uniqueness

[bo198214]

To be simple we have two sets of methods: those applicable for lower bases (and they are equal) and those applicable for higher bases (I suppose they also equal with each other). Probably we would not need a proof for equivalence of these two sets of methods because their areas of applicability do not overlap.

They do overlap, Matrix-Power and Intuitive Abel are applicable to all bases \( b>1 \).
And even Cauchy-Integral we saw that there are similar methods for for both base ranges.

P.S. Where I could learn about matrix powers method?

See the pdf Gottfried posted here.

[bo198214]

Does not it use Carleman matrices whic also used for partial iteration formula?

What is partial iteration?

Indeed the matrix power method coincides with regular iteration if applied to a fixed point.

[bo198214]

You derived your Newton formula using Carleman matrices, and they also use Carleman matrices of arbitrary iteration of exp function.

thats exactly what i said: at fixed points both methods coincide.

[tommy1729]

f(x) = (x + q(x)) ^2 = x^2 + 2q(x)x + q(x)^2

f(f(x)) = ( x^2 + 2q(x)x + q(x)^2 + q(x^2 + 2q(x)x + q(x)^2) ) ^2

q must appear inside q.

so ??

thats no proof or disproof of anything ?


regards

tommy1729

[tommy1729]

here are the equations that make half-iterate of exp(x) unique :

(under condition f(x) maps reals to reals and f(x) > x )

exp(x)

= f(f(x))

= D f(f(x)) = f ' (f(x)) * f ' (x)

= D^2 f(f(x)) = f '' (f(x)) * f ' (x)^2 + f ' (f(x)) * f '' (x)

= D^3 f(f(x)) = D^4 f(f(x))


regards

tommy1729

is there a solution where f ' (x) is not strictly rising but f(x) is Coo ??

[tommy1729]

here are the equations that make half-iterate of exp(x) unique :

(under condition f(x) maps reals to reals and f(x) > x )

exp(x)

= f(f(x))

= D f(f(x)) = f ' (f(x)) * f ' (x)

= D^2 f(f(x)) = f '' (f(x)) * f ' (x)^2 + f ' (f(x)) * f '' (x)

= D^3 f(f(x)) = D^4 f(f(x))


regards

tommy1729

is there a solution where f ' ' (x) is not strictly rising but f(x) is Coo ??

*corrected*

[tommy1729]

here are the equations that make half-iterate of exp(x) unique :

(under condition f(x) maps reals to reals and f(x) > x )

exp(x)

= f(f(x))

= D f(f(x)) = f ' (f(x)) * f ' (x)

= D^2 f(f(x)) = f '' (f(x)) * f ' (x)^2 + f ' (f(x)) * f '' (x)

= D^3 f(f(x)) = D^4 f(f(x))


regards

tommy1729

is there a solution where f ' ' (x) is not strictly rising but f(x) is Coo ??

*corrected*

ok.

i see now.

i was a bit confused ...

the new uniqueness conditions , i think ,

(under condition f(x) maps reals to reals and f(x) > x )

f ' (0) > 0

f '' (0) > 0

f ''' (0) > 0

...



and f(x) is Coo


regards

tommy1729

[bo198214]

Before uniqueness, one should show however existence, because a uniqueness criterion makes no sense if there are no functions satisfying this criterion ...

[tommy1729]

Before uniqueness, one should show however existence, because a uniqueness criterion makes no sense if there are no functions satisfying this criterion ...

isnt that already done ?

by kneser ?

by robbins ?

doesnt robbins solution satisfy my uniqueness criterions ?

regards

tommy1729