Proving Taylor’s Theorem via Indeterminate Coefficients I’m going over a book and it shows a rather neat way of proving Taylor’s theorem.$$f(x+h)=f(x)+hf’(x)+\frac {h^2}{1.2}f’’(x)+\ldots+\frac {h^n}{n!}f^n(x+\theta h)\tag1$$But given how ancient it is, I’m having some trouble understanding what to do. Here’s what the book says

Proof: (i) Assume $f(x+h)=A+Bh+Ch^2+\&\text c.$ Differentiate both sides of this equation, first for $x$ and again for $h$, and equate coefficients in the two results.
(ii) Lemma. If $f(x)$ vanishes when $x=a$ and also when $x=b$, and if $f(x)$ and $f’(x)$ are continuous functions between the same limits; then $f’(x)$ vanishes for some value of $x$ between $a$ and $b$. For $f’(x)$ must change sign somewhere between the assigned limits, and being continuous, it must vanish in passing from plus to minus.
(iii) Now, the expression$$\begin{multline}f(a+x)-f(a)-xf’(x)-\ldots-\frac {x^n}{n!}f^n(a)-\frac {x^{n+1}}{(n+1)!}\frac {(n+1)!}{h^{n+1}}\left\{f(a+h)-f(a)-hf’(a)-\ldots-\frac {h^n}{n!}f^n(a)\right\}\end{multline}$$vanishes when $x=0$ and when $x=h$. Therefore the differential coefficient vanishes when $x=0$ for some value of $x$ between $0$ and $h$ by the lemma. Let $\theta h$ be this value. Differentiate and apply the lemma to the resulting expression, which vanishes when $x=0$ and when $x=\theta\theta h$. Perform this same process $n+1$ times successively, writing $\theta h$ for $\theta\theta h$, &c., since $\theta$ merely stands for some quantity less than unity. The result shews that$$f^{n+1}(a+x)-\frac {(n+1)!}{h^{n+1}}\left\{f(a+h)-f(a)-hf’(a)-\ldots-\frac {h^n}{n!}f^n(a)\right\}$$vanishes when $x=\theta h$. Substituting $\theta h$ and equating to zero, the theorem is proven.

I’m able to prove the first part quite easily, establishing an infinite expansion for $f(x+h)$, but I’m not sure how they’re getting both expressions for (iii) and what they mean by ‘differentiating successively for $\theta h$ for $\theta\theta h$.’
Any help?
 A: First of all, let me point out that (i) does not establish an infinite expansion for $f(x + h)$. It shows that if $f(x+h)$ has an expansion, then that expansion is given by the Taylor series. But there are infinitely differentiable functions for which the Taylor series does not give the function. $e^{-1/x^2}$ expanded about $0$ is the most commonly given example. All of its derivatives at $0$ are $0$. 
(ii) seems to be a rather awkward proof of Rolles' theorem, taking for granted that if $f(x) = 0$ for two values of $x$, then of course the derivative must take on both positive and negative values between. The common proof given in calculus courses is much more robust (other than needing the extreme value theorem, which is always stated without proof in those courses).
(iii) addresses the deficit in (i) to actually prove the result. The "$\theta h$ notation is rather poor, IMO, as it makes it seem like the same multiplier $\theta$ works at each step, which is not the case. The expression (which I will call $F(x)$) was chosen so that $F(h) = 0$ and for all $k \le n, F^{(k)}(0) = 0$. What he notes is that


*

*Since $F(0) = F(h) = 0$, there is some point $h_1 \in [0, h]$, which he calls $\theta h$, such that $F'(h_1) = 0$.

*Since $F'(0) = F'(h_1) = 0$, there is some point $h_2 \in [0, h_1]$ (which he calls $\theta\theta h$) such that $F''(h_2) = 0$.

*Since $F''(0) = F''(h_2) = 0$, there is some point $h_3 \in [0, h_2]$ such that $F'''(h_3) = 0$.
$$\vdots$$

*Since $F^{(n)}(0) = F^{(n)}(h_n) = 0$, there is some point $h_{n+1}
       \in [0, h_n]$ such that $F^{(n+1)}(h_{n+1}) = 0$


If you do the calculation, you will see that 
$$F^{(n+1)}(x) = f^{(n+1)}(a+x)-\frac {(n+1)!}{h^{n+1}}\left\{f(a+h)-f(a)-hf’(a)-\ldots-\frac {h^n}{n!}f^{(n)}(a)\right\}$$
Therefore since $F^{(n+1)}(h_{n+1}) = 0$, we have 
$$f^{(n+1)}(a+h_{n+1})=\frac {(n+1)!}{h^{n+1}}\left\{f(a+h)-f(a)-hf’(a)-
\ldots-\frac {h^n}{n!}f^{(n)}(a)\right\}$$
or turning it around,
$$f(a + h) = f(a)+hf’(a)+
\ldots+\frac {h^n}{n!}f^{(n)}(a) + \frac {h^{n+1}}{(n+1)!}f^{(n+1)}(a+h_{n+1})$$
