Constructing a Power Series using a geometric series as a base.

i am to use $\displaystyle \frac{1}{1-x}=\sum{x^n}$

to find a power series representation for $\displaystyle \frac{1+x}{1-x}$

additionally I need to find the interval of convergence.

since, $\displaystyle \frac{1}{1-x}=\sum{x^n}$

and

$\displaystyle (1+x)\frac{1}{1-x}=\frac{1+x}{1-x}$

then

$\displaystyle (1+x)*\sum{x^n}$

the book says $\displaystyle 1+2\sum{x^n}$ is a power series representation of the original function $\displaystyle \frac{1+x}{1-x}$ can someone point out my error?

Re: Constructing a Power Series using a geometric series as a base.

$\displaystyle (1+x)*\sum{x^n}$ is correct, but it is not in the form $\displaystyle \sum c_nx^n$.

Re: Constructing a Power Series using a geometric series as a base.

we'll i don't think i can use the distributive property to make it

$\displaystyle \sum{x^n}+\sum{x^{n+1}} $

since this is basically an polynomial of infinitely many terms being multiplied by a binomial.

i don't really see anything to do here.

Re: Constructing a Power Series using a geometric series as a base.

i don't know of a way to dump out the parentheses and make a "$\displaystyle c_n$"

Re: Constructing a Power Series using a geometric series as a base.

Quote:

Originally Posted by

**bkbowser** we'll i don't think i can use the distributive property to make it

$\displaystyle \sum{x^n}+\sum{x^{n+1}} $

since this is basically an polynomial of infinitely many terms being multiplied by a binomial.

This depends on the level of rigor with which you need to "find" the series for $\displaystyle \frac{1+x}{1-x}$. Do you need an accurate proof that it converges to the function inside the interval of convergence? I agree that infinite sums require special care, but using distributivity in the way you did is fine as the first approximation.

Re: Constructing a Power Series using a geometric series as a base.

$\displaystyle \frac{1+x}{1-x}$

$\displaystyle \frac{1}{1-x} + \frac{x}{1-x}$

$\displaystyle (1 + x + x^2 + x^3 + ... ) + (x + x^2 + x^3 + x^4 + ...)$

$\displaystyle 1 + 2x + 2x^2 + 2x^3 + ...$

$\displaystyle 1 + 2(x + x^2 + x^3 + ... ) = 1 + 2\sum_{n=1}^\infty x^n$