フェルマーの最終定理の証明 (846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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832: １３２人目の素数さん [] 09/13(土)11:15 ID:tt8WnsBt(1/3)
B(p,q)=∫_0^1?x^(p-1) (1-x)^(q-1) dx
B(p,q)=∫[0→π/2](sinθ)^(2p-1)(cosθ)^(2q-1)dθ
x=sin^2θ?=(sinθ)?^2 dx=2sinθcosθdθ
x:0→1 θ:0→π/2
B(p,q)=∫[0→1]x^(p-1)(1-x)^(q-1) dx
=∫[0→π/2](sin^2θ)^(p-1)(1-sin^2θ)^(q-1) ? 2sinθcosθdθ
=2∫[0→π/2](sinθ)^(2p-2) sinθ(cosθ)^(2q-2) cosθ? dθ
=2∫[0→π/2](sinθ)^(2p-1)(cosθ)^(2q-1)dθ
∫[0→π/2]tanθ)^(1/n)dθ
t=sin^2θ=(sinθ)^2
sin^2θ=1-cos^2θ
cos^2θ=1-t
dt=2sinθcosθdθ=2√t √(1-t) dθ
dθ=dt/(2√t √(1-t))=(t^(-1/2)(1-t)^(1/2))/2 dt
(sinθ)^(1/n)=(√t)^(1/n)=t^(1/2n)
(cosθ)^(1/n)=(√(1-t))^(1/n)=(1-t)^(1/2n)
∫[0→π/2]tanθ)^(1/n)dx
=∫[0→π/2] (sinθ)^(1/n))/( (cosθ)^(1/n) ) dθ
=∫[0→π/2] t^(1/2n))/(1-t)^(1/2n)
(t^(-1/2)(1-t)^(1/2))/2 dt
=(1/2)∫[0→π/2]?t^(1/2n)(1-t)^(-1/2n) t?^(-1/2)(1-t)^(-1/2)dt
=(1/2)∫[0→π/2]t^(1/2n-1/2)(1-t)^(-1/2n-1/2)dt
=(1/2)∫[0→π/2]t^(1/2+1/2n-1)(1-t)^(1/2-1/2n-1)dt
=(1/2)∫[0→π/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)
B(p,q)=∫[0→π/2](sinθ)^(2p-1)(cosθ)^(2q-1)dθ
(1/2)∫[0→π/2]t^(1/2+1/2n-1)(1-t)^(1/2-1/2n-1)dt
=∫[0→π/2](sinθ)^(2p-1)(cosθ)^(2q-1)dθ

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833: １３２人目の素数さん [] 09/13(土)11:16 ID:tt8WnsBt(2/3)
y_s=1/(D+i) (2i/(e^2ix+1)^2 )=e^(-ix) 1/D e^ix 2i/(e^2ix+1)^2 =e^(-ix) ∫(2ie^2ix)/(e^2ix+1)^2 dx
t=e^2ix+1 dt=2ie^2ix dx dx=dt/(2ie^2ix )
∫?(2ie^2ix)/(e^2ix+1)^2 dx?=∫?(2ie^2ix)/t^2 dt/(2ie^2ix )?=∫t^(-2) dt=-1/t=-1/(e^2ix+1)
y_s=e^(-ix) ∫(2ie^2ix)/(e^2ix+1)^2 dx=-e^(-ix)/(e^2ix+1)
=(- e^(-ix) (e^(-ix)+e^ix-e^ix ))/(e^(-ix) (e^2ix+1) ) =(- e^(-ix) (e^(-ix)+e^ix )+1)/(e^ix+e^(-ix) )
=- e^(-ix)+1/(e^ix+e^(-ix) )=- e^(-ix)+1/2cos(x)

y=C_1 cos(x)+C_2 sin(x)- e^(-ix)+1/2cos(x)
=C_1 cos(x)+C_2 sin(x)- cos(x)+isin(x)+1/2cos(x)
=(C_1-1)cos(x)+(C_2+i)sin(x)+1/2cos(x)
=Acos(x)+Bsin(x)+1/2cos(x)
y_s=1/2cos(x)
y=C_2 cos(x)+C_1 sin(x)- 1/2 cos(2x) 1/cos(x)
=C_2 cos(x)+C_1 sin(x)- 1/2 (2?cos?^2 (x)-1) 1/cos(x)
=C_2 cos(x)+C_1 sin(x)- (?cos?^2 (x)-1/2)/cos(x)
=C_2 cos(x)+C_1 sin(x)- cos(x)+1/2 1/cos(x)
=(C_2-1)cos(x)+C_1 sin(x)+1/2cos(x)
=Acos(x)+Bsin(x)+1/2cos(x)

