Answer to Question #188492 in Calculus for Amit

Question

1-Calculate the third-degree Taylor polynomial about x₀=0 for f (x) =(1+ x)^1/2

2-Use the polynomial in part (i) to approximate (1.1)^1/2 and find a bound for the

error involved.

3.Use the polynomial in part (i) to approximate integration[0,0.1]

(1 +x)^1/2dx .

Answer 1

1.

$f(x)=(1+x)^{\frac{1}{2}} \newline f'(x)=\frac{1}{2}(1+x)^{-\frac{1}{2}}\newline f''(x)=-\frac{1}{4}(1+x)^{-\frac{3}{2}}\newline f'''(x)=\frac{3}{8}(1+x)^{-\frac{5}{2}}\newline Taylor series,\newline f(x)=\sum_{n=0}^{3}\frac{f^n(x_{0})}{n!}(x-x_{0})^n\newline f(x)=f(x_{0})+\frac{f'(x_{0})}{1!}(x-x_{0})+\frac{f''(x_{0})}{2!}(x-x_{0})^2+\frac{f'''(x_{0})}{3!}(x-x_{0})^3\newline f(x)=f(0)+\frac{f'(0)}{1!}(x-0)+\frac{f''(0)}{2!}(x-0)^2+\frac{f'''(0)}{3!}(x-0)^3\newline f(x)=f(0)+\frac{f'(0)}{1!}(0)+\frac{f''(0)}{2!}(0)^2+\frac{f'''(0)}{3!}(0)^3\newline f(x)=1+\frac{1}{2}x-\frac{1}{8}x^2+\frac{3}{48}x^3$

2.

Given, (1.1)^1/2.

Actual value of (1.1)^1/2=1.048808848. (9 decimal places)

$f(0.1)=(1.1)^{\frac{1}{2}} \newline f(0.1)=1+\frac{1}{2}(0.1)-\frac{1}{8}(0.1)^2+\frac{3}{48}(0.1)^3\newline f(0.1)=1.048812500$

Error=$\begin{vmatrix} approx -exact \\ \end{vmatrix} =\begin{vmatrix} 1.0488125 -1.048808848 \\ \end{vmatrix} =0.000003652$

3.

Approximate $\int_{0}^{0.1}(1+x)^\frac{1}{2}dx$ ,

$\int_{0}^{0.1}f(x)dx=\int_{0}^{0.1}(1+\frac{1}{2}x-\frac{1}{8}x^2+\frac{3}{48}x^3)dx=0.102459896$