The function f(x) = \frac{6}{1+49x^2} is represented as a power series: \sum_{n=0}^{\infty}...

Question:

The function {eq}f(x) = \frac{6}{1+49x^2} {/eq} is represented as a power series:

{eq}f(x) = \sum_{n=0}^{\infty} c_nx^n {/eq}

(a) Find the first five coefficients of the power series.

(b) Find the radius of convergence R of the series.

Infinite Geometric Series Sum:

Sum of infinite geometric series with initial term {eq}a{/eq} and common ratio {eq}r{/eq} is given by,

{eq}\displaystyle \sum_{n=0}^{\infty} ar^n = \frac{a}{1-r} {/eq}

A special case of this is with a=1,r=x

{eq}\displaystyle \sum_{n=0}^{\infty} x^n = \frac{1}{1-x} {/eq}

Interval of convergence of this series is given by,

{eq}|x|<1 {/eq}

Consider the geometric series,

{eq}\begin{align} \displaystyle \frac{1}{1-x} &= \sum_{n=0}^{\infty} x^n \\ \displaystyle \frac{1}{1-(-49x^2)} &= \sum_{n=0}^{\infty} (-49x^2)^n && \cdots \text{Substituting }-49x^2 \text{ in place of }x \\ \displaystyle \frac{1}{1+49x^2} &= \sum_{n=0}^{\infty} (-1)^n49^nx^{2n} \\ \displaystyle \frac{6}{1+49x^2} &= \sum_{n=0}^{\infty} 6(-1)^n49^nx^{2n} && \cdots \text{Multiplying by }6 \\ \displaystyle \sum_{n=0}^{\infty} c_nx^n &= \sum_{n=0}^{\infty} 6(-1)^n49^nx^{2n} \\ \displaystyle \sum_{n=0}^{\infty} c_nx^n &= 6(-1)^049^0x^{2(0)}+6(-1)^149^1x^{2(1)}+6(-1)^249^2x^{2(2)}+... \\ \displaystyle \sum_{n=0}^{\infty} c_nx^n &= 6-294x^2+14406x^4+... \\ \displaystyle \color{blue}{ c_0 =6, c_1=0, c_2} &\color{blue}{ =-294,c_3=0,c_4=14406} &&....\text{Comparing corresponding coefficients on both sides} \end{align} {/eq}

Interval of convergence is given by,

{eq}\begin{align} |-49x^2| &<1 \\ |49x^2| &<1 \\ \displaystyle |x^2| &<\frac{1}{49} \\ \displaystyle |x| &<\frac{1}{7} \\ \displaystyle \frac{-1}{7} &< x <\frac{1}{7} \end{align} {/eq}

Hence,

Radius of convergence, {eq}\displaystyle \color{blue}{R = \frac{1}{7}} {/eq} 