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

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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｡

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673: 08/20(水)10:07 ID:kS5YreVJ(1/9) AAS
1/m+1/n=3/77 (m>n)を満たす自然数の組(m,n)
1/m+1/n=(m+n)/mn=3/77 77(m+n)=3mn
3mn-77m-77n=0
9mn-231m-231n=0
(3m-77)(3n-77)-77^2=0
(3m-77)(3n-77)=77^2=7^2･11^2
(7^2･11^2,1), (7･11^2,7),(7^2･11,11),(11^2･7,7),(11^2,7^2 )
?(7^2･11^2,1)のとき
3m-77=7^2･11^2 3m=7^2･11^2+77=6006 ∴m=2002
3n-77=1 3n=78 ∴n=26
?(7･11^2,7)のとき
3m-77=7･11^2 3m=7･11^2+77=924 ∴m=308
3n-77=7 3n=7+77=84 ∴n=28
?(7^2･11,11)のとき
3m-77=11･7^2 3m=11･7^2+77=616
?(11^2,7^2 )のとき
3m-77=11^2 3m=11^2+77=198 ∴m=66
3n-77=7^2 3n=7^2+77=84 ∴n=42
(m,n)=(2002,26), (308,28), (66,42)

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674: 08/20(水)10:09 ID:kS5YreVJ(2/9) AAS
e^(-1.822r)( 0.223 + 0.223e^r + 0.223e^(2r) + 0.331e^(3r) ) = 1
0.223e^(-1.822r) + 0.223e^(-0.822r) + 0.223e^(0.178r) + 0.331e^(1.178r) = 1
f(r) = 0.223e^(-1.822r) + 0.223e^(-0.822r) + 0.223e^(0.178r) + 0.331e^(1.178r)

f'(r) = -0.406306e^(-1.822r) - 0.183306e^(-0.822r) + 0.039694e^(0.178r) + 0.389918e^(1.178r)
f'(0) = -0.16
f(r) は下に凸で f(0) = 1,f(1) ≒ 1.5 なので 0 < r < 1 の範囲に r = 0 以外の解がある。
a[0] = 1
a[n]-{0.223e^(-1.822a[n])+0.223e^(-0.822a[n])+0.223e^(0.178a[n])+0.331e^(1.178a[n])-1}
a[n+1] = ───────────────────────────────────────────
{-0.406306e^(-1.822a[n])-0.183306e^(-0.822a[n])+0.039694e^(0.178a[n])+0.389918e^(1.178a[n])}
a[1] = 0.5926787635…
a[2] = 0.3745976647…
a[3] = 0.2787742875…
a[4] = 0.2500774128…
a[5] = 0.2469209086…
a[6] = 0.2468816358…
a[7] = 0.2468816298…
a[8] = 0.2468816298…

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675: 08/20(水)10:10 ID:kS5YreVJ(3/9) AAS
{! 123456789 <─-i
101110011 1*2^8+1*2^7+･････+1*2^0 i
x := 2*x+v 2*0+1 1
2*(2*0+1)+0 2
2*(2*(2*0+1)+0)+1 3
2*(2*(2*(2*0+1)+0)+1)+1 4
2*(2*(2*(2*(2*0+1)+0)+1)+1)+1 5
2*(2*(2*(2*(2*(2*0+1)+0)+1)+1)+1)+0 6
2*(2*(2*(2*(2*(2*(2*0+1)+0)+1)+1)+1)+0)+0 7
2*(2*(2*(2*(2*(2*(2*(2*0+1)+0)+1)+1)+1)+0)+0)+1 8
2*(2*(2*(2*(2*(2*(2*(2*(2*0+1)+0)+1)+1)+1)+0)+0)+1)+1 9 !}
function BinToDec(const S: string):string;
var
i,x,v,n: Integer;
begin
Result := '';
x := 0;
n := Length(S);
for i := 1 to n do
begin
v := Ord(S[i]) - Ord('0');
x := 2*x+v;
end;
Result := IntToStr(x);
end;

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676: 08/20(水)10:17 ID:kS5YreVJ(4/9) AAS
∫[a→b]dx/√((x-a)(x-b))
=∫[a→b]dx/√{-x^2+(a+b)x-ab}
=∫[a→b]dx/√{(1/4)(a+b)^2-ab-{x-(a+b)/2}^2}
=∫[a→b]dx/√{(1/4)(a-b)^2-{x-(a+b)/2}^2}
= {2/(b-a)}∫[a→b]dx/√{1-{2{x-(a+b)/2}/(b-a)}^2}
= [arcsin{2{x-(a+b)/2}/(b-a)}][a→b]
=π

