Science Forums

# Iterated Exponents And Nested Logarithms

## Recommended Posts

I was looking into operators beyond exponents when I discovered a few properties of iterated exponents and nested logarithms. 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 iterated exponentiation and nested logarithms anywhere.

Iterated Exponentiation

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

$\left(a^b\right)^c=a^{b\, c}$

So, I began to wonder about iterated exponents and if I could derive any properties associated with them. Now, I do realize that iterated exponentiation is well known. However, let's define iterated exponentiation as

$a^{\left \langle n \right \rangle} = \underbrace{(((a)^a)^a)^{\cdot^{\cdot^{a}}}}_n = a^{a^{n-1}}$

such that

$a^{\left \langle 1 \right \rangle} = a = a^{a^{(1-1)}}$,

$a^{\left \langle 2 \right \rangle} = (a)^a = a^{a^{(2-1)}}$,

$a^{\left \langle 3 \right \rangle} = ((a)^a)^a = a^{a^{(3-1)}}$,

$a^{\left \langle 4 \right \rangle} = (((a)^a)^a)^a = a^{a^{(4-1)}}$,

etc...

Now that we have the generalized form, $a^{a^{n-1}}$, that predicts the outcome of iterated exponentiation, we can derive the derivatives / integrals using Calculus and define an algorithm for determining iterated roots using the Newton-Rhapson method. However, the operation that I believe is new is what I call the nested logarithm.

Nested Logarithms

The nested log is just like normal logarithms except it determines the value of the iterated exponent:

$\text{nLog}_{\, a}\left(a^{\left \langle b \right \rangle}\right)=b$

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

$\text{nLog}_{\, a}\left(b\right)=1+\frac{\text{ln}\left(\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}$

$\text{nLog}_{\, a}\left(b\right)=1-\frac{\text{ln}\left(\frac{\text{ln}\left(a\right)}{\text{ln}\left(b\right)}\right)}{\text{ln}\left(a\right)}$

Proof that - $\text{nLog}_{\, a}\left(b\right)=1+\text{ln}(\text{ln}\left(b\right) / \text{ln}\left(a\right)) / \text{ln}\left(a\right)$

Let $a$ equal the base of the iterated exponent and let $b$ equal $a^{\left \langle n \right \rangle}$, which is also equal to $a^{a^{n-1}}$.

$\text{nLog}_{\, a}\left(b\right)=\text{nLog}_{\, a}\left(a^{\left \langle n \right \rangle}\right)=\text{nLog}_{\, a}\left(a^{a^{n-1}}\right)=n$

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

$\text{ln}\left(a^{a^{n-1}}\right)=a^{n-1} \, \text{ln}\left(a\right)$

Divide both sides by $\text{ln}\left(a\right)$:

$\frac{\text{ln}\left(a^{a^{n-1}}\right)}{\text{ln}\left(a\right)}=a^{n-1}$

Take the natural log of both sides:

$\text{ln}\left(\frac{\text{ln}\left(a^{a^{n-1}}\right)}{\text{ln}\left(a\right)}\right)=\left(n-1\right) \, \text{ln}\left(a\right)$

Again, divide both sides by $\text{ln}\left(a\right)$:

$\frac{\text{ln}\left(\frac{\text{ln}\left(a^{a^{n-1}}\right)}{\text{ln}\left(a\right)}\right)}{ \text{ln}\left(a\right)}=n-1$

$1+\frac{\text{ln}\left(\frac{\text{ln}\left(a^{a^{n-1}}\right)}{\text{ln}\left(a\right)}\right)}{ \text{ln}\left(a\right)}=n$

Substitute $a^{\left \langle n \right \rangle}$ in place of $a^{a^{n-1}}$:

$\text{nLog}_{\, a}\left(a^{\left \langle n \right \rangle}\right)=1+\frac{\text{ln}\left(\frac{\text{ln}\left(a^{\left \langle n \right \rangle}\right)}{\text{ln}\left(a\right)}\right)}{ \text{ln}\left(a\right)}=n$

Substitute $b$ in place of $a^{\left \langle n \right \rangle}$:

$\text{nLog}_{\, a}\left(b\right)=1+\frac{\text{ln}\left(\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)}{ \text{ln}\left(a\right)}$

Q.E.D.

