Periodic question?

  • #1
periodic question??

is cos (x^2) a periodic function,
also tell about cos([tex]\sqrt[2]{x}[/tex])

and what's its period if it is?
 

Answers and Replies

  • #2
mathman
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Neither are periodic.
 
  • #3
Defennder
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In general, how do we tell if a given function is periodic?
 
  • #4
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If for some [tex]\alpha[/tex] and any [tex]x[/tex], [tex]f(x+\alpha)=f(x)[/tex], then [tex]f[/tex] is periodic with period [tex]\alpha[/tex]
 
  • #5
Defennder
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Yes, I'm pretty sure anyone who understands what a periodic function is would know that. I was asking how do we know if for some function f(x), there exists [tex]\alpha[/tex] such that [tex]f(x+\alpha) = f(x)[/tex]?
 
  • #6
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Yes, I'm pretty sure anyone who understands what a periodic function is would know that. I was asking how do we know if for some function f(x), there exists [tex]\alpha[/tex] such that [tex]f(x+\alpha) = f(x)[/tex]?
I guess we try it. Say, take one of your functions

[tex]f(x)=cosx^2, f(x+a)=cos(x+a)^2[/tex] then we want to show whether f(x)=f(x+a) or not.

[tex]cosx^2=cos(x+a)^2=>cosx^2-cos(x+a)^2=0=>-2sin\frac{x^2+(a+x)^2}{2}sin\frac{x^2-(a+x)^2}{2}=0[/tex]

Now this is zero when

[tex] sin\frac{x^2+(a+x)^2}{2}=0, or:sin\frac{x^2-(a+x)^2}{2}=0[/tex]

From the first, it is zero when

[tex]x^2+(a+x)^2=2k\pi[/tex]

But from the second one we get

[tex]x^2-(a+x)^2=2k\pi=>x^2-a^2-2ax-x^2=2k\pi=>-a^2-2ax=2k\pi[/tex]

[tex]a^2+2ax+2k\pi=0[/tex] since

[tex] D=b^2-4ac=4a^2x^2-8k^2\pi^2[/tex]

Then i guess we conclude that this function is not periodic since we cannot locate a without expressing it in terms of x. So, it means that a is a function of x, which means that a changes as x changes, but is not rather a fixed number that is valid for all x-es, i don't know though my reasoning might be totally wrong.
 
  • #7
matt grime
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That is overkill. If any function is periodic, the values of x for which it takes a given value are an arithmetic progression. E.g. cos (x)=1 for x =0, 2pi, 4pi, 6pi etc.

Now, where is cos(x^2) equal to 1? at the square roots of those values, and they aren't an arithmetic progression.
 
  • #8
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Is it possible to construct an example x = f(x)+g(x), where f and g are periodic functions?
 
  • #9
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That is overkill. If any function is periodic, the values of x for which it takes a given value are an arithmetic progression. E.g. cos (x)=1 for x =0, 2pi, 4pi, 6pi etc.

Now, where is cos(x^2) equal to 1? at the square roots of those values, and they aren't an arithmetic progression.
I see your point, but it is hard to make it precise.

For example,
  • f(x) = { 0 if x is integer; a random value between 0.1 and 0.2 if x is non-integer } takes a value of 0 for x in an arithmetic progression. Is it periodic?
  • g(x) = sin(x) + sin(2x) takes a value of 0 for x in an arithmetic progression, but also on other values of x, yet it is periodic.
  • h(x) = { 1 if exp(x) divides pi, 0 otherwise } takes a value of 0 for x in an arithmetic progression, but also on other values of x, and it is not periodic.

Surely something must be said for all values of the function in a range, not only for one value.
 
  • #10
matt grime
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What do you mean 'isn't precise'? On reflection I'd say what I wrote is wrong. I should have said the sequence of zeroes is a union of arithmetic sequences.

Let's take your examples.

1: isn't even a function

2: sorry, don't see the point here. (actually, I think I do).

3: what does it mean for something to divide pi? Divides in what ring?

Perhaps you want extra detail, but all that detail is is that the sequence of the square roots of 2pi, 4pi etc contains no infinite arithmetic progression, which doesn't need any justification, really. I do get your point, though, and I shouldn't have left the details to the reader.
 
Last edited:
  • #11
thank you very much for the help mates....
i got it
they r not periodic
 

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