10
votes
How to obtain the asymptotics of Legendre polynomials directly from their generating function
As described in Analytic Combinatorics by Flajolet and Sedgewick, page 4, the pole $t_0$ of the generating function $F(t)$ of smallest absolute value governs the exponential asymptotics $P_n\sim (1/...
- 188k
8
votes
Accepted
How to obtain the asymptotics of Legendre polynomials directly from their generating function
You may write $2x=a+1/a$ for certain $a$, $|a|>1$ (I guess you mean $|x|>1$), then
$$
\frac1{\sqrt{1-2tx+t^2}}=
\frac1{\sqrt{(1-at)(1-a^{-1}t)}}\\= \sum (-1)^n{-1/2\choose n}a^nt^n\cdot \sum (-1)...
- 109k
6
votes
Accepted
Convergence of the series of Legendre polynomials
The answer is yes. By a theorem of Fatou
Theorem [Fatou]
If $a_n\to0$ and the function $f(z)=\sum_{n=0}^\infty a_nz^n$ is analytic at
the point $z=1$, then the series $\sum_{n=0}^\infty a_n$ ...
- 7,024
6
votes
Accepted
Clausen’s identity for associated Legendre polynomials
There is a similar formula
$$
\small{\left(P_n^m(\cos\theta)\right)^2=(\sin{\theta})^{2m}\frac{(m+n)!}{(n-m)!}\sum_{k=0}^{n-m}\frac{(-1)^k}{4^{k+m}}\binom{n+m}{k+2m}\binom{n+k+m}{n+m}\binom{2k+2m}{k+...
- 5,624
6
votes
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Legendre Polynomial Integral over half space
Integration of Equation (34) in MathWorld gives the integral $I_{nm}$ as a sum
$$I_{nm}=\sum _{q=0}^m \frac{2^{-q}}{q+1} \binom{-m-1}{q} \binom{m}{q} \, _3F_2\left(-n,n+1,q+1;1,q+2;\tfrac{1}{2}\right)...
- 188k
6
votes
Accepted
Looking for bound in integral involving Legendre polynomial
We have
$$
\int_{0}^1 \int_{0}^1 \frac{p_n(x) p_n(y)}{1-xy} dx dy=
\int_{0}^1 \int_{0}^1 p_n(x) p_n(y)\sum_{k=0}^\infty(xy)^k dx dy=
\sum_{k=0}^\infty \left(\int_0^1 p_n(x)x^kdx\right)^2.
$$
Next, ...
- 109k
5
votes
Accepted
Expansion of the associated Legendre polynomials $P^m_l(\cos\vartheta)$ for $\vartheta \rightarrow 0$
In view the Rodrigues formula for the associated Legendre polynomials, one finds
$$a_{nm}=\lim_{\vartheta\rightarrow 0} \vartheta^{-m}P^m_n(\cos\vartheta)= (-1)^m\frac{1}{2^n n!} \lim_{x\rightarrow 1}\...
- 188k
5
votes
Accepted
Generating function of the product of Legendre polynomials
Using the recursion relation
$$
P_{n-1} (x) = x P_n (x) - \frac{x^2 -1}{n} \frac{d}{dx} P_n (x) \ ,
$$
you can reduce your expression to a sum of the generating function you quote and a combined ...
- 6,696
5
votes
Proof of spherical harmonic addition theorem
Like most such things, this was shown by Ferrers (1877), in Chapter IV, Art. 14, in very elementary (and therefore not very compact, but still readable) fashion.
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5
votes
Integral formula involving Legendre polynomial
Using the relation
$$ P_n(x)=\frac{1}{2n+1}\left(P'_{n+1}(x)-P'_{n-1}(x)\right)$$
and integration by parts, we get for your integral the expression
$$ \frac{1}{2n+1}\left(-\int_{-1}^1 \frac{P_{n+1}(x)}...
- 1,549
4
votes
Accepted
Computing the integral $\int_{-1}^1 dx \, |x| J_0(\alpha \sqrt{1 - x^2}) P_\ell(x)$
Thanks to the comment by Johannes, the solution can indeed be obtained by using the following identities:
\begin{equation}
P_\ell(z)
=
\frac{1}{2^\ell}
\sum\limits_{k=0}^{\left\lfloor \frac{\ell}{2}\...
- 159
4
votes
Legendre equation: An interpretation
Legendre's equation with arbitrary $\ell$ has infinitely many solutions on $(-1,1)$.
Solutions make a vector space of dimension $2$. There is a subspace $V_1$ of dimension $1$ which consists of ...
- 91.8k
3
votes
Proof of spherical harmonic addition theorem
The OP asks for a group theoretic derivation that is also elementary. I have not found one which combines these two properties (unless one considers the rotation operator as "elementary"). ...
- 188k
3
votes
Accepted
Legendre equation: An interpretation
I am not sure if this is the qualitative/geometric interpretation -of the integrality of the $l$ parameter- you are looking for, but if the parameter $l$ is a non-negative integer then the Legendre ...
