tF}[ΜΕIθΜΨΎ (843Ϊ½)
γΊO1-V
oπ Ϊ½x
41: PRQlΪΜf³ρ [] 04/28()21:42:10.15 ID:0uxBMTmL(1)
κκΖ©’Δ'ΪΪ'ΖΗήB
137: ^μ [] 05/25(ϊ)12:07:12.15 ID:4EyabO0p(3/3)
n=2ΜΖ«AX^n+Y^n=Z^nΝ©Rππ³ΙΒB
X^2+Y^2=Z^2πy^2=(x+1)^2-x^2c(1)Ζ¨B(y,xΝL)
(1)π(y-1)(y+1)=2xc(2)Ζ¨B
(2)Ν(y-1)=2ΜΖ«A(y+1)=xͺ¬§ΒB
(2)π(y-1)(y+1)=k2x/kc(3)Ζ¨B
(3)Νk/k=1ΘΜΕA(y-1)=k2ΜΖ«A(y+1)=x/kͺ¬§ΒB
n=2ΜΖ«AX^n+Y^n=Z^nΝ©Rππ³ΙΒB
369: ^μ [] 07/17(Ψ)12:00:52.15 ID:4J9At0pY(3/17)
n=2ΜΖ«AX^n+Y^n=Z^nΝ©Rππ³ΙΒB
X^2+Y^2=Z^2πy^2=(x+1)^2-x^2c(1)Ζ¨B(y,xΝL)
(1)π(y-1)(y+1)=k2x/kc(2)Ζ¨B
(2)Νk=1ΜΖ«A(y-1)=2A(y+1)=xΖΘιB
(2)ͺk=1ΜΖ«A¬§ΒΘηΞAk=1ΘOΕଧΒB
n=2ΜΖ«AX^n+Y^n=Z^nΝ©Rππ³ΙΒB
388: PRQlΪΜf³ρ [] 07/17(Ψ)15:29:19.15 ID:88t231TB(14/15)
f*g(t)= η_(-)^?f(Ρ)g(t-Ρ) dΡ
F[f*g(t)]=η_(-)^?(η_(-)^?f(Ρ)g(t-Ρ) dΡ) e^(-jΦt) dt
=η_(-)^?(η_(-)^?f(Ρ)g(t-Ρ) e^(-jΦt) dt) dΡ
=η_(-)^?f(Ρ)(η_(-)^?g(t-Ρ) e^(-jΦt) dt) dΡ??¦
664: PRQlΪΜf³ρ [] 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/Σ(D) = e^iat/Σ(ia) cc (2)
@(1)ΜΑκππ v Ζ·ιΖ
(D^2+2D+5)v = 10e^it
@(2)πgΑΔ
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: PRQlΪΜf³ρ [] 09/07(ϊ)11:02:46.15 ID:g2aKRGvd(3/4)
η[0¨Ξ/2]( tan(x) )^(1/n) dx@@(n2)
η_0^(Ξ/2)?(tan(x))^(1/n) dx πίιB
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=2γt γ(1-t) dx
dx=dt/(2γt 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^(Ξ/2)?(tan(x))^(1/n) dx=η_0^(Ξ/2)?( (sin(x))^(1/n))/( (cos(x))^(1/n) ) dx=η_0^(Ξ/2)?( t^(1/2n))/(1-t)^(1/2n) (t^(-1/2) (1-t)^(1/2))/2 dt
=1/2 η_0^(Ξ/2)???t^(1/2n) (1-t)^(-1/2n) t?^(-1/2) (1-t)^(-1/2) ? dt
=1/2 η_0^(Ξ/2)??t^(1/2n-1/2) (1-t)^(-1/2n-1/2) ? dt
=1/2 η_0^(Ξ/2)??t^(1/2+1/2n-1) (1-t)^(1/2-1/2n-1) ? dt
=1/2 η_0^(Ξ/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) ‘( 1/2+1/(2n) ) ‘( 1/2-1/(2n) ) / ‘( 1/2+1/(2n) + 1/2-1/(2n) )
= (1/2) ‘(z) ‘(1-z) / ‘(1)
= (1/2) ( Ξ/sin(Ξz) ) / 0I
= Ξ/( 2 sin(Ξz) )
= Ξ/( 2 sin(Ξ/2+Ξ/(2n)) )
= Ξ/( 2 cos(Ξ/(2n)) ).
γΊO1-VΦΚΒυέxπ
½Ϊξρ ΤΪ½o ζΪ½o πΜ’Η½Ϊ AA»ΡΘ²Ω
Κ±Μθ Κ±TOP 1.578s*