Pentru ca o funcție f să fie dezvoltabilă în serie Taylor este necesar ca f să fie indefinit derivabilă .
Această restricţie este severă.
Chiar dacă se foloseşte formula Taylor cu restcontinuitatea funcţiei.
Multe funcţii însă, care descriu fenomene fizice importante, nu sunt continue şi nu pot fi reprezentate nici cu formula Taylor.
De exemplu, funcţia care descrie evoluţia în timp a tensiunii într-un circuit electric în care apar comutări de contacte, nu este continuă.
Seriile Fourier oferă posibilitatea de reprezentare a funcţiilor continue şi continue pe porţiuni.
Aceasta întrucât pentru a construi seria Fourier, funcţia trebuie să fie doar integrabilă Riemann-Darboux.
Definiţie .
Fie
f
:
[
−
π
,
π
]
→
R
{\displaystyle f:[-\pi ,\pi ]\rightarrow \mathbb {R} \!}
funcție continuă pe porţiuni.
Seria Fourier a lui f este, prin definiţie, seria de funcţii :
f
(
x
)
∼
a
0
2
+
∑
n
=
1
∞
(
a
n
⋅
cos
n
x
+
b
n
⋅
sin
n
x
)
.
{\displaystyle f(x)\sim {\frac {a_{0}}{2}}+\sum _{n=1}^{\infty }(a_{n}\cdot \cos nx+b_{n}\cdot \sin nx).\!}
în care coeficienţii Fourier
a
n
,
b
n
{\displaystyle a_{n},b_{n}\!}
a
n
=
1
π
∫
−
π
π
f
(
x
)
⋅
cos
n
x
d
x
n
=
0
,
1
,
2
,
⋯
{\displaystyle a_{n}={\frac {1}{\pi }}\int _{-\pi }^{\pi }f(x)\cdot \cos nx\;dx\;\;n=0,1,2,\cdots \!}
a
n
=
1
π
∫
−
π
π
f
(
x
)
⋅
sin
n
x
d
x
n
=
0
,
1
,
2
,
⋯
{\displaystyle a_{n}={\frac {1}{\pi }}\int _{-\pi }^{\pi }f(x)\cdot \sin nx\;dx\;\;n=0,1,2,\cdots \!}
Observaţie .
În mod tradiţional, până când problema convergenţei seriei Fourier nu este tranşată, relaţia dintre funcţia f şi seria Fourier a lui f se notează cu semnul
∼
{\displaystyle \sim \!}
Rezultatul fundamental care va fi stabilit în această secţiune se referă la convergenţa seriei Fourier a unei funcţii continue pe porţiuni şi la suma seriei.
Acest rezultat se bazează pe două leme ce vor fi prezentate ulterior.
Lema 1 . [de reprezentare integrală a sumei parţiale
S
n
(
x
)
{\displaystyle S_{n}(x)\!}
n ,
S
n
(
x
)
,
{\displaystyle S_{n}(x),\!}
f
:
[
−
π
,
π
]
→
R
{\displaystyle f:[-\pi ,\pi ]\rightarrow \mathbb {R} \!}
R
{\displaystyle \mathbb R \!}
S
n
(
x
)
=
1
π
∫
−
π
π
f
(
x
−
u
)
⋅
sin
(
n
+
1
2
)
u
2
sin
1
2
u
d
u
,
x
∈
R
.
{\displaystyle S_{n}(x)={\frac {1}{\pi }}\int _{-\pi }^{\pi }f(x-u)\cdot {\frac {\sin(n+{\frac {1}{2}})u}{2\sin {\frac {1}{2}}u}}du,\;\;x\in \mathbb {R} .\!}
Demonstraţie .
Considerăm suma parţială
S
n
(
x
)
{\displaystyle S_{n}(x)\!}
f :
S
n
(
x
)
=
a
0
2
+
∑
k
=
1
n
(
a
k
⋅
cos
k
x
+
b
k
⋅
sin
k
x
)
.
{\displaystyle S_{n}(x)={\frac {a_{0}}{2}}+\sum _{k=1}^{n}(a_{k}\cdot \cos kx+b_{k}\cdot \sin kx).\!}
În virtutea definiţiei coeficienţilor Fourier
a
k
,
b
k
,
{\displaystyle a_{k},b_{k},\!}
S
n
(
x
)
=
1
2
π
∫
−
π
π
f
(
t
)
d
t
+
1
π
∑
k
=
1
n
[
cos
k
x
∫
−
π
π
cos
k
t
d
t
+
sin
k
x
∫
−
π
π
f
(
t
)
sin
k
t
d
t
]
{\displaystyle S_{n}(x)={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(t)dt+{\frac {1}{\pi }}\sum _{k=1}^{n}\left[\cos kx\int _{-\pi }^{\pi }\cos ktdt+\sin kx\int _{-\pi }^{\pi }f(t)\sin ktdt\right]\!}
Introducând
cos
k
x
{\displaystyle \cos kx\!}
sin
k
x
{\displaystyle \sin kx\!}
cos
k
(
x
−
t
)
=
cos
k
x
⋅
cos
k
t
+
sin
k
x
⋅
sin
k
t
,
{\displaystyle \cos k(x-t)=\cos kx\cdot \cos kt+\sin kx\cdot \sin kt,\!}
obţinem:
S
n
(
x
)
=
1
π
∫
−
π
π
f
(
t
)
1
2
+
∑
k
=
1
n
cos
k
(
x
−
t
)
d
t
.
