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?

  • 2
    $\begingroup$ Did you write the geometric series of $\dfrac{1}{1+4x^2}$.? $\endgroup$
    – Nosrati
    Feb 13, 2017 at 15:39
  • $\begingroup$ what do u mean?? @MyGlasses $\endgroup$ Feb 13, 2017 at 15:39
  • $\begingroup$ Try substituting $u=2x$ for the first one (because then $u^2=4x^2$), and $u=3x$ for the second one. $\endgroup$
    – Théophile
    Feb 13, 2017 at 15:45
  • $\begingroup$ @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 $\endgroup$ Feb 13, 2017 at 15:51
  • $\begingroup$ how can i check my answers???? $\endgroup$ Feb 13, 2017 at 15:52

1 Answer 1


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: $$ \frac{1}{1+9x^2}=\sum_{n=0}^{\infty}(-1)^n\,(3x)^{2n} $$

  • $\begingroup$ thank you.. but when u derive dont u add a +1 to the n, so now the series will start at n=1 not n=0... also so the answer for 9 would be (-1)^n(3x)^2n correct? also, how can i possibly check my answers? $\endgroup$ Feb 13, 2017 at 16:00
  • $\begingroup$ Yes, you are right!! Try doing the same process for the other expression! =) $\endgroup$ Feb 13, 2017 at 16:03
  • $\begingroup$ mathworld.wolfram.com/MaclaurinSeries.html $\endgroup$ Feb 13, 2017 at 16:05
  • $\begingroup$ Out of curiosity, if you take the derivative of a series of functions, shouldn't the uniform convergence be checked? $\endgroup$ Feb 13, 2017 at 16:06
  • $\begingroup$ Yes! You need to justify why you can do what I did! $\endgroup$ Feb 13, 2017 at 16:07

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