Question

In the Taylor series expansion of $exp\left(x\right)+sin\left(x\right)$ about the point $x=\pi$, the coefficient of $(x-\pi {)}^2$ is

• A. $exp\left(\pi \right)$
• B. $0.5exp\left(\pi \right)$
• C. $exp\left(\pi \right)+1$
• D. $exp\left(\pi \right)-1$

Answer 1

The correct option is B $0.5exp\left(\pi \right)$

$f\left(x\right)=e^x+sinx$
sin x contains odd powers x only. So the cofficients of even parts will be zero in the expansion of sin x so only $e^x$ has the even powers which is as follows.
$e^x=e^{\pi }{e}^{x-\pi }=e^{\pi }[1+(x-\pi )+\frac{(x-\pi {)}^2}{2!}+...]$
Cofficient of $(x-\pi {)}^2=\frac{1}{2}{e}^{\pi }$