# cyclic function proof..

• Dec 13th 2008, 11:29 PM
transgalactic
cyclic function proof..
prove that function f(x)=sinx +sinax
is cyclic if and only if "a" is rational
??
• Dec 13th 2008, 11:31 PM
Jhevon
Quote:

Originally Posted by transgalactic
prove that function f(x)=sinx +sinax
is cyclic if and only if "a" is rational
??

what is your definition of "cyclic"?
• Dec 13th 2008, 11:32 PM
transgalactic
cyclic is a function which goes in cycles
like sinx cosx etc..
• Dec 14th 2008, 12:41 AM
Opalg
Quote:

Originally Posted by transgalactic
prove that function f(x)=sinx +sinax
is cyclic [the usual word is "periodic"] if and only if "a" is rational
??

If a is rational, a=m/n, then it's easy to see that f is periodic, with period 2nπ.

It's harder to prove the converse. Suppose that f(x) is periodic. Then $\displaystyle f''(x) = -\sin(x) -a^2\sin(ax)$ will also be periodic, with the same period. But when x=0, f(0)=f''(0)=0. If there is another point, say x=c, at which f(c)=f''(c)=0, and a≠±1, then you should be able to show that sin(c) = sin(ac) = 0. With c≠0, this is only possible if a is rational. But if there is no such point c then f cannot be periodic.
• Dec 14th 2008, 01:58 AM
transgalactic
you said that if f(0)=f'(0)=f''(0)=0
but when there is another point which gives the same result
then its not periodic
but why should differ +1 and -1

how to use this definition a=m/n of a rational number in this proof
?
• Dec 14th 2008, 09:17 AM
Mathstud28
Quote:

Originally Posted by Opalg
If a is rational, a=m/n, then it's easy to see that f is periodic, with period 2nπ.

It's harder to prove the converse. Suppose that f(x) is periodic. Then $\displaystyle f''(x) = -\sin(x) -a^2\sin(ax)$ will also be periodic, with the same period. But when x=0, f(0)=f''(0)=0. If there is another point, say x=c, at which f(c)=f''(c)=0, and a≠±1, then you should be able to show that sin(c) = sin(ac) = 0. With c≠0, this is only possible if a is rational. But if there is no such point c then f cannot be periodic.

Question? Maybe I am misremembering but doesnt the period of $\displaystyle \sin(ax)+\sin(bx)$ have something to do with $\displaystyle \text{gcf}(a,b)$ or $\displaystyle \text{gcd}(a,b)$ or something? If so wouldn't that suffice to show that because $\displaystyle a$ is irrational that this doesnt make sense?
• Dec 14th 2008, 01:45 PM
transgalactic
i dont know how to prove this thing
• Dec 14th 2008, 02:23 PM
chiph588@
Quote:

Originally Posted by Opalg
If there is another point, say x=c, at which f(c)=f''(c)=0, and a≠±1, then you should be able to show that sin(c) = sin(ac) = 0. With c≠0, this is only possible if a is rational. But if there is no such point c then f cannot be periodic.

How do you know this?
• Dec 16th 2008, 02:08 AM
transgalactic
i know that i need to use
the definition that if a number is rational
then it ca be represented by two whole numbers a/b

what to do here?
• Dec 16th 2008, 02:56 AM
Opalg
Quote:

Originally Posted by transgalactic
i know that i need to use
the definition that if a number is rational
then it ca be represented by two whole numbers a/b

what to do here?

See my previous comment (#4 above). If $\displaystyle f(x)=\sin x +\sin ax$ is periodic, with period c, then $\displaystyle \sin c +\sin ac = f(c) = f(0) = 0$. The second derivative $\displaystyle f''(x)=-\sin x - a^2\sin ax$ will also be periodic with period c, and therefore $\displaystyle -\sin c - a^2\sin ac = 0$. Add those two equations together, and you see that $\displaystyle (1-a^2)\sin ac = 0$. Assuming that $\displaystyle 1-a^2\ne0$, it follows that $\displaystyle \sin ac=0$. Since $\displaystyle \sin c +\sin ac = 0$, it also follows that $\displaystyle \sin c=0$.

But if $\displaystyle \sin x = 0$ then x must be a multiple of π. Therefore both c and ac are multiples of π, say c=nπ and ac=mπ. Then a=m/n, which is rational.