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Obtain the Maclaurin's series expansion for:$\;\large\frac{1}{1+x}$

Note: This is part 3rd of a 4 part question, split as 4 separate questions here.

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  • 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 $\large\frac{1}{1+x}$
Step 1:
$f(x)=\large\frac{1}{1+x}$               $f(0)=1$
$f'(x)=\large\frac{-1}{(1+x)^2}$       $f'(0)=-1$
$f''(x)=\large\frac{2}{(1+x)^3}$        $f''(0)=2$
$f'''(x)=\large\frac{-6}{(1+x)^4}$      $f'''(0)=-6$
$f^{iv}(x)=\large\frac{24}{(1+x)^4}$    $f^{iv} (0)=-24$
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:
$\large\frac{1}{1+x}=1- \large\frac{1}{1!}x+\frac{2}{2 !}$$x^2+\large\frac{6}{3 !}$$x^3+\large\frac{24}{4 !}$$x^4.....$
$\qquad=1-x+x^2-x^3+x^4+.......$

 

answered Jul 25, 2013 by meena.p
 
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