Is it possible to find F(n), summation of the function f(k), where k=1,2...,n ( Is there some formula, general formula, for any f(k), of course if k is in domain of function)
sum_(k=1, to n) f(k) = F(n)
If it is not possible, why?
Is it possible to find F(n), summation of the function f(k), where k=1,2...,n ( Is there some formula, general formula, for any f(k), of course if k is in domain of function)
sum_(k=1, to n) f(k) = F(n)
If it is not possible, why?
No. The sum $\displaystyle f(1) + f(2) + \dots + f(n)$ is sometimes known as a Riemann sum. There is no general formula for a Riemann sum, other than approximating that sum with the integral of the function f(n) (assuming f is integrable).
There are definitely formulas for specific functions f, e.g. $\displaystyle f(n) = n$, $\displaystyle f(n) = n^2$, $\displaystyle f(n) = 3n+1$, $\displaystyle f(n) = \frac{1}{n(n+1)}$ etc. but there is no general formula.
But then how to solve this: sum_(k=1, to n) sin(1/k)
Is it possible, I put it into wolfram alpha, but I didn t get the solution.
Why is not possible to get it? Do exist conditions when the sum is possible to calculate, or is it just technical problem?
The sum $\displaystyle \sum_{k=1}^n\sin(1/k)$ is, of course, a well-defined function of n. However, in general there is no reason why some function has to be represented as a finite composition of a few basic functions and operations: addition, multiplication, sine, square root, etc. For example, the function that, given five coefficients of a quintic equation with the leading coefficient 1, returns the least real root of this equation, is also well-defined for all arguments. However, it is proven that this function cannot be expressed as a finite composition of the four arithmetic operations and roots of any degree.
I would also like to know sufficient, or, better, necessary and sufficient conditions for when a recurrence relation $\displaystyle x_{n+1}=g(x_n)$ (note that the sum $\displaystyle x_n=\sum_{k=1}^nf(k)$ satisfies $\displaystyle x_{n+1}=x_n+f(n+1))$ has a closed-form solution. Well, there are sufficient conditions, such as when a relation is a linear homogeneous recurrence relation with constant coefficients, but they are not very general.