Question

How do you find the maclaurin series of $f\left( x \right)={{e}^{-2x}}?$

Answer 1

The maclaurin series or taylor series of a real complex-valued function $f\left( x \right)$ that is initially differentiable at a real or complex number $a$ is the power series.
$f\left( a \right)+\dfrac{f'\left( a \right)}{1!}(x-a)+\dfrac{f''\left( a \right)}{2!}{{\left( x-a \right)}^{2}}+f'''\dfrac{\left( a \right)}{3!}{{\left( x-a \right)}^{3}}+........$
Where $n!$ denotes the factorial of $n$ in more compact sigma notation this can be written as $\sum\limits_{n=0}^{0}{\dfrac{{{f}^{\left( n \right)}}\left( a \right)}{n!}{{\left( x-a \right)}^{n}}}.$

Complete step by step solution:
The maclaurin series of $f\left( x \right)={{e}^{-2xq}}s.$
$f\left( x \right)=1+\left( -2x \right)+\dfrac{{{\left( -2x \right)}^{2}}}{2!}+\dfrac{{{\left( -2x \right)}^{3}}}{3!}+.........$
First solution method: the maclaurin series of $y={{e}^{x}}ps.$
$y=1+z+\dfrac{{{z}^{2}}}{2!}+\dfrac{{{z}^{3}}}{3!}+\dfrac{{{z}^{4}}}{4!}+...........$
Let $z=-2x$
Then $f\left( x \right)={{e}^{-2x}}={{e}^{z}}$ and $f\left( x \right)$ has the same maclaurin series as the one. Above except use set $z=-2x$ and get.
$f\left( x \right)=1+\left( -2x \right)+\dfrac{{{\left( -2x \right)}^{2}}}{2!}+\dfrac{{{\left( -2x \right)}^{3}}}{3!}+.........$
The maclaurin series of $f\left( x \right)\text{Ps}.$
$f\left( x \right)=f\left( x=0 \right)+\dfrac{f'\left( x=0 \right)x}{1!}+\dfrac{f''\left( x=0 \right){{x}^{2}}}{2!}+\dfrac{f'''\left( x=0 \right){{x}^{3}}}{3!}+..........$
The first term is $f\left( x=0 \right)$ Here $f\left( x=0 \right)={{e}^{-2(0)=1}}$.
The second term is $\dfrac{f'\left( x=0 \right)x}{1!}=\dfrac{-2{{e}^{-2\left( 0 \right)}}x}{1}=-2x$.
The third term is $\dfrac{f'\left( x=0 \right){{x}^{2}}}{2!}=\dfrac{{{\left( -2 \right)}^{2}}{{e}^{-2\left( 0 \right)}}{{x}^{2}}}{1}=\dfrac{{{\left( -2x \right)}^{2}}}{2!}$.
There are the same terms as in the maclaurin series written above.
By observing a pattern the ${{n}^{th}}$ term of the the series is $\dfrac{{{\left( -2x \right)}^{n}}}{n!}$.
Using a summation sign, the maclaurin series at $f\left( x \right)$ can be written instead as $f\left( x \right)=\sum\limits_{n=0}^{n=\infty }{\left[ \dfrac{{{\left( -2x \right)}^{n}}}{n!} \right]}$.

Note: Do not get confused between taylor series and maclaurin series both are the same.
Derivation at ${{e}^{-2x}}$ is $(-2){{e}^{-2x}}$ do not write ${{e}^{-2x}}$ many at us make these mistakes.