ƒtƒFƒ‹ƒ}[‚̍ŏI’藝‚̏ؖΎ (843Ϊ½)
γ‰Ί‘OŽŸ1-V
’Šo‰πœ Ϊ½žx

41: ‚P‚R‚Ql–Ϊ‚Μ‘f”‚³‚ρ [] 04/28(ŒŽ)21:42:10.15 ID:0uxBMTmL(1)
•κ•κ‚Ζ‚©‚’‚Δ'‚Ϊ‚Ϊ'‚ƓǂށB
137: —^μ [] 05/25(“ϊ)12:07:12.15 ID:4EyabO0p(3/3)
n=2‚Μ‚Ζ‚«AX^n+Y^n=Z^n‚ΝŽ©‘R”‰π‚𖳐”‚ΙŽ‚B
X^2+Y^2=Z^2‚πy^2=(x+1)^2-x^2c(1)‚Ζ‚¨‚­B(y,x‚Ν—L—”)
(1)‚π(y-1)(y+1)=2xc(2)‚Ζ‚¨‚­B
(2)‚Ν(y-1)=2‚Μ‚Ζ‚«A(y+1)=x‚ͺ¬—§‚B
(2)‚π(y-1)(y+1)=k2x/kc(3)‚Ζ‚¨‚­B
(3)‚Νk/k=1‚Ȃ̂ŁA(y-1)=k2‚Μ‚Ζ‚«A(y+1)=x/k‚ͺ¬—§‚B
ˆn=2‚Μ‚Ζ‚«AX^n+Y^n=Z^n‚ΝŽ©‘R”‰π‚𖳐”‚ΙŽ‚B
369: —^μ [] 07/17(–Ψ)12:00:52.15 ID:4J9At0pY(3/17)
n=2‚Μ‚Ζ‚«AX^n+Y^n=Z^n‚ΝŽ©‘R”‰π‚𖳐”‚ΙŽ‚B
X^2+Y^2=Z^2‚πy^2=(x+1)^2-x^2c(1)‚Ζ‚¨‚­B(y,x‚Ν—L—”)
(1)‚π(y-1)(y+1)=k2x/kc(2)‚Ζ‚¨‚­B
(2)‚Νk=1‚Μ‚Ζ‚«A(y-1)=2A(y+1)=x‚Ζ‚Θ‚ιB
(2)‚ͺk=1‚Μ‚Ζ‚«A¬—§‚Β‚Θ‚η‚΁Ak=1ˆΘŠO‚Ε‚ΰ¬—§‚B
ˆn=2‚Μ‚Ζ‚«AX^n+Y^n=Z^n‚ΝŽ©‘R”‰π‚𖳐”‚ΙŽ‚B
388: ‚P‚R‚Ql–Ϊ‚Μ‘f”‚³‚ρ [] 07/17(–Ψ)15:29:19.15 ID:88t231TB(14/15)
f*g(t)= η_(-‡)^‡?f(ƒΡ)g(t-ƒΡ) dƒΡ
F[f*g(t)]=η_(-‡)^‡?(η_(-‡)^‡?f(ƒΡ)g(t-ƒΡ) dƒΡ) e^(-jƒΦt) dt
=η_(-‡)^‡?(η_(-‡)^‡?f(ƒΡ)g(t-ƒΡ) e^(-jƒΦt) dt) dƒΡ
=η_(-‡)^‡?f(ƒΡ)(η_(-‡)^‡?g(t-ƒΡ) e^(-jƒΦt) dt) dƒΡ??¦
664: ‚P‚R‚Ql–Ϊ‚Μ‘f”‚³‚ρ [] 08/19(‰Ξ)19:51:43.15 ID:UNSSr5hH(10/12)
y''+ 2y' + 5y = 10cos(t)

y''+ 2y' + 5y = 0
y0 = e^(-t)( C1cos(2t) + C2sin(2t) )

e^iat/ƒΣ(D) = e^iat/ƒΣ(ia) cc (2)
@(1)‚Μ“ΑŽκ‰π‚π v ‚Ζ‚·‚ι‚Ζ
(D^2+2D+5)v = 10e^it
@(2)‚πŽg‚Α‚Δ
10e^it/(D^2+2D+5)
2cos(t) - icos(t) - 2isin(t) + sin(t)
v = 2cos(t) + sin(t)
y = e^(-t)( C1cos(2t) + C2sin(2t) ) + 2cos(t) + sin(t)
797: ‚P‚R‚Ql–Ϊ‚Μ‘f”‚³‚ρ [] 09/07(“ϊ)11:02:46.15 ID:g2aKRGvd(3/4)
η[0¨ƒΞ/2]( tan(x) )^(1/n) dx@@(n†2)
η_0^(ƒΞ/2)?(tan(x))^(1/n) dx ‚π‹‚ί‚ιB
t=?sin?^2 x=(sin(x))^2
?sin?^2 x=1-?cos?^2 x ?cos?^2 x=1-t
dt=2sin(x)cos(x)dx=2γt γ(1-t) dx
dx=dt/(2γt γ(1-t))=(t^(-1/2) (1-t)^(1/2))/2 dt
(sin(x))^(1/n)=(γt)^(1/n)=t^(1/2n) (cos(x))^(1/n)=(γ(1-t))^(1/n)=(1-t)^(1/2n)
η_0^(ƒΞ/2)?(tan(x))^(1/n) dx=η_0^(ƒΞ/2)?( (sin(x))^(1/n))/( (cos(x))^(1/n) ) dx=η_0^(ƒΞ/2)?( t^(1/2n))/(1-t)^(1/2n) (t^(-1/2) (1-t)^(1/2))/2 dt

=1/2 η_0^(ƒΞ/2)???t^(1/2n) (1-t)^(-1/2n) t?^(-1/2) (1-t)^(-1/2) ? dt
=1/2 η_0^(ƒΞ/2)??t^(1/2n-1/2) (1-t)^(-1/2n-1/2) ? dt
=1/2 η_0^(ƒΞ/2)??t^(1/2+1/2n-1) (1-t)^(1/2-1/2n-1) ? dt
=1/2 η_0^(ƒΞ/2)??t^(1/2+1/2n-1) (1-t)^(1/2-1/2n-1) ? dt
(1/2) B(1/2+1/(2n), 1/2-1/(2n))
= (1/2) ƒ‘( 1/2+1/(2n) ) ƒ‘( 1/2-1/(2n) ) / ƒ‘( 1/2+1/(2n) + 1/2-1/(2n) )
= (1/2) ƒ‘(z) ƒ‘(1-z) / ƒ‘(1)
= (1/2) ( ƒΞ/sin(ƒΞz) ) / 0I
= ƒΞ/( 2 sin(ƒΞz) )
= ƒΞ/( 2 sin(ƒΞ/2+ƒΞ/(2n)) )
= ƒΞ/( 2 cos(ƒΞ/(2n)) ).
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