Expanding to power series, and finding the Laurent Series

In summary, the conversation focuses on the process of expanding a logarithm to a power series and the use of series expansions for natural logarithms. The first question is about the technique for expanding a log to a power series, while the second question involves setting a variable and manipulating the function to find its series expansion. The conversation also touches on the derivation of the series expansion for ln(1+x) and its use in integration.
  • #1
nacho-man
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0
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?
 

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  • #2
nacho said:
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)... $\displaystyle \ln (1 + i\ z) - \ln (1-i\ z) = \sum_{n=1}^{\infty} \{1- (-1)^{n}\}\ \frac{(i\ z)^{n}}{n} = 2\ i\ \sum_{n=1}^{\infty} (-1)^{n-1}\ \frac{z^{2n-1}}{2n-1} = 2\ i\ \tan^{-1} z\ (5)$

Kind regards$\chi$ $\sigma$
 
  • #3
nacho said:
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
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
 
  • #5
nacho said:
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
chisigma said:
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
 

FAQ: Expanding to power series, and finding the Laurent Series

What is a power series?

A power series is an infinite series of the form ∑n=0∞ cn(x-a)n, where cn are constants, x is the variable, and a is the center of the series. This series is used to represent a function as a polynomial, which can be useful in solving problems in calculus and other areas of mathematics.

How do you find the power series expansion of a function?

To find the power series expansion of a function, you can use the Taylor series formula, which is given by f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)2/2! + f'''(a)(x-a)3/3! + ..., where f'(a), f''(a), and f'''(a) are the derivatives of the function evaluated at the center a. This formula can be used to find the coefficients cn in the power series.

What is a Laurent series?

A Laurent series is a type of power series that includes negative powers of the variable, such as ∑n=-∞∞ cn(x-a)n. This type of series is useful for representing functions with singularities, such as poles or branch points, and is commonly used in complex analysis.

How do you find the Laurent series of a function?

To find the Laurent series of a function, you can use the formula ∑n=-∞∞ cn(z-a)n, where cn are the coefficients and z is the complex variable. The coefficients can be found by evaluating the function at the center a and its derivatives, similar to the process for finding a power series.

What are some applications of power and Laurent series?

Power and Laurent series have many applications in mathematics, physics, and engineering. They can be used to approximate functions, solve differential equations, and study the behavior of functions near singularities. They are also commonly used in numerical analysis and computer simulations to model complex systems and phenomena.

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