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 any $n$? 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.