{d}{dx}x^2=2x$. What I am showing by these two examples are two questions that are from completely different areas of math. However, they both have have only one solution. Problems with multiple answers doesn't necessarily mean they are subjective though, such as $|x|=2,$ which has two solutions. My question is that are any such problems that depend entirely on perspective? If all subjective math problems follow a certain pattern, please tell me what that pattern is. I really have no idea of any examples of this and I would really be interested to see one. Thank you very much.

There's plenty of room for subjective opinion in mathematics. It usually doesn't concern questions of the form Is this true? since we have a good consensus how to recognize an acceptable proof and which assumptions for such a proof you need to state explicitly.

As soon as we move onwards to Is this useful? and Is this interesting?, or even Is this likely to work?, subjectivity hits us in full force. Even in pure mathematics, it's easy to choose a set of axioms and derive consequences from them, but if you want anyone to spend time reading your work, you need to tackle the subjective questions and have an explanation why what you're doing is either useful or interesting, or preferably both.

In applied math, these questions are accompanied by Is this the best way to model such-and-such real-world problem? -- where "best" again comes down to usefulness (does the model answer questions we need to have answered?) and interest (does the model give us any insight about the situation we wouldn't have without it?).

The subjective questions are important in research, but can also arise at more elementary level. The high-school teacher who chooses to devote several lessons to presenting Cardano's method for solving the generic third-degree equation will certainly have to answer his students' questions why this is useful or interesting. Perhaps he has an answer. Perhaps he has an answer that the students don't agree with. In that case, he cannot look for a deductive argument concluding that Cardano's formula is interesting -- he'll have to appeal to emotions, curiosity, all of those fluffy touchy-feely considerations that we need to use to tackle subjective questions.

Yeah, it can:

Consider that polynomials of degree 4 or lower are said to be solvable (or solvable "in radicals" or whatever) whereas the higher-order ones are said to be unsolvable. But this is quite subjective and arbitrary. The definition of $\sqrt{2}$ is literally "the positive real number $x$ that satisfies $x^2 = 2$". What makes solving $x^2 = 2$ any different from solving $x^5 + x + 10 = 0$? I really don't see a meaningful distinction -- both are real numbers and both are irrational. To write either of them out as decimals, you have to use the same kinds of root-finding algorithms for both. To me, the distinction that one of them can be written in terms of radicals and the other one can't seems pretty arbitrary and subjective.
Just look at the definition of e.g. "closed-form solution". It's completely subjective.

etc.


Why is $ \lim_{x \to 0} \frac{x^2}{1-\cos(x)} $ not equal to 0? I was watching Calculus 1 lectures on Youtube, particularly Professor Leonard's, and I thought I would be able to answer a question on the board very simply, but I've reached quite an odd answer and I can't identify the faulty logic in my way of solving. Here is the question and my steps to solve it:
Evaluate: $$ \lim_{x \to 0} \frac{x^2}{1-\cos(x)} $$
My approach to solving this was to take $ x $ out as a factor, then deal with the two expressions as two limits, and eventually treat the latter as a reciprocal of the fundamental $\lim_{x \to 0} \frac{1-\cos x}{x}= 0$
Steps:
$$ \lim_{x \to 0} \left[x \cdot \frac{x}{1-\cos(x)}\right] $$
$$ \lim_{x \to 0} x \cdot \lim_{x \to 0} \frac{x}{1-\cos(x)} $$
$$ \lim_{x \to 0} x \cdot \lim_{x \to 0} \left(\frac{1-\cos(x)}{x}\right)^{-1} $$
$$ \lim_{x \to 0} x \cdot \left[\lim_{x \to 0} \frac{1-\cos(x)}{x}\right]^{-1} $$
$$ 0 \cdot 0^{-1} = 0 $$
Now I am aware that my answer is mathematically wrong and that the correct limit is 2, but I can't deduce why by using my current knowledge of calculus. I would appreciate it if someone pointed out the reason I can not evaluate this limit this way.

Generally speaking the "intuitive" rules for calculating limits such as
$$
\lim f(x)g(x)=\lim f(x) \cdot\lim g(x)
$$
do only work if all limits exist. This is not the case here since the limit
$$
\lim_{x\to0} \frac{x}{1-\cos(x)}
$$
does not exist.
That being said if you extend the real numbers with plus and minus infinity you can also extend the rules to some more cases e.g. $\infty \cdot a=\infty$ for $a\neq0$. However that is not the case for your specific example which reads $0\cdot\infty$ which is often referred to as "indeterminate form".
Please refer to the chapter "properties" in the following link:
https://en.m.wikipedia.org/wiki/Limit_of_a_function

"My approach to solving this was to take x out as a factor, then deal with the two expressions as two limits"
You can not EVER do that.  EVER.  It's a rookie mistake but the values of $x$ and the values of $1-\cos x$ are not independent.
If you look at the behavior of $x^2$ and $1-\cos x$ where $x$ is near zero you can't consider the behavior and nearness of one without also considering the behavior and nearness of the other.  So you can not considered thir limits separately.
Here is an example of way it doesn't work.
Take $\lim_{x\to 0} \frac {x}{2x}$. If you consider the limits separately you can try "doing the denominator first".  SO as $x \to 0$ we have $\frac {x}{2x}\to \frac 0{2x} = 0$. Then after we do that we consider $2x\to 0$ we get $\frac 0{2x} = 0$ cosistantly so $0\to 0$ as $2x\to 0$.  So $\lim_{x\to 0} \frac x{2x}= \lim_{\color{green}x\to 0}\lim_{\color{red}x \to 0}\frac {\color{red}x}{\color{green}{2x}}=\lim_{\color{green}x \to 0}\frac 0{2x}=\lim_{\color{green}x \to 0} 0 = 0$.
Or we not $2x \to 0$ and we do that "first"  $\frac {x}{2x} =\frac {x}0 = \infty$ and $\infty \to \infty$ as $x^2 \to 0$ so $\lim_{x\to 0} \frac x{2x}= \lim_{\color{red}x\to 0}\lim_{\color{green}x \to 0}\frac {\color{red}x}{\color{green}{2x}}=\lim_{\color{red}x \to 0}\frac x{0}=\lim_{\color{green}x \to 0} \infty = \infty$.
But if we do them both "at the same time" we get $\lim_{x\to 0}\frac x{2x} = \lim_{x\to 0} \frac 12 = \frac 12$.
So which is it.  Well whenever we reduce $\color{red} x$ to so something small such as $\color{red}x = \delta$ then we must also reduce $\color{green}{2x}$ to something equivalently small such as $\color{green}{2x} = 2\delta$.  We can just make one of the small and leave the other one hanging around till we are ready for it.
So the first two are wrong.
.....
Okay.... you are asking "but what about the theorem that says if $\lim_{x\to a} f(x) = M$ and $\lim_{x\to a} g(x) = N$ then $\lim_{x\to a} f(x)g(x) = MN$.  Isn't that actually saying we can do just said we can't do".
Welllll.... sort of.  But not really.  In this case $\lim_{x\to a}f(x) = M$ and $\lim_{x\to a} g(x) =N$ must actually be things where $MN$ is a value that makes sense.   If $\lim_{x \to a}f(x) = 0$ and $\lim_{x\to a}g(x) =\infty$ then "$0\cdot \infty$" is meaningless garbage and it doesn't make any sense to talk of it.
This is what we can "indeterminate form".
(SUBTLE thing to note:  When we say $\lim x f(x)g(x) = ML = \lim f(x)\lim g(x)$ we aren't "separating" the limits until after we get the result.  We are not taking the limits one after another, we are still playing with the limits both and the same time.  We just know ahead of time the result will hold. So we can focus on one of the other so long as we are aware the limits themselves are both occuring simultaneous.)
.....
Fortunately, for limits in indeterminate form we have l'hopitals rule.
If $\lim \frac {x}{1-\cos x}$ yiels $\frac {\lim x}{\lim (1-\cos x)} = \frac 00$ is in indeterminate form, we can sleep easy knowing
$\lim \frac {x}{1 - \cos x} = \lim\frac {derivative\ of\ x^2}{derivative\ of\ 1-\cos x} = \lim \frac {2x}{\sin x}= \frac {2}{1} = 2$.


Matrix Multiplication Rules Unfortunately I cannot post up the full question, but I need this little idea (whether its true or false to answer another question)

Let's say we have $v, w \in \mathbb{R}^n$ and $A,B$ be matrices with $B$ being symmetric. Then if

$v^tA^tBw = v^tBAw$

Is there a rule that can help me argue that $A^tB = BA$?  $(*)$

Somehow if I can show this then this means,

$A^tB^t = BA$ iff $(BA)^t = BA$. Hence $BA$ is symmetric.

