Le të jenë f:A→B, g:B→C f x
∈
{\displaystyle \scriptstyle \in }
f(x) B
Fig. 1.14.
kurse me pasgyrimin g g(f (x)) C f:A→B g:B→C x
∈
{\displaystyle \scriptstyle \in }
g ( f (x)) C A C shumëzim (kompozim, superpozicion) i pasqyrimeve f g g
∘
{\displaystyle \circ }
g
∗
{\displaystyle *}
g (f (x)) shumëzimi i pasqyrimeve f dhe g ). Pra, shumëzimi i funksioneve f, g
Fig. 1.16.
(
∀
{\displaystyle \scriptstyle {\forall }}
∈
{\displaystyle \scriptstyle \in }
∃
!
{\displaystyle \scriptstyle {\exists !}}
∈
{\displaystyle \scriptstyle \in }
∘
{\displaystyle \circ }
=
{\displaystyle \scriptstyle {=}}
Siç shihet në shumëzimin g
∘
{\displaystyle \circ }
f, g f
∘
{\displaystyle \circ }
Përkufizimi i shumëzimit [ redakto ] Të caktohet shumëzimi i pasqyrimeve f :A→B g :B→C A
=
{\displaystyle \scriptstyle {=}}
=
{\displaystyle \scriptstyle {=}}
f
=
(
1
2
3
4
a
b
c
d
)
,
g
=
(
a
b
c
d
α
γ
δ
β
)
{\displaystyle f={\bigl (}{\begin{smallmatrix}1&2&3&4\\a&b&c&d\\\end{smallmatrix}}{\bigr )},\ g={\bigl (}{\begin{smallmatrix}a&b&c&d\\\alpha &\gamma &\delta &\beta \\\end{smallmatrix}}{\bigr )}}
Zgjidhje : Prodhimi
g
∘
f
:
A
→
C
{\displaystyle g\circ f:A\rightarrow C}
(
g
∘
f
)
(
1
)
=
g
(
f
(
1
)
)
=
g
(
b
)
=
γ
{\displaystyle (g\circ f)(1)=g(f(1))=g(b)=\gamma }
(
g
∘
f
)
(
2
)
=
g
(
f
(
2
)
)
=
g
(
c
)
=
δ
{\displaystyle (g\circ f)(2)=g(f(2))=g(c)=\delta }
(
g
∘
f
)
(
3
)
=
g
(
f
(
3
)
)
=
g
(
a
)
=
α
{\displaystyle (g\circ f)(3)=g(f(3))=g(a)=\alpha }
(
g
∘
f
)
(
4
)
=
g
(
f
(
4
)
)
=
g
(
d
)
=
β
{\displaystyle (g\circ f)(4)=g(f(4))=g(d)=\beta }
pra :
g
∘
f
=
(
a
b
c
d
α
γ
δ
β
)
∘
(
1
2
3
4
a
b
c
d
)
=
(
1
2
3
4
α
γ
δ
β
)
{\displaystyle g\circ f={\bigl (}{\begin{smallmatrix}a&b&c&d\\\alpha &\gamma &\delta &\beta \\\end{smallmatrix}}{\bigr )}\circ {\bigl (}{\begin{smallmatrix}1&2&3&4\\a&b&c&d\\\end{smallmatrix}}{\bigr )}={\bigl (}{\begin{smallmatrix}1&2&3&4\\\alpha &\gamma &\delta &\beta \\\end{smallmatrix}}{\bigr )}}
Këtu shumëzimi f
∘
{\displaystyle \circ }
Veprimi jokomutative [ redakto ] Të caktohen shumëzimet (1) g
∘
{\displaystyle \circ }
∘
{\displaystyle \circ }
-1
∘
{\displaystyle \circ }
f, g
f
(
x
)
=
x
2
+
2
x
−
1
{\displaystyle f(x)=x^{2}+2x-1}
g
(
x
)
=
3
x
+
2
{\displaystyle g(x)=3x+2}
Zgjidhje :
(1)
(
g
∘
f
)
(
x
)
=
g
(
f
(
x
)
)
=
g
(
x
2
+
2
x
−
1
)
=
3
(
x
2
+
2
x
−
1
)
+
2
=
3
x
2
+
6
x
−
1
{\displaystyle (g\circ f)(x)=g(f(x))=g(x^{2}+2x-1)=3(x2+2x-1)+2=3x2+6x-1}
(2)
(
f
∘
g
)
(
x
)
=
f
(
g
(
x
)
)
=
f
(
3
x
+
2
)
=
(
3
x
+
2
)
2
+
2
(
3
x
+
2
)
−
1
=
9
x
2
+
18
x
+
7
{\displaystyle (f\circ g)(x)=f(g(x))=f(3x+2)=(3x+2)^{2}+2(3x+2)-1=9x^{2}+18x+7}
(3) Meqë
g
−
1
(
x
)
=
(
x
−
2
)
3
,
(
g
−
1
∘
g
)
(
x
)
=
g
−
1
(
3
x
+
2
)
=
x
{\displaystyle g^{-1}(x)={\frac {(x-2)}{3}},(g^{-1}\circ g)(x)=g^{-1}(3x+2)=x}
Siç shihet pra, edhe kur ekzistojnë dy shumëzime g
∘
{\displaystyle \circ }
f
∘
{\displaystyle \circ }
f, g f, g
g
∘
{\displaystyle \circ }
≠
{\displaystyle \scriptstyle {\neq }}
∘
{\displaystyle \circ }
g (f (x))
≠
{\displaystyle \scriptstyle {\neq }}
Nga relacioni përkufizues (39) dhe fig. 1.16. del se për shumëzimin e pasqyrimeve f, g
(g
∘
{\displaystyle \circ }
-1
=
{\displaystyle \scriptstyle {=}}
-1
∘
{\displaystyle \circ }
-1
sepse : (g
∘
{\displaystyle \circ }
-1
=
{\displaystyle \scriptstyle {=}}
-1 -1
∀
{\displaystyle \scriptstyle {\forall }}
∈
{\displaystyle \scriptstyle \in }
Pasqyrimi identik [ redakto ] Nëse f A B f -1
∘
{\displaystyle \circ }
pasqyrimin identik të bashkësisë A
f:x→y
=
{\displaystyle \scriptstyle {=}}
∀
{\displaystyle \scriptstyle {\forall }}
∈
{\displaystyle \scriptstyle \in }
f-1
=
{\displaystyle \scriptstyle {=}}
-1
∀
{\displaystyle \scriptstyle {\forall }}
∈
{\displaystyle \scriptstyle \in }
dhe
(f-1
∘
{\displaystyle \circ }
=
{\displaystyle \scriptstyle {=}}
∀
{\displaystyle \scriptstyle {\forall }}
∈
{\displaystyle \scriptstyle \in }
Teorema për n pasqyrime [ redakto ] Le të jenë f:A→B g:B→C h:C→D f, g h :
(h o g)
∘
{\displaystyle \circ }
=
{\displaystyle \scriptstyle {=}}
∘
{\displaystyle \circ }
∘
{\displaystyle \circ }
Vërtetim : Transformojmë anën e djathtë të formulës (43) :
[h
∘
{\displaystyle \circ }
∘
{\displaystyle \circ }
=
{\displaystyle \scriptstyle {=}}
∘
{\displaystyle \circ }
=
{\displaystyle \scriptstyle {=}}
=
{\displaystyle \scriptstyle {=}}
∘
{\displaystyle \circ }
∘
{\displaystyle \circ }
∘
{\displaystyle \circ }
Relacioni i fundit vërteton pohimin e teoremës.
