How to simplify a partial sum obtained by Legendre polynomials

Question

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 even number $2k$.

$$S{(2k)}=\frac1{2^{2k}}\sum_{m=0}^{k}(-1)^m\binom{2k}{m}\binom{4k-2m}{2k}\,\frac{1}{2k-2m+1}.$$

The boundary problem that this question originates from has odd symmetry. Therefore, I expect all even Fourier coefficients--except potentially $2k=0$--to be null. This expectation is verified by the correct answer below.

Question

Can the above expression be simplified further?

Answer 1

We can rewrite the sum as $$ \eqalign{ & S(k) = {1 \over {2^{\,2k} }}\sum\limits_{m = 0}^k {\left( { - 1} \right)^{\,m} \left( \matrix{ 2k \cr m \cr} \right)\left( \matrix{ 4k - 2m \cr 2k \cr} \right) {1 \over {2k - 2m + 1}}} = \cr & = {1 \over {2^{\,2k} }}\sum\limits_{m = 0}^k {\left( { - 1} \right)^{\,m} \left( \matrix{ 2k \cr m \cr} \right){{\left( {4k - 2m} \right)^{\,\underline {\,2k\,} } } \over {\left( {2k} \right)!}}{1 \over {\left( {2k - 2m + 1} \right)}}} = \cr & = {1 \over {2^{\,2k} \left( {2k} \right)!}}\sum\limits_{m = 0}^k {\left( { - 1} \right)^{\,m} \left( \matrix{ 2k \cr m \cr} \right) \left( {2\left( {2k - m} \right)} \right)^{\,\underline {\,2k - 1\,} } } \cr} $$

Now the Falling factorial $$ p(m,k) = \left( {2\left( {2k - m} \right)} \right)^{\,\underline {\,2k - 1\,} } $$ is a polynomial in $m$ with the following characteristics.
$$ \left\{ {\matrix{ {k = 0\quad \Rightarrow \quad } & \matrix{ p(m,0) = \left( { - 2m} \right)^{\,\underline {\, - 1\,} } = {1 \over {\left( { - 2m + 1} \right)^{\underline {\,1\,} } }} = {1 \over { - 2m + 1}}\quad \Rightarrow \hfill \cr \Rightarrow \quad p(0,0) = 1 \hfill \cr} \cr {1 \le k\quad \Rightarrow \quad } & \matrix{ p(m,k) = \left( {2\left( {2k - m} \right)} \right)^{\,\underline {\,2k - 1\,} } \hfill \cr {\rm polynomial}\,{\rm in}\,{\rm m}\,{\rm ofdegree}\;2k - 1 \hfill \cr {\rm with}\,{\rm zeros} \in \left[ {k + 1,\;2k} \right] \hfill \cr} \cr } } \right. $$

Therefore we have $$ S(0) = 1 $$ and for $1 \le k$ $$ S(k)\quad \left| {\;1 \le k} \right. = {1 \over {2^{\,2k} \left( {2k} \right)!}}\sum\limits_{m = 0}^k {\left( { - 1} \right)^{\,m} \left( \matrix{ 2k \cr m \cr} \right)p(m,k)} $$ Since $p(m,k) = 0$ for $k+1 \le m \le 2m$ we can extend the sum to $2k$ and multiply the summand by $1 = (-1)^{2k}$ $$ \eqalign{ & S(k)\quad \left| {\;1 \le k} \right.\quad = {1 \over {2^{\,2k} \left( {2k} \right)!}}\sum\limits_{m = 0}^{2k} {\left( { - 1} \right)^{\,m} \left( \matrix{ 2k \cr m \cr} \right)p(m,k)} = \cr & = {1 \over {2^{\,2k} \left( {2k} \right)!}}\sum\limits_{m = 0}^{2k} {\left( { - 1} \right)^{2k - \,m} \left( \matrix{ 2k \cr m \cr} \right)p(m,k)} \cr} $$ Here we recognize that the sum represents the finite difference (unitary step) of order $2k$ of $p(m,k)$, and it is known (re. to the Newton series) that the difference of a polynomial , of order greater than the degree of the same is identically null.

In conclusion $$S(k)= \delta _{\,k, \, 0} =\binom {0}{k}$$

Answer 2

Since (a known formula which easily follows from e.g. Rodrigues') $$P_n(x)=\frac{1}{2^n}\sum_{m=0}^{\lfloor n/2\rfloor}(-1)^m\binom{n}{m}\binom{2n-2m}{n}x^{n-2m},$$ the given expression is equal to $\int_0^1 P_{2k}(x)\,dx=\frac12\int_{-1}^1 P_{2k}(x)\,dx$.

This is $1$ for $k=0$ and $0$ for $k>0$.

Answer 3

We use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ of a series. This way we can write for instance \begin{align*} \binom{n}{k}=[z^k](1+z)^n\tag{1} \end{align*}

We obtain for $k>0$ \begin{align*} \color{blue}{S(2k)}&=\color{blue}{\frac{1}{2^{2k}}\sum_{m=0}^k(-1)^m\binom{2k}{m}\binom{4k-2m}{2k}\frac{1}{2k-2m+1}}\\ &=\frac{1}{k2^{k+1}}\sum_{m\geq 0}(-1)^m\binom{2k}{m}\binom{4k-2m}{2k-2m+1}\tag{2}\\ &=\frac{1}{k2^{k+1}}\sum_{m\geq 0}(-1)^m\binom{2k}{m}[z^{2k-2m+1}](1+z)^{4k-2m}\tag{3}\\ &=\frac{1}{k2^{k+1}}[z^{2k+1}](1+z)^{4k}\sum_{m\geq 0}(-1)^m\binom{2k}{m}\left(\frac{z}{1+z}\right)^{2m}\tag{4}\\ &=\frac{1}{k2^{k+1}}[z^{2k+1}](1+z)^{4k}\left(1-\left(\frac{z}{1+z}\right)^2\right)^{2k}\tag{5}\\ &=\frac{1}{k2^{k+1}}[z^{2k+1}](1+2z)^{2k}\tag{6}\\ &\,\,\color{blue}{=0}\tag{7} \end{align*}

Comment:

• In (2) we use the binomial identity $\binom{4k-2m}{2k}\frac{1}{2k-2m+1}=\binom{4k-2m}{2k-2m}\frac{1}{2k-2m+1} =\binom{4k-2m}{2k-2m-1}\frac{1}{2k}$.

• In (3) we use the coefficient of operator according to (1). We also set the upper limit of the sum to $\infty$ which doesn't change the sum since we are adding zeros only.

• In (4) we use the linearity of the coefficient of operator and apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.

• In (5) we apply the binomial theorem.

• In (6) we do some simplifications.

• In (7) we select the coefficient of $z^{2k+1}$.