order of growth of the error function erf

Question

The error function is defined as

$$ erf (x )= \frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2} dt . $$

But expanding the integrand $e^{-t^2}$ into Taylor series, it is easy to get the Taylor series of erf. We then see that it defines an entire function.

The question is, what is it order of growth?

Answer 1

For this kind of problem with $\text{erf}(x)$, an approximation such as $$\text{erf}(x)\sim \text{sgn}(x)\sqrt{1-e^{-\frac{4 }{\pi } x^2}} $$ could be sufficient.

Answer 2

From the links I gave you, you can conclude that $$ \operatorname{erf}(z) = 1 - \frac{e^{ - z^2 } }{z\sqrt \pi }\left( 1 + \mathcal{O}\left( \frac{1}{z} \right) \right), $$ as $z\to \infty$ in $\mathbb{C}$ with $\Re(z)\geq 0$ and $$ \operatorname{erf}(z) = -1 - \frac{e^{ - z^2 } }{z\sqrt \pi }\left( 1 + \mathcal{O}\left( \frac{1}{z} \right) \right), $$ as $z\to \infty$ in $\mathbb{C}$ with $\Re(z)\leq 0$. Note that there is no ambiguity on the imaginary axis since the second terms are exponentially large compared to the constants $\pm 1$.

Addendum:

You may write the above in one formula as $$ \operatorname{erf}(z) = \operatorname{sgn}(\Re(z)) - \frac{e^{ - z^2 } }{z\sqrt \pi }\left( 1 + \mathcal{O}\left( \frac{1}{z} \right) \right), $$ as $z\to \infty$ in $\mathbb{C}$.