- #1

- 25

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Can someone help explain this? Wolfram says it is zero but I don't know why?

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- Thread starter powerplayer
- Start date

- #1

- 25

- 0

Can someone help explain this? Wolfram says it is zero but I don't know why?

- #2

- 650

- 3

Perhaps this will help: http://en.wikipedia.org/wiki/Euler's_formula

- #3

- 25

- 0

Ok I know eulers but how does 1^x - (-1)^x = 0?

- #4

- 650

- 3

That's ##\left(-1\right)^x-\left(-1\right)^x##, after some simplification.

- #5

Mentallic

Homework Helper

- 3,798

- 94

Ok I know eulers but how does 1^x - (-1)^x = 0?

[tex]e^{ix}=\cos(x)+i \sin(x)[/tex] hence after plugging [itex]x=-\pi[/itex] we get [tex]e^{-i\pi}=\cos(-\pi) +i \sin(-\pi)[/tex] and recall that [itex]\cos(-x)=\cos(x)[/itex] and [itex]\sin(-x)=-\sin(x)[/itex] thus we have [tex]e^{-i\pi}=\cos(\pi)-i\sin(\pi)=-1-0i=-1[/tex] while similarly, [tex]e^{i\pi}=\cos(\pi)+i\sin(\pi)=-1+0i=-1[/tex]

- #6

- 25

- 0

Ok I see now thx[tex]e^{ix}=\cos(x)+i \sin(x)[/tex] hence after plugging [itex]x=-\pi[/itex] we get [tex]e^{-i\pi}=\cos(-\pi) +i \sin(-\pi)[/tex] and recall that [itex]\cos(-x)=\cos(x)[/itex] and [itex]\sin(-x)=-\sin(x)[/itex] thus we have [tex]e^{-i\pi}=\cos(\pi)-i\sin(\pi)=-1-0i=-1[/tex] while similarly, [tex]e^{i\pi}=\cos(\pi)+i\sin(\pi)=-1+0i=-1[/tex]

- #7

- 36,396

- 6,936

Ok I know eulers but how does 1^x - (-1)^x = 0?

((e^(i*pi))^x)-((e^(-i*pi))^x) does not reduce to 1^x - (-1)^x.

It reduces to (-1)^x - (-1)^x.

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