## Tetration and other large number sequences

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Whoa, that is getting complicated. I'm not sure about what to do to simplify that. However, I would suggest using your computer to do some of those long and tedious derivatives. I'm using Mathematica, but I'm sure there are some web pages that will differentiate functions for you for free. It would save a lot of time.

Unfortunately, I have no real progress to report on this post. Currently, there are two things I want to do: 1) Prove that (sqtrt(2))^2 is irrational (if it is) and 2) See if there are any other complex values for sqtrt(2) besides 1.5596104... (I know that sqrt's always have 2 complex answers (except when you're taking the square root of 0)).
jinydu
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Using Mathematica, I have discovered more square tetraroots of 2. Their approximate values are:

2.4694707146516336821689110092445532382691 - 3.0139944067814401210239704716263122994094i

and

2.4694707146516336821689110092445532382691 + 3.0139944067814401210239704716263122994094i

and

3.252323305365845283819667616801169642309 + 5.144480323313504290103555493388109472269i

and

3.252323305365845283819667616801169642309 - 5.144480323313504290103555493388109472269i

and

5.067894839159135336012016491758015553827895124350966366809 + 10.484181198728861852212771924010665497316768426408241810963i

and

5.067894839159135336012016491758015553827895124350966366809 - 10.484181198728861852212771924010665497316768426408241810963i

I'm starting to suspect that there may be infinitely many square tetraroots of two. I'll have to think more about that.
jinydu
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Ooh, that makes sense actually. I think I can prove that there are infinitely many. I'll use both rectangular and exponential form (with t instead of theta)

let z = r e[sup]i*t[/sup]

then z^z = e[sup]z ln z[/sup]

z^z = e[sup]([a+bi]*ln[re^ti])[/sup]

z^z = e[sup]([a+bi]*[ln(r)+ti])[/sup]

z^z = e[sup](aln(r) + ibln(r) + ati - bt)[/sup]

z^z = e[sup](aln(r) - bt)[/sup] e[sup](ibln(r) + ati)[/sup]

Now, in exponential form, you can add any multiple of "2pi" to the angle, and get the same answer. i.e.

z^z = e[sup](aln(r) - bt)[/sup] e[sup](i[bln(r) + at + 2*pi*k])[/sup]
for any integer k

Now, suppose z^z = 2
There are infinitely many expressions for z^z, so there are infinitely many expressions for 2, in terms of the components of z.

I don't know if it's possible to solve this equation. Hopefully it isn't, because otherwise the roots of tetration would be algebraic numbers.

I do think that it may be possible to express any trtroot in terms of the primary trtroot. It might be useful to try to find a kind of DeMoivre's theorum for tetration.

PWrong
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I've found a simple method for finding the n'th derivative of a function, in terms of the function itself, using first principles. It's not exactly what I was looking for, but it might help.

by the definition of a derivative: (as h approaches zero)

f'(x) = [f(x+h) - f(x)]/h

Now, to find f''(x):

f''(x) = [f'(x+h) - f'(x)]/h

Now we can simply substitute the orininal definition into this.

f''(x) = ([f(x+2h) - f(x+h)] - [f(x+h) - f(x)])/h^2

f''(x) = (f(x+2h) - 2f(x+h) + f(x)])/h^2

We do the same for the next derivative:

f'''(x) = [f(x+3h) - 3f(x+2h) + 3f(x+h) - f(x)]/h^3

And the next...

f''''(x) = [f(x+4h) - 4f(x+3h) + 6f(x+2h) - 4f(x+h) + f(x)]/h^4

It's Pascal's triangle again. Unfortunately, I can't seem to use it to find the n'th derivative of e^f(x) in terms of successive derivatives of f(x). But anyway, it might come in useful somewhere.

It might be easier to have a program that could do more than a few derivatives, so I could try to find a pattern. But in this case, I don't think I'll find one. The rule is probably too complicated to find just by inspecting the numbers.[/img]

PWrong
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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.

I'm still working on "digesting" your previous post, on the infinitude of solutions to x^x = 2.

Anyway, I now think that the reason for all the difficulties with tetration is that it is based on a "noncommutative chain". Let me try to explain this more clearly:

Multiplication is based on a chain of addition. That is it can be written as:

x*a = x(1) + x(2) + x(3) + ... + x(k-1) + x(k) + x(k+1) + ... + x(a)
where x(1) = x(2) = x(3) = ... = x(a) (= x).

