##\int \frac{dx}{x} = ln(x) + C##, assuming x > 0.Jbreezy said:Homework Statement
I have to find the power series representation for integral (1/x) dx
Homework Equations
ln (1+x)
The Attempt at a Solution
This is very similar to ln(1+x) but I don't know if this helps me.
Jbreezy said:Is this ln(x) shifted one to the right? So maybe I can use what is already the power series for ln(1+x) = Ʃ (-1)^(n-1) (x^n)/n from n = 1 to ∞
so could I do ln(x) = [(-1)^(n) (x^(n+1)] / (n+1)
NO? Maybe I shifted it wrong?
Instead of writing "answers" show me some mathematics reasoning.Jbreezy said:OK so this
Ʃ (-1)^(n-1) (x^n)/n from n = 1 to ∞
Should be Ʃ (-1)^(n-2) (x^(n-1))/(n-1) from n = 1 to ∞
Right?
A power series is an infinite series of the form ∑an(x-c)n, where a and c are constants and x is a variable. It is a mathematical concept used to represent a function as a sum of terms with increasing powers of x.
An integral is a mathematical concept that represents the area under a curve on a graph. It is used to find the total quantity or value of a function over a given interval.
A power series can be used to represent an integral by using the Taylor series expansion of a function. This involves expressing the function as a sum of terms with increasing powers of x, which can then be integrated term by term to find the integral.
The power series for the integral of (1/x) dx is ∑(1/n)(x-c)n+1 + C, where C is a constant and n is a whole number. This series is derived from the Taylor series expansion of ln(x).
The power series for the integral of (1/x) dx converges for values of x within the radius of convergence, which is x > 0. This means that the power series is a valid representation of the integral for values of x greater than 0.