User:Dmitrii Kouznetsov/loginal

Template:Under construction; Name of article is temporal. Loginal of function $g$ at some space S is function $K$ such tat

(1) $f^{a}(K(t))=K(a+t)$ for all $t\in {\rm {S}}$

Loginal allow the solution $f$ of equation

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

in form

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

Loginal should be invertable

(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, at the substitution to the initial equation (1)

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

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

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

Special cases

Summation

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

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

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, $g(x)=x*c$, then

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

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

Exponentiation

For exponentiation, $K$ is tetration,

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

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

In particular, I 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) $f(f(x))=F\!\left(1+F^{-1}(x)\right)=\exp(F(F^{-1}(x))=\exp(x)$

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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