# how many points do I need to uniquely determine a nonlinear monotonically increasing function?

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 $$i=1,\dots, k$$.

• 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. Oct 31 '18 at 22:45
• 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 $a_2$. Oct 31 '18 at 22:47
• 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. Oct 31 '18 at 22:50
• @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.
– Iván
Oct 31 '18 at 23:15
$$f(x,a_1,a_2,a_3)=x+a_1+a_2+a_3$$ 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.
• 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. Nov 1 '18 at 0:35