A dihedral projection

[Milo]

Please wait a while as I would like to derive the C3 class solution on my own.

Have I waited long enough? Because I'm still hoping to find the C^∞ solution, and I did some more work on this myself.

I figured out how to code Faà di Bruno's formula in Maxima:

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halfangles: true$
g[n] := at(diff(asin(cos(a)/sqrt(2)), a, n), a=acos(1/3)/2)$
bell[n,k] := if k=1 then g[n] else fullratsimp(sum(binomial(n,j) * g[j] * bell[n-j,k-1], j, 1, n-1) / k)$
f[n] := fullratsimp(sum(f[k] * bell[n,k], k, 1, n-1) / (1 - bell[n,n]))$ /* Note that this solves f(a) = f(g(a)), given g(a) = a, ignoring constant and linear terms in the actual formula we want to solve, and so is valid only for n >= 2.  n = 1 is manually patched in below. */
f[1]: %pi/acos(1/3) / 3$

Producing:
f[1] = %pi/acos(1/3) * 1/3
f[2] = %pi/acos(1/3) * -1/3 / sqrt(2)
f[3] = %pi/acos(1/3) * 1/9
f[4] = %pi/acos(1/3) * -1/5 / sqrt(2)
f[5] = %pi/acos(1/3) * -29/9
f[6] = %pi/acos(1/3) * 439/77 / sqrt(2)
f[7] = %pi/acos(1/3) * -42979/4257
f[8] = %pi/acos(1/3) * -65951/3655 / sqrt(2)
f[9] = %pi/acos(1/3) * 883085651/1375011

Unfortunately these follow no obvious pattern, although they do appear to stay rational multiples of either %pi/acos(1/3) (odd derivatives) or %pi/acos(1/3)/sqrt(2) (even derivatives).

Furthermore this computes the derivatives at a = acos(1/3)/2, whereas a Taylor series would be much simpler (containing only odd-exponent terms) around a = %pi/4. Plus, given the non-analytic nature of the function, the Taylor series around a = acos(1/3)/2 probably doesn't actually converge.

