prove that function f(x)=sinx +sinax

is cyclic if and only if "a" is rational

??

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- December 14th 2008, 12:29 AMtransgalacticcyclic function proof..
prove that function f(x)=sinx +sinax

is cyclic if and only if "a" is rational

?? - December 14th 2008, 12:31 AMJhevon
- December 14th 2008, 12:32 AMtransgalactic
cyclic is a function which goes in cycles

like sinx cosx etc.. - December 14th 2008, 01:41 AMOpalg
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 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. - December 14th 2008, 02:58 AMtransgalactic
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

? - December 14th 2008, 10:17 AMMathstud28
- December 14th 2008, 02:45 PMtransgalactic
i dont know how to prove this thing

- December 14th 2008, 03:23 PMchiph588@
- December 16th 2008, 03:08 AMtransgalactic
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? - December 16th 2008, 03:56 AMOpalg
See my previous comment (#4 above). If is periodic, with period c, then . The second derivative will also be periodic with period c, and therefore . Add those two equations together, and you see that . Assuming that , it follows that . Since , it also follows that .

But if 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.