Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 17

The Taylor formula

Brook Taylor (1685-1731)

So far, we have only considered power series of the form ${}\sum _{k=0}^{\infty }c_{k}x^{k}$. Now we allow that the variable ${}x$ may be replaced by a "shifted variable“ ${}x-a$, in order to study the local behavior in the expansion point ${}a$. Convergence means, in this case, that some ${}\epsilon >0$ exists, such that for

${}x\in ]a-\epsilon ,a+\epsilon [\,,$

the series converges. In this situation, the function, presented by the power series, is again differentiable, and its derivative is given as in Theorem 16.1 . For a convergent power series

${}f(x):=\sum _{k=0}^{\infty }c_{k}(x-a)^{k}\,,$

the polynomials ${}\sum _{k=0}^{n}c_{k}(x-a)^{k}$ yield polynomial approximations for the function ${}f$ in the point ${}a$. Moreover, the function ${}f$ is arbitrarily often differentiable in ${}a$, and the higher derivatives in the point ${}a$ can be read of from the power series directly, namely

${}f^{(n)}(a)=n!c_{n}\,.$

We consider now the question whether we can find, starting with a differentiable function of sufficiently high order, approximating polynomials (or a power series). This is the content of the Taylor expansion.

Definition

Let ${}I\subseteq \mathbb {R}$ denote an interval,

$f\colon I\longrightarrow \mathbb {R}$

an ${}n$-times differentiable function, and ${}a\in I$. Then

${}T_{a,n}(f)(x):=\sum _{k=0}^{n}{\frac {f^{(k)}(a)}{k!}}(x-a)^{k}\,$

is called the Taylor polynomial of degree ${}n$ for ${}f$ in the point ${}a$.

So

${}T_{a,0}(f)(x):=f(a)\,$

is the constant approximation,

${}T_{a,1}(f)(x):=f(a)+f'(a)(x-a)\,$

is the linear approximation,

${}T_{a,2}(f)(x):=f(a)+f'(a)(x-a)+{\frac {f^{\prime \prime }(a)}{2}}(x-a)^{2}\,$

is the quadratic approximation,

${}T_{a,3}(f)(x):=f(a)+f'(a)(x-a)+{\frac {f^{\prime \prime }(a)}{2}}(x-a)^{2}+{\frac {f^{\prime \prime \prime }(a)}{6}}(x-a)^{3}\,$

is the approximation of degree ${}3$, etc. The Taylor polynomial of degree ${}n$ is the (uniquely determined) polynomial of degree ${}\leq n$ with the property that its derivatives and the derivatives of ${}f$ at ${}a$ coincide up to order ${}n$.

Theorem

Let ${}I$ denote a real interval,

$f\colon I\longrightarrow \mathbb {R}$

an ${}(n+1)$-times differentiable function, and ${}a\in I$ an inner point of the interval. Then for every point ${}x\in I$, there exists some ${}c\in I$ such that

$f(x)=\sum _{k=0}^{n}{\frac {f^{(k)}(a)}{k!}}(x-a)^{k}+{\frac {f^{(n+1)}(c)}{(n+1)!}}(x-a)^{n+1}.$

Here, ${}c$ may be chosen between

${}a$ and ${}x$.

Proof

This proof was not presented in the lecture.

$\Box$

The real sine function, together with several approximating Taylor polynomials (of odd degree).

Corollary

Suppose that ${}I$ is a bounded closed interval,

$f\colon I\longrightarrow \mathbb {R}$

is an ${}(n+1)$-times continuously differentiable function, ${}a\in I$ an inner point, and ${}B:={\max {\left(\vert {f^{(n+1)}(c)}\vert ,c\in I\right)}}$. Then, between ${}f(x)$ and the ${}n$-th Taylor polynomial, we have the estimate

${}\vert {f(x)-\sum _{k=0}^{n}{\frac {f^{(k)}(a)}{k!}}(x-a)^{k}}\vert \leq {\frac {B}{(n+1)!}}\vert {x-a}\vert ^{n+1}\,.$

Proof

The number ${}B$ exists due to Theorem 11.13 , since the ${}(n+1)$-th derivative ${}f^{(n+1)}$ is continuous on the compact interval ${}I$. The statement follows, therefore, directly from Theorem 17.2 .

$\Box$

Criteria for extrema

We have seen in the 15th lecture that it is a necessary condition for a differentiable function to have a local extremum at a point, that its derivative equals ${}0$ at this point. We give now an important sufficient criterion, which relies on higher derivatives.

Theorem

Let ${}I$ denote a real interval,

$f\colon I\longrightarrow \mathbb {R}$

an ${}(n+1)$-times continuously differentiable function, and ${}a\in I$ an inner point of the interval. Suppose that

$f'(a)=f^{\prime \prime }(a)=\ldots =f^{(n)}(a)=0{\text{ and }}f^{(n+1)}(a)\neq 0$

is fulfilled. Then the following statements hold.

