Solve Inverse Problem: Find f(k) for sum_(k=1, to n) f(k) = F(n)

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Discussion Overview

The discussion revolves around the inverse problem of finding a function f(k) such that the sum from k=1 to n of f(k) equals a given function F(n). Participants explore specific examples, general methods, and the implications of uniqueness in the inverse function.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents the problem of finding f(k) given F(n) = n^2 and suggests that f(k) could be 2k-1, as this satisfies the equation when summed from k=1 to n.
  • Another participant questions the uniqueness of the inverse function, suggesting that without constraints, multiple functions could yield the same sum, complicating the search for a unique f(k).
  • Concerns are raised about more complex functions, such as F(n) = sin(n)*n^2, and whether a general formula exists to derive f(k) from such functions.
  • Examples are provided to illustrate how different functions can yield the same sum, emphasizing the challenge of finding a unique inverse function.

Areas of Agreement / Disagreement

Participants express differing views on the uniqueness of the inverse function, with some suggesting that without constraints, a unique solution may not exist. The discussion remains unresolved regarding the existence of a general method for finding f(k) for arbitrary F(n).

Contextual Notes

Participants note the importance of defining constraints on F(n) to potentially identify a unique f(k). The implications of degrees of freedom in function selection are also discussed, indicating that more complex functions may lead to multiple valid inverses.

Emilijo
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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?
 
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Emilijo said:
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?

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?
 
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)
 
Emilijo said:
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)

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.
 

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