How Do You Calculate the Power Series of (1-x)/(1-2x)^3?

[dan]

Hi, I am stuck on this problem!

Given the function, find the first four terms of the power series representation and find the domain of convergence?

function is;
(1-x)/(1-2x)^3

my calculation;
find derivatives of the function.

y'=(1-x)/(1-2x)^4
y''=(1-x)/(1-2x)^5

I am having difficulty in knowing if what I did was correct.

I'm not sure if this is how you are meant to approach it and if what I did was the correct way to go about solving the problem.

If anyone can help me in solving my problem your help will be appreciated.

Dj

 

[arcnets]

Hi dan,
I think you should use Taylor's formula.
It's true you need the derivatives.
But I think your results are wrong. I get, for instance,

y'=(5-4x)/(1-2x)^4

 

[mathman]

Use the binomial expansion to get a series for (1-2x)^-3, multiply the first few terms by (1-x). It will converge for |x|<1/2.

 

[Imperial]

Your derivatives are not exactly correct. You need to use the quotient rule and the chain rule in this case.
Remember the quotient rule y' = (vu' - uv')/v^2.
Now let u = (1-x) and v = (1-2x)^3.
u' = -1, v' = -6(1-2x)^2 (chain rule).

therefore y' = (-(1-2x)^3-(-6)(1-x)(1-2x)^2)/(1-2x)^6
= ((1-2x)^2(5-4x))/(1-2x)^6
= (5-4x)/(1-2x)^4