TITLE: $\|T\|= \sup_{\| f\|=1} |\langle Tf,f \rangle|$ for complex Hilbert spaces
QUESTION [1 upvotes]: Let $H$ be a hilbert space over the complex numbers. Let $T:H\to H$ be a linear bounded operator prove that $$\| T\|= \sup_{\| f\|=1} |\langle Tf,f \rangle|.$$

Obviously $\sup_{\| f\|=1} |\langle Tf,f\rangle| \le \| T\|$ by C-S inequality. But how about the other inequality?

REPLY [3 votes]: As user8268 points out, this is wrong even in complex hilbert spaces.
But there is at least the following substitute:
Define $B(u,v) := \langle Tu, v\rangle$. Then $B$ is sesquilinear (i.e. linear in the first argument and antiliear in the second). Define $Q(u) := B(u,u)$. The sesquilinearity of $B$ implies the polarization identity
$$B(u,v) = \frac{1}{4} \sum_{j=0}^3 i^j Q(u + i^j v),$$
where $i = \sqrt{-1}$.
We thus derive
$$|\langle Tu, v\rangle| = |B(u,v)| \leq \frac{1}{4} \sum_{j=0}^3 |Q(u + i^j v)|.$$
The right hand side can be estimated by $4 \cdot \sup_{\Vert f \Vert = 1}|\langle Tf, f\rangle|$, because $\Vert u + i^j v\Vert \leq 2$ if $\Vert u \Vert, \Vert v \Vert \leq 1$.
Then use $\Vert Tu \Vert = \sup_{\Vert v \Vert \leq 1} |\langle Tu, v\rangle|$.


TITLE: Prove $A$ and $B$ are equivalent if and only if $\text{rank}(A) =\text{rank}(B)$
QUESTION [6 upvotes]: Knowing that $A$ is equivalent to $B$ if there exists an invertible $m\times m$ matrix $P$ and an invertible $n\times n$ matrix $Q$ such that $PAQ = B$, how can I prove that $A$ and $B$ are equivalent iff $\text{rank}(A) =\text{rank}(B)$?
I've managed to solve the forward direction of the iff and am confident it is correct:
Suppose $A$ and $B$ are equivalent. Then, $PAQ = B$. Knowing this, we can assume
$$
\text{rank}(PAQ) \leq \text{rank}(A) = \text{rank}(P^{-1} B Q^{-1}) \leq \text{rank}(B)
$$
As $\text{rank}(PAQ) = \text{rank}(B)$, all inequalities must be equalities, so $\text{rank}(A) =\text{rank}(B)$.
I am not sure how to prove this statement in the reverse direction. I think that the invertible matrix theorem could be useful for this problem

REPLY [0 votes]: For one way, the answer is correct.
The other way, let $A$ and $B$ be matrices which are equivalent.
Note that the rank of $A$ is equal to the rank of $B$, therefore the dimensions of the image of $A$ and the image of $B$ are the same. Let $k$ be the rank of $A$(and of $B$). Let $\{v_i\}_{i=1,...,k}$ and $\{w_i\}_{i=1,...,k}$  be bases for the images of $A$ and $B$ respectively. Note that $k \leq m,n$ by the fact that row rank equals column rank. So,  we complete the bases of the images, to bases for $\mathbb R^m$, and to avoid confusion, the completed bases are $\{v_i\}_{i=1,...,n}$ and $\{w_i\}_{i=1,...,n}$.
Now, we want matrices $P,Q$ such that $PAQ = B$. Think of it this way : $Q$ rewrites the input vector of $B$ in a manner convenient for $A$. Then $A$ does its job on the rewritten vector which it finds easy to work with, and then $P$ rewrites the output of $A$ in a way which $B$ would have written it. That is the break up of the jobs of $P$ and $Q$ : they are one way translators from the input/output language of $A$ to the language of $B$ and vice-versa, if you like.
For each $v_i$, pick a single preimage $e_i$, and for each $w_i$, pick a preimage $f_i$. Now, $\{e_i\}_{i=1,...,k}$ and $\{f_i\}_{i=1,...,k}$ are linearly independent sets (check!) so they can be completed to bases of $\mathbb R^n$. Without confusion, we will call these bases as $\{e_i\}_{i=1,...,n}$ and $\{f_i\}_{i=1,...,n}$ respectively.
Now, what has to be thought here, is that $\{e_i\}$ is like $A$'s mother tongue and $\{f_i\}$ is like $B$'s mother tongue.
Therefore, the task of $Q$, when it receives a vector that is $B$'s mother tongue, is to convert it into $A$'s mother tongue. That leads to a very simple answer : $Q$ is the basis conversion matrix from $f_i$ to $e_i$. That is, $Q$ is the matrix of the unique linear transformation that satisfies $Q(f_i) = e_i$ for all $i = 1,...,n$.
Now that $Q$ has done its job, $A$ receives input in its mother tongue, so it outputs some vector whose entries are in the basis of $\{v_i\}$.
But $B$ outputs in the basis $\{w_i\}$! Obviously, it is clear that $P$, then, must be the unique linear transformation with $P(v_i) = w_i$.
So $Q$ and $P$ are just basis transformation matrices.
Finally, we can provide a proof that $PAQ = B$. Let $x \in \mathbb R^n$.
Then, $x = \sum_{i=1}^n x_if_i$, so $Bx = \sum_{i=1}^k x_iw_i$.
Alternately, by the nature of $Q$, $Qx = \sum_{i=1}^n x_ie_i$. Now, given what $A$ does, $AQx = \sum_{i=1}^k x_iv_i$, and then by what $P$ does, $PAQx = \sum_{i=1}^k x_iw_i$.
Hence $B = PAQ$. Since $P,Q$ are change of basis matrices, they are clearly invertible. They are also seen to be of the right dimension.
This proves the proposition.


