Measure: dm= x3 dx
Appendix 2: Gaussian Quadrature Rule with a non-standard measure
The following examples were solved using the Basic Gaussian quadrature Rule with measures dm = x3dx, and dm = ©-xdx.
Measure = x3 dx
Interval: [0,4]
f(x)= x/(5 - 4 x - x2)(3/2)
∫[0,4]f(x)dm = 383.816
[pic]
|M |G |
|3 |383.719 |
|5 |383.811 |
|7 |383.816 |
By using M = 7, we were able to get a 6 place approximation.
Measure = x3 dx
Interval: [7,9]
f(x)= x/(5 - 4 x - x2)(3/2)
∫[7,9]f(x)dm = 14817.6
[pic]
|M |G |
|3 |14817.6 |
Using the same function as in the last example, but a different interval, we got a 6 place approximation by using M = 3. A small value of M worked well because f(x) is very regular in the interval [7,9].
Measure = x3 dx
Interval: [0.5,5]
f(x)= √[1+3x2+3x4+x6]
∫[0,4]f(x)dm = 12113.6
[pic]
|M |G |
|3 |12113.5 |
|5 |12113.5 |
|7 |12113.6 |
By using M = 7, the integral approximation agreed to six places with Mathematica.
Measure = x3 dx
Interval: [6,9]
f(x)= √[1+3x2+3 x4+x6]
∫[6,9]f(x) dm = 658736
[pic]
|M |G |
|3 |658736 |
Using a different interval, where the graph is more regular, we only had to use M= 3 to be able to get a six place approximation.
Measure = ©-x dx
Interval: [1,3]
f(x)= x/Sqrt[5+2 x+x^2]
∫[1,3]f(x) dm = 0.156019
[pic]
|M |G |
|3 |0.156055 |
|5 |0.156018 |
|7 |0.156019 |
After increasing M to 7, we got a 6 place agreement with Mathematica.
Measure = ©-x dx
Interval: [0,7]
f(x)= 6xx3Cos[x]
∫[0,7]f(x) dm = 55193.4
[pic]
|M |G |
|3 |0.156055 |
|5 |0.156018 |
|7 |6785.8 |
|13 |55258.9 |
|15 |55193.5 |
|17 |55193.5 |
|19 |55047.6 |
|21 |53498.1 -0.0000390941 ™ |
By using M = 19, we were only able to get a 2 place approximation.
Measure = ©-x dx
Interval: [0,1]
f(x)= x/(1 + Sin[4x])
∫[0,1]f(x) dm = 0.269956 -2.48412×10-15 ™
[pic]
|M |G |
|3 |0.214661 |
|5 |0.252846 |
|9 |0.268829 |
|13 |0.269896 |
|17 |0.269927 |
|17 |0.271829 +3.89613×10-13 ™ |
|21 |0.263063 -2.09768×10-12 ™ |
Surprisingly, Mathematica was not able to integrate f(x) within the interval [0,1]. By using Gaussian Quadrature Rule with a non-standard measure, we were able to get 0.269927 as the interregnal approximation for f(x).
Measure = ©-x dx
Interval: [0.2,1]
f(x)= 1/Sqrt[-25+4 x2]
∫[0.2,1]f(x) dm = 0.0311716
[pic]
|M |G |
|7 |0.022763 |
|9 |0.0251327 |
|15 |0.0269181 |
|17 |0.0245198 +3.53511×10-11 ™ |
After increasing M to 17, we were not able to approximate the integral for f(x).
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