User:MathFacts/Tetration Summary

This is a summary of properties of tetration. $a$ tetrated to the $x$ is represented by $\operatorname {sexp} _{a}(x)$ or $f_{a}(x)$. The functions $\operatorname {log} _{a}$ and $\operatorname {exp} _{a}$ denote logarithm and exponentiation to base $a$, respectively. $f^{[k]}(1)$ is the result of applying the function $f$ to 1 $k$ times in succession. So $\exp _{a}^{[k]}(1)$, for example, is equal to $\operatorname {sexp} _{a}(k)$, for integer $k$.

The symbol $C_{m}^{k}$ denotes the binomial coefficient, equal to ${\frac {\Gamma (k+1)}{\Gamma (m+1)\Gamma (k-m+1)}}$. $B_{n}(x)$ is the degree $n$ Bernoulli polynomial of x, while $B_{n}$ without an argument is the $n$th Bernoulli number, equal to $B_{n}(0)$.

Evaluation methods

1. Newton method. It can be derived from the Newton polynomial interpolation formula or from the partial iteration theorem.

$\operatorname {sexp} _{a}(x)=\sum _{m=0}^{\infty }{\binom {x}{m}}\sum _{k=0}^{m}\,{\binom {m}{k}}\,(-1)^{m-k}\,\operatorname {exp} _{a}^{[k]}(1)$

This method works for any positive base $a\leq e^{1/e}\,$.

2. Lagrange method. It can be derived from the Lagrange polynomial interpolation formula.

$\operatorname {sexp} _{a}(x)=\lim _{n\to \infty }{\binom {x}{n}}\sum _{k=0}^{n}{\frac {x-n}{x-k}}{\binom {n}{k}}(-1)^{n-k}\exp _{a}^{[k]}(1)$

This method works for positive real bases $a\leq e^{1/e}\,$.

3. Rational interpolation method

$\operatorname {sexp} _{a}(x)=\lim _{n\to \infty }{\frac {\sum _{k=0}^{2n}{\frac {(-1)^{k}(k+2)\exp _{a}^{[k]}(1)}{(x-k)\Gamma (k+1)\Gamma (2n-k+1)}}}{\sum _{k=0}^{2n}{\frac {(-1)^{k}(k+2)}{(x-k)\Gamma (k+1)\Gamma (2n-k+1)}}}}$

This method works for positive real bases $a\leq e^{1/e}\,$.

4. Regular iteration method. It is derived using the technique of regular iteration at the fixed point of the exponential function.

$\operatorname {sexp} _{a}(x)=\lim _{n\to \infty }\log _{a}^{[n]}\left(\left(1-\left(\ln \left({\frac {W(-\ln a)}{-\ln a}}\right)\right)^{x}\right){\frac {W(-\ln a)}{-\ln a}}+\ln \left({\frac {W(-\ln a)}{-\ln a}}\right)\exp _{a}^{[n]}(1)\right)$

$\operatorname {sexp} _{a}(x)=\left(\lim _{n\to \infty }{\frac {\ln \left({\frac {{\frac {W(-\ln a)}{\ln a}}+\exp _{a}^{[n]}(x)}{{\frac {W(-\ln a)}{\ln a}}+\exp _{a}^{[n]}(1)}}\right)}{\ln \ln \left({\frac {W(-\ln a)}{-\ln a}}\right)}}\right)^{[-1]}$

Here the W function is the product logarithm or Lambert W function, which is defined by $W(x)\exp(W(x))=x\,$. The expression ${\frac {W(-\ln a)}{-\ln a}}$ is the principal fixed point of the function $a^{x}$.

This method works for real bases $e^{-1/e}<a\leq e^{1/e}\,$.

5. Matrix power method. One can use Carleman matrices to find the iterates.

This method works for real bases $a>1\,$.

6. Cauchy integral iteration method: A method developed by Dmitriy Kouznetsov .

One should apply the Cauchy integral formula to the loop around the rectangle from $n-1-Ki$ to $n+1+Ki$, where $n$ is the integer part of $x$, and $K$ is substantially larger than the real or imaginary parts of $x$. The value of the integrands along the imaginary-axis edges of this rectangle can be determined from their value along the imaginary line through $n$, while the value along the real-axis edges of the rectangle can be estimated under the assumption that the function tends to the two (complex) fixed points of $x=a^{x}$ at infinite imaginary values.

To obtain the value of the function along the imaginary line through $n$, we start with a guess value for this function, and recursively correct it using the Cauchy formula.

This method works for real bases $a\geq e^{1/e}\,$.

7. Intuitive iteration method (formerly called "natural" method, or matrix inverse method). A method first developed by Peter Walker, then rediscovered by Andrew Robbins, which uses matrices to solve the Abel functional equation for exponentials:

$\alpha (b^{x})=\alpha (x)+1\,$

applying the Carleman matrix to both sides, and simplifying the matrices a bit, we are left with the matrix equation:

$(C[b^{x}]^{T}-I)D[\alpha ]=D[1]\,$

where C is a Carleman matrix, and D is a Taylor coefficient column vector. Since this equation is linear, we can solve for the Taylor coefficients of the Abel function $\alpha (x)$ making $(C[b^{x}]^{T}-I)$ invertible (by truncating the first column and last row). This truncation is called the Abel matrix:

$A[b^{x}]=J(C[b^{x}]^{T}-I)K\,$

where J is the identity matrix without the last row, and K is the identity matrix without the first column. The importance of the Abel matrix is that it is often invertible, in which case we can solve for the Taylor coefficients of $\alpha (x)\,$ as $D[\alpha ]=A[b^{x}]^{-1}D[1]\,$.

