## Nested Exponentiation

The Next Operation 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.

Nested Exponentiation:

Most of us are aware of nested exponents and how to simplify the mathematics associated with them:

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

This led me to inquire about series of nested exponents and if I could derive any properties of continued nested exponentiation. I developed the following notation to simplify the respresentation of this operation:

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

We can demonstrate this as follows:

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

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

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

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

Now that we have a generalized equation that represents nested exponentiation, we can derive the nested logarithm. We can also use Calculus to derive the derivatives / integrals and define nested roots using the Newton-Rhapson method.

Nested Logarithms:

The nested logarithm is similar to normal logarithms in that it returns the value of the exponent. However, the similarities end there and nested logarithms define a whole new suit of identities that are different from normal logarithms. The following notation is used to denote nested logarithms and shows their relationship to nested exponentiation:

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

The natural nested logarithm is notated as follows:

$\text{nLog}_{\, e}\left(e^{\left \langle n \right \rangle}\right)=\text{nLn}\left(e^{\left \langle n \right \rangle}\right)=n$

The above definitions keeps things simple, but they do not allow us to do calculations involving nested logarithms. We will need to use the definition for nested exponentiation to derive an equation that represents the nested logarithm. This can be accomplished as follows:

Definition for nested exponentiation as previously defined:

$a^{\left \langle n \right \rangle} = a^{a^{n-1}}$

Let $a$ equal the base of the nested exponential and let $b$ equal $a^{\left \langle n \right \rangle}$ such that:

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

Take the natural log of the generalized equation:

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

Divide both sides by ln(a):

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

Take the natural logarithm 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 ln(a):

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

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

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 $a^{\left \langle n \right \rangle}$:

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

As we can see from the above proof, nested logarithms are derived from natural logarithms. If we analyze the equation that we just derived for nested logarithms, we can use the “changing base” identity for normal logarithms and see that the nested logarithm is truly nesting normal logarithms.

Changing the base of normal exponentials:

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

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

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

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

There is also another form that the equation for nested logarithms can take. The equation seems counterintuitive, but we will see that this second form allows us to yield some interesting identities for nested logarithms and normal logarithms. To prove the definition, we must first define the equation and work backwards to yield the nested exponential. This second equation for nested logarithms is as follows:

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

We can see that the fraction in the numerator has been inverted such that the natural log of $a$ is now being divided by the natural log of $b$. Now that we have defined the second equation for nested logarithms, we can work backwards from this form and derive nested exponentiation such that:

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

Swap the variable $n$ and the term using the natural logarithms:

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

Multiply both sides by ln(a):

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

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

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

Raise $e$ to the power of each side:

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

Simplify the result:

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

Now that we have the definitions for nested exponentiation and nested logarithms, we can define identities associated with these operations.

Identities for Nested Exponentiation:

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

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

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

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

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

Derivatives for Nested Exponentiation (Derived using Mathematica):

Nested exponential function that has a variable as the base and a constant as the nested exponent:

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

The derivative of the natural log of the nested exponential:

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

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

Nested exponential function that has a constant as the base and a variable as the nested exponent:

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

The derivative of the natural log of the nested exponential:

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

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

Nested exponential function that has a variable as the base and as the nested exponent:

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

The derivative of the natural log of the nested exponential:

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

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 for Nested Exponentiation (Derived using Mathematica):

The integrals for nested exponentiation are based on the exponential integral or they are unkown.

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

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

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

The exponential integral:

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

Identities for Nested Logarithms:

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

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

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

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

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

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

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

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

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

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

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

Identities for Nested Logarithms Similiar to Normal Logarithms:

Multiplication:

$\text{log}_{\, a}\left(x \ \times \ y\right)\ =\ \big(\text{nLog}_{\, a}\left(a^x\right)-1\big)\ +\ \big(\text{nLog}_{\, a}\left(a^y\right)-1\big)\ =\ \text{nLog}_{\, a}\left(a^x\right)\ +\ \text{nLog}_{\, a}\left(a^y\right)\ -\ 2$