Proof that - $\text{nLog}_{\, a}\left(b\right)=1-\text{ln}(\text{ln}\left(a\right) / \text{ln}\left(b\right)) / \text{ln}\left(a\right)$

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

$1-\frac{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(a^{\left \langle n \right \rangle}\right)}\right)}{\text{ln}\left(a\right)}=1-\frac{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(a^{a^{n-1}}\right)}\right)}{\text{ln}\left(a\right)}=n$

Move $n$ and the term using the natural logarithms to the opposite side of the equation:

$1-n=\frac{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(a^{a^{n-1}}\right)}\right)}{\text{ln}\left(a\right)}$

Multiply both sides by $\text{ln}\left(a\right)$:

$(1-n)\, \text{ln}\left(a\right)=\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(a^{a^{n-1}}\right)}\right)$

Raise $e$ to the power of each side:

$e^{(1-n)\, \text{ln}\left(a\right)}=e^{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(a^{a^{n-1}}\right)}\right)}$

Simplify the result:

$a^{1-n}=\frac{\text{ln}\left(a\right)}{\text{ln}\left(a^{a^{n-1}}\right)}$

Multiply both sides by $\text{ln}\left(a^{a^{n-1}}\right)$:

$a^{1-n}\, \text{ln}\left(a^{a^{n-1}}\right)=\text{ln}\left(a\right)$

Divide both sides by $a^{1-n}$:

$\text{ln}\left(a^{a^{n-1}}\right)=\frac{\text{ln}\left(a\right)}{a^{1-n}}$

Raise $e$ to the power of each side:

$e^{\text{ln}\left(a^{a^{n-1}}\right)}=e^{\left(\frac{\text{ln}\left(a\right)}{a^{1-n}}\right)}$

Simplify the result:

$a^{a^{n-1}}=a^{a^{n-1}}$

Q.E.D.

Trivial Identities

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

$a^{\left \langle 1 \right \rangle}=a^{a^{(1-1)}}=a^{a^0}=a$

$a^{\left \langle 2 \right \rangle}=a^{a^{(2-1)}}=a^{a^1}=a^a$

$\left(a^{\left \langle b \right \rangle}\right)^{a^c}=\left(a^{a^{b-1}}\right)^{a^c}=a^{a^{b-1} a^{c}}=a^{a^{b+c-1}}=a^{\left \langle b+c \right \rangle}$

$\left(a^{\left \langle b \right \rangle}\right)^{a^{-c}}=\left(a^{a^{b-1}}\right)^{a^{-c}}=a^{a^{b-1} a^{-c}}=a^{a^{b-c-1}}=a^{\left \langle b-c \right \rangle}$

$\text{nLog}_{\, a}\left(a\right)=1+\frac{\text{ln}\left(\frac{\text{ln}\left(a\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}=1+\frac{\text{ln}\left(1\right)}{\text{ln}\left(a\right)}=1+\frac{0}{\text{ln}\left(a\right)}=1$

Both, $\text{nLog}_{\, a}\left(0\right)$ and $\text{nLog}_{\, a}\left(1\right)$, are undefined:

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

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

Properties of Nested Logarithms

$\text{nLog}_{\, a}\left(b\right)=1+\frac{\text{ln}\left (\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)} \ \ \ \ \ \ \ \text{nLog}_{\, a}\left(b\right)=1-\frac{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(b\right)}\right)}{\text{ln}\left(a\right)}$

$\text{nLog}_{\, a}\left(b\right)=1+\text{log}_{a}\left(\text{log}_{a}\left(b\right)\right) \ \ \ \ \ \ \ \text{nLog}_{\, e}\left(b\right)=1+\text{ln}\left(\text{ln}\left(b\right)\right)$

