A conflicting result of Euler's formula

Undoubtedly0

It can be shown from Euler's formula that

[tex] \left(-1\right)^x = \cos(\pi x) + i\sin(\pi x) [/tex]

However, consider [itex] x = 2/3 [/itex]. The left expression gives

[tex] \left(-1\right)^\frac{2}{3} = \left(\left(-1\right)^2\right)^\frac{1}{3} = \left(1\right)^\frac{1}{3} = 1 [/tex]

while the right expression gives

[tex] \cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right) = -\frac{1}{2} + \frac{\sqrt{3}}{2}i [/tex]

What has gone wrong here? Thanks all.

andrien

because you have not done it in general.
(-1)^x=cos(n*pi*x)+isin(n*pi*x) where n is odd integer.now put x=2/3 and n=3.
you get it equal to 1.also check out that -1 has three cube roots so just check out that that complex number is arising because of other roots.

DonAntonio

Complexpowers, roots don't behave as real ones. Read about multivaluate complex functions...or wait until the appropiate course at university.

DonAntonio

Undoubtedly0

How does one derive [itex] (-1)^x=\cos(n\pi x)+i\sin(n\pi x) [/itex]?

To derive [itex] (-1)^x=\cos(\pi x)+i\sin(\pi x) [/itex] I simply used
[tex] \ln\left((-1)^x)\right)=x\ln(-1)=x\pi i [/tex]
therefore [itex] (-1)^x = e^{x\pi i} = \cos(\pi x)+i\sin(\pi x) [/itex].

Where did I miss the general case? Thanks again.

andrien

can't you just verify that -1 is equal to exp(i*n*pi) when n is odd.

Techian

The equation, (-1)^x=cos(pi*x)+i*sin(pi*x) is derived from Euler's identity based on the assumption that 'x' is an integer. Hence substituting fractional value for 'x' in the equation will lead to erroneous results. The derivation is as follows,
e^(i*y)=cos(y)+i*sin(y).....(Original Euler's identity)
Now replacing 'y' with 'pi*x', you will end up with
e^(i*pi*x)=cos(pi*x)+i*sin(pi*x)......(1)
Assuming that 'x' is an integer, cos(pi*x)=(-1)^x and sin(pi*x)=0.
substituting them in (1)
e^(i*pi*x)=(-1)^x.......(2)
substituting (2) in (1), you will end up with (-1)^x=cos(pi*x)+i*sin(pi*x) for x is an integer.