Fie
(
A
,
+
,
⋅
)
{\displaystyle (A,+,\cdot )}
inel integru .
Se numește serie formală , într-o variabilă cu coeficienți în inelul
A
,
{\displaystyle A,}
funcție
f
:
N
→
A
.
{\displaystyle f:\mathbb {N} \to A.}
Fie mulțimea valorilor lui
f
:
I
m
f
=
{
a
0
,
a
1
,
a
2
,
⋯
,
a
n
,
⋯
}
.
{\displaystyle f:\;\;Im\;f=\{a_{0},a_{1},a_{2},\cdots ,a_{n},\cdots \}.}
a
0
+
a
1
T
+
a
2
T
2
+
⋯
a
n
T
n
+
⋯
,
{\displaystyle a_{0}+a_{1}T+a_{2}T^{2}+\cdots a_{n}T^{n}+\cdots ,}
unde
T
=
{
0
,
1
,
0
,
0
,
⋯
,
0
,
⋯
}
∈
A
N
{\displaystyle T=\{0,1,0,0,\cdots ,0,\cdots \}\in A^{\mathbb {N} }}
șirul de numere reale care are al doilea element 1, iar celelalte elemente zero.
Deoarece seriei formale i se poate asocia expresia de mai sus, în loc de
f
:
N
→
A
{\displaystyle f:\mathbb {N} \to A}
f
=
a
0
+
a
1
T
+
a
2
T
2
+
⋯
a
n
T
n
+
⋯
,
{\displaystyle f=a_{0}+a_{1}T+a_{2}T^{2}+\cdots a_{n}T^{n}+\cdots ,}
a
0
,
a
1
,
a
2
,
⋯
,
a
n
,
⋯
{\displaystyle a_{0},a_{1},a_{2},\cdots ,a_{n},\cdots }
f .
Mulțimea seriilor formale într-o variabilă cu coeficienți în inelul integru
A
{\displaystyle A}
A
[
[
T
]
]
.
{\displaystyle A[[T]].}
1
)
{\displaystyle 1)}
A
{\displaystyle A}
f
=
a
0
+
a
1
T
+
⋯
+
a
n
T
n
+
⋯
∈
A
[
[
T
]
]
,
{\displaystyle f=a_{0}+a_{1}T+\cdots +a_{n}T^{n}+\cdots \in A[[T]],}
astfel încât există
k
∈
N
{\displaystyle k\in \mathbb {N} }
a
k
≠
0
,
{\displaystyle a_{k}\neq 0,}
a
k
+
1
=
a
k
+
2
=
⋯
=
a
k
+
s
=
0
,
∀
s
≥
1.
{\displaystyle a_{k+1}=a_{k+2}=\cdots =a_{k+s}=0,\;\forall s\geq 1.}
k se numește gradul polinomului f , iar f se mai scrie sub forma:
f
=
a
0
+
a
1
T
+
a
2
T
2
+
⋯
+
a
n
T
n
.
{\displaystyle f=a_{0}+a_{1}T+a_{2}T^{2}+\cdots +a_{n}T^{n}.}
Dacă
a
k
=
0
,
∀
k
∈
N
,
{\displaystyle a_{k}=0,\;\forall k\in \mathbb {N} ,}
f se consideră a fi
−
∞
.
{\displaystyle -\infty .}
2
)
{\displaystyle 2)}
f
=
1
+
T
+
T
2
+
⋯
+
T
n
+
⋯
∈
Z
[
[
T
]
]
.
{\displaystyle f=1+T+T^{2}+\cdots +T^{n}+\cdots \in \mathbb {Z} [[T]].}
3
)
{\displaystyle 3)}
f
=
1
+
T
3
+
T
2
9
+
T
3
27
+
⋯
+
T
n
3
n
+
⋯
∈
Q
[
[
T
]
]
.
{\displaystyle f=1+{\frac {T}{3}}+{\frac {T^{2}}{9}}+{\frac {T^{3}}{27}}+\cdots +{\frac {T^{n}}{3^{n}}}+\cdots \in \mathbb {Q} [[T]].}
4
)
{\displaystyle 4)}
f
=
T
2
−
2
T
3
+
3
T
4
−
⋯
+
(
n
−
1
)
(
−
1
)
n
T
n
+
⋯
∈
Z
[
[
T
]
]
.
{\displaystyle f=T^{2}-2T^{3}+3T^{4}-\cdots +(n-1)(-1)^{n}T^{n}+\cdots \in \mathbb {Z} [[T]].}
O
b
s
e
r
v
a
t
′
i
e
.
{\displaystyle Observa{\underset {'}{t}}ie.}
1
{\displaystyle 1}
număr natural bine determinat, gradul acestuia.
Pentru o serie formală oarecare
f
=
a
0
+
a
1
T
+
a
2
T
2
+
⋯
∈
A
[
[
T
]
]
{\displaystyle f=a_{0}+a_{1}T+a_{2}T^{2}+\cdots \in A[[T]]}
k
{\displaystyle k}
a
k
+
l
=
0
{\displaystyle a_{k+l}=0}
l
≥
1.
