What does $2^π$ : the multiplication of $2 \pi$ many times?

Question

This is rather an intuitive question, in the sense that indeed a real number raised to the power of an irrational doesn't make sense but I wanted to know what does it mean intuitively. If $2^7=2*2*2*2*2*2*2, 2^{\frac{1}{2}}=\frac{1}{\sqrt{2}}$ and $a^b=a*a*a*a*a*...*a\, \, b$ many times, I'm curious as to what does $2^π$ mean. Obviously it's equal to $8.8249778271...$ but what's the meaning in the aforementioned scenario?

Answer 1

There is no meaning, if you use that definition of exponentiation (not that they teach you that in high shcool when they introduce irrational exponents; they usually just power through and expect you to accept it, because there is so much curriculum to get through and so little time. Kudos to you for being sceptical and actually thinking about it).

But that's an important part of what mathematics is about: Take something that seems like it ought to have a meaning, but doesn't quite have one, and see if you can't force it to work anyways (it's called "generalization"). There are three common ways to mend your issue:

1. Instead define $a^b$ as $$\exp(b\ln a):= 1 + (b\ln a) + \frac12(b+\ln a)^2 + \frac16(b+\ln a)^3+\cdots$$ where the denominators of the fractions are the factorial numbers. The logarithm of $a$, $\ln a$ is the one number such that $\exp(\ln a) = a$, and $\pi\cdot \ln 2$ makes sense, since (I hope) you know how to make sense of multiplication of two real numbers. To prove: $\ln 2$ and the $\exp$ function actually exist and are well-defined.

2. Define the value of irrational exponents as the limit of a sequence using rational exponents. In other words, take the sequence $$ 2^3, 2^{3.1}, 2^{3.14},2^{3.141},\ldots $$ and define $2^\pi$ to be the limit of this sequence. To prove: Different sequences of rational exponents that all converge to $\pi$ give the same result when you apply the above.

3. Take a function $f: \Bbb R$ to $\Bbb R$ such that

   • $f$ is continuous
   • $f(a + b) = f(a)\cdot f(b)$
   • $f(1) = 2$

   then define $2^x := f(x)$, and evaluate $f(\pi)$. To prove: $f$ exists and is unique.

So, all of these approaches have things that need to be proven before they can be used effectively, and those things are by no means easy (and proving that they give the same result is not easy either). But as I said above, that's what mathematics is about.

Answer 2

$a^{k}$ is $a*a*...*a$, k times.
$a^{1/n}$ is the n-th root of $a$.
If $r$ is irrational, then $a^r=\lim\limits_{x \to r}a^x=\lim\limits_{k/n \to r} a^{k/n}$.

Answer 3

We can make sense of numbers on the form $2^x$ when $x$ is rational using the way you describe: if $x = \frac{m}{n}$ with $m,n$ integers then $2^x = (2^{m})^{1/n}$ so it's the $n$'th root of $2\cdot 2\cdots 2\cdot 2$ ($m$ factors).

This way of looking at it does not apply to irrational $x$. But any irrational $x$ can be approximated arbitrarily well by rational numbers, i.e. we can find integer sequences $m_k$ and $n_k$ such that $x = \lim_{k\to \infty} \frac{m_k}{n_k}$ (for example for $\pi$ we can for example take $m_k=\{3, 31, 314, 3141,\cdots\}$ and $n_k = \{1,10,100,\cdots\}$). Now the usual way of defining $2^x$ is simply as the limit $2^x \equiv \lim_{k\to\infty}2^{\frac{m_k}{n_k}}$.

What this procedure is doing is to extend the power-function $2^x$ from the rationals (where we know how to compute it) to all the reals by demanding it to be a continuous function.

Answer 4

$2^\pi = e^{\pi \ln(2)} = \sum_{n=0}^{\infty}\frac{(\pi\ln(2))^n}{n!}$...

Answer 5

You know how to calculate powers when exponents are rational. Namely $$2^{\frac{m}{n}} = \sqrt[n]{2^m}$$

$2^\pi$ can then be defined as:

$$2^\pi = \sup\{2^q : q \in \mathbb{Q}, 0 < q < \pi\}$$

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An equivalent definition would be $$2^\pi = \exp(\pi \ln2) = \lim_{n\to\infty} \left(1 + \frac{\pi \ln 2}{n}\right)^n$$

where $\exp$ is defined to be $$\exp x = \lim_{n\to\infty}\left(1 + \frac{x}{n}\right)^n$$ for $x \in \mathbb{R}$.

Answer 6

Start with $2^3=8$. Then think about what $2^{3.1}$ is:

$$2^{3.1} = 2^{31/10} = \sqrt[10]{2^{31}} \approx 8.574187700.$$

Then consider

$$2^{3.14} = 2^{314/100} = \sqrt[100]{2^{314}} \approx 8.815240927.$$

Since the sequence $3, 3.1, 3.14, 3.141, \ldots$ approaches $\pi,$, the sequence

$$2^3, 2^{3.1}, 2^{3.14}, 2^{3.141}, \ldots$$

should approach $2^\pi.$ So it's probably best to think of $2^\pi$ as a limit.

Answer 7

Consider the sequence $2^3, 2^{3.1}, 2^{3.14}, \dots$. We define $2^\pi$ to be the limit of this sequence.

We can compute, for example, $2^{3.14} = 2^{314/100}=\sqrt[100]{2^{314}}$. Practically, this is rediculous since it means we have to compute the $100^{\text{th}}$ root of

$$33374797436264220037422214158899251790667258161822699530422525122222183215322508594108782608384$$

But, theoretically, it makes perfect sense and there are tricks that might make such a calculation more reasonable to compute.