Bi-ons = Complexified Anions for the HP-41

Overview
 

 0°)  M-Code Routines
 1°)  Multiplication

  a)  General Case
  b)  Bi-ons with proportional imaginary parts ( focal program & M-code routine B*B1 )

 2°)  Miscellaneous Functions + 2 M-Code routines:  Z*B & Z*IMB
 3°)  Polynomials with complex coefficients
 4°)  Bionic Equations F(b) = b
 

-In "Anions for the HP-41", we have used the Cayley-Dickson formula

         ( a , b ) ( c , d ) = ( a c - d* b , d a + b c* )       where   * = conjugate

 to construct the complexes from the real numbers, regarding a complex as a pair of real numbers,
 and then the quaternions from the complexes, the octonions from the quaternions, the sedenions from the octonions,
 the 32-ons from the sedenions, the 64-ons from the 32-ons ... and so on ... for the 2^n-ons

-But if we start with the complexes, we gradually get the bi-complexes, bi-quaternions, bi-octonions, bi-sedenions ...

-The programs listed in this page calculate elementary functions of these "bi-ons"
 

0°)  M-Code Routines
 

-Unfortunately, I've not found a good way to avoid using M-code routines, especially to compute complex functions, namely:

  Z*Z  Z^2  SQRTZ  1/Z  Z/Z   Z+Z         cf "A few M-Code Routines for the HP-41"

and also  X+1   SINH   COSH                  -------------------------------------------

-Several M-Code routines that are useful for anions may also be employed for bi-ons:

  A+A  A-A  ST*A  ST/A  AMOVE  ASWAP   cf "Anions" and "Anionic Functions for the HP-41"

-Of course, you can use your own routines, provided  Z*Z  does not alter synthetic register Q
 

1°)  Multiplication
 

     a)  General Case
 

-This program employs the Cayley-Dickson formula recursively.
-Unfortunately, there are not enough registers for bi-64-ons.

-The 2n components of the 1st bion are to be sored into R01-R02 , R03-R04 , ............... , R2n-1-R2n   ( 2 real numbers for each b_j )
-Likewise for the 2n components of the 2nd bion into  R2n+1-R2n+2 , ............................ , R4n-1-R4n   ( 2 real numbers for each b'_j )

     b = b₀ + b₁ e₁ + ............... + b_n-1 e_n-1
    b' = b'₀ + b'₁ e₁ + ............... + b'_n-1 e_n-1

-"B*B" does not use any M-Code routine !
 

Data Registers:           •  R00 = 2n ( 4 , 8 , 16 , 32 or 64 )                  ( Registers R00 thru R4n are to be initialized before executing "B*B" )

                                      •  R01   ......  •  R2n = the components of the 1st bi-ion
                                      •  R2n+1 ....  •  R4n = the components of the 2nd bi-ion

  >>>  When the program stops:     R01   ......  R2n = the 2n components of the product

-During the calculations,

    R01 to R04 = bicomplex1    R09 to R16 = biquaternion1  R25 to R40 = bioctonion1  R57 to R88 = bisedenion1     R121 to R184 = bi-32-on1
    R05 to R08 = bicomplex2    R17 to R24 = biquaternion2  R41 to R56 = bioctonion2  R89 to R120 = bisedenion2   R185 to R248 = bi-32-on2

Flags: /
Subroutines:  /
 
 
 

 
       ( 378 bytes / SIZE 8n-7 )
 
 

   where  1.2n  is the control number of the result.

Example:   Find the product of the bisedenions:

  b = ( 25 + 108 i ) + ( 105 + 113 i ) e₁ + ( 48 + 3 i ) e₂ + ( 123 + 65 i ) e₃ + ( 45 + 11 i ) e₄ + ( 58 + 20 i ) e₅ + ( 34 + 84 i ) e₆ + ( 38 + 117 i ) e₇
    + ( 81 + 46 i ) e₈ + ( 52 + 36 i ) e₉ + ( 35 + 125 i ) e₁₀ + ( 16 + i ) e₁₁ + ( 41 + 109 i ) e₁₂ + ( 15 + 91 i ) e₁₃ + ( 63 + 94 i ) e₁₄ + ( 55 + 28 i ) e₁₅

  b' = ( 100 + 39 i ) + ( 27 + 59 i ) e₁ + ( 61 + 12 i ) e₂ + ( 99 + 129 i ) e₃ + ( 49 + 44 i ) e₄ + ( 101 + 80 i ) e₅ + ( 5 + 74 i ) e₆ + ( 21 + 75 i ) e₇
     + ( 62 + 53 i ) e₈  + ( 77 + 13 i ) e₉ + ( 9 + 107 i ) e₁₀ + ( 64 + 4 i ) e₁₁ + ( 33 + 43 i ) e₁₂ + ( 60 + 102 i ) e₁₃ + ( 121 + 114 i ) e₁₄ + ( 89 + 112 i ) e₁₅