1. If ${}n$ is even, then ${}f$ does not have a local extremum in ${}a$.
2. Suppose that ${}n$ is odd. In case ${}f^{(n+1)}(a)>0$, the function ${}f$ has an isolated local minimum in ${}a$.
3. Suppose that ${}n$ is odd. In case ${}f^{(n+1)}(a)<0$, the function ${}f$ has an isolated local maximum in ${}a$.

Proof

Under the given conditions, the Taylor formula becomes

${}f(x)-f(a)={\frac {f^{(n+1)}(c)}{(n+1)!}}(x-a)^{n+1}\,,$

with some ${}c$ (depending on ${}x$) between ${}a$ and ${}x$. Depending on whether ${}f^{(n+1)}(a)>0$ or ${}f^{(n+1)}(a)<0$ holds, we have (due to the continuity of the ${}(n+1)$-th derivative) ${}f^{(n+1)}(x)>0$ or ${}f^{(n+1)}(x)<0$ for ${}x\in [a-\epsilon ,a+\epsilon ]$, for a suitable ${}\epsilon >0$. For these ${}x$, we have ${}c\in [a-\epsilon ,a+\epsilon ]$, so that the sign of ${}f^{(n+1)}(c)$ depends on the sign of ${}f^{(n+1)}(a)$.
For ${}n$ even, ${}n+1$ is odd and therefore the sign of ${}(x-a)^{n+1}$ changes at ${}x=a$ (for ${}x<a$, the sign is negative, and for ${}x>a$, the sign is positive). Since the sign of ${}f^{(n+1)}(c)$ does not change, the sign of ${}f(x)-f(a)$ is changing. This means that there can not be an extremum.
Suppose now that ${}n$ is odd. Then ${}n+1$ is even, hence ${}(x-a)^{n+1}>0$ for all ${}x\neq a$ in the neighborhood. This means, in the neighborhood, in case ${}f^{(n+1)}(a)>0$, that ${}f(x)>f(a)$ holds, and we have an isolated minimum in ${}a$. If ${}f^{(n+1)}(a)<0$, then ${}f(x)<f(a)$ holds, and we have an isolated maximum in ${}a$.

$\Box$

A special case of this is that in case ${}f'(a)=0$ and ${}f^{\prime \prime }(a)>0$, then we have an isolated minimum, and in case ${}f'(a)=0$ and ${}f^{\prime \prime }(a)<0$ we have an isolated maximum.

The Taylor series

Definition

Let ${}I\subseteq \mathbb {R}$ denote an interval,

$f\colon I\longrightarrow \mathbb {R}$

an infinitely often differentiable function, and ${}a\in I$. Then

$\sum _{k=0}^{\infty }{\frac {f^{(k)}(a)}{k!}}(x-a)^{k}$

is called the Taylor series of ${}f$ in the point ${}a$.

Theorem

Let ${}\sum _{n=0}^{\infty }c_{n}x^{n}$ denote a power series which converges on the interval ${}]-r,r[$, and let

$f\colon ]-r,r[\longrightarrow \mathbb {R}$

denote the function defined via Theorem 12.2 . Then ${}f$ is infinitely often differentiable, and the Taylor series of ${}f$ in ${}0$ coincides with the given power series.

Proof

That ${}f$ is infinitely often differentiable, follows directly from Theorem 16.1 by induction. Therefore, the Taylor series exists in particular in the point ${}0$. Hence, we only have to show that the ${}n$-th derivative has ${}c_{n}n!$ as its value. But this follows also from Theorem 16.1 .

$\Box$

Example

We consider the function

$f\colon \mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto f(x),$

given by

${}f(x):={\begin{cases}0,\,{\text{ if }}x\leq 0\,,\\e^{-{\frac {1}{x}}},\,{\text{ if }}x>0\,.\end{cases}}\,$

We claim that this function is infinitely often differentiable, which is only in ${}0$ not directly clear. We first show, by induction, that all derivatives of ${}e^{-{\frac {1}{x}}}$ have the form ${}p{\left({\frac {1}{x}}\right)}e^{-{\frac {1}{x}}}$ with certain polynomials ${}p\in \mathbb {R} [Z]$, and that therefore the limit for ${}x\rightarrow 0,\,x>0$ equals ${}0$ (see Exercise 17.16 and Exercise 17.17 ). Therefore, the limit exists for all derivatives and is ${}0$. So all derivatives in ${}0$ have value ${}0$, and therefore the Taylor series in ${}0$ is just the zero series. However, the Function ${}f$ is in no neighborhood of ${}0$ the zero function, since ${}e^{-{\frac {1}{x}}}>0$.

Power series ansatz

Taylor series/R/Power series/Section

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