[Side note: I have used $v^t$ to denote the transpose]. Again sorry I cannot give more context I cannot post up the question due to reasons. So my main point is $(*)$

Since you're trying to show $A^TB^T = BA \iff (BA)^T = BA$: I think you would be able to get by with the general statement
$$
(AB)^T = B^TA^T
$$
which holds for any matrices $A,B$.  Here's a proof.

I'm not quite sure on your wording, but assuming you mean $v^tA^tBw = v^tBAw$ holds for all $v,w\in\mathbb{R}^n$ and you want to prove that $A^tB=BA$, then you can write $v^tA^tBw = v^tBAw$ as $v^tC=v^tD$, where $C=A^tBw,D=BAw$. Thus we see that $C$ has to equal $D$. Now repeat this process.


How to evaluate $\int \frac{dx}{\sin(\ln(x))}$? I am wondering how to evaluate the indefinite integral
$$\int \frac{dx}{\sin(\ln(x))} \quad (1)$$

Attempt 1
I tried using Weierstrass substitution.

The Weierstrass substitution, (named after K.Weierstrass (1815)), is a substitution used in order to convert trigonometric functions rational expressions to polynomial rational expressions. Integrals of this type are usually easier to evaluate.

This substitution is constructed by letting:
$$t = \tan\left(\frac{x}{2}\right) \iff x = 2\arctan(t)  \iff dx = \frac{2}{t^2+1}$$
Using basic trigonometric identities it is easy to prove that:
$$\cos x = \dfrac{1 - t^2}{1 +  t^2}$$
$$\sin x = \dfrac{2t}{1 + t^2}$$
But I couldn't express $\ln(x)$ in terms of $t$.

Attempt 2
I tried using integration by parts but I couldn't find a workaround, it gets more complicated, really fast.
$$ \int \frac{dx}{\sin(\ln(x))} \ = x \sin(\ln(x)) - \int \frac{\cot \left(\ln \left(x\right)\right)}{x\sin \left(\ln \left(x\right)\right)} $$

Attempt 3
The most logical substitution I could think of. It doesn't seem to lead anywhere though.
Let, $\ln(x) = u \iff dx = \, e^u du$
$$ (1) \iff \int \frac{dx}{\sin(\ln(x))} = \int \frac{e^u}{\sin(u)} du = \int \frac{(e^u)'}{\sin(u)} du = $$
$$ \frac{(e^u)'}{\sin(u)} - \int e^u \left(\frac{1}{\sin(u)}\right)' = \frac{(e^u)'}{\sin(u)} - \int e^u \frac{\cos(u)}{\sin^2(u)} =  ?$$

Attempt 4
A combination of attempts 1,2, 3.
Let $\ln(x) = t$ then $dx = e^t dt$, therefore,
$$\int \frac{dx}{\sin(\ln(x))} dx = \int \frac{e^t }{\sin(t)}dt \quad (1)$$
Let's first evaluate $$ \int \frac{1\:}{\sin\left(t\right)}dt \quad (2)$$
Using the Weierstrass substitution $$ t = \arctan(\frac{x}{2})$$
it is easy to prove that
$$ (2) = \int \frac{1\:}{\sin\left(t\right)}dt= \ln \left|\tan \left(\frac{t}{2}\right)\right|+C$$
Therefore,
$$ (1) \iff I = \int e^x\left(\ln \:\left|\tan \:\left(\frac{t}{2}\right)\right|\right)'dt = e^x \ln \:\left|\tan \:\left(\frac{t}{2}\right)\right| - \int (e^x)' \ln \:\left|\tan \:\left(\frac{t}{2}\right)\right|dt = $$
$$  e^x \ln \:\left|\tan \:\left(\frac{t}{2}\right)\right| - \left( e^x \ln \:\left|\tan \:\left(\frac{t}{2}\right)\right|  - \int e^x \left(\ln \:\left|\tan \:\left(\frac{t}{2}\right)\right|\right)'dt \right) $$
$$ I = 0 + I \iff 0=0$$
Tautology. No answer here.

Attempt 5
Ask a question on MathExchange: Any ideas?
Note: A complex-plane solution was proposed in the comments, but I am evaluating this on $\mathbb{R}$

By Euler's formula,
$$\sin(\ln(x))=\frac{e^{i\ln(x)}-e^{-i\ln(x)}}{2i}=\frac{x^i-x^{-i}}{2i}$$
In the integral, this works out to give us
$$\int\frac{\mathrm dx}{\sin(\ln(x))}=\int\frac{2i~\mathrm dx}{x^i-x^{-i}}=2i\int\frac{x^i~\mathrm dx}{x^{2i}-1}=-2i\int\frac{x^i~\mathrm dx}{1-x^{2i}}$$
By expanding with geometric series, this then becomes
$$\int\frac{x^i~\mathrm dx}{1-x^{2i}}=\sum_{k=0}^\infty\int x^{(2k+1)i}~\mathrm dx=\sum_{k=0}^\infty\frac{x^{1+(2k+1)i}}{1+(2k+1)i}$$
Observe that the ratio of consecutive terms in this series is given by
$$\frac{x^{1+(2k+3)i}/(1+(2k+3)i)}{x^{1+(2k+1)i}/(1+(2k+1)i)}=\frac{(2k+1)i+1}{(2k+3)i+1}x^{2i}=\frac{(k+\color{#3377cc}{\frac{1+i}2})(k+\color{#3377cc}1)}{k+\color{#339999}{\frac{1+3i}2}}\frac{\color{#dd3333}{x^{2i}}}{k+1}$$
which implies the series is a hypergeometric function:
$$\sum_{k=0}^\infty\frac{x^{1+(2k+1)i}}{1+(2k+1)i}=x^{1+i}{}_2F_1\left(\color{#3377cc}{\frac{1+i}2},\color{#3377cc}1;\color{#339999}{\frac{1+3i}2};\color{#dd3333}{x^{2i}}\right)$$
and altogether,
$$\int\frac{\mathrm dx}{\sin(\ln(x))}=-2ix^{1+i}{}_2F_1\left(\frac{1+i}2,1;\frac{1+3i}2;x^{2i}\right)\color{#999999}{{}+C}$$
which likely cannot be simplified further, though can be rewritten using various hypergeometric identities.

Note: The above manipulations require that the series converges, but the end results in terms of hypergeometric functions hold everywhere they both exist, as they are defined through the use of analytic continuation.

I used MMA to calculate this indefinite integral and found that his result contains a hypergeometric function, so the primitive function of this indefinite integral cannot be expressed by elementary functions.


Trig substitution problem. Stuck but close. I have this problem:

$$\int_0^2 \sqrt{1 + 4x^2}$$

Here's how I start. I focus on finding the antiderivative:
$$ \int \sqrt{4 * (\frac{1}{4} + x^2)}$$

$$ \int 2* \sqrt{\frac{1}{4} + x^2)} dx$$

so let $x = \frac{1}{2} * tan{\theta}$ so $dx = \frac{1}{2} * sec^2\theta d\theta$

so...

$$ \int 2 \sqrt{\frac{1}{4} + \frac{1}{4}tan^2{\theta}} * \frac{1}{2}sec^2{\theta} \cdot d\theta$$

and by $sec^2{\theta} = 1 + tan^2{\theta}$:

$$ \int 2 \cdot \frac{1}{2} sec{\theta} \cdot \frac{1}{2} sec^2{\theta} \cdot d\theta$$

$$ \int \frac{1}{2} sec{\theta} \cdot sec^2{\theta} \cdot d\theta$$ Let's call the above line : (1)

and now integration by parts:

$u = sec{\theta}$ and so $du = sec{\theta}tan{\theta}$

$dv = sec^2{\theta}$ and so $v = tan{\theta}$

so = $$\frac{1}{2} (\cdot sec{\theta}tan{\theta} - \int tan^2{\theta} sec{\theta} \cdot d\theta)$$

and so since $tan^2{\theta} = sec^2{\theta} - 1$

$$\frac{1}{2} (\cdot sec{\theta}tan{\theta} - \int sec^3{\theta} - \int sec{\theta} \cdot d\theta)$$

So using (1) we have:

$$(\frac{1}{2}  \cdot sec{\theta} \cdot tan{\theta} - \frac{1}{2} \int sec^3{\theta} - \frac{1}{2}  \int sec{\theta} \cdot d\theta) = \int \frac{1}{2} sec{\theta} \cdot sec^2{\theta} \cdot d\theta = \int \frac{1}{2} sec^3{\theta} d\theta$$

combining like sec^3{\theta} terms:

$$ (\frac{1}{2}  \cdot sec{\theta} \cdot tan{\theta} - \frac{1}{2}  \int sec{\theta} \cdot d\theta) = sec^3{\theta} $$

and now multiplying by $\frac{1}{2}$ again:

$$ (\frac{1}{4} \cdot sec{\theta} \cdot tan{\theta} - \frac{1}{4}  \int sec{\theta} \cdot d\theta) = \frac{1}{2} sec^3{\theta} $$

But what's the antiderivative of $\int sec{\theta} d\theta$ ?