The important point to note is that this is a very "flexible" chain. I can state this more precisely in several ways:

1) Its possibly to break off the chain at any number of arbitrary point, reassemble it any way you want, and still get the same answer. More importantly, this would be true even if the individual terms were different. In other words a+b+c+d = d+c+b+a = any other ordering. This property is very important because it allows us to prove many properties. For instance, if we break the chain after x(k), we will have two separate chains. However, combining the two chains will still lead to the same answer. Furthermore, it doesn't matter what order in which order we combine it. In equation form:

[x(1) + x(2) + x(3) + ... + x(k-1) + x(k)] + [x(k+1) + ... + x(a)] = [x(k+1) + ... + x(a)] + [x(1) + x(2) + x(3) + ... + x(k-1) + x(k)]

Using multiplicative notation: x*k + x*(a-k) = x*(a-k) + x*k = x*a
Letting m = k and n = a-k, we have:
x*m + x*n = x*n + x*m = x*(m+n)

2) A simple modification to the chain will lead to a simple overall expression for the chain. For instance, let's change x(k) from x to x+u. Then, we have:

x(1) + x(2) + x(3) + ... + x(k-1) + (x+u) + x(k+1) + ... + x(a)

Since this chain is so flexible, we can move things around easily:

x(1) + x(2) + x(3) + ... + x(k-1) + x(k+1) + ... + x(a) + (u+x)
x(1) + x(2) + x(3) + ... + x(k-1) + x(k+1) + ... + x(a) + (x+u)
x(1) + x(2) + x(3) + ... + x(k-1) + x(k+1) + ... + x(a) + x + u
x*(a-1) + x + u
x*(a-1+1) + u
x*a + u

See how the answer (xa+u) is only slightly more complicated than the original answer, before the modification (xa).

Similar properties hold for exponentiation, which is a chain of multiplication:

x^a = x(1) * x(2) * x(3) * ... * x(k-1) * x(k) * x(k+1) * ... * x(a)
where x(1) = x(2) = x(3) = ... = x(a) (= x).

This is also a very "flexible" chain:

1) Its possibly to break off the chain at any number of arbitrary point, reassemble it any way you want, and still get the same answer. More importantly, this would be true even if the individual terms were different. In other words a*b*c*d = d*c*b*a = any other ordering. This property is very important because it allows us to prove many properties. For instance, if we break the chain after x(k), we will have two separate chains. However, combining the two chains will still lead to the same answer. Furthermore, it doesn't matter what order in which order we combine it. In equation form:

[x(1) * x(2) * x(3) * ... * x(k-1) * x(k)] * [x(k+1) * ... * x(a)] = [x(k+1) * ... * x(a)] * [x(1) * x(2) * x(3) * ... * x(k-1) * x(k)]

Using multiplicative notation: x^k * x^(a-k) = x^(a-k) + x^k = x^a
Letting m = k and n = a-k, we have:
x^m * x^n = x^n + x^m = x^(m+n)

2) A simple modification to the chain will lead to a simple overall expression for the chain. For instance, let's change x(k) from x to u*(x+v). Then, we have:

x(1) * x(2) * x(3) * ... * x(k-1) * (u*(x+v)) * x(k+1) * ... * x(a)

Since this chain is so flexible, we can move things around easily:

x(1) * x(2) * x(3) * ... * x(k-1) * x(k+1) * ... * x(a) * (u*(x+v))
x(1) * x(2) * x(3) * ... * x(k-1) * x(k+1) * ... * x(a) * ((x+v)*u)
x(1) * x(2) * x(3) * ... * x(k-1) * x(k+1) * ... * x(a) * (x*u + v*u)
x^(a-1) * (x*u + v*u)
(x^(a-1)*(x*u)) + (x^(a-1)*(v*u))
(x^(a-1+1)*u) + (x^(a-1)*v*u)
((x^a)*u) + (u*v*(x^(a-1)))
u*(x^a) + u*v*(x^(a-1))

Well, the final expression, u*(x^a) + u*v*(x^(a-1)), is a bit more complicated this time. But this is at least in part due to the fact that our modification to the original chain was more complex than last time. Thus, simple expressions for slightly modified chains still exist.

