Aptarimas : Matematika/Furje eilutės
b
n
=
∫
−
π
π
f
(
x
)
sin
(
n
x
)
d
x
=
1
π
[
a
0
2
∫
−
π
π
sin
(
n
x
)
d
x
+
∑
m
=
1
∞
(
a
m
∫
−
π
π
sin
(
n
x
)
cos
(
m
x
)
d
x
+
b
m
∫
−
π
π
sin
(
n
x
)
sin
(
m
x
)
d
x
)
]
=
{\displaystyle b_{n}=\int _{-\pi }^{\pi }f(x)\sin(nx){\text{d}}x={\frac {1}{\pi }}\left[{\frac {a_{0}}{2}}\int _{-\pi }^{\pi }\sin(nx){\text{d}}x+\sum _{m=1}^{\infty }\left(a_{m}\int _{-\pi }^{\pi }\sin(nx)\cos(mx){\text{d}}x+b_{m}\int _{-\pi }^{\pi }\sin(nx)\sin(mx){\text{d}}x\right)\right]=}
=
1
π
[
0
+
∑
m
=
1
∞
(
0
+
b
m
∫
−
π
π
sin
(
n
x
)
sin
(
m
x
)
d
x
)
]
=
b
n
π
∫
−
π
π
sin
2
(
n
x
)
d
x
=
b
n
π
(
∫
−
π
0
sin
2
(
n
x
)
d
x
+
∫
0
π
sin
2
(
n
x
)
d
x
)
=
{\displaystyle ={\frac {1}{\pi }}\left[0+\sum _{m=1}^{\infty }\left(0+b_{m}\int _{-\pi }^{\pi }\sin(nx)\sin(mx){\text{d}}x\right)\right]={\frac {b_{n}}{\pi }}\int _{-\pi }^{\pi }\sin ^{2}(nx){\text{d}}x={\frac {b_{n}}{\pi }}\left(\int _{-\pi }^{0}\sin ^{2}(nx){\text{d}}x+\int _{0}^{\pi }\sin ^{2}(nx){\text{d}}x\right)=}
=
b
n
π
(
1
2
∫
−
π
0
(
1
−
cos
(
2
n
x
)
)
d
x
+
1
2
∫
0
π
(
1
−
cos
(
2
n
x
)
)
d
x
)
=
b
n
π
[
1
2
(
x
−
1
2
n
sin
(
2
n
x
)
)
|
−
π
0
+
1
2
(
x
−
1
2
n
sin
(
2
n
x
)
)
|
0
π
]
=
{\displaystyle ={\frac {b_{n}}{\pi }}\left({\frac {1}{2}}\int _{-\pi }^{0}(1-\cos(2nx)){\text{d}}x+{\frac {1}{2}}\int _{0}^{\pi }(1-\cos(2nx)){\text{d}}x\right)={\frac {b_{n}}{\pi }}\left[{\frac {1}{2}}\left(x-{\frac {1}{2n}}\sin(2nx)\right)|_{-\pi }^{0}+{\frac {1}{2}}\left(x-{\frac {1}{2n}}\sin(2nx)\right)|_{0}^{\pi }\right]=}
=
b
n
π
[
x
2
|
−
π
0
+
x
2
|
0
π
]
=
b
n
2
π
(
x
|
−
π
0
+
x
|
0
π
)
=
b
n
2
π
[
(
0
−
(
−
π
)
)
+
(
π
−
0
)
]
=
b
n
2
π
[
π
+
π
]
=
b
n
.
