In a program I'm writing, I need to evaluate an expression of the form

z = e^x / (e^x + e^y)

Because x and y may be *very* large numbers, computing e^x and e^y may
result in overflows.  Is there a way to approximate z instead of
getting an exact figure?

Alternatively, is there a way to get an exact or approximate value of
z if we use the following form instead?

e^z = e^x / (e^x + e^y)

Hello gw

 1)          Simple way would be to calculate value   of u=1/z

 1/z = 1 + e^(y-x)

 That has the advantage that y-x may be small even if each is large.
 After you have value of 1/z, you can test it's value and invert it is
 not too close to zero.

2) In the second case  

 z2 = x - lg ( e^x + e^y) 

That may work if x and y are both in range of the domain, but 1 method
looks better.

For more info, you would have to specify language (c, fortran ..)
 actual meaning of large (how large is large) and precision you need.

Hedgie

z=e^x/(e^x-e^y) = [e^x/e^x]/[(e^x+e^y)/(e^x)] = 1/[1+e^(y-x)]

so if both become too big y-x will stay small 
This formula has problem only if y is increasing and/or x is decreasing

agmpinia-ga, thanks, that seems to be a workable solution. :)

I would agree with simplifying to 1/[1+e^(y-x)]. However, you won't
only have problems with this in the case y increasing, x decreasing.
You will have problems any time x and y are too far apart (try it for
y = 2x at x = 1000 and you'll see what I mean). If y is more than a
little bit bigger than x the result will come back as 0, and if x is
more than a little bit bigger than y it will come back as 1.

If that's OK for you, then fine. Otherwise, if y is significantly
greater than x (and note that what counts is y-x, not y/x),
approximate as z = e^(x-y) (this is just ignoring the e^x term in the
denominator, since it's going to be immaterial). If x is significantly
greater than y, z will be very close to 1, so we need an approximation
for z - 1 = -e^y/(e^x + e^y), which is the original formula with x and
y reversed; so we approximate this as z - 1 = -e^(y-x).

For my purposes, x and y will be constrained such that the equation
e^x / (e^x + e^y) will always fall between 0 and 1.

I need to "approximately" sort some 9,000 items based on this
fraction, and I'm starting to think that I might be able to get away
with simply sorting by (y-x)...