{\displaystyle l\geq 1.}
l
{\displaystyle l}
a
l
≠
0
{\displaystyle a_{l}\neq 0}
l
=
0
{\displaystyle l=0}
Fie
f
=
a
0
+
a
1
T
+
⋯
+
a
n
T
n
+
⋯
{\displaystyle f=a_{0}+a_{1}T+\cdots +a_{n}T^{n}+\cdots }
g
=
b
0
+
b
1
T
+
⋯
+
b
n
T
n
+
⋯
{\displaystyle g=b_{0}+b_{1}T+\cdots +b_{n}T^{n}+\cdots }
A
.
{\displaystyle A.}
f
+
g
=
a
0
+
b
0
+
(
a
1
+
b
1
)
T
+
⋯
+
(
a
n
+
b
n
)
T
+
⋯
{\displaystyle f+g=a_{0}+b_{0}+(a_{1}+b_{1})T+\cdots +(a_{n}+b_{n})T+\cdots }
f
g
=
a
0
b
0
+
(
a
0
b
1
+
a
1
b
0
)
T
+
⋯
+
(
a
0
b
i
+
a
1
b
i
−
1
+
⋯
+
a
i
−
1
b
1
+
a
i
b
0
)
T
i
+
⋯
{\displaystyle fg=a_{0}b_{0}+(a_{0}b_{1}+a_{1}b_{0})T+\cdots +(a_{0}b_{i}+a_{1}b_{i-1}+\cdots +a_{i-1}b_{1}+a_{i}b_{0})T^{i}+\cdots }
P
r
o
p
o
z
i
t
′
i
e
.
{\displaystyle Propozi{\underset {'}{t}}ie.}
(
A
,
+
,
⋅
)
{\displaystyle (A,+,\cdot )}
inel comutativ , atunci și
(
A
,
[
[
T
]
]
,
+
,
⋅
)
{\displaystyle (A,[[T]],+,\cdot )}
D
e
m
o
n
s
t
r
a
t
′
i
e
.
{\displaystyle Demonstra{\underset {'}{t}}ie.}
A
{\displaystyle A}
asociative și comutative deoarece "adunarea" și "înmulțirea" din inelul
A
{\displaystyle A}
Seria formală
0
=
0
+
0
⋅
T
+
⋯
+
0
⋅
T
n
+
⋯
{\displaystyle 0=0+0\cdot T+\cdots +0\cdot T^{n}+\cdots }
element neutru pentru adunarea seriilor formale.
Dacă
f
=
a
0
+
a
1
T
+
⋯
+
a
n
T
n
+
⋯
∈
A
[
[
T
]
]
,
{\displaystyle f=a_{0}+a_{1}T+\cdots +a_{n}T^{n}+\cdots \in A[[T]],}
−
f
=
−
a
0
+
(
−
a
1
)
T
+
⋯
+
(
−
a
n
)
T
n
{\displaystyle -f=-a_{0}+(-a_{1})T+\cdots +(-a_{n})T^{n}}
f
{\displaystyle f}
f
+
(
−
f
)
=
(
−
f
)
+
f
=
0.
{\displaystyle f+(-f)=(-f)+f=0.}
1
=
1
+
0
⋅
T
+
⋯
+
0
⋅
T
n
+
⋯
{\displaystyle 1=1+0\cdot T+\cdots +0\cdot T^{n}+\cdots }
P
r
o
p
o
z
i
t
′
i
e
.
{\displaystyle Propozi{\underset {'}{t}}ie.}
A
:
{\displaystyle A:}
f
=
a
0
+
a
1
T
+
a
2
T
2
+
⋯
+
a
n
T
n
+
⋯
{\displaystyle f=a_{0}+a_{1}T+a_{2}T^{2}+\cdots +a_{n}T^{n}+\cdots }
este inversabilă în
A
[
[
T
]
]
{\displaystyle A[[T]]}
dacă și numai dacă elementul
a
0
{\displaystyle a_{0}}
A
.
{\displaystyle A.}
D
e
m
o
n
s
t
r
a
t
′
i
e
.
{\displaystyle Demonstra{\underset {'}{t}}ie.}
f
=
a
0
+
a
1
T
+
a
2
T
2
+
⋯
+
a
n
T
n
+
⋯
{\displaystyle f=a_{0}+a_{1}T+a_{2}T^{2}+\cdots +a_{n}T^{n}+\cdots }
A
[
[
T
]
]
,
{\displaystyle A[[T]],}
a
0
{\displaystyle a_{0}}
A
.
