# Exploring Maclaurin Expansions: Log(x) vs Log(x+1)

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• Mr Davis 97

#### Mr Davis 97

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?

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?

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.

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.

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

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.

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