[ keisti ] Eilutė pavidalo
a
0
2
+
a
1
cos
x
+
b
1
sin
x
+
a
2
cos
(
2
x
)
+
b
2
sin
(
2
x
)
+
a
3
cos
(
3
x
)
+
b
3
sin
(
3
x
)
+
.
.
.
+
a
n
cos
(
n
x
)
+
b
n
sin
(
n
x
)
+
.
.
.
=
{\displaystyle {\frac {a_{0}}{2}}+a_{1}\cos x+b_{1}\sin x+a_{2}\cos(2x)+b_{2}\sin(2x)+a_{3}\cos(3x)+b_{3}\sin(3x)+...+a_{n}\cos(nx)+b_{n}\sin(nx)+...=}
=
a
0
2
+
∑
n
=
1
∞
(
a
n
cos
(
n
x
)
+
b
n
sin
(
n
x
)
)
(
1
)
{\displaystyle ={\frac {a_{0}}{2}}+\sum _{n=1}^{\infty }(a_{n}\cos(nx)+b_{n}\sin(nx))\quad (1)}
vadinasi trigonometrine eilute, o skaičiai
a
0
,
a
1
,
b
1
,
a
2
,
b
2
,
.
.
.
,
a
n
,
b
n
,
.
.
.
−
{\displaystyle a_{0},\;a_{1},\;b_{1},\;a_{2},\;b_{2},\;...,\;a_{n},\;b_{n},\;...\;-}
koeficientais trigonometrinės eilutės. Skirtumas nuo laipsninės eilutės yra, kad trigonometrinėje eilutėje vietoje paprasčiausių funkcijų 1, x,
x
2
{\displaystyle x^{2}}
x
n
{\displaystyle x^{n}}
1
2
,
cos
x
,
sin
x
,
cos
(
2
x
)
,
sin
(
2
x
)
,
.
.
.
,
cos
(
n
x
)
,
sin
(
n
x
)
,
.
.
.
,
(
2
)
{\displaystyle {\frac {1}{2}},\;\cos x,\;\sin x,\;\cos(2x),\;\sin(2x),\;...,\;\cos(nx),\;\sin(nx),\;...,\quad (2)}
kurios taip pat gerai išnagrinėtos.
Visų pirma pažymėsime, kad visos funkcijios sistemos (2) yra periodinės su periodu
2
π
{\displaystyle 2\pi }
1
2
{\displaystyle {\frac {1}{2}}}
sin
(
n
x
)
{\displaystyle \sin(nx)}
cos
(
n
x
)
{\displaystyle \cos(nx)}
n
=
1
,
2
,
.
.
.
{\displaystyle n=1,2,...}
2
π
n
{\displaystyle {\frac {2\pi }{n}}}
sin
[
n
(
x
±
2
π
n
)
]
=
sin
(
n
x
±
2
π
)
=
sin
(
n
x
)
{\displaystyle \sin[n(x\pm {\frac {2\pi }{n}})]=\sin(nx\pm 2\pi )=\sin(nx)}
2
π
=
n
{\displaystyle 2\pi =n}
(
2
π
n
)
{\displaystyle ({\frac {2\pi }{n}})}
2
π
{\displaystyle 2\pi }
2
π
{\displaystyle 2\pi }
x iki
x
+
2
π
{\displaystyle x+2\pi }
[
−
π
;
π
]
{\displaystyle [-\pi ;\;\pi ]}
2
π
{\displaystyle 2\pi }
Kita svarbia savybe funkcijos (2) yra jų statmenumas atkarpoje
[
−
π
;
π
]
:
{\displaystyle [-\pi ;\;\pi ]:}
[
−
π
;
π
]
{\displaystyle [-\pi ;\;\pi ]}
[
−
π
;
π
]
{\displaystyle [-\pi ;\;\pi ]}
Iš tiesų,
∫
−
π
π
1
2
cos
(
k
x
)
d
x
=
∫
−
π
π
1
2
k
cos
(
k
x
)
d
(
k
x
)
=
1
2
k
sin
(
k
x
)
|
−
π
π
=
0
;
(
3
)
{\displaystyle \int _{-\pi }^{\pi }{\frac {1}{2}}\cos(kx)\;{\text{d}}x=\int _{-\pi }^{\pi }{\frac {1}{2k}}\cos(kx)\;{\text{d}}(kx)={\frac {1}{2k}}\sin(kx)|_{-\pi }^{\pi }=0;\quad (3)}
∫
−
π
π
1
2
sin
(
k
x
)
d
x
=
∫
−
π
π
1
2
k
sin
(
k
x
)
d
(
k
x
)
=
−
1
2
k
cos
(
k
x
)
|
−
π
π
=
−
1
2
k
(
cos
(
k
π
)
−
cos
(
−
k
π
)
)
=
0
;
(
3
)
{\displaystyle \int _{-\pi }^{\pi }{\frac {1}{2}}\sin(kx)\;{\text{d}}x=\int _{-\pi }^{\pi }{\frac {1}{2k}}\sin(kx)\;{\text{d}}(kx)=-{\frac {1}{2k}}\cos(kx)|_{-\pi }^{\pi }=-{\frac {1}{2k}}(\cos(k\pi )-\cos(-k\pi ))=0;\quad (3)}
čia
d
(
k
x
)
=
k
⋅
d
x
;
d
(
k
x
)
k
=
d
x
.
{\displaystyle {\text{d}}(kx)=k\cdot {\text{d}}x;\;\;{\frac {{\text{d}}(kx)}{k}}={\text{d}}x.}
Toliau,
∫
−
π
π
cos
(
k
x
)
cos
(
n
x
)
d
x
=
1
2
∫
−
π
π
[
cos
(
(
k
+
n
)
x
)
+
cos
(
(
k
−
n
)
x
)
]
d
x
=
{\displaystyle \int _{-\pi }^{\pi }\cos(kx)\cos(nx){\text{d}}x={\frac {1}{2}}\int _{-\pi }^{\pi }[\cos((k+n)x)+\cos((k-n)x)]{\text{d}}x=}
=
1
2
[
∫
−
π
π
cos
(
(
k
+
n
)
x
)
k
+
n
d
(
(
k
+
n
)
x
)
+
∫
−
π
π
cos
(
(
k
−
n
)
x
)
k
−
n
d
(
(
k
−
n
)
x
)
]
=
{\displaystyle ={\frac {1}{2}}\left[\int _{-\pi }^{\pi }{\frac {\cos((k+n)x)}{k+n}}{\text{d}}((k+n)x)+\int _{-\pi }^{\pi }{\frac {\cos((k-n)x)}{k-n}}{\text{d}}((k-n)x)\right]=}
=
1
2
[
sin
(
(
k
+
n
)
x
)
k
+
n
|
−
π
π
+
sin
(
(
k
−
n
)
x
)
k
−
n
|
−
π
π
]
=
0
,
kai
k
≠
n
,
(
4
)
{\displaystyle ={\frac {1}{2}}\left[{\frac {\sin((k+n)x)}{k+n}}|_{-\pi }^{\pi }+{\frac {\sin((k-n)x)}{k-n}}|_{-\pi }^{\pi }\right]=0,\;{\text{kai}}\;k\neq n,\quad (4)}
čia
d
(
(
k
+
n
)
x
)
=
(
k
+
n
)
d
x
,
d
(
(
k
+
n
)
x
)
k
+
n
=
d
x
{\displaystyle {\text{d}}((k+n)x)=(k+n){\text{d}}x,\;{\frac {{\text{d}}((k+n)x)}{k+n}}={\text{d}}x}
d
(
(
k
−
n
)
x
)
=
(
k
−
n
)
d
x
,
d
(
(
k
−
n
)
x
)
k
−
n
=
d
x
{\displaystyle {\text{d}}((k-n)x)=(k-n){\text{d}}x,\;{\frac {{\text{d}}((k-n)x)}{k-n}}={\text{d}}x}
cos
(
A
)
⋅
cos
(
B
)
=
1
2
[
cos
(
A
+
B
)
+
cos
(
A
−
B
)
]
.
{\displaystyle \cos(A)\cdot \cos(B)={\frac {1}{2}}[\cos(A+B)+\cos(A-B)].}
Analogiškai randame
∫
−
π
π
sin
(
k
x
)
sin
(
n
x
)
d
x
=
−
1
2
∫
−
π
π
[
cos
(
(
k
+
n
)
x
)
−
cos
(
(
k
−
n
)
x
)
]
d
x
=
{\displaystyle \int _{-\pi }^{\pi }\sin(kx)\sin(nx){\text{d}}x=-{\frac {1}{2}}\int _{-\pi }^{\pi }[\cos((k+n)x)-\cos((k-n)x)]{\text{d}}x=}
=
−
1
2
[
∫
−
π
π
cos
(
(
k
+
n
)
x
)
k
+
n
d
(
(
k
+
n
)
x
)
−
∫
−
π
π
cos
(
(
k
−
n
)
x
)
k
−
n
d
(
(
k
−
n
)
x
)
]
=
{\displaystyle =-{\frac {1}{2}}\left[\int _{-\pi }^{\pi }{\frac {\cos((k+n)x)}{k+n}}{\text{d}}((k+n)x)-\int _{-\pi }^{\pi }{\frac {\cos((k-n)x)}{k-n}}{\text{d}}((k-n)x)\right]=}
=
−
1
2
[
sin
(
(
k
+
n
)
x
)
k
+
n
|
−
π
π
−
sin
(
(
k
−
n
)
x
)
k
−
n
|
−
π
π
]
=
0
,
kai
k
≠
n
,
(
5.1
)
{\displaystyle =-{\frac {1}{2}}\left[{\frac {\sin((k+n)x)}{k+n}}|_{-\pi }^{\pi }-{\frac {\sin((k-n)x)}{k-n}}|_{-\pi }^{\pi }\right]=0,\;{\text{kai}}\;k\neq n,\quad (5.1)}
čia
d
(
(
k
+
n
)
x
)
=
(
k
+
n
)
d
x
,
d
(
(
k
+
n
)
x
)
k
+
n
=
d
x
{\displaystyle {\text{d}}((k+n)x)=(k+n){\text{d}}x,\;{\frac {{\text{d}}((k+n)x)}{k+n}}={\text{d}}x}
d
(
(
k
−
n
)
x
)
=
(
k
−
n
)
d
x
,
d
(
(
k
−
n
)
x
)
k
−
n
=
d
x
{\displaystyle {\text{d}}((k-n)x)=(k-n){\text{d}}x,\;{\frac {{\text{d}}((k-n)x)}{k-n}}={\text{d}}x}
sin
(
A
)
⋅
sin
(
B
)
=
−
1
2
[
cos
(
A
+
B
)
−
cos
(
A
−
B
)
]
;
{\displaystyle \sin(A)\cdot \sin(B)=-{\frac {1}{2}}[\cos(A+B)-\cos(A-B)];}
∫
−
π
π
sin
(
k
x
)
cos
(
n
x
)
d
x
=
1
2
∫
−
π
π
[
sin
(
(
k
+
n
)
x
)
+
sin
(
(
k
−
n
)
x
)
]
d
x
=
−
1
2
(
cos
(
(
k
+
n
)
x
)
k
+
n
+
cos
(
(
k
−
n
)
x
)
k
−
n
)
|
−
π
π
=
0.
(
5.2
)
{\displaystyle \int _{-\pi }^{\pi }\sin(kx)\cos(nx){\text{d}}x={\frac {1}{2}}\int _{-\pi }^{\pi }[\sin((k+n)x)+\sin((k-n)x)]{\text{d}}x=-{\frac {1}{2}}{\Big (}{\frac {\cos((k+n)x)}{k+n}}+{\frac {\cos((k-n)x)}{k-n}}{\Big )}|_{-\pi }^{\pi }=0.\quad (5.2)}
Kai (5.2) integrale
k
=
n
=
p
,
{\displaystyle k=n=p,}
∫
−
π
π
sin
(
p
x
)
cos
(
p
x
)
d
x
=
∫
−
π
π
sin
(
p
x
)
cos
(
p
x
)
d
(
sin
(
p
x
)
)
p
cos
(
p
x
)
=
1
p
∫
−
π
π
sin
(
p
x
)
d
(
sin
(
p
x
)
)
=
1
2
p
sin
2
(
p
x
)
|
−
π
π
=
0
;
{\displaystyle \int _{-\pi }^{\pi }\sin(px)\cos(px){\text{d}}x=\int _{-\pi }^{\pi }\sin(px)\cos(px){\frac {{\text{d}}(\sin(px))}{p\cos(px)}}={\frac {1}{p}}\int _{-\pi }^{\pi }\sin(px){\text{d}}(\sin(px))={\frac {1}{2p}}\sin ^{2}(px)|_{-\pi }^{\pi }=0;}
d
(
sin
(
p
x
)
)
=
p
cos
(
p
x
)
d
x
.
{\displaystyle {\text{d}}(\sin(px))=p\cos(px)\;dx.}
Pagaliau,
∫
−
π
π
cos
2
(
k
x
)
d
x
=
1
2
∫
−
π
π
(
1
+
cos
(
2
k
x
)
)
d
x
=
1
2
(
x
+
1
2
k
sin
(
2
k
x
)
)
|
−
π
π
=
1
2
(
π
−
(
−
π
)
)
=
π
,
(
6
)
{\displaystyle \int _{-\pi }^{\pi }\cos ^{2}(kx)\;{\text{d}}x={\frac {1}{2}}\int _{-\pi }^{\pi }(1+\cos(2kx)){\text{d}}x={\frac {1}{2}}\left(x+{\frac {1}{2k}}\sin(2kx)\right)|_{-\pi }^{\pi }={\frac {1}{2}}(\pi -(-\pi ))=\pi ,\quad (6)}
∫
−
π
π
sin
2
(
k
x
)
d
x
=
1
2
∫
−
π
π
(
1
−
cos
(
2
k
x
)
)
d
x
=
1
2
(
x
−
1
2
k
sin
(
2
k
x
)
)
|
−
π
π
=
1
2
(
π
−
(
−
π
)
)
=
π
,
(
6
)
{\displaystyle \int _{-\pi }^{\pi }\sin ^{2}(kx)\;{\text{d}}x={\frac {1}{2}}\int _{-\pi }^{\pi }(1-\cos(2kx)){\text{d}}x={\frac {1}{2}}\left(x-{\frac {1}{2k}}\sin(2kx)\right)|_{-\pi }^{\pi }={\frac {1}{2}}(\pi -(-\pi ))=\pi ,\quad (6)}
∫
−
π
π
(
1
2
)
2
d
x
=
1
4
x
|
−
π
π
=
1
4
(
π
−
(
−
π
)
)
=
1
4
⋅
2
π
=
π
2
,
{\displaystyle \int _{-\pi }^{\pi }\left({\frac {1}{2}}\right)^{2}{\text{d}}x={\frac {1}{4}}x|_{-\pi }^{\pi }={\frac {1}{4}}(\pi -(-\pi ))={\frac {1}{4}}\cdot 2\pi ={\frac {\pi }{2}},}
ką ir reikėjo parodyti. Analogiškai laipsninei eilutei, trigonometrinė eilutė turi tokią teoremą.
Teorema 1. Jeigu funkcija
f
(
x
)
{\displaystyle f(x)}
[
−
π
;
π
]
,
{\displaystyle [-\pi ;\pi ],}
išsiskaido į trigonometrinę eilutę
f
(
x
)
=
a
0
2
+
∑
n
=
1
∞
(
a
n
cos
(
n
x
)
+
b
n
sin
(
n
x
)
)
(
7
)
{\displaystyle f(x)={\frac {a_{0}}{2}}+\sum _{n=1}^{\infty }(a_{n}\cos(nx)+b_{n}\sin(nx))\quad (7)}
kurią galima integruoti panariui, tai šitas išskaidymas vienintelis. Įrodymas . Integruodami (7), gauname
∫
−
π
π
f
(
x
)
d
x
=
∫
−
π
π
a
0
2
d
x
+
∑
n
=
1
∞
(
a
n
∫
−
π
π
cos
(
n
x
)
d
x
+
b
n
∫
−
π
π
sin
(
n
x
)
d
x
)
,
{\displaystyle \int _{-\pi }^{\pi }f(x){\text{d}}x=\int _{-\pi }^{\pi }{\frac {a_{0}}{2}}{\text{d}}x+\sum _{n=1}^{\infty }\left(a_{n}\int _{-\pi }^{\pi }\cos(nx){\text{d}}x+b_{n}\int _{-\pi }^{\pi }\sin(nx){\text{d}}x\right),}
Iš kur, atsižvelgę į (3), randame
∫
−
π
π
f
(
x
)
d
x
=
a
0
2
⋅
x
|
−
π
π
+
∑
n
=
1
∞
(
a
n
sin
(
n
x
)
n
|
−
π
π
−
b
n
cos
(
n
x
)
n
|
−
π
π
)
,
{\displaystyle \int _{-\pi }^{\pi }f(x){\text{d}}x={\frac {a_{0}}{2}}\cdot x|_{-\pi }^{\pi }+\sum _{n=1}^{\infty }\left(a_{n}{\frac {\sin(nx)}{n}}|_{-\pi }^{\pi }-b_{n}{\frac {\cos(nx)}{n}}|_{-\pi }^{\pi }\right),}
∫
−
π
π
f
(
x
)
d
x
=
a
0
2
⋅
(
π
−
(
−
π
)
)
,
{\displaystyle \int _{-\pi }^{\pi }f(x){\text{d}}x={\frac {a_{0}}{2}}\cdot (\pi -(-\pi )),}
∫
−
π
π
f
(
x
)
d
x
=
a
0
2
⋅
2
π
,
{\displaystyle \int _{-\pi }^{\pi }f(x){\text{d}}x={\frac {a_{0}}{2}}\cdot 2\pi ,}
a
0
π
=
∫
−
π
π
f
(
x
)
d
x
,
{\displaystyle a_{0}\pi =\int _{-\pi }^{\pi }f(x){\text{d}}x,}
a
0
=
1
π
∫
−
π
π
f
(
x
)
d
x
.
(
8
)
{\displaystyle a_{0}={\frac {1}{\pi }}\int _{-\pi }^{\pi }f(x){\text{d}}x.\quad (8)}
Kad nustatyti koeficientą
a
k
{\displaystyle a_{k}}
cos
(
k
x
)
{\displaystyle \cos(kx)}
−
{\displaystyle -\;}
cos
(
k
x
)
{\displaystyle \cos(kx)}
x nuo
−
π
{\displaystyle -\pi }
π
{\displaystyle \pi }
−
{\displaystyle -\;}
∫
−
π
π
f
(
x
)
cos
(
k
x
)
d
x
=
a
0
2
∫
−
π
π
cos
(
k
x
)
d
x
+
∑
n
=
1
∞
(
a
n
∫
−
π
π
cos
(
k
x
)
cos
(
n
x
)
d
x
+
b
n
∫
−
π
π
cos
(
k
x
)
sin
(
n
x
)
d
x
)
=
{\displaystyle \int _{-\pi }^{\pi }f(x)\cos(kx){\text{d}}x={\frac {a_{0}}{2}}\int _{-\pi }^{\pi }\cos(kx){\text{d}}x+\sum _{n=1}^{\infty }\left(a_{n}\int _{-\pi }^{\pi }\cos(kx)\cos(nx){\text{d}}x+b_{n}\int _{-\pi }^{\pi }\cos(kx)\sin(nx){\text{d}}x\right)=}
=
a
0
2
k
sin
(
k
x
)
|
−
π
π
+
a
1
∫
−
π
π
cos
(
k
x
)
cos
(
x
)
d
x
+
b
1
∫
−
π
π
cos
(
k
x
)
sin
(
x
)
d
x
+
{\displaystyle ={\frac {a_{0}}{2k}}\sin(kx)|_{-\pi }^{\pi }+a_{1}\int _{-\pi }^{\pi }\cos(kx)\cos(x){\text{d}}x+b_{1}\int _{-\pi }^{\pi }\cos(kx)\sin(x){\text{d}}x+}
+
a
2
∫
−
π
π
cos
(
k
x
)
cos
(
2
x
)
d
x
+
b
2
∫
−
π
π
cos
(
k
x
)
sin
(
2
x
)
d
x
+
.
.
.
+
a
k
∫
−
π
π
cos
(
k
x
)
cos
(
k
x
)
d
x
+
b
k
∫
−
π
π
cos
(
k
x
)
sin
(
k
x
)
d
x
+
.
.
.
+
{\displaystyle +a_{2}\int _{-\pi }^{\pi }\cos(kx)\cos(2x){\text{d}}x+b_{2}\int _{-\pi }^{\pi }\cos(kx)\sin(2x){\text{d}}x+...+a_{k}\int _{-\pi }^{\pi }\cos(kx)\cos(kx){\text{d}}x+b_{k}\int _{-\pi }^{\pi }\cos(kx)\sin(kx){\text{d}}x+...+}
+
a
n
∫
−
π
π
cos
(
k
x
)
cos
(
n
x
)
d
x
+
b
n
∫
−
π
π
cos
(
k
x
)
sin
(
n
x
)
d
x
+
.
.
.
=
{\displaystyle +a_{n}\int _{-\pi }^{\pi }\cos(kx)\cos(nx){\text{d}}x+b_{n}\int _{-\pi }^{\pi }\cos(kx)\sin(nx){\text{d}}x+...=}
=
0
+
0
+
0
+
.
.
.
+
a
k
∫
−
π
π
cos
(
k
x
)
cos
(
k
x
)
d
x
+
b
k
∫
−
π
π
cos
(
k
x
)
sin
(
k
x
)
d
x
+
.
.
.
+
0
+
0
+
.
.
.
