# Approximating the erf function

I was trying to find an approximate solution to the following:

$\DeclareMathOperator\erf{erf}$

$$\frac12 \sqrt{\pi} \erf\left (\frac{x-2}{\sqrt{10}}\right) + \frac12 \sqrt{\pi} \erf \left(\frac{x+2}{\sqrt{10}}\right) = \frac25 \sqrt{\pi}$$

This naturally equals

$$\int_0^{\frac{x-2}{\sqrt{10}}} e^{-t^2} dt+ \int_0^{\frac{x+2}{\sqrt{10}}} e^{-t^2} dt = \frac25 \sqrt\pi$$

What I tried then was to use the taylor approximation and then solve the polynomial equations.. but the results I obtained were not consistently getting close to the correct solution ($x \sim 1.71$), obtained with wolfram alpha

In fact, using $\displaystyle \int_0^x e^{-t^2}dt = x - \frac{x^3}{3} + \frac{x^5}{10} - \frac{x^7}{42} + \dots$

I obtained the following approximations ($x_i$ indicates the result obtained considering the terms until $x^i$) $$x_1 \sim 1.12\ \ \ \ x_3 = -4.975 \ \ \ x_5 = 1.59 \ \ \ x_7 = -4.718 | 3.729 | 1.755 \ \ \ \ x_9 = 1.70$$

Questions

1) If I take into account the whole series, what assures me that only one solution will be found (with $x_7$, for example, I find $3$ real solutions)

2) I understand that as long as I am near $0$, the solution will be good approximation. But a priori I don't know what values $x$ is going to take, so the error can be large as large as $x$ and the taylor approximation is useless.

3) Why is (eventually) the error going to $0$ if one takes the whole series?

I don't think I have a clear understanding of the passage between taylor polynomial (valid only near a point $x_0$) and series (why exactly they are convergent everywhere (well, at least in the case of $e^x$) if we consider an an infinite amount of terms)

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Yes, but the error in taylor series is $o((x-x_0)^n)$. If $x >> x_0$, the error can become huge.. So why does it go to $0$ even when $x >> x_0$? – Ant May 5 '14 at 19:53