Suppose we try to define f (more accurately, f_2 from before, valid between acos(1/3)/2 and pi/2-acos(1/3)/2) as
f(a) := sum(c[m] / (2*m-1)! * (a-%pi/4)^(2*m-1), m, 1, inf) + %pi/4
Then c[m] need to satisfy the following relationship relative to the previously-computed f[n]:
f[n] = sum(c[m] / (2*m-1-n)! * (acos(1/3)/2-%pi/4)^(2*m-1-n), m, ceiling((n+1)/2), inf)
Actually solving this to infinity is difficult, but if we limit the summation to, say, five terms, then we get:
c[1] = -(%pi*(145*2^(9/2)*acos(1/3)^4-145*2^(11/2)*%pi*acos(1/3)^3-44352*acos(1/3)^3+435*2^(7/2)*%pi^2*acos(1/3)^2+66528*%pi*acos(1/3)^2-4785*2^(13/2)*acos(1/3)^2-145*2^(7/2)*%pi^3*acos(1/3)-33264*%pi^2*acos(1/3)+4785*2^(13/2)*%pi*acos(1/3)-5892480*acos(1/3)+145*sqrt(2)*%pi^4+5544*%pi^3-4785*2^(9/2)*%pi^2+2946240*%pi-495*2^(31/2)))/(1485*2^(31/2)*acos(1/3))
c[2] = (%pi*(1160*acos(1/3)^3-1740*%pi*acos(1/3)^2-297*2^(11/2)*acos(1/3)^2+870*%pi^2*acos(1/3)+297*2^(11/2)*%pi*acos(1/3)-100320*acos(1/3)-145*%pi^3-297*2^(7/2)*%pi^2+50160*%pi-17325*2^(11/2)))/(380160*acos(1/3)*(2*acos(1/3)-%pi))
c[3] = -(%pi*(232*acos(1/3)^3-348*%pi*acos(1/3)^2-99*2^(9/2)*acos(1/3)^2+174*%pi^2*acos(1/3)+99*2^(9/2)*%pi*acos(1/3)-12320*acos(1/3)-29*%pi^3-99*2^(5/2)*%pi^2+6160*%pi-1155*2^(11/2)))/(264*acos(1/3)*(2*acos(1/3)-%pi)^3)
c[4] = (8*%pi*(1160*acos(1/3)^3-1740*%pi*acos(1/3)^2-99*2^(13/2)*acos(1/3)^2+870*%pi^2*acos(1/3)+99*2^(13/2)*%pi*acos(1/3)-36960*acos(1/3)-145*%pi^3-99*2^(9/2)*%pi^2+18480*%pi-3465*2^(11/2)))/(33*acos(1/3)*(2*acos(1/3)-%pi)^5)
c[5] = -(1792*%pi*(1160*acos(1/3)^3-1740*%pi*acos(1/3)^2-297*2^(9/2)*acos(1/3)^2+870*%pi^2*acos(1/3)+297*2^(9/2)*%pi*acos(1/3)-26400*acos(1/3)-145*%pi^3-297*2^(5/2)*%pi^2+13200*%pi-2475*2^(11/2)))/(33*acos(1/3)*(2*acos(1/3)-%pi)^7)
...Yeah. Numerically, this gives:
c[1] = 0.778240456947945
c[2] = 7.423254731907882
c[3] = -1488.676285956136
c[4] = 619910.1478436295
c[5] = -214811061.0840599
Compare the earlier C² version, which had:
c[1] = 0.799609546057049
c[2] = 3.540212531753472
Or the even earlier C¹ version, which had:
c[1] = 0.8507165520199232
Note that c[m] grows even faster than (2*m-1)! does, so even if we redefined the Taylor summation to omit that part, they'd still get pretty high.
Going the other way, a C²⁰ version has:
c[1] = 0.764495900777279 = 0.764495900777279 * 1!
c[2] = 16.2937621351 = 2.715627022516667 = 3!
c[3] = -20022.43327616207 = -166.8536106346839 * 5!
c[4] = 7.076072603208672E+7 = 14039.82659366799 * 7!
c[5] = -3.92158350587567E+11 = -1080683.285349336 * 9!
c[6] = 2.852412677766556E+15 = 71458951.56341581 * 11!
c[7] = -2.489653753710991E+19 = -3998145876.935229 * 13!
c[8] = 2.466375940530462E+23 = 188607806414.5754 * 15!
c[9] = -2.650236337102935E+27 = -7451026175678.147 * 17!
c[10] = 3.14376953136687E+31 = 258437826168181.4 * 19!
c[11] = -4.142165954208365E+35 = -8107437009650615 * 21!
c[12] = 8.454279234842538E+36 = 327025907504003 * 23!
c[13] = -1.603867150320828E+39 = -103400517808757.8 * 25!
c[14] = 1.686363403437531E+41 = 15487038494825.02 * 27!
c[15] = -6.217355432072262E+44 = -70318059188591.4 * 29!
c[16] = -1.637231301705888E+45 = -199107798481.9793 * 31!
c[17] = 1.401911494166907E+48 = 161448832774.4398 * 33!
c[18] = 9.29084015974988E+49 = 8991296930.976983 * 35!
c[19] = -3.282134015861778E+51 = -238462139.948875 * 37!
c[20] = -8.760449740820171E+52 = -4294783.99078278 * 39!
This does feel like it converges to something, but...

Oh, the command for computing those:

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coefficients(continuity) := solve(makelist(f[n] = sum(c[m] / (2*m-1-n)! * (acos(1/3)/2-%pi/4)^(2*m-1-n), m, ceiling((n+1)/2), continuity), n, 1, continuity), makelist(c[n], n, 1, continuity))[1]$

[dummy_index]

I'm sorry, I just couldn't find the time.

(I was trying to figure out which symbol to use for acos(1/3)/2 and pi/4-acos(1/3)/2 avoiding a, b, c and d.
m = acos(-1/3) (Maraldi angle)
m_1 = pi - m = acos(1/3)
m_2 = pi/2 - m_1 = pi/2 - acos(1/3)
(m_3 = pi/4 - m_2 ?)
h_1 = m_1 / 2 = acos(1/3)/2
h_2 = m_2 / 2 = pi/4 - acos(1/3)/2
)

Suppose we try to define f

Hmmm, wouldn't that method end up trying to make an analytic continuation between f_1 and f_2? If you can analytic continuation and it becomes one analytic function, it will not be possible to have y = f(a) - d/3 be linear only up to the middle.
I think you need to consider other basis functions. For example, the Fourier series can finally reproduce a trigonometric wave by adding infinitely many more, and not just matching the height of the vertices (i.e. C^1) nor C^20 in the middle step.

