Discoveries by Daedalus

[Daedalus]

The Next Operator Beyond Exponents:

I was looking into operations greater than exponents when I discovered a few properties of nested exponentiation. I'm not claiming that I am the first to discover these properties. I am well aware of work that has been done on tetration and the Ackermann function. However, I have not found these properties of nested exponentiation anywhere. Also, according to Wikipedia, nested exponentation is not even listed as a hyperoperation:

 

Shown here are examples of the first four hyper operators, with tetration as the fourth (and succession, the unary operation denoted a' = a + 1 taking a and yielding the number after a, as the 0th):

• Addition [math]a + n = a\!\underbrace{''{}^{\cdots}{}'}_n[/math] [math]a[/math] succeeded [math]n[/math] times.
• Multiplication [math]a \times n = \underbrace{a + a + \cdots + a}_n[/math] [math]a[/math] added to itself, [math]n[/math] times.
• Exponentiation [math]a^n = \underbrace{a \times a \times \cdots \times a}_n[/math] [math]a[/math] multiplied by itself, [math]n[/math] times.
• Tetration [math]{^{n}a} = \underbrace{a^{a^{\cdot^{\cdot^{a}}}}}_n[/math] [math]a[/math] exponentiated by itself, [math]n-1[/math] times.

Another clue is that my version of Mathematica does not have any of these properties listed as operators and does not know how to simplify equations using the new operations I have derived. I fully understand that this does not mean that I am truly the first one to discover these properties, but I have been searching for a reference to them ever since I discovered these properties in 2005. Also, I have shown these to a few math professors at UCO and OU. They were completely unaware of these operations themselves and told me to do further research in the math library, which I have done. If you are aware of these properties, then please provide a link so that I can read up on who discovered them and how they are applied.

Nested Exponentiation:

Most of us are aware of nested exponents and how to simplify the math that uses them:

[math](a^b)^c=a^{b\, c}[/math]

So I began to wonder about series of nested exponents and if I could derive any properties associated with them. Now I do realize that this next part may be known, but here is the relationship for a series of nested exponents along with how I denote nested exponentiation:

[math]a^{\left \langle n \right \rangle} = \underbrace{(((a)^a)^a)^{\cdot^{\cdot^{a}}}}_n = a^{a^{n-1}}[/math]

We can demonstrate this by the following:

[math]a^{\left \langle 1 \right \rangle} = a = a^{a^{(1-1)}}[/math]

[math]a^{\left \langle 2 \right \rangle} = (a)^a = a^{a^{(2-1)}}[/math]

[math]a^{\left \langle 3 \right \rangle} = ((a)^a)^a = a^{a^{(3-1)}}[/math]

[math]a^{\left \langle 4 \right \rangle} = (((a)^a)^a)^a = a^{a^{(4-1)}}[/math]

Now that we have the generalized form, [math]a^{a^{n-1}}[/math], that predicts the outcome of continued nested exponentiation, we can derive the derivatives / integrals using Calculus and define an algorithm for determining nested roots using the Newton-Rhapson method. However, the operation that I believe is new is what I call the nested logarithm. The nested log is just like normal logarithms except it determines the value of the nested exponent:

[math]{}_n\, \!\text{Log}_{\, a}\left(a^{\left \langle b \right \rangle}\right)=b[/math]

I've also managed to work out a few generalized forms of the nested logarithm:

[math]{}_n\, \!\text{Log}_{\, a}\left(b\right)=1+\frac{\text{ln}\left (\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}[/math]

[math]{}_n\, \!\text{Log}_{\, a}\left(b\right)=1-\frac{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(b\right)}\right)}{\text{ln}\left(a\right)}[/math]


Proof that [math]{}_n\, \!\text{Log}_{\, a}\left(b\right)=1+\text{ln}(\text{ln}(b) / \text{ln}(a)) / \text{ln}(a)[/math] :


Let [math]a[/math] equal the base of the nested exponent and let [math]b[/math] equal [math]a^{\left \langle n \right \rangle}[/math] which is also equal to [math]a^{a^{n-1}}[/math] :

[math]{}_n\, \!\text{Log}_{\, a}\left(b\right)={}_n\, \!\text{Log}_{\, a}\left(a^{\left \langle n \right \rangle}\right)={}_n\, \!\text{Log}_{\, a}\left(a^{a^{n-1}}\right)=n[/math]

We can now use the general form, [math]a^{a^{n-1}}[/math], to derive the nested logarithm by using natural logarithms:

[math]\text{ln}\left(a^{a^{n-1}}\right)=a^{n-1} \, \text{ln}\left(a\right)[/math]

Divide both sides by [math]\text{ln}\left(a\right)[/math]:

[math]\frac{\text{ln}\left(a^{a^{n-1}}\right)}{\text{ln}\left(a\right)}=a^{n-1}[/math]

Take the natural log of both sides:

[math]\text{ln}\left(\frac{\text{ln}\left(a^{a^{n-1}}\right)}{\text{ln}\left(a\right)}\right)=\left(n-1\right) \, \text{ln}\left(a\right)[/math]

Again, divide both sides by [math]\text{ln}\left(a\right)[/math]:

[math]\frac{\text{ln}\left(\frac{\text{ln}\left(a^{a^{n-1}}\right)}{\text{ln}\left(a\right)}\right)}{ \text{ln}\left(a\right)}=n-1[/math]

Add one to both sides:

