jinydu wrote:P.S. PWrong, you're 12 posts away from something big!
Hey, so I am.
I wonder what it'll be like...
jinydu wrote:It turns out that when working with complex number, the natural logarithm of a number has an infinite number of answers. See Equation 12 at
http://mathworld.wolfram.com/NaturalLogarithm.html (r represents the absolute value of z and theta is the argument).
As can be seen, the natural log is usually taken by setting n = 0, but in this case, doing so causes big problems. Thus, I instead tried taking n = 1. Now, I have a workable formula for tetrating to negative integers.
Hey, that's great! I'd pretty much given up on negative tetration until now. Using the other values of ln(0) is a great idea. So that would mean there are infinitely many values of x tetra -2, depending on the value of n.
I might have to alter my "top-function" ideas to account for this later. I think the idea will still work though.
Unbelievably, numerical evidence from Mathematica seems to indicate that e tetra -n converges to a finite complex number as n approaches infinity! This number has the approximate value:
0.318131505205 + 1.3372357014307i
That can be a new goal: Prove that (e tetra -n) really does converge, and try to find an exact expression for it!
Hey, I've seen that number before. I found it ages ago by taking the natural log of a number over and over. It makes sense according to your formula for e tetra -n
I think the simplest way to express it would be to say that it's a solution to e^x = x
this equation has more than one solution though, so e tetra -n can converge onto more than one spot. This also makes sense, because there are infinitely many values of e tetra n anyway.
jinydu wrote:Whoa, nice work. I've been able to confirm your formula for f''(x) for the case of f(x) = x^3. But why do you say that you can't find the nth derivative of e^f(x) in terms of f(x) itself? It seems like it should be workable.
Well, it might be possible to convert f(x+nh) into derivatives of f(x) with some juggling, but it seems like a roundabout way to do it. But anyway, I've found a formula on mathworld that will do the job.
http://mathworld.wolfram.com/FaadiBrunosFormula.html
It's quite complicated (it took me a while to figure out the whole "partition" thing), but it does what I wanted. It's also more general, as it finds the n'th derivative of f(g(t)) while I only wanted the n'th derivative of z^f(t). Having two different f(t)'s is potentially confusing, so I'll let the top-function be F(t). So we substitute f(u) =z^u, and g(u) = F(u), then use Faá di Bruno to expand z^F(0).
By the way, apparently D[sup]n[/sup] is shorthand for "the n'th derivative of". I hadn't encountered this notation before, so I thought I'd mention it.
Now I can theoretically write out an infinite system of equations, using the condition I already have for the top-function.
The condition as it stands is this:
D[sup]n[/sup](F(1)) = D[sup]n[/sup](z^F(0))
for all positive integers n
I can expand the derivatives of F(1) and F(0) using something like Taylor's theorum.
f(t) = a(0) + a(1) t + a(2) t^2 + ...
D[sup]n[/sup](f(t)) = sum [ k!/(k-n)! a[sub]k[/sub]t[sub]k-n[/sub] ]
Subsituting t=1, we get:
D[sup]n[/sup](f(1)) = sum [ k!/(k-n)! a[sub]k[/sub] ]
Subsituting t=0:
D[sup]n[/sup](f(0)) = n!a[sub]n[/sub]
Now, the left side is a sum to infinity, and the right side is in terms of the derivatives of f(0). Using Taylor's theorum again, but with t=0, we get
D[sup]n[/sup](f(0)) = n! a[sub]n[/sub]
Before I use this, I'll expand the right side using Faá di Bruno's formula. Unfortunately, Faá di Bruno's formula is way too big to put it here and simplify it, but it cancels down a bit once you subsitute the functions in. So I'll just show the final result, and you can try it yourself, or take my word for it, or I'll go through it in my next post if you want.
Faá di Bruno's formula cancels down to:
Sigma (sum over all all partitions of n) :
[ n! ln[sup]k[/sup]z ] / [ k[sub]1[/sub]! k[sub]2[/sub]! ...k[sub]n[/sub]! ]
* (a[sub]1[/sub])^(k[sub]1[/sub])
* (a[sub]2[/sub])^(k[sub]2[/sub])
:
* (a[sub]n[/sub])^(k[sub]n[/sub])
(where k1, k2... make up each partition, and k is simply the sum of k1, k2,... )
So, I now have the infinite system of equations I needed
.
We have sigma[ k!/(k-n)! a[sub]k[/sub] ] on the left side, and the chunky expression above on the right. Note that the k's on the left aren't the same as the ones on the right. I should have used different variables.
Granted it's a bit complicated, and it's going to be really hard to solve, but at least its there. The "umbral calculus" bit is over now, and I can start solving the equations.
My plan is now to solve the system bit by bit, for increasing values of n, up to about n=5, then try to go for the whole thing somehow. I'll work out how to use the "greedy algorithm" to find partitions, and I might have a go at solving it with matrices.