Using the first fundamental theorem of calculus to derive Taylor expansion/series

Question

Appendix C of my textbook, Algorithms for Optimization, by Kochenderfer and Wheeler, begins to derive the Taylor expansion/series as follows:

From the first fundamental theorem of calculus, we know that

$$f(x + h) = f(x) + \int_0^h f'(x + a) \ da$$

Something about this application of the first fundamental theorem seemed wrong to me, so I looked at the Wikipedia article. The article presents the relevant equation as

$$F(x_{1}+\Delta x)-F(x_{1})=\int _{x_{1}}^{x_{1}+\Delta x}f(t)\,dt.\qquad$$

If we set $\Delta x = h$, $t = x + a$, and $x_1 = 0$, we get

$$F(0 + h)-F(0)=\int _{0}^{0+h}f(x + a)\,dt,\qquad$$

which is a different result from what the author has.

I'm confused as to how the author's result is valid. I would greatly appreciate it if people would please take the time to clarify this.

Answer 1

For a fixed $x$, let $g(h)=f(x+h)$. Then $g'(h)=f'(x+h)$ and\begin{align}f(x+h)-f(x)&=g(h)-g(0)\\&=\int_0^hg'(a)\,\mathrm da\text{ (by the fundamental theorem of Calculus)}\\&=\int_0^hf'(x+a)\,\mathrm da.\end{align}So, yes, that textbook of yours is correct.

Note that your equality$$F(0+h)-F(0)=\int _0^{0+h}f(x+a)\,dt,$$is weird, since the RHS depends on $x$, whereas the LHS doesn't.

Answer 2

Set not $x_1=0$, but $x_1=x$. Also, what Wikipedia calls $F$ would in your example be $f$. Then Wikipedia's $f$ would become $f'$, and we would have:$$f(x + h)-f(x)=\int _{x}^{x+h}f'(t)\,dt,\qquad$$ From there, a substitution $t=x+a, dt=da$ would give the author's claim.