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)
Best Answer
$\DeclareMathOperator\erf{erf}$Approximations of $\erf$ according to the Handbook of Mathematical Functions:
http://people.math.sfu.ca/~cbm/aands/page_299.htm