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 fractional powers of function inversion (was: changing terminology) Gottfried Ultimate Fellow Posts: 757 Threads: 116 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 « Next Oldest | Next Newest »

 Messages In This Thread fractional powers of function inversion (was: changing terminology) - by Base-Acid Tetration - 08/10/2009, 06:14 PM RE: fractional powers of function inversion (was: changing terminology) - by Base-Acid Tetration - 08/11/2009, 02:07 AM RE: fractional powers of function inversion (was: changing terminology) - by bo198214 - 08/11/2009, 08:49 AM RE: fractional powers of function inversion (was: changing terminology) - by Gottfried - 08/11/2009, 12:02 PM RE: fractional powers of function inversion (was: changing terminology) - by bo198214 - 08/11/2009, 01:07 PM

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