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 ifd \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)

Note that, by dividing both sides of the binomial formula by n!,

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