175 – Quiz 8

Here is quiz 8.

Problem 1 asks to find the power series {a_0+a_1x+a_2x^2+a_3x^3+\cdots} for {(2-x)^3}.

Method 1: Using basic algebra, expand the cube:

\displaystyle  \begin{array}{rcl} (2-x)^3&=&2^3-3\cdot2^2\cdot x+3\cdot 2\cdot x^2-x^3\\ &=&8-12x+6x^2-x^3. \end{array}

This gives {(2-x)^3} as a sum of powers of {x}. Since there is only one way of doing this, this is the series we were looking for (and we have {a_4=a_5=a_6=\dots=0}).

Method 2: Let {f(x)=(2-x)^3}. Then we have:

\displaystyle  \begin{array}{ccc} f(x)=(2-x)^3&\mbox{ and }&f(0)=8\\ f'(x)=-3(2-x)^2&\mbox{ and }&f'(0)=-12\\ f''(x)=6(2-x)&\mbox{ and }&f''(0)=12\\ f'''(x)=-6&\mbox{ and }&f'''(0)=-6\\ f^{(4)}(x)=0&\mbox{ and }&f^{(4)}(0)=f^{(5)}(0)=\dots=0. \end{array}

Using that {f(x)=}

\displaystyle  f(0)+f'(0)x+\frac{f''(0)}2 x^2+\frac{f'''(0)}6 x^3+\frac{f^{(4)}(0)}{4!}x^4+\dots,

we have

\displaystyle  f(x)=8-12x+6x^2-x^3+0+0+0+\dots\,.

Problem 2 asks to find the power series {a_0+a_1x+a_2x^2+a_3x^3+\cdots} for {\displaystyle \frac{x}{1+x^2}}.

Perhaps the easiest way to proceed is to note that

\displaystyle  \begin{array}{rcl} \displaystyle \frac1{1+x^2}&=&\displaystyle \frac1{1-(-x^2)}\\ &=&\displaystyle 1+(-x^2)+(-x^2)^2+(-x^2)^3+\dots\\ &=&\displaystyle 1-x^2+x^4-x^6+x^8-\dots \end{array}

and therefore

\displaystyle  \frac x{1+x^2}=x-x^3+x^5-x^7+x^9-\dots

Problem 3 asks to find the radius {R} of convergence of the series

\displaystyle \sum_{n=0}^\infty \frac{(2x)^{2n+1}}{n^2+1}=2x+4x^3+\frac{32}5 x^5+\frac{64}{5}x^7+\cdots\,.

As usual, we look at {\displaystyle b_n=\left|\frac{(2x)^{2n+1}}{n^2+1}\right|} and {b_{n+1}}, which we find by replacing {n} with {n+1} in the previous formula, so we have

\displaystyle  b_{n+1}=\left|\frac{(2x)^{2(n+1)+1}}{(n+1)^2+1}\right|.

We then take the quotient {b_{n+1}/b_n} and simplify:

\displaystyle  \begin{array}{rcl} \displaystyle \frac{b_{n+1}}{b_n}&=&\displaystyle \left|\frac{(2x)^{2(n+1)+1}}{(n+1)^2+1}\cdot\frac{n^2+1}{(2x)^{2n+1}}\right|\\ &=&\displaystyle |(2x)^2|\frac{n^2+1}{(n+1)^2+1}, \end{array}

and this last expression converges to {|(2x)^2|} as {n\rightarrow\infty}. To see this, note that

\displaystyle  \begin{array}{rcl} \displaystyle \frac{n^2+1}{(n+1)^2+1}&=&\displaystyle \frac{n^2+1}{n^2+2n+2}\\ &=&\displaystyle \frac{1+\displaystyle \frac1{n^2}}{\displaystyle 1+\frac2n+\frac2{n^2}}\\ &\rightarrow& 1. \end{array}

By the ratio test, the series converges if {|(2x)^2|<1} and diverges if {|(2x)^2|>1}. Note that {|(2x)^2|<1} is the same as {|2x|<1} or {|x|<1/2,} i.e., {R=1/2}.

Typeset using LaTeX2WP. Here is a printable version of this post.

Advertisements

One Response to 175 – Quiz 8

  1. There is a small typo in the pdf file: In method 1 for problem 1, “Since here” should be “Since there”.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: