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Here is an elementary derivation (without calculus) of Euler's formula $e^{ix}=\cos{x}+i\sin{x}$, although not a strict proof or derivation. My question is:

Can we estimate the value of e in an elementary way? I mean without calculus, just like we estimate $3<\pi<2\sqrt{3}$ (or more precise).

Step 1

For any positive real number y ≠ 1 (e.g. y = 10), we have $y^i=a+bi$, and we also have $y^{-i}=a-bi$. [1]

Because $1=y^{i}y^{-i}=(a+bi)(a-bi)=a^2+b^2$, there should be a real number x matching $a=\cos{x}$ and $b=\sin{x}$. Then we get $y^i=\cos{x}+i\sin{x}$.

This also means, for any real number x ≠ 0, there should be a real number y matching $y^i=\cos{x}+i\sin{x}$.

So we can get the real number $e=y^{1/x}$ matching $e^{ix}=y^i=\cos{x}+i\sin{x}$. (Although we don't know the exact value of e here, and e may be dependent on x.)

Step 2

Make x = 1, we get $e^i=\cos{1}+i\sin{1}$.

Then we generalize de Moivre's formula $(\cos\theta+i\sin\theta)^n=\cos{n}\theta+i\sin{n}\theta$ to non-integer powers $(\cos\theta+i\sin\theta)^x=\cos{x}\theta+i\sin{x}\theta$. [2]

Make θ = 1, we get $(\cos{1}+i\sin{1})^x=\cos{x}+i\sin{x}$, and finally we get $e^{ix}=\cos{x}+i\sin{x}$.

This means, the unique real number e (although we still don't know the value here) matches x = 1 and all other values.

Notes

[1] Here I use a rule of Complex conjugate:

In general, if $\varphi$ is a holomorphic function whose restriction to the real numbers is real-valued, and $\varphi (z)$ and $\varphi (\overline {z})$ are defined, then

$\varphi({\overline z})=\overline{\varphi(z)}$

This rule doesn't look elementary. However, we can accept it intuitively. According to the definition of imaginary number i2 = −1, i can be replaced with -i in any equation, and vice versa.

[2] The generalization doesn't look elementary (may use some complex analysis) and may fail in some cases. See here.

I raised this question because I read The Feynman Lectures on Physics Vol. I Ch. 22: Algebra recently, where Feynman mentioned finally:

We summarize with this, the most remarkable formula in mathematics:

$e^{i\theta}=\cos\theta+i\sin\theta$ (22.9)

This is our jewel.

I think Feynman's derivation is redundant so I tried to derive it in a simple way beginning from de Moivre's formula.

auntyellow
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    Define elementary, would you be satisfied with the series definitions of $\cos(1)$ and $\sin(1)$? You could use those to define $e^{i}$ – QC_QAOA Jun 02 '21 at 03:30
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    You could define $e^{ix}$ as the solution to the IVP $$y''+y=0\ y(0)=1 \y'(0)=i$$ – Matthew H. Jun 02 '21 at 03:39
  • For an alternative viewpoint, see this answer. – user2661923 Jun 02 '21 at 03:43
  • @MatthewPilling You don't have to define that, you can simply DERIVE that by stating $y=e^{ix}$ and then take a derivative twice. This would indeed be my alternative approach for arriving at the OP's result – imranfat Jun 02 '21 at 03:53
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    I don't think it can be done without calculus. It seems unreasonable to expect that topics from the Wiki link would suffice when complex numbers themselves aren't even included the link. – Stuck Jun 02 '21 at 04:46
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    How do you even define $e$ without calculus? – Zeekless Jun 02 '21 at 06:50
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    You can't even formulate the claim using the mathematics from https://en.wikipedia.org/wiki/Elementary_mathematics There are no complex numbers in there, let alone exponentiation with complex powers. – Magdiragdag Jun 02 '21 at 06:50

1 Answers1

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You can define a function $f(x)=\dfrac{\cos x+i\sin x}{e^{ix}}$. If you know the quotient rule, then $$f'(x)=\frac{e^{ix}(-\sin x+i\cos x)-(\cos x+i\sin x)\cdot ie^{ix}}{e^{2ix}}=0$$. Since the derivative of the function is $0$, therefore the function must be equal to some constant $C$. Therefore, $$\frac{\cos x+i\sin x}{e^{ix}}=C$$, where we can find the constant by putting $x=0$. $$C=\frac{\cos 0+i\sin 0}{e^{0i}}=1$$. Hence, the Euler's formula is proved.

I don't know if this matches your 'elementary' definition but this is an alternative way to using Maclaurin series.