[math]1+\frac{\text{ln}\left(\frac{\text{ln}\left(a^{a^{n-1}}\right)}{\text{ln}\left(a\right)}\right)}{ \text{ln}\left(a\right)}=n[/math]

Substitute [math]a^{\left \langle n \right \rangle}[/math] in place of [math]a^{a^{n-1}}[/math]:

[math]{}_n\, \!\text{Log}_{\, a}\left(a^{\left \langle n \right \rangle}\right)=1+\frac{\text{ln}\left(\frac{\text{ln}\left(a^{\left \langle n \right \rangle}\right)}{\text{ln}\left(a\right)}\right)}{ \text{ln}\left(a\right)}=n[/math]

Substitute [math]b[/math] in place of [math]a^{\left \langle n \right \rangle}[/math]:

[math]{}_n\, \!\text{Log}_{\, a}\left(b\right)=1+\frac{\text{ln}\left(\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)}{ \text{ln}\left(a\right)}[/math]

Q.E.D.


Proof that [math]{}_n\, \!\text{Log}_{\, a}\left(b\right)=1-\text{ln}(\text{ln}(a) / \text{ln}(b)) / \text{ln}(a)[/math] :


Working backwards, we can prove that the second form is true. Let [math]a[/math] equal the base of the nested exponent and let [math]b[/math] equal [math]a^{\left \langle n \right \rangle}[/math] which is also equal to [math]a^{a^{n-1}}[/math] such that:

[math]1-\frac{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(a^{\left \langle n \right \rangle}\right)}\right)}{\text{ln}\left(a\right)}=1-\frac{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(a^{a^{n-1}}\right)}\right)}{\text{ln}\left(a\right)}=n[/math]

Swap the variable [math]n[/math] and the term using the natural logarithms:

[math]1-n=\frac{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(a^{a^{n-1}}\right)}\right)}{\text{ln}\left(a\right)}[/math]

Multiply both sides by [math]\text{ln}\left(a\right)[/math]:

[math](1-n)\, \text{ln}\left(a\right)=\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(a^{a^{n-1}}\right)}\right)[/math]

Raise [math]e[/math] to the power of each side:

[math]e^{(1-n)\, \text{ln}\left(a\right)}=e^{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(a^{a^{n-1}}\right)}\right)}[/math]

Simplify the result:

[math]a^{1-n}=\frac{\text{ln}\left(a\right)}{\text{ln}\left(a^{a^{n-1}}\right)}[/math]

Multiply both sides by [math]\text{ln}\left(a^{a^{n-1}}\right)[/math]:

[math]a^{1-n}\, \text{ln}\left(a^{a^{n-1}}\right)=\text{ln}\left(a\right)[/math]

Divide both sides by [math]a^{1-n}[/math]:

[math]\text{ln}\left(a^{a^{n-1}}\right)=\frac{\text{ln}\left(a\right)}{a^{1-n}}[/math]

Raise [math]e[/math] to the power of each side:

[math]e^{\text{ln}\left(a^{a^{n-1}}\right)}=e^{\frac{\text{ln}\left(a\right)}{a^{1-n}}}[/math]

Simplify the result:

[math]a^{a^{n-1}}=a^{a^{n-1}}[/math]

Q.E.D.


Properties of nested exponentiation:


[math]{}_n\, \!\text{Log}_{\, a}\left(b\right)=1+\frac{\text{ln}\left (\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)} \ \ \ \ \ \ \ {}_n\, \!\text{Log}_{\, a}\left(b\right)=1-\frac{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(b\right)}\right)}{\text{ln}\left(a\right)}[/math]


[math]{}_n\, \!\text{Log}_{\, a}\left(b\right)=1+\text{log}_{a}\left(\text{log}_{a}\left(b\right)\right) \ \ \ \ \ \ \ {}_n\, \!\text{Log}_{\, e}\left(b\right)=1+\text{ln}\left(\text{ln}\left(b\right)\right)[/math]


[math]{}_n\, \!\text{Log}_{\, a}\left(b\right)=1-\frac{\text{ln}\left(\text{ln}\left(a\right)\right)}{\text{ln}\left(a\right)}+\frac{\text{ln}\left(\text{ln}\left(b\right)\right)}{\text{ln}\left(a\right)} \ \ \ \ \ \ \ \text{log}_{\, a}\left(b\right)=\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}=a^{\left({}_n\, \!\text{Log}_{\, a}\left(b\right)-1\right)}[/math]


[math]\text{ln}\left(\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)=\text{ln}\left(a\right) \left({}_n\, \!\text{Log}_{\, a}\left(b\right) - 1\right) \ \ \ \ \ \ \ \text{ln}\left(\frac{\text{ln}\left(a\right)}{\text{ln}\left(b\right)}\right)=\text{ln}\left(a\right) \left(1-{}_n\, \!\text{Log}_{\, a}\left(b\right)\right)[/math]


Mathematica 5.2 cannot determine that this identity equals zero:


[math]\left(1+\frac{\text{ln}\left (\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}\right)-\left(1-\frac{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(b\right)}\right)}{\text{ln}\left(a\right)}\right)=0[/math]


It produces the following:


[math]\frac{\text{ln}\left (\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right) + \text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(b\right)}\right)}{\text{ln}\left(a\right)}[/math]


Every number I plug in produces zero, which is what is expected as determined by the above proofs. The following identity is true as determined by the proof below:


[math]\frac{\text{ln}\left (\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right) + \text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(b\right)}\right)}{\text{ln}\left(a\right)} = 0[/math]


[math]\text{ln}\left (\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right) + \text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(b\right)}\right) = 0[/math]


Proof:


Raise [math]e[/math] to the above identity:


[math]e^{\text{ln}\left (\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right) + \text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(b\right)}\right)}=1[/math]


This is equal to:


[math]e^{\text{ln}\left (\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)} \times e^{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(b\right)}\right)}=1[/math]


Which simplifies to:


[math]\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)} \times \frac{\text{ln}\left(a\right)}{\text{ln}\left(b\right)} = 1[/math]


Q.E.D.


