# Write out the first three nonzero terms of the Taylor series for the following functions centered at the given point a. Then write the series using summation notation. f(x)=tan^{-1}4x,a=0

Write out the first three nonzero terms of the Taylor series for the following functions centered at the given point a. Then write the series using summation notation.
$$f(x)=\tan^{-1}4x,a=0$$

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Corben Pittman
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
$$f(x)=f(0)=xf'(0)+\frac{x^2}{2!}f''(0)+\frac{x^3}{3!}f'''(0)+...$$
$$\tan^{-1}(4x)=0+x(4)+\frac{x^2}{2!}\times0+\frac{x^3}{3!}\times(-128)+...$$
$$\tan^{-1}(4x)=4x-128\frac{x^3}{3!}+...$$
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