approximating alternating series of the gaussian integral

Question

$$f(t) = \int_{0}^{t}e^{-x^2}dx $$

so I'm suppose to approximate f(3/5) to within 1/129.

I know the answer is about 15/28. However, I don't know the work to prove it. All I have is that the Gaussian integral is equal to $$\sqrt{\pi}$$

But I don't know how to do it. I've been reading math chapters after chapters. This is way too way for math education. Please help and show me how to do it.

Answer 1

You have the tag power series, so use it! $$ e^{-x^2}=1-x^2+\frac{x^4}{2}+.... $$ The above is easy enough to integrate, and to insure error, look up the various forms of Taylor's formula for the remainder, and pick the most convenient to get the desired accuracy, remembering to remember to multiply by the length of the interval of integration, $3/5$.

Also, that the integral from 0 to infinity is $\frac{\sqrt{\pi}}{\sqrt{2}}$ is of no help here.

Answer 2

Note that we can represent the exponential $e^{-x^2}$ as the power series

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

which converges uniformly on any closed and bounded interval since the radius of convergence is $\infty$.

Hence we can integrate term by term to obtain

$$\bbox[5px,border:2px solid #C0A000]{\int_0^t e^{-x^2}\,dx=\sum_{n=0}^\infty \frac{(-1)^nt^{2n+1}}{(2n+1)n!}}$$

Finally, recall that for a convergent alternating series, the truncation error $E_N(t)$ is bounded as

$$E_N(t)=\left|\sum_{N+1}^\infty \frac{(-1)^nx^{2n}}{n!} \right|\le \frac{t^{2N+3}}{(2N+3)(N+1)!} \tag 1$$

Finally, find the smallest integer $N$ such that the right-hand side of $(1)$ is less than the tolerance $1/129$.

NOTE:

The integral of interest can be represented by its Burmann Series of one-half odd-integer powers of $\sqrt{1-e^{-t^2}}$. That series converges must more rapidly than the corresponding power series.