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8/20/2019 Furije Jun Jul Fizika
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8/20/2019 Furije Jun Jul Fizika
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n 1
an = 1
π
π
−π
2 −
|x|
π
cos(nx) dx =
2
π
π
−π
2 −
x
π
cos(nx) dx = =
2
π2n2(1 − (−1)n).
S (x) f (x) S (x) = a0
2
+∞n=1
an cos(nx) +
bn sin(nx)
S (x) = 3
2 +
2
π2
∞n=1
(1 − (−1)n)
n2 cos(nx) =
3
2 +
4
π2
∞n=1
1
(2n − 1)2 cos((2n − 1)x),
n 1 − (−1)n = 0
∞n=1
cos(10n − 5)
(2n − 1)2 S (5)
x = 5 S (5) = 3
2 +
4
π2
∞
n=1
cos(10n − 5)
(2n − 1)2
∞
n=1
cos(10n − 5)
(2n − 1)2 =
π2
4 S (5) − 3
2
S (5)
S (5) = f̃ (5)
f̃ 2π f
S (5) = f (5) = 2 − 5
π 5 /∈ [−π, π]
f̃ (5) = f (5)
f̃ (5)
f
f̃ (5) = f̃ (5−2π) = f (5−2π) 5−2π ∈ [−π, π]
f̃
f̃
f̃ = f
S (5) = f (5 − 2π) = 2 − 5 − 2π
π = 2 −
2π − 5
π =
5
π.
∞n=1
cos(10n − 5)
(2n − 1)2 =
π2
4
S (5) −
3
2
=
π2
4 5
π −
3
2
=
π(10 − 3π)
8 .
(2n−1) 4
f (2n − 1) 2 2 · 2 = 4
secret
m
a2n b2n 2m
1
π
π
−π(f (x))2 dx =
a202
+∞n=1
(a2n + b2n).
1
π
π
−π
2 −
|x|
π
2dx =
2
π
π
0
2 −
x
π
2dx = =
14
3 ,
32
2 + 4
π2
2 ∞n=1
1
(2n − 1)4 =
9
2 +
16
π4
∞n=1
1
(2n − 1)4,
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14
3 =
9
2 +
16
π4
∞n=1
1
(2n − 1)4
∞
n=1
1
(2n − 1)4 =
π4
1614
3 −
9
2 =
π4
96.
https://www.youtube.com/watch?v=Kz3UIYmfhS4
f (x) =
−x2, x ∈ (0, 1)
0, x ∈ [1, 2) (0, 2)
∞n=1
2(−1)n+1 − (2n − 1)π
(2n − 1)3
an = 0 n 0 an
f an = 0 (0, 2)
(−2, 0]
f̃
f̃ (x) =
0, x ∈ (−2, −1],
x2, x ∈ (−1, 0),
0, x = 0,
−x2 (= f (x)), x ∈ (0, 1),
0 (= f (x)), x ∈ [1, 2).
(−2, 2) 2l = 4
2l = 4
f̃
bn = 1
2
2
−2f̃ (x)sin
nπx
2
dx =
2
0
f̃ (x)sinnπx
2
dx =
2
0
f (x)sinnπx
2
dx
=
1
0
(−x2)sinnπx
2
dx +
2
1
0 · sinnπx
2
dx =
1
0
(−x2)sinnπx
2
dx
= = 2
π3
n3 (π2n2 − 8)cosnπ
2− 4πn sinnπ
2+ 8.
f
S (x) = 2
π3
∞n=1
(π2n2 − 8)cosnπ
2
− 4πn sin
nπ
2
n3
sinnπx
2
.
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nπ
2
S (x) =
∞k=1
”2k” +
∞k=1
”2k − 1” = 2
π3
∞k=1
(4π2k2 − 8) cos(kπ) − 8πk sin(kπ) + 8
8k3 sin(kπx)
+∞k=1
(π2(2k − 1)2 − 8) cos(kπ − π
2 ) − 4π(2k − 1) sin(kπ − π
2 ) + 8(2k − 1)3
sin
(kπ − π2
)x
= 2
π3
∞k=1
(4π2k2 − 8)cos(kπ) − 8πk sin(kπ) + 8
8k3 sin(kπx)
+∞k=1
(π2(2k − 1)2 − 8) sin(kπ) + 4π(2k − 1) cos(kπ) + 8
(2k − 1)3 sin
(kπ −
π
2)x
= ().
cos(kπ) = (−1)k sin(kπ) = 0
() = 2
π3
∞k=1
(4π2k2 − 8)(−1)k + 8
8k3 sin(kπx) +
∞k=1
4π(2k − 1)(−1)k + 8
(2k − 1)3 sin
(kπ −
π
2)x
.
x = 1
S (1) = 2
π3
∞k=1
(4π2k2 − 8)(−1)k + 8
8k3 sin(kπ) +
∞k=1
4π(2k − 1)(−1)k + 8
(2k − 1)3 sin
(kπ −
π
2)
= 2
π3
∞k=1
4π(2k − 1)(−1)k + 8
(2k − 1)3 (− cos(kπ)) =
2
π3
∞k=1
4π(2k − 1)(−1)k + 8
(2k − 1)3 (−1)k+1
= 2
π3
∞k=1
4π(2k − 1)(−1) + 8(−1)k+1
(2k − 1)3 = 8
π3
∞k=1
2(−1)k+1 − (2k − 1)π
(2k − 1)3 ,
π3
8 S (1)
x = 1
f̃
f̃ (1) = f (1) 1 ∈ (0, 2)
S (1) = limx→1− f (x) + limx→1+ f (x)
2 =
−1 + 0
2 = −
1
2,
∞
n=1
2(−1)n+1 − (2n − 1)π
(2n − 1)3 =
π3
8−1
2 = −
π3
16.
THE END
