推 jacky7987 :喔感謝大大QAQ 08/31 22:32
推 keroro321 :推 09/01 00:26
※ 引述《jacky7987 (憶)》之銘言：
: 其實我完全沒學過這類方法所以來求救QAQ
: 1 1 1 1
: a=----- + ----- + ------ + ------ + ...
: 2^2 6^2 10^2 14^2
: Find the value of a by the method of the Fourier series
: 感謝大家QAQ
1 1 1 1 1
a = ----- ( ----- + ----- + ----- + ----- + ... )
2^2 1^2 3^2 5^2 7^2
1 1 1 1 1
= ----- [ ( ----- + ----- + ----- + ----- + ... )
2^2 1^2 2^2 3^2 4^2
1 1 1 1
- ( ----- + ----- + ----- + ----- + ... ) ]
2^2 4^2 6^2 8^2
1 1 1 1 1
= --- [ ( ----- + ----- + ----- + ----- + ... )
4 1^2 2^2 3^2 4^2
1 1 1 1 1
- ----- ( ----- + ----- + ----- + ----- + ... ) ]
2^2 1^2 2^2 3^2 4^2
1 1 1 1 1 1
= --- ( 1 - - ) ( ----- + ----- + ----- + ----- + ... )
4 4 1^2 2^2 3^2 4^2
3 1 1 1 1 ∞ 1
= ---- ( ----- + ----- + ----- + ----- ... ) = (3/16) Σ ---
16 1^2 2^2 3^2 4^2 k=1 k^2
∫x^2 cos(ωx) dx = (2x/ω^2)cos(ωx) + [(x^2/ω) - (2/ω^3)]sin(ωx) + C
∫(x/2)^2 cos(ωx) dx = (x/2ω^2)cos(ωx) + [(x^2/4ω) - (1/2ω^3)]sin(ωx) + C
∞ kπx ∞ kπx
f(x) = a0 + Σ ak cos(----) + Σ bk sin(----)
k=1 L k=1 L
1 L 1 L kπx 1 L kπx
a0 = --- ∫ f(x) dx，ak = - ∫ f(x) cos(----) dx，bk = - ∫ f(x) sin(----) dx
2 L -L L -L L L -L L
此時 ω = kπ/L，在此設 L = π，f(x) = (x/2)^2 = x^2/4 => ω = k
則 a0 = π^2/12，ak = (1/k^2) cos(kπ) + (π/2k) sin(kπ) - (1/k^3π) sin(kπ)
bk = 0
又因為 k 為整數 => ak = (1/k^2) (-1)^k
x^2 π^2 ∞ (-1)^k
f(x) = --- = ------ + Σ ------ cos(kx)
4 12 k=1 k^2
π^2 π^2 ∞ (-1)^k π^2 ∞ (-1)^k
f(π) = ---- = ---- + Σ ------ cos(kπ) = ---- + Σ ------ (-1)^k
4 12 k=1 k^2 12 k=1 k^2
π^2 ∞ 1 ∞ 1 π^2 π^2 3π^2 π^2 2π^2 π^2
= ---- + Σ --- => Σ --- = ---- - ---- = ----- - ---- = ----- = ----
12 k=1 k^2 k=1 k^2 4 12 12 12 12 6
∞ 1 3 π^2 π^2
a = (3/16) Σ --- = -- ---- = ----
k=1 k^2 16 6 32
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