TAOCP 1.2.11.3 Exercise 1
Equation (5) states that f(x) = \sum_{k=0}^{n}\frac{f^{(k)}(0)}{k!
Section 1.2.11.3: Some Asymptotic Calculations
Exercise 1. [HM20] Prove Eq. (5) by induction on $n$.
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Solve time: 1m21s
Solution
Equation (5) states that
$$ f(x) = \sum_{k=0}^{n}\frac{f^{(k)}(0)}{k!}x^k + \frac1{n!}\int_0^x f^{(n+1)}(t)(x-t)^n,dt. \tag{5} $$
We prove this formula by induction on $n$.
For $n=0$,
$$ f(x) = f(0)+\int_0^x f'(t),dt, $$
by the fundamental theorem of calculus. This is exactly (5) when $n=0$.
Assume that (5) holds for some $n\ge0$:
$$ f(x) = \sum_{k=0}^{n}\frac{f^{(k)}(0)}{k!}x^k + \frac1{n!}\int_0^x f^{(n+1)}(t)(x-t)^n,dt. \tag{1} $$
Consider the remainder integral in (1). Integrating by parts with
$$ u=f^{(n+1)}(t), \qquad dv=\frac{(x-t)^n}{n!},dt, $$
gives
$$ du=f^{(n+2)}(t),dt, \qquad v=-\frac{(x-t)^{n+1}}{(n+1)!}. $$
Hence
\begin{align*}
\frac1{n!}\int_0^x f^{(n+1)}(t)(x-t)^n,dt
&=
\left[
-\frac{f^{(n+1)}(t)(x-t)^{n+1}}{(n+1)!}
\right]_{0}^{x}
&\qquad + \frac1{(n+1)!} \int_0^x f^{(n+2)}(t)(x-t)^{n+1},dt . \end{align*} At $t=x$ the boundary term is $0$; at $t=0$ it is $$ -\frac{f^{(n+1)}(0)x^{n+1}}{(n+1)!}. $$ Therefore $$ \frac1{n!}\int_0^x f^{(n+1)}(t)(x-t)^n,dt = \frac{f^{(n+1)}(0)x^{n+1}}{(n+1)!} + \frac1{(n+1)!} \int_0^x f^{(n+2)}(t)(x-t)^{n+1},dt. \tag{2} $$ Substituting (2) into (1), we obtain $$ f(x) = \sum_{k=0}^{n+1}\frac{f^{(k)}(0)}{k!}x^k + \frac1{(n+1)!} \int_0^x f^{(n+2)}(t)(x-t)^{n+1},dt. $$ This is equation (5) with $n$ replaced by $n+1$. Since the formula holds for $n=0$ and the truth for $n$ implies the truth for $n+1$, equation (5) holds for all $n\ge0$. This completes the proof. ∎ ::contentReference[oaicite:0]{index=0}