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This question already has an answer here:

$1$ square is $1$, so is raised $1$ to $123434234$.

My maths teacher claims that $1$ raised to infinity is not $1$, but not defined. Is there any reason for this?

I know that any number raised to infinity is not defined, but shouldn't $1$ be an exception?

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marked as duplicate by Ilmari Karonen, Davide Giraudo, Lord_Farin, Emily, Julian Kuelshammer May 1 '13 at 13:32

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Dr. Math has a not half bad answer to this: see here. $\endgroup$ – user1729 May 1 '13 at 9:34
  • $\begingroup$ See this. $\endgroup$ – David Mitra May 1 '13 at 9:50
  • $\begingroup$ Infinite exponentiation means that you take 1 and multiply it by a scaler, $a$, infitnite number of times. Scaling by a > 1 yields greater and greater number. So, $(a > 1)^\infty = \infty$. Downscaling infinitly by a<1 yields 0: $(a < 1)^\infty = 0$. Scaling by 1 does not change the number, so $1^\infty = 1$. So, 1 is a clear border between inf and 0 and your teacher is wrong. $\endgroup$ – Val May 1 '13 at 10:22
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What $1^\infty$ is, or is not, is merely a matter of definition. Normally, one would only define $a^b$ for some specific class of pairs of $a,b$ - say $b$ - positive integer, $a$ - real number.

When extending the definition of exponentiation to more general pairs, the key thing people keep in mind is that various nice properties are preserved. For instance, for $ b$ - positive integer, you want to put $a^{-b} = \frac{1}{a^b}$ so that the rule $a^ba^c = a^{b+c}$ is preserved.

It may make sense in some context to speak of infinities in the context of limits, but this is usually more a rule of thumb than rigorous mathematics. This may be seen as extending the rule that $(a,b) \mapsto a^b$ is continuous (i.e. if $\lim_n a_n = a$ and $\lim_n b_n = b$, then $\lim_n a_n^{b_n} = a^b$) to allow for $b_n \to \infty$. For instance, you may risk saying that: $$\lim_{n} (2+\frac{1}{n})^n = 2^{\infty} = \infty$$ If you agree to use rules of this kind, you might be tempted to also say: $$\lim_{n} (1+\frac{1}{n})^n = 1^{\infty} = 1$$ but this would lead you astray, since in reality: $$\lim_{n} (1+\frac{1}{n})^n = e \neq 1$$ Thus, it is safer to leave $1^\infty$ undefined.


A more thorough discussion can be found on Wikipedia.

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  • $\begingroup$ 1^infinity as 1 * 1 * 1 * 1 * 1 ... is still 1 in its determinate form. 1^infinity would be undetermined in the example you provided because it is in an indeterminate form. Correct? $\endgroup$ – Gary S. Weaver May 1 '13 at 13:16
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    $\begingroup$ @GaryS.Weaver: I'm not quite sure how to understand your question. If you consider $1^\infty$ as something which arises as a product of infinitely many $1$'s, then it is $1$. If you consider $1^\infty$ as a shorthand for $a^b$ with $a \simeq 1$ and $b$ is extremely large, then it is not determined, in the sense you need to know more about $a$ and $b$ to say anything about the value. I am not sure to what extent this answers your question. $\endgroup$ – Jakub Konieczny May 1 '13 at 14:04
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    $\begingroup$ Thanks! You understood me. $1^\infty$ can have a determinant form that equates to 1, e.g. if you know it comes from the result of multiplying 1 by itself an infinite number of times. So, $1^\infty$ = $1^\infty$ or sometimes $1^\infty$ != $1^\infty$ depending on how you produced $1^\infty$. In other words, if you don't know how you produced $1^\infty$, you don't know which form it is: determinate or indeterminate? $\endgroup$ – Gary S. Weaver May 1 '13 at 14:32
  • $\begingroup$ That sounds about right. The expression $1^\infty$ can stand for many different things, and hence it might be that $1^\infty \neq 1^\infty$, if these come from different directions. $\endgroup$ – Jakub Konieczny May 1 '13 at 14:49
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When your teacher talks about $0/0$ or $\infty/\infty$ or $1^\infty$ he/she's not talking about numbers, but about functions, more precisely about limits of functions.

It's just a convenient expression, but it should not be confused with computations on simple numbers (which $\infty$ isn't, by the way).

When $1^\infty$ is referred to, it is to mean the following situation: there are two functions $f$ and $g$ defined in a neighborhood of $c$, with the properties

  1. $\lim\limits_{x\to c} f(x)=1$

  2. $\lim\limits_{x\to c} g(x)=\infty$ (or $-\infty$)

(of course, $c$ can also be $\infty$ or $-\infty$).

Saying that $1^\infty$ is an indeterminate form is just a mnemonic way to say that you cannot compute

$$\lim_{x\to c}f(x)^{g(x)}$$

just by saying “the base goes to $1$, so the limit is $1$ because $1^t=1$”. Indeed this can be grossly wrong as the fundamental example

$$\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{x}=e$$

shows.

Why is that? It's easy if you always write $f(x)^{g(x)}$ as $\exp(g(x)\log f(x))$ and compute the limit of $g(x)\log f(x)$, then applying the properties of the exponential function.

In the case above we'd have

  1. $\lim\limits_{x\to c} \log f(x)=0$

  2. $\lim\limits_{x\to c} g(x)=\infty$ (or $-\infty$)

so the limit

$$\lim_{x\to c}g(x)\log f(x)$$

is in the other $\infty\cdot0$ indeterminate form (that you should know). Why is it “indeterminate”? Because we have many instances of that form where the limit is not predictable by simply doing a (nonsense) multiplication:

\begin{gather} \lim_{x\to 0+}x\cdot\frac{1}{x}=1\\ \lim_{x\to 0+}x^2\cdot\frac{1}{x}=0\\ \lim_{x\to 0+}x\cdot\frac{1}{x^2}=\infty \end{gather}

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By the way, just to give a much simpler answer (which indeed does not really explain the issue but might help if you're not studying calculus yet):

The problem here is that, in reality, $\infty$ is not a number. It is used to represent an unimaginably big number, but you obviously can't tell which. Therefore, infinity itself is not a defined number.

That's the background reason of why having $1^a$ with $a$ being a single, precise number does give you

$$ 1^a=1 $$

while with $a=\infty$, or any other undefined number, it generally gives you $undefined$.

It's not the only case, actually. This is one example of indeterminate forms. Other similar examples are $\infty^0$ and $0*\infty$, which are also undefined.

To fully demonstrate it or show practical examples, you would actually have to go into functions and theory of limits (which are, by the way, not the actual values that functions take but where they converge, or approach, to). And the other answers have already done that, so I'll stop here.

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