New Taylor Series From Old

Section: 
10.3
Date: 
Monday, October 25, 2010 - 14:00 - 15:00
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fall2010math2300_10-3-taylor-series-notes.pdf58.76 KB

Math 2300 Section 005 – Calculus II – Fall 2010

Notes on Obtaining New Taylor Series from Old – October 19, 2010

Taylor Series We Currently Know

\displaystyle\sin x \displaystyle=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}+\cdots=\sum _{{k=0}}^{\infty}\frac{(-1)^{k}}{(2k+1)!}x^{{2k+1}}
\displaystyle\cos x \displaystyle=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\frac{x^{6}}{6!}+\cdots=\sum _{{k=0}}^{\infty}\frac{(-1)^{k}}{(2k)!}x^{{2k}}
\displaystyle e^{x} \displaystyle=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\cdots=\sum _{{k=0}}^{\infty}\frac{x^{k}}{k!}
\displaystyle(1+x)^{p} \displaystyle=1+px+\frac{p(p-1)}{2!}x^{2}+\frac{p(p-1)(p-2)}{3!}x^{3}+\cdots
\displaystyle\frac{1}{1+x} \displaystyle=(1+x)^{{-1}}=1-x+x^{2}-x^{3}+x^{4}-\cdots=\sum _{{k=0}}^{\infty}(-1)^{k}x^{k}
\displaystyle\ln(x+1) \displaystyle=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\cdots=\sum _{{k=1}}^{\infty}\frac{(-1)^{{k+1}}}{k}x^{k}

There are several ways we could go about obtaining new series from these ones. First, we can substitute inside an existing series. Then, we could differentiate or integrate existing series. We'll consider these with some examples.

Substituting Inside a Taylor Series

If we're given one series for a function f(x) and we wish to create the series for f(g(x)), then we can replace g(x) for x in the Taylor series of f(x).

Examples:

  1. Find the Taylor series for \sin(2x) about x=0.

    Solution: We plug in 2x everywhere we see x in the Taylor series expansion of \sin(x) about x=0. That is, the desired Taylor series is

    \displaystyle sin(2x) \displaystyle=(2x)-\frac{(2x)^{3}}{3!}+\frac{(2x)^{5}}{5!}-\frac{(2x)^{7}}{7!}+\cdots
    \displaystyle=2x-\frac{2^{3}x^{3}}{3!}+\frac{2^{5}x^{5}}{5!}-\frac{2^{7}x^{7}}{7!}+\cdots
    \displaystyle=\sum _{{k=0}}^{\infty}\frac{(-1)^{k}2^{{2k+1}}}{(2k+1)!}x^{{2k+1}}
  2. Find the Taylor series for \frac{1}{1-x} about x=0.

    Solution: We know that

    \frac{1}{1+x}=1-x+x^{2}-x^{3}+x^{4}-\cdots=\sum _{{k=0}}^{\infty}(-1)^{k}x^{k}.

    If we replace the x in this series by -x, we will then have the desired series. So, we have

    \displaystyle\frac{1}{1-x} \displaystyle=\frac{1}{1+(-x)}
    \displaystyle=1-(-x)+(-x)^{2}-(-x)^{3}+(-x)^{4}-\cdots=\sum _{{k=0}}^{\infty}(-1)^{k}(-x)^{k}
    \displaystyle=1+x+x^{2}+x^{3}+x^{4}+\cdots=\sum _{{k=0}}^{\infty}(-1)^{k}(-1)^{k}x^{k}
    \displaystyle=1+x+x^{2}+x^{3}+x^{4}+\cdots=\sum _{{k=0}}^{\infty}x^{k}
  3. Example 2 on page 520 provides a much more complicated example.

Differentiation and Integration

If you know a Taylor series for a function, you can find the Taylor series for its derivative or integral by differentiating or integrating the terms of the Taylor series. A few examples make this more obvious.

Examples:

  1. Find the Taylor series for \displaystyle\frac{1}{(1+x)^{2}} around zero.

    Solution: We know the Taylor series for \displaystyle\frac{1}{1+x} and we know that

    \frac{d}{dx}\frac{1}{1+x}=-\frac{1}{(1+x)^{2}}.

    Now, we have

    \displaystyle\frac{1}{1+x} \displaystyle=1-x+x^{2}-x^{3}+\cdots=\sum _{{k=0}}^{\infty}(-1)^{k}x^{k}
    \displaystyle\frac{d}{dx}\frac{1}{1+x} \displaystyle=\frac{d}{dx}\left(1-x+x^{2}-x^{3}+\cdots\right)=\frac{d}{dx}\left(\sum _{{k=0}}^{\infty}(-1)^{k}x^{k}\right)
    \displaystyle-\frac{1}{(1+x)^{2}} \displaystyle=-1+2x-3x^{2}+4x^{3}-\cdots=\sum _{{k=0}}^{\infty}(-1)^{{k+1}}(k+1)x^{k}
    \displaystyle\frac{1}{(1+x)^{2}} \displaystyle=1-2x+3x^{2}-4x^{3}+\cdots=\sum _{{k=0}}^{\infty}(-1)^{{k}}(k+1)x^{k}
  2. Find the Taylor series for \arctan x about x=0. Use the resulting Taylor series to approximate \pi.

    Solution: We know that

    \frac{d}{dx}\arctan x=\frac{1}{1+x^{2}}.

    Thus, if we can find the Taylor series for this function, we can integrate it and get back to \arctan x. We can perform a substitution and find

    \displaystyle\frac{1}{1+x} \displaystyle=1-x+x^{2}-x^{3}+\cdots=\sum _{{k=0}}^{\infty}(-1)^{k}x^{k}
    \displaystyle\frac{1}{1+(x^{2})} \displaystyle=1-(x^{2})+(x^{2})^{2}-(x^{3})^{2}+\cdots=\sum _{{k=0}}^{\infty}(-1)^{k}(x^{2})^{k}
    \displaystyle\frac{1}{1+x^{2}} \displaystyle=1-x^{2}+x^{4}-x^{6}+\cdots=\sum _{{k=0}}^{\infty}(-1)^{k}x^{{2k}}

    Then, we integrate and find

    \displaystyle\arctan x \displaystyle=\int\frac{1}{1+x^{2}}\, dx
    \displaystyle=\int 1-x^{2}+x^{4}-x^{6}+\cdots\, dx
    \displaystyle=x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}+\cdots
    \displaystyle=\sum _{{k=0}}^{\infty}\frac{(-1)^{k}}{2k+1}x^{{2k+1}}.

    So, we now have a Taylor series for \arctan x around x=0. Approxmating \pi is then a simple trick. We recall that

    \frac{\pi}{4}=\arctan 1\qquad\text{and so}\qquad\pi=4\arctan 1.

    Using our Taylor series we then have

    \displaystyle\pi \displaystyle=4\arctan(1)
    \displaystyle=4\left(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\cdots\right)
    \displaystyle=4\sum _{{k=0}}^{\infty}\frac{(-1)^{k}}{2k+1}

© 2011 Jason B. Hill. All Rights Reserved.