Whats the best way to express this...?

[NotLKH]

I'm at a loss here.

I'm trying to describe the following, but I'll use an example to illustrate.

Given variables X, Y, and Z.

You have an equation, E1, such that

E1 = X²Y + X²Z + Y²X + Y²Z + Z²X + Z²Y

As you can see, the right hand side is the summation of all combinations of 2 elements taken from the group (X,Y,Z) Where one element is raised to the 2nd power, while the second element is to the first.

So, I'd like to ascribe a symbolic description to this, something along the lines of

E1 = Sum{X,Y,Z}^2,1,0
Of course, the 0 is superfluous, so this could also be said to be identical to:
E1 = Sum{X,Y,Z}^2,1

However, is this the best way? How would you do it?


Next:

lets say you had the following:

E1 = Sum{X₁,X₂,X₃}^1,1

and E2 = E1, thus
E1 x E2 = Sum{X₁,X₂,X₃}^2,2 + 2*Sum{X₁,X₂,X₃}^2,1,1

How would you approach developing a program to input E1 and E2, and outputing their product?

[kedaman]

your generic polynom form would be
Sum(x=0 to infinity)Sum(y=0 to infinity)Sum(z=0 to infinity) X^x + Y^y + Z^z
for instance X^2Y can be written X^2+Y^1+Z^0, 0'th power is not superfluous.

multiplying the polynoms, you have the cartesian product ex, (a+b) (c+d) = a(c+d) + b(c+d) = ac + ad + bc + bd, using distributivity * over +

[NotLKH]

Originally posted by kedaman
for instance X^2Y can be written X^2+Y^1+Z^0, 0'th power is not superfluous.

I agree in your instance, you are right. However, in the system of equations I'm building, X²+Y¹+Z⁰ never occurs.



Where I said:

E1 = Sum{X,Y,Z}^2,1,0
Of course, the 0 is superfluous, so this could also be said to be identical to:
E1 = Sum{X,Y,Z}^2,1

Sum{X,Y,Z}^2,1,0 expands to:

X²Y¹Z⁰+X²Z¹Y⁰+Y²X¹Z⁰+Y²Z¹X⁰+Z²X¹Y⁰+Z²Y¹X⁰

and

Sum{X,Y,Z}^2,1 expands to:

X²Y¹+X²Z¹+Y²X¹+Y²Z¹+Z²X¹+Z²Y¹

Which we see then that Sum{X,Y,Z}^2,1,0 = Sum{X,Y,Z}^2,1, when any general K⁰ is defined to equal 1.

This was why I said that the 0 was superfluous.

-Lou

[NotLKH]

Of course, this might be a better notation:

if we have a set of numbers : {0, 1, 2}, and all the permutations
of this set of 3 elements taken 3 at a time can be noted as:

₃{0,1,2}₃

which physically expands to:

{0,1,2}
{0,2,1}
{1,0,2}
{1,2,0}
{2,0,1}
{2,1,0}


then perhaps I could say that,
Sum(X,Y,Z)^2,1,0 should really be expressed as:

[XYZ]^{(₃{0,1,2}₃)}

?
-Lou

[kedaman]

I see now, I wasn't familiar with your notation so i just assumed you wanted to have a generic polynom form

You seem to have the summed the permutations of (X,Y,Z) passed to a function F(x,y,z) = X^x Y^y Z^z
if this is so you could write it
f(i,j,k)=å"sÎ{(x,y,z) Î Sym({i,j,k})| (X^x Y^y Z^z)})

[NotLKH]

Originally posted by kedaman
I see now, I wasn't familiar with your notation so i just assumed you wanted to have a generic polynom form

You seem to have the summed the permutations of (X,Y,Z) passed to a function F(x,y,z) = X^x Y^y Z^z
if this is so you could write it
f(i,j,k)=å"sÎ{(x,y,z) Î Sym({i,j,k})| (X^x Y^y Z^z)})

Now we're getting somewhere.


Alrighty then. Useing your notation, how would you express
the expansion of:

(X₁ + X₂ + X₃)³

The notation I used to use can be seen here:



And goal was to have a notation that I could verbally and consistantly explain, and which would reduce the amount of pencil lead used to notate such formulas.

As you can see, my old notation Reduced the amount written out significantly {If I didn't add a mapping for everything I wrote :laugh: } , but I always had a tough time trying to express what it meant, {still do!}

[kedaman]

damn, that's some way to make it more complicated but make it look less complicated
did you come up with the ¥ notation just for your own use? In that case you could stick to it, I think its quite consise

[A$$Bandit]

You could (and I would normally) say that:

(x + y + z)³ = Sum(x³) + Sum(x²y) + Sum(xyz)

where Sum() can be considered to be the greek capital sigma. This can be literally translated as:

The cube of (x + y + z) = the sum of all things like x³ added to the sum of all things like x²y added to the sum of all things like xyz.

In fact it's fair to say that:

{a + b + c + d + ...)³ = Sum(a³) + Sum(a²b) + Sum(abc).

Now whether that helps anyone at all I don't know...

[Alphanos]

It looks to me like you're just expanding trinomials, unless I'm missing something... a generic trinomial expansion equation you can use is:

(a+b+c)^n = SUM ( (n!/(x!y!z!)) * a^x * b^y * c^z)

for each x+y+z=n;
0<=x,y,z<=n;
x,y,z,n are elements of the whole number set.


You can also extend this for any desired polynomial expansion, just like this:

(a + b + c + d)^n = SUM ( (n!/(w!x!y!z!)) * a^w * b^x * c^y *d^z)

There are lots of interesting things about trinomials... instead of pascal's triangle, you have a tetrahedron, with three pascal's triangles running down the three sides (not including bottom) and interesting stuff in the center.

[NotLKH]

It sucks, that weekends are seperated by 5 days of actual work!


Bear with me, I've got a reason for this. 3 more days! { hopefully less}