December 2019

[0scar]

Originally Posted by yae9911

f(x) = c3 + x * (x / (c0 / (c1 - x*x) + c2) + x / (x*x / (d0 / (d1 - x*x) + d2) + d3 ))

c0= 1.3555986881256104
c1= -0.51428204774856567
c2= 2.8334267139434814
c3= 4.7942781820893288E-003
d0= 0.33841833472251892
d1= -0.51365238428115845
d2= 0.70969653129577637
d3= 2.6548692956566811E-002

MSE ~= 1.8215E-7.

Using Newton's method, I found these coefficients:

c0 = 1.316653736569374
c1 = -0.5038314121686646
c2 = 2.809476206784873
c3 = 4.927057764206701E-003
d0 = 0.3287562928937044
d1 = -0.5033390767749206
d2 = 0.7036393598362972
d3 = 2.681336409906903E-002

MSE ~= 1.803235712965E-007

I suspect that this value is the actual record MSE quoted by IBM, as f(x) can be computed using 15 operations.
Rewrite f(x) as a quotient of even polynomials P8(x)/Q6(x), then compute its (real) partial fraction decomposition, with three quadratic denominators.

[SmartMersenne]

Originally Posted by 0scar

Using Newton's method, I found these coefficients:

c0 = 1.316653736569374
c1 = -0.5038314121686646
c2 = 2.809476206784873
c3 = 4.927057764206701E-003
d0 = 0.3287562928937044
d1 = -0.5033390767749206
d2 = 0.7036393598362972
d3 = 2.681336409906903E-002

MSE ~= 1.803235712965E-007

I suspect that this value is the actual record MSE quoted by IBM, as f(x) can be computed using 15 operations.
Rewrite f(x) as a quotient of even polynomials P8(x)/Q6(x), then compute its (real) partial fraction decomposition, with three quadratic denominators.

After the split, the roots will fall into range [-1.1], so the mse integrals will blow up.

[Dr Sardonicus]

Originally Posted by SmartMersenne

After the split, the roots will fall into range [-1.1], so the mse integrals will blow up.

I don't understand. Call the rational function P/Q where P is even of degree 8 and Q is even of degree 6. With the given coefficients, Q has no real zeroes. Replacing x^2 with x in Q gives a cubic with three real zeroes, all negative, so of the form -k^2. Thus Q is the product of three quadratic polynomials, all of the form x^2 + k^2. Nothing in (x - P/Q)^2 is going to "blow up" anywhere.

[SmartMersenne]

Originally Posted by Dr Sardonicus

I don't understand. Call the rational function P/Q where P is even of degree 8 and Q is even of degree 6. With the given coefficients, Q has no real zeroes. Replacing x^2 with x in Q gives a cubic with three real zeroes, all negative, so of the form -k^2. Thus Q is the product of three quadratic polynomials, all of the form x^2 + k^2. Nothing in (x - P/Q)^2 is going to "blow up" anywhere.

Then I must be making a mistake somewhere while converting this into fractions.

[Dr Sardonicus]

Here's my check of the proffered answer.

The function polsturm() counts the number of real zeroes.

? f = c3 + x * (x / (c0 / (c1 - x*x) + c2) + x / (x*x / (d0 / (d1 - x*x) + d2) + d3 ));

? pg=numerator(f);qg=denominator(f);

? p=substvec(pg,[c0,c1,c2,c3,d0,d1,d2,d3],[1.316653736569374, -0.5038314121686646, 2.809476206784873, .004927057764206701, 0.3287562928937044, -0.5033390767749206, 0.7036393598362972, .02681336409906903]);

? q=substvec(qg,[c0,c1,c2,c3,d0,d1,d2,d3],[1.316653736569374, -0.5038314121686646, 2.809476206784873, .004927057764206701, 0.3287562928937044, -0.5033390767749206, 0.7036393598362972, .02681336409906903]);

? polsturm(q)
%5 = 0

? factor(q)
%6 = 
[x^2 + 0.E-38*x + 0.0013081357715409056020674263005885229664 1]

[x^2 + 0.E-38*x + 0.035184006154098301694382918810177883871 1]

[x^2 + 0.E-38*x + 0.52089787935310618048251629609512747703 1]

? qr=substpol(q,x^2,x)
%7 = 2.8094762067848730000000000000000000000*x^3 + 1.5659740026819492386683626944779388983*x^2 + 0.053533739760533625202130885426609124581*x + 0.000067355964222339810991940300871368415406

? factor(qr)
%8 = 
[x + 0.52089787935310618048251629609512747703 1]

[x + 0.035184006154098301694382918810177883871 1]

[x + 0.0013081357715409056020674263005885229664 1]

The MSE is equal to the integral of (x - p/q)^2 over [0,1]. Numerical integration gave the value 0.00000018032357129653610602755968291726293260, approximately. Check!