-You can use the short routine below to store these coefficients into R01 thru R64
 
 

 
-Bisedenions so n = 16 and 2n = 32  STO 00

-SIZE 121 at least

  XEQ "B*B"  >>>>  1.032             in about 6 minutes

-And we get   R01 = 22450  R02 = -93103  .......................  R31 = -14895  R32 = 36483

-Thus,

   b.b' = ( 22450 - 93103 i ) + ( -3665 + 12974 i ) e₁ + ( 32291 - 31812 i ) e₂ + ( 18482 + 29668 i ) e₃ + ( -7840 + 8985 i ) e₄ + ( 1001 + 48348 i ) e₅
          + ( -10494 + 6090 i ) e₆ + ( -1657 + 25709 i ) e₇ + ( -14243 - 595 i ) e₈  + ( 9295 + 1562 i ) e₉ + ( -19125 + 35197 i ) e₁₀ + ( 18449 - 20036 i ) e₁₁
          + ( -3315 + 35478 i ) e₁₂ + ( -22176 + 9676 i ) e₁₃ + ( 11818 + 23190 i ) e₁₄ + ( -14895 + 36483 i ) e₁₅

Note:

-The product of 2 bi-32-ons should last about 24 minutes.
 

     b)  Bi-ons with proportional imaginary parts
 

-Fortunately, the formulas are much simpler if the 2 bi-ons have the same imaginary direction, I mean if, given  b & b'

     b = b₀ + b₁ e₁ + ............... + b_n-1 e_n-1
    b' = b'₀ + b'₁ e₁ + ............... + b'_n-1 e_n-1

  there exist a complex number c such that, for all  i > 0 ,  b'_i = c b_i  or  b_i = c b'_i

-The program hereunder is useful in this case, and it uses only 4n+1 registers
 

Data Registers:           •  R00 = 2n > 3                                   ( Registers R00 thru R4n are to be initialized before executing "B*B1" )

                                      •  R01   ......  •  R2n = the n components of the 1st bi-on
                                      •  R2n+1 ....  •  R4n = the n components of the 2nd bi-on

  >>>  When the program stops:     R01   ......  R2n = the n components of the product,   R2n+1  ........ R4n  are unchanged.

Flags: /
Subroutine:  Z*Z
 
 
 

 
      ( 124 bytes / SIZE 4n+1 )
 
 

   where  1.2n  is the control number of the product.

Example:   Find the product of the biquaternions

  b = ( 6 - 9 i ) + ( -4 + 2 i ) e₁ + ( -3 + i ) e₂ + ( -4 + 7 i ) e₃                            here,   Im(b') = ( 2 + 3i ) Im(b)
  b' = ( 4 + 7 i ) + ( -14 - 8 i ) e₁ + ( -9 - 7 i ) e₂ + ( -29 + 2 i ) e₃

-> biquaternions, so  8  STO 00

-Store the coefficients into R01 thru R16

    6  STO 01   -9  STO 02   -4  STO 03   2  STO 04  ..........................  -29  STO 15    2  STO 16

    XEQ "B*B1"  >>>>  1.008                 ---Execution time = 6 seconds---

  R01 = -121    R02 = 201
  R03 = -186    R04 =  58
  R05 = -136    R06 =  22
  R07 = -221    R08 = 273

-Thus,  b b' = ( -121 + 201 i ) + ( -186 + 58 i ) e₁ + ( -136 + 22 i ) e₂ + ( -221 + 273 i ) e₃

Notes:

-"B*B1" does not check that the imaginary parts are proportional.
-This routine uses the synthetic registers M , N , O , P & Q
-Most of the "bionic" functions deal with such hypercomplexes and "B*B1" is sufficient.

-So, an M-code routine is quite useful and will be much faster:

Initialization

-This routine is alread listed in "Anions for the HP-41"
-Please take a look at this page for more details.
 

260  SETHEX                   @EFC8  in my ROM
378  C=c
03C  RCR 3
106  A=C S&X
130  LDI S&X
200  200h                           correct value for an HP-41CV/CX or an HP-41C with a QUAD memory module
306  ?A<C S&X
381  ?NCGO
00A  NEXIST
0A6  A<>C S&X
270   RAMSELECT
038   READATA
38D  ?NCXQ
008   N->S&X
2E6  ?C#0 S&X
0B5  ?NCGO
0A2  ERROR
05A  C=0 M
23A  C=C+1 M
106  A=C S&X
1BC  RCR 11
0A6  A<>C S&X
1E6  C=C+C S&X
10E  A=C ALL
378  C=c
03C  RCR 3
0E6  B<>C S&X
378  C=c
0C6  C=B S&X
1BC  RCR 11
0C6  C=B S&X
20E  C=A+C ALL
24C  ?FSET 9
013  JNC+02
03C  RCR 3
070  N=C ALL
10E  A=C ALL
130  LDI S&X
200  200h
306  ?A<C S&X
381  ?NCGO
00A  NEXIST
3E0  RTN                        @EFF2  in my ROM

  ( 43 words )
 

-To call this subroutine:    ?NCXQ  EFC8  =  321   3BC    in my ROM
-Change these 2 words if you have loaded this routine at another address...