I'm told to do this but I don't really get it:

\begin{align}
\int \sec \theta \, d\theta = \int \frac{\sec^2 \theta + \sec \theta \tan \theta}{\sec \theta + \tan \theta} \, d\theta=\ln (\sec \theta + \tan \theta) + c
\end{align}

Note that this is due to the numerator is the derivative of the denominator.

$$\int \frac{f'(t)}{f(t)} \, dt=\ln f(t) + c$$

Using a double-angle formula:

$$\sqrt{17}+\frac{1}{4} \sinh ^{-1}(4)$$


Every odd composite $=$ prime ${}+ 2x^2$ I was looking through some project-euler questions and I came across one that said

  Every odd composite number can be written as the sum of a prime and twice a square...This was proven false.

So, I solved the problem through brute-force in Java but I was wondering how I can do this on paper? Please don't give me the answer, just some hints

Thanks.

Some proofs that a number-theoretic statement like that isn't true basically amount to finding a counterexample. That isn't to say there isn't an elegant analytic approach, but the claim that this was proven false might simply be a reference to the fact that someone has found a number which can't be written in that way, and shown that it fits by brute force as well.

Starting from $p_{n}=3$

Every odd composite $N$ can be written as $N= p^2_{n}+ 2p_{n}c$

Ex for $p_{n}=5$

$25 + 10c =N$

For $c$ from 0 to 7 we have the set of odd composites $[25,35,45,55,65,75,85]$

Intuitively $N= p^2_{n}+ 2p_{n}c$ can't be equal to $N= p_{n}+ 2x^2$


Number of picks to end the process I am encounter this question, I don't know how to start. Can you guys help me on this?
$N$ is a positive number. I randomly pick a number, $K$, from $[0, N-1]$. If $K = 0$, the process ends, otherwise we keep picking a number from $[0, K-1]$ until we pick $0$. What is the expected number of picks to end the game?

Let $T_n$ be the number of picks to end when we are picking from $[0,n-1]$.
Note that $E[T_1] = 1$ and if we pick $i$, then we need to look at $T_i$. Then $$E[T_n] = \frac{E[T_1] + \sum_{i=2}^{n-1} (E[T_i] + 1)}{n} = 1 + \frac{\sum_{i=2}^{n-1}E[T_i]}{n} = 1 + \frac{1}{n} + \frac{\sum_{i=2}^{n-2}E[T_i]}{n-1} = ... =\sum_{i=1}^{n}\frac{1}{i} = H(n)$$


Minimum vertices set bipartite graph covering-special case I was wondering if anyone here could give me any pointers as to how to solve the following problem.

Let B=(L,R,E) be an undirected bipartite graph, ∀u∈L, ∃ s= {ei(u,wi)} ∈E; i=1,2.....n connect u to wi, where w∈R.

The problem is to find a minimum set K from L covering all R in B, K⊆L , ∑u∈K is minimal.

To clarify what I mean by covering: all vertices of R should should have at least one edge to any u∈K.

My intuition is that it's NP-Hard. If that is the case, any idea of what would be the best way to approximate the result (ie a minimum set K of L covering R)? Such that:

Each vertex from L could cover multi-vertices of R
The vertex from L which has maximum edges toward R will be selected first
Each vertex of R is connected with at least one vertex of L

Edit: Here is an example, consider the following bipartite graph: G={L∪R,E},
L={1,2,3,4,5,6},
R={A,B,C,D} ,
E={1A,1B,2A,2B,2C,3A,3C,4A,4B,4D,5A,5B,6A,6D}

And here is a covering minimum set will be {2,4}

I have read many algorithms and solutions like maximum matching, complete matching, stable marriage, set cover problem, vertex cover in hypergraphs ...etc.
Unfortunately no one match my case.

So i am here asking your help guys cause i about to fed up!!! SOS!!

This is just the set cover problem in disguise, to each vertex of $L$ assign the set of vertices of $R$ in its neighborhood. We now have transformed $L$ into a set of subsets of $R$ so now we want to find the set covering with the fewest sets.

It is true, as you suspected that this problem is $NP$ complete, but a lot of research on it exists.

I recently stumbled upon this problem as well and I thought about it.

I believe your problem to be easily solvable in polynomial time and not being related to the classical NP-complete set vertex cover as @jorge-fernández suggested, since you only care about vertices on one side of the bipartite graph and not on both. This difference makes everything way more simple.

The greedy algorithm is quite trivial:

Chose a vertex A from v1 which covers "more" nodes left in v2 (i.e the vertex from v1 with more edges)

Add the vertex A the mininum cover set
Remove all the nodes in v2 having an edge to A
Remove the vertex A from v1

In case v2 is not empty, go to point 1

The algorithm produces an optimal solution and terminates in polynomial time.

The optimal solution is returned because picking up the vertex with more edges from v1 does not prevent any remaining vertex from being picked up later on (i.e. taking this step does exclude any other vertices nearby to be selected later in the process. This instead happens in the classic NP-Complete problem). The descent is convex and we will not land in a local minumum
In the worst case, the algorithm will take |v2| loops to complete (every loop will just remove a single vertex from v2)


Sequences of pairs where adjacent elements differ in exactly one component I have a finite alphabet A and a set S of $n$ ordered pairs $(a, b)$ with $a,b\in A.$  I want to find out how many sequences $(a_1, b_1),\ldots,(a_n,b_n)$ there are where adjacent elements have either $a_{k-1}=a_k$ or $b_{k-1}=b_k.$ (Since S is a set, they can't have both.)

Any ideas on how to go about finding this?  Formulas, generating functions, algorithms, or whatever you have.  In my particular application I have $|A|=10$ and $|S|=21$, but the question seems general.

Also, in my application $(a_1,b_1)$ and $(a_k,b_k)$ are not adjacent. (They can have, but need not have, one member different.)

Edit: For clarification, I'm not allowing all elements from $A\times A$, just those in $S$.  No valid answer can ignore S.  For example, I might have S = {(1, 1), (3, 3)} in which case there are no valid orderings.

For those interested in using a matrix approach (not sure if this can work; the path needs to go through each pair exactly once), here's the adjacency matrix:

[0 1 1 1 0 0 1 0 1 0 0 0 0 1 0 1 0 0 0 0 0]
[1 0 1 1 1 0 0 0 0 1 0 1 0 0 0 0 1 0 1 0 0]
[1 1 0 1 0 0 0 1 0 0 1 0 0 0 1 0 0 0 0 0 1]
[1 1 1 0 0 1 0 0 0 0 0 0 1 0 0 0 0 1 0 1 0]
[0 1 0 0 0 1 0 0 0 1 0 1 0 0 0 0 1 0 1 0 0]
[0 0 0 1 1 0 0 0 0 0 0 0 1 0 0 0 0 1 0 1 0]
[1 0 0 0 0 0 0 1 1 0 0 0 0 1 0 1 0 0 0 0 0]
[0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 1]
[1 0 0 0 0 0 1 0 0 1 1 0 0 1 0 1 0 0 0 0 0]
[0 1 0 0 1 0 0 0 1 0 1 1 0 0 0 0 1 0 1 0 0]
[0 0 1 0 0 0 0 1 1 1 0 0 0 0 1 0 0 0 0 0 1]
[0 1 0 0 1 0 0 0 0 1 0 0 1 0 0 0 1 0 1 0 0]
[0 0 0 1 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 1 0]
[1 0 0 0 0 0 1 0 1 0 0 0 0 0 1 1 0 0 0 0 0]
[0 0 1 0 0 0 0 1 0 0 1 0 0 1 0 0 0 0 0 0 1]
[1 0 0 0 0 0 1 0 1 0 0 0 0 1 0 0 1 1 0 0 0]
[0 1 0 0 1 0 0 0 0 1 0 1 0 0 0 1 0 1 1 0 0]
[0 0 0 1 0 1 0 0 0 0 0 0 1 0 0 1 1 0 0 1 0]
[0 1 0 0 1 0 0 0 0 1 0 1 0 0 0 0 1 0 0 1 0]
[0 0 0 1 0 1 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0]
[0 0 1 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0 0 0 0]

Kudos to anyone who recognizes it (and thus my goal).

Let $m = \lvert A\rvert$, and denote by $L_{m,m}$ what is variously called the $m\times m$ rook's graph or lattice graph: it has $m\times m$ vertices arranged in a square, with an edge connecting two vertices if they are in the same row or column. There is a natural mapping between the vertices of $L_{m,m}$ and ordered pairs of elements from $A$, of which $S$ is a subset. Then a sequence of length $n$ in your problem is equivalent to a Hamiltonian path in the subgraph of $L_{m,m}$ induced by $S$.

The number of such paths will certainly depend on the elements of $S$, and not just their cardinality. For example, let $m = 2$ and $n = \lvert S\rvert = 2$; if the two elements of $S$ are adjacent, then there is one Hamiltonian path, otherwise none.

