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"erf"
- Thread starter Wilmer
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quantaentangled
New member
- Apr 2, 2012
- 9
This is the 'inverse error function'. It even has a Taylor series. Google 'inverse erf' and you will find info about it.
- Feb 5, 2012
- 1,621
Hi Wilmer,
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 not true in general that,
\[f^{-1}(x)=\frac{1}{f(x)}\]
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,
\[\mbox{erf}(z)= \frac{2}{\sqrt{\pi}}\sum_{n=0}^\infty\frac{(-1)^n z^{2n+1}}{n! (2n+1)} \]
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.
Last edited:
- Thread starter
- #4
- Feb 5, 2012
- 1,621
You are welcome.
Yes in a way, but note that the trigonometric functions are periodic and the error function is not. See this.Kind Regards,
Sudharaka.
CaptainBlack
Well-known member
- Jan 26, 2012
- 890
You should also be aware of the realtionship between the error function and the cumulative normal distribution:
\( \displaystyle \Phi(x)=\frac{1}{2}\left[1+{\rm{erf}} \left( \frac{x}{\sqrt{2}} \right) \right] \)
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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- #7
chisigma
Well-known member
- Feb 13, 2012
- 1,704
The function erf(*) is an entire function that can be alternatively defined by its McLaurin series...
$\displaystyle \text{erf}\ (z)= \frac{2}{\sqrt{\pi}}\ \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!\ (2n+1)}\ z^{2n+1}$ (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
$\chi$ $\sigma$
$\displaystyle \text{erf}\ (z)= \frac{2}{\sqrt{\pi}}\ \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!\ (2n+1)}\ z^{2n+1}$ (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
$\chi$ $\sigma$