🚀 go-pugleaf

In article <4b38d0$2c3@ayame.ipc.chiba-u.ac.jp> hagiwara@yuri.ipc.chiba-u.ac.jp (ManAV) writes:
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	...
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>(x + 1/x)^2 = x^2 + 1/x^2 + 2 $B$N7W;;$r$9$k>e$G(J
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	...
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$B!V(J(x+1)/x=1 $B$N$H$-(J 1/x $B$r5a$a$h!W$K$7$^$7$g$&!#(J

$B!!2r(J 1$B!'(J (x+1)/x = 1+1/x =1 $B$@$+$i!"(J1/x = 0 $B$G$"$k!#(J
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$B!!2r(J 2$B!'(J (x+1)/x = 1 $B$@$+$i(J x/(x+1)=1/1=1 $B$G$"$k!J(Jx=-1 $B$G$O$"$j$($J$$$+$i!K!#(J
	1/x = 1/x+(x-x) = (x + 1/x)-x = (x^2+1)/x - x
	= (x^2+2x+1)/x - (2x)/x - x
	= ((x+1)^2/x) * (x/(x+1)) - 2 - x	(x/(x+1)=1 $B$r;H$C$?(J)
	= (x+1)^2/(x+1) - 2 - x = (x+1) - 2 - x = 1-2 = -1
	$B$D$^$j(J 1/x = -1 $B!J!)!)!)!K(J
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* 1-1+1-1+... = 1/2
$B!!2r(J 1$B!'(J 1/(1-x) = 1+x+x^2+...$B!"$=$3$G(J x=-1 $B$H$9$k!#(J
$B!!2r(J 2$B!'(J x=1-1+1-1+... $B$H$*$/$H!"(Jx=1-(1-1+1-...) = 1-x $B$7$?$,$C$F(J x=1/2

* log(-1) = 0 $B!J%i%$%W%K%C%D!"%Y%k%L!<%$!"%*%$%i!<O@Ah!K(J
$B!!$?$H$($P(J d(-x)/(-x) = dx/x $B$H$+(J 2log(-1) = log(-1)^2 = log 1 $B$J$I$+$i!#(J

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* 1=-1
  -1 = i^2 = sqrt(-1)*sqrt(-1) = sqrt((-1)*(-1)) = sqrt(1) = 1
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* sin x = 2 $B$N$H$-(J cos2x $B$r5a$a$h!#(J
  cos2x = 1-2sin^2 x = 1-2*2^2 = -7
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