Why is 0 multiplied by infinity considered an indeterminate form?

[Mk]

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 1^infinity? 1? Guess not.

infinity / infinity? 1? Guess not.

 

[matt grime]

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]

my book states that in some cases you can calculate 0*infinity by using l'hopital rule when 0*\infty=(1/\infty)*\infty.

 

[matt grime]

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]

0 * infinity
How about 1^infinity? 1? Guess not.

I've always wondered why 1^\infty is considered indeterminate. Why? How many ever times you multiply one by itself it's going to stay one, won't it?

 

[Hurkyl]

So, does

<br /> e = \lim_{x \rightarrow \infty} \left(1 + \frac{1}{x} \right)^x =<br /> \left( \lim_{x \rightarrow \infty} 1 + \frac{1}{x} \right)^<br /> {\lim_{x \rightarrow \infty} x}<br /> = 1^\infty = 1<br />

?

 

[matt grime]

I've always wondered why 1^\infty 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]

I've always wondered why 1^\infty 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]

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]

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]

Hurkyl's post is a reductio ad absurdem. It is a proof that

\lim (a\circ b)\neq \lim a \circ \lim b

where o is some operation.

 

[Hurkyl]

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 1^{\infty} is a determinate form.

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


(Those should all be positive infinities, but it doesn't typeset well)

 

[dasmurph]

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]

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]

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]

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]

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:

<br /> \lim_{x \rightarrow \infty} 1^x =1<br />

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

<br /> \lim_{x \rightarrow \infty}{\left(1+\frac{1}{x}\right)^{x}} = e<br />

why not define 1^{\infty} as e? Or for that matter, as any arbitrarily chosen number, since:<br /> \lim_{x \rightarrow \infty} \left(1+\frac{1}{x}\right)^{n x} = e^{n}<br />

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.

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:

\lim_{m\rightarrow\infty}\lim_{n\rightarrow\infty}\frac{m}{m+n}

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]

So, does

<br /> e = \lim_{x \rightarrow \infty} \left(1 + \frac{1}{x} \right)^x =<br /> \left( \lim_{x \rightarrow \infty} 1 + \frac{1}{x} \right)^<br /> {\lim_{x \rightarrow \infty} x}<br /> = 1^\infty = 1<br />

?

no but what about <br /> not_e = \lim_{x \rightarrow \infty} \left(1 \right)^x =<br /> \left( \lim_{x \rightarrow \infty} 1 \right)^<br /> {\lim_{x \rightarrow \infty} x}<br /> = 1^\infty = 1<br />?

 

[makc]

since I replied to something on 1st page, I missed this:

The reason you would say this is:

<br /> \lim_{x \rightarrow \infty} 1^x =1<br />

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

<br /> \lim_{x \rightarrow \infty}{\left(1+\frac{1}{x}\right)^{x}} = e<br />

why not define 1^{\infty} as e?

well, because we were talking about 1^ something, and with e limit it is (something that gradually approaches 1)^ something. It is worse as definition of 1^{\infty} for that reason.

 

[Montebianco]

well, because we were talking about 1^ something, and with e limit it is (something that gradually approaches 1)^ something. It is worse as definition of 1^{\infty} for that reason.

Well, when there is ambiguity, I suppose one could always arbitrarily pick one definition and declare it to be the correct one. But this allegedly better definition fails an important test: it doesn't preserve the rules of arithmetic. No definition of 1^{\infty} does, which is why conventional mathematics refers to such constructions as indeterminate. Any attempt to define it as some specific quantity breaks arithmetic as we know it.

By the way, if we talk about 0/0, are we talking about 0/something, or something/0?

PS - my apologies if resurrecting a long dead thread is considered bad form. Someone at another forum pointed this thread out to me, and I posted here without noticing that it was very old.

 

[Gib Z]

no but what about <br /> not_e = \lim_{x \rightarrow \infty} \left(1 \right)^x =<br /> \left( \lim_{x \rightarrow \infty} 1 \right)^<br /> {\lim_{x \rightarrow \infty} x}<br /> = 1^\infty = 1<br />?

\lim_{x\to a} f(x)g(x) = \lim_{x\to a} f(x) \cdot \lim_{x\to a} g(x) if and only if each individual limit exists.

And if f(x) is not equal to zero, the following holds.
\log \lim_{x\to a} f(x)^{g(x)} = \lim_{x\to a} g(x) \log (f(x))
We can take the log out because the log is defined as the integral of 1/x, which is continuous everywhere except x=0, and also because the integral of a continuous function is another continuous function, so the Log function is continuous.

Now we have another limit of a product, and since we know that the limit of a product is the product of the each individual limit, if and if only both limits exist, we have shown that the exponential limits must obey the same conditions.

PS. The limit is not considered to exist if the sequence of terms does not tend to any finite real number, of which infinity is not.

 

[Algr]

How does one then justify .333... = 1/3?

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.

Well this is curious. How does one then justify .333... = 1/3? If an infinite number of zeros is undefined, then how can an infinite number of positive value .x3 have a rational value?

 

[Diffy]

This post was started 2 years ago, how did it start up?

There are so many posts regarding the properties of arithmetic and infinity, why not make complete stickied thread? If Hurkyl or Gib_z could right an detailed post explaining why you can't use binary operators on infinity then perhaps it would reduce the number of posts we see regarding this topic in here.

 

[Diffy]

Well this is curious. How does one then justify .333... = 1/3? If an infinite number of zeros is undefined, then how can an infinite number of positive value .x3 have a rational value?

How do you know infinity has zeros in it?

 

[Algr]

Well the opening question to this thread was 0 x infinity, so I was working from that idea. All repeating decimals are an infinite number of positive values added together. How can a value defined by infinite addition be rational?

 

[arildno]

This post was started 2 years ago, how did it start up?

There are so many posts regarding the properties of arithmetic and infinity, why not make complete stickied thread? If Hurkyl or Gib_z could right an detailed post explaining why you can't use binary operators on infinity

It doesn't need to be more detailed than just saying that "infinity" is not a real (or complex) number, yet the standard binary operations explicitly assumes its two arguments to be such numbers.
Hence, you can't use "infinity" as one of those numbers.

 

[HallsofIvy]

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

That is the limit for that very simple function. But if you did the same with [(x-1)/x]^x, (x-1)/x goes to 1 as x goes to infinity so it has every bit as good a claim to represent "1^infinity". But it does not go to 1 as x goes to infinity. One example does not prove anything.