=
{\displaystyle =0+0+0+...+a_{k}\int _{-\pi }^{\pi }\cos(kx)\cos(kx){\text{d}}x+b_{k}\int _{-\pi }^{\pi }\cos(kx)\sin(kx){\text{d}}x+...+0+0+...=}
=
a
k
∫
−
π
π
cos
2
(
k
x
)
d
x
+
b
k
∫
−
π
π
1
2
[
sin
(
(
k
+
k
)
x
)
−
sin
(
(
k
−
k
)
x
)
]
d
x
=
a
k
∫
−
π
π
cos
2
(
k
x
)
d
x
+
b
k
2
∫
−
π
π
sin
(
(
k
+
k
)
x
)
d
x
=
{\displaystyle =a_{k}\int _{-\pi }^{\pi }\cos ^{2}(kx){\text{d}}x+b_{k}\int _{-\pi }^{\pi }{\frac {1}{2}}[\sin((k+k)x)-\sin((k-k)x)]{\text{d}}x=a_{k}\int _{-\pi }^{\pi }\cos ^{2}(kx){\text{d}}x+{\frac {b_{k}}{2}}\int _{-\pi }^{\pi }\sin((k+k)x){\text{d}}x=}
=
a
k
∫
−
π
π
1
2
(
1
+
cos
(
2
k
x
)
)
d
x
−
b
k
2
⋅
cos
(
2
k
x
)
2
k
|
−
π
π
=
a
k
2
(
x
+
1
2
k
sin
(
2
k
x
)
)
|
−
π
π
−
b
k
4
k
(
cos
(
2
k
π
)
−
cos
(
−
2
k
π
)
)
=
{\displaystyle =a_{k}\int _{-\pi }^{\pi }{\frac {1}{2}}(1+\cos(2kx)){\text{d}}x-{\frac {b_{k}}{2}}\cdot {\frac {\cos(2kx)}{2k}}|_{-\pi }^{\pi }={\frac {a_{k}}{2}}(x+{\frac {1}{2k}}\sin(2kx))|_{-\pi }^{\pi }-{\frac {b_{k}}{4k}}(\cos(2k\pi )-\cos(-2k\pi ))=}
=
a
k
2
⋅
x
|
−
π
π
=
a
k
2
⋅
(
π
−
(
−
π
)
)
=
a
k
π
,
{\displaystyle ={\frac {a_{k}}{2}}\cdot x|_{-\pi }^{\pi }={\frac {a_{k}}{2}}\cdot (\pi -(-\pi ))=a_{k}\pi ,}
iš kur
a
k
=
1
π
∫
−
π
π
f
(
x
)
cos
(
k
x
)
d
x
.
(
9
)
{\displaystyle a_{k}={\frac {1}{\pi }}\int _{-\pi }^{\pi }f(x)\cos(kx){\text{d}}x.\quad (9)}
Analogiškaim padauginę lygybę (7) iš
sin
(
k
x
)
{\displaystyle \sin(kx)}
−
π
{\displaystyle -\pi }
π
{\displaystyle \pi }
∫
−
π
π
f
(
x
)
sin
(
k
x
)
d
x
=
a
0
2
∫
−
π
π
sin
(
k
x
)
d
x
+
∑
n
=
1
∞
(
a
n
∫
−
π
π
sin
(
k
x
)
cos
(
n
x
)
d
x
+
b
n
∫
−
π
π
sin
(
k
x
)
sin
(
n
x
)
d
x
)
=
{\displaystyle \int _{-\pi }^{\pi }f(x)\sin(kx){\text{d}}x={\frac {a_{0}}{2}}\int _{-\pi }^{\pi }\sin(kx){\text{d}}x+\sum _{n=1}^{\infty }\left(a_{n}\int _{-\pi }^{\pi }\sin(kx)\cos(nx){\text{d}}x+b_{n}\int _{-\pi }^{\pi }\sin(kx)\sin(nx){\text{d}}x\right)=}
=
b
k
∫
−
π
π
sin
2
(
k
x
)
d
x
=
b
k
π
,
{\displaystyle =b_{k}\int _{-\pi }^{\pi }\sin ^{2}(kx){\text{d}}x=b_{k}\pi ,}
iš kur randame
b
k
=
1
π
∫
−
π
π
f
(
x
)
sin
(
k
x
)
d
x
.
(
10
)
{\displaystyle b_{k}={\frac {1}{\pi }}\int _{-\pi }^{\pi }f(x)\sin(kx){\text{d}}x.\quad (10)}
Tokiu budu, koeficientai
a
0
,
a
k
{\displaystyle a_{0},a_{k}\;}
b
k
{\displaystyle b_{k}\;}
−
{\displaystyle -\;}
Šita teorema duoda pagrindą įvesti tokį apibrežimą.
Apibrėžimas . Tegu
f
(
x
)
−
{\displaystyle f(x)\;-}
funkcija, apibrėžta ir integruojama atkarpoje
[
−
π
;
π
]
{\displaystyle [-\pi ;\pi ]}
Tada skaičiai
a
0
,
a
n
,
b
n
,
{\displaystyle a_{0},\;a_{n},\;b_{n},}
rasti pagal formules (8)
−
{\displaystyle -\;}
vadinasi koeficientais Furje, o eilutė
a
0
2
+
∑
n
=
1
∞
(
a
n
cos
(
n
x
)
+
b
n
sin
(
n
x
)
)
{\displaystyle {\frac {a_{0}}{2}}+\sum _{n=1}^{\infty }(a_{n}\cos(nx)+b_{n}\sin(nx))}
su šitais koeficientais vadinasi eilute Furje funkcijos f(x) .[ keisti ] Įvesime sąvoka periodinio pratesimo funkcijos
f
(
x
)
,
{\displaystyle f(x),\;}
[
−
π
;
π
]
.
{\displaystyle [-\pi ;\;\pi ].}
Sakysime, kad funkcija
F
(
x
)
,
{\displaystyle F(x),\;}
2
π
{\displaystyle 2\pi }
f
(
x
)
,
{\displaystyle f(x),\;}
[
−
π
;
π
]
F
(
x
)
=
f
(
x
)
.
{\displaystyle [-\pi ;\;\pi ]\;F(x)=f(x).}
Akivaizdu, kad jeigu atkarpoje
[
−
π
;
π
]
{\displaystyle [-\pi ;\;\pi ]}
f
(
x
)
,
{\displaystyle f(x),\;}
Nustatysime kokiomis sąlygomis Furje eilutė funkcijos
f
(
x
)
{\displaystyle f(x)\;}
Teorema 2 . Tegu funkcija
f
(
x
)
{\displaystyle f(x)\;}
ir jos išvestinė
f
′
(
x
)
−
{\displaystyle f'(x)\;-}
netrūkios funkcijos atkarpoje
[
−
π
;
π
]
{\displaystyle [-\pi ;\;\pi ]}
arba turi atkarpoje
[
−
π
;
π
]
{\displaystyle [-\pi ;\;\pi ]}
baigtinį skaičių trūkių 1-ojo tipo. Tada Furje eilutė funkcijos
f
(
x
)
{\displaystyle f(x)\;}
konverguoja visoje skaičių tiesėje, be kita ko kiekviename taške
x
∈
(
−
π
;
π
)
,
{\displaystyle x\in (-\pi ;\pi ),}
kuriame
f
(
x
)
{\displaystyle f(x)\;}
netruki, suma eilutės lygi
f
(
x
)
,
{\displaystyle f(x),\;}
o kiekviename trūkio taške
x
0
{\displaystyle x_{0}}
funkcijos suma eilutės lygi
f
(
x
0
−
)
+
f
(
x
0
+
)
2
,
{\displaystyle {\frac {f(x_{0}-)+f(x_{0}+)}{2}},}
kur
f
(
x
0
−
)
=
lim
x
→
x
0
−
f
(
x
)
{\displaystyle f(x_{0}-)=\lim _{x\to x_{0}-}f(x)}
f
(
x
0
+
)
=
lim
x
→
x
0
+
f
(
x
)
.
{\displaystyle f(x_{0}+)=\lim _{x\to x_{0}+}f(x).}
Ant kraštų atkapros
[
−
π
;
π
]
{\displaystyle [-\pi ;\;\pi ]}
suma eilutės lygi
f
(
−
π
)
+
f
(
π
)
2
.
{\displaystyle {\frac {f(-\pi )+f(\pi )}{2}}.}
Bet kuriame taške
x
∈
[
−
π
;
π
]
{\displaystyle x\in [-\pi ;\pi ]}
suma Furje eilutės lygi
F
(
x
)
,
{\displaystyle F(x),\;}
jeigu
x
{\displaystyle x}
−
{\displaystyle -\;}
netrūkio taškas
F
(
x
)
,
{\displaystyle F(x),\;}
ir lygi
F
(
x
−
)
+
F
(
x
+
)
2
,
{\displaystyle {\frac {F(x-)+F(x+)}{2}},}
x
{\displaystyle x}
−
{\displaystyle -\;}
trūkio taškas
F
(
x
)
,
{\displaystyle F(x),\;}
F
(
x
)
−
{\displaystyle F(x)\;-}
f
(
x
)
.
{\displaystyle f(x).\;}
[ keisti ] Tegu funkcija
f
(
x
)
{\displaystyle f(x)\;}
[
−
π
;
π
]
{\displaystyle [-\pi ;\;\pi ]}
f
(
−
x
)
=
f
(
x
)
.
{\displaystyle f(-x)=f(x).\;}
b
n
{\displaystyle b_{n}}
b
n
=
1
π
∫
−
π
π
f
(
x
)
sin
(
n
x
)
d
x
=
1
π
[
∫
−
π
0
f
(
x
)
sin
(
n
x
)
d
x
+
∫
0
π
f
(
x
)
sin
(
n
x
)
d
x
]
.
{\displaystyle b_{n}={\frac {1}{\pi }}\int _{-\pi }^{\pi }f(x)\sin(nx)\;{\text{d}}x={\frac {1}{\pi }}\left[\int _{-\pi }^{0}f(x)\sin(nx)\;{\text{d}}x+\int _{0}^{\pi }f(x)\sin(nx)\;{\text{d}}x\right].}
Pirmame integrale kvadratiniuose skliaustuose padarysime pakeitimą kintamojo. Tarsime
x
=
−
t
.
{\displaystyle x=-t.}
d
x
=
d
t
;
{\displaystyle {\text{d}}x={\text{d}}t;}
x
=
−
π
{\displaystyle x=-\pi }
t
=
π
{\displaystyle t=\pi }
x
=
0
{\displaystyle x=0}
t
=
0.
{\displaystyle t=0.}
f
(
x
)
{\displaystyle f(x)\;}
sin
x
−
{\displaystyle \sin x\;-}
∫
−
π
0
f
(
x
)
sin
(
n
x
)
d
x
=
−
∫
π
0
f
(
−
t
)
sin
(
n
(
−
t
)
)
d
t
=
−
∫
0
π
f
(
t
)
sin
(
n
t
)
d
t
.
{\displaystyle \int _{-\pi }^{0}f(x)\sin(nx){\text{d}}x=-\int _{\pi }^{0}f(-t)\sin(n(-t)){\text{d}}t=-\int _{0}^{\pi }f(t)\sin(nt){\text{d}}t.}
Todėl,
b
n
=
1
π
[
−
∫
0
π
f
(
t
)
sin
(
n
t
)
d
t
+
∫
0
π
f
(
x
)
sin
(
n
x
)
d
x
]
=
0
{\displaystyle b_{n}={\frac {1}{\pi }}\left[-\int _{0}^{\pi }f(t)\sin(nt){\text{d}}t+\int _{0}^{\pi }f(x)\sin(nx){\text{d}}x\right]=0}
(priminsime, kad apibrėžtinis integralas nepriklauso nuo pažymėjimo kintamojo integravimo).
