Wikipedia:Reference desk/Archives/Mathematics/2012 August 10

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August 10

Kernel

See here. I've been waiting months for some insight on Problem 10; now August 8 has passed, I have you guys! I reduced Part a) to showing that

$(k-1)D_{k}=x_{1}D_{k-1}+\dots +x_{k-1}D_{1},\qquad D_{1}=1$

has solution $D_{k}=\Delta (x_{k})$, but I have no idea how to do that. Better methods or a more fleshed out approach to the whole problem would be very much appreciated. Cheers. —Anonymous Dissident^Talk 14:21, 10 August 2012 (UTC)

I would approach the problem by trying to understand the summation

$\sum _{m_{1}+2m_{2}+\cdots +rm_{r}=r}{\frac {x_{1}^{m_{1}}\cdots x_{r}^{m_{r}}}{1^{m_{1}}m_{1}!\cdots r^{m_{r}}m_{r}!}}.$

If the $x_{i}$ are the power sum symmetric polynomials in some set of auxiliary variables, then this summation gives the complete homogeneous symmetric polynomials (up to fiddling with the signs). Sławomir Biały (talk) 15:35, 10 August 2012 (UTC)

Name of squashed sphere?

I'm looking for the mathematically correct name (or 3D primative name) of a squashed sphere. Not an ovoid. (which is egg shaped). A square is to a rectangle as a sphere is to a ______? --206.209.221.240 (talk) 18:06, 10 August 2012 (UTC)

oblate spheroid. Looie496 (talk) 18:10, 10 August 2012 (UTC)

Well, a square is to a rectangle as a circle is to an ellipse. If you want something in three dimensions then I'd recommend the ellipsoid article. — Fly by Night (talk) 17:44, 11 August 2012 (UTC)

Dirac delta function

Can ${\frac {1}{x}}{\frac {d}{dx}}(x{\frac {1}{x}})$ be expressed in terms of the Dirac delta function? Widener (talk) 19:17, 10 August 2012 (UTC)

No. If you want a more informative answer, you'll have to ask a more informative question. Looie496 (talk) 21:46, 10 August 2012 (UTC)

$x{\frac {1}{x}}$ isn't really a meaningful expression. It is difficult to know how that should be interpreted as a function. Should you simplify it and then it is just 1 everywhere? Or should it be left as undefined at x=0? You probably want the latter, but in that case the natural way to integrate would just give zero (since an integral isn't affected by individual points). The Dirac delta function only has a non-zero integral because it is explicitly defined that way. --Tango (talk) 17:18, 11 August 2012 (UTC)

I think that x·(1/x) is a perfectly meaningful expression: you multiply a number by its reciprocal. Although it is indeterminate at x = 0 but has a removable singularity. — Fly by Night (talk) 17:52, 11 August 2012 (UTC)

I think Widener is considering something similar to the Laplacian of the Coulomb potential. For a sherically symmetric potential V(r), we have:

$\nabla ^{2}V={\frac {1}{r}}{\frac {\partial ^{2}}{\partial r^{2}}}\left(rV\right)$

For V(r) = 1/r, the Laplacian is -4 pi delta(r), but then the above exression is not valid at r = 0. But you can ask if you can still extract the delta function from that expression in some way. Count Iblis (talk) 17:48, 11 August 2012 (UTC)

In the context of Fourier analysis, the ambiguity in defining the distribution 1/x is precisely a delta function. (Usually we define 1/x via the principal value; see Hilbert transform). However, since 1/x is multiplied by x, this ambiguity disappears, and you're left with the constant distribution 1. The derivative of a constant distribution is equal to the zero distribution. Sławomir Biały (talk) 14:24, 12 August 2012 (UTC)

