Question

In the Taylor series expansion of $ \exp \left( x \right)+\sin \left( x \right) $ about the point $ x=\pi $ , the coefficient of $ {{\left( x=\pi \right)}^{2}} $ is
A. $ \exp \left( \pi \right) $
B. $ 0.5\exp \left( \pi \right) $
C. $ \exp \left( \pi \right)+1 $
D. $ \exp \left( \pi \right)-1 $

Answer 1

Hint: Our function $ f\left( x \right)=\exp \left( x \right)+\sin \left( x \right) $ , so we will directly use the Taylor series expansion, which is given as, $ f\left( x \right)=f\left( a \right)+\left( x-a \right)f'\left( a \right)+\dfrac{{{\left( x-a \right)}^{2}}}{2!}f''\left( a \right) $ . We will put $ a=\pi $ in the series and solve accordingly to get the desired answer.

Complete step-by-step answer:
It is given in the question that we have to find the coefficient of $ {{\left( x=\pi \right)}^{2}} $ in the Taylor series expansion of $ \exp \left( x \right)+\sin \left( x \right) $ about the point $ x=\pi $ . To start solving this question, let us assume that $ f\left( x \right)={{e}^{x}}+\sin \left( x \right) $ . Now, we know that the expansion of the Taylor series is given as, $ f\left( x \right)=f\left( a \right)+\left( x-a \right)f'\left( a \right)+\dfrac{{{\left( x-a \right)}^{2}}}{2!}f''\left( a \right) $ . In the question that has been given to us, we have $ a=\pi $ . So, we will put the value of $ a=\pi $ in the Taylor series expansion. So, we will get as,
  $ f\left( x \right)=f\left( \pi \right)+\left( x-\pi \right)f'\left( \pi \right)+\dfrac{{{\left( x-\pi \right)}^{2}}}{2!}f''\left( \pi \right) $
Now, from the Taylor expansion, we get the coefficient of $ {{\left( x=\pi \right)}^{2}} $ is $ \dfrac{f''\left( \pi \right)}{2} $ . We know that $ f''\left( \pi \right)=\exp \left( x \right)-\sin x $ , where $ x=\pi $ . So, we get,
 $ \begin{align}
  & f''\left( \pi \right)={{\left[ \exp \left( x \right)-\sin x \right]}_{x=\pi }} \\
 & \Rightarrow f''\left( \pi \right)=\exp \pi -\sin \pi \\
\end{align} $
Now, we also know that $ \sin \pi =0 $ , therefore, we will get,
 $ \begin{align}
  & f''\left( \pi \right)=\exp \pi -0 \\
 & \Rightarrow f''\left( \pi \right)=\exp \pi \\
\end{align} $
Now, we can substitute it in the coefficient of $ {{\left( x=\pi \right)}^{2}} $ , which we have as $ \dfrac{f''\left( \pi \right)}{2} $ . So, we get the coefficient as,
 $ {{\left( x=\pi \right)}^{2}}=\dfrac{\exp \pi }{2} $
Thus, we will get the coefficient of $ {{\left( x=\pi \right)}^{2}}=0.5\exp \left( \pi \right) $ .
Therefore, we get the correct answer as option B.

Note: Majority of the students make a mistake in the last step. They may take the coefficient of $ {{\left( x=\pi \right)}^{2}} $ as $ 0.5\exp \left( \pi \right)+\sin \left( \pi \right) $ , but as we know that $ \sin \pi =0 $ , so it does not need to be included in the final answer, thus the correct answer is $ 0.5\exp \left( \pi \right) $ .