Given monotonically increasing function $f(x,a_1,a_2, \dots, a_n)$ for $x\geq0$ where $a_i$ are the coefficients, and the function verifies that $$f(x,a_1,a_2, \dots, a_n)|_{x=0} = 0$$

Is there any rule to determine how many points $y_i$, $$y_i = f(x_i,a_1,a_2, \dots, a_n)$$ do I need to uniquely determine the function $f(x,a_1,a_2, \dots, a_n)$?

Is there any relationship between the number of coefficients and the number of points? The functional form is defined, but what I don't know is which are the coefficients. To make the situation even more complex the coefficients $a_i$ can be nonlinear (or the dependence of the function with the coefficients can be nonlinear).

To give two clear examples:

  • if $f(x) = ax$, with only one point $(x_1, y_1)$ I can determine the whole behaviour of $g(x)$ ($a=y_1/x_1$).
  • if $f(x) = e^{ax}-1$, again one points is enough $(x_1, y_1)$ ($a=\log((y_1+1)/x_1)$).
  • if $f(x)=ax+bx^2$ I know that two points $(x_1, y_1)$, $(x_2, y_2)$ are enough.

I do not pretend to find the function, but given a definition of a function $f(x)$ in terms of the variable $x$ and of the coefficients $a_i$ I want to say something like that if the function $f(x)$ is defined at let's say $k$ points then there is no other possible function $f(x)$ with the same functional form but different coefficients that verifies that $f(x_i)=y_i$ for $i=1,\dots, k$.

  • $\begingroup$ If you know the functional form and it uses $n$ arbitrary parameters then usually knowing the value at $n$ points is enough.The details can be subtle in each case. $\endgroup$ Oct 31 '18 at 22:45
  • $\begingroup$ What if $f(x;a_1,a_2)=a_1-a_2$? Then any number of points is not sufficient to recover $a_1$ and $a_2$. $\endgroup$
    – Federico
    Oct 31 '18 at 22:47
  • $\begingroup$ You are basically asking when a system of nonlinear equations $f(x_1;a_1,\dots,a_n)=y_1$, ..., $f(x_k;a_1,\dots,a_n)=y_k$ has a unique solution in $a_1,\dots,a_n$. This doesn't have a general answer. $\endgroup$
    – Federico
    Oct 31 '18 at 22:50
  • $\begingroup$ @Federico $f(x;a_1, a_2)$ should verify that $f(0)=0$ since I also asked for that, and in that case I think that one point is enough to point will give the value of $a_1=a_2$ and then f(x)=0. However a constant function is not an increasing function I think. $\endgroup$
    – Iván
    Oct 31 '18 at 23:15
  • $\begingroup$ @EthanBolker Where I can find more info about your answer? $\endgroup$
    – Iván
    Oct 31 '18 at 23:17

$$f(x,a_1,a_2,a_3)=x+a_1+a_2+a_3$$ where $a_1+a_2+a_3=0$ is not uniquely determined by $a_1, a_2, a_3$. It is increasing and linear, and no matter how many points you are given you can't determine the parameters.

  • 1
    $\begingroup$ The OP wants $f(0) = 0$ so this example doesn't work. But the idea does: set $f(x) = x + a + b + c$ with $a + b + c = 0$; you can't find the separate values of the three parameters. $\endgroup$ Nov 1 '18 at 0:35

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