Question

Question: In Problems 29–34, determine the Taylor series about the point x₀for the given functions and values of x₀.

29. f(x)= cosx, x_{0 =$\pi$}

Short answer

The required expression is $\sum _{n-0}^{\infty }\frac{{(-1)}^{n+1}}{\left(2n\right)!}.{(x-\mathrm{\pi })}^{2n}$.

Step-by-step solution

Taylor series

For a function ^{$f\left(x\right)$} the Taylor series expansion about a point ^$x_0$ is given by,

$f(x-x0)=f\left(x0\right)+f\text{'}\left(x0\right)$.$\frac{{(x-x_0)}^2}{2!}+f\text{'}\text{'}\text{'}\left(x_0\right).\frac{{(x-x_0)}^3}{3!}+...$

Derivatives of function at x0

We have to calculate the Taylor series expansion for, f(x) = cos x at x₀ = $\mathrm{\pi }$.

Calculating the derivatives of function at x₀ .

f(x)= cos x then f(x₀) = -1

f'(x)= -sin x then f'(x₀) = 0

f'' (x) = -cos x then f''(x₀) =1

f'''(x) =sin x then f'''(x₀) =0

f''''(x) = cos x then f''''(x₀) =-1

Substitute the derivatives in Taylor series

Substituting the above derivatives in Taylor series expansion for the function at

x0 = 'then,

$$\begin{gathered} \cos \left(x-\mathrm{\pi }\right)=-1+0.(x-\mathrm{\pi })+1\frac{{(\mathrm{x}-\mathrm{\pi })}^2}{2!}+0\frac{{(\mathrm{x}-\mathrm{\pi })}^3}{3!}-1\frac{{(\mathrm{x}-\mathrm{\pi })}^4}{4!} \\ =-1+\frac{{(\mathrm{x}-\mathrm{\pi })}^2}{-2!}-\frac{{(\mathrm{x}-\mathrm{\pi })}^4}{4!}+... \\ =\sum _{\mathrm{n}-0}^{\infty }\frac{{(-1)}^{\mathrm{n}+1}}{\left(2\mathrm{n}\right)!}.{(\mathrm{x}-\mathrm{\pi })}^{2\mathrm{n}} \end{gathered}$$

Hence, the required expression is$\sum _{\mathrm{n}-0}^{\infty }\frac{{(-1)}^{\mathrm{n}+1}}{\left(2\mathrm{n}\right)!}.{(\mathrm{x}-\mathrm{\pi })}^{2\mathrm{n}}$