## Tetration and other large number sequences

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I could collapse the Triadic Function into a single-variable function. I'll call it the hyperfactorial (I know there's already a relatively obscure function called the hyperfactorial, but this has nothing to do with it). I denote the hyperfactorial of x as hfac(x). The function is defined as:

hfac(x) = hy(x,x,x)

For example:

hfac(1) = hy(1,1,1) = 1 + 1 = 2
hfac(2) = hy(2,2,2) = 2 * 2 = 4
hfac(3) = hy(3,3,3) = 3 ^ 3 = 27
hfac(4) = hy(4,4,4) = 4 tetra 4 = 4^(4^(4^(4))) = 4^(4^256)
hfac(5) = hy(5,5,5) = 5 penta 5 = 5tetra(5tetra(5tetra(5tetra(5))))

As you can see, this function grows very fast. hfac(4) is already larger than a googolplex.
jinydu
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MY GOD. That's a BIG number. But what about hfac(3.5)? Keiji

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PWrong wrote:i.e. what is x if x^x=2?

X=logx2
RQ
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bobxp wrote:MY GOD. That's a BIG number. But what about hfac(3.5)?

Well, hfac(3.5) = hy(3.5,3.5,3.5)

That means you would have to use some operation halfway between exponentiation and tetration. I have no clue how to do that.

How about starting with a more modest goal at the beginning:

Define hy(p,1.5,q), where p and q are positive integers. We're looking for an operation halfway between addition and multiplication. Any ideas?
jinydu
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RQ wrote:
PWrong wrote:i.e. what is x if x^x=2?

X=logx2

AH, now I know, that fits in the series subtract, divide, root, log. Why didn't I think of that?

We're looking for an operation halfway between addition and multiplication. Any ideas?

This, perhaps?

Examples:
hy(p,1.5,q) = (p + q) * 0.5 + (p * q) * 0.5
hy(p,2.3,q) = (p * q) * 0.7 + (p ^ q) * 0.3
hy(p,1.999,q) = (p + q) * 0.001 + (p * q) * 0.999

Get it? :wink:

Edit: I think this might not be very good at all. Look at the graph I produced, and you'll see why.   Keiji

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bobxp wrote:Edit: I think this might not be very good at all. Look at the graph I produced, and you'll see why. Let me guess, its because there's a sharp corner (i.e. discontinuous derivative) at n = 2. This is caused by multiplication suddenly appearing, out of "nowhere". Also, the use of + between the two terms and * as a weighing factor seems arbitrary, since + and * are just two of an infinitude of operations in the "hyper-system".

I've also had some ideas on inverse tetration.

Suppose x^x = c

Then define the "square tetraroot", sqtrt function so that:

sqtrt(c) = x

Similarly, the "cube tetraroot" (cbtrt) function is the inverse of

f(x) = x^(x^x)

In general, the "nth tetraroot" is the inverse of

g(x) = x tetra n

I think that none of the nth tetraroot functions can be expressed in terms of "lower" functions (those consisting of addition, multiplication and exponentiation). (Proof?) Also, numerical examples seem to show that the sqtrt function grows a slightly more slowly than the logarithm. That is:

sqtrt(c*n)/sqtrt(n)

seems to decrease slowly, at least when c*n is less than a trillion. Still, it would be nice if someone could prove this. Also, I think that higher-order tetraroots should grow even more slowly (Proof?).

Note that sqtrt(x) has 2 answers when 0 < x < 1. This is due to the shape of the function f(x) = x^x. When working only with real numbers, that function is defined and continuous only when x > 0. Using differentiation, it can be shown that the function decreases from 0 < x < 1/e, has a minimum at (1/e,[1/e]^[1/e]), then increases from there, passing through (1,1). Thus, it would seem that sqtrt(x) is undefined for x < (1/e)^(1/e) ~= 0.6922. But:

(-1)^(-1) = -1
(-2)^(-2) = 1/4
(-3)^(-3) = -1/27
(-4)^(-4) = 1/256

Thus, it seems that the sqtrt function cannot be used with only the real numbers, complex numbers will probably be needed. This means that a complete graph will need (ouch!) 4 dimensions...
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Well, I've thought about it more and it seems I was wrong to say that the tetraroot is the inverse of tetration. Think back to the inverses of "lower" operations:

Multiplication -----> Division
Exponentiation -----> Root and Log

Notice that exponentiation has 2 inverse operations. This is because exponentiation is not commutative. That is, x^y does not, in general, equal y^x. There are two ways to ask, "What is the inverse of exponentiation?" One way is to have a variable base raised to a fixed exponent. Another way is to have a fixed base raised to a variable exponent. In both cases, we are trying to find the operation that gives us the variable quantity. In the first case, the operation is taking the nth root (where n is the fixed exponent). In the second case, the operation is taking the logarithm base n (where n is the fixed base).

Now, tetration is also not commutative. For example:

3 tetra 2 = 3^3 = 27
2 tetra 3 = 2^(2^2) = 16

Thus, we should expect to see two inverse operations, corresponding to each of the 2 following cases:

1. (variable) tetra (constant)
2. (constant) tetra (variable)

Case 1: Let the variable = x and the constant = n. Then:

nth tetraroot (x tetra n) = x.

x is an nth tetraroot of y if and only if:

(x tetra n) = y

A number can have more than one tetraroot of a particular order. For example, 0.030065368 and 0.8881353288 are both (approximately) square tetraroots of 0.9. I can't prove that those are the only square tetraroots. Please let me know if you can find more (all complex number answers are acceptable).

Case 2: Let the variable = x and the constant = n. Then:

nth tetralog (n tetra x) = x

Unfortunately, the only values this function that output at the moment are positive integers, since I don't know how to calculate (n tetra x) when x is not a positive integer.

x is an nth tetralog of y if and only if:

(n tetra x) = y

I'm think, based on the definition of tetration, that the nth tetralog of x is equivalent to taking the nested logarithm (base n) of x, x times. I would be grateful if someone could check on that.

