# Power series representatin of a function

• Jul 17th 2009, 09:39 AM
calc101
Power series representatin of a function
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?
• Jul 17th 2009, 10:06 AM
skeeter
Quote:

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.
• Jul 17th 2009, 10:32 AM
calc101
And I calculated the radius of convergence to be: 1, with the following interval: (-1,1). Correct?
• Jul 17th 2009, 10:35 AM
DeMath
Quote:

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!
• Jul 17th 2009, 10:42 AM
Bruno J.
Quote:

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 $\frac{3}{1-2^2}=3\sum_{j=1}^\infty2^{2j}$?
• Jul 17th 2009, 10:53 AM
Plato
Quote:

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
Quote:

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!