Točno
28. lipnja 2013. 12:03 (11 godine)
Let
![g: \mathbb{C} \rightarrow \mathbb{C}](/media/m/7/9/b/79ba0b7485f3de0891ba44fb849a82c8.png)
,
![\omega \in \mathbb{C}](/media/m/9/7/6/976b43a8e16f1d8dbaae1336fbbf27c8.png)
,
![a \in \mathbb{C}](/media/m/d/7/4/d741f48125f557742a174d23f430e615.png)
,
![\omega^3 = 1](/media/m/b/1/a/b1aabe639240abc28e6ed2cdd8ff60f9.png)
, and
![\omega \ne 1](/media/m/0/5/0/0500b80de85096dbcc5733e473770dda.png)
. Show that there is one and only one function
![f: \mathbb{C} \rightarrow \mathbb{C}](/media/m/5/3/9/539719be7762fcf680cd4ded72425ae6.png)
such that
%V0
Let $g: \mathbb{C} \rightarrow \mathbb{C}$, $\omega \in \mathbb{C}$, $a \in \mathbb{C}$, $\omega^3 = 1$, and $\omega \ne 1$. Show that there is one and only one function $f: \mathbb{C} \rightarrow \mathbb{C}$ such that
$$f(z) + f(\omega z + a) = g(z),z\in \mathbb{C}$$
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%V0
$$f(z) + f(\omega z + a) = g(z)$$ $$f(z) = g(z) - f(\omega z + a)$$ $$f(\omega z + a) = g(\omega z + a) - f(\omega^2 z + \omega a + a)$$ $$f(\omega^2 z + \omega a + a) = g(\omega^2 z + \omega a + a) - f(\omega^3 z + \omega^2 a + \omega a + a)$$
Sada imamo $\omega^3 = 1$, $\omega \neq 1$ i $1 + \omega + \omega^2 = \dfrac{1 - \omega^3}{1 - \omega} = 0$, pa je $$f(\omega^2 z + \omega a + a) = g(\omega^2 z + \omega a + a) - f(z)$$
Uvrštavamo u početnu jednadžbu:
$$f(z) + \!\left( g(\omega z + a) - \left( g(\omega^2 z + \omega a + a) - f(z) \right) \right) = g(z)$$ $$f(z) = \frac12 \left( g(z) - g(\omega z + a) + g(\omega^2 z + \omega a + a) \right)
31. kolovoza 2015. 13:45 | grga | Točno |