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- Thread starter Wilmer
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- Apr 2, 2012

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- Feb 5, 2012

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Hi Wilmer,What does erf^(-1)(x) mean?

erf^(-1)(.6) = ?

is erf^(-1) same as 1/erf?

I found out erf = error function....hmmm....

THANKS for any explanations.

As you have correctly stated, \(\mbox{erf }\) denotes the error function, which is defined by the integral,

\[\mbox{erf }(x) = \frac{2}{\sqrt{\pi}}\int_{0}^x e^{-t^2} dt\]

The inverse error function is denoted by \(\mbox{erf}^{-1}\). If we consider the power series representation of the inverse error function an approximate value for \(\mbox{erf}^{-1}(0.6)\) can be found out to any given precision.

\[\mbox{erf}^{-1}(0.6)\approx 0.5951160814499948500193\]

The inverse of a function \(f:X\rightarrow Y\) is defined as a function \(f^{-1}:Y\rightarrow X\) such that,

\[f(x) = y\,\,\text{if and only if}\,\,f^{-1}(y) = x\]

However it is

\[f^{-1}(x)=\frac{1}{f(x)}\]

For an example in the case of the error function,

\[\mbox{erf}(0.6)\approx 0.60385609084793\Rightarrow \frac{1}{\mbox{erf}(0.6)}=1.656023703587745\neq \mbox{erf}^{-1}(0.6)\]

So it is clear that generally,

\[\mbox{erf}^{-1}(x)\neq\frac{1}{\mbox{erf}(x)}\]

The Taylor expansion of the error function is given by,Is there a value for erf, like there is for pi and e?

\[\mbox{erf}(z)= \frac{2}{\sqrt{\pi}}\sum_{n=0}^\infty\frac{(-1)^n z^{2n+1}}{n! (2n+1)} \]

By the Alternating Series Test this series converges. Therefore the error function has a specific value at each point. However closed form expressions for these values may or may not exist. But there are closed form approximations of the error function so that values of certain accuracy can be found.

Kind Regards,

Sudharaka.

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- Feb 5, 2012

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You are welcome. Yes in a way, but note that the trigonometric functions are periodic and the error function is not. See this.Thanks a lot, Sudharaka; very clear.

So, as far as the "mechanics" go, it works a bit likeSIN or COS functions, right?

I mean every case is different....

Kind Regards,

Sudharaka.

- Jan 26, 2012

- 890

You should also be aware of the realtionship between the error function and the cumulative normal distribution:Thanks a lot, Sudharaka; very clear.

So, as far as the "mechanics" go, it works a bit likeSIN or COS functions, right?

I mean every case is different....

\( \displaystyle \Phi(x)=\frac{1}{2}\left[1+{\rm{erf}} \left( \frac{x}{\sqrt{2}} \right) \right] \)

and hence the inverse functions:

\( \displaystyle {\rm{erf}}^{-1} (x) = \frac{1}{\sqrt{2}} \Phi^{-1}\left( \frac{x-1}{2}\right) \)

CB

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- Feb 13, 2012

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$\displaystyle \text{erf}\ (z)= \frac{2}{\sqrt{\pi}}\ \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!\ (2n+1)}\ z^{2n+1}$ (1)

It belongs to the family of functions having McLaurin series of the type...

$\displaystyle w=f(z)= c_{1}\ z + c_{2}\ z^{2} + c_{3}\ z^{3} + ...\ ,\ c_{1} \ne 0$ (2)

... and for them the coefficients inverse function McLaurin expansion...

$\displaystyle z= f^{-1}(w) = d_{1}\ w + d_{2}\ w^{2} + d_{3}\ w^{3}+...$ (3)

... can be computed with the formula...

$\displaystyle d_{n}=\frac{1}{n!}\ \lim_{z \rightarrow 0} \frac{d^{n-1}}{d z^{n-1}} \{\frac{z}{f(z)}\}^{n}$ (4)

The computation of the $d_{n}$ for the function $\text{erf}^{-1}(w)$ using (4) is tedious but not very difficult and it will performed in a successive post...

Kind regards

$\chi$ $\sigma$