If that reasoning is correct, we can expect the tetraroot function to give non-real answers often, because it will often involve taking the logarithm of a negative number.
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uh oh, looks like the tetraroot can't be used to extend tetration to reals. http://users.forthnet.gr/ath/jgal/math/exponents4.html

Not only is tetration not commutative, it's also not associative.

From the above website:
The main problem showing up after one has defined 1/nm to be the n-th order hyperroot of the real number x, is that there are two "natural" ways to define [sup]m/n[/sup]x, either as [sup]m[/sup]([sup]1/n[/sup]x) or [sup]1/n[/sup]([sup]m[/sup]x).

He goes on to describe a version of tetration that is continuous everywhere. I'm currently trying to create a similar version that is also differentiable everywhere. It's slow work, but I'm nearly there. PWrong
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I have been able to confirm that:

(nth tetraroot x) tetra m doesn't always = nth tetraroot (x tetra m)

Let x = 5, n = 2, m = 3

Then:

(nth tetraroot x) tetra m = (sqtrt (5)) tetra 3 ~= 43.7782288861192637

nth tetraroot (x tetra m) = sqtrt (5 tetra 3) ~= 758.45020035537072545

However, it looks like exponentiation isn't associative either.

http://en.wikipedia.org/wiki/Exponentiation

Quoting from that page:

Whereas addition or multiplication are commutative (for example, 2+3 = 5 = 3+2 and 2×3 = 6 = 3×2), this is not true of exponentiation: 2^3 = 8 while 3^2 = 9. Similarly, whereas addition or multiplication are associative (for example, (2+3)+4 = 9 = 2+(3+4) and (2×3)×4 = 24 = 2×(3×4)), this is not true of exponentiation either: 2^3 to the 4th power is 8^4 or 4,096 ,while 2 to the 3^4 power is 2^81 or 2,417,851,639,229,258,349,412,352.

I've also been thinking about whether its possible to solve equations with tetration in them. For instance, how about solving a "quadratic tetranomial"? That equation would involve an x^x term, an x term, and a constant term. There would be coefficients (possibly more than one per term, since its possibly to multiply or exponentiate a term by a constant). Preferably, we could try to solve this equation using just the elementary functions + the sqtrt function. So far, I've tried without success to solve the equation:

x^x + x - 3 = 0 (my graphing calculator assures me that a real number solution does exist).
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jinydu wrote:However, it looks like exponentiation isn't associative either.

Maybe associative isn't the right word:.? The important thing is that

x[sup]m/n[/sup] = (x[sup]m[/sup])[sup]1/n[/sup] = (x[sup]1/n[/sup])[sup]m[/sup]

But the same isn't true for tetration.

I was just about to try a tetration quadratic before I came on here. I can easily solve it using Newton's algorithm, but that won't give us the answer in elementary functions.

I've also thought about the standard quadratic formula a bit. The formula is based on the idea of factoring the quadratic into two parts.
ax^2 + bx + c = (x-d)(x-e)

I'm not sure if it's possible to express x^x+x-3 any differently. It's very difficult to find any simple rules for combining tetration with other operators.

One problem is that you often end up with something like (x+a)^x
The only thing you can do with that is use the binomial theorum to expand it, which requires x!, which only works for integers.

Another problem is x^x is both a polynomial and an exponential function. I don't know if there's a method for solving equations with exponentials in it. Maybe we should first try solving equations like x^2 - e^x = 0

If tetration became more widely used, it's possible that they might count the tetraroot as an elementary function. That would open a lot of new numbers up that are currently transcendential (cannot be expressed in elementary functions).

Presumably there is also a "pentaroot" and "hexaroot" and so on. If we redefined elementary functions to include all of these roots, down to infinity, I wonder if we could possibly express pi or e using these functions. PWrong
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Well... There is actually no real need for all the names addition, multiplication, exponentiation, tetration, etc. Surely you could just use the hy(a,n,b) function for everything?

Also, another thing to bring up. There are the triangle numbers, 1, 3, 6, 10, 15, etc., where they are formed by 1+2+3+4+...+n. Then there are the factorial numbers formed by 1*2*3*4*...*n. What about the third series - 1^(2^(3^(4^(...))))? What about the fourth series - 1#(2#(3#(4#(...)))) using # to mean the tetra operation?

Now couldn't we define all these serieses (whatever the plural is) as a simple function - iinc(n,s) where n is the number of numbers there are, and s is the series? For example, iinc(4,2) = 4! = 1*2*3*4 = 24.

Also, there is a need for a lower bound. I've seen some places write "incomplete factorials" as a·b which would represent b! / a! or a*a+1*a+2*...*b-2*b-1*b. So 3·5 would mean 3*4*5 = 60.

So we could extend the iinc function to iinc(l,n,s) where l is the lower bound and n is the upper bound. Any thoughts on this? Keiji

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Thanks for the feedback. I really appreciate it.

To PWrong: When I asked if anyone could solve:

x^x + x - 3 = 0

using elementary functions, I meant to include tetration and the square tetraroot (I don't think you would need cube tetraroots) as elementary functions (not in general, just for the purposes of solving that equation).

After all, the sqrt function was defined so that equations like:

x^2 = c

could be solved. I meant to define the sqtrt function analogously, to allow equations like:

x tetra 2 (= x^x) = c

can be solved. I have little reason to believe that sqtrt can be expressed using the preexisting elementary functions, since sqrt cannot be expressed in terms of +, - , * and /. Now, my question is, do you need any more new functions to solve

x^x + x - 3 = 0

or is the sqtrt + the preexisting elementary functions enough?

Also, you're right to say that Newton's method could be used to solve that equation. However, that method may have difficulties with other "tetrapolynomial" equations, such as:

x tetra 1000 = giggol (for example)

You would have to expand that (x tetra 1000) into exponential form, and differentiate that ! Unless, of course, you're smart enough to differentiate that function in tetra form. Just for starters:

Derivative of (x tetra 2) = (x tetra 2)*(1+ln x)
Derivative of (x tetra 3) = (x tetra 3)*(x^(x-1) + (x^x)*ln x*(1+ln x))
I got those answers from a mathematics software called Mathematica. It would be great if someone could differentiate (x tetra n) and express the answer simply enough to allow practical computation using the Newton method for the above example.

