设关于x∈[-1, 1]的多项式 T(n, x)= cos(n*arccos x),对任何整数n
T(n+1, x) = cos((n+1)*arccos x)
= cos(n*arccos x + arccos x)
= cos(n*arccos x) * cos(arccos x) - sin(n*arccos x)*sin(arccos x)
= T(n, x)*x - sin(n*arccos x)*sin(arccos x)
T(n-1, x) = = cos((n-1)*arccos x)
= cos(n*arccos x - arccos x)
= cos(n*arccos x) * cos(arccos x) + sin(n*arccos x)*sin(arccos x)
= T(n, x)*x + sin(n*arccos x)*sin(arccos x)
所以 T(n+1, x) + T(n-1, x) = 2T(n, x)*x
再由T(0, x)=1, T(1, x)=x,可以得到T(2, x) = 2x²-1
因为ℓ2x²-1ℓ≤1,arccos(2x²-1)= 2arccos x,所以T(2n, x)= T(n, 2x²-1)
由于可以归纳出T(n, x)是整系数多项式,所以只要x²是有理数,T(2n, x)=T(n, 2x²-1)就也是有理数
T(3, x)= 2x(2x²-1)-x = 4x³-3x
T(4, x)= 2x(4x³-3x)-(2x²-1) = 8x⁴-8x²+1
T(5, x)= 2x(8x⁴-8x²+1)-(4x³-3x) = 16x^5-20x^3+5x
T(10, 1/√6) = T(5, -2/3) = -16*32/243+20*8/27-5*2/3 = 118/243
T(n+1, x) = cos((n+1)*arccos x)
= cos(n*arccos x + arccos x)
= cos(n*arccos x) * cos(arccos x) - sin(n*arccos x)*sin(arccos x)
= T(n, x)*x - sin(n*arccos x)*sin(arccos x)
T(n-1, x) = = cos((n-1)*arccos x)
= cos(n*arccos x - arccos x)
= cos(n*arccos x) * cos(arccos x) + sin(n*arccos x)*sin(arccos x)
= T(n, x)*x + sin(n*arccos x)*sin(arccos x)
所以 T(n+1, x) + T(n-1, x) = 2T(n, x)*x
再由T(0, x)=1, T(1, x)=x,可以得到T(2, x) = 2x²-1
因为ℓ2x²-1ℓ≤1,arccos(2x²-1)= 2arccos x,所以T(2n, x)= T(n, 2x²-1)
由于可以归纳出T(n, x)是整系数多项式,所以只要x²是有理数,T(2n, x)=T(n, 2x²-1)就也是有理数
T(3, x)= 2x(2x²-1)-x = 4x³-3x
T(4, x)= 2x(4x³-3x)-(2x²-1) = 8x⁴-8x²+1
T(5, x)= 2x(8x⁴-8x²+1)-(4x³-3x) = 16x^5-20x^3+5x
T(10, 1/√6) = T(5, -2/3) = -16*32/243+20*8/27-5*2/3 = 118/243