Multiplication Routine:

- B*B1 below does the same calculations as "B*B1" above
 

0B1  "1"
002   "B"
02A  "*"
002   "B"
244   CLRF 9
321   ?NCXQ                   executes the initialization
3BC  EFC8                       routine above
378   C=c
03C  RCR 3
226   C=C+1 S&X
226   C=C+1 S&X
226   C=C+1 S&X
106   A=C S&X
0B0  C=N ALL
306   ?A<C S&X
0B5   ?NCGO
0A2   DATAERROR         There will be a DATA ERROR message if  R00 = 2n < 4
03C   RCR 3
106   A=C S&X
17C  RCR 6
0A6  A<>C S&X
226   C=C+1 S&X
226   C=C+1 S&X
0BC  RCR 5
070   N=C ALL
27C  RCR 9
10E  A=C ALL
270  RAMSLCT
038  READATA
0AE  A<>C ALL
266   C=C-1 S&X
270  RAMSLCT
038  READATA
0EE  B<>C ALL
046  C=0 S&X
270  RAMSLCT
0EE  B<>C ALL
068  Z=C
0AE A<>C ALL
028  T=C
0B0  C=N ALL
17C  RCR 6
270  RAMSLCT
038  READATA
10E  A=C ALL
0B0  C=N ALL
17C  RCR 6
226  C=C+1 S&X
270  RAMSLCT
038  READATA
0EE  B<>C ALL
046  C=0 S&X
270  RAMSLCT
0EE  B<>C ALL
0A8  Y=C
0AE  A<>C ALL
0E8  X=C
0B0  C=N ALL
268   Q=C
259   ?NCXQ            The 1st executable word of my Z*Z routine is @F696 in my ROM
3D8   Z*Z                  Change these 2 words according to your own ROM
0F8   C=X
168   M=C
0B8  C=Y                  Here, M and N are synthetic registers ( not CPU registers )
1A8  N=C
278  C=Q
070  N=C ALL
260  SETHEX
0B0  C=N ALL          LOOP    @FD79 in my ROM
03C  RCR 3
10E  A=C ALL
270  RAMSLCT
038  READATA
0AE  A<>C ALL
266  C=C-1 S&X
270  RAMSLCT
038  READATA
0EE  B<>C ALL
046  C=0 S&X
270  RAMSLCT
0EE  B<>C ALL
068  Z=C
0AE  A<>C ALL
028  T=C
0B0  C=N ALL
27C  RCR 9
10E  A=C ALL
270  RAMSLCT
038  READATA
0AE  A<>C ALL
266  C=C-1 S&X
270  RAMSLCT
038  READATA
0EE  B<>C ALL
046  C=0 S&X
270  RAMSLCT
0EE  B<>C ALL
0E8  X=C
0AE  A<>C ALL
0A8  Y=C
0B0  C=N ALL
268  Q=C
259   ?NCXQ            The 1st executable word of my Z*Z routine is @F696 in my ROM
3D8   Z*Z                  Change these 2 words according to your own ROM
0F8  C=X
1E8  O=C
0B8  C=Y
228   P=C
046  C
270  =
038  T
0A8  Y=C
078  C=Z
0E8  X=C
278  C=Q
070  N=C ALL
10E  A=C ALL
270  RAMSLCT
038  READATA
0AE  A<>C ALL
260  SETHEX
266  C=C-1 S&X
270  RAMSLCT
038  READATA
0EE  B<>C ALL
046  C=0 S&X
270  RAMSLCT
0EE  B<>C ALL
068  Z=C
0AE  A<>C ALL
028   T=C
259   ?NCXQ            The 1st executable word of my Z*Z routine is @F696 in my ROM
3D8   Z*Z                  Change these 2 words according to your own ROM
0F8  C=X
2BE  C=-C
10E  A=C ALL
178  C=M
01D  C=
060  A+C
168  M=C
0B8  C=Y
2BE  C=-C
10E  A=C ALL
1B8  C=N
01D  C=
060  A+C
1A8  N=C
278  C=Q
070  N=C ALL
17C  RCR 6
10E  A=C ALL
270  RAMSLCT
038  READATA
0AE A<>C ALL
260  SETHEX
226  C=C+1 S&X
270  RAMSLCT
038  READATA
0EE  B<>C ALL
046  C=0 S&X
270  RAMSLCT
0EE  B<>C ALL
0A8  Y=C
0AE  A<>C ALL
0E8  X=C
259   ?NCXQ            The 1st executable word of my Z*Z routine is @F696 in my ROM
3D8   Z*Z                  Change these 2 words according to your own ROM
0B8  C=Y
10E  A=C ALL
238  C=P
01D  C=
060  A+C
10E  A=C ALL
278  C=Q
03C  RCR 3
070  N=C ALL
270  RAMSLCT
0AE A<>C ALL
2F0  WRITDATA
046  C=0 S&X
270  RAMSLCT
0F8  C=X
10E  A=C ALL
1F8  C=O
01D  C=
060  A+C
10E  A=C ALL
0B0  C=N ALL
260  SETHEX
266  C=C-1 S&X
270  RAMSLCT
0AE  A<>C ALL
2F0  WRITDATA
0B0  C=N ALL
266  C=C-1 S&X
266  C=C-1 S&X
1BC  RCR 11
266  C=C-1 S&X
266  C=C-1 S&X
070  N=C ALL
106  A=C S&X
27C  RCR 9
226  C=C+1 S&X
306  ?A<C S&X
1E5   ?NCXQ              GOTO LOOP @FD79 in my ROM
3F6   FD79                  Change these 2 words according to your own ROM
046  C=0 S&X
270  RAMSLCT
1B8  C=N
10E  A=C ALL
178  C=M
0EE  B<>C ALL
0B0  C=N ALL
17C  RCR 6
270  RAMSLCT
0EE  B<>C ALL
2F0  WRITDATA
0EE  B<>C ALL
226  C=C+1 S&X
270  RAMSLCT
0AE A<>C ALL
2F0  WRITDATA
0B0  C=N ALL
17C  RCR 6
266  C=C-1 S&X
270  RAMSLCT
038  READATA
2A0  SETDEC
266  C=C-1 S&X
266  C=C-1 S&X
266  C=C-1 S&X
10E  A=C ALL
046  C=0 S&X
270  RAMSLCT
04E  C
35C  =
050   1
01D  C=
060  A+C
0E8  X=C
345  ?NCGO
042    CLA

