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.

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

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.

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.

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.

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