Periodic Solutions: Are All Valid?

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Homework Statement


Hi guys

Say I have an equation of the form

f(x) = cos(x)+cos(0.2x),

and I wish to find the solutions x. When I plot this graph, I see multiple solutions, but there is no apparent period for the solutions. Are all the solutions equally valid, or can some be discarded?
 
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By "solutions", do you mean "zeroes", i.e., solutions of the equation [tex]f(x) = 0[/tex]?

If this is the case, there is no sense in which any solution is less "equally valid" than any other. A solution of this equation is a solution of the equation.

However, if it seems to you that the solutions of this particular equation are not periodic, you need to look more closely -- at the equation, not at a graph. The function [tex]f(x) = \cos x + \cos (0.2x)[/tex] is periodic indeed.
 
Hmm, let's say that a solution x0 to f(x) is to be used in another function g(x)=sin(x). Since the zeroes of f(x) are not periodic with 2π, it matters which zero I choose. Is there a way to determine which one?
 
One can't answer this meaningfully without more context.
 
Niles said:
Hmm, let's say that a solution x0 to f(x) is to be used in another function g(x)=sin(x). Since the zeroes of f(x) are not periodic with 2π, it matters which zero I choose. Is there a way to determine which one?

Your terminology is very imprecise. There is no such thing as a solution to a function. There is the idea of a solution to an equation, so maybe you mean solutions to the equation f(x) = 0. IOW the x-intercepts of this function.

If you look at the graph of the function f(x) = cos(x) + cos(.2x), you should be able to see that it is periodic. In fact, its period is exactly the same as the period of cos(.2x). Once you figure out what the period is, it's straightforward to prove that f is periodic with that period, by showing that f(x + P) = f(x) for all real x.
 
tiny-tim said:
Hi Niles! :smile:

Use one of the standard trigonometric identities to get f(x) as a product of two sines.

That should give you a period, and slso a pattern within that period. :wink:

You mean two cosines, right? But I understand the thing with the period; my problem is that within that period, there are solutions that are not 2π-periodic with each other. So when I have a harmonic function g(x), which takes the zeroes of f(x) as arguments, then I am not sure which one to pick.
Mark44 said:
Your terminology is very imprecise. There is no such thing as a solution to a function. There is the idea of a solution to an equation, so maybe you mean solutions to the equation f(x) = 0. IOW the x-intercepts of this function.

If you look at the graph of the function f(x) = cos(x) + cos(.2x), you should be able to see that it is periodic. In fact, its period is exactly the same as the period of cos(.2x). Once you figure out what the period is, it's straightforward to prove that f is periodic with that period, by showing that f(x + P) = f(x) for all real x.

Yeah, I agree. It is very imprecise, but that is exactly what I mean. Thanks for claryfing that.