- #1
Mr Davis 97
- 1,462
- 44
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?
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)
A Maclaurin expansion is a mathematical series that represents a function as an infinite sum of terms, starting with the constant term and including all of the derivatives of the function evaluated at 0. It is a special case of a Taylor series, where the center of the series is at 0.
A Maclaurin expansion is a special case of a Taylor series, where the center of the series is at 0. This means that the derivatives of the function at 0 are used to calculate the terms in the series. In a Taylor series, the center can be any value, not just 0.
The Maclaurin expansion of log(x) is log(x) = (x - 1) - \frac{1}{2}(x - 1)^2 + \frac{1}{3}(x - 1)^3 - \frac{1}{4}(x - 1)^4 + \cdots
The Maclaurin expansion of log(x) and log(x+1) are similar, as they both involve the logarithm function. However, the center of the series is different. The Maclaurin expansion of log(x) has a center of 0, while the Maclaurin expansion of log(x+1) has a center of 1. This means that the terms in the series will be calculated using the derivatives of the function at 0 for log(x), and at 1 for log(x+1).
The Maclaurin expansion of log(x+1) can be useful in situations where we need to approximate the value of log(x) for values of x close to 1. By using the Maclaurin expansion, we can calculate the value of log(x+1) and then subtract 1 to get an approximation for log(x). This can be particularly useful in calculus and other areas of mathematics where log(x) is commonly used.