PWrong wrote:Presumably there is also a "pentaroot" and "hexaroot" and so on. If we redefined elementary functions to include all of these roots, down to infinity, I wonder if we could possibly express pi or e using these functions.

Whoa! I never thought of that! An exact and finite formula for pi and e, instant Fields Medal!

To bobxp: Yes, you're right, there is no logical need for the names addition, multiplication, etc. However, the first three are already deeply embedded in our language (can you imagine telling a 3-year old: "hyper 2, 1, 2 equals 4" instead of "2 plus 2 equals 4"?). As for tetration, well, look at the name of this forum .

I guess its similar with other things in math too. For example, we don't need to say "squaring" and "cubing". We could just say "raising to the second power" and "raising to the third power".

As for the iinc (by the way, why did you choose that name) function, I think its a great idea. Now, I'll try to post what I can think of at the moment, and ask some of my own questions.

It turns out that 1+2+3+...+n can be expressed simply using multiplication:

1+2+3+...+n = n*(n+1)/2

However, I don't know of a way to express 1*2*3*...*n simply in terms of exponentiation. Still, I do know of this: http://mathworld.wolfram.com/StirlingsA ... ation.html
Admittedly, its nowhere near as crisp and elegant as n*(n+1)/2, and its not even an exact answer. Its possible that an exact answer is not possible in terms of exponentiation, but perhaps tetration can do the job!

Let's take a look at the third series:

iinc(1,3) = 1
iinc(2,3) = 2
iinc(3,3) = 8
iinc(4,3) ~= 2.41785*10^24

I'm quite sure that iinc(n,3) also grows faster than (n tetra c). The tower of exponents gets taller and taller with iinc, while it stays at the same height for (n tetra c). Numerical examples seem to show that the height of the power tower is much more important than the numbers in the tower. I'm also quite sure iinc(n,3) grows faster than (c tetra n) because they both have n levels of exponents, but the exponents grow in the iinc function while the exponents remain constant in the (c tetra n) function. This is analogous to the factorial case. The factorial function grows faster than any polynomial (analogue to n tetra c) or exponential (analogue to c tetra n) function.

However, iinc(n,3) grows slower than (n tetra n). This is obvious when I put them into exponential form, bearing in mind that the heights of the power towers are equal (I assume that you know what order to perform the exponentiations, so I leave out the parentheses):

iinc(n,3) = 1^2^3^4^5^...^n
n tetra n = n^n^n^n^n...^n

We can express all this using what I think are elegant inequalities. Here, I use the "<" symbol to mean "grow more slowly than", rather than its usual meaning:

(n^c) < (c^n) < iinc(n,2) < (n tetra 2)
(n tetra c) < (c tetra n) < iinc(n,3) < (n penta 2)

I could probably do a similar analysis for the 1#2#3#4#..., but I'm feeling tired right now. I think the results should be similar. Maybe you could try that.

There shouldn't be any problem with setting a lower bound other than 1 for any of these functions, as far as I can tell.
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Sorry for the double post, that last one was really getting too long. I have some other ideas about tetration too.

First, I have found a property that applies to addition, multiplication and exponentiation, but does not apply to tetration:

(a o b) o c = (a o c) o b where o represents the operation

Addition: (a+b)+c = a+b+c = a+c+b = (a+c)+b

Multiplication: (a*b)*c = a*b*c = a*c*b = (a*c)*b

Exponentiation: (a^b)^c = a^(b*c) = a^(c*b) = (a^c)^b

To show that this is not true for tetration, it will suffice to show a single example: a = 2, b = 3, c = 2

(a tetra b) tetra c = (2 tetra 3) tetra 2 = 16 tetra 2 ~= 1.84467440737096 * 10^19

(a tetra c) tetra b = (2 tetra 2) tetra 3 = 4 tetra 3 ~= 1.34078079299426 * 10^154

In order to help simplify expressions containing tetration, it would help if we could find some property that is true for tetration. But so far, I haven't found any that are worth mentioning.

There are also other issues with tetration.

One problem is that computers, the way they are currently designed, often do a poor job of calculating tetration for "reasonably large" input values. For example, 4 tetra 3 overloads most handheld calculators (which can only deal with numbers less than a googol, 10^100). Big, heavy, desktop computers can improve on this, but not by much. My big P4 3.055 GHz computer at home, armed with Mathematica 5, can handle numbers beyond 10^(~600 million, I don't remember the exact number), but gets overflowed by 10^(700 million). I would guess that supercomputers could calculate up to 10^(10^12). I think we can be quite sure that using the current system of number representation, we will never have a computer that can calculate up to a googolplex, 10^(10^100), which is less than (4 tetra 4). If I'm not mistaken, computers store and manipulate numbers in binary representation. That means that the amount of memory needed to store a number is proportional to the number of digits that number has. And the number of digits a number has is proportional to the log of the number. Putting this in a more mathematical form:

M = c * log (N)

where M is the memory needed and N is the number to be stored. Rearranging the equation:

N = 10^(M/c)

Herein lies the problem. If you add in a constant amount of memory to your computer, the maximum number you can store grows exponentially. If your memory grows exponentially (like you would expect from Moore's Law), the number grows like 10^(10^x), with x increasing linearly. That function grows more slowly than (x tetra 3). Thus, you have to wait longer and longer for new computer technology every time you increase x by 1 (and this is assuming Moore's Law continues to hold indefinitely). The situation is even more hopeless for (x tetra 4), and "infinitely" more hopeless for (2 tetra x). Basically, my point is that we would need a new standard system of representation, and this ties in with my next point:

We need a standard form, which is capable of easily comparing numbers, for expressing tetrated numbers. For most numbers we deal with in the everyday world, scientific notation suffices:

a * 10^n where 1 <= a < 10 and n is a positive number small enough to be written in decimal form.

I think that a useful form for writting tetrated numbers is:

(10 tetra n)^a where 0.1 < a <= 1 and n is a positive number small enough to be written in decimal form.