However, the picture changes dramatically for tetration (i.e. repeated exponentiation). We have the chain:

x tetra a = x(1)^x(2)^x(3)^...^x(k-1)^x(k)^x(k+1)^....^x(a)
where x(1) = x(2) = x(3) = ... = x(a) (= x), and I assume you know to go from right to left in performing the exponentiations.

However, in general, a^(b^(c^d)) does not = d^(c^(b^a)), let alone all the other possible permutations. Thus, it is not possible to break off "segements" from a tetration chain, and move them around. This makes it hard to prove properties about tetration.

Also, it is much more difficult to derive formulas for slightly modified chains. For example, let's try what's probably the simplest modification possible, changing x(k) from x to x+1:

x(1)^x(2)^x(3)^...^x(k-1)^(x+1)^x(k+1)^....^x(a)
x(1)^x(2)^x(3)^...^x(k-1)^((x+1)^(x tetra a-k))

Because exponentiation is noncommutative, it is NOT possible to change x(1)^x(2)^x(3)^...^x(k-1) to (x tetra k-1). For a simple example of this, try x = 2, a = 5 and k = 3. Trying to perform the simplification yields the incorrect answer, 4^81. The real answer is much larger, 2^(2^81).

In fact, I don't know of any to contract x(1)^x(2)^x(3)^...^x(k-1) into tetration form. This, and the lack of "Laws of Tetration" (analogous to the Laws of Exponentiation) makes many problems very difficult.

All these difficulties stem from the fact that I can't move individual "segements" around the tetration chain. This, in turn, is caused by the fact that exponentiation is not commutative. A logical question from this would be "why are addition and multiplication commutative, while exponentiation is not"?

Of course, this is definitely NOT an excuse to give up! I'll keep looking for a way to express all the solutions to x^x = 2 in terms of sqtrt(2) [from now on, sqtrt(2) refers to the principal, i.e. real, tetraroot of 2]. Also, in my next post, I'll talk about an idea I've had for extending (x tetra y) to negative values of y.

P.S. PWrong, you're 12 posts away from something big!
jinydu
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Ok, this is what I think about tetrating to negative integers. log (a, b) = "log base a of b".

Problem:

(x tetra n) = x^(x tetra n-1)

(x tetra n-1) = log (x, x tetra n)

x tetra 1 = x

x tetra 0 = log (x, x) = 1

x tetra -1 = log (x, 1) = 0

The computer balks at my attempt to calculate x tetra -2, log (x, 0)

Unfortunately, I wasn't able to come up with a "natural" way to fix the problem, so I decided to do it in the least unnatural way I can think of. I introduced a "discontinuity" after x tetra 0. 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.

First, go back to the formula: (x tetra n-1) = log (x, x tetra n)

The change of base formula states that: log (a, b) = [log (c, b)]/[log (c, a)]

Therefore, (x tetra n-1) = (ln (x tetra n))/(ln x), taking c = e

Taking the natural log of x shouldn't lead to any problems (for now, I'm restricting myself to keeping x a positive number). The numerator may cause difficulties if (x tetra n) is not positive. This is where using a complex answer for the natural log comes in handy. Let's go back to our initial problem, but this time use our new technique:

x tetra 0 = 1

x tetra -1 = (ln 1)/(ln x) = (0 + i*(0 + 2pi))/(ln x) = (2pi*i)/(ln x)

Notice that the answer above is a pure imaginary number (i.e. the real part is equal to 0). I simply took the real number answer for (ln x), for the sake of convenience. Now, we can continue onwards (keep in mind that theta now equals pi/2):

x tetra -2 = (ln ((2pi*i)/(ln x)))/(ln x) = (ln (2pi*i) - ln (ln x))/(ln x)

I could keep going and calculate x tetra -3, but now things get considerably more complicated. ln (ln x) is an complex number with both real and imaginary parts not = 0 when 0<x<1. Taking the natural log of this complex number (after it has been divided by ln x, of course) is even more complicated, involving the arctan function. Thus, its probably best left to computers, except in special (simplified) cases.

One such case is obtained by setting x = e. Using the method above, I can tetrate e to negative integers.

e tetra 0 = 1

e tetra -1 = 2pi*i

e tetra -2 = ln (2pi*i) = ln (2pi) + (pi*i)/2

e tetra -n = ln (e tetra -(n-1))

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!
jinydu
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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.

