power series centered at zero

Apr 2010
21
0
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
Feb 2009
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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.
 
Nov 2009
202
20
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
Aug 2008
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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 \(\displaystyle \ln{(3 + x)}\).
 
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Nov 2009
202
20
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)
 
Nov 2009
202
20
\(\displaystyle \frac{d}{dx}[\ln{(3 + x)}] = \frac{1}{3 + x}\).


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!!
 
Nov 2009
202
20
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.
 

matheagle

MHF Hall of Honor
Feb 2009
2,763
1,146
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.
 
Apr 2010
21
0
thank you!! i understand all of them except the integral one.
can i get more help on that one please!
Thanx so much!