b8-taylor and maclaurin series: problem 9\n(3 points)\nthe taylor series of function ( f(x)=ln (x) ) at ( a…

b8-taylor and maclaurin series: problem 9\n(3 points)\nthe taylor series of function ( f(x)=ln (x) ) at ( a = 6 ) is given by:\n f(x)=sum_{n = 0}^{infty} c_{n}(x - 6)^{n} \nfind the following coefficients:\n( c_{0}= )\n( c_{1}= )\n( c_{2}= )\n( c_{3}= )\n( c_{4}= )\ndetermine the interval of convergence:\nnote: give your answer in help (intervals)\nnote: you can earn partial credit on this problem.

b8-taylor and maclaurin series: problem 9\n(3 points)\nthe taylor series of function ( f(x)=ln (x) ) at ( a = 6 ) is given by:\n f(x)=sum_{n = 0}^{infty} c_{n}(x - 6)^{n} \nfind the following coefficients:\n( c_{0}= )\n( c_{1}= )\n( c_{2}= )\n( c_{3}= )\n( c_{4}= )\ndetermine the interval of convergence:\nnote: give your answer in help (intervals)\nnote: you can earn partial credit on this problem.

Answer

Answer:

(c_0=\ln(6)), (c_1=\frac{1}{6}), (c_2 =-\frac{1}{72}), (c_3=\frac{1}{648}), (c_4=-\frac{1}{6912}), Interval of convergence: ((0,12))

Explanation:

Step1: Recall the Taylor series formula

The Taylor series of a function (f(x)) about (x = a) is (f(x)=\sum_{n = 0}^{\infty}\frac{f^{(n)}(a)}{n!}(x - a)^n), where (f^{(n)}(a)) is the (n) - th derivative of (f(x)) evaluated at (x=a)

Step2: Find the derivatives of (y = f(x)=\ln(x))

  • (f(x)=\ln(x)), so (f(6)=\ln(6)), then (c_0=\frac{f(6)}{0!}=\ln(6))
  • (f^{\prime}(x)=\frac{1}{x}), (f^{\prime}(6)=\frac{1}{6}), then (c_1=\frac{f^{\prime}(6)}{1!}=\frac{1}{6})
  • (f^{\prime\prime}(x)=-\frac{1}{x^{2}}), (f^{\prime\prime}(6)=-\frac{1}{6^{2}}), then (c_2=\frac{f^{\prime\prime}(6)}{2!}=-\frac{1}{2\times6^{2}}=-\frac{1}{72})
  • (f^{(3)}(x)=\frac{2}{x^{3}}), (f^{(3)}(6)=\frac{2}{6^{3}}), then (c_3=\frac{f^{(3)}(6)}{3!}=\frac{2}{6\times6^{3}}=\frac{1}{648})
  • (f^{(4)}(x)=-\frac{6}{x^{4}}), (f^{(4)}(6)=-\frac{6}{6^{4}}), then (c_4=\frac{f^{(4)}(6)}{4!}=-\frac{6}{24\times6^{4}}=-\frac{1}{6912})

Step3: Find the interval of convergence

We use the ratio test. Let (u_n=\frac{f^{(n)}(6)}{n!}(x - 6)^n)

(f^{(n)}(x)=(-1)^{n - 1}\frac{(n-1)!}{x^{n}}), so (u_n=\frac{(-1)^{n - 1}(n - 1)!}{n!6^{n}}(x - 6)^n=\frac{(-1)^{n - 1}}{n6^{n}}(x - 6)^n)

(\lim_{n\rightarrow\infty}\left|\frac{u_{n + 1}}{u_n}\right|=\lim_{n\rightarrow\infty}\left|\frac{\frac{(-1)^{n}}{(n+1)6^{n+1}}(x - 6)^{n+1}}{\frac{(-1)^{n - 1}}{n6^{n}}(x - 6)^n}\right|=\lim_{n\rightarrow\infty}\left|\frac{-n(x - 6)}{6(n + 1)}\right|=\left|\frac{x - 6}{6}\right|)

For convergence, (\left|\frac{x - 6}{6}\right|\lt1), i.e., (- 6\lt x - 6\lt6), so (0\lt x\lt12)

When (x = 0): (\sum_{n = 1}^{\infty}\frac{(-1)^{n - 1}}{n6^{n}}(-6)^n=\sum_{n = 1}^{\infty}\frac{-1}{n}), which diverges (harmonic - like series)

When (x = 12): (\sum_{n = 1}^{\infty}\frac{(-1)^{n - 1}}{n6^{n}}(6)^n=\sum_{n = 1}^{\infty}\frac{(-1)^{n-1}}{n}), which converges (alternating harmonic series)

So the interval of convergence is ((0,12))