Chapter 1.2, Problem 15E

a.

To determine

To show: The error function is equal to $erf\left(x\right)=\frac{2}{\sqrt{\pi }}{\displaystyle \sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k{x}^{2k+1}}{\left(2k+1\right)k!}}$ when the Maclaurin series for $e^{-x^2}$ is integrated.

b.

To determine

To show: The error function can also be expressed in the form $erf\left(x\right)=\frac{2}{\sqrt{\pi }}{e}^{-x^2}{\displaystyle \sum _{k=0}^{\infty }\frac{2^k{x}^{2k+1}}{1\cdot 3\cdot 5\dots \left(2k+1\right)}}$ .

c.

To determine

To find: The approximate value of $erf\left(1\right)$ to within ${10}^{-7}$ using the series $erf\left(x\right)=\frac{2}{\sqrt{\pi }}{\displaystyle \sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k{x}^{2k+1}}{\left(2k+1\right)k!}}$ .

d.

To determine

To find: The approximate value of $erf\left(1\right)$ for $k=11$ using the series $erf\left(x\right)=\frac{2}{\sqrt{\pi }}{e}^{-x^2}{\displaystyle \sum _{k=0}^{\infty }\frac{2^k{x}^{2k+1}}{1\cdot 3\cdot 5\dots \left(2k+1\right)}}$ .

e.

To determine

To explain: The reason why difficulties occur using the series $erf\left(x\right)=\frac{2}{\sqrt{\pi }}{e}^{-x^2}{\displaystyle \sum _{k=0}^{\infty }\frac{2^k{x}^{2k+1}}{1\cdot 3\cdot 5\dots \left(2k+1\right)}}$ to approximate $erf\left(x\right)$ .