Power series

What is a power series?

A power series is $C(x)=\sum_{n=0}^{\infty}c_{n}x^{n}$ where x is a variable and $c_0,c_1,c_2,…$ are constants called coefficients of the power series.

What does absolute convergence for smaller modulus mean with relation to a power series?

Suppose for a given $x_0\in\mathbb{R}$ the power series $C(x_0)$ is convergent. Then for all y with |y|<|x_0|

1. $C(y)$ is absolutely convergent

2. For any fixed $p\ge0$ we have \sum_{n=1}^{\infty}\left|n^{p}c_{n}y^{n}\right|<+\infty

Whats the interval of convergence of the power series?

The set of points where the power series is convergent.

What’s the radius of convergence of the power series?

$R\in [0,+\infty]$ is the radius of convergence. If $R=+\infty$ then the seres is convergent for all $x\in\mathbb{R}$.

If 0<R<+\infty at $x=R$ and $x=-R$ the series may be

• absolutely convergent

• conditionally convergent

• divergent

How do you find the interval of convergence?

1. Apply ratio test to the non-negative series $\sum_{n=0}^{\infty}\left|a_{n}\right|\left|x\right|^{n}$

2. Based on the test, determine R such that $\sum_{n=0}^{\infty}\left|a_{n}\right|\left|x\right|^{n}$ is absolutely convergent when |x|<R and is not absolutely convergent when |x|>R. This is the radius of convergence.

3. Use further tests to check convergence of$\sum_{n=0}^{\infty}a_{n}x^{n}$ at $x=-R$ and $x=R$

What is Lipschitz Continuous?

Let $L$ be a non-negative constant. A function $f:[a,b]→\mathbb{R}$ is called L-Lipschitz continuous iff $\left|f\left(y\right)-f\left(x\right)\right|\le L\left|y-x\right|$ for all $x,y\in[a,b]$.

Proof that a function defined by a power series is continuous

Let $C(x)$ denote the sum of a power series with radius of convergence $R$. Then $C(x)$ is a continuous function on the interval $(-R,R)$.

Proof: $\forall$ closed interval $[-S,S]$ inside $(-R,R)$, we show that $C(x)$ is continuous on $[-S,S]$ for a suitable $S$ (namely if |x|<S<R) this will prove continuity of $C(x)$ at all points of $(-R,R)$.

If $x,y\in[-S,S]$ then by the Infinite Triangle Inequality,

$$|C(y)-C(x)|=\left|\sum_{n=1}^{\infty}c_{n}\left(y^{n}-x^{n}\right)\right|\le\sum_{n=1}^{\infty}\left|c_{n}\Vert y^{n}-x^{n}\right|$$

Since $x^n$ is Lipschits continuous on a bounded interval ($\forall x,y\in [-S,S]$ $|y^n-x^n|\le nS^{n-1}|y-x|$ ).

$$|C(y)-C(x)|\le\frac{1}{S}\left(\sum_{n=1}^{\infty}n\left|c_{n}\right|S^{n}\right)\left|y-x\right|$$

Since $\sum_{n=1}^{\infty}n\left|c_{n}\right|S^{n}$ is a convergent series, so the factor $L=\frac{1}{S}\sum_{}^{}n\left|c_{n}\right|S^{n}$ is finite. Hence the function $C(x)$ is L-lipschiotz continuous and so continuous, on $[-S,S]$.

What is the exponetnial in terms of a power series?

$$\exp\left(x\right)=1+\frac{x}{1!}+\frac{x^2}{2!}+\cdots=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}$$

$$e=\exp(1)$$

Proof that exp(x) is continuous.

If $x=0$ the series is 1+0+0+… which is convergent with sum 1. If $x\ne 0$, apply the ratio test to the series $\sum_{n=0}^{\infty}\frac{\left|x\right|^{n}}{n!}$ to find $l=\lim_{n\to\infty}\frac{\frac{x^{n+1}}{\left(n+1\right)!}}{\frac{\left|x\right|^{n}}{n!}}=\lim_{n\to\infty}\frac{\vert x\vert}{\left(n+1\right)}=0,\forall\vert x\vert$

Since 0<1 the power series exp(x) is absolutely convergent.

Thus, exp(x) is well defined for all x so it follows that exp(x) is a continuous function on all of $\mathbb{R}$.

Proof that exp(x+y)=exp(x)exp(y)

Recall that $\exp(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!}$. By the binomial formula, $\frac{1}{n!}(x+y)^n=\frac{x^n}{n!}+\frac{x^{n-1}}{(n-1)!}\frac{y}{1!}+\frac{x^{n-2}}{(n-2)!}\frac{y^2}{2!}+\cdots+\frac{x}{1!}\frac{y^{n-1}}{(n-1)!}+\frac{y^n}{n!}.$

For $n=0,1,2,\dots$ write the terms of these expansions in the $n$th column of a double series, filling the rest of the column with zeros.

Then the $n$th column sum is $\text{ColSum}_n=\frac{1}{n!}(x+y)^n,$ so $\sum_{n=0}^{\infty}\text{ColSum}_n=\sum_{n=0}^{\infty}\frac{(x+y)^n}{n!}=\exp(x+y).$

On the other hand, the $m$th row has common factor $\frac{y^m}{m!}$, so $\text{RowSum}_m=\left(1+\frac{x}{1!}+\frac{x^2}{2!}+\cdots\right)\frac{y^m}{m!}=\exp(x)\frac{y^m}{m!}.$

Hence $\sum_{m=0}^{\infty}\text{RowSum}_m=\exp(x)\sum_{m=0}^{\infty}\frac{y^m}{m!}=\exp(x)\exp(y).$

Since the double series is absolutely convergent, we may change the order of summation. Therefore $\exp(x+y)=\exp(x)\exp(y).$"