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Question: In the Taylor series expansion of \( \exp \left( x \right)+\sin \left( x \right) \) about the point ...

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

Explanation

Solution

Hint: Our function f(x)=exp⁡(x)+sin⁡(x)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(x)=f(a)+(x−a)f′(a)+(x−a)22!f′′(a)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=πa=\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 (x=π)2{{\left( x=\pi \right)}^{2}} in the Taylor series expansion of exp⁡(x)+sin⁡(x)\exp \left( x \right)+\sin \left( x \right) about the point x=πx=\pi . To start solving this question, let us assume that f(x)=ex+sin⁡(x)f\left( x \right)={{e}^{x}}+\sin \left( x \right) . Now, we know that the expansion of the Taylor series is given as, f(x)=f(a)+(x−a)f′(a)+(x−a)22!f′′(a)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=πa=\pi . So, we will put the value of a=πa=\pi in the Taylor series expansion. So, we will get as,
f(x)=f(π)+(x−π)f′(π)+(x−π)22!f′′(π)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 (x=π)2{{\left( x=\pi \right)}^{2}} is f′′(π)2\dfrac{f''\left( \pi \right)}{2} . We know that f′′(π)=exp⁡(x)−sin⁡xf''\left( \pi \right)=\exp \left( x \right)-\sin x , where x=πx=\pi . So, we get,
f′′(π)=[exp⁡(x)−sin⁡x]x=π ⇒f′′(π)=exp⁡π−sin⁡π \begin{aligned} & f''\left( \pi \right)={{\left[ \exp \left( x \right)-\sin x \right]}_{x=\pi }} \\\ & \Rightarrow f''\left( \pi \right)=\exp \pi -\sin \pi \\\ \end{aligned}
Now, we also know that sin⁡π=0\sin \pi =0 , therefore, we will get,
f′′(π)=exp⁡π−0 ⇒f′′(π)=exp⁡π \begin{aligned} & f''\left( \pi \right)=\exp \pi -0 \\\ & \Rightarrow f''\left( \pi \right)=\exp \pi \\\ \end{aligned}
Now, we can substitute it in the coefficient of (x=π)2{{\left( x=\pi \right)}^{2}} , which we have as f′′(π)2\dfrac{f''\left( \pi \right)}{2} . So, we get the coefficient as,
(x=π)2=exp⁡π2{{\left( x=\pi \right)}^{2}}=\dfrac{\exp \pi }{2}
Thus, we will get the coefficient of (x=π)2=0.5exp⁡(π){{\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 (x=π)2{{\left( x=\pi \right)}^{2}} as 0.5exp⁡(π)+sin⁡(π)0.5\exp \left( \pi \right)+\sin \left( \pi \right) , but as we know that sin⁡π=0\sin \pi =0 , so it does not need to be included in the final answer, thus the correct answer is 0.5exp⁡(π)0.5\exp \left( \pi \right) .