Question

How do you find the maclaurin series of \[{{e}^{\sin x}}\] ?

Answer 1

Hint: To solve this question first we need to know the maclaurin series. In the process we first take the \[{{e}^{x}}\] series and replace \[x\] with \[\sin x\] so that we get a series which is required to get the answer .Simplify the series we got until we arrive at the answer.

Complete step by step answer:
First we should know about the Maclaurin series. It is derived from the Taylor series.
So, we should start with the taylor series.
We can write taylor expansion of function \[f\left( x \right)\] about \[x=a\] as
\[f\left( x \right)=\sum\limits_{n=0}^{\infty }{\dfrac{{{f}^{\left( n \right)}}\left( a \right)}{n!}}{{\left( x-a \right)}^{n}}\]
\[\Rightarrow f\left( a \right)+f'\left( a \right)\left( x-a \right)+\dfrac{f''\left( a \right)}{2!}{{\left( x-a \right)}^{2}}+\dfrac{f'''\left( a \right)}{3!}{{\left( x-a \right)}^{3}}+.....\]
If the function \[f\left( x \right)\] is about then it is called as Maclaurin Series
The Maclaurin series is given by
\[f\left( x \right)=\sum\limits_{n=0}^{\infty }{\dfrac{{{f}^{\left( n \right)}}\left( 0 \right)}{n!}}{{\left( x \right)}^{n}}\]
\[\Rightarrow f\left( 0 \right)+f'\left( 0 \right)x+\dfrac{f''\left( 0 \right)}{2!}{{\left( x \right)}^{2}}+\dfrac{f'''\left( 0 \right)}{3!}{{\left( x \right)}^{3}}+.....\]
So here we should use the maclaurin series formula to solve the question
First we take maclaurin series formula
\[f\left( x \right)=f\left( 0 \right)+f'\left( 0 \right)x+\dfrac{f''\left( 0 \right)}{2!}{{\left( x \right)}^{2}}+\dfrac{f'''\left( 0 \right)}{3!}{{\left( x \right)}^{3}}+.....\]
Let \[f\left( x \right)={{e}^{\sin x}}\]
Now we have to derive the maclaurin series using the formula
First we have to calculate the derivatives of the given \[f\left( x \right)\]
\[f\left( x \right)={{e}^{\sin x}}\]
\[f'\left( x \right)={{e}^{\sin x}}\cos x\]
\[f''\left( x \right)={{e}^{\sin x}}{{\cos }^{2}}x-{{e}^{\sin x}}\sin x\]
Now we have to write the maclaurin series
\[f\left( x \right)={{e}^{\sin x}}+x{{e}^{\sin x}}\cos x+\dfrac{1}{{{x}^{2}}}\left( {{e}^{\sin x}}{{\cos }^{2}}x-{{e}^{\sin x}}\sin x \right)\]
According to the maclaurin series formula we have to take 0 in place of x. so we are substituting the 0 in the above equation
\[\Rightarrow {{e}^{\sin \left( 0 \right)}}+x\cos \left( 0 \right){{e}^{\sin \left( 0 \right)}}+\dfrac{1}{2}{{x}^{2}}\left( {{\cos }^{2}}\left( 0 \right)-\cos \left( 0 \right)\sin \left( 0 \right) \right){{e}^{\sin \left( 0 \right)}}+....\]
As we already know the values of \[\sin 0\] and \[\cos 0\], and also we know the value of \[{{e}^{0}}\]
\[\sin 0=0\]
\[\cos 0=1\]
\[{{e}^{0}}=1\]
Using these we can rewrite the equation as
 \[\Rightarrow {{e}^{0}}+x{{e}^{0}}\times 1+\dfrac{1}{{{x}^{2}}}\left( {{e}^{0}}\times 1-{{e}^{0}}\times 0 \right)\]
From this we can write
\[\Rightarrow 1+x+\dfrac{{{x}^{2}}}{2}+.....\]

So the maclaurin series of \[{{e}^{\sin x}}\] is
\[{{e}^{\sin x}}=1+x+\dfrac{{{x}^{2}}}{2}+.....\]

Note: we have to make sure that we write the formula correctly as we will have a confusion between taylor and maclaurin series. We had done infinite series here as it is the most generalized method but we can also do it until the highest degree asked in the question.