Subsection 3.4.10 (Optional) — Derivation of the Error Formulae
¶In this section we will derive the formula for the error that we gave in equation 3.4.33 — namely
\begin{align*}
R_n(x) = f(x) - T_n(x) &= \frac{1}{(n+1)!}f^{(n+1)}(c)\cdot (x-a)^{n+1}
\end{align*}
for some \(c\) strictly between \(a\) and \(x\text{,}\) and where \(T_n(x)\) is the \(n^\mathrm{th}\) degree Taylor polynomial approximation of \(f(x)\) about \(x=a\text{:}\)
\begin{align*}
T_n(x) &= \sum_{k=0}^n \frac{1}{k!} f^{(k)}(a).
\end{align*}
Recall that we have already proved a special case of this formula for the constant approximation using the Mean-Value Theorem (Theorem 2.13.5). To prove the general case we need the following generalisation 22 of that theorem:
Theorem 3.4.38 Generalised Mean-Value Theorem
Let the functions \(F(x)\) and \(G(x)\) both be defined and continuous on \(a\le x\le b\) and both be differentiable on \(a \lt x \lt b\text{.}\) Furthermore, suppose that \(G'(x)\ne 0\) for all \(a \lt x \lt b\text{.}\) Then, there is a number \(c\) obeying \(a \lt c \lt b\) such that
\begin{gather*}
\frac{F(b)-F(a)}{G(b)-G(a)}=\frac{F'(c)}{G'(c)}
\end{gather*}
Notice that setting \(G(x) = x\) recovers the original Mean-Value Theorem. It turns out that this theorem is not too difficult to prove from the MVT using some sneaky algebraic manipulations:
Proof
- First we construct a new function \(h(x)\) as a linear combination of \(F(x)\) and \(G(x)\) so that \(h(a)=h(b)=0\text{.}\) Some experimentation yields
\begin{gather*}
h(x)=\big[F(b)-F(a)\big]\cdot \big[G(x)-G(a)\big]-
\big[G(b)-G(a)\big] \cdot \big[F(x)-F(a)\big]
\end{gather*}
- Since \(h(a)=h(b)=0\text{,}\) the Mean-Value theorem (actually Rolle's theorem) tells us that there is a number \(c\) obeying \(a \lt c \lt b\) such that \(h'(c)=0\text{:}\)
\begin{align*}
h'(x) &= \big[F(b)-F(a)\big] \cdot G'(x) - \big[G(b)-G(a)\big] \cdot F'(x) & \text{
so}\\
0 &= \big[F(b)-F(a)\big] \cdot G'(c) - \big[G(b)-G(a)\big] \cdot F'(c)
\end{align*}
Now move the \(G'(c)\) terms to one side and the \(F'(c)\) terms to the other:
\begin{align*}
\big[F(b)-F(a)\big] \cdot G'(c) &= \big[G(b)-G(a)\big] \cdot F'(c).
\end{align*}
- Since we have \(G'(x) \neq 0\text{,}\) we know that \(G'(c) \neq 0\text{.}\) Further the Mean-Value theorem ensures 23 that \(G(a) \neq G(b)\text{.}\) Hence we can move terms about to get
\begin{align*}
\big[F(b)-F(a)\big] &= \big[G(b)-G(a)\big] \cdot \frac{F'(c)}{G'(c)}\\
\frac{F(b)-F(a)}{G(b)-G(a)} &= \frac{F'(c)}{G'(c)}
\end{align*}
as required.
Armed with the above theorem we can now move on to the proof of the Taylor remainder formula.
Proof of equation 3.4.33
We begin by proving the remainder formula for \(n=1\text{.}\) That is
\begin{align*}
f(x) - T_1(x) &= \frac{1}{2}f''(c) \cdot(x-a)^2
\end{align*}
- Start by setting
\begin{align*}
F(x) &= f(x)-T_1(x) &
G(x) &= (x-a)^2
\end{align*}
Notice that, since \(T_1(a)=f(a)\) and \(T'_1(x) = f'(a)\text{,}\)
\begin{align*}
F(a) &= 0 & G(a)&=0\\
F'(x) &= f'(x)-f'(a) & G'(x) &= 2(x-a)
\end{align*}
- Now apply the generalised MVT with \(b=x\text{:}\) there exists a point \(q\) between \(a\) and \(x\) such that
\begin{align*}
\frac{F(x)-F(a)}{G(x)-G(a)} &= \frac{F'(q)}{G'(q)}\\
\frac{F(x)-0}{G(x) - 0} &= \frac{f'(q)-f'(a)}{2(q-a)}\\
2 \cdot \frac{F(x)}{G(x)} &= \frac{f'(q)-f'(a)}{q-a}
\end{align*}
- Consider the right-hand side of the above equation and set \(g(x) = f'(x)\text{.}\) Then we have the term \(\frac{g(q)-g(a)}{q-a}\) — this is exactly the form needed to apply the MVT. So now apply the standard MVT to the right-hand side of the above equation — there is some \(c\) between \(q\) and \(a\) so that
\begin{align*}
\frac{f'(q)-f'(a)}{q-a} &= \frac{g(q)-g(a)}{q-a} = g'(c) = f''(c)
\end{align*}
Notice that here we have assumed that \(f''(x)\) exists.
- Putting this together we have that
\begin{align*}
2 \cdot \frac{F(x)}{G(x)} &= \frac{f'(q)-f'(a)}{q-a} = f''(c)\\
2 \frac{f(x)-T_1(x)}{(x-a)^2} &= f''(c)\\
f(x) - T_1(x) &= \frac{1}{2!} f''(c) \cdot (x-a)^2
\end{align*}
as required.
