Simpson’s rule

Simpson’s rule is a numerical method that uses quadratic functions to approximate definite integrals.

With reference to the diagram above, $\int_{-h}^hf(x)dx$ is the area under the cubic curve demarcated by two equal intervals, each of magnitude h. The area is estimated using 3 points on the quadratic function $P(x)=Ax^2+Bx+C$ , where

$\int_{-h}^hf(x)dx\approx \int_{-h}^hP(x)dx=\frac{h}{3}(2Ah^2+6C)\;\;\;\;\;\;\;\;(23)$

Substitute $P(0)=y_1$ , $P(-h)=y_0$ and $P(h)=y_2$ in $P(x)$ , we have $C=y_1$ , $Ah^2-Bh+C=y_0$ and $Ah^2+Bh+C=y_2$ . Combining the last two equations, $2Ah^2+2C=y_0+y_2$ . Substitute this equation and $C=y_1$ in eq23,

$\int_{-h}^hf(x)dx\approx\frac{h}{3}(y_0+4y_1+y_2)$

Similarly, if we consider the domain from h to h‘ (see above diagram),

$\int_{h}^{h'}f(x)dx\approx\frac{h}{3}(y_2+4y_3+y_4)$

Therefore, using 5 points on $f(x)$ , we have

$\int_{-h}^{h'}f(x)dx\approx\frac{h}{3}(y_0+4y_1+2y_2+4y_3+y_4)$

Extending the logic for n points on $f(x)$ , we have,

$\int_{a}^{b}f(x)dx\approx\frac{h}{3}\{f(a)+4[f(x_1)+f(x_3)+\cdots+f(x_{n-1})]$
$+2[f(x_2)+f(x_4)+\cdots+f(x_{n-2})]+f(b)\}\;\;\;\;\;\;\;\;(24)$

where $a=x_0$ and $b=x_n$ .

This implies that there must be an even number of equal intervals for the rule to work, i.e. n must be odd.

Question

Evaluate
$DI=\int_0^\infty\phi_i(x_i)\left(\int_0^\infty f(x_i,x_j)dx_j\right)\phi_i(x_i)dx_i$

Answer

Substituting eq24 in the above equation,

$DI=\int_0^{\infty}\phi_i^2(x_i)\frac{h}{3}\{f(x_i,x_0)+4[f(x_i,x_1)+f(x_i,x_3)+\cdots+f(x_i,x_{\infty-1})]$ $+2[f(x_i,x_2)+f(x_i,x_4)+\cdots+f(x_i,x_{\infty-2})]+f(x_i,x_{\infty})\}dx_i$

So,

$DI=\frac{h}{3}[A+B+C+\cdots+D]\;\;\;\;\;\;\;\;(25)$

where (form 1)

$A=\int_0^{\infty}\phi_i^2(x_i)f(x_i,x_0)dx_i=\frac{h}{3}[\phi_0^2(x_0)f(x_0,x_0)+4\phi_1^2(x_1)f(x_1,x_0)$ $+2\phi_2^2(x_2)f(x_2,x_0)+\cdots+\phi_{\infty}^2(x_{\infty})f(x_{\infty},x_0)]$

$B=4\int_0^{\infty}\phi_i^2(x_i)f(x_i,x_1)dx_i=\frac{4h}{3}[\phi_0^2(x_0)f(x_0,x_1)+4\phi_1^2(x_1)f(x_1,x_1)$ $+2\phi_2^2(x_2)f(x_2,x_1)+\cdots+\phi_{\infty}^2(x_{\infty})f(x_{\infty},x_1)]$

$C=2\int_0^{\infty}\phi_i^2(x_i)f(x_i,x_2)dx_i=\frac{2h}{3}[\phi_0^2(x_0)f(x_0,x_2)+4\phi_1^2(x_1)f(x_1,x_2)$ $+2\phi_2^2(x_2)f(x_2,x_2)+\cdots+\phi_{\infty}^2(x_{\infty})f(x_{\infty},x_2)]$

$D=\int_0^{\infty}\phi_i^2(x_i)f(x_i,x_{\infty})dx_i=\frac{h}{3}[\phi_0^2(x_0)f(x_0,x_{\infty})+4\phi_1^2(x_1)f(x_1,x_{\infty})$ $+2\phi_2^2(x_2)f(x_2,x_{\infty})+\cdots+\phi_{\infty}^2(x_{\infty})f(x_{\infty},x_{\infty})]$

or equivalently (form 2),

$A=\int_0^{\infty}\phi_i^2(x_i)f(x_i,x_0)dx_i=\frac{h}{3}[\phi_0^2(x_0)f(x_0,x_0)+4\phi_0^2(x_0)f(x_0,x_1)$ $+2\phi_0^2(x_0)f(x_0,x_2)+\cdots+\phi_0^2(x_0)f(x_0,x_{\infty})]$

$B=4\int_0^{\infty}\phi_i^2(x_i)f(x_i,x_1)dx_i=\frac{4h}{3}[\phi_1^2(x_1)f(x_1,x_0)+4\phi_1^2(x_1)f(x_1,x_1)$ $+2\phi_1^2(x_1)f(x_1,x_2)+\cdots+\phi_1^2(x_1)f(x_1,x_{\infty}]$

$C=2\int_0^{\infty}\phi_i^2(x_i)f(x_i,x_2)dx_i=\frac{2h}{3}[\phi_2^2(x_2)f(x_2,x_0)+4\phi_2^2(x_2)f(x_2,x_1)$ $+2\phi_2^2(x_2)f(x_2,x_2)+\cdots+\phi_2^2(x_2)f(x_2,x_{\infty})]$

$D=\int_0^{\infty}\phi_i^2(x_i)f(x_i,x_{\infty})dx_i=\frac{h}{3}[\phi_{\infty}^2(x_{\infty})f(x_{\infty},x_0)+4\phi_{\infty}^2(x_{\infty})f(x_{\infty},x_1)$ $+2\phi_{\infty}^2(x_{\infty})f(x_{\infty},x_2)+\cdots+\phi_{\infty}^2(x_{\infty})f(x_{\infty},x_{\infty})]$

These integrals are useful for computing the numerical solution of the Hartree self-consistent field method for helium.

Question

Answer

Next article: Slater-type orbitals

Previous article: Proof of the Hartree equations

Content page of computational chemistry

Content page of advanced chemistry

Main content page

Leave a Reply Cancel reply