How can one find the dervative of a function

[epkid08]

How can I find the derivative of a function say, f(n) = ln(n), where n is a natural number?

In other words, how can find g(x) in the following equation:
f(x) + g(x) = f(x+1)

I know there isn't a general formula for this, but are there some techniques to find g(x) for a specific equation f(x)?

 

[HallsofIvy]

How can I find the derivative of a function say, f(n) = ln(n), where n is a natural number?

In other words, how can find g(x) in the following equation:
f(x) + g(x) = f(x+1)

I know there isn't a general formula for this, but are there some techniques to find g(x) for a specific equation f(x)?

If you mean the domain of f is the set of natural numbers, then there is no derivative. The domain has to be a continuous set in order to define a derivative.

As far as finding g(x) in f(x)+ g(x)= f(x+1) is concerned, what's wrong with the obvious g(x)= f(x+1)- f(x)? Unless you have more information on f(x) I don't know what else you want.

Are you, perhaps, talking about "finite differences"? The finite difference \Delta f is defined as f(n+1)- f(n) for functions defined only on the natural numbers or, more generally, as (f(x+h)- f(x))/h for h a fixed, non-zero, real number. Still the only way to calculste f(n+1)- f(n) is to evaluate f(n+1) and f(n) and actually do that algebra.

 

[epkid08]

If you mean the domain of f is the set of natural numbers, then there is no derivative. The domain has to be a continuous set in order to define a derivative.

As far as finding g(x) in f(x)+ g(x)= f(x+1) is concerned, what's wrong with the obvious g(x)= f(x+1)- f(x)? Unless you have more information on f(x) I don't know what else you want.

Are you, perhaps, talking about "finite differences"? The finite difference \Delta f is defined as f(n+1)- f(n) for functions defined only on the natural numbers or, more generally, as (f(x+h)- f(x))/h for h a fixed, non-zero, real number. Still the only way to calculste f(n+1)- f(n) is to evaluate f(n+1) and f(n) and actually do that algebra.

I guess technically I was talking about "finite differences," but I wanted to find it in terms of the actual derivative of the function. Here's an easy example:

f(n) = n^2, n ∈ ℕ
f(n+1)=f(n)+\frac{f'(n)+f'(n+1)}{2}

The finite difference here is expressed through the derivative of the function. I realize that this would readily cancel out, just like f(n) - f(n) + f(n+1) = f(n+1) would, as it would with most polynomial functions, but I'm wondering that for a function like ln(n), there exists the finite difference in terms of f'(n) where it wouldn't readily cancel out.

If we take for example, f(n) = ln(n), a simple guess could be made about the finite difference in terms of f'(n):

f(n+1)=f(n)+\frac{f'(n)}{f'(f(n))}

though probably not true.