Derivatives for nested exponents (Found by using Mathematica):


[math]\frac{dy}{dx} \ x^{\left \langle a \right \rangle} = x^{x^{(a-1)}}\left(x^{(a-2)}\left(1+\left(a-1\right)\, \text{ln}\left(x\right)\right)\right)=x^{\left \langle a \right \rangle}\left(x^{(a-2)}\left(1+\left(a-1\right)\, \text{ln}\left(x\right)\right)\right)[/math]


[math]\frac{dy}{dx} \ a^{\left \langle x \right \rangle} = a^{a^{(x-1)}} \, a^{(x-1)} \, \text{ln}\left(a\right)^2=a^{\left \langle x \right \rangle} \left(a^{(x-1)} \, \text{ln}\left(a\right)^2\right)[/math]


[math]\frac{dy}{dx} \ x^{\left \langle x \right \rangle} = x^{x^{(x-1)}}\left(x^{(x-2)}+x^{(x-1)}\, \text{ln}\left(x\right)\left(\text{ln}\left(x\right)+\frac{x-1}{x}\right)\right)=x^{\left \langle x \right \rangle}\left(x^{(x-2)}+x^{(x-1)}\, \text{ln}\left(x\right)\left(\text{ln}\left(x\right)+\frac{x-1}{x}\right)\right)[/math]


[math]\frac{dy}{dx} \ {}_n\, \!\text{Log}_{\, a}\left(x\right) = \frac{1}{x\, \text{ln}\left(x\right)\, \text{ln}\left(a\right)}=\frac{{}_n\, \!\text{Log}_{\, a}\left(x\right)}{x\, \text{ln}\left(x\right) \left(\text{ln}\left(a\right) +\text{ln}\left(\frac{\text{ln}\left(x\right)}{\text{ln}\left(a\right)}\right)\right)}[/math]


[math]\frac{dy}{dx} \ {}_n\, \!\text{Log}_{\, x}\left(a\right) = \frac{\text{ln}\left(\frac{\text{ln}\left(x\right)}{\text{ln}\left(a\right)}\right)-1}{x\, \text{ln}\left(x\right)^2}=\frac{\text{ln}\left(x\right) \left(1-{}_n\, \!\text{Log}_{\, x}\left(a\right)\right)-1}{x\, \text{ln}\left(x\right)^2}[/math]


Properties of Derivatives for nested exponents (Found by using Mathematica):


Nested exponential function that has the constant as the nested exponent, [math]x^{\left \langle a \right \rangle}[/math]:

[math]\frac{dy}{dx} \ \text{ln}\left(x^{\left \langle a \right \rangle}\right) = \left(x^{(a-2)}\left(1+\left(a-1\right)\, \text{ln}\left(x\right)\right)\right)[/math]

Which allows us to define the following by substitution:

[math]\frac{dy}{dx} \ x^{\left \langle a \right \rangle} = x^{\left \langle a \right \rangle} \ \frac{dy}{dx} \ \text{ln}\left(x^{\left \langle a \right \rangle}\right)[/math]

Such that:

[math]x^{\left \langle a \right \rangle}=\frac{\frac{dy}{dx} \ x^{\left \langle a \right \rangle}}{\frac{dy}{dx} \ \text{ln}\left(x^{\left \langle a \right \rangle}\right)}[/math]



Nested exponential function that has the variable as the nested exponent, [math]a^{\left \langle x \right \rangle}[/math]:

[math]\frac{dy}{dx} \ \text{ln}\left(a^{\left \langle x \right \rangle}\right) = \left(a^{(x-1)} \, \text{ln}\left(a\right)^2\right)[/math]

Which allows us to define the following by substitution:

[math]\frac{dy}{dx} \ a^{\left \langle x \right \rangle} = a^{\left \langle x \right \rangle} \ \frac{dy}{dx} \ \text{ln}\left(a^{\left \langle x \right \rangle}\right)[/math]

Such that:

[math]a^{\left \langle x \right \rangle}=\frac{\frac{dy}{dx} \ a^{\left \langle x \right \rangle}}{\frac{dy}{dx} \ \text{ln}\left(a^{\left \langle x \right \rangle}\right)}[/math]



Nested exponential function that has the variable as the base and as the nested exponent, [math]x^{\left \langle x \right \rangle}[/math]:

[math]\frac{dy}{dx} \ \text{ln}\left(x^{\left \langle x \right \rangle}\right) = \left(x^{(x-2)}+x^{(x-1)}\, \text{ln}\left(x\right)\left(\text{ln}\left(x\right)+\frac{x-1}{x}\right)\right)[/math]

Which allows us to define the following by substitution:

[math]\frac{dy}{dx} \ x^{\left \langle x \right \rangle} = x^{\left \langle x \right \rangle} \ \frac{dy}{dx} \ \text{ln}\left(x^{\left \langle x \right \rangle}\right)[/math]