For example:
2 ~= (10 tetra 1)^0.3010299957
10 = (10 tetra 1)^1
100 = (10 tetra 2)^0.2
500 ~= (10 tetra 2)^0.269897
7000 ~= (10 tetra 2)^0.384509804
800 million ~= (10 tetra 2)^0.890308999
(3 tetra 3) ~= (10 tetra 3)^(1.28822739 * 10^-9)

As you can see, that form has its flaws, and it would likely need modifications. But its the best I can think of now.
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I remember posting a new notation on this forum capable of storing huge numbers with erratic lengths required for each. Maybe that could be used to store numbers in a computer? And also, to display numbers that are too big to be calculated. Also, I made a program in C++ a while ago that can do addition, subtraction and multiplication (that's all so far) on HUGE whole numbers (any number of bytes, only limited by the width of the screen) but only in hexadecimal. Keiji

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bobxp wrote:I remember posting a new notation on this forum capable of storing huge numbers with erratic lengths required for each. Maybe that could be used to store numbers in a computer? And also, to display numbers that are too big to be calculated. Also, I made a program in C++ a while ago that can do addition, subtraction and multiplication (that's all so far) on HUGE whole numbers (any number of bytes, only limited by the width of the screen) but only in hexadecimal.

I meant a standard (and hopefully accurate) "tetration notation" that can represent and operate on numbers far larger than computers can currently handle.

Suppose your computer has 1 TB (one terabyte) of storage space, which is far beyond what most PCs have at the moment. 1 byte = 8 bits, therefore, you have about 8 * 10^12 bits of memory. Now, suppose you could use all of that memory to represent a single large number, using the current system. Each bit could represent one binary digit of the number. Thus, the largest number you could represent would be about 2^(8*10^12), or if you prefer base ten, 10^(2.41*10^12). While that number may seem large, it is in fact less than (11 tetra 3), hardly a giant in the universe of numbers that can be conveniently represented using tetration. In practice, no current PC can store numbers anywhere close to that size due to things like program limitations (and the fact that most people want their PC to have an operating system and other software).

So I was thinking about a new system of number representation using tetration. It should be able to:

1) Handle numbers as large as a giggol, (10 tetra 100)
2) Allow simple comparison of two numbers, so that a user can easily see which one is larger
3) Perform computations on numbers (basic operations, as well as algorithms like the Newton Method)
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To PWrong: When I asked if anyone could solve:

x^x + x - 3 = 0

using elementary functions, I meant to include tetration and the square tetraroot (I don't think you would need cube tetraroots) as elementary functions (not in general, just for the purposes of solving that equation).

I did understand that, but I can't solve the equation. The best I could do algebraically is this:
x^x + x - 3 = 0
x^x = 3-x
Taking tetraroots of both sides:
x=sqrtrt(3-x)

But I have no idea where to go from there.

In order to help simplify expressions containing tetration, it would help if we could find some property that is true for tetration. But so far, I haven't found any that are worth mentioning.

I tried for a few weeks and couldn't find anything decent either. It would be really nice to have something like
(a+b)^2 = a^2 + 2ab + b^2,
for tetration.

One rule I do have is this.
(a+b)^^2 = (a+b)[sup](a+b)[/sup]
= (a+b)[sup]a[/sup](a+b)[sup]b[/sup]

But you can't do much else other than use the binomial theorum. I really don't want to do that, as you'd end up with a huge pile of ugly factorials, and you probably wouldn't find an answer anyway.

bobxp wrote:There are the triangle numbers, 1, 3, 6, 10, 15, etc., where they are formed by 1+2+3+4+...+n. Then there are the factorial numbers formed by 1*2*3*4*...*n. What about the third series - 1^(2^(3^(4^(...))))?

You'd have to go the other way round, otherwise it would always equal 1.
Try something like
iinc(4) = 4^(3^(2^(1)))=26144

Jinydu wrote:Its possible that an exact answer is not possible in terms of exponentiation, but perhaps tetration can do the job!

I don't know if you noticed equation 9 on the page you linked to, about Stirling's approximation:
n! = n^n * e^-n * sqrt(2 pi n)
Looks like tetration has already done the job!

And have a look at this limit for e. It's equation 5 on
http://mathworld.wolfram.com/e.html
Not only does it use tetration 4 times, but the limit is apparently connected to prime numbers somehow!

And speaking of primes, I've considered a version of the prime numbers that involves exponentiation instead of multiplication. A prime number can't be expressed as a product of more than two factors, i.e. it can't be expressed as a*b for integers other than 1 and itself. What about numbers that can't be expressed as a^b, with similar restrictions? Obviously, there are many more "exponential primes" than "multiplicative primes". However I did some graphs a while back, and they appear to be scattered at random. I also tried, "tetrational primes", but nearly every number is a tetrational prime so it's difficult to look for patterns.

Jinydu wrote:You would have to expand that (x tetra 1000) into exponential form, and differentiate that ! Unless, of course, you're smart enough to differentiate that function in tetra form.

I know how to differentiate x^^n in terms of the derivative of x^^(n-1). You can do this over and over until you get a proper expression.

By the definition of tetration:
x^^n = x[sup](x^^(n-1))[/sup]

Now we turn this into an exponential function by taking e to the log:

= e ^ln(x[sup](x^^(n-1))[/sup])
= e [sup]x^^(n-1) * ln(x)[/sup]

Now we differentiate the top part using the product rule, and multiply by the original function.

= [ d/dx(x^^(n-1))*lnx + x^^(n-1)*d/dx(lnx) ]e[sup]x^^(n-1) * ln(x)[/sup]

= [ d/dx(x^^(n-1))*lnx + x^^(n-1)/x ]x^^n

Hope that helps. You should be able to do that for x^^1000, provided you have a supercomputer handy.

On the subject of large numbers, apparently someone recently broke a record for the largest finite number ever used in physics (or rather, the smallest number). Using quantum physics, they calculated the approximate probability that quantum fluctuations will result in the big bang happening, again. The probability is 1:10^10^100, but they didn't say which order to read it in! Apparently it's so small, that it doesn't really matter what units you measure in. So the chance of it happening within a hundred years, somewhere in this galaxy, is almost the same as the chance that there's another universe brewing in your fridge, right now. Creepy stuff. If there's too many replies to this, it should probably go in a different topic. But it might put into perspective how useless tetration is when only applied to the integers.