TITLE: Effects of terrain on outdoor sound propagation?
QUESTION [1 upvotes]: I'm looking for a computationally fast model for capturing the effects of varying terrain topology in far-field outdoor acoustic propagation, and was wondering if anyone had any insight or could point me in the right direction. I'm currently aware of Rasmussen's terrain profile methodology, as well as various PE solvers for capturing these effects, but I was hoping for something much faster, even if it's just a really rough first-order model. Even if it's as simple as applying function weightings based on terrain gradient or something similar, that would be excellent.
Do any such models even exist? I'm greatly valuing speed over accuracy, so any low fidelity model would do.

REPLY [0 votes]: Answering my own question: it seems as if ISO 9613 might be the best way to accomplish this. In it, an algebraic methodology for calculating ground attenuation based on average heights AGL from source to receiver is given. Comparisons of this methodology with measurements and more complex PE solvers can also be easily found in literature.


TITLE: Question about proof of Chebychev's inequality
QUESTION [0 upvotes]: I am completely new to measure theory and probability theory, my question may be silly but I am confused. Let me first state the theorem:
Let $X$ and $Y$ be two random variables on a probability space $(\Omega, B, P)$ where $\Omega$ is a sample space, $B$ is a Borel set and $P$ is a probability measure. Then for all $\lambda>0$,
$P(|X|>\lambda)\leq\frac{E(|X|)}{\lambda}$.
The proof goes like this:
As $|X|$ is a non-negative random variable, $E(|X|)$ makes sense. Note that
$$
\begin{aligned}
E(|X|) &= \int_{|X|>\lambda}|X|dP + \int_{|X|<\lambda}|X|dP \\
&\geq \int_{|X|>\lambda}|X|dP \\
&\geq \int_{|X|>\lambda}\lambda dP \\
&\geq \lambda P(|X|>\lambda)
\end{aligned}
$$
My questions are:
(1) Why there is a random variable $Y$ in the statement of the theorem?
(2) I know that $E(X)$=$\int_{\Omega}XdP$, But I don't understand how they have written the integral over $|X|>\lambda $ and $|X|<\lambda$; maybe I am missing something here.
What I have understood is that basically $|X|>\lambda$ means $\{\omega\in \Omega$ such that $X(\omega)>\lambda\}$ so with the help of this and $|X|<\lambda$ we are covering all of $\Omega$. I would be thankful if someone answered these two questions.

REPLY [1 votes]: To answer your questions:

I don't know why there is a random variable $Y$ in the statement of the theorem; it should not be there at all. Also, if $(\Omega, B, P)$ is your probability space, $B$ should not be a Borel set, but it should be a event space, i.e., $\sigma$-algebra.

They can not write the integral like this. You were right about them splitting up the sample space $\Omega$ into two sections, the problem is that they forgot where $|X|=\lambda$. This is not like Riemann integration, where we can forget about "singleton values" (loosely speaking), because here we are talking about values in the range, not the domain.

Here is a correct statement of the theorem and modified version of your proof (the inequality of your statement was correct, but I strengthened it):
Theorem (Chebychev). If $X$ is a random variable on a probability space $(\Omega,B,P)$, then $P(|X|\geq \lambda)\leq\frac{E(|X|)}{\lambda}$.
Proof. We have $$
\begin{aligned}
E(|X|)&=\int_{|X| \geq \lambda} |X| dP + \int_{|X| < \lambda} |X| dP \\
&\geq \int_{|X| \geq \lambda} |X| dP \\
&\geq  \int_{|X| \geq \lambda} \lambda dP \\
&\geq \lambda P(|X| \geq \lambda),
\end{aligned}$$
so divide through by $\lambda$ to get the result.