Such that:

[math]x^{\left \langle x \right \rangle}=\frac{\frac{dy}{dx} \ x^{\left \langle x \right \rangle}}{\frac{dy}{dx} \ \text{ln}\left(x^{\left \langle x \right \rangle}\right)}[/math]



Integrals for nested exponents (Found by using Mathematica):

There are no general expressions for the integrals. They are based on the exponential integral, logarithmic integral, or unknown. Here is what Mathematica derived:


[math]\int x^{\left \langle a \right \rangle} \, dx = \int x^{x^{a-1}} \, dx = \text{unknown}[/math]


[math]\int a^{\left \langle x \right \rangle} \, dx = \int a^{a^{x-1}} \, dx = \frac{\text{Ei}\left(a^{x-1}\, \text{ln}(a)\right)}{\text{ln}(a)}+C[/math]


[math]\int x^{\left \langle x \right \rangle} \, dx = \int x^{x^{x-1}} \, dx = \text{unknown}[/math]


[math]\int {}_n\, \!\text{Log}_{\, a}\left(x\right) \, dx = \int 1\, +\, \frac{\text{ln}\left (\frac{\text{ln}\left(x\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)} \, dx = x \, + \, \frac{x \, \text{ln}\left(\frac{\text{ln}(x)}{\text{ln}(a)}\right)}{\text{ln}(a)}\, -\, \frac{\text{li}(x)}{\text{ln}(a)}\, +\, C = x\, \left({}_n\, \!\text{Log}_{\, a}\left(x\right)\right)\, -\, \frac{\text{li}(x)}{\text{ln}(a)}\, +\, C[/math]


[math]\int {}_n\, \!\text{Log}_{\, x}\left(a\right) \, dx = \int 1+\frac{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(x\right)}\right)}{\text{ln}\left(x\right)} \, dx = \text{unknown}[/math]


The exponential integral, [math]\text{Ei}(x)[/math]:

[math]\text{Ei}(x)=\int \limits_{-\infty}^{x} \frac{e^t}{t} \, dt[/math]


The logarithmic integral, [math]\text{li}(x)[/math]:

[math]\text{li}(x)=\int \limits_{0}^{x} \frac{dt}{\text{ln}(t)}[/math]


Finding nested roots and nested logarithms using the Newton-Rhapson method:


First it is important to note that the Newton-Rhapson method may fail to converge to the desired result. One must read and understand the proof of quadratic convergence for Newton's iterative method which is described in Wikipedia. With that being said I have not had a problem finding nested roots or nested logarithms using this approach.


[math]x_{n+1}=x_{n} - \frac{f(x_{n})}{f'(x_{n})}[/math]


Nested Roots:

To find the nested root we must use the nested exponential function that has the constant as the nested exponent, [math]x^{\left \langle a \right \rangle}[/math]. Using this function with Newton's method yields the following algorithm for nested roots (Repeat the process until you have obtained the desired result):


[math]\sqrt[\left \langle a \right \rangle]{b}\ \ \ \ \text{approximated by} \ \ \ \ x_{n+1}=x_{n} - \frac{\left(x_{n}\right)^{\left(x_{n}\right)^{(a-1)}}-b}{\left(x_{n}\right)^{\left(x_{n}\right)^{(a-1)}}\left(\left(x_{n}\right)^{(a-2)}\left(1+\left(a-1\right)\, \text{ln}\left(x\right)\right)\right)}[/math]


Nested Logarithms:

To find the nested logarithm we must use the nested exponential function that has the variable as thenested exponent, [math]a^{\left \langle x \right \rangle}[/math]. Using this function with Newton's method yields the following algorithm for nested logarithms (Repeat the process until you have obtained the desired result):

[math]{}_n\, \!\text{Log}_{\, a}\left(b\right)\ \ \ \text{approximated by} \ \ \ x_{n+1}=x_{n} - \frac{a^{a^{\left(x_{n}\right)-1}}-b}{a^{a^{\left(x_{n}\right)-1}} a^{\left(x_{n}\right)-1} \text{ln}\left(a\right)^2}[/math]



I'm sorry about all of the edits. I was making sure that I had all of the properties that I have derived posted and I had to correct a few things. This sums it up for now. I will continue to post to this thread whenever I find something new. Also, I have a lot of mathematics which might be new that I will be posting here. I hope you all have enjoyed reading this thread : )

If anyone can suggest a few good places to publish my work, I'd be most appreciative. Otherwise, I will consult with one of my professors this Thursday unless someone can show me that this work has already been done.

 

(bugfix)

Edited October 1, 2013 by Cap'n Refsmmat

[Daedalus]

 

Notation

I've been thinking about the notation for nested exponentiation / nested logarithms and was wondering if anyone had some thoughts on how they could be written which would distinguish the operations from others that are similiar in nature.

