# Taylor series

On a fixed interval $$I$$ in the real line, we have a point $$a\in I$$. The $$N$$th-order Taylor polynomial for $$f : I\to\mathbb{R}$$ at $$a$$ is

$P_N(x) = \sum\limits_{n=0}^N\frac{f^{(n)}(a)}{n!}(x-a)^n$

The constant $$\frac{f^{(n)}(a)}{n!}$$ of the polynomal is chosen so that the nth derivative of $$P_N(x)$$ and $$f(x)$$ are equal at $$a$$:

$\begin{array}{rl}P_N^{(0)}(a) &= f^{(0)}(a)\\P_N^{(1)}(a) &= f^{(1)}(a)\\P_N^{(2)}(a) &= f^{(2)}(a)\\&= \vdots\\P_N^{(N)}(a) &= f^{(N)}(a)\\\end{array}$

Taylor series of a real or complex-valued function $$f(x)$$ that is infinitely differentiable at a real or complex number $$a$$ is the power series

$P_\infty(x)=\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n$

where $$f^{(n)}(a)$$ denotes the $$n$$th derivative of $$f(x)$$ evaluated at point $$a$$. The derivative of order zero of $$f(x)$$ is defined to be $$f(x)$$ itself.

The Taylor series for any polynomial is the polynomial itself.

## Maclaurin series

When $$a=0$$, the Taylor series is also called a Maclaurin series

$\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n$

### Geometric Series

We know the geometric series $$\sum\limits_{i=0}^\infty x^i$$ converges to $$\frac{1}{1-x}$$ for $$|x|<1$$. Reversing this convergence with Maclaurin series is possible by determining the $$n$$th derivative

$\frac{\partial ^n}{\partial x^n} \frac{1}{1-x} = \frac{n!}{(1 - x)^{n + 1}}$

From which we can find the original geometric series as a Maclaurin series:

$\frac{1}{1-x} =\sum _{n=0}^{\infty }x^n = 1+x+x^2+x^3+\dots \forall x\in(-1,1)$

### Exponential function

$e^{x}=\exp(x)=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}=1+x+{\frac {x^{2}}{2!}}+{\frac {x^{3}}{3!}}+\dotsb$

This way $$e:=\exp(1)$$ can be defined.

### Natural logarithms

$\ln(1+x) = \sum \limits_{n=1}^{\infty }(-1)^{n+1}\frac{x^n}{n} = x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots\forall x\in(-1,1]$

$\ln(1-x) = \sum \limits_{n=1}^{\infty }(-1)^{n}\frac{x^n}{n} = -x+\frac{x^2}{2}-\frac{x^3}{3}+\frac{x^4}{4}-\dots\forall x\in(-1,1]$

For the boundary on $$x=1$$ we see that $$\ln(1+1) = \sum \limits_{n=1}^{\infty }(-1)^{n+1}\frac{1}{n}$$ and therefore

$\ln(2) = 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots$

$\log\left(\frac{1+x}{1-x}\right) = \sum\limits_{n=1}^\infty \frac{2x^{2n-1}}{2n-1}$

### Cosine and Sine function

$\cos(x)=\sum _{n=0}^{\infty }(-1)^{n}{\frac {x^{2n}}{(2n)!}}={\frac {x^{0}}{0!}}-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-\frac{x^6}{6!}+ \dotsb$

$\sin(x)=\sum _{n=0}^{\infty }(-1)^{n}{\frac {x^{2n+1}}{(2n+1)!}}={\frac {x^1}{1!}}-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}- \frac{x^7}{7!}+\dotsb$

It is interesting to see that the polynomial of sine has all the odd exponents divided by odd factorials and the cosine has all even exponents divided by even factorials, so they are kind of complementary.

### Tangent function

$\tan(x) = x + \frac{1}{3}x^3 + \frac{2}{15}x^5 + \frac{17}{315}x^7 + \dotsb$

### Arcus Tangent function

$\tan^{-1}(x) = \sum _{n=0}^{\infty }(-1)^{n}\frac{x^{2n+1}}{2n+1} = \frac{x^1}{1!} -\frac{x^3}{3!}+\frac{x^5}{5!} -\frac{x^7}{7!}+\dots\forall x\in[-1,1]$

### Arcus Cotangent function

$\begin{array}{rl}\cot^{-1}(x) &= \frac{\pi}{2} - \tan^{-1}(x)\\&= \frac{\pi}{2} - \sum _{n=0}^{\infty }(-1)^{n}\frac{x^{2n+1}}{2n+1}\end{array}$

### Hyperbolic functions

{\displaystyle {\begin{aligned}\cosh(z)&=1+{\frac {z^{2}}{2!}}+{\frac {z^{4}}{4!}}+{\frac {z^{6}}{6!}}+\dots =\sum _{n=0}^{\infty }{\frac {z^{2n}}{(2n)!}}\\\sinh(z)&=z+{\frac {z^{3}}{3!}}+{\frac {z^{5}}{5!}}+{\frac {z^{7}}{7!}}+\dots =\sum _{n=0}^{\infty }{\frac {z^{2n+1}}{(2n+1)!}}\end{aligned}}}

## Derivation of complex numbers

When we add the Maclaurin series of $$\cos(x)$$ and $$\sin(x)$$, we get

$\cos(x) + \sin(x) = {\frac {x^{0}}{0!}}+{\frac {x}{1!}}-{\frac {x^{2}}{2!}}-{\frac {x^{3}}{3!}}+{\frac {x^{4}}{4!}}+{\frac {x^{5}}{5!}}-\frac{x^6}{6!}- \frac{x^7}{7!}+\dotsb$

which is pretty similar to the Maclaurin series of $$e^x$$. The only difference is the signs. In order to properly add the series, we need something that is negative when squared. The problem is that $$i^2\geq 0\forall i\in\mathbb{R}$$. The solution is the clever trick of defining $$i^2=-1$$, which gives the famous equation

$\begin{array}{rl}e^{ix} &= \frac{(ix)^0}{0!} + \frac{(ix)^1}{1!} + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \frac{(ix)^5}{5!} + \frac{(ix)^6}{6!} + \dotsb\\&= \left({\frac {x^{0}}{0!}}-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-\frac{x^6}{6!}+ \dotsb\right)+ i\left({\frac {x}{1!}}-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}- \frac{x^7}{7!}+\dotsb\right)\\&= \cos(x) + i\sin(x)\end{array}$

Plugging in $$x:= \pi$$ results in what some call the most beautiful equation in mathematics, Euler’s identity:

$e^{i\pi} + 1 = 0$

But what is $$i$$ anyway? Instead of seeing numbers on a line, this opens a whole plane of numbers, the Complex Numbers.

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