[AccessD] complex query!!

To raise a quantity to an arbitrary power use:

X^Y

The ^ operator does exactly that.

?2^3
 8 
?1.2^2
 1.44 
?2^1.2
 2.29739670999407 

John W. Colby
www.ColbyConsulting.com 

Contribute your unused CPU cycles to a good cause:
http://folding.stanford.edu/

-----Original Message-----
From: accessd-bounces at databaseadvisors.com
[mailto:accessd-bounces at databaseadvisors.com] On Behalf Of Steve Erbach
Sent: Tuesday, January 18, 2005 9:21 AM
To: Access Developers discussion and problem solving
Subject: Re: [AccessD] complex query!!


Stuart,

O!, Ancient One! We are honored and humbled that you have descended from
your extended care facility to help us in our dire need!

>> I can see a possible use for it, it  shows how alike the items are
in each pair. <<

...and one could graph the "alikeness", too, I suppose.

I was going to modify the SQL statement I came up with for Pedro a few
messages ago, but I discovered that Access has very poor math
function. It has the exp function (e^x) but no function for raising a
quantity to an arbitrary power. It has the sqr (square root) function
but no function for squaring a number! And the StDev function doesn't
seem to allow me to feed it a range of data points. What it wants is a
field name and that's that. I could fiddle around with logs and such
to get the result that Pedro wants, but sheesh! It's like going back
to using a slide rule! I've decided that I've had enough of that kind
of pie. I don't feel particularly clever this morning.

Steve Erbach
Neenah, WI

On Tue, 18 Jan 2005 11:44:50 +1000, Stuart McLachlan
<stuart at lexacorp.com.pg> wrote:
> On 17 Jan 2005 at 19:35, Steve Erbach wrote:
> 
> > Pedro,
> >
> > Just so I'm clear (I'm 52 years old; have a little pity for the aged),
> 
> You young pups don't know nuttin.  When you get to my age you have a
better
> chance of understanding.  :-)
> 
> > do you want what Stuart suggested? That is:
> >
> > For each pair of values that are used to generate an average, you want
> > the standard deviation of that pair of data points? For two points the
> > standard deviation is equal to the difference between one of the
> > points and the average. So if the points are 8.4 and 9.3, the average
> > is 8.85. The "standard deviation" would be 9.3 - 8.85 = 0.45. Is that
> > what you want? I mean, it's easy enough to compute, but I'm having a
> > hard time seeing how useful it is.
> >
> 
> I can see a possible use for it, it  shows how alike the items are  in
each
> pair.
> 
> --
> Stuart
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