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The error function is defined as, $$\text{erf}\bigg(\frac{(ax-b)}{\sqrt{2}}\bigg) = \frac{2}{\sqrt{\pi}}\int_{0}^{\frac{(ax-b)}{\sqrt{2}}} e^{-t^{2}/2}dt$$. My question is how to expand the above function. Am I able to write it as, $$\text{erf}\bigg(\frac{(ax-b)}{\sqrt{2}}\bigg) = \text{erf}\bigg(\frac{ax}{\sqrt{2}}\bigg) + \text{some_value}$$. I did a transformation $u = \frac{t}{\sqrt{2}}$, but it didn't help.

Thank you in advance.

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To answer your question directly...

\begin{align*} \text{erf}\left(c_1 x + c_2\right) &= \frac{2}{\sqrt{\pi}}\int_0^{c_1x+c_2}e^{-t^2/2}dt \\[1.2ex] &= \frac{2}{\sqrt{\pi}}\int_0^{c_1x}e^{-t^2/2}dt + \frac{2}{\sqrt{\pi}}\int_{c_1x}^{c_1x+c_2}e^{-t^2/2}dt \\[1.2ex] &= \text{erf}(c_1 x) + \frac{2}{\sqrt{\pi}}\int_{c_1x}^{c_1x+c_2}e^{-t^2/2}dt. \end{align*}

This implies that "some_value" is equal to $\frac{2}{\sqrt{\pi}}\int_{c_1x}^{c_1x+c_2}e^{-t^2/2}dt$, which unfortunately can't be simplified in any particularly useful way (at least that I can see).

In fact, if I was given this second term by itself, I would immediately re-write it as $\text{erf}(c_1x+c_2) - \text{erf}(c_1 x)$ which gets us right back to the original expression. Can you give more details on what you are hoping to accomplish?

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  • $\begingroup$ Actually, I am simplifying a CDF $\Phi(ax-b)$ into $\Phi(ax) + Some_value$. Then I realized that I can write the CDF $\Phi(x) = 1/2[1 + errf(x/\sqrt{2}]$. That is the reason, I need to expand $erf$ in order to expand $\Phi(ax-b)$ $\endgroup$ – score324 Feb 16 at 2:02
  • $\begingroup$ @score324 Why do you need to expand $\Phi(ax+b)$? What is the larger goal here? As the answer shows, you (unfortunately) can't expand the error function in a tractable way. $\endgroup$ – knrumsey Feb 16 at 4:49

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