 

 

I used the following notation in the original post:

 

Nested exponentiation:

 

[math]a^{\left \langle b \right \rangle}[/math]

 

[math]e^{\left \langle x \right \rangle}={}_n\, \!\text{Exp}\left(x\right)[/math]

 

Nested logs:

 

[math]{}_n\, \!\text{Log}_{\, a}\left(b\right)[/math]

 

Natural nested logs:

 

[math]{}_n\, \!\text{Log}_{\, e}\left(b\right)={}_n\, \!\text{Ln}\left(b\right)[/math]

 

 

 

But, this notation might be better because it doesn't imply that the [math]n[/math] is some sort of variable and requires less LaTeX syntax to display it:

 

Nested Exponentiation:

 

[math]e^{\left \langle x \right \rangle}=\text{nExp}\left(x\right)[/math]

 

Nested logs:

 

[math]\text{nLog}_{\, a}\left(b\right)[/math]

 

Natural nested logs:

 

[math]\text{nLog}_{\, e}\left(b\right)=\text{nLn}\left(b\right)[/math]

 

 

 

The [math]n[/math] is being used to let us know that we are dealing with "nested" exonents and the capitalized [math]\text{L}[/math] and [math]\text{E}[/math] are meant to suggest that we are dealing with operations bigger than normal exponents. Both symbols are meant to distinguish these operators from the traditional logarithm and natural logarithm operations. I suppose both options would be acceptable. I am open to suggestions.

 

 

Identities of Nested Exponentiation and Nested Logarithms

 

I have derived some new properties of nested exponentiation and nested logarithms. I have listed these properties along with some trivial identities below.

 

Trivial Identities:

 

 

[math]a^{\left \langle 0 \right \rangle}=a^{a^{(0-1)}}=a^{a^{-1}}=a^{\frac{1}{a}}=\sqrt[a]{a}[/math]

 

[math]a^{\left \langle 1 \right \rangle}=a^{a^{(1-1)}}=a^{a^0}=a[/math]

 

[math]a^{\left \langle 2 \right \rangle}=a^{a^{(2-1)}}=a^{a^1}=a^a[/math]

 

 

[math]\text{nLog}_{\, a}\left(a\right)=1+\frac{\text{ln}\left(\frac{\text{ln}\left(a\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}=1+\frac{\text{ln}\left(1\right)}{\text{ln}\left(a\right)}=1+\frac{0}{\text{ln}\left(a\right)}=1[/math]

 

 

Both [math]\text{nLog}_{\, a}\left(0\right)[/math] and [math]\text{nLog}_{\, a}\left(1\right)[/math] are undefined:

 

 

[math]\lim_{b \to 0}\big(\text{nLog}_{\, a}\left(b\right)\big) =\lim_{b \to 0} \left(1+\frac{\text{ln}\left(\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}\right)=\infty[/math]

 

 

[math]\lim_{b \to 1}\big(\text{nLog}_{\, a}\left(b\right)\big) =\lim_{b \to 1} \left(1+\frac{\text{ln}\left(\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}\right)=-\infty[/math]

 

 

 

Relationship to normal exponentials:

 

 

[math]a^b=a\ \frac{\text{ln}\left(a^{\left \langle b \right \rangle}\right)}{\text{ln}\left(a\right)}=a\ \text{log}_{\, a}\left(a^{\left \langle b \right \rangle}\right)[/math]

 

 

Proof:

 

 

Nested exponentiation by definition:

 

 

[math]a^{\left \langle b \right \rangle}=a^{a^{(b-1)}}[/math]

 

 

Take the natural log of both sides:

 

 

[math]\text{ln}\left(a^{a^{(b-1)}}\right)=\text{ln}\left(a^{\left \langle b \right \rangle}\right)[/math]

 

 

Simplify the result:

 

 

[math]a^{(b-1)}\, \text{ln}\left(a\right)=\text{ln}\left(a^{\left \langle b \right \rangle}\right)[/math]

 

 

Divide both sides by [math]\text{ln}\left(a\right)[/math]:

 

 

[math]a^{(b-1)}=\frac{\text{ln}\left(a^{\left \langle b \right \rangle}\right)}{\text{ln}\left(a\right)}[/math]

 

 

Multiply both sides by [math]a[/math]:

 

 

[math]a^{b}=a\ \frac{\text{ln}\left(a^{\left \langle b \right \rangle}\right)}{\text{ln}\left(a\right)}[/math]

 

 

Convert [math]\text{ln}\left(a^{\left \langle b \right \rangle}\right) / \text{ln}\left(a\right)[/math] to a logarithm by the "changing the base" identity:

 

 

[math]a^{b}=a\ \text{log}_{\, a}\left(a^{\left \langle b \right \rangle}\right)[/math]

 

 

Q.E.D.

 

 

Relationship between nested logarithms and natural nested logarithms:

 

 

[math]\text{nLog}_{\, a}\left(b\right) = 1+\frac{\text{nLn}\left(b\right) - \text{nLn}\left(a\right)}{\text{ln}\left(a\right)}[/math]

 

 

Proof:

 

 

Natural nested logarithm by definition from the previous post:

 

 

[math]\text{nLn}\left(x\right)=1+\text{ln}\left(\text{ln}\left(x\right)\right)[/math]

 

 

Nested logarithm by definition from the previous post:

 

 

[math]\text{nLog}_{\, a}\left(b\right) =1+\frac{\text{ln}\left(\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}= 1+\frac{\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)}{\text{ln}\left(a\right)}[/math]

 

 

Such that:

 

 

[math]1+\frac{\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)}{\text{ln}\left(a\right)}=1+\frac{\left(1+\text{ln}\left(\text{ln}\left(b\right)\right)\right)-\left(1+\text{ln}\left(\text{ln}\left(a\right)\right)\right)}{\text{ln}\left(a\right)}[/math]

 

 

Simplify the right side by cancelling the ones in the numerator:

 

 

[math]1+\frac{\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)}{\text{ln}\left(a\right)}=1+\frac{\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)}{\text{ln}\left(a\right)}[/math]

 

 

Q.E.D.

