find the taylor series for f(x) centered at the given value of a. assume that f has a power series…

find 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. f(x)=7/x, a = - 3 find the associated radius of convergence r. step 1 the taylor series formula is given. f(a)+f(a)(x - a)+f(a)/2!(x - a)^2+f(a)/3!(x - a)^3+f^(4)(a)/4!(x - a)^4+... the 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.

find 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. f(x)=7/x, a = - 3 find the associated radius of convergence r. step 1 the taylor series formula is given. f(a)+f(a)(x - a)+f(a)/2!(x - a)^2+f(a)/3!(x - a)^3+f^(4)(a)/4!(x - a)^4+... the 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 (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+\cdots)

Step2: Find (f(a))

Given (f(x)=\frac{7}{x}=7x^{-1}) and (a=-3). Then (f(a)=f(-3)=\frac{7}{-3}=-\frac{7}{3})

Step3: Find the first - derivative (f^{\prime}(x))

Using the power rule ((x^n)^\prime=nx^{n - 1}), if (f(x)=7x^{-1}), then (f^{\prime}(x)=7(-1)x^{-2}=-\frac{7}{x^{2}}), and (f^{\prime}(a)=f^{\prime}(-3)=-\frac{7}{(-3)^{2}}=-\frac{7}{9})

Step4: Find the second - derivative (f^{\prime\prime}(x))

(f^{\prime\prime}(x)=7(-1)(-2)x^{-3}=\frac{14}{x^{3}}), and (f^{\prime\prime}(a)=f^{\prime\prime}(-3)=\frac{14}{(-3)^{3}}=-\frac{14}{27})

Step5: Find the third - derivative (f^{(3)}(x))

(f^{(3)}(x)=7(-1)(-2)(-3)x^{-4}=-\frac{42}{x^{4}}), and (f^{(3)}(a)=f^{(3)}(-3)=-\frac{42}{(-3)^{4}}=-\frac{42}{81})

Step6: Generalize the (n) - th derivative (f^{(n)}(x))

(f^{(n)}(x)=7(-1)^{n}n!x^{-(n + 1)}), so (f^{(n)}(a)=7(-1)^{n}n!(-3)^{-(n + 1)})

Step7: Write the Taylor series

[ \begin{align*} f(x)&=\sum_{n = 0}^{\infty}\frac{f^{(n)}(a)}{n!}(x - a)^n\ &=-\frac{7}{3}-\frac{7}{9}(x + 3)-\frac{14/27}{2}(x + 3)^2-\frac{42/81}{6}(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 + 1}}{(-3)^{n+1}}(x + 3)^n \end{align*} ]

Step8: Find the radius of convergence using the ratio test

Let (u_n=\frac{7(-1)^{n + 1}}{(-3)^{n+1}}(x + 3)^n). Then (u_{n+1}=\frac{7(-1)^{n+2}}{(-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+2}}{(-3)^{n + 2}}(x + 3)^{n+1}}{\frac{7(-1)^{n + 1}}{(-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), which implies (|x+3|\lt3). So the radius of convergence (R = 3)

Answer:

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