I was wondering what differences and relations are between indeterminate equation and functional equation? Are they the same concept? Thanks and regards!

  • $\begingroup$ What is an indeterminate equation? $\endgroup$ – Qiaochu Yuan May 8 '11 at 2:38
  • $\begingroup$ @Qiaochu: Links added. $\endgroup$ – Tim May 8 '11 at 2:41

They are very different concepts. An indeterminate equation is just one with infinitely many solutions. A functional equation is an equation where the thing you want to solve for is a function, rather than a variable.

"Indeterminate equation" doesn't strike me as particularly useful terminology anyway. My guess is it originates from linear algebra, where if a system of linear equations doesn't have a unique solution, it has infinitely many solutions. But for non-linear equations it can happen that there are finitely many solutions, but more than one. For example, the equation $(x - 1)(x - 2) = 0$ has two solutions $x = 1, x = 2$.

The main difference between a functional equation and an ordinary equation is what the quantifiers are. In an ordinary equation, the claim is something like "decide whether there exists $x$ such that (some equation involving $x$)." In a functional equation, the question is "decide whether there exists a function $f$ such that (some equation involving $f$ and some other variables) for all values of the variables." For example,

$$f(x) = f(x + 1)$$

is the functional equation describing functions with period $1$. The equation is not just a collection of symbols: it comes with a quantifier requiring that the above relation hold for all values of $x$.

  • $\begingroup$ Thanks! I think they are very similar. They both seem to have functions as their solutions essentially. $\endgroup$ – Tim May 8 '11 at 2:49
  • $\begingroup$ Sorry, Qia0chu, I was composing while you posted, otherwise I would have only commented with the link I give in my answer. So an indeterminate function may not necessarily have an infinite number of solutions, but more generally, it encompasses functions that have no unique solution. In the example I give, that would mean that e.g. $f(x) = x^2 = 4$ is still indeterminate, since the solution (for x) is not unique? $\endgroup$ – Namaste May 8 '11 at 2:57
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    $\begingroup$ @Tim: I don't understand your comment. For example, the equation $x^2 + y^2 = 1$ has infinitely many pairs $(x, y)$ of real solutions, but $x$ and $y$ are variables, not functions. $\endgroup$ – Qiaochu Yuan May 8 '11 at 3:28
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    $\begingroup$ @Tim: in general you can't solve functional equations this way. Solving functional equations requires very different techniques. $\endgroup$ – Qiaochu Yuan May 8 '11 at 4:33
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    $\begingroup$ @Tim: sure, if you want to think about it that way, but I don't really see the point of this particular change in perspective. Functional equations are generally harder to solve than ordinary equations. $\endgroup$ – Qiaochu Yuan May 8 '11 at 5:24

No, they aren't the same concept.

An indeterminate equation is an equation for which there are an infinite number of solutions; that is, there is not enough information within the equation itself to solve the equation, were it not for a "given" value at which to evaluate the function.

E.g. $y = x^2$ is satisfied by ordered pairs $(0,0),(1,1),(-2,4),(2,4)\dots (x,x^2)$.

A solution to a system of equations can also be indeterminate: e.g., the system of two equations, each in three variables, say $x$, $y$, and $z$, is indeterminate.

A functional equation is a function which is defined implicitly in terms of a function or in terms of the function at some value. For a more thorough explanation and examples, reread the links you provide in your question.

[Edited:] Seriously, as Qiaochu states, they are very different. In a functional equation, you need to solve for a function . The functional equation may very well also be an indeterminate equation, in that it admits of more than one solutions (all functions). [Thanks to Qiaochu for pointing out my earlier mis-statement of a functional equation; I've since replaced "equation" with "function"].

  • $\begingroup$ In a functional equation, you need to solve for a function. $\endgroup$ – Qiaochu Yuan May 8 '11 at 3:37
  • $\begingroup$ @Qiaochu: Yes, of course: one indeed aims to solve for a function when given a functional equation; all the "equation" talk -> "slip of the tongue". I've corrected my answer. $\endgroup$ – Namaste May 8 '11 at 3:53

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