APM2611
ASSIGNMENT 2 2021
QUESTION 1
𝑻𝒉𝒆 𝑭𝒐𝒖𝒓𝒊𝒆𝒓 𝑺𝒆𝒓𝒊𝒆𝒔 𝒅𝒆𝒇𝒊𝒏𝒆𝒅 𝒐𝒏 𝒕𝒉𝒆 𝒊𝒏𝒕𝒆𝒓𝒗𝒂𝒍 (−𝒑, 𝒑):
∞
𝒂𝒐 𝒏𝝅𝒙 𝒏𝝅𝒙
𝒇(𝒙) = + ∑ (𝒂𝒏 𝐜𝐨𝐬 ( ) +𝒃𝒏 𝐬𝐢𝐧 ( ))
𝟐 𝒑 𝒑
𝒏=𝟏
𝟏 𝒑
𝒘𝒉𝒆𝒓𝒆 ∶ 𝒂𝒐 = ∫ 𝒇(𝒙) 𝒅𝒙
𝒑 −𝒑
𝟏 𝒑 𝒏𝝅𝒙
𝒂𝒏 = ∫ 𝒇(𝒙) 𝐜𝐨𝐬 ( ) 𝒅𝒙
𝒑 −𝒑 𝒑
𝟏 𝒑 𝒏𝝅𝒙
𝒃𝒏 = ∫ 𝒇(𝒙) 𝐬𝐢𝐧 ( ) 𝒅𝒙
𝒑 −𝒑 𝒑
0, − 1 ≤ 𝑥 ≤ 0
𝑓(𝑥) = {
𝑥, 0≤𝑥≤1
𝑇ℎ𝑒 𝑓𝑜𝑢𝑟𝑖𝑒𝑟 𝑠𝑒𝑟𝑖𝑒𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 [−1,1]
1 1
𝑎𝑜 = ∫ 𝑓(𝑥) 𝑑𝑥
1 −1
0 1
𝑎𝑜 = ∫ 0 𝑑𝑥 + ∫ 𝑥 𝑑𝑥
−1 0
1
𝑥2
𝑎𝑜 = 0 + [ ]
2 0
(1)2 (0)2
𝑎𝑜 = [ − ]
2 2
1
𝑎𝑜 =
2
1 𝑝 𝑛𝜋𝑥
𝑎𝑛 = ∫ 𝑓(𝑥) cos ( ) 𝑑𝑥
𝑝 −𝑝 𝑝
1 1 𝑛𝜋𝑥
𝑎𝑛 = [∫ 𝑓(𝑥) cos ( ) 𝑑𝑥]
1 −1 𝑝
, 1 0 𝑛𝜋𝑥 0
𝑛𝜋𝑥
𝑎𝑛 = [∫ 0 ∙ cos ( ) 𝑑𝑥 + ∫ 𝑥 cos ( ) 𝑑𝑥 ]
1 −1 1 −1 1
0 1
𝑎𝑛 = ∫ 0 ∙ cos(𝑛𝜋𝑥) 𝑑𝑥 + ∫ 𝑥 cos(𝑛𝜋𝑥) 𝑑𝑥
−1 0
1
𝑎𝑛 = ∫ 𝑥 cos(𝑛𝜋𝑥) 𝑑𝑥
0
𝑆𝑜𝑙𝑣𝑒 ∫ 𝑥 cos(𝑛𝜋𝑥) 𝑑𝑥 𝑏𝑦 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛 𝑏𝑦 𝑝𝑎𝑟𝑡𝑠
∫ 𝑥 cos(𝑛𝜋𝑥) 𝑑𝑥
𝐿𝑒𝑡: 𝑢 = 𝑥 𝑎𝑛𝑑 𝑑𝑣 = cos(𝑛𝜋𝑥)
sin(𝑛𝜋𝑥)
𝑑𝑢 = 1 𝑎𝑛𝑑 𝑣 =
𝑛𝜋
∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢
sin(𝑛𝜋𝑥) sin(𝑛𝜋𝑥)
∫ 𝑥 cos(𝑛𝜋𝑥) 𝑑𝑥 = 𝑥 −∫ 𝑑𝑥
𝑛𝜋 𝑛𝜋
𝑥 sin(𝑛𝜋𝑥) 1
= − ∫ sin(𝑛𝜋𝑥) 𝑑𝑥
𝑛𝜋 𝑛𝜋
𝑥 sin(𝑛𝜋𝑥) 1 cos(𝑛𝜋𝑥)
= − (− )
𝑛𝜋 𝑛𝜋 𝑛𝜋
𝑥 sin(𝑛𝜋𝑥) cos(𝑛𝜋𝑥)
= +
𝑛𝜋 𝑛2 𝜋 2
1 1