 

 

This bring us to another interesting relationship:

 

 

[math]\text{nLn}\left(b\right) - \text{nLn}\left(a\right)=\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)[/math]

 

 

Such that:

 

 

[math]e^{\text{nLn}\left(b\right) - \text{nLn}\left(a\right)}=e^{\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)}=\text{log}_{\, a}\left(b\right)[/math]

 

 

Proof:

 

 

[math]1+\frac{\text{nLn}\left(b\right) - \text{nLn}\left(a\right)}{\text{ln}\left(a\right)}=1+\frac{\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)}{\text{ln}\left(a\right)}[/math]

 

 

Subtract one from both sides:

 

 

[math]\frac{\text{nLn}\left(b\right) - \text{nLn}\left(a\right)}{\text{ln}\left(a\right)}=\frac{\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)}{\text{ln}\left(a\right)}[/math]

 

 

Multiply both sides by [math]\text{ln}\left(a\right)[/math]:

 

 

[math]\text{nLn}\left(b\right) - \text{nLn}\left(a\right)=\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)[/math]

 

 

Raise [math]e[/math] to the power of both sides:

 

 

[math]e^{\text{nLn}\left(b\right) - \text{nLn}\left(a\right)}=e^{\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)}[/math]

 

 

Undo the subtraction in the exponent:

 

 

[math]\frac{e^{\text{nLn}\left(b\right)}}{e^{\text{nLn}\left(a\right)}}=\frac{e^{\text{ln}\left(\text{ln}\left(b\right)\right)}}{e^{\text{ln}\left(\text{ln}\left(a\right)\right)}}[/math]

 

 

Simplify the result:

 

 

[math]\frac{e^{1+\text{ln}\left(\text{ln}\left(b\right)\right)}}{e^{1+\text{ln}\left(\text{ln}\left(a\right)\right)}}=\frac{e\ \text{ln}\left(b\right)}{e\ \text{ln}\left(a\right)}=\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}[/math]

 

 

Cancel out the [math]e[/math] and change the base:

 

 

[math]e^{\text{nLn}\left(b\right) - \text{nLn}\left(a\right)}=\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}=\text{log}_{\, a}\left(b\right)[/math]

 

 

Q.E.D.

 

 

The nested logarithm of a double nested exponential with the same base (or something like that lol):

 

 

[math]\text{nLog}_{\, a}\left(\left(a^{\left \langle b \right \rangle}\right)^{\left \langle c \right \rangle}\right)=\left(c-1\right)\, a^{(b-1)}+b[/math]

 

 

Proof:

 

 

[math]\left(a^{\left \langle b \right \rangle}\right)^{\left \langle c \right \rangle}=\left(a^{\left \langle b \right \rangle}\right)^{\left(a^{\left \langle b \right \rangle}\right)^{(c-1)}}=\left(a^{a^{(b-1)}}\right)^{\left(a^{a^{(b-1)}}\right)^{(c-1)}}=[/math]

 

[math]\left(a^{a^{(b-1)}}\right)^{\left(a^{(c-1)\, a^{(b-1)}}\right)}=a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}[/math]

 

 

Take the natural log of the simplified version:

 

 

[math]\text{ln}\left(a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}\right)=\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right) \text{ln}\left(a\right)[/math]

 

 

Divide both sides by [math]\text{ln}\left(a\right)[/math]:

 

 

[math]\frac{\text{ln}\left(a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}\right)}{\text{ln}\left(a\right)}=\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)[/math]

 

 

Take the natural log of both sides:

 

 

[math]\text{ln}\left(\frac{\text{ln}\left(a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}\right)}{\text{ln}\left(a\right)}\right)=\text{ln}\left(\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)\right)[/math]

 

 

Simplify the right hand side:

 

 

[math]\text{ln}\left(\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)\right)=\text{ln}\left(a^{(b-1)}\right) + \text{ln}\left(a^{(c-1)\, a^{(b-1)}}\right)=[/math]

 

[math](b-1) \, \text{ln}\left(a\right)\, + \, (c-1)\, a^{(b-1)} \, \text{ln}\left(a\right)=\text{ln}\left(a\right)\, \left((b-1)\, +\, (c-1)\, a^{(b-1)}\right)[/math]

 

 

Such that:

 

 

[math]\text{ln}\left(\frac{\text{ln}\left(a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}\right)}{\text{ln}\left(a\right)}\right)=\text{ln}\left(a\right)\, \left((b-1)\, +\, (c-1)\, a^{(b-1)}\right)[/math]

 

 

Divide both sides by [math]\text{ln}\left(a\right)[/math]:

 

 

[math]\frac{\text{ln}\left(\frac{\text{ln}\left(a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}=(b-1) \, + \, (c-1)\, a^{(b-1)}[/math]

 

 

Complete the nested logarithm by adding one to both sides:

 

 

[math]1+\frac{\text{ln}\left(\frac{\text{ln}\left(a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}=1+(b-1) \, + \, (c-1)\, a^{(b-1)}[/math]

 

 

Simplify the result:

 

 

[math]\text{nLog}_{\, a}\left(\left(a^{\left \langle b \right \rangle}\right)^{\left \langle c \right \rangle}\right)=1+\frac{\text{ln}\left(\frac{\text{ln}\left(a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}=(c-1)\, a^{(b-1)}\, +\, b[/math]

 

 

Q.E.D.

