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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