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677: 08/20(水)10:23 ID:kS5YreVJ(5/9) AAS
f(z)=1/(1-z) z=i で展開
?@) |z-i|<√2
(1-i)(1-(z-i)/(1-i))=1-i+(1-i) (z-i)/(1-i)
1/(1-z)=1/(1-i-(z-i) )=1/(1-i)?1/(1-(z-i)/(1-i))
=1/(1-i) (1+((z-i)/(1-i))+((z-i)/(1-i))^2+((z-i)/(1-i))^3+?)
=(z-i)^0/(1-i)+(z-i)^1/(1-i)^2 +(z-i)^2/(1-i)^3 +?
=((1+i) (z-i)^0)/2+((1+i)^2 (z-i)^1)/2^2 +((1+i)^3 (z-i)^2)/2^3 +?
=?[n=0→∞]((1+i)/2)^(n+1) (z-i)^n
※(1 )/(1-i)^2 =(1/(1-i))(1/(1-i))=(1+i)/((1-i)(1+i))((1+i)/(1-i)(1+i)) =(1+i)^2/2^2

?A) |z-i|>√2の場合
|z-i|/√2=|(z-i)/(1-i)|>1
すなわち、0<|(1-i)/(z-i)|<1となるから((1-i)/(z-i))^n の級数展開を考える。
1/(1-z)=1/(1-i-(z-i) )=-1/(z-i)?1/(1-(1-i)/(z-i))
=-1/(z-i) (1+((1-i)/(z-i))+((1-i)/(z-i))^2+((1-i)/(z-i))^3+?)
=-(1/(z-i)+(1-i)/(z-i)^2 +(1-i)^2/(z-i)^3 +?)
=-(1/(z-i)+2/(1+i)(z-i)^2 +2^2/?(1+i)^2 (z-i)?^3 +?)
=-((2^0 (z-i)^(-1))/(1+i)^0 +(2^1 (z-i)^(-2))/(1+i)^1 +(2^2 (z-i)^(-3))/(1+i)^2 +?)
=-(?(1+i)^0 (z-i)?^(-1)/2^0 +?(1+i)^(-1) (z-i)?^(-2)/2^(-1) +((1+i)^(-2) (z-i)^(-3))/2^(-2) +?)
=-?[n=1→∞]((1+i)/2)^(1-n) (z-i)^(-n)
※(1-i)^2=(1-i)(1-i)=(1-i)(1+i)/(1+i)?(1-i)(1+i)/(1+i)=2^2/(1+i)^2
(1-i)^n=2^n/(1+i)^n

()^?

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681: 08/20(水)18:15 ID:kS5YreVJ(6/9) AAS
y''(t) - 3y'(t) + 2y(t) = e^(-t) ･･･････?（初期条件）y(0) = 1/6, y'(0) = 5/6

L[y''(t)] = s^2Y(s) - sy(0) - y'(0) = s^2Y(s) - s/6 - 5/6
L[3y'(t)] = 3( sY(s) - y(0) ) = 3sY(s) - 1/2
L[2y(t)] = 2Y(s)
L[e^(-t)] = 1/(s + 1)

s^2Y(s) - s/6 - 5/6 - (3sY(s) -1/2) + 2Y(s) = 1/(s+1)
Y(s)(s^2 - 3s + 2) - s/6 -1/3 = 1/(s+1)
s 1 1
Y(s)(s-1)(s-2) = ─ + ─ + ──
6 3 s+1

s(s+1) + 2(s+1) + 6 s^2 + 3s + 8
= ────────── = ───────
6(s+1) 6(s+1)

s^2 + 3s + 8 A B C
Y(s) = ──────── = ── + ── + ──
6(s+1)(s-1)(s-2) s+1 s-1 s-2

s^2 + 3s + 8 = 6( A(s-1)(s-2) + B(s+1)(s-2) + C(s+1)(s-1) )
s = -1 のとき 1 - 3 + 8 = 6A(-2)(-3) 36A = 6 A = 1/6
s = 1 のとき 1 + 3 + 8 = 6B(2)(-1) -12B = 12 B = -1
s = 2 のとき 4 + 6 + 8 = 6C(3)(1) 18C = 18 C = 1