{\displaystyle S_{n}(x)={\frac {1}{\pi }}\int _{-\pi }^{\pi }f(t){\frac {1}{2}}+\sum _{k=1}^{n}\cos k(x-t)dt.\!}
Aplicând identitatea:
1
2
+
∑
k
=
1
n
cos
k
(
x
−
t
)
=
sin
(
n
+
1
2
)
(
x
−
t
)
2
sin
1
2
(
x
−
t
)
.
{\displaystyle {\frac {1}{2}}+\sum _{k=1}^{n}\cos k(x-t)={\frac {\sin(n+{\frac {1}{2}})(x-t)}{2\sin {\frac {1}{2}}(x-t)}}.\!}
şi introducând
u
=
x
−
t
,
S
n
(
x
)
{\displaystyle u=x-t,\;\;S_{n}(x)\!}
S
n
(
x
)
=
1
π
∫
x
−
π
x
+
π
f
(
x
−
u
)
⋅
sin
(
n
+
1
2
)
u
2
sin
1
2
u
d
u
.
{\displaystyle S_{n}(x)={\frac {1}{\pi }}\int _{x-\pi }^{x+\pi }f(x-u)\cdot {\frac {\sin(n+{\frac {1}{2}})u}{2\sin {\frac {1}{2}}u}}du.\!}
Integrandul este o funcţie de u , periodică de perioadă
2
π
{\displaystyle 2 \pi \!}
S
n
(
x
)
=
1
π
∫
−
π
π
f
(
x
−
u
)
⋅
sin
(
n
+
1
2
)
u
2
sin
1
2
u
d
u
.
{\displaystyle S_{n}(x)={\frac {1}{\pi }}\int _{-\pi }^{\pi }f(x-u)\cdot {\frac {\sin(n+{\frac {1}{2}})u}{2\sin {\frac {1}{2}}u}}du.\!}
Lema 2 .
Dacă
f
:
[
−
π
,
π
]
→
R
{\displaystyle f:[-\pi ,\pi ]\rightarrow \mathbb {R} \!}
funcție continuă pe porţiuni atunci sunt adevărate următoarele egalităţi:
a)
lim
n
→
∞
a
n
=
lim
n
→
∞
b
n
=
0.
{\displaystyle \lim _{n\to \infty }a_{n}=\lim _{n\to \infty }b_{n}=0.\!}
b)
lim
n
→
∞
f
(
x
)
sin
(
n
+
1
2
)
x
d
x
=
0
{\displaystyle \lim _{n\to \infty }f(x)\sin(n+{\frac {1}{2}})x\;dx=0\!}
−
π
≤
a
<
b
≤
π
{\displaystyle -\pi \leq a<b\leq \pi \!}
unde
a
n
{\displaystyle a_n \!}
b
n
{\displaystyle b_n\!}
f .
Demonstraţie .
Se consideră identitatea:
∫
−
π
π
|
f
(
x
)
−
S
n
(
x
)
|
2
d
x
=
∫
−
π
π
f
2
(
x
)
d
x
−
2
∫
−
π
π
f
(
x
)
S
n
(
x
)
d
x
+
∫
−
π
π
|
S
n
(
x
)
|
2
d
x
{\displaystyle \int _{-\pi }^{\pi }|f(x)-S_{n}(x)|^{2}dx=\int _{-\pi }^{\pi }f^{2}(x)dx-2\int _{-\pi }^{\pi }f(x)S_{n}(x)dx+\int _{-\pi }^{\pi }|S_{n}(x)|^{2}dx\!}
Prin calcul se arată că avem:
∫
−
π
π
|
S
n
(
x
)
|
2
d
x
=
∫
−
π
π
f
(
x
)
S
n
(
x
)
d
x
=
π
[
a
0
2
2
+
∑
k
=
1
n
(
a
k
2
+
b
k
2
)
]
{\displaystyle \int _{-\pi }^{\pi }|S_{n}(x)|^{2}dx=\int _{-\pi }^{\pi }f(x)S_{n}(x)dx=\pi \left[{\frac {a_{0}^{2}}{2}}+\sum _{k=1}^{n}(a_{k}^{2}+b_{k}^{2})\right]\!}
Ţinând seama de acestă egalitate din urmă, identitatea considerată devine:
∫
−
π
π
|
f
(
x
)
−
S
n
(
x
)
|
2
d
x
=
∫
−
π
π
|
f
(
x
)
|
2
d
x
−
π
[
a
0
2
2
+
∑
k
=
1
n
(
a
k
2
+
b
k
2
)
]
.
