# Thread: power series centered at zero

1. ## 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

2. 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.

3. Originally Posted by matheagle
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.
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).

4. 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).
$\displaystyle \frac{d}{dx}[\ln{(3 + x)}] = \frac{1}{3 + x}$.

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 $\displaystyle \ln{(3 + x)}$.

5. 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)

6. Originally Posted by Prove 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 $\displaystyle \ln{(3 + x)}$.
that makes a lot of sense!!!thanks!!

7. 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.

8. 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

$\displaystyle (1 + 2x)^{-\frac{1}{4}}$.

9. 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.

10. thank you!! i understand all of them except the integral one.
can i get more help on that one please!
Thanx so much!

11. 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.