- 7,746
3
votes
Relation between Legendre and Chebyshev polynomials
On pp.13~15 of Fox, L. Parker. Chebyshev polynomials in numerical analysis. No. 519.4 F6. 1968., especially (64)(65), we can see the arguement. As an approach to the minimax solution to the function $\...
- 8,071
2
votes
Accepted
Bounds on Legendre polynomials on the complex plane
Setting $Q_{n}(z)=(n+1/2)^{1/2}P_{n}(z)$ for the orthonormalized Legendre polynomial, and
$$\kappa_{n}(z)=\sum_{j=0}^{n}|Q_{j}(z)|^{2},$$
for the inverse of the Christoffel function, it is known that ...
- 4,034
2
votes
Accepted
Relation between Legendre and Chebyshev polynomials
Both the Legendre and Chebyshev polynomials are particular cases of Jacobi polynomials $P_n^{(\alpha,\beta)}(x)$. A general connection formula of the type $$P_n^{(\gamma,\delta)}(x)=\sum_{k=0}^nc_{n,k}...
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2
votes
Accepted
A recurrence formula for the Legendre function $P_\mu^\nu(x)$
Let's take relation 14.10.3 from the NIST Handbook, which, after renaming $\mu \leftrightarrow \nu $ and shifting the new $\mu \rightarrow \mu -2$ reads
$$
(\mu -\nu ) P_{\mu }^{\nu } (x) - (2\mu -1)x ...
- 6,696
2
votes
Legendre Polynomial Integral over half space
I found the following answer, based on the idea by
Dougall, John, The product of two Legendre polynomials, Proc. Glasg. Math. Assoc. 1, 121-125 (1953). ZBL0052.06404.
Expand the product in Legendre ...
- 213
2
votes
Accepted
Prove the orthogonality of vector spherical harmonics
Here is a proof that the last integral is zero, from which the orthogonality follows.
We have the definition
$$Y_{lm}(\theta,\varphi) = \sqrt{\frac{2l+1}{4\pi}\frac{(l-m)!}{(l+m)!}}P_l^m(\cos\theta)e^...
- 136
2
votes
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How are the Legendre Polynomials of second kind for negative degrees defined?
It helps to rewrite the expression from Gradshteyn,
$$Q_\nu^0(z)=\frac{ \Gamma \left(\frac{1}{2}\right) \Gamma (\nu+1)\, _2F_1\left(\frac{\nu}{2}+1,\frac{\nu}{2}+\frac{1}{2};\nu+\frac{3}{2};\frac{1}{z^...
- 188k
1
vote
Properties of a function $C_\ell(\ell)$ which checks an inequality in ideal case (decreasing assumption) and after estimating impact in general case
It is not enough to assume that $Y=C_l$ is decreasing in $l$.
Indeed, if your inequality were true for all such $Y$, then its non-strict version would be true for all constant $Y$, say for $Y=1$ for ...
- 128k
1
vote
Defining Legendre polynomials in terms of a sinusoidal function for $|x| \leq 1$
No doubt that $x=\cos\theta$ is a meaningful substitution for the Legendre polynomials. The functions $P_n(\cos\theta)$ were already considered by Legendre in the spherical harmonic expansion of the ...
- 60.5k
1
vote
How to obtain the asymptotics of Legendre polynomials directly from their generating function
One of the possibiliteis is a Liouville–Steklov method (see P. K. Suetin, “Classical Orthogonal Polynomials,” Nauka, Moscow, 1974 or 2005). It gives (Theorem 4.5)
$$
(\sin \theta)^{1 / 2} P_{n}(\cos \...
- 12.3k
1
vote
Closed form for the integral of a squared Legendre function
The best evaluation, a single sum, that I can derive is
$$ \int_0^{\pi/2} \big( P_n^m(\cos{\theta}) \big)^2\ d\theta = \frac{\pi}{2} \frac{(2m)!}{m!^4} \Big( \frac{(n+m)!}{(n-m)!} \Big)^2
{}_4F_3
\...
- 894
1
vote
Bounds for associated Legendre polynomials
Please have a look at the following paper
https://www.sciencedirect.com/science/article/pii/S0021904598932075
In particular, Eq. (6) to get the answer to your question.
Lohöfer, G., Inequalities for ...
- 31
1
vote
A recurrence formula for the Legendre function $P_\mu^\nu(x)$
See equation 37 here. And now some more characters.
- 96.4k
1
vote
Gegenbauer's addition theorem for Jacobi polynomials
Too long for a comment: Since the Legendre polynomials define the zonal harmonics on the sphere, a similar formula should hold at least for the Jacobi polynomials which correspond to harmonics for the ...
- 3,209
1
vote
Accepted
Generate a two-variable polynomial from its "roots
There a number of very well researched techniques that can be used to solve your problem in a practical sense. Most of these have come from computer graphics and computer vision where taking a set of ...
- 4,862
1
vote
Integral with Legendre polynomial
I don't have a general expression as a function of $n$, but there is one in terms of harmonic numbers $H_k$ as a function of $k\in\mathbb{R}^+$ for given $n$; for example
$$I_{2,k}=\tfrac{1}{2}\Gamma(...
- 188k
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