Mathematics - 2017-06-28 (page 4 of 7)

Hey if I just read that the vector space $H$ of all homogenous solutions of a linear differential equations system fulfils: $dim H \le dim \mathbb{C}^n$. When having constant coefficients supposedly we have $dim H = dim \mathbb{C}^n$, so it just with non-constant coefficients that we get a lower dimension? Is there aneasy example?

Astyx

Example of what ?

In dimension one, you can easily take $y'=y$ and $y' = 2x y$

Felix.C

of a linear differential equation where $dim H < dim \mathbb{C}^n$

Astyx

In both case $H$ is of dimension 1

I guess you take the coefficient of the derivative of highest order to be 1

Otherwise, for instance $xy' =1$ has no solutions on $\Bbb R$

Dattier

that's not a linear equation

Astyx

However if you write $y^{(n)} + \sum_{k=0}^{n-1}f_k(x) y^{(k)} = 0$, then by Cauchy Lipschit the space of solutions is of dimension $n$ (isomorphic to $\Bbb C^n$)

My bad, it isn't

Dattier

2:42 PM

no y

all constante

Felix.C

@Astyx You example is of first order and has one solution, but I want like a linear equation of n-th order that has k<n linear independant solutions

Dattier

dim 1

Astyx

I'm confusing myself

Felix.C

and I'm somewhat sure that this is possible only if all coefficients are non-constant, but I'd would like to see an example

but thank you

Astyx

Yeah, if you take $xy' = (x+2)y$, when $x\ne 0$ you get $y' = (1+{2\over x})y$ of which a solution on $\Bbb R_+^*$ (and $\Bbb R_-^*$) is $y(x) = x^2e^x$

So the dimension of the space of solutions is $2$

Because any function $$f_{\alpha, \beta} : x\mapsto\begin{cases}\alpha x^2e^x \text{ if }x\ge0\\\beta x^2e^x \text{ otherwise} \end{cases}$$ works

@Felix.C

So $\dim H\gt 1$ which makes me doubt your statement

Sha Vuklia

2:55 PM

Guys, consider
$$
a=b\cdot\frac{n}{d},
$$
where $\gcd(b,d)=1$ and $1\leq b\leq d$, and $1\leq a\leq n$. Does it follow from here that $\gcd(a,n)=n/d$? It’s obvious that $n/d$ is a divisor of both $a$ and $n$. My guess is that if $\gcd(a,n)>n/d$, then we would have a contradiction with $\gcd(b,d)=1$. I know I can prove it the other way around, for if $\gcd(b,d)>1$, then $\gcd(a,n)>n/d$, which is a contradiction.
Say $\gcd(a,n)=k>n/d$. Then we can write $a=b’k$, and we have $b’<b$. But yea, I’m not sure what to do from here on.

user193319

If $(x_n)$ is an increasing sequence of positive numbers, then can I say that $\frac{x_n}{x_{n+1}}$ is monotonic? I know I cannot generally say that the ratio it is increasing, as $x_n = n$ would provide a counterexample to this. But can I say that it will be generally monotonic?

Astyx

No

Take any sequence of numbers $u_n \ge 1$, and define $x_n = \prod_{k=1}^n u_k$

user193319

@Astyx Hmm...So if $(x_n)$ is an increasing sequence of positive numbers, will $\frac{x_n}{x_{n+1}}$ generally converge?

Felix.C

@Astyx I have to think again some minutes about my question...to make an at least in my opninion correct statement ;)

Astyx

No @user193319 with the same argument

@Felix.C Take your time :p

Abhishekstudent

3:09 PM

hi

Chris Nguyen

hi

Abhishekstudent

howdy

Astyx

How do I compute efficiently the determinant of the matrix that has random diagonal coefficients $\alpha_i$, all $a$ above the main diagonal and all $b$ under ?

Felix.C

@Astyx First of all nice example, but you just found one single solution. You mentioned that we have the same solution for $\Bbb R_+^*$ and $\Bbb R_-^*$ that are two differant intervalls, so that doesn't contradict my claim.

Astyx

No, these are solutions in $\Bbb R$

So you have two linearily independant solutions

Semiclassical

3:19 PM

haha, today's xkcd is apropos given our 'threat' conversation earlier:

Real analysis is way realer than I expected.

2

Astyx

Ohi Semi, any idea about my determinant question ? :)

Semiclassical

didn't see it.

Astyx

like, 7 messages above

Semiclassical

Ahh.

Astyx

I believe xkcd spies on this chatroom to get inspiration

It's a classic, but I don't remember it

Semiclassical

3:20 PM

Hmmmm

SteamyRoot

If $a = b = 1$ I know a good solution :P

Semiclassical

so a matrix $L+U+D$ where $L$ is constant lower-triangular, $U$ is constant upper-triangular, and $D$ is some arbitrary diagonal matrix.

Astyx

Right

SteamyRoot

You can, of course, assume either $a$ or $b$ is $1$.

