Expansion of 1/f(z)

**dudyu** · May 8th 2011, 11:40 PM

Hello,
If I have an expansion of a certain function- $f(z) = \sum_{n = 0}^\infty a_n z^n$ , how do I derive the expantion of 1/f(z) ?
Namely, I am looking for a series $b_m$ such that:

$\sum_{m = -\infty}^\infty b_m z^m = \frac{1}{\sum_{n = 0}^\infty a_n z^n}$

And I'm looking to express it as a function of the given series of $a_n$

Thank you

**Prove It** · May 8th 2011, 11:52 PM

I don't know if this would work, but since f(z) is a complex function, you should be able to write it in terms of its real and imaginary parts, and from there be able to find $\overline{f(z)}$ .

You should know that $\frac{1}{f(z)} = \frac{\overline{f(z)}}{f(z)\overline{f(z)}} = \frac{\overline{f(z)}}{\left|f(z)\right|^2}$

**chisigma** · May 9th 2011, 06:22 AM

Originally Posted by dudyu

Hello,
If I have an expansion of a certain function- $f(z) = \sum_{n = 0}^\infty a_n z^n$ , how do I derive the expantion of 1/f(z) ?
Namely, I am looking for a series $b_m$ such that:

$\sum_{m = -\infty}^\infty b_m z^m = \frac{1}{\sum_{n = 0}^\infty a_n z^n}$

And I'm looking to express it as a function of the given series of $a_n$

Thank you

If

, then is...

(1)

... and the b coefficients are found as follows...

(2)

Kind regards

$\chi$ $\sigma$

**dudyu** · May 9th 2011, 06:35 AM

Originally Posted by chisigma

If

, then is...

(1)

... and the b coefficients are found as follows...

(2)

Kind regards

$\chi$ $\sigma$

Thank you!
and if $a_0$ is equal to zero, do I simply start from $a_1$ ?
Namely-
$b_n = - \frac{1}{a_1} \sum_{k = 2}^n a_k b_{n-k}$
?

**chisigma** · May 9th 2011, 08:01 AM

Originally Posted by dudyu

Thank you!
and if $a_0$ is equal to zero, do I simply start from $a_1$ ?
Namely-
$b_n = - \frac{1}{a_1} \sum_{k = 2}^n a_k b_{n-k}$
?

If

and

You can write...

(1)

(2)

... and then compute the b coefficients imposing that...

(3)

Kind regards

$\chi$ $\sigma$

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