# Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 16/refcontrol

    Derivative of power series
Many important functions, like the exponential function or the trigonometric functions, are represented by a power series. The following theorem shows that these functions are differentiable, and that the derivative of a power series is itself a power series, given by differentiating the individual terms of the series.

## TheoremTheorem 16.1 change

Let
    $g(x):=\sum _{n=0}^{\infty }a_{n}x^{n}\,$
denote a power seriesMDLD/power series (R) which convergesMDLD/converges (power series R) on the open intervalMDLD/open interval (R) $]-r,r[$, and represents there a function $f\colon ]-r,r[\rightarrow \mathbb {R}$. Then the formally differentiated power series
    ${\tilde {g}}(x):=\sum _{n=1}^{\infty }na_{n}x^{n-1}\,$
is convergent on $]-r,r[$. The function $f$ is differentiableMDLD/differentiable (R) in every point of the interval, and
    $f'(x)={\tilde {g}}(x)\,$

### Proof

$\Box$
In the formulation of the theorem, we have distinguished between $g$ for the power series and $f$ for the function, defined by the series, in order to stress the roles they play. This distinction is now not necessary anymore.

## CorollaryCreate referencenumber
### Proof

This follows immediately from Theorem 16.1 .
$\Box$

## TheoremTheorem 16.3 change

The exponential functionMDLD/exponential function (R)
    $\mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto \exp x,$
differentiableMDLD/differentiable (R) with
    $\exp \!'(x)=\exp x\,.$

### Proof

Due to Theorem 16.1 , we have
    ${\begin{aligned}\exp \!'(x)&={\left(\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}\right)}'\\&=\sum _{n=1}^{\infty }{\left({\frac {x^{n}}{n!}}\right)}'\\&=\sum _{n=1}^{\infty }{\frac {n}{n!}}x^{n-1}\\&=\sum _{n=1}^{\infty }{\frac {1}{(n-1)!}}x^{n-1}\\&=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}\\&=\exp x.\end{aligned}}$
$\Box$

## TheoremTheorem 16.4 change

The exponential functionMDLD/exponential function (real base)
    $\mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto a^{x},$
with base $a>0$, is differentiableMDLD/differentiable (R) with
    ${\left(a^{x}\right)}'={\left(\ln a\right)}a^{x}\,.$

### Proof

By definition,MDLD/definition we have
    $a^{x}=\exp {\left(x\,\ln a\right)}\,.$
The derivativeMDLD/derivative (R) with respect to $x$ equals
    ${\left(a^{x}\right)}'={\left(\exp {\left(x\,\ln a\right)}\right)}'={\left(\ln a\right)}\exp '(x\,\ln a)={\left(\ln a\right)}\exp {\left(x\,\ln a\right)}={\left(\ln a\right)}a^{x}\,,$
due to Theorem 16.3
and the chain rule.
$\Box$

## Remark

For a real exponential functionMDLD/real exponential function
    $y(x)=a^{x}\,,$
the relation
    $y'={\left(\ln a\right)}y\,$
holds, due to Theorem 16.4 . Hence, there is a proportional relationship between the function $y$ and its derivative $y'$, and $\ln a$ is the factor. This is still true if $a^{x}$ is multiplied with a constant. If we consider $y$ as a function depending on time $x$, then $y'(x)$ describes the growing behavior at that point of time. The equation $y'={\left(\ln a\right)}y$ means that the instantaneous growing rate is always proportional with the magnitude of the function. Such an increasing behavior (or decreasing behavior, if $a<1$) occurs in nature for a population, if there is no competition for resources, and if the dying rate is neglectable (the number of mice is then proportional with the number of mice born). A condition of the form
    $y'=by\,$
is an example of a differential equation. This is an equation for a function, which expresses a condition for the derivative. A solution for such a differential equation is a differentiable function which fulfills the condition on its derivative. The differential equation just mentioned are fulfilled by the functions
    $y(x)=ce^{bx}\,.$
We will study differential equations in the second semester.

## CorollaryCorollary 16.6 change

The derivativeMDLD/derivative (R) of the natural logarithmMDLD/natural logarithm
    $\ln \colon \mathbb {R} _{+}\longrightarrow \mathbb {R} ,x\longmapsto \ln x,$
    $\ln \!'\colon \mathbb {R} _{+}\longrightarrow \mathbb {R} ,x\longmapsto {\frac {1}{x}}.$

### Proof

As the logarithm is the inverse function of the exponential function, we can apply Theorem 14.9
and get
    $\ln '(x)={\frac {1}{\exp '(\ln x)}}={\frac {1}{\exp(\ln x)}}={\frac {1}{x}}\,,$
using Theorem 16.3 .
$\Box$

## CorollaryCreate referencenumber

Let $\alpha \in \mathbb {R}$. Then the functionMDLD/function
    $f\colon \mathbb {R} _{+}\longrightarrow \mathbb {R} _{+},x\longmapsto x^{\alpha },$
is differentiable,MDLD/differentiable (R) and its derivativeMDLD/derivative (R) is
    $f'(x)=\alpha x^{\alpha -1}\,.$

