Question

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

Answer 1

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,

& \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{aligned}$$ Continuing in this way we get $${{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{aligned} & \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{aligned}$$ Thus $$\text{e}{{\text{q}}^{\text{n}}}\left( 1 \right)$$ becomes, $$\begin{aligned} & \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{aligned}$$ **Hence the Taylor’s series expansion of $${{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{aligned} & \Rightarrow f'\left( x \right)={{e}^{x}} \\\ & \Rightarrow f''\left( x \right)={{e}^{x}} \\\ & \Rightarrow f'''\left( x \right)={{e}^{x}} \\\ \end{aligned}$$ In fact the $${{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{aligned} & \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{aligned}$$ Hence, the Maclaurin’s series for $$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{aligned} & \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{aligned}$$ Hence the Taylor’s series expansion of $${{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!}+......$$