answer: 9/10 x^2(1 + x^3)^(-1) #____ - find the fourth term of the taylor series for f(x)=x^2/(1 + x^3)…

answer: 9/10 x^2(1 + x^3)^(-1) #____ - find the fourth term of the taylor series for f(x)=x^2/(1 + x^3) centered at x = 0. p_4x=f(0)+f(0)+f(0)x^2/2!+f(0)x^3/3! f^(4)(0)x^4/4!=24/24 x^4=x^4 answer: 4x^3 #____ - the taylor series for a function f about x = 0 is given by sum_(n = 1)^(infinity) ((-1)^(n + 1)x^n)/n! and converges to f for all real numbers x. if the fourth - degree taylor polynomial for f about x = 0 is used to approximate f(1/3), what is the alternating series error bound? answer: 4(x - 3)^3/3! #____ - suppose that the function f(x) is approximated near x = 5 by a third - degree taylor polynomial 7-3(x - 5)+8(x - 5)^2-10(x - 5)^3. find the value of f(5).

answer: 9/10 x^2(1 + x^3)^(-1) #____ - find the fourth term of the taylor series for f(x)=x^2/(1 + x^3) centered at x = 0. p_4x=f(0)+f(0)+f(0)x^2/2!+f(0)x^3/3! f^(4)(0)x^4/4!=24/24 x^4=x^4 answer: 4x^3 #____ - the taylor series for a function f about x = 0 is given by sum_(n = 1)^(infinity) ((-1)^(n + 1)x^n)/n! and converges to f for all real numbers x. if the fourth - degree taylor polynomial for f about x = 0 is used to approximate f(1/3), what is the alternating series error bound? answer: 4(x - 3)^3/3! #____ - suppose that the function f(x) is approximated near x = 5 by a third - degree taylor polynomial 7-3(x - 5)+8(x - 5)^2-10(x - 5)^3. find the value of f(5).

Answer

Explanation:

Step1: Recall Taylor - series formula

The Taylor series of a function (f(x)) centered at (a) is given by (P_n(x)=\sum_{k = 0}^{n}\frac{f^{(k)}(a)}{k!}(x - a)^k=f(a)+f^{\prime}(a)(x - a)+\frac{f^{\prime\prime}(a)}{2!}(x - a)^2+\cdots+\frac{f^{(n)}(a)}{n!}(x - a)^n).

Step2: Solve the first problem

For (f(x)=\frac{x^{2}}{1 + x^{3}}), centered at (x = 0), the (n) - th term of the Taylor series is (\frac{f^{(n)}(0)}{n!}x^{n}). We know that (f^{(4)}(0)=24), so the fourth - term is (\frac{f^{(4)}(0)}{4!}x^{4}=\frac{24}{24}x^{4}=x^{4}).

Step3: Solve the second problem

The Taylor series (\sum_{n = 1}^{\infty}\frac{(-1)^{n + 1}x^{n}}{n!}) is an alternating series. For an alternating series (\sum_{n = 1}^{\infty}(-1)^{n+1}a_n) ((a_n\gt0), (a_{n+1}\leq a_n) and (\lim_{n\rightarrow\infty}a_n = 0)), the error bound (E_N) for the sum (S_N=\sum_{n = 1}^{N}(-1)^{n + 1}a_n) is (|E_N|\leq a_{N+1}). When using the fourth - degree Taylor polynomial ((N = 4)) to approximate (f(\frac{1}{3})), we substitute (x=\frac{1}{3}) into the next term ((n = 5)) of the series. (a_n=\frac{|x|^{n}}{n!}), so (a_5=\frac{(\frac{1}{3})^{5}}{5!}=\frac{1}{3^5\times120}=\frac{1}{29160}).

Step4: Solve the third problem

The third - degree Taylor polynomial of (f(x)) centered at (a = 5) is (P_3(x)=f(5)+f^{\prime}(5)(x - 5)+\frac{f^{\prime\prime}(5)}{2!}(x - 5)^2+\frac{f^{(3)}(5)}{3!}(x - 5)^3). Given (P_3(x)=7-3(x - 5)+8(x - 5)^2-10(x - 5)^3), we know that (\frac{f^{(3)}(5)}{3!}=-10). Then (f^{(3)}(5)=-10\times3!=-60). Since the coefficient of ((x - 5)^4) in a third - degree Taylor polynomial is (0), (f^{(4)}(5)=0).

Answer:

  1. (x^{4})
  2. (\frac{1}{29160})
  3. (0)