 

 

Graph of [math]\text{nLog}_{\, a}\left(x\right)[/math]:

 

This graph shows both the real part (blue) and the imaginary part (green)

 

 

 

Graph of [math]\text{nLog}_{\, x}\left(a\right)[/math]:

 

This graph shows both the real part (blue) and the imaginary part (green)

 

 

 

That's it for now. I hope you all enjoy reading this post : )

 

I just realized that on the graph of [math]\text{nLog}_{\, x}\left(a\right)[/math], located at the bottom, I did not plot the correct equation for the imaginary part. I will post the correct graph when I get home tomorrow.

 

(bugfix?)

Edited October 1, 2013 by Cap'n Refsmmat

[michel123456]

oops sorry.

again.

I made a bad comment caused by confusion of your [math] a^{\left \langle n \right \rangle}[/math] notation

different from [math] a^n[/math]

Edited November 10, 2011 by michel123456

[Daedalus]

oops sorry.

again.

I made a bad comment caused by confusion of your [math] a^{\left \langle n \right \rangle}[/math] notation

different from [math] a^n[/math]

 

Yeah, that is why I asked if anyone had some thoughts on how nested exponentiation / nested logarithms could be written which would distinguish the operations from others that are similiar in nature. I like the notation that I developed. I just don't want it to be confusing.

 

 

A few more trivial identities:

 

 

[math]\left(a^{\left \langle b \right \rangle}\right)^{a^c}=\left(a^{a^{b-1}}\right)^{a^c}=a^{a^{b-1} a^{c}}=a^{a^{b+c-1}}=a^{\left \langle b+c \right \rangle}[/math]

 

 

[math]\left(a^{\left \langle b \right \rangle}\right)^{a^{-c}}=\left(a^{a^{b-1}}\right)^{a^{-c}}=a^{a^{b-1} a^{-c}}=a^{a^{b-c-1}}=a^{\left \langle b-c \right \rangle}[/math]

 

 

Graph of [math]x^{\left \langle a \right \rangle}[/math]:

This graph shows both the real part (blue) and the imaginary part (green).

 

 

 

Graph of [math]a^{\left \langle x \right \rangle}[/math]:

 

This graph shows both the real part (blue) and the imaginary part (green).

 

 

 

Graph of [math]x^{\left \langle x \right \rangle}[/math]:

 

This graph shows both the real part (blue) and the imaginary part (green).

 

 

 

Graph of [math]\text{nLog}_{\, a} \left(x\right)[/math]:

 

This graph shows both the real part (blue) and the imaginary part (green).

 

 

 

Graph of [math]\text{nLog}_{\, x} \left(a\right)[/math]:

 

This graph shows both the real part (blue) and the imaginary part (green).

 

 

 

Graph of [math]\text{nLog}_{\, x} \left(x\right)[/math]:

 

This graph shows both the real part (blue) and the imaginary part (green).

 

[Cool_Rocket_Scientist]

This is excellent work, and fascinating. I actually used it to help me check my work on a calculus problem. I would be interested in seeing any and all (more polished) publications.

Have you explored any variations, such as when the exponents are different from the "base" number. For example: n ^ a ^ <b> where n is not equal to a?

[Daedalus]

This is excellent work, and fascinating. I actually used it to help me check my work on a calculus problem. I would be interested in seeing any and all (more polished) publications.

Have you explored any variations, such as when the exponents are different from the "base" number. For example: n ^ a ^ <b> where n is not equal to a?

Thank you. It brings a smile to my face knowing that my work actually proved to be useful . I do have a blog here at SFN where I posted a polished version of this thread. However, there must have been a software update because the page is not rendering correctly. Summer session is over and I have a three week break from my academic studies before fall session begins. With that being said, I will work towards updating my blog and post a link here once it is fixed.

 

As for your questions, I have explored similar variations to the one that you suggested. So far, I have not found any other useful properties other than the ones I posted in this thread. On a separate note, I am curious as to how you used my equations to check your work. If you are not comfortable sharing your work publicly in this thread, then you can always send me a private message.

[Daedalus]

After looking into the problem with my blog not rendering certain equations correctly, I have determined that latex can no longer be successfully embedded within a HTML table. I am currently trying to find a workaround for this problem.

Edited August 9, 2012 by Daedalus

[studiot]

This is the first I have seen of this thread and it looks most interesting.

 

Thanks Daedalus

[Daedalus]

This is the first I have seen of this thread and it looks most interesting.

 

Thanks Daedalus

 

Thank you studiot.

[Amaton]

Hey Daedalus

 

Great work. I saw this referenced in a recent thread, and despite its inactivity as of late, I must say this is fantastic.

 

You're pioneering a relatively dark corner of mathematics (at least from a popularized scope). Unlike fractals and such, hyper-operations are a little harder to love for layman like me. But I've always thought that a large cooperative insight into this subject might open up a lot of doors, both in the pure and applied mathematics.

 

Though I might not be able to contribute much, I'd love to learn and discuss this topic.

[Daedalus]

Hey Daedalus

 

Great work. I saw this referenced in a recent thread, and despite its inactivity as of late, I must say this is fantastic.

 

You're pioneering a relatively dark corner of mathematics (at least from a popularized scope). Unlike fractals and such, hyper-operations are a little harder to love for layman like me. But I've always thought that a large cooperative insight into this subject might open up a lot of doors, both in the pure and applied mathematics.

 

Though I might not be able to contribute much, I'd love to learn and discuss this topic.