The simplest algorithm I can think of is to simply generate all permutations of the $n$ vertices and check whether each one is a valid path. Finding a single Hamiltonian path is NP-complete, so I'm not sure you can do much better. You could do a search for "count number of hamiltonian paths" and see what you get.

Perhaps I'm misunderstanding your problem. But if you want to count how many sequences of pairs there are which where each two conscutive pairs have at least one element in common, the answer seems easy. Just count how many sequences there are that have not that property, and substract from the total.

I get, then, $a^{2n} - a^2 (a-1)^{2(n-1)} = a^2 [ a^{2(n-1)} - (a-1)^{2(n-1)}]$ (where $a = |A|$)


Why MLR (monotone likelihood ratio) implies stochastic increasing? the following argument holds:
for $\theta_1<\theta_2$, $\dfrac{f(x\mid\theta_2)}{f(x\mid\theta_1)}$ is increasing in $x$. Then, $F(x\mid\theta_2)\leq F(x\mid\theta_1)$ for all $x$.

Intuitively, this seems to be natural but I could not find a way.
Would you let me know this or have some hints?

Let's suppose all is one-dimensional, and denote $g(x) = \dfrac{f(x\mid\theta_2)}{f(x\mid\theta_1)}$, then $g$ is increasing.

Define $t = \sup\{x: g(x) \leq 1\}$, then obviously for all $x \leq t$, $F(x\mid\theta_2)\leq F(x\mid\theta_1)$ since on $(-\infty,t)$ $f(x\mid\theta_2) \leq f(x\mid\theta_1)$ and $F(x\mid\theta_i) = \int_\infty^x f(y\mid\theta_i)dy, i = 1,2$

And for $x > t$, remark that $F(x\mid\theta_i) = 1 - \int^\infty_x f(y\mid\theta_i)dy, i = 1,2$ and on $(t, +\infty)$ we have $f(x\mid\theta_2) \geq f(x\mid\theta_1)$

Sometimes it is good learning about doubts. This relation is well known from the beginning of mathematical statistics. In the classic book of Lehmann "Testing Statistical Hypothesis" you can find this implication and more.


matrix structures obtainable by multiplication by unitary matrices Out of "Numerical Linear Algebra" by Trefethen and Bau:

  Suppose we have $3\times 3$ matrices and wish to introduce zeroes by left- and/or right-multiplications by unitary matrices $Q_j$ such as Householder reflectors or Givens rotations. Consider the following matrix structures:
  $$
(a)\begin{bmatrix}
\times & \times & 0\\
0 & \times & \times\\
0 & 0 & \times\\
\end{bmatrix}\ \ \ \ \ \ \ \ (b)\begin{bmatrix}
\times & \times & 0\\
\times & 0 & \times\\
0 & \times & \times\\
\end{bmatrix}\ \ \ \ \ \ \ \ (c)\begin{bmatrix}
\times & \times & 0\\
0 & 0 & \times\\
0 & 0 & \times\\
\end{bmatrix}
$$

  (where $\times$ "represents an entry that is not necessarily zero")

  For each one, decide which of the following situations olds and justify your claim

  (i) Can be obtained by a sequence of left-multiplications by matrices $Q_j$

  (ii) Not (i), but can be obtained by a sequence of left- and right-multiplication by matrices $Q_j$

  (iii) Cannot be obtained by any sequence of left- and right-multiplications by matrices $Q_j$.

Not sure how to argue, here, and not sure how $Q_j$'s being unitary factors in.

Some help would be hot.

Hints.

(a) Can every rank-1 matrix be reduced to this form?

(b) Handle the first column by Householder reflection. Then try to handle the $2\times2$ submatrix at the top right corner by singular value decomposition. Note that in such a decomposition $USV^\ast$, the unitary matrices $U,V$ may have determinants $-1$. How do you remedy for this?

(c) What is the rank of a matrix in this pattern?


How to derive that for every real $y > 0$, for every positive real $z \neq 1$, there is a $x \in \mathbb{R}$ such that $y=z^x$. I am not sure on how to derive the following statement concerning the reals (that I think should be true).

  For every real $y > 0$, for every positive real $z \neq 1$, there is a $x \in \mathbb{R}$ such that $y=z^x$.

QUESTIONS:

Is this part of the definition of exponentiation, or it is a result that can be proven?
If it can be proven, how (maybe by using the Archimedean Property)?

Thank you for your time.

PS: Few days ago I asked almost the same question, but a user pointed out (rightly) that the question I really wanted to ask is the one here.

Since the exponential function $\exp(t) = e^t$ strictly increases on $\mathbb{R}$, (Check this in your analysis textbook) its inverse, the logarithmic function $\log t$ on $\mathbb{R}^+$ is well-defined. If we take $x = \log y / \log z$, $ z^x = e^{\log y} = y$ clearly.


Equal polynomials of equal degree at equal x must have equal coefficients? does anyone know if the following is true?

I'm looking at something like
$$2^{s_1+s_2+\Delta_2}+2^{s_2+\Delta_2}3-2^{s_1}3^2=2^{s_1+s_2+\Delta_1}+2^{s_2}3-2^{s_1+\Delta_1}3^2$$

And what I notice is that these look like two polynomials (quadratic) at the same $x$ value (if $x=3$) are equal, have the same degree and so I would like to say therefore their coefficients must be equal.

In general maybe
$$2^{s_1+s_2+\Delta_2}+2^{s_2+\Delta_2}x-2^{s_1}x^2=2^{s_1+s_2+\Delta_1}+2^{s_2}x-2^{s_1+\Delta_1}x^2$$

I don't think it works for any $x$, because any two graphs could be equal at an intersection, but maybe the argument works because $3$ is coprime to $2$?

Is it safe to say that
$$\therefore s_1+s_2+\Delta_2= s_1+s_2+\Delta_1$$
$$\therefore s_2+\Delta_2= s_2$$
$$\therefore s_1= s_1+\Delta_1$$
and so at the end of the day
$$\Delta_1=\Delta_2=0$$
because that result would really make me scream with joy.

Also, could this same argument be applied for similar polynomials of higher degree, maybe something like
$$2^{s_2+s_3+\Delta_3}+2^{s_1+s_3+\Delta_3}3+2^{s_3+\Delta_3}3^2-2^{s_2}3^3-2^{s_1}3^4$$ $$=2^{s_2+s_3+\Delta_2}+2^{s_1+s_3+\Delta_1}3+2^{s_3}3^2-2^{s_2+\Delta_2}3^3-2^{s_1+\Delta_1}3^4$$
allow me to conclude the similar result that
$$\therefore \Delta_1=\Delta_2=\Delta_3=0\mbox{ ?}$$

Any help will be greatly appreciated.

         Edit:

Does this work for $s_1,s_2,\Delta_1,\Delta_2\in\mathbb{R}$? if not,

Does this work if either $s_n\in\mathbb{N}$ or $(s_n+\Delta_n)\in\mathbb{N}$?

Suppose $s_1 = s_2 = 0$, then the equation can be simplified to:

$$
8*2^{\Delta_1} + 4 * 2^{\Delta_2} = 12
$$

Clearly we have the solution of $\Delta_1 = \Delta_2 = 0$.

But, if we substitute $x=2^{\Delta_1},y=2^{\Delta_2}$, then we have a linear equation in two unknowns:

$$
8x + 4y = 12
$$

This equation gives us a line of solutions in the $x,y$-plane. In particular it has infinite solutions where $x,y$ are positive so they can be written as powers of two.

For example, we can see that $x=\frac{1}{2}, y=2$ is a solution corresponding to $\Delta_1 = -1, \Delta_2 = 1$.

So, even if we assume that $s_1=s_2=0$, there will still be infinite solutions for $\Delta_1, \Delta_2$.

The situation will be similar for other choices of $s_1,s_2$.


Rotation Matrix and Reflection Matrix, Unitary Matrix I am working on a problem that asks me to show a $2 \times 2$ orthogonal matrix is reflection matrix or the rotation matrix and I do not know the definitions of the rotation matrix and the reflection matrix, the unitary matrix and some good examples of them and could anyone tell me something about them?

Unitary matrices matrices $A$ are invertible matrices that have property that $A^T = A^{-1}$. This leads to a very neat geometric characterisation: a matrix $A$ is unitary if and only if the corresponding linear map $v \mapsto Av$ is an isometry, meaning that it preserves lengths and distances. In other words, $\|Av\| = \|v\|$ for all $v$.

To give you an idea of why this is true, consider the identity $\|w\|^2 = w^T w$, where $w$ is an $n \times 1$ column vector. Then, if $A$ is unitary,
$$\|Aw\|^2 = (Aw)^TAw = w^T A^T A w = w^T I w = w^T w = \|w\|^2.$$
Proving the other direction is a little more involved, so I won't get into it, but this should give you an idea as to how these concepts are connected.

If you think about rotations and reflection transformations, they also preserve lengths and distances, so their matrices should indeed be unitary.