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.

PWrong
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Whoa, a bit complicated for me, a first quarter undergraduate freshman. Especially on Midterm exam week. I'll try to look into that in-depth when I've got a lot of free time. Let me guess; once all those equations are solved, your top function will be an infinite polynomial that will (hopefully) converge in all cases where we expect tetration to work?

First, an interesting observation is that e^x always has only one possible value, even when we're working with complex numbers. However, that doesn't seem to be the case with e tetra n for negative values of n. Things do seem to get more complicated if you want to have an infinite number of possible values. Its easy to show that:

e tetra -1 = 2*n*pi*i

However, caluclating e tetra -2 is much harder. Each possible value of e tetra -1 (except n = 0) leads to an infinite number of values for e tetra -2. And then, each value for e tetra -2 leads to an infinitely many values for e tetra -3. So in a very unrigorous sense:

# of values for e tetra -1 = infinity
# of values for e tetra -2 = infinity^2
# of values for e tetra -3 = infinity^3
etc.

Of course, Cantor assures us that all of these infinities are countable (except maybe if we try to calculate (e tetra -infinity)), but that still doesn't stop my head from hurting. I just think its more convenient to work with a single value for (x tetra -n). Thus, I think I can define the Principle Value of (x tetra -p) as:

x tetra -p = (ln abs(x tetra -(p-1)) + i*(arg(x tetra -(p-1)) + 2*n*pi))/(ln x)
where n = 1 if p = 1 and n = 0 if p>1.

On the other hand, defining x tetra n so that it has an infinite number of values when n is positive is less elegant, I think. We already defined:

x tetra 1 = x

To derive other possible values of x tetra 1, we would have to first compute x tetra 2 (= x^x), then work backwards by taking a log. But once we've computed x tetra 3, we'll find that the value we took for x tetra 2 was only one of an infinite number of values, the rest of which can be computed by taking the log of x tetra 3. And then, when we calculate x tetra 4, we'll see that we made an arbitrary choice with x tetra 3, and...

Well, I think you see the problem with that...
jinydu
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jinydu wrote:Whoa, a bit complicated for me, a first quarter undergraduate freshman. Especially on Midterm exam week. I'll try to look into that in-depth when I've got a lot of free time.

, don't worry about it if you're not bothered. I just included what I found for completeness. We're getting close to the final exams now, so I should probably do a bit less of this and a bit more studying.

jinydu wrote:Let me guess; once all those equations are solved, your top function will be an infinite polynomial that will (hopefully) converge in all cases where we expect tetration to work?

Yep, pretty much. Although converge might not be the right word. Infinitely differentiable is what I'm looking for. An annoying thing about differentiable tetration is that it produces very ugly graphs, while ordinary continuous tetration is much more elegant, even though it's not differentiable.

However, caluclating e tetra -2 is much harder. Each possible value of e tetra -1 (except n = 0) leads to an infinite number of values for e tetra -2. And then, each value for e tetra -2 leads to an infinitely many values for e tetra -3.

Are all the infinities really neccessary? It seems like what we're doing here is a bit like that proof of 1=2, except more complicated. With the extra "2*k*pi*i", it's possible to prove many false results:

y=x
e^y=e^x
e^y=e^(x+2*k*pi*i)

ln(e^y)=ln(e^(x+2*k*pi*i))
y=x+2*k*pi*i

Therefore x=x+2*k*pi*i
i.e. there are infinitely many values of any number, x.
Clearly there must be a problem with this proof. You could say that x is being treated as both a number and an angle at the same time (because exponential form only works in radians), so the result isn't valid.

Thus, I think I can define the Principle Value of (x tetra -p) as:

x tetra -p = (ln abs(x tetra -(p-1)) + i*(arg(x tetra -(p-1)) + 2*n*pi))/(ln x)
where n = 1 if p = 1 and n = 0 if p>1.

Should "2*n*pi" be "2*n*pi*i"? I made the same mistake a few posts ago, and only just noticed it.

Instead of defining principle values for tetration, would it be easier to redefine ln(x), so that it can only have one value for any given x, and so that ln(1)=2*pi*i?[/quote]

PWrong
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PWrong wrote:y=x
e^y=e^x
e^y=e^(x+2*k*pi*i)

ln(e^y)=ln(e^(x+2*k*pi*i))
y=x+2*k*pi*i

Therefore x=x+2*k*pi*i
i.e. there are infinitely many values of any number, x.
Clearly there must be a problem with this proof. You could say that x is being treated as both a number and an angle at the same time (because exponential form only works in radians), so the result isn't valid.

