# Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 12

Power series

## Definition

Let ${\displaystyle {}{\left(c_{n}\right)}_{n\in \mathbb {N} }}$ be a sequence of real numbers and ${\displaystyle {}x}$ another real number. Then the series

${\displaystyle \sum _{n=0}^{\infty }c_{n}x^{n}}$
is called the power series in ${\displaystyle {}x}$ for the coefficients ${\displaystyle {}{\left(c_{n}\right)}_{n\in \mathbb {N} }}$.

For a power series, it is important that ${\displaystyle {}x}$ varies and that the power series represents in some convergence interval a function in ${\displaystyle {}x}$. Every polynomial is a power series, but one for which all coefficients starting with a certain member are ${\displaystyle {}0}$. In this case, the convergence is everywhere.

We have encountered an important power series earlier, the geometric series ${\displaystyle {}\sum _{n=0}^{\infty }x^{n}}$ (here all coefficients equal ${\displaystyle {}1}$), which converges for ${\displaystyle {}\vert {x}\vert <1}$ and represents the function ${\displaystyle {}1/(1-x)}$, see Theorem 9.13 . Another important power series is the exponential series, which for every real number converges and represents the real exponential function. Its inverse function is the natural logarithm.

The behavior of convergence of a power series is given by the following theorem.

## Theorem

Let

${\displaystyle {}f(x):=\sum _{n=0}^{\infty }c_{n}x^{n}\,}$

be a power series and suppose that there exists some ${\displaystyle {}x_{0}\neq 0}$ such that ${\displaystyle {}\sum _{n=0}^{\infty }c_{n}x_{0}^{n}}$ converges. Then there exists a positive ${\displaystyle {}R}$ (where ${\displaystyle {}R=\infty }$ is allowed) such that for all ${\displaystyle {}x\in \mathbb {R} }$ fulfilling ${\displaystyle {}\vert {x}\vert the series converges absolutely. On such an (open) interval of convergence, the power series ${\displaystyle {}f(x)}$ represents a continuous function.

### Proof

The proof needs a systematic study of power series and of limits of sequences of functions. We will not do this here.
${\displaystyle \Box }$

If two functions are given by power series, then their sum is simply given by the (componentwise defined) sum of the power series. It is not clear at all by which power series the product of two power series is described. The answer is given by the Cauchy-product of series.

## Definition

For two series ${\displaystyle {}\sum _{i=0}^{\infty }a_{i}}$ and ${\displaystyle {}\sum _{j=0}^{\infty }b_{j}}$ of real numbers, the series

${\displaystyle \sum _{k=0}^{\infty }c_{k}{\text{ with }}c_{k}:=\sum _{i=0}^{k}a_{i}b_{k-i}}$
is called the Cauchy-product of the series.

Also, for the following statement we do not provide a proof.

## Lemma

Let

${\displaystyle \sum _{k=0}^{\infty }a_{k}{\text{ and }}\sum _{k=0}^{\infty }b_{k}}$

be absolutely convergent series of real numbers. Then also the Cauchy product ${\displaystyle {}\sum _{k=0}^{\infty }c_{k}}$ is absolutely convergent and for its sum the equation

${\displaystyle {}\sum _{k=0}^{\infty }c_{k}={\left(\sum _{k=0}^{\infty }a_{k}\right)}\cdot {\left(\sum _{k=0}^{\infty }b_{k}\right)}\,}$

holds.

From this we can infer that the product of power series is given by the power series whose coefficients are those which arise by the multiplication of polynomials, see Exercise 12.3 .

Exponential series and exponential function

We discuss another important power series, the exponential series and the exponential function defined by it.

## Definition

For every ${\displaystyle {}x\in \mathbb {R} }$, the series

${\displaystyle \sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}}$
is called the exponential series in ${\displaystyle {}x}$.

