辛普森的 1/3 法则和 3/8 法则的区别
在辛普森的 1/3 规则中,我们基于二次逼近来逼近多项式。在这种情况下,每个近似值实际上涵盖了两个子区间。这就是为什么我们要求子区间的数量是偶数的。有些近似看起来更像一条线而不是二次曲线,但它们确实是二次曲线。
辛普森的¹/₃ 规则公式
ₐ∫ᵇ f (x) dx = h/₃ [(y₀ + yₙ) + 4 (y₁ + y₃ + ..) + 2(y₂ + y₄ + ..)]
where,
a, b is the interval of integration
h = (b – a)/ n
y₀ means the first terms and yₙ means last terms.
(y₁ + y₃ + ..) means the sum of odd terms.
(y₂ + y₄ + …) means sum of even terms.
示例:使用辛普森的 1/3 规则找到解决方案。x f(x) 0.0 1.0000 0.1 0.9975 0.2 0.9900 0.3 0.9776 0.4 0.8604
解决方案:
Using Simpson’s 1/3 rule
ₐ∫ᵇ f (x) dx = h/₃ [(y₀ + yₙ) + 4 (y₁ + y₃ + …) + 2 (y₂ + y₄ + …)]
h = 0.1
ₐ∫ᵇ f (x) dx = 0.1/3 [(1+0.8604)+4×(0.9975+0.9776)+2×(0.99)]
ₐ∫ᵇ f (x) dx = 0.1/3 [(1+0.8604)+4×(1.9751)+2×(0.99)]
ₐ∫ᵇ f (x) dx = 0.39136
Solution of Simpson’s 1/3 rule = 0.39136
在辛普森的 3/8 规则中,我们基于二次逼近来逼近多项式。然而,每个近似实际上覆盖了三个子区间而不是两个。
辛普森 3/8 法则的公式
ₐ∫ᵇ f (x) dx = 3h/8[(y₀ + yₙ) + 3(y₁ + y₂ + y₄ + …) + 2(y₃ + y₆ +…)]
where,
a, b is the interval of integration
h = (b – a )/ n
y₀ means the first terms and yₙ means the last terms.
( y₁ + y₂ + y₄ + … ) means the sum of remaining terms.
( y₃ + y₆ +…) means the multiples of 3.
示例:使用辛普森的 1/3 规则找到解决方案。x f(x) 0.0 1.0000 0.1 0.9975 0.2 0.9900 0.3 0.9776 0.4 0.8604
解决方案:
Using Simpson’s 3/8 rule:
ₐ∫ᵇ f (x) dx = 3h/8[(y₀ + yₙ) + 3(y₁ + y₂ + y₄ + …) + 2(y₃ + y₆ +…)]
h = 0.1
ₐ∫ᵇ f (x) dx = 3h/8 [(y0+y4)+2(y3)+3(y1+y2)]
ₐ∫ᵇ f (x) dx = 3* 0.1/8 [(1+0.8604)+2×(0.9776)+3×(0.9975+0.99)]
ₐ∫ᵇ f (x) dx = 3* 0.1/8 [(1+0.8604)+2×(0.9776)+3×(1.9875)]
ₐ∫ᵇ f (x) dx = 0.36668
Solution of Simpson’s 3/8 rule = 0.36668
以下是辛普森 1/3 规则和辛普森 3/8 规则之间的差异表 Simpson’s 1/3 rule Simpson’s 3/8 rule Estimation in truncation error in Simpson ‘s 1/3 rule is E< -nh5/180 y iv(x) where h = (b-a)/n Estimation in truncation error in Simpson ‘s 3/8 rule is E< -nh5/90 y iv(x) where h = (b-a)/nSr. No 1. It approximates function y = f(x) by a parabola i.e. by 2nd order polynomial. It approximates the function y = f(x) by a parabola i.e. by 3rd order polynomial. 2. In this, the chances of error are more than Simpson’s 3/8 rule. In this, the chances of error are less. 3. The integral function can be calculated as = h/3 [(sum of 1st and last ordinates) + 4 (sum of odd ordinates) + 2 (sum of even ordinates)]. The integral function can be calculated as = 3h/8 [(sum of 1st and last ordinates) + 2 (sum of multiple of 3 ordinates) + 3 (sum of remaining ordinates)]. 4. 5. This rule is applied where N is an even number. This rule is applied where N is a multiple of 3.