R help - Jun 2004 - erf function documentation

Hi all.  I may be wrong, (and often am), but in trying
to determine how to calculate the erf function, the
documentation for 'pnorm' states:

## if you want the so-called 'error function'
erf <- function(x) 2 * pnorm(x * sqrt(2)) - 1
## and the so-called 'complementary error function'
erfc <- function(x) 2 * pnorm(x * sqrt(2),
lower=FALSE)

Should, instead, it read:
## if you want the so-called 'error function'
erf <- function(x) 2 * pnorm(x / sqrt(2)) - 1
## and the so-called 'complementary error function'
erfc <- function(x) 2 * pnorm(x / sqrt(2),
lower=FALSE)

I've looked at a couple references and they all show
that 'x' should be divided by, not multiplied by,
'sqrt(2)'.

Again, I may be incorrect.  If so, kindly let me know.
 But, if I am correct, perhaps the documentation could
be corrected in a subsequent R release.


Thanks,
Charles

On Wed, 16 Jun 2004, Charles Maner wrote:
> 
> Hi all.  I may be wrong, (and often am), but in trying
> to determine how to calculate the erf function, the
> documentation for 'pnorm' states:
> 
> ## if you want the so-called 'error function'
> erf <- function(x) 2 * pnorm(x * sqrt(2)) - 1
> ## and the so-called 'complementary error function'
> erfc <- function(x) 2 * pnorm(x * sqrt(2),
> lower=FALSE)
> 
> Should, instead, it read:
> ## if you want the so-called 'error function'
> erf <- function(x) 2 * pnorm(x / sqrt(2)) - 1
> ## and the so-called 'complementary error function'
> erfc <- function(x) 2 * pnorm(x / sqrt(2),
> lower=FALSE)
> 
> I've looked at a couple references and they all show
Which are?  `Both', surely.
> that 'x' should be divided by, not multiplied by,
> 'sqrt(2)'.
> 
> Again, I may be incorrect.  If so, kindly let me know.
>  But, if I am correct, perhaps the documentation could
> be corrected in a subsequent R release.
According to Abramowitz and Stegun section 7.1, 

erf z = \frac{2}{\sqrt{pi} \int^z_0 e^_{-t^2} dt

Now pnorm(x) = \frac{1}{sqrt{2\pi}} \int^x_{-\infty} e^_{-u^2/2} du

so pnorm(x) - 1/2 = \frac{1}{sqrt{2\pi}} \int^x_0 e^_{-u^2/2} du

Now substitute t = u/sqrt{2}

pnorm(x) - 1/2 = \frac{1}{sqrt{2\pi}} \int^{x\sqrt{2}}_0 e^_{-t^2} dt\sqrt{2}

so pnorm(x) - 1/2 = 1/2 erf x\sqrt{2} or erf z = 2 pnorm(x/\sqrt{2} - 1

as you suggest.


-- 
Brian D. Ripley,                  ripley at stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford,             Tel:  +44 1865 272861 (self)
1 South Parks Road,                     +44 1865 272866 (PA)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595

On Wed, 16 Jun 2004 14:14:16 -0700 (PDT), Charles Maner
<ckjmaner at yahoo.com> wrote:
>Hi all.  I may be wrong, (and often am), but in trying
>to determine how to calculate the erf function, the
>documentation for 'pnorm' states:
>
>## if you want the so-called 'error function'
>erf <- function(x) 2 * pnorm(x * sqrt(2)) - 1
>## and the so-called 'complementary error function'
>erfc <- function(x) 2 * pnorm(x * sqrt(2),
>lower=FALSE)
>
>Should, instead, it read:
>## if you want the so-called 'error function'
>erf <- function(x) 2 * pnorm(x / sqrt(2)) - 1
>## and the so-called 'complementary error function'
>erfc <- function(x) 2 * pnorm(x / sqrt(2),
>lower=FALSE)
>
>I've looked at a couple references and they all show
>that 'x' should be divided by, not multiplied by,
>'sqrt(2)'.
The "Mathworld" web page at 

http://mathworld.wolfram.com/NormalDistributionFunction.html

shows

pnorm(x) - 1/2 = (1/2)*erf(x/sqrt(2))

(note that their Phi(x) = pnorm(x) - 1/2).  This agrees with the pnorm
documentation once you rearrange things.

Duncan Murdoch