To summarize, the super-logarithm is the Abel function of exponentials, so the Taylor coefficients of the super-logarithm can be found in the first column of the inverse of the Abel matrix.

This method works for real bases $a>1\,$.

Functional and differential equations

1. Main functional equation

$f_{a}(x+1)=a^{f_{a}(x)}$

2. Main functional equation for inverse function

$f_{a}^{[-1]}(b)=1+f_{a}^{[-1]}(\log _{a}b)\,$

3. Differential-difference equation

$f'_{a}(x+1)=f'_{a}(x)f_{a}(x+1)\ln a\,$

Periodicity

Superexponential with fixed base b is periodic with period

$T={\frac {2\pi i}{\ln \ln {\frac {W(-\ln b)}{-\ln b}}}}$

Values in fixed points and asymptotic properties

• $f_{a}(0)=1\,$
• $f_{a}(-1)=0\,$
• $f_{a}(1)=a\,$
• $f_{a}(-2)=-{\frac {\infty }{\ln(a)}}$
• $f_{a}(+\infty )=-{\frac {\mathrm {W} (-\ln {a})}{\ln {a}}}$
• $f'_{e}(-1)=f'_{e}(0)\,$

Other properties

$\log _{a}{\frac {f'_{a}(x)}{f'_{a}(0)(\ln a)^{x}}}=\sum _{k=0}^{x-1}f_{a}(k)$

${\frac {f'_{a}(x)}{f'_{a}(0)(\ln a)^{x}}}=\prod _{k=0}^{x}f_{a}(k)$

$\log _{a}{\frac {f'_{a}(x)}{f'_{a}(0)(\ln a)^{x}}}=\sum _{n=1}^{\infty }{\frac {f^{(n-1)}(0)}{n!}}(B_{n}(x)-B_{n})$

Differentiation rules

1. Differentiating tetration with fixed height

$(\operatorname {sexp} _{x}(1))'=1$

$(\operatorname {sexp} _{x}(n))'=\left((\operatorname {sexp} _{x}(n-1))'\ln x+{\frac {\operatorname {sexp} _{x}(n-1)}{x}}\right)\operatorname {sexp} _{x}(n)$

Generalized,

$({\text{sexp}}_{x}(n))'={\frac {1}{x}}\sum _{k=1}^{n}(\ln x)^{k-1}\prod _{j=n-k}^{n}{\text{sexp}}_{x}(j)$

2. Differentiating tetration with fixed base

$\operatorname {sexp} '_{a}(x)=\operatorname {sexp} '_{a}(x-1)\operatorname {sexp} _{a}(x)\ln a\,$

Generalized,

$(\operatorname {sexp} _{a}(x))'=\operatorname {sexp} '_{a}(0)(\ln a)^{x}\prod _{j=1}^{x}\operatorname {sexp} _{a}(j)$

$(\operatorname {sexp} _{a}(x))'=\operatorname {sexp} '_{a}(c)(\ln a)^{x-c}\!\prod _{j=c+1}^{x}\!\operatorname {sexp} _{a}(j)$

Approximation methods

In light of the main equation above, it suffices to define an approximation function on an interval of unit length.

1. Linear approximation: On the interval [-1, 0], we have $\operatorname {sexp} _{a}(x)\approx x+1$. This approximation is continuous everywhere, but generally not differentiable at integers.
2. Quadratic approximation: On the interval [-1, 0], we have $\operatorname {sexp} _{a}(x)\approx {\log(a)-1 \over 1+\log(a)}x^{2}+{2\log(a) \over 1+\log(a)}x+1$. This approximation is continuously differentiable everywhere, but its second derivative is discontinuous at each integer.
3. Shifted linear approximation: Developed by Jay D. Fox at [1]. On some interval [c, c+1], we have $\operatorname {sexp} _{a}(x)\approx {1+\log(\log(a)) \over \log(a)}(x-c)-{\log(\log(a)) \over \log(a)}$. The value c is chosen so that the approximation gives $\operatorname {sexp} _{a}(0)\approx 1$; if $a\geq e$, this means setting $c=-1-{\log(\log(a)) \over 1+\log(\log(a))}$, while if $a\leq e\leq a^{a}$, it means setting $c={-\log(a)-\log(\log(a)) \over 1+\log(\log(a))}$. This approximation is continuously differentiable everywhere, but its second derivative is discontinuous at c+n for any integer n. It is only defined for $a>e^{1/e}$.

If $a=e$ all three of the above approximations are equivalent.

Approximate values

• For base 10:

Approximation	$\operatorname {sexp} _{10}(1/2)$	$\operatorname {slog} _{10}(2)$
Linear: $\,{}^{x}10\approx 10^{x}$ for $0<x<1$	3.162776601...	0.301029995...
Quadratic: $\,{}^{x}10\approx {}^{x}a\approx 10^{x+{\frac {-1+\log(10)}{1+\log(10)}}(x^{2}-x)}=10^{x+0.39441379...(x^{2}-x)}$ for $0<x<1$	2.5199768...
Shifted Linear: $\,{}^{x}10\approx .79651...(x+1)$ for $-1.454753...<x<-0.454753...$	2.50181...	0.377936...