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2020-01-16, 03:34   #56
0scar

Jan 2020

22·3 Posts

Quote:
 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.

2020-01-17, 00:26   #57
SmartMersenne

Sep 2017

7710 Posts

Quote:
 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.

2020-01-18, 02:43   #58
Dr Sardonicus

Feb 2017
Nowhere

23·3·5·29 Posts

Quote:
 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.

2020-01-18, 03:16   #59
SmartMersenne

Sep 2017

7·11 Posts

Quote:
 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.

 2020-01-18, 13:36 #60 Dr Sardonicus     Feb 2017 Nowhere 23·3·5·29 Posts Here's my check of the proffered answer. The function polsturm() counts the number of real zeroes. Code: ? 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!

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