A nice series for b^^h , base sqrt(2), by diagonalization

[Gottfried]

Fun...

I now could make use of the upper (repelling) fixpoint with this type of series.

Here we have the upper fixpoint t=4, u=log(4) ~ 1.3862943611... for the same base b = sqrt(2), and we can compute the fractional heights for some appropriate initial-value x, say x=5:

to get context I quote the previous:

Here for base b = sqrt(2) and for brevity v = u^h:
\(
\begin{tabular}{llll}
\exp_{\sqrt{2}}^h(1)
& = 2 \\
& - 0.6321 v \\
& - 0.2253 v^2 \\
& - 0.08541 v^3+ \cdots
\end{tabular}
\)

and again v for u^h and f(h) for the longish exp-expression:

\(
\begin{tabular}{llll}
f(h) = \exp_{\sqrt{2}}^h(5)
& = 4 \\
& + ??? v \cdots
\end{tabular}
\)

with different coefficients. (We have to compute the appropriate Ut-matrix and also W-matrix now)


For x=5 we have by fixpoint-shift x1 = x/t - 1 = 5/4 - 1 = 0.25 and the evaluation of the schröder-function

Code:

´   S~ = V(5/4-1)~ * W = V(0.25)~ * W

gives, by the summation in each column, a set of series, which converge good with 64 terms(n=64 is my selected vector/matrix-dimension) and give the vector S containing all results. Then the coefficients from WI (which represent the inverse of schröder-function) are multiplied into to get the constant vector M, which has now the coefficients which are independent from u^h:

Code:

´   M~ = S ~ * diag(WI[,1])

Then the intermediate value y as in the previous msg is again:

\( \hspace{24} y = M\sim * V(u^h) = \sum_{k=0}^{\infty} m_k * (u^h)^k \)

Thus, having the fixpoint t=4 here, we get the new coefficients for the series from the vector M:

Code:

´  f(h) = 4 + 4* y
         = 4 + 4* (M~ * V(u^h))

which can be evaluated for some height h, as long as the occuring series (by M~*V(u^h)) converges or can be Euler-summed.
This gives the powerseries:
\(
\begin{tabular}{llll}
f(h) = \exp_{\sqrt{2}}^h(5)
& = 4 \\
& + 0.694707147143 v \\
& + 0.216496377971 v^2\\
& + 0.0638276619589 v^3\\
& + 0.0181280769146 v^4 \\
& + 0.00500577162906 v^5 \\
& + 0.00135142475056 v^6 \\
& + 0.000358055753514 v^7 \\
& + 0.0000933533652020 v^8 \\
& + 0.0000240008373227 v^9 \\
& + 0.00000609458207498 v^{10} \\
& + 0.00000153056011763 v^{11} \\
& + 0.000000380551461011 v^{12} \\
& + 0.0000000937615006278 v^{13} \\
& + 0.0000000229095405831 v^{14} \\
& + 0.00000000555487772536 v^{15} \\
& + 0.00000000133735491281 v^{16} \\
& + 0.000000000319852431282 v^{17} \\
& + 7.60281626580E+11 v^{18} \cdots
\end{tabular}
\)



For the heights h=0..-2 in 1/32-steps I get the following values:

Code:

´   h    |  f(h) = exp_sqrt(2)°h(5)=
-----------------------------    
       0  |  5.00000000000
   -1/32  |  4.98555564349
   -1/16  |  4.97137811129
   -3/32  |  4.95746089099
    -1/8  |  4.94379767428
   -5/32  |  4.93038234919
   -3/16  |  4.91720899253
   -7/32  |  4.90427186283

    -1/4  |  4.89156539347
   -9/32  |  4.87908418618
   -5/16  |  4.86682300480
  -11/32  |  4.85477676933
    -3/8  |  4.84294055021
  -13/32  |  4.83130956288
   -7/16  |  4.81987916254
  -15/32  |  4.80864483915

    -1/2  |  4.79760221266
  -17/32  |  4.78674702837
   -9/16  |  4.77607515259
  -19/32  |  4.76558256839
    -5/8  |  4.75526537155
  -21/32  |  4.74511976670
  -11/16  |  4.73514206358
  -23/32  |  4.72532867348

    -3/4  |  4.71567610582
  -25/32  |  4.70618096483
  -13/16  |  4.69683994642
  -27/32  |  4.68764983512
    -7/8  |  4.67860750119
  -29/32  |  4.66970989776
  -15/16  |  4.66095405821
  -31/32  |  4.65233709352

      -1  |  4.64385618977
  -33/32  |  4.63550860581
  -17/16  |  4.62729167087
  -35/32  |  4.61920278239
    -9/8  |  4.61123940385
  -37/32  |  4.60339906273
  -19/16  |  4.59567934854
  -39/32  |  4.58807791086

    -5/4  |  4.58059245755
  -41/32  |  4.57322075293
  -21/16  |  4.56596061612
  -43/32  |  4.55880991933
   -11/8  |  4.55176658633
  -45/32  |  4.54482859088
  -23/16  |  4.53799395524
  -47/32  |  4.53126074878

    -3/2  |  4.52462708658
  -49/32  |  4.51809112807
  -25/16  |  4.51165107581
  -51/32  |  4.50530517419
   -13/8  |  4.49905170827
  -53/32  |  4.49288900258
  -27/16  |  4.48681542007
  -55/32  |  4.48082936094

    -7/4  |  4.47492926166
  -57/32  |  4.46911359397
  -29/16  |  4.46338086382
  -59/32  |  4.45772961051
   -15/8  |  4.45215840574
  -61/32  |  4.44666585274
  -31/16  |  4.44125058537
  -63/32  |  4.43591126737

      -2  |  4.43064659147

For instance, f(-1) should be log(5)/log(b) and we have from the table at h=-1 the value 4.64385618977 which agrees with direct computation log(5)/log(sqrt(2)) = 4.64385618977 , and it should be b^f(-1.5) = f(-0.5), which can be checked easily using values from the table.

Don't know yet, whether this has some benefit so far.