Power series representatin of a function

**calc101** · July 17th 2009, 08:39 AM

Question:

Find a power series representation for the function.

$f(x) = \frac{3}{1-x^2}$

My solution:

Since, $\frac{1}{1-x} = 1 + x + x^2 + x^3 + .... = \sum_{n=0}^{\infty} x^n$

Then,

$ \frac{3}{1-x^2} = \sum_{n=0}^{\infty} 3x^{2n} $

Am I correct?

**skeeter** · July 17th 2009, 09:06 AM

Originally Posted by calc101

Question:

Find a power series representation for the function.

$f(x) = \frac{3}{1-x^2}$

My solution:

Since, $\frac{1}{1-x} = 1 + x + x^2 + x^3 + .... = \sum_{n=0}^{\infty} x^n$

Then,

$ \frac{3}{1-x^2} = \sum_{n=0}^{\infty} 3x^{2n} $

Am I correct?

looks fine.

**calc101** · July 17th 2009, 09:32 AM

And I calculated the radius of convergence to be: 1, with the following interval: (-1,1). Correct?

**DeMath** · July 17th 2009, 09:35 AM

Originally Posted by calc101

Question:

Find a power series representation for the function.

$f(x) = \frac{3}{1-x^2}$

My solution:

Since, $\frac{1}{1-x} = 1 + x + x^2 + x^3 + .... = \sum_{n=0}^{\infty} x^n$

Then,

$ \frac{3}{1-x^2} = \sum_{n=0}^{\infty} 3x^{2n} $

Am I correct?

Sorry!

**Bruno J.** · July 17th 2009, 09:42 AM

Originally Posted by DeMath

A small clarification: for $x \in \left( { - \infty ;{\text{ }} - 1} \right) \cup \left( { - 1;{\text{ }}1} \right) \cup \left( {1;{\text{ }}\infty } \right)$ .

Are you saying for example that $\frac{3}{1-2^2}=3\sum_{j=1}^\infty2^{2j}$ ?

**Plato** · July 17th 2009, 09:53 AM

Originally Posted by calc101

And I calculated the radius of convergence to be: 1, with the following interval: (-1,1). Correct?

Yes this correct.

Whereas

Originally Posted by DeMath

A small clarification: for $x \in \left( { - \infty ;{\text{ }} - 1} \right) \cup \left( { - 1;{\text{ }}1} \right) \cup \left( {1;{\text{ }}\infty } \right)$ .

Is grossly misguided!

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