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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?

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    $\begingroup$ Can you give the lead-up to this $S(k)$? That may provide for a more direct explanation of why $S(k)$ vanishes for $k>0$. $\endgroup$ Jan 16, 2021 at 18:33
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    $\begingroup$ 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.) $\endgroup$
    – metamorphy
    Jan 16, 2021 at 20:25
  • $\begingroup$ For a proof of orthogonality via the Legendre ODE, see for instance math.stackexchange.com/questions/2003754/… $\endgroup$ Jan 16, 2021 at 20:45

3 Answers 3

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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 $z^{2k+1}$.

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

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    $\begingroup$ @MichaelLevy: added explanation for each step $\endgroup$
    – G Cab
    Jan 16, 2021 at 12:29
  • $\begingroup$ @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. $\endgroup$
    – G Cab
    Jan 16, 2021 at 16:02
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    $\begingroup$ @metamorphy: I amended my answer by recasting it as the difference of order $2k$ of a polynomial of degree $2k-1$. $\endgroup$
    – G Cab
    Jan 16, 2021 at 18:27
  • $\begingroup$ @MichaelLevy: cannot grasp what you mean to do: I don't see any change in the sum .. $\endgroup$
    – G Cab
    Jan 16, 2021 at 19:35
  • $\begingroup$ @MichaelLevy: so meaning that $k$ could also be a multiple of $1/2$ ? and the sum going from $0$ to $floor (k/2)$ ? $\endgroup$
    – G Cab
    Jan 16, 2021 at 19:48
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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 $k>0$.

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  • $\begingroup$ @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.) $\endgroup$
    – metamorphy
    Jan 16, 2021 at 11:49
  • $\begingroup$ 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.) $\endgroup$ Jan 16, 2021 at 18:46
  • $\begingroup$ @Semiclassical: this is addressed to OP rather than me, right? ;) $\endgroup$
    – metamorphy
    Jan 16, 2021 at 19:43
  • $\begingroup$ 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.) $\endgroup$ Jan 16, 2021 at 19:57

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