If $\sum_{s=0}^\infty \sum_{m=0}^\infty b_{s,m}x^{2s}(1-x^2)^{m}=0$ for all $x\in [-1,1]$ what can we say about coefficients $b_{s,m}$?

Question

Consider the following power series: \begin{align} f(x)=\sum_{s=0}^\infty \sum_{m=0}^{\infty} b_{s,m}x^{2s}(1-x^2)^{m}. \end{align} Suppose that we know that $f(x)=0$ for all $x\in [-1,1]$. My question is, under the above assumption, what can we conclude about the coefficients $b_{s,m}$?

Some Thoughts $f$ can be re-written in terms of a single sum as \begin{align} \sum_{s=0}^\infty \sum_{m=0}^\infty b_{s,m}x^{2s}(1-x^2)^{m}= \sum_{k=0}^\infty a_k x^{2k}. \end{align} The expression for $a_k$ in terms $b_{s,m}$ was found here and is given by \begin{align} a_k =\sum_{s=0}^k \sum_{m \geq (k-s)} b_{s,m} {m \choose (k-s)} (-1)^{k-s} . \end{align}

Since $f(x)=0,x \in [-1,1]$, this implies that $a_k=0$ for all $k$. However, since the mapping between $b_{s,m}$ and $a_k$ is not simple I am not sure what we can conclude about the coefficients $b_{s,m}$ and what $a_k$'s equal zero imply.

Edit 1: As suggest in the comment section, it might be usefull to transform the problem by setting $x=\sin(t)$ in which case we get \begin{align} 0= \sum_{s=0}^\infty \sum_{m=0}^\infty b_{s,m}\sin^{2s}(t)\cos^{2m}(t), \end{align} for all $t\in [-\pi,\pi]$.

Edit 2: If assumptions are needed, then we can assume that $\sum b_{s,m}^2 <\infty$.

Answer 1

Here's some work on a Fourier expansion that could maybe lead to some conditions on $b_{s,m}$. Let $f(t) = \sin^{2s}(t) \cos^{2m}(t)$, for a fixed $s$ and $m$ (apologies for the inconsistent notation). First note that there is no sine term in the Fourier transform of $f(t)$ with a non-zero coefficient, since $f(t)$ is an even function and $\sin(nt)$ is an odd function for every integer $n$. So, let its Fourier transform be $$ f(t) = \sum_{n=0}^{\infty} \tilde f_n \cos(nt). $$ Note that $$ \tilde f_n = \frac{1}{\pi} \int_{-\pi}^\pi \sin^{2s}(t) \cos^{2m}(t) \cos(nt) \, dt = \frac{1}{2\pi} \int_{0}^\pi (1-\cos(2t))^s (1+\cos(2t))^m \cos(nt) \, dt $$ must be 0 for every odd value of $n$. To see this, define $u = 2t-\pi$, and the integral becomes $$ \tilde f_n = \frac{1}{4\pi} \int_{-\pi}^\pi (1+\cos(u))^s (1-\cos(u))^m \cos \left( \frac{nu}{2} + \frac{n\pi}{2} \right) \, du $$ The left term in the integrand is an even function of $u$. When $n$ is odd, the right term is odd, hence making the integral zero. Otherwise, let $n=2k$, making the (even) right term $\cos(ku+k\pi) = (-1)^k \cos(ku)$: $$ \tilde f_{2k} = \frac{(-1)^k}{4\pi} \int_{-\pi}^\pi (1+\cos(u))^s (1-\cos(u))^m \cos(ku) \, du $$ But since the integrand is still even, the same game yields $$ \tilde f_{2k} = \frac{(-1)^k}{4\pi} \int_{-\pi}^\pi \left(1-\sin\left(\frac{v}{2}\right)\right)^s \left(1+\sin\left(\frac{v}{2}\right)\right)^m \cos\left(\frac{kv}{2}+\frac{k\pi}{2}\right) \, dv$$ and thus $$ \tilde f_n = \left\{ \begin{array}{l|l} \frac{(-1)^\ell}{4\pi} \int_{-\pi}^\pi \left(1-\sin\left(\frac{v}{2}\right)\right)^s \left(1+\sin\left(\frac{v}{2}\right)\right)^m \cos\left(\ell v\right) \, dv & n = 4 \ell\\ \frac{(-1)^\ell}{4\pi} \int_{-\pi}^\pi \left(1-\sin\left(\frac{v}{2}\right)\right)^s \left(1+\sin\left(\frac{v}{2}\right)\right)^m \sin\left(\left(\ell+\frac{1}{2}\right)v\right) \, dv & n = 4 \ell + 2 \\ 0 & \textrm{otherwise} \end{array} \right. $$ where $v = 2u-\pi$. Now the left side of the integrand is neither even nor odd, so the game stops here.

