[dundee] How to tell if you are a techie, part 2 (or 0010)

[Robert Ladyman]

If you laugh at the last comment for this discussion*, and your wife / blokie 
/ significant other doesn't understand why it's funny and you can't talk for 
laughing.

Or maybe that's just me.

* not to mention having followed the discussion, of course.


[Quote...]
 
No. g(x)=1/x does not 'attain' infinity at x=0. It is undefined at x=0. (To 
quickly see why, look at what happens when x approaches 0 from below and what 
happens when x approaches 0 from above, you get two different results and 
there is no canonical choice, so g(x)=1/x is undefined at x=0). 
 
As g(x) is undefined at 0, then for all x>0 there is a real number k with 
g(x)<k. This is not a useful characterisation to make. 
 
The correct way to understand the 'infinite' behaviour of g(x)=1/x is to say 
that it is unbounded in any neighbourhood of 0. Or in other words, for any 
number N there exists a (probably small) number e such that for all 0<x<e we 
have g(x)>N. 
 
This statement is the mathematically rigorous way of saying g(x) tends to 
infinity as x tends to 0 (from above). 
 
Exponential functions behave in a similar way: f(x)=2^x is not defined at 
x=infinity in the same way that g(x) is not defined at x=0. But again for any 
number N there exists a (probably large) number e such that for all 
e<x<infinity we have f(x)>N.

 
Of course you are right that 1/x does not "attain" infinity; Sorry, my maths 
vocabulary is slighty rusty. However, there is still an important distinction 
between the behaviour of 1/x and 2^x; unlike 1/x, 2^x is defined for any 
finite x. The question of it being defined or not at infinity is a moot point; 
to my knowledge, infinity is not generally considered part of the domain of 
any function. What's more, you cannot say that for any real number x there is 
a k such that k > 1/x, since this has no sense for x=0. However for k > 2^x it 
does have sense and is true. Consequently, this is a useful characterisation 
to make. 
 
To get back on topic, my point was that the id generation will take inifinite 
time for a finite number of occupied IDs (1000000), so the execution time 
cannot depend exponentially on the number of occupied IDs (at least it cannot 
be dominated by an exponential term near where it matters). 
 
 
This discussion is getting less and less funny as it goes along. It is, 
however, still a little funny. It appears to be asymptotic humour.


[End Quote]

-- 

Robert Ladyman
File-Away Limited, 32 Church Street, Newtyle
Perthshire, PH12 8TZ SCOTLAND
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