Let me have a go at it again. If you look at the vector field ${\vec {v}}={\frac {1}{r}}{\hat {\theta }}$ and consider a disk $D$ of radius 1 then we know $\int _{C}{\frac {1}{1}}{\hat {\theta }}\cdot dC=\int _{D}{\frac {1}{r}}{\frac {d}{dr}}(r{\frac {1}{r}}){\hat {z}}\cdot dD$ by Stokes' theorem. This implies $2\pi =\int _{D}{\frac {1}{r}}{\frac {d}{dr}}(r{\frac {1}{r}}){\hat {z}}\cdot dD.$ $dD$ is in the direction of z and ${\frac {1}{r}}{\frac {d}{dr}}(r{\frac {1}{r}})$ is zero for all $r\neq 0$ so my logic would dictate that ${\frac {1}{x}}{\frac {d}{dx}}(x{\frac {1}{x}})=2\pi \delta (x)$. What do you think? --Widener (talk) 02:04, 12 August 2012 (UTC)

The Dirac delta "function" is kind of dodgy anyway. --Widener (talk) 02:25, 12 August 2012 (UTC)

If you're interested, the question arose when trying to dot-integrating this particular vector field: $k(\phi /r){\hat {\phi }}$ over a circle of radius R. I get $2\pi ^{2}k$ for the line integral. Widener (talk) 02:25, 12 August 2012 (UTC)

It's actually the curl of your ${\hat {\theta }}$ that gives the delta function, not the scalar factor that multiplies it. This is invisible in your calculation since the identity expressing the curl in cylindrical coordinates is only valid away from zero. Sławomir Biały (talk) 14:24, 12 August 2012 (UTC)

I wonder whether hyperfunctions might not be a suitable context in which to consider this. As described in The Road to Reality, it provides a formal structure within which to use path-dependent integrals of holomorphic functions on C for functions on R. This would generalize to dot-integrating curl-free 3D fields containing 1D curves of infinite curl. — Quondum^☏ 19:21, 12 August 2012 (UTC)

I don't think this is necessary, as conventional distributions are already a proper subset of hyperfunctions. Really all that's being asserted is that $d(d\theta )=2\pi \delta$ is the delta measure. This is easily checked by integrating against any compactly supported test function. Sławomir Biały (talk) 14:50, 13 August 2012 (UTC)

p^2 | numerator of H_{p-1} for p a prime number larger or equal to 5

I have a simple proof of this, let's see if the Ref Desk is up to task. The harmonic numbers are defined as

$H_{n}=\sum _{k=1}^{n}{\frac {1}{k}}$

Expressing this as a fraction r/s with GCD(r,s) = 1, you can look at the integer factorization of the numerator r. It turns out that for p a prime number, the numerator of $H_{p-1}$ is divisible by p, and for $p\geq 5$, it is divisible by $p^{2}$.

To not spoil this problem for others, don't post the answer until everyone has had a chance to try it!

Count Iblis (talk) 20:58, 10 August 2012 (UTC)

The divisibility by p part is just Wilson's Theorem and some basic results, I feel like the second part is on the tip of my brain, but I've never been that into number theory, so I'm missing some crucial (obvious) connection. Sorry, by the way, if this spoils anything, I'm not good with N.theory and am assuming everyone's already got the easy part:-) Phoenixia1177 (talk) 10:21, 11 August 2012 (UTC)

I don't know much about harmonic numbers, but what I find interesting is the fact that H_p-1 being divisible by p³ is equivalent to p being a Wolstenholme prime. That's not an answer to your problem of course, just saying. -- Toshio Yamaguchi (tlk−ctb) 10:56, 11 August 2012 (UTC)

Hint: You can start with divisibility by p. Per the discussion below this, it should be clear that $H_{p-1}$ modulo p is well defined... Count Iblis (talk) 17:49, 12 August 2012 (UTC)

For the $p^{2}$ case, consider how you can compute $a^{-1}{\bmod {\ }}p^{2}$ from $a^{-1}{\bmod {\ }}p$ using Hensel lifting... Count Iblis (talk) 20:44, 12 August 2012 (UTC)

Interesting lemma, thanks for showing it to me; but since I can see the gist of this, I'm going to stop where I am and be satisfied (sorry, this just isn't my kind of "for fun" problem.) Thanks, though, it is still interesting to learn something new. As an aside, I worked the easy part out in my head, realized later you don't really need Wilson's theorem for anything, just made it easier to keep track of stuff in my brain. This posting of problems thing is interesting, looking forward to more (supposing that this is in the spirit of the reference desk and no one objects):-) Phoenixia1177 (talk) 21:39, 12 August 2012 (UTC)