Kitoks paaiškinimas. Kadangi
g
(
x
)
=
sin
x
{\displaystyle g(x)=\sin x}
x kinta nuo
−
π
{\displaystyle -\pi }
π
{\displaystyle \pi }
sin
(
30
∘
)
=
1
2
{\displaystyle \sin(30^{\circ })={\frac {1}{2}}}
sin
(
−
30
∘
)
=
−
1
2
,
{\displaystyle \sin(-30^{\circ })=-{\frac {1}{2}},}
f
(
x
)
,
{\displaystyle f(x),\;}
f
(
x
)
{\displaystyle f(x)\;}
f
(
1
)
⋅
sin
(
1
)
=
f
(
−
1
)
⋅
sin
(
−
1
)
.
{\displaystyle f(1)\cdot \sin(1)=f(-1)\cdot \sin(-1).}
f
(
x
)
{\displaystyle f(x)\;}
f
(
1
)
⋅
sin
(
1
)
≠
f
(
−
1
)
⋅
sin
(
−
1
)
.
{\displaystyle f(1)\cdot \sin(1)\neq f(-1)\cdot \sin(-1).}
f
(
x
)
{\displaystyle f(x)\;}
f
(
−
x
)
=
−
f
(
x
)
{\displaystyle f(-x)=-f(x)\;}
g
(
x
)
=
sin
x
{\displaystyle g(x)=\sin x}
x
=
t
{\displaystyle x=t}
x
=
−
t
{\displaystyle x=-t}
f
(
t
)
sin
t
=
u
{\displaystyle f(t)\sin t=u}
f
(
−
t
)
sin
(
−
t
)
=
u
{\displaystyle f(-t)\sin(-t)=u}
f
(
x
)
{\displaystyle f(x)\;}
f
(
−
x
)
=
f
(
x
)
{\displaystyle f(-x)=f(x)\;}
g
(
x
)
=
sin
x
{\displaystyle g(x)=\sin x}
x
=
t
{\displaystyle x=t}
x
=
−
t
{\displaystyle x=-t}
f
(
t
)
sin
t
=
u
{\displaystyle f(t)\sin t=u}
f
(
−
t
)
sin
(
−
t
)
=
−
u
{\displaystyle f(-t)\sin(-t)=-u}
f
(
x
)
{\displaystyle f(x)\;}
∫
−
π
0
f
(
x
)
sin
(
n
x
)
d
x
=
−
∫
0
π
f
(
x
)
sin
(
n
x
)
d
x
(
jeigu
f
(
−
x
)
=
f
(
x
)
)
.
{\displaystyle \int _{-\pi }^{0}f(x)\sin(nx){\text{d}}x=-\int _{0}^{\pi }f(x)\sin(nx){\text{d}}x\quad ({\text{jeigu}}\;\;f(-x)=f(x)).}
Palyginimui, kai funkcija
f
(
x
)
{\displaystyle f(x)\;}
∫
−
π
0
f
(
x
)
sin
(
n
x
)
d
x
=
∫
0
π
f
(
x
)
sin
(
n
x
)
d
x
(
jeigu
f
(
−
x
)
=
−
f
(
x
)
)
.
{\displaystyle \int _{-\pi }^{0}f(x)\sin(nx){\text{d}}x=\int _{0}^{\pi }f(x)\sin(nx){\text{d}}x\quad ({\text{jeigu}}\;\;f(-x)=-f(x)).}
Todėl, kai funkcija
f
(
x
)
{\displaystyle f(x)\;}
b
n
=
1
π
∫
−
π
π
f
(
x
)
sin
(
n
x
)
d
x
=
1
π
[
∫
−
π
0
f
(
x
)
sin
(
n
x
)
d
x
+
∫
0
π
f
(
x
)
sin
(
n
x
)
d
x
]
=
{\displaystyle b_{n}={\frac {1}{\pi }}\int _{-\pi }^{\pi }f(x)\sin(nx)\;{\text{d}}x={\frac {1}{\pi }}\left[\int _{-\pi }^{0}f(x)\sin(nx)\;{\text{d}}x+\int _{0}^{\pi }f(x)\sin(nx)\;{\text{d}}x\right]=}
=
1
π
[
−
∫
0
π
f
(
x
)
sin
(
n
x
)
d
x
+
∫
0
π
f
(
x
)
sin
(
n
x
)
d
x
]
=
0
(
jeigu
f
(
−
x
)
=
f
(
x
)
)
.
{\displaystyle ={\frac {1}{\pi }}\left[-\int _{0}^{\pi }f(x)\sin(nx){\text{d}}x+\int _{0}^{\pi }f(x)\sin(nx){\text{d}}x\right]=0\quad ({\text{jeigu}}\;\;f(-x)=f(x)).}
Pažynėsime, kad reikia sudėti daug dalių funkcijos kai reikšmė x padalinta į daug mažų intervalų. Pavyzdžiui, jei funkcija lyginė
f
(
x
)
=
x
2
{\displaystyle f(x)=x^{2}}
∫
−
π
0
x
2
d
x
=
x
3
3
|
−
π
0
=
0
3
3
−
(
−
π
)
3
3
=
π
3
3
{\displaystyle \int _{-\pi }^{0}x^{2}{\text{d}}x={\frac {x^{3}}{3}}|_{-\pi }^{0}={\frac {0^{3}}{3}}-{\frac {(-\pi )^{3}}{3}}={\frac {\pi ^{3}}{3}}}
∫
0
π
x
2
d
x
=
x
3
3
|
−
π
0
=
π
3
3
−
0
3
3
=
π
3
3
{\displaystyle \int _{0}^{\pi }x^{2}{\text{d}}x={\frac {x^{3}}{3}}|_{-\pi }^{0}={\frac {\pi ^{3}}{3}}-{\frac {0^{3}}{3}}={\frac {\pi ^{3}}{3}}}
f
(
x
)
=
x
3
{\displaystyle f(x)=x^{3}}
∫
−
π
0
x
3
d
x
=
x
4
4
|
−
π
0
=
0
4
4
−
(
−
π
)
4
4
=
−
π
4
4
{\displaystyle \int _{-\pi }^{0}x^{3}{\text{d}}x={\frac {x^{4}}{4}}|_{-\pi }^{0}={\frac {0^{4}}{4}}-{\frac {(-\pi )^{4}}{4}}=-{\frac {\pi ^{4}}{4}}}
∫
0
π
x
3
d
x
=
x
4
4
|
0
π
=
π
4
4
−
0
4
4
=
π
4
4
.
{\displaystyle \int _{0}^{\pi }x^{3}{\text{d}}x={\frac {x^{4}}{4}}|_{0}^{\pi }={\frac {\pi ^{4}}{4}}-{\frac {0^{4}}{4}}={\frac {\pi ^{4}}{4}}.}
∫
−
10
0
x
2
dx
=
x
3
3
|
−
10
0
=
0
3
3
−
(
−
10
)
3
3
=
1000
3
=
333.
(
3
)
≈
{\displaystyle \int _{-10}^{0}x^{2}{\text{dx}}={\frac {x^{3}}{3}}|_{-10}^{0}={\frac {0^{3}}{3}}-{\frac {(-10)^{3}}{3}}={\frac {1000}{3}}=333.(3)\approx }
≈
(
−
1
)
2
+
(
−
2
)
2
+
(
−
3
)
2
+
(
−
4
)
2
+
(
−
5
)
2
+
(
−
6
)
2
+
(
−
7
)
2
+
(
−
8
)
2
+
(
−
9
)
2
+
(
−
10
)
2
=
385
,
{\displaystyle \approx (-1)^{2}+(-2)^{2}+(-3)^{2}+(-4)^{2}+(-5)^{2}+(-6)^{2}+(-7)^{2}+(-8)^{2}+(-9)^{2}+(-10)^{2}=385,}
∫
0
10
x
2
dx
=
x
3
3
|
0
10
=
10
3
3
−
0
3
3
=
1000
3
=
333.
(
3
)
≈
{\displaystyle \int _{0}^{10}x^{2}{\text{dx}}={\frac {x^{3}}{3}}|_{0}^{10}={\frac {10^{3}}{3}}-{\frac {0^{3}}{3}}={\frac {1000}{3}}=333.(3)\approx }
≈
1
2
+
2
2
+
3
2
+
4
2
+
5
2
+
6
2
+
7
2
+
8
2
+
9
2
+
10
2
=
385
{\displaystyle \approx 1^{2}+2^{2}+3^{2}+4^{2}+5^{2}+6^{2}+7^{2}+8^{2}+9^{2}+10^{2}=385}
bei
∫
−
10
0
x
3
dx
=
x
4
4
|
−
10
0
=
0
4
4
−
(
−
10
)
4
4
=
−
10000
4
=
−
2500
≈
{\displaystyle \int _{-10}^{0}x^{3}{\text{dx}}={\frac {x^{4}}{4}}|_{-10}^{0}={\frac {0^{4}}{4}}-{\frac {(-10)^{4}}{4}}=-{\frac {10000}{4}}=-2500\approx }
≈
(
−
1
)
3
+
(
−
2
)
3
+
(
−
3
)
3
+
(
−
4
)
3
+
(
−
5
)
3
+
(
−
6
)
3
+
(
−
7
)
3
+
(
−
8
)
3
+
(
−
9
)
3
+
(
−
10
)
3
=
−
3025
,
{\displaystyle \approx (-1)^{3}+(-2)^{3}+(-3)^{3}+(-4)^{3}+(-5)^{3}+(-6)^{3}+(-7)^{3}+(-8)^{3}+(-9)^{3}+(-10)^{3}=-3025,}
∫
0
10
x
3
dx
=
x
4
4
|
0
10
=
10
4
4
−
0
4
4
=
10000
4
=
2500
≈
{\displaystyle \int _{0}^{10}x^{3}{\text{dx}}={\frac {x^{4}}{4}}|_{0}^{10}={\frac {10^{4}}{4}}-{\frac {0^{4}}{4}}={\frac {10000}{4}}=2500\approx }
≈
1
3
+
2
3
+
3
3
+
4
3
+
5
3
+
6
3
+
7
3
+
8
3
+
9
3
+
10
3
=
3025.
{\displaystyle \approx 1^{3}+2^{3}+3^{3}+4^{3}+5^{3}+6^{3}+7^{3}+8^{3}+9^{3}+10^{3}=3025.}
Taigi, mes iš karto matome, kad galime iškelti minuso ženklą nelyginės (kuri yra
sin
x
{\displaystyle \sin x}
x
∈
(
−
π
;
π
)
{\displaystyle x\in (-\pi ;\pi )}
f
(
x
)
sin
x
{\displaystyle f(x)\sin x}
π
{\displaystyle \pi }
Analogiškai, atsižvelgiant, kad funkcijos
f
(
x
)
{\displaystyle f(x)\;}
cos
(
x
)
{\displaystyle \cos(x)}
cos
(
x
)
{\displaystyle \cos(x)}
x
∈
[
−
π
;
π
]
,
{\displaystyle x\in [-\pi ;\pi ],}
cos
(
−
150
∘
)
=
cos
(
150
∘
)
=
−
3
2
{\displaystyle \cos(-150^{\circ })=\cos(150^{\circ })=-{\frac {\sqrt {3}}{2}}}
a
n
{\displaystyle a_{n}}
a
0
=
2
π
∫
0
π
f
(
x
)
d
x
,
a
n
=
2
π
∫
0
π
f
(
x
)
cos
(
n
x
)
d
x
.
