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.

$$ \operatorname{sexp}_a(x)=\sum_{m=0}^{\infty} \binom xm \sum_{k=0}^m\,\binom mk\,(-1)^{m-k}\,\operatorname{exp}_a^{[k]}(1) $$

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

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

$$ \operatorname{sexp}_a(x)=\lim_{n\to\infty}\binom xn\sum_{k=0}^n\frac{x-n}{x-k}\binom nk(-1)^{n-k}\exp_a^{[k]}(1) $$

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

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 $$a\le e^{1/e} \,$$.

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 $$e^{-1/e} < a\le e^{1/e} \,$$.

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 $$ n-1 - K i $$ to $$ n+1 + K i $$, 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 $$ a \ge e^{1/e} \,$$.

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

$$f^{[-1]}_a (b) = 1 + f^{[-1]}_a (\log_a b) \,$$

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,
 * $$(\text{sexp}_x(n))'=\frac1x \sum _{k=1}^n (\ln x)^{k-1}\prod _{j=n-k}^n \text{sexp}_x(j)$$

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 . 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 $$ a \ge e $$, this means setting $$ c = -1 - {\log(\log(a)) \over 1 + \log(\log(a))} $$, while if $$ a \le e \le a^a $$, 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:
 * {| class="wikitable"

! 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...
 * }
 * Shifted Linear: $$\,{}^{x}10 \approx .79651... (x+1) $$ for $$-1.454753...<x<-0.454753...$$
 * 2.50181...
 * 0.377936...
 * }