$\text{nLog}_{\, a}\left(b\right)=1-\frac{\text{ln}\left(\text{ln}\left(a\right)\right)}{\text{ln}\left(a\right)}+\frac{\text{ln}\left(\text{ln}\left(b\right)\right)}{\text{ln}\left(a\right)} \ \ \ \ \ \ \ \text{log}_{\, a}\left(b\right)=\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}=a^{\left(\text{nLog}_{\, a}\left(b\right)-1\right)}$

$\text{ln}\left(\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)=\text{ln}\left(a\right) \left(\text{nLog}_{\, a}\left(b\right) - 1\right) \ \ \ \ \ \ \ \text{ln}\left(\frac{\text{ln}\left(a\right)}{\text{ln}\left(b\right)}\right)=\text{ln}\left(a\right) \left(1-\text{nLog}_{\, a}\left(b\right)\right)$

Relationship to Exponentials

$a^b=a\ \frac{\text{ln}\left(a^{\left \langle b \right \rangle}\right)}{\text{ln}\left(a\right)}=a\ \text{log}_{\, a}\left(a^{\left \langle b \right \rangle}\right)$

Proof:

Iterated exponentiation by definition:

$a^{\left \langle b \right \rangle}=a^{a^{(b-1)}}$

Take the natural log of both sides:

$\text{ln}\left(a^{a^{(b-1)}}\right)=\text{ln}\left(a^{\left \langle b \right \rangle}\right)$

Simplify the result:

$a^{(b-1)}\, \text{ln}\left(a\right)=\text{ln}\left(a^{\left \langle b \right \rangle}\right)$

Divide both sides by $\text{ln}\left(a\right)$:

$a^{(b-1)}=\frac{\text{ln}\left(a^{\left \langle b \right \rangle}\right)}{\text{ln}\left(a\right)}$

Multiply both sides by $a$:

$a^{b}=a\ \frac{\text{ln}\left(a^{\left \langle b \right \rangle}\right)}{\text{ln}\left(a\right)}$

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

$a^{b}=a\ \text{log}_{\, a}\left(a^{\left \langle b \right \rangle}\right)$

Q.E.D.

Relationship between Nested Logarithms and Natural Nested Logarithms

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

Proof:

Natural nested logarithm by definition:

$\text{nLn}\left(x\right)=1+\text{ln}\left(\text{ln}\left(x\right)\right)$

Nested logarithm by definition:

$\text{nLog}_{\, a}\left(b\right) =1+\frac{\text{ln}\left(\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}= 1+\frac{\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)}{\text{ln}\left(a\right)}$

such that

$1+\frac{\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)}{\text{ln}\left(a\right)}=1+\frac{\left(1+\text{ln}\left(\text{ln}\left(b\right)\right)\right)-\left(1+\text{ln}\left(\text{ln}\left(a\right)\right)\right)}{\text{ln}\left(a\right)}$

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

$1+\frac{\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)}{\text{ln}\left(a\right)}=1+\frac{\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)}{\text{ln}\left(a\right)}$

Q.E.D.

Relationship between the Difference of Natural Nested Logarithms

$\text{nLn}\left(b\right) - \text{nLn}\left(a\right)=\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)$

such that

$e^{\text{nLn}\left(b\right) - \text{nLn}\left(a\right)}=e^{\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)}=\text{log}_{\, a}\left(b\right)$

Proof:

$1+\frac{\text{nLn}\left(b\right) - \text{nLn}\left(a\right)}{\text{ln}\left(a\right)}=1+\frac{\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)}{\text{ln}\left(a\right)}$

Subtract one from both sides:

$\frac{\text{nLn}\left(b\right) - \text{nLn}\left(a\right)}{\text{ln}\left(a\right)}=\frac{\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)}{\text{ln}\left(a\right)}$

Multiply both sides by $\text{ln}\left(a\right)$:

$\text{nLn}\left(b\right) - \text{nLn}\left(a\right)=\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)$

Raise $e$ to the power of both sides:

$e^{\text{nLn}\left(b\right) - \text{nLn}\left(a\right)}=e^{\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)}$

Undo the subtraction in the exponent:

$\frac{e^{\text{nLn}\left(b\right)}}{e^{\text{nLn}\left(a\right)}}=\frac{e^{\text{ln}\left(\text{ln}\left(b\right)\right)}}{e^{\text{ln}\left(\text{ln}\left(a\right)\right)}}$

Simplify the result:

$\frac{e^{1+\text{ln}\left(\text{ln}\left(b\right)\right)}}{e^{1+\text{ln}\left(\text{ln}\left(a\right)\right)}}=\frac{e\ \text{ln}\left(b\right)}{e\ \text{ln}\left(a\right)}=\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}$

Cancel out the $e$ and change the base:

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

Q.E.D.

