/sci/ - Science & Math

>>4340470
I don't see any reason you can assume <span class="math">K(x,y)=A(x)B(y)[/spoiler]. Not all functions of two variables separate in this way (<span class="math">\cos(x+y)[/spoiler], for instance).


>>4339176
>Hey, looks like you got yourself a tripcode :)
Yes, and you can probably guess where I got the idea to enclose a short, relevant word between two /'s.

>>4338837
Sorry, I'm getting myself confused with something else (it's late — that's my excuse). You can show that <span class="math">\mathcal{K}[/spoiler] is the limit of finite rank operators. I'll omit the details, but any book that discusses compact operators will include this as its first or second example. All that's required is that <span class="math">||K||_{L^2([0,1]\times[0,1])}<\infty[/spoiler].

>>4338813
<span class="math">\mathcal{K}[/spoiler] maps <span class="math">L^2[/spoiler] bounded sets to precompact sets:

Suppose <span class="math">u_n[/spoiler] is a bounded sequence in <span class="math">L^2(\Omega)[/spoiler], <span class="math">||u_n||_{L^2}\leq C[/spoiler].

Using the Arzelà–Ascoli theorem, you can show that <span class="math">Ku_n[/spoiler] must have an <span class="math">L^2[/spoiler]-convergent subsequence.

>>4338797
>Note that K^2 is compact, self-adjoint and positive
oops. Disregard the "self-adjoint". It isn't needed, and it would only obviously be true if <span class="math">K(x,y)=K(y,x)[/spoiler] for all <span class="math">x,y\in[0,1][/spoiler].

>>4338805
The composition of a bounded operator and a compact operator is compact.

>>4338788
[cont'd]

Note that <span class="math">\mathcal{K}^2[/spoiler] is compact, self-adjoint and positive with <span class="math">\mathcal{K}^2(f)=f[/spoiler] and <span class="math">\mathcal{K}^2(g)=g[/spoiler]. Since <span class="math">f[/spoiler] and <span class="math">g[/spoiler] are positive, by (c) it follows that 1 must be the spectral radius of <span class="math">\mathcal{K}^2[/spoiler]. By (a), the corresponding eigenspace must have dimension one. Hence, <span class="math">f[/spoiler] and <span class="math">g[/spoiler] must be a multiples of each other: <span class="math">f= \alpha g[/spoiler] for some <span class="math">\alpha>0[/spoiler].

Applying <span class="math">\mathcal{K}[/spoiler] to both sides yields <span class="math">g = \alpha f[/spoiler]. Consequently, <span class="math">\alpha^2=1[/spoiler]. Since <span class="math">\alpha[/spoiler] must be positive, we conclude <span class="math">\alpha=1[/spoiler].

>>4338788
[cont'd]


Note that <span class="math">\mathcal{K}^2[/spoiler] is compact, self-adjoint and positive-definite with <span class="math">\mathcal{K}^2(f)=f[/spoiler] and <span class="math">\mathcal{K}^2(g)=g[/spoiler]. Since <span class="math">f[/spoiler] and <span class="math">g[/spoiler] are positive, by (c) it follows that 1 must be the spectral radius of <span class="math">\mathcal{K}^2[/spoiler]. By (a), the corresponding eigenspace must have dimension one. Hence, <span class="math">f[/spoiler] and <span class="math">g[/spoiler] must be a multiples of each other: <span class="math">f= \alpha g[/spoiler] for some <span class="math">\alpha>0[/spoiler].

Applying <span class="math">\mathcal{K}[/spoiler] to both sides yields <span class="math">g = \alpha f[/spoiler]. Consequently, <span class="math">\alpha^2=1[/spoiler]. Since <span class="math">\alpha[/spoiler] must be positive, we conclude <span class="math">\alpha=1[/spoiler].

[oops. I posted a solution and then deleted it after I realized it was incomplete.]

This solution is pretty straightforward — if you've had a graduate course in functional analysis. I realize there must be a more elementary solution, but since I deal with integral operators so often these days, this is what came to me first:

Let <span class="math">\mathcal{K}:L^2([0,1])\to L^2([0,1])[/spoiler] be the compact, self-adjoint operator <span class="math">\mathcal{K}(u) =\int_0^1 u(y)K(x,y)\,dy[/spoiler] (self-adjointness requires that <span class="math">K[/spoiler] be symmetric). <span class="math">\mathcal{K}[/spoiler] is also a positive operator in the sense that <span class="math">u>0\mathrm{ a.e. }\Rightarrow \mathcal{K}u>0[/spoiler] a.e.

