answer: 9/10 x^2(1 + x^3)^(-1) #____ - find t...
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. \(0\)