{\displaystyle A.}
g
=
b
0
+
b
1
T
+
b
2
T
2
+
⋯
+
b
n
T
n
+
⋯
∈
A
[
[
T
]
]
{\displaystyle g=b_{0}+b_{1}T+b_{2}T^{2}+\cdots +b_{n}T^{n}+\cdots \in A[[T]]}
(
a
0
+
a
1
T
+
a
2
T
2
+
⋯
+
a
n
T
n
+
⋯
)
⋅
(
b
0
+
b
1
T
+
b
2
T
2
+
⋯
+
b
n
T
n
+
⋯
)
=
1.
{\displaystyle (a_{0}+a_{1}T+a_{2}T^{2}+\cdots +a_{n}T^{n}+\cdots )\cdot (b_{0}+b_{1}T+b_{2}T^{2}+\cdots +b_{n}T^{n}+\cdots )=1.}
a
0
b
0
=
1
,
{\displaystyle a_{0}b_{0}=1,}
a
0
{\displaystyle a_{0}}
A
.
{\displaystyle A.}
Reciproc, acum se presupune că elementul
a
0
{\displaystyle a_{0}}
A
{\displaystyle A}
f
=
a
0
+
a
1
T
+
a
2
T
2
+
⋯
+
a
n
T
n
+
⋯
{\displaystyle f=a_{0}+a_{1}T+a_{2}T^{2}+\cdots +a_{n}T^{n}+\cdots }
A
[
[
T
]
]
.
{\displaystyle A[[T]].}
g
=
b
0
+
b
1
T
+
b
2
T
2
+
⋯
+
b
n
T
n
+
⋯
∈
a
[
[
T
]
]
{\displaystyle g=b_{0}+b_{1}T+b_{2}T^{2}+\cdots +b_{n}T^{n}+\cdots \in a[[T]]}
f
g
=
1.
{\displaystyle fg=1.}
b
0
,
b
1
,
b
2
,
⋯
,
b
n
,
⋯
∈
A
{\displaystyle b_{0},b_{1},b_{2},\cdots ,b_{n},\cdots \in A}
a
0
b
0
=
1
{\displaystyle a_{0}b_{0}=1}
a
0
b
1
+
a
1
b
0
=
0
{\displaystyle a_{0}b_{1}+a_{1}b_{0}=0}
a
0
b
2
+
a
1
b
1
+
a
2
b
0
=
0
{\displaystyle a_{0}b_{2}+a_{1}b_{1}+a_{2}b_{0}=0}
⋯
⋯
⋯
⋯
⋯
⋯
{\displaystyle \cdots \cdots \cdots \cdots \cdots \cdots }
a
0
b
i
+
a
1
b
i
−
1
+
⋯
+
a
i
−
1
b
1
+
a
i
b
0
=
0.
{\displaystyle a_{0}b_{i}+a_{1}b_{i-1}+\cdots +a_{i-1}b_{1}+a_{i}b_{0}=0.}
⋯
⋯
⋯
⋯
⋯
⋯
{\displaystyle \cdots \cdots \cdots \cdots \cdots \cdots }
Din
a
0
b
0
=
1
{\displaystyle a_{0}b_{0}=1}
b
0
=
a
0
−
1
.
{\displaystyle b_{0}=a_{0}^{-1}.}
Din
a
0
b
1
+
a
1
b
0
=
0
{\displaystyle a_{0}b_{1}+a_{1}b_{0}=0}
b
1
=
a
0
−
1
(
−
a
1
b
0
)
.
{\displaystyle b_{1}=a_{0}^{-1}(-a_{1}b_{0}).}
Din
a
0
b
2
+
a
1
b
1
+
a
2
b
0
=
0
{\displaystyle a_{0}b_{2}+a_{1}b_{1}+a_{2}b_{0}=0}
b
2
=
a
0
−
1
(
−
a
1
b
1
−
a
2
b
0
)
.
{\displaystyle b_{2}=a_{0}^{-1}(-a_{1}b_{1}-a_{2}b_{0}).}
Dacă se presupune că sunt determinați
b
0
,
b
1
,
⋯
,
b
i
−
1
,
{\displaystyle b_{0},b_{1},\cdots ,b_{i-1},}
a
0
b
i
+
a
1
b
i
−
1
+
⋯
+
a
i
−
1
b
1
+
a
i
b
0
=
0
{\displaystyle a_{0}b_{i}+a_{1}b_{i-1}+\cdots +a_{i-1}b_{1}+a_{i}b_{0}=0}
b
i
=
a
0
−
1
(
−
a
1
b
i
−
1
−
⋯
−
a
i
−
1
b
1
−
a
i
b
0
)
.
{\displaystyle b_{i}=a_{0}^{-1}(-a_{1}b_{i-1}-\cdots -a_{i-1}b_{1}-a_{i}b_{0}).}
Deci există o serie formală
g
=
b
0
+
b
1
T
+
⋯
+
b
n
T
n
+
⋯
{\displaystyle g=b_{0}+b_{1}T+\cdots +b_{n}T^{n}+\cdots }
f
g
=
1.
{\displaystyle fg=1.}
1.
{\displaystyle 1.}
f
=
1
−
X
∈
Z
[
[
X
]
]
.
