Taylor Series of log(1-t)

vjj

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Mar 3, 2012
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Hello
How do you take the geometric/taylor (unsure which it is) series of log(1-t)?
I do already have a result given to me of 0 -t -(t^2)/2 -(t^3)/3 -(t^4)/4 -...
But don't know where its come from?
Please help
Also, I need to find log(1-t)-log(1-2t) in that form but getting different answers with my friend so would like some reassurement!
is it just (-t -(t^2)/2 -(t^3)/3 -(t^4)/4 -...) - (-2t -((2t)^2)/2 -((2t)^3)/3 -((2t)^4)/4 -... ?
 
Hello
How do you take the geometric/taylor (unsure which it is) series of log(1-t)?
I do already have a result given to me of 0 -t -(t^2)/2 -(t^3)/3 -(t^4)/4 -...
But don't know where its come from?
Please help
Also, I need to find log(1-t)-log(1-2t) in that form but getting different answers with my friend so would like some reassurement!
is it just (-t -(t^2)/2 -(t^3)/3 -(t^4)/4 -...) - (-2t -((2t)^2)/2 -((2t)^3)/3 -((2t)^4)/4 -... ?

Do you know the expansion of \(\displaystyle \frac{1}{1-t}\)?

Can you see a way to get from that expansion to expansion of loge(1-t)?
 
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