All Questions
Tagged with legendre-polynomials summation
25
questions
1
vote
1
answer
62
views
How to prove this summation equation? [duplicate]
I'm looking for some hints on proving the following (either directly or by induction):
$$
\sum_{k={0}}^{l/2} \frac{(-1)^k(2l-2k)!}{k!(l-k)!(l-2k)!} =2^l
$$
I do know it is actually true from various ...
- 13
0
votes
0
answers
83
views
Re-writing a sum of binomial coefficients as an integral of shifted Legendre polynomials
This is a question regarding the answer presented here.
In order to make this post self-contained, I am wondering if someone can explain why the sum
$$
\sum_{k=0}^{n}(-1)^{n+k}\binom{n}{k}\binom{n+k}{...
user1211323
2
votes
1
answer
96
views
How to evaluate the following sum: $\sum_{n=1}^\infty \frac{P_n(x) - P_{n-1}(x)}{ n } \cos(nt)$
I am trying to find a closed form expression for the following sum,
$$
F(x,t)= \frac{1}{\log\left(\frac{1+x}{2}\right)}\sum_{n=1}^\infty \frac{P_n(x) - P_{n-1}(x)}{ n } \cos(nt) ~,
$$
where $P_n(x)$ ...
- 139
0
votes
1
answer
101
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summation of Legendre polynomials over a power law
I am trying to find a closed-form for the summation:
$$\sum_{n=0}^{\infty}\frac{P_{n}(x)}{(n+k)^{\alpha}}$$
where $P_{n}(x)$ denote the Legendre polynomials, k is a constant, and $\alpha$ is positive ...
- 461
2
votes
3
answers
162
views
How to simplify a partial sum obtained by Legendre polynomials
Context
I am working on an electrostatics problem. I have undertaken Fourier analysis. By $k$, I denote a natural number $k=0,1,2,\ldots$. I have obtained the following partial sum in terms of the ...
- 1,112
1
vote
0
answers
172
views
Spherical Harmonics Sum Identity
I'm taking a course in Quantum Mechanics and this problem is causing me some struggles. Can someone help me prove this identity?
$$\sum_{m = -l}^l m^2 |Y_{l}^{m}(\theta, \phi)|^2 = \frac{l(l+1)(2l+1)}{...
6
votes
2
answers
419
views
Proof for a combinatorial identity
I have the following formula, which I believe it's true since it works in Mathematica for all values of $N$ I have tried, but I don't know how to prove it:
$$\sum_{q=0}^{N} {N \choose q}^2 x^{q} = \...
- 759
1
vote
1
answer
95
views
Why swapping between the derivative operator and this infinite sum leads to different results?
While working on a mathematical physical problem, i came across seemingly contradictory results.
Notations
Let's consider $\mathbf{x}_1$ to be the origin of a spherical coordinate system and $\...
1
vote
1
answer
2k
views
Legendre polynomial recurrence relation proof using the generation function
I want to prove the following recurrence relation for Legendre polynomials:
$$P'_{n+1}(x) − P'_{n−1}(x) = (2n + 1)P_n(x)$$
Using the generating function for the Legendre polynomials which is,
$$(1-...
- 519
2
votes
0
answers
712
views
Use Rodrigues’formula to generate the Legendre Polynomials
may any one tell how the circled step was given
1
vote
1
answer
642
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Sum of associated Legendre functions
I want to find the sums of two expressions involving the Schmidt-normalized associated Legendre functions. They are defined by
\begin{align}
S_l^0(x) &= P_l^0(x) \\
S_l^m(x) &= \sqrt{2 \frac{(...
- 1,026
1
vote
0
answers
103
views
Simplifying with spherical harmonics
I originally asked this on the physics Stack Exchange site, but perhaps it could be more easily answered here.
Given the definition of the correlation function for CMB temperature fluctuations as
$$ ...
- 205
1
vote
1
answer
91
views
Evaluating $\sum\limits_{k=0}^n \frac{(-1)^k (4 n - 2 k)!}{k! (2 n - k)! (2 n - 2 k)! (2 n + 2 r + 1 - 2 k)}$
I'm trying to prove an property of Legendre polynomial, namely $\int_{-1}^1x^{2r}P_{2n}(x)dx=\frac{2^{2n+1}(2r)!(r+n)!}{(2r+2n+1)!(r-n)!}$
I'm required to use the general formula for Legendre ...
- 471
0
votes
1
answer
810
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Convergence of Series involving Legendre Polynomials
I have this following series:
\begin{equation}
G = \frac{-1}{4\pi}\sum_{l=0}^{\infty}\frac{2l+1}{\frac{l(l+1)}{R^2}+\frac{1}{\alpha}}P_l(\cos(\gamma))
\end{equation}
where $R= 6371$, $\alpha>0$ ...
- 488
2
votes
0
answers
220
views
Double summation over a product of combinations
Would there be any simplification of the following double sum?
$$\sum^{n-1}_{k=1}\sum^n_{j=k}{n-1\choose k-1}{n\choose j}a^{k-1}(1-a)^{n-k}b^j(1-b)^{n-j}$$.
or, equivalently,
$$(1-a)^{n-1}(1-b)^n\...
- 69