De la Wikipedia, enciclopedia liberă
În matematică o ecuație funcțională exponențială este ansamblul tuturor funcțiilor
f
:
R
→
R
{\displaystyle f:\mathbb {R} \;\rightarrow \;\mathbb {R} \;}
x
,
y
∈
R
{\displaystyle x,y\in \mathbb {R} \;}
f
(
x
+
y
)
=
f
(
x
)
⋅
f
(
y
)
{\displaystyle f(x+y)=f(x)\cdot f(y)}
Exemplul 1 . Funcțiile continue
f
:
R
→
R
{\displaystyle f:\mathbb {R} \;\rightarrow \;\mathbb {R} \;}
x
,
y
∈
R
{\displaystyle x,y\in \mathbb {R} \;}
f
(
x
+
y
)
=
f
(
x
)
⋅
f
(
y
)
{\displaystyle f(x+y)=f(x)\cdot f(y)}
Soluție: Fie
t
∈
R
{\displaystyle t\in \mathbb {R} \;}
x
=
t
2
=
y
{\displaystyle x={\frac {t}{2}}=y}
f
(
t
)
=
f
(
t
2
+
t
2
)
=
f
(
t
2
)
2
≥
0
{\displaystyle f(t)=f\left({\frac {t}{2}}+{\frac {t}{2}}\right)={f\left({\frac {t}{2}}\right)}^{2}\geq 0}
s
∈
R
{\displaystyle s\in \mathbb {R} \;}
f
(
s
)
=
0
{\displaystyle f(s)=0}
t
∈
R
{\displaystyle t\in \mathbb {R} \;}
f
(
t
)
=
f
(
t
−
s
+
s
)
=
f
(
t
−
s
)
⋅
f
(
s
)
=
0
{\displaystyle f(t)=f(t-s+s)=f(t-s)\cdot f(s)=0}
f
:
R
→
R
{\displaystyle f:\mathbb {R} \;\rightarrow \;\mathbb {R} \;}
f
(
t
)
>
0
{\displaystyle f(t)>0}
t
∈
R
{\displaystyle t\in \mathbb {R} \;}
f
(
t
)
>
0
{\displaystyle f(t)>0}
t
∈
R
{\displaystyle t\in \mathbb {R} \;}
g
:
R
→
R
{\displaystyle g:\mathbb {R} \;\rightarrow \;\mathbb {R} \;}
x
∈
R
{\displaystyle x\in \mathbb {R} \;}
g
(
x
)
=
ln
f
(
x
)
{\displaystyle g(x)=\ln {f(x)}}
x
,
y
∈
R
{\displaystyle x,y\in \mathbb {R} \;}
g
(
x
+
y
)
=
ln
f
(
x
)
f
(
y
)
=
ln
f
(
x
)
+
ln
f
(
y
)
=
g
(
x
)
+
g
(
y
)
{\displaystyle g(x+y)=\ln {f(x)f(y)}=\ln {f(x)}+\ln {f(y)}=g(x)+g(y)}
g
:
R
→
R
{\displaystyle g:\mathbb {R} \;\rightarrow \;\mathbb {R} \;}
g
(
x
)
=
x
⋅
g
(
1
)
{\displaystyle g(x)=x\cdot g(1)}
x
∈
R
{\displaystyle x\in \mathbb {R} \;}
g
(
1
)
=
ln
f
(
1
)
{\displaystyle g(1)=\ln {f(1)}}
f
(
1
)
=
e
(
g
(
1
)
)
{\displaystyle f(1)=e^{(g(1))}}
x
,
y
∈
R
{\displaystyle x,y\in \mathbb {R} \;}
f
(
x
)
=
e
g
(
x
)
=
e
x
⋅
g
(
1
)
=
(
e
g
(
1
)
)
x
=
f
(
1
)
x
{\displaystyle f(x)=e^{g(x)}=e^{x\cdot g(1)}=(e^{g(1)})^{x}={f(1)}^{x}}
Exemplul 2 . Funcțiile continue
f
:
R
→
R
{\displaystyle f:\mathbb {R} \;\rightarrow \;\mathbb {R} \;}
x
,
y
∈
R
{\displaystyle x,y\in \mathbb {R} \;}
f
(
x
+
y
)
=
f
(
x
)
⋅
f
(
y
)
−
x
⋅
f
(
y
)
−
y
⋅
f
(
x
)
+
x
y
+
x
+
y
{\displaystyle f(x+y)=f(x)\cdot f(y)-x\cdot f(y)-y\cdot f(x)+xy+x+y}
Soluție: Fie
x
,
y
∈
R
{\displaystyle x,y\in \mathbb {R} \;}
f
(
x
+
y
)
−
(
x
+
y
)
=
(
f
(
x
)
−
x
)
⋅
(
f
(
y
)
−
y
)
{\displaystyle f(x+y)-(x+y)=(f(x)-x)\cdot (f(y)-y)}
g
:
R
→
R
{\displaystyle g:\mathbb {R} \;\rightarrow \;\mathbb {R} \;}
t
∈
R
{\displaystyle t\in \mathbb {R} \;}
g
(
t
)
=
f
(
t
)
−
t
{\displaystyle g(t)=f(t)-t}
g
(
x
+
y
)
=
g
(
x
)
⋅
g
(
y
)
{\displaystyle g(x+y)=g(x)\cdot g(y)}
Dacă
g
(
1
)
=
0
{\displaystyle g(1)=0}
x
∈
R
{\displaystyle x\in \mathbb {R} \;}
g
(
x
)
=
g
(
1
+
x
−
1
)
=
g
(
1
)
⋅
g
(
x
−
1
)
=
0
⋅
g
(
x
−
1
)
=
0
{\displaystyle g(x)=g(1+x-1)=g(1)\cdot g(x-1)=0\cdot g(x-1)=0}
f
(
t
)
=
t
{\displaystyle f(t)=t}
t
∈
R
{\displaystyle t\in \mathbb {R} \;}
g
(
1
)
≠
0
{\displaystyle g(1)\neq \;0}
0
≠
g
(
1
)
=
g
(
1
2
+
1
2
)
=
g
(
1
2
)
⋅
g
(
1
2
)
=
g
(
1
2
)
2
>
0
{\displaystyle 0\neq \;g(1)=g\left({\frac {1}{2}}+{\frac {1}{2}}\right)=g\left({\frac {1}{2}}\right)\cdot g\left({\frac {1}{2}}\right)={g\left({\frac {1}{2}}\right)}^{2}>0}
f
:
R
→
R
{\displaystyle f:\mathbb {R} \;\rightarrow \;\mathbb {R} \;}
g
:
R
→
R
{\displaystyle g:\mathbb {R} \;\rightarrow \;\mathbb {R} \;}
t
∈
R
{\displaystyle t\in \mathbb {R} \;}
g
(
t
)
=
g
(
1
)
t
=
(
f
(
1
)
−
1
t
{\displaystyle g(t)={g(1)}^{t}={(f(1)-1}^{t}}
f
(
t
)
=
t
+
f
(
t
)
=
t
+
f
(
1
)
−
1
t
{\displaystyle f(t)=t+f(t)=t+{f(1)-1}^{t}}
M. O. Drimbe, 200 de ecuații funcționale pe N,Z,Q , Editura GIL, Zalău, 2003, ISBN 973-9417-10-8
A. Engel, Probleme de matematică. Strategii de rezolvare , Editura GIL, Zalău, 2006, ISBN 973-9417-65-5 
