Question

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

Answer 1

Hint: 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.