(cache)組立除法でa^n+b^n=P(a+b,ab)やcos(nx),sin(nx)を解く

組立除法でa^n+b^n=P(a+b,ab)やcos(nx),sin(nx)を解く
3項間漸化式xn+1=pxn−qxn−1x_{n+1} = p x_{n} - q x_{n-1}xn+1​=pxn​−qxn−1​を解くプロセスは漸化式をそれ以上代入できなくなるまで再帰的に代入する操作なのだが、漸化式を多項式xn+1=pxn−qxn−1x^{n+1} = p x^{n} - q x^{n-1}xn+1=pxn−qxn−1に置き換えると、この再帰的な代入の操作は多項式x2−px+q(=0)x^2-px+q(=0)x2−px+q(=0)で割り算して剰余を求める操作に置き換わる。だから組立除法に乗せられる。

p:=a+bp:=a+bp:=a+b,q:=abq:=abq:=ab,xn:=an+bnx_n:=a^n+b^nxn​:=an+bnとする。
このときx0=2x_0 = 2x0​=2,x1=px_1 = px1​=pであり、
x2=px1−qx0=p2−2qx_2 = p x_1 -q x_0=p^2-2qx2​=px1​−qx0​=p2−2qよりa2+b2=(a+b)2−2aba^2+b^2=(a+b)^2-2aba2+b2=(a+b)2−2ab

1000ppp2−q−q−pq1p ∣∣p2−q−pq\begin{array}{r|llll} & 1 & 0 & 0 & 0\\\hline p & & p & p^2\\ -q & & & -q & -pq\\\hline & 1 & p \ ||& p^2-q & -pq \end{array}p−q​11​0pp ∣∣​0p2−qp2−q​0−pq−pq​​
x3=(p2−q)x1−pqx0=(p2−q)p−2pqx_3 = (p^2 - q) x_1 - p q x_0 = (p^2 - q) p - 2 p qx3​=(p2−q)x1​−pqx0​=(p2−q)p−2pq
a3+b3=(a+b)3−3ab(a+b)a^3+b^3=(a+b)^3-3ab(a+b)a3+b3=(a+b)3−3ab(a+b)

10000ppp2p3−pq−q−q−pqq2−p2q1pp2−q ∣∣p3−2pqq2−p2q\begin{array}{r|lllll} & 1 & 0 & 0 & 0 & 0\\\hline p & & p & p^2 &p^3-pq\\ -q & & & -q & -pq & q^2-p^2q\\\hline & 1 & p & p^2-q\ || & p^3-2pq & q^2-p^2q \end{array}p−q​11​0pp​0p2−qp2−q ∣∣​0p3−pq−pqp3−2pq​0q2−p2qq2−p2q​​
x4=(p3−2pq)x1+(q2−p2q)x0=(p3−2pq)p+2(q2−p2q)x_4 = (p^3 - 2 p q) x_1 + (q^2 - p^2 q) x_0 = (p^3 - 2 p q) p + 2 (q^2 - p^2 q)x4​=(p3−2pq)x1​+(q2−p2q)x0​=(p3−2pq)p+2(q2−p2q)
a4+b4=(a+b)4−4ab(a+b)2+2(ab)2a^4+b^4=(a+b)^4-4ab(a+b)^2 + 2(ab)^2a4+b4=(a+b)4−4ab(a+b)2+2(ab)2

100000ppp2p3−pqp4−2p2q−q−q−pqq2−p2q2pq2−p3q1pp2−qp3−2pq ∣∣p4−3p2q+q22pq2−p3q\begin{array}{r|llllll} & 1 & 0 & 0 & 0 & 0 & 0 \\\hline p & & p & p^2 & p^3-pq & p^4-2p^2q \\ -q & & & -q & -pq & q^2-p^2q & 2pq^2-p^3q \\\hline & 1 & p & p^2-q & p^3-2pq\ || & p^4 - 3 p^2 q + q^2 & 2pq^2-p^3q \end{array}p−q​11​0pp​0p2−qp2−q​0p3−pq−pqp3−2pq ∣∣​0p4−2p2qq2−p2qp4−3p2q+q2​02pq2−p3q2pq2−p3q​​
x5=(p4−3p2q+q2)x1+(2pq2−p3q)x0=(p4−3p2q+q2)p+2(2pq2−p3q)x_5 = (p^4 - 3 p^2 q + q^2) x_1 + (2 p q^2 - p^3 q) x_0 = (p^4 - 3 p^2 q + q^2) p + 2 (2 p q^2 - p^3 q)x5​=(p4−3p2q+q2)x1​+(2pq2−p3q)x0​=(p4−3p2q+q2)p+2(2pq2−p3q)
a5+b5=(a+b)5−5ab(a+b)3+5(ab)2(a+b)a^5+b^5=(a+b)^5-5ab(a+b)^3 + 5(ab)^2(a+b)a5+b5=(a+b)5−5ab(a+b)3+5(ab)2(a+b)

問題：なぜ解ける？

cos⁡nx, sin⁡nx\cos nx,\ \sin nxcosnx, sinnxを組立除法でcos⁡x, sin⁡x\cos x,\ \sin xcosx, sinxの式にする
3項間漸化式cos⁡(n+1)x⏟xn+1=2cos⁡xcos⁡nx⏟xn−cos⁡(n−1)x⏟xn−1=x1Pn(cos⁡x)+x0Qn(cos⁡x)\underbrace{\cos(n+1)x}_{x_{n+1}} = 2\cos x\underbrace{\cos nx}_{x_{n}} - \underbrace{\cos(n-1)x}_{x_{n-1}} = x_1 P_n(\cos x)+x_0 Q_n(\cos x)xn+1​cos(n+1)x​​=2cosxxn​cosnx​​−xn−1​cos(n−1)x​​=x1​Pn​(cosx)+x0​Qn​(cosx)を再帰的に解いているだけ