### Proof

By definition,MDLD/definition we have
    $x^{\alpha }=\exp {\left(\alpha \,\ln x\right)}\,.$
The derivativeMDLD/derivative (R) with respect to $x$ equals
    ${\left(x^{\alpha }\right)}'={\left(\exp {\left(\alpha \,\ln x\right)}\right)}'={\frac {\alpha }{x}}\cdot \exp {\left(\alpha \,\ln x\right)}={\frac {\alpha }{x}}x^{\alpha }=\alpha x^{\alpha -1}\,$
using Theorem 16.3 , Corollary 16.6
and the chain rule.
$\Box$

## TheoremTheorem 16.8 change

The sine functionMDLD/sine function (R)
    $\mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto \sin x,$
is differentiable,MDLD/differentiable (R) with
    $\sin \!'(x)=\cos x\,,$
and the cosine functionMDLD/cosine function (R)
    $\mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto \cos x,$
is differentiable, with
    $\cos \!'(x)=-\sin x\,.$

### Proof

$\Box$

## TheoremCreate referencenumber

The tangent functionMDLD/tangent function (R)
    $\mathbb {R} \setminus {\left({\frac {\pi }{2}}+\mathbb {Z} \pi \right)}\longrightarrow \mathbb {R} ,x\longmapsto \tan x,$
is differentiable,MDLD/differentiable (R) with
    $\tan \!'(x)={\frac {1}{\cos ^{2}x}}\,,$
and the cotangent functionMDLD/cotangent function (R)
    $\mathbb {R} \setminus \mathbb {Z} \pi \longrightarrow \mathbb {R} ,x\longmapsto \cot x,$
is differentiable, with
    $\cot \!'(x)=-{\frac {1}{\sin ^{2}x}}\,.$

### Proof

Using the quotient rule, Theorem 16.8 , and the circle equation, we get
    ${\begin{aligned}(\tan x)^{\prime }&={\left({\frac {\sin x}{\cos x}}\right)}^{\prime }\\&={\frac {(\cos x)(\cos x)-(\sin x)(-\sin x)}{\cos ^{2}x}}\\&={\frac {1}{\cos ^{2}x}}.\end{aligned}}$
The derivative of the cotangent function follows in the same way.
$\Box$
    The number $\pi$
The number $\pi$ is the area and half of the circumference of a circle with radius $1$. But, in order to build a precise definition for this number on this, we would have first to establish measure theory or the theory of the length of curves. Also, the trigonometric functions have an intuitive interpretation at the unit circle, but also this requires the concept of the arc length. An alternative approach is to define the functions sine and cosine by their power series, and then to define the number $\pi$ with the help of them, and establishing finally the relation with the circle.

## LemmaLemma 16.10 change

The cosine functionMDLD/cosine function (R) has, within the real intervalMDLD/real interval $[0,2]$, exactly one

### Proof

We consider the cosine seriesMDLD/cosine series (R)
    $\cos x=\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{2n}}{(2n)!}}\,.$
For $x=0$, we have $\cos 0=1$. For $x=2$, one can write
    ${\begin{aligned}\cos 2&=1-{\frac {2^{2}}{2!}}+{\frac {2^{4}}{4!}}-{\frac {2^{6}}{6!}}+{\frac {2^{8}}{8!}}-\ldots \\&=1-{\frac {2^{2}}{2!}}{\left(1-{\frac {4}{3\cdot 4}}\right)}-{\frac {2^{6}}{6!}}{\left(1-{\frac {4}{7\cdot 8}}\right)}-\ldots \\&=1-2(2/3)-\ldots \\&\leq -1/3.\end{aligned}}$
Hence, due to the intermediate value theorem, there exists at least one zero in the given interval. To prove uniqueness, we consider the derivativeMDLD/derivative (R) of cosine, which is
    $\cos 'x=-\sin x\,,$
due to Theorem 16.8 . Hence, it is enough to show that sine is positive in the interval $]0,2[$, because then cosine is strictly decreasingMDLD/strictly decreasing by Theorem 15.7
in the interval and there is only one zero. Now, for $x\in0,2]}$, we have
    ${\begin{aligned}\sin x&=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+\ldots \\&=x{\left(1-{\frac {x^{2}}{3!}}\right)}+{\frac {x^{5}}{5!}}{\left(1-{\frac {x^{2}}{6\cdot 7}}\right)}+\ldots \\&\geq x{\left(1-{\frac {4}{3!}}\right)}+{\frac {x^{5}}{5!}}{\left(1-{\frac {4}{6\cdot 7}}\right)}+\ldots \\&\geq x/3\\&>0.\end{aligned}}$
$\Box$

## Definition

Let $s$ denote the unique (according to Lemma 16.10 ) realMDLD/real zeroMDLD/zero (function) of the cosine functionMDLD/cosine function (R) in the intervalMDLD/interval (R) $[0,2]$. Then the number $\pi$ is defined by
    $\pi :=2s\,.$

