Showing $F(x+h)=F(x)+hF'(x)+\frac{h^2}{2}F''(x)+h^2\phi (h)$, not with Taylor series. The book I am studying, has this question, show $F(x+h)=F(x)+hF'(x)+\frac{h^2}{2}F''(x)+h^2\phi (h)$ where $\phi(h) \to 0$ as $h\to 0$. It is noted that this is a Taylor expansion, but the method suggested was to use $F(x+h)-F(x)=\int_x^{x+h}F'(y)dy$ and $F'(y)=F'(x)+(y-x)F''(x)+(y-x)\psi(y-x)$ where $\psi(h) \to 0$ as $h\to 0$. I tried it and am having trouble at the last step.\begin{align*} F(x+h)-F(x)&=\int_x^{x+h}F'(y)dy\\ &= \int_x^{x+h}F'(x)+(y-x)F''(x)+(y-x)\psi(y-x)dy\\ &=hF'(x)+\frac{h^2}{2}F''(x)+\int_x^{x+h}(y-x)\psi(y-x)dy\\&=hF'(x)+\frac{h^2}{2}F''(x)+\int_0^{h}k\psi(k)dk  \end{align*}
How do I finish it off, i.e. how do I show $\int_0^{h}k\psi(k)dk=h^2\phi (h)$. Thanks
 A: We need to find $\phi$ so $\phi(h)\to0$ as $h\to0^+$ and $\int_0^h k\psi(k)\,dk=h^2\phi(h).$ So we define $\phi$ in the only way possible: $$\phi(h)=h^{-2}\int_0^h k\psi(k)\,dk.$$
Defining $M(h)=\sup_{0<t<h}|\psi(t)|$ we have $$|\phi(h)|\le M(h)h^{-2}\int_0^h k\,dk=\frac12 M(h),$$so we need only show that $M(h)\to0$, which is clear; in fact if $|\psi(t)|<\epsilon$ for $0<t<\delta$ then $M(h)\le\epsilon$ for $0<h<\delta$.
A: I have to answer your question "how do we know that it will be equal to $h^2ϕ(h)$" here as I still do not have enough reputation to comment.
If I didn't guess wrongly, the book mentioned is actually the Fourier Analysis: An introduction by Elias M. Stein and Rami Shakarchi. I am also reading the book currently.
In that question, we hope to use the result to deduce that
$$
\frac{F(x+h)+F(x-h)-2F(x)}{h^2} \to F''(x) \text{  as } h\to 0,
$$
which seems to be intuitively true.
Then, from this final result we aim to prove, we can speculate that there must exist an error term like $\varepsilon(h) $, s.t.
$$
F(x+h)+F(x-h)-2F(x)=h^2(F''(x)+\varepsilon(h)) \text{ with } \varepsilon(h)\to 0 \text{  as } h\to 0.  
$$
Then, manipulate the expression above a bit and use the given hints we can see that the motivation for $h^2\phi(h)$.
I hope this partly answers your question.
