The Taylor expansion for ln(1+z) can be derived using the geometric series and integration techniques. The series expansion is given by the formula: ln(1+z) = ∑ (-1)^(n-1) * z^n / n for |z| < 1. Participants in the discussion emphasized the importance of evaluating derivatives at z = 0 and integrating the geometric series to achieve the correct expansion. The conversation highlighted common misunderstandings regarding the application of Taylor series and the geometric series method.
PREREQUISITESStudents and educators in calculus, mathematicians interested in series expansions, and anyone looking to deepen their understanding of logarithmic functions and Taylor series.
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.