Sequence Interpolation

[MphLee]

I see... this is really interesting.

Definition. Let \(H_n(b,x)=y\) be a real arguments extension of an hyperoperation family \(H_n:\mathbb R\times\mathbb R\). Define \(L_n(b,y)\) as the solution \(x\) to the equation \[H_n(b,x)=y\]

a) The difference operator. We want to solve the generalize the functional equation so that \(F\) is the discrete difference of \(f\)
\[f(x)+F(x)=f(x+1)\]
This is the same as solving \(H_1(f(x),F(x))=f(H_0(x))\) assuming \(H=goodstein\) This gives \(F=\Delta f\).

b) The forward difference operator. Consider now the forward difference operator with delay \(h\in\mathbb R\). It is the solution of \[f(x)+F(x)=f(h+x)\] i.e.\(H_1(f(x),F(x))=f(H_0^h(x))=f(H_1(x,h))\)  and \(F\) measures the additive \(H_1\) perturbation on the output of \(f\) given an iterated \(H_0\) successor perturbation of the input of \(f\).

c) How to measure the perturbation. The perturbation in the output can be measured in different ways. I believe, based on some heuristics on differentiation, that the perturbation in the output must be measured as quantity of times we have to iterate \(H_{0}(f(x),-)\) i.e. we measure it "hyper-logarithmically"
\[H_{0}^{F(x)}(f(x),H_1(f(x),0))=f(H_{0}^h(x,H_1(x,0)))\]
but by definition of goodstein \(H_n^h(b,H_{n+1}(b,0))=H_{n+1}(b,h)\) and we can consider instead the functional equation
\[H_1(f(x),F(x))=f(H_1(x,h))\]

d) Generalizing to higher ranks. We follow this schema by looking at the solutions of the following functional equation
\[H_n(f(x),F(x))=f(H_n(x,h)) \]

Definition (hyperdifference). We define the Hyperdifference at \(x\) of \(f\) as the \(H_n\) perturbation of the output of \(f\) given an \(H_n\) perturbation of the input of \(f\) at \(x\). Let \(f:\mathbb R\to\mathbb R\). Define the partial function \(\Delta^H_{n,h}[f]:\mathbb R\to\mathbb R\) as
\[\Delta^H_{n,h}[f] (x):=L_n(f(x),f(H_n(x,h)))\]

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e) The differentiation schema consider now the functional equation \(H_{n+1}(\Delta^H_{n,h}[{\rm id}] (x),F(x))=\Delta^H_{n,h}[f] (x)\) to be solved in the \(F\) we call the solution \(D^H_{n,h}[f]\)

Definition. Let \(f:\mathbb R\to\mathbb R\). Define the partial function \(D^H_{n,h}[f]:\mathbb R\to\mathbb R\) as

\[D^H_{n,h}[f] (x):=L_{n+1}(\Delta^H_{n,h}[{\rm id}] (x),\Delta^H_{n,h}[f] (x))\]

Lets take the limit now and define the hyperderivative of rank \(n\), it may not exists globally, as the previous operators.

Definition (hyperderivative). Let \(f:\mathbb R\to\mathbb R\). Define the partial function \(D^H_{n}[f]:\mathbb R\to\mathbb R\) as

\[D^H_{n,h}[f] (x):=\lim_{h\to\varepsilon_n}D^H_{n,h}[f] (x)\]
where \(\varepsilon_n\) is the left unit element \(H_n{b,\varepsilon_n}=b\).

Question. is it possible to extend Newton, Taylor formalism and calculus in general to this? Is it useful?

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example. Unwind the definition and observe that \(D^H_{n,h}[f] (x)=L_{n+1}(L_n(x,H_n(x,h)),L_n(f(x),f(H_n(x,h)))) =L_{n+1}(h,L_n( f(x),f(H_n(x,h)) )  )  \)

Assume \(H\) is the standard Goodstein family.

• \(D^H_{-1,h}[f] (x)=L_{0}(h,L_0( f(x),f(H_{0}(x,h)) ) ) = f(h+1)-2 \);
  
• \(D^H_{0,h}[f] (x)=L_{1}(h,L_0( f(x),f(H_0(x,h)) )  ) = f(h+1)-(h+1) \) but \(\varepsilon_0\) doesn't exists;
  
• \(D^H_{1,h}[f] (x)=L_{2}(h,L_1( f(x),f(H_1(x,h)) )  ) =\frac{f(x+h)-f(x)}{h}  \);
  
• \(D^H_{2,h}[f] (x)=L_{3}(h,L_2( f(x),f(H_2(x,h)) )  ) =\frac{\ln f(hx)-\ln f(x)}{\ln h}\);
  
• \(D^H_{3,h}[f] (x)=L_{4}(h,L_3( f(x),f(H_3(x,h)) )  )  ={\rm slog}_h (\frac{\ln f(x^h)}{\ln f(x)})\).
  
• ...
  

Assume \(B\) is the Bennet family. (What follows was already considered by Rubtsov 20-30 years ago)

• \(D^B_{-1,h}[f] (x)=L_{0}(h,L_{-1}( f(x),f(B_{-1}(x,h)) )  )= \ln(e^{f(\ln (e^x+e^h))}-e^{f(x)})-h =\ln(\frac{e^{f(\ln (e^x+e^h))}-e^{f(x)}}{e^h})\) obviously \(D^B_{-1}f(x)=\lim_{h\to -\infty}\ln(\frac{e^{f(\ln (e^x+e^h))}-e^{f(x)}}{e^h})\);
  
• \(D^B_{0,h}[f] (x)=L_{1}(h,L_0( f(x),f(B_0(x,h)) )  )  =\frac{f(x+h)-f(x)}{h}\)
  
• \(D^B_{1,h}[f] (x)=L_{2}(h,L_1( f(x),f(B_1(x,h)) )  )  =\frac{f(hx)}{f(x)}\ominus_2 h=\exp( \frac{\ln(\frac{f(hx)}{f(x)})}{\ln h})=\sqrt[\ln h]{\frac{f(xh)}{f(x)
  }}\) note also that we can express it as \(  \exp( \frac{  \ln f(hx)-\ln f(x)}  {\ln h}  )  \). The Bennet derivative of rank \(1\) is \(D^B_1 f(x)=\lim_{h\to 1}e^{D^H_{2,h} f(x)}=e^{D^H_{2} f(x)}\)
  
• ...