Yeah, I was aware of a direct relation to differentiation:
\( \frac{d}{dx} f(x) = e^{(\bigtriangleup \frac{d}{dx} f(x)) + x - f(x)} \)
And I've extended the definition of the series to:
\( \ln(x) = a + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{ne^{an}}(x-e^a)^n \)
And I've found a beautiful return:
\( \ln(x) = e + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{ne^{ne}}(x-e^e)^n \)
Converges for values as high as 30. My computer overflows before it stops converging.
I think my assumptions were correct. If this is true, I might find an infinite convergent power series for ln(x)
edit:
sadly, doesn't converge for values less than 1
\( \frac{d}{dx} f(x) = e^{(\bigtriangleup \frac{d}{dx} f(x)) + x - f(x)} \)
And I've extended the definition of the series to:
\( \ln(x) = a + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{ne^{an}}(x-e^a)^n \)
And I've found a beautiful return:
\( \ln(x) = e + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{ne^{ne}}(x-e^e)^n \)
Converges for values as high as 30. My computer overflows before it stops converging.
I think my assumptions were correct. If this is true, I might find an infinite convergent power series for ln(x)
edit:
sadly, doesn't converge for values less than 1
