Question

What is the Taylor series for $\sin \left( 2x \right)$ ?

Answer 1

Hint: To calculate the Taylor series of any function, we will first understand what does a Taylor series mean and what is its representation. Taylor series gives the value of any function as an infinite sum of terms. That is, it gives us the value of a function $f\left( x \right)$ in the neighborhood of a point ‘a’ in the form of derivative of the function at point ‘a’.

Complete step by step answer:
The general formula for writing the Taylor series of a function in the neighborhood of a point ‘a’ is given by:
$\Rightarrow f\left( x \right)=\sum\limits_{n=0}^{\infty }{\dfrac{{{f}^{n}}\left( a \right)}{n!}{{\left( x-a \right)}^{n}}}$
Where,
‘n!’ is the factorial of ‘n’.
‘a’ is any real or complex number. And,
${{f}^{n}}\left( a \right)$ is the ${{n}^{th}}$ derivative of the function at ‘a’.
In our problem, we have been given the function (say, $f\left( x \right)$) for which Taylor series is to be calculated as:
$\Rightarrow f\left( x \right)=\sin 2x$
Now, we will first calculate the Taylor series around a general point ‘a’ and then find the standard Taylor series for, $a=0$.
In the first case, our function will be written as:
$\Rightarrow \sin 2x=\dfrac{{{\left[ \sin 2x \right]}^{0}}\left( a \right)}{0!}{{\left( 2x-a \right)}^{0}}+\dfrac{{{\left[ \sin 2x \right]}^{1}}\left( a \right)}{1!}{{\left( 2x-a \right)}^{1}}+\dfrac{{{\left[ \sin 2x \right]}^{2}}\left( a \right)}{2!}{{\left( 2x-a \right)}^{2}}+.....$
Where, the terms: ${{\left[ \sin 2x \right]}^{0}}\left( a \right),{{\left[ \sin 2x \right]}^{1}}\left( a \right),{{\left[ \sin 2x \right]}^{2}}\left( a \right)...........$are not powers but zeroth, first, second and so on differential of the sine function evaluated at the point ‘a’.
Thus, our expression reduces to:
$\Rightarrow \sin 2x=\sin 2a+\dfrac{{{2}^{1}}\cos 2a}{1!}\left( 2x-a \right)-\dfrac{{{2}^{2}}\sin 2a}{2!}{{\left( 2x-a \right)}^{2}}-\dfrac{{{2}^{3}}\cos 2a}{3!}{{\left( 2x-a \right)}^{3}}+.....$
Here, we get the Taylor series of $\sin 2x$in the neighborhood of point ‘a’. For, $a=0$, we can see that all our sine terms are equal to zero and all of our cosine terms are equal to one. And, the new series has an alternating sign for each successive term. This new series could be written as:
$\begin{align}
  & \Rightarrow \sin 2x=\dfrac{{{2}^{1}}\cos 2a}{1!}\left( 2x \right)-\dfrac{{{2}^{3}}\cos 2a}{3!}{{\left( 2x \right)}^{3}}+\dfrac{{{2}^{5}}\cos 2a}{5!}{{\left( 2x \right)}^{5}}..... \\
 & \Rightarrow \sin 2x=\dfrac{{{2}^{1}}}{1!}\left( 2x \right)-\dfrac{{{2}^{3}}}{3!}{{\left( 2x \right)}^{3}}+\dfrac{{{2}^{5}}}{5!}{{\left( 2x \right)}^{5}}..... \\
\end{align}$
Thus, we can write the generalized form of this Taylor series, for $a=0$ as:
$\Rightarrow \sin 2x=\sum\limits_{k=0}^{\infty }{\dfrac{{{\left( -1 \right)}^{k}}}{\left( 2k+1 \right)!}{{\left( 2x \right)}^{k+1}}}$
Hence, the standard Taylor series for $\sin \left( 2x \right)$ at, $a=0$, comes out to be $\sum\limits_{k=0}^{\infty }{\dfrac{{{\left( -1 \right)}^{k}}}{\left( 2k+1 \right)!}{{\left( 2x \right)}^{k+1}}}$.

Note: In numerical, when we use the Taylor series to evaluate the value of a function at a certain point, we get an approximate value of the function and not the absolute value. To verify our solution, we should calculate the absolute value with the help of a scientific calculator and check if our answer is close to the absolute value or not. If it is close to the absolute value, then our answer is correct.