Domanda

Please provide me with a method to prove Euler's formula

Pensiero

Exploring Euler’s formula proofs

The user wants a method to prove Euler's formula, but the term is ambiguous. It could mean either the complex exponential identity $e^{i\theta} = \cos\theta + i\sin\theta$ or the polyhedron formula $V - E + F = 2$ . I’ll tackle the first formula by providing a necessary proof using the power series expansion. I'll break it down clearly.

Using power series proof

To prove $e^{i\theta} = \cos\theta + i\sin\theta$ , I’ll use the power series approach. First, write the power series for $e^x$ , $\sin x$ , and $\cos x$ , then substitute $x \rightarrow i\theta$ and separate real and imaginary parts. I’ll keep the explanation concise, without extra details, but mention that other methods, like differential equations, exist.

Power series proof

I’ll start by using power series for $e^x$ , $\sin x$ , and $\cos x$ , and substitute $x = i\theta$ . By separating the even and odd powers, I show that $e^{i\theta} = \cos\theta + i\sin\theta$ . I’ll justify the use of power series based on absolute convergence, but also mention the alternative differential equation method for completeness.

Choosing the power series method

Since the user asked for a method, I’ll focus on the power series approach and keep explanations clear and concise. I’ll start by defining the Maclaurin series for $e^x$ , $\sin x$ , and $\cos x$ for real $x$ , noting their convergence. Then I'll substitute $x = i\theta$ , step by step.

Thought for 39s

Risposta

Here’s a clean, standard way—using power series—to prove Euler’s formula

e^{i\theta}=\cos\theta+i\sin\theta.

Power-series method (rigorous and short)

Start from the Maclaurin series (radius of convergence is ∞):

e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!},\quad \cos x=\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{(2n)!},\quad \sin x=\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!}.

Substitute $x=i\theta$ into the exponential series (this is valid because the series converges absolutely for all complex $x$ ):

e^{i\theta}=\sum_{n=0}^{\infty}\frac{(i\theta)^{n}}{n!} =\sum_{n=0}^{\infty}\frac{i^{n}\theta^{n}}{n!}.

Split into even and odd powers, using $i^{2n}=(-1)^n$ and $i^{2n+1}=i(-1)^n$ :

e^{i\theta} =\sum_{n=0}^{\infty}\frac{(-1)^n\theta^{2n}}{(2n)!} +i\sum_{n=0}^{\infty}\frac{(-1)^n\theta^{2n+1}}{(2n+1)!}.

Recognize the two series:

e^{i\theta}=\cos\theta+i\sin\theta.

That’s it. (As a corollary, set $\theta=\pi$ to get $e^{i\pi}+1=0$ .)

If you’d like a different “method”: there’s also the ODE approach—define $f(\theta)=\cos\theta+i\sin\theta$ . Then $f'(θ)=i f(θ)$ with $f(0)=1$ . The function $g(θ)=e^{iθ}$ satisfies the same differential equation and initial condition; by uniqueness of solutions, $f\equiv g$ , giving Euler’s formula.

Please provide me with a method to prove Euler's f...