{\displaystyle \int _{-\pi }^{\pi }|f(x)-S_{n}(x)|^{2}dx=\int _{-\pi }^{\pi }|f(x)|^{2}dx-\pi \left[{\frac {a_{0}^{2}}{2}}+\sum _{k=1}^{n}(a_{k}^{2}+b_{k}^{2})\right].\!}
De aici se deduce inegalitatea:
a
0
2
2
+
∑
k
=
1
n
(
a
k
2
+
b
k
2
)
≤
1
π
∫
−
π
π
f
2
(
x
)
d
x
{\displaystyle {\frac {a_{0}^{2}}{2}}+\sum _{k=1}^{n}(a_{k}^{2}+b_{k}^{2})\leq {\frac {1}{\pi }}\int _{-\pi }^{\pi }f^{2}(x)dx\!}
valabilă pentru orice n şi cunoscută sub denumirea de inegalitatea lui Bessel .
Din inegalitatea lui Bessel rezultă că seria numerică:
a
0
2
2
+
∑
k
=
1
∞
(
a
k
2
+
b
k
2
)
{\displaystyle {\frac {a_{0}^{2}}{2}}+\sum _{k=1}^{\infty }(a_{k}^{2}+b_{k}^{2})\!}
este convergentă şi, prin urmare,
a
k
⟶
k
→
∞
0
{\displaystyle a_{k}{\underset {k\to \infty }{\longrightarrow }}0\!}
b
k
⟶
k
→
∞
0.
{\displaystyle b_{k}{\underset {k\to \infty }{\longrightarrow }}0.\!}
Remarcăm aici faptul că dacă:
lim
n
→
∞
∫
−
π
π
|
f
(
x
)
−
S
n
(
x
)
|
2
d
x
=
0.
{\displaystyle \lim _{n\to \infty }\int _{-\pi }^{\pi }|f(x)-S_{n}(x)|^{2}dx=0.\!}
atunci are loc egalitatea:
a
0
2
2
+
∑
k
=
1
∞
(
a
n
2
+
b
n
2
)
=
1
π
∫
−
π
π
f
2
(
x
)
d
x
{\displaystyle {\frac {a_{0}^{2}}{2}}+\sum _{k=1}^{\infty }(a_{n}^{2}+b_{n}^{2})={\frac {1}{\pi }}\int _{-\pi }^{\pi }f^{2}(x)dx\!}
numită egalitatea lui Parseval .
Convergenţa:
lim
n
→
∞
|
f
(
x
)
−
S
n
(
x
)
|
2
d
x
=
0
{\displaystyle \lim _{n\to \infty }|f(x)-S_{n}(x)|^{2}dx=0\!}
se numeşte convergenţa în medie de sume parţiale
S
n
(
x
)
{\displaystyle S_{n}(x)\!}
f
(
x
)
.
{\displaystyle f(x).\!}
Acum să arătăm că pentru
a
<
b
,
a
,
b
∈
[
−
π
,
π
]
{\displaystyle a<b,\;a,b\in [-\pi ,\pi ]\!}
lim
n
→
∞
∫
a
b
f
(
x
)
sin
(
n
+
1
2
)
x
d
x
=
0.
{\displaystyle \lim _{n\to \infty }\int _{a}^{b}f(x)\sin(n+{\frac {1}{2}})x\;dx=0.\!}
Remarcăm la început că pentru
α
<
β
{\displaystyle \alpha <\beta \!}
|
∫
α
β
sin
(
n
+
1
2
)
x
d
x
|
=
|
cos
(
n
+
1
2
)
α
−
cos
(
n
+
1
2
)
β
n
+
1
2
|
≤
2
n
+
1
2
{\displaystyle \left|\int _{\alpha }^{\beta }\sin(n+{\frac {1}{2}})x\;dx\right|=\left|{\frac {\cos(n+{\frac {1}{2}})\alpha -\cos(n+{\frac {1}{2}})\beta }{n+{\frac {1}{2}}}}\right|\leq {\frac {2}{n+{\frac {1}{2}}}}\!}
Fie
a
=
x
0
<
x
1
<
⋯
<
x
P
=
b
{\displaystyle a=x_{0}<x_{1}<\cdots <x_{P}=b\!}
[
a
,
b
]
{\displaystyle [a, b] \!}
∫
a
b
f
(
x
)
sin
(
n
+
1
2
)
x
d
x
=
∑
i
=
0
p
−
1
∫
x
i
x
i
+
1
f
(
x
)
sin
(
n
+
1
2
)
x
d
x
{\displaystyle \int _{a}^{b}f(x)\sin(n+{\frac {1}{2}})x\;dx=\sum _{i=0}^{p-1}\int _{x_{i}}^{x_{i+1}}f(x)\sin(n+{\frac {1}{2}})x\;dx\!}
Notăm:
m
i
=
i
n
f
{
f
(
x
)
|
x
∈
[
x
