1
$\begingroup$

At the start of my first year of university in a Foundations module, my lecturer stated Euler's formula for complex numbers, $e^{i \theta} = \cos\theta + i \sin\theta$. In passing he mentioned this is just a definition which conveniently works when you plug the numbers into the Taylor expansions and that you could actually define it differently (a number other than $e$). What would be the consequences of this? Is there any other number which gives interesting results?

$\endgroup$
  • 2
    $\begingroup$ Einstein first tried $E=m\,a^2$, then $E=m\,b^2$, and when he arrived at $E=m\,c^2$ he cried: Heureka! $\endgroup$ – Christian Blatter May 19 '18 at 18:31
2
$\begingroup$

Suppose you want the following desirable properties:

  1. $\operatorname{exp}:\theta \mapsto e^{i\theta}$ is smooth

  2. $\exp(x+y)=\exp(x)\cdot\exp(y)$

the first requirement is reasonable since we want the function to be nice, and the second is desirable, because it says most essentially that if we add angles, we get

$(\cos(x)+i\sin(x))(\cos(y)+i\sin(y))$

which after working out with some trigonometric identities reduces to $\cos(x+y)+\sin(x+iy)$

so that adding angles does what we would expect on the unit circle.

For any function with these two properties: $f(0)=1$ and differentiating with respect to $x$ and letting $x=0$ gives $$f^{\prime}(y) =f(y)\cdot f^{\prime}(0)$$

which solving returns an exponential function $Ce^{a \theta}$ where $a=f^{\prime}(0)$, although the requirement that $f(0)=1$ forces that $C=1$.

In this case, one could in principle define exponential functions with different bases, but this seems to me inconvenient.

The special thing about base $e$ is that the derivative at $0$ is precisely $1$, which is why we get the clean formulas.

$\endgroup$
1
$\begingroup$

The simplest possible answer is that your question doesn't make any sense. If we take the point of view that Euler's formula is a definition, then the $e$ in $e^{i\theta}$ isn't really Euler's number anymore, it's just a symbol. The notation $e^{i\theta}$, at that point, is no more meaningful than $f(\theta)$ - it's just a shorthand for $\cos\theta+i\sin\theta$. Thus it doesn't make sense to talk about "replacing $e$ with a different number".

For a slightly more nuanced answer, we can look in more detail at where Euler's formula comes from. In particular, most people would not take it as a definition. The spirit of what your professor said is true, since there are arbitrary (for the purposes of this conversation) definitions involved here, but they happen a few steps before you get to Euler's formula.

First, we define the exponential of real numbers. We say that $a^n$ is $a$ multiplied by itself $n$ times, and then extend that to rational exponents via roots and to irrational numbers by continuity. Once we've done all that, we find that there's an infinite series formula for exponentiation:

$$a^x=\sum_{n=0}^\infty\lambda_a^n\frac{x^n}{n!}$$

where $\lambda_a$ is some peculiar number that depends on $a$ in a complicated way. Now we ask which value of $a$ we would need to pick to make the $\lambda$ go away, and we find the answer is

$$a=e\approx2.718$$

Now, if we want to raise a number to a complex exponent, we can just try setting $x$ to a complex number in the power series. This is just a definition. Another thing that's just a definition is that, since we also have power series for $\sin(x)$ and $\cos(x)$, we can use the same idea to define $\sin(x)$ and $\cos(x)$ for complex $x$. However, given all of those definition, Euler's formula is true. It is a theorem, not a definition. The formula states:

$$e^{i\theta}=\cos(\theta) + i\sin(\theta)$$

Expanding each of those functions in terms of our power series definitions, we get:

$$\sum_{i=0}^\infty\frac{(i\theta)^n}{n!}=\sum_{\text{even }n}\left(\pm\frac{\theta^n}{n!}\right)+i\sum_{\text{odd }n}\left(\pm\frac{\theta^n}{n!}\right)$$

where the $\pm$ symbols mean the signs of the terms alternate (starting with $+$). Even just glancing at this equation, you should be able to see how the two right hand sums will "slot together" (or maybe "zip together" would be more descriptive) to form the sum on the left. If you work out the details, you'll see that the various $i$-s and $\pm$-s will cancel out in the right way to make the equation true.

If we replaced $e$ with a different number $a$, then on the left we would have the $\lambda_a$ factors, and the equation would not be correct at all.

$\endgroup$

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.