In the below routine, are the instances where i=0 due to not being able to
express certain values exactly in binary?

And can it be reliably solved for any 'r' by ROUNDing to 15 places?

Or is there something else I'm missing?

Thanks.

===================
Sub foo()
Dim r
Dim i

For r = 2 To 16

i = Fix(Log(r) / Log(r))
Debug.Print ("For r = " & r & ", i = " & i)
Next r

End Sub
================


--ron

It is a little more complex than just rounding error. Try
this:

Public Sub foo()
Dim r
Dim i

For r = 2 To 16

i = Fix(Log(r) / Log(r))
Debug.Print "r:" & r & " i:" & i _
& " Log(r):" & Log(r) & " Log(r)/Log(r):" _
& Log(r) / Log(r) & " Fix:" & Fix(Log(r) / Log(r))
Next r


End Sub

You will see that when the results of the formula Fix(Log
(r) / Log(r)) are printed directly it gives the result 1
but when it is assigned to the variable i it gives the
result 0! This seems to occur even if I explicitly define
r and i to be either long integers or a double precision
floating point variables!


-----Original Message-----
In the below routine, are the instances where i=0 due to
not being able to
express certain values exactly in binary?

And can it be reliably solved for any 'r' by ROUNDing to
15 places?

Or is there something else I'm missing?

Thanks.

===================
Sub foo()
Dim r
Dim i

For r = 2 To 16

i = Fix(Log(r) / Log(r))
Debug.Print ("For r = " & r & ", i = " & i)
Next r

End Sub
================


--ron
.