フェルマーの最終定理の証明 (692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
ﾘﾛｰﾄﾞ規制です｡10分ほどで解除するので､他のﾌﾞﾗｳｻﾞへ避難してください｡
393: 07/18(金)07:29 ID:QNG/Z1cz(1/12) AA×
394: 07/18(金)07:30 ID:QNG/Z1cz(2/12) AA×
395: 07/18(金)07:31 ID:QNG/Z1cz(3/12) AA×
396: 07/18(金)07:32 ID:QNG/Z1cz(4/12) AAS
e^z=e^(x+iy)=e^x e^iy=e^x (cos(y)+isin(y))
=e^x cos(y)+ie^x sin(y)
u(x,y)=e^x cos(y)
u_x=e^x cos(y), u_y=-e^x sin(y)
v(x,y)=e^x sin(y)
v_x=e^x sin(y), v_y=e^x cos(y)
したがってC-R の方程式
u_x=v_y, v_x=-u_y
が成り立つ。
(e^z )^'=e^z
より
(e^iz )^'=e^iz (iz)^'=ie^iz
(e^(-iz) )^'=e^(-iz) (-iz)^'=-ie^(-iz)
(sin(z))^'=((e^iz-e^(-iz))/2i)^'=(ie^iz+?ie?^(-iz))/2i=(e^iz+e^(-iz))/2=cos(z)
(cos(z))^'=((e^iz+e^(-iz))/2)^'=(ie^iz-?ie?^(-iz))/2=(e^iz-e^(-iz))/2i=sin(z)
(tan(z))^'=(sin(z)/cos(z) )^'=(cos(z)cos(z)+sin(z)sin(z))/(?cos?^2 (z) )=1/(?cos?^2 (z) )
397: 07/18(金)07:34 ID:QNG/Z1cz(5/12) AA×
398: 07/18(金)07:35 ID:QNG/Z1cz(6/12) AA×
406: 07/18(金)16:31 ID:QNG/Z1cz(7/12) AAS
F(ω)=∫_(-∞)^∞??f(t) e^(-jωt) ? dt フーリエ変換 ???
f(t)= F^(-1) [F(ω)]=1/2π ∫_(-∞)^∞??F(ω) e^jωt ? dω 逆フーリエ変換???
f(t)=1, f(t)=t, f(t)= sin(ωt)
g(t)={■(0&t<0@f(t) e^(-σt)&t?0)┤
G(ω)=∫_(-∞)^∞??g(t) e^(-jωt) ? dt=∫_0^∞??g(t) e^(-jωt) ? dt=∫_0^∞??f(t) e^(-σt) e^(-jωt) ? dt
=∫_0^∞??f(t) e^(-(σ+jω)t) ? dt
s=σ+jω
F(s)=∫_0^∞??f(t) e^(-st) ? dt ラプラス変換 ???
407: 07/18(金)16:32 ID:QNG/Z1cz(8/12) AAS
s=σ+jω ds=jdω ω: -∞ → ∞
s? σ-j∞→σ+j∞
g(t)=1/2π ∫_(-∞)^∞??F(s) e^jωt ? dω=1/2πj ∫_(σ-j∞)^(σ+j∞)??F(s) e^jωt ? ds
f(t) e^(-σt)=f(t)/e^σt =1/2πj ∫_(σ-j∞)^(σ+j∞)??F(s) e^jωt ? ds
f(t)=1/2πj ∫_(σ-j∞)^(σ+j∞)??F(s) e^σt e^jωt ? ds=1/2πj ∫_(σ-j∞)^(σ+j∞)??F(s) e^(σ+jω)t ? ds
∴f(t)=1/2πj ∫_(σ-j∞)^(σ+j∞)??F(s) e^st ? ds ラプラス逆変換 ???
408: 07/18(金)16:33 ID:QNG/Z1cz(9/12) AAS
∫_0^1?x^m (1-x)^n dx=1/(m+n+1)!
I(m,n)=∫_0^1?x^m (1-x)^n dx
I(m,n)=∫_0^1?x^m (1-x)^n dx=∫_0^1??(x^(m+1)/(m+1))^' (1-x)^n ? dx
=[?( @x^(m+1)/(m+1)@ )(1-x)^n ]_0^1-∫_0^1?x^(m+1)/(m+1) (-n) (1-x)^(n-1) dx
=n/(m+1) ∫_0^1?x^(m+1) (1-x)^(n-1) dx=n/(m+1) I(m+1,n-1)
I(m,n)=n/(m+1) I(m+1,n-1)
=n/(m+1)?(n-1)/(m+2) I(m+2,n-2)
=n/(m+1)?(n-1)/(m+2)?(n-2)/(m+3) I(m+3,n-3)
=n/(m+1)?(n-1)/(m+2)?(n-2)/(m+3)? ??? ?1/(m+n) I(m+n,0)
n/(m+1)?(n-1)/(m+2)?(n-2)/(m+3)? ??? ?1/(m+n)
=(1?2?(m-1)m?n(n-1)(n-2)? ??? ?1)/(1?2?(m-1)m(m+1)(m+2)? ??? ?(m+n) )=m!n!/(m+n)!
