$f(a+h)= \sum_{k=0}^{n} (1/k!)f^{(k)}(a) h^k$ for any polynomial f(x) (Comparing summations)

Question

My final goal is to show that for any polynomial $f(x)∈F[x]$ ,

$f(a+h)= \sum_{k=0}^{n} (1/k!)f^{(k)}(a) h^k$ (where $F$:field and $a, h∈F$)

I expanded $LHS$ and $RHS$ terms and arrived to the point that, for polynomial $f(x)=\sum_{i=0}^{n} b_i x^i $,

$LHS= \sum_{i=0}^{n} \sum_{k=0}^{i} {i \choose k} b_i a^{i-k} h^k$ ------(A)

and

$RHS = \sum_{k=0}^{n} \sum_{i=k}^{n} {i \choose k} b_i a^{i-k} h^k$ ------(B)

but am stuck and do not see how I could proceed.

How do I know that the two summations (A) and (B) are equal?

Please help! Thank you so much in advance.

Answer 1

As the formula $$f(a+h)= \sum_{k=0}^{n} \frac{1}{k!}f^{(k)}(a) h^k$$ is linear in $f$, it is sufficient to prove it for monomials. And for a monomial $p(x) = x^n$ of degree $n$ and $0 \le k \le n$

$$p^{(k)}(a) = \frac{n!}{(n-k)!}a^{n-k}.$$ Therefore

$$\sum_{k=0}^{n} \frac{1}{k!}f^{(k)}(a) h^k= \sum_{k=0}^{n}\frac{n!}{(n-k)!k!}a^{n-k}h^k = p(a+h)$$ according to the binomial theorem

$$(a+h)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k}h^k.$$

Note: your formula is only valid if $n$ is greater of equal to the degree of the polynomial.