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Numerical integration vs. integrals that are zero
03-06-2019, 05:23 PM
Post: #10
RE: Numerical integration vs. integrals that are zero
(03-06-2019 02:19 PM)Thomas Klemm Wrote:  
(03-06-2019 12:55 PM)Albert Chan Wrote:  What is the formula for fourth, fifth ... iterations ?

Cf. About the Algorithm

4th iteration
  4.04296875
 33.22265625
 83.49609375
146.42578125
213.57421875
276.50390625
326.77734375
355.95703125

5th iteration
  1.03271484375
  8.89892578125
 23.62060546875
 44.14306640625
 69.41162109375
 98.37158203125
129.96826171875
163.14697265625
196.85302734375
230.03173828125
261.62841796875
290.58837890625
315.85693359375
336.37939453125
351.10107421875
358.96728515625

So if I understand correctly, \(v\) is chosen uniformly from the interval \([-1,1]\) (using \(\{0\}\) for the first iteration, \(\{-\frac{1}{2}, \frac{1}{2}\}\) for the second iteration, \(\{-\frac{3}{4}, -\frac{1}{4}, \frac{1}{4}, \frac{3}{4}\}\) for the third iteration, etc.), converted to \(u\) via \(u=\frac{3}{2}v-\frac{1}{2}v^3\) and then converted to \(x\) via \(x=\frac{b-a}{2}u+\frac{b+a}{2}\) (where \(a\) is the lower limit and \(b\) is the upper limit)?

— Ian Abbott
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RE: Numerical integration vs. integrals that are zero - ijabbott - 03-06-2019 05:23 PM



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