[Milo]

(I was trying to figure out which symbol to use for acos(1/3)/2 and pi/4-acos(1/3)/2 avoiding a, b, c and d.)

Well, it's traditional for angles to be assigned to Greek letters, so just pick one that isn't used much, I guess.

Hmmm, wouldn't that method end up trying to make an analytic continuation between f_1 and f_2?

No. Presumably, the resulting Taylor series would have a finite radius of convergence. Ideally, the radius of convergence would be pi/4 - acos(1/3)/2, so convergence fails at the same point analyticity does.

This is another reason why I want the Taylor series to be centered on pi/4. That way, it has the largest possible radius of convergence, with the unavoidable disanalyticities (is that what you call the point where a function isn't analytic?) being equally far in either direction.

[dummy_index]

No. Presumably, the resulting Taylor series would have a finite radius of convergence. Ideally, the radius of convergence would be pi/4 - acos(1/3)/2, so convergence fails at the same point analyticity does.

Is it possible to control the radius of convergence in this way? It may be possible, but it does not seem to be working at present.

For simplicity, let f_1 be a quadratic function and
f[0] is not constrained,
f[1] = 1 or something,
f[2] = 1 or something,
f[3] = f[4] = ... = 0.
Then
c[3] = c[4] = ... = 0,
and a cubic function (c[2]*x^3 + c[1]*x) connects between f_1 and f_3. This is not non-analytic, but non-smooth.

Actually in this case, solution is impossible because f[3]=... and f[2]=... require different c[2]. However, even in such basic case, we want to output an infinite sum representing a non-analytic function.

[Milo]

Is it possible to control the radius of convergence in this way?

I think so. For example, consider the function f(x)=exp(-1/x^2) (with f(0)=0). I had Maxima compute and plot its Taylor expansions to 100 terms, using:

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plot2d([exp(-1/x^2), taylor(exp(-1/x^2), x, 1, 100), taylor(exp(-1/x^2), x, 2, 100), taylor(exp(-1/x^2), x, 3, 100)], [x,-6,+6], [y,-0.5,1.5], [legend, "e^{-1/x^2}", "Taylor expansion around x=1", "Taylor expansion around x=2", "Taylor expansion around x=3"])$

The result is

roc.png (6.51 KiB) Viewed 10912 times

You can see from where they veer off that the expansion around x=1 seems to converge on approximately ⟨0,2⟩, the expansion around x=2 seems to converge on approximately ⟨0,4⟩, and the expansion around x=3 seems to converge on approximately ⟨0,6⟩. This supports the idea that the radius of convergence is equal in either direction from the center, and as large as it can be without hitting the disanalyticity. (Actually, the expansions veer off somewhat short of these bounds, but this is presumably due to me only summing the Taylor expansion to 100 terms. It makes sense that near the point where convergence fails, convergence would be very slow, so even 100 terms isn't enough.)

(Note that while this function is infinitely-real-differentiable for all real x, including x=0, it is not complex-differentiable or even complex-continuous at x=0, explaining the analyticity failure.)

The alternative would be that the function we're looking for isn't analytic at all, even away from the guaranteed disanalyticity points of a=acos(1/3)/2 and a=pi/2-acos(1/3)/2. Everywhere-smooth but nowhere-analytic functions do exist...

[dummy_index]

nonanalytic points are at acos(1/3)/2 and pi/2-acos(1/3)/2. any smooth transition functions could be used, but p1 and p2 need to be adjusted to gain resolution-efficiency (consider off the apothem).

h_1=acos(1.0/3)/2
h_2=pi/4-acos(1.0/3)/2
p1=0.6482 (?)
p2=6.0 (?)
f_2m(x)=p1+f[1]/1!*(x-h_1)+f[2]/2!*(x-h_1)**2+f[3]/3!*(x-h_1)**3+f[4]/4!*(x-h_1)**4+f[5]/5!*(x-h_1)**5+f[6]/6!*(x-h_1)**6+f[7]/7!*(x-h_1)**7+f[8]/8!*(x-h_1)**8+f[9]/9!*(x-h_1)**9 (+O(x**11))
logistic(x)=x<-300?0.0:x>300?1.0:1/(1+exp(-x))
f_2(x)=f_2m(x)*logistic(-p2*tan((x-pi/4)/h_2*pi/2))+(pi/2-f_2m(pi/2-x))*logistic(p2*tan((x-pi/4)/h_2*pi/2))