TITLE: Show $\sqrt{-n}$ is irreducible in $R = \mathbb{Z}[\sqrt{-n}]$
QUESTION [2 upvotes]: Let $R = \mathbb{Z}[\sqrt{-n}]$ where $n$ is a square free integer greater or equal than $3$. Prove that $2$ and $\sqrt{-n}$ are irreducible in $R$.

Proof. Recall that if $R$ is an integral domain, and $r \in R$ and is not a unit, then $r$ is called irreducible in $R$ if whenever $r = ab$ with $a,b\in R$, at least one $a$ or $b$ must be a unit in $R$.
Then let $2 =  xy$. Then taking the norm, we have $4 = N(x)N(y)$.
Then we need to show that $N(x)=1$ and $N(y) = 4$ or $N(y)=1$ and $N(x) = 4$.
Suppose $x = a + b\sqrt{-n}$. Then assume $N(x) = 2$. Then $x = a + b\sqrt{-n}$ implies $N(x)= 2 = a^2 + nb^2$  . Since $n \ge 3$, then equation only makes sense if $b = 0$ and $a = \sqrt2$ . So we have a contradiction, so $N(x) $ =  $4$ and $N(y)$ must be $1$, so that $y$ is a unit.
So $2$ is irreducible.
Now for $\sqrt{-n}$, suppose $\sqrt{-n} = xy$. Then taking the norm, we have, $n = N(x)N(y)$. Then let $\sqrt{-n} = xy = (a + b\sqrt{-n})(c + d\sqrt{-n})$. Then $n = (a^2 + nb^2)(c^2 + nd^2)$ .
Can someone please help me for $\sqrt{-n}$ ? I am stuck.
Thank you!

REPLY [3 votes]: If $b\neq 0$, then $a^2+nb^2 \geq n$. The equality occurs only when $a=0$ and $b=\pm 1$. In which case the factorization you have is the trivial one $\sqrt{-n}=\pm\sqrt{-n}(c+d\sqrt{-n})$. Thus $c=\pm 1,d=0$.
If $b=0$, then $n=(ac)^2+(ad)^2n$, using the same reasoning as above, we can conclude that $ac=0$ and $ad=1$. Consequently $c=0$ and $a=d=\pm 1$. Which again gives a trivial factorization.
Note: In the second case $d \neq 0$, otherwise $n$ will not be square free.


TITLE: Fourier series with positive indices defines a complex function?
QUESTION [0 upvotes]: Consider the following Fourier series
$ f\left(t\right)=\sum_{k=1}^{N}c_{k}e^{i\frac{2\pi}{T}kt} $
Can we prove that the function which its Fourier series is $ f $ is a complex function due to the fact that its Fourier series only contains positive indices?
Also, are there other cool properties of a function that we can tell just from looking at its Fourier series? I know that for real functions we can tell if the function is odd/even by looking at the Fourier series in the form of the $ \sin + \cos $ orthogonal system. Are there any other properties?
Thanks in advance.

REPLY [2 votes]: If $f(t)$ is real valued then $$\sum_{k=1}^{N}c_{k}e^{i\frac{2\pi}{T}kt} =\sum_{k=1}^{N}c_{k}e^{-i\frac{2\pi}{T}kt} $$ and orthogonality of the functiosn $e^{i\frac{2\pi}{T}kt}$, $k \in \mathbb Z$ on $[0,T]$ show that each $c_k$ must be $0$.


TITLE: Number of partitions of a number
QUESTION [2 upvotes]: I've been trying to find out how many partitions of a number I can make if I restrict the partition size, The specific problem I am trying to tackle is,
How many ways can you partition the number '9' when each partition has a size less than or equal 6 but greater than 0 (note: all partitions have a discrete size ( can't have a decimal or fractional size)
My attempt:
let A+B+C=9 and let $(A,B,C)\subset S^3$ where $S= (x: where \ 6\ge x\ge 0)$
and then I did the stars and bars approach to finding the number of partitions $\binom{11}{2} $ but I am stuck on how to remove the partitions of where one of the partitions has is greater than 6. Is there any way to directly arrive at the answer required or more preferably a way to fix this formula?

REPLY [0 votes]: You want the number of partitions of $9$ into any parts, minus those with a part of size $7$, $8$ and $9$. So the number of partitions of $9$ is simply $30$. The number of partitions with a part of size at least $7$ is clearly
$$(9),\;\;(8,1),\;\;(7,2),\;\;(7,1,1)$$
so four. Thus there are $26$ partitions remaining.


text:
{\bf Problem.} How many of the first $1000$ positive integers can be expressed in the form
\[\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor\]where $x$ is a real number, and $\lfloor z \rfloor$ denotes the greatest integer less than or equal to $z$?
{\bf Level.} Level 4
{\bf Type.} Intermediate Algebra
{\bf Solution.} Let $f(x)$ be the given expression. We first examine the possible values of $f(x)$ for $x$ in the interval $(0, 1].$ Note that $f(0) = 0,$ while $f(1) = 2 + 4 + 6 + 8 = 20.$

As we increase $x$ from $0$ to $1,$ each of the four floor functions "jumps up" by $1$ at certain points. Furthermore, if multiple floor functions "jump up" at the same value of $x,$ then some integers will be skipped.

