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**: $$ e^{i\pi} + 1 = 0 $$