A well-known result from functional analysis (Krein-Rutman):
If <span class="math">\mathcal{A}[/spoiler] is a compact, positive operator on <span class="math">L^2(\Omega)[/spoiler], then
(a) the spectral radius <span class="math">\rho_\mathcal{A}[/spoiler] is a simple eigenvalue of <span class="math">\mathcal{A}[/spoiler] (i.e., the corresponding eigenspace has dimension one),
(b) <span class="math">\rho_\mathcal{A}[/spoiler] has a corresponding eigenfunction that is positive a.e. <span class="math">(\Omega)[/spoiler],
(c) <span class="math">\rho_\mathcal{A}[/spoiler] is the _only_ eigenvalue that corresponds to a positive (a.e.) eigenfunction.

[Here's a link to a proof: http://www.worldscibooks.com/etextbook/5999/5999_chap1.pdf ]

I love this problem. The solution is pretty straightforward — if you've had a graduate course in functional analysis. I realize there must be a more elementary solution, but since I deal with integral operators so often these days, this is what came to me first:

Let <span class="math">\mathcal{K}:L^2([0,1])\to L^2([0,1])[/spoiler] be the compact, self-adjoint, positive-definite operator <span class="math">\mathcal{K}(u) =\int_0^1 u(y)K(x,y)\,dy[/spoiler] (positive-definiteness follows easily from the fact that <span class="math">K[/spoiler] is strictly positive and continuous on its compact domain, and therefore bounded away from 0).

A well-known result from functional analysis:
If <span class="math">\mathcal{A}[/spoiler] is a compact, self-adjoint, positive definite operator on <span class="math">L^2(\Omega)[/spoiler], then
(a) the spectral radius <span class="math">\rho_\mathcal{A}[/spoiler] is a simple eigenvalue of <span class="math">\mathcal{A}[/spoiler] (i.e., the corresponding eigenspace has dimension one),
(b) <span class="math">\rho_\mathcal{A}[/spoiler] has a corresponding eigenfunction that is positive a.e. <span class="math">(\Omega)[/spoiler],
(c) <span class="math">\rho_\mathcal{A}[/spoiler] is the _only_ eigenvalue that corresponds to a positive (a.e.) eigenfunction (this is a consequence of <span class="math">\mathcal{A}[/spoiler] being self-adjoint, and therefore having orthogonal eigenfunctions: two positive functions cannot possibly be orthogonal).


Note that <span class="math">\mathcal{K}^2[/spoiler] is compact with <span class="math">\mathcal{K}^2(f)=f[/spoiler] and <span class="math">\mathcal{K}^2(g)=g[/spoiler]. Since <span class="math">f[/spoiler] and <span class="math">g[/spoiler] are positive, by (c) it follows that 1 must be the spectral radius of <span class="math">\mathcal{K}^2[/spoiler]. By (a), the corresponding eigenspace must have dimension one. Hence, <span class="math">f[/spoiler] and <span class="math">g[/spoiler] must be a multiples of each other: <span class="math">f= \alpha g[/spoiler] for some <span class="math">\alpha>0[/spoiler].

Applying <span class="math">\mathcal{K}[/spoiler] to both sides yields <span class="math">g = \alpha f[/spoiler]. Consequently, <span class="math">\alpha^2=1[/spoiler]. Since <span class="math">\alpha[/spoiler] must be positive, we conclude <span class="math">\alpha=1[/spoiler].

I love this problem. The solution is pretty straightforward — if you've had a graduate course in functional analysis. I realize there must be a more elementary solution, but since I deal with integral operators so often these days, this is what came to me first:

Let <span class="math">\mathcal{K}:L^2([0,1])\to L^2([0,1])[/spoiler] be the compact, self-adjoint, positive-definite operator <span class="math">\mathcal{K}(u) =\int_0^1 u(y)K(x,y)\,dy[/spoiler] (positive-definiteness follows easily from the fact that <span class="math">K[/spoiler] is strictly positive and continuous on its compact domain, and therefore bounded away from 0).

