
#1
September 26th, 2013,
07:46
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?
Last edited by nacho; September 26th, 2013 at 08:15.

September 26th, 2013 07:46
# ADS
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MHB Master
#2
September 26th, 2013,
08:58
Originally Posted by
nacho
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?
(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)...
$\displaystyle \ln (1 + i\ z)  \ln (1i\ z) = \sum_{n=1}^{\infty} \{1 (1)^{n}\}\ \frac{(i\ z)^{n}}{n} = 2\ i\ \sum_{n=1}^{\infty} (1)^{n1}\ \frac{z^{2n1}}{2n1} = 2\ i\ \tan^{1} z\ (5)$
Kind regards
$\chi$ $\sigma$

MHB Master
#3
September 26th, 2013,
09:36
Originally Posted by
nacho
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?
(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
$\chi$ $\sigma$

#4
September 26th, 2013,
10:44
Thread Author
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

MHB Master
#5
September 26th, 2013,
12:14
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
$\chi$ $\sigma$

#6
September 26th, 2013,
12:15
Thread Author
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