Think periodically, Part 2
What is the longest possible fundamental period of a function that satisfies the following equation for all real x ? f ( x + 8 ) + f ( x + 5 ) + f ( x + 3 ) + f ( x ) = f ( x + 7 ) + f ( x + 4 ) + f ( x + 1 )
The answer is 15.
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1 solution
You show that 15 is always a period. But you did not convince us that there does exist a function for which the fundamental period is indeed 15. Can you give an example?
You claim that 1 5 / 3 = 5 can also be a period, but, in light of your equation f ( x ) + f ( x + 5 ) + f ( x + 1 0 ) = 0 , that will be the case only for the zero function.
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Ya. Sorry, I wasn't thinking correctly. How about sin(2*pi/15 x)? It's fundamental period is 15.
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Yes, that is the natural candidate, but how do you convince us that sin ( 1 5 2 π x ) solves our functional equation f ( x + 8 ) . . . = . . . + f ( x + 1 ) ? As I showed, sin ( 5 2 π x ) is not a solution, for example.
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@Otto Bretscher – Then you're asking if there's a way to work backwards from f(x+15)=f(x) to your equation. I don't know for sure but I checked the graphs of your equation with sin(2*pi/15 x) in the place of f(x) and the RHS and LHS both produced the same graph. So there must be a way.
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@Vishnu C – Obviously, there is no way to work backwards from f ( x + 1 5 ) = f ( 0 ) to our functional equation f ( x + 8 ) + . . . , since a nonzero function with period 5 satisfies f ( x + 1 5 ) = f ( x ) but does not satisfy our functional equation.
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@Otto Bretscher – If you can't work back, then how do you verify?
Questions like these (especially in trigonometry) bother me when they appear in tests. If you play around with the equations, you're bound to get the answer; but in a test, you don't have time to play around to see what works. I haven't figured out (yet) a general algorithm for solving such questions. Do you have one figured out?
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As you know, questions of periodicity are closely related to roots of unity... look at my solution to the "Inspiration" problem. When I have a little more time, I will write something up... or maybe you can figure it out yourself.
The basic idea is this: When dealing with a functional equation (including differential and difference equations), look for exponential solutions first.
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@Otto Bretscher – Although problems concerning periods of functions are a part of this play-with-equations-till-you-get-the-solution genre, there are also various other problems in algebra and trigonometry that are in this genre.
I took a look at your solution and I couldn't understand most of it. I've only heard of characteristic equations in matrices and in differential equations with order>1. Under which category do difference equations fall? Numerical Analysis? Interpolation? Well, I've not read up on those yet.
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@Vishnu C – Roughly speaking, a characteristic equation is an equation you get when you use an exponential function as a trial solution for a functional equation.
If we use the trial solution f ( x ) = c x for our given equation, we get an expression that looks like c 8 c x + . . . = . . . + c × c x . We can divide by c x and we end up with the polynomial equation c 8 − c 7 + c 5 − c 4 + c 3 − c + 1 = 0 ... that's the characteristic equation in this case. Now comes the difficult step... this is a Level 5 question after all ;)
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@Otto Bretscher – Why choose an exponential function? Why not some other function?
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@Vishnu C – It depends on the problem you are working on. Remember, our goal here is to show that sin ( 1 5 2 π x ) is a solution of our functional equation. Now, sine and cosine are just the real and imaginary parts of a complex exponential... that's why we are looking for exponential solutions.
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@Otto Bretscher – Ya. I solved the characteristic equation and I got 8 of the 15 roots of unity and exp(2*pi/15) was a part of the solution set.
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@Vishnu C – Are you sure e 2 π i / 5 is part of the solution set? According to what we said earlier , it shouldn't be...
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@Otto Bretscher – Damn it! I put 5 instead of 15.
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@Vishnu C – Typos happen to all of us ;)
So, you figured it out, congrats! The roots of the characteristic polynomial are the primitive 15th roots of unity, e 2 k π i / 1 5 , for k = 1 , 2 , 4 , 7 , 8 , 1 1 , 1 3 , 1 4 , since the polynomial is a factor of c 1 5 − 1 . (This is called the cyclotomic polynomial for n = 1 5 .)The real and imaginary parts of e 2 k π i x / 1 5 are solutions as well, for example, sin ( 2 k π x / 1 5 ) . For k = 1 you get the function we want, with fundamental period 15.
Good luck with your exams! I can see that you are a very talented young man... you will be fine.
Solved exactly the same way!
This is what I did:
f(x+9)+f(x+6)+f(x+4)+f(x+1)=f(x+8)+f(x+5)+f(x+2)
Substituting f(x+4)+f(x+1) from the equation in the question we get
f(x+9)+f(x+6)+f(x+3)+f(x)=f(x+2)+f(x+7).-----------------(1)
From the equation in the question, if we substitute x+2 in the place of x,
f(x+10)+f(x+7)+f(x+5)+f(x+2)=f(x+3)+f(x+6)+f(x+9)
The RHS is the same as f(x+2)+f(x+7)-f(x), from (1).
So we get
f(x+10)+f(x+5)=-f(x)
So, f(x+15)+f(x+10)+f(x+5)=0
Therefore, f(x+15)=f(x) for all real x. As the question asks for the longest possible fundamental period, we can say that 15 is the largest possible fundamental period for such functions, although 15/2, etc. can also be periods of functions where f(x)=f(x+15).