Točno
19. kolovoza 2017. 18:43 (6 godine, 11 mjeseci)
Find all functions
![f \colon \mathbb R \to \mathbb R](/media/m/4/2/7/4274589bc617b7d9cd7565e617fd02cf.png)
such that the equality
![y^2f(x) + x^2f(y) + xy = xyf(x + y) + x^2 + y^2](/media/m/3/f/8/3f81b0867f6373582d7375b4185fc8b1.png)
holds for all
![x, y \in \Bbb R](/media/m/c/3/8/c386d592c03d772e719bb0956ead7aaa.png)
, where
![\Bbb R](/media/m/0/c/a/0ca3deced1777f29d31572a0e19d110a.png)
is the set of real numbers.
%V0
Find all functions $f \colon \mathbb R \to \mathbb R$ such that the equality $$y^2f(x) + x^2f(y) + xy = xyf(x + y) + x^2 + y^2$$ holds for all $x, y \in \Bbb R$, where $\Bbb R$ is the set of real numbers.
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Uvrstimo li u početnu jednadžbu vrijednost
![y=0](/media/m/3/1/3/313c785b69a2965d5eb1b358bea8e41c.png)
dobijamo
![\Rightarrow f(0) = 1](/media/m/3/6/1/36103188dbe312658d2f196a3c81471a.png)
Uvedimo funkciju
![m(x) = f(x) - 1](/media/m/0/3/0/0305e1352e59c86ae064c92e891dfb0e.png)
, tada vrijedi
![m(0) = 0](/media/m/4/6/1/4614cbc73f84548ae86066df6acce074.png)
, a početna jednakost postaje:
![y^2m(x) + x^2m(x) = xy \cdot m(x+y) \qquad \textbf{(*)}](/media/m/b/b/c/bbc382f0b5490c7c7ea2fe82ff5b1792.png)
Neka
![P(x,y)](/media/m/0/2/1/0216a09a9765bb0d1de05e7905a94648.png)
označava uvrštavanje varijabli
![x,y](/media/m/f/b/6/fb60533620f22cd699e5b58ce9a646a4.png)
u
![\textbf{(*)}](/media/m/8/b/f/8bf725205c81cf2d9a0b5a79cdebca6f.png)
.
![\Rightarrow -m(x) = m(-x)](/media/m/0/4/3/04368e4a2f1aa95db1e4b4843c5fc944.png)
![P(x,1) \Rightarrow m(x) + x^2 m(1) = xm(x+1) \qquad \textbf{(1)}](/media/m/7/c/e/7ce886c519f03c348eb04c5dd955945a.png)
![P(x, -1) \Rightarrow m(x) + x^2 m(-1) = -x m(x-1) \qquad \textbf{(2)}](/media/m/3/2/5/3250ba209fa5719d27ed89b9f5dfd8ef.png)
Ako u
![\textbf{(2)}](/media/m/a/1/8/a1895833fbdee2b5fc426076104bd4b9.png)
zamijenimo
![x \rightarrow x+1](/media/m/4/5/9/4597cfa1e14a2f8a01a4b64e7081dc5d.png)
dobivamo:
![m(x+1) + (x+1)^2 \cdot m(-1) = -(x+1) m(x) \qquad \textbf{(3)}](/media/m/e/5/b/e5bb96958eab8aac0069bb22084db724.png)
Ako iz
![\textbf{(1)}](/media/m/3/6/0/360636f6a99eec69bd849eaed0bfe0eb.png)
i
![\textbf{(3)}](/media/m/a/e/9/ae9cbb63d5f061bfe726362534760cc5.png)
izjednačimo
![m(x+1)](/media/m/c/a/6/ca6dde81d0629e9856585b588ee6ea7e.png)
i iskoristimo
![m(-1) = -m(1)](/media/m/8/0/2/80262a158401910eb0bcdf5417246fd5.png)
dobivamo:
![m(x+1) = \frac{m(x) + x^2 m(1)}{x}](/media/m/3/c/f/3cf139ceba7d370f6152ba90c2a84846.png)
i
![m(x+1) = (x+1)^2 m(1) - (x+1)m(x)](/media/m/0/c/9/0c9d23de77d9ebe5824ede1ca81d79c5.png)
odakle sređivanjem slijedi
![m(x) (x^2 + x + 1) = x m(1)(x^2 + x + 1)](/media/m/8/e/5/8e57ce7db6cb7c19ea37856fef24a097.png)
a budući da je
![x^2 + x + 1 \geq 0](/media/m/3/8/9/389183cc896c4795a51fe53dfdd11d2f.png)
za svaki
![x \in \mathbb{R}](/media/m/c/1/6/c162a96a3e70076e1031361250564464.png)
vrijedi
![m(x) = m(1)x \Rightarrow f(x) = x m(1) +1](/media/m/5/2/7/527ac7288130db263b5c98c7cc590e01.png)
. Ako
![m(1) = c \Rightarrow f(x) = cx + 1](/media/m/5/d/2/5d27370601c8ed5a7143551d1731a6cd.png)
što uvrštavanjem u početnu jednakost zaista jest rješenje.
%V0
Uvrstimo li u početnu jednadžbu vrijednost $y=0$ dobijamo $x^2 f(0) = x^2$ $ \Rightarrow f(0) = 1$
Uvedimo funkciju $ m(x) = f(x) - 1$ \, tada vrijedi $m(0) = 0$ \, a početna jednakost postaje: $$ y^2m(x) + x^2m(x) = xy \cdot m(x+y) \qquad \textbf{(*)}$$
Neka $P(x,y)$ označava uvrštavanje varijabli $x,y$ u $\textbf{(*)}$.
$P(x,-x) \Rightarrow x^2 m(x) + x^2 m(-x) = 0$ $ \Rightarrow -m(x) = m(-x)$
$P(x,1) \Rightarrow m(x) + x^2 m(1) = xm(x+1) \qquad \textbf{(1)}$
$P(x, -1) \Rightarrow m(x) + x^2 m(-1) = -x m(x-1) \qquad \textbf{(2)}$
Ako u $ \textbf{(2)}$ zamijenimo $x \rightarrow x+1$ dobivamo:
$m(x+1) + (x+1)^2 \cdot m(-1) = -(x+1) m(x) \qquad \textbf{(3)}$
Ako iz $\textbf{(1)}$ i $\textbf{(3)}$ izjednačimo $m(x+1)$ i iskoristimo $ m(-1) = -m(1)$ dobivamo:
$m(x+1) = \frac{m(x) + x^2 m(1)}{x}$ i $m(x+1) = (x+1)^2 m(1) - (x+1)m(x)$ odakle sređivanjem slijedi $$m(x) (x^2 + x + 1) = x m(1)(x^2 + x + 1)$$ a budući da je $x^2 + x + 1 \geq 0$ za svaki $ x \in \mathbb{R}$ vrijedi $m(x) = m(1)x \Rightarrow f(x) = x m(1) +1$. Ako $m(1) = c \Rightarrow f(x) = cx + 1$ što uvrštavanjem u početnu jednakost zaista jest rješenje.