# 4.8 Taylor polynomials and taylor's remainder theorem

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This module contains taylor's remainder theorem, some remarks about Taylor series, and a test for local maxima and minima.

Let $f$ be in ${C}^{n}\left({B}_{r}\left(c\right)\right)$ for $c$ a fixed complex number, $r>0,$ and $n$ a positive integer. Define the Taylor polynomial of degree $n$ for $f$ at $c$ to be the polynomial ${T}^{n}\equiv {T}_{\left(f,c\right)}^{n}$ given by the formula:

$\left({T}_{\left(f,c\right)}^{n}\right)\left(z\right)=\sum _{j=0}^{n}{a}_{j}{\left(z-c\right)}^{j},$

where ${a}_{j}={f}^{\left(j\right)}\left(c\right)/j!.$

REMARK If $f$ is expandable in a Taylor series on ${B}_{r}\left(c\right),$ then the Taylor polynomial for $f$ of degree $n$ is nothing but the $n$ th partial sum of the Taylor series for $f$ on ${B}_{r}\left(c\right).$ However, any function that is $n$ times differentiable at a point $c$ has a Taylor polynomial of order $n.$ Functions that are infinitely differentiable have Taylor polynomials of all orders, and we might suspect that these polynomials are some kind of good approximation to the function itself.

Prove that $f$ is expandable in a Taylor series function around a point $c$ (with radius of convergence $r>0$ ) if and only if the sequence $\left\{{T}_{\left(f,c\right)}^{n}\right\}$ of Taylor polynomials converges pointwise to $f;$ i.e.,

$f\left(z\right)=lim\left({T}_{\left(f,c\right)}^{n}\right)\left(z\right)$

for all $z$ in ${B}_{r}\left(c\right).$

Let $f\in {C}^{n}\left({B}_{r}\left(c\right)\right).$ Prove that ${f}^{\text{'}}\in {C}^{n-1}\left({B}_{r}\left(c\right)\right).$ Prove also that ${\left({T}_{\left(f,c\right)}^{n}\right)}^{\text{'}}={T}_{\left({f}^{\text{'}},c\right)}^{n-1}.$

The next theorem is, in many ways, the fundamental theorem of numerical analysis. It clearly has to do withapproximating a general function by polynomials. It is a generalization of the Mean Value Theorem, and as in that casethis theorem holds only for real-valued functions of a real variable.

## Taylor's remainder theorem

Let $f$ be a real-valued function on an interval $\left(c-r,c+r\right),$ and assume that $f\in {C}^{n}\left(\left(c-r,c+r\right)\right),$ and that ${f}^{\left(n\right)}$ is differentiable on $\left(c-r,c+r\right).$ Then, for each $x$ in $\left(c-r,c+r\right)$ there exists a $y$ between $c$ and $x$ such that

$f\left(x\right)-\left({T}_{\left(f,c\right)}^{n}\right)\left(x\right)=\frac{{f}^{\left(n+1\right)}\left(y\right)}{\left(n+1\right)!}{\left(x-c\right)}^{n+1}.\phantom{\rule{2.em}{0ex}}\left(4.7\right)$

REMARK If we write $f\left(x\right)={T}_{f,c}^{n}\right)\left(x\right)+{R}_{n+1}\left(x\right),$ where ${R}_{n+1}\left(x\right)$ is the error or remainder term, then this theorem gives a formula, and hence an estimate, for that remainder term.This is the evident connection with Numerical Analysis.

