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I was wondering, can we have an equation for any graph even the most zigzagged one?

Ben Millwood
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Bonjour
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  • define equation – Dan Rust Sep 04 '13 at 22:14
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    It depends on exactly what you mean by graph and equation. – mrf Sep 04 '13 at 22:14
  • There is someone that pick up a number randomly (1,20,8,-5,7..), we graph it, can we find its equation? Merci. – Bonjour Sep 04 '13 at 22:16
  • If we pick even a single number randomly (say by throwing a dart at a one-dimensional dartboard and measuring where it hits) can we describe with complete accuracy with some (finite) expression? Probably not. So I think one may as well ask the question in one dimension, and the answer will still be "no". – Trevor Wilson Sep 04 '13 at 22:31
  • Graph as in map into an xy plane, xyz plane, a 9 dimension space, or what? There is also the question of how well defined over what domain and range are you plotting here. If you are plotting over the Real or some other complete number system there are a lot of points to define here. – JB King Sep 04 '13 at 22:35

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There are many, many more graphs in the plane than there are formulas we can write down. In fact, since equations must be written in finitely many characters from a finite alphabet, the number of possible equations - or definite descriptions of any kind - we can write is a countable infinity. However, there are uncountably many functions with different graphs on the plane.

I would have a hard time, of course, describing to you in a definite way one of the uncountable multitude of graphs for which there exists no definite description. They're everywhere, though.

G Tony Jacobs
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  • In some sense, if you can describe their graph entirely, you probably have a formula. – Dan Rust Sep 04 '13 at 22:23
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    Yeah, and zig-zags really aren't the issue. We can't even describe all the horizontal lines, i.e., graphs of constant functions. – G Tony Jacobs Sep 04 '13 at 22:26
  • It seems quite unfair to say "I have a graph here, but I can't finitely describe it" but then say "I don't accept equations that can't be finitely described as candidates for matching this graph". – Ben Millwood Sep 04 '13 at 22:31
  • @BenMillwood What if the graph is a graph of some random physical process that we can approximate by measurement? (In this case I think we can still pretend that the "real" graph exists in some sense, even though with our finitary measurements we can only approximate it.) – Trevor Wilson Sep 04 '13 at 22:33
  • @BenMillwood, I'm not sure I understand what you mean by "unfair". When you say "equations that can't be finitely described," what kind of thing are you thinking of? Isn't an equation nothing more than a finite description of a relation among numbers? – G Tony Jacobs Sep 04 '13 at 22:36
  • @TrevorWilson: Then you'll have difficulty, but it won't be difficulty with turning graphs into equations: whatever you decide the graph is, you can turn it into an equation. I mean, no-one says "you can't add integers because you can't tell me what $2 + x$ is, where $x$ is $1$ if Goldbach's conjecture is true and $0$ otherwise." – Ben Millwood Sep 04 '13 at 22:37
  • @GTonyJacobs: I'm saying that I don't see the justification for restricting equations to having finite descriptions but not restricting graphs in this way. – Ben Millwood Sep 04 '13 at 22:38
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    @BenMillwood To me, a graph means the graph of a function (set theoretically, this is the same as a function itself.) That is, any subset of the plane that passes the vertical line test. There is no definability requirement. On the other hand, "equation" usually means something that can be expressed in a finite language. – Trevor Wilson Sep 04 '13 at 22:40
  • @BenMillwood, my "justification" is that I think I'm just using both words in the senses in which they're usually used. It's what Trevor just said. A "graph", the way we usually use the word, is just a subset of the plane. The collection of these has a cardinality. An "equation", the way we usually use the word, is a statement of finite length in a finite language. The collection of these has a cardinality. The cardinality is greater for the collection of graphs than for the collection of equations. That's really all I'm saying. – G Tony Jacobs Sep 04 '13 at 22:44
  • Neither of you are wrong to use words in this way, but I don't think you are using them in the way the question meant them. Moreover, I think the difference in describability requirements is purely conventional/notational, a reflection of the contexts in which we're interested in graphs and equations rather than any intrinsic properties of those things, and thus the objection here is not, in my view, mathematically enlightening. – Ben Millwood Sep 04 '13 at 22:52
  • What the OP meant is ultimately a question we can't settle without input from the OP. I like the question about cardinality that I tried to answer, but I also understand the question that your answer addresses. I guess that's why there's more than one of us, huh? :) – G Tony Jacobs Sep 04 '13 at 22:59
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    There's an interesting question in the comments, too - does $\forall x\exists! y R(x,y)\implies\exists F(x)(\forall x R(x,F(x)))$? This is a core foundational question, and not one that has a unique answer; different formal systems provide different answers, and some versions of this question are unknown (for instance, if $R$ is a polynomial-time predicate and $F$ is intended to be as well, then the question is a slight variation on the P=NP question...) – Steven Stadnicki Sep 05 '13 at 22:22
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If you specify any finite collection of points with distinct $x$ co-ordinates and say you want a graph that will go through them, there is an equation, a polynomial even, that will go through all your points. This is called an interpolating polynomial and is not too hard to construct. I wrote an article on interpolating polynomials that explains the process a bit.