So this is just the series

${\displaystyle 1+x+{\frac {x^{2}}{2}}+{\frac {x^{3}}{6}}+{\frac {x^{4}}{24}}+{\frac {x^{5}}{120}}+\ldots .}$

## Theorem

For every ${\displaystyle {}x\in \mathbb {R} }$, the exponential series

${\displaystyle \sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}}$
is

### Proof

For ${\displaystyle {}x=0}$, the statement is clear. Else, we consider the fraction

${\displaystyle {}\vert {\frac {\frac {x^{n+1}}{(n+1)!}}{\frac {x^{n}}{n!}}}\vert =\vert {\frac {x}{n+1}}\vert ={\frac {\vert {x}\vert }{n+1}}\,.}$

This is, for ${\displaystyle {}n\geq 2\vert {x}\vert }$, smaller than ${\displaystyle {}1/2}$. By the ratio test, we get convergence.

${\displaystyle \Box }$

Due to this property, we can define the real exponential function.

## Definition

The function

${\displaystyle \mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto \exp x:=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}},}$

is called the (real)

exponential function.

The following statement is called the functional equation for the exponential function.

## Theorem

For real numbers ${\displaystyle {}x,y\in \mathbb {R} }$, the equation

${\displaystyle {}\exp {\left(x+y\right)}=\exp x\cdot \exp y\,}$

holds.

### Proof

The Cauchy product of the two exponential series is

${\displaystyle \sum _{n=0}^{\infty }c_{n},}$

where

${\displaystyle {}c_{n}=\sum _{i=0}^{n}{\frac {x^{i}}{i!}}\cdot {\frac {y^{n-i}}{(n-i)!}}\,.}$

This series is due to Lemma 12.4 absolutely convergent and the limit is the product of the two limits. Furthermore, the ${\displaystyle {}n}$-th summand of the exponential series of ${\displaystyle {}x+y}$ equals

${\displaystyle {}{\frac {(x+y)^{n}}{n!}}={\frac {1}{n!}}\sum _{i=0}^{n}{\binom {n}{i}}x^{i}y^{n-i}=c_{n}\,,}$

so that both sides coincide.

${\displaystyle \Box }$

## Corollary

${\displaystyle \mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto \exp x,}$
fulfills the following properties.
1. ${\displaystyle {}\exp 0=1}$.
2. For every ${\displaystyle {}x\in \mathbb {R} }$, we have ${\displaystyle {}\exp {\left(-x\right)}=(\exp x)^{-1}}$. In particular ${\displaystyle {}\exp x\neq 0}$.
3. For integers ${\displaystyle {}n\in \mathbb {Z} }$, the relation ${\displaystyle {}\exp n=(\exp 1)^{n}}$ holds.
4. For every ${\displaystyle {}x}$, we have ${\displaystyle {}\exp x\in \mathbb {R} _{+}}$.
5. For ${\displaystyle {}x>0}$ we have ${\displaystyle {}\exp x>1}$, and for ${\displaystyle {}x<0}$ we have ${\displaystyle {}\exp x<1}$.
6. The real exponential function is strictly increasing.

### Proof

(1) follows directly from the definition.
(2) follows from

${\displaystyle {}\exp x\cdot \exp {\left(-x\right)}=\exp {\left(x-x\right)}=\exp 0=1\,}$

using Theorem 12.8 .
(3) follows for ${\displaystyle {}n\in \mathbb {N} }$ from Theorem 12.8 by induction, and from that it follows with the help of (2) also for negative ${\displaystyle {}n}$.
(4). Nonnegativity follows from

${\displaystyle {}\exp x=\exp {\left({\frac {x}{2}}+{\frac {x}{2}}\right)}=\exp {\frac {x}{2}}\cdot \exp {\frac {x}{2}}={\left(\exp {\frac {x}{2}}\right)}^{2}\geq 0\,.}$

