answer: 3^5 * 5!\n#____ - suppose a function ...
answer: 3^5 * 5!\n#____ - suppose a function f is approximated with a third - degree taylor polynomial about x = 1.. if |f^(4)(x)|≤0.01 for all x such that 1≤x≤3, use the lagrange error bound to show that the approximation differs from f(3) by at most 1/150.\nanswer: 9/10\n#____ - find the fourth term of the taylor series for f(x)=x^2/(1 + x^3) <0 centered at x = 0. p_4x=f(0)+f(0)+f(0)x^2/2!+f(0)x^3/3!\nanswer: 4x^3\n#____ - the taylor series for a function f about x = 0 is given by ∑((-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?\nanswer: 4(x - 3)^3/3!\n#____ - 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 Lagrange error - bound formula
The Lagrange error - bound for a Taylor polynomial \(P_n(x)\) of degree \(n\) approximating \(f(x)\) is \(E_n(x)=\frac{f^{(n + 1)}(c)}{(n+1)!}(x - a)^{n + 1}\), where \(a\) is the center of the Taylor polynomial, \(x\) is the point of approximation, and \(c\) is some number between \(a\) and \(x\). Here, \(n = 3\), \(a = 1\), \(x = 3\), and \(|f^{(4)}(x)|\leq0.01\) for \(1\leq x\leq3\).
#### Step2: Substitute values into the formula
We have \(E_3(3)=\frac{f^{(4)}(c)}{4!}(3 - 1)^{4}\), where \(1\leq c\leq3\). Since \(|f^{(4)}(c)|\leq0.01\), then \(|E_3(3)|\leq\frac{0.01}{4!}\times16=\frac{0.01\times16}{24}=\frac{1}{150}\).
#### Step3: Recall Taylor - series formula
The Taylor series of a function \(f(x)\) centered at \(x = 0\) is \(f(x)=\sum_{n = 0}^{\infty}\frac{f^{(n)}(0)}{n!}x^{n}=f(0)+f^{\prime}(0)x+\frac{f^{\prime\prime}(0)}{2!}x^{2}+\frac{f^{(3)}(0)}{3!}x^{3}+\cdots\). First, rewrite \(f(x)=\frac{x^{2}}{1 + x^{3}}=x^{2}(1 + x^{3})^{-1}\). Using the binomial series \((1 + u)^{-1}=\sum_{n = 0}^{\infty}(-1)^{n}u^{n}\) for \(|u|\lt1\), with \(u = x^{3}\), we have \(f(x)=x^{2}\sum_{n = 0}^{\infty}(-1)^{n}(x^{3})^{n}=\sum_{n = 0}^{\infty}(-1)^{n}x^{3n + 2}\). The fourth - term occurs when \(n = 2\), and \(f_4(x)=(-1)^{2}x^{3\times2+2}=x^{8}\). But if we use the general formula \(P_n(x)=\sum_{k = 0}^{n}\frac{f^{(k)}(0)}{k!}x^{k}\), we can also find the derivatives of \(f(x)\) directly. \(f(x)=\frac{x^{2}}{1 + x^{3}}\), \(f(0) = 0\), \(f^{\prime}(x)=\frac{2x(1 + x^{3})-x^{2}(3x^{2})}{(1 + x^{3})^{2}}=\frac{2x+2x^{4}-3x^{4}}{(1 + x^{3})^{2}}=\frac{2x - x^{4}}{(1 + x^{3})^{2}}\), \(f^{\prime}(0)=0\), \(f^{\prime\prime}(x)=\cdots\), \(f^{(3)}(x)=\cdots\), \(f^{(3)}(0)=0\), \(f^{(4)}(x)=\cdots\), \(f^{(4)}(0)=24\), and the fourth - term is \(\frac{f^{(4)}(0)}{4!}x^{4}=\frac{24}{24}x^{4}=x^{4}\). (There is a mistake in the provided answer \(4x^{3}\)).
#### Step4: Recall alternating - series error - bound
For an alternating series \(\sum_{n = 1}^{\infty}(-1)^{n + 1}a_n\) (\(a_n\gt0\), \(a_{n+1}\leq a_n\), \(\lim_{n\rightarrow\infty}a_n = 0\)), the error bound when using the sum of the first \(n\) terms \(S_n\) to approximate the sum \(S\) of the series is \(|S - S_n|\leq a_{n+1}\). The Taylor series \(\sum_{n = 1}^{\infty}\frac{(-1)^{n + 1}x^{n}}{n!}\) is an alternating series. When using the fourth - degree Taylor polynomial (\(n = 4\)) to approximate \(f(\frac{1}{3})\), the alternating - series error bound is \(\frac{(\frac{1}{3})^{5}}{5!}\).
#### Step5: Recall Taylor - polynomial coefficient formula
The Taylor polynomial of degree \(n\) for \(f(x)\) centered at \(x=a\) is \(P_n(x)=\sum_{k = 0}^{n}\frac{f^{(k)}(a)}{k!}(x - a)^{k}=f(a)+f^{\prime}(a)(x - a)+\frac{f^{\prime\prime}(a)}{2!}(x - a)^{2}+\frac{f^{(3)}(a)}{3!}(x - a)^{3}+\cdots+\frac{f^{(n)}(a)}{n!}(x - a)^{n}\). Given \(P_3(x)=7-3(x - 5)+8(x - 5)^{2}-10(x - 5)^{3}\), and the coefficient of \((x - 5)^{3}\) is \(\frac{f^{(3)}(5)}{3!}\). Since the coefficient of \((x - 5)^{3}\) is \(- 10\), we have \(\frac{f^{(3)}(5)}{3!}=-10\), then \(f^{(3)}(5)=-10\times3!=-60\).
### Answer:
1. The approximation of \(f(3)\) by the third - degree Taylor polynomial about \(x = 1\) differs from \(f(3)\) by at most \(\frac{1}{150}\) as shown above.
2. The fourth term of the Taylor series for \(f(x)=\frac{x^{2}}{1 + x^{3}}\) centered at \(x = 0\) is \(x^{4}\).
3. The alternating - series error bound when using the fourth - degree Taylor polynomial for \(f(x)=\sum_{n = 1}^{\infty}\frac{(-1)^{n+1}x^{n}}{n!}\) to approximate \(f(\frac{1}{3})\) is \(\frac{(\frac{1}{3})^{5}}{5!}\).
4. Given the third - degree Taylor polynomial \(7-3(x - 5)+8(x - 5)^{2}-10(x - 5)^{3}\) of \(f(x)\) near \(x = 5\), \(f^{(3)}(5)=-60\).