approximate the integral below so that the magnitude of the error is at most 1/216

the integral from 0 to 1 of e^-x^2

I am pretty sure this is a series problem but I cannot figure out how to approach it?

thanks in advance

Printable View

- Apr 7th 2009, 10:04 PMcalculushelptxmagnitude of error
approximate the integral below so that the magnitude of the error is at most 1/216

the integral from 0 to 1 of e^-x^2

I am pretty sure this is a series problem but I cannot figure out how to approach it?

thanks in advance - Apr 8th 2009, 01:12 AMchisigma
The Taylor expansion of the 'gaussian function' is...

$\displaystyle e^{-x^{2}}= \sum_{n=0}^{\infty} (-1)^{n}\cdot \frac{x^{2n}}{n!}$ (1)

... and integrating it 'term by term'...

$\displaystyle \int e^{-x{2}}\cdot dx = \sum_{n=0}^{\infty} (-1)^{n}\cdot \frac {x^{2n+1}}{(2n+1)\cdot n!} + c$ (2)

The (2) is an 'alternating terms' series, so that the error can't be greater than the first term you neglect. The condition for error then becomes...

$\displaystyle \frac {1}{(2n+1)\cdot n!} \le \frac {1}{216} \rightarrow (2n+1)\cdot n! \ge 216$ (3)

... and that happens just for $\displaystyle n=4$ so that the first 4 tems of (2) will give the requested accuracy...

$\displaystyle \int_{0}^{1} e^{-x{2}}\cdot dx \approx 1 - \frac{1}{3} + \frac {1}{10} - \frac {1}{42} = .74285714\dots$

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Apr 8th 2009, 05:24 AMcalculushelptx
Now I see it. Thank you so much now I see where I went wrong.