This post is getting really long too, so I'll do a double post to explain my continuous/differentiable tetration. PWrong
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PWrong wrote:

I've been exposed! Ok, I cheated a bit:

iinc(2) = 2
iinc(3) = 2^3
iinc(4) = 2^(3^4)

As you can see, I ignored the initial 1, to avoid always getting the answer 1. Still, its interesting to note that

1 + 2 + 3 + ... + n = n + (n-1) + (n-2) + ... + 1
1 * 2 * 3 * ... * n = n * (n-1) * (n-2) * ... * 1

but

1^(2^(3^(...^n))))) not = n^((n-1)^((n-2)^...^(1)))))

Thus, it seems your way of defining iinc(n,3) is consistent, while mine isn't.

PWrong wrote:
I don't know if you noticed equation 9 on the page you linked to, about Stirling's approximation:
n! = n^n * e^-n * sqrt(2 pi n)
Looks like tetration has already done the job!

Not exactly, its only an approximation. For example, when n = 2, the formula gives approximately 1.919, instead of the correct answer, 2. However, the approximation gets better as n increases. For n = 20, the formula is off by about 0.41577% (you can try it yourself).

PWrong wrote:
And have a look at this limit for e. It's equation 5 on
http://mathworld.wolfram.com/e.html
Not only does it use tetration 4 times, but the limit is apparently connected to prime numbers somehow!

You have got a point there. But do you think its possible to express e exactly in closed form (no limits as n approaches infinity) using tetration? That would really be interesting.

PWrong wrote:
You should be able to do that for x^^1000, provided you have a supercomputer handy.

Maybe I'm asking for too much, but is it possible to get a simple, closed formula for the derivative. I mean, the derivative of x^n is simply n*x^(n-1). The reason is to allow efficient use of Newton's Method to solve equations with tetration. For instance

x^1000 = c is easy to solve numerically

but

x tetra 1000 = c would be a real pain

Also, I've found another difficulty with extending tetration, but maybe your idea will do the trick. This problem follows quickly from the definition fo tetration. Note: log (p,q) = log (base p of q)

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

taking the log (base x) of both sides:

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

x tetra 1 = x

Thus:

x tetra 0 = log (x,x) = 1
x tetra -1 = log (x,1) = 0
x tetra -2 = log (x,0) which is undefined
jinydu
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I've split out these posts about tetration and large number sequences from the original topic. What about 5.5! for example? :? Keiji

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bobxp wrote:What about 5.5! for example? :?

That's already been taken care of: See http://mathworld.wolfram.com/GammaFunction.html.

5.5! = Gamma(6.5) = Integral (((ln(1/t))^(5.5))dt,0,1) ~= 288

If you scroll down to the middle of the page, Equation 52 gives an exact formula Gamma(n/2), where n is odd. It turns out that there is an exact answer for 5.5!:

5.5! = 10395pi/64

Take a look at http://mathworld.wolfram.com/Superfactorial.html for two definitions of "Superfactorial". The second definition (by Sloane and Plouffe) evidently grows more slowly, so I'll analyze it first:

1\$ = 1! = 1
2\$ = 1! * 2! = (1)*(2*1) = (2) * (1)^2 = 2
3\$ = 1! * 2! * 3! = (1)*(2*1)*(3*2*1) = 3 *(2)^2 * (1)^3 = 12
n\$ = 1! * 2! * ... * n! = (1)*(2*1)*...*(n*(n-1)*(n-2)...)
= n * (n-1)^2 * (n-2)^3 * ... (2)^(n-1) * (1)^n

n\$ < (n^n)*n = (n tetra 2)*n

Now, I'll look at the first definition, by Pickover (I'm guessing on what order I should exponentiate. I chose the one that gives the highest value).

1\$ = 1! = 1
2\$ = 2!^2! = 2^2 = 4
3\$ = 3!^3!^3!^3!^3!^3! = 6^6^6^6^6^6
n\$ = (n! tetra n!) = (n! penta 2)
jinydu
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It looks like some people are WAY ahead of us:

jinydu
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LOL, a bit TOO far ahead for me, I'll just stick here  Keiji

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I don't think we'd fit in on that forum. They all sound like real pure mathematicians, so they might not appreciate amatuers like us.

I did notice that they mention Ioannis Galidakis a few times.
Galidakis was presented as the leading researcher in tetration.

I found this page a while back, and it sounds like noone else has improved on it yet.
http://users.forthnet.gr/ath/jgal/math/exponents4.html
This website is a bit complicated, so I'll give you the gist of it.

Galidakis found a continuous extension for n^^x. Instead of using the hyperroot function, he simply adds an extra n, and puts the fractional part of x at the top of the "power tower".

n^^2.0 = n^(n^(n^0))
since n^0 = 1:
n^^2.0 = n^n

n^^2.5 = n^n^(n^0.5) = n^n^(n^0.5)

The function is continuous because as x approaches an integer, it approaches the same value from either direction.

2^^2.9999 = 2^2^(2^.9999) = 15.99573
2^^3.0001 = 2^2^2^(2^.0001) = 16.00295

His function is continuous for all complex n, and real x. Because the base is complex, I tend to use z instead of n, and t instead of x.

Anyway, when I found this I realised that his function wasn't differentiable. I've been working on a differentiable version for a while, and I've just finished. All I did was put an arbitrary function of t at the top of the tower, instead of t.

i.e.
z^^t = z^z^z^...^(z^f(t))

Then I found some conditions for the "top-function" as I call it, so that z^^t is differentiable for all z.

I won't got through the whole process I went through, but I'll tell you the main points.

The derivative respect to t is:
[z^^t * z^^(t-1) * z^^(t-2) * ... * z^z^f(t) * z^f(t)] * f'(t) * (ln(z))^int(t)
(how about that for a superfactorial?)

The conditions I found basically ensure that the limits are the same on both sides.
1. Tetration is continuous iff
f(0) = 0
f(1) = 1

2. The derivative is continuous iff
f'(1) = f'(0)*ln(z)
note that this condition is in terms of z. This is so that

I also found a condition for the second order derivative.

