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Here is a nice graph representing the solution: wolframalpha. I wish to draw such a graph myself but don't have any idea which methods exist and which of them are more appropriate for equations of this kind.

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we can write it as sin(x)-sin(y)-sin(x*y)=0 and use some numerical approximation method newton's method or something like this. first trivial solution is x=y=0 –  dato datuashvili Aug 25 '11 at 6:35
    
@user3196, nice! But to draw a graph I need many solutions. How can I find them? By choosing different starting points? –  max taldykin Aug 25 '11 at 8:03
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A very nice graph. Whose artist's hand is this? (Well, I'm joking... but not only joking...) –  Gottfried Helms Aug 25 '11 at 10:45
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You'll want to look into Jeff Tupper's thesis on the matter of graphing implicit Cartesian equations; his software GrafEq is a particular implementation of his ideas. –  J. M. Aug 25 '11 at 16:52
    
@J. M.: thanks, it seems quite useful (but first I need to read hundred of pages to understand interval arithmetic). I wish I could accept your answer. –  max taldykin Aug 26 '11 at 6:33

3 Answers 3

up vote 3 down vote accepted

The standard numerical method for plotting an implicit function graph, say $f(x,y) = 0$, is to evaluate $f$ at points on a triangular grid, see which cells of the grid have a change of sign, and use linear interpolation to approximate the function in each cell. Thus if you have three grid points $A$, $B$, $C$ with $f(A) = 1$, $f(B) = -1$ and $f(C) = 2$, you would draw a line segment from the midpoint of $AB$ to the point $1/3$ of the way from $B$ to $C$.
More sophisticated software might sample more points in a cell once a sign change has been detected, or use quadratic interpolation.

Another approach would be to numerically solve a system of differential equations such as $\frac{dx}{dt} = \frac{\partial f}{\partial y}$, $\frac{dy}{dt} = - \frac{\partial f}{\partial x}$, starting at an initial solution. This would run into trouble at points where both partial derivatives are $0$ (which typically happens at self-intersections of the curve). Also, it's hard to know a priori how many initial solutions to look for (one in each component of the curve).

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thanks! can you give some references to more detailed description of these methods, or maybe their names (which I can google for)? –  max taldykin Aug 25 '11 at 8:23
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@max, see math.stackexchange.com/questions/32528/… –  lhf Aug 26 '11 at 18:59
    
@lhf, great links! (definitely I need to understand interval arithmetic) –  max taldykin Aug 26 '11 at 20:00

Note that the plot from Wolfram Alpha isn't particularly good; you can see a lot of jagged artifacts all over the place, presumably due to insufficient initial sampling. Here's a better one from Maple, made using

plots:-implicitplot(sin(x)+sin(y)-sin(x*y), x=-5..15, y=-10..10, grid=[100,100], gridrefine=2);

Implicit plot of sin(x)+sin(y) = sin(xy)

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The figure shows that there is an arc through the origin which near $(0,0)$ is well approximated by the hyperbola $x+y=x y$. For this arc (a tiny portion of the whole image!) you can recursively compute a Taylor expansion as follows: Write $$x(t):=t +\sum_{k=0}^\infty a_{2k} t^{2k}\ ,\qquad y(t):=x(-t)$$ (note that I have chosen a particular parametrization) and plug this "Ansatz" into the equation $\sin\bigl(x(t)\bigr)+\sin\bigl(y(t)\bigr)=\sin\bigl(x(t)y(t)\bigr)$. Using the Taylor expansion of $\sin$ and comparing coefficients for all even powers of $t$ the $a_{2k}$ can now be determined one for one. Doing this with Mathematica I have obtained the following first coefficients: $$x(t)=t - {t^2\over2} - {t^4\over8} + {t^6\over12} - {319 t^8\over5760} - {541 t^{10}\over26880}+\ldots\quad .$$ Now make a parametric plot of $t\mapsto\bigl(x(t),x(-t)\bigr)$ for $|t|\leq 1$, say, and you will see the arc in question.

I have told this story because it gives an "analytical" approach to such problems and is radically different from the ideas brought forward in the other comments and answers.

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sorry, but I don't see how this answers my question. As I understand you are trying to represent original equation in parametric form by comparing two Taylor expansions. But wolframalpha draws something completely different for resulting equation. –  max taldykin Aug 26 '11 at 6:20
    
@max taldykin: I have edited my answer. Perhaps the idea has become clear now. –  Christian Blatter Aug 26 '11 at 13:36
    
it seems that this method is somehow similar to the second method suggested by Robert Israel. thanks for your answer, it gives some perspective from analytical viewpoint. –  max taldykin Aug 26 '11 at 20:05

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