(
11
)
{\displaystyle a_{0}={\frac {2}{\pi }}\int _{0}^{\pi }f(x){\text{d}}x,\quad a_{n}={\frac {2}{\pi }}\int _{0}^{\pi }f(x)\cos(nx){\text{d}}x.\quad (11)}
Kai abi funkcijos
f
(
x
)
{\displaystyle f(x)\;}
cos
(
x
)
{\displaystyle \cos(x)}
a
n
=
1
π
∫
−
π
π
f
(
x
)
cos
(
n
x
)
d
x
=
1
π
(
∫
−
π
0
f
(
x
)
cos
(
n
x
)
d
x
+
∫
0
π
f
(
x
)
cos
(
n
x
)
d
x
)
=
2
π
∫
0
π
f
(
x
)
cos
(
n
x
)
d
x
.
{\displaystyle a_{n}={\frac {1}{\pi }}\int _{-\pi }^{\pi }f(x)\cos(nx){\text{d}}x={\frac {1}{\pi }}\left(\int _{-\pi }^{0}f(x)\cos(nx){\text{d}}x+\int _{0}^{\pi }f(x)\cos(nx){\text{d}}x\right)={\frac {2}{\pi }}\int _{0}^{\pi }f(x)\cos(nx){\text{d}}x.}
Nes integruojant lyginę funkciją (pvz.,
x
2
{\displaystyle x^{2}}
x
6
=
x
2
⋅
x
4
{\displaystyle x^{6}=x^{2}\cdot x^{4}}
x kinta nuo
−
π
{\displaystyle -\pi }
x kinta nuo 0 iki
π
{\displaystyle \pi }
∫
−
π
0
f
(
x
)
cos
(
n
x
)
d
x
=
∫
0
π
f
(
x
)
cos
(
n
x
)
d
x
(
kai
f
(
−
x
)
=
f
(
x
)
ir
cos
(
−
t
)
=
cos
(
t
)
)
.
{\displaystyle \int _{-\pi }^{0}f(x)\cos(nx){\text{d}}x=\int _{0}^{\pi }f(x)\cos(nx){\text{d}}x\quad ({\text{kai}}\;\;f(-x)=f(x)\;\;{\text{ir}}\;\;\cos(-t)=\cos(t)).}
a
0
=
1
π
∫
−
π
π
f
(
x
)
d
x
=
1
π
(
∫
−
π
0
f
(
x
)
d
x
+
∫
0
π
f
(
x
)
d
x
)
=
2
π
∫
0
π
f
(
x
)
d
x
,
kai
f
(
−
x
)
=
f
(
x
)
.
{\displaystyle a_{0}={\frac {1}{\pi }}\int _{-\pi }^{\pi }f(x){\text{d}}x={\frac {1}{\pi }}\left(\int _{-\pi }^{0}f(x){\text{d}}x+\int _{0}^{\pi }f(x){\text{d}}x\right)={\frac {2}{\pi }}\int _{0}^{\pi }f(x){\text{d}}x,\;\;{\text{kai}}\;\;f(-x)=f(x).}
a
0
=
1
π
∫
−
π
π
f
(
x
)
d
x
=
1
π
(
∫
−
π
0
f
(
x
)
d
x
+
∫
0
π
f
(
x
)
d
x
)
=
1
π
(
−
∫
0
π
f
(
x
)
d
x
+
∫
0
π
f
(
x
)
d
x
)
=
0
,
kai
f
(
−
x
)
=
−
f
(
x
)
.
{\displaystyle a_{0}={\frac {1}{\pi }}\int _{-\pi }^{\pi }f(x){\text{d}}x={\frac {1}{\pi }}\left(\int _{-\pi }^{0}f(x){\text{d}}x+\int _{0}^{\pi }f(x){\text{d}}x\right)={\frac {1}{\pi }}\left(-\int _{0}^{\pi }f(x){\text{d}}x+\int _{0}^{\pi }f(x){\text{d}}x\right)=0,\;{\text{kai}}\;f(-x)=-f(x).}
Tegu, dabar funkcija
f
(
x
)
,
{\displaystyle f(x),\;}
[
−
π
;
π
]
,
{\displaystyle [-\pi ;\pi ],}
f
(
x
)
=
−
f
(
−
x
)
.
{\displaystyle f(x)=-f(-x).}
a
n
{\displaystyle a_{n}}
b
n
{\displaystyle b_{n}}
b
n
=
2
π
∫
0
π
f
(
x
)
sin
(
n
x
)
d
x
.
(
12
)
{\displaystyle b_{n}={\frac {2}{\pi }}\int _{0}^{\pi }f(x)\sin(nx){\text{d}}x.\quad (12)}
Nes tuomet, kai abi funkcijos nelyginės
f
(
−
x
)
sin
(
−
x
)
=
−
f
(
x
)
⋅
(
−
sin
(
x
)
)
=
f
(
x
)
sin
(
x
)
{\displaystyle f(-x)\sin(-x)=-f(x)\cdot (-\sin(x))=f(x)\sin(x)}
b
n
=
1
π
∫
−
π
π
f
(
x
)
sin
(
n
x
)
d
x
=
1
π
(
∫
−
π
0
f
(
x
)
sin
(
n
x
)
d
x
+
∫
0
π
f
(
x
)
sin
(
n
x
)
d
x
)
=
2
π
∫
0
π
f
(
x
)
sin
(
n
x
)
d
x
,
kai
f
(
−
x
)
=
−
f
(
x
)
;
{\displaystyle b_{n}={\frac {1}{\pi }}\int _{-\pi }^{\pi }f(x)\sin(nx){\text{d}}x={\frac {1}{\pi }}\left(\int _{-\pi }^{0}f(x)\sin(nx){\text{d}}x+\int _{0}^{\pi }f(x)\sin(nx){\text{d}}x\right)={\frac {2}{\pi }}\int _{0}^{\pi }f(x)\sin(nx){\text{d}}x,\;{\text{kai}}\;f(-x)=-f(x);}
1
π
∫
−
π
0
f
(
x
)
sin
(
n
x
)
d
x
=
1
π
∫
0
π
f
(
x
)
sin
(
n
x
)
d
x
,
kai
f
(
−
x
)
=
−
f
(
x
)
.
{\displaystyle {\frac {1}{\pi }}\int _{-\pi }^{0}f(x)\sin(nx){\text{d}}x={\frac {1}{\pi }}\int _{0}^{\pi }f(x)\sin(nx){\text{d}}x,\;{\text{kai}}\;f(-x)=-f(x).}
Tas pats kas integruojant
x
3
⋅
x
5
=
x
8
{\displaystyle x^{3}\cdot x^{5}=x^{8}}
∫
−
π
0
x
3
dx
=
−
∫
0
π
x
3
dx
;
{\displaystyle \int _{-\pi }^{0}x^{3}{\text{dx}}=-\int _{0}^{\pi }x^{3}{\text{dx}};}
∫
−
π
0
x
5
dx
=
−
∫
0
π
x
5
dx
;
{\displaystyle \int _{-\pi }^{0}x^{5}{\text{dx}}=-\int _{0}^{\pi }x^{5}{\text{dx}};}
∫
−
π
0
x
3
⋅
x
5
dx
=
∫
0
π
x
3
⋅
x
5
dx
.
{\displaystyle \int _{-\pi }^{0}x^{3}\cdot x^{5}{\text{dx}}=\int _{0}^{\pi }x^{3}\cdot x^{5}{\text{dx}}.}
Tokiu budu, jeigu funkcija
f
(
x
)
{\displaystyle f(x)\;}
f
(
x
)
{\displaystyle f(x)\;}
Vaizdas:219apav.jpg a). Panagrinėkime funkcija
f
(
x
)
=
x
.
{\displaystyle f(x)=x.\;}
teoremą 2 ir todėl gali būti išdeliota į eilutę Furjė. Kadangi jinai nelyginė, tai jos koeficientai Furjė
a
n
=
0
,
{\displaystyle a_{n}=0,}
b
n
{\displaystyle b_{n}}
b
n
=
2
π
∫
0
π
x
sin
(
n
x
)
d
x
=
2
π
[
−
1
n
x
cos
(
n
x
)
|
0
π
+
1
n
∫
0
π
cos
(
n
x
)
d
x
]
=
{\displaystyle b_{n}={\frac {2}{\pi }}\int _{0}^{\pi }x\sin(nx){\text{d}}x={\frac {2}{\pi }}\left[-{\frac {1}{n}}x\cos(nx)|_{0}^{\pi }+{\frac {1}{n}}\int _{0}^{\pi }\cos(nx){\text{d}}x\right]=}
=
2
π
[
−
1
n
(
π
cos
(
n
π
)
−
0
⋅
cos
(
n
⋅
0
)
)
+
1
n
2
sin
(
n
x
)
|
0
π
]
=
{\displaystyle ={\frac {2}{\pi }}\left[-{\frac {1}{n}}(\pi \cos(n\pi )-0\cdot \cos(n\cdot 0))+{\frac {1}{n^{2}}}\sin(nx)|_{0}^{\pi }\right]=}
=
2
π
[
−
π
cos
(
n
π
)
n
+
1
n
2
(
sin
(
n
π
)
−
sin
(
n
⋅
0
)
)
]
=
{\displaystyle ={\frac {2}{\pi }}\left[-{\frac {\pi \cos(n\pi )}{n}}+{\frac {1}{n^{2}}}(\sin(n\pi )-\sin(n\cdot 0))\right]=}
=
2
π
[
−
π
cos
(
n
π
)
n
]
=
−
2
cos
(
n
π
)
n
=
−
2
⋅
(
−
1
)
n
n
=
(
−
1
)
n
+
1
2
n
.
{\displaystyle ={\frac {2}{\pi }}\left[-{\frac {\pi \cos(n\pi )}{n}}\right]=-{\frac {2\cos(n\pi )}{n}}=-{\frac {2\cdot (-1)^{n}}{n}}=(-1)^{n+1}{\frac {2}{n}}.}
Tokiu budu, gauname eilutę Furjė duotos funkcijos
x
=
2
(
sin
x
1
−
sin
2
x
2
+
sin
3
x
3
−
sin
4
x
4
+
.
.
.
+
(
−
1
)
n
+
1
sin
n
x
n
+
.
.
.
)
.
{\displaystyle x=2\left({\frac {\sin x}{1}}-{\frac {\sin 2x}{2}}+{\frac {\sin 3x}{3}}-{\frac {\sin 4x}{4}}+...+(-1)^{n+1}{\frac {\sin nx}{n}}+...\right).}
Šita lygybė teisinga betkuriam
x
∈
(
−
π
;
π
)
.
{\displaystyle x\in (-\pi ;\;\pi ).}
x
=
±
π
{\displaystyle x=\pm \pi }
teoremą 2 nesutampa su reikšmėmis funkcijos
f
(
x
)
=
x
,
{\displaystyle f(x)=x,}
f
(
−
π
)
+
f
(
π
)
2
=
−
π
+
π
2
=
0.
{\displaystyle {\frac {f(-\pi )+f(\pi )}{2}}={\frac {-\pi +\pi }{2}}=0.}
[
−
π
;
π
]
{\displaystyle [-\pi ;\;\pi ]}
f
(
x
)
=
x
{\displaystyle f(x)=x}
Vaizdas:219bpav.jpg b). Panagrinėkime funkciją
f
(
x
)
=
x
2
.
