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For me, multiplication is a binary operation so it can be applied only on a finite sequence of numbers. but $1^{\infty}$ requires that we apply multiplication infinitly which is not defined as multiplication is a binary operation.

Is that a good reason? If not, what is the reason?

If my reason is ok, So similarly, $5^{\infty}$ is indeterminate ?

ِAdded:

I noticed that all answers are in context of "limits". Algebraically, multiplication is a binary operation, So it ONLY can be used to define multiplication of finite sequence of numbers as not a infinite sequence. So algebraically, what does $1^{\infty}$ even mean?

Najib Idrissi
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FNH
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    E.g., compare $1^n$ and $(1+1/n)^n$, for large $n$. – David Mitra Oct 05 '14 at 09:16
  • Let $A=1^{\infty}, \ln A=\infty\ln1=\dfrac00$

    Similarly, $B=5^{\infty}, \ln b=\infty\ln5=\infty$

     – lab bhattacharjee Oct 05 '14 at 09:16
  • @RainiervanEs , My question is different! I'm asking from an algebraic point of view as binary operation see the added part not in context of limits. – FNH Oct 05 '14 at 09:23
  • @MathsLover: Any time you mention $\infty$, you are talking about limits. Without speaking of convergence, $1^\infty$ is unlikely to mean anything. – Eric Stucky Oct 05 '14 at 09:23
  • @labbhattacharjee,$ln(a^n)=n ln(a)$ only true when $n$ is real number not infinity. – FNH Oct 05 '14 at 09:24
  • @DavidMitra, Yes, I recognised that before you added your comment so I deleted it. – FNH Oct 05 '14 at 09:42
  • About what you added - we actually need limits to make exponentiation defined everywhere. For example, how would you go about defining $2^\pi$ if you couldn't use limits? Similar thing applies when we try to do it for infinite exponent. – Wojowu Oct 05 '14 at 09:48
  • @Wojowu, We can define $2^\pi$ to be the least upper bound of the set ${ 2^x : x\in \mathbb{R} & x<\pi}$ because this set is bounded above by say $2^5$ so it has a least upper bound property :) – FNH Oct 05 '14 at 09:58
  • You'd have to use $x\in\Bbb Q$, but that's about right. However, using l.u.b. is an analytic property of real numbers in my dictionary, not algebraic. – Wojowu Oct 05 '14 at 10:01
  • @Wojowu, yes, rational , I've only miswritten that :). But also, definition using limits is not algebraic, I think. afterall, what the wrong with analytic definition? – FNH Oct 05 '14 at 10:29
  • You are the one who wanted to know what this means algebraically. – Wojowu Oct 05 '14 at 10:30

2 Answers2

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The reason on why this is indeterminate is because of how it behaves when we go to limits. For example, if you look at $1^n$ as $n\rightarrow\infty$ we would get that $1^\infty=1$, while by looking at $(1+\frac{1}{n})^n$ we get that $1^\infty=e$. We don't get such thing with $5^\infty$ because, no matter what, it always diverges to infinity.

Wojowu
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  • Why not to define $1^\infty$ as $lim_{x\rightarrow \infty} 1^x$ neglecting the other limit? Why should we consider the other limit at all? I knowthat the two expressions get closer to each other as $x$ goes larger but in any given real $n$ ,they still are not the same! – FNH Oct 05 '14 at 10:03
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    You can define it like that, but this is one of infinitely many possible definitions. Remember that for determinate forms, like $2^4$, the result of $\lim_{x\rightarrow 2,y\rightarrow 4}x^y$ does not depend on the way that $x,y$ tend to their limits. – Wojowu Oct 05 '14 at 10:06
  • So, Can we say that we define $a^n$ to be the limit $lim_{x\rightarrow a,y\rightarrow n}x^y$? So if the limit exists then it's a determinate form otherwise indetermindate form? Is that the standard definition? – FNH Oct 05 '14 at 10:32
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    I believe this is a definition most people would agree with. – Wojowu Oct 05 '14 at 10:36
  • But according to that definition, $5^\infty$ is indeterminate as in the link: http://www.wolframalpha.com/input/?i=lim+x%5Ey+%28x+approachs+5+%2C+y+approachs++infinity%29 – FNH Oct 05 '14 at 11:00
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Because $1=a^0$, and $0\cdot\infty$ is $($also$)$ undetermined.

Because all convergent infinite products are of the form $1^\infty$, since their general term tends to $1$, and the number of terms is infinite, but they don't all converge to the same value. Furthermore, there are also divergent infinite products whose general term also tends to $1$.

Etc.

Lucian
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  • I don't uderstand the paragraph "because all .... ", could you clarify plz? – FNH Oct 05 '14 at 09:43
  • @MathsLover: Were you taught infinite products ? – Lucian Oct 05 '14 at 09:48
  • I never knew that $1^\infty$ can be interpreted as infinite product. It's quite unnatural for me, because some of the terms in such product can be quite far from 1. – Wojowu Oct 05 '14 at 09:57
  • @Lucian , No but Whenever I was seing thing like $a^{infty}$ i was interpreting it according to this definition. – FNH Oct 05 '14 at 10:09
  • @Wojowu: Infinity minus some is still infinity. The same reasoning applies to infinite sums, where the general term tends to $0$. Though for $0\cdot\infty$ we have simpler ways of showing it to be undetermined. – Lucian Oct 05 '14 at 10:20
  • But if we have, say, a sum of form $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...$ then it isn't of the form $0\cdot\infty$, because all terms are strictly greater than 0. That the terms tend to 0 doesn't matter here. – Wojowu Oct 05 '14 at 10:29
  • @Wojowu: If you were asked to make a brute approximation of the portion of the harmonic series starting at $\dfrac1{10^9}$, would you not be tempted, at least a bit, to express it as $0\cdot\infty$ ? Furthermore, aren't all definite integrals limits of the form $0\cdot\infty$ ? An area being expressed as the sum of an infinite number of terms, whose individual area is nil, since the width of each line segment is zero. Yet, we know that the value of such an integral could be any finite quantity, as well as $\pm\infty$. – Lucian Oct 05 '14 at 11:01