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16-Point Gaussian Quadrature - Gamo - 04-09-2021 05:17 AM Recently I try this program called 16-Point Gaussian Quadrature found in the HP-11C Solutions Handbook on page 30 So this program do the Integration problem. I have try diffrent problem and the answer come out very well except one. This one doesn't give enough accuracy when compare to the Integration Function on the HP-15C Here is the problem: ∫ 0 to 2 √4 - X² dx = Pi = 3.141592654 On the HP-11C and HP-15 Equation is [X²] [4] [X<>] [-] [√] Answer on HP-11C is 3.141722674 Answer on HP-15C is 3.141592653 So is program provided on the HP-11C Solutions Book only give three decimal place accuracy or it only happen to certain problem. Gamo RE: 16-Point Gaussian Quadrature - robve - 04-09-2021 11:53 AM (04-09-2021 05:17 AM)Gamo Wrote: Here is the problem: ∫ 0 to 2 √4 - X² dx = Pi = 3.141592654 This looks right and not bad for 16 points. 16 points will only give you a (rough) approximation of \( \int_0^2 \sqrt{4-x^2}\,dx \). Gausss 10 point: 3.14209916979661 Gauss Kronrod 21 point: 3.14161975053083 By comparison (edited to add more examples): Romberg 524289 points (1e-9 error threshold) gives 3.14159265256818 Romberg 16385 points (1e-7 error threshold) gives 3.14159265256818 Tanh-Sinh 56 points (1e-9 error threshold) gives 3.14159265358672 Tanh-Sinh 30 points (1e-7 error threshold) gives 3.14159265358975 Adaptive Simpson 417 points (1e-9 error threshold) gives 3.14159111522653 Adaptive Simpson 153 points (1e-7 error threshold) gives 3.14159111489381 - Rob RE: 16-Point Gaussian Quadrature - Albert Chan - 04-09-2021 04:41 PM (04-09-2021 05:17 AM)Gamo Wrote: Here is the problem: ∫ 0 to 2 √4 - X² dx = Pi = 3.141592654 Both Gaussian and Romberg quadrature are based on fitting the curve to polynomial. When X ≈ 2, the curve fall off a cliff, which is hard to curve fit by polynomial. HP-15C does a u-substitution, which in this case, removed the cliff entirely. int(√(4-x^2), x=0..2.) = int(4*√(1-x^2), x=0..1.) = int(24*u*(1-u) * √(1 - (u*u*(3-2*u))^2), u = 0..1.) Try this u-transformed function, 16-points should converge to pi pretty good. --- Another way is to avoid the cliff. y = √(4-x²), x≥0 → x = √(4-y²) Note the x/y symmetry, instead of integrating from 0 to 2, we can do 0 to √2 With limit = [0,√2], doubled the result over-counted by a square, area = (√2)² = 2 int(√(4-x^2), x=0..2.) = int(√(4-x^2), x=0..√(2.)) * 2 - 2 RE: 16-Point Gaussian Quadrature - Gamo - 04-10-2021 02:54 AM The HP-15C Integration Function is excellent. I will go back to read more information about this great integration function in the Advance Functions Handbook. Gamo RE: 16-Point Gaussian Quadrature - Wes Loewer - 04-10-2021 03:50 PM (04-09-2021 05:17 AM)Gamo Wrote: Here is the problem: ∫ 0 to 2 √4 - X² dx = Pi = 3.141592654 I put that exact integral on my last calculus test. :-) (04-09-2021 04:41 PM)Albert Chan Wrote: Try this u-transformed function, 16-points should converge to pi pretty good. The TI-Nspire calculator uses this u-transformation with a 15 point Gauss-Kronrod method. |