# 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 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?

• Can you give the lead-up to this $S(k)$? That may provide for a more direct explanation of why $S(k)$ vanishes for $k>0$. Jan 16, 2021 at 18:33
• My conclusion (well, suspicion) to "shout it loud". If the original problem is to show that $\int_0^1 P_{2k}(x)\,dx=0$ for $k>0$ (armed with a suitable definition of $P_{2k}$), and your attempt is to put the explicit expression for $P_{2k}(x)$ in powers of $x$ here, then this is a detour. Any known idea of proving orthogonality of $P_*$, like using Rodrigues' formlua, is a great deal faster. (But, if we're assumed to know nothing about $P_{2k}$ but the mentioned expression, the answer by @GCab is perhaps the way to go.) Jan 16, 2021 at 20:25
• For a proof of orthogonality via the Legendre ODE, see for instance math.stackexchange.com/questions/2003754/… Jan 16, 2021 at 20:45

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 $$z^{2k+1}$$.

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

• @MichaelLevy: added explanation for each step Jan 16, 2021 at 12:29
• @metamorphy: oohps .. bad sign mistake, sorry and thanks for signalling: I ask MLevy to cancel the acceptance so that I may delete the answer while rethinking the conclusion , which any way is right. Jan 16, 2021 at 16:02
• @metamorphy: I amended my answer by recasting it as the difference of order $2k$ of a polynomial of degree $2k-1$. Jan 16, 2021 at 18:27
• @MichaelLevy: cannot grasp what you mean to do: I don't see any change in the sum .. Jan 16, 2021 at 19:35
• @MichaelLevy: so meaning that $k$ could also be a multiple of $1/2$ ? and the sum going from $0$ to $floor (k/2)$ ? Jan 16, 2021 at 19:48

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 $$k>0$$.

• @MichaelLevy: Then it depends on how do you define the Legendre polynomials. With Rodrigues' formula $P_n(x)=\frac{1}{2^n n!}\frac{d^n}{dx^n}(x^2-1)^n$ used as a definition, $\int_{-1}^1 P_n(x)P_m(x)\,dx=0$ for $n\neq m$ (and particularly the case $m=0$) is very easy to prove using integration by parts, as well as the "known formula" above (expand $(x^2-1)^n$ and differentiate termwise). So I'm not sure whether it is a "backwards" way. (Unless you prefer a "Legendre-free" approach.) Jan 16, 2021 at 11:49
• To justify the integral over the Legendre polynomials, note that $\int_{-1}^1 P_m (x)\,dx=\int_{-1}^1 P(m)P_0(x)\,dx$ since $P_0(x)=1$. So this is just a direct consequence of the orthogonality of Legendre polynomials. (And if you want to prove said orthogonality, there's better ways to do that than considering particular sums.) Jan 16, 2021 at 18:46
• @Semiclassical: this is addressed to OP rather than me, right? ;) Jan 16, 2021 at 19:43
• It's addressed to anyone who hasn't seen the $P_0(x)=1$ trick :). (Though I see now I typoed $P_m(x)$ as $P(m)$, silly me.) Jan 16, 2021 at 19:57