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x1(t), x2(t) are periodic
( x1(t) + x2(t) ) is aperiodic

2007-01-20 16:53:37 · 5 answers · asked by thawhizzhardhofhoz 1 in Science & Mathematics Mathematics

5 answers

Good question! Let c be any irrational number (for instance c = pi or c = square root of 2). Let
x1(t) = cos(t) and x2(t) = cos(c*t)
Then f(t) = x1(t) + x2(t) is aperiodic. To see this, suppose that f were periodic with period p, i.e., f(t) = f(t+p) for all t. In particular, f(n*p) = f(0) = 2 for all integers n. Since cos(x) is never more than one, this means that cos(n*p) = 1 and cos(c*n*p) = 1 for all n. But cos(x) = 1 if and only if x = m*2*pi for some integer m. Hence since cos(p) = 1, we have p = m*2*pi for some m. Using cos(c*p) = 1 gives that c*p = c*m*2*pi is an integer multiple of 2*pi as well, which is impossible because c*m is not an integer, because we chose c to be irrational.

2007-01-20 17:13:11 · answer #1 · answered by bobqwatson 2 · 1 0

Aperiodic Function

2017-01-19 11:21:18 · answer #2 · answered by ? 4 · 0 0

The sum of two periodic functions is always periodic, not aperiodic.

2007-01-20 17:02:35 · answer #3 · answered by adrian b 3 · 0 1

thats an easy one,

sin(x) + sin( sqrt[2]*x)

This is a nonperiodic function, because over NO interval L does it EXACTLY repeat itself. It might qualify as "semiperiodic", as over any L it can be represented as a sum of periodic functions.

2007-01-20 17:17:47 · answer #4 · answered by WOMBAT, Manliness Expert 7 · 0 0

I'm not buying bobqwatso's argument using the irrationality of c because pi is irrational too and doesn't prevent periodicity.

2007-01-20 17:18:34 · answer #5 · answered by modulo_function 7 · 0 1

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