Find the taylor series of ln(1+x)

AI Thread Summary
The discussion focuses on finding the Taylor series of ln(1+x) centered at zero. The derivatives of the function are calculated, revealing values at zero that include f(0) = 0, f'(0) = 1, f''(0) = -1, f'''(0) = 2, and f''''(0) = -6. Participants express confusion about the coefficients in the series, particularly the values 2 and 6, and seek to identify a general term for the nth expansion. The general term is suggested to be xn(-1)n+1/n!, with a hint that the coefficients follow a factorial pattern. The discussion concludes with a resolution to the confusion regarding the coefficients.
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Homework Statement


find the taylor series of ln(1+x) centered at zero



Homework Equations



from 0 to infinity ∑ cn(x-a)n

cn = f(n)(a)/n!



The Attempt at a Solution



f(x) = ln(1+x)
f'(x) = 1/(1+x)
f''(x) = -1/(1+x)2
f'''(x) = 2/(1+x)3
f''''(x) = -6/(1+x)4


f(0) = 0
f'(0) = 1
f''(0) = -1
f'''(0) = 2
f''''(0) = -6

^ I feel like there is a mistake somewhere in the drivative, because one i set out the taylor series, it doesn't make much sense to me

0 + x/1! -x2/2! +2x3/3! -6x4/4!
 
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There's no mistake. Can you see a pattern? Can you find the general equation for the nth term in the expansion?
 
ideasrule said:
There's no mistake. Can you see a pattern? Can you find the general equation for the nth term in the expansion?

I tried to develop an nth term, but the 2 and 6 are messing me up, that why I thought it was wrong.

general term xn(-1)n+1/n! starting at n=1

It should resemble something like that, but I am not too sure, again I cannot figure out how to develop a general term that will explain the 2 and 6
 
Note that 2=2*1 and 6=3*2*1. I bet the next coefficient will be 24=4*3*2*1 (check this!)
 
2=2!

6=3!

Look at the next couple of derivatives of f if you need to.
 
okay I got it thanks
 
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