User:Dmitrii Kouznetsov/loginal

Loginal of function $g$ at some space S is function $K$ such that

(0) $g(K(t))=K(1+t)$ for all $t\in {\rm {S}}$

Loginal allow the solution $f$ of equation

(1) $f^{n}(x)=g(x)$

in form

(2) $f(x)=K\!\left({\frac {1}{n}}+K^{-1}(x)\right)$

Generalization

The straightforward generalization of equaiton (0) can be written in form

(3) $g^{a}(K(t))=K(a+t)$ for all $t\in {\rm {S}}$ and any $a$ from some set A that includes integers</math>

In some cases, it is possible to extend the set A to complex numbers.

Loginal should be invertable

As loginal $K$ of function $g$ is implemented, together with its inverse function $K^{-1}$ , the solution of equaiton (1) becomes straightforward:

(4) $f(x)=g^{1/n}(x)=g^{1/n}{\Big (}K\!{\Big (}K^{-1}(x){\Big )}{\Big )}$$=K\!\left({\frac {1}{n}}+K^{-1}(x)\right)$

Then, for the initial equation (1)

(5) $f^{2}(x)=f(f(x))=f\!\left(K\!\left({\frac {1}{n}}+K^{-1}(x)\right)\right)$

(6) $f^{2}(x)=K\!\left({\frac {1}{n}}+K^{-1}\left(K\!\left({\frac {1}{n}}+K^{-1}(x)\right)\right)\right)$

(7) $f^{2}(x)=K\left({\frac {1}{n}}+{\frac {1}{n}}+K^{-1}(x)\right)=K\left({\frac {2}{n}}+K^{-1}(x)\right)$

Similarly, for any $m$

(8) $f^{m+1}(x)=f(f^{m}(x))=K\left({\frac {1}{n}}+{\frac {m}{n}}+K^{-1}(x)\right)=K\left({\frac {m+1}{n}}+K^{-1}(x)\right)$

Special cases

For simple function $g$, it is easy to find its loginal.

Summation

In particular, if $g$ means addition a constant $c$, id est, $g(x)=x+c$, then

(8) $nc+K(t)=K(t+n)$

means that $K(t)=t=K^{-1}(t)$

In such a way, this case is trivial.

Multiplication

If $g$ means multiplication by a constant $c$, id est, $g(x)=cx$, then

(9) $c^{n}K(t)=K(t+n)$

means that $K(t)=c^{t}$ and $K^{-1}(t)=\log _{c}(t)$.

Exponentiation

For exponentiation, $K$ is tetration,

(10) $\exp(K(x))=K(x+1)$;

or $\exp ^{n}(K(x))=K(x+n)$

In particular, one can extract the square root of exponential, id est, to find finction $f={\sqrt {\exp }}=\exp ^{1/2}$ such that

(12) $f(f(x))=\exp(x)$

The calculation is straightforward:

(13) $f(x)=\exp ^{1/2}(x)=\exp ^{1/2}\left(F(F^{-1}(x))\right)=F\left({\frac {1}{2}}+F^{-1}(x)\right)$

Checkback:

(14) $f(f(x))=F\!\left({\frac {1}{2}}+F^{-1}\!\left(F\!\left({\frac {1}{2}}+F^{-1}(x)\right)\right)\right)$

(15) $f(f(x))=F\!\left({\frac {1}{2}}+{\frac {1}{2}}+F^{-1}(x)\right)$

(16) $f(f(x))=F\!\left(1+F^{-1}(x)\right)=\exp(F(F^{-1}(x))=\exp(x)$

Possible application

In the case when a signal is supposed to pass through a set of $N$ identical elements, and the transfer function of the integral cirquit is known, the loginal of this transfer function allows to calculate the response function of each indifidual element, extracting root of power $N$ from the integral response function.

The elements have no need to be discreet, formula (4) can be applied for real values of $n$ as well. At least tetration (case of exponential function $g$) seems to be naturally extendable for the complex values. The continuous case may refer to the nonlinear optical fiber cirquit.

Conclusion

Roughly, loginal of a funciton allows to count, how many times the function should be applied to get the given function; this allows to apply a function some "fractal number of times. For summation and multiplication, loginal is easy to express. For exponential, loginal is operation of tetration. In general case, finding of loginal $K$ of a heneral function $g$ is not trivial.

References

(needs to be cleaned up)

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