{\displaystyle f(x)=x^{2}.\;}
teoremos 2 ir todėl gali būti išdeliota į eilutę Furjė. Kadangi ji lyginė, tai jos koeficientai Furjė
b
n
=
0
,
{\displaystyle b_{n}=0,}
a
n
{\displaystyle a_{n}}
a
0
=
2
π
∫
0
π
x
2
d
x
=
2
π
⋅
π
3
3
=
2
π
2
3
;
{\displaystyle a_{0}={\frac {2}{\pi }}\int _{0}^{\pi }x^{2}{\text{d}}x={\frac {2}{\pi }}\cdot {\frac {\pi ^{3}}{3}}={\frac {2\pi ^{2}}{3}};}
a
n
=
2
π
∫
0
π
x
2
cos
(
n
x
)
d
x
=
2
π
[
x
2
sin
(
n
x
)
n
|
0
π
−
2
n
∫
0
π
x
sin
(
n
x
)
d
x
]
=
{\displaystyle a_{n}={\frac {2}{\pi }}\int _{0}^{\pi }x^{2}\cos(nx){\text{d}}x={\frac {2}{\pi }}\left[{\frac {x^{2}\sin(nx)}{n}}|_{0}^{\pi }-{\frac {2}{n}}\int _{0}^{\pi }x\sin(nx){\text{d}}x\right]=}
=
−
4
n
π
∫
0
π
x
sin
(
n
x
)
d
x
=
−
4
n
π
(
−
x
cos
(
n
x
)
n
|
0
π
−
−
1
n
∫
0
π
cos
(
n
x
)
d
x
)
=
{\displaystyle =-{\frac {4}{n\pi }}\int _{0}^{\pi }x\sin(nx){\text{d}}x=-{\frac {4}{n\pi }}\left(-{\frac {x\cos(nx)}{n}}|_{0}^{\pi }-{\frac {-1}{n}}\int _{0}^{\pi }\cos(nx){\text{d}}x\right)=}
=
−
4
n
π
(
−
π
cos
(
n
π
)
n
+
1
n
2
sin
(
n
x
)
|
0
π
)
=
{\displaystyle =-{\frac {4}{n\pi }}\left(-{\frac {\pi \cos(n\pi )}{n}}+{\frac {1}{n^{2}}}\sin(nx)|_{0}^{\pi }\right)=}
=
4
n
2
cos
(
n
π
)
=
4
n
2
⋅
(
−
1
)
n
=
(
−
1
)
n
4
n
2
;
{\displaystyle ={\frac {4}{n^{2}}}\cos(n\pi )={\frac {4}{n^{2}}}\cdot (-1)^{n}=(-1)^{n}{\frac {4}{n^{2}}};}
čia pasinaudojome integravimu dalimis
∫
u
(
x
)
v
′
(
x
)
d
x
=
u
(
x
)
v
(
x
)
−
∫
u
′
(
x
)
v
(
x
)
d
x
.
{\displaystyle \int u(x)v'(x){\mathsf {d}}x=u(x)v(x)-\int u'(x)v(x){\mathsf {d}}x.}
Reiškia, eilutė Furjė duotos funkcijos turi pavidalą
x
2
=
π
2
3
−
4
(
cos
x
1
−
cos
(
2
x
)
2
2
+
cos
(
3
x
)
3
2
−
.
.
.
)
.
{\displaystyle x^{2}={\frac {\pi ^{2}}{3}}-4\left({\frac {\cos x}{1}}-{\frac {\cos(2x)}{2^{2}}}+{\frac {\cos(3x)}{3^{2}}}-...\right).}
Šita lygybė teisinga betkuriam
x
∈
[
−
π
;
π
]
,
{\displaystyle x\in [-\pi ;\pi ],}
x
=
±
π
{\displaystyle x=\pm \pi }
f
(
x
)
=
x
2
,
{\displaystyle f(x)=x^{2},}
f
(
−
π
)
+
f
(
π
)
2
=
π
2
+
π
2
2
=
π
2
=
f
(
π
)
=
f
(
−
π
)
.
{\displaystyle {\frac {f(-\pi )+f(\pi )}{2}}={\frac {\pi ^{2}+\pi ^{2}}{2}}=\pi ^{2}=f(\pi )=f(-\pi ).}
f
(
x
)
=
x
2
{\displaystyle f(x)=x^{2}\;}
Panagrinėkime funkcija
f
(
x
)
=
x
3
.
{\displaystyle f(x)=x^{3}.\;}
teoremą 2 ir todėl gali būti išdeliota į eilutę Furjė. Kadangi jinai nelyginė, tai jos koeficientai Furjė
a
n
=
0
,
{\displaystyle a_{n}=0,}
b
n
{\displaystyle b_{n}}
b
n
=
2
π
∫
0
π
x
3
sin
(
n
x
)
d
x
=
2
π
[
(
3
x
2
n
2
−
6
n
4
)
sin
(
n
x
)
−
(
x
3
n
−
6
x
n
3
)
cos
(
n
x
)
]
|
0
π
=
{\displaystyle b_{n}={\frac {2}{\pi }}\int _{0}^{\pi }x^{3}\sin(nx){\text{d}}x={\frac {2}{\pi }}\left[\left({\frac {3x^{2}}{n^{2}}}-{\frac {6}{n^{4}}}\right)\sin(nx)-\left({\frac {x^{3}}{n}}-{\frac {6x}{n^{3}}}\right)\cos(nx)\right]|_{0}^{\pi }=}
=
2
π
[
−
(
x
3
n
−
6
x
n
3
)
cos
(
n
x
)
]
|
0
π
=
−
2
π
(
π
3
n
−
6
π
n
3
)
cos
(
n
π
)
=
(
12
n
3
−
2
π
2
n
)
cos
(
n
π
)
=
(
−
1
)
n
(
12
n
3
−
2
π
2
n
)
.
{\displaystyle ={\frac {2}{\pi }}\left[-\left({\frac {x^{3}}{n}}-{\frac {6x}{n^{3}}}\right)\cos(nx)\right]|_{0}^{\pi }=-{\frac {2}{\pi }}\left({\frac {\pi ^{3}}{n}}-{\frac {6\pi }{n^{3}}}\right)\cos(n\pi )=\left({\frac {12}{n^{3}}}-{\frac {2\pi ^{2}}{n}}\right)\cos(n\pi )=(-1)^{n}\left({\frac {12}{n^{3}}}-{\frac {2\pi ^{2}}{n}}\right).}
Tokiu budu, gauname eilutę Furjė duotos funkcijos
x
3
=
−
(
12
1
3
−
2
π
2
1
)
sin
x
+
(
12
2
3
−
2
π
2
2
)
sin
2
x
−
(
12
3
3
−
2
π
2
3
)
sin
3
x
+
(
12
4
3
−
2
π
2
4
)
sin
4
x
−
⋯
.
{\displaystyle x^{3}=-\left({\frac {12}{1^{3}}}-{\frac {2\pi ^{2}}{1}}\right)\sin x+\left({\frac {12}{2^{3}}}-{\frac {2\pi ^{2}}{2}}\right)\sin 2x-\left({\frac {12}{3^{3}}}-{\frac {2\pi ^{2}}{3}}\right)\sin 3x+\left({\frac {12}{4^{3}}}-{\frac {2\pi ^{2}}{4}}\right)\sin 4x-\cdots .}
[ keisti ] Tegu funkcija
f
(
x
)
{\displaystyle f(x)\;}
[
−
l
;
l
]
{\displaystyle [-l;\;l]}
l
−
{\displaystyle l\;-}
teoremos 2 . Išdeliosimę ją į Furjė eilutę.
Įvesime naują kintamajį
ξ
{\displaystyle \xi }
x
=
l
ξ
π
{\displaystyle x={\frac {l\xi }{\pi }}}
ir panagrinėsime funkciją
ϕ
(
ξ
)
=
f
(
l
ξ
π
)
=
f
(
x
)
.
{\displaystyle \phi (\xi )=f\left({\frac {l\xi }{\pi }}\right)=f(x).}
Akivaizdu, funkcija
ϕ
(
ξ
)
{\displaystyle \phi (\xi )}
[
−
π
;
π
]
{\displaystyle [-\pi ;\pi ]}
teoremos 2 .
Išdeliosime funkciją
ϕ
(
ξ
)
{\displaystyle \phi (\xi )}
[
−
π
;
π
]
{\displaystyle [-\pi ;\pi ]}
ϕ
(
ξ
)
=
a
0
2
+
∑
n
=
1
∞
(
a
n
cos
(
n
ξ
)
+
b
n
sin
(
n
ξ
)
)
,
(
13
)
{\displaystyle \phi (\xi )={\frac {a_{0}}{2}}+\sum _{n=1}^{\infty }(a_{n}\cos(n\xi )+b_{n}\sin(n\xi )),\quad (13)}
kur
a
0
=
1
π
∫
−
π
π
ϕ
(
ξ
)
d
ξ
;
a
n
=
1
π
∫
−
π
π
ϕ
(
ξ
)
cos
(
n
ξ
)
d
ξ
;
b
n
=
1
π
∫
−
π
π
ϕ
(
ξ
)
sin
(
n
ξ
)
d
ξ
.
{\displaystyle a_{0}={\frac {1}{\pi }}\int _{-\pi }^{\pi }\phi (\xi ){\text{d}}\xi ;\;a_{n}={\frac {1}{\pi }}\int _{-\pi }^{\pi }\phi (\xi )\cos(n\xi ){\text{d}}\xi ;\;b_{n}={\frac {1}{\pi }}\int _{-\pi }^{\pi }\phi (\xi )\sin(n\xi ){\text{d}}\xi .}
Grįšime dabar prie senojo kintamojo x :
x
=
l
π
ξ
,
ξ
=
x
π
l
,
d
ξ
=
d
(
x
π
l
)
=
π
l
d
x
,
d
x
=
l
π
d
ξ
.
{\displaystyle x={\frac {l}{\pi }}\xi ,\;\xi =x{\frac {\pi }{l}},\;{\text{d}}\xi ={\text{d}}\left({\frac {x\pi }{l}}\right)={\frac {\pi }{l}}{\text{d}}x,\;{\text{d}}x={\frac {l}{\pi }}{\text{d}}\xi .}
f
(
x
)
=
a
0
2
+
∑
n
=
1
∞
(
a
n
cos
n
π
x
l
+
b
n
sin
n
π
x
l
)
,
(
14
)
{\displaystyle f(x)={\frac {a_{0}}{2}}+\sum _{n=1}^{\infty }(a_{n}\cos {\frac {n\pi x}{l}}+b_{n}\sin {\frac {n\pi x}{l}}),\quad (14)}
kur
a
0
=
1
π
∫
−
π
π
ϕ
(
ξ
)
d
ξ
=
1
π
∫
−
l
l
f
(
x
)
π
l
d
x
=
1
l
∫
−
l
l
f
(
x
)
d
x
;
{\displaystyle a_{0}={\frac {1}{\pi }}\int _{-\pi }^{\pi }\phi (\xi ){\text{d}}\xi ={\frac {1}{\pi }}\int _{-l}^{l}f(x){\frac {\pi }{l}}{\text{d}}x={\frac {1}{l}}\int _{-l}^{l}f(x){\text{d}}x;}
a
n
=
1
l
∫
−
l
l
f
(
x
)
cos
n
π
x
l
d
x
;
{\displaystyle a_{n}={\frac {1}{l}}\int _{-l}^{l}f(x)\cos {\frac {n\pi x}{l}}{\text{d}}x;}
b
n
=
1
l
∫
−
l
l
f
(
x
)
sin
n
π
x
l
d
x
.