𝑥 sin(𝑛𝜋𝑥) cos(𝑛𝜋𝑥)
𝑎𝑛 = ∫ 𝑥 cos(𝑛𝜋𝑥) 𝑑𝑥 = [ + ]
0 𝑛𝜋 𝑛2 𝜋 2 0
(1) sin(𝑛𝜋(1)) cos(𝑛𝜋(1)) (0) sin(𝑛𝜋(0)) cos(𝑛𝜋(0))
𝑎𝑛 = [( + 2 2 )−( + )]
𝑛𝜋 𝑛 𝜋 𝑛𝜋 𝑛2 𝜋 2
sin(𝑛𝜋) cos(𝑛𝜋) sin(0) cos((0))
𝑎𝑛 = [( + 2 2 )−( + )] ∴ sin(𝑛𝜋) = 0 𝑎𝑛𝑑 cos(𝑛𝜋) = (−1)𝑛
𝑛𝜋 𝑛 𝜋 𝑛𝜋 𝑛2 𝜋 2
0 (−1)𝑛 0 1
𝑎𝑛 = [( + 2 2 ) − ( + 2 2 )]
𝑛𝜋 𝑛 𝜋 𝑛𝜋 𝑛 𝜋
(−1)𝑛 1
𝑎𝑛 = 2 2
− 2 2
𝑛 𝜋 𝑛 𝜋
, (−1)𝑛 − 1
𝑎𝑛 =
𝑛2 𝜋 2
1 𝑝 𝑛𝜋𝑥
𝑏𝑛 = ∫ 𝑓(𝑥) sin ( ) 𝑑𝑥
𝑝 −𝑝 𝑝
1 1 𝑛𝜋𝑥
𝑏𝑛 = ∫ 𝑓(𝑥) sin ( ) 𝑑𝑥
1 −1 1
1
𝑏𝑛 = ∫ 𝑓(𝑥) sin(𝑛𝜋𝑥) 𝑑𝑥
−1
0 1
𝑏𝑛 = ∫ 0 sin(𝑛𝜋𝑥) 𝑑𝑥 + ∫ 𝑥 sin(𝑛𝜋𝑥) 𝑑𝑥
−1 0
1
𝑏𝑛 = ∫ 𝑥 sin(𝑛𝜋𝑥) 𝑑𝑥
0
𝑆𝑜𝑙𝑣𝑒 ∫ 𝑥 sin(𝑛𝜋𝑥) 𝑑𝑥 𝑏𝑦 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛 𝑏𝑦 𝑝𝑎𝑟𝑡𝑠:
∫ 𝑥 sin(𝑛𝜋𝑥) 𝑑𝑥
𝐿𝑒𝑡: 𝑢 = 𝑥 𝑎𝑛𝑑 𝑑𝑣 = sin(𝑛𝜋𝑥)
cos(𝑛𝜋𝑥)
𝑑𝑢 = 1 𝑎𝑛𝑑 𝑣 = −
𝑛𝜋
∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢
𝑥 cos(𝑛𝜋𝑥) cos(𝑛𝜋𝑥)
∫ 𝑥 sin(𝑛𝜋𝑥) 𝑑𝑥 = − +∫ 𝑑𝑥
𝑛𝜋 𝑛𝜋
𝑥 cos(𝑛𝜋𝑥) 1
=− + ∫ cos(𝑛𝜋𝑥) 𝑑𝑥
𝑛𝜋 𝑛𝜋
𝑥 cos(𝑛𝜋𝑥) 1 sin(𝑛𝜋𝑥)
=− + ( )
𝑛𝜋 𝑛𝜋 𝑛𝜋
𝑥 cos(𝑛𝜋𝑥) sin(𝑛𝜋𝑥)
=− +
𝑛𝜋 𝑛2 𝜋 2
1 1
𝑥 cos(𝑛𝜋𝑥) sin(𝑛𝜋𝑥)
𝑏𝑛 = ∫ 𝑥 sin(𝑛𝜋𝑥) 𝑑𝑥 = [− + ]
0 𝑛𝜋 𝑛2 𝜋 2 0
(1) cos(𝑛𝜋(1)) sin(𝑛𝜋(1)) (0) cos(𝑛𝜋(0)) sin(𝑛𝜋(0))
𝑏𝑛 = [(− + 2 2 ) − (− + )]
𝑛𝜋 𝑛 𝜋 𝑛𝜋 𝑛2 𝜋 2
