009C Sample Final 2, Problem 7 Detailed Solution

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

(b) Find its radius of convergence.

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Background Information:
1. The Taylor polynomial of  $f(x)$  at  $a$  is
$\sum _{n=0}^{\infty }c_{n}(x-a)^{n}$  where  $c_{n}={\frac {f^{(n)}(a)}{n!}}.$
2. Ratio Test
Let  $\sum a_{n}$  be a series and  $L=\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}.$  Then,
If  $L<1,$  the series is absolutely convergent.
If  $L>1,$  the series is divergent.
If  $L=1,$  the test is inconclusive.

Solution:

(a)

Step 1:
We begin by finding the coefficients of the Maclaurin series for  $f(x)={\frac {1}{(1-{\frac {1}{2}}x)^{2}}}.$
We make a table to find the coefficients of the Maclaurin series.
$n$ $f^{(n)}(x)$ $f^{(n)}(0)$ ${\frac {f^{(n)}(0)}{n!}}$ $0$ ${\frac {1}{(1-{\frac {1}{2}}x)^{2}}}$ $1$ $1$ $1$ ${\frac {1}{(1-{\frac {1}{2}}x)^{3}}}$ $1$ $1$ $2$ ${\frac {\frac {3}{2}}{(1-{\frac {1}{2}}x)^{4}}}$ ${\frac {3}{2}}$ ${\frac {3}{4}}$

Step 2:
So, the first three terms of the Binomial Series are
$1+x+{\frac {3}{4}}x^{2}.$

(b)

Step 1:
By taking the derivative of the known series
${\frac {1}{1-x}}\,=\,1+x+x^{2}+\cdots ,$
we find that the Maclaurin series of  ${\frac {1}{(1-x)^{2}}}$  is
$\sum _{n=0}^{\infty }(n+1)x^{n}.$
Letting  ${\frac {x}{2}}$  play the role of  $x,$  the Maclaurin series of  ${\frac {1}{(1-{\frac {1}{2}}x)^{2}}}$  is
$\sum _{n=0}^{\infty }(n+1){\bigg (}{\frac {1}{2}}x{\bigg )}^{n}=\sum _{n=0}^{\infty }{\frac {(n+1)x^{n}}{2^{n}}}.$

Step 2:
Now, we use the Ratio Test to determine the radius of convergence of this power series.
We have
${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {(n+2)x^{n+1}}{2^{n+1}}}{\frac {2^{n}}{(n+1)x^{n}}}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {|x|}{2}}{\frac {n+2}{n+1}}}\\&&\\&=&\displaystyle {{\frac {|x|}{2}}\lim _{n\rightarrow \infty }{\frac {n+2}{n+1}}}\\&&\\&=&\displaystyle {{\frac {|x|}{2}}.}\end{array}}$
Now, the Ratio Test says this series converges if  ${\frac {|x|}{2}}<1.$
So,  $|x|<2.$
Hence, the radius of convergence is  $R=2.$

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

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