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834: １３２人目の素数さん [] 09/13(土)11:17 ID:tt8WnsBt(3/3)
f^((k) ) (z)=(n!/2πi)?_Cf(ζ)/(ζ-z)^(k+1)dζ
?@)n=1のとき
f(z)=1/( 2πi) ?_Cf(ζ)/((ζ-z) ) dζ
f(z+h)=1/( 2πi) ?_Cf(ζ)/(ζ-(z+Δz) ) dζ
f(z+h)-f(z)=1/( 2πi) ?_Cf(ζ)/(ζ-(z+h) )-f(ζ)/((ζ-z) ) dζ
=1/( 2πi) ?_Cf(ζ)((ζ-z)-(ζ-z-h))/(ζ-z-h)(ζ-z)dζ
=1/( 2πi) ?_Cf(ζ)(ζ-z-ζ+z+h)/(ζ-z-h)(ζ-z)dζ
=1/( 2πi) ?_Cf(ζ)h/(ζ-z-h)(ζ-z)dζ
=h/( 2πi) ?_Cf(ζ)/(ζ-z-h)(ζ-z)dζ
( f(z+h)-f(z))/h=1/( 2πi) ?_Cf(ζ)/(ζ-z-h)(ζ-z)dζ
　h→0
f'(z)= f^((1)) (z)=1/2πi ?_C(f(ζ))/(ζ-z)^2dζ
?A)n=k(k=1,2,3,…)のとき
f^((k)) (z)=k!/2πi ?_C(f(ζ))/(ζ-z)^(k+1)dζ ⇒f^((k+1)) (z)=(k+1)!/( 2πi) ?_Cf(ζ)/(ζ-z)^(k+2)dζ
f^((k)(z+h)- f^((k) ) (z))/h
=k!/( 2πih) ?_Cf(ζ)/(ζ-(z+h))^(k+1) -f(ζ)/(ζ-z)^(k+1)dζ
=k!/( 2πih) ?_C((ζ-z)^(k+1)-(ζ-z-h)^(k+1))/((ζ-z-h)^(k+1) (ζ-z)^(k+1) ) f(ζ)dζ??※
(a+b)^(k+1)
=(_k+1^ )C_0 a^n b^0+(_k+1^ )C_1 a^(k+1-1) b^1+(_k+1^ )C_2 a^(k+1-2) b^2+?+(_k+1^ )C_r a^(k+1-r) b^r+?+b^(k+1)
=a^(k+1)+(k+1) a^k b+(_k+1^ )C_2 a^(k-1) b^2+?+(_k+1^ )C_r a^(k+1-r) b^r+? +b^(k+1)
(ζ-z-h)^(k+1)
=(ζ-z)^(k+1)-(k+1) (ζ-z)^k h + (_k+1^ )C_2 (ζ-z)^(k-1) h^2-?+h^(k+1)
(ζ-z)^(k+1)-(ζ-z-h)^(k+1)
=(k+1) (ζ-z)^k h-(_k+1^ )C_2 (ζ-z)^(k-1) h^2+?-h^(k+1)
( f^((k) ) (z+h)- f^((k) ) (z))/h
=k!/( 2πih) ?_C((k+1) (ζ-z)^k h-(_k+1^ )C_2 (ζ-z)^(k-1) h^2+?-h^(k+1))/((ζ-z-h)^(k+1) (ζ-z)^(k+1) ) f(ζ)dζ
=(k+1)!/( 2πi) ?_Cf(ζ)/((ζ-z-h)^(k+1) (ζ-z) ) dζ-k!/( 2πi) ?_C((_k+1^ )C_2 (ζ-z)^(k-1) h-?+h^k)/((ζ-z-h)^(k+1) (ζ-z)^(k+1) ) f(ζ)dζ
　h→0
f^((k+1)) (z)

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