Please Register or Log in to view the hidden image! so e exp (i Pi) = -1 Cos Pi =-1 e exp (i Pi) = cos Pi is this true?
From e^(i*pi) = -1 you can write exp(i*pi) = -1, since exp(z) is a synonym for e^z. But e * exp(i*pi) = -1 doesn't follow, unless you mean exp(i*pi) with incorrect notation. But yes, since exp(i*pi) = -1, it's also equal to cos(pi), or 1-2, or 47-48, or anything else that evaluates to -1.
The \(\cos\) function is defined in analysis as: \(\cos x \equiv \frac{e^{ix} + e^{-ix}}{2} = \frac{1 + e^{2 i x}}{2 e^{ix}}\) [from which a proof that \(\cos ( x ) = \cos (-x)\) comes automatically]. So \(\cos \pi = \frac{e^{i\pi} + e^{-i\pi}}{2} = \frac{1 + \left(e^{i \pi}\right)^2}{2 e^{i\pi}}\) But since \(e^{i \pi} + 1 = 0\) it follows that \(e^{i \pi} = -1\) and therefore \(\cos \pi = \frac{1 + \left(-1\right)^2}{2 \, \times \, -1} = \frac{2}{-2} = -1\) just by happenstance.* They are related by their use of the complex exponential (or similarly, by the respective Taylor expansions of \(e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}\) and \(\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}\)). Related to this, is the equation \(\frac{1 + x^2}{2 x} = x\) which has solutions only at \(x = -1, \; x =1\). So \(\forall n,m \in \mathbb{Z} \quad \cos (2n \pi) = e^{2 i m \pi}, \; \cos ((2n + 1) \pi) = e^{(2 m + 1 ) i \pi}\) are more general observations. * Or is it entirely happenstance? \(\cos x \equiv \frac{e^{ix} + e^{-ix}}{2} = \frac{1 + e^{2 i x}}{2 e^{ix}}\) \(\sin x \equiv \frac{e^{ix} - e^{-ix}}{2 i} = - \frac{1 - e^{2 i x}}{2 e^{ix}}\) and \(\pi\) can be defined as the first positive root of \(\sin x = 0\). Therefore \(2 i \sin \pi = e^{i\pi} - e^{-i\pi} = e^{i\pi} - \frac{1}{e^{i\pi}}= 0\). Therefore it follows that \(e^{i\pi} = \frac{1}{e^{i\pi}} = e^{-i\pi}\) and so \(\cos \pi = \frac{e^{i\pi} + e^{-i\pi}}{2} = \frac{2 e^{i\pi} }{2} = e^{i\pi}\) from the definitions of \(\pi, \; \cos, \; \textrm{and} \; \sin\) without using Euler's formula or calculating exactly what the value of \(e^{i\pi}\) is.
The connection between \(e^{ix}\) and \(\cos x\) is the real point. Everyone who takes a course in analysis learns these complex number identities and most come into class knowing them: \(e^x \quad \equiv \quad \sum_{k=0}^{\infty} \frac{x^k}{k!} \quad = \quad \cosh \, x \; + \; \sinh \, x \quad = \quad \cos (ix) \; - \; i \, \sin (ix) \\ e^{ix} \quad = \quad \cosh \, ix \; + \; \sinh \, ix \quad = \quad \cos \, x \; + \; i \, \sin \, x \\ e^{-ix} \quad = \quad \cosh \, ix \; - \; \sinh \, ix \quad = \quad \cos \, x \; - \; i \, \sin \, x \\ \cosh \, x \quad \equiv \quad \frac{e^{x} + e^{-x}}{2} \quad = \quad \sum_{k=0}^{\infty} \frac{x^{2k}}{(2k)!} \quad = \quad \cos (ix) \\ \sinh \, x \quad \equiv \quad \frac{e^{x} - e^{-x}}{2} \quad = \quad \sum_{k=0}^{\infty} \frac{x^{2k+1}}{(2k+1)!} \quad = \quad -i \, \sin (ix) \\ \cos \, x \quad \equiv \quad \frac{e^{ix} + e^{-ix}}{2} \quad = \quad \sum_{k=0}^{\infty} \frac{(-1)^k \, x^{2k}}{(2k)!} \quad = \quad \cosh (ix) \\ \sin \, x \quad \equiv \quad \frac{e^{ix} - e^{-ix}}{2i} \quad = \quad \sum_{k=0}^{\infty} \frac{(-1)^k \, x^{2k+1}}{(2k+1)!} \quad = \quad -i \sinh (ix)\) Since for any real number x, \(\cos x = \Re \, e^{ix}\), then it follows that if \(\Im \, e^{ix} = \sin x = 0\) then \(\cos x = e^{ix}\). This is hardly surprising to a student of analysis.
Excellent as always, rpenner. Fully wired, to say the least! Please Register or Log in to view the hidden image! Locust, The identity I committed to memory long ago is the third line of rpenner's post immediately above* e[sup]-iθ[/sup] = cos(θ) - i sin(θ) From this you have e[sup]iπ[/sup] = cos(π) + i sin(π) = -1 + 0 = -1 ∴ e[sup]iπ[/sup] + 1 = 0 *good way to remember it, since it eliminates possible confusion over how to treat the minus sign.
You seem like a pretty smart guy rpenner mayby you can help me out? Check out my thread it's in the math and physics section under help needed.
