Question

How do you find the Maclauren series for f(x) = sin (2x)?

Answer 1

Hint: Maclauren series of any function is given by $$f\left(x\right)=f\left(0\right)+{\displaystyle \frac{f^{\prime }\left(0\right)}{1!}}x+{\displaystyle \frac{f^{\prime }\left(0\right)}{2!}}{x}^2+{\displaystyle \frac{f^{‴}\left(0\right)}{3!}}{x}^3+{\displaystyle \frac{f^{⁗}\left(0\right)}{4!}}{x}^4+...+{\displaystyle \frac{f^n\left(0\right)}{n!}}{x}^n+...\left(1\right)$$ Here the given function is sin (2x). To find Maclauren series we will find the first, second, third and higher derivatives of the given function and find the values of derivatives at zero. Hence we will substitute the values in the Maclauren series expansion and hence find the Maclauren series for the given function.

Complete step-by-step solution:
Now let us first understand the concept of Taylor series and Maclauren series.
Taylor series is expansion of the function at any point.
Taylor series expansion of the function f(x) about any point a is given by
$$f\left(x\right)=f\left(a\right)+{\displaystyle \frac{f^{\prime }\left(a\right)}{1!}}{(x-a)}^1+{\displaystyle \frac{f^{″}\left(a\right)}{2!}}{(x-a)}^2+{\displaystyle \frac{f^{‴}\left(a\right)}{3!}}{(x-a)}^3+{\displaystyle \frac{f^{⁗}\left(a\right)}{4!}}{(x-a)}^4+...+{\displaystyle \frac{f^n\left(a\right)}{n!}}{(x-a)}^n+...$$
Maclauren series is nothing but Taylor series expansion of a function about point zero.
Hence to get the expansion for Maclaurin series we substitute a = 0 in the expansion of Taylor series.
Hence we get the expansion of Maclaurin series as,
$$f\left(x\right)=f\left(0\right)+{\displaystyle \frac{f^{\prime }\left(0\right)}{1!}}x+{\displaystyle \frac{f^{\prime }\left(0\right)}{2!}}{x}^2+{\displaystyle \frac{f^{‴}\left(0\right)}{3!}}{x}^3+{\displaystyle \frac{f^{⁗}\left(0\right)}{4!}}{x}^4+...+{\displaystyle \frac{f^n\left(0\right)}{n!}}{x}^n+...\left(1\right)$$
Now Consider the given function f(x) = sin (2x)
Let us take some derivatives of f(x) = sin (2x).
$\begin{array}{rl}& \Rightarrow f^{\prime }\left(x\right)=2\cos \left(2x\right)\\ & \Rightarrow f^{″}\left(x\right)=-4\sin \left(2x\right)\\ & \Rightarrow f^{‴}\left(x\right)=-8\cos \left(2x\right)\\ & \Rightarrow f^{⁗}\left(x\right)=16\sin \left(2x\right)\\ & \Rightarrow f^{\prime \prime \prime \prime \prime }\left(x\right)=32\cos \left(2x\right)\end{array}$
And so on.
Now let us calculate the values of these about zero.
$\begin{array}{rl}& f\left(0\right)=\cos \left(0\right)=1\\ & f^{\prime }\left(0\right)=2\cos \left(0\right)=2\\ & f^{″}\left(0\right)=-4\sin \left(0\right)=0\end{array}$
$f^{‴}\left(0\right)=-8\cos \left(0\right)=-8$
$\begin{array}{rl}& f^{⁗}\left(0\right)=16\sin \left(0\right)=0\\ & f^{\prime \prime \prime \prime \prime }\left(0\right)=32\cos \left(0\right)=32\end{array}$
Now let us substitute these values in equation (1) Hence, we get
$$\begin{array}{rl}& f\left(x\right)=0+{\displaystyle \frac{2}{1!}}x+{\displaystyle \frac{0}{2!}}{x}^2-{\displaystyle \frac{8}{3!}}{x}^3+{\displaystyle \frac{0}{4!}}{x}^4+{\displaystyle \frac{32}{5!}}{x}^5...\\ & \Rightarrow \sin \left(2x\right)=0+{\displaystyle \frac{\left(2x\right)}{1}}-{\displaystyle \frac{{\left(2x\right)}^3}{3!}}+{\displaystyle \frac{{\left(2x\right)}^5}{5!}}+....\end{array}$$
Hence the Maclauren series of sin (2x) is $$\sin \left(2x\right)=0+{\displaystyle \frac{\left(2x\right)}{1}}-{\displaystyle \frac{{\left(2x\right)}^3}{3!}}+{\displaystyle \frac{{\left(2x\right)}^5}{5!}}+....$$

Note: Now note that if we know the Maclauren series expansion of sin(x) which is $$\sin \left(x\right)=0+{\displaystyle \frac{\left(x\right)}{1}}-{\displaystyle \frac{{\left(x\right)}^3}{3!}}+{\displaystyle \frac{{\left(x\right)}^5}{5!}}+....$$ then we can easily find the Maclauren series expansion for the function sin (2x) by substituting 2x instead of x and hence we get the Maclaurin series of sin (2x) as $$\sin \left(2x\right)=0+{\displaystyle \frac{\left(2x\right)}{1}}-{\displaystyle \frac{{\left(2x\right)}^3}{3!}}+{\displaystyle \frac{{\left(2x\right)}^5}{5!}}+....$$