Find the period of the function

  • Thread starter Thread starter utkarshakash
  • Start date Start date
  • Tags Tags
    Function Period
AI Thread Summary
The function f(x) = e^(3(x - [x])) involves the greatest integer function, which leads to the periodicity of the fractional part of x. The discussion focuses on finding the period T by setting f(x) = f(x + T) and analyzing the equation derived from the properties of the greatest integer function. It is established that the fractional part of x, denoted as {x}, repeats every 1 unit, indicating that the period of the function e^(x - [x]) is also 1. Graphing the function can provide visual confirmation of this periodicity. Ultimately, the period of the function is confirmed to be 1.
utkarshakash
Gold Member
Messages
852
Reaction score
13

Homework Statement


f(x)=e3(x-[x])
[] denotes greatest integer function

Homework Equations



The Attempt at a Solution


f(x)=f(x+T)

e3{x}=e{x+T}
Taking ln of both sides
3{x}=3{x+T}
{x}={x+T}

x-[x]=x+T-[x+T]
T=[x+T]-[x]
 
Physics news on Phys.org
Now you have to find T which satisfy this equation.
Alternatively, graph x-[x], the period is obvious in the graph.
 
mfb said:
Now you have to find T which satisfy this equation.
Alternatively, graph x-[x], the period is obvious in the graph.

T=[x]+t-[x]
0=0

How to find T?
Also why you are asking me to draw the graph of x-[x] when the question says e raised to x-[x]?
 
utkarshakash said:
Also why you are asking me to draw the graph of x-[x] when the question says e raised to x-[x]?

Give it a try at least. ;)

You see that x-[x] which is equivalent to fractional part of x repeats itself after 1. The period of e^(x-[x]) is obvious.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

Similar threads

Back
Top