Simple approximations for $\operatorname{erf}(x)$

Question

I am looking for some potentially unexpected approximations of $\operatorname{erf}(x)$, by which I mean not a really long series expansion, and something that resembles the correct shape for an arbitrary amount of the real domain. Here are some I have come up with myself: $$ \begin{matrix} \tanh(1.182\cdot x) & \text{the constant is a trade-off}\\ \tanh(1.152\cdot x)+\dfrac{e^{-|x-1.5|}\cos(x-1.5)}{20} & \text{smaller error but not symmetrical}\\ \tanh\left(\frac{x|x|}{4}+x\right) & \text{constant not optimised for this yet} \end{matrix} $$ I also read this article which listed a method for assuming solutions of the form: $$\tanh(f(x))$$ Just like I initially thought. My question is really, does anyone have any surprisingly accurate short approximations for it? Thanks

Answer 1

You can easily build your own approximations using as a basis $$\text{erf}(x) \sim \sqrt{1-\exp\Big[ -\frac {4x^2} {\pi} P_{n}(x) \Big]}$$ when $P_n(x)$ is the ratio of two polynomials of degree $n$ in $x^2$ or just a polynomial in $x^2$.

These are quite easy to build writing $$P=-\frac{\pi \log \left(1-\text{erf}(x)^2\right)}{4 x^2}$$ and developing as as series around $x=0$ $$P=1-\frac{2 (\pi -3) x^2}{3 \pi }+\frac{2 \left(120-60 \pi +7 \pi ^2\right) x^4}{45 \pi ^2}-\frac{4 \left(-420+280 \pi -56 \pi ^2+3 \pi ^3\right) x^6}{105 \pi ^3}+\frac{2 \left(120960-100800 \pi +28560 \pi ^2-3040 \pi ^3+83 \pi ^4\right) x^8}{4725 \pi ^4}+O\left(x^{10}\right)$$ which can easily be transformed into a Padé approximant.

For example $$P_0(x)=1 \qquad \qquad P_1(x)=\frac {1+\frac{\left(10-\pi ^2\right) }{5 (\pi -3) \pi }x^2 } {1+\frac{\left(120-60 \pi +7 \pi ^2\right) }{15 (\pi -3) \pi }x^2 }$$

To give you an idea, computing the norm $$\Phi_n=\int_0^\infty \left(\text{erf}(x)-\sqrt{1-\exp\Big[ -\frac {4x^2} {\pi} P_{n}(x) \Big]}\right) ^2\,dx $$ we have $$\Phi_0=3.23\times 10^{-5}\qquad \Phi_1=3.04\times 10^{-8}\qquad\Phi_2=1.20\times 10^{-10}\qquad\Phi_3=3.97\times 10^{-12}$$

Edit

Using the ideas from the linked paper, we could do basically the same expanding $$\tanh ^{-1}(\text{erf}(x))=\frac{2 x}{\sqrt{\pi }}-\frac{2 (\pi -4) x^3}{3 \pi ^{3/2}}+\frac{\left(96-40 \pi +3 \pi ^2\right) x^5}{15 \pi ^{5/2}}+\frac{\left(5760-3360 \pi +532 \pi ^2-15 \pi ^3\right) x^7}{315 \pi ^{7/2}}+O\left(x^9\right)$$ and use, as another approximation, $$\text{erf}(x)\sim \tanh \left(\frac{2 x}{\sqrt{\pi }} Q_n(x)\right)$$ and, similar to $P_1(x)$, write $$Q_1(x)=\frac{1+\frac{128-40 \pi -\pi ^2}{30 (\pi -4) \pi } x^2} {1+\frac{96-40 \pi +3 \pi ^2}{10 (\pi -4) \pi } x^2}$$ which is much better than what is given in the paper.

$$\Psi_1=\int_0^\infty\left(\text{erf}(x)-\tanh \left(\frac{2 x}{\sqrt{\pi }} Q_n(x)\right)\right) ^2 \,dx =2.22\times 10^{-7} $$

Even if the two formalisms look similar in terms of accuracy, I would favor the first one because it makes the inversion of the error function easier ; with $P_1(x)$ you just need to solve a quadratic equation in $x^2$ while with $Q_1(x)$ you just need to solve a cubic equation in $x$.

Comparing with Taylor series, the difference between $\text{erf}(x)$ and its approximations is $1.56\times 10^{-4}x^7$ for $P_1(x)$ and $5.916\times 10^{-4}x^7$ for $Q_1(x)$.

Answer 2

I had made a similar question here, and I found experimentally this simple approximation to the $\text{erf}(x)$ function:

$$z(x)=\tanh\left(\frac{2}{\sqrt{\pi}}\left(x+\frac{11}{123}x^3\right)\right)$$

it keeps the difference $|\text{erf}(x)-z(x)|<0.00036$.

Also I compared in Wolfram-Alpha using $z(x)$ for taking probabilities of the standard Gaussian distribution, and the max mistake looks its below $0.018\%$, quite accurate in my opinion.

The formula is a modification by trial and error so the last fraction $11/123$ surely could be improved. It comes from the series expansion of $\tanh^{-1}(\text{erf}(x))$ shown in Wolfram-Alpha, and just the first 2 terms makes already simple approximation than works quite good: $$f(x)=\tanh\left(\frac{2}{\sqrt{\pi}}\left(x+\frac{(4-\pi)x^3}{3\pi}\right)\right)$$

Here you could see them in Desmos.

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Added later

In the following question I found another approximate much better than the previous, a little more complicated but still simpler than many of the approximations displayed on Wikipedia page for the Error function: Numerical approximations:

$$\tanh\left(\frac{2}{\sqrt{\pi}}\left(x+\frac{28}{305}x^3\sqrt{1-\frac{x^2}{33}}\right)\right)\to \text{max error}=0.0000551322$$

Again, be aware the constants were find by trial and error in Desmos, there nothing special about them a surely could be improved.