次数n3211002C2C−1−11 ∣∣2C−1\begin{array}{r|llll} 次数n & 3 & 2 & 1 \\ & 1 & 0 & 0 \\\hline 2C & & 2C & \\ -1 & & & -1 \\\hline & 1 \ ||& 2C & -1 \end{array}次数n2C−1​311 ∣∣​202C2C​10−1−1​​
∴cos⁡3x=2cos⁡x⏟2C(2cos⁡2x−1)⏟cos⁡2x−⏟−1cos⁡x\therefore \cos3x = \underbrace{2\cos x}_{2C}\underbrace{(2\cos^2x - 1)}_{\cos2x} \underbrace{-}_{-1} \cos x∴cos3x=2C2cosx​​cos2x(2cos2x−1)​​−1−​​cosx
∴sin⁡3x=2cos⁡x⏟2C(2cos⁡xsin⁡x)⏟sin⁡2x−⏟−1sin⁡x\therefore \sin3x = \underbrace{2\cos x}_{2C}\underbrace{(2\cos x\sin x)}_{\sin2x} \underbrace{-}_{-1} \sin x∴sin3x=2C2cosx​​sin2x(2cosxsinx)​​−1−​​sinx

次数n432110002C2C4C2−1−1−2C12C ∣∣4C2−1−2C\begin{array}{r|llll} 次数 n & 4 & 3 & 2 & 1 \\ & 1 & 0 & 0 & 0\\\hline 2C & & 2C & 4C^2 \\ -1 & & & -1 & -2C \\\hline & 1 & 2C \ ||& 4C^2-1 & - 2C \end{array}次数n2C−1​411​302C2C ∣∣​204C2−14C2−1​10−2C−2C​​
∴cos⁡4x=(4cos⁡2x−1)(2cos⁡2x−1)−2cos⁡xcos⁡x\therefore \cos4x = (4\cos^2x - 1)(2\cos^2x -1) - 2\cos x\cos x∴cos4x=(4cos2x−1)(2cos2x−1)−2cosxcosx
∴sin⁡4x=(4cos⁡2x−1)(2cos⁡xsin⁡x)−2cos⁡xsin⁡x\therefore \sin4x = (4\cos^2x - 1)(2\cos x\sin x) - 2\cos x\sin x∴sin4x=(4cos2x−1)(2cosxsinx)−2cosxsinx

次数n54321100002C2C4C28C3−2C−1−1−2C−4C2+112C4C2−1 ∣∣8C3−4C−4C2+1\begin{array}{r|llll} 次数 n & 5 & 4 & 3 & 2 & 1 \\ & 1 & 0 & 0 & 0 & 0 \\\hline 2 C & & 2 C & 4 C^2 & 8 C^3 - 2 C \\ -1 & & & -1 & -2 C & -4 C^2 + 1 \\\hline & 1 & 2C & 4C^2-1 \ ||& 8C^3 - 4C & -4 C^2 + 1 \end{array}次数n2C−1​511​402C2C​304C2−14C2−1 ∣∣​208C3−2C−2C8C3−4C​10−4C2+1−4C2+1​​
∴cos⁡5x=(8cos⁡3x−4cos⁡x)(2cos⁡2x−1)+(−4cos⁡2x+1)cos⁡x\therefore \cos5x = (8\cos^3x - 4\cos x)(2\cos^2x-1)+( - 4\cos^2x + 1)\cos x∴cos5x=(8cos3x−4cosx)(2cos2x−1)+(−4cos2x+1)cosx
∴sin⁡5x=(8cos⁡3x−4cos⁡x)(2cos⁡xsin⁡x)+(−4cos⁡2x+1)sin⁡x\therefore \sin5x = (8\cos^3x - 4\cos x)(2\cos x\sin x) + (- 4\cos^2x + 1)\sin x∴sin5x=(8cos3x−4cosx)(2cosxsinx)+(−4cos2x+1)sinx

sin⁡1°\sin1\degreesin1°は無理数か？
sin⁡(2n+1)=2cos⁡1sin⁡2n−sin⁡(2n−1)\sin(2n+1) = 2 \cos1 \sin2n - \sin(2n-1)sin(2n+1)=2cos1sin2n−sin(2n−1)
=2cos⁡1[2cos⁡1sin⁡(2n−1)−sin⁡(2n−2)]−sin⁡(2n−1)= 2 \cos1 [2 \cos1 \sin(2n-1) - \sin(2n-2)] - \sin(2n-1)=2cos1[2cos1sin(2n−1)−sin(2n−2)]−sin(2n−1)
=(4cos⁡21−1)sin⁡(2n−1)−2cos⁡1sin⁡(2n−2)= (4 \cos^21 - 1)\sin(2n-1) - 2 \cos1 \sin(2n-2)=(4cos21−1)sin(2n−1)−2cos1sin(2n−2)
=(4cos⁡21−1)sin⁡(2n−1)−(sin⁡(2n−1)+sin⁡(2n−3))= (4 \cos^21 - 1)\sin(2n-1) - (\sin(2n-1) + \sin(2n-3))=(4cos21−1)sin(2n−1)−(sin(2n−1)+sin(2n−3))
=(4cos⁡21−2)sin⁡(2n−1)−sin⁡(2n−3)= (4 \cos^21 - 2)\sin(2n-1) - \sin(2n-3)=(4cos21−2)sin(2n−1)−sin(2n−3)
=(2−4sin⁡21)sin⁡(2n−1)−sin⁡(2n−3)= (2 - 4 \sin^21) \sin(2n-1) - \sin(2n-3)=(2−4sin21)sin(2n−1)−sin(2n−3)