{\displaystyle f=1-X\in \mathbb {Z} [[X]].}
(
1
−
X
)
(
1
+
X
+
⋯
+
X
n
+
⋯
)
=
1
,
{\displaystyle (1-X)(1+X+\cdots +X^{n}+\cdots )=1,}
(
1
−
X
)
−
1
=
1
+
X
+
X
2
+
⋯
+
X
n
+
⋯
{\displaystyle (1-X)^{-1}=1+X+X^{2}+\cdots +X^{n}+\cdots }
Se observă că
1
−
X
{\displaystyle 1-X}
Z
[
[
X
]
]
,
{\displaystyle \mathbb {Z} [[X]],}
Z
[
X
]
.
{\displaystyle \mathbb {Z} [X].}
2.
{\displaystyle 2.}
f
=
1
2
−
1
2
T
+
1
2
T
2
∈
Q
[
[
T
]
]
.
{\displaystyle f={\frac {1}{2}}-{\frac {1}{2}}T+{\frac {1}{2}}T^{2}\in \mathbb {Q} [[T]].}
Elementul
1
2
∈
Q
{\displaystyle {\frac {1}{2}}\in \mathbb {Q} }
f
{\displaystyle f}
Q
[
[
T
]
]
.
{\displaystyle \mathbb {Q} [[T]].}
f
−
1
=
b
0
+
b
1
T
+
⋯
+
b
n
T
n
+
⋯
∈
Q
[
[
T
]
]
.
{\displaystyle f^{-1}=b_{0}+b_{1}T+\cdots +b_{n}T^{n}+\cdots \in \mathbb {Q} [[T]].}
Se obține:
(
1
2
−
1
2
T
+
1
2
T
2
)
⋅
(
b
0
+
b
1
T
+
⋯
+
b
n
T
n
+
⋯
)
=
1.
{\displaystyle \left({\frac {1}{2}}-{\frac {1}{2}}T+{\frac {1}{2}}T^{2}\right)\cdot (b_{0}+b_{1}T+\cdots +b_{n}T^{n}+\cdots )=1.}
Prin identificarea coeficienților, se obține:
1
2
b
0
=
1
,
{\displaystyle {\frac {1}{2}}b_{0}=1,}
deci
b
0
=
2
{\displaystyle b_{0}=2}
1
2
b
1
−
1
2
b
0
=
1
2
b
1
−
1
=
0
,
{\displaystyle {\frac {1}{2}}b_{1}-{\frac {1}{2}}b_{0}={\frac {1}{2}}b_{1}-1=0,}
deci
b
1
=
2
{\displaystyle b_{1}=2}
1
2
b
2
−
1
2
b
1
+
1
2
b
0
=
1
2
b
2
−
1
+
1
=
1
2
b
2
=
0
,
{\displaystyle {\frac {1}{2}}b_{2}-{\frac {1}{2}}b_{1}+{\frac {1}{2}}b_{0}={\frac {1}{2}}b_{2}-1+1={\frac {1}{2}}b_{2}=0,}
deci
b
2
=
0
{\displaystyle b_{2}=0}
1
2
b
3
−
1
2
b
2
+
1
2
b
1
=
1
2
b
3
+
1
=
0
,
{\displaystyle {\frac {1}{2}}b_{3}-{\frac {1}{2}}b_{2}+{\frac {1}{2}}b_{1}={\frac {1}{2}}b_{3}+1=0,}
deci
b
3
=
−
2
{\displaystyle b_{3}=-2}
1
2
b
4
−
1
2
b
3
+
1
2
b
2
=
1
2
b
4
+
1
=
0
,
{\displaystyle {\frac {1}{2}}b_{4}-{\frac {1}{2}}b_{3}+{\frac {1}{2}}b_{2}={\frac {1}{2}}b_{4}+1=0,}
deci
b
4
=
−
2
{\displaystyle b_{4}=-2}
1
2
b
5
−
1
2
b
4
+
1
2
b
3
=
1
2
b
5
+
1
−
1
=
1
2
b
5
=
0
,
{\displaystyle {\frac {1}{2}}b_{5}-{\frac {1}{2}}b_{4}+{\frac {1}{2}}b_{3}={\frac {1}{2}}b_{5}+1-1={\frac {1}{2}}b_{5}=0,}
deci
b
5
=
0
{\displaystyle b_{5}=0}
1
2
b
6
−
1
2
b
5
+
1
2
b
4
=
1
2
b
6
−
1
=
0
,
{\displaystyle {\frac {1}{2}}b_{6}-{\frac {1}{2}}b_{5}+{\frac {1}{2}}b_{4}={\frac {1}{2}}b_{6}-1=0,}
deci
b
6
=
2.
