20200116, 03:34  #56  
Jan 2020
2^{2}×3^{2} Posts 
Quote:
c0 = 1.316653736569374 c1 = 0.5038314121686646 c2 = 2.809476206784873 c3 = 4.927057764206701E003 d0 = 0.3287562928937044 d1 = 0.5033390767749206 d2 = 0.7036393598362972 d3 = 2.681336409906903E002 MSE ~= 1.803235712965E007 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. 

20200117, 00:26  #57  
Sep 2017
10000101_{2} Posts 
Quote:


20200118, 02:43  #58 
Feb 2017
Nowhere
7·29^{2} Posts 
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.

20200118, 03:16  #59  
Sep 2017
7·19 Posts 
Quote:


20200118, 13:36  #60 
Feb 2017
Nowhere
7·29^{2} 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.E38*x + 0.0013081357715409056020674263005885229664 1] [x^2 + 0.E38*x + 0.035184006154098301694382918810177883871 1] [x^2 + 0.E38*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] 
Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
December 2018  Xyzzy  Puzzles  6  20190106 23:07 
December 2017  Batalov  Puzzles  4  20180104 04:33 
December 2016  Xyzzy  Puzzles  11  20170124 12:27 
December 2015  Xyzzy  Puzzles  15  20160106 10:23 
December 2014  Xyzzy  Puzzles  13  20150102 19:41 