B^A = e^(A * lnB)
A^B = e^(B * lnA)
B^A = e^(lnB * A)
A^B = e^(lnA * B)
houserichichi wrote:I haven't done the actual working out of the matrices lnA and lnB, but I suspect these results to be accurate as matrix multiplication is noncommutative in general.
ln([ a b ] )
[ c d ]
=
p [(a-d) 2b ] +q [1 0]
[ 2c (d-a)] [0 1]
where p = ln{ (a+d+delta)/(2k) } / delta
q= ln(k)
delta = sqrt( (a-d)^2 +4bc )
k= sqrt( ad - bc )
houserichichi wrote:Since you have found two different values for A^B and B^A each you've just shown that the operation (^) isn't well-defined in this context so it's really of no use, unfortunately
A binary operation f(x,y) on a nonempty set A is a map f: AxA->A such that
1. f is defined for every pair of elements in A, and
2. f uniquely associates each pair of elements in A to some element of A.
= sum k [ sum j [ (k^j )/( j! k! ) *x^j *(lnx)^(k+j) ]]
x tetra n
= sum k1 [ sum k2 [ ... sum kn [
(k1^k2 k2^k3 ... k(n-1) ^kn) / (k1! k2!...kn! )
*x^kn*(lnx)^(k1+k2+...kn)
houserichichi wrote:[...]Anywho, it's not a well-defined binary operation because it (generally) gives two different answers for any particular A and B, although when A and B commute, that is [A,B] = 0, I suspect (and again, I'm too lazy to work out the arithmetic, but be my guest) that the results would be unique.
If it helps any, the square root function is not well-defined in general either - any real number has two square roots (a positive one and a negative one), but when working with just the positive reals it's all fine and dandy. There are tricks to get out of that mess, but they're not important in this discussion. Just wanted to throw another example at you in case it helped in understanding what "well-defined" meant.
jinydu wrote:Actually, I don't think its really correct to talk about the "period" of the natural logarithm, because I think "period" normally refers to the independent variable (x), not the dependent variable (f(x) or y, depending on your notation).
If z = log x, then z + 2n*pi*i is also = log x, for any integer value of n. This follows from Euler's famous formula
e^(iy) = cos y + i * sin y
and the fact that both cos and sin repeat with a period of 2pi.
In fact, even when you take the log of a positive real number, you still get infinitely many values. Its only when you completely restrict yourself to real numbers, not allowing non-real numbers in either the domain or the range, that the natural log is a single-valued function.
jinydu wrote:That's true in general for any complex function; you can see the whole thing if you make a 4D graph. Making a computer projection of the 4D graph is probably not difficult (in fact, I would think that someone has already done it). The problem is that our brains will probably just interpret it as a jumble of lines.
PWrong wrote:Can't you view a complex function as a 2D surface in a 3D graph? :?
PWrong wrote:Can't you view a complex function as a 2D surface in a 3D graph? :?
Anyway, I think I've found another potentially useful idea.
Think of tetration as a reccurence relation. (Sorry about the lack of subscripts)
y_(n+1) = e^(y_n)
The difference between terms is y_(n+1) - y_n = e^y - y
Now since we're trying to extend tetration to reals, I thought it might be useful to turn this into a differential equation.
dy/dx = e^y - y
I don't think the solution for this is the same as tetration, but it's interesting anyway.
I haven't formally learned how to solve differential equations, but I've read about seperable equations from my textbook. It turns into:
integral dy/(e^y - y) = x + c
integrating 1/ (e^y - y) is the hardest part. Dividing the top and bottom by e^y, we get e^-y / ( 1-ye^-y)
let u = y e^-y, then it becomes e^-y * 1/(1-u), which turns into
e^-y * sum[ u^n ], for n=0 to infinity
= e^-y * sum[y^n e^-ny ]
= sum[y^n e^-(n+1)y ]
Now I didn't know how to integrate this, but the Integrator website came up with this:
-(n+1)^-(n+1) * Gamma[n+1, (n+1)y]
where gamma is the incomplete gamma function.
So we sum from n=1 to infinity instead of n=0, and get:
-sum [ Gamma [n,ny] / n^n ] = x
I like this formula, but unfortunately it gives x in terms of y, and I have no idea how to invert it.
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