Question

In the Taylor series expansion of $e^x$ about $x=2$, the coefficient of $(x-2{)}^4$

• A. $1/4!$
• B. $2^4/4!$
• C. $e^2/4!$
• D. $e^4/4!$

Answer 1

The correct option is C $e^2/4!$

Let $y=e^{x}then,y^{\prime }=e^x,y^{\prime \prime }=e^x,y^{\prime \prime \prime }=e^x,y^{\prime \prime \prime \prime }=e^x$
According to Taylor series expansinson,
$y\left(x\right)=y\left(2\right)+(x-2)y^{\prime }\left(2\right)+\frac{(x-2{)}^2}{2!}{y}^{\prime \prime }\left(2\right)+\frac{(x-2{)}^3}{3!}{y}^{\prime \prime \prime }\left(2\right)+\frac{(x-2{)}^4}{4!}{y}^{\prime \prime \prime \prime }\left(2\right)+...$
$\Rightarrow y\left(x\right)=e^2+(x-2).e^2+\frac{(x-2{)}^2}{2!}{e}^2+\frac{(x-2{)}^3}{3!}{e}^2+\frac{(x-2{)}^4{e}^2}{4!}+...$
$\because Cofficientof(x-2{)}^{4}is\frac{e^2}{4!}$
Alternative Solution
$e^x=e^2{e}^{x-2}$
$=e^2[1+(x-2)+\frac{(x-2{)}^2}{2!}+\frac{(x-2{)}^3}{3!}+\frac{(x-2{)}^4}{4!}+....]$
coefficient of $(x-2{)}^4=\frac{e^2}{4!}$