Undergrad Why Is There No Maclaurin Expansion for Log(x) Compared to Log(x+1)?

[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?

 

[FactChecker]

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.

 

[Mr Davis 97]

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]

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]

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]

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

 

[FactChecker]

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]

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