What happens to the remainder when differentiating a Taylor polynomial

Question

Say we have $F(x+h) = F(x) + hf(x) + \frac{h^2}{2}f'(x) + \frac{h^3}{6}f''(x) + \frac{h^4}{24}f'''(\xi)$ and when we differentiate we get $f(x+h) = f(x) + hf'(x) + \frac{h^2}{2}f''(x) + \frac{h^3}{6}f'''(x)$, but we expect a remainder for this aswell. Is there a correlation between the remainder $\xi$ from $F(x+h)$ and the missing remainder $f(x)$?

Will $f(x+h) = f(x) + hf'(x) + \frac{h^2}{2}f''(x) + \frac{h^3}{6}f'''(x) + \frac{h^4}{24}f''''(\xi)$ or will this be another number at the place for $\xi$.

Answer 1

No, these numbers will be, in the general case, be close, but not equal.

The remainder of the Taylor series expansion for $F(x+h)$ starts with $$ \frac{h^4}{24}f'''(x)+\frac{h^5}{5!}f^{(4)}(x)+O(h^6) = \frac{h^4}{4!}\left(f'''(x)+\frac{h}{5}f^{(4)}(x)+O(h^2)\right) $$ This can now be interpreted as Taylor expansion of $f'''$, $$ \frac{h^4}{4!}f'''\left(x+\frac{h}{5}+O(h^2)\right)+O(h^6), $$ and this midpoint estimate contains no reference to the expanded function. So the midpoints will be the same up to $O(h^2)$. But when including the next term this reduction gives $$ \frac{h^4}{4!}f'''\left(x+\frac{h}{5}\right) +\left(\frac{h^6}{6!}-\frac{h^6}{50\cdot 4!}\right)f^{(5)}(x)+O(h^7) =\frac{h^4}{4!}f'''\left(x+\frac{h}{5}+\frac{h^2}{75}\frac{f^{(5)}(x)}{f^{(4)}(x)}\right)+O(h^7) $$ This now clearly depends on $f$ and its derivatives, so replacing $F$ with $f$ will in general give a different midpoint.

Answer 2

If you want to differentiate that first expression with respect to $h$ (as it appears you are), then your last term should be "$... + \frac{h^3}{6}f'''(\xi) \cdot \xi'(h)$.

As to whether or not $\xi$ is a differentiable function of $h$, or what further conditions might ensure that it is, I'm not sure.