Wikipedia:Reference desk/Archives/Mathematics/2011 June 13

= June 13 =

Ogden's lemma examples
Our article on Ogden's lemma (and many other pages I've found on the web) use the example {aibjckdl : i = 0 or j = k = l} but this language can be proven non-context-free using the pumping lemma and closure of CFLs under intersection by regular languages (take ab*c*d*). Are there examples of languages which can be proven to be non-CFLs with Ogden's lemma, but not with the pumping lemma and closure of CFLs under regular intersection and gsm mappings? --146.141.1.96 (talk) 10:51, 13 June 2011 (UTC)

nilpotent cone of a Lie algebra
The article nilpotent cone claims that "The nilpotent cone...is invariant under the adjoint action of $$\mathfrak{g}$$ on itself." Take the example of $$\mathfrak{g}= \mathfrak{sl}_2 = \langle E,F,H \rangle$$ as in the article, so the nilcone N is spanned by E and F. Then $$ \operatorname{ad}(E)(F) = H $$ which is not in the nilcone, so N is not ad-invariant. Have I misunderstood something, or is the article wrong? Tinfoilcat (talk) 11:15, 13 June 2011 (UTC)


 * If g is a Lie algebra and x is an element of g, then the adjoint action adx : g → g is defined by adx(y) = [x,y] for all y in g. To prove that the nilpotent cone, say N, is invariant under the adjoint action you just need to show that for all x and y in N, the Lie bracket [x,y] is again in N. — Fly by Night  ( talk )  14:47, 13 June 2011 (UTC)


 * FbN, I know what the adjoint map is and what it means to be ad-invariant (it's not quite what you say - ad-invariant means a Lie ideal, not a Lie subalgebra). The example I gave in my post seems to show that the nilcone is not ad-invariant since $$[E,F]=H$$ which is not in the nilcone, whereas E and F are. Tinfoilcat (talk) 14:54, 13 June 2011 (UTC)


 * I don't recall mentioning subalgebras or ideas. I'm sorry I wasn't able to help. — Fly by Night  ( talk )  15:50, 13 June 2011 (UTC)


 * FbN, I was very bitey above. Sorry.  When you say "for all x and y in N, the Lie bracket [x,y] is again in N " that's equivalent to it being a subalgebra.  Being an ideal (= being ad-invariant) is that for n in N, x in the Lie algebra, $$[x,n] \in N$$ - a little stronger. Tinfoilcat (talk) 15:59, 13 June 2011 (UTC)
 * No problem. Proving it's a subalgebra means that it's automatically an ideal too, by definition (because of the way the nilpotent cone is defined in the first place). Or at least I think so&hellip; — Fly by Night  ( talk )  16:06, 13 June 2011 (UTC)

The statement is wrong. The nilpotent cone is invariant under the action on Int(g) on g (interior auromorphisms). Sławomir Biały (talk) 15:53, 13 June 2011 (UTC)


 * Thanks. I'll fix the page (if you haven't got there first). Tinfoilcat (talk) 15:59, 13 June 2011 (UTC)

Normalizing Associated Legendre Polynomials to get Spherical Harmonics
Hi all. I've been working on this problem for like weeks, and I can't seem to figure it out. I'm trying to normalize Associated Legendre polynomials to turn them into Spherical harmonics. The integral comes out to:

$$\int_{-1}^1 P_{\ell_1}^m(x) P_{\ell_2}^m(x) dx = \frac{1}{A^2}$$

where $$A$$ is the normalization constant. $$A$$ can be found in Spherical harmonics. I know that it involves integrating by parts $$\ell+m$$ times, and that the boundary terms vanish in each case, but I'm not sure why they vanish. Can anyone point me to a (very detailed) discussion of how to actually do the integral, or maybe a better way than by parts? Thanks!--Dudemanfellabra (talk) 23:32, 13 June 2011 (UTC)


 * Hi, I had a quick look in Boas - 'Mathematical Methods in the Physical Sciences'. She has a problem (prob 10.10 in second edition) about the normalisation, first some results:


 * a) $$ P^m_l(x) = \frac{1}{2^l l!} (1-x^2)^{m/2} \frac{d^{l+m}}{dx^{l+m}} (x^2-1)^l

$$


 * This comes from substituting in Rodriques' formula (see Legendre_polynomials) for the Legendre polynomials. Next Boas says derive


 * b) $$ P^m_l(x) = (-1)^m \frac{(l+m)!}{(l-m)!}\frac{(1-x^2)^{-m/2}}{2^l l!} \frac{d^{l-m}}{dx^{l-m}} (x^2-1)^l $$


 * Multiply these two results together (this forms your normalisation integrand) then integrate by parts repeatedly until both l+m and l-m derivatives are just l derivatives. Then use
 * $$\int^1_{-1} [P_l(x)]^2 dx = \frac{2}{2l+1} $$ - I assume because you end up having an integrand that looks like two Legendre polynomials multiplied together.