The Nested Logarithm of a Doubly Iterated Exponential with the Same Base (or something like that lol)

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

Proof:

$\left(a^{\left \langle b \right \rangle}\right)^{\left \langle c \right \rangle}=\left(a^{\left \langle b \right \rangle}\right)^{\left(a^{\left \langle b \right \rangle}\right)^{(c-1)}}=\left(a^{a^{(b-1)}}\right)^{\left(a^{a^{(b-1)}}\right)^{(c-1)}}=$

$\left(a^{a^{(b-1)}}\right)^{\left(a^{(c-1)\, a^{(b-1)}}\right)}=a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}$

Take the natural log of the simplified version:

$\text{ln}\left(a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}\right)=\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right) \text{ln}\left(a\right)$

Divide both sides by $\text{ln}\left(a\right)$:

$\frac{\text{ln}\left(a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}\right)}{\text{ln}\left(a\right)}=\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)$

Take the natural log of both sides:

$\text{ln}\left(\frac{\text{ln}\left(a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}\right)}{\text{ln}\left(a\right)}\right)=\text{ln}\left(\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)\right)$

Simplify the right hand side:

$\text{ln}\left(\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)\right)=\text{ln}\left(a^{(b-1)}\right) + \text{ln}\left(a^{(c-1)\, a^{(b-1)}}\right)=$

$(b-1) \, \text{ln}\left(a\right)\, + \, (c-1)\, a^{(b-1)} \, \text{ln}\left(a\right)=\text{ln}\left(a\right)\, \left((b-1)\, +\, (c-1)\, a^{(b-1)}\right)$

such that

$\text{ln}\left(\frac{\text{ln}\left(a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}\right)}{\text{ln}\left(a\right)}\right)=\text{ln}\left(a\right)\, \left((b-1)\, +\, (c-1)\, a^{(b-1)}\right)$

Divide both sides by $\text{ln}\left(a\right)$:

$\frac{\text{ln}\left(\frac{\text{ln}\left(a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}=(b-1) \, + \, (c-1)\, a^{(b-1)}$

Complete the nested logarithm by adding one to both sides:

$1+\frac{\text{ln}\left(\frac{\text{ln}\left(a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}=1+(b-1) \, + \, (c-1)\, a^{(b-1)}$

Simplify the result:

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

Q.E.D.

Derivatives of Iterated Exponents and Nested Logarithms (Found by using Mathematica)

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

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

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

$\frac{dy}{dx} \ \text{nLog}_{\, a}\left(x\right) = \frac{1}{x\, \text{ln}\left(x\right)\, \text{ln}\left(a\right)}=\frac{\text{nLog}_{\, a}\left(x\right)}{x\, \text{ln}\left(x\right) \left(\text{ln}\left(a\right) +\text{ln}\left(\frac{\text{ln}\left(x\right)}{\text{ln}\left(a\right)}\right)\right)}$

$\frac{dy}{dx} \ \text{nLog}_{\, x}\left(a\right) = \frac{\text{ln}\left(\frac{\text{ln}\left(x\right)}{\text{ln}\left(a\right)}\right)-1}{x\, \text{ln}\left(x\right)^2}=\frac{\text{ln}\left(x\right) \left(1-\text{nLog}_{\, x}\left(a\right)\right)-1}{x\, \text{ln}\left(x\right)^2}$

Properties of Derivatives for Iterated Exponents (Found by using Mathematica)

An iterated exponential function that has the constant as the nested exponent, $x^{\left \langle a \right \rangle}$:

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

Which allows us to define the following by substitution:

