Question

# Calculate the following:a. Find the Maclaurin series of cos(x) and find the radius of this series, without using any known power or Maclaurin series, besides geometric.b. Find exactly the series of cos(-2x)

Series

Calculate the following:
a. Find the Maclaurin series of $$\cos(x)$$ and find the radius of this series, without using any known power or Maclaurin series, besides geometric.
b. Find exactly the series of $$\cos(-2x)$$

2021-02-05

(a) The Maclaurin series of a function f(x) is given by
$$f(x)=f(0)+f'(0)x+f''(0)\frac{x^2}{2!}+f'''(0)\frac{x^3}{3!}+f^{(4)}(0)\frac{x^4}{4!}+...$$
We have $$f(x)=\cos x$$
Compute the first few derivatives of $$\cos x$$.
Now, $$f(0)=1$$
$$f'(0)=0$$
$$f''(0)=-1$$
$$f'''(0)=0$$
$$f^{(4)}(0)=1$$
So, the Maclaurin series is
$$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}\pm...=1+(-1)\frac{x^{2\cdot1}}{2!}+(-1)^2\frac{x^{2\cdot2}}{4!}+...$$
$$\Rightarrow\cos x=\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n)!}$$ is the required Maclaurin series.
Use Ratio Test for radius of convergence:
$$\lim_{n\rightarrow\infty}|\frac{(-1)^{n+1}\frac{x^{2(n+1)}}{(2(n+1))!}}{(-1)^n\frac{x^{2n}}{(2n)!}}|=\lim_{n\rightarrow\infty}|\frac{x^{2n+2}}{x^{2n}}\cdot\frac{(2n)!}{(2n+2)!}|$$
$$=x^2\lim_{n\rightarrow\infty}|\frac{(2n)!}{(2n+2)\cdot(2n+1)\cdot(2n)!}|$$
$$=x^2\lim_{n\rightarrow\infty}|\frac{1}{(2n+2)\cdot(2n+1)}|=0<1$$
The series converges independently of the value of x. Hence, the radius of the series is $$\infty$$ (b) We have $$f(x)=\cos(-2x)$$
Compute the first few derivatives of cos x.
$$f'(x)=2\sin(-2x)$$
$$f''(x)=-2^2\cos(-2x)$$
$$f'''(x)=-2^3\sin(-2x)$$
$$f^{(4)}(x)=2^4\cos(-2x)$$
Now,
$$f(0)=1$$
$$f'(0)=0$$
$$f''(0)=-2^2$$
$$f'''(0)=0$$
$$f^{(4)}(x)=2^4$$
So, the Maclaurin series is
$$\cos(-2x)=1-\frac{2^2x^2}{2!}+\frac{2^4x^4}{4!}\pm...=1+(-1)\frac{2^{2\cdot1}x^{2\cdot1}}{2!}+(-1)^2\frac{2^{2\cdot2}x^{2\cdot2}}{4!}+...$$
$$\Rightarrow\cos(-2x)=\sum_{n=0}^\infty(-1)^n\frac{2^{2n}x^{2n}}{(2n)!}$$ is the required Maclaurin series