# New Series from Old 1

There are three common ways to get new series from old series without calculating all the derivatives and substituting into the Taylor Series general term.

• Substituting into known series
• Differentiating and integrating
• Approaching rational functions as geometric series

This will be  the subject of this and my next two posts.

Substituting

As you might expect the Taylor/Maclaurin series for the “parent” functions are known. We’ve seen the series for ln(x) and sin(x) and you can develop or look up series for ex, cos(x) and so on. The power series for compositions involving these series can be found by substituting.

Example 1: To find the Maclaurin series for sin(3x), substitute 3x for x in the sin(x) series

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

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

Substituting along with some algebra also works, as the next example shows.

Example 2: To find the series for multiply the last answer by x2

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

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

Example 3: You must be a little careful sometimes. A recent AP Calculus exam asked for the first 4 terms of the Maclaurin series for $\sin \left( 5x+\tfrac{\pi }{4} \right)$. Substituting $\sin \left( 5x+\tfrac{\pi }{4} \right)$ into the sin(x) series gives a series centered at $x=-\tfrac{\pi }{20}$ but the questions required a series centered at x = 0. Almost all students calculated derivatives and used them in the general form of a Maclaurin series. But with a little trigonometry you can substitute after some rewriting:

$\sin \left( 5x+\tfrac{\pi }{4} \right)=\sin \left( 5x \right)\cos \left( \tfrac{\pi }{4} \right)+\cos \left( 5x \right)\sin \left( \tfrac{\pi }{4} \right)$

$\sin \left( 5x+\tfrac{\pi }{4} \right)=\tfrac{\sqrt{2}}{2}\left( \sin \left( 5x \right)+\cos \left( 5x \right) \right)$

Now we can substitute into the series for the sin(x) and cos(x):

$\sin \left( 5x+\tfrac{\pi }{4} \right)=\tfrac{\sqrt{2}}{2}\left( \left( 5x-\frac{{{\left( 5x \right)}^{3}}}{3!} \right)+\left( 1-\frac{{{\left( 5x \right)}^{2}}}{2!} \right) \right)$

$\sin \left( 5x+\tfrac{\pi }{4} \right)=\tfrac{\sqrt{2}}{2}+\tfrac{5\sqrt{2}}{2}x-\tfrac{25\sqrt{2}}{4}{{x}^{2}}-\tfrac{125\sqrt{2}}{12}{{x}^{3}}$

As you can see from the last two examples, any sort of “legal” algebra or trigonometry can be used to find a new series from one you already know.

Next post: Differentiating and Integrating