0
$\begingroup$

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?

$\endgroup$
5
  • 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

1
$\begingroup$

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

$\endgroup$
7
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.