ASSIGNMENT 2 2021
QUESTION 1
𝑻𝒉𝒆 𝑭𝒐𝒖𝒓𝒊𝒆𝒓 𝑺𝒆𝒓𝒊𝒆𝒔 𝒅𝒆𝒇𝒊𝒏𝒆𝒅 𝒐𝒏 𝒕𝒉𝒆 𝒊𝒏𝒕𝒆𝒓𝒗𝒂𝒍 (−𝒑, 𝒑):
∞
𝒂𝒐 𝒏𝝅𝒙 𝒏𝝅𝒙
𝒇(𝒙) = + ∑ (𝒂𝒏 𝐜𝐨𝐬 ( ) +𝒃𝒏 𝐬𝐢𝐧 ( ))
𝟐 𝒑 𝒑
𝒏=𝟏
𝟏 𝒑
𝒘𝒉𝒆𝒓𝒆 ∶ 𝒂𝒐 = ∫ 𝒇(𝒙) 𝒅𝒙
𝒑 −𝒑
𝟏 𝒑 𝒏𝝅𝒙
𝒂𝒏 = ∫ 𝒇(𝒙) 𝐜𝐨𝐬 ( ) 𝒅𝒙
𝒑 −𝒑 𝒑
𝟏 𝒑 𝒏𝝅𝒙
𝒃𝒏 = ∫ 𝒇(𝒙) 𝐬𝐢𝐧 ( ) 𝒅𝒙
𝒑 −𝒑 𝒑
0, − 1 ≤ 𝑥 ≤ 0
𝑓(𝑥) = {
𝑥, 0≤𝑥≤1
𝑇ℎ𝑒 𝑓𝑜𝑢𝑟𝑖𝑒𝑟 𝑠𝑒𝑟𝑖𝑒𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 [−1,1]
1 1
𝑎𝑜 = ∫ 𝑓(𝑥) 𝑑𝑥
1 −1
0 1
𝑎𝑜 = ∫ 0 𝑑𝑥 + ∫ 𝑥 𝑑𝑥
−1 0
1
𝑥2
𝑎𝑜 = 0 + [ ]
2 0
(1)2 (0)2
𝑎𝑜 = [ − ]
2 2
1
𝑎𝑜 =
2
1 𝑝 𝑛𝜋𝑥
𝑎𝑛 = ∫ 𝑓(𝑥) cos ( ) 𝑑𝑥
𝑝 −𝑝 𝑝
1 1 𝑛𝜋𝑥
𝑎𝑛 = [∫ 𝑓(𝑥) cos ( ) 𝑑𝑥]
1 −1 𝑝
, 1 0 𝑛𝜋𝑥 0
𝑛𝜋𝑥
𝑎𝑛 = [∫ 0 ∙ cos ( ) 𝑑𝑥 + ∫ 𝑥 cos ( ) 𝑑𝑥 ]
1 −1 1 −1 1
0 1
𝑎𝑛 = ∫ 0 ∙ cos(𝑛𝜋𝑥) 𝑑𝑥 + ∫ 𝑥 cos(𝑛𝜋𝑥) 𝑑𝑥
−1 0
1
𝑎𝑛 = ∫ 𝑥 cos(𝑛𝜋𝑥) 𝑑𝑥
0
𝑆𝑜𝑙𝑣𝑒 ∫ 𝑥 cos(𝑛𝜋𝑥) 𝑑𝑥 𝑏𝑦 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛 𝑏𝑦 𝑝𝑎𝑟𝑡𝑠
∫ 𝑥 cos(𝑛𝜋𝑥) 𝑑𝑥
𝐿𝑒𝑡: 𝑢 = 𝑥 𝑎𝑛𝑑 𝑑𝑣 = cos(𝑛𝜋𝑥)
sin(𝑛𝜋𝑥)
𝑑𝑢 = 1 𝑎𝑛𝑑 𝑣 =
𝑛𝜋
∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢
sin(𝑛𝜋𝑥) sin(𝑛𝜋𝑥)
∫ 𝑥 cos(𝑛𝜋𝑥) 𝑑𝑥 = 𝑥 −∫ 𝑑𝑥
𝑛𝜋 𝑛𝜋
𝑥 sin(𝑛𝜋𝑥) 1
= − ∫ sin(𝑛𝜋𝑥) 𝑑𝑥
𝑛𝜋 𝑛𝜋
𝑥 sin(𝑛𝜋𝑥) 1 cos(𝑛𝜋𝑥)
= − (− )
𝑛𝜋 𝑛𝜋 𝑛𝜋
𝑥 sin(𝑛𝜋𝑥) cos(𝑛𝜋𝑥)
= +
𝑛𝜋 𝑛2 𝜋 2
1 1