$\text{log}_{\, a}\big( \text{log}_{\, a}\left(x\right) \ \times \ \text{log}_{\, a}\left(y\right)\big)\ =\ \big(\text{nLog}_{\, a}\left(x\right)-1\big)\ +\ \big(\text{nLog}_{\, a}\left(y\right)-1\big)\ =\ \text{nLog}_{\, a}\left(x\right)\ +\ \text{nLog}_{\, a}\left(y\right)\ -\ 2$

Division:

$\text{log}_{\, a}\left(\frac{x}{y}\right)\ =\ \big(\text{nLog}_{\, a}\left(a^x\right)-1\big)\ -\ \big(\text{nLog}_{\, a}\left(a^y\right)-1\big)\ =\ \text{nLog}_{\, a}\left(a^x\right)\ -\ \text{nLog}_{\, a}\left(a^y\right)$

$\text{log}_{\, a}\left( \frac{\text{log}_{\, a}\left(x\right)}{\text{log}_{\, a}\left(y\right)}\right)\ =\ \big(\text{nLog}_{\, a}\left(x\right)-1\big)-\big(\text{nLog}_{\, a}\left(y\right)-1\big)\ =\ \text{nLog}_{\, a}\left(x\right)\ -\ \text{nLog}_{\, a}\left(y\right)$

Exponents:

$\text{nLog}_{\, a}\left(x^{y}\right)=\text{log}_{\, a}\left(y\right)\ +\ \text{nLog}_{\, a}\left(x\right)$

Derivatives for Nested Logarithms (Derived using Mathematica):

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

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

Integrals for Nested Logarithms (Derived using Mathematica):

The integrals for nested logarithms are based on the logarithmic integral or they are unkown.

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

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

The logarithmic integral:

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

Interesting Relationships for Nested Exponentials and Nested Logarithms:

The following relationships and their proofs are just a few of the more interesting properties that can be derived by manipulating the equations. Even though most of these relationships may not have a usage in standard mathematics, some may prove to be useful when dealing with large numbers.

Relationship between nested exponentials and normal exponentials:

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

Nested 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 ln(a):

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

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 to a standard logarithm by “changing the base”:

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

Relationship between logarithms with the base and input swapped:

$\text{ln}\left (\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right) + \text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(b\right)}\right) = 0$

This is derived by subtracting both equations for nested logarithms:

$\left(1+\frac{\text{ln}\left (\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}\right)-\left(1-\frac{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(b\right)}\right)}{\text{ln}\left(a\right)}\right)=0$

Simplify the left side:

$\frac{\text{ln}\left (\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right) + \text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(b\right)}\right)}{\text{ln}\left(a\right)} = 0$

Multiply both sides by ln(a) to derive the relationship:

$\text{ln}\left (\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right) + \text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(b\right)}\right) = 0$

We can prove this by raising $e$ to the above identity:

$e^{\text{ln}\left (\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right) + \text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(b\right)}\right)}=e^0$

This is equal to:

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

Which simplifies to:

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

Relationship between natural logarithms and natural nested logarithms:

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

Natural nested logarithm by definition:

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

Subtract the natural nested log of $a$ from the natural nested log of $b$:

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

Simplify the right hand side:

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

Simplify the division of the natural logarithms:

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

We can see that they are both equal which brings us to the following relationship by raising $e$ to the power of the above identity:

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

The nested logarithm of a double nested exponential with the same base:

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

Expand the double nested exponential using the general definition:

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

Take the natural log of the expanded 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 ln(a):

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

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 ln(a):

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

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:

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

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.

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

Nested Roots:

To find the nested root we must use the nested exponential function that has the constant as the nested exponent, $x^{\left \langle a \right \rangle}$. Using this function with Newton’s method yields the following algorithm for nested 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 nested exponential function that has the variable as thenested 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):

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

Graphs of nested exponentials and nested logarithms

$\text{The graph of } x^{\left \langle a \right \rangle}$:

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

$\text{The graph of } a^{\left \langle x \right \rangle}$:

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

$\text{The graph of } x^{\left \langle x \right \rangle}$:

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

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

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

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

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

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

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

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