The discussion revolves around developing the Taylor expansion for the function ln(1+z). Participants are exploring the necessary components of the Taylor series, particularly the evaluation point and the derivatives involved.
The conversation is ongoing, with participants providing different methods and questioning the validity of their approaches. Some guidance has been offered regarding the use of geometric series, but there is no clear consensus on the best method to proceed.
There are indications that some participants feel the problem is poorly worded, which adds to the confusion. Additionally, there are references to topics not covered in class, suggesting a gap in foundational knowledge that may be affecting the discussion.
Physics Monkey said:z = 0 provided you're expanding about zero.
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?
d_leet said:Well you can derive a taylor expansion for it by using the geometric series and a bit of integration.
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.
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.
Wishbone said:ya but what do i use as a? I end up with the same problem...
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)}.
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.
Wishbone said:wait won't the nth derivative on ln (1+z) where z=0 always equal 1?
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.
Wishbone said:yes but isn't it 1 to a negative power?
d_leet said:What is the derivative of something to a negative power though?
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
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
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!
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.
Wishbone said:well I don't know how to show natural logs as a geometric series, that's why i couldn't try that
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.
Wishbone said:well it depends on what you start out with right?
d_leet said:I don't know what you mean.