{\displaystyle b_{6}=2.}
Deci coeficienții se repetă. Prin urmare:
(
1
2
−
1
2
T
+
1
2
T
2
)
−
1
=
{\displaystyle \left({\frac {1}{2}}-{\frac {1}{2}}T+{\frac {1}{2}}T^{2}\right)^{-1}=}
=
2
+
2
T
−
2
T
3
−
2
T
4
+
⋯
−
2
T
6
k
−
2
+
2
T
6
k
+
2
T
6
k
+
1
−
2
T
6
k
+
3
−
2
T
6
k
+
4
+
⋯
{\displaystyle =2+2T-2T^{3}-2T^{4}+\cdots -2T^{6k-2}+2T^{6k}+2T^{6k+1}-2T^{6k+3}-2T^{6k+4}+\cdots }
Fie
f
=
1
+
T
+
T
2
+
⋯
∈
Z
[
[
T
]
]
{\displaystyle f=1+T+T^{2}+\cdots \in \mathbb {Z} [[T]]}
g
=
(
1
+
T
)
(
1
+
T
2
)
(
1
+
T
4
)
⋯
∈
Z
[
[
T
]
]
.
{\displaystyle g=(1+T)(1+T^{2})(1+T^{4})\cdots \in \mathbb {Z} [[T]].}
f
=
g
.
{\displaystyle f=g.}
(
1
−
T
)
f
=
(
1
−
T
)
(
1
+
T
+
T
2
+
⋯
)
=
1
{\displaystyle (1-T)f=(1-T)(1+T+T^{2}+\cdots )=1}
(
1
−
T
)
g
=
(
1
−
T
)
(
1
+
T
)
(
1
+
T
2
)
(
1
+
T
4
)
⋅
⋯
=
(
1
−
T
2
)
(
1
+
T
2
)
(
1
+
T
4
)
⋅
⋯
=
{\displaystyle (1-T)g=(1-T)(1+T)(1+T^{2})(1+T^{4})\cdot \cdots =(1-T^{2})(1+T^{2})(1+T^{4})\cdot \cdots =}
=
(
1
−
T
4
)
(
1
+
T
4
)
⋅
⋯
=
(
1
T
2
k
)
(
1
+
T
2
k
)
(
1
+
T
2
k
+
1
)
⋅
⋯
{\displaystyle =(1-T^{4})(1+T^{4})\cdot \cdots =(1_{T}^{2k})(1+T^{2k})(1+T^{2k+1})\cdot \cdots }
⋯
⋯
⋯
⋯
⋯
=
1
{\displaystyle \cdots \cdots \cdots \cdots \cdots =1}
T
i
{\displaystyle T_{i}}
Prin urmare
(
1
−
T
)
f
=
(
1
−
T
)
g
.
{\displaystyle (1-T)f=(1-T)g.}
(
1
−
T
)
−
1
(
1
−
T
)
f
=
(
1
−
T
)
−
1
(
1
−
T
)
g
,
{\displaystyle (1-T)^{-1}(1-T)f=(1-T)^{-1}(1-T)g,}
f
=
g
.
{\displaystyle f=g.}
Se vor defini cu ajutorul seriilor formale unele funcții elementare care sunt utilizate frecvent.
Pentru a demonstra unele proprietăți ale acestor funcții, se va utiliza noțiunea de derivată a unei serii formale.
D
e
f
i
n
i
t
′
i
e
.
{\displaystyle Defini{\underset {'}{t}}ie.}
A
:
{\displaystyle A:}
f
=
a
0
+
a
1
T
+
⋯
+
a
n
T
n
+
⋯
{\displaystyle f=a_{0}+a_{1}T+\cdots +a_{n}T^{n}+\cdots }
seria formală:
f
′
=
a
1
+
2
a
2
T
+
⋅
+
n
a
n
T
n
−
1
+
⋯
{\displaystyle f^{\prime }=a_{1}+2a_{2}T+\cdot +na_{n}T^{n-1}+\cdots }
Derivata unei serii formale
f
{\displaystyle f}
d
f
{\displaystyle df}
d
f
/
d
T
.
{\displaystyle df/dT.}
Se remarcă faptul că dacă
f
{\displaystyle f}
d
f
{\displaystyle df}
f
:
R
→
R
,
f
(
x
)
=
a
0
+
a
1
x
+
⋯
+
a
n
x
n
+
⋯
{\displaystyle f:\mathbb {R} \to \mathbb {R} ,\;f(x)=a_{0}+a_{1}x+\cdots +a_{n}x^{n}+\cdots }
O
b
s
e
r
v
a
t
′
i
e
.
d
f
=
0
⇔
o
r
d
f
=
0.
{\displaystyle Observa{\underset {'}{t}}ie.\;\;df=0\;\Leftrightarrow \;ord\;f=0.}
D
e
f
i
n
i
t
′
i
e
.
{\displaystyle Defini{\underset {'}{t}}ie.}
sinus formal următoarea serie formală într-o variabilă cu coeficienți reali:
sin
T
=
T
−
T
3
3
!
+
T
5
5
!
−
⋯
+
(
−
1
)
k
T
2
k
+
1
(
2
k
+
1
)
!