We prove this theorem by induction on $n.$ For $n=0,$ this is precisely the Mean Value Theorem. Thus,

$f\left(x\right)-{T}_{f,c}^{0}\left(x\right)=f\left(x\right)-f\left(c\right)={f}^{\text{'}}\left(y\right)\left(x-c.$

Now, assuming the theorem is true for all functionsin ${C}^{n-1}\left(\left(c-r,c+r\right)\right),$ let us show it is true for the given function $f\in {C}^{n}\left(\left(c-r,c+r\right)\right).$ Set $g\left(x\right)=f\left(x\right)-\left({T}_{\left(f,c\right)}^{n}\right)\left(x\right)$ and let $h\left(x\right)={\left(x-c\right)}^{n+1}.$ Observe that both $g\left(c\right)=0$ and $h\left(c\right)=0.$ Also, if $x\ne c,$ then $h\left(x\right)\ne 0.$ So, by the Cauchy Mean Value Theorem, we have that

$\frac{g\left(x\right)}{h\left(x\right)}=\frac{g\left(x\right)-g\left(c\right)}{h\left(x\right)-h\left(c\right)}=\frac{{g}^{\text{'}}\left(w\right)}{{h}^{\text{'}}\left(w\right)}$

for some $w$ between $c$ and $x.$ Now

${g}^{\text{'}}\left(w\right)={f}^{\text{'}}\left(w\right)-{\left({t}_{\left(f,c\right)}^{n}\right)}^{\text{'}}\left(w\right)={f}^{\text{'}}\left(w\right)-\left({T}_{\left({f}^{\text{'}},c\right)}^{n-1}\right)\left(w\right)$

(See the preceding exercise.), and ${h}^{\text{'}}\left(w\right)=\left(n+1\right){\left(w-c\right)}^{n}.$ Therefore,

$\begin{array}{ccc}\hfill \frac{f\left(x\right)-\left({T}_{\left(f,c\right)}^{n}\right)\left(x\right)}{{\left(x-c\right)}^{n+1}}& =& \frac{g\left(x\right)}{h\left(x\right)}\hfill \\ & =& \frac{{g}^{\text{'}}\left(w\right)}{{h}^{\text{'}}\left(w\right)}\hfill \\ & =& \frac{{f}^{\text{'}}\left(w\right)-\left({T}_{\left({f}^{\text{'}},c\right)}^{n-1}\right)\left(w\right)}{\left(n+1\right){\left(w-c\right)}^{n}}.\hfill \end{array}$

We apply the inductive hypotheses to the function ${f}^{\text{'}}$ (which is in ${C}^{n-1}\left(\left(c-r,c+r\right)\right)\right)$ and obtain

$\begin{array}{ccc}\hfill \frac{f\left(x\right)-\left({T}_{\left(f,c\right)}^{n}\right)\left(x\right)}{{\left(x-c\right)}^{n+1}}& =& \frac{{f}^{\text{'}}\left(w\right)-\left({T}_{\left({f}^{\text{'}},c\right)}^{n-1}\right)\left(w\right)}{\left(n+1\right){\left(w-c\right)}^{n}}\hfill \\ & =& \frac{\frac{{{f}^{\text{'}}}^{\left(n\right)}\left(y\right)}{n!}{\left(w-c\right)}^{n}}{\left(n+1\right){\left(w-c\right)}^{n}}\hfill \\ & =& \frac{{{f}^{\text{'}}}^{\left(n\right)}\left(y\right)}{\left(n+1\right)!}\hfill \\ & =& \frac{{f}^{\left(n+1\right)}\left(y\right)}{\left(n+1\right)!}\hfill \end{array}$

for some $y$ between $c$ and $w.$ But this implies that

$f\left(x\right)-\left({T}_{\left(f,c\right)}^{n}\right)\left(x\right)=\frac{{f}^{\left(n+1\right)}\left(y\right){\left(x-c\right)}^{n+1}}{\left(n+1\right)!},$

for some $y$ between $c$ and $x,$ which finishes the proof of the theorem.

Define $f\left(x\right)=0$ for $x\le 0$ and $f\left(x\right)={e}^{-1/x}$ for $x>0.$ Verify that $f\in {C}^{\infty }\left(R\right),$ that ${f}^{\left(n\right)}\left(0\right)=0$ for all $n,$ and yet $f$ is not expandable in a Taylor series around $0.$ Interpret Taylor's Remainder Theorem for this function. That is, describe the remainder ${R}_{n+1}\left(x\right).$

As a first application of Taylor's Remainder Theorem we give the following result, which should be familiar from calculus.It is the generalized version of what's ordinarily called the “second derivative test.”