{\displaystyle b_{n}={\frac {1}{l}}\int _{-l}^{l}f(x)\sin {\frac {n\pi x}{l}}{\text{d}}x.}
Formulė (14) ir yra Furjė eilutė su periodu
2
l
.
{\displaystyle 2l.\;}
Vaizdas:Fourier220.jpg Pav. 2. Išdelioti į Furjė eilutę su periodu
2
l
{\displaystyle 2l\;}
f
(
x
)
,
{\displaystyle f(x),\;}
[
−
l
;
l
]
{\displaystyle [-l;\;l]}
f
(
x
)
=
|
x
|
.
{\displaystyle f(x)=|x|.\;}
Sprendimas . Kadangi funkcija
f
(
x
)
=
|
x
|
{\displaystyle f(x)=|x|\;}
b
n
=
0
;
{\displaystyle b_{n}=0;}
a
0
=
2
l
∫
0
l
x
d
x
=
2
l
⋅
x
2
2
|
0
l
=
2
l
⋅
l
2
2
=
l
;
{\displaystyle a_{0}={\frac {2}{l}}\int _{0}^{l}x{\text{d}}x={\frac {2}{l}}\cdot {\frac {x^{2}}{2}}|_{0}^{l}={\frac {2}{l}}\cdot {\frac {l^{2}}{2}}=l;}
a
n
=
2
l
∫
0
l
x
cos
n
π
x
l
d
x
=
2
l
(
l
⋅
x
sin
n
π
x
l
n
π
|
0
l
−
l
n
π
∫
0
l
sin
n
π
x
l
d
x
)
=
{\displaystyle a_{n}={\frac {2}{l}}\int _{0}^{l}x\cos {\frac {n\pi x}{l}}{\text{d}}x={\frac {2}{l}}\left({\frac {l\cdot x\sin {\frac {n\pi x}{l}}}{n\pi }}|_{0}^{l}-{\frac {l}{n\pi }}\int _{0}^{l}\sin {\frac {n\pi x}{l}}{\text{d}}x\right)=}
=
2
l
(
l
2
sin
n
π
l
l
n
π
+
l
n
2
π
2
cos
n
π
x
l
|
0
l
)
=
2
l
(
l
2
sin
(
n
π
)
n
π
+
l
n
2
π
2
(
cos
n
π
l
l
−
cos
n
π
⋅
0
l
)
)
=
{\displaystyle ={\frac {2}{l}}\left({\frac {l^{2}\sin {\frac {n\pi l}{l}}}{n\pi }}+{\frac {l}{n^{2}\pi ^{2}}}\cos {\frac {n\pi x}{l}}|_{0}^{l}\right)={\frac {2}{l}}\left({\frac {l^{2}\sin(n\pi )}{n\pi }}+{\frac {l}{n^{2}\pi ^{2}}}(\cos {\frac {n\pi l}{l}}-\cos {\frac {n\pi \cdot 0}{l}})\right)=}
=
2
l
⋅
l
n
2
π
2
(
cos
(
n
π
)
−
1
)
=
2
n
2
π
2
(
cos
(
n
π
)
−
1
)
=
2
n
2
π
2
(
(
−
1
)
n
−
1
)
=
{
0
,
kai
n
lyginis
,
−
4
l
n
2
π
2
,
kai
n
nelyginis
.
{\displaystyle ={\frac {2}{l}}\cdot {\frac {l}{n^{2}\pi ^{2}}}(\cos(n\pi )-1)={\frac {2}{n^{2}\pi ^{2}}}(\cos(n\pi )-1)={\frac {2}{n^{2}\pi ^{2}}}((-1)^{n}-1)={\begin{cases}0,\quad {\text{kai}}\;n\;{\text{lyginis}},&\\-{\frac {4l}{n^{2}\pi ^{2}}},\quad {\text{kai}}\;n\;{\text{nelyginis}}.&\end{cases}}}
Furjė eilutė funkcijos
f
(
x
)
{\displaystyle f(x)\;}
|
x
|
=
l
2
−
4
l
π
2
(
cos
π
x
l
+
1
3
2
cos
3
π
x
l
+
1
5
2
cos
5
π
x
l
+
.
.
.
)
.
{\displaystyle |x|={\frac {l}{2}}-{\frac {4l}{\pi ^{2}}}(\cos {\frac {\pi x}{l}}+{\frac {1}{3^{2}}}\cos {\frac {3\pi x}{l}}+{\frac {1}{5^{2}}}\cos {\frac {5\pi x}{l}}+...).}
Funkcija
|
x
|
{\displaystyle |x|}
teoremos 2 ir gauta lygybė teisinga bet kokiam
x
∈
[
−
l
;
l
]
,
{\displaystyle x\in [-l;l],}
Pažymėsime, kad Furjė eilutės plačiai taikomos tiek teoriniuose tyrimuose, tiek ir praktiniuose uždaviniuose.
Išdelioti į Furjė eilutę su periodu
2
l
{\displaystyle 2l\;}
f
(
x
)
,
{\displaystyle f(x),\;}
[
−
l
;
l
]
{\displaystyle [-l;\;l]}
f
(
x
)
=
x
2
.
{\displaystyle f(x)=x^{2}.\;}
Sprendimas . Įvedame keitinį
x
=
ξ
l
π
;
ξ
=
π
x
l
;
d
ξ
=
π
l
d
x
.
{\displaystyle x={\frac {\xi l}{\pi }};\;\xi ={\frac {\pi x}{l}};\;{\text{d}}\xi ={\frac {\pi }{l}}{\text{d}}x.}
ϕ
(
ξ
)
{\displaystyle \phi (\xi )}
[
−
π
;
π
]
.
{\displaystyle [-\pi ;\;\pi ].}
f
(
x
)
=
x
2
{\displaystyle f(x)=x^{2}\;}
b
n
=
0
;
{\displaystyle b_{n}=0;}
a
0
=
1
π
∫
−
π
π
ϕ
(
ξ
)
d
ξ
=
2
π
∫
0
π
ϕ
(
ξ
)
d
ξ
=
2
π
∫
0
l
f
(
x
)
⋅
π
l
d
x
=
2
l
∫
0
l
x
2
d
x
=
2
l
x
3
3
|
0
l
=
2
l
⋅
l
3
3
=
2
l
2
3
;
{\displaystyle a_{0}={\frac {1}{\pi }}\int _{-\pi }^{\pi }\phi (\xi ){\text{d}}\xi ={\frac {2}{\pi }}\int _{0}^{\pi }\phi (\xi ){\text{d}}\xi ={\frac {2}{\pi }}\int _{0}^{l}f(x)\cdot {\frac {\pi }{l}}{\text{d}}x={\frac {2}{l}}\int _{0}^{l}x^{2}{\text{d}}x={\frac {2}{l}}{\frac {x^{3}}{3}}|_{0}^{l}={\frac {2}{l}}\cdot {\frac {l^{3}}{3}}={\frac {2l^{2}}{3}};}
a
n
=
1
π
∫
−
π
π
ϕ
(
ξ
)
cos
(
n
ξ
)
d
ξ
=
2
π
∫
0
π
ϕ
(
ξ
)
cos
(
n
ξ
)
d
ξ
=
2
π
⋅
π
l
∫
0
l
f
(
x
)
cos
n
π
x
l
d
x
=
{\displaystyle a_{n}={\frac {1}{\pi }}\int _{-\pi }^{\pi }\phi (\xi )\cos(n\xi ){\text{d}}\xi ={\frac {2}{\pi }}\int _{0}^{\pi }\phi (\xi )\cos(n\xi ){\text{d}}\xi ={\frac {2}{\pi }}\cdot {\frac {\pi }{l}}\int _{0}^{l}f(x)\cos {\frac {n\pi x}{l}}{\text{d}}x=}
=
2
l
∫
0
l
x
2
cos
n
π
x
l
d
x
=
2
l
[
x
2
sin
n
π
x
l
n
π
l
|
0
l
−
∫
0
l
2
x
⋅
sin
n
π
x
l
n
π
l
d
x
]
=
{\displaystyle ={\frac {2}{l}}\int _{0}^{l}x^{2}\cos {\frac {n\pi x}{l}}{\text{d}}x={\frac {2}{l}}\left[{\frac {x^{2}\sin {\frac {n\pi x}{l}}}{\frac {n\pi }{l}}}|_{0}^{l}-\int _{0}^{l}2x\cdot {\frac {\sin {\frac {n\pi x}{l}}}{\frac {n\pi }{l}}}{\text{d}}x\right]=}
=
2
l
⋅
(
−
2
n
π
l
)
∫
0
l
x
sin
n
π
x
l
d
x
=
−
4
n
π
(
−
x
cos
n
π
x
l
n
π
l
|
0
l
−
−
1
n
π
l
∫
0
l
cos
n
π
l
d
x
)
=
{\displaystyle ={\frac {2}{l}}\cdot \left(-{\frac {2}{\frac {n\pi }{l}}}\right)\int _{0}^{l}x\sin {\frac {n\pi x}{l}}{\text{d}}x=-{\frac {4}{n\pi }}\left(-{\frac {x\cos {\frac {n\pi x}{l}}}{\frac {n\pi }{l}}}|_{0}^{l}-{\frac {-1}{\frac {n\pi }{l}}}\int _{0}^{l}\cos {\frac {n\pi }{l}}{\text{d}}x\right)=}
=
−
4
n
π
(
−
l
cos
(
n
π
)
n
π
l
+
1
(
n
π
l
)
2
sin
n
π
x
l
|
0
l
)
=
4
l
2
n
2
π
2
cos
(
n
π
)
=
(
−
1
)
n
4
l
2
n
2
π
2
;
{\displaystyle =-{\frac {4}{n\pi }}\left(-{\frac {l\cos(n\pi )}{\frac {n\pi }{l}}}+{\frac {1}{({\frac {n\pi }{l}})^{2}}}\sin {\frac {n\pi x}{l}}|_{0}^{l}\right)={\frac {4l^{2}}{n^{2}\pi ^{2}}}\cos(n\pi )=(-1)^{n}{\frac {4l^{2}}{n^{2}\pi ^{2}}};}
čia pasinaudojome integravimu dalimis
∫
u
(
x
)
v
′
(
x
)
d
x
=
u
(
x
)
v
(
x
)
−
∫
u
′
(
x
)
v
(
x
)
d
x
{\displaystyle \int u(x)v'(x){\mathsf {d}}x=u(x)v(x)-\int u'(x)v(x){\mathsf {d}}x}
Furjė eilutė funkcijos
f
(
x
)
{\displaystyle f(x)\;}
x
2
=
2
l
2
3
2
+
(
−
4
l
2
π
2
cos
π
x
l
+
4
l
2
2
2
π
2
cos
2
π
x
l
−
4
l
2
3
2
π
2
cos
3
π
x
l
+
.
.
.
