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+--- Thread: fractional powers of function inversion (was: changing terminology) (/showthread.php?tid=321)
fractional powers of function inversion (was: changing terminology) - Base-Acid Tetration - 08/10/2009
(08/10/2009, 11:32 AM)Ansus Wrote: Heh it would be a good idea to introduce an 'arc' or 'inv' operator instead of ugly f^-1, commonly used.
so you might want to consider fractional iterates of functional inversion operator inv[]?
such that inv^2[f] = f (hopefully)
can we assume this f^a)^b = f^(ab) for most cases?
can real or complex iterates of functional inversion be associated with powers of -1?
is []^i=inv^(1/2)[], so that (f^i)^i = f^-1?
how are these complex iterate thingies numerically computed anyway?
RE: fractional powers of function inversion (was: changing terminology) - Base-Acid Tetration - 08/11/2009
but if we consider the half-iterate of inversion, what is the half iterate of inverting a function? an i-th iterate?
RE: fractional powers of function inversion (was: changing terminology) - bo198214 - 08/11/2009
(08/10/2009, 06:14 PM)Tetratophile Wrote: so you might want to consider fractional iterates of functional inversion operator inv[]?
slowly, slowly, I can not remember that we anywhere on the forum already considered extended iterates of an *operator*.
Perhaps you can make an infinite matrix that maps the powerseries of the input function to the powerseries of the output function. Then one could take matrix powers to define the fractional iterates, however I didnt see this anywhere done.
I dont even know whether inv is expressible as such a matrix.
Quote:but if we consider the half-iterate of inv, it will be a hopefully.
you mean i?
Quote:can we assume this f^a)^b = f^(ab) for most cases?
Yes I think so (for real a,b), it should be derivable from f^(a+b)=f^a o f^b, when taking into account cancelability, i.e. g^n=h^n should imply g=h for g and h being an iterate of f.
Quote:how are these complex iterate thingies numerically computed anyway?
for *functions* you just plug i into the formula
I.e. into a regular iteration formula.
RE: fractional powers of function inversion (was: changing terminology) - Gottfried - 08/11/2009
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(08/10/2009, 06:14 PM)Tetratophile Wrote:(08/10/2009, 11:32 AM)Ansus Wrote: Heh it would be a good idea to introduce an 'arc' or 'inv' operator instead of ugly f^-1, commonly used.
so you might want to consider fractional iterates of functional inversion operator inv[]?
such that inv^2[f] = f (hopefully)
can we assume this f^a)^b = f^(ab) for most cases?
can real or complex iterates of functional inversion be associated with powers of -1?
is []^i=inv^(1/2)[], so that (f^i)^i = f^-1?
how are these complex iterate thingies numerically computed anyway?
If you only mean inverse of f like inv(f) or half-step or even complex-step of this then I think, this is the question of powers of the iterator-parameter:
Code:
inv(f) = f°[-1](x)
inv(inv(f)) = f°[-1](f°[-1](x)) = f°[(-1)*(-1)](x) = f°[1](x) = f(x)inv(inv(...(inv(f(x)))) = f°[(-1)^h](x) \\ where then "inv" occurs h-times
and fractional "iterates of inversion" is then multivalued with complex heights according to the complex roots of -1
inv°[s](f(x)) = f°[(-1)^s](x)
But we already have a concept of complex heights, at least with functions, which can be represented by Bell-matrices: just compute the s'th power of the Bell-matrix and use its entries for the coefficients of the Taylor-series of the new function.
For an easier example than ours (which is the exponential f(x) = exp(x)) you can look at the function f(x) = x+1 and the fractional and complex powers of the Pascal-matrix.
Say , with the vandermonde(column-)vector V(x)= [1,x,x^2,x^3,...]~ and the (lower triangular) pascalmatrix P
Code:
´
P * V(x) = V(x+1) implements f(x) = x+1
P^-1 * V(x) = V(x-1) implements inv(f(x)) = x - 1
Generally, using the matrix-logarithm and -exponential
PL = Log(P)
P^s = EXP( PL * s) // for all complex s
Then also
P^((-1)^s) = EXP( PL * (-1)^s) // for complex s
which is what you asking for, and practically
P^((-1)^s) * V(x) = V(x+(-1)^s) implements inv^[s](f(x)) = x + (-1)^sIf you define the vandermonde-vector V(x) as diagonal-matrix dV(x), then
Code:
´
P^s = dV(s)* P * dV(1/s)
and
P^s * V(x) = dV(s)*P*dV(1/s) * V(x)
= dV(s)* P *V(x/s)
= dV(s) * V(x/s+1)
= V(s*(x/s + 1)
= V(x+s)Code:
´
PI = P^i = dV(i)*P*dV(1/i)
PI * V(x) = V(x+i)Code:
´
PI^2* V(x) = PI * V(x+i) = V(x+2i)Thus
Code:
´
PII = (P^i)^i = P^(i^2) = P^-1
= dV(i) * PI * dV(1/i)
PII * dV(x) = dV(i) * PI * dV(1/i) *V(x)
= dV(i)* dV(i)*P*dV(1/i) *dV(1/i) *V(x)
= dV(-1) *P* dV(-1) *V(x)
= dV(-1) *P * V(-x)
= dV(-1) * V(-x+1)
= V(-(-x+1))
= V(x - 1)However, the latter nice and easy computation of arbitrary powers of P by simply multiplication with diagonal-vectors is not available for our exponential-iteration, here we need the matrix-log or eigensystem-decomposition of the bell matrix to get fractional powers and then fractional iterates, or even complex powers steming from complex unit-roots to implement "fractional-step-inversion"...
But I can provide a picture, where I plotted complex heights for the base b =sqrt(2) such that we have the curves for b^^h, where h=(-1)^m, where m is real, thus the "inversion in fractional steps". The graph has four curves, the relevant is the blue one: for h=1 the curve is on the real axis at x=(real,imag)=(sqrt(2),0), for h=-1 is x=(log(1)/log(b), 0) = (0,0), and for the "half-inverse" (having h=(-1)^0.5=I) it is at the thick blue point.
RE: fractional powers of function inversion (was: changing terminology) - bo198214 - 08/11/2009
(08/11/2009, 12:02 PM)Gottfried Wrote: and fractional "iterates of inversion" is then multivalued with complex heights according to the complex roots of -1
inv°[s](f(x)) = f°[(-1)^s](x)
Ya true, for iteration operators \( S[f]:=f^{\circ s} \) we dont need a new technique for general operator iteration, it can be reduced to powers of the exponent \( S^{\circ x}[f]=f^{\circ s^x} \).
RE: fractional powers of function inversion (was: changing terminology) - bo198214 - 08/11/2009
(08/11/2009, 01:17 PM)Tetratophile Wrote: @bo: ...not an operator, sorry. i meant a property. is the i-th power one of them. that was what i meant to ask.
Inv is an operator.
Most of the on the forum featured iteration extensions allow to plug in a complex iteration exponent.