# power series centered at zero

• May 15th 2010, 06:03 PM
twofortwo
power 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 PM
matheagle
You can use Taylor's formula, but I'd use the geometric series for the first one.

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

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

now integrate wrt x.
• May 16th 2010, 03:49 AM
Pulock2009
Quote:

Originally Posted by matheagle
You can use Taylor's formula, but I'd use the geometric series for the first one.

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

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

now integrate wrt x.

although i didnot have mathematics as honours in college i didnot understand one thing: How can ln(3+x) be expressed as 1/(3+x).
• May 16th 2010, 03:52 AM
Prove It
Quote:

Originally Posted by Pulock2009
although i didnot have mathematics as honours in college i didnot understand one thing: How can ln(3+x) be expressed as 1/(3+x).

$\frac{d}{dx}[\ln{(3 + x)}] = \frac{1}{3 + x}$.

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

So if you have a power series for $\frac{1}{3 + x}$, integrate it, then you will have a power series for $\ln{(3 + x)}$.
• May 16th 2010, 03:52 AM
Pulock2009
Quote:

Originally Posted by twofortwo
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

a)F(x)=ln(3+x)
=ln(3(1+x/3))
=ln3+ln(1+x/3)
=ln3+(the standard logarithmic expansion)
• May 16th 2010, 03:56 AM
Pulock2009
Quote:

Originally Posted by Prove It
$\frac{d}{dx}[\ln{(3 + x)}] = \frac{1}{3 + x}$.

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

So if you have a power series for $\frac{1}{3 + x}$, integrate it, then you will have a power series for $\ln{(3 + x)}$.

that makes a lot of sense!!!thanks!!
• May 16th 2010, 04:16 AM
Pulock2009
Quote:

Originally Posted by twofortwo
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

b)G(x) can be expanded binomially to get a power series if i am not wrong.
• May 16th 2010, 04:28 AM
Prove It
Quote:

Originally Posted by Pulock2009
b)G(x) can be expanded binomially to get a power series if i am not wrong.

Yes, as long as you write it as

$(1 + 2x)^{-\frac{1}{4}}$.
• May 16th 2010, 06:40 AM
matheagle
Quote:

Originally Posted by Pulock2009
although i didnot have mathematics as honours in college i didnot understand one thing: How can ln(3+x) be expressed as 1/(3+x).

I clearly said you should now integrate.
• May 16th 2010, 05:48 PM
twofortwo
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 PM
Prove It
Quote:

Originally Posted by matheagle
I clearly said you should now integrate.

Yes you did, the OP just didn't understand WHY you should integrate, hence my fill-in post. :)