You can look up formulas for rotation and reflection matrices, but it's also possible to derive them. Let's begin with a rotation matrix. We want, for any angle $\theta$, to find the matrix corresponding to the (linear) rotation map that rotates the space anti-clockwise (as is usually the convention).

The way I usually like to do this is with complex numbers. If we take an arbitrary vector $(x, y) \in \mathbb{R}^2$, we can identify it with a complex number $x + iy \in \mathbb{C}$. In $\mathbb{C}$, rotation anti-clockwise by $\theta$ can be done by multiplying $x + iy$ by the complex number $e^{i\theta} = \cos(\theta) + i \sin(\theta)$, which has modulus $1$ and argument $\theta$. The resulting complex number is:
$$e^{i\theta}(x + iy) = x\cos(\theta) - y \sin(\theta) + i(x \sin(\theta) + y \cos(\theta)).$$
If we bring this back to $\mathbb{R}^2$, then the transformation, in total, does the following:
$$(x, y) \mapsto (x\cos(\theta) - y \sin(\theta), x \sin(\theta) + y \cos(\theta)).$$
Putting this as a matrix gives us
$$\begin{pmatrix}
\cos(\theta) & -\sin(\theta) \\
\sin(\theta) & \cos(\theta)
\end{pmatrix}.$$
I'll leave it to you to verify that $A^T A = A A^T = I$, and hence $A^{-1} = A^T$.

As for reflections, again, I prefer to do it with complex numbers. Recall that the complex conjugate reflects about the $x$-axis. Let's say that we want to reflect about a line that makes an angle $\theta$ with the positive $x$-axis (again, anti-clockwise). Our plan is to rotate the space so that this line becomes the $x$-axis, reflect in the $x$-axis using the conjugate, and rotate the space back.

Take an arbitrary $z = x + iy \in \mathbb{C}$. First, to rotate the space, we rotate everything clockwise (not anti-clockwise) by $\theta$. Our point is now at $e^{-i\theta}z$. We now take the complex conjugate, giving us,
$$\overline{e^{-i\theta}z} = \overline{e^{-i\theta}}\overline{z} = e^{i\theta}\overline{z}.$$
Finally, we rotate back, that is anti-clockwise by $\theta$. This gives us our final total of
$$e^{i\theta}e^{i\theta}\overline{z} = (e^{i\theta})^2\overline{z} = e^{i2\theta}\overline{z}.$$
Putting $z = x + iy$, we get,
$$e^{i2\theta}\overline{z} = (\cos(2\theta) + i \sin(2\theta))(x - iy) = (x\cos(2\theta)+ y\sin(2\theta)) + i(x \sin(2\theta) - y \cos(2\theta)).$$
As a transformation on $\mathbb{R}^2$,
$$(x, y) \mapsto (x\cos(2\theta)+ y\sin(2\theta), x \sin(2\theta) - y \cos(2\theta)).$$
This gives us a matrix
$$\begin{pmatrix}
\cos(2\theta) & \sin(2\theta) \\ \sin(2\theta) & -\cos(2\theta)
\end{pmatrix}.$$
Again, show that $A^T A = A A^T = I$, and you're done.

Try drawing in the coordinate system the vector $e_1=(1,0)$ and another vector rotated slightly upwards by an angle $\phi$. This represents the same vector both with length $1$.

The projection of the new vector onto $e_1$, and $e_1$ itself form a right-angled triangle from which you can find the coordinates of $x$ and $y$ of the new rotated vector using trigonometry with $\phi$ and length $1$.

Do the same with $e_2=(0,1)$ by rotating it slightly left from the $y$ line.


Equivalent way of writing the norm of Lp Given a measurable $E\subset \Bbb R^d $  and a measurable function $f:E\rightarrow \Bbb R^d $, prove that :

$$
\int (\left\lvert f \right\rvert)^r d\mu  = r\int_{0}^\infty t^{r-1} \mu(\{x  \in  E \mid  \left\lvert f(x) \right\rvert>t\})\,dt
$$
where $r   \ge  1$

The $r=1$ case is simple; however, I can't prove the general case. Thank you in advance

If you know how to prove the case $r=1$, it's easy :

let $g(t)= |f(t)|^r$. You have (assuming you can prove the result for $r=1$)
$$\int_E |f(t)|^r dt = \int_E |g(t)| dt = \int_0^{\infty} \mu(\{ x\in E \ :\ |g(x)| > t\}) dt$$

it follow

$$= \int_0^{\infty} \mu(\{ x\in E \ :\ |f(x)|^r > t\}) dt = \int_0^{\infty} \mu(\{ x\in E \ :\ |f(x)| > t^{\frac{1}{r}}\}) dt $$

Now, you do the change of variable $y= t^\frac{1}{r}$, and it gives you the result :

$$= \int_0^{\infty} ry^{1-r} \mu(\{ x\in E \ :\ |f(x)| > y \}) dy $$

You can derive the general case from the case $r=1$. Because $|f|>t$ if and only if $|f|^r>t^r$.


Adjustable Sigmoid Curve (S-Curve) from $(0,0)$ to $ (1,1)$ I feel like this is such a simple question but I am at such a loss. I currently have a set of values that I would like to weigh by an S Curve. My data ranges from $0$ to $1$ and never leaves those bounds, but nearly every version of a Sigmoid I see assumes $x = 0$ is the point of inflection. Furthermore many of the Sigmoids reach y=0 at approximately $x = -6$, and $y=1$ at approx $x= 6$. However in my case I need to reach $y=0$ at $x=0$ and $y=1$ at $x=1$, and the point to of inflection to be at $x = 0.5$.

To summarize I need a S Curve that meets the following requirements..

$x$ from $0$ to $1$
$y$ from $0$ to $1$
$0$ remains $0$ and $1$ remains $1$
Adjustable shape for different Curves
Bonus If I can adjust the point of inflection while maintaining the above

Here is an example of the above (except for a movable inflection)
http://www.guillermoluijk.com/misc/contrast.gif

I have looked at Sigmoids, Logistics, tan and sing approximations, contrast and AI equations, and nothing quite gets me what I need. Any explanation to go along with an answer would be much appreciated!

I put three in a chain: I use the positive domain of the NTS to 'skew' the input, the second is the basic sigmoid, and then I use the positive domain to skew the output. The result is an extremely flexible curve which I tend to use everywhere now. Here is a Desmos sheet, I think it is what you were looking for, although the other answers seem good too. https://www.desmos.com/calculator/tswgrnoosy

Here is a variant of sigmoid described in question I have made through arctan() function

$$y=\left(\frac{-1}{2\cdot\arctan(-\frac{a}{2\pi})}\right)\cdot\arctan\left(\frac{a\cdot(x-0.5)}{\pi}\right)+0.5$$

Those big left multiplier depends only on parameter and can be calculated separately so basically it is:
$$K=\frac{-1}{2\cdot\arctan(-\frac{a}{2\pi})}$$
$$y=K\cdot\arctan\left(\frac{a\cdot(x-0.5)}{\pi}\right)+0.5$$

arctan parametric sigmoid picture

Pseudocode:

a = 100.0;
K = -1.0 / (2.0 * atan(-a * 0.5 / pi));

for(x = 0.0; x <= 1.0; x += 0.01)
{
    y = K * atan(a * (x - 0.5) / pi) + 0.5;
}


Given an arbitrary sequence {$x_n$} in $\Bbb{R}$, find a test function $f$ with $f^{(n)}(0)=x_n$ Given an arbitrary sequence {$x_n$} in $\Bbb{R}$, can I find a test function having the $n$-derivative equal to $x_n$ at $0$?

Taking the $n^{th}$ derivative of an infinite polynomials gives:

$
f^{(n)}(z) = n!(a_n) +  ((n+1)!/1!)(a_{n+1})z + ((n+2)!/2!)(an_{n+2})z^2 + ...$

$f^{(n)}(z) = n!(a_n) + O{(z)}
$

So in general we need $a_n = x_n/n!$

Hence the function is $f(z) = x_0 + x_1z + (x_2/2)z^2 + ... + (x_n/n!)z^n + ...$


How to show that a mapping is linear Let $T: \mathbb{R}^2 \rightarrow \mathbb{R}^3$ be defined by $T\begin{pmatrix}u_1\\u_2\end{pmatrix} = \begin{pmatrix}u^2_2\\2\\u_1-u_2\end{pmatrix}$
Show T is linear.

