Question

$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{c}\mathit{o}\mathit{t}\mathit{x}\mathbf{,}\mathit{a}\mathbf{=}\frac{\mathbf{\pi }}{\mathbf{2}}$

Short answer

The first few terms of Taylor’s series expansion are:

$\cot \left(x\right)=\left(x-\frac{\pi }{2}\right)-\frac{{\left(x-\frac{\pi }{2}\right)}^3}{3}-\frac{2{\left(x-\frac{\pi }{2}\right)}^5}{15}-\dots$

Step-by-step solution

Given Information

The given function

$f\left(x\right)=cot\hspace{0.17em}x$

Definition of Taylor’s series

The Taylor’s series is a power series that yields the expansion of a function in the vicinity of a point if the function is continuous, all of its derivatives exist, and the series converges.

Rewrite cotxin required form

Rewrite and use $\cot \left(\frac{\pi }{2}+x\right)=-\tan \left(x\right)$

$\cot \left(x\right)=\cot \left(\frac{\pi }{2}+\left(x-\frac{\pi }{2}\right)\right)$

$\cot \left(\frac{\pi }{2}+\left(x-\frac{\pi }{2}\right)\right)=-\tan \left(x-\frac{\pi }{2}\right)$

Find the Taylor's series forcotx .

Use the Maclaurin series of $\tan x$and replace $x$by $(x-\frac{\pi }{2})$,

$\tan \left(x\right)=x+\frac{x^3}{3}+\frac{2x^5}{15}+\dots$

$-\tan \left(x\right)=-x-\frac{x^3}{3}-\frac{2x^5}{15}-\dots$

$-\tan \left(x-\frac{\pi }{2}\right)=-\left(x-\frac{\pi }{2}\right)-\frac{{\left(x-\frac{\pi }{2}\right)}^3}{3}-\frac{2{\left(x-\frac{\pi }{2}\right)}^5}{15}-\dots$

$\cot \left(x\right)=\left(x-\frac{\pi }{2}\right)-\frac{{\left(x-\frac{\pi }{2}\right)}^3}{3}-\frac{2{\left(x-\frac{\pi }{2}\right)}^5}{15}-\dots$

Write first few terms of Taylor’s series about $a=\frac{\mathrm{\pi }}{2}$,

role="math" localid="1657419575245" $\cot \left(x\right)=\left(x-\frac{\pi }{2}\right)-\frac{{\left(x-\frac{\pi }{2}\right)}^3}{3}-\frac{2{\left(x-\frac{\pi }{2}\right)}^5}{15}-\dots$

Hence, first few terms of Taylor’s series expansion are:

$\cot \left(x\right)=\left(x-\frac{\pi }{2}\right)-\frac{{\left(x-\frac{\pi }{2}\right)}^3}{3}-\frac{2{\left(x-\frac{\pi }{2}\right)}^5}{15}-\dots$