i
,
x
i
+
1
]
}
{\displaystyle m_{i}=inf\{f(x)\;|\;x\in [x_{i},x_{i+1}]\}\!}
şi reprezentăm integrala
∫
a
b
f
(
x
)
sin
(
n
+
1
2
)
x
d
x
{\displaystyle \int _{a}^{b}f(x)\sin(n+{\frac {1}{2}})x\;dx\!}
∫
a
b
f
(
x
)
sin
(
n
+
1
2
)
x
d
x
=
∑
i
=
0
p
−
1
∫
x
i
x
i
+
1
[
f
(
x
)
−
m
i
]
sin
(
n
+
1
2
)
x
d
x
+
∑
i
=
0
p
−
1
∫
x
i
x
i
+
1
sin
(
n
+
1
2
)
x
d
x
{\displaystyle \int _{a}^{b}f(x)\sin(n+{\frac {1}{2}})x\;dx=\sum _{i=0}^{p-1}\int _{x_{i}}^{x_{i+1}}[f(x)-m_{i}]\sin(n+{\frac {1}{2}})x\;dx+\sum _{i=0}^{p-1}\int _{x_{i}}^{x_{i+1}}\sin(n+{\frac {1}{2}})x\;dx\!}
Pentru
ω
i
=
M
i
−
m
i
,
{\displaystyle \omega _{i}=M_{i}-m_{i},\!}
M
i
=
s
u
p
{
f
(
x
)
|
x
∈
[
x
i
,
x
i
+
1
]
}
,
{\displaystyle M_{i}=sup\{f(x)\;|\;x\in [x_{i},x_{i+1}]\},\!}
f
(
x
)
−
m
i
≤
M
i
−
m
i
=
ω
i
∀
x
∈
[
x
i
,
x
i
+
1
]
{\displaystyle f(x)-m_{i}\leq M_{i}-m_{i}=\omega _{i}\;\;\forall x\in [x_{i},x_{i+1}]\!}
i
=
0
,
1
,
⋯
,
p
−
1.
{\displaystyle i=0,1,\cdots ,p-1.\!}
De aici rezultă:
|
∫
a
b
f
(
x
)
sin
(
n
+
1
2
)
x
d
x
|
≤
∑
i
=
0
p
−
1
ω
i
Δ
x
i
+
2
n
+
1
2
∑
i
=
0
p
−
1
|
m
i
|
{\displaystyle \left|\int _{a}^{b}f(x)\sin(n+{\frac {1}{2}})x\;dx\right|\leq \sum _{i=0}^{p-1}\omega _{i}\Delta x_{i}+{\frac {2}{n+{\frac {1}{2}}}}\sum _{i=0}^{p-1}|m_{i}|\!}
Pentru
ε
>
0
{\displaystyle \varepsilon >0 \!}
∑
i
=
0
p
−
1
ω
i
Δ
x
i
<
ε
2
{\displaystyle \sum _{i=0}^{p-1}\omega _{i}\Delta x_{i}<{\frac {\varepsilon }{2}}\!}
Acest lucru este posibil pentru că funcţia f este continuă pe porţiuni şi este integrabilă.
Acum putem lua
n
>
4
ε
M
(
b
−
a
)
,
{\displaystyle n>{\frac {4}{\varepsilon }}M(b-a),\!}
M
=
s
u
p
{
f
(
x
)
|
x
∈
[
a
,
b
]
}
{\displaystyle M=sup\{f(x)\;|\;x\in [a,b]\}\!}
n obţinem inegalitatea:
|
∫
a
b
f
(
x
)
sin
(
n
+
1
2
x
d
x
)
|
≤
ε
∀
ε
>
0.
{\displaystyle \left|\int _{a}^{b}f(x)\sin(n+{\frac {1}{2}}x\;dx)\right|\leq \varepsilon \;\;\forall \varepsilon >0.\!}
Teoremă (Fourier ).
Fie
f
:
[
−
π
,
π
]
→
R
{\displaystyle f:[-\pi ,\pi ]\rightarrow \mathbb {R} \!}
reală .
Dacă f are derivate laterale finite în punctele ei de discontinuitate atunci următoarele afirmaţii sunt adevărate:
a) dacă
x
0
{\displaystyle x_0 \!}
f atunci:
lim
n
→
∞
S
n
(
x
0
)
=
f
(
x
0
)
.
{\displaystyle \lim _{n\to \infty }S_{n}(x_{0})=f(x_{0}).\!}
b) dacă
x
0
{\displaystyle x_0 \!}
f atunci:
lim
n
→
∞
S
n
(
x
0
)
=
1
2
[
f
(
x
0
+
)
+
f
(
x
0
−
)
]
{\displaystyle \lim _{n\to \infty }S_{n}(x_{0})={\frac {1}{2}}[f(x_{0}^{+})+f(x_{0}^{-})]\!}
Demonstraţie .
Facem demonstraţia în cazul în care f nu este continuă într-un punct
x
0
.