$\frac{dy}{dx} \ x^{\left \langle a \right \rangle} = x^{\left \langle a \right \rangle} \ \frac{dy}{dx} \ \text{ln}\left(x^{\left \langle a \right \rangle}\right)$

such that:

$x^{\left \langle a \right \rangle}=\frac{\frac{dy}{dx} \ x^{\left \langle a \right \rangle}}{\frac{dy}{dx} \ \text{ln}\left(x^{\left \langle a \right \rangle}\right)}$

An iterated exponential function that has the variable as the nested exponent, $a^{\left \langle x \right \rangle}$:

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

Which allows us to define the following by substitution:

$\frac{dy}{dx} \ a^{\left \langle x \right \rangle} = a^{\left \langle x \right \rangle} \ \frac{dy}{dx} \ \text{ln}\left(a^{\left \langle x \right \rangle}\right)$

such that:

$a^{\left \langle x \right \rangle}=\frac{\frac{dy}{dx} \ a^{\left \langle x \right \rangle}}{\frac{dy}{dx} \ \text{ln}\left(a^{\left \langle x \right \rangle}\right)}$

An iterated exponential function that has the variable as the base and as the nested exponent, $x^{\left \langle x \right \rangle}$:

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

Which allows us to define the following by substitution:

$\frac{dy}{dx} \ x^{\left \langle x \right \rangle} = x^{\left \langle x \right \rangle} \ \frac{dy}{dx} \ \text{ln}\left(x^{\left \langle x \right \rangle}\right)$

such that:

$x^{\left \langle x \right \rangle}=\frac{\frac{dy}{dx} \ x^{\left \langle x \right \rangle}}{\frac{dy}{dx} \ \text{ln}\left(x^{\left \langle x \right \rangle}\right)}$

Integrals of Iterated Exponents and Nested Logarithms (Found by using Mathematica)

There are no general expressions for the integrals. They are based on the exponential integral, logarithmic integral, or unknown:

$\int x^{\left \langle a \right \rangle} \, dx = \int x^{x^{a-1}} \, dx = \text{unknown}$

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

$\int x^{\left \langle x \right \rangle} \, dx = \int x^{x^{x-1}} \, dx = \text{unknown}$

$\int \text{nLog}_{\, a}\left(x\right) \, dx = \int 1\, +\, \frac{\text{ln}\left (\frac{\text{ln}\left(x\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)} \, dx = x \, + \, \frac{x \, \text{ln}\left(\frac{\text{ln}(x)}{\text{ln}(a)}\right)}{\text{ln}(a)}\, -\, \frac{\text{li}(x)}{\text{ln}(a)}\, +\, C = x\, \left(\text{nLog}_{\, a}\left(x\right)\right)\, -\, \frac{\text{li}(x)}{\text{ln}(a)}\, +\, C$

$\int \text{nLog}_{\, x}\left(a\right) \, dx = \int 1+\frac{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(x\right)}\right)}{\text{ln}\left(x\right)} \, dx = \text{unknown}$

The exponential integral, $\text{Ei}\left(x\right)$:

$\text{Ei}(x)=\int \limits_{-\infty}^{x} \frac{e^t}{t} \, dt$

The logarithmic integral, $\text{li}\left(x\right)$:

$\text{li}(x)=\int \limits_{0}^{x} \frac{dt}{\text{ln}(t)}$

Finding Iterated 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 iterated roots or nested logarithms using this approach.

$x_{n+1}=x_{n} - \frac{f(x_{n})}{f'(x_{n})}$

Iterated Roots:

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

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

Nested Logarithms:

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

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

Graph of $x^{\left \langle a \right \rangle}$:

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

Graph of $a^{\left \langle x \right \rangle}$:

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

Graph of $x^{\left \langle x \right \rangle}$:

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

Graph of $\text{nLog}_{\, a} \left(x\right)$:

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

Graph of $\text{nLog}_{\, x} \left(a\right)$:

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

Graph of $\text{nLog}_{\, x} \left(x\right)$:

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

Edited by DaedalusSFN

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

×   Pasted as rich text.   Paste as plain text instead

Only 75 emoji are allowed.

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.