Euler’s Formula Without Calculus, pt. 1 – Exponentiation

part two

I took a break from studying today to see whether I could prove
Euler’s Formula
without using any calculus concepts – Taylor series, differential equations and the like. Incidentally, according to MathWorld, the formula first appeared in 1714, when Euler was seven (or possibly six) years old! Euler was good, but not that good. It was published by a different guy (R. Coles, in case in comes up on Jeopardy or during sex or something).

Our goal will ultimately be to discover the value of

(a + b\imath)^{c+d\imath} .

Also, we’ll stop along the way to figure out what that \imath thing is. Unfortunately the expression we’re investigating has no meaning until you define it. But fortunately, most definitions would suck. That means we can narrowly restrict our choices for the definition by cutting out all the sucky ones. Let’s get started not sucking.

If a is a real number and b is a positive integer, then define

a^b \equiv a*a*a*\ldots*a \,\,\,\,\, \left[b \textrm{ times}\right] .

This yields two important properties. If a is a real number and b, c are positive integers, then by our definition

a^b*a^c = (a*a*\ldots*a)*(a*a*\ldots*a) = (a*a*a*a*\ldots*a*a) = a^{b+c}

(a^b)^c = a^b*\ldots*a^b = (a*a*\ldots*a)*\ldots*(a*a*\ldots*a) = a^{bc}

That makes exponentiation by integers easier, but the story is just beginning. Now let’s make a new definition. Let

a^{-b} = \frac{1}{a^b} .

Why? Because if you do that

a^{-b}*a^c = \frac{a^c}{a^b} = \frac{a*a*\ldots*a}{a*a*\ldots*a} = a^{c-b} ,

which is nice because it means our property

a^b*a^c = a^{b+c}

still works, even for negative numbers. That’s progress, but we don’t yet know how to do something like

4^{\frac{1}{2}}

because we only defined exponentiation for integers. If we want to evaluate fractional exponents, we first need to say what that means. But let’s figure out a definition that doesn’t suck by demanding that the two properties we found about integer exponents hold for all fractions, too. If this is true, then

(4^{\frac{1}{2}})^2 = 4^{2*\frac{1}{2}} = 4^1 = 4 .

So if

4^{\frac{1}{2}} = x ,

then x is a number such that

x^2 = 4

which means

x=\pm 2 .

That doesn’t give us a good definition, because we wanted to make a function (well, I did, at least). A function gobbles up one thing (at a time) and spits out one thing. The function can’t be wishy-washy and tell us the answer is positive two or minus two. Let’s choose. We’ll make it positive two because, come on, did you seriously think I would pick minus two? If you did, you are probably an electron.

Generalizing,

(a^{\frac{1}{b}})^b = a

so

a^\frac{1}{b}

is the number x such that

x^b = a ,

which is commonly called the bth root of a.

There could be a problem here. If a is positive, it has a square root, a cube root, a fourth root, and all the other all the way down (they get closer and closer to one whether a itself is less than or greater than one). But if a is negative, it doesn’t have a square root. It has a cube root, but let’s ignore that fact for now. At the moment, we’re only going to define the exponential of positive numbers.

We can continue generalizing using the rules.

a^{\frac{p}{q}} = \left(a^{\frac{1}{q}}\right)^p for a > 0 .

This definition could have gone the other way. What if we defined

a^{\frac{p}{q}} = \left(a^p\right)^{1/q} ?

Would these be equivalent? Yes. It’s obvious, but can you prove it? I’m not going to do everything here. We’re almost 500 words in and I still haven’t gotten to the interesting part.

That gives us exponentials with a base that’s a positive real and an exponent that’s a rational. To take a number to a real, non-rational exponent (like \pi), you’d need to use the same sort of methods that are used to define the reals (limits of a sequence of rationals or repeated fractions or something funny like that. Having never studied analysis I don’t know precisely how it’s supposed to work.)

You may be somewhat bothered by the requirement that the base a is required to be positive. Well we can let it be zero by defining

0^x = 0 for all x \neq 0

and leaving 0^0 as “undefined” or “indeterminate”.

But you really wanted to know, what is {-9}^{\frac{1}{2}}? For everything we’ve done so far, the following rule holds:

a^x*b^x = (ab)^x .

The rule is obvious for repeated multiplication, and kind of cascades down from there until it works for any exponent x we want. So demand that the rule continues to hold for negative bases.

{-9}^{\frac{1}{2}} = (-1*9)^{\frac{1}{2}} = {-1}^{\frac{1}{2}}*9^{\frac{1}{2}} = {-1}^\frac{1}{2}*3

The same would be true for any negative number. If we could only figure out what {-1}^{\frac{1}{2}} is, we’d be all set to take the square root for any real base (i.e. all your base would be belong to us).

There are no real numbers x such that x^2 = -1, because all positive numbers square to positive numbers, and negative numbers square to positive numbers, and zero squares to itself. That pretty much exhausts the choices.

So we’ll just invent a new symbol, called the imaginary unit, which looks like \imath. Then we’ll assert \imath^2 = -1. This is totally sleazy, because it doesn’t make a damn bit of sense to say some random symbol you just invented squares to a number. It isn’t a number, so how can two of them multiplied together make a number?

Never fear; it’s possible to build the complex numbers out of the real numbers. Define a complex number as an ordered pair of real numbers (a,b), with addition defined by

(a,b) + (c,d) = (a+c, b+d)

and multiplication defined by

(a,b)*(c,d) = (ac - bd, bc+ad) .

Then the ordered pairs satisfy the axioms for a field, since you can find multiplicative and additive identities and inverses. The only non-trivial one is the multiplicative inverse, which comes from playing around with the algebra until you discover

(a,b)*(\frac{a}{a^2+b^2},\frac{-b}{a^2+b^2}) = (1,0) ,

with (1,0) the multiplicative identity. There appears to be some significant Pythagorean Theorem stuff going on there. That suggests a geometric interpretation to the complex numbers that involves some right angles. There is one (at least). Watch this movie. I certify it to be very very nice. (Complex numbers are movie 5, but they’re all worth your time. That’s probably not saying too much, though, since you apparently spend all your time reading blogs.)

Associating the real numbers with a complex number by

x \to (x,0) ,

The ordinary rules for addition and multiplication of real numbers still work (and, because exponentiation was based on those rules, it holds, too). The complex numbers have subsumed the reals.

In practice, nobody really bothers with the ordered pair business, and a general complex number (a,b) is just written as a + b\imath, where the \imath is treated as a variable you have to carry around, until you square it, at which point you can make the substitution

\imath^2 = -1 .

This reproduces the structure of the ordered pair definition of complex numbers, and has the bonus of being easier for humans to work with.

The next task will be to examine

x^{a+b\imath}

with x a real number. But that’s the next post, and for now I’m leaving you hangi

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2 Responses to “Euler’s Formula Without Calculus, pt. 1 – Exponentiation”

  1. Euler’s Formula Without Calculus, pt. 2 - « Arcsecond Says:

    [...] Arcsecond As Simple As Improbable, But No Less Unconfusing « Euler’s Formula Without Calculus, pt. 1 – Exponentiation [...]

  2. Euler’s Formula without Calculus pt. 3 - The Natural Logarithm « Arcsecond Says:

    [...] pt. 3 – The Natural Logarithm By meichenl Let’s recapitulate the previous two posts (one and [...]

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