Astyx

I know the answer btw

Just not how to find it :p

Semiclassical

3:23 PM

Think I see a good route via the matrix determinant lemma.

SteamyRoot

Wild guess: $$a^n \left(\prod_{i=1}^n (\alpha_i - b/a) + \sum_{i=1}^n \prod_{j \neq i} (\alpha_j - b/a)\right)$$

Astyx

There surely is a polynomial argument there

Semiclassical

Namely, we can write the matrix as $A+buu^T$ where $u$ is a column vector of ones. (uu^T will be a matrix of ones)

Astyx

Solution is $$b\prod_{i=1}^n(\alpha_i - a) - a\prod_{i=1}^n(\alpha_i-b)\over b-a$$

Semiclassical

consequently the matrix determinant lemma tells us that $\det(A+buu^T)=(1+bu^T A^{-1}u)\det A$.

But $A$ is our initial matrix with every element decremented by B. So that'll be $\alpha_i-b$ for the diagonal entries, $a-b$ above the diagonal, and 0 below the diagonal.

The determinant of $A$ is then immediately $\prod_i^n (\alpha_i-b)$. That leaves the first term to compute.

SteamyRoot

3:28 PM

@Astyx In that case it's pretty easy, no?

Induction

Astyx

Yeah but that's not beautiful :(

Semiclassical

Also, note that $u^T M u$ will be the sum of all the entries of $M$.

SteamyRoot

Why not?

Astyx

I was thinking of considering $M-XJ$ where J is all 1

SteamyRoot

If you just do one row operation it gets really easy.

Semiclassical

3:29 PM

@Astyx That's basically what I'm doing :)

Astyx

This values $\prod \alpha_i$ when $X = a,b$

Semiclassical

with $X=bI$, specifically.

Astyx

Oh, I don't have the insight to realise that

Yeah, I meant X to be the unknow of a polynomial

Semiclassical

Sure.

Alas, I think my approach isn't as good as I'd like---I don't remember how to do matrix inversion in this case.

user193319

@Astyx I don't think that example you gave always works. If $u_n = n \ge 1$ for every $n$, then $x_ n = n!$ and $\frac{x_n}{x_{n+1}} = \frac{1}{n+1} \to 0$.

Astyx

3:33 PM

So ?

user193319

@Astyx You said that any sequence satisfying the condition would work.

Semiclassical

Yeah, I no longer trust my approach.

Astyx

So I said that was an equivalent definition of your sequences. You're asking wether ${1\over u_n}$ always converges.

user193319

@Astyx I am not sure I follow. I was asking whether $\frac{x_n}{x_{n+1}}$ converges, provided $x_n$ is an increasing sequence of positive numbers.

Semiclassical

Good news is, I remembered how to make such a matrix efficiently in Mathematica

SparseArray[{{i_, j_} /; i < j->b, {i_, j_}/;i > j ->a, {i_, i_} -> \[Alpha][i]}, {n, n}]

Astyx

3:37 PM

Yes, I'm saying that's equivalent to asking wether $1\over u_n$ always converges for any $u_n\ge 1$

Semiclassical

where n is the matrix size.

Works efficiently for arbitrarily large $n$ :)

SteamyRoot

Anyway, the induction argument boils down to rewriting
$$(\alpha_n - b)\frac{b\prod_{i=1}^{n-1}(\alpha_i-a) - a\prod_{i=1}^{n-1}(\alpha_i-b)}{b-a} + (\alpha_{n-1}-a) \frac{b\prod_{i=1}^{n-2}(\alpha_i-a)(b-a)}{b-a}$$

Semiclassical

Of course, Array works well here too:

Array[Which[#1 > #2, a, #1 < #2, b, #1 == #2, \[Alpha][#1]] &, {n, n}]

SteamyRoot

Which you can almost do on sight :P

Sha Vuklia

Guys, if $\gcd(m,n)=k$, how can I show that $\gcd(m,n/k)=1$? I tried using contradiction: assume $\gcd(m,n/k)=l>1$. Then we know that $l\mid m$ and $lk\mid n$. It seems to me I should be able to get that some number $r>k$ divides $m$, but I don't know how. Any ideas?

Astyx

3:42 PM

Oh right, polynomial approach destroys the problem

You can prove $\det (M-XI)$ is affine

And you can easilly compute it for $X = a,b$

So you have the value for $X=0$

Semiclassical

If gcd(m,n)=k then that means k|m, k|n, right?

Martin Sleziak

@ShaVuklia This is not true, try m=3 and n=9.

Semiclassical

(This is not a socratic question, this is me genuinely not remembering things)

Sha Vuklia

oh huh

I got it from here

Martin Sleziak

Probably you meant $\gcd(m/k,n/k)=1$?

Semiclassical

3:44 PM

heh.

