0
n = 2k,
=
2 π
0 −π
f
(x)
sin
nxdx,
n = 2k − 1.
4
ÓnŒ
a2n = 0.
K0.1.4. an, bn´±Ï•2π ŒÈ½ýéŒÈ¼êf Fp“Xê,y²²£ ¼êf (x + h) Fp“Xê´
An = an cos nh + bn sin nh,
Bn = bn cos nh + an sin nh.
bn
=
1 π
π −π
f
(x)
sin
nxdx
=
1 π
0 −π
f (x)
sin
nxdx
+
1 π
0πf (x) sin nxdx
=
1 π
0 −π
f (x)
sin
nxdx
+
1 π
0 −π
f
(t
+
π)
sin
n(t
+
π)dt
=
1 π
ห้องสมุดไป่ตู้
0 −π
f
(x)
cos
nxdx
+
(−1)n+1
1 π
0 −π
f
(t)
sin
ntdt
§0.1 13.1 Fp“?ê Ä Vg
1
§0.1 13.1 Fp“?ê Ä Vg
K0.1.1. y²V4 (Legendre)õ‘ª
´[−1, 1]þ
p0(x) = 1
pn(x)
=
1 2nn!
dn
(x2−1)n dxn
,
n
=
1, 2, · · ·
¼êX.
Proof. V4 õ‘ª´[−1, 1]þ
)
dx
dn−1(x2 − 1)n F (x) = ( dxn−1 )
= (x2 − 1)g(x)
¤±F (1) = F (−1) = 0,
1 −1
dn(x2−1)n dxn
dm
(x2−1)m dxm
dx
=
−
1 −1
dn−1 (x2 −1)n dxn−1
(
dm
(x2−1)m dxm
)
dx
=
−
( 1 d(n−2)(x2−1)n
−1
dx(n−2)
)
( dm+1(x2−1)m
dxm+1
)
dx
=
·
·
·
=
(−1)m
( 1 d(n−m)(x2−1)n
−1
dx(n−m)
)
(
dm+m (x2 −1)m dxm+m
)
dx
=
(−1)m(2m)!
dx 1 d(n−m)(x2−1)n
−1 dx(n−m)
0,
n > m,
=
(−1)n
2
(2n)!! (2n+1)!!
=
1 π
0 −π
f (x)
cos
nxdx
+
1 π
0 −π
f
(t
+
π)
cos
n(t
+
π)dt
=
1 π
0 −π
f
(x)
cos
nxdx
+
(−1)n
1 π
0 −π
f
(t)
cos
ntdt
0
n = 2k − 1,
=
2 π
0 −π
f
(x)
cos
nxdx,
n = 2k.
ÓnŒ
b2n−1 = 0.
(2) é?¿ n > 0,
k=1
=
∞
A0 2
+
(Ak cos kx + Bk sin kx)
k=1
¤±n ¼êTn Fp“?ê´§ .
K0.1.3. f (x)´±Ï•2π ŒÈ¼ê§y²µ(1) XJf 3[π, π]¥÷vf (x + π) = f (x), @of 4|“Xê÷v
a2n−1 = b2n−1 = 0, n ≥ 1.
,
n = m.
2
K0.1.2. y²n õ‘ª
n
Tn(x) = (Ak cos kx + Bk sin kx)
k=0
Fp“?êÒ´§gC.
Proof. Äk£Á˜en ¼êX
5µ
π
= 0, n = m
cos nx cos mxdx
−π
= π, n = m.
π
= 0, n = m
sin nx sin mxdx
¼êX⇔
1
= 0, n = m,
pn(x) · pm(x)dx
−1
= 0, n = m.
1 dn(x2 − 1)n dm(x2 − 1)m
= 0, n = m,
⇔
−1
dxn
dxm
dx = 0, n = m.
e¡b
1
1
whenn = m = 0, p0(x) · p0(x)dx = 1dx = x|1−1 = 2.
k=1
= Am
∀m > n,Ó dn ¼êX
55Υam = 0. =
am =
Am, 0 ≤ m ≤ n 0, m > n.
§0.1 13.1 Fp“?ê Ä Vg
3
bm =
Bm, 1 ≤ m ≤ n 0, m > n.
¤±Tn
Fp“?ê•
∞
a0 2
+
(ak cos kx + bk sin kx)
k=1
= 2A0
ùp^ n ¼êX
5.
éu?¿ n ≥ m ≥ 1,k
am
=
1 π
π −π
Tn
cos
mxdx
=
1 π
π −π
A0 2
cos
mxdx
+
1 π
π −π
n
(Ak cos kx + Bk sin kx) cos mxdx
k=1
=
1 π
π −π
n
(Ak cos kx cos mx + Bk sin kx cos mx)dx
Proof.
An
=
1 π
π −π
f
(x
+
h)
cos
nxdx,
=
1 π
π+h −π+h
f
(t)
cos
n(t
−
h)dx(-t
=
x
+
h)
=
1 π
π+h −π+h
f (t)
cos
nx cos
nhdx
+
1 π
π+h −π+h
f
(t)
sin
nx
sin
nhdt
=
cos
nh
1 π
π −π
f
(t)
cos
nxdx
+
sin
−1
−1
n ≥ m,
1 −1
dn(x2−1)n dxn
dm(x2−1)m dxm
dx
=
−11(
dn−1 (x2 −1)n dxn−1
)
dm
(x2−1)m dxm
dx
=
(
dn−1 (x2 −1)n dxn−1
)
dm
(x2−1)m dxm
|1−1
−
1 −1
dn−1 (x2 −1)n dxn−1
(
dm
(x2−1)m dxm
nh
1 π
π −π
f
(t)
sin
nxdt
= an cos nh + bn sin nh.
Bn
=
1 π
π −π
f
(x
+
h)
sin
nxdx,
=
1 π
π+h −π+h
f
(t)
sin
n(t
−
h)dx(-t
=
x
+
h)
=
1 π
π+h −π+h
f (t)
sin
nx cos
nhdx
−
1 π
π+h −π+h
(2) XJf 3[π, π]¥÷vf (x + π) = −f (x), @of 4|“Xê÷v
a2n = b2n = 0, n ≥ 1.
Proof. (1) é?¿ n > 0,
an
=
1 π
π −π
f
(x)
cos
nxdx
=
1 π
0 −π
f (x)
cos
nxdx
+
1 π
0πf (x) cos nxdx
−π
= π, n = m.
π
cos nx sin mxdx = 0.
−π
Tn(x)´±Ï•2π ¼ê§cos kx, sin kx3«m[−π, π]þŒÈ©§l ¤±Tn(x) Fp“XêŒd½Â¦Ñ.
Tn(x)ŒÈ§
a0
=
1 π
π −π
Tndx
=