## Tetration and other large number sequences

Other scientific, philosophical, mathematical etc. topics go here.
I've just found a startling and unexpected result from the book I mentioned.

Integral from 0 to 1 of (1/(x^x)) dx = 1 + 1/(2^2) + 1/(3^3) + 1/(4^4) + ...

!!!

The proof involves applying integration by parts repeatedly
jinydu
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Mathworld has the same integral, and also a similar definite integral of x^x as well, which alternates between positive and negative terms. I hadn't realised how simple they are. It's called "a sophomore's dream".
http://mathworld.wolfram.com/SophomoresDream.html

There isn't a finite expression for the indefinite integral, but by following the same process, I just found the large double sum formula for it. It's more complicated than the integral from 0 to 1, but still more elegant than I expected. It's a bit big to post here, however.

I can't say I've found anything on interpolation, but I've had an idea about extending tetration to matrices.

I found a definition of the exponential of a square matrix, using the infinite sum:
e^[X]= sum( (1/n!)[X]^n )
http://mathworld.wolfram.com/MatrixExponential.html
It can also be defined without the sum, but the formula is huge.
(equation 14)

If I can somehow find the inverse of that, and call it ln(X), then I can define a matrix to the power of another matrix.
A^B = e^(B ln(A)) for matrices A and B.
Then A tetra n = A^(A tetra (n-1)) just like the usual definition.

Although it's not very helpful for extending it to reals, it might help establish some more rules for tetration.

PWrong
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I've managed to invert the formula, but it's even bigger than e^X, so matrix powers are extremely unwieldy.

The interesting about the matrix powers is that it's even less commutative than multiplication. For two matrices, A and B, there are four different powers.

B^A = e^(A * lnB)
A^B = e^(B * lnA)
B^A = e^(lnB * A)
A^B = e^(lnA * B)

I've confirmed that for most matrices, all four are different. I'll have to make up some notation for this. I haven't found a useful link with tetration yet, but while I was finding ln(X), the equations looked very similar to the Lambert W function, which we already know is important to tetration. I was hoping I could define the log of a matrix using the W function, but it didn't work.

I've also found a way you could potentially use the gamma function in tetration. The basic definition is
z tet (x+1) = z^ (z tet x)

We want a function of x that is defined for reals and fits these values. Just as e^x is simpler than a^x, e tetra x is simpler than z tetra x, so let z = e.
The function we want satisfies
f(x+1) = e^ f(x)

Now use the infinite series for e:
f(x+1) = 1/0! + f(x)/1! + f(x)^2/2! + f(x)^3/3! + ...
And we can simply replace the factorial with the gamma function.

Maybe you could then define f(x+1.5) by using values in between. It won't guarantee anything useful, but it's worth a try.

f(x+1.5) = f(x)^.5 /G(.5) + f(x) /G(1) + f(x)^1.5 /G(1.5) + f(x)^2[/sup]/G(2) + f(x)^2.5 /G(2.5) + ...
(G =Gamma)

For integers, this would become an integral, rather than a sum.
i.e.
Integral( f(x)^t / G(t) dt )

And now I'm stuck. I'll try to integrate it, but I doubt I'll get very far.
The definition I've come up with doesn't seem to work anyway, but the general idea might come in handy. You might be able to combine this with whatever insights you gain from your book.

PWrong
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Actually, gamma(n) does not equal factorial (n), even when n is an integer. Instead, gamma(n) = (n-1)!. This seems to be due to some property of the integral that is used to define gamma.
jinydu
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You're right, I forgot about that. I've tried it the correct way, however, and it still doesn't work.

The integral would actually be simpler if gamma had been defined the intuitive way.

(n-1)! =gamma(n) = integral( t^(n-1) e^-t dt
The gamma seems to be the odd one out here. Maybe it's just convention.

By the way, I've almost found a quadruple sum formula for the integral of x tetra 3.
It's too complicated to actually use, but it's nice to know that it's possible.

PWrong
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B^A = e^(A * lnB)
A^B = e^(B * lnA)
B^A = e^(lnB * A)
A^B = e^(lnA * B)

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. 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 (and technically not even a binary operation at all) in this context so it's really of no use, unfortunately - though kudos to you for working through the arithmetic to find lnA and lnB...I don't have that kind of patience at this time of day!
Last edited by houserichichi on Tue Feb 08, 2005 6:57 pm, edited 1 time in total.
houserichichi
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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.