The problem is that its possible for:

e^x = e^y

when x is not equal to y if x and y are complex numbers. That is, the exponential function is a many-to-one function in the complex plane. Your contradiction is only a more sophisticated version of this one:

1 = 1
(-1)^2 = (1)^2
-1 = 1

The problem arises because the square root of 1 can have two answers. Things are much "worse" with the logarithm, because it can have infinitely many answers.

As for my definition of the principle value, no, I didn't miss an i. There's already an i outside the bracket.

Hopefully, I'll have new thoughts to post soon.
jinydu
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Ok, I've managed to find the derivative of sqtrt(x). I did it using the Inverse Function Derivative formula at http://www.sosmath.com/calculus/diff/der08/der08.html. That webpage didn't supply a proof, so I had to prove it myself (I used the chain rule), but I'll omit that in this post.

Let f(x) = x^x = x tetra 2. As we all know, f'(x) = (x tetra 2)*(1 + ln x)Applying the formula, I get:

(sqtrt x)' = 1/(f'(sqtrt x))

(sqtrt x)' = 1/(x*(1 + ln (sqtrt x))

As a corollary, note that for x>= 1, sqtrt x >=. Therefore, ln (sqtrt x) >= 0, and 1 + ln (sqtrt x) >= 1. Hence, the denominator must be >= x. Finally, the entire expression must be <= 1/x. Thus, sqtrt x grows more slowly than ln x. This isn't too surprising, since we know that x^x grows faster than e^x.
jinydu
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ooh, that could be useful. I'm starting to think that trtroots are probably the best way to define continuous tetration after all.

I haven't found anything new yet, but I'm going to start thinking about different top-functions I could use instead of a polynomial, possibly a top-function including the tetraroot itself. What we really need is a new way to define the trtroot. Something like "sqrt(x) = x^(1/2)" except up a level.

I know exactly how to calculate the polynomial now, but I can't do it without something like mathematica. Unfortunately, it's hard to find any simple functions to even obey them even for n=0 and n=1. I can't find any exponential or trig functions, so my main hope now is to use roots, logs or possibly trtroots. It could be anything, but I'm hoping for something relatively simple.

Although the whole polynomial excercise seemed pointless, at least it tells us that a top-function does exist. Also, the infinite polynomial might be an expansion of a simpler function anyway. Many ordinary functions can be represented as an infinite polynomial by taking the derivative repeatedly:

Start with f(x) = a0 + a1 x + a2 x^2 + a3 x^3 + ...
then find all the derivatives of f(x). By letting x=0 each time, you get a value for a0, a1, a2, e.t.c.
This was an extension in my calculus textbook. It shows how to find a power series for e^x and sin(x) e.t.c.
I've used it for f(x) = x^x, f(x) = sqrt(1-x^2) (half a circle), and a few others. I think it works for most functions as long as all derivatives are all defined for x=0.
I think this is similar to "expanding the taylor series".

It's very interesting to do this with iterated functions like tetration.
For instance, you can expand e^(e^x) by substituting y=e^x and expanding e^y. Then you get an infinite polynomial of infinite polynomials!

I haven't done much recently apart from dabble with these taylor series, and with the "partitions" that I mentioned earlier, which are also very interesting. I think I'll come back to tetration (and the rest of the forum, I haven't forgotten everyone) more seriously after my exams are over (they start in about a week). Wish me luck, and good luck on your own exams.

PWrong
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Yes, good luck on your exams.

Unfortunately, I haven't been taught Taylor's Theorem yet, so I'm not as clear on it as I would be had I been taught. I'm quite sure that taking x = 0 will not work for tetration because 0^0 is undefined. Surprisingly, Mathematica says that x^x = 0 does have a solution, but I disagree, as I will show by contradiction:

x^x = 0
e^(x * lnx) = 0
Let x * lnx = a + bi
e^(a + bi) = 0
(e^a)*(e^bi) = 0
(e^a)*(cos b + i*sin b) = 0

In order for that equation to be true, either the first or the second parenthesis must be equal to zero. The first can never be zero, since we know that e^a > 0 for any real a. Therefore, the second parenthesis must be zero. That is:

cos b = 0 and i*sin b = 0
cos b = 0 and sin b = 0
b = (pi/2) + n*pi and b = n*pi

These equations can't both be true, so there is no solution to x^x = 0.

