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 fractional powers of function inversion (was: changing terminology) Base-Acid Tetration Fellow Posts: 94 Threads: 15 Joined: Apr 2009 08/10/2009, 06:14 PM (This post was last modified: 08/11/2009, 02:08 AM by Base-Acid Tetration.) (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? Base-Acid Tetration Fellow Posts: 94 Threads: 15 Joined: Apr 2009 08/11/2009, 02:07 AM (This post was last modified: 08/11/2009, 02:08 AM by Base-Acid Tetration.) but if we consider the half-iterate of inversion, what is the half iterate of inverting a function? an i-th iterate? bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 08/11/2009, 08:49 AM (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. Gottfried Ultimate Fellow Posts: 758 Threads: 117 Joined: Aug 2007 08/11/2009, 12:02 PM (This post was last modified: 08/11/2009, 12:53 PM by Gottfried.) r (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)```As you want to do arithmethic with the "number of parts of inv"-operations, then I think, that this is 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)^s```One nice property of the Pascal-matrix is, that you even don't need the LOG and EXP for fractional powers. If 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)```So, for the "half-inverse" in this sense, we need P to the (-1)^0.5 = I 'th power Code:```´   PI  = P^i = dV(i)*P*dV(1/i)   PI * V(x) = V(x+i)```Now we cannot simply use the iterate Code:```´   PI^2* V(x) = PI * V(x+i) = V(x+2i)```because it were in fact P^(i+i) = P^(2i) but need the i'th power of PI, such that P^(i^2) = P^(-1) is the result. 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.         Gottfried Helms, Kassel bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 08/11/2009, 01:07 PM (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 $S^{\circ x}[f]=f^{\circ s^x}$. bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 08/11/2009, 01:26 PM (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. « Next Oldest | Next Newest »

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