Oof! We have now proved the cases \(n=1\) (and we did \(n=0\) earlier).
To proceed — assume we have proved our result for \(n=1,2,\cdots, k\text{.}\) We realise that we haven't done this yet, but bear with us. Using that assumption we will prove the result is true for \(n=k+1\text{.}\) Once we have done that, then
- we have proved the result is true for \(n=1\text{,}\) and
- we have shown if the result is true for \(n=k\) then it is true for \(n=k+1\)
Hence it must be true for all \(n \geq 1\text{.}\) This style of proof is called mathematical induction. You can think of the process as something like climbing a ladder:
- prove that you can get onto the ladder (the result is true for \(n=1\)), and
- if I can stand on the current rung, then I can step up to the next rung (if the result is true for \(n=k\) then it is also true for \(n=k+1\))
Hence I can climb as high as like.
- Let \(k \gt 0\) and assume we have proved
\begin{align*}
f(x) - T_k(x) &= \frac{1}{(k+1)!} f^{(k+1)}(c) \cdot (x-a)^{k+1}
\end{align*}
for some \(c\) between \(a\) and \(x\text{.}\)
- Now set
\begin{align*}
F(x) &= f(x) - T_{k+1}(x) & G(x) &= (x-a)^{k+1}\\
\end{align*}
and notice that, since \(T_{k+1}(a)=f(a)\text{,}\)
\begin{align*}
F(a) &= f(a)-T_{k+1}(a)=0 & G(a) &= 0 & G'(x) &= (k+1)(x-a)^k
\end{align*}
and apply the generalised MVT with \(b=x\text{:}\) hence there exists a \(q\) between \(a\) and \(x\) so that
\begin{align*}
\frac{F(x)-F(a)}{G(x)-G(a)} &= \frac{F'(q)}{G'(q)} &\text{which becomes}\\
\frac{F(x)}{(x-a)^{k+1}} &= \frac{F'(q)}{(k+1)(q-a)^k} & \text{rearrange}\\
F(x) &= \frac{(x-a)^{k+1}}{(k+1)(q-a)^k} \cdot F'(q)
\end{align*}
- We now examine \(F'(q)\text{.}\) First carefully differentiate \(F(x)\text{:}\)
\begin{align*}
F'(x) &= \diff{}{x} \bigg[f(x) - \bigg( f(a) + f'(a)(x-a) + \frac{1}{2}
f''(a)(x-a)^2 + \cdots \\
\amp\hskip2.5in+ \frac{1}{k!}f^{(k)}(x-a)^k \bigg) \bigg]\\
&= f'(x) - \bigg( f'(a) + \frac{2}{2} f''(a)(x-a) + \frac{3}{3!} f'''(a)(x-a)^2 +
\cdots \\
\amp\hskip2.5in+ \frac{k}{k!}f^{(k)}(a) (x-a)^{k-1} \bigg)\\
&= f'(x) - \bigg( f'(a) + f''(a)(x-a) + \frac{1}{2} f'''(a)(x-a)^2 +\cdots \\
\amp\hskip2.5in+ \frac{1}{(k-1)!}f^{(k)}(a)(x-a)^{k-1} \bigg)
\end{align*}
Now notice that if we set \(f'(x) = g(x)\) then this becomes
\begin{align*}
F'(x) &= g(x) - \bigg( g(a) + g'(a)(x-a) + \frac{1}{2} g''(a)(x-a)^2 +
\cdots \\
\amp\hskip2.5in+ \frac{1}{(k-1)!}g^{(k-1)}(a)(x-a)^{k-1} \bigg)
\end{align*}
So \(F'(x)\) is then exactly the remainder formula but for a degree \(k-1\) approximation to the function \(g(x) = f'(x)\text{.}\)
- Hence the function \(F'(q)\) is the remainder when we approximate \(f'(q)\) with a degree \(k-1\) Taylor polynomial. The remainder formula, equation 3.4.33, then tells us that there is a number \(c\) between \(a\) and \(q\) so that
\begin{align*}
F'(q) &= g(q) - \bigg( g(a) + g'(a)(q-a) + \frac{1}{2} g''(a)(q-a)^2 + \cdots \\
\amp\hskip2.5in + \frac{1}{(k-1)!}g^{(k-1)}(a)(q-a)^{k-1} \bigg)\\
&= \frac{1}{k!} g^{(k)}(c) (q-a)^k = \frac{1}{k!} f^{(k+1)}(c)(q-a)^k
\end{align*}
Notice that here we have assumed that \(f^{(k+1)}(x)\) exists.
- Now substitute this back into our equation above
\begin{align*}
F(x) &= \frac{(x-a)^{k+1}}{(k+1)(q-a)^k} \cdot F'(q)\\
&= \frac{(x-a)^{k+1}}{(k+1)(q-a)^k} \cdot \frac{1}{k!} f^{(k+1)}(c)(q-a)^k\\
&= \frac{1}{(k+1)k!} \cdot f^{(k+1)}(c) \cdot
\frac{(x-a)^{k+1}(q-a)^k}{(q-a)^k}\\
&= \frac{1}{(k+1)!} \cdot f^{(k+1)}(c) \cdot(x-a)^{k+1}
\end{align*}
as required.
So we now know that
- if, for some \(k\text{,}\) the remainder formula (with \(n=k\)) is true for all \(k\) times differentiable functions,
- then the remainder formula is true (with \(n=k+1\)) for all \(k+1\) times differentiable functions.
Repeatedly applying this for \(k=1,2,3,4,\cdots\) (and recalling that we have shown the remainder formula is true when \(n=0,1\)) gives equation 3.4.33 for all \(n=0,1,2,\cdots\text{.}\)