# Thread: Taylor Series and Approximation

1. ## Taylor Series and Approximation

Hello Guys,

I need help solving these two problem, in the first one i dont really know what im supposed to do with the x = 9... Please i need help...

Problem 1

Find the Taylor series for f(x) = √x using reference point
x = 9. For what values of x does it converge to √x?

Problem 2

Estimate ln(1/2) to within 1/1000.

2. Originally Posted by tapiaghector
Hello Guys,

I need help solving these two problem, in the first one i dont really know what im supposed to do with the x = 9... Please i need help...

Problem 1

Find the Taylor series for f(x) = √x using reference point
x = 9. For what values of x does it converge to √x?

Problem 2

Estimate ln(1/2) to within 1/1000.
$\displaystyle f'(x)=\frac{1}{2}x^{-1/2}$
$\displaystyle f''(x)=\frac{-1}{2^2}x^{-3/2}$
$\displaystyle f'''(x)=\frac{3}{2^3}x^{-5/2}$
$\displaystyle f^{4}(x)=\frac{-3\cdot 5}{2^4}x^{-7/2}$

for $\displaystyle n \ge 2$
$\displaystyle f^{n}(x)=\frac{(-1)^{n+1}1\cdot 3 \cdots (2n-3)}{2^n}x^{\frac{-(2n-1)}{2}}$

$\displaystyle f^{n}(9)=\frac{(-1)^{n+1}1\cdot 3 \cdots (2n-3)}{2^n}(9)^{\frac{-(2n-1)}{2}}$

$\displaystyle f^{n}(9)=\frac{(-1)^{n+1}1\cdot 3 \cdots (2n-3)}{3^{2n-1}2^n}$

$\displaystyle T(x)=3+\frac{1}{6}(x-3)+\sum_{n=2}^{\infty}\frac{(-1)^{n+1}1\cdot 3 \cdots (2n-3)}{2^n 3^{2n-1}n!}(x-9)^n$

$\displaystyle f(x)=ln(x+1)$

$\displaystyle f'(x)=\frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^nx^n$

Now we can integrate to get

$\displaystyle f(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{n+1}}{n+1}$

You want

$\displaystyle f\left( \frac{1}{2}\right)=\sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n+1}n+1}$

This is an alternating series so the error cannot be bigger then the next term in the series n=10 will work because $\displaystyle 2^{10}=1024$

3. Hey TheEmptySet,