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

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    What is an indeterminate equation?2011-05-08
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    @Qiaochu: Links added.2011-05-08

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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 $x$.

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    Thanks! I think they are very similar. They both seem to have functions as their solutions essentially.2011-05-08
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    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?2011-05-08
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    @Amy: Did you forget the link? I don't think the example is indeterminate, because there are not infinitely many solutions.2011-05-08
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    @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.2011-05-08
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    @Tim: the link I was going to add was one of your links (in your question). Are you going to argue that only a finite number of points lie on the parabola given by $f(x) = y = x^2$? And, as Qiaochu suggests, any function which has no *unique* solution (barring any additional information or constraints) is indeterminate (meaning the solution cannot be determined by the function alone). Note that it is possible for a functional equation to be indeterminate (a functional equation for which there is no unique function that satisfies it).2011-05-08
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    Just to be more rigorous (on my part)...lacking a unique solution isn't sufficient for an equation to be indeterminate. I should say that an equation is an indeterminate equation if there exist more than one solution. If no solution exists, it's unsolvable...or rather, fails to be an equation! E.g. $$-|y| = x^2: x,y \in \mathbb{R}$$ then I'm not sure what it's called, but certainly not an "indeterminate" equation. Intractable?2011-05-08
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    Actually, the example immediately above does evaluate to a unique (Boolean) value: *false*!2011-05-08
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    @Amy: The solutions of indeterminate equation $y=x^2$ and of functional equation $f(x)=x^2$ are essentially the same, except that the former is represented as a set of pairs, and the latter as a function.2011-05-08
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    @Qiaochu: The solutions of $x^2+y^2=1$ as an indeterminate equation and of $x^2+f(x)^2=1$ as a functional equation, are similar, although the former is represented as a set of pairs and the latter as a set of functions. For functional equation $f(x)=f(x+1)$, I can't find a corresponding indeterminate equation yet.2011-05-08
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    @Tim: the solutions to a particular equation may be packaged into a single function, but the solutions to a particular functional equation form a _set of functions._ For example, the functional equation $f(x) = f(x + 1)$ has solutions $f(x) = \sin 2 \pi x$ as well as $f(x) = \cos 2 \pi x$, which are two _different collections of $(x, f(x))$ values_ (and of course there are many other solutions as well). You are conflating solutions with sets of solutions. The set of solutions to an equation may well form something similar to a function, but for a functional equation the _individual_ solutions2011-05-08
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    Tim: $f(x) = x^2$ is simply different notation: it express the same thing as $y = x^2$. The solutions to the former are also ordered pairs (x, x^2), which are identical to the ordered pairs satisfying the latter (x, y). Simple substitution: (If are equal to the same thing, then they are equal to each other.) Since $$f(x) = x^2 \land y = x^2 \rightarrow f(x) = y$$ Seriously, if you cannot understand this, you need to be asking different questions.2011-05-08
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    ... are _themselves_ functions.2011-05-08
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    @Qiaochu: From the beginning, I have already understood the differences in what a solution is to each type of equation. I have been asking about connection between them, like what I did in my last two comments, finding the corresponding equation of the other type for an equation of one type, so that their solutions can be derived from each other in that obvious way.2011-05-08
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    @Tim: in general you can't solve functional equations this way. Solving functional equations requires very different techniques.2011-05-08
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    @Qiaochu: Can I say an indeterminate equation always has a functional equation corresponding to it in the way above, although not vice versa?2011-05-08
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    @Tim: no. Even in the example $x^2 + y^2 = 1$, it is not true that the set of all solutions forms a function of $y$ in terms of $x$, since $y$ is not uniquely determined by the value of $x$ except when $x = 1, -1$.2011-05-08
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    @Qiaochu: In that case, **the set of function solutions** to the functional equation can still be derived from those to the indeterminate equation.2011-05-08
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    @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.2011-05-08
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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"].

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    In a functional equation, you need to solve for a function.2011-05-08
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    @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.2011-05-08