(5). For real ${\displaystyle {}x}$ we have ${\displaystyle {}\exp x\cdot \exp {\left(-x\right)}=1}$, so that because of (4), one factor must be ${\displaystyle {}\geq 1}$ and the other factor must be ${\displaystyle {}\leq 1}$. For ${\displaystyle {}x>0}$, we have

${\displaystyle {}\exp x=\sum _{n=0}^{\infty }{\frac {1}{n!}}x^{n}=1+x+{\frac {1}{2}}x^{2}+\ldots >1\,}$

as only positive numbers are added.
(6). For real ${\displaystyle {}y>x}$, we have ${\displaystyle {}y-x>0}$, and therefore, because of (5) ${\displaystyle {}\exp {\left(y-x\right)}>1}$, hence

${\displaystyle {}\exp y=\exp {\left(y-x+x\right)}=\exp {\left(y-x\right)}\cdot \exp x>\exp x\,.}$

${\displaystyle \Box }$

With the help of the exponential series, we also define Euler's number.

## Definition

The real number

${\displaystyle {}e:=\sum _{k=0}^{\infty }{\frac {1}{k!}}\,}$
is called Euler's number.

So we have ${\displaystyle {}e=\exp 1}$. Its numerical value is

${\displaystyle {}e=1+1+{\frac {1}{2}}+{\frac {1}{6}}+{\frac {1}{24}}+\ldots \cong 2,71...\,.}$

## Remark

For Euler's number there is also the description

${\displaystyle {}e=\lim _{n\rightarrow \infty }{\left(1+{\frac {1}{n}}\right)}^{n}\,,}$

so that ${\displaystyle {}e}$ can also be introduced as the limit of this sequence. However, the convergence in the exponential series is much faster.

We will write also ${\displaystyle {}e^{x}}$ instead of ${\displaystyle {}\exp x}$. This is, for ${\displaystyle {}x\in \mathbb {Z} }$, compatible with the usual meaning of powers in the sense of the fourth lecture due to Corollary 12.9 . The compatibility with arbitrary roots (if the exponents are rational) follows from Remark 12.17 and Exercise 12.17 .

## Theorem

${\displaystyle \mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto \exp x,}$
is

continuous and defines a bijection between ${\displaystyle {}\mathbb {R} }$ and ${\displaystyle {}\mathbb {R} _{+}}$.

### Proof

The continuity follows from Theorem 12.2 , since the exponential function is defined with the help of a power series. Due to Corollary 12.9 , the image lies in ${\displaystyle {}\mathbb {R} _{+}}$, and the image is, because of the intermediate value theorem, an interval. The unboundedness of the image follows from Corollary 12.9 . This implies, because of Corollary 12.9 , that also arbitrary small positive real numbers are obtained. Thus the image is ${\displaystyle {}\mathbb {R} _{+}}$. Injectivity follows from Corollary 12.9 , in connection with Exercise 5.38 .

${\displaystyle \Box }$

Logarithms

## Definition

The natural logarithm

${\displaystyle \ln \colon \mathbb {R} _{+}\longrightarrow \mathbb {R} ,x\longmapsto \ln x,}$

is defined as the inverse function of the

real exponential function.

## Theorem

${\displaystyle \ln \colon \mathbb {R} _{+}\longrightarrow \mathbb {R} ,x\longmapsto \ln x,}$
is a

continuous strictly increasing function, which defines a bijection between ${\displaystyle {}\mathbb {R} _{+}}$ and ${\displaystyle {}\mathbb {R} }$. Moreover, the functional equation

${\displaystyle {}\ln(x\cdot y)=\ln x+\ln y\,}$

holds for all ${\displaystyle {}x,y\in \mathbb {R} _{+}}$.

