Find the fourier sin series for the function f(x) =x²in (0,pi)
Answers
Answer:
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Step-by-step explanation:
Assume that f(x) is an odd function on the interval [−L/2, L/2].
a_m=\dfrac{1}{L}\displaystyle\int_{-L}^{L}f(x)\cos(\dfrac{m\pi x}{L})dx=0, m=0,1,2,...a
m
=
L
1
∫
−L
L
f(x)cos(
L
mπx
)dx=0,m=0,1,2,...
b_n=\dfrac{1}{L}\displaystyle\int_{-L}^{L}f(x)\sin(\dfrac{n\pi x}{L})dx, n=1,2,...b
n
=
L
1
∫
−L
L
f(x)sin(
L
nπx
)dx,n=1,2,...
b_n=\dfrac{1}{L}\displaystyle\int_{-L}^{L}f(x)\sin(\dfrac{n\pi x}{L})dx=b
n
=
L
1
∫
−L
L
f(x)sin(
L
nπx
)dx=
=\dfrac{2}{\pi/2}\displaystyle\int_{0}^{\pi/2}\cos x\sin(\dfrac{n\pi x}{\pi/2})dx==
π/2
2
∫
0
π/2
cosxsin(
π/2
nπx
)dx=
=\dfrac{4}{\pi}\displaystyle\int_{0}^{\pi/2}\cos x\sin(2nx)dx=
π
4
∫
0
π/2
cosxsin(2nx)dx
\int\cos x\sin (2nx)dx=∫cosxsin(2nx)dx=
=\dfrac{1}{2}\int(\sin ((2n+1)x)+\sin ((2n-1)x))dx==
2
1
∫(sin((2n+1)x)+sin((2n−1)x))dx=
=-\dfrac{1}{2(2n+1)}\cos((2n+1)x)-\dfrac{1}{2(2n-1)}\cos((2n-1)x)+C=−
2(2n+1)
1
cos((2n+1)x)−
2(2n−1)
1
cos((2n−1)x)+C
b_n=\dfrac{4}{\pi}\displaystyle\int_{0}^{\pi/2}\cos x\sin(2nx)dx=b
n
=
π
4
∫
0
π/2
cosxsin(2nx)dx=
=-\dfrac{2}{\pi(2n+1)}(\cos((2n+1)\dfrac{\pi}{2})+\dfrac{2}{\pi(2n+1)}\cos((2n+1)(0))-=−
π(2n+1)
2
(cos((2n+1)
2
π
)+
π(2n+1)
2
cos((2n+1)(0))−
-\dfrac{2}{\pi(2n-1)}(\cos((2n-1)\dfrac{\pi}{2})+\dfrac{2}{\pi(2n-1)}\cos((2n-1)(0))=−
π(2n−1)
2
(cos((2n−1)
2
π
)+
π(2n−1)
2
cos((2n−1)(0))=
=\dfrac{2}{\pi(2n+1)}\sin(\pi n)+\dfrac{2}{\pi(2n+1)}+=
π(2n+1)
2
sin(πn)+
π(2n+1)
2
+
+\dfrac{2}{\pi(2n-1)}\sin(\pi n)+\dfrac{2}{\pi(2n-1)}=\dfrac{8n}{\pi(4n^2-1)}+
π(2n−1)
2
sin(πn)+
π(2n−1)
2
=
π(4n
2
−1)
8n
f(x)\sim\displaystyle\sum_{n=1}^\infin\dfrac{8n}{\pi(4n^2-1)}\sin(2nx)f(x)∼
n=1
∑
∞
π(4n
2
−1)
8n
sin(2nx)