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# mcclaurin series fo log(1-x) Watch

1. I can do it for log(1+x) just fine.

for the first 3 terms, I found the coefficients were 0, -1 and -1

when i plug this into the taylor series formula

I end up with -x -(x^2)/2

apparently it's meant to be like

-1-x-x^2-x^3 etc
2. (Original post by f45)
I can do it for log(1+x) just fine.
then substitute -x into x
log(1-x)=log(1+(-x))
for the first 3 terms, I found the coefficients were 0, -1 and -1
I think it to be 0, +1, -1/2 because the nth coefficient is
3. (Original post by ztibor)
then substitute -x into x
log(1-x)=log(1+(-x))

I think it to be 0, +1, -1/2 because the nth coefficient is
I make it 0,-1,-1/2,-1/3,...

That is etc.
4. (Original post by Get me off the £\?%!^@ computer)
I make it 0,-1,-1/2,-1/3,...

That is etc.
Yep, I agree:
5. (Original post by Get me off the £\?%!^@ computer)
I make it 0,-1,-1/2,-1/3,...

That is etc.
OK, but I wrote the coefficients of log(1+x)
as OP wrote he could work out this.
6. (Original post by ztibor)
OK, but I wrote the coefficients of log(1+x)
as OP wrote he could work out this.
Surely.

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