Perhaps there is then hope for evaluating each of these cases, even if just through the messy binomial expansion that expresses each as a sum of terms proportional to $$ \int_{-\pi}^\pi \sin\left(\frac{v}{2} \right)^w \cos (\ell v) \, dv = \left\{ \begin{array}{l|l} \frac{(-1)^{\ell}\pi}{2^{w-1}} \binom{w}{w/2-\ell} & \textrm{even } w \\ 0 & \textrm{otherwise} \end{array} \right. $$ (note that the cosine is always an even function while the sine power is only even when its exponent is even) and $$ \int_{-\pi}^\pi \sin\left(\frac{v}{2} \right)^w \sin \left(\left(\ell+\frac{1}{2} \right) v\right) \, dv = \left\{ \begin{array}{l|l} \frac{(-1)^{\ell}\pi}{2^{w-1}} \binom{w}{(w-1)/2-\ell} & \textrm{odd } w \\ 0 & \textrm{otherwise} \end{array} \right. $$ where $w$ is an index in the summation. These integrals were derived using the expansions given in nth power of sine as sum of sine and cosine terms.

Next, using the series expansion $$ (1-S)^s (1+s)^m = \sum_{a=0}^s \sum_{b=0}^m (-1)^a \binom{s}{a} \binom{m}{b} S^{a+b} $$ we arrive finally at $$ \tilde f_n = \left\{ \begin{array}{l|l} \sum_{a=0}^s \sum_{b=0}^m I(a+b \textrm{ even}) \, \frac{(-1)^a}{2^{a+b+1}} \binom{s}{a} \binom{m}{b} \binom{a+b}{(a+b)/2-\ell} & n = 4 \ell\\ \sum_{a=0}^s \sum_{b=0}^m I(a+b \textrm{ odd}) \, \frac{(-1)^a}{2^{a+b+1}} \binom{s}{a} \binom{m}{b} \binom{a+b}{(a+b-1)/2-\ell} & n = 4 \ell + 2 \\ 0 & \textrm{otherwise} \end{array} \right. $$

Since the function in the question was a sum of my $f(t)$ over all non-negative values of $s$ and $m$ with the coefficients $b_{s,m}$, and is identically zero, the above Fourier coefficients could in principle lead to some restrictions on $b_{s,m}$. However, they might be rather messy, unless there is some magical cancellation of some of the binomial and power-of-two coefficients. Rearranging summations, one class of these restrictions would be, for arbitrary $\ell$, $$ 0 = \sum_{a=0}^\infty \sum_{b=0}^\infty I(a+b \textrm{ even}) \, \frac{(-1)^a}{2^{a+b+1}} \binom{a+b}{(a+b)/2-\ell} \, \left( \sum_{s=a}^\infty \sum_{m=b}^\infty b_{s,m} \binom{s}{a} \binom{m}{b} \right) $$

Answer 2

Let $\;y=x^2,\;$ then from the identity $$g(y)=\sum\limits_{s=0}^\infty\,\sum\limits_{m=0}^\infty\,b_{s,m}y^s(1-y)^m=f(x)=0,$$ should $$g^{(n)}(0)=g^{(n)}(1)=0,\qquad(n=0,1,2\dots)$$ or \begin{cases} \sum\limits_{m=0}^\infty\,b_{0,m} = \sum\limits_{s=0}^\infty\,b_{s,0} = 0\\[4pt] \sum\limits_{m=0}^\infty\,(-m\,b_{0,m}+s\,b_{1,m}) = \sum\limits_{s=0}^\infty\,(s\,b_{s,0}-mb_{s,1}) = 0\\[4pt] \sum\limits_{m=0}^\infty\,(m(m-1)b_{0,m}-sm\,b_{1,m}+s(s-1)\,b_{2,m}) \\[4pt] = \sum\limits_{s=0}^\infty\,(s(s-1)b_{s,0}-sm\,b_{s,1}+m(m-1)\,b_{s,2}) = 0\\[4pt] \dots\\[4pt] \sum\limits_{m=0}^\infty\,\sum\limits_{k=0}^n (-1)^k (m-k+1)_{(k)}(s-n+k-1)_{(n-k)} b_{k,m}\\[4pt] = \sum\limits_{s=0}^\infty\,\sum\limits_{k=0}^n (-1)^k (m-k+1)_{(k)} (s-n+k-1)_{(n-k)} b_{s,n-k} = 0\\[4pt] \dots, \end{cases} where $\;a_{(p)} = a(a+1)\dots(a+p-1)\;$ is the Pochhammer symbol.