I am trying to understand the most significant jewel in mathematics - the Euler's formula. But first I try to re-catch my understanding of exponent function.

At the very beginning, exponent is used as a shorthand notion of multiplying several identical number together. For example, $5*5*5$ is noted as $5^3$. In this context, the exponent can only be $N$.

Then the exponent extends naturally to $0$, negative number, and fractions. These are easy to understand with just a little bit of reasoning. Thus the exponent extends to $Q$

Then it came to irrational number. I don't quite understand what an irrational exponent means? For example, how do we calculate the $5^{\sqrt{2}}$? Do we first get an approximate value of $\sqrt{2}$, say $1.414$. Then convert it to $\frac{1414}{1000}$. And then multiply 5 for 1414 times and then get the $1000^{th}$ root of the result?


Thanks to the replies so far.

In the thread recommended by several comments, a function definition is mentioned as below:

$$ ln(x) = \int_1^x \frac{1}{t}\,\mathrm{d}t $$

And its inverse function is intentionally written like this:

$$ exp(x) $$

And it implies this is the logarithms function because it abides by the laws of logarithms.

I guess by the laws of logarithms that thread means something like this:

$$ f(x_1*x_2)=f(x_1)+f(x_2) $$

But that doesn't necessarily mean the function $f$ is the logarithms function. I can think of several function definitions satisfying the above law.

So what if we don't explicitly name the function as $ln(x)$ but write it like this:

$$ g(x) = \int_1^x \frac{1}{t}\,\mathrm{d}t $$

And its reverse as this:

$$ g^{-1}(x) $$

How can we tell they are still the logarithm/exponent function as we know them?

  • 3
    $\begingroup$ Check this mse question $\endgroup$ – user378947 Oct 31 '16 at 2:24
  • 1
    $\begingroup$ You should check for similar questions before asking a new one. $\endgroup$ – user378947 Oct 31 '16 at 2:25
  • 1
    $\begingroup$ @mathbeing Thanks. But that thread may not fully solve my confusion. I updated my question. $\endgroup$ – smwikipedia Oct 31 '16 at 3:14
  • 1
    $\begingroup$ Well, much to Pytagoras' shock two millennia ago, there is no rational number whose square is 2. So again, its definition is through successive approximation like 1.4, 1.41, etc. Such a sequence is called Cauchy Sequence. $\endgroup$ – Momo Oct 31 '16 at 3:33
  • 2
    $\begingroup$ If you have time, I recommend you to study the axioms of real numbers and constructions of reals from rationals of Cauchy, Dedekind, Weierstrass. It's not trivial; that is why it has not been formalized earlier. If you don't have time to waste, you may think at $5^\sqrt{2}$ as something mysterious between $5^{1.41}$ and $5^{1.42}$, that you can get with as many decimals as you need, but never exactly :) $\endgroup$ – Momo Oct 31 '16 at 3:49

Yes, you can approximate the result by approximating the irrational exponent with a rational number and proceed with computing integer powers and integer roots. But this does not give you much insight into what an irrational exponent might mean, and I think this is what you mostly care about.

The best insightful explanation I've seen comes from Khalid at BetterExplained.com.

The short summary is that we have to stop seeing exponents as repeated multiplication and start seeing them as continuous growth functions, where $e$ plays a central role.

So $5^{\sqrt2}$ can be written as $(e^{ln(5)})^\sqrt2 = e^{\sqrt2\cdot ln(5)}$. This can be interpreted as continuous growth for $1$ unit of time at a rate of $\sqrt2\cdot ln(5)$, or continuous growth for $\sqrt2$ units of time at a rate of $ln(5)$, or continuous growth for $ln(5)$ units of time at a rate of $\sqrt2$. They are all equivalent.

Check out these links for a much more detailed explanation:

| cite | improve this answer | |
  • 1
    $\begingroup$ The continuous growth explanation of $e$ is unbelievably amazing! I have never viewed $e$ like this! Thanks. Hope it can help me solve my confusion. $\endgroup$ – smwikipedia Oct 31 '16 at 8:01
  • $\begingroup$ And a useful link about various growth functions. betterexplained.com/articles/… $\endgroup$ – smwikipedia Oct 31 '16 at 8:02
  • $\begingroup$ Regarding the continuous compound growth, I have another question posted here: math.stackexchange.com/questions/1992743/… $\endgroup$ – smwikipedia Oct 31 '16 at 8:38

One way of defining the real numbers is as equivalence classes of "the collection of all Cauchy sequences (or, equivalently, the collection of all increasing sequences with upper bound) of rational numbers" with "$\{a_n\}$ equivalent to $\{b_n\}$ if and only if $\{a_n- b_n\}$ converges to 0. The essentially says that, for example, $\pi$ is "represented" by the infinite decimal 3.1415926.... From that definition, if a is an irrational number then there exist a sequence of rational numbers $r_1, r_2, r_3, ...$ that converges to a. We then define $x^a$ to be the limit of the sequence $x^{r_1}, x^{r_2}, x^{r_3}, ...$. Using the same example as before, $2^\pi$ is defined as the limit of the sequence $2^3, 2^{3.1}, 2^{3.14}, 2^{3.141}, 2^{3.1415}, 2^{3.14159}, 2^{3.141592}, 2^{31415926}, ...$.

| cite | improve this answer | |
  • 1
    $\begingroup$ One key point is that the limit of $x^{r_{\large n}}$ does not depend on which sequence $r_n$ converging to $a$ is chosen. $\endgroup$ – anon Nov 5 '17 at 21:41

My two cents:

The definition has to do with the fact that $\mathbb R$ is by definition the completion of $\mathbb Q$:

$5^\sqrt{2}=\sup\{5^a:a\in \mathbb Q \wedge a<\sqrt 2\}$

You may show that the set on the RHS is bounded and nonempty, so $\sup$ exists.

So your approach is right (calculate $5^a$ for successive approximations of $\sqrt 2$ like 1.4, 1.41 and so on, and it will converge to the result).

I've seen the theory developed the other way around (first study the series, then define elementary functions as series), but this approach seems very "unnatural" to me, as it is not how this functions were developed historically.

| cite | improve this answer | |

The exponential function is an order-preserving bijection over the rationals. Filling in the holes gives an exponential function over the reals that is an order-preserving bijection. This is done by letting a^i be the supremum of {a^(p/q) | (p/q)

| cite | improve this answer | |
  • 1
    $\begingroup$ If you used the proper TeX style formatting, your answer would likely be a bit clearer. $\endgroup$ – rajb245 Oct 31 '16 at 2:49
  • $\begingroup$ @rajb245 yeah i typed it clear but the damn compiler or whatever was written by a faschist. $\endgroup$ – Jacob Wakem Oct 31 '16 at 2:55
  • $\begingroup$ You just need to put dollar signs to make math Jax read it. You will also need to use \ to escape your curly braces for set notation. $\endgroup$ – Ian Oct 31 '16 at 2:56
  • $\begingroup$ @Alephnull: please use polite words, and don't call names. $\endgroup$ – P Vanchinathan Oct 31 '16 at 3:49
  • $\begingroup$ Especially if you don't know how to spell the names! $\endgroup$ – user247327 Nov 5 '17 at 21:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.