I think that's a warning to me, don't trust Mathematica too much.

I think that its possible to get a Taylor expansion of a function about any point, so long as all the derivatives are known. I tried this for the function f(x) = x tetra 2 = x^x. I found the first four derivatives (by hand) and my calculations were verified by Mathematica (after some further simplication of the Mathematica output, it doesn't simplify all the way). The formulas get quite long, so I'll only post the first few.

f(x) = x^x
f'(x) = (x^x)*(1 + lnx)
f''(x) = (x^x)*((1 + lnx)^2 + 1/x)

Here are the values of the function and its derivatives at x = 1 (which I thought was the most convenient point):

f(1) = 1
f'(1) = 1
f''(1) = 2
f'''(1) = 3
f''''(1) = 8

Mathematica can go much farther, but I haven't checked these by hand yet (again, these are all evaluated at x = 1):

5th derivative = 10
6th derivative = 54
7th derivative = -42
8th derivative = 944
9th derivative = -5112
10th derivative = 47160
11th derivative = -419760
12th derivative = 4297512

Not the most orderly sequence I've ever seen.
jinydu
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Topic split

Keiji

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This post is copied from the 0/0 thread:

Getting back to tetration, here are the first few terms of the Taylor series for x^x about x = 1:

x^x ~= 1 + (x-1) + (x-1)^2 + (1/2)(x-1)^3 + (1/3)(x-1)^4

Expanding and simplifying:

x^x ~= (1/3)x^4 - (5/6)x^3 + (3/2)x^2 - (5/6)x + (5/6)

For x = 1.1, this approximation gives x^x =~ 1.110533333333333333..., quite close to the actual (provably irrational) value of 1.11053424...
jinydu
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Looks like I'm not the first to think about differentiating x^x:

http://www.research.att.com/cgi-bin/acc ... um=A005727

The values at the first 20 derivatives (including the value of x^x itself) at 1 are:

1, 1, 2, 3, 8, 10, 54, -42, 944, -5112, 47160, -419760, 4297512, -47607144, 575023344, -7500202920, 105180931200, -1578296510400, 25238664189504, -428528786243904, 7700297625889920

I couldn't understand their description of a formula:

For n>0, a(n)=sum(b(n,k),k=0..n), where b(n,k) is a
Lehmer-Comtet number of the first kind (see A008296).
E.g.f.: (1+x)^(1+x). a(n) = Sum_{k=0..n}

Here is the Mathematica code to calculate more derivatives (change n into the number of derivatives you want):

NestList[ Factor[ D[ #1,x ] ]&, x^x, n ] /. (x->1)

The page also says that these numbers are known as "Lehmer-Comtet" numbers. Unfortunately, a Google search for Lehmer +Comtet didn't turn up anything particularly relevant.

For the sake of comparison, here are the first 10 derivatives of x tetra 3, also at x = 1. Unfortunately, my computer stalled when I tried to calculate 15 derivatives.

1, 1, 2, 9, 32, 180, 954, 6524, 45016, 360144, 3023640

For x tetra 4 (again, at x = 1, and you'll notice my computer is hanging up faster and faster):

1, 1, 2, 9, 56, 360, 2934, 26054, 269128

x tetra 5:

1, 1, 2, 9, 56, 360, 2934, 26054

Interesting... More and more terms in common. Maybe the limit as n approaches infinity of (x tetra n) has finite derivatives at x = 1.
jinydu
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houserichichi
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Thanks for the link. I've read through it and found that it covers only some of the things mentioned in this thread. This means that we've still found a lot of new things!

What I've noticed is that they've focused a lot less on lower "tetra-powers" than I have. Admittedly, most of my efforts have been on x^x and its inverse, sqtrt(x). Apparently, things are different for odd "tetra-powers". Also, it seems that infinite tetration is possible, though only in an interval, and its been studied quite heavily.

