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

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The discussion revolves around developing the Taylor expansion for ln(1+z). Participants express confusion about the problem's wording and the appropriate point of expansion, with suggestions to use z=0 for simplicity. The use of derivatives to find the series terms is debated, with one participant suggesting deriving the expansion from the geometric series through integration. The conversation highlights the importance of understanding Taylor series and the conditions for convergence, particularly when applying these concepts to complex variables. Ultimately, the thread emphasizes the need for clarity in mathematical problems and the application of foundational concepts.
Wishbone
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the problem reads develop expansion of ln(1+z)

of course I just tried throwing it into the formula for taylor expansions, however I do not know what F(a) is, the problem doesn't specify, so how can I use a taylor series?
 
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What exactly did you find difficult about using the formula? You just take derivatives of your function and evaluate them at z = 0 provided you're expanding about zero. The terms in the series are then \frac{1}{n!} f^{(n)}(0) z^n where in your case f(z) = \ln{(1+z)}.
 
Physics Monkey said:
z = 0 provided you're expanding about zero.

the problem doesn't say that, see that's what my question is... Am I supposed to assume that?
 
Wishbone said:
the problem reads develop expansion of ln(1+z)

of course I just tried throwing it into the formula for taylor expansions, however I do not know what F(a) is, the problem doesn't specify, so how can I use a taylor series?

Well you can derive a taylor expansion for it by using the geometric series and a bit of integration.
 
d_leet said:
Well you can derive a taylor expansion for it by using the geometric series and a bit of integration.

well it says develop the taylor expansion, I think it means use the taylor expansion.. I don't know tho, its a really stupid, poorly worded question.
 
Wishbone said:
well it says develop the taylor expansion, I think it means use the taylor expansion.. I don't know tho, its a really stupid, poorly worded question.

Well you would end up with the same answer, or at least you should, so it may just be easier to derive it from the sum of an infinite geometric series.
 
d_leet said:
Well you would end up with the same answer, or at least you should, so it may just be easier to derive it from the sum of an infinite geometric series.

ya but what do i use as a? I end up with the same problem...
 
Wishbone said:
ya but what do i use as a? I end up with the same problem...

Well if you do it the way I suggested you could just use a, leave it general and just make sure to note the rquirements on a.
 
Physics Monkey said:
What exactly did you find difficult about using the formula? You just take derivatives of your function and evaluate them at z = 0 provided you're expanding about zero. The terms in the series are then \frac{1}{n!} f^{(n)}(0) z^n where in your case f(z) = \ln{(1+z)}.


ya I must have to do it the other way becausethis wouldn't even work. If I take z=0, then df(0)/dz= undefined
 
  • #10
You must have made a mistake, Wishbone. \frac{d}{dz} \ln{(1+z)} = \frac{1}{1+z} which is perfectly well defined at z = 0.

Also, you can expand about whatever point you want, you just have to worry about the radius of convergence. I suggested z = 0 because that is what is usually done in practice with this function.
 
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  • #11
Physics Monkey said:
You must have made a mistake, Wishbone. \frac{d}{dz} \ln{(1+z)} = \frac{1}{1+z} which is perfectly well defined at z = 0.

Also, you can expand about whatever point you want, you just have to worry about the radius of convergence. I suggested z = 0 because that is what is usually done in practice with this function.


oh geez, you I was taking the derivate of ln z. whoops.
 
  • #12
wait won't the nth derivative on ln (1+z) where z=0 always equal 1?
 
  • #13
Wishbone said:
wait won't the nth derivative on ln (1+z) where z=0 always equal 1?

No... because ypu end up with a number raised to a negative power and so it will osscilate between positive and negative values.
 
  • #14
d_leet said:
No... because ypu end up with a number raised to a negative power and so it will osscilate between positive and negative values.

yes but isn't it 1 to a negative power?
 
  • #15
Wishbone said:
yes but isn't it 1 to a negative power?

What is the derivative of something to a negative power though?
 
  • #16
d_leet said:
What is the derivative of something to a negative power though?

oh duh! man I do not know what is wrong with me tonight
 
  • #17
hmmmm I did it out, and I didnt get the answer in the book, it says it should look like:

(-1)^{n-1}*z^n/n
 
  • #18
Wishbone said:
hmmmm I did it out, and I didnt get the answer in the book, it says it should look like:

(-1)^{n-1}*z^n/n

And what did you get?
 
  • #19
\sum 0 + (1)z + \frac{\frac{z^2}{z(1+z)^3}}{2} + \frac{\frac{z^3}{-6(1+z)}{3*2}
 
  • #20
\sum 0 + (1)z + \frac{(z^2)}{(2(1+z)^3)*2} + \frac{(z^3)}{(-6(1+z)^3)*3*2 }
 
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  • #21
Wishbone said:
\sum 0 + (1)z + frac{1}{2(1+z)^3}(z^2)/2 + frac{1}{-6(1+z)^3}(z^3)/3*2

Is this what you meant? The formatting is horrible but it isn't right that much I can tell, how did you get that?
 
  • #22
ug, you sorry about that is it clearer now?
 
  • #23
It's ugly as hell, how did you get that?
 
  • #24
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!
 
  • #25
Wishbone said:
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!

Well you need to evaluate the f primes, and I still say it would be much eaiser to derive this from the geometric series.
 
  • #26
d_leet said:
Well you need to evaluate the f primes, and I still say it would be much eaiser to derive this from the geometric series.


well I don't know how to show natural logs as a geometric series, that's why i couldn't try that
 
  • #27
Wishbone said:
well I don't know how to show natural logs as a geometric series, that's why i couldn't try that

I never said to show a natural log as a geometric series, I said to derive it from one, what happens if you integrate the formula for the sum of an infinite geometric series.
 
  • #28
d_leet said:
I never said to show a natural log as a geometric series, I said to derive it from one, what happens if you integrate the formula for the sum of an infinite geometric series.


well it depends on what you start out with right?
 
  • #29
Wishbone said:
well it depends on what you start out with right?

I don't know what you mean.
 
  • #30
d_leet said:
I don't know what you mean.

well i guess, no, i don't know how to
 

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