### Proof

${\displaystyle \Box }$

## Definition

For a positive real number ${\displaystyle {}b>0}$, the exponential function for the base ${\displaystyle {}b}$ is defined as

${\displaystyle {}b^{x}:=\exp(x\ln b)\,.}$

## Theorem

Let ${\displaystyle {}b}$ denote a positive real number. Then the exponential function

${\displaystyle f\colon \mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto b^{x},}$
fulfills the following properties.
1. We have ${\displaystyle {}b^{x+x'}=b^{x}\cdot b^{x'}}$ for all ${\displaystyle {}x,x'\in \mathbb {R} }$.
2. We have ${\displaystyle {}b^{-x}={\frac {1}{b^{x}}}}$.
3. For ${\displaystyle {}b>1}$ and ${\displaystyle {}x>0}$, we have ${\displaystyle {}b^{x}>1}$.
4. For ${\displaystyle {}b<1}$ and ${\displaystyle {}x>0}$, we have ${\displaystyle {}b^{x}<1}$.
5. For ${\displaystyle {}b>1}$, the function ${\displaystyle {}f}$ is strictly increasing.
6. For ${\displaystyle {}b<1}$, the function ${\displaystyle {}f}$ is strictly decreasing.
7. We have ${\displaystyle {}(b^{x})^{x'}=b^{x\cdot x'}}$ for all ${\displaystyle {}x,x'\in \mathbb {R} }$.
8. For ${\displaystyle {}a\in \mathbb {R} _{+}}$, we have ${\displaystyle {}(ab)^{x}=a^{x}\cdot b^{x}}$.

### Proof

${\displaystyle \Box }$

## Remark

There is another way to introduce the exponential function ${\displaystyle {}x\mapsto a^{x}}$ to base ${\displaystyle {}a>0}$. For a natural number ${\displaystyle {}n\neq 0}$, one takes the ${\displaystyle {}n}$th product of ${\displaystyle {}a}$ with itself as definition for ${\displaystyle {}a^{n}}$. For a negative integer ${\displaystyle {}x}$, one sets ${\displaystyle {}a^{x}:=(a^{-x})^{-1}}$. For a positive rational number ${\displaystyle {}x=r/s}$, one sets

${\displaystyle {}a^{x}:={\sqrt[{s}]{a^{r}}}\,,}$

where one has to show that this is independent of the chosen representation as a fraction. For a negative rational number, one takes again the inverse. For an arbitrary real number ${\displaystyle {}x}$, one takes a sequence ${\displaystyle {}q_{n}}$ of rational numbers converging to ${\displaystyle {}x}$, and defines

${\displaystyle {}a^{x}:=\lim _{n\rightarrow \infty }a^{q_{n}}\,.}$

For this, one has to show that these limits exist and that they are independent of the chosen rational sequence. For the passage from ${\displaystyle {}\mathbb {Q} }$ to ${\displaystyle {}\mathbb {R} }$, the concept of uniform continuity is crucial.

## Definition

For a positive real number ${\displaystyle {}b>0}$, ${\displaystyle {}b\neq 1}$, the logarithm to base ${\displaystyle {}b}$ of ${\displaystyle {}x\in \mathbb {R} _{+}}$ is defined by

${\displaystyle {}\log _{b}x:={\frac {\ln x}{\ln b}}\,.}$

## Theorem

${\displaystyle {}b}$ fulfill the following rules.
1. We have ${\displaystyle {}\log _{b}(b^{x})=x}$ and ${\displaystyle {}b^{\log _{b}(y)}=y}$, this means that the logarithm to Base ${\displaystyle {}b}$ is the inverse function for the exponential function to base ${\displaystyle {}b}$.
2. We have ${\displaystyle {}\log _{b}(y\cdot z)=\log _{b}y+\log _{b}z}$.
3. We have ${\displaystyle {}\log _{b}y^{u}=u\cdot \log _{b}y}$ for ${\displaystyle {}u\in \mathbb {R} }$.
4. We have
${\displaystyle {}\log _{a}y=\log _{a}{\left(b^{\log _{b}y}\right)}=\log _{b}y\cdot \log _{a}b\,.}$

### Proof

${\displaystyle \Box }$