3. The second derivative is continuous iff
f''(1) = f''(0)*ln(z) +(f'(0)*ln(z))^2

I've yet to find conditions for the 3rd order derivatives, but it's

Then, I found a suitable polynomial top-function to satisfy the conditions.

Interestingly, the simplest top-function for continuity is linear, the top-function for the 1st derivative is a quadratic, and the top-function for the 2nd derivative is a cubic.

This would all be great, but it turns out that the functions themselves are huge and inelegant. They're definitely correct; I've checked them by graphing.

It's not neccessary to post the whole thing, but the coefficients, especially for the 2nd derivative, are ugly combinations of ln(z), full of square roots and a lot of unexpected numbers. It's also very difficult to simplify, even though there are a few repeating bits. It just strikes me as odd that a polynomial with a special property looks so arbitrary. PWrong
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PWrong wrote:
Interestingly, the simplest top-function for continuity is linear, the top-function for the 1st derivative is a quadratic, and the top-function for the 2nd derivative is a cubic.

I guess this means that if you want the first n derivatives to be continuous, you would need a polynomial of degree n+1.

Here's an important question: Do you always get exactly the same answer for (x tetra y), regardless of whether you use the linear, quadratic or cubic "top-function"? This is important because we (x tetra y) should have a unique value.

Also, does your extension of tetration satisfy:

(x tetra 1/2) = sqtrt(x)?
(x tetra 1/3) = cbtrt(x)?
etc.

I would also like to know, how do you calculate (x tetra -1) and (x tetra -2)?
jinydu
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jinydu wrote:I would also like to know, how do you calculate (x tetra -1) and (x tetra -2)?

x tetra 0 = x
x tetra -1 = sqrt(x) (NOT SQTRT!!!!)
x tetra -2 = rt[sub]3[/sub](x)
x tetra -3 = rt[sub]4[/sub](x)
x tetra -n = rt[sub]n[/sub](x)

I don't know if that is right though. Keiji

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bobxp wrote:x tetra 0 = x
x tetra -1 = sqrt(x) (NOT SQTRT!!!!)
x tetra -2 = rt[sub]3[/sub](x)
x tetra -3 = rt[sub]4[/sub](x)
x tetra -n = rt[sub]n[/sub](x)

I don't know if that is right though.

That would lead to the strange property:

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

Then,

x tetra 1 = x tetra -(-1) = x^(1/-1) = x^-1 = 1/x
x tetra 2 = x tetra -(-2) = x^(1/-2) = 1/(x^(1/2)) = 1/s(qrt(x))

Thus, it doesn't seem to be consistent with the definition of tetration for positive integers.

-----------------------------------------------------------------------------------

Now, today I had an interesting talk with my Math teaching assistant. He told the class about how to differentiate the funcion, f(x) = x^x.

First, he quickly brushed through the "conventional" way of doing it:

f(x) = x^x = e^(x*ln x)
f'(x) = (x*ln x)' * e^(x*ln x)
f'(x) = (1+ln x)*(x^x)

Then, he talked about his friend's approach to the problem. According to his friend, people commonly make 2 kinds of mistakes when trying to differentiate f(x):

1) They think of the exponent as a constant (that is, the problem can be thought of as x^a). Therefore:

f'(x) = x*(x^(x-1)) = x^x

2) They think of the base as a constant (that is, the problem can be thought of as a^x). Therefore:

f'(x) = (x^x)*(ln x)

However, something very interesting happens when you add both incorrect answers together:

x^x + (x^x)*(ln x) = (1+ln x)*(x^x) = f'(x)

In this case, two wrongs do make a right!

I was very surprised when I heard this, and asked my TA to try differentiating something much more general:

h(x) = f(x)^g(x)

First, he did it the "conventional way" (I admit its quite hard to read using text box formatting):

Let y = h(x)
ln y = g(x)*ln(f(x))
(1/y)*h'(x) = g'(x)*ln(f(x)) + g(x)*(f'(x)/f(x))
h'(x) = (f(x)^g(x))*((f'(x)*g(x)/f(x)) + ln (f(x))*g'(x))

Then, he tried making both types of "mistakes":

1) g(x) is a constant. Let g(x) = a
(f(x)^a)' = a*(f(x)^(a-1))*f'(x) = (f(x)^g(x))*((f'(x)*g(x))/f(x))

2) f(x) is a constant. Let f(x) = a
(a^g(x))' = g'(x)*(a^g(x))*(ln a) = (f(x)^g(x))*(ln (f(x))*g'(x))

Adding them together (and using the distributive property):

(f(x)^g(x))*((f'(x)*g(x))/f(x) + ln (f(x))*g'(x))

Remarkably, this is exactly the correct answer. Furthermore, I have verified everything using the Mathematica.

I've also tried applying this "technique" to addition (h(x) = f(x) + g(x)) and multiplication (h(x) = f(x)*g(x)). The technique correctly gave the Sum and Product Rules.

The obvious thing to do with the (proven) Exponentiation Rule is to apply it to tetration.

Let f(x) = x, g(x) = (x tetra n-1)
Then h(x) = (x tetra n)

(x tetra n)' = (x tetra n)*((x tetra n-1)/x + (ln x)*(x tetra n-1)')

x tetra 1 = x. So we already know that (x tetra 1)' = 1. Thus, we can derive the derivatives for the next few values of n:

(x tetra 2)' = (x tetra 2)*(x/x + ln (x)*1) = (x tetra 2)*(1+ln x)
(x tetra 3)' = (x tetra 3)*((x tetra 2)/x + (ln x)*(x tetra 2)') = (x tetra 3)*((x tetra 2)/x + (ln x)*(x tetra 2)*(1+ln x))

This is not as good as a fully fledged formula for (x tetra n)' in terms of x, n and tetration + "lower" operations, but it's still progess.
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jinydu wrote:I guess this means that if you want the first n derivatives to be continuous, you would need a polynomial of degree n+1.