Sha Vuklia

I guess I misinterpreted it

but I try to show that $\#S(k)=\phi(n/k)$

Semiclassical

Ah. Yeah, the greatest common divisor of m,n is by definition a divisor of n :)

Astyx

Actually that's nuts

Martin Sleziak

And you use there exactly that $\gcd(m,n)=k$ $\Leftrightarrow$ $\gcd(m/k,n/k)=1$.

Astyx

$\det(M + XJ)$ is always always affine

Sha Vuklia

3:46 PM

I thought that $S(k)$ consists of the elements $m$ such that $\gcd(m,n)=k$ and $\phi(n/k)$ counts the positive elements $m$ such that $\gcd(m,n/k)=1$

Astyx

Cause $\det(C_1 + X, \dots, C_n + X) = \det(C_1 + X, C_2-C_1, \dots, C_n - C_1)$

Semiclassical

I forget what it means for a determinant to be affine.

Astyx

Anyway, I need to go, thanks for your time

Martin Sleziak

The map $m\mapsto m/k$ is the bijection between $S(k)$ and the $\{j \le n/k; \gcd(j,k)=1\}$.

Astyx

I mean $\det(M+xJ) = \alpha x + \beta$ for some constants $\alpha, \beta$

affine here is about the polynomial

Semiclassical

3:47 PM

Huh.

Astyx

I don't know anything about affine meaning something general for determinants

Semiclassical

Okay.

Sha Vuklia

@Martin but how do we know that $m\mid k$?

Martin Sleziak

@ShaVuklia $\phi(n/k)$ counts integers such that $\gcd(m,n/k)=1$ and *additionally $m\le n/k$.

@ShaVuklia Do you mean $k\mid m$?

Sha Vuklia

oh yea sorry that's indeed what I meant

Martin Sleziak

3:50 PM

If $m\in S(k)$, then you know that $\gcd(m,n)=k$ and, consequently, $k$ is a divisor of $m$.

Sha Vuklia

ohh right

Akiva Weinberger

Happy Tau Day!

7

Martin Sleziak

BTW the top voted answer in this question gives basically the same proof: Is there a direct, elementary proof of $n = \sum_{k|n} \phi(k)$?

SteamyRoot

Ew, tau.

Leaky Nun

@AkivaWeinberger it's my last 9 minutes of tau day

and I never noticed

Just win baby

3:51 PM

Martin Sleziak

D'oh! Your screenshot is from that answer, right?

Sha Vuklia

yes that is correct

alright I'll try to understand that answer @Martin

SteamyRoot

@Astyx And now, try to find the determinant when $a = b$? :P

Akiva Weinberger

@MartinSleziak Write the fractions $\{\frac kn:1\le k\le n\}$ in reduced form

For 6, for example, you get $\frac16$, $\frac13$, $\frac12$, $\frac23$, $\frac56$, $\frac11$

Note that $\phi(6)=2$, $\phi(3)=2$, $\phi(2)=1$, and $\phi(1)=1$

Liad

could someone explain to me what is the exact definition of $L \ ^ 2 [0,1]$ ? im not sure about it.

Sha Vuklia

3:54 PM

I find it so hard, I have to read it 10 times over, so that's why you don't hear from me

Akiva Weinberger

and also that there are two fractions with denominator $6$, two with denominator $3$, one with denominator $2$, and one with denominator $1$.

Liad

is it defined to be the complement of $C[0,1]$ to a complete metric space?

Akiva Weinberger

I don't remember, but I think it's the ones whose squares are integratable?

Semiclassical

Should be square-integrable functions on $[0,1]$, yeah.

(finite L^2-norm)

Liad

so am i correct?

Sha Vuklia

3:56 PM

@Martin how do we know that $t$ divides $n$, in $s/t$?

Akiva Weinberger

In any case, $|\{\frac kn:1\le k\le n\}|=n$, clearly, and counting by the ones of each denominator (when written in reduced form) gives you $\sum_{k|n}\phi(k)$.

Thus $n=\sum_{k|n}\phi(k)$, QED.

SteamyRoot

$L^2([0,1])$ is the Hilbert space of square-integrable functions equipped with the inner product $\langle f,g \rangle = \int_0^1 f(x)\overline{g(x)} dx $

Liad

i also need to prove that if $\{c_n \} \subset \Bbb C$ s.t $\sum |c_n| \ ^ 2 \lt \infty$ then there is $F \in L \ ^ 2 [0,1]$ s.t the fourier's coefficients are $c_n$ . i thought defining $F(x) = \sum c_n e_n(x)$ , and then $d_n = <F , e_n>$ would be $c_n$ because $\{e_n\}$ is ortonormal basis.

Sha Vuklia

@Akiva what do you mean by "counting the ones of each denominator"? You mean that we're going to count the ones with the same denominator?

apparently that gives $\phi(k)$?

for denominator $k$, I suppose?

Akiva Weinberger

@ShaVuklia Did you see the example above with $n=6$?

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