What are the methods for finding the points of symmetry/antisymmetry of a function?

I think this question needs to be restricted in scope for any hope of a complete response. I think we should first consider only 2D functions on the x,y plane, and consider only 1) mirror symmetry about any line, and 2) symmetry about any point, which is the same as self-congruence about a 180 degree rotation. Once we have the basics, we can try to generalize to higher order functions.

The idea of "local symmetry" is too unclear at this point to go further with this, and maybe deserves its own Q&A. Any point of any function can be approximated by a power series, and so if the coefficients disappear for the x, x^2, x^3,.. terms, we then end up with either a mirror or rotation symmetry, depending on the remaining dominant term's power being even or odd. But I am not sure if this is good enough for you. For example, the function x + x^2, would you say that it has both "local symmetry" and "local antisymmetry" about the point (0,0)? Clarify?

I'll get back to this later.

Edit: Back. Okay, like I said, let's break this problem down into smaller pieces, instead of all at once. Right now, I'll just address the question of existence of symmetries for 2D functions, which I will express in implicit form, f(x,y) = 0, which is the most general way. There are only 2 kinds of 2D symmetries, which is 1) point symmetry, and 2) line symmetry.

1) Point symmetry: This is actually very easy. If for any particular (a,b) (which is the point of the symmetry):

f( x, y ) = f( -(x-a), -(y-b) )

then there is a point symmetry about (a,b).

2) Line symmetry: This is more complicated, because the line of symmetry can be at any angle and anyplace. If for any particular a and b, a being angle and b being displacement:

f( x, y ) = f( b + bCos(2a) - Cos(2a)x + Sin(2a)y, bSin(2a) - Sin(2a)x + Cos(2a)y )

then there is a line symmetry about the line at angle a from the x axis, passing through it at x = b. This is a general formula that will work for even simple symmetries such as for y = x², where a = b = 0. As an example, consider the implicit expression:

x² - 3x + y² + y - 2xy + 2 = 0

Letting a = - 45° and b = 1 and working it all out, you'll get exactly the same expression as before. This proves that there is a line of symmetry at 45°, intersecting the x axis at x = 1.

Unfortunately, FINDING the values of (a, b) in the case of point symmetries, and a, b in the case of line symmetries is altogether another matter. There is no systematic way of determining them that will work for any and all implicit functions.

Generalizing this to higher dimensions brings in many more forms of symmetries that are possible, which enormously complicates the job, and probably is worthy of a paper by itself. I think there is a version which is x, y symmetric, but that's another chore to work out.

Edit 2: It should be noted that the expression given for testing line symmetry isn't the only suitable one. There are others similar to it that will work just as well. I think there's a version which is x, y symmetric, but that's another chore to work out.

Edit 3: How to derive "the formula" for line symmetry testing: 1) shift the function along x axis by b, so that line of symmetry passes through origin. 2) rotate function counterclockwise by angle a so that line of symmetry is coincident with y axis. 3) flip function about y axis 4) rotate function clockwise by angle a. 5) shift funciton back along x axix by b. Combine the following steps:

1)..x2 = x1 - b

.....y2 = y1

2)...x3 = Cos(a) x2 + Sin(a) y2

......y3 = -Sin(a) x2 + Cos(a) y2

3)...x4 = -x3

......y4 = y3

4)...x5 = Cos(a) x4 - Sin(a) y4

......y5 = Sin(a) x4 + Cos(a) y4

5)...x6 = x5 + b

......y6 = y5

to get the final formula. Like I said, this isn't the only possible test for line symmetry, but it does generally look like this anyway. Steps 2) and 4) are the classic rotation matrixes, which are of the form:

....(Cos(a)....Sin(a))

....(-Sin(a)...Cos(a))

Look up "Rotation Matrix". As a matter of fact, these are all part of "symmetry transformations", the study of which is a huge part of not only mathematics, but physics as well. Symmetry plays a central role in the laws of physics, where for every symmetry there is, there is an associated conservation law, such as conservation of energy. Look up Noether's Theorem. Gauge symmetry is the foundation of the Standard Model of fundamental particles. The concept of symmetry is now generalized beyond basic geometric transformations, to wherever there exists a transform H() such that for any arbitrary function or object A, if H(A) = A, i.e., leaves it unaffected, then it's said that there exists a symmetry. This is an important and extremely powerful idea in both mathematics and physics, in widespread use. Also, look up "Lie Groups" as one example of "local symmetry". Y!A is cutting me off here now.

Could you clarify exactly what you mean by a symmetric function in n variables? In n variables, a symmetric function usually means once that is equal to itself if you permute the variables in any way. (Presumably, an antisymmetric function according to this definition would be one that is equal to ±1 times itself depending on whether the permutation of the variables is "even" or "odd".)

But you're grouping this discussion together with a question about functions of one variable being symmetric/asymmetric about a point, and this is related to the discussion in your other question, but is not related to the concept above.

Anyway... with one variable, an antisymmetric function has to be equal to 0 at the point of antisymmetry. Indeed, if c is the point of asymmetry of f(x), we then have for all x (or, in the locally antisymmetric case, all x close enough to c):

f(x) = -f(2c-x)

So:

f(c) = -f(2c-c) = -f(c) => f(c) = 0

So if you can find the zeros of f, this gives you a starting point.

Similarly, if f(x) is symmetric about c, then:

f(x) = f(2c-x)

Taking derivatives, we have:

f'(x) = -f'(2c-x)

i.e., f'(x) is antisymmetric about c. So if you can find the zeros of the derivative, you get a starting point for finding the points of symmetry.

Let me think about this some more.

because of the fact the x values are calmly spaced, there's a shortcut to looking the coefficient A in the quadratic equation y = Ax^2 + Bx + C. After that the gadget of equations is rather consumer-friendly to resolve for B and C. Take the modifications between the y-values (a million, 2) Take the version between those modifications (a million) Divide by way of the undemanding distinction between the x values (a million). that provides you A. So instead of 0 = A + B + C a million = 4A + 2B + C 3 = 9A + 3B + C ... you at present have: 0 = a million + B + C a million = 4 + 2B + C (you do not elect a 0.33 equation for the reason that there are purely 2 variables) At this element you will get rid of C by way of subtraction.