Maybe I can start looking at (x tetra 3) and perhaps even cbtrt(x).
jinydu
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hmm, that is interesting. It might be difficult to find a simple formula for it though. If you really want the n'th derivative of x^x, you probably have to use Faa Di Bruno's formula for the n'th derivative of f(g(x)), http://mathworld.wolfram.com/FaadiBrunosFormula.html.
It uses partitions a lot, which makes it difficult to predict.

First of all, we have to express x^x in the form, f(g(x)).
x^x= e^(xlnx),
so let f(x)=e^x
and g(x)=xlnx

Any derivative of e^x is of course, e^x, and the first few derivatives of xlnx are:

g(x) = xlnx
g'(x) = lnx+1
g''(x) = 1/x
g'''(x) = -1x^-2
g''''(x) = 2x^-3

From then on:
g[sup]n+1[/sup](x) = (-1)^n*n!*x^-(n+1)

The functions themselves don't matter much, as they follow simple patterns, especially for x=0.

But the partitions are really hard to sort out, even though they're based on a pretty simple idea.
http://mathworld.wolfram.com/Partition.html

I've thought of a new way to represent them using matrices, but I have no idea how to go about writing a matrix here.

Anyway, I'm going to try to get mathematica in the near future. I've thought about it for a while, and it seems almost impossible to do any maths without it recently.

PWrong
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I've managed to confirm that the 4th and 5th derivatives are 8 and 10, using Faa Di Bruno.

I've also written a program on my calculator to find partitions, and I'm working on one that will automatically find the n'th derivative of x^x at x=1. A program for variable x will be more difficult.

I reckon it'll be much harder to find the n'th derivative of x tetra m, because

x tetra m = e^ (x tetra (m-1) * lnx)

which involves both a chain and a product.

This means we need to combine Faa di Bruno's formula for f(g(x)), with the "Leibniz Identity" for f(x)*g(x). http://mathworld.wolfram.com/LeibnizIdentity.html
It's possible, but it will take a long time to work out.

PWrong
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Yes, I've had a look at (x tetra 3):

Function: x^x^x
First Derivative: (x^x^x)*(x^x)*((log x)^2 + (log x) + (1/x))

I also calculated the second derivative, but it is extremely complicated (about two lines when written out completely!). You can derive it yourself using the Product Rule.

x tetra 3 has a limit of 0 as x approaches 0. It is always positive and defined for x > 0.

By rewriting the first derivative, we can see that x tetra 3 is always strictly increasing (I omit the factors x^x^x and x^x because clearly, they are always positive when x is positive):

(log x)^2 + (log x) + (1/x) = ((log x) + 1/2)^2 + (1/x - 1/4)

If 0 < x < 4, the first term is positive while the second term is also positive, so the entire expression is positive.

If x = 4, the first term is positive while the second term is zero, so the entire expression is still positive.

If x > 4, we have to look a little more closely

log x > 0
(log x) + 1/2 > 1/2
((log x) + 1/2)^2 > 1/4

Since the second term must be greater than -1/4, the entire expression is still positive.

Thus, since the first derivative is always positive, x tetra 3 is a strictly increasing function.

Now, for the second derivative, things get even more interesting.

If x = 0.5, (x tetra 3)'' ~= -0.851252887
If x = 1, (x tetra 3)'' ~= 2

By the Intermediate Value Theorem, there must be some value of x such that

(x tetra 3)'' = 0

i.e. a point of inflection. The equation for the second derivative is fiendishly complicated, so Mathematica balks at my attempts to calculate it exactly. But Mathematica can solve it numerically. The approximate solution is:

x = 0.6676572765017400132511172567

So it seems that tetration does have something in common with exponentiation.

x^2 and x tetra 2 both have a single global minimum point.
x^3 and x tetra 3 both have a single point of inflection
jinydu
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It might be easier to differentiate x^f(x) first. Then we can substitute functions in for x tetra 3 and greater.

i.e. suppose we want the derivatives of x tetra n.
Then, let f(x) = x tetra (n-1)

y = x[sup]f(x)[/sup]
y' = ( f'(x)lnx +f(x)/x ) x[sup]f(x)[/sup]

Now, clearly f(x) and x[sup]f(x)[/sup] are always positive when x>0
Also, while x>1, lnx > 0.
So, if f'(x)>0, then y'>0

i.e. if x tetra (n-1) is strictly increasing after x>1, then x tetra n is strictly increasing when x>1. And we already know that x tetra 3 is strictly increasing.
Thus, by induction, x tetra n is strictly increasing for x>1, for all n.

