New Series from Old 2

Differentiating and integrating a known series can help you find other series. Since \frac{d}{dx}\sin \left( x \right)=\cos \left( x \right) we can find the series for cos(x) this way

\frac{d}{dx}\sin \left( x \right)=\frac{d}{dx}(x-\frac{{{x}^{3}}}{3!}+\frac{{{x}^{5}}}{5!}-\frac{{{x}^{7}}}{7!}+\cdots +\frac{{{\left( -1 \right)}^{n-1}}{{x}^{2n-1}}}{\left( 2n-1 \right)!}+\cdots )

\frac{d}{dx}\sin \left( x \right)=1-\frac{3{{x}^{2}}}{3!}+\frac{5{{x}^{4}}}{5!}-\frac{7{{x}^{6}}}{7!}+\cdots \frac{{{\left( -1 \right)}^{n-1}}\left( 2n-1 \right){{x}^{2n}}}{\left( 2n-1 \right)!}+\cdots

\cos \left( x \right)=1-\frac{{{x}^{2}}}{2!}+\frac{{{x}^{4}}}{4!}-\frac{{{x}^{6}}}{6!}+\cdots \frac{{{\left( -1 \right)}^{n-1}}{{x}^{2n}}}{\left( 2n \right)!}+\cdots

 Of course we could also have integrated the series for sin(x) to get the series for  –cos(x) and then changed the signs.

In out next post we will find that

 \frac{1}{1+{{x}^{2}}}=1-{{x}^{2}}+{{x}^{4}}-{{x}^{6}}+\cdots +{{\left( -1 \right)}^{2n+1}}{{x}^{2n-2}}+\cdots

Recall that \frac{d}{dx}\arctan \left( x \right)=\frac{1}{1+{{x}^{2}}}, so we can integrate the series above to find the series for arctan(x).

\arctan \left( x \right)=\int_{{}}^{{}}{(1-{{x}^{2}}+{{x}^{4}}-{{x}^{6}}+\cdots +{{\left( -1 \right)}^{2n-1}}{{x}^{2n-2}}+\cdots )dx}

\arctan \left( x \right)=C+x-\tfrac{1}{3}{{x}^{3}}+\tfrac{1}{5}{{x}^{5}}-\tfrac{1}{7}{{x}^{7}}+\cdots +\tfrac{{{\left( -1 \right)}^{2n+1}}}{2n-1}{{x}^{2n-1}}+\cdots

Since \arctan \left( 0 \right)=0 it follows that the constant  of integration C=0 so

 \arctan \left( x \right)=x-\tfrac{1}{3}{{x}^{3}}+\tfrac{1}{5}{{x}^{5}}-\tfrac{1}{7}{{x}^{7}}+\cdots +\tfrac{{{\left( -1 \right)}^{2n+1}}}{2n-1}{{x}^{2n-1}}+\cdots

When differentiating or integrating the interval of convergence cannot get any larger, but it can get smaller. If the endpoints are included before differentiating or integrating then you must check to see if they are included in the new series.

Next post: Rational functions and geometric series.


2 thoughts on “New Series from Old 2

  1. Often the AP exam asks for a series and the general term. Does the general term always need to start with n=1 or is okay to start with n=0? I saw this asked last year on the AP Calculus forum but there was no answer.


    • A series is determined by the terms. There is always more then one way to write the general terms. You can pretty much start anywhere you want provided that your expression gives the correct terms. For example, \sum\limits_{n=1}^{\infty }{n}=\sum\limits_{n=11}^{\infty }{\left( n-10 \right)} That said, 1 is usually a good place to start.


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