(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

\(f(x)=f(0)+f'(0)x+f''(0)\frac{x^2}{2!}+f'''(0)\frac{x^3}{3!}+f^{(4)}(0)\frac{x^4}{4!}+...\)

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