)
=
{\displaystyle x^{2}={\frac {\frac {2l^{2}}{3}}{2}}+\left(-{\frac {4l^{2}}{\pi ^{2}}}\cos {\frac {\pi x}{l}}+{\frac {4l^{2}}{2^{2}\pi ^{2}}}\cos {\frac {2\pi x}{l}}-{\frac {4l^{2}}{3^{2}\pi ^{2}}}\cos {\frac {3\pi x}{l}}+...\right)=}
=
l
2
3
−
4
l
2
π
2
(
cos
π
x
l
1
2
−
cos
2
π
x
l
2
2
+
cos
3
π
x
l
3
2
−
.
.
.
)
.
{\displaystyle ={\frac {l^{2}}{3}}-{\frac {4l^{2}}{\pi ^{2}}}\left({\frac {\cos {\frac {\pi x}{l}}}{1^{2}}}-{\frac {\cos {\frac {2\pi x}{l}}}{2^{2}}}+{\frac {\cos {\frac {3\pi x}{l}}}{3^{2}}}-...\right).}
![{\displaystyle \sin[n(x\pm {\frac {2\pi }{n}})]=\sin(nx\pm 2\pi )=\sin(nx)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5c2235822d1ad21691648e6865bf692b5e791630)
![{\displaystyle [-\pi ;\;\pi ]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f6df219fde2f828a2ff0af793fe836b55b19e754)
![{\displaystyle [-\pi ;\;\pi ]:}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9e1a00a61cb24bf8a8fa7cf84940b86ff364193c)
![{\displaystyle \int _{-\pi }^{\pi }\cos(kx)\cos(nx){\text{d}}x={\frac {1}{2}}\int _{-\pi }^{\pi }[\cos((k+n)x)+\cos((k-n)x)]{\text{d}}x=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9b0bca719414e4ba6e8eb67a7146a109cbc14f23)
![{\displaystyle ={\frac {1}{2}}\left[\int _{-\pi }^{\pi }{\frac {\cos((k+n)x)}{k+n}}{\text{d}}((k+n)x)+\int _{-\pi }^{\pi }{\frac {\cos((k-n)x)}{k-n}}{\text{d}}((k-n)x)\right]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/93f700a39d9202bc047481ced82f7be53c1b0b7f)
![{\displaystyle ={\frac {1}{2}}\left[{\frac {\sin((k+n)x)}{k+n}}|_{-\pi }^{\pi }+{\frac {\sin((k-n)x)}{k-n}}|_{-\pi }^{\pi }\right]=0,\;{\text{kai}}\;k\neq n,\quad (4)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4fd2e8c14f71c74f14fea3e3d281a9e1cd58b1cb)
![{\displaystyle \cos(A)\cdot \cos(B)={\frac {1}{2}}[\cos(A+B)+\cos(A-B)].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/20ef44deebd5cf0869fabbbb0476185c9faab535)
![{\displaystyle \int _{-\pi }^{\pi }\sin(kx)\sin(nx){\text{d}}x=-{\frac {1}{2}}\int _{-\pi }^{\pi }[\cos((k+n)x)-\cos((k-n)x)]{\text{d}}x=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2f793febc373091acfb0014c9ad3de9b5dc97e89)
![{\displaystyle =-{\frac {1}{2}}\left[\int _{-\pi }^{\pi }{\frac {\cos((k+n)x)}{k+n}}{\text{d}}((k+n)x)-\int _{-\pi }^{\pi }{\frac {\cos((k-n)x)}{k-n}}{\text{d}}((k-n)x)\right]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1e4542eb84177347afc7d28d5a61a64696f1a743)
![{\displaystyle =-{\frac {1}{2}}\left[{\frac {\sin((k+n)x)}{k+n}}|_{-\pi }^{\pi }-{\frac {\sin((k-n)x)}{k-n}}|_{-\pi }^{\pi }\right]=0,\;{\text{kai}}\;k\neq n,\quad (5.1)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c4728a4a3f7f23ec0e8adb581c2580ddb4bf4394)
![{\displaystyle \sin(A)\cdot \sin(B)=-{\frac {1}{2}}[\cos(A+B)-\cos(A-B)];}](https://wikimedia.org/api/rest_v1/media/math/render/svg/47a9370c8c1c1ab2e819615a0c4059ade80512fc)
![{\displaystyle \int _{-\pi }^{\pi }\sin(kx)\cos(nx){\text{d}}x={\frac {1}{2}}\int _{-\pi }^{\pi }[\sin((k+n)x)+\sin((k-n)x)]{\text{d}}x=-{\frac {1}{2}}{\Big (}{\frac {\cos((k+n)x)}{k+n}}+{\frac {\cos((k-n)x)}{k-n}}{\Big )}|_{-\pi }^{\pi }=0.\quad (5.2)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eefcb1d0b93974bbc81f051c57edda6f26166266)
![{\displaystyle [-\pi ;\pi ],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/be3bb6a7405877090d192b894618273824135a6a)
![{\displaystyle =a_{k}\int _{-\pi }^{\pi }\cos ^{2}(kx){\text{d}}x+b_{k}\int _{-\pi }^{\pi }{\frac {1}{2}}[\sin((k+k)x)-\sin((k-k)x)]{\text{d}}x=a_{k}\int _{-\pi }^{\pi }\cos ^{2}(kx){\text{d}}x+{\frac {b_{k}}{2}}\int _{-\pi }^{\pi }\sin((k+k)x){\text{d}}x=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/66df71b4e7fc28cdc701e61ac834b24f4972322b)
![{\displaystyle [-\pi ;\pi ]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1e483c5368694d4985a9181e5288540173d602c4)
![{\displaystyle [-\pi ;\;\pi ].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8e6cfd2393e995e7c43edf90debacebce850c8ac)
![{\displaystyle [-\pi ;\;\pi ]\;F(x)=f(x).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3307de6eea634f0b663ca38570d1fec0310618bb)
![{\displaystyle x\in [-\pi ;\pi ]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b1380c33dc47f81124bf40b0d0631106a330101c)
![{\displaystyle b_{n}={\frac {1}{\pi }}\int _{-\pi }^{\pi }f(x)\sin(nx)\;{\text{d}}x={\frac {1}{\pi }}\left[\int _{-\pi }^{0}f(x)\sin(nx)\;{\text{d}}x+\int _{0}^{\pi }f(x)\sin(nx)\;{\text{d}}x\right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4a4f7c1b52aef2dde4ca2dadf6772cc51cd262e0)
![{\displaystyle b_{n}={\frac {1}{\pi }}\left[-\int _{0}^{\pi }f(t)\sin(nt){\text{d}}t+\int _{0}^{\pi }f(x)\sin(nx){\text{d}}x\right]=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9397b3e4daa384296931db81089dde24de660137)
![{\displaystyle b_{n}={\frac {1}{\pi }}\int _{-\pi }^{\pi }f(x)\sin(nx)\;{\text{d}}x={\frac {1}{\pi }}\left[\int _{-\pi }^{0}f(x)\sin(nx)\;{\text{d}}x+\int _{0}^{\pi }f(x)\sin(nx)\;{\text{d}}x\right]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/835d957a1b21de401cf51617e71db3e761c4d957)
![{\displaystyle ={\frac {1}{\pi }}\left[-\int _{0}^{\pi }f(x)\sin(nx){\text{d}}x+\int _{0}^{\pi }f(x)\sin(nx){\text{d}}x\right]=0\quad ({\text{jeigu}}\;\;f(-x)=f(x)).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6005a389c4cf5e27f3344f885e36762590eaa0c4)
![{\displaystyle x\in [-\pi ;\pi ],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a13f723553224078019ea1d77d799cf37fbe5d19)
![{\displaystyle b_{n}={\frac {2}{\pi }}\int _{0}^{\pi }x\sin(nx){\text{d}}x={\frac {2}{\pi }}\left[-{\frac {1}{n}}x\cos(nx)|_{0}^{\pi }+{\frac {1}{n}}\int _{0}^{\pi }\cos(nx){\text{d}}x\right]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/65950b6c205f47e2659b529ca117d6a6128fa200)
![{\displaystyle ={\frac {2}{\pi }}\left[-{\frac {1}{n}}(\pi \cos(n\pi )-0\cdot \cos(n\cdot 0))+{\frac {1}{n^{2}}}\sin(nx)|_{0}^{\pi }\right]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f65619d0e925148de72b4b691207cac98bf92f6e)
![{\displaystyle ={\frac {2}{\pi }}\left[-{\frac {\pi \cos(n\pi )}{n}}+{\frac {1}{n^{2}}}(\sin(n\pi )-\sin(n\cdot 0))\right]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1b7680badae5f9f8bb6559e1b10ae4eb9ecf7dc5)
![{\displaystyle ={\frac {2}{\pi }}\left[-{\frac {\pi \cos(n\pi )}{n}}\right]=-{\frac {2\cos(n\pi )}{n}}=-{\frac {2\cdot (-1)^{n}}{n}}=(-1)^{n+1}{\frac {2}{n}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dffcf986c16cd34337b80f1101d6f7c2511dd281)
![{\displaystyle a_{n}={\frac {2}{\pi }}\int _{0}^{\pi }x^{2}\cos(nx){\text{d}}x={\frac {2}{\pi }}\left[{\frac {x^{2}\sin(nx)}{n}}|_{0}^{\pi }-{\frac {2}{n}}\int _{0}^{\pi }x\sin(nx){\text{d}}x\right]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0852e8856f08a9021cca38cd634125cd01c148e8)
![{\displaystyle b_{n}={\frac {2}{\pi }}\int _{0}^{\pi }x^{3}\sin(nx){\text{d}}x={\frac {2}{\pi }}\left[\left({\frac {3x^{2}}{n^{2}}}-{\frac {6}{n^{4}}}\right)\sin(nx)-\left({\frac {x^{3}}{n}}-{\frac {6x}{n^{3}}}\right)\cos(nx)\right]|_{0}^{\pi }=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/79d0cf9124e5b312053609be04c8334c4a45d230)
![{\displaystyle ={\frac {2}{\pi }}\left[-\left({\frac {x^{3}}{n}}-{\frac {6x}{n^{3}}}\right)\cos(nx)\right]|_{0}^{\pi }=-{\frac {2}{\pi }}\left({\frac {\pi ^{3}}{n}}-{\frac {6\pi }{n^{3}}}\right)\cos(n\pi )=\left({\frac {12}{n^{3}}}-{\frac {2\pi ^{2}}{n}}\right)\cos(n\pi )=(-1)^{n}\left({\frac {12}{n^{3}}}-{\frac {2\pi ^{2}}{n}}\right).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ca9d17a0571be1ff51dbe116d1bc7b96efb9267e)
![{\displaystyle [-l;\;l]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9da9174e553377dec3e7af4240d98289d038547f)
![{\displaystyle x\in [-l;l],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f3aa5a7c1a68ac98c35171954148703eee1b9a6f)
![{\displaystyle ={\frac {2}{l}}\int _{0}^{l}x^{2}\cos {\frac {n\pi x}{l}}{\text{d}}x={\frac {2}{l}}\left[{\frac {x^{2}\sin {\frac {n\pi x}{l}}}{\frac {n\pi }{l}}}|_{0}^{l}-\int _{0}^{l}2x\cdot {\frac {\sin {\frac {n\pi x}{l}}}{\frac {n\pi }{l}}}{\text{d}}x\right]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/79c3800d6f734438a95b148bb8ea7c66f90ddac1)