A well-known result from functional analysis:
If <span class="math">\mathcal{A}[/spoiler] is a compact, self-adjoint, positive definite operator on <span class="math">L^2(\Omega)[/spoiler], then
(a) the spectral radius <span class="math">\rho_mathcal{A}[/spoiler] is a simple eigenvalue of <span class="math">\mathcal{A}[/spoiler] (i.e., the corresponding eigenspace has dimension one),
(b) <span class="math">\rho_\mathcal{A}[/spoiler] has a corresponding eigenfunction that is positive a.e. <span class="math">(\Omega)[/spoiler],
(c) <span class="math">\rho_\mathcal{A}[/spoiler] is the _only_ eigenvalue that corresponds to a positive (a.e.) eigenfunction (this is a consequence of <span class="math">\mathcal{A}[/spoiler] being self-adjoint, and therefore having orthogonal eigenfunctions: two positive functions cannot possibly be orthogonal).


Note that <span class="math">\mathcal{K}^2[/spoiler] is compact with <span class="math">\mathcal{K}^2(f)=f[/spoiler] and <span class="math">\mathcal{K}^2(g)=g[/spoiler]. Since <span class="math">f[/spoiler] and <span class="math">g[/spoiler] are positive, by (c) it follows that 1 must be the spectral radius of <span class="math">\mathcal{K}^2[/spoiler]. By (a), the corresponding eigenspace must have dimension one. Hence, <span class="math">f[/spoiler] and <span class="math">g[/spoiler] must be a multiples of each other: <span class="math">f= \alpha g[/spoiler] for some <span class="math">\alpha>0[/spoiler].

Applying <span class="math">\mathcal{K}[/spoiler] to both sides yields <span class="math">g = \alpha f[/spoiler]. Consequently, <span class="math">\alpha^2=1[/spoiler]. Since <span class="math">\alpha[/spoiler] must be positive, we conclude <span class="math">\alpha=1[/spoiler].

>>4333106
>both sides of the inequality
I meant inequalitIES.

>>4332850
You don't really need to consider the ratio distribution for this problem. All you need to know is that <span class="math">(X,Y)[/spoiler] is uniformly distributed in the unit square <span class="math">(0,1)\times(0,1)[/spoiler]. You eliminate the pesky ratio by multiplying both sides of the inequality <span class="math"> 2k -1/2 \leq x/y < 2k+1/2[/spoiler] by <span class="math">y[/spoiler].

For <span class="math">x,y\in(0,1)[/spoiler] and <span class="math">k\in\mathbb{Z}[/spoiler],
<span class="math">x/y\in[2k-1/2,2k+1/2)[/spoiler] iff <div class="math">0<x<y/2,\quad k=0,</div>
<div class="math">(2k-1/2)y \leq x < (2k+1/2)y,\quad k\geq1.\quad\quad\quad (I_k)</div> (The inequality cannot hold for negative <span class="math">k[/spoiler].)
We must find <span class="math">P(I_0\cup I_1\cup\cdots)[/spoiler]. These events are mutually
exclusive, so <div class="math"> P(I_0\cup I_1\cup\cdots)= P(I_0)+\sum_{k=1}^\infty
P(I_k).</div> <span class="math">P(I_0) = \int_0^1 y/2\,dy=1/4[/spoiler]. <div class="math">P(I_k) =
\int_0^1\frac{x}{2k-1/2} - \frac{x}{2k+1/2}= \frac{1}{4k-1} -
\frac{1}{4k+1}.</div> So
<span class="math">P(I_0\cup I_1\cup\cdots)= 1/4 + (1/3 - 1/5) + (1/7 - 1/9) +\cdots[/spoiler]
<span class="math">= 5/4 - (1 -1/3 + 1/5 - 1/7 + \cdots)[/spoiler].

I couldn't at first remember if the series in parentheses is the series for <span class="math">tan^{-1}(1)[/spoiler] or <span class="math">sin^{-1}(1)[/spoiler]. But if it were the latter, the probability would be negative ( <span class="math">5/4 - \pi/2 <0[/spoiler]), so that makes me confident it's the former.

