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Fantezi Cebir => Analiz-Cebir => Konuyu başlatan: stuart clark - Temmuz 15, 2011, 12:39:45 ös

Başlık: period of f(x)
Gönderen: stuart clark - Temmuz 15, 2011, 12:39:45 ös

If (http://latex.codecogs.com/gif.latex?f(x)) be a real value function which satisfies

(http://latex.codecogs.com/gif.latex?f\left(x+\frac{3}{2}\right)+f(x)=f(x+1)+f\left(x+\frac{1}{2}\right)) and

(http://latex.codecogs.com/gif.latex?|f(x)|\leq%202,%20\forall%20x%20\in%20R,). then period of (http://latex.codecogs.com/gif.latex?f(x))  is

Başlık: Ynt: period of f(x)
Gönderen: senior - Temmuz 29, 2011, 08:16:24 öö

f(x+3/2) - f(x+1/2) = f(x+1) - f(x) --> T = 1/2;
however, if you are asking for fundamental period, it needs further proving, I guess.

Başlık: Ynt: period of f(x)
Gönderen: stuart clark - Temmuz 30, 2011, 06:10:32 öö

Yes i am asking for fundamental period.

Başlık: Ynt: period of f(x)
Gönderen: senior - Ağustos 01, 2011, 12:50:07 ös

f(x+3/2) - f(x+1/2) = f(x+1) - f(x) --> f(k) - f(k-1) = f(1) - f(0) = M and
the difference between f(k) and f(k-1) is increasing as we move toward infinity. So M must be zero to satistfy |f(x)| <= 2
Then, f(x) = f(x-1) for all x. And T = 1.
Sorry for previous answer.

Başlık: Ynt: period of f(x)
Gönderen: stuart clark - Ağustos 02, 2011, 07:13:47 öö

Thanks Moderator

please I want more explanation for last post.

Başlık: Ynt: period of f(x)
Gönderen: senior - Ağustos 02, 2011, 09:22:28 öö

We have f(x+3/2) - f(x+1/2) = f(x+1) - f(x),
Let M = f(1) - f(0), Then for x=0, we have f(3/2) - f(1/2) = f(1) - f(0) = M
So, we move 1/2 forward, and now have f(3/2) - f(1/2) = M
x = 1/2 --> we have f(2) - f(1) = f(3/2) - f(1/2) = M, we moved 1/2 forward again.
x = 1 --> we have f(5/2) - f(3/2) = f(2) - f(1) = M ....
f(x+3/2) - f(x+1/2) = M always...
So, if f(x+3/2) =  f(x+1/2) + M and if M is non-zero, eventually |f(x)| will be greater than 2. So, M = 0, and we have
f(x+3/2) =  f(x+1/2) which  means T = 1