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Zero times infinity

by Mk
Tags: infinity, times
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Mk
#1
Apr27-06, 04:11 AM
P: 2,056
0 * infinity
What is up? I read that it this expression is called an "indeterminate form." Why isn't zero multiplied by infinity equal to zero? Even if infinity is really big, if there are zero amounts of infinity, that would make zero.

How about 1infinity? 1? Guess not.

infinity / infinity? 1? Guess not.
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matt grime
#2
Apr27-06, 04:17 AM
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Infinity, not being part of the natural, rational, real or complex numbers, does not come with the innate possibility to multiply it by anything in those sets. We can define a consistent notion of arithmetic on the extended numbers (gotten by adding in a symbol for infinity) in many cases. To be consistent with these definitions then we cannot define the multiplication 0*infinity. It is no big deal, and nothing to do with 'zero amounts of anything'. You must not think of the mathematical notion of infinity as being the same as your 'really big number, bigger than you can imagine' unmathematical notion.
MathematicalPhysicist
#3
Apr27-06, 04:53 AM
P: 3,215
my book states that in some cases you can calculate 0*infinity by using l'hopital rule when 0*[tex]\infty[/tex]=(1/[tex]\infty[/tex])*[tex]\infty[/tex].

matt grime
#4
Apr27-06, 05:07 AM
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Zero times infinity

that is analysis, that is about taking limits of things *that tend to infinity*. It is lazy short hand to actually say it 'is' a statement about infinity, rather than what it actually is: a statement about limits.

As a demonstration, suppose g(x)=0 at x=0 and suppose f(x) tends to infinity as x tends to zero (lazy short hand would be f(0)=infinity).

Then f(x)g(x), as x went to zero, if it were to exist, would be '0*infinity' in the lazy short hand. Picking f(x)=1/x and g(x)=nx for any real number n, then the result is n, hence indeterminate.
neutrino
#5
Apr27-06, 07:16 AM
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Quote Quote by Mk
0 * infinity
How about 1infinity? 1? Guess not.
I've always wondered why [tex]1^\infty[/tex] is considered indeterminate. Why? How many ever times you multiply one by itself it's going to stay one, won't it?
Hurkyl
#6
Apr27-06, 07:22 AM
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So, does

[tex]
e = \lim_{x \rightarrow \infty} \left(1 + \frac{1}{x} \right)^x =
\left( \lim_{x \rightarrow \infty} 1 + \frac{1}{x} \right)^
{\lim_{x \rightarrow \infty} x}
= 1^\infty = 1
[/tex]

?
matt grime
#7
Apr27-06, 08:28 AM
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Quote Quote by neutrino
I've always wondered why [tex]1^\infty[/tex] is considered indeterminate. Why? How many ever times you multiply one by itself it's going to stay one, won't it?
Since infinity is not a natural number your argument is not valid. Why do people think that their reasoning about things of type Y applies to things of type X where anything of type X is strictly different from anything of type Y?
DaveC426913
#8
Apr27-06, 08:34 AM
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Quote Quote by neutrino
I've always wondered why [tex]1^\infty[/tex] is considered indeterminate. Why? How many ever times you multiply one by itself it's going to stay one, won't it?
Will it? Prove it.

Start multiplying them, and I'll ask you in a few hours if you have reached an answer.

Your answer, of course, will be "the answer has not been determined yet, as I am not finished the calculation."

And then I'll come back in a few months and ask again. What do you suppose your answer will be then?

Remember, there is no guessing or anticipating in math...
neutrino
#9
Apr27-06, 10:17 AM
P: 2,047
Thanks for all the comments, guys.

Your answer, of course, will be "the answer has not been determined yet, as I am not finished the calculation."
Actually my reply will be something like "I know the answer will be one whenever you ask me." j/k But that appears to be the "intuitive" reply a non-mathematician like me would give.


Btw, how does one resolve the apparent contradiction in the Hurkyl's post? That's the first time that I've seen a limit applied to the power.
matt grime
#10
Apr27-06, 10:44 AM
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What contradiction in Hurkyl's post? There is no reason to suppose the limits behave like that, and that demonstrates that they do not. Just because you can manipulate some symbols does not mean you may do so and have the result make sense.
krab
#11
Apr27-06, 01:53 PM
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Hurkyl's post is a reductio ad absurdem. It is a proof that

[tex]\lim (a\circ b)\neq \lim a \circ \lim b[/tex]

where o is some operation.
Hurkyl
#12
Apr27-06, 06:06 PM
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Btw, how does one resolve the apparent contradiction in the Hurkyl's post? That's the first time that I've seen a limit applied to the power.
Exponentiation of positive real numbers is a continuous function, just like addition, multiplication, division, logarithms, et cetera.

The contradiction is resolved by discarding the hypothesis that [itex]1^{\infty}[/itex] is a determinate form.

As krab said, it's a proof by contradiction. If [itex]1^{\infty}[/itex] is determinate, then my manipulations are justified. But, my manipulations derive a contradiction, and thus [itex]1^{\infty}[/itex] is indeterminate.


