# power series centered at zero

#### 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

#### matheagle

MHF Hall of Honor
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.

#### Pulock2009

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).

#### Prove It

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

Pulock2009

#### Pulock2009

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)

#### Pulock2009

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

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!!

#### Pulock2009

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.

#### Prove It

MHF Helper
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

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

#### matheagle

MHF Hall of Honor
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.

#### twofortwo

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