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

[Wishbone]

im sorry, these topics were not covered in class.

 

[d_leet]

im sorry, these topics were not covered in class.

Well do you know what the formula for the sum of an infinite geometric series? Maybe try it this way, take the derivative of the ln(1 + z) and then try to find a geometric series for that and then integrate it to find one for the log...

 

[VietDao29]

Well do you know what the formula for the sum of an infinite geometric series? Maybe try it this way, take the derivative of the ln(1 + z) and then try to find a geometric series for that and then integrate it to find one for the log...

d_leet, I think it might be better if you hint him more clearly. AFAIK, he has tried his best to understand the problem. I haven't seen such a 3-page thread that reaches nowhere like this one for a long time.
Wishbone, I think you should re-read your textbook to understand the concept of Taylor's series a little bit more clearer. And try again to see if you can do this problem.
You, however can do it by using geometric series as d_leet has pointed out.
For a common ratio |z| < 1, we have:
\frac{a}{1 - z} = \sum_{k = 0} ^ {\infty} az ^ k, so apply it here, we have:
\frac{1}{1 + z} = \sum_{k = 0} ^ {\infty} (-z) ^ k, \ \forall |z| &lt; 1 (a geometric series with a common ratio -z, and the first term is 1).
Integrate both sides gives:
\int \frac{dz}{1 + z} = \int \left( \sum_{k = 0} ^ {\infty} (-z) ^ k \right) dz
Now, hopefully, you can go from here, right? :)

 

[assyrian_77]

I haven't seen such a 3-page thread that reaches nowhere like this one for a long time.

Then take a look at this thread below: "Urgend Geometric series question".

Sorry for the off-topic.

 

[shmoe]

ok i took

F'(z)= 1/(1+z)^2
F''(z) = 1/2(1+z)^3
F'''(z) = 1/-6(1+z)^4

so I do F(0) + F'(z)*(z-0) + F''(z)*(z-0)^2/2! + F'''(z)*(z-0)^3/3!

These derivatives are incorrect. If F(z)=ln(1+z) then:

F&#039;(z)=(1+z)^{-1}

try finding F'', F''', etc. again. Carefully.

 

[Axiom_137]

For a common ratio |z| < 1, we have:
\frac{a}{1 - z} = \sum_{k = 0} ^ {\infty} az ^ k, so apply it here, we have:
\frac{1}{1 + z} = \sum_{k = 0} ^ {\infty} (-z) ^ k, \ \forall |z| &lt; 1 (a geometric series with a common ratio -z, and the first term is 1).
Integrate both sides gives:
\int \frac{dz}{1 + z} = \int \left( \sum_{k = 0} ^ {\infty} (-z) ^ k \right) dz

This seems a little too simple for the complex plane.

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

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

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

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?)

 

[d_leet]

This seems a little too simple for the complex plane.

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

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

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

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?)

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.

 

[Axiom_137]

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

 

[d_leet]

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)

 

[HallsofIvy]

Please, tell us what you think the derivative of 1/(1+ z) is!

 

[Hurkyl]

This seems a little too simple for the complex plane.

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

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

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

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

You forgot the constant of integration.