23. Taylor Series

e. Taylor Remainder

The Taylor remainder of degree \(k\) centered at \(x=a\) for the function \(f(x)\) is \[ R_k f(x)=f(x)-T_k f(x) \] where the Taylor polynomial of degree \(k\) is \[ T_k f(x)=\sum_{n=0}^k \dfrac{\,f^{(n)}(a)}{n!}(x-a)^n \]

1. Formula and Inequality

If the \((k+1)^\text{st}\) derivative exists on an open interval containing \(a\) and \(x\), then there is a number \(c\) between \(a\) and \(x\) such that the Taylor remainder of degree \(k\) is \[ R_k f(x)=\dfrac{f^{(k+1)}(c)}{(k+1)!}(x-a)^{k+1} \]

The the proof is similar to that for the Mean Value Theorem and is also based on Rolle's Theorem.

If \(g(x)\) is a function on the interval \([a,b]\) with a continuous derivative, \(g'(x)\), and \(g(a)=g(b)\) then there is a number \(c\) in \((a,b)\) such that \[ g'(c)=0. \]

a. Proof

To simplify notation, we replace \(x\) by \(b\) and assume \(b>a\). Then we want to prove there is a number \(c\) in \((a,b)\) such that: \[ f(b)-T_kf(b)=\dfrac{f^{(k+1)}(c)}{(k+1)!}(b-a)^{k+1} \]

Let \(S\) be defined by \[\begin{aligned} f(b)-T_kf(b)&=\dfrac{S}{(k+1)!}(b-a)^{(k+1)} \qquad (*) \end{aligned}\] (We are trying to prove \(S=f^{(k+1)}(c)\).) We now, define \[\begin{aligned} g(x)&=f(x)+f'(x)(b-x)+\dfrac{1}{2}f''(x)(b-x)^2+\cdots \\ &\quad+\dfrac{1}{k!}f^{(k)}(x)(b-x)^k+\dfrac{S}{(k+1)!}(b-x)^{k+1} \qquad \qquad \end{aligned}\] We compute the values of \(g(x)\) at the endpoints: \[\begin{aligned} g(a)&=f(a)+f'(a)(b-a)+\dfrac{1}{2}f''(a)(b-a)^2+\cdots \\ &\quad+\dfrac{1}{k!}f^{(k)}(a)(b-a)^k+\dfrac{S}{(k+1)!}(b-a)^{k+1}=f(b) \qquad \text{by}\,(*) \\ g(b)&=f(b)+f'(b)(b-b)+\dfrac{1}{2}f''(b)(b-b)^2+\cdots \\ &\quad+\dfrac{1}{k!}f^{(k)}(b)(b-b)^k+\dfrac{S}{(k+1)!}(b-b)^{k+1}=f(b) \end{aligned}\] Since \(g(a)=g(b)\), we can apply Rolle's Theorem to conclude there is a number \(c\) in \((a,b)\) such that \(g'(c)=0\). However: \[\begin{aligned} g'(x)&=\dfrac{d}{dx}\left[f(x)+f'(x)(b-x)+\dfrac{1}{2}f''(x)(b-x)^2+\cdots\right. \\ &\left.\qquad\quad+\dfrac{1}{k!}f^{(k)}(x)(b-x)^k+\dfrac{S}{(k+1)!}(b-x)^{k+1}\right] \\ \end{aligned}\] We need the Product Rule on all but the first and last terms. Then terms cancel in pairs, except for the last two terms: \[\begin{aligned} g'(x)&=f'(x)-f'(x)+f''(x)(b-x)-f''(x)(b-x)+\dfrac{1}{2}f'''(x)(b-x)^2+\cdots \\ &\quad-\dfrac{1}{(k-1)!}f^{(k)}(x)(b-x)^{k-1} +\dfrac{1}{k!}f^{(k+1)}(x)(b-x)^k-\dfrac{S}{k!}(b-x)^k \\ &=\dfrac{1}{k!}f^{(k+1)}(x)(b-x)^k-\dfrac{S}{k!}(b-x)^k \\ &=\dfrac{1}{k!}\left[f^{(k+1)}(x)-S\right](b-x)^{k} \end{aligned}\] So \[ g'(c)=\dfrac{1}{k!}\left[f^{(k+1)}(c)-S\right](b-c)^{k}=0 \] Since \(c\ne b\), we conclude \(S=f^{(k+1)}(c)\).

Note: This proof is adapted from one by James Keesling posted at https://people.clas.ufl.edu/kees/files/TaylorRemainderProof.pdf