Evaluating a sinusoidal series

Question

Define the sequence of functions

$$f_n(x)=\sum_{m=n}^\infty(-1)^m\frac{x^{2m}}{(2m+1)!} {m \choose n} $$

Is there a closed form expression for arbitrary $n$? It is clear that the result should assume the form

$$n!f_n(x)=P_n(x)\cos x+Q_n(x)\frac{\sin x}{x}$$

where $P_n, Q_n$ are $n$-th degree polynomials, but I haven't been able to calculate the coefficients of the polynomials. The coefficients may be connected to Stirling numbers, if one could express the falling factorial in terms of the following linear combination of falling factorials:

$$(m)_n=\sum_{k=0}^n A_{nk}(2m+1)_k$$

Then the infinite sums would become simpler, but this requires an explicit calculation of the $A_{nk}$'s.

For up to $n=4$, the expressions read $$f_0(x)=\frac{\sin x}{x}$$ $$f_1(x)=\frac{1}{2}\cos x-\frac{1}{2}\frac{\sin x}{x}$$ $$f_2(x)=-\frac{3}{8} \cos x + \frac{3 - x^2}{8 } \frac{\sin x}{x}$$ $$f_3(x)=\frac{1}{48} (15 - x^2) \cos x + \frac{2 x^2 - 5}{16 } \frac{\sin x}{x}$$ $$f_4(x)=\frac{5}{384} (-21 + 2 x^2) \cos x + \frac{105 - 45 x^2 + x^4}{384 } \frac{\sin x}{x}$$

EDIT: I am mostly interested in finding the value $f_n(\pi/2)$ explicitly, so even though the expression provided in the answers below is technically a closed form, it does not help evaluate the function at $x=\pi/2$ directly.

Answer 1

Those are very close to the spherical Bessel functions of the first kind $$f_{n}=\left(\frac{-z}{2}\right)^{n}j_{n}(z)=(-1)^{n}\frac{\sqrt{\pi}}{2}\left(\frac{z}{2}\right)^{n-\frac{1}{2}}J_{n+\frac{1}{2}}(z).$$

Answer 2

Letting $y:=-x^2$, we reduce the calculation to $$g_n(y):=\sum_{m=n}^\infty \frac{y^{m-n}}{(2m+1)!}\,m(m-1)\cdots(m-(n-1)) \\ =\partial_y^n\sum_{m=0}^\infty \frac{y^m}{(2m+1)!} =\frac12\,\partial_y^n h_+(y)+\frac12\,\partial_y^n h_-(y),$$ where $$h_\pm(y):=\frac{e^{\pm\sqrt y}}{\pm\sqrt y}.$$ Note that $$\partial_y h_\pm(y)=\frac12\,h_\pm(y)(\pm y^{1/2}-y^{-1}).$$ So, by the Leibnitz differentiation rule,
$$\partial_y^n h_\pm(y)=h_\pm(y)P_{n,\pm}(y^{-1/2})$$ where $P_{n,\pm}$ are certain polynomials satisfying a certain recurrence.

Answer 3

Here's a formula for $A_{nk}$: $$A_{nk} = (-1)^{n-k}2^{k-2n} \frac{(2n-k)!}{k!\,(n-k)!}.$$ This can be proved by standard methods for proving binomial coefficient identities.