𝑥 sin(𝑛𝜋𝑥) cos(𝑛𝜋𝑥)
𝑎𝑛 = ∫ 𝑥 cos(𝑛𝜋𝑥) 𝑑𝑥 = [ + ]
0 𝑛𝜋 𝑛2 𝜋 2 0
(1) sin(𝑛𝜋(1)) cos(𝑛𝜋(1)) (0) sin(𝑛𝜋(0)) cos(𝑛𝜋(0))
𝑎𝑛 = [( + 2 2 )−( + )]
𝑛𝜋 𝑛 𝜋 𝑛𝜋 𝑛2 𝜋 2
sin(𝑛𝜋) cos(𝑛𝜋) sin(0) cos((0))
𝑎𝑛 = [( + 2 2 )−( + )] ∴ sin(𝑛𝜋) = 0 𝑎𝑛𝑑 cos(𝑛𝜋) = (−1)𝑛
𝑛𝜋 𝑛 𝜋 𝑛𝜋 𝑛2 𝜋 2
0 (−1)𝑛 0 1
𝑎𝑛 = [( + 2 2 ) − ( + 2 2 )]
𝑛𝜋 𝑛 𝜋 𝑛𝜋 𝑛 𝜋
(−1)𝑛 1
𝑎𝑛 = 2 2
− 2 2
𝑛 𝜋 𝑛 𝜋
, (−1)𝑛 − 1
𝑎𝑛 =
𝑛2 𝜋 2
1 𝑝 𝑛𝜋𝑥
𝑏𝑛 = ∫ 𝑓(𝑥) sin ( ) 𝑑𝑥
𝑝 −𝑝 𝑝
1 1 𝑛𝜋𝑥
𝑏𝑛 = ∫ 𝑓(𝑥) sin ( ) 𝑑𝑥
1 −1 1
1
𝑏𝑛 = ∫ 𝑓(𝑥) sin(𝑛𝜋𝑥) 𝑑𝑥
−1
0 1
𝑏𝑛 = ∫ 0 sin(𝑛𝜋𝑥) 𝑑𝑥 + ∫ 𝑥 sin(𝑛𝜋𝑥) 𝑑𝑥
−1 0
1
𝑏𝑛 = ∫ 𝑥 sin(𝑛𝜋𝑥) 𝑑𝑥
0
𝑆𝑜𝑙𝑣𝑒 ∫ 𝑥 sin(𝑛𝜋𝑥) 𝑑𝑥 𝑏𝑦 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛 𝑏𝑦 𝑝𝑎𝑟𝑡𝑠:
∫ 𝑥 sin(𝑛𝜋𝑥) 𝑑𝑥
𝐿𝑒𝑡: 𝑢 = 𝑥 𝑎𝑛𝑑 𝑑𝑣 = sin(𝑛𝜋𝑥)
cos(𝑛𝜋𝑥)
𝑑𝑢 = 1 𝑎𝑛𝑑 𝑣 = −
𝑛𝜋
∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢
𝑥 cos(𝑛𝜋𝑥) cos(𝑛𝜋𝑥)
∫ 𝑥 sin(𝑛𝜋𝑥) 𝑑𝑥 = − +∫ 𝑑𝑥
𝑛𝜋 𝑛𝜋
𝑥 cos(𝑛𝜋𝑥) 1
=− + ∫ cos(𝑛𝜋𝑥) 𝑑𝑥
𝑛𝜋 𝑛𝜋
𝑥 cos(𝑛𝜋𝑥) 1 sin(𝑛𝜋𝑥)
=− + ( )
𝑛𝜋 𝑛𝜋 𝑛𝜋
𝑥 cos(𝑛𝜋𝑥) sin(𝑛𝜋𝑥)
=− +
𝑛𝜋 𝑛2 𝜋 2
1 1
𝑥 cos(𝑛𝜋𝑥) sin(𝑛𝜋𝑥)
𝑏𝑛 = ∫ 𝑥 sin(𝑛𝜋𝑥) 𝑑𝑥 = [− + ]
0 𝑛𝜋 𝑛2 𝜋 2 0
(1) cos(𝑛𝜋(1)) sin(𝑛𝜋(1)) (0) cos(𝑛𝜋(0)) sin(𝑛𝜋(0))
𝑏𝑛 = [(− + 2 2 ) − (− + )]
𝑛𝜋 𝑛 𝜋 𝑛𝜋 𝑛2 𝜋 2
