Wave Function in specific range

In summary, the function given by f(x) = x/2∏ for 0<x<2∏ is not defined at x = 0 or x = 2∏. If the function is defined at these points, then f(x) would also have to be defined for all values of x in order for the function to be periodic.
  • #1
ZedCar
354
1
I obtained the following from a book.

Question is:

Periodic Sawtooth described by the following;

f(x) = x/2∏ for 0<x<2∏
f(x+2∏) = f(x) for -∞<x<+∞


The solution is:

If x = 0
y = 0

If x = 2∏
y = 2∏/2∏ = 1

If x = 4∏
y = f(2∏+2∏) = 2∏ = 1

Can anyone explain to me why when x = 4∏ y = 1 ? I'm not clear on that bit.

I would just think if you're putting x = 4∏ into the 2nd equation in the question you would get y = f(4∏+4∏) = f(8∏)

I know a full rotation is 2 ∏, so I can see how 8∏ would be the same as 2∏, but then how did they go from 2∏ to 1 in the part above which I have emboldened?

Thank you
 
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  • #2
You were given f(x) = x/2∏ for 0<x<2∏. Your function is not defined at x = 0 or at x = 2∏. So you don't have f(0) = 0 and f(2∏) = 1. The value of a function like that at the two end points is irrelevant in calculating things like the Fourier Series so they are usually left undefined. If you insist on defining f(0)= 0, then you have no choice but to also choose f(2∏) = 0 if you want the function to be periodic.
 

Related to Wave Function in specific range

1. What is a wave function in a specific range?

A wave function in a specific range is a mathematical representation of the probability of finding a particle in a certain location within a given range. It is often used in quantum mechanics to describe the behavior of particles on a microscopic level.

2. How is a wave function in a specific range calculated?

A wave function in a specific range is calculated using the Schrödinger equation, which takes into account the properties of the particle and its surroundings. The resulting equation is a complex function that describes the probability of finding the particle in different locations within the specified range.

3. What is the significance of a wave function in a specific range?

The wave function in a specific range is significant because it allows scientists to make predictions about the behavior of particles in quantum systems. It also provides insight into the nature of matter and the fundamental principles of the universe.

4. How does a wave function in a specific range relate to the uncertainty principle?

The uncertainty principle states that it is impossible to know both the position and momentum of a particle with absolute certainty. The wave function in a specific range represents the probability of finding the particle in a certain location, which is a manifestation of this uncertainty principle.

5. Can a wave function in a specific range change over time?

Yes, a wave function in a specific range can change over time as the particle moves and interacts with its surroundings. This change is described by the time-dependent Schrödinger equation, which takes into account the changing conditions of the system.

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