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Obtain the Maclaurin's series expansion for:$\;e^{2x}$

Note: This is part 1st 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 $e^{2x}$
Step 1:
$f(x)=e^{2x}$                    $f(0)=e^0=1$
$f'(x)=2e^{2x}$                 $f'(0)=2$
$f''(x)=4e^{2x}$                $f''(0)=4$
$f'''(x)=8e^{2x}$                $f'''(0)=8$
$f^{iv}(x)=16e^{2x}$         $f^{iv} (0)=16$
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:
$e^{2x}=1+ \large\frac{2x}{1!}+\frac{4x^2}{2!}+\frac{8x^3}{3 !}+\frac{16x^4}{4 !}.....$


answered Jul 25, 2013 by meena.p
edited Jul 25, 2013 by meena.p
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