# Calculus-taylor series

1. Nov 26, 2006

### 413

f(x) =ln (1-x)

a) Compute f'(x), f''(x), f'''(x). Spot the pattern and give an expression for f ^(n) (x) [the n-th derivative of f(x)]

b) Compute the MacLaurin series of f(x) (i.e. the Taylor series of f(x) around x=0)

c) Compute the radius of convergence and determine the interval of convergence of the series in b).

d) Determine the Taylor series of f'(x) around x=0. Can you do so without using b)?

e) How would you have computed part b) if you had first done part d)?

for part a) i got, please check for me.

f'(x) = -1/(1-x)
f''(x) = -1/(1-x)^2
f'''(x) = -2/(1-x)^3

f^(n) (x) = -((n-1)!)/(1-x)^n
for n = 1,2,3,...

am i right so far, how do i do the other ones? Thanks.

2. Nov 26, 2006

### d_leet

You're off by a sing on the second derivative, and your patter will for the same reason be off by a sign for all even order derivatives. What problems are you having with the rest of the question, it is pretty straightforward to use the derivatives at 0 to find the MacLaurin series for the original function, and then use the ratio test to find the radius of convergence.

3. Nov 26, 2006

### 413

oh the second derivative is f(x)=1/(1-x)^2 correct?

What is the expression for the n-th derivative now? What do i change to show that the signs is negative for all even order?

4. Nov 26, 2006

### d_leet

Yes, that is correct. What is the sign of (-1)n when n is even? When n is odd?

5. Nov 26, 2006

### 413

so that means i put a (-1)^n in front of my orginal equation?

f^(n) (x) = (-1)^n((n-1)!)/(1-x)^n like this?

6. Nov 26, 2006

### d_leet

Yes. /*extra characters*/

7. Nov 26, 2006

### 413

thanks, let me give part b) a try.

equation for Taylor series is f(x)= f(0) + [f'(0)/1!]*x + [f''(0)/2!]*x^2 + ....

I did...
f(0)=0
f'(0)=-1 thus f'(0)/1!=-1
f''(0)=1 thus f"(0)/2!=1/2
f'''(0)=-1 thus f'''(0)/3! =-1/3

therefore answer for part b is... taylor series of f(x)= (-x/1) +(x^2/2) - (x^3/3) +...
seems right, any mistakes?

8. Nov 26, 2006

### d_leet

That looks right, except that f'''(0) is not -1.

9. Nov 26, 2006

### 413

oh, its -2, then f'''(0)/3! would be -2/3?
taylor series become (x)= (-x/1) +(x^2/2) - 2(x^3/3) +...

10. Nov 26, 2006

### d_leet

No, it was -2! not 2. So the coefficient would be -2!/3! which is -1/3, I don't think there was anything wrong with your Taylor series at all, just that you stated the value of the derivative incorrectly.

11. Nov 26, 2006

### 413

i see, thanks for catching that. Usually for these questions, do i need to put it in the general forumla like with the reimann sums symbol? if so, how would i do it?

12. Nov 26, 2006

### d_leet

You can if you want to, but if you do want to all you need is a general form for your coeffiecient of xn. In your case this is pretty easy because you can fairly easily see that the nth derivative at x=0 is (-1)n(n-1)! and then the coefficient of xn is just f(n)(0)/n!, so you can just substitute for the nth derivative at zero there and pretty easily write an infinite sum for the Taylor series.

13. Nov 26, 2006

### 413

for finding the radius and interval of convergence you mention the ratio test, the ration test is ︱an+1/an︱, in my case, what is an?

14. Nov 26, 2006

### d_leet

an is the coefficient of xn in the Taylor series, I told you how you could find this fairly eaily in my previous post, and it should also be fairly evident from the terms in your series.

15. Nov 26, 2006

### 413

ok, i got that figured out.

now part d) Determine the Taylor series of f'(x) around x=0. Can you do so without using b)?
do i just repeat part b again?

16. Nov 26, 2006

### d_leet

Part d can be done in one of two ways, both of which are fairly simple. If you use part b then you can differentiate the series found in part b term by term, or you could just as well just take the derivative of the original function and expand that in a Taylor series which is fairly easy since it is pretty much teh closed form for the geometric series. And this made me notice a mistake of my own, your original derivatives were correct, they should all be negative, not alternating I'm sorry about that.

17. Nov 27, 2006

### 413

why? why isn't the 2nd derivative positive?

18. Nov 27, 2006

### HallsofIvy

Staff Emeritus
Actually, the "an" to use in the ratio test to determine radius of convergence is the full term, Anxn. Of course, then the fraction |an+1/an| is just |(An+1/An)| |x|. That's why the ratio test works so nicely.

19. Nov 28, 2006

### 413

so you mean that my derivatives are all positive? which means
f'(x) = -1/(1-x)
f''(x) = -1/(1-x)^2
f'''(x) = -2/(1-x)^3

20. Nov 28, 2006

### d_leet

Negative, not positive, but what you have here is correct.