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I have a little question regarding the error function that disturbs me a lot:

Let us consider the classical error function defined as:

erf($z$) $= \frac{2}{\sqrt{\pi}}\int_{0}^{z} e^{-t^2} dt$

It is well-known [1] that for $z= \infty,$ erf($z$) = 1. My question may be idiotic but is the following one: is the error function only equals to 1 at $z= \infty$ ? In other words, is the error function never reaches the value of 1 before the strict value $z= \infty$ ? Don't tell me to refer to tables of values of erf($z$) because the latter are based on numerical integrations and thus subjected to numerical uncertainties.

Thank you very much for your help,

Regards

[1] http://mathworld.wolfram.com/Erf.html

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    $\begingroup$ Yes for the question in title, as the error function is strictly increasing since the integrand is strictly positive. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 May 10 '18 at 0:26
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    $\begingroup$ The value of the integral starts at zero and grows toward the asymptote one. It reaches one only when z reaches infinity (so to speak, b/c/ infinity is not a number). $\endgroup$ – Christopher Marley May 10 '18 at 0:26
  • $\begingroup$ It does take on the value $1$ in the complex plane, however: the roots with smallest absolute value are at roughly $-1.35 \pm 1.99i$. $\endgroup$ – Chappers May 10 '18 at 0:46
  • $\begingroup$ @GNUSupporter I wanted to write an answer, but it is basically your comment. Maybe you should answer? $\endgroup$ – SK19 May 10 '18 at 0:51
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    $\begingroup$ @SK19 By writing $z$, I think OP wants the domain to be complex. As there's already a answer for the complex case, I think there's no need to add one for the real case. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 May 10 '18 at 3:49
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If you see that: $$\frac{d}{dz} \text{erf}(z) = \frac{2e^{-z^2}}{\sqrt{\pi}}>0$$ Therefore the function is strictly increasing.

  • As $\text{erf}(0)=0$ then $\text{erf}(x)>0$.
  • $\frac{2}{\sqrt{\pi}}\int_{0}^{\infty} e^{-t^2} dt=1$

Take $x>0$.

$$\frac{2}{\sqrt{\pi}}\int_{0}^{\infty} e^{-t^2} dt = \frac{2}{\sqrt{\pi}}\int_{0}^{x} e^{-t^2} dt + \frac{2}{\sqrt{\pi}}\int_{x}^{\infty} e^{-t^2} dt $$ $$1 = \text{erf}(x) + \frac{2}{\sqrt{\pi}}\int_{x}^{\infty} e^{-t^2} dt$$ $$1- \text{erf}(x) = \frac{2}{\sqrt{\pi}}\int_{x}^{\infty} e^{-t^2} dt >0$$

$$1>\text{erf}(x)$$

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Probably not very useful.

In the linked page, they give the asymptotic series $$\text{erf}(x)=1-\frac{e^{-x^2}}{\sqrt{\pi }}\sum_{n=0}^\infty (-1)^n \frac{(2n-1)!!}{2^n} x^{-(2n+1)}$$ and a very accurate asymptotics would be $$\text{erf}(x)=1-e^{-x^2} \left(\frac{1}{\sqrt{\pi } x}+O\left(\frac{1}{x^2}\right)\right)$$

For $x=10$, the "exact" value is $$0.99999999999999999999999999999999999999999999791151$$ while the above truncated expression gives $$0.99999999999999999999999999999999999999999999790117$$

Edit

After GNU Supporter's comment, consider $$a_n=(-1)^n \frac{(2n-1)!!}{2^n} x^{-(2n+1)}$$ This gives $$\left|\frac{a_{n+1}}{a_n}\right|=\frac{2 n+1}{2 x^2}\approx \frac n {x^2}$$ which decreases very fast.

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  • $\begingroup$ I find it useful as it gives a second-order estimation. To answer OP's question, it remains to set $x$ large enough so that $$\frac{1}{\sqrt{\pi } x}+O\left(\frac{1}{x^2}\right) > 0.$$ $\endgroup$ – GNUSupporter 8964民主女神 地下教會 May 10 '18 at 3:41

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