((e^(i*pi))^x)-((e^(-i*pi))^x)=0? how?

  • Thread starter powerplayer
  • Start date
In summary, Wolfram says that the equation 1^x - (-1)^x = 0 is equal to e^{ix}-e^{-ix}, which reduces to (-1)^x - (-1)^x after simplification.
  • #1
Can someone help explain this? Wolfram says it is zero but I don't know why?
 
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  • #3
Ok I know eulers but how does 1^x - (-1)^x = 0?
 
  • #4
That's ##\left(-1\right)^x-\left(-1\right)^x##, after some simplification.
 
  • #5
powerplayer said:
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
Mentallic said:
[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]
Ok I see now thx
 
  • #7
powerplayer said:
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.
 

1. What is the equation ((e^(i*pi))^x)-((e^(-i*pi))^x)=0?

The equation ((e^(i*pi))^x)-((e^(-i*pi))^x)=0 is a complex equation that involves the imaginary number i, the mathematical constant e, and the irrational number pi. It is commonly known as Euler's identity and is used in many branches of mathematics.

2. What does the equation represent?

The equation represents the equalities between two complex numbers with different exponents. The left side of the equation is a combination of two complex numbers raised to the power of x, while the right side is simply 0. This equation is commonly used in trigonometry and complex analysis.

3. How is this equation derived?

The equation ((e^(i*pi))^x)-((e^(-i*pi))^x)=0 can be derived from Euler's formula, e^(ix)=cos(x)+i*sin(x). By substituting pi for x in the formula, we get e^(i*pi)=-1 and e^(-i*pi)=1. Plugging these values into the equation, we get (-1)^x-1^x=0, which simplifies to (-1)^x-1=0. Finally, by applying the exponent rule for negative exponents, we get (-1)^x-1=(-1)^0, which equals 0.

4. What is the significance of this equation?

The equation ((e^(i*pi))^x)-((e^(-i*pi))^x)=0 is significant because it is a fundamental result in complex analysis and is used in many areas of mathematics, including Fourier analysis, differential equations, and number theory. It also has connections to other important mathematical concepts, such as the Riemann zeta function and the prime number theorem.

5. How is this equation used in real-world applications?

The equation ((e^(i*pi))^x)-((e^(-i*pi))^x)=0 has practical applications in engineering, physics, and signal processing. It is used in the analysis of alternating current circuits, quantum mechanics, and the study of wave phenomena. Additionally, it has been used in the creation of digital signal processing algorithms for various applications, such as audio and image processing.

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