前>>538訂正。
>>433
直角三角形の相似比より、
x：√(x^2+1/x^2)=R：1/x-R
x(1/x-R)=R√(x^2+1/x^2)
1-xR=R√(x^2+1/x^2)
1-2xR+x^2R^2=x^2R^2+R^2/x^2
x^2-2x^3R=R^2
R^2+2x^3R-x^2=0
R=-x^3+√(x^6+x^2)
=x√(x^4+1)-x^3
R^2=2x^6+x^2-2x^4√(x^4+1)
一方1/x-2R=r+r√(x^2+1/x^2)/x
x-2x^2R=r｛x^2+√(x^4+1)｝
R=x√(x^4+1)-x^3を代入し、
x-2x^3√(x^4+1)+2x^5=r｛x^2+√(x^4+1)｝
辺々√(x^4+1)-x^2を掛け、
r=｛2x^5+x-2x^3√(x^4+1)｝｛√(x^4+1)-x^2｝
=(4x^5+x)√(x^4+1)-x^3(2x^4+1)-2x^3(x^4+1)
=(4x^5+x)√(x^4+1)-4x^7-3x^3
r^2=(16x^10+8x^6+x^2)(x^4+1)+(4x^7+3x^3)^2-2x^4(16x^8+16x^4+3)√(x^4+1)
=32x^14+48x^10+18x^6+x^2-(32x^12+32x^8+6x^4)√(x^4+1)
R^2+r^2
=2x^6+x^2-2x^4√(x^4+1)
+32x^14+48x^10+18x^6+x^2
-(32x^12+32x^8+6x^4)√(x^4+1)
=32x^14+48x^10+20x^6+2x^2
-(32x^12+32x^8+8x^4)√(x^4+1)=f(x)とおくと、
f'(x)=448x^13+480x^9+120x^5+4x
-(384x^11+256x^7+32x^3)√(x^4+1)
-｛(32x^12+32x^8+8x^4)4x^3｝/｛2√(x^4+1)｝=0
-の項を移項し、辺々√(x^4+1)/4xを掛け、
(112x^12+120x^8+30x^4+1)√(x^4+1)
=(96x^10+64x^6+8x^2)(x^4+1)
+16x^14+16x^10+4x^6
(112x^12+120x^8+30x^4+1)√(x^4+1)
=112x^14+176x^10+76x^6+8x^2
辺々2乗し、
(112x^12+120x^8+30x^4+1)^2(x^4+1)
=(112x^14+176x^10+76x^6+8x^2)^2
12544x^24+14400x^16+900x^8+1
26880x^20+6720x^16+224x^12
+7200x^12+240x^8+60x^4
+12544x^28+14400x^20+900x^12+x^4
26880x^24+6720x^20+224x^16
+7200x^16+240x^12+60x^8
=12544x^28+30976x^20+5776x^12+64x^4
+39424x^24+17024x^20+1792x^16
+26752x^16+2816x^12+1216x^8
12544x^28+39424x^24+48000x^20+28544x^16
+8564x^12+1200x^8+61x^4+1
=12544x^28+39424x^24+48000x^20+28544x^16
+8592x^12+1216x^8+64x^4
28x^12+16x^8+3x^4-1=0
x^4=Xとおくと、
28X^3+16X^2+3X-1=0
X=0.1595819881204……
x^2=0.3994771434……
x=0.6320420424…… やっとできたみたい。
∴S=π(R^2+r^2)=0.6970328246……