 

Hi Amaton. Thanks for your interest in my work. I'm not sure if hyper operators are a relatively dark corner of mathematics as you suggest, but they do not get much attention due to a lack of practical application. It would be nice to discover a physical application that involves such operations. Most people think of very large numbers or operations that grow very quickly when dealing with hyper operators. However, the inverse operations, nested logarithms for instance, grow slower than traditional operators such as logarithms. Perhaps one day, we will come to learn how such operations plays their part in the scheme of things.

Edited January 22, 2013 by Daedalus

[Amaton]

Hi Amaton. Thanks for your interest in my work. I'm not sure if hyper operators are a relatively dark corner of mathematics as you suggest, but they do not get much attention due to a lack of practical application. It would be nice to discover a physical application that involves such operations. Most people think of very large numbers or operations that grow very quickly when dealing with hyper operators.

 

Thanks for the reply, Daedalus

 

I was thinking "dark" in respect to both attention and understanding. Like I said, it's far from well-known in a popularized scope (as opposed to fractals, game theory, mandelbulbs, etc.). Operations beyond exponentiation are also not as established or well-formulated as the preceding iterations. But they follow the same thought train which gives us these predecessors (well, your iterated powers can be extended more easily as given the identity [math]x^{<n>}=x^{x^{(n-1)}}[/math]). But I hold doubts to the range of their applicability nonetheless.

 

Most people think of very large numbers or operations that grow very quickly when dealing with hyper operators. However, the inverse operations, nested logarithms for instance, grow slower than traditional operators such as logarithms. Perhaps one day, we will come to learn how such operations plays their part in the scheme of things.

 

Possibly and hopefully. Maybe future technology will depend upon such operations for design and computation.

 

One problem in developing an understanding however seems to be the increasing numerity of operations. The sequence of hyper-operations only deals with right-associative operations. Thus, tetration --> pentation --> etc. Simple.

 

However, disregard the right-associative restriction, and it seems the numerity of operations per level increases geometrically by a factor of two.

Edited January 23, 2013 by Amaton

[pantheory]

Hey Cleve,

 

Haven't talked to you for a while now. How is school going? That was a great posting in the News here that I saw today concerning fluid dynamics as a future way to model the quantum world. I hadn't seen this info anywhere else yet. If you get a chance give me a hollar at forrest_forrest@netzero.net.

 

Hope all is well with you in Oklahoma. This summer has been pretty cool so far this summer here in Southern California.

 

Did you ever publish your paper. In any event I would like to read it if you could send it to me

 

take care my friend, pantheory (Forrest Noble)

[Daedalus]

Hey Cleve,

 

Haven't talked to you for a while now. How is school going? That was a great posting in the News here that I saw today concerning fluid dynamics as a future way to model the quantum world. I hadn't seen this info anywhere else yet. If you get a chance give me a hollar at forrest_forrest@netzero.net.

 

Hope all is well with you in Oklahoma. This summer has been pretty cool so far this summer here in Southern California.

 

Did you ever publish your paper. In any event I would like to read it if you could send it to me

 

take care my friend, pantheory (Forrest Noble)

 

Hi Pantheory!!! I was just thinking about you the other day. School is going great, and I still have a 4.0 GPA As for the paper, I've been so busy with school and getting ready for fall session that I haven't had much time to finalize it. I also have a friend, J.M. Jennings, who is a published author that is helping me write a book on temporal uniformity. I've expanded on the concept since we've last talked, and he helps me proof read everything so that it is much easier to read and understand. Once I find the time, I will send you what I have so far on nested / iterated exponentiation, and we can catch up on current events (especially concerning fluid dynamics in regards to modeling QM). I'll send you an e-mail around the middle of next week after I take my kiddos back to their mom. I've been spending a lot of time with them the past few weeks since school is about to start back up

Edited August 11, 2013 by Daedalus

[Daedalus]

The algorithm for nested roots in the first post is no longer being displayed properly by LaTeX. I am re-posting it so that everyone can see the algorithm again:

Nested Roots:

To find the nested root we must use the nested exponential function that has the constant as the nested exponent, [math]x^{\left\langle a\right\rangle}[/math]. Using this function with Newton's method yields the following algorithm for nested roots (Repeat the process until you have obtained the desired result):

[math]\sqrt[\left\langle a\right\rangle]{b}[/math] [math]\text{approximated by}[/math] [math]x_{n+1}=x_{n}-\frac{\left(x_{n}\right)^{\left(x_{n}\right)^{(a-1)}}-b}{\left(x_{n}\right)^{\left(x_{n}\right)^{(a-1)}}\left(\left(x_{n}\right)^{(a-2)}\left(1+\left(a-1\right)\,\text{ln}\left(x\right)\right)\right)}[/math]

Nested Logarithms:

To find the nested logarithm we must use the nested exponential function that has the variable as thenested exponent, [math]a^{\left\langle x\right\rangle}[/math]. Using this function with Newton's method yields the following algorithm for nested logarithms (Repeat the process until you have obtained the desired result):

[math]\text{nLog}_{\, a}\left(b\right)[/math] [math]\text{approximated by}[/math] [math]x_{n+1}=x_{n}-\frac{a^{a^{\left(x_{n}\right)-1}}-b}{a^{a^{\left(x_{n}\right)-1}}a^{\left(x_{n}\right)-1}\text{ln}\left(a\right)^2}[/math]

Edited October 1, 2013 by Daedalus