Question

How do I use Taylor expansion to show that \[{{e}^{i\theta }}=\cos \left( \theta \right)+i\sin \left( \theta \right)\]?
Using this, what is the value of \[{{e}^{i\pi }}+1\]?

Answer 1

Hint: In this problem there is no line \[x=a\] is given so we assume that the given equation will pass through the origin.So we will use maclaurin series to expand \[{{e}^{i\theta }}\].After expanding \[{{e}^{i\theta }}\] we will simplify or rearrange the terms in such a way that we will arrive at the solution . Here we have to write the series for \[\cos \theta \] and \[\sin \theta \] so that if we arrive at that series we will substitute with them.

Complete step by step answer:
First we have to know both Taylor expansion and maclaurin 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
\[f\left( x \right)=\sum\limits_{n=0}^{\infty }{\dfrac{{{f}^{\left( n \right)}}\left( 0 \right)}{n!}}{{\left( x \right)}^{n}}\]
\[\Rightarrow f\left( 0 \right)+f'\left( 0 \right)x+\dfrac{f''\left( 0 \right)}{2!}{{\left( x \right)}^{2}}+\dfrac{f'''\left( 0 \right)}{3!}{{\left( x \right)}^{3}}+.....\]
Now as our equation is passing through we write the maclaurin series for \[{{e}^{i\theta }}\].
As we already said we must know the series for both \[\cos \theta \]and \[\sin \theta \].
So the maclaurin series are
\[{{e}^{x}}=1+x+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{4}}}{4!}+\dfrac{{{x}^{5}}}{5!}+....\]
\[\sin x=x-\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}+\dfrac{{{x}^{7}}}{7!}+\dfrac{{{x}^{9}}}{9!}+.....\]
\[\cos x=1-\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{4}}}{4!}+\dfrac{{{x}^{6}}}{6!}+\dfrac{{{x}^{8}}}{8!}+.....\]
So, these are the three maclaurin series that we will use in our problem. These series are obtained by substituting the values in the above formulas given.
 Now , our problem is to prove \[{{e}^{i\theta }}=\cos \left( \theta \right)+i\sin \left( \theta \right)\]
We already have the series of \[{{e}^{x}}\] so the series of \[{{e}^{i\theta }}\] will be obtained by substituting the \[i\Theta \] in place of \[x\] then the series will look like
\[{{e}^{i\theta }}=1+i\theta +\dfrac{i{{\theta }^{2}}}{2!}+\dfrac{i{{\theta }^{3}}}{3!}+\dfrac{i{{\theta }^{4}}}{4!}+\dfrac{i{{\theta }^{5}}}{5!}+....\]
As we all know that \[{{i}^{2}}=1\] so \[i\] in even terms will become 1 then
\[\Rightarrow 1+i\theta -\dfrac{{{\theta }^{2}}}{2!}-\dfrac{i{{\theta }^{3}}}{3!}+\dfrac{{{\theta }^{4}}}{4!}+\dfrac{i{{\theta }^{5}}}{5!}-\dfrac{{{\theta }^{6}}}{6!}+.....\]
\[\Rightarrow \left\{ 1-\dfrac{{{\theta }^{2}}}{2!}+\dfrac{{{\theta }^{4}}}{4!}-\dfrac{{{\theta }^{6}}}{6!}+... \right\}+\left\{ i\theta -\dfrac{i{{\theta }^{3}}}{3!}+\dfrac{i{{\theta }^{5}}}{5!}+.... \right\}\]
Now taking \[i\]common from second term then
\[\Rightarrow \left\{ 1-\dfrac{{{\theta }^{2}}}{2!}+\dfrac{{{\theta }^{4}}}{4!}-\dfrac{{{\theta }^{6}}}{6!}+... \right\}+i\left\{ \theta -\dfrac{{{\theta }^{3}}}{3!}+\dfrac{{{\theta }^{5}}}{5!}+.... \right\}\]
We already have \[\cos x\] and \[\sin x\] series by substituting \[\theta \] in the series
\[\sin \theta =\theta -\dfrac{{{\theta }^{3}}}{3!}+\dfrac{{{\theta }^{5}}}{5!}+\dfrac{{{\theta }^{7}}}{7!}+\dfrac{{{\theta }^{9}}}{9!}+.....\]
\[\cos \theta =1-\dfrac{{{\theta }^{2}}}{2!}+\dfrac{{{\theta }^{4}}}{4!}+\dfrac{{{\theta }^{6}}}{6!}+\dfrac{{{\theta }^{8}}}{8!}+.....\]
Here we can observe the \[\cos \Theta \] and \[\sin \Theta \] series same as in our equation so we substitute these in that terms place
\[\Rightarrow \cos \theta +i\sin \theta \]
Hence we have proved
 \[{{e}^{i\theta }}\Rightarrow \cos \theta +i\sin \theta \]
Now using this we have to get the value of \[{{e}^{i\pi }}+1\]
So we get
\[{{e}^{i\pi }}=\cos \pi +i\sin \pi \]
As we already know the values of \[\cos \pi =-1\] and \[\sin \pi =0\]
\[\Rightarrow -1+0\]
So, we get
\[{{e}^{i\pi }}+1=0\]
The value of \[{{e}^{i\pi }}+1\] is 0

Note:
We can solve this question in this way or we directly have an identity called Euler’s identity. Using that we can solve the problem easily. We also have to know the taylor and maclaurin series formulas to solve this question.