find a power series for the given function, centered at a=0.

a)F(x)=ln(3+x)

b) G(x)=1/(1+2x)^(1/4)

c) Integral, 0 to x, (1-cos(t)/t^2) dt

Printable View

- May 15th 2010, 06:03 PMtwofortwopower series centered at zero
find a power series for the given function, centered at a=0.

a)F(x)=ln(3+x)

b) G(x)=1/(1+2x)^(1/4)

c) Integral, 0 to x, (1-cos(t)/t^2) dt - May 15th 2010, 06:14 PMmatheagle
You can use Taylor's formula, but I'd use the geometric series for the first one.

$\displaystyle {1\over 3+x}={1/3\over 1-(-x/3)}={1\over 3}\sum_{n=0}^{\infty}(-x/3)^n$

$\displaystyle =\sum_{n=0}^{\infty}{(-1)^nx^n\over 3^{n+1}}$

now integrate wrt x. - May 16th 2010, 03:49 AMPulock2009
- May 16th 2010, 03:52 AMProve It
$\displaystyle \frac{d}{dx}[\ln{(3 + x)}] = \frac{1}{3 + x}$.

So $\displaystyle \int{\frac{1}{3 + x}\,dx} = \ln{(3 + x)}$.

So if you have a power series for $\displaystyle \frac{1}{3 + x}$, integrate it, then you will have a power series for $\displaystyle \ln{(3 + x)}$. - May 16th 2010, 03:52 AMPulock2009
- May 16th 2010, 03:56 AMPulock2009
- May 16th 2010, 04:16 AMPulock2009
- May 16th 2010, 04:28 AMProve It
- May 16th 2010, 06:40 AMmatheagle
- May 16th 2010, 05:48 PMtwofortwo
thank you!! i understand all of them except the integral one.

can i get more help on that one please!

Thanx so much! - May 16th 2010, 07:04 PMProve It