I hope so. I'm having trouble finding the n'th derivative of tetration, but I think I've found a common pattern in the conditions for the top-function. I'd really like tetration to be continuous for all derivatives, so for that I'd need an infinite polynomial.

jinydu wrote:Here's an important question: Do you always get exactly the same answer for (x tetra y), regardless of whether you use the linear, quadratic or cubic "top-function"? This is important because we (x tetra y) should have a unique value.

Nope. It's different for each top-function, unless x is an integer. I'd like tetration to be continuous for every derivative. If it turns out that you need a polynomial of degree n+1 for the first n derivatives to be continuous, then I'll need an infinite polynomial. That's not a problem, as I'd just need a sum to infinity.

My eventual plan was to find the n'th derivative of z^^t, then use it to find an infinite polynomial for the top-function, (using a sum to infinity to express it). If I could do that, I would define x^^y using only this function. Unfortunately, it turns out to be impossible. It's not really a matter of a single polynomial. The top-function has to be in terms of the ln of z.

f(x) = ax^2 +bx + c

From my conditions, you can show that

c=0
a+b= 1
2a + b=blnz

From these,

b = 2/(lnz+1)
a = 1 - 2/(lnz+1)

So the function itself is
f(x) = [ 1 - 2/(lnz+1) ]x^2 + [ 2/(lnz+1) ]x

This is complicated enough, but the cubic is enormous. To make it more simple, I let w = lnz.

From the conditions, you can find similar equations to those above, and then end up with a quadratic involving c. But each coefficient is a polynomial of w. So you can find c in terms of w with the quadratic formula, but the expression is huge. Once you do that, you can find expressions for b and a.

I just had a go at the quartic, and naturally it's even worse. I ended up with a cubic involving d, again with the coefficients in terms of w. I know vaguely how to solve cubics, so I'll have a go after the final exams, when I'll have plenty of time. Besides, even if I do that, and the next one, I'd then have to solve a quintic, which Godel proved is impossible using elementary functions. But I have no intention of going that far for a few years anyway. Maybe I'll be able to do it with "non-elementary" functions one day. jinydu wrote:Also, does your extension of tetration satisfy:

(x tetra 1/2) = sqtrt(x)?
(x tetra 1/3) = cbtrt(x)?
etc.

Not at the moment. At least, it doesn't satisfy them using the quadratic or cubic top-function. If it does for any other, I'll be very surprised. Maybe I should use that as a starting point, rather than assuming that tetration is differentiable.

jinydu wrote:I would also like to know, how do you calculate (x tetra -1) and (x tetra -2)?

Frankly, I have no idea. Galidakis only defined his function for reals. Since the top-function uses the fraction part, it doesn't really have any meaning for values outside 1 and 0.

This is just speculation, but I think tetration might be connected with probability. For one thing, we have numbers that have to be between 1 and 0. Like probabilities, they have no meaning otherwise. Furthermore, one equation that keeps popping up for all the top functions is that the coefficients all sum to 1 (just like probabilities sum to one).

That derivative technique is very interesting. Do you think that would work for any operation? PWrong
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PWrong wrote:Besides, even if I do that, and the next one, I'd then have to solve a quintic, which Godel proved is impossible using elementary functions. But I have no intention of going that far for a few years anyway. Maybe I'll be able to do it with "non-elementary" functions one day. Well, first of all, it was Abel, not Godel, who proved that the general quintic is unsolvable by radicals. Also, not all quintics are unsolvable by radicals. For example:

x^5 -32 = 0

is solvable (obviously, x = 2). What Abel proved is that not all quintics can be solved using the following operations: +, -, x, /, ^2, ^3, ^4, ^5, sqrt, cbrt, 4th rt and 5th rt. Thus, its still possible that your quintic could be a solvable quintic. However, I realize I'm being a nitpicker. Even if you could solve the quintic, you would then have to deal with a 6th degree polynomial, and even then, that would be a long way from a general formula.

PWrong wrote:That derivative technique is very interesting. Do you think that would work for any operation?

I don't know. My teaching assistant just told me about it yesterday. I guess it might depend on what you mean by an "operation". According to Mathworld (http://mathworld.wolfram.com/Operation.html):

"Let A be a set. An operation on A is a function from a power of A into A. More precisely, given an ordinal number (alpha), a function from (A^alpha) into A is an (alpha)-ary operation on A. If (alpha=n) is a finite ordinal, then the n-ary operation f is a finitary operation on A."

Do you I understand the above? No. The way I have used the word "operation" so far is eerily similar to what a function is. After all, addition, multiplication, exponentiation and tetration can be thought of as (2-variable) functions:

f(x,y) = x+y
g(x,y) = x*y
h(x,y) = x^y
k(x,y) = x tetra y

Anyway, about the exponentiation rule, I can also use it to try and differentiate (n tetra x). Keep in mind that in order for (n tetra x) to be differentiable at all, it must be a continuous function (at least in a certain interval)! In other words, if this derivative is correct, then it is possible to define (a tetra b) for all real values of b within a certain (possibly infinite!) interval. Ok, here goes:

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

(n tetra x)' = (n tetra x)*(ln n)*(n tetra x-1)'

This equation implies that if (n tetra x-1)' exists, then so does (n tetra x)'. I could rearrange the equation and solve for (n tetra x-1)':

(n tetra x-1)' = (n tetra x)'/((n tetra x)*(ln n))

If I could just show that (n tetra x)' exists at one value of x, that would show that (n tetra x)' is defined for infinitely many values of x. Since we seem to be best at tetrating to positive integers, I would recommend trying to differentiate (n tetra x) for a positive integer value of x. Then, we would know the derivative for all integer values!

