Proving Euler's identity, $e^{i\pi} + 1 = 0$, can be demonstrated using LaTeX to typeset the mathematical steps clearly. The proof relies on the Maclaurin series expansions of the exponential, sine, and cosine functions. ** Maclaurin Series Expansions First, we recall the Maclaurin series for $e^x$, $\sin(x)$, and $\cos(x)$: $$ \begin{align} e^x &= 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots \\ \sin(x) &= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots \\ \cos(x) &= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots \end{align} $$ * Deriving Euler's Formula Next, we substitute $x = i\theta$ into the Maclaurin series for $e^x$, where $i$ is the imaginary unit ($i^2 = -1$). $$ \begin{align} e^{i\theta} &= 1 + (i\theta) + \frac{(i\theta)^2}{2!} + \frac{(i\theta)^3}{3!} + \frac{(i\theta)^4}{4!} + \dots \\ &= 1 + i\theta - \frac{\theta^2}{2!} - \frac{i\theta^3}{3!} + \frac{\theta^4}{4!} + \dots \end{align} $$ Now, we separate the real and imaginary terms: $$ e^{i\theta} = \left( 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \dots \right) + i \left( \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \dots \right) $$ By comparing this with the Maclaurin series for $\cos(\theta)$ and $\sin(\theta)$, we can see that: $$ e^{i\theta} = \cos(\theta) + i\sin(\theta) $$ This is Euler's formula. Proving Euler's Identity To prove Euler's identity, we set $\theta = \pi$ in Euler's formula: $$ e^{i\pi} = \cos(\pi) + i\sin(\pi) $$ We know that $\cos(\pi) = -1$ and $\sin(\pi) = 0$. Substituting these values gives us: $$ e^{i\pi} = -1 + i(0) $$ $$ e^{i\pi} = -1 $$ Finally, adding $1$ to both sides yields Euler's identity**: $$ e^{i\pi} + 1 = 0 $$
