I'll derive Simpson's 1/3 rule using this method to show what I mean:
Simpson’s 1/3 rule involves fitting a quadratic through three points, so put:
$$I_2=\int_{x_0}^{x_2} f(x) \space dx = c_0y_0 + c_1y_1 + c_2y_2$$
With $\space x_0=-h,\space \space x_1=0, \space \space x_2=h \space$, $\space f(x_n)=y_n \space$ and $\space c_0,c_1,c_2\in\mathbb{R} \space$ being values we will have to solve for.
Let $\space I \space$ be exact for $f(x)=1,x,x^2 \space$ and hence for any quadratic:
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(1) $f(x)=1$:
$$\int_{x_0}^{x_2} \space dx = (x_2-x_0)=2h=c_0+c_1+c_2$$
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(2) $f(x)=x$:
$$\int_{x_0}^{x_2} x \space dx = {1\over 2}((x_2)^2-(x_0)^2)={1\over 2}(h^2-h^2)=c_0x_0+c_1x_1+c_2x_2=0$$
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(3) $f(x)=x$:
$$\int_{x_0}^{x_2} x^2 \space dx = {1\over 3}((x_2)^3-(x_0)^3)={1\over 3}(h^3-(-h)^3)=c_0x_0+c_1x_1+c_2x_2={2\over 3}h^3$$
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Then equations (1), (2) and (3) reduce to:
(1): $ \space c_0+c_1+c_2 = 2h \space$
(2): $\space -hc_0+hc_2=0 \space $ OR $\space c_2=c_0 \space $
(3): $ \space c_0h^2+c_2h^2 = {2\over 3}h^3 \quad$ OR $ \quad 2c_0h^2= {2\over 3}h^3 \quad $ OR $ \quad c_0= {1\over 3}h \space$
So we now have from (1):
${2\over 3}h + c_1 = 2h \quad$ OR $\quad c_1={4 \over 3}h$
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This tells us that:
$$I_2=\int_{x_0}^{x_2} f(x) \space dx = {h\over 3}y_0 + {4\over 3}y_1 + {h\over 3}y_2$$
$$I_2=\int_{x_0}^{x_2} f(x) \space dx = {h\over 3}(y_0 + 4y_1 + y_2)$$
Thus, generally speaking, for $\space n\in\mathbb{N} \space$ is EVEN:
$$I_n=\int_{x_0}^{x_2} f(x) \space dx + \int_{x_2}^{x_4} f(x) \space dx + … + \int_{x_{n-2}}^{x_n} f(x) \space dx$$
OR
$$I_n = {h\over 3}(y_0 + 4y_1 + y_2) + {h\over 3}(y_2 + 4y_3 + y_4) +…+ {h\over 3}(y_{n-2} + 4y_{n-1} + y_n)$$
OR better yet
$$I_n = {h\over 3}(y_0 + 4y_1 + 2y_2 + 4y_3 + 2y_4 +…+ 2y_{n-2} + 4y_{n-1} + y_n)$$
Now, Simpson's 3/8 rule states the following equality:
$$S_3 = \int_{x_0}^{x_3} f(x) \space dx = {3\over 8}h (y_0+3y_1+3y_2+y_3)$$
OR more generally:
$$S_n = \int_{x_0}^{x_n} f(x) \space dx = {3\over 8}h (y_0+3y_1+3y_2+3y_3+3y_4+…+3y_{n-1}+y_n)$$
Where this time, $\space n\in\mathbb{N} \space$ is a multiple of 3.
I can't seem to derive this integration rule using the method I used for the 1/3 rule.
Can someone please help shed some light?
Thank you.
Best Answer
Yes. Simpson's $3/8$-rule (the simpler form $$S_3 = \frac{3}{8} h (y_0 + 3y_1 + 3y_2 + y_3)$$ is the closed Newton-Cotes formula of order $3$, which means you calculate the interpolating cubic through the points $(x_k,y_k)$ and integrate it exactly. (Equivalently, you can use the method of undetermined coefficients.) So $$S_3 = \sum_{k=0}^3 h y_k \int_0^3 \prod_{j \ne k} \frac{t-j}{k-j} \, \mathrm{d}t.$$ For example, the coefficient of $y_0$ is $$h \int_0^3 \frac{(t-1)(t-2)(t-3)}{-6} \, \mathrm{d}t = \frac{3}{8}h.$$