{\displaystyle x_0. \!}
f
(
x
0
−
)
=
lim
x
→
x
0
−
f
(
x
)
{\displaystyle f(x_{0}^{-})=\lim _{x\to x_{0}^{-}}f(x)\!}
şi
f
(
x
0
+
)
=
lim
x
→
x
0
+
f
(
x
)
{\displaystyle f(x_{0}^{+})=\lim _{x\to x_{0}^{+}}f(x)\!}
Conform lemei 1, avem:
S
n
(
x
0
)
=
1
π
∫
−
π
π
f
(
x
0
−
u
)
⋅
sin
(
n
+
1
2
)
u
2
sin
1
2
u
d
u
{\displaystyle S_{n}(x_{0})={\frac {1}{\pi }}\int _{-\pi }^{\pi }f(x_{0}-u)\cdot {\frac {\sin(n+{\frac {1}{2}})u}{2\sin {\frac {1}{2}}u}}du\!}
Avem de asemenea:
1
2
⋅
f
(
x
0
+
)
=
1
π
∫
−
π
0
f
(
x
0
+
)
⋅
sin
(
n
+
1
2
)
u
2
sin
1
2
u
d
u
{\displaystyle {\frac {1}{2}}\cdot f(x_{0}^{+})={\frac {1}{\pi }}\int _{-\pi }^{0}f(x_{0}^{+})\cdot {\frac {\sin(n+{\frac {1}{2}})u}{2\sin {\frac {1}{2}}u}}du\!}
1
2
⋅
f
(
x
0
−
)
=
1
π
∫
0
π
f
(
x
0
−
)
⋅
sin
(
n
+
1
2
)
u
2
sin
1
2
u
d
u
{\displaystyle {\frac {1}{2}}\cdot f(x_{0}^{-})={\frac {1}{\pi }}\int _{0}^{\pi }f(x_{0}^{-})\cdot {\frac {\sin(n+{\frac {1}{2}})u}{2\sin {\frac {1}{2}}u}}du\!}
şi, prin urmare:
S
n
(
x
0
)
−
1
2
[
f
(
x
0
+
)
+
f
(
x
0
−
)
]
=
1
π
∫
−
π
0
[
f
(
x
0
−
u
)
−
f
(
x
0
+
)
]
⋅
sin
(
n
+
1
2
)
u
2
sin
1
2
u
d
u
+
{\displaystyle S_{n}(x_{0})-{\frac {1}{2}}[f(x_{0}^{+})+f(x_{0}^{-})]={\frac {1}{\pi }}\int _{-\pi }^{0}[f(x_{0}-u)-f(x_{0}^{+})]\cdot {\frac {\sin(n+{\frac {1}{2}})u}{2\sin {\frac {1}{2}}u}}du+\!}
+
1
π
∫
0
π
[
f
(
x
0
−
u
)
−
f
(
x
0
−
)
]
⋅
sin
(
n
+
1
2
)
u
2
sin
1
2
u
d
u
{\displaystyle +{\frac {1}{\pi }}\int _{0}^{\pi }[f(x_{0}-u)-f(x_{0}^{-})]\cdot {\frac {\sin(n+{\frac {1}{2}})u}{2\sin {\frac {1}{2}}u}}du\!}
Integranzii sunt bine definiţi peste tot, cu excepţia lui
u
=
0
{\displaystyle u=0\!}
F
1
(
u
)
⋅
sin
(
n
+
1
2
)
u
{\displaystyle F_{1}(u)\cdot \sin(n+{\frac {1}{2}})u\!}
unde
F
1
(
u
)
=
f
(
x
0
−
u
)
−
f
(
x
0
+
)
u
⋅
1
2
u
sin
1
2
u
{\displaystyle F_{1}(u)={\frac {f(x_{0}-u)-f(x_{0}^{+})}{u}}\cdot {\frac {{\frac {1}{2}}u}{\sin {\frac {1}{2}}u}}\!}
Pentru
u
→
0
−
,
{\displaystyle u\to 0^{-},\!}
1 şi produsul tinde la
−
f
′
(
x
0
+
)
.
{\displaystyle -f'(x_{0}^{+}).\!}
F
1
(
0
)
=
−
f
′
(
x
0
+
)
{\displaystyle F_{1}(0)=-f'(x_{0}^{+})\!}
u
=
0.
{\displaystyle u=0. \!}
În mod similar avem:
F
2
(
u
)
=
f
(
x
0
−
u
)
−
f
(
x
0
−
)
u
⋅
−
1
2
u
sin
1
2
u
{\displaystyle F_{2}(u)={\frac {f(x_{0}-u)-f(x_{0}^{-})}{u}}\cdot {\frac {-{\frac {1}{2}}u}{\sin {\frac {1}{2}}u}}\!}
are limita
+
f
′
(
x
0
−
)
{\displaystyle +f'(x_{0}^{-})\!}
F
2
(
0
)
=
+
f
′
(
x
0
−
)
{\displaystyle F_{2}(0)=+f'(x_{0}^{-})\!}
Prin urmare avem:
S
n
(
x
0
)
−
1
2
[
f
(
x
0
+
)
+
f
(
x
0
−
)
]
=
1
π
∫
−
π
0
F
1
(
u
)
⋅
sin
(
n
+
1
2
)
u
d
u
+
1
π
∫
0
π
⋅
sin
(
n
+
1
2
)
u
d
u
.