If you draw a line on a page with a pencil (so the line has nonzero thickness), there is even a polynomial that stays within that pencil line, as long as it's reasonable (no gaps in the line, no going back on yourself, etc.). That is to say, any continuous graph can be approximated as close as you like (e.g. within a pencil stroke's thickness) with polynomials; one method of doing so is using Bernstein polynomials.

Hence, I suspect the answer to the question you were trying to ask is "yes".

But that's sort of besides the point. The important thing to understand is that "equations" can really be whatever you like. I can write an equation, if it pleases me, of the following form:

\[y = \text{the vertical distance of Graph A from the $x$-axis, at the point distance $x$ from the $y$-axis}\]

As long as I also show you what Graph A is, this is a statement of equality between two well-defined numeric expressions: an equation. So by definition, there is an equation associated with every graph.

You may be more interested in questions like "suppose I have a graph with property $P$, can I match it with an equation with property $Q$". The answers to these questions of course depend on what $P$ and $Q$ are, but as shown in the beginning of my answer, there are a lot of good results you can find.

Ben Millwood
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I would say the apt way of representing graphs are not equations, but functions (a function can reduce to an equation in x for 2d, if the function is smooth and continuous, by using some sort of spline interpolation or using nth order polynomial fit). On the other hand any curves in 2d plane atleast, you can write for the worst case, as discontinuous functions representing the graph, even if the function has a limit going to infinity at some point. But one may not want to do this if the graph is too complicated.

Vaidyanathan
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  • So are you saying the answer is yes or no? – Trevor Wilson Sep 04 '13 at 22:35
  • In principle as I said, you should be able to fit the curve to a nth degree polynomial. I hope I am not overlooking anything here. – Vaidyanathan Sep 04 '13 at 22:38
  • The equation $x^2 + y^2 = 1$ has a graph, but it's not a function, in the sense of $y$ being a function of $x$. – G Tony Jacobs Sep 04 '13 at 22:39
  • Also, there are graphs of functions - most functions in fact - that are discontinuous at every point in the domain. No amount of polynomial interpolation to any degree will get you that. – G Tony Jacobs Sep 04 '13 at 22:40
  • yes, i agree discontinuous functions cannot be represented by interpolation. There you can write it as piecewise continuous functions. My point was just to say that, equation may not be a right word to represent a graph. Also can you not represent a circle by 4 piecewise continuous functions, one in each quadrant? – Vaidyanathan Sep 04 '13 at 22:45
  • also can you not use fourier series to represent the kind of functions which are discontinuous?? – Vaidyanathan Sep 04 '13 at 22:47
  • You only need 2 functions to represent the circle. I'm not sure you got what I was saying about functions that are discontinuous everywhere. I'm talking about functions where every single real number is a point of discontinuity. There are no continuous bits anywhere, just a dot here, a dot there, a dot there. This can pass the vertical line test, no problem, but it's not piecewise continuous in any sense. – G Tony Jacobs Sep 04 '13 at 22:48
  • even piecewise continuous won't get you very far. most functions (in the technical sense) are not continuous. – Dan Rust Sep 04 '13 at 22:48
  • hmm.. okay.. sorry that i did not understand your argument earlier. yes, these cases like a set of random numbers which has no "correlation" with each other cannot be represented. – Vaidyanathan Sep 04 '13 at 22:50