+
⋯
{\displaystyle \sin T=T-{\frac {T^{3}}{3!}}+{\frac {T^{5}}{5!}}-\cdots +(-1)^{k}{\frac {T^{2k+1}}{(2k+1)!}}+\cdots }
Se numește funcția cosinus formal următoarea serie formală într-o variabilă cu coeficienți reali:
sin
T
=
1
−
T
2
2
!
+
T
4
4
!
−
⋯
+
(
−
1
)
k
T
2
k
(
2
k
)
!
+
⋯
{\displaystyle \sin T=1-{\frac {T^{2}}{2!}}+{\frac {T^{4}}{4!}}-\cdots +(-1)^{k}{\frac {T^{2k}}{(2k)!}}+\cdots }
T
e
o
r
e
m
a
˘
.
{\displaystyle Teorem{\breve {a}}.}
funcțiile trigonometrice formale există relațiile:
i
)
sin
(
−
x
)
=
−
sin
x
,
cos
(
−
x
)
=
cos
x
{\displaystyle i)\;\sin(-x)=-\sin x,\;\cos(-x)=\cos x}
i
i
)
sin
′
x
=
cos
x
,
cos
′
x
=
−
sin
x
.
{\displaystyle ii)\;\sin 'x=\cos x,\;\cos 'x=-\sin x.}
D
e
m
o
n
s
t
r
a
t
′
i
e
.
{\displaystyle Demonstra{\underset {'}{t}}ie.}
i
)
sin
(
−
x
)
=
−
x
+
x
3
3
!
−
⋯
+
(
−
1
)
k
+
1
+
⋯
=
−
sin
x
.
{\displaystyle i)\;\sin(-x)=-x+{\frac {x^{3}}{3!}}-\cdots +(-1)^{k+1}+\cdots =-\sin x.}
cos
(
−
x
)
=
1
−
x
2
2
!
+
x
4
4
!
−
⋯
+
(
−
1
)
k
x
2
k
(
2
k
)
!
+
⋯
=
cos
x
.
{\displaystyle \cos(-x)=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-\cdots +(-1)^{k}{\frac {x^{2k}}{(2k)!}}+\cdots =\cos x.}
i
i
)
sin
′
x
=
1
−
3
x
3
3
!
+
⋯
+
(
−
1
)
k
(
2
k
+
1
)
x
2
k
(
2
k
+
1
)
!
+
⋯
=
cos
x
{\displaystyle ii)\;\sin 'x=1-3{\frac {x^{3}}{3!}}+\cdots +(-1)^{k}{\frac {(2k+1)x^{2k}}{(2k+1)!}}+\cdots =\cos x}
cos
′
x
=
−
2
x
2
!
+
⋯
+
(
−
1
)
k
2
k
x
2
k
−
1
(
2
k
)
!
+
⋯
=
−
sin
x
.
{\displaystyle \cos 'x=-{\frac {2x}{2!}}+\cdots +(-1)^{k}{\frac {2kx^{2k-1}}{(2k)!}}+\cdots =-\sin x.}
De remarcat faptul că:
sin
0
=
0
,
cos
0
=
1.
{\displaystyle \sin 0=0,\;\cos 0=1.}
T
e
o
r
e
m
a
˘
.
{\displaystyle Teorem{\breve {a}}.}
a
,
x
∈
R
{\displaystyle a,x\in \mathbb {R} }
sin
(
a
+
x
)
=
sin
a
⋅
cos
x
+
cos
a
⋅
sin
x
{\displaystyle \sin(a+x)=\sin a\cdot \cos x+\cos a\cdot \sin x}
cos
(
a
+
x
)
=
cos
a
⋅
cos
x
−
sin
a
⋅
sin
x
.
{\displaystyle \cos(a+x)=\cos a\cdot \cos x-\sin a\cdot \sin x.}
D
e
m
o
n
s
t
r
a
t
′
i
e
.
{\displaystyle Demonstra{\underset {'}{t}}ie.}
x
{\displaystyle x}
F
1
(
x
)
=
sin
(
a
+
x
)
−
sin
a
⋅
cos
x
−
cos
a
⋅
sin
x
{\displaystyle F_{1}(x)=\sin(a+x)-\sin a\cdot \cos x-\cos a\cdot \sin x}
F
2
(
x
)
=
cos
(
a
+
x
)
−
cos
a
⋅
cos
x
+
sin
a
⋅
sin
x
{\displaystyle F_{2}(x)=\cos(a+x)-\cos a\cdot \cos x+\sin a\cdot \sin x}
F
(
x
)
=
F
1
2
(
x
)
=
F
1
2
(
x
)
+
F
2
2
(
x
)
.
{\displaystyle F(x)=F_{1}^{2}(x)=F_{1}^{2}(x)+F_{2}^{2}(x).}
Se remarcă faptul că:
F
1
(
0
)
=
F
2
(
0
)
=
0.