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682: 08/20(水)18:16 ID:kS5YreVJ(7/9) AAS
M(θ)=E[e^θX ]=∫_(-∞)^∞??e^θx f(x)dx?
M(θ)=E[e^θX ]=1/(√2π σ) ∫_(-∞)^∞??e^θx e^(-(x-μ)^2/(2σ^2 )) ? dx=1/(√2π σ) ∫_(-∞)^∞?e^(θx-(x-μ)^2/(2σ^2 )) dx
θx-(x-μ)^2/(2σ^2 )=1/(2σ^2 ) (2σ^2 θx-(x-μ)^2 )=-1/(2σ^2 ) (? (x-μ)?^2-2σ^2 θx )
=-1/(2σ^2 ) (? x?^2+μ^2-2μx-2σ^2 θx )
=-1/(2σ^2 ) (? x?^2-2(μ+σ^2 θ)x+μ^2 )
=-1/(2σ^2 ) ((x-(μ+σ^2 θ))^2-(μ+σ^2 θ)^2+μ^2 )
=-1/(2σ^2 ) ((x-(μ+σ^2 θ))^2-(μ^2+2μσ^2 θ+σ^4 θ^2 )+μ^2 )
=-1/(2σ^2 ) ((x-(μ+σ^2 θ))^2-(2μσ^2 θ+σ^4 θ^2 ) )
=-(x-(μ+σ^2 θ))^2/(2σ^2 )+μθ+(σ^2 θ^2)/2
M(θ)=1/(√2π σ) ∫_(-∞)^∞?e^(θx-(x-μ)^2/(2σ^2 )) dx
=1/(√2π σ) ∫_(-∞)^∞?e^((-(x-(μ+σ^2 θ))^2/(2σ^2 )+μθ+(σ^2 θ^2)/2) ) dx
=1/(√2π σ) e^(μθ+(σ^2 θ^2)/2) ∫_(-∞)^∞?e^((-(x-(μ+σ^2 θ))^2/(2σ^2 )) ) dx
t=(x-(μ+σ^2 θ))/(√2 σ) x=√2 σt+μ+σ^2 θ dx=√2 σdt
(x-(μ+σ^2 θ))^2/(2σ^2 )=((x-(μ+σ^2 θ))/(√2 σ))^2=t^2
-∞<x?∞ ⇒-∞<t?∞

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683: 08/20(水)18:25 ID:kS5YreVJ(8/9) AAS
y''(t) - 3y'(t) + 2y(t) = e^(-t) ･･･････?（初期条件）y(0) = 1/6, y'(0) = 5/6
L[y''(t)] = s^2Y(s) - sy(0) - y'(0) = s^2Y(s) - s/6 - 5/6
L[3y'(t)] = 3( sY(s) - y(0) ) = 3sY(s) - 1/2
L[2y(t)] = 2Y(s)
L[e^(-t)] = 1/(s + 1)

s^2Y(s) - s/6 - 5/6 - (3sY(s) -1/2) + 2Y(s) = 1/(s+1)
Y(s)(s^2 - 3s + 2) - s/6 -1/3 = 1/(s+1)
Y(s)(s-1)(s-2) = s/6+1/3+1/(s+1) = (s(s+1)+2(s+1)+6)/6(s+1) = (s^2 + 3s + 8)/6(s+1)
Y(s) = (s^2 + 3s + 8)/6(s+1)(s-1)(s-2) = A/(s+1) + B/(s-1) + C/(s-2)
s^2 + 3s + 8 = 6( A(s-1)(s-2) + B(s+1)(s-2) + C(s+1)(s-1) )
s = -1 のとき 1 - 3 + 8 = 6A(-2)(-3) 36A = 6 A = 1/6
s = 1 のとき 1 + 3 + 8 = 6B(2)(-1) -12B = 12 B = -1
s = 2 のとき 4 + 6 + 8 = 6C(3)(1) 18C = 18 C = 1

Y(s) = 1/6(s+1) - 1/(s-1) + 1/(s-2)
y(t) = -e^t + e^(2t) + (1/6)e^(-t)

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684: 08/20(水)18:25 ID:kS5YreVJ(9/9) AAS
y''(t) - 3y'(t) + 2y(t) = e^(-t) ･･･････?（初期条件）y(0) = 1/6, y'(0) = 5/6
k^2 -3k + 2 = (k-1)(k-2) = 0 k = 1, 2
y''(t) - 3'y(t) + 2y(t) = 0
y0 = C1e^t + C2e^(2t)
v(t) = 1/(D-1)(D-2)*e^(-t)
= 1/(D-2)*e^(-t) - 1/(D-1)*e^(-t)
= (-1/3)e^(-t) + (1/2)e^(-t) = (1/6)e^(-t)
y(t) = C1e^t + C2e^(2t) + (1/6)e^(-t)
y(0) = C1 + C2 + 1/6 = 1/6
C1 + C2 = 0 …… ?
y'(t) = C1e^t + C2*2e^(2t) - (1/6)e^(-t)
y'(0) = C1 + C2*2 - 1/6 = 5/6
C1+ 2C2 = 1……?
　??より
C1 = -1, C2= 1
y(t) = -e^t +e^(2t) + (1/6)e^(-t)

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上下前次1-新書関写板覧索設栞歴

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