Given,

We have to find the Taylor series for the function \(f(x)=\tan^{-1}(4x)\) centered at a=0

Formula Used:

The general term for the Taylor Series \(f(x)=\sum_{n=0}^\infty\frac{f^n(a)}{n!}(x-a)^n\) centered at origin a.

Calculation: \(f(x)=\sum_{n=0}^\infty\frac{f^n(0)}{n!}(x)^n\)

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

Given the function \(f(x)=\tan^{-1}(4x)\)

\(f(x)=\tan^{-1}(4x)\Rightarrow f(0)=0\)

\(f'(x)=\frac{4}{1+16x^2}\Rightarrow f'(0)=4\)

\(f''(x)=-\frac{128}{(1+16x^2)^2}\Rightarrow f''(0)=0\)

\(f'''(x)=-\frac{128(1-64x+16x^2-1024x^3)}{(1+16x^2)^3}\Rightarrow f'''(0)=-128\)

...

Hence the Taylor series of the function \(f(x)=\tan^{-1}(4x)\) centered at 0 then

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

\(\tan^{-1}(4x)=0+x(4)+\frac{x^2}{2!}\times0+\frac{x^3}{3!}\times(-128)+...\)

\(\tan^{-1}(4x)=4x-128\frac{x^3}{3!}+...\)

We have to find the Taylor series for the function \(f(x)=\tan^{-1}(4x)\) centered at a=0

Formula Used:

The general term for the Taylor Series \(f(x)=\sum_{n=0}^\infty\frac{f^n(a)}{n!}(x-a)^n\) centered at origin a.

Calculation: \(f(x)=\sum_{n=0}^\infty\frac{f^n(0)}{n!}(x)^n\)

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

Given the function \(f(x)=\tan^{-1}(4x)\)

\(f(x)=\tan^{-1}(4x)\Rightarrow f(0)=0\)

\(f'(x)=\frac{4}{1+16x^2}\Rightarrow f'(0)=4\)

\(f''(x)=-\frac{128}{(1+16x^2)^2}\Rightarrow f''(0)=0\)

\(f'''(x)=-\frac{128(1-64x+16x^2-1024x^3)}{(1+16x^2)^3}\Rightarrow f'''(0)=-128\)

...

Hence the Taylor series of the function \(f(x)=\tan^{-1}(4x)\) centered at 0 then

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

\(\tan^{-1}(4x)=0+x(4)+\frac{x^2}{2!}\times0+\frac{x^3}{3!}\times(-128)+...\)

\(\tan^{-1}(4x)=4x-128\frac{x^3}{3!}+...\)