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)

Question
Series
asked 2021-02-04
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 \(\cos(-2x)\)

Answers (1)

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:
\(\lim_{n\rightarrow\infty}|\frac{(-1)^{n+1}\frac{x^{2(n+1)}}{(2(n+1))!}}{(-1)^n\frac{x^{2n}}{(2n)!}}|=\lim_{n\rightarrow\infty}|\frac{x^{2n+2}}{x^{2n}}\cdot\frac{(2n)!}{(2n+2)!}|\)
\(=x^2\lim_{n\rightarrow\infty}|\frac{(2n)!}{(2n+2)\cdot(2n+1)\cdot(2n)!}|\)
\(=x^2\lim_{n\rightarrow\infty}|\frac{1}{(2n+2)\cdot(2n+1)}|=0<1\)</span>
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
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