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This question is a about (if not proving) at least guessing the Euler's formula. I don't want the proof using the infinite sums.

We can guess by logic that for example that the equation $x^2+1=\sqrt{x}$ has no real solutions because $x^2=\sqrt{x}$ has 2 solutions $x=0, x=1$ but by adding 1 on the left side, we cancel these 2 solutions, so there are no solutions.

I want to know if there is a way to guess by logic that $e^{ix}=\cos(x)+i\sin(x)$. I guess that the most important here here will be $\frac{d}{dx}e^x=e^x$. And suggestions?

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  • $\begingroup$ I'm not sure what you mean by "logical" here. Does it have to do with mathematical logic (i.e. formulating the argument in a certain logic and language, and some particular theory), or are you just looking for an explanation appealing intuitive logic about what's correct? $\endgroup$
    – Asaf Karagila ♦
    May 5, 2013 at 17:28
  • $\begingroup$ You need to choose a definition for the exponential. You don't want to us the series definition, so which would you like? $\endgroup$
    – vadim123
    May 5, 2013 at 17:29
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    $\begingroup$ @AsafKaragila It's actually your second statement explanation appealing intuitive logic about what's correct. $\endgroup$
    – moray95
    May 5, 2013 at 17:32
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    $\begingroup$ The equation $x^2+1 = \sqrt{x}$ does have solutions. Squaring both sides and rearranging gives $x^4 + 2x^2 -x + 1 = 0$. By the fundamental theorem of algebra, this has - when counted with multiplicity - four solutions over the complex plane. The solutions to $x^2 + 1 = \sqrt{x}$ will be solutions of $x^4 + 2x^2 -x + 1 = 0$, although the converse need not be true. $\endgroup$
    – Fly by Night
    May 5, 2013 at 17:41
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    $\begingroup$ I think a nice (and rigorous) way is showing that $f(\theta)=\cos\left(\theta\right)+i\sin\left(\theta\right)$ satifies both $f'\left(\theta\right)=i\cdot f\left(\theta\right)$ and $f\left(0\right)=1$. Showing that $\left(f\left(\theta\right)\right)^n=f\left(n\theta\right)$ also feeds intuitions. $\endgroup$
    – Ian Mateus
    May 5, 2013 at 18:19

2 Answers 2

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The power series argument, while simple, is indeed unenlightening. You can easily show that $f(\theta)=\cos\left(\theta\right)+i\sin\left(\theta\right)$ satifies both $f'\left(\theta\right)=i\cdot f\left(\theta\right)$ and $f\left(0\right)=1$, and it looks remarkably similar to one definition of $\gamma(t)=e^{\alpha t}$: the function $\gamma\,\colon\mathbb{C}\rightarrow\mathbb{C}$ that both $\left(e^{\alpha t}\right)'=\alpha e^{\alpha t}$ and $\gamma(0)=1$.

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This isn't a proof, but an illustration as why the formula is defined in that way:

Consider $h(z) = A^z$, A > 0. This is well-defined for z real, and you want to extend it to when z is complex. It's enough to define it whenever z is pure imaginary.
Define $f(x) := Re(h(ix))$, $g(x) := Im(h(ix))$ So $h(ix) == f(x) + i*g(x)$

1) $h(0) == 1$, so $f(0) == 1$, $g(0) == 0$.

2) The distinction between i and -i is arbitrary, so when extending real-valued functions to the complex numbers, you usually want $h(\bar{z}) == \bar{h(z)}$, so $f(-x) == f(x)$, $g(-x) == -g(x)$

3) $A^m * A^n == A^{m+n}$, so $h(i(x+y)) == h(ix) * h(iy)$ $f(x+y) + i*g(x+y) == (f(x) + i*g(x)) (f(y) + i*g(y))$ $f(x+y) + i*g(x+y) == f(x)*f(y) + i*g(x)*f(y) + i*f(x)*g(y) - g(x)*g(y)$ $f(x+y) + i*g(x+y) == (f(x)*f(y) - g(x)*g(y)) + i*(g(x)*f(y) + f(x)*g(y))$

Equating real and imaginary parts: $f(x+y) == f(x)*f(y) - g(x)*g(y)$ $g(x+y) == f(x)*g(h) + f(y)*g(x)$

Those formulas should be very familiar, and they suggest that $A^{ix}$ behaves very much like $cos(x) + i*sin(x)$.

($cos(kx) + i*sin(kx)$) satisfies all the formulas for any $k$. Determining which value of $k$ corresponds to a particular value of $A$ requires calculus or the equivalent.)

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