The discussion centers around the absence of a Maclaurin expansion for the function log(x) compared to the presence of such an expansion for log(x+1). Participants explore the implications of the function's definition and the conditions under which Maclaurin series can be applied.
Participants generally agree that log(x) cannot be expanded using a Maclaurin series due to its definition at x=0. However, there are multiple competing views regarding the choice of log(x+1) for expansion and the implications of using other forms.
Limitations include the dependence on the definition of the logarithmic function and the specific conditions under which the Maclaurin series is applicable. The discussion does not resolve the implications of radius and speed of convergence for the series.
So why does it have log(x+1), and not log(x+1/2) or log(x+2), for example?FactChecker said: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.
Mostly convenience, I suppose. Each of the other two expressions could also be expanded as Maclaurin series.Mr Davis 97 said:So why does it have log(x+1), and not log(x+1/2) or log(x+2), for example?
Expanding the Maclaurin series at x=0 would be trying to evaluate the log at negative numbers.mathman said:Log((x+1)/(x-1)) gives a series that can be used for any y=(x+1)/(x-1)