Chapter 5: Problem 2
The complementary error function is defined as $$ \operatorname{erfc}(x)=\frac{2}{\sqrt{\pi}} \int_{x}^{\infty} \exp \left(-u^{2}\right) \mathrm{d} u $$ Show that the second term in the expansion of \(1 / r\) is given by $$ \frac{2}{\sqrt{\pi}} \int_{g}^{\infty} \exp \left(-r^{2} \rho^{2}\right) \mathrm{d} \rho=\frac{1}{r} \operatorname{erfc}(g r) $$ Use a spreadsheet or other programming tool to show that this function falls to zero much faster than \(1 / r\). A good numerical approximation is $$ \operatorname{erfc}(x)=\exp \left(-x^{2}\right) \sum_{n=1}^{5} a_{n} t^{n} $$ where \(t=(1+b x)^{-1}\), and \(a_{1}=0.25483, a_{2}=\) \(-0.28450, a_{3}=1.42141, a_{4}=1.45315, a_{5}=1.06141\) and \(b=0.32759\).
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