Left associative exponentiation- an iteration exercise Gottfried Ultimate Fellow     Posts: 889 Threads: 130 Joined: Aug 2007 09/12/2008, 08:17 PM (This post was last modified: 09/25/2008, 03:17 PM by Gottfried.) A nice accidental result - although it is nearly trivial... I tried, which matrix-operator would perform the left-associative exponentiation - well, this can easily be solved, but the appearance of this operator was much surprising to me. First assume the factorially similarity-scaled matrices of Stirling numbers 2nd and 1st kind (as I described them many times). If S2 is that lower-triangle containing Stirling-numbers 2nd kind and S1 that containing 1st kind, then fS2F and fS1F are the factorially similarity scalings: fS1F = dF^-1 * S1 * dF fS2F = dF^-1 * S2 * dF for instance: Code:´      fS2F = fS1F^-1      V(x)~ * fS1F = V(log(1+x))~ Now consider the matrix Code:´     DPW_m = fS1F * dV(m) * fS2F If we use it as matrix-operator we have the following transform of powerseries: Code:´ V(x-1)~        * fS1F  = V(log(x))~ V(log(x))~     * dV(m) = V(m*log(x))~ = V(log(x^m))~ V(log(x^m))~   * fS2F  = V(exp(log(x^m))-1) ~                         = V(x^m -1)~so Code:´ V(x - 1)~ * DPW_m  = V(x^m - 1)~ and integer iterations by integer powers of DPW_m Code:´ V(x - 1)~ * DPW_m^k = V(x^(m^k) - 1)~ If we interpret V(x-1) as a binomial-transform of V(x) then we can note this at the beginning and at end we have: Code:´ (V(x)~ * P^-1~)  * DPW_m^k   =  (V(x^(m^k))~ * P^-1 ~) then V(x)~ * (P^-1~  * DPW_m^k * P~) = V(x^(m^k))~ and denote the new product of matrices in the parenthese formally Code:´    PW_m = P^-1 ~ * (DPW_m * P~) then we have the operator for x-> x^m or, when k-times iterated, x -> x^(m^k), whenever the product P^-1~ * () is computable. The much surprising aspect is the shape of DPW_m * P~: this is just the binomial-matrix where the columns represent the binomial-coefficients for the orders according to m - which also can be fractional. For instance, m=2, DPW_2*P~ = Code:´   1  1  1   1   1    1    1     1   .  2  4   6   8   10   12    14   .  1  6  15  28   45   66    91   .  .  4  20  56  120  220   364   .  .  1  15  70  210  495  1001   .  .  .   6  56  252  792  2002   .  .  .   1  28  210  924  3003   .  .  .   .   8  120  792  3432 which gives, premultiplied by P^-1~ (here the multiplication is possible) Code:´   1  .  .  .  .  .  .  .   .  .  .  .  .  .  .  .   .  1  .  .  .  .  .  .   .  .  .  .  .  .  .  .   .  .  1  .  .  .  .  .   .  .  .  .  .  .  .  .   .  .  .  1  .  .  .  .   .  .  .  .  .  .  .  . the "quasi"-diagonal matrix, which simply shifts all powers of x in V(x) to their 2*power-position V(x^2). For instance, the even-indexed powers in [(1),x,(x^2),x^3,(x^4)...] get simply shifted to the consecutive positions [(1),(x^2),(x^4),(x^6),...], so we have a transform of x->x^2 and of all consecutive powers. For fractional "bases of iteration" ( this means exponents here) m=1/2 Code:´   1         1  1        1  1        1  1        1   .       1/2  1      3/2  2      5/2  3      7/2   .      -1/8  .      3/8  1     15/8  3     35/8   .      1/16  .    -1/16  .     5/16  1    35/16   .    -5/128  .    3/128  .   -5/128  .   35/128   .     7/256  .   -3/256  .    3/256  .   -7/256   .  -21/1024  .   7/1024  .  -5/1024  .   7/1024   .   33/2048  .  -9/2048  .   5/2048  .  -5/2048the premultiplication by P^-1~ gives divergent results, cannot be done and must applied to the V(x)~-parameter instead. The most interesting aspect is, that with integer m the column-vectors of PW_m are finite - so we can safely define the version with P^-1~ premultiplied - which gives, for integer m (and heights) - nicely the quasi-diagonal-matrices, which transform a vandermondevector V(x) into one of V(x^2), V(x^3) or the like... And finally, the aspect, which really pleases me, is, that for fractional m we just get the correct column-vectors of fractional binomials - a smooth procedure, which I was looking for over the last monthes (unsatisfied with the hardcoded binomial-expressions otherwise) Here another example with m=2/3, which means, each multiple of third power of x in the result is composed by smaller integer powers of x, and only the other powers are composed by the infinite sequences of according fractional binomially weighted powers of x: Code:´   1          1          1  1         1          1  1          1   .        2/3        4/3  2       8/3       10/3  4       14/3   .       -1/9        2/9  1      20/9       35/9  6       77/9   .       4/81      -4/81  .     40/81     140/81  4     616/81   .     -7/243      5/243  .   -10/243     35/243  1    770/243   .     14/729     -8/729  .     8/729    -14/729  .    308/729   .   -91/6561    44/6561  .  -28/6561    35/6561  .  -154/6561   .  208/19683  -88/19683  .  40/19683  -40/19683  .   88/19683 so, the spin-off-result here is: the fractional m in fS1F*dV(m)*fS2F*P~ gives the correct composition for the matrix-operator, which performs the binomial-theorem for fractional exponents, and its (possibly fractional) powers give the expected formal powerseries for the expected binomial composition. Gottfried Helms, Kassel « Next Oldest | Next Newest »

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