■正葉曲線の等分点(その3)
P1(x)=1,Q1(x)=1
sl(2u)=2sl(u)sl'(u)/(1+sl^4(u))より
P2(x)=2,Q2(x)=1+x
Q3(x)=Q1(x){Q2^2(x)+x(1−x)P2^2(x)}
=(1+x)^2+4x(1−x)=1+6x−3x^2
P3(x)=2(1−x)P2(x)Q2(x)Q1(x)−P1(x){Q2^2(x)+x(1−x)P2^2(x)}
=4(1−x)(1+x)−{(1+x)^2+4x(1−x)}
=4−4x^2−{1+6x−3x^2}
=3−6x−x^2
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Q4(x)=Q2(x){Q3^2(x)+xP3^2(x)}
=(1+x){(1+6x−3x^2)^2+x(3−6x−x^2)^2}
=(1+x){1+36x^2+9x^4+12x−36x^3−6x^2
+9x+36x^3+x^5−36x^2+12x^4−6x^3}
=(1+x){1+21x−6x^2−6x^3+21x^4+x^5}
=(1+x)^2{1+20x−26x^2+20x^3+x^4}
P4(x)=2P3(x)Q3(x)Q2(x)−P2(x){Q3^2(x)+xP3^2(x)}
=2(3−6x−x^2)(1+6x−3x^2)(1+x)−2(1+x){1+20x−26x^2+20x^3+x^4}
=2(1+x){3+12x−46x^2+12x^3+3x^4−1−20x+26x^2−20x^3−x^4}
=2(1+x){2−8x−20x^2−8x^3+2x^4}
=2(1+x)^2{2−10x−10x^2+2x^3)
=4(1+x)^3{1−6x+x^2)
最大公約数因子で割って
Q4(x)={1+20x−26x^2+20x^3+x^4}
P4(x)=4(1+x){1−6x+x^2)
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Q5(x)=(1−2x+5x^2)(1+52x−26x^2−12x^3+x^4}
P5(x)=(5−2x+x^2)(1−12x−26x^2+52x^3+x^4}
係数の表現の対称性がおもしろい.
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