# Maclaurin series therom question of artan

find the Maclaurin series of $$\frac{1}{1+4x^2}$$ and $$\frac{1}{1+9x^2}$$ using $$\arctan x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1}$$

Firstly I derived the formula they gave us, then substituted $x \rightarrow 9x$. Is this the right approach?

• Did you write the geometric series of $\dfrac{1}{1+4x^2}$.? Feb 13, 2017 at 15:39
• what do u mean?? @MyGlasses Feb 13, 2017 at 15:39
• Try substituting $u=2x$ for the first one (because then $u^2=4x^2$), and $u=3x$ for the second one. Feb 13, 2017 at 15:45
• @Théophile yes i did that and found that 1/3arctan (3x) = 1/1+9x^2 ..... so then i substiuted and found that 1/3arctan (3x) = sigma from n=1 is [(-1)^n 9^n x^(2n+1)] / 2n+1... using the thing thing they gave us in the question. so then i derived both and found 1/1+9x^2 to be sigm from n=1 is (-1)^n 9^n x^2n, which i think is the wrong answer Feb 13, 2017 at 15:51
• how can i check my answers???? Feb 13, 2017 at 15:52

As you know that: $$\arctan(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1} \quad\Rightarrow\quad \arctan (2x) = \sum_{n=0}^{\infty} (-1)^n \frac{(2x)^{2n+1}}{2n+1}$$ Also, you have: $$\frac{d}{dx}\left[\arctan(x)\right] = \frac{1}{1+x^2} \quad\Rightarrow\quad \frac{d}{dx}\left[\arctan(2x)\right] = \frac{2}{1+4x^2}$$ Thus, \begin{align} \color{red}{\frac{1}{1+4x^2}} &= \frac{1}{2}\,\frac{d}{dx} \left[\color{white}{\frac{}{}}\arctan(2x)\color{white}{\frac{}{}}\right] = \frac{1}{2}\,\frac{d}{dx} \left[\,\sum_{n=0}^{\infty}(-1)^n\,\frac{(2x)^{2n+1}}{2n+1}\,\right] \\[2mm] &= \frac{1}{2}\,\sum_{n=0}^{\infty}(-1)^n\,\frac{\left(\,(2x)^{2n+1}\,\right)'}{2n+1} \\[2mm] &= \frac{1}{2}\,\sum_{n=0}^{\infty}(-1)^n\,\frac{{2\,(2n+1)}\,(2x)^{2n}}{2n+1} \\[2mm] &= \color{red}{\sum_{n=0}^{\infty}(-1)^n\,(2x)^{2n}} \end{align} And with similar analysis, you should get: $$\frac{1}{1+9x^2}=\sum_{n=0}^{\infty}(-1)^n\,(3x)^{2n}$$