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Nowhere analytic superexponential convergence
(01/20/2011, 05:23 PM)sheldonison Wrote: .... In particular, my hypothesis is that is also a nowhere analytic function...
The goal of this post is to show that f(x), the superexponential reciprocal sum is not analytic on the complex plane, even though it converges superexponentially fast at the real axis. This post assumes that f(x) is , and that all of the derivatives of f(x) converge via the limit equation. First, assume that the sum is analytic. Then, around any point "x" along the real axis, there must exist r, which is the radius of a circle, for which all points inside that circle, such that , for the nth approximation term, |1/sexp(x+n+z)|<1. After choosing any radius, no matter how small, we can choose a value of n, such that the approximation of f(x) has n terms, and show that for the nth term, , there exists a counter example, with |z|<r, and . Moreover, for the n+1 term, the radius will be superexponentially smaller still.

Even though the superexponential reciprocal summation converges superexponentially quickly at the real axis, it does not converge in the complex plane, which means the function is only defined at the real axis. Also, each individual approximation is analytic, and there are no singularities, so long as . But as n increases, gets arbitrarily small, and the function converges at the real axis. But, as n increases, the radius for which holds also gets arbitrarily small, which is the exact opposite from what would be expected if the function were defined in the complex plane.

Here I give an empirical (unproven) estimate of what the radius of z is. There exists a value of z, where , and where , as long as x>3.5 or so. This gives a way to calculate the upper bounds for the integer value of n to use for any particular radius r. n>slog(1/r)+2-x will work, as long as slog(1/r)>=1.5. Summation terms with n greater than this value will not be stable in the complex plane within the radius r, because the term will be greater than 1, which is contrary to the assumption. Moreover, for any value of r, no matter how small, there is a value of n for which the summation not converging in the complex plane.

Now, a few graphs. The first graph is a graph of the superexponential reciprocal summation, and its first three derivatives. For x>2.5, f(x)<2E-78. Beginning with the third derivatives, oscillation is visible. For higher derivatives, the oscillation becomes more and more pronounced, and larger in magnitude.
The next chart, is a contour plot of where where |sexp(z+1)|=1. If |sexp(z+1)|=1 then it is also true that the absolute value of the reciprocal=1, |1/sexp(z+1)|=1. This contour is also the contour where . This chart has . This contour is also where , where . If z is chosen from a point on this curve, then . Between this curve and the real axis, |1/sexp(z+1)|<1. For every point on the real axis, if the radius is bigger than some value, the circle tangentially intersects the curve, and the radius of that circle gets arbitrarily small as x increases.

Then, going back to the super-exponential reciprocal summation, in the complex plane the individual terms don't converge, because there isn't a radius for which all of the approximation terms of . This curve tracks very closely to the singularity points for the base change sexp(z), which tracks the 2sinh(z) singularities, and I could post some similarities later. I suspect there is some way to show that all three nowhere analytic functions behave in some underlying similar manner, and that perhaps this can help to prove that they are all nowhere analytic.

As an example, consider f(0.5). The n=3 approximation term is 1/sexp(3.5+z). If z=0, 1/sexp(3.5)=1.6E-78, which is very small, and convergence looks very good at the real axis. But, in the complex plane, the curve 1/sexp(3.5+z), is not so well behaved. In particular 1/sexp(3.5794+0.163i)=1.005-0.174i, which is>1. This is approximately the closest point. The radius is given by r=|abs(3.5794+0.163i-3.5)|=0.1813. The slog(1/r)+2 approximation equation gives 3.538, which is pretty good, compared to 3.5. Here is a graph of the contour of the n=3 term, a circle of radius r=0.181, for f(z)=1/sexp(3.5+z). For abs(z)<0.181, the |1/sexp(3.5+z)|<1. First, I graph the circle from z=r*exp(0) to z=r*exp(Pi). Then I zoom in on z=r*exp(0.9...1.4). Finally, I graph the absolute value, |1/sexp(x+z)| where z=r*exp(0.9..1.4). These are contour graphs, where red=real and green=imaginary. The third graph shows the approximately Gaussian decay of the |1/sexp(3.5+z)|, on the contour of the circle. The abs(1/sexp(3.5+z) quickly decays to 1E-10, and will continue to decay from there. In some ways, the rapid growth from 1/sexp(3.5)=1E-78 resembles the way a singularity might behave. As 1/sexp(z) super-exponentially decays, the Gaussian distribution falls over a smaller and smaller region of the circle.
Also, here are the first 50 derivatives, for 1/sexp(2.5+z) compared with 1/sexp(3.5+z). For the first 43 derivatives, the 1/sexp(3.5+z) terms are smaller, but after that, the 1/sexp(3.5+z) terms take over. Notice that 1/sexp(3.5) is only 1.6E-78 at the real axis; which is a very small correction.
n   1/sexp(2.5)     1/sexp(3.5)
0   0.00558298914   1.625991332E-78
1  -0.0456240094   -2.380006310E-75
2   0.2563009796    3.458149315E-72
3  -0.6054787553   -4.987380173E-69
4  -2.741862125     7.138720290E-66
5   8.757799374    -1.014009718E-62
6   157.8507067     1.429193425E-59
7   491.3929443    -1.998567535E-56
8  -9026.157526     2.772527041E-53
9  -153649.9284    -3.815139828E-50
10 -787222.9479     5.206785176E-47
11  13796409.21    -7.046904390E-44
12  402118496.3     9.456701274E-41
13  4664848650.    -1.258156134E-37
14 -9014484432.     1.659288212E-34
15 -1.700819629E12 -2.168892791E-31
16 -4.135791005E13  2.809424917E-28
17 -4.796963471E14 -3.605715028E-25
18  4.244145460E15  4.584466348E-22
19  3.817993655E17 -5.773450813E-19
20  1.088186879E19  7.200359831E-16
21  1.655763911E20 -8.891282534E-13
22 -8.910968566E20  0.000000001086880942
23 -1.582505635E23 -0.000001314979395
24 -6.115215612E24  0.001574284791
25 -1.400241746E26 -1.864569574
26 -9.368959331E26  2184.256553
27  9.368403311E28 -2530184.310
28  5.823757822E30  2897425490.
29  2.043837038E32 -3.279186635E12
30  4.246059314E33  3.666817736E15
31 -2.095476396E34 -4.049955874E18
32 -7.113541073E36  4.416819823E21
33 -4.085907645E38 -4.754661211E24
34 -1.505115263E40  5.050363334E27
35 -3.226262684E41 -5.291161995E30
36  4.065047260E42  5.465445548E33
37  8.703655084E44 -5.563575680E36
38  5.486386165E46  5.578658876E39
39  2.313255539E48 -5.507193104E42
40  6.041186242E49  5.349517047E45
41 -3.838827157E50 -5.110001027E48
42 -1.700626152E53  4.796939500E51
43 -1.303482620E55 -4.422133083E54
44 -6.682907802E56  4.000180567E57
45 -2.333449528E58 -3.547534383E60
46 -2.282950303E59  3.081401865E63
47  4.588400766E61 -2.618595179E66
48  4.904697005E63  2.174441199E69
49  3.205058435E65 -1.761857223E72
50  1.513287370E67  1.390679176E75
- Sheldon Levenstein

Messages In This Thread
RE: Nowhere analytic superexponential convergence - by sheldonison - 01/26/2011, 12:49 AM

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