b implies a, so you could shorten your list, if you want.
In general, aren't all decimal expansions infinite sums? Decimal representation - Wikipedia, the free encyclopedia
What makes equations that involve infinity wrong?
b implies a, so you could shorten your list, if you want.
In general, aren't all decimal expansions infinite sums? Decimal representation - Wikipedia, the free encyclopedia
What makes equations that involve infinity wrong?
Any number multiplied by its inverse is 1 since N*(1/N)=1 because the N's cancel on each other. Now, since 0 is the inverse of infinity, is it safe so assume that 0*infinity=1? Methinks so.
0 isn't the reciprocal of infinity.
inf/inf = 1
1/inf = 0
0inf = 0
All of these are made rigorous by the limit concept.
Think of it this way:
Think infinity as the universe ( and ignore my signature ).
The universe is constantly expanding. Likewise, two infinities may not be the same...
I know... I think... I guess... who am I kidding?
Devoted my life to programming...
Except 0. Otherwise...
0 * (1/0) = 1
Since the above shows you, I hope, that 0 can't be the inverse of infinity (infinity has no inverse), 0 * infinity must be constructed differently. The current agreement is that infinity multiplied by any number, including 0 is undefined.because the N's cancel on each other. Now, since 0 is the inverse of infinity, is it safe so assume that 0*infinity=1? Methinks so.
Originally Posted by brewbuck:
Reimplementing a large system in another language to get a 25% performance boost is nonsense. It would be cheaper to just get a computer which is 25% faster.
It depends. Do you mean 0 multiplied by something that approaches infinity or something that approaches 0 multiplied by something that approaches infinity? If the former, no, it would be 0. If the latter, sometimes; It could approach anything, because it is an indeterminate form(Indeterminate form - Wikipedia, the free encyclopedia).
lim(x->inf)x/lim(x->inf)x^2 = lim(x->inf)x/x^2 = lim(x->inf)1/x = 0
The second is correct. (I even used it above.)
lim(x->inf)1/x * lim(x->inf)x = lim(x->inf)x/x = lim(x->inf)1 = 1
The first and third are indeterminate, and can't be given a general rule.
Last edited by User Name:; 04-01-2011 at 12:11 PM.