Numerical studies                    To the theory　　To sample expansions of 1/(1±x)
                   To the calculation form for 1/(1-x)  To the calculation form for 1/(1+x)
 As examples, we try to expand 1/(1±x) as p=2 and q=5, that is, p＜x＜q. We can see smaller
errors for larger n or for x around the center of the interval. Figures in ( ) of the tables 
are for p=2 or q=5. They are not valid but shown to see limits.

x
Sum of
series
True
value
Error(r1)
2(=p)
(-0.64)
(-1.00)
(0.36)
2.5
-0.56
-0.67
0.11
3.0
-0.48
-0.50
0.02
3.5
-0.40
-0.40
0.00
4.0
-0.32
-0.33
0.01
4.5
-0.24
-0.29
0.05
5(=q)
(-0.16)
(-0.25)
(0.09)

1. Expansion of 1/(1-x) 

First degree
　From (8), a10=-2{1+(5+2)/(5+2-2)}/(5+2-2)=-0.96 ,
 a11=4/(5+2-2)2=0.16 ,
 Hence,
 1/(1-x)=-0.96+0.16x+r1　　             ..........(18)
 See table of right side.

x
Sum of
series
True
value
Error(r2)
2(=p)
(-0.784)
(-1.0000)
(0.216)
2.5
-0.624
-0.6667
0.0427
3.0
-0.496
-0.5000
0.0040
3.5
-0.4000
-0.4000
0.0000
4.0
-0.336
-0.3333
-0.0027
4.5
-0.304
-0.2857
-0.0183
5(=q)
(-0.304)
(-0.2500)
(-0.0540)

Second degree
　From (10), a20=-2[1+(5+2)/(5+2-2)
 +{(5+2)/(5+2-2)}2]/(5+2-2)=-1.744 ,
 a21=4{1+2(5+2)/(5+2-2)}/(5+2-2)2=0.608 ,
 a22=-8/(5+2-2)3=-0.064
 Hence,
 1/(1-x)=-1.744+0.608x-0.064x2+r2　　..........(19)
 See table of right side.

x
Sum of
series
True
value
Error(r3)
2(=p)
(-0.8704)
(-1.0000)
(0.1296)
2.5
-0.6496
-0.6667
0.0171
3.0
-0.4992
-0.5000
0.0008
3.5
-0.4000
-0.4000
0.0000
4.0
-0.3328
-0.3333
0.0005
4.5
-0.2784
-0.2857
0.0073
5(=q)
(-0.2176)
(-0.2500)
(0.0324)

Third degree
　Substituting p=2 and q=5 in (12), a30=-2.8416 ,
 a31=1.5488 , a32=-0.3328 , a33=0.0256, 
 Hence,
 1/(1-x)=-2.8416+1.5488x-0.3328x2+0.0256x3+r3....(20)
 See table of right side.

x
Sum of
series
True
value
Error(r4)
2(=p)
(-0.92224)
(-1.0000)
(0.07776)
2.5
-0.65984
-0.66667
0.00683
3.0
-0.49984
-0.50000
0.00016
3.5
-0.40000
-0.40000
0.00000
4.0
-0.33344
-0.33333
-0.00011
4.5
-0.28864
-0.28571
-0.00293
5(=q)
(-0.26944)
(-0.25000)
(-0.01944)

Fourth degree
　Substituting p=2 and q=5 in (14), a40=-4.37824 , 
 a41=3.30496 , a42=-1.08544 , a43=0.16896 ,
 a44=-0.01024,
 Hence,
 1/(1-x)=-4.37824+3.30496x-1.08544x2+0.16896x3
 -0.01024x4+r4　                ..........(21)
 See table of right side.


2. Expansion of 1/(1+x)

x
Sum of
series
True
value
Error(r1)
2(=p)
(0.29630)
(0.33333)
(-0.03704)
2.5
0.27160
0.28571
-0.01411
3.0
0.24691
0.25000
-0.00309
3.5
0.22222
0.22222
0.00000
4.0
0.19753
0.20000
-0.00247
4.5
0.17284
0.18182
-0.00898
5(=q)
(0.14815)
(0.16667)
(-0.01852)

First degree
　From (17), b10=2{1+(5+2)/(5+2+2)}/(5+2+2)
  =0.39506 , b11=-4/(5+2+2)2=-0.04938 ,
  Hence,
　1/(1+x)=0.39506-0.04938x+r1　　   ..........(22)
  See table of right side.

x
Sum of
series
True
value
Error(r2)
2(=p)
(0.32099)
(0.33333)
(-0.01234)
2.5
0.28258
0.28571
-0.00314
3.0
0.24966
0.25000
-0.00034
3.5
0.22222
0.22222
0.00000
4.0
0.20027
0.20000
0.00027
4.5
0.18381
0.18182
0.00200
5(=q)
(0.17284)
(0.16667)
(0.00617)

Second degree
　From (17), b20=2[1+(5+2)/(5+2+2)
  +{(5+2)/(5+2+2)}2]/(5+2+2)=0.52949 ,
  b21=-4{1+2(5+2)/(5+2+2)}/(5+2+2)2=-0.12620 ,
  b22=8/(5+2+2)3=0.01097 ,
  Hence,
　1/(1+x)=0.52949-0.12620x+0.01097x2+r2......(23)
  See table of right side.

x
Sum of
series
True
value
Error(r3)
2(=p)
(0.32922)
(0.33333)
(-0.00412)
2.5
0.28502
0.28571
-0.00070
3.0
0.24996
0.25000
-0.00004
3.5
0.22222
0.22222
0.00000
4.0
0.19997
0.20000
-0.00003
4.5
0.18137
0.18182
-0.00044
5(=q)
(0.16461)
(0.16667)
(-0.00206)

Third degree
　Substituting p=2 and q=5 in (17),
  b30=0.63405 , b31=-0.21582 , b32=0.03658 ,
  b33=-0.00244 , 
  Hence,
  1/(1+x)=0.63405-0.21582x+0.03658x2
  -0.00244x3+r3                       ........(24)
  See table of right side.

x
Sum of
series
True
value
Error(r4)
2(=p)
(0.33196)
(0.33333)
(-0.00137)
2.5
0.28556
0.28571
-0.00015
3.0
0.25000
0.25000
-0.00000
3.5
0.22222
0.22222
0.00000
4.0
0.20000
0.20000
0.00000
4.5
0.18192
0.18182
0.00010
5(=q)
(0.16735)
(0.16667)
(0.00069)

Fourth degree
　Substituting p=2 and q= in (17),
  b40=0.71537 , b41=-0.30876 , b42=0.07641 ,
  b43=-0.01003 , b44=0.00054 ,
  Hence,
　1/(1+x)=0.71537-0.30876x+0.07641x2-0.01003x3
  +0.00054x4+r4　                   ..........(25)
  See table of right side.