§ 133. — Le serie di Fourier,
Sia una funzione
f
(
x
)
{\displaystyle f(x)}
2
π
{\displaystyle 2\pi }
2
π
{\displaystyle 2\pi }
f
(
x
)
{\displaystyle f(x)}
f
(
x
)
=
∑
n
=
0
∞
(
a
n
cos
n
x
+
b
n
sen
n
x
)
,
{\displaystyle f(x)=\sum _{n=0}^{\infty }(a_{n}\,{\text{cos}}\,nx+b_{n}\,{\text{sen}}\,nx),\,\,\,\,\,\,\,\,\,\,}
dove
n
{\displaystyle n}
0
,
1
,
2
,
3.....
{\displaystyle 0,1,2,3.....}
a
n
,
b
n
{\displaystyle a_{n},b_{n}}
n
=
0
{\displaystyle n=0}
a
0
{\displaystyle a_{0}}
f
(
x
)
=
a
0
+
∑
1
n
(
a
n
cos
n
x
+
b
n
sen
n
x
)
.
{\displaystyle f(x)=a_{0}+\sum _{1}^{n}(a_{n}\,{\text{cos}}\,nx+b_{n}\,{\text{sen}}\,nx).\,\,\,\,\,}
bis
Ricordando che, se
α
{\displaystyle \alpha }
∫
0
2
π
cos
α
x
d
x
{\displaystyle \int _{0}^{2\pi }\,{\text{cos}}\,\alpha \,xdx}
α
≠
0
{\displaystyle \alpha \not =0}
2
π
{\displaystyle 2\pi }
α
=
0
{\displaystyle \alpha =0}
∫
0
2
π
cos
n
x
cos
m
x
d
x
=
1
2
∫
0
2
π
cos
(
n
+
m
)
x
d
x
+
1
2
∫
0
2
π
cos
(
n
−
m
)
x
d
x
{\displaystyle \int _{0}^{2\pi }\,{\text{cos}}\,nx\,{\text{cos}}\,mxdx={\frac {1}{2}}\int _{0}^{2\pi }\,{\text{cos}}(n+m)xdx+{\frac {1}{2}}\int _{0}^{2\pi }\,{\text{cos}}\,(n-m)xdx}
troviamo, se
n
,
m
{\displaystyle n,m}
∫
0
2
π
cos
n
x
cos
m
x
d
x
=
{
2
π
se
n
=
m
=
0
π
se
n
=
m
≠
0
0
se
n
≠
m
.
{\displaystyle \int _{0}^{2\pi }\,{\text{cos}}\,nx\,{\text{cos}}\,mx\,dx=\left\{{\begin{matrix}2\pi \,{\text{se}}\,n=m=0\\\pi \,{\text{se}}\,n=m\not =0\\0\,{\text{se}}\,n\not =m.\end{matrix}}\right.}
In modo simile si prova:
∫
0
2
π
cos
n
x
sen
m
x
d
x
=
0
{\displaystyle \int _{0}^{2\pi }\,{\text{cos}}\,nx\,{\text{sen}}\,mx\,dx=0}
∫
0
2
π
sen
n
x
sen
m
x
d
x
=
{
0
se
n
=
m
=
0
0
se
n
≠
m
π
se
n
=
m
≠
0.
{\displaystyle \int _{0}^{2\pi }\,{\text{sen}}\,nx\,{\text{sen}}\,mx\,dx=\left\{{\begin{matrix}0\,{\text{se}}\,n=m=0\\0\,{\text{se}}\,n\not =m\\\pi \,{\text{se}}\,n=m\not =0.\end{matrix}}\right.}