{\displaystyle ={\frac {b_{n}}{\pi }}\left[{\frac {x}{2}}|_{-\pi }^{0}+{\frac {x}{2}}|_{0}^{\pi }\right]={\frac {b_{n}}{2\pi }}(x|_{-\pi }^{0}+x|_{0}^{\pi })={\frac {b_{n}}{2\pi }}[(0-(-\pi ))+(\pi -0)]={\frac {b_{n}}{2\pi }}[\pi +\pi ]=b_{n}.}
Kitoks paaiškinimas (taikomas tokioms lyginėms arba nelyginėms funkcijoms kaip
f
(
x
)
=
x
2
{\displaystyle f(x)=x^{2}}
f
(
x
)
=
x
3
{\displaystyle f(x)=x^{3}}
f
(
x
)
=
x
4
{\displaystyle f(x)=x^{4}}
f
(
x
)
=
x
5
{\displaystyle f(x)=x^{5}}
f
(
x
)
=
−
|
x
|
,
{\displaystyle f(x)=-|x|,}
f
(
x
)
=
−
x
2
,
{\displaystyle f(x)=-x^{2},}
f
(
x
)
=
−
|
x
3
|
{\displaystyle f(x)=-|x^{3}|}
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}
f
(
x
)
⋅
sin
x
=
|
u
|
{\displaystyle f(x)\cdot \sin x=|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
>
0
{\displaystyle x>0}
x
<
0
{\displaystyle x<0}
f
(
x
)
⋅
sin
x
=
u
{\displaystyle f(x)\cdot \sin x=u}
f
(
x
)
{\displaystyle f(x)\;}
∫
−
π
0
f
(
x
)
sin
(
n
x
)
d
x
=
−
∫
0
π
f
(
x
)
sin
(
n
x
)
d
x
.
{\displaystyle \int _{-\pi }^{0}f(x)\sin(nx){\text{d}}x=-\int _{0}^{\pi }f(x)\sin(nx){\text{d}}x.}
Grįžti į "Matematika/Furje eilutės" puslapį.
![{\displaystyle b_{n}=\int _{-\pi }^{\pi }f(x)\sin(nx){\text{d}}x={\frac {1}{\pi }}\left[{\frac {a_{0}}{2}}\int _{-\pi }^{\pi }\sin(nx){\text{d}}x+\sum _{m=1}^{\infty }\left(a_{m}\int _{-\pi }^{\pi }\sin(nx)\cos(mx){\text{d}}x+b_{m}\int _{-\pi }^{\pi }\sin(nx)\sin(mx){\text{d}}x\right)\right]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0614a445d69555271607d61f97e9ba4f917682df)
![{\displaystyle ={\frac {1}{\pi }}\left[0+\sum _{m=1}^{\infty }\left(0+b_{m}\int _{-\pi }^{\pi }\sin(nx)\sin(mx){\text{d}}x\right)\right]={\frac {b_{n}}{\pi }}\int _{-\pi }^{\pi }\sin ^{2}(nx){\text{d}}x={\frac {b_{n}}{\pi }}\left(\int _{-\pi }^{0}\sin ^{2}(nx){\text{d}}x+\int _{0}^{\pi }\sin ^{2}(nx){\text{d}}x\right)=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/98c4f526aa9067c5a31130ac1b743a353325ef72)
![{\displaystyle ={\frac {b_{n}}{\pi }}\left({\frac {1}{2}}\int _{-\pi }^{0}(1-\cos(2nx)){\text{d}}x+{\frac {1}{2}}\int _{0}^{\pi }(1-\cos(2nx)){\text{d}}x\right)={\frac {b_{n}}{\pi }}\left[{\frac {1}{2}}\left(x-{\frac {1}{2n}}\sin(2nx)\right)|_{-\pi }^{0}+{\frac {1}{2}}\left(x-{\frac {1}{2n}}\sin(2nx)\right)|_{0}^{\pi }\right]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/22d18777b219a5841b5233969d8810803bff2bad)
![{\displaystyle ={\frac {b_{n}}{\pi }}\left[{\frac {x}{2}}|_{-\pi }^{0}+{\frac {x}{2}}|_{0}^{\pi }\right]={\frac {b_{n}}{2\pi }}(x|_{-\pi }^{0}+x|_{0}^{\pi })={\frac {b_{n}}{2\pi }}[(0-(-\pi ))+(\pi -0)]={\frac {b_{n}}{2\pi }}[\pi +\pi ]=b_{n}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/694e73b8000d954b0003d4e218f1fd008e44ef25)
