# 009C Sample Final 2, Problem 7

(a) Consider the function  ${\displaystyle f(x)={\bigg (}1-{\frac {1}{2}}x{\bigg )}^{-2}.}$  Find the first three terms of its Binomial Series.

(b) Find its radius of convergence.

Foundations:
1. The Taylor polynomial of   ${\displaystyle f(x)}$   at   ${\displaystyle a}$   is

${\displaystyle \sum _{n=0}^{\infty }c_{n}(x-a)^{n}}$ where ${\displaystyle c_{n}={\frac {f^{(n)}(a)}{n!}}.}$

2. Ratio Test
Let  ${\displaystyle \sum a_{n}}$  be a series and  ${\displaystyle L=\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}.}$
Then,

If  ${\displaystyle L<1,}$  the series is absolutely convergent.

If  ${\displaystyle L>1,}$  the series is divergent.

If  ${\displaystyle L=1,}$  the test is inconclusive.

Solution:

(a)

Step 1:
We begin by finding the coefficients of the Maclaurin series for  ${\displaystyle f(x)={\frac {1}{(1-{\frac {1}{2}}x)^{2}}}.}$
We make a table to find the coefficients of the Maclaurin series.
 ${\displaystyle n}$ ${\displaystyle f^{(n)}(x)}$ ${\displaystyle f^{(n)}(0)}$ ${\displaystyle {\frac {f^{(n)}(0)}{n!}}}$ ${\displaystyle 0}$ ${\displaystyle {\frac {1}{(1-{\frac {1}{2}}x)^{2}}}}$ ${\displaystyle 1}$ ${\displaystyle 1}$ ${\displaystyle 1}$ ${\displaystyle {\frac {1}{(1-{\frac {1}{2}}x)^{3}}}}$ ${\displaystyle 1}$ ${\displaystyle 1}$ ${\displaystyle 2}$ ${\displaystyle {\frac {\frac {3}{2}}{(1-{\frac {1}{2}}x)^{4}}}}$ ${\displaystyle {\frac {3}{2}}}$ ${\displaystyle {\frac {3}{4}}}$
Step 2:
So, the first three terms of the Binomial Series is
${\displaystyle 1+x+{\frac {3}{4}}x^{2}.}$

(b)

Step 1:
By taking the derivative of the known series
${\displaystyle {\frac {1}{1-x}}\,=\,1+x+x^{2}+\cdots ,}$
we find that the Maclaurin series of  ${\displaystyle {\frac {1}{(1-x)^{2}}}}$  is
${\displaystyle \sum _{n=0}^{\infty }(n+1)x^{n}.}$
Letting ${\displaystyle x/2}$   play the role of ${\displaystyle x,}$ the Maclaurin series of  ${\displaystyle {\frac {1}{(1-{\frac {1}{2}}x)^{2}}}}$  is
${\displaystyle \sum _{n=0}^{\infty }(n+1){\bigg (}{\frac {1}{2}}x{\bigg )}^{n}=\sum _{n=0}^{\infty }{\frac {(n+1)x^{n}}{2^{n}}}.}$
Step 2:
Now, we use the Ratio Test to determine the radius of convergence of this power series.
We have
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {(n+2)x^{n+1}}{2^{n+1}}}{\frac {2^{n}}{(n+1)x^{n}}}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {|x|}{2}}{\frac {n+2}{n+1}}}\\&&\\&=&\displaystyle {{\frac {|x|}{2}}\lim _{n\rightarrow \infty }{\frac {n+2}{n+1}}}\\&&\\&=&\displaystyle {{\frac {|x|}{2}}.}\end{array}}}$
Now, the Ratio Test says this series converges if  ${\displaystyle {\frac {|x|}{2}}<1.}$  So,  ${\displaystyle |x|<2.}$
Hence, the radius of convergence is  ${\displaystyle R=2.}$

(a)    ${\displaystyle 1+x+{\frac {3}{4}}x^{2}}$
(b)    The radius of convergence is  ${\displaystyle R=2.}$