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$\DeclareMathOperator\cis{cis}$ I define $e^x$ to be $$\lim_{n\to\infty} \left( 1+\frac{x}{n} \right)^n,$$ so $$e^{i\theta} = \lim_{n\to\infty} \left( 1+\frac{i\theta}{n} \right)^n.$$

Now I express $1+\dfrac{i\theta}{n}$ in polar form. The magnitude will be $$\sqrt{1^2+ \left( \frac{\theta}{n} \right)^2} = \sqrt{1+ \frac{\theta^2}{n^2} },$$ the argument will be $$\tan^{-1} \left( \dfrac{\theta/n}{1} \right) = \tan^{-1} \left( \dfrac{\theta}{n} \right).$$ So $$e^{i\theta} = \lim_{n\to\infty} \left( \sqrt{1+ \frac{\theta^2}{n^2} } \left( \cis \left( \tan^{-1} \left( \frac{\theta}{n} \right) \right) \right) \right)^n.$$ By De Moivre’s theorem we get \begin{align} e^{i\theta} &= \lim_{n\to\infty} \left( \left( \sqrt{1+ \frac{\theta^2}{n^2} } \right)^n \left( \cis \left( n\tan^{-1} \left( \frac{\theta}{n} \right) \right) \right) \right) \\ &= \left( \lim_{n\to\infty} \left( \sqrt{1+ \frac{\theta^2}{n^2} } \right)^n \right) \cis \left( \lim_{n\to\infty} \left(n\tan^{-1} \left( \frac{\theta}{n} \right) \right) \right). \end{align} Using L’Hopital’s rule you can find $$\lim_{n\to\infty} \left( \sqrt{1+ \frac{\theta^2}{n^2} } \right)^n = 1$$ and $$\lim_{n\to\infty} \left(n\tan^{-1} \left( \frac{\theta}{n} \right) \right) = \theta,$$ so \begin{align} e^{i\theta} &= 1 \cdot \cis(\theta) \\ &= \cos\theta+i\sin\theta, \end{align} which is Euler’s identity. However, I am now questioning whether this is valid or not, as I can not find similar proofs online. This proof is the easiest to understand for me, so considering I can’t find much like it, it is probably wrong.

Any help would be appreciated.

Rócherz
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Jaden Amatuzzo
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1 Answers1

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If I understand your posting, you are attempting to prove Euler's assertion that

$$e^{i\theta} = \cos(\theta) + i\sin(\theta). \tag1 $$

In fact, it is impossible for your attempt to be valid, because the underlying idea is not an assertion. Instead, the underlying idea, as represented in (1) above, is nothing more than syntactic sugar.

You are starting from the premise that for any Real number $x$, that

$$e^x = \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n. \tag2 $$

The assertion in (2) above is absolutely true (i.e. has been proven), in Real Analysis, for any Real number $x$. However, without some arbitrary definition (i.e. syntactic sugar), any expression of the form

$$e^{(x + iy)} ~: x,y \in \Bbb{R}$$

is gibberish whenever $y \neq 0.$

Therefore, the assertion can't be proven, because the assertion, in an of itself, is gibberish.

Euler decided to arbitrarily define

$$e^{(x + iy)} = e^x \times [\cos(y) + i\sin(y)].$$

Euler noticed that if he did this, that the exponent law (for example) :

$$e^{x_1} \times e^{x_2} = e^{(x_1 + x_2)}$$

still held, when complex numbers $z_1, z_2$ were substituted for $x_1, x_2.$

He then noticed that the behavior of certain limits, when the Real variable $x$ was replaced with the complex variable $z$ were well behaved.

So, the attack on certain Complex Analysis Math problems was facilitated.

However, the idea that you are attempting to prove, can't be proven. That is, without the arbitrary definition, dipped in syntactic sugar, the idea is gibberish.

user2661923
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  • I don't see anything wrong or unjustified with taking $\lim (1+i\theta / n)^n$ as a definition. All we need is complex number arithmetic (which immediately grants us integer powers) and to evaluate the limit as $|(1+i\theta/n)^n-L|<\varepsilon$ using the complex modulus. We don't need foreknowledge of the complex exponential (nor its properties) for that. – Jam Oct 13 '22 at 09:44
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    @Jam If you define $$e^{i\theta} ~: ~\theta \in \Bbb{R}$$ as $$\lim_{n \to \infty} \left(1 + \frac{i\theta}{n}\right)^n,$$ then yes, this definition can be shown to be equivalent to defining $$e^{i\theta} = \cos(\theta) + i\sin(\theta).$$ However, this is all cart before the horse. You are still forced to assume some arbitrary definition of $$e^{i\theta} ~: ~\theta \in \Bbb{R},$$ without which, you are left with gibberish. What you can not do is justify the arbitrary definition by noticing that the corresponding assertion is true when $(i\theta)$ is replaced by $x \in \Bbb{R}.$ – user2661923 Oct 13 '22 at 09:50
  • @user26661923 thanks for the answer, i just have two questions, 1), you say that euler arbitrarily decided to define e^(x+iy) as e^x*(cos(y)+isin(y), but i heard that euler discovered it via a taylor maclaurin series, so that doesn't seen arbitrary. 2) if it is impossible to prove eulers identity, and my proof proved eulers identity, then there must be something wrong with my proof, do you know what my error is? – Jaden Amatuzzo Oct 13 '22 at 09:53
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    @JadenAmatuzzo Euler discovered that if he made the corresponding arbitrary definition, that the Real Analysis Taylor MacLaurin Series could be consistently extended into the realm of Complex Analysis. The facto that this extension dovetailed nicely with Real Analysis and facilitated the attack of Complex Analysis Math problems does not imply that the Taylor Macluren Series was proven to be extendable into Complex Analysis. Instead, it merely showed that if you adopted an arbitrary convenient definition, nice things happened. ...see next comment – user2661923 Oct 13 '22 at 09:58
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    @JadenAmatuzzo As far as finding something wrong with your proof, it is impossible for your proof or any corresponding proof to be valid, without some arbitrary definition of $$e^{i\theta} ~: ~\theta \in \Bbb{R}.$$ There is nothing in Real Analysis that leads to such an arbitrary definition, which is nothing more than consistently extending Real Analysis ideas into the realm of Complex Analysis. Without any such arbitrary definition of $e^{i\theta}$, any assertions that involve $e^{i\theta}$ must be regarded as gibberish, because $e^{i\theta}$ is itself gibberish. – user2661923 Oct 13 '22 at 10:01
  • @user2661923 thank you for answering, i think i misinterpreted your answer as saying that eulers identity was arbitrary, rather than the definition of e^x for complex numbers was arbitrary and eulers identity was a result of that arbitrary choice. one final question if I arbitrarily assume e^z = (1+z/n)^n where z is a complex number, is this proof correct? – Jaden Amatuzzo Oct 13 '22 at 10:09
  • @JadenAmatuzzo I know that if you start with the definition of $$e^z = \lim_{n\to \infty} \left(1 + \frac{z}{n}\right)^n ~: ~z = (x + iy), ~x,y \in \Bbb{R},$$ then you can prove that $$e^z = e^x \times [\cos(y) + i\sin(y)].$$ Whether each step in your posting happens to be valid, or whether there is some subtle analytical omission in your posting, I can't really say. My knowledge of Complex-Analysis/Real-Analysis is not deep enough to have an opinion on that specific issue. – user2661923 Oct 13 '22 at 10:16