Let's consider divided difference:
$$f[a_1, a_1, \dots,a_1, a_2]$$
where we have $n$ occurrences of $a_1$ and function $f$ is $n$ times differentiable on segment $K = \mathbb C$. I want to find closed form solution for function $f$.
My problem is that when we consider $n = 1$ we have:
$$f[a_1, a_2] = \frac{f(a_2) - f(a_1)}{a_2 - a_1}$$
And for $n = 2$:
$$f[a_1, a_1, a_2] = \frac{f[a_1, a_2] - f[a_1, a_1]}{a_2 - a_1}$$
but here, the problem is that divided difference $f[a_1, a_1] = \frac{f(a_1) - f(a_1)}{a_1 - a_1} = \frac 0 0$ is problematic to calculate.
Could you please give me a hand in solving this problem?
You can understand $f[a,a,...,a]$ as the limit of $f[a,a+h,a+2h,...,a+nh]$ for $h\to 0$. It should be almost intuitive that this evaluates to $$ f[\underbrace{a,a,...,a}_{n+1~{\rm times}}]=\frac1{n!}f^{(n)}(a). $$ For your expression one gets thus $$ f[\underbrace{a,a,...,a}_{n~{\rm times}},b] =\frac{f[\overbrace{a,a,...,a}^{n-1},b]-f[\overbrace{a,a,...,a}^{n}]}{b-a} =\frac{f[\overbrace{a,a,...,a}^{n-1},b]-\frac1{(n-1)!}f^{(n-1)}(a)}{b-a} $$ which iterates out to $$ =\frac{f(b)-f(a)-f'(a)(b-a)-...-\frac1{(n-1)!}f^{(n-1)}(a)(b-a)^{n-1}}{(b-a)^n}. $$ So the left side is the coefficient of the remainder term of the Taylor polynomial in $a$ of degree $n-1$.