# Taylor series

## Main Question or Discussion Point

I am looking at examples of Maclaurin expansions for different functions, such as e^x, and sinx. But there is no expansion for log(x), only log(x+1). Why is that?

## Answers and Replies

FactChecker
Science Advisor
Gold Member
The Maclaurin series uses the values of the function and it's derivatives at x=0. But log(x) is not defined at x=0.

The Maclaurin series uses the values of the function and it's derivatives at x=0. But log(x) is not defined at x=0.
So why does it have log(x+1), and not log(x+1/2) or log(x+2), for example?

Mark44
Mentor
So why does it have log(x+1), and not log(x+1/2) or log(x+2), for example?
Mostly convenience, I suppose. Each of the other two expressions could also be expanded as Maclaurin series.

FactChecker
Science Advisor
Gold Member
Issues to consider are the radius of convergence and the speed of convergence of the series. An expansion of log(x+1) around x=0 will give values for log(y), y=x+1 from y=0 to y=2.

mathman
Science Advisor
Log((x+1)/(x-1)) gives a series that can be used for any y=(x+1)/(x-1)

FactChecker
Science Advisor
Gold Member
Log((x+1)/(x-1)) gives a series that can be used for any y=(x+1)/(x-1)
Expanding the Maclaurin series at x=0 would be trying to evaluate the log at negative numbers.

mathman
Science Advisor
MacLaurin series at x = 0 has minus infinity as the constant term.