Anyway, I'll give the second derivative of x^f(x), and try to prove some more complicated stuff later.

y'' = [ ( f''(x)lnx+2f'(x)/x - f(x)/x[sup]2[/sup] ) + ( f'(x)lnx +f(x)/x )[sup]2[/sup] ] x[sup]f(x)[/sup]

PWrong
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PWrong wrote:Thus, by induction, x tetra n is strictly increasing for x>1, for all n.

Actually, this is not the case. In one of my earlier posts on this thread, I showed that x tetra 2 is in fact decreasing when 0 < x < 1/e
jinydu
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yes, that's true, but I specified that x>1.

PWrong
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do you think it's too soon to start investigating x^x for x<0?

the result is usually a complex number, that spirals outward from (0,0), but the imaginary part keeps mysteriously jumping to 0. It's hard to explain without a graph, but you can probably see it for yourself with mathematica.

This isn't unique to tetration though, the jumping also happens for (-1)[sup]x[/sup].
It's hard to describe when the jumps occur, because with higher resolution the density increases. They always appear evenly spaced though.

PWrong
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Things do start to get much more "interesting" when you let x be a negative number. In general, raising a number to a non-integer power results in more than one possible complex answer. By restricting tetration to positive real numbers, we can sidestep this problem because there is only one possible positive real answer, which we can define to be the "principal" value.

We can compute x^x using DeMoivre's Theorem, but be warned; if x is an irrational number, there will be an infinite number of solutions. Here's an example:

-1/2 tetra 2 = (-1/2)^(-1/2) = 1/((-1/2)^(1/2)) = 1/(+- i*sqrt(2) /2) = +- i*sqrt(2)
jinydu
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I've made a bit more progress with the top-function. I found a much faster way to calculate it, and I've now found the 7-degree polynomial for e tetra x.

Interestingly, the top-functions for e tetra x are very close to y=x, but not exactly.

Recently, I redefined the top-function using the idea I had before about how the first principles method following pascals triangle. This helped me indirectly prove that the top-function can work for complex numbers.

It turns out that f(|x|+1)=z^f(|x|), which is basically the definition for tetration. Unfortunately, it doesn't help find the function much. Also, it's not particularly exciting for complex numbers, since it's just an absolute value.

PWrong
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Currently, I'm waiting for a book to get back to my university's library (somebody keeps renewing it ).

It seems to me that the greatest problem is finding a way to define:

x tetra a

for real a. This is essentially an interpolation problem:

Given a function that is defined for natural numbers and/or integers, find a "natural" way to extend it to real numbers, while preserving the important properties of the original function.

I know that Euler was a master at this type of problem, so I'm planning to borrow a book that details how he managed to do this for the Gamma function, an extension of the factorial function.

Its easy to find sources that show that the Gamma function does preserve the properties of the factorial function. Unfortunately, I've yet to find any sources that do the reverse, how to go from the properties that you want to preserve to the extension function.

Hopefully, the process that Euler took will give some insights into how to attack the problem of extending tetration.
jinydu
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Hey, I've found a stronger definition of the trtroot.

Have a look at the Lambert W function. http://mathworld.wolfram.com/LambertW-Function.html
Basically, W(a) is the solution to the (xe^x) = a

Consider this equation:
e^x = trtroot(e)

i.e. (e^x)^(e^x) = e
e^(xe^x) = e
xe^x=1

The solution is W(1) = 0.56714...
Therefore, the trtroot of e is simply ln(W(1))

Now, let's find x, where 2^x = trtroot(2)
(2^x)^(2^x) = 2
2^(x2^x) = 2

taking logs
x2^x = 1
xe^(xln2) = 1

* both sides by ln2

(xln2)*e^(xln2) = ln2
xln2 = W(ln2)
x = W(ln2)/ln2
Therefore,

trtroot(2) = 2^[W(ln2)/ln2]
This simplifies to
trtroot(2) = e^W(ln2)

In general:
Trtroot(a) = e^ W(ln a)

The Lambert W function has a few different expansions, but I can't get any of them to work. I'm finding it difficult to extend this new definition for anything other than the square trtroot. However, if I can do that, the extension might apply to the reals as well.

PWrong
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Interesting idea.

Have you been able to find anything about function interpolation?
jinydu
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