{\displaystyle S_{n}(x_{0})-{\frac {1}{2}}[f(x_{0}^{+})+f(x_{0}^{-})]={\frac {1}{\pi }}\int _{-\pi }^{0}F_{1}(u)\cdot \sin(n+{\frac {1}{2}})u\;du+{\frac {1}{\pi }}\int _{0}^{\pi }\cdot \sin(n+{\frac {1}{2}})u\;du.\!}
şi aplicând lema 2, concluzionăm că:
lim
n
→
∞
S
n
(
x
0
)
=
1
2
[
f
(
x
0
+
)
+
f
(
x
0
−
)
]
{\displaystyle \lim _{n\to \infty }S_{n}(x_{0})={\frac {1}{2}}[f(x_{0}^{+})+f(x_{0}^{-})]\!}
Se arată uşor că dacă f este continuă în
x
0
{\displaystyle x_0 \!}
lim
n
→
∞
S
n
(
x
0
)
=
f
(
x
0
)
.
{\displaystyle \lim _{n\to \infty }S_{n}(x_{0})=f(x_{0}).\!}
Diferite forme ale seriei Fourier [ ] Teoremă . [schimbarea originii intervalului fundamental
[
−
π
,
π
]
{\displaystyle [-\pi, \pi]\!}
]
Dacă
f
:
[
−
π
,
π
]
→
R
{\displaystyle f:[-\pi ,\pi ]\rightarrow \mathbb {R} \!}
funcție continuă pe porţiuni şi este prelungită pe
R
{\displaystyle \mathbb R \!}
α
,
{\displaystyle \alpha, \!}
a
n
,
b
n
{\displaystyle a_{n},b_{n}\!}
f verifică relaţiile:
a
n
=
1
π
∫
α
−
π
α
+
π
f
(
x
)
⋅
cos
n
x
d
x
n
=
0
,
1
,
2
,
⋯
{\displaystyle a_{n}={\frac {1}{\pi }}\int _{\alpha -\pi }^{\alpha +\pi }f(x)\cdot \cos nx\;dx\;\;\;n=0,1,2,\cdots \!}
b
n
=
1
π
∫
α
−
π
α
+
π
f
(
x
)
⋅
sin
n
x
d
x
n
=
0
,
1
,
2
,
⋯
{\displaystyle b_{n}={\frac {1}{\pi }}\int _{\alpha -\pi }^{\alpha +\pi }f(x)\cdot \sin nx\;dx\;\;\;n=0,1,2,\cdots \!}
şi
f
(
x
)
=
a
0
2
+
∑
n
=
1
∞
(
a
n
⋅
cos
n
x
+
b
n
⋅
sin
n
x
)
{\displaystyle f(x)={\frac {a_{0}}{2}}+\sum _{n=1}^{\infty }(a_{n}\cdot \cos nx+b_{n}\cdot \sin nx)\!}
în orice punct de continuitate
x
∈
[
α
−
π
,
α
+
π
]
.
{\displaystyle x\in [\alpha -\pi ,\alpha +\pi ].\!}
Demonstraţie .
Funcţiile
f
(
x
)
⋅
cos
n
x
,
f
(
x
)
⋅
sin
n
x
{\displaystyle f(x)\cdot \cos nx,\;f(x)\cdot \sin nx\!}
periodice de perioadă
2
π
.
{\displaystyle 2 \pi. \!}
integralele lor sunt aceleaşi pe orice interval de lungime
2
π
.
{\displaystyle 2 \pi. \!}
1
π
∫
α
−
π
α
+
π
f
(
x
)
⋅
cos
n
x
d
x
=
1
π
∫
−
π
π
f
(
x
)
⋅
cos
n
x
d
x
=
a
n
{\displaystyle {\frac {1}{\pi }}\int _{\alpha -\pi }^{\alpha +\pi }f(x)\cdot \cos nxdx={\frac {1}{\pi }}\int _{-\pi }^{\pi }f(x)\cdot \cos nxdx=a_{n}\!}
1
π
∫
α
−
π
α
+
π
f
(
x
)
⋅
sin
n
x
d
x
=
1
π
∫
−
π
π
f
(
x
)
⋅
sin
n
x
d
x
=
b
n
{\displaystyle {\frac {1}{\pi }}\int _{\alpha -\pi }^{\alpha +\pi }f(x)\cdot \sin nxdx={\frac {1}{\pi }}\int _{-\pi }^{\pi }f(x)\cdot \sin nxdx=b_{n}\!}
f
(
x
)
=
a
0
2
+
∑
n
=
1
∞
(
a
n
⋅
cos
n
x
+
b
n
⋅
sin
n
x
)
{\displaystyle f(x)={\frac {a_{0}}{2}}+\sum _{n=1}^{\infty }(a_{n}\cdot \cos nx+b_{n}\cdot \sin nx)\!}
considerăm șirul
y
k
=
x
−
2
k
π
,
k
∈
Z
.
