tutorial exercise\nfind the taylor series for f(x) centered at the given value of a. assume that f has a…

tutorial exercise\nfind the taylor series for f(x) centered at the given value of a. assume that f has a power - series expansion. do not show that rn(x)→0.\nf(x)=7/x, a = - 3\nfind the associated radius of convergence r.\nstep 1\nthe taylor series formula is given.\nf(a)+f(a)(x - a)+f(a)/2!(x - a)^2 + f(a)/3!(x - a)^3 + f^(4)(a)/4!(x - a)^4+...\nthe function f(x)=7/x can also be written as f(x)=(7)1/x, which has derivatives f(x)=(7) - 1/x^2, f(x)=(7)2/x^3, f(x)=(7) - 6/x^4, and f^(4)(x)=(7)24/x^5.

tutorial exercise\nfind the taylor series for f(x) centered at the given value of a. assume that f has a power - series expansion. do not show that rn(x)→0.\nf(x)=7/x, a = - 3\nfind the associated radius of convergence r.\nstep 1\nthe taylor series formula is given.\nf(a)+f(a)(x - a)+f(a)/2!(x - a)^2 + f(a)/3!(x - a)^3 + f^(4)(a)/4!(x - a)^4+...\nthe function f(x)=7/x can also be written as f(x)=(7)1/x, which has derivatives f(x)=(7) - 1/x^2, f(x)=(7)2/x^3, f(x)=(7) - 6/x^4, and f^(4)(x)=(7)24/x^5.

Answer

Explanation:

Step1: Recall Taylor - series formula

The Taylor series of a function (f(x)) centered at (x = a) is (f(x)=\sum_{n = 0}^{\infty}\frac{f^{(n)}(a)}{n!}(x - a)^{n}=f(a)+f^{\prime}(a)(x - a)+\frac{f^{\prime\prime}(a)}{2!}(x - a)^{2}+\frac{f^{(3)}(a)}{3!}(x - a)^{3}+\frac{f^{(4)}(a)}{4!}(x - a)^{4}+\cdots) For (f(x)=\frac{7}{x}=7x^{-1}) and (a=-3), first find the (n) - th derivative of (f(x)). The first - derivative: (f^{\prime}(x)=7(-1)x^{-2}=-\frac{7}{x^{2}}) The second - derivative: (f^{\prime\prime}(x)=7(-1)(-2)x^{-3}=\frac{7\times2!}{x^{3}}) The third - derivative: (f^{(3)}(x)=7(-1)(-2)(-3)x^{-4}=-\frac{7\times3!}{x^{4}}) The (n) - th derivative: (f^{(n)}(x)=7(-1)^{n}n!x^{-(n + 1)})

Step2: Evaluate (f^{(n)}(a))

Substitute (x = a=-3) into (f^{(n)}(x)). (f^{(n)}(-3)=7(-1)^{n}n!(-3)^{-(n + 1)})

Step3: Write the Taylor series

The Taylor series of (f(x)=\frac{7}{x}) centered at (a=-3) is: [ \begin{align*} f(x)&=\sum_{n = 0}^{\infty}\frac{f^{(n)}(-3)}{n!}(x + 3)^{n}\ &=\frac{7}{-3}+\frac{7(-1)}{(-3)^{2}}(x + 3)+\frac{7\times2!}{2!(-3)^{3}}(x + 3)^{2}+\frac{7\times(-3!)}{3!(-3)^{4}}(x + 3)^{3}+\cdots\ &=-\frac{7}{3}-\frac{7}{9}(x + 3)-\frac{7}{27}(x + 3)^{2}-\frac{7}{81}(x + 3)^{3}-\cdots\ &=\sum_{n = 0}^{\infty}\frac{7(-1)^{n}}{(-3)^{n+1}}(x + 3)^{n} \end{align*} ]

Step4: Find the radius of convergence

Use the ratio test. Let (u_{n}=\frac{7(-1)^{n}}{(-3)^{n + 1}}(x + 3)^{n}). Then (u_{n+1}=\frac{7(-1)^{n+1}}{(-3)^{n+2}}(x + 3)^{n+1}) [ \begin{align*} \lim_{n\rightarrow\infty}\left|\frac{u_{n + 1}}{u_{n}}\right|&=\lim_{n\rightarrow\infty}\left|\frac{\frac{7(-1)^{n+1}}{(-3)^{n+2}}(x + 3)^{n+1}}{\frac{7(-1)^{n}}{(-3)^{n+1}}(x + 3)^{n}}\right|\ &=\lim_{n\rightarrow\infty}\left|\frac{(-1)(x + 3)}{-3}\right|\ &=\left|\frac{x + 3}{3}\right| \end{align*} ] For convergence, (\left|\frac{x + 3}{3}\right|\lt1), i.e., (|x+3|\lt3). So the radius of convergence (R = 3)

Answer:

The Taylor series of (f(x)=\frac{7}{x}) centered at (a=-3) is (\sum_{n = 0}^{\infty}\frac{7(-1)^{n}}{(-3)^{n+1}}(x + 3)^{n}), and the radius of convergence (R = 3)