Taylor Series and Approximation

**tapiaghector** · June 10th 2009, 07:13 PM

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.

**TheEmptySet** · June 10th 2009, 07:43 PM

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.

$f'(x)=\frac{1}{2}x^{-1/2}$
$f''(x)=\frac{-1}{2^2}x^{-3/2}$
$f'''(x)=\frac{3}{2^3}x^{-5/2}$
$f^{4}(x)=\frac{-3\cdot 5}{2^4}x^{-7/2}$

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

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

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

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

For the 2nd start with

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

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

Now we can integrate to get

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

You want

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

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

**tapiaghector** · June 10th 2009, 08:30 PM

Hey TheEmptySet,

Thanks for your help, know i understand more about taylor series...

Hava a great night and thanks again!

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