# Taylor expansion for ln(1+z)

1. Feb 23, 2006

### Wishbone

the problem reads develop expansion of ln(1+z)

of course I just tried throwing it in to 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?

2. Feb 23, 2006

### Physics Monkey

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 $$f(z) = \ln{(1+z)}$$.

3. Feb 23, 2006

### Wishbone

the problem doesn't say that, see thats what my question is... Am I supposed to assume that?

4. Feb 23, 2006

### d_leet

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

5. Feb 23, 2006

### Wishbone

well it says develop the taylor expansion, I think it means use the taylor expansion.. I dont know tho, its a really stupid, poorly worded question.

6. Feb 23, 2006

### d_leet

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.

7. Feb 23, 2006

### Wishbone

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

8. Feb 23, 2006

### d_leet

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.

9. Feb 23, 2006

### Wishbone

ya I must have to do it the other way becausethis wouldnt even work. If I take z=0, then df(0)/dz= undefined

10. Feb 23, 2006

### Physics Monkey

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.

Last edited: Feb 23, 2006
11. Feb 23, 2006

### Wishbone

oh geez, ya I was taking the derivate of ln z. whoops.

12. Feb 23, 2006

### Wishbone

wait wont the nth derivative on ln (1+z) where z=0 always equal 1?

13. Feb 23, 2006

### d_leet

No... because ypu end up with a number raised to a negative power and so it will osscilate between positive and negative values.

14. Feb 23, 2006

### Wishbone

yes but isnt it 1 to a negative power?

15. Feb 23, 2006

### d_leet

What is the derivative of something to a negative power though?

16. Feb 23, 2006

### Wishbone

oh duh! man I do not know what is wrong with me tonight

17. Feb 23, 2006

### Wishbone

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. Feb 23, 2006

### d_leet

And what did you get?

19. Feb 23, 2006

### Wishbone

$$\sum 0 + (1)z + \frac{\frac{z^2}{z(1+z)^3}}{2} + \frac{\frac{z^3}{-6(1+z)}{3*2}$$

20. Feb 23, 2006

### Wishbone

$$\sum 0 + (1)z + \frac{(z^2)}{(2(1+z)^3)*2} + \frac{(z^3)}{(-6(1+z)^3)*3*2 }$$

Last edited: Feb 23, 2006