Question

What is a Taylor’s expansion of \[{{e}^{-2x}}\] centered at \[x=0\]?

Answer 1

Hint: This type of question depends on the concept of Taylor’s series expansion of a function at a particular point. We know that the Taylor’s series expands any function till an infinite sum of terms which are expressed in terms of the derivatives of the function at a point. We know that the Taylor’s series expansion of a function centered at \[x=0\] is known as Maclaurin’s series. The general formula for Maclaurin’s series is \[f\left( x \right)=\sum\limits_{n=0}^{\infty }{{{f}^{n}}\left( 0 \right)\dfrac{{{x}^{n}}}{n!}}\]

Complete step by step solution:
Now, we have to find Taylor’s series expansion of \[{{e}^{-2x}}\] centered at \[x=0\].
We know that Taylor’s series expansion at \[x=0\] is known as Maclaurin’s series which is given by,
\[\Rightarrow f\left( x \right)=\sum\limits_{n=0}^{\infty }{{{f}^{n}}\left( 0 \right)\dfrac{{{x}^{n}}}{n!}}\]
Which we can also write as
\[\Rightarrow f\left( x \right)=f\left( 0 \right)+f'\left( 0 \right)x+f''\left( 0 \right)\dfrac{{{x}^{2}}}{2!}+f'''\left( 0 \right)\dfrac{{{x}^{3}}}{3!}+.........+{{f}^{n}}\left( 0 \right)\dfrac{{{x}^{n}}}{n!}+.......\text{ e}{{\text{q}}^{\text{n}}}\left( 1 \right)\]
Let us consider,
\[\Rightarrow f\left( x \right)={{e}^{-2x}}\]
On taking derivatives, we get,
\[\begin{align}
  & \Rightarrow f'\left( x \right)=-2{{e}^{-2x}} \\
 & \Rightarrow f''\left( x \right)=4{{e}^{-2x}} \\
 & \Rightarrow f'''\left( x \right)=-8{{e}^{-2x}} \\
\end{align}\]
Continuing in this way we get \[{{n}^{th}}\] order derivative,
\[\Rightarrow {{f}^{n}}\left( x \right)={{\left( -2 \right)}^{n}}{{e}^{-2x}}\]
Now, by substituting \[x=0\] we can write
\[\begin{align}
  & \Rightarrow f\left( 0 \right)=1 \\
 & \Rightarrow f'\left( 0 \right)=-2 \\
 & \Rightarrow f''\left( 0 \right)=4 \\
 & \Rightarrow f'''\left( 0 \right)=-8 \\
 & \Rightarrow {{f}^{n}}\left( 0 \right)={{\left( -2 \right)}^{n}} \\
\end{align}\]
Thus \[\text{e}{{\text{q}}^{\text{n}}}\left( 1 \right)\] becomes,
\[\begin{align}
  & \Rightarrow {{e}^{-2x}}=1+\left( -2 \right)x+4\dfrac{{{x}^{2}}}{2!}+\left( -8 \right)\dfrac{{{x}^{3}}}{3!}+.........+{{\left( -2 \right)}^{n}}\dfrac{{{x}^{n}}}{n!}+....... \\
 & \Rightarrow {{e}^{-2x}}=1-2x+2{{x}^{2}}-\dfrac{4}{3}{{x}^{3}}+............+{{\left( -2 \right)}^{n}}\dfrac{{{x}^{n}}}{n!}+...... \\
\end{align}\]
Hence the Taylor’s series expansion of \[{{e}^{-2x}}\] centered at \[x=0\] is given by, \[{{e}^{-2x}}=1-2x+2{{x}^{2}}-\dfrac{4}{3}{{x}^{3}}+............+{{\left( -2 \right)}^{n}}\dfrac{{{x}^{n}}}{n!}+......\]

Note: This type of question can also be solved in another way. One of the students may solve this as first finding the Maclaurin’s series for the function \[f\left( x \right)={{e}^{x}}\] and then to obtain series expansion of \[f\left( x \right)={{e}^{-2x}}\] only replace each of \[x\] by \[-2x\]. Hence, if we consider \[f\left( x \right)={{e}^{x}}\] then
\[\begin{align}
  & \Rightarrow f'\left( x \right)={{e}^{x}} \\
 & \Rightarrow f''\left( x \right)={{e}^{x}} \\
 & \Rightarrow f'''\left( x \right)={{e}^{x}} \\
\end{align}\]
In fact the \[{{n}^{th}}\] order derivative of \[{{e}^{x}}\] is \[{{e}^{x}}\]itself.
\[\Rightarrow {{f}^{n}}\left( x \right)={{e}^{x}}\]
Now, by substituting \[x=0\] we can write
\[\begin{align}
  & \Rightarrow f'\left( 0 \right)=1 \\
 & \Rightarrow f''\left( 0 \right)=1 \\
 & \Rightarrow f'''\left( 0 \right)=1 \\
 & \Rightarrow {{f}^{n}}\left( 0 \right)=1 \\
\end{align}\]
Hence, the Maclaurin’s series for \[f\left( x \right)={{e}^{x}}\] is given by,
\[\Rightarrow {{e}^{x}}=1+x+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{3}}}{3!}+.........+\dfrac{{{x}^{n}}}{n!}+.......\]
Now we will replace each of x’s by (-2x) to obtain Maclaurin’s series for \[f\left( x \right)={{e}^{x}}\].
\[\begin{align}
  & \Rightarrow {{e}^{-2x}}=1+\left( -2x \right)+\left( \dfrac{{{\left( -2x \right)}^{2}}}{2!} \right)+\left( \dfrac{{{\left( -2x \right)}^{3}}}{3!} \right)+.........+\left( \dfrac{{{\left( -2x \right)}^{3}}}{n!} \right)+..... \\
 & \Rightarrow {{e}^{-2x}}=1+\left( -2 \right)x+4\dfrac{{{x}^{2}}}{2!}+\left( -8 \right)\dfrac{{{x}^{3}}}{3!}+.........+{{\left( -2 \right)}^{n}}\dfrac{{{x}^{n}}}{n!}+....... \\
 & \Rightarrow {{e}^{-2x}}=1-2x+2{{x}^{2}}-\dfrac{4}{3}{{x}^{3}}+............+{{\left( -2 \right)}^{n}}\dfrac{{{x}^{n}}}{n!}+...... \\
\end{align}\]
Hence the Taylor’s series expansion of \[{{e}^{-2x}}\] centered at \[x=0\] is given by, \[{{e}^{-2x}}=1-2x+2{{x}^{2}}-\dfrac{4}{3}{{x}^{3}}+............+{{\left( -2 \right)}^{n}}\dfrac{{{x}^{n}}}{n!}+......\]