Integration?

[bo198214]

Unfortunately, all those summations make calculating even \( Ti_{3}(1) \) extraordinarily difficult. Can anyone calculate this series with Maple or Mathematica? (I know that the integral approximation is ~0.573... but I want to check my series with that)

I use sage with my formal_powerseries package. With a code like this

Code:

sage: load formal_powerseries.py
sage: F = FormalPowerSeriesRing(QQ)
sage: pt3 = F.Exp(F.Log_inc * F.Exp(F.Log_inc * F.Inc))
sage: pt3
[1, 1, 1, 3/2, 4/3, 3/2, 53/40, 233/180, 5627/5040, 2501/2520, 8399/10080, ...]
sage: #or the derivatives
sage: pt3.mul_fact()
[1, 1, 2, 9, 32, 180, 954, 6524, 45016, 360144, 3023640, 27617832, ...]

we get the Taylor coefficients of \( \text{pt3}(x)=(x+1)^{(x+1)^{x+1}} \) at 0 which is the same as the Taylor coefficients of \( x^{x^x} \) at 1.

We can now integrate this:

Code:

sage: pt3i = pt3.integral()
sage: pt3i
[0, 1, 1/2, 1/3, 3/8, 4/15, 1/4, 53/280, 233/1440, 5627/45360, 2501/25200, ...]
sage: pt3i.mul_fact()
[0, 1, 1, 2, 9, 32, 180, 954, 6524, 45016, 360144, 3023640, 27617832, ...]

I always include the derivatives which is the powerseries coefficients multiplied by n! because this is an integer sequence, which you possibly can look up at Sloane's Online Encypclopedia of Integer Sequences. And now I see that the sequence for the 3rd powertower (or its integral which is just the series shifted to the right by one) isnt found there (the second powertower is of course contained). I will filed a new submission.

And now we can compute the integral pt3i(1-1) - pt3i(0-1) = -pt3i(-1) by truncating the infinite series, say to 100. To have an idea about the accuracy we also calculate the 101-truncation:

Code:

sage: p = pt3i.polynomial(100)
sage: p2 = pt3i.polynomial(101)
sage: RR(-p(-1))
0.573116317784245
sage: RR(-p2(-1))
0.573116465191639

This means \( \text{Ti}_3(1)\approx 0.573116 \).

Edit: If we improve precision a bit taking a polynomial with 200 terms which takes a bit longer to compute we see that the last digits are not accurate:

Code:

sage: p3 = pt3i.polynomial(200)
sage: p3x = pt3i.polynomial(201)
sage: RR(-p3(-1))
0.573120826997061
sage: RR(-p3x(-1))
0.573120837278469

This time a bit more careful with the last digits I would say: \( \text{Ti}_3(1)\approx 0.573121 \).

Caution: actually one need to be careful with this kind of evaluation of a powerseries. Because we know that the convergence radius of the powertowers are at most 1. That means that -1 is at boundary of the disk of convergence of pt3, so it may or may not converge. And I guess the same is true for pt3i.