Exercise 8.3.9

Question

\begin{enumerate}[label=(\alph*)] 
\item Show $$f(x) = f(0) +\int_0^x f'(t) dt\ .$$
\item Now use a previous result from this section to show $$f(x) = f(0) + f'(0)x + \int_0^x f''(t)(x-t)dt \ .$$
\item Continue in this fashion to complete the proof of the theorem.
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
\item This is a simple application of the Fundamental Theorem of Calculus.
\item Use Exercise 8.3.2 (integration by parts), with $h(t) = f'(t)$ and $k(t) = t-x$:
$$\int_0^x f'(t)dt = f'(x)\cdot 0 - f'(0) \cdot (-x) - \int_0^x f''(t) (t-x) dt = f'(0)x + \int_0^x f''(t)(x-t)dt$$
\item In general, with $h = f^{(n+1)}$ and $k(t) = -\frac{(x-t)^{n+1}}{n+1}$,
$$\int_0^x f^{(n+1)}(t)(x-t)^n = f^{(n+1)}(x) \cdot 0 + f^{(n+1)}(0) \cdot \left(\frac{x^{n+1}}{n+1}\right) + \frac{1}{n+1}\int_0^x f^{(n+2)}(t)(x-t)^{n+1} dt$$
$$\frac{1}{n!} \int_0^x f^{(n+1)}(t) (x-t)^n = \frac{f^{(n+1)}}{(n+1)!}x^{n+1} + \frac{1}{(n+1)!} \int_0^x f^{(n+2)}(t) (x-t)^{n+1} dt$$
Armed with this, we can show inductively that, for any $N \in \mathbf{N}$,
$$f(x) = \frac{1}{N!}\int_0^x f^{(N+1)}(t) (x-t)^{N} dt + \sum^N_{i=0} \frac{f^{(i)}(x)}{i!}x^i = E_N(x) + S_N(x)$$
 \end{enumerate}

Exercise 8.3.9

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Exercise 8.3.9

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