フェルマーの最終定理の証明 (577ﾚｽ)
上下前次1-新
抽出解除 必死ﾁｪｯｶｰ(本家) (べ) 自ID ﾚｽ栞 あぼーん

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546: 08/01(金)17:03 ID:2hip4JpQ(1/8) AAS
∫_0^∞?(sin(x))/x dx
∂/∂s (e^(-sx) (sin(x))/x)=-xe^(-sx) (sin(x))/x=-e^(-sx) sin(x)

F(s)=∫_0^∞??e^(-sx) (sin(x))/x? dx (s?0)

dF(s)/ds=d/ds ∫_0^∞??e^(-sx) sin?(x)/x? dx
=∫_0^∞??∂/ds e^(-sx) sin?(x)/x? dx
=∫_0^∞??-xe^(-sx) sin?(x)/x? dx=-∫_0^∞??e^(-sx) sin?(x) ? dx
=-∫_0^∞??-1/s (e^(-sx) )^' sin(x)? dx
省7

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547: 08/01(金)17:04 ID:2hip4JpQ(2/8) AAS
Q? √(6&2^(2x^2+15)/x^(4x+30) ) (x=√2n, n?5) ･････(#12)
x=e^logx 2=e^log2
2^(2x^2+15) = ?(e^log2)?^(2x^2+15)=e^((2x^2+15)log2)
x^(4x+30)=?(e^logx)?^(4x+30)=e^((4x+30)logx)
　ここで
(2x^2+15)log2 >(4x+30)logx (x?12) ･････(#14)
2^(2x^2+15)/x^(4x+30) =e^((2x^2+15)log2)/e^((4x+30)logx) =e^((2x^2+15)log2-(4x+30)logx)>e^0
省11

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548: 08/01(金)17:05 ID:2hip4JpQ(3/8) AAS
x≡1 (mod 3)
x≡2 (mod 5)
x≡3 (mod 11)

3*5*11 = 165
x≡1 (mod 3) ……?
x≡2 (mod 5) ……?
x≡3 (mod 11) ……?
省9

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549: 08/01(金)17:05 ID:2hip4JpQ(4/8) AAS
6x≡3 (mod 15)
2x≡1 (mod 5)
6x≡3 (mod 5)
6≡1, 6x≡x (mod 5)
x≡3≡8≡13 (mod 5)
∴x≡3, 8, 13 (mod 15)

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550: 08/01(金)17:07 ID:2hip4JpQ(5/8) AAS
74x≡117 (mod 403)
74x≡1 (mod 403)

403=5*74+33
74=2*33+8
33=4*8+1

1=33-4*8=33-4(74-2*33)
=33-4*74+8*33=9*33-4*74
省5

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554: 08/01(金)21:29 ID:2hip4JpQ(6/8) AAS
∫_α^β??(x-α)^m (β-x)^n ? dx=m!n!/(m+n+1)! (β-α)^(m+n+1)
t=(β-α)x+α dt=(β-α)dx dx=dt/(β-α)
x:0→1 t:α→β
x=(t-α)/(β-α) 1-x=(β-α-(t-α))/(β-α)=(β-t)/(β-α)
∫_0^1?x^m (1-x)^n dx
=∫_α^β??((t-α)/(β-α))^m ((β-t)/(β-α))^n ? dt/(β-α)=∫_α^β?((t-α)^m (β-t)^m)/(β-α)^(m+n+1) dt
=1/(β-α)^(m+n+1) ∫_α^β??(t-α)^m (β-t)^m ? dt=m!n!/(m+n+1)!
省7

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555: 08/01(金)21:30 ID:2hip4JpQ(7/8) AAS
∂u/∂t=(∂u^2)/(∂x^2 ) (0<x<1, t>0)
u_x (0,t)=u_x (1,t)=0 境界条件（断熱条件）
u(x,0)=δ(x-1/2) 初期条件

u(x,t)=X(x)T(t)
∂u/∂t=XT^'
∂u/∂x=TX^' (∂u^2)/(∂x^2 )=∂/∂x TX^'=TX^''
XT^'= TX^'' T^'/T=X^''/X
省4

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556: 08/01(金)21:31 ID:2hip4JpQ(8/8) AAS
x ?+ax ?+bx=0 ???
λ^2+aλ+b=0
λ=α, β ⇒ x= C_1 e^αt+C_2 e^βt
λ=α (重解) ⇒ x= C_1 e^αt+C_2 te^βt
λ=α±βi ⇒ x= e^αt (C_1 cos?(βt)+C_2 cos?(βt))
λ^2-μ=0
0^2-4(-μ)=4μ
省18

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