# Thread: Expanding to power series, and finding the Laurent Series

1. Please refer to attached image.

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?

2.

3. Originally Posted by nacho

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

$\chi$ $\sigma$

4. Originally Posted by nacho

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

$\chi$ $\sigma$

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

6. Originally Posted by nacho
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

$\chi$ $\sigma$

Originally Posted by chisigma
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

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