## TheoremCreate referencenumber

$\mathbb {R}$
* We have $\cos {\left(x+2\pi \right)}=\cos x$ and $\sin {\left(x+2\pi \right)}=\sin x$ for all $x\in \mathbb {R}$.
* We have $\cos {\left(x+\pi \right)}=-\cos x$ and $\sin {\left(x+\pi \right)}=-\sin x$ for all $x\in \mathbb {R}$.
* We have $\cos {\left(x+\pi /2\right)}=-\sin x$ and $\sin {\left(x+\pi /2\right)}=\cos x$ for all $x\in \mathbb {R}$.
* We have $\cos 0=1$, $\cos \pi /2=0$, $\cos \pi =-1$, $\cos 3\pi /2=0$, and $\cos 2\pi =1$.
* We have $\sin 0=0$, $\sin \pi /2=1$, $\sin \pi =0$, $\sin 3\pi /2=-1$, and $\sin 2\pi =0$.

### Proof

Due to the circle equation
    $(\cos z)^{2}+(\sin z)^{2}=1\,,$
we have ${\left(\sin {\frac {\pi }{2}}\right)}^{2}=1$, hence $\sin {\frac {\pi }{2}}=1$, because of the reasoning in the proof of Lemma 16.10 . From that we deduce, with the help of the addition theorems, the relations between sine and cosine as mentioned in (3), e.g.
    $\cos {\left(z+{\frac {\pi }{2}}\right)}=\cos z\cos {\left({\frac {\pi }{2}}\right)}-\sin z\sin {\left({\frac {\pi }{2}}\right)}=-\sin z\,.$
Hence it is enough to prove the statements for cosine. All statements follow from the definition of $\pi$ and from (3).
$\Box$

## Definition

A functionMDLD/function $f\colon \mathbb {R} \rightarrow \mathbb {R}$ is called periodic with period length $L>0$, if the equality
    $f(x)=f(x+L)\,$
holds for all
$x\in \mathbb {R}$
The trigonometric functions sin and cosine are periodic functions with the period length $2\pi$.
    The inverse trigonometric functions

## CorollaryCorollary 16.14 change

The real sine functionMDLD/real sine function induces a bijective,MDLD/bijective strictly increasingMDLD/strictly increasing function
    $[-\pi /2,\pi /2]\longrightarrow [-1,1],$
and the real cosine functionMDLD/real cosine function induces a bijective, strictly decreasing function
    $[0,\pi ]\longrightarrow [-1,1].$

### Proof

$\Box$

## CorollaryCreate referencenumber

The real tangent functionMDLD/real tangent function induces a bijective,MDLD/bijective strictly increasingMDLD/strictly increasing function
    $]-\pi /2,\pi /2[\longrightarrow \mathbb {R} ,$
and the real cotangent functionMDLD/real cotangent function induces a bijective strictly decreasing function
    $[0,\pi ]\longrightarrow \mathbb {R} .$

### Proof

$\Box$
Due to the bijectivity of sine, cosine, tangent and cotangent on suitable interval, there exist the following inverse functions.

## Definition

The inverse functionMDLD/inverse function of the real sine functionMDLD/sine function (R) is
    $[-1,1]\longrightarrow [-{\frac {\pi }{2}},{\frac {\pi }{2}}],x\longmapsto \arcsin x,$

## Definition

The inverse functionMDLD/inverse function of the real cosine functionMDLD/cosine function (R) is
    $[-1,1]\longrightarrow [0,\pi ],x\longmapsto \arccos x,$

## Definition

The inverse functionMDLD/inverse function of the real tangent functionMDLD/tangent function (R) is
    $\mathbb {R} \longrightarrow ]-{\frac {\pi }{2}},{\frac {\pi }{2}}[,x\longmapsto \arctan x,$

## Definition

The inverse functionMDLD/inverse function of the real cotangent functionMDLD/cotangent function (R) is
    $\mathbb {R} \longrightarrow ]0,\pi [,x\longmapsto \operatorname {arccot} x,$

## TheoremTheorem 16.20 change

The inverse trigonometric functions have the following
    ${\left(\arcsin x\right)}'={\frac {1}{\sqrt {1-x^{2}}}}\,.$
    ${\left(\arccos x\right)}'=-{\frac {1}{\sqrt {1-x^{2}}}}\,.$
    ${\left(\arctan x\right)}'={\frac {1}{1+x^{2}}}\,.$
    ${\left(\operatorname {arccot} x\right)}'=-{\frac {1}{1+x^{2}}}\,.$

### Proof

For example, for the arctangent, we have, due to Theorem 14.9 ,
    ${\begin{aligned}(\arctan x)^{\prime }&={\frac {1}{\frac {1}{\cos ^{2}(\arctan x)}}}\\&={\frac {1}{\frac {\cos ^{2}(\arctan x)+\sin ^{2}(\arctan x)}{\cos ^{2}(\arctan x)}}}\\&={\frac {1}{1+\tan ^{2}(\arctan x)}}\\&={\frac {1}{1+x^{2}}}.\end{aligned}}$
$\Box$