How can I do this using the rules I know
$i)  \, T(u +v) = T(u) + T(v)$
$ii) \,  T(ru) = rT(u)$ ?
For example, in this case what is $u$ and $v$ for me?
UPDATE

Let $S: \mathbb{R}^2 \rightarrow \mathbb{R}^3$ be defined by $T(\begin{pmatrix}u_1\\u_2\end{pmatrix}) = \begin{pmatrix}u_2^2\\2\\u_1-u_2\end{pmatrix}$. Show $S$ is not linear

So I think I've got a solution but want to double check that this answer will suffice. (By the way I let $T = S$
I know that I can use the property
$T(ru) = rT(u)$ and let $u = 0$ to get $T(0) = \begin{bmatrix}0\\2\\0\end{bmatrix} \ne 0 \implies$ not linear. Isn't that all I really need to prove that $S$ is not linear

To answer your question:

$u,v$ are just elements of a vector space, in your case it is $\mathbb R^2$, that means
$$
u=(u_1,u_2),v=(v_1,v_2) \text{ with } u_1,u_2,v_1,v_2\in \mathbb R
$$
$r$ on the other hand is called a scalar and in your case the corresponding filed (over which your vector space is defined) is again $\mathbb R$, so $r\in \mathbb R$.

The following was an answer to the question before editing:

Now just check for example what @Bye_World already mentioned in the comments to see that
$$
T(ru) = rT(u)
$$
fails (or check whether the $0$ is mapped to $0$) or just check your first property with
$$
u=(1,0),v=(0,1)
$$
which gives
$$
T(u+v)=T((1,1))=(1,2,0)\neq(0,2,1)+(1,2,-1)=(1,4,0)=T(u)+T(v)
$$
So $T$ is not a linear mapping.

The following is an answer to the question after editing

Here $T$ is indeed linear, the second property is obvious and for the first one we just check
\begin{align}
T(u+v)=T((u_1+v_1,u_2+v_2))&=(u_2+v_2,0,u_1+v_1-u_2-v_2)\\&=(u_2,0,u_1-u_2)+(v_2,0,v_1-v_2)=T(u)+T(v)
\end{align}

In this case, your $u$ and $v$ will be vector in the form $\begin{pmatrix} a\\b\end{pmatrix}$.

To check the first property, let $u = \begin{pmatrix} a\\b \end{pmatrix},v=\begin{pmatrix} c\\d \end{pmatrix}.$ Then we have

$$T(u+v)=T(\begin{pmatrix} a\\b \end{pmatrix}+\begin{pmatrix} c\\d \end{pmatrix})=T(\begin{pmatrix} a+c \\b+d \end{pmatrix})  \text{......(why?)}$$

You then compute $T(u) + T(v)$. If these two are equal then you are done for the first property. However, I am not saying that $T$ is a linear transformation since I haven't done the computation. But in fact you can tell that $T$ is not a linear transformation. This is a consequence of the following fact:

https://proofwiki.org/wiki/Linear_Transformation_Maps_Zero_Vector_to_Zero_Vector


Distributing n distinct objects to r different people - Why is my method wrong? I saw this problem:
How do I distribute 5 different candies to 3 different students?
I decided to try and make a modified Stars and Bars formula to handle the different candies.
My Method:
Suppose I had 5 different candies:
A B C D E
and 2 bars for 3 different students: ||
combined: A B C D E | |
To find all the different combinations of candies and students, I need to do
$${\frac{7!}{2!}}$$
However, my answer was wrong. I'm not sure what the correct answer is, because the quiz I took did not show it.
Why is my method wrong?

You just have to affect one student to each candy. For each candy, you have $3$ different possibilities, so this gives
$$3^5=243$$
possibilities.


$\lim_{x \to \infty } {1.0000000000001^x}/{x^{100}} = 0$ I am confused as to how Wolfram alpha got the answer to be $0$, because I am fairly sure that the answer should be infinity, can somebody explain this to me?

Since $e^x\ge1+x$,
$$
\begin{align}
\left(1+10^{-13}\right)^x
&=e^{x\log\left(1+10^{-13}\right)}\\[6pt]
&=e^{\left(\left.x\log\left(1+10^{-13}\right)\middle/200\right.\right)\cdot200}\\[6pt]
&\ge\left(1+\frac{x\log\left(1+10^{-13}\right)}{200}\right)^{200}\\
&\ge x^{200}\,\left(\frac{\log\left(1+10^{-13}\right)}{200}\right)^{200}
\end{align}
$$
Therefore,
$$
\frac{\left(1+10^{-13}\right)^x}{x^{100}}
\ge x^{100}\,\left(\frac{\log\left(1+10^{-13}\right)}{200}\right)^{200}\\
$$
and the limit is obviously infinite.

As mentioned in comments, the problem is most likely round-off error due to finite precision arithmetic.

As I mentioned in comments, it takes $x\gt38181117481549541$ to even get $\frac{\left(1+10^{-13}\right)^x}{x^{100}}\gt1$.

If
$0 < c < 1$
and
$a > 0$
then

$\begin{array}\\
f(x)
&=\ln(\dfrac{(1+c)^x}{x^a})\\
&=x\ln(1+c)-a\ln(x)\\
&\gt xc/2-a\ln(x)
\qquad\text{since } \ln(1+c) > c/2\\
&\gt xc/2-2a\sqrt{x}
\qquad\text{since } \ln(x) < 2\sqrt{x}\\
&=\sqrt{x}(\sqrt{x}c/2-2a)\\
\end{array}
$

so
$f(x) > 0$
for
$x > (4a/c)^2
$
and
$f(x) > \sqrt{x}$
for
$x > (2(2a+1)/c)^2
$.

In particular,
$f(x) \to \infty$.

These bounds, of course,
are much worse than
the best bounds.


What is the intuitive relationship between SVD and PCA? Singular value decomposition (SVD) and principal component analysis (PCA) are two eigenvalue methods used to reduce a high-dimensional data set into fewer dimensions while retaining important information. Online articles say that these methods are 'related' but never specify the exact relation.

What is the intuitive relationship between PCA and SVD? As PCA uses the SVD in its calculation, clearly there is some 'extra' analysis done. What does PCA 'pay attention' to differently than the SVD? What kinds of relationships do each method utilize more in their calculations? Is one method 'blind' to a certain type of data that the other is not?

(I assume for the purposes of this answer that the data has been preprocessed to have zero mean.)

Simply put, the PCA viewpoint requires that one compute the eigenvalues and eigenvectors of the covariance matrix, which is the product $\frac{1}{n-1}\mathbf X\mathbf X^\top$, where $\mathbf X$ is the data matrix. Since the covariance matrix is symmetric, the matrix is diagonalizable, and the eigenvectors can be normalized such that they are orthonormal:

$\frac{1}{n-1}\mathbf X\mathbf X^\top=\frac{1}{n-1}\mathbf W\mathbf D\mathbf W^\top$

On the other hand, applying SVD to the data matrix $\mathbf X$ as follows:

$\mathbf X=\mathbf U\mathbf \Sigma\mathbf V^\top$

and attempting to construct the covariance matrix from this decomposition gives
$$
\frac{1}{n-1}\mathbf X\mathbf X^\top
=\frac{1}{n-1}(\mathbf U\mathbf \Sigma\mathbf V^\top)(\mathbf U\mathbf \Sigma\mathbf V^\top)^\top
= \frac{1}{n-1}(\mathbf U\mathbf \Sigma\mathbf V^\top)(\mathbf V\mathbf \Sigma\mathbf U^\top)
$$

and since $\mathbf V$ is an orthogonal matrix ($\mathbf V^\top \mathbf V=\mathbf I$),

$\frac{1}{n-1}\mathbf X\mathbf X^\top=\frac{1}{n-1}\mathbf U\mathbf \Sigma^2 \mathbf U^\top$

and the correspondence is easily seen (the square roots of the eigenvalues of $\mathbf X\mathbf X^\top$ are the singular values of $\mathbf X$, etc.)

In fact, using the SVD to perform PCA makes much better sense numerically than forming the covariance matrix to begin with, since the formation of $\mathbf X\mathbf X^\top$ can cause loss of precision. This is detailed in books on numerical linear algebra, but I'll leave you with an example of a matrix that can be stable SVD'd, but forming $\mathbf X\mathbf X^\top$ can be disastrous, the Läuchli matrix:

$\begin{pmatrix}1&1&1\\ \epsilon&0&0\\0&\epsilon&0\\0&0&\epsilon\end{pmatrix}^\top,$

where $\epsilon$ is a tiny number.

There is a way to do an SVD on a sparse matrix that treats missing features as missing (using gradient search). I don't know any way to do PCA on a sparse matrix except by treating missing features as zero.


Differentiability of piecewise function using the definition analytically In the following example particularly
$$
f(x)= \begin{cases}  (x-2)^2  + 5 \quad\text{when $x\geq 2$} \\
   (x-2)^2  + 4 \quad\text{when $x<2$} \end{cases}
$$
for the above function I know quite well graphically that the function is not differentiable as the when lim $h\to 0^-$ of difference quotient is approaching infinity.

But my question now is how to prove that analytically without graph (animation). In other words if we differentiate LHS and RHS at $x=2$ we will find the same value which is $2(x-2)=0$

For example for $f(x)=|x|$ in many videos on the internet, the tutor differentiate both sides independently but if we do this in my example it will yield the same real number which is zero though the left hand limit shouldn't exist?