(Those should all be positive infinities, but it doesn't typeset well)
dasmurph
#13
May9-06, 11:16 PM
P: 2
It is true that the actual number infinity is unlimited, however the only way to perform any mathematical calculation is one step at a time. But imagine one^infinity as a line, as a graph of y=1^x. That is, of course, a straight line. Since said line will continue forever, we can see that it can "reach" infinity, although infinity can never be reached in the real sense. Basically, it will go on forever, so 1^infinity =1
jbusc
#14
May10-06, 12:01 AM
P: 212
Consider the function f(x) = (1+x)^(1/x)

Clearly the "limit" of this function as x approaches 0 is 1^infinity, yet try to graph it.

The actual limit is e, 2.71828, which is then "equal" to 1^infinity. Clearly e is not equal to 1 so therefore 1^infinity doesn't always have to equal 1 (i.e., it's indeterminate)

The "one step at a time" thing is irrelevant; when dealing with infinities, you have to accept that there can be "supertasks", i.e. tasks that are doable but unable to complete in a finite series of steps.
mathwonk
#15
May10-06, 12:50 AM
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I guess what they are saying is that the limit of a^b, where a approaches 1 and b approaches infinity, is hard to determine.


this should not be too shocking, since if we take the log, we are back to matts exampels of how the limit of a product can be anything, when one factor approaches zero and the other approaches infinity.

for instance it is not hard to calculate the limit of log(1 + 1/x) as x goes to infinity from the integral definition of log, and estimate it as between 1/x+1 and 1/x. hence x times it goes to 1, which gives the limit of xlog(1 + 1/x), i.e. the log of (1 + 1/x)^x. since the log goes to 1, the original limit is e.
matt grime
#16
May10-06, 03:15 AM
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Quote Quote by dasmurph
It is true that the actual number infinity
what is an 'actual number'? Certainly not a 'real number' or a 'complex number' or a 'rational number'.


is unlimited, however the only way to perform any mathematical calculation is one step at a time.
not strictly true. The only way to physically perform any mathematical calculation, perhaps, using a computer, but even that is moot.

we can see that it can "reach" infinity, although infinity can never be reached in the real sense. Basically, it will go on forever, so 1^infinity =1
What does 'reach' mean? In its " " sense or its 'real' sense? 'it' will go on forever? what is 'it'?
Montebianco
#17
Jan30-08, 04:00 PM
P: 2
Quote Quote by neutrino View Post
Actually my reply will be something like "I know the answer will be one whenever you ask me." j/k But that appears to be the "intuitive" reply a non-mathematician like me would give.
The reason you would say this is:

[tex]
\lim_{x \rightarrow \infty} 1^x =1
[/tex]

Certainly you can take the next step and define [tex]1^{\infty}[/tex] to be one if you like. But this definition is only one of many inconsistent definitions that could be made. Since

[tex]
\lim_{x \rightarrow \infty}{\left(1+\frac{1}{x}\right)^{x}} = e
[/tex]

why not define [tex]1^{\infty}[/tex] as e? Or for that matter, as any arbitrarily chosen number, since:


[tex]
\lim_{x \rightarrow \infty} \left(1+\frac{1}{x}\right)^{n x} = e^{n}
[/tex]

The crux of the matter is, infinity doesn't obey the rules that numbers obey. It can't - this is provable. A great deal of trouble occurs when people take laws of arithmetic and try to apply them to infinity. If you assume that infinity obeys all the same laws of arithmetic, then you have assumed a contradiction, and are able to "prove" anything.

Quote Quote by neutrino View Post
Btw, how does one resolve the apparent contradiction in the Hurkyl's post? That's the first time that I've seen a limit applied to the power.
The apparent contradiction comes about because the very first split (the splitting of the limit into two limits, taken in a particular order) is not a valid operation. You can't manipulate limits in this way. To take a similar example from Rudin's book Principles of Mathematical Analysis, what is:

[tex]\lim_{m\rightarrow\infty}\lim_{n\rightarrow\infty}\frac{m}{m+n}[/tex]

The answer you get depends on the order in which you take the limits. You have to be very careful when working with limits to make sure each step in a derivation is justified. The first step in Hurkyl's isn't.
makc
#18
Feb1-08, 04:58 AM
P: 61
Quote Quote by Hurkyl View Post
So, does

[tex]
e = \lim_{x \rightarrow \infty} \left(1 + \frac{1}{x} \right)^x =
\left( \lim_{x \rightarrow \infty} 1 + \frac{1}{x} \right)^
{\lim_{x \rightarrow \infty} x}
= 1^\infty = 1
[/tex]

?
no but what about [tex]
not_e = \lim_{x \rightarrow \infty} \left(1 \right)^x =
\left( \lim_{x \rightarrow \infty} 1 \right)^
{\lim_{x \rightarrow \infty} x}
= 1^\infty = 1
[/tex]?


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