What is the Taylor expansion for ln(1+z)?

In summary, The conversation discusses how to develop the taylor expansion of ln(1+z) and the difficulty of using the formula. It is suggested to use the geometric series and integration to derive the taylor expansion. There is also a discussion about the nth derivative and how to evaluate it at z = 0. The conversation ends with a suggestion to derive the taylor expansion from the sum of an infinite geometric series.
  • #36
VietDao29 said:
For a common ratio |z| < 1, we have:
[tex]\frac{a}{1 - z} = \sum_{k = 0} ^ {\infty} az ^ k[/tex], so apply it here, we have:
[tex]\frac{1}{1 + z} = \sum_{k = 0} ^ {\infty} (-z) ^ k, \ \forall |z| < 1[/tex] (a geometric series with a common ratio -z, and the first term is 1).
Integrate both sides gives:
[tex]\int \frac{dz}{1 + z} = \int \left( \sum_{k = 0} ^ {\infty} (-z) ^ k \right) dz[/tex]

This seems a little too simple for the complex plane.

What about ln(i-z)? hmmm...

[tex]\int \left( \frac{1}{i} \right)\frac{dz}{1 + iz} = \int \left( -i\sum_{k = 0} ^ {\infty} (-iz) ^ k \right) dz[/tex]

[tex]ln(i-z) = \sum_{k = 0} ^ {\infty} \frac{(-iz)^{k+1}}{(k+1)}[/tex]

Plug in z=0 and the sum is zero...My calculator says ln(i) is not zero. ln(i)= i*(pi/2)

So is there seems to be a problem using this method with complex variables.

(Or have I made a mistake with my math?)

:rolleyes:
 
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  • #37
Axiom_137 said:
This seems a little too simple for the complex plane.

What about ln(i-z)? hmmm...

[tex]\left( \frac{1}{i} \right)\frac{dz}{1 + iz} = \int \left( -i\sum_{k = 0} ^ {\infty} (-iz) ^ k \right) dz[/tex]

[tex]ln(i-z) = \sum_{k = 0} ^ {\infty} \frac{(-iz)^{k+1}}{(k+1)}[/tex]

Plug in z=0 and the sum is zero...


My calculator says ln(i) is not zero. ln(i)= i*(pi/2)

So is there seems to be a problem using this method with complex variables.

(Or have I made a mistake with my math?)

:rolleyes:


I think that sum only holds for |z|<1, that is strictly less than one, whereas |i|=1 so one should not expect the equation to hold for z=i since i does not satisfy the necessary inequality.
 
  • #38
d_leet said:
I think that sum only holds for |z|<1, that is strictly less than one, whereas |i|=1 so one should not expect the equation to hold for z=i since i does not satisfy the necessary inequality.


For the geometric series, the inequality for this problem would be :

|-iz|<1 or simply |z|<1

z=0 satisfies |z|<1
 
  • #39
Axiom_137 said:
For the geometric series, the inequality for this problem would be :

|-iz|<1 or simply |z|<1

z=0 satisfies |z|<1

Yes you're absolutely right, I have no idea what I was thinking, and I cannot think of a reason as to why the sum does not work other than the fact that the series does, at least, correctly calculate the real part of ln(i):rolleyes:
 
  • #40
Please, tell us what you think the derivative of 1/(1+ z) is!
 
  • #41
Axiom_137 said:
This seems a little too simple for the complex plane.

What about ln(i-z)? hmmm...

[tex]\int \left( \frac{1}{i} \right)\frac{dz}{1 + iz} = \int \left( -i\sum_{k = 0} ^ {\infty} (-iz) ^ k \right) dz[/tex]

[tex]ln(i-z) = \sum_{k = 0} ^ {\infty} \frac{(-iz)^{k+1}}{(k+1)}[/tex]

Plug in z=0 and the sum is zero...
You forgot the constant of integration.
 
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