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+--- Thread: Kneser tetration Polynomial approximation for base e (/showthread.php?tid=1827)
Kneser tetration Polynomial approximation for base e - RaeesHarris - 09/02/2026
I found a polynomial approximation of Kneser tetration without using Schroder, Abel, Carlemann matrices or interpolation. However the method to get a polynomial approximation is quite limited because it is only best at an 8th-order, and higher orders diverges. The method to calculate this polynomial approximation is:
\[\operatorname{f}_{1}(x) = x\]
\[\operatorname{f}_{n}(x) = \underset{\text{extend sum}}{\sum^{x}_{k=1}}\sum^{n}_{m=2} \operatorname{B}_{n,m}(\operatorname{f}_{1}(k-1), \dots, \operatorname{f}_{n-m+1}(k-1))\]
Where \(\operatorname{B}_{n,k}\) are the partial bell polynomials. Then the approximation of Kneser tetration is:
\[\sum_{k=1}^{8} \frac{\operatorname{f}_{k}(x+1)}{k!}\]
I truncated it to only 8 since it has the most accuracy. The Sum gives a 15 degree polynomial:
\[\operatorname{pol}(x) = \sum_{k=0}^{15} a_{k}x^{k}\]
Where the values of \(a_{k}\) are:
\[a_{0} = 1\]
\[a_{1} = \frac{87159883}{79833600}\]
\[a_{2} = \frac{3944583499}{14529715200}\]
\[a_{3} = \frac{17362542271}{81729648000}\]
\[a_{4} = \frac{199863943}{2874009600}\]
\[a_{5} = \frac{254779193}{5748019200}\]
\[a_{6} = \frac{15390481}{1045094400}\]
\[a_{7} = \frac{5637281}{653184000}\]
\[a_{8} = \frac{763207}{270950400}\]
\[a_{9} = \frac{5989049}{3657830400}\]
\[a_{10} = \frac{69841}{149299200}\]
\[a_{11} = \frac{4043953}{14370048000}\]
\[a_{12} = \frac{1095811}{11496038400}\]
\[a_{13} = \frac{66061}{1437004800}\]
\[a_{14} = \frac{320791}{43589145600}\]
\[a_{15} = \frac{929569}{81729648000}\]
We can use some simple PARI/GP code to generate these polynomials:
Code:
bellpepper(n,k,V)={if(n==0&&k==0,return(1));if(n==0||k==0||k>n,return(0));my(m=matrix(n+1,k+1));m[1,1]=1;for(i=1,n,for(j=max(1,k-n+i),min(i,k),m[i+1,j+1]=sum(m_iter=1,min(i-j+1,#V),binomial(i-1,m_iter-1)*V[m_iter]*m[i-m_iter+1,j])));return(m[n+1,k+1]);}
C = [x];
for(i=2,8,C=concat(C,sum(k=0,2*i - 1,polcoeff(sum(k=2,i,bellpepper(i,k,subst(C[1..i-k+1],x,x-1))),k)*((bernpol(k+1,x+1)-bernpol(k+1,1))/(k+1)))))
print(sum(k=1,8,subst(C[k],x,x+1)/k!))Anyways some numerical tests across \(\mathbb{R}(x) = [-1,0]\) and some more regions where \(\operatorname{sexp}(x)\) is the Kneser tetration function and \(\operatorname{pol}(x)\) is my polynomial approximation of Kneser tetration:
\[\int_{-1}^{0} \lvert\operatorname{sexp}(x) - \operatorname{pol}(x)\rvert dx \approx\ 0.00000055214213417011314579275348914020\]
\[\int_{-1}^{0}\left(\int_{-1}^{1} \lvert\operatorname{sexp}(x+ti) - \operatorname{pol}(x+ti)\rvert dx\right)dt \approx\ 0.000064043929412371895800823868927624\]
This method is quite limited, but the polynomial I got out of the method is quite useful for low lever approximations. I might explain this method further and how I derived it in a future math paper.