- #1
Jhenrique
- 685
- 4
Accidentally I wrote in the wolfram f(x) = f(1/x) the the wolfram give me the solution for this equation (f(x) = Abs(log(x))). Hummmm, nice! Thus I thought: given the definition of derivative, ##f'(x) = \frac{f(x+dx)-f(x)}{dx}##, is possible to isolate f(x) in this equation? If yes, how?
I realized that exist a pattern for the expression of derivative: [tex]\\ f^{(1)}(x) = \frac{f(x+dx)-f(x)}{dx} \\ \\ f^{(2)}(x) = \frac{f(x+2dx) - 2 f(x+dx) + f(x)}{dx^2} \\ \\ f^{(3)}(x) = \frac{f(x+3dx) - 3 f(x+2dx) + 3 f(x+dx) - f(x)}{dx^3} \\ \\ f^{(4)}(x) = \frac{f(x+4dx) - 4 f(x+3dx) + 6 f(x+2dx) - 4 f(x+dx) + f(x)}{dx^4}[/tex] So maybe could exist a pattern too for f(-1), f(-2), f(-3)... But I haven't sure if is possible to develop this pattern retroactively, I think that would be necessary to define binomial coefficients for negative argument... In the end, what you think about?
I realized that exist a pattern for the expression of derivative: [tex]\\ f^{(1)}(x) = \frac{f(x+dx)-f(x)}{dx} \\ \\ f^{(2)}(x) = \frac{f(x+2dx) - 2 f(x+dx) + f(x)}{dx^2} \\ \\ f^{(3)}(x) = \frac{f(x+3dx) - 3 f(x+2dx) + 3 f(x+dx) - f(x)}{dx^3} \\ \\ f^{(4)}(x) = \frac{f(x+4dx) - 4 f(x+3dx) + 6 f(x+2dx) - 4 f(x+dx) + f(x)}{dx^4}[/tex] So maybe could exist a pattern too for f(-1), f(-2), f(-3)... But I haven't sure if is possible to develop this pattern retroactively, I think that would be necessary to define binomial coefficients for negative argument... In the end, what you think about?