{\displaystyle F_{1}(0)=F_{2}(0)=0.}
F
1
′
(
x
)
=
cos
(
a
+
x
)
+
sin
a
⋅
sin
x
−
cos
a
⋅
cos
x
=
F
2
(
x
)
{\displaystyle F'_{1}(x)=\cos(a+x)+\sin a\cdot \sin x-\cos a\cdot \cos x=F_{2}(x)}
F
2
′
(
x
)
=
−
sin
(
a
+
x
)
+
cos
a
⋅
sin
x
+
sin
a
⋅
cos
x
=
−
F
1
(
x
)
.
{\displaystyle F'_{2}(x)=-\sin(a+x)+\cos a\cdot \sin x+\sin a\cdot \cos x=-F_{1}(x).}
Derivând și seria formală
F
(
x
)
{\displaystyle F(x)}
F
′
(
x
)
=
2
F
1
(
x
)
F
1
′
(
x
)
+
2
F
2
(
x
)
F
2
′
(
x
)
=
2
F
1
(
x
)
F
2
(
x
)
−
2
F
2
(
x
)
F
1
(
x
)
=
0.
{\displaystyle F'(x)=2F_{1}(x)F'_{1}(x)+2F_{2}(x)F'_{2}(x)=2F_{1}(x)F_{2}(x)-2F_{2}(x)F_{1}(x)=0.}
Dacă
F
(
x
)
{\displaystyle F(x)}
F
(
x
)
=
c
,
c
∈
R
.
{\displaystyle F(x)=c,\;c\in \mathbb {R} .}
F
(
0
)
=
0
,
{\displaystyle F(0)=0,}
F
(
x
)
=
0.
{\displaystyle F(x)=0.}
F
1
(
x
)
=
F
2
(
x
)
=
0
,
{\displaystyle F_{1}(x)=F_{2}(x)=0,}
sin
(
a
+
x
)
=
sin
a
⋅
cos
x
+
cos
a
⋅
sin
x
{\displaystyle \sin(a+x)=\sin a\cdot \cos x+\cos a\cdot \sin x}
cos
(
a
+
x
)
=
cos
a
⋅
cos
x
−
sin
a
⋅
sin
x
.
{\displaystyle \cos(a+x)=\cos a\cdot \cos x-\sin a\cdot \sin x.}
D
e
f
i
n
i
t
′
i
e
.
{\displaystyle Defini{\underset {'}{t}}ie.}
funcție exponențială
exp
:
R
→
R
{\displaystyle \exp :\mathbb {R} \to \mathbb {R} }
exp
T
=
1
+
T
1
!
+
⋯
+
T
n
n
!
+
⋯
{\displaystyle \exp T=1+{\frac {T}{1!}}+\cdots +{\frac {T^{n}}{n!}}+\cdots }
P
r
o
p
o
z
i
t
′
i
e
.
(
exp
x
)
′
=
exp
x
,
∀
x
∈
R
.
{\displaystyle Propozi{\underset {'}{t}}ie.\;(\exp x)'=\exp x,\;\forall x\in \mathbb {R} .}
D
e
m
o
n
s
t
r
a
t
′
i
e
.
(
exp
x
)
′
=
1
+
2
x
2
!
+
⋯
+
n
x
n
−
1
n
!
+
⋯
=
exp
x
.
{\displaystyle Demonstra{\underset {'}{t}}ie.\;(\exp x)'=1+{\frac {2x}{2!}}+\cdots +{\frac {nx^{n-1}}{n!}}+\cdots =\exp x.}
T
e
o
r
e
m
a
˘
.
exp
(
x
+
y
)
=
exp
x
⋅
exp
y
,
∀
x
,
y
∈
R
.
{\displaystyle Teorem{\breve {a}}.\;\exp(x+y)=\exp x\cdot \exp y,\;\forall x,y\in \mathbb {R} .}
D
e
m
o
n
s
t
r
a
t
′
i
e
.
{\displaystyle Demonstra{\underset {'}{t}}ie.}
exp
x
⋅
exp
y
=
(
1
+
x
1
!
+
⋯
+
x
n
n
!
+
⋯
)
⋅
(
1
+
y
1
!
+
⋯
+
y
n
n
!
+
⋯
)
=
{\displaystyle \exp x\cdot \exp y=\left(1+{\frac {x}{1!}}+\cdots +{\frac {x^{n}}{n!}}+\cdots \right)\cdot \left(1+{\frac {y}{1!}}+\cdots +{\frac {y^{n}}{n!}}+\cdots \right)=}
=
a
0
+
a
1
t
+
⋯
+
a
n
t
n
+
⋯
,
{\displaystyle =a_{0}+a_{1}t+\cdots +a_{n}t^{n}+\cdots ,}
unde
a
n
t
n
=
∑
p
=
0
n
x
p
y
n
−
p
p
!
(
n
−
p
)
!
=
1
n
!
∑
p
=
0
n
n
!
p
!
(
n
−
p
)
!
x
p
y
n
−
p
=
1
n
!
