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Obtain the Maclaurin's series expansion for:$\;\tan x,- \large\frac{\pi}{2} \lt \normalsize x \lt \large\frac{\pi}{2}$

Note: This is part 4th of a 4 part question, split as 4 separate questions here.

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Toolbox:
  • Maclaurin's series :If $ f(x)$ and its derivatives (a sufficiently large number of derivatives ) are continuous on [o x], then Maclaurin series for the expansion of $f(x)$ is given by $f(x)=f(0)+\large\frac{f'(0)}{1!} x+\frac{f''(0)}{2 !}x^2+............$
Given $\tan x\;\large \frac{-\pi}{2} < x <\large\frac{\pi}{2}$
Step 1:
$f(x)=\tan x$                                                                                            $f(0)=0$
$f'(x)=1+\tan ^2 x$                                                                                   $f'(0)=1$
$f''(x)=2 \tan x \sec^2 x $
$\qquad=2\tan x(1+\tan ^2 x)$
$\qquad=2\tan x+2\tan ^3 x$                                                                     $f''(0)=0$
$f'''(x)=2 \sec^2 x +6 \tan2 x \sec^2 x$
$\qquad= 2+ 8 \tan ^2 x +6 \tan ^4 x$                                                        $f'''(0)=2$
$f^{iv}(x)=16 \tan x \sec^2 x +24 \tan^3 x \sec^2 x$
$\qquad=16 \tan x (1+\tan ^2 x)+24 \tan ^3 x (1+\tan ^2 x) $
$\qquad=16 \tan x +40 \tan ^3 x +24 \tan ^5 x $                                         $f^{iv} (0)=0$
$f^{v}(x)=16 \sec^2 x +120 \tan^2 x \sec^2 x+120 \tan^4 x \sec^2 x$      $f^{v} (0)=16$
Step 2:
Maclaurin's series for $f(x)$ is given by
$f(x)=f(0)+\large\frac{f'(0)}{1!} x+\frac{f''(0)}{2 !}x^2+............$
Step 3:
$\tan x=x+ \large\frac{2x^3}{3!}+\frac{16x^5}{5 !}+.....$
$\qquad=x+\large\frac{x^3}{3}+\frac{2x^5}{15}....$

 

answered Jul 26, 2013 by meena.p
 
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