# 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.

• 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.