 * Now, the derivation of result (b) is a task in itself. Apparently one can show that
 * $$ \frac{d^{l-m}}{dx^{l-m}} (x^2-1)^l = \frac{(l-m)!}{(l+m)!}(x^2 - 1)^m \frac{d^{l+m}}{dx^{l+m}} (x^2-1)^l

$$ by starting with $$ (x^2 - 1)^l = (x-1)^l (x+1)^l $$ and finding derivatives using Leibniz' rule. Good luck, I'd find a copy of Boas and work from that if I was you. Christopherlumb (talk) 19:55, 14 June 2011 (UTC)


 * I found a copy of Boas, but unfortunately the method for part b) is not given. After applying Leibniz' rule, I get
 * $$\frac{d^{l-m}}{dx^{l-m}} (x^2-1)^l = \sum_{i=0}^{l-m} \binom{l-m}{i} \frac{d^{l-m-i}}{dx^{l-m-i}} (x+1)^l \frac{d^i}{dx^i} (x-1)^l$$
 * I have no idea where to go from here...--Dudemanfellabra (talk) 23:13, 14 June 2011 (UTC)


 * I'm really struggling myself, I think you have to get expressions for the $$l+m$$th and $$l-m$$th derivatives and compare them term-by-term, so
 * $$\frac{d^{l+m}}{dx^{l+m}} (x^2-1)^l = \sum_{i=0}^{l+m} \binom{l+m}{i} \frac{d^{l+m-i}}{dx^{l+m-i}} (x+1)^l \frac{d^i}{dx^i} (x-1)^l$$
 * we note that the derivatives on the right-hand side are just acting on $$ (x \pm 1)^l $$ so that any derivative higher than $$l$$ differentiates to zero. This means the sum gets truncated: we must have $$i \leq l$$ (else the d/dx^i goes to zero) and $$i \geq m$$ (else the d/dx^(l+m-i) will go to zero). So:
 * $$\frac{d^{l+m}}{dx^{l+m}} (x^2-1)^l = \sum_{i=m}^{l} \binom{l+m}{i} \frac{d^{l+m-i}}{dx^{l+m-i}} (x+1)^l \frac{d^i}{dx^i} (x-1)^l$$
 * now we have no-more terms in this sum than for the $$l-m$$ sum, next evaluate the derivatives:
 * $$\frac{d^{l+m}}{dx^{l+m}} (x^2-1)^l = \sum_{i=m}^{l} \binom{l+m}{i} l(l-1)\ldots(1-l-m+i) \frac{(x+1)^l}{(x+1)^{l+m-i}} $$
 * $$\times l(l-1)\ldots(1-i)\frac{(x-1)^l}{(x-1)^i}$$
 * I've written the derivatives as fractions where you'd normally just write $$(x^2-1)^{l-i}$$. Next job is to make this look like the l-m derivative:
 * $$\frac{d^{l-m}}{dx^{l-m}} (x^2-1)^l = \sum_{j=0}^{l-m} \binom{l-m}{j} \frac{d^{l-m-j}}{dx^{l-m-j}} (x+1)^l \frac{d^j}{dx^j} (x-1)^l$$
 * $$\frac{d^{l-m}}{dx^{l-m}} (x^2-1)^l = \sum_{j=0}^{l-m} \binom{l-m}{j} l(l-1)\ldots(1-l+m+j) \frac{(x+1)^l}{(x+1)^{l-m-j}} $$
 * $$\times l(l-1)\ldots(1-j)\frac{(x-1)^l}{(x-1)^j}$$
 * Let $$j=k-m$$ then our sum runs from $$k=m$$ to $$k=l$$:


 * $$\frac{d^{l-m}}{dx^{l-m}} (x^2-1)^l = \sum_{k=m}^{l} \binom{l-m}{k-m} l(l-1)\ldots(1-l+k) \frac{(x+1)^l}{(x+1)^{l-k}} $$
 * $$\times l(l-1)\ldots(1-k+m)\frac{(x-1)^l}{(x-1)^{k-m}}$$
 * Pull a factor of (x^2-1)^m out the front:


 * $$\frac{d^{l-m}}{dx^{l-m}} (x^2-1)^l = (x^2-1)^{m} \sum_{k=m}^{l} \binom{l-m}{k-m} l(l-1)\ldots(1-l+k) \frac{(x+1)^l}{(x+1)^{l+m-k}} $$
 * $$\times l(l-1)\ldots(1-k+m)\frac{(x-1)^l}{(x-1)^{k}}$$
 * It's looking very similar to the l+m version above, I haven't the energy to go through to check the factorials all work out but I think this is on the right track 213.249.173.33 (talk) 21:09, 15 June 2011 (UTC)


 * AH! THANK YOU THANK YOU THANK YOU THANK YOU! Haha I had figured out what the (l+m-i)th and (l-m-i)th derivatives were (btw, the factorials in your above derivation are a little off. I'll show the correct one below.), but I didn't think about shifting the index of one of the sums so that they can be compared properly. By shifting that index and using falling factorials for the derivatives, I was able to show the correct relationship. I'll show it below and continue to work on the rest of the problem. Thanks for all your help!

$$\frac{d^{l+m}}{dx^{l+m}} (x^2-1)^l = \sum_{i=0}^{l+m} \binom{l+m}{i} \frac{d^{l+m-i}}{dx^{l+m-i}} (x+1)^l \frac{d^i}{dx^i} (x-1)^l$$
 * Derivation showing alternative way to write Plm(x)

$$(x \pm 1)^l$$ has only l non-vanishing derivatives, so $$i \geq m, i \leq l$$. Therefore, the sum can be written from i=m to l. Using falling factorial notation for the derivatives and expanding the binomial coefficients in factorial form yields

$$\frac{d^{l+m}}{dx^{l+m}} (x^2-1)^l = \sum_{i=m}^{l} \frac{(l+m)!}{i!(l+m-i)!} \frac{l!}{(i-m)!} (x+1)^{i-m} \frac{l!}{(l-i)!} (x-1)^{l-i}$$

Doing the Leibniz formula for the l-m derivative yields

$$\frac{d^{l-m}}{dx^{l-m}} (x^2-1)^l = \sum_{j=0}^{l-m} \binom{l-m}{j} \frac{d^{l-m-j}}{dx^{l-m-j}} (x+1)^l \frac{d^j}{dx^j} (x-1)^l$$

Again, expanding the coefficients and using falling factorials,

$$\frac{d^{l-m}}{dx^{l-m}} (x^2-1)^l = \sum_{j=0}^{l-m} \frac{(l-m)!}{j!(l-m-j)!} \frac{l!}{(m+j)!} (x+1)^{m+j} \frac{l!}{(l-j)!} (x-1)^{l-j}$$

Let $$k=m+j \Rightarrow k \in [m,l]$$ Plugging in:

$$\frac{d^{l-m}}{dx^{l-m}} (x^2-1)^l = \sum_{k=m}^{l} \frac{(l-m)!}{(k-m)!(l-k)!} \frac{l!}{k!} (x+1)^{k} \frac{l!}{(l-k+m)!} (x-1)^{l-k+m}$$

Now that the two sums have the same indices (i and k are just dummy variables, so we may as well say they're the same thing; let's choose i), we can divide the two sums to find the ratio of the l+m and l-m derivatives. That is,

$$\frac{\frac{d^{l+m}}{dx^{l+m}} (x^2-1)^l}{\frac{d^{l-m}}{dx^{l-m}} (x^2-1)^l} = \sum_{i=m}^l \frac{(l+m)!}{i!(l+m-i)!} \frac{l!}{(i-m)!} \frac{l!}{(l-i)!} \frac{(i-m)!(l-i)!}{(l-m)!} \frac{i!}{l!} \frac{(l-i+m)!}{l!} (x+1)^{-m} (x-1)^{-m}$$

which, after cancelling and observing that all i's disappear so the sum can be dropped, yields:

$$\frac{\frac{d^{l+m}}{dx^{l+m}} (x^2-1)^l}{\frac{d^{l-m}}{dx^{l-m}} (x^2-1)^l} = \frac{(l+m)!}{(l-m)!} (x^2-1)^{-m}$$

This gives the relationship desired in part b) of Christopherlumb's comment above (with the exception of the $$(-1)^m$$ factor... I'm hoping that doesn't matter though haha). I'll now attempt to work through the integral using this identity. Again, thanks for all the help!--Dudemanfellabra (talk) 23:34, 15 June 2011 (UTC)