Whether or not this will help you depends on what the exact posing of the problem is, but you might find it helpful to recall that a function that is not continuous at a point cannot be differentiable at that point.


discontinuity of an operator I want to show that if $X=Y$ is the subspace of $L^1(0,1)$ over $\mathbb C $ consisting of all polynomials, then $T:X \times Y \rightarrow \mathbb C$ given by

$T(f,g)=\int_0^1 f(t)g(t) dt$

is not continuous.

I have tried the following:

$T$ is surely bilinear, hence if it is continuous it is bounded.
I am not given the norm in $X\times Y$ but I then proceed saying take $x_n=(n+1)x^n$

$|T(x_n,x_n)|\rightarrow \infty $ for $n\rightarrow \infty$

But if T were bounded, $|T(x_n,x_n)|\leq M ||(x_n,x_n)||_{X\times Y}$

now I would imagine we can bound the norm in the product space by a sum of the norms in $L^1$ but I am not sure how to go about it.
If we can get this bound then we are done since $||x_n||_X=1 \ \ \ \forall n$

If you could help filling the gap or find a mistake in my reasoning it would be extremely helpful!

For normed spaces $X,Y$, the continuity or boundedness of a bilinear form

$$T \colon X\times Y \to \mathbb{K}$$

is equivalent to the existence of a constant $M \geqslant 0$ with

$$\lvert T(x,y)\rvert \leqslant M\cdot \lVert x\rVert\cdot \lVert y\rVert\tag{1}$$

for all $x\in X$, $y\in Y$. That $(1)$ implies continuity is easily verified, and conversely, by continuity, there is  a neighbourhood $W$ of $(0,0)$ such that $\lvert T(x,y)\rvert  \leqslant 1$ for all $(x,y) \in W$. By the definition of the product topology, there are $\alpha, \beta > 0$ such that $B_\alpha \times B_\beta \subset W$, where $B_r$ is the (closed, but that's not important) ball with radius $r$ and centre $0$ in the respective space. Then one can verify that

$$\lvert T(x,y)\rvert \leqslant \frac{1}{\alpha \beta}\lVert x\rVert\cdot \lvert y\rVert$$

for all $x,y$.

Thus, the chosen $x_n(t) = (n+1)t^n$ show that the given $T$ is not continuous.

You are right, the norm on the product space is not given, so one would assume that it is a $p$-norm:
$$
|| (x,y) || =\left(||x||^p + ||y||^p\right)^{\frac 1 p}
$$
which are all equivalent each other. Hence you can assume $p=1$ even if usually $p=2$ is the preferred choice.


Sin properties - Why is this undefined? Why is sin(-π/2)^(1/2) undefined? sin(-π/2) is -1, so I thought the the answer would be the same, -1^(1/2)=-1?  But my calculator and the answers sheet both say undefined...

Why do you think the principal square root of $-1$ is $-1$? That would suggest that $(-1)^2=(-1)\times(-1)=-1$ when clearly $(-1)\times(-1)=+1$. Instead, we have $(-1)^\frac12=i$ -- have you learned about imaginary numbers yet?

What everyone else is saying is -1^(1/2) is undefined, because anything ^(1/2) means it's the square root. (AKA x^(1/2) = sqrt(x)). (-pi/2) is negative so trying to get a square root of it doesn't work, and is therefore undefined.


Inverse of a matrix having zeroes in diagonal and one elsewhere Could any one help me to find inverse of such matrix? I observed that $A=  J-I$, where J is a matrix having all entries 1.
Thanks for helping.

Note that the rank of $J$ is 1 because its range is spanned by $(1, \cdots, 1)$. Moreover $(1, \cdots, 1)$ is an eigenvector with eigenvalue $n$. Thus  eigenvalues of $J$ are $n$ and $0$ (with multiplicity $n-1$). So the eigenvalues of $A$ are $n-1$ and $-1$ (with multiplicity $n-1$). The minimal polynomial of $A$ is then
\begin{align*} (A-(n-1)I)(A+I)&=0\\
 A^2-(n-2)A-(n-1)I&=0\\
 A(A-(n-2)I)&=(n-1)I \\
 A^{-1} &=\frac{1}{n-1}(A-(n-2)I)\end{align*}

If your $A$ is an $n\times n$ matrix, for $n\geq2$, the inverse is
$$
\frac{1}{n-1}(A-(n-2)I)
$$


Minimum radius of N congruent circles on a sphere, placed optimally, such that the sphere is covered by the circles? What is the minimum (radius/radii/range of such) of N congruent circles that are placed (optimally) on a sphere in such a way that they cover the entire surface of the sphere?

For 2 circles, a easier visualization would be having two points, growing them into circles and to keep growing them until they touch each other... Then keep growing them until the... 'tunnel-thing' of area covered by the circles is as... wide as the circumference of the sphere that you want, or something...

Tarnai and Gaspar (1991) state that

  How must a sphere be covered by $n$ equal circles so that the angular radius of the circles will be as small as possible?  In this paper, conjectured solutions of this problem for $n=15$ to $20$ are given and some sporadic results for $n>20$ $(n=22,26,38,42,50)$ are presented.

and

  Solutions have been established for $n=2$ to $7$ and for $n=10,12,14$; conjectured solutions exist for $n=8,9,11,13,16,20,32,72,122,132$

and they provide the references.

Reference:

T. Tarnai and Zs. Gáspár, Covering a sphere by equal circles, and the rigidity of its graph, Mathematical Proceedings of the Cambridge Philosophical Society, Volume 110, Issue 01, July 1991, pp. 71-89.

Start with in-circles of faces of the only possible 5 Platonic solids. What you say "Circles" are spherical caps. Unless the regular polygons grow at their corners the sphere cannot be fully covered.


Stochastic Calculus - Ito decomposition I have got one question about Ito decomposition. Suppose $W_t$ is a Brownian Motion:

$X_t = W_t^2 + \int_0^t(W_t^3-1)du$

How to get $dX_t$? I am quited comfused by the integral. Should we calculate the integral first since it is just a normal integral. Or we take differential on both sides like:
$dX_t=dW_t^2+(W_t^3-1)du$ and use Ito's formula to calculate $dW_t^2$? If it is the latter case, why can't we do the integral first? If it is the former case, why can we do the differential first? Thanks so much!

By the very nature of paths of unbounded variation (such as those of Brownian motion), an expression such as $dW_t$ must be interpreted in integral form. In other words, when someone writes an expression of the form
$$dX_t = b dt + \sigma dW_t, \quad t \geqslant 0,$$
what they really mean is
$$ X_t - X_0 = \int_0^t b ds + \int_0^t \sigma dW_s, \quad t\geqslant 0.$$
The differential notation is shorthand for the integral form and is defined to be the process $X_t$ which satisfies the integrated form in a suitable sense.

To answer your question, you can do either. That is, you can compute the integral $\int_0^t (W_s^3 - 1) ds$ directly, and then apply Itô's formula when applying the differential (as you must do with the $W_t^2$ term). However, if all you want is to write the process in differential form, there is no need to do this. Indeed,

\begin{align*}
d(X_t) &= d(W_t^2) + (W_t^3 - 1)dt \\
&= 2W_t dW_t + dt + (W_t^3 - 1)dt \\
&= 2W_t dW_t + W_t^3 dt.
\end{align*}
Again, in the above, the first equality only makes sense because
$$W_t^2 - W_0^2 = \int_0^t 2W_s dW_s + \int_0^t ds.$$

X(t) = F(X,t) = X^2 + (X^3-1)t

Now you can use taylor expansion in terms of t and x and 1/2x^2
thats it.


For $a>1$, show that $\frac{1}{1+x}-\frac{1}{1+ax} \leq \frac{\sqrt{a}-1}{\sqrt{a}+1}$, $x \geq 1$ I'm self-learning the analysis I "by Herbert Amann" and got stuck in this problem. It's in Chapter IV Taylor's theorem.

For $a>1$ and $x\geq 1$ show that $$\frac{1}{1+x}-\frac{1}{1+ax} \leq \frac{\sqrt{a}-1}{\sqrt{a}+1}$$

This is what I've tried:
Let $f(t)=\frac{1}{1+tx}$ which is convex and,
$f(a) \geq f(1)+f'(1)(a-1)$
so $\frac{1}{1+x}-\frac{1}{1+ax} \leq \frac{x}{(1+x)^2}(a-1) \leq \frac{a-1}{4}$

Isn't Taylor an overkill here? Given is equivalent to
$$\frac{x(a-1)}{(1+x)(1+ax)} \leq \frac{a-1}{(\sqrt{a}+1)^2}$$ or
$$x(\sqrt{a}+1)^2\leq {(1+x)(1+ax)}$$ or
$$xa+2x\sqrt{a}+x \leq 1+x+ax+ax^2$$
or $$2x\sqrt{a} \leq 1+ax^2$$
which is true since $$0\leq (1-x\sqrt{a})^2$$ and we didn' t even use $x>1$, but it must be $> -{1\over a}$.