(
x
+
y
)
n
.
{\displaystyle a_{n}t^{n}=\sum _{p=0}^{n}{\frac {x^{p}y^{n-p}}{p!(n-p)!}}={\frac {1}{n!}}\sum _{p=0}^{n}{\frac {n!}{p!(n-p)!}}x^{p}y^{n-p}={\frac {1}{n!}}(x+y)^{n}.}
Prin urmare:
exp
x
⋅
exp
y
=
1
+
x
+
y
1
!
+
⋯
+
(
x
+
y
)
n
n
!
+
⋯
=
exp
(
x
+
y
)
.
{\displaystyle \exp x\cdot \exp y=1+{\frac {x+y}{1!}}+\cdots +{\frac {(x+y)^{n}}{n!}}+\cdots =\exp(x+y).}
Pentru funcția exponențială
exp
x
{\displaystyle \exp x}
e
x
.
{\displaystyle e^{x}.}
Se remarcă faptul că
exp
0
=
1
,
exp
(
−
x
)
=
(
exp
x
)
−
1
,
exp
x
≠
0
,
∀
x
∈
R
.
{\displaystyle \exp 0=1,\;\exp(-x)=(\exp x)^{-1},\;\exp x\neq 0,\;\forall x\in \mathbb {R} .}
Miron Nicolescu - Analiză matematică , vol. I, 1957;P. Samuel, O. Zariski - Comutative algebra , vol. I, 1959;
N. Radu, I. D. Ion - Algebra , 1970. 
![{\displaystyle A[[T]].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e316b955931f74385e0a9422d830ec3449899623)
![{\displaystyle f=a_{0}+a_{1}T+\cdots +a_{n}T^{n}+\cdots \in A[[T]],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/403c592ee6c186fea308fbb4860b8d0c24ed927d)
![{\displaystyle f=1+T+T^{2}+\cdots +T^{n}+\cdots \in \mathbb {Z} [[T]].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/84783a293f5d84f38b5cdd85008aaebe4bdcbfd4)
![{\displaystyle f=1+{\frac {T}{3}}+{\frac {T^{2}}{9}}+{\frac {T^{3}}{27}}+\cdots +{\frac {T^{n}}{3^{n}}}+\cdots \in \mathbb {Q} [[T]].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/710fbadfad70f55d711e8bdcf1d796cc49b9b5bb)
![{\displaystyle f=T^{2}-2T^{3}+3T^{4}-\cdots +(n-1)(-1)^{n}T^{n}+\cdots \in \mathbb {Z} [[T]].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f51da7cb2252f4791103be811492438c9925fad6)
![{\displaystyle f=a_{0}+a_{1}T+a_{2}T^{2}+\cdots \in A[[T]]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8348f72ddb1d0ee07752dc77810f55ccf59a712d)
![{\displaystyle (A,[[T]],+,\cdot )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/93491c7952ab1df481c1dc4009098e6523914390)
![{\displaystyle A[[T]]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4a21a6184c65fc40563cad8990c44d304b1fae88)
![{\displaystyle A[[T]],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6970187aa8e4d942e069419003519f18d9f2b2dc)
![{\displaystyle g=b_{0}+b_{1}T+b_{2}T^{2}+\cdots +b_{n}T^{n}+\cdots \in A[[T]]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d587bbd8bc7016654974a955a99d7cd28960866d)
![{\displaystyle g=b_{0}+b_{1}T+b_{2}T^{2}+\cdots +b_{n}T^{n}+\cdots \in a[[T]]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/600880f17ef3058036636593e58b28fb6d04873c)
![{\displaystyle f=1-X\in \mathbb {Z} [[X]].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/87c51225ca0ddb8e94994ffa1e24bd50420120aa)
![{\displaystyle \mathbb {Z} [[X]],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8cbcc2a9dd4557e1a88962deb55a3d7cdfc84f50)
![{\displaystyle \mathbb {Z} [X].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6dff5f5cc8f92c9047be4851007baca114ceb3e2)
![{\displaystyle f={\frac {1}{2}}-{\frac {1}{2}}T+{\frac {1}{2}}T^{2}\in \mathbb {Q} [[T]].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/119970bc14bf7f045668f5af528dcc842a9b0062)
![{\displaystyle \mathbb {Q} [[T]].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fe38f8cb031d9dd870a1eb5768be88eea6a97708)
![{\displaystyle f^{-1}=b_{0}+b_{1}T+\cdots +b_{n}T^{n}+\cdots \in \mathbb {Q} [[T]].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/93f934e4b7220de2063e9240f3ac4fed4495699a)
![{\displaystyle f=1+T+T^{2}+\cdots \in \mathbb {Z} [[T]]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/714c5ca36ba00d71e7f82f7c9534674cdf61f53e)
![{\displaystyle g=(1+T)(1+T^{2})(1+T^{4})\cdots \in \mathbb {Z} [[T]].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a451852bbb4cc8e4934e347f8f3390c7c156165e)