{\displaystyle y_{k}=x-2k\pi ,\;k\in \mathbb {Z} .\!}
k
0
∈
Z
{\displaystyle k_{0}\in \mathbb {Z} \!}
y
k
0
∈
[
−
π
,
π
]
{\displaystyle y_{k_{0}}\in [-\pi ,\pi ]\!}
k
0
{\displaystyle k_{0}\!}
f
(
y
k
0
)
=
f
(
x
−
2
k
0
π
)
=
f
(
x
)
,
{\displaystyle f(y_{k_{0}})=f(x-2k_{0}\pi )=f(x),\!}
precum şi:
f
(
y
k
0
)
=
a
0
2
+
∑
n
=
1
∞
(
a
n
⋅
cos
n
y
k
0
+
b
n
⋅
sin
n
y
k
0
)
=
a
0
2
+
∑
n
=
1
∞
(
a
n
⋅
cos
n
x
+
b
n
⋅
sin
n
x
)
.
{\displaystyle f(y_{k_{0}})={\frac {a_{0}}{2}}+\sum _{n=1}^{\infty }(a_{n}\cdot \cos ny_{k_{0}}+b_{n}\cdot \sin ny_{k_{0}})={\frac {a_{0}}{2}}+\sum _{n=1}^{\infty }(a_{n}\cdot \cos nx+b_{n}\cdot \sin nx).\!}
QED .
Observaţie .
În condiţiile teoremei 1, dacă x este un punct de discontinuitate a prelungirii lui f prin periodicitate, atunci are loc:
1
2
[
f
(
x
+
)
+
f
(
x
−
)
]
=
a
0
2
+
∑
n
=
1
∞
(
a
n
⋅
cos
n
x
+
b
n
⋅
sin
n
x
)
.
{\displaystyle {\frac {1}{2}}[f(x^{+})+f(x^{-})]={\frac {a_{0}}{2}}+\sum _{n=1}^{\infty }(a_{n}\cdot \cos nx+b_{n}\cdot \sin nx).\!}
Această egalitate se demonstrează asemănător cu egalitatea din teorema 1.
Teorema 2 .
[schimbarea lungimii intervalului ]
Dacă
f
:
[
−
L
,
L
]
→
R
{\displaystyle f:[-L,L]\rightarrow \mathbb {R} \!}
funcție continuă pe porţiuni, atunci pentru orice
x
∈
[
−
L
,
L
]
{\displaystyle x\in [-L,L]\!}
f
(
x
)
=
a
0
2
+
∑
n
=
1
∞
(
a
n
⋅
cos
n
π
x
L
+
b
n
⋅
sin
n
π
x
L
)
{\displaystyle f(x)={\frac {a_{0}}{2}}+\sum _{n=1}^{\infty }(a_{n}\cdot \cos {\frac {n\pi x}{L}}+b_{n}\cdot \sin {\frac {n\pi x}{L}})\!}
dacă f este continuă în x .
1
2
[
f
(
x
+
)
+
f
(
x
−
)
]
=
a
0
2
+
∑
n
=
1
∞
(
a
n
⋅
cos
n
π
x
L
+
b
n
⋅
sin
n
π
x
L
)
{\displaystyle {\frac {1}{2}}[f(x^{+})+f(x^{-})]={\frac {a_{0}}{2}}+\sum _{n=1}^{\infty }(a_{n}\cdot \cos {\frac {n\pi x}{L}}+b_{n}\cdot \sin {\frac {n\pi x}{L}})\!}
dacă f nu este continuă în x , unde:
a
n
=
1
L
∫
−
L
L
f
(
x
)
⋅
cos
n
π
x
L
d
x
n
=
0
,
1
,
2
,
⋯
{\displaystyle a_{n}={\frac {1}{L}}\int _{-L}^{L}f(x)\cdot \cos {\frac {n\pi x}{L}}dx\;\;n=0,1,2,\cdots \!}
b
n
=
1
L
∫
−
L
L
f
(
x
)
⋅
sin
n
π
x
L
d
x
n
=
0
,
1
,
2
,
⋯
{\displaystyle b_{n}={\frac {1}{L}}\int _{-L}^{L}f(x)\cdot \sin {\frac {n\pi x}{L}}dx\;\;n=0,1,2,\cdots \!}
Demonstraţie .
