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I realized something the other day that's got my mind in a knot again. Given some constant $ n $, I know that: $$ n^2 = n \cdot n$$ $$ n^3 = n \cdot n \cdot n$$ And so on so forth. I also know that: where, $ n^{1/2} = k $, then: $ k \cdot k = n $ and so on and so forth. Things get a little complicated when $ n^{2/3} = k $, but with some work I understand that $ n \cdot n = k \cdot k \cdot k $. An intuitive explanation might be "the length of one side of that square is equal to the length of one side of that cube." Higher dimensions, of course, break this "intuitiveness" but I can still process mathematically what these rational powers mean.

But what does it mean when you're given $ n^\pi $? How would I visualize this? Is it really as simple as just approximating to the degree of accuracy you need?

$$ n^{31415} = k^{10000} $$

What's a more rigorous definition (or alternative explanation) of $ n^\pi $ than my understanding of powers mentioned above?

I ask this question because I read this thread: Real Numbers to Irrational Powers

npengra317
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  • The standard definition is that it's $n^\pi=\mathrm e^{\pi\log n}$. That's the way hand-held computers calculate $x^y$. – Bernard Feb 13 '18 at 21:03
  • @Bernard: this doesn't address the question, does it ? $\pi \log n$ is still an irrational number, not counting that the question relates to a stage of learning where logarithms are not available yet. –  Feb 13 '18 at 21:05
  • @Bernard Interesting. I didn't know that. But my question still stands, what does $ e^\pi $ mean? Do I calculate that value with just an approximation technique I suggested in my question? – npengra317 Feb 13 '18 at 21:06
  • You are right, this is essentially the way exponentials of a real can be defined: by means of a sequence of rationals converging to a real (you should read about the Dedekind cuts). Such a definition preserves the properties of exponentiation with rational exponents, such as the law $a^ba^c=a^{b+c}$. –  Feb 13 '18 at 21:07
  • @YvesDaoust: I'm not so sure logarithms are not available to the O.P. Anyway I don't see how one can answer this question in a simple way without logarithms. – Bernard Feb 13 '18 at 21:08
  • @Bernard: check the answers: no logarithms. And remember that logarithms come *after* exponentiation. –  Feb 13 '18 at 21:10
  • @npengra317: Yes. All one really knows is that $\mathrm e^\pi$ is the infinite sum $1+\pi+\frac{\pi^2}2+\frac{\pi^3}{3!}+\dotsm$. – Bernard Feb 13 '18 at 21:10
  • @Bernard I suspect a computer does something with power series to find $x^y$. Anyone know the truth? – Jonathan Feb 13 '18 at 21:12
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    Basically, it computes $\mathrm e^{y\ln x}$, with some sophistication like the CORDIC algorithm, I believe. – Bernard Feb 13 '18 at 21:14
  • @YvesDaoust, I understand logs so I don't mind if the answer is explained using logs. Seems that most the answers are explained in limits, which occur after logs anyways. – npengra317 Feb 13 '18 at 21:15
  • @JonathanBrown: general exponentiation is done by logarithms/antilogarithms ($e^{y\log x}$). Logarithms and anti- can be computed by CORDIC or by series or by optimal approximation polynomials. When done using the floating-point representation, processing the exponent can be trivial. –  Feb 13 '18 at 21:16
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    @npengra317: you are asking about irrational exponents. Logarithms are most of the time irrational, so the argument is irrelevant. If you swallowed the logarithms, why are you asking about irrational powers ? –  Feb 13 '18 at 21:22

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The most standard definition uses the idea of a limit: basically, the limit of a function $f(x)$ "as $x$ goes to $c$" is $L$ if you can get $f(x)$ to be as close as you want to $L$ by taking $x$ to be close enough to $c$. For example, the limit of $\frac{1}{x}$ as $x$ goes to $2$ is $\frac{1}{2}$, because no matter how precise I want to measure, I can always choose an $x$ close enough to $2$ that I won't be able to measure the difference between $\frac{1}{2}$ and $\frac{1}{x}$. We write this $\lim_{x \to c}f(x) = L$.

Formally, when $r$ is irrational, we define $a^r$ as $\lim_{x \to a}a^x$. In other words, we say that $a^r$ is the number you get closer and closer to when you raise $a$ to rational powers that are closer and closer to $r$.

Thinking of $n^{\pi}$ as looking at things like $n^{31415} = k^{10000}$ is a pretty good way of thinking about it, but it has the disadvantage of using unreasonably large numbers. I would think of it more as "$n^{\pi}$ is somewhere between $n^{3.1415}$ and $n^{3.1416}$".

Reese Johnston
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  • Interesting. Will the limit yield an irrational number most of the time? Or are there some cases where the limit will yield a number with a finite number of decimal places? – npengra317 Feb 13 '18 at 21:36
  • @npengra317 In a very precise sense, yes, "most of the time" the result is irrational. But there are many cases in which it isn't - for example, there's a particular number $z$ (usually called $\log_23$) so that $2^z = 3$. Because no power of $2$ is also a power of $3$, this $z$ has to be irrational. In fact, for any two positive numbers $a$ and $b$, there is a number $z$ with $a^z = b$; if $a$ and $b$ are integers and aren't powers of the same integer, then this $z$ has to be irrational. – Reese Johnston Feb 14 '18 at 01:18
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Let $\{p_1, p_2, \dots \}$ be a sequence of rational numbers such that $\lim_{i \to \infty} p_i =\pi$. Then one can define $n^\pi = \lim_{i \to \infty} n^{p_i}$

Jonathan
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