So the answer is: <span class="math">5/4 - \pi/4[/spoiler].

For <span class="math">x,y\in(0,1)[/spoiler] and <span class="math">k\in\mathbb{Z}[/spoiler], <span class="math">x/y\in[2k-1/2,2k+1/2)[/spoiler] iff <div class="math">0<x<y/2,\quad k=0,</div> <div class="math">(2k-1/2)y \leq x < (2k+1/2)y.\quad k\geq1.\quad\quad\quad (I_k)</div> (The inequality cannot hold for negative <span class="math">k[/spoiler].) We must find <span class="math">P(I_0\cup I_1\cup\cdots)[/spoiler]. These events are mutually exclusive, so <div class="math"> P(I_0\cup I_1\cup\cdots)= P(I_0)+\sum_{k=1}^\infty P(I_k).</div> <span class="math">P(I_0) = \int_0^1 y/2\,dy=1/4[/spoiler]. <span class="math">P(I_k) = \int_0^1\frac{x}{2k-1/2} - \frac{x}{2k+1/2}= (\frac{1}{4k-1} - \frac{1}{4k+1}[/spoiler].

So
<span class="math">P(I_0\cup I_1\cup\cdots)= 1/4 + (1/3 - 1/5) + (1/7 - 1/9) +\cdots[/spoiler]
<span class="math">= 5/4 - (1 -1/3 + 1/5 - 1/7 + \cdots)[/spoiler]. I couldn't at first remember if the series in parentheses is the series for <span class="math">tan^{-1}(1)[/spoiler] or <span class="math">sin^{-1}(1)[/spoiler]. But if it were the latter, the probability would be negative ( <span class="math">5/4 - \pi/2 <0[/spoiler]), so that makes me confident it's the former.

So the answer is: <span class="math">5/4 - \pi/4[/spoiler].

>>4330763
It's a very nice solution. It's certainly not quite a putnam-like problem, because you can't be expected to completely answer it by hand. But it is still pretty nice.

By the way, you can say "factor" or "divisor" — they mean the same thing.

>>4330694
lol
>>4330712
I had actually forgotten that the question was whether it was possible to find such a sequence.

>>4330574
>The relationship is this: if there is no such sequence, there are no odd perfects.

Is everything below that statement a proof of the statement? can't follow it, if it is.

>>4330474
>I can buy that.
I meant: I can believe it, but I'm not going to bother to think about it right now. I guess the poster of >>4330452 retracted his claim anyway.

>>4330450
Doing the same for all other <span class="math">a_k[/spoiler] follows that <span class="math">b_ka_k = \prod_{j\neq k} a_j[/spoiler] for some odd <span class="math">b_k[/spoiler].

Therefore, <div class="math">b_ka_k^2 = b_ma_m^2 </div> for all <span class="math">k,m[/spoiler].

>>4330452
I can buy that.

>>4330445
Doh. I meant: <span class="math">a_1[/spoiler] divides <span class="math">\prod_{j\geq2}a_j[/spoiler].

>>4330312
Another observation resulting from the equation I posted:

Since <span class="math">a_1[/spoiler] is a factor of all terms on the rhs, and since <span class="math">a_1[/spoiler] and <span class="math">a_1 - 1[/spoiler] share no nontrivial factors,
<span class="math">a_1[/spoiler] divides <span class="math">\prod_{j\geq1}a_j[/spoiler].

>>4330346
I don't see this. Starting with the assumption that last equation holds for some nondecreasing finite sequence of odd <span class="math">d_k>1[/spoiler], why must it be the case that <span class="math">\{d_0,\ldots,d_{n-1}\}[/spoiler] includes ALL proper divisors of <span class="math">d_n[/spoiler]?

>>4330312
[ahem]
<span class="math">k=1,\ldots,n[/spoiler].

>>4330251
First observation: if <span class="math">n[/spoiler] such odd numbers <span class="math">a_k[/spoiler], <span class="math">n=1,\ldots,n[/spoiler], exist, there must be an odd number of them. <div class="math"> (a_1-1)\prod_{j\neq1}a_j = \sum_{k=2}^n \prod_{j\neq k}a_k. </div> So (even number) = (sum of <span class="math">n-1[/spoiler] odd numbers).