# Expanding to power series, and finding the Laurent Series

#### nacho

##### Active member

Hi,
I'm a bit lost here with the first question. Unfortunately the online lecture covering this material isn't available due to their having been made some technical difficulties, and I find our textbook difficult to comprehend!
My lecture notes are pretty ambiguous in relation to these two questions.

Firstly, how exactly does one expand a log to a power series? Is there some trick required here, like converting the given logs to it's equivalent exponential, and then using the polar form?

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#### chisigma

##### Well-known member

Hi,
I'm a bit lost here with the first question. Unfortunately the online lecture covering this material isn't available due to their having been made some technical difficulties, and I find our textbook difficult to comprehend!
My lecture notes are pretty ambiguous in relation to these two questions.

Firstly, how exactly does one expand a log to a power series? Is there some trick required here, like converting the given logs to it's equivalent exponential, and then using the polar form?
(i) Is...

$\displaystyle \ln (1 + s) = - \sum_{n=1}^{\infty} (-1)^{n} \frac{s^{n}}{n}\ (1)$

$\displaystyle \ln (1 - s) = - \sum_{n=1}^{\infty} \frac{s^{n}}{n}\ (2)$

... and setting $\displaystyle s = i\ z$ You obtain...

$\displaystyle \ln (1 + i\ z) = - \sum_{n=1}^{\infty} (-1)^{n} \frac{(i\ z)^{n}}{n}\ (3)$

$\displaystyle \ln (1 - i\ z) = - \sum_{n=1}^{\infty} \frac{(i\ z)^{n}}{n}\ (4)$

From (3) and (4)...

$\displaystyle \ln (1 + i\ z) - \ln (1-i\ z) = \sum_{n=1}^{\infty} \{1- (-1)^{n}\}\ \frac{(i\ z)^{n}}{n} = 2\ i\ \sum_{n=1}^{\infty} (-1)^{n-1}\ \frac{z^{2n-1}}{2n-1} = 2\ i\ \tan^{-1} z\ (5)$

Kind regards

$\chi$ $\sigma$

#### chisigma

##### Well-known member

Hi,
I'm a bit lost here with the first question. Unfortunately the online lecture covering this material isn't available due to their having been made some technical difficulties, and I find our textbook difficult to comprehend!
My lecture notes are pretty ambiguous in relation to these two questions.

Firstly, how exactly does one expand a log to a power series? Is there some trick required here, like converting the given logs to it's equivalent exponential, and then using the polar form?
(ii) For semplicity we set $\displaystyle s = z - 1$ so that the function becomes...

$\displaystyle f(s) = \frac{1}{s}\ \frac{1 + s}{2 + s} = \frac{1}{2}\ \frac{1}{s}\ \frac{1 + s}{1 + \frac{s}{2}} = \frac{1}{2}\ \frac{1}{s}\ (1 + s)\ (1 - \frac{s}{2} + \frac{s^{2}}{4} - \frac{s^{3}}{8} + ...)= \frac{1}{2}\ (\frac{1}{s} + \frac{1}{2} - \frac{s}{4} + \frac{s^{2}}{8} - ...)\ (1)$

Kind regards

$\chi$ $\sigma$

#### nacho

##### Active member
thanks for the response both of you.

curiously, for

i) when you said

$\displaystyle \ln (1 + s) = - \sum_{n=1}^{\infty} (-1)^{n} \frac{s^{n}}{n}\ (1)$

Is that simply the definition for a power series of natural logs, or did you do some quick manipulation otherwise?

Thank you very much, the rest of it makes perfect sense, I was just getting stuck on how to start it.

I'm having a look at ii) now again

#### chisigma

##### Well-known member
thanks for the response both of you.

curiously, for

i) when you said

$\displaystyle \ln (1 + s) = - \sum_{n=1}^{\infty} (-1)^{n} \frac{s^{n}}{n}\ (1)$

Is that simply the definition for a power series of natural logs, or did you do some quick manipulation otherwise?

Thank you very much, the rest of it makes perfect sense, I was just getting stuck on how to start it.

I'm having a look at ii) now again
The series expansion of $\ln (1 + x)$ derives from the well know expansion...

$\displaystyle \frac{1}{1+x} = \sum_{n=0}^{\infty} (-1)^{n}\ x^{n}\ (1)$

... and integrating (1) 'term by term' ...

$\displaystyle \int \frac{d x}{1+x} = \ln (1+x) = - \sum_{n=1}^{\infty} (-1)^{n} \frac{x^{n}}{n}\ (2)$

Kind regards

$\chi$ $\sigma$

#### nacho

##### Active member
The series expansion of $\ln (1 + x)$ derives from the well know expansion...

$\displaystyle \frac{1}{1+x} = \sum_{n=0}^{\infty} (-1)^{n}\ x^{n}\ (1)$

... and integrating (1) 'term by term' ...

$\displaystyle \int \frac{d x}{1+x} = \ln (1+x) = - \sum_{n=1}^{\infty} (-1)^{n} \frac{x^{n}}{n}\ (2)$

Kind regards

$\chi$ $\sigma$
oh wow, this has changed my perspective of series completely!

thanks for that, i'll keep it in mind