I've just gone over it again, and I've simplified lnA enough to present it here, if anyone's interested. Sorry about all the extra variables though.

Code: Select all
`ln([ a  b ] )    [ c  d ]=p [(a-d)  2b  ] +q [1 0]   [  2c  (d-a)]      [0 1]where p = ln{ (a+d+delta)/(2k) } / deltaq= ln(k)delta = sqrt( (a-d)^2 +4bc )k= sqrt( ad - bc )`

I'm think that's right.

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

I just looked up "well-defined", but the expression doesn't seem to be very well defined itself. It says it requires a unique interpretation or value, but the only examples it gives are non-associative numbers.

The two different answers look like opposites of each other. I thought you could just use both answers, and call them conjugates of each other. A bit like 4^0.5 = both 2 and -2. Even if it is an ambiguous definition, surely some interesting maths could come out of it anyway? [/quote]

PWrong
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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.

This part is taken right out of mathworld:

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.

Your ^ operation, which we could just as easily have written as ^(A,B) for square matrices A and B, is not binary as it doesn't satisfy condition 2.

Hope that helped, but sorry to be the bringer of bad news!!! All good ideas come from failed attempts though.
houserichichi
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Getting back to tetration, I've started making some fractals based on tetration.

I used UltraFractal to iterate the formula: z = z tetra n + c, for a given n.
For n=2, this is similar to the mandelbrot set, although the picture doesn't actually look like the mandelbrot. You'd think it would be more complex, but it's actually very simple and boring.

I also made some julia sets from the same formula. The julias are much more interesting than the mandelbrot version. I'll post some pictures on here when I finish them.[/img]

PWrong
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http://mathworld.wolfram.com/i.html

They have a point about i^i and i^(i^(i^(i^... near the bottom. They've even got a joke about i
jinydu
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I had an idea after noticing how n tetra x alternates for even and odd x. Look at the graphs for small n, and you'll see what I mean. If you look at only the even numbers, they seem to follow a smooth curve.

Maybe we should try to find two interpolating functions, one that preserves the values for even x, and one for odd x.

PWrong
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I just found a very nice way to integrate x tetra n for any n, using n sums to infinity.

Let f(x,a,b) be the integral of x^a * (ln x)^b
You can get this function by integrating by parts repeatedly.

Now x^(x^x) = e^(x^x lnx)
Using the taylor series, (read "sum k" as "sum k=0 to k=infinity"
= sum k[ (x^(kx) (lnx)^k /k! ]
= sum k[ sum j[ k^j x^j (lnx)^j *1/j!] * (lnx)^k *1/k! ]

Now we just put everything inside the last sum.
Code: Select all
`= sum k [ sum j [ (k^j )/( j! k! ) *x^j *(lnx)^(k+j) ]]`

For x tetra n, we use the same process for k1, k2, k3, ... kn (read as k superscript n), and we get n infinite sums, ending in an interesting looking fraction, and something of the form x^a * (lnx)^b

Code: Select all
`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) `

Now we can simply integrate each term using the function above. Replace
x^kn*(lnx)^(k1+k2+...kn) with f(x, kn, k1+k2+...kn )

The same method might be useful for extending tetration. The formula above shows that x tetra n is a multiple sum formula, and at the centre of it is a function of n dimensions, (not including x). To represent x tetra 2.5, we really need a 2.5 dimensional function. I'm starting to wonder whether the concept of fractal dimension might be useful for this somehow.

PWrong
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Hmm, that's interesting. Keep up the good work, PWrong .

(Someday, we'll be granted tetraspace citizenship)
jinydu
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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.

Actually, I don't see anything wrong about having a logarithm function that does not give a unique answer. Logarithms are unique only when applied to positive real numbers, anyway, so I don't see it as a hindrance.

