Originally posted by Acolytei suppose, when you put it that way...that you could prove that it is equal to a lot of things-i got 1...
Or: lim(x->0+) sqrt(-x)*sqrt(1/x)
= i
as x->infinity, 1/x->0 and x/x=1 so 0*infinity=1...
but not 0 as 1/0 is infinty, so you have infinite sets of 0 to make a whole number...hey-that kinda proves that it's one too!...in fact...2/0=infinity....3/0 etc....wawzer!...
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Originally posted by geniusI think there are (infinitely!) many functions on x that evaluate to infinity * zero as x approaches some limit. A lot of these functions will have a definite value at that point, but different functions will give different values. So the value of infinity * zero depends on what function produced it and is undefined without that information.
i suppose, when you put it that way...that you could prove that it is equal to a lot of things-i got 1...
as x->infinity, 1/x->0 and x/x=1 so 0*infinity=1...
but not 0 as 1/0 is infinty, so you have infinite sets of 0 to make a whole number...hey-that kinda proves that it's one too!...in fact...2/0=infinity....3/0 etc....wawzer!...
0*infinity has never been defined, because there is no logical answer.
by definition of infinity: x*inf = inf for x >0
and easy to proof is 0*x = 0 for x a real (yes, even this has a proof)
this would give lim{n->0} n*inf =/= lim{k->inf} 0*k
so neither of those limits would be a good definition of 0*inf
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Originally posted by FiathahelPerhaps it should be defined to have a random value somewhere between + and - infinity!
0*infinity has never been defined, because there is no logical answer.
by definition of infinity: x*inf = inf for x >0
and easy to proof is 0*x = 0 for x a real (yes, even this has a proof)
this would give lim{n->0} n*inf =/= lim{k->inf} 0*k
so neither of those limits would be a good definition of 0*inf
Originally posted by iamatigeri don't understand that, having not taken calc, but it seems to me that multiplication is similer to exponenta in that
What about this!
0*I = lim(x -> 0) x * tan(x + pi()/2))
= -1
5^3=5*5*5
5*3=5+5+5
multiplying anything by one makes no diference, and adding 0 makes no diference, thus it is also threw that
5^3=5*5*5*1*1*1*1......
5*3=5+5+5+0+0+0+0......
thus
N^0=1*1*1*1*1....=1
N*0=0+0+0+0+0.....=0
thus
I^0=1
I*0=0
infinity times zero is undefined.
However, the multiple of two functions, one evaluating to infinity, the other to zero can sometimes be resolved using their derivatives. (As an example of the form, consider L'Hopital's rule; where derivitives of functions are used to avoid a division by zero. But that is a seperate question too...)
It can't be defined in any real terms, any calculation must use values tending towards infinity, and is therefore conceptually wrong, I agree that in principle it would appear that I*0=0 but that is an 'engineers view' as my math teacher always taught me. Meaning that it is good enough to function in reality, but doesn't follow pure mathematics. Hmmmm, I never really saw the insult in being able to relate to reality. It's cetainly something I'd like to do someday.😛
you might consider it this way.
we know that 1/0 is infinite, so we can "define" infinity as 1/0 (or any other positive finite number divided by zero).
now, 0 times that gives 0/0, which is undefined since it could be anything.
these are two "indefinite" forms (0*infinity and 0/0). a (the?) third is 1^infinity. i'm sure most of you have seen something like this: lim(x->0) (1+x)^(1/x), which is the transcendental number e. another way this can be written is to substitute y for 1/x and take the limit as y goes to +infinity.
my favorite equation, relating five universal constants, is based on the above: e^(2*pi*i) = 1.
Originally posted by BarefootChessPlayerThe above is indeed a beautiful equation. A more motivated proof might result from using series representations of circular functions.
you might consider it this way.
we know that 1/0 is infinite, so we can "define" infinity as 1/0 (or any other positive finite number divided by zero).
now, 0 times that gives 0/0, which is undefined since it could be anything.
these are two "indefinite" forms (0*infinity and 0/0). a (the?) third is 1^infinity. i'm sure most of you have s ...[text shortened]... ation, relating five universal constants, is based on the above: e^(2*pi*i) = 1.
Originally posted by BarefootChessPlayer0/0 is undifined, and it can be argued that so is 0*infinity, but 1^infinity is definetly 1. the higher the exponent, the sharper the curve. thus the curve for infinity is infinetly sharpe.
you might consider it this way.
we know that 1/0 is infinite, so we can "define" infinity as 1/0 (or any other positive finite number divided by zero).
now, 0 times that gives 0/0, which is undefined since it could be anything.
these are two "indefinite" forms (0*infinity and 0/0). a (the?) third is 1^infinity. i'm sure most of you have s ...[text shortened]... ation, relating five universal constants, is based on the above: e^(2*pi*i) = 1.
0^i=0
1^i=(0^i)i=0*i=1 as i proved earlyer, but you seem to not agree.
2^i=((0^i)i)i=1*i=infinity
Originally posted by fearlessleaderwell, consider it this way. take the original formula i gave, and substitute: (1 + 2/x)^x or the other way, (1 + 2*y)^(1/y) (i may have these reversed; i can't see what i put in before from this page).
0/0 is undifined, and it can be argued that so is 0*infinity, but 1^infinity is definetly 1. the higher the exponent, the sharper the curve. thus the curve for infinity is infinetly sharpe.
0^i=0
1^i=(0^i)i=0*i=1 as i proved earlyer, but you seem to not agree.
2^i=((0^i)i)i=1*i=infinity
you won't get 1 as the limit is evaluated! (if i've done it right, it will be exp(2).)
so, since this is a form of 1^inf, that must be an indefinite form.
l'hopital's rule with logarithmic differentiation is a good way to find this. going back to the original formula, ln (1+ x)^(1/x) = (ln (1 + x))/x. for x=0, this is 0/0, an indeterminate form, so the result at this point would be exp(0/0). differentiation of the logarithm gives (x/(1 + x) - ln (1 + x))/x^2. it's been too long since i did one of these to find the limit, but i know it will come out to e if it's done right.
so, 1^inf cannot have a single value.