# Sum operator

1. Jun 15, 2012

### Emilijo

I have general function
sum_(k=1, to n) f(k) = F(n)

ex.) sum_(k=1, to n) k = F(n)
solution: F(n) = (1+n)n/2

But If I have inverse problem?:
ex.) sum_(k=1, to n) f(k) = n^2
how to get f(k) ?

Generaly, If I have F(n), how to get f(k)
for sum_(k=1, to n) f(k) = F(n)

Is there some formula or method to solve this?

2. Jun 15, 2012

### chiro

Hey Emilijo and welcome to the forums.

The first thing you should ask yourself is whether a unique inverse for a function is guaranteed to exist. If you do not place constraints on your function, do you think that you will have a unique function? What does this imply about the inverse?

3. Jun 16, 2012

### Emilijo

For F(n)=n^2 in my example f(k)=2k-1,
because sum_(k=1,to n) (2k-1)= n^2 (wolfram alpha)
but what if I had a more complicated function, ex. sin(n)*n^2
Is there some formula to get f(k)?
In other words what function I have to sum when k goes from 1 to n
to get for example sin(n)*n^2, or generally to get function F(n)

4. Jun 16, 2012

### chiro

The reason I asked you to consider whether the function is unique is because if its not, then you can never actually find a unique inverse for F(n). In general the answer is no which means that you won't be able to find a unique F(n) given the sum and the number of parameters in general.

As an example think of F(N) = ((N-1) MOD 3)-1 and G(N) = 1 - ((N-1) MOD 3). Now Calculate F(6) and G(6). F(6) = -1 + 0 + 1 + -1 + 0 + 1 = 0 and G(6) = 1 + 0 + -1 + 1 + 0 + -1 = 0. Both functions are completely different yet give the same answer.

Basically this has to do with the degrees of freedom you are given: the more degrees of freedom (in other words the higher the N) the more choices you can have and that translates eventually if you don't restrict things yourself to lots more functions.