For example, the complex logarithm ln|z| is a bizarre function with many branches and a cyclic period of pi (or was it pi/2 ?) along the imaginary axis. Yet it's still well-defined, at least as a multivalued function, and analytic, even. The Lambert W function is also multivalued for x<0. I guess 'functions' like these aren't really 'functions' in the strictest sense of the word, but mathematicians seem to have no trouble talking about the 'branches of the W function', etc., so I don't see why a matrix exponential/logarithm couldn't be treated the same way.
quickfur
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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
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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.

True, so it's more that the exponential function is periodic along the imaginary axis, and it is this periodicity that gives rise to the multi-branchedness of the logarithm. I got the two mixed up, oops.

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.

That's what I meant, when I said 'complex logarithm' I mean ln|z| for any complex number z, which includes the real numbers. The logarithm function as defined on the real numbers has only real values, but the generalized logarithm (extended to the complex plane) will give multiple values for the same real numbers.

Speaking of which, wouldn't it be neat if we could see a 4D graph of the complex logarithm? Then we'd immediately realize that the complex exponential is just a reflection of the graph along some plane, and we'd be able to pick out the sine and cosine components of the two functions. After playing with John McIntosh's 4D maze game, I think I'm actually starting to understand how to visualize 4D via 3D projections. I wonder what the projected graph of the complex logarithm looks like. Anyone dare to write an applet or something that performs this projection? :-)
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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.
jinydu
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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.

But wouldn't the graph be a 3-brane in 4-space, and so would (usually) project to a 3D volume? The key to 4D visualization is to think in terms of 3D volumes that fill the 3D retina space you project the 4D object onto. Of course, this is not always easy, esp. when the 3D volumes are not purely realmar (not confined to a flat realm in the 4D space). Plus, if displayed by a computer program, you'd have to do another projection from 3D to 2D, which makes it even harder to visualize. But I think with practice it shouldn't be too hard. You can use a wire mesh to represent the projected 3D volumes.

Maybe one of these days I should try doing some 4D graphs this way. Of course, the program would have to allow free 4D rotation so that you can view it from various different angles. And probably a way to selectively display the constituent volumes of the graph so that you can get a better idea of the shape the projected volume rather than to try to deduce it from a mess of lines.

A lot of flexibility would be needed so that the viewer can get a good idea of what the 4D shape is really like. There has to be a way to clip by 4D distance, for example, so that when you look at, say, the spherical projection of a hypersphere, you have a way of knowing that the volume around its origin is bulging "towards" you in the 4th direction, and not merely flat. Or some other way of indicating this will be necessary, maybe by longitudinal/latitudinal/tetritudinal lines so that you can see that the sub-volumes around the center of the sphere are larger, implying that they are closer in the 4th direction.

Clipping of hidden volumes would be a must, since otherwise you end up with an incomprehensible tangle of lines. The free 4D rotation will make up for this, in that the user can fly around the graph in 4D to look at the obscured parts from another angle.

Non-trivial, but I still think it's doable.
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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.

PWrong
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PWrong wrote:Can't you view a complex function as a 2D surface in a 3D graph? :?

The problem with this is that you can only see the real/imaginary parts of the function separately, and it takes a lot of mental effort to put the two together to get an idea of what it looks like on the complex plane. Well actually, you won't be able to get a good idea of how the function changes across the domain, because it has 4D curvature which individual slices of the function (which is all you can get in a 3D graph) are unable to adequately represent.
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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.

Interesting. Well, I do know that Gamma is a generalization of the factorial to real numbers. More precisely:

Gamma(n) = (n-1)!

But I don't know how to solve that differential equation either...
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http://mathworld.wolfram.com/PowerTower.html
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I normally designate CE2 as the cartesian product of complex numbers, CE2 stands for complex-euclidean 2d.

The real representation of this is E4 (ie x, xi, y, yi), but unlike 2D, the thing is a proper subset of E4.

In CE2, we can solve, for example the line-equation through any two points, ie there is a unique line through x1,y1 and x2, y2. In the E4 representation of this, it suggests that there is only one plane E2 through the two points. In practice, there is a whole sphere of such things, but only one set of points on this sphere makes for a CE1.

An equation like the exponential function in CE2, then is an E2 (hedrix), that looks like (in the x,0,y,0) plane, a normal exponential plane, but as you drag it through yi axis, the thing rotates around the circle in the x, xi plane: ie it's not sinosoidal.
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