Welcome to our community

Be a part of something great, join today!

"erf"

Wilmer

In Memoriam
Mar 19, 2012
376
What does erf^(-1)(x) mean?

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

Is there a value for erf, like there is for pi and e?

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

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

THANKS for any explanations.
 

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.
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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.
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,

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

Is there a value for erf, like there is for pi and e?
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:

Wilmer

In Memoriam
Mar 19, 2012
376
Thanks a lot, Sudharaka; very clear.

So, as far as the "mechanics" go, it works a bit like SIN or COS functions, right?
I mean every case is different....
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Thanks a lot, Sudharaka; very clear.

So, as far as the "mechanics" go, it works a bit like SIN or COS functions, right?
I mean every case is different....
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
Thanks a lot, Sudharaka; very clear.

So, as far as the "mechanics" go, it works a bit like SIN or COS functions, right?
I mean every case is different....
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
 

Wilmer

In Memoriam
Mar 19, 2012
376
Merci beaucoup, ya'll !
 

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

$\chi$ $\sigma$