I(m,n)=n/(m+1) I(m+1,n-1)
=m!n!/(m+n)! I(m+n,0) =m!n!/(m+n)! ∫_0^1?x^(m+n) (1-x)^0 dx
=m!n!/(m+n)! ∫_0^1?x^(m+n) dx=m!n!/(m+n)! [?( @x^(m+n+1)/(m+n+1)@ )]_0^1
=m!n!/(m+n+1)!
∴∫_0^1?x^m (1-x)^n dx=1/(m+n+1)!
409: 07/18(金)16:34 ID:QNG/Z1cz(10/12) 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)!
∴∫_α^β??(x-α)^m (β-x)^n ? dx=m!n!/(m+n+1)! (β-α)^(m+n+1)
m=1,n=1⇒∫_α^β?(x-α)(x-β) dx=-∫_α^β?(x-α)(β-x) dx
=-1/6 (β-α)^3
m=2,n=1⇒∫_α^β?(x-α)(x-β) dx=-∫_α^β??(x-α)^2 (β-x) ? dx
=-1/12 (β-α)^4
m=2,n=2⇒∫_α^β??(x-α)^2 (x-β)^2 ? dx=∫_α^β??(x-α)^2 (β-x)^2 ? dx
=(2?2)/(5?4?3?2?1) (β-α)^5=1/30 (β-α)^5
411: 07/18(金)21:14 ID:QNG/Z1cz(11/12) AAS
x^'' (t)+3x^' (t)+2x(t)=e^2t , x(0)=0, x^' (0)=1
L[x^'' (t)]=s^2 X(s)-sx(0)-x^' (0)=s^2 X(s)-1
3L[x^' (t)]=3(sX(s)-x(0))=3sX(s)
2L[x(t)]=2X(s)
L[x^'' (t)]+2L[x^' (t)]+ 5L[x(t)]
=s^2 X(s)-1+3sX(s)+2X(s)
=X(s)(s^2+3s+2)-1=X(s)(s+1)(s+2)-1
L[e^2t ]=1/(s-2) (L[e^at ]=1/(s-a))
X(s)(s+1)(s+2)-1=1/(s-2)
X(s)(s+1)(s+2)=1/(s-2)+1=(1+s-2)/(s-2)=(s-1)/(s-2)
X(s)=(s-1)/(s+1)(s+2)(s-2) =A/(s+1)+B/(s+2)+C/(s-2)
s-1=A(s+2)(s-2)+B(s+1)(s-2)+C(s+1)(s+2)
s=-1⇒-2=A(-3) A=2?3
s=-2⇒-3=B(-1)(-4) 4B=-3 B=-3?4
s=2⇒1=C(3)(4) 12C=1 C=1?12
X(s)=2/3 1/(s+1)-3/4 1/(s+2)+1/12 1/(s-2)
L^(-1) [2/3 1/(s-(-1))-3/4 1/(s-(-2))+1/12 1/(s-2)]=2/3 e^(-t)-3/4 e^(-2t)+1/12 e^2t
412: 07/18(金)21:15 ID:QNG/Z1cz(12/12) AAS
L[y^'' (t)]=s^2 Y(s)-sy(0)-y^' (0) =s^2 Y(s)-2s-4
L[?4y?^' (t)]=4(sY(s)-y(0))=4sY(s)-8
L[4y(t)]=4Y(s)
L[y^'' (t)]-L[?4y?^' (t)]+ L[4y(t)]
=s^2 Y(s)-2s-4-4sY(s)+8+4Y(s)
=Y(s)(s^2-4s+4)-2s+4
L[6te^2t ]=6L[t^1 e^2t ]=6 1!/(s-2)^2 =6/(s-2)^2 ( L[t^n e^at ]=n!/(s-a)^(n+1) )
Y(s)(s^2-4s+4)-2s+4=6/(s-2)^2
Y(s) (s-2)^2-2s+4=6/(s-2)^2
Y(s) (s-2)^2=6/(s-2)^2 +2(s-2)
Y(s)=6/(s-2)^4 +2/(s-2)
Y(s)= F(s-2)とおくと
F(s-2)=6/(s-2)^4 +2/(s-2)
∴F(s)=6/s^4 +2/s=3!/s^(3+1) +2/s
y(t)=L^(-1)[F(s-2)]=e^2t L^(-1) [F(s)] ( L^(-1) [F(s-a)]=e^at L^(-1) [F(s)])
=e^2t L^(-1) [3!/s^(3+1) +2/s] (L[t^n ]=n!/s^(n+1) )
=e^2t (t^3+2)
上下前次1-新書関写板覧索設栞歴
ｽﾚ情報 赤ﾚｽ抽出 画像ﾚｽ抽出 歴の未読ｽﾚ
ぬこの手 ぬこTOP 0.071s