Vezi şi [ ] Resurse [ ] ![{\displaystyle f:[-\pi ,\pi ]\rightarrow \mathbb {R} \!}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/5fba6d5c15df864a2f710f97af2317fdffdb9912)
![{\displaystyle S_{n}(x)={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(t)dt+{\frac {1}{\pi }}\sum _{k=1}^{n}\left[\cos kx\int _{-\pi }^{\pi }\cos ktdt+\sin kx\int _{-\pi }^{\pi }f(t)\sin ktdt\right]\!}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/9d38cffa558d6eed3832f34144ebf2bbf532ed57)
![{\displaystyle \int _{-\pi }^{\pi }|S_{n}(x)|^{2}dx=\int _{-\pi }^{\pi }f(x)S_{n}(x)dx=\pi \left[{\frac {a_{0}^{2}}{2}}+\sum _{k=1}^{n}(a_{k}^{2}+b_{k}^{2})\right]\!}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/82e12fffc2286905fde481945521568b1e9fa152)
![{\displaystyle \int _{-\pi }^{\pi }|f(x)-S_{n}(x)|^{2}dx=\int _{-\pi }^{\pi }|f(x)|^{2}dx-\pi \left[{\frac {a_{0}^{2}}{2}}+\sum _{k=1}^{n}(a_{k}^{2}+b_{k}^{2})\right].\!}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/8ab42e61f5f9473f688f643be9ab8c9f1653c163)
![{\displaystyle a<b,\;a,b\in [-\pi ,\pi ]\!}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/753e2841285e4febdaf350716ede59cafc2856c4)
![{\displaystyle [a,b]\!}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/f819c7db66b223d5d6748df5905edff33082ca1b)
![{\displaystyle m_{i}=inf\{f(x)\;|\;x\in [x_{i},x_{i+1}]\}\!}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/d28a8c4bae7c93b7adfd2c89014ad2c35337cf41)
![{\displaystyle \int _{a}^{b}f(x)\sin(n+{\frac {1}{2}})x\;dx=\sum _{i=0}^{p-1}\int _{x_{i}}^{x_{i+1}}[f(x)-m_{i}]\sin(n+{\frac {1}{2}})x\;dx+\sum _{i=0}^{p-1}\int _{x_{i}}^{x_{i+1}}\sin(n+{\frac {1}{2}})x\;dx\!}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/17b6b75e5c3e33f08fc504d197685a13e3afad1c)
![{\displaystyle M_{i}=sup\{f(x)\;|\;x\in [x_{i},x_{i+1}]\},\!}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/c60f99805047b85d013702a7bb5762d08cb2118b)
![{\displaystyle f(x)-m_{i}\leq M_{i}-m_{i}=\omega _{i}\;\;\forall x\in [x_{i},x_{i+1}]\!}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/661504ff52d02c50b107bfb138dc49181405ca05)
![{\displaystyle M=sup\{f(x)\;|\;x\in [a,b]\}\!}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/8dc3f3060f3cacd25f88384aa6b9e6167b72ad33)
![{\displaystyle \lim _{n\to \infty }S_{n}(x_{0})={\frac {1}{2}}[f(x_{0}^{+})+f(x_{0}^{-})]\!}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/4141489a9047ec102681c53caa559516bcf6d22d)
![{\displaystyle S_{n}(x_{0})-{\frac {1}{2}}[f(x_{0}^{+})+f(x_{0}^{-})]={\frac {1}{\pi }}\int _{-\pi }^{0}[f(x_{0}-u)-f(x_{0}^{+})]\cdot {\frac {\sin(n+{\frac {1}{2}})u}{2\sin {\frac {1}{2}}u}}du+\!}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/fe840fe94f38d0518d58eb4be9262d78cb3e8d16)
![{\displaystyle +{\frac {1}{\pi }}\int _{0}^{\pi }[f(x_{0}-u)-f(x_{0}^{-})]\cdot {\frac {\sin(n+{\frac {1}{2}})u}{2\sin {\frac {1}{2}}u}}du\!}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/1cce23c8e9ee3d7797f38aaf377b78e695114aa5)
![{\displaystyle S_{n}(x_{0})-{\frac {1}{2}}[f(x_{0}^{+})+f(x_{0}^{-})]={\frac {1}{\pi }}\int _{-\pi }^{0}F_{1}(u)\cdot \sin(n+{\frac {1}{2}})u\;du+{\frac {1}{\pi }}\int _{0}^{\pi }\cdot \sin(n+{\frac {1}{2}})u\;du.\!}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/550ee37ebf488d419d6e850eb5777b1453d4ea9e)
![{\displaystyle [-\pi ,\pi ]\!}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/09fee10ccc83d7d32ca494f7d29841589dbccbc5)
![{\displaystyle x\in [\alpha -\pi ,\alpha +\pi ].\!}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/4234d9dbacb774341416b125b9653457de81ed8e)
![{\displaystyle y_{k_{0}}\in [-\pi ,\pi ]\!}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/ec99221d2291f0b766e6efada043ae3bba19bc50)
![{\displaystyle {\frac {1}{2}}[f(x^{+})+f(x^{-})]={\frac {a_{0}}{2}}+\sum _{n=1}^{\infty }(a_{n}\cdot \cos nx+b_{n}\cdot \sin nx).\!}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/9b2df3bf6476593e383869745b7b4da08bbe0e30)
![{\displaystyle f:[-L,L]\rightarrow \mathbb {R} \!}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/09efb898708a21dd0cfc8aed6ca664b2daefc9b3)
![{\displaystyle x\in [-L,L]\!}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/77d8a70494361217cf12e5996f3490f198586115)
![{\displaystyle {\frac {1}{2}}[f(x^{+})+f(x^{-})]={\frac {a_{0}}{2}}+\sum _{n=1}^{\infty }(a_{n}\cdot \cos {\frac {n\pi x}{L}}+b_{n}\cdot \sin {\frac {n\pi x}{L}})\!}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/d9640282a0e70d4914a9ff3f48a795bd94f15973)
