How Do You Calculate the Taylor Series for ln(1-x)?

[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.

 

[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.

 

[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?

 

[d_leet]

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?

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

 

[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?

 

[d_leet]

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?

Yes. /*extra characters*/

 

[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?

 

[d_leet]

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?

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

 

[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) +...

 

[d_leet]

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

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.

 

[413]

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

 

[d_leet]

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

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.

 

[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?

 

[d_leet]

a_n 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.

 

[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?

 

[d_leet]

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?

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 the 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.

 

[413]

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

 

[HallsofIvy]

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

 

[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

 

[d_leet]

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

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

 

[413]

really? i thought the negatives actually cancels out.

 

[d_leet]

really? i thought the negatives actually cancels out.

No, they don't cancel out.

 

[jkh4]

i don't get this one.

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

if you plug in n =1, the equation wouldn't just = 0?

 

[d_leet]

i don't get this one.

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

if you plug in n =1, the equation wouldn't just = 0?

No, you get 0! which is not 0.

 

[jkh4]

No, you get 0! which is not 0.

but 0!/(1-x)^n is still not same as -1/(1-x)^1 when n = 1 right?

 

[d_leet]

but 0!/(1-x)^n is still not same as -1/(1-x)^1 when n = 1 right?

Well you get -0!/(1-x) which is the same thing as -1/(1-x).

 

[jkh4]

Well you get -0!/(1-x) which is the same thing as -1/(1-x).

o...thanks

just curious, what's 0! when expand?

 

[d_leet]

o...thanks

just curious, what's 0! when expand?

0! = 1
/*extra characters*/

 

[jkh4]

0! = 1
/*extra characters*/

thanks

by the way, for part c) Compute the radius of convergence and determine the interval of convergence of the series in b).

i do the radius of convergence with the equation x^n/n! that i figure from part b), just wondering if that's correct

 

[d_leet]

i do the radius of convergence with the equation x^n/n! that i figure from part b), just wondering if that's correct

That isn't the correct general term in the first place, and to find the radius of convergence the best way for this problem is either to notice the relation between the series you found and the geometric series or just use the ratio test.