One more thing; if we accept the unproven method for differentiating operations, we could come up with the derivative of (f(x) tetra g(x)). However, this formula would be recursive, and thus not nearly as powerful as the Sum, Product and Exponentiation Rules. I'll try it anyway:

Assume g(x) is constant. Let a = g(x)
Use the Chain Rule
f'(x)*(f(x) tetra a)*((f(x) tetra a-1)/f(x) + (ln f(x))(f(x) tetra a-1)')
f'(x)*(f(x) tetra g(x))*((f(x) tetra g(x)-1)/f(x) + (ln f(x))(f(x) tetra g(x)-1)')
(What a mess)

Assume f(x) is constant, Let a = f(x)
Use the Chain Rule
g'(x)*(a tetra u)*(ln a)*(a tetra u-1)'
g'(x)*(f(x) tetra g(x))*(ln f(x))*(f(x) tetra g(x)-1)'

Now, for the biggest mess of all:

(f(x) tetra g(x))' = f'(x)*(f(x) tetra g(x))*((f(x) tetra g(x)-1)/f(x) + (ln f(x))(f(x) tetra g(x)-1)') + g'(x)*(f(x) tetra g(x))*(ln f(x))*(f(x) tetra g(x)-1)'

(f(x) tetra g(x))' = (f(x) tetra g(x))(f'(x)((f(x) tetra g(x)-1)/f(x) + (ln f(x))(f(x) tetra g(x)-1)') + g'(x)*(ln f(x))*(f(x) tetra g(x)-1)')

If it wasn't for those recursive derivatives (and the fact that it is unproven), we could use the above formula to launch a frontal assault on the differentiation of pentation. For example, substituting in f(x) = g(x) = x, we have the formula for (x tetra x)' = (x penta 2)'

(x penta 2)' = (x penta 2)((x tetra x-1)/x + 2(ln x)(x tetra x-1)'))

However, it is a long formula, a text box is not the easily place to type a formula, and I cannot use the computer to verify my result (since it obviously doesn't recognize tetration); so please check the calculations if you can, and forgive me if I made any mistakes.
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Well, first of all, it was Abel, not Godel, who proved that the general quintic is unsolvable by radicals.

Oops, I was thinking of Galois, not Godel. I think Galois and Abel proved it at the same time. Anyway, this quintic would almost certainly be a more general quintic, since the coefficients are all functions of w.

Keep in mind that in order for (n tetra x) to be differentiable at all, it must be a continuous function (at least in a certain interval)! In other words, if this derivative is correct, then it is possible to define (a tetra b) for all real values of b within a certain (possibly infinite!) interval.

Differentiation implies that the function is already defined for reals. I don't think you could find the derivative first, and use that to define it for reals.

Specifically:
(n^^x)' = the limit of [n^^(x+h) - n^^x] / h
as h approaches 0
Maybe that would be a better starting point. PWrong
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PWrong wrote:Differentiation implies that the function is already defined for reals. I don't think you could find the derivative first, and use that to define it for reals.

Specifically:
(n^^x)' = the limit of [n^^(x+h) - n^^x] / h
as h approaches 0
Maybe that would be a better starting point.

Right back to that problem again. I don't know how to define (n tetra (x+h)) - (n tetra x) unless both x and h are positive integers.

However, I have managed to get something done: Prove that the square tetraroot of an integer is either an integer or irrational.

Assume that:

sqtrt(n) = p/q where p and q are integers with no common factors

By definition of sqtrt:

n = (p/q)^(p/q)

n^q = (p/q)^p

n^q = (p^p)/(q^p)

Now, the left hand side is clearly an integer. However, since p and q have no common factors, p^p and q^p cannot have any common factors either, since raising an integer to an integer power only repeats the existing factors. Since the numerator and denominator have no common factors, there is no hope of cancelling out all of (q^p)'s factors, unless q^p = 1. There are two ways this equation might be satisfied:

p = 0. However, plugging this into the equation "n = (p/q)^(p/q)", leads to n = 0^0, an unacceptable result.

q = 1. Plugging this into the original equation "sqtrt(n) = p/q" yields sqtrt(n) = p, which basically says that sqtrt(n) can be an integer. If this is not the case, then the original assumption that "sqtrt(n) = p/q" must be false, and hence sqtrt(n) is irrational.

This proof establishes that (unsurprisingly) sqtrt(2), sqtrt(3), sqtrt(5), sqtrt(6), etc. are all irrational.

Maybe I'll try extending this to cube tetraroots later. A proof for nth tetraroots would probably be a real challenge.

Also, if I could establish that sqtrt(2) is transcendental, that would be a significant acheivement, since it add one more entry to the following "sequence":

Natural numbers, Integers, Rational Numbers, Radical Numbers, Algebraic Numbers, ...

Note that the integers are obtained using subtraction, rationals using division, radicals using (exponential) roots and algebraics using polynomials (which are based on exponentiation).
jinydu
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That's a nice, elegant proof. At least not everything in tetration is ridiculously complicated.

As for my continuous expression, I think I've found a way to find the conditions for n'th order differentiability, without having to find the n'th derivative of the entire tetration function.

I looked at my list of "conditions", and I found a pattern that makes perfect sense.

My conditions were f(1)=1 and f(0)=0

However, since z^0=1, you can rearrange this as,

f(1) = z^f(0) and f(0)=0
Interestingly, when you differentiate both sides, you get,

f'(1) = z^f(0) f'(0) ln(z)
this gives the next condition,
f'(1) = f'(0) ln(z)

If you differentiate again, you get the next condition, and so forth. I haven't proved this rigidly yet, but it seems to work. Up until now I had to differentiate z^^t entirely n times, to get the n'th condition. But now it should be easier.

The only problem is, the n'th derivative of z^f(x), or even e^f(x) is actually quite difficult.

A similar problem, the n'th derivative of f(x)*g(x) is relatively simple. I tend to ignore the (x) for simplicity, and just use f and g.

fg

f'g + fg'

f''g + 2f'g' + fg''

f'''g + 3f''g' + 3f'g'' + g'''

This is just like the expansion of (a+b)^n.

I was hoping there might be a similar expansion of e^f, but I simply can't find one.

e^f

f'e^f

f'^2 e^f + f'' e^f

f'^3e^f + 2 f' f'' e^f + f'' f' e^f + f''' e^f
=f'^3e^f + 3f' f'' e^f + f''' e^f

f'^4 e^f + 6f'^2f''e^f + 3f''^2e + 4f'f''e^f + ef''''

The main problem is that it implements the product rule, power rule and e^f(x) rule all at once, so you get a lot more complexity. I've tried a lot of ways to figure it out, but I can't seem to get my head around the problem. Any suggestions? PWrong
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