2 fixed typo

This is a repost, and partial rewrite of an earlier deleted answer by Anixx. If you want to discuss the wisdom of that deletion, take it to the meta thread; let's keep this post focused on math only. This answer is commuunity community wiki, so that others can improve it.

If $a_k$ is any sequence of real numbers, indexed by the nonnegative integers, then define $\Delta^m(a) = \sum_{k=0}^{m} (-1)^k \binom{m}{k} a_k$. Then, for integer $n$, we have $a_n = \sum_{m=0}^{\infty} \binom{n}{m} \Delta^m(a)$. Note that the sum is finite, because all but finitely many binomial coefficients vanish. One can then try defining $$A(x) = \sum_{m=0}^{\infty} \binom{x}{m} \Delta^m(a).$$ If this sum converges, it defines a function $A$ which interpolates $a_n$. This is sometimes called Newton's interpolation formula.

Anixx points out that, $a_n = \sin^{[n]}(x)$ this method appears to give a good answer, but for $\cos^{[n]}(x)$, it appears not to.

If $a_k$ is any sequence of real numbers, indexed by the nonnegative integers, then define $\Delta^m(a) = \sum_{k=0}^{m} (-1)^k \binom{m}{k} a_k$. Then, for integer $n$, we have $a_n = \sum_{m=0}^{\infty} \binom{n}{m} \Delta^m(a)$. Note that the sum is finite, because all but finitely many binomial coefficients vanish. One can then try defining $$A(x) = \sum_{m=0}^{\infty} \binom{x}{m} \Delta^m(a).$$ If this sum converges, it defines a function $A$ which interpolates $a_n$. This is sometimes called Newton's interpolation formula.
Anixx points out that, $a_n = \sin^{[n]}(x)$ this method appears to give a good answer, but for $\cos^{[n]}(x)$, it appears not to.