Show that the difference quotient of $1/x^n$ exists Let $n>0$ be a positive integer. For all $x\not=0$, prove that $f(x) = 1/x^n$ is differentiable at $x$ with $f^\prime(x) = -n/x^{n+1}$ by showing that the limit of the difference quotient exists.

I am having trouble seeing how I can manipulate the difference quotient in order to get a limit that exists

so far I have

$$f^\prime(x)= \lim_{h\rightarrow 0} \frac{f(x+h) - f(x)}{h} =\lim_{h\rightarrow 0} \frac{1/(x+h)^n - 1/x^n}{h}$$

All help is much appreciated.

Let $x\not=0$. Use the binomial theorem:

\begin{align}
\frac{\frac{1}{(x+h)^n}-\frac{1}{x^n}}{h} &=-\frac{(x+h)^n - x^n}{(x+h)^n x^n h}\\
&=-\frac{1}{h (x+h)^n x^n} \left(\sum_{k=0}^n {n\choose k} x^k h^{n-k} - x^n\right)\\
&=-\frac{1}{h (x+h)^n x^n} \sum_{k=0}^{n-1} {n\choose k} x^k h^{n-k}\\
&=-\frac{1}{(x+h)^n x^n} \sum_{k=0}^{n-1} {n\choose k} x^k h^{n-k-1}
\end{align}

Now we can safely let $h\rightarrow 0$. This makes the sum collapse to the summand with $k=n-1$ and we get

$$f^\prime(x)=-\frac{1}{x^{2n}} {n\choose n-1} x^{n-1}=-\frac{n}{x^{n+1}}$$

Collect fractions and use the binomial theorem.


Simple group with Klein four Sylow If $G$ is a simple group, with a Sylow $2$-subgroup isomorphic to the Klein four group $\mathbb{Z}_2 \times \mathbb{Z}_2$, then I want to show that any two involutions in a given Sylow $2$-subgroup are conjugate in $G$.

Any help would be appreciated.

Note: A solution/method in relatively elementary terms would be appreciated — I am alright with some machinery as long as you explain it.

I think you gate this from Sylow's 2nd theorem


Prove that the dual (ie, "midpoint polygon") of an isosceles trapezoid is a rhombus. Prove that the dual of an isosceles trapezoid is a rhombus.

Here, the "dual" of any polygon is where its sides intersects the midpoint of each side of the "outer" figure. (In other words, it's the "midpoint polygon".)

I'm aware of the properties of an isosceles trapezoid:

By definition, the legs are congruent.
The lower and upper base angles are congruent.
Any lower base angle is supplementary to any upper base angle.

Here is my drawing of the dual polygon in the isosceles trapezoid:

I showed that $\triangle FAE\cong\triangle HDE$ by SAS, and that $\triangle GBF\cong\triangle GCH$ by SAS. So by CPCTC, $\overline{FE}\cong\overline{HE}$, and $\overline{FG}\cong\overline{GH}$.

  I'm unsure of how to show that all of the sides of the dual quadrilateral are congruent, making it a rhombus.

By properties of the midpoint polygon, each side of the midpoint polygon is parallel and equal to half the length of a diagonal (why?). An isosceles trapezoid has diagonals of equal length, hence the result.

   Consider the triangle $\Delta ABC$. As $AF=FB$ and $BG=GC$, we get $\Delta BFG\sim\Delta BAC$ and the similarity ratio is a half. The other sides of the midpoint polygon follows somewhat similarly.

Compare with Varignon's theorem.

Since G and E are the base midpoints of the isosceles trapezoid, BC $\perp$ GE. Since  F and H are the side midpoints of the isosceles trapezoid, BC || FH. Therefore, GE $\perp$ FH.

Since GOHC and GEDC are similar, GO:OE = CH:HD = 1:1. Also, FO = OH = $\frac14$ (BC + AD). Therefore, O bisects the perpendicular GE and FH, which leads to that the four right triangles are congruent, hence a rhombus.


How can a finite set not have a maximum or minimum? Consider a set of real numbers such that $0\le x<1$. My book (Tom Apostol) says that this set do not have any maximum. How is it possible? Isn't the number just below $1$ the maximum of the set? Isn't that the least upper bound (supremum)? Why is $1$ the supremum?

The set of real numbers in $[0,1)$ isn't finite, it is bounded.
There is no real number "just below" 1. Give any real number $\alpha < 1$
and $\alpha + \frac{(1-\alpha)}{2}$ is also strictly less than 1, and strictly greater than $\alpha$.

For any real number a which is in S.
There exists another number b which is larger than a: b=(a+1)/2.
So there is no maximum.


Class of $p_i$ such that $\sum_{i=1}^n \frac{1}{p_i} = l$ for some $l$. Let $l$ be any integer and $p_i$ belong to natural numbers for $1 \leq i \leq n$. Suppose they satisfy the following equation $$\sum_{i=1}^n \frac{1}{p_i} = l$$ Suppose we are given $l$ and $n$ is fixed. What can we conculde about $p_i$? More specifically, can we say any class of $p_i$ which statisfies the above equation. Or Is there any we can say explicitly which $p_i$ satifies the equation.

Given $n,l$ enumerate all the $p_1\le p_2\le \ldots \le p_n$ such that $$l=\sum_{i=1}^n \frac1{p_i}$$

Due to the $\le$ restriction given $p_1\ldots p_{j-1}$ we must have $p_j\le \frac{n-j}{l-\sum_{i=1}^{j-1} \frac1{p_i}}$. This gives the naive algorithm :

Starting with $j=1$
If $j\le n$ let $r=\sum_{i=1}^{j-1} \frac1{p_i}$

If $r=l$ then return
Otherwise for each $p_j\le \frac{n-j}{l-r}$, enumerate recursively the solutions $p_{j+1}\le \ldots \le p_n$

If $j=n+1$

if $\sum_{i=1}^n \frac1{p_i} \ne l$ then return
otherwise enumerate all the permutations of $p_1,\ldots,p_n$


How to solve $\lim\limits_{n \to \infty} \frac{\sum\limits_{k=1}^{n} k^m}{n^{m+1}}$ I'm trying to get the limit of $\lim\limits_{n \to \infty} \frac{\sum\limits_{k=1}^{n} k^m}{n^{m+1}}$
I know that $\sum\limits_{k=1}^{n} k^m$ is a polynom of $\deg(m +1)$ by the Euler-MacLaurin formula. And that I may apply L'hôspitals rule $m+1$ times.
So there will be some constant in the denominator and the numerator I guess? Maybe something like $(m+1)!$
But I can't find out what exactly yields the $m+1$ the derivitive of that sum. Has anybody a tipp on how to solve this?
Edit: I forgot the $+1$, sry

Hint:
$$\frac{1}{n}\sum_{k=1}^n\left(  \frac{k}{n} \right)^m \approx  \int_{0}^{1}x^m\text{d}x $$
The above is true due to Riemann sum & how we define the integral. Therefore: $$ \lim_{n \to  \infty }   \frac{1}{n} \sum_{k=1}^{n} \left(  \frac{k}{n} \right)^m= \int_{0}^{1}x^m\text{d}x= \frac{1}{m+1} $$


Is the following set connected given that the union and intersection is connected Suppose $U_1, U_2$ are open sets in a space $X$. Suppose $U_1 \cap U_2$ and $U_1 \cup U_2$ are connected. Can we conclude that $U_1$ must be connected??

I am trying to find a counterexample, but I failed. PErhaps it is true? Can someone help me find a counter example? thanks

Suppose $A$ and $B$ form a separation of $U_1$, i.e., $A$ and $B$ are disjoint nonempty open sets such that $A\cup B = U_1$. Because $U_1 \cap U_2$ is a connected subset of $U_1$, it must be entirely contained in either $A$ or $B$ (else we would get a separation of $U_1 \cap U_2$ by intersecting $A$ and $B$ with $U_1 \cap U_2$); WLOG let $U_1 \cap U_2 \subset A$. Then, $A\cup U_2$ and $B$ forms a separation of $U_1 \cup U_2$, a contradiction.

I'm assuming all of the above are non-empty.

If we can't separate $U_1 \cap U_2$ into disjoint sets, how could we possible separate $U_1$ into disjoint sets?

That is, suppose for contradiction that $U_1 = S_1 \sqcup S_2$ (which is to say $U_1 = S_1 \cup S_2$ with $S_1 \cap S_2 = \emptyset$). Then we have

$$U_1 \cap U_2 = (S_1 \sqcup S_2) \cap U_2 = (S_1 \cap U_2) \sqcup (S_2 \cap U_2).$$ But then we've just written $U_1 \cap U_2$ as a disjoint union, contradicting our assumption that $U_1 \cap U_2$ is connected.