天才置換を合理的に①　を読んで

次が成り立つ。
$\phi ([0,1])=[0,1]$
$\phi ={\phi }^{-1}$
$1+\phi =\frac{2}{1+t}$
$1-\phi =\frac{2t}{1+t}$
$1+{\phi }^2=\frac{2(1+t^2)}{(1+t{)}^2}$

次が成り立つ。
$1-{\phi }^2=\frac{4t}{(1+t{)}^2}$
$1+{\phi }^3=\frac{2(1+3t^2)}{(1+t{)}^3}$
$1-{\phi }^3=\frac{2t(3+t^2)}{(1+t{)}^3}$
$1+{\phi }^4=\frac{2(1+6t^2+t^4)}{(1+t{)}^4}$
$1-{\phi }^4=\frac{8t(1+t^2)}{(1+t{)}^4}$
$1+{\phi }^5=\frac{2(1+10t^2+5t^4)}{(1+t{)}^5}$
$1-{\phi }^5=\frac{2t(5+10t^2+t^4)}{(1+t{)}^5}$
$1+{\phi }^6=\frac{2(1+15t^2+15t^4+t^6)}{(1+t{)}^6}$
$1-{\phi }^6=\frac{2t(6+20t^2+6t^4)}{(1+t{)}^6}$

より一般に
$1+{\phi }^{2n-1}=\frac{2\sum _{i=0}^{n-1}{}_{2n-1}{\mathrm{C}}_{2k}{t}^{2k}}{(1+t{)}^{2n-1}}$
$1-{\phi }^{2n-1}=\frac{2t\sum _{i=0}^{n-1}{}_{2n-1}{\mathrm{C}}_{2k+1}{t}^{2k}}{(1+t{)}^{2n-1}}$
$1+{\phi }^{2n}=\frac{2\sum _{i=0}^{n-1}{}_{2n}{\mathrm{C}}_{2k}{t}^{2k}}{(1+t{)}^{2n}}$
$1-{\phi }^{2n}=\frac{2t\sum _{i=0}^{n-1}{}_{2n}{\mathrm{C}}_{2k+1}{t}^{2k}}{(1+t{)}^{2n}}$

次が成り立つ。
${\int }_0^1\frac{f(x)}{1+x}dx={\int }_0^1\frac{f\circ \phi }{1+t}dt$
${\int }_0^1\frac{f(x)}{1-x}dx={\int }_0^1\frac{f\circ \phi }{t(1+t)}dt$
${\int }_0^1\frac{f(x)}{(1+x{)}^2}dx=\frac{1}{2}{\int }_0^{1}f\circ \phi dt$
${\int }_0^1\frac{f(x)}{(1-x{)}^2}dx=\frac{1}{2}{\int }_0^1\frac{f\circ \phi }{t^2}dt$
${\int }_0^1\frac{f(x)}{1+x^2}dx={\int }_0^1\frac{f\circ \phi }{1+t^2}dt$
${\int }_0^1\frac{f(x)}{1-x^2}dx=\frac{1}{2}{\int }_0^1\frac{f\circ \phi }{t}dt$
${\int }_0^1\frac{f(x)}{1-x^4}dx=\frac{1}{4}{\int }_0^1\frac{f\circ \phi }{t}dt+\frac{1}{2}{\int }_0^1\frac{f\circ \phi }{1+t^2}dt=\frac{1}{4}{\int }_0^1\frac{f\circ \phi }{t}dt+\frac{1}{2}{\int }_0^1\frac{f(x)}{1+x^2}dx$