For each $k,$ the function $\lfloor kx \rfloor$ "jumps up" at $x = \tfrac{1}{k}, \tfrac{2}{k}, \ldots, \tfrac{k-1}{k}, \tfrac{k}{k}.$ Therefore, we see that at $x = \tfrac{1}{2}$ and $x = 1,$ all four of the given functions "jump up," so that three integers are skipped. Also, for $x = \tfrac{1}{4}$ and $x =\tfrac{3}{4},$ the functions $\lfloor 4x \rfloor$ and $\lfloor 8x \rfloor$ both "jump up," skipping one integer.

Thus, for $0 < x \le 1,$ $f(x)$ takes $20 - 3 - 3 - 1 - 1 = 12$ positive integer values. Notice that \[\begin{aligned} f(x+1) &= \lfloor 2(x+1) \rfloor + \lfloor 4(x+1) \rfloor + \lfloor 6(x+1) \rfloor + \lfloor 8(x+1) \rfloor \\ &= \left(\lfloor 2x \rfloor+2\right) + \left(\lfloor 4x \rfloor +4\right)+ \left(\lfloor 6x\rfloor+6 \right)+ \left(\lfloor 8x \rfloor +8\right) \\ &= f(x) + 20. \end{aligned}\]Therefore, in the interval $1 < x \le 2,$ $f(x)$ takes $12$ more integer values between $21$ and $40,$ respectively. In general, $f(x)$ takes $12$ out of every $20$ positive integer values from the list $20a, 20a+1, \ldots, 2a+19.$

Since $20$ is a divisor of $1000,$ exactly $\tfrac{12}{20} = \tfrac{3}{5}$ of the first $1000$ positive integers are possible values for $f(x).$ Thus the answer is $1000 \cdot \tfrac{3}{5} = \boxed{600}.$

set_name: MATH


TITLE: ODE - how to solve this ODE obtained from a geodesic problem?
QUESTION [0 upvotes]: I am stuck at solving this ODE that I obtained by trying to compute geodesics:
$$
\begin{equation}
g(\theta) \frac{d^2 \theta}{dt^2} + \frac{1}{2} \frac{\partial g(\theta)}{\partial \theta} \left( \frac{d\theta}{dt} \right)^2 = 0.
\end{equation}
$$
where
$$g(\theta) = \frac{3(\theta^2+1)}{(\theta-1)^2(\theta+1)^2},$$
$$\frac{\partial g(\theta)}{\partial \theta} = \frac{-6\theta(\theta^2 + 3)}{(\theta-1)^3 (\theta+1)^3}$$
Any help? (It has been several years that I have not been confronted to differential equations... Thank you)
We can assume $\theta(0) = \theta_0$ and $\theta(1) = \theta_1$ as initial conditions.
In SAGEMath, I obtain:
-1/2*(sqrt(2)*arcsinh(2*(theta(t) - 1)/abs(2*theta(t) + 2)) + sqrt(2)*arcsinh(2*(theta(t) + 1)/abs(2*theta(t) - 2)) - 2*arcsinh(theta(t)))/_K1 == _K2 + t

REPLY [2 votes]: Sage gave you the solution (in implicit form).  Maple's is essentially the same:
$$ {\rm arcsinh} \left(\theta\right)-\frac{\sqrt {2}}{2}{\rm arctanh} \left(
\,{\frac {1+\theta}{\sqrt {2\,{\theta}^{2}+2}}}\right)+\frac{\sqrt
{2}}{2}{\rm arctanh} \left({\frac {1-\theta}{\sqrt {2\,{\theta}^{2}
+2}}}\right)-{c_1}\,t-{c_2}=0
$$
Write this as $f(\theta) = c_1 t + c_2$.  The boundary conditions give you the values of $c_1$ and $c_2$: $c_2 = f(\theta_0)$, $c_1 = f(\theta_1) - f(\theta_0)$.
The graph of $f(\theta)$ looks like this:

Note the singularities at $\theta = \pm 1$, where $f(\theta) \to \pm \infty$.
On each of the intervals $(-\infty, -1)$, $(-1,1)$ and $(1,+\infty)$ the function is monotone, so there will be a unique solution to your initial value problem if $\theta_0$ and $\theta_1$ are both in one of those intervals.  If $\theta_0$ and $\theta_1$ are in different intervals, there is no solution.  You might have guessed that from the fact that $g(\theta)$ is singular at $\theta = \pm 1$.