## Test for local maxima and minima

Let $f$ be a real-valued function in ${C}^{n}\left(c-r,c+r\right),$ suppose that the $n+1$ st derivative ${f}^{\left(n+1\right)}$ of $f$ exists everywhere on $\left(c-r,c+r\right)$ and is continuous at $c,$ and suppose that ${f}^{\left(k\right)}\left(c\right)=0$ for all $1\le k\le n$ and that ${f}^{\left(n+1\right)}\left(c\right)\ne 0.$ Then:

1. If $n$ is even, $f$ attains neither a local maximum nor a local minimum at $c.$ In this case, $c$ is called an inflection point.
2. If $n$ is odd and ${f}^{\left(n+1\right)}\left(c\right)<0,$ then $f$ attains a local maximum at $c.$
3. If $n$ is odd and ${f}^{\left(n+1\right)}\left(c\right)>0,$ then $f$ attains a local minimum at $c.$

Since ${f}^{\left(n+1\right)}$ is continuous at $c,$ there exists a $\delta >0$ such that ${f}^{\left(n+1\right)}\left(y\right)$ has the same sign as ${f}^{\left(n+1\right)}\left(c\right)$ for all $y\in \left(c-\delta ,c+\delta \right).$ We have by Taylor's Theorem that if $x\in \left(c-\delta ,c+\delta \right)$ then there exists a $y$ between $x$ and $c$ such that

$f\left(x\right)=\left({T}_{\left(f,c\right)}^{n}\right)\left(x\right)+\frac{{f}^{\left(n+1\right)}\left(y\right)}{\left(n+1\right)!}{\left(x-c\right)}^{n+1},$

from which it follows that

$\begin{array}{ccc}\hfill f\left(x\right)-f\left(c\right)& =& \sum _{k=1}^{n}{f}^{\left(k\right)}\left(c\right)k!{\left(x-c\right)}^{k}+\frac{{f}^{\left(n+1\right)}\left(y\right)}{\left(n+1\right)!}{\left(x-c\right)}^{n+1}\hfill \\ & =& \frac{{f}^{\left(n+1\right)}\left(y\right)}{\left(n+1\right)!}{\left(x-c\right)}^{n+1}.\hfill \end{array}$

Suppose $n$ is even. It follows then that if $x the sign of ${\left(x-c\right)}^{n+1}$ is negative, so that the sign of $f\left(x\right)-f\left(c\right)$ is the opposite of the sign of ${f}^{\left(n+1\right)}\left(c\right).$ On the other hand, if $x>c,$ then ${\left(x-c\right)}^{n+1}>0,$ so that the sign of $f\left(x\right)-f\left(c\right)$ is the same as the sign of ${f}^{\left(n+1\right)}\left(c\right).$ So, $f\left(x\right)>f\left(c\right)$ for all nearby $x$ on one side of $c,$ while $f\left(x\right) for all nearby $x$ on the other side of $c.$ Therefore, $f$ attains neither a local maximum nor a local minimum at $c.$ This proves part (1).

Now, if $n$ is odd, the sign of $f\left(x\right)-f\left(c\right)$ is the same as the sign of ${f}^{\left(n+1\right)}\left(y\right),$ which is the same as the sign of ${f}^{\left(n+1\right)}\left(c\right),$ for all $x\in \left(c-\delta ,c+\delta \right).$ Hence, if ${f}^{\left(n+1\right)}\left(c\right)<0,$ then $f\left(x\right)-f\left(c\right)<0$ for all $x\in \left(c-\delta ,c+\delta \right),$ showing that $f$ attains a local maximum at $c.$ And, if ${f}^{\left(n+1\right)}\left(c\right)>0,$ then the sign of $f\left(x\right)-f\left(c\right)$ is positive for all $x\in \left(c-\delta ,c+\delta \right),$ showing that $f$ attains a local minimum at $c.$ This proves parts (2) and (3).

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