( 242 words )
 
 

   where  1.2n  is the control number of the product.
 

Note:

-The product of 2 bisedenions ( 2n = 32 ) is about 7 seconds
 

2°)  Miscellaneous Functions
 

-These programs compute elementary functions of a "bi-on":

    •  "1/B"    inverse

         B^-1 = B* / | B |²    where B* is the conjugate of  B   and   | B |² = b₀² + b₁² + ................. + b_N-1²

-Note that here  | B |  may equal 0 even if b # 0 since the components b_j are complexes !
-So, b # 0 may not have an inverse !
 

    •  "E^B"   exponential

        exp( b₀ + b₁ e₁ + .... + b_N-1 b_N-1 ) = e^b0 [ cos µ + ( sin µ ). I ]

       where  µ = ( b₁² + ............... + b_N-1² )^1/2    and    I = ( b₁ e₁ + ............. + b_N-1 e_N-1 ) / µ

   •   "LNB"  natural logarithm

       Ln( b₀ + b₁ e₁ + .... + b_N-1 b_N-1 )   = Ln ( b₀² + b₁² + ................. + b_N-1² )^1/2 + A(µ,b₀). I

       where  µ = ( b₁² + ............... + b_N-1² )^1/2    and    I = ( b₁ e₁ + ............. + b_N-1 e_N-1 ) / µ

>>> Here,  A(µ,b₀) generalizes the ATAN2 function to complexes:

-We have to solve  Sin Z = µ / ( µ² + b₀² )^1/2  ,  Cos Z = b₀ / ( µ² + b₀² )^1/2

  it yields  Z = A(µ,b₀) = - i Ln ( b₀ + i µ ) / ( µ² + b₀² )^1/2

>>>  If µ = 0  and  Im(b) # 0 ,  Ln b = Ln b₀ + ( b₁ e₁ + ............. + b_N-1 e_N-1 ) / b₀    ( if  b₀ = 0 , Ln b  does not exist )
>>>  If µ = 0  and  Im(b) = 0 ,  Ln b = Ln b₀
 

    •  "B^X"  raising a bion to a real power         B^X = exp ( X Ln B )

    •  "X^B"  raising a real to a bionic power       X^B = exp ( B Ln X )

    •  "B^Z"  raising a bion to a complex power         B^Z = exp ( Z Ln B )

    •  "Z*B"  multiplies a bi-on by the complex x + i.y

    •  "SHB"  hyperbolic sine

        Sinh b = ( exp(b) - exp(-b) ) / 2

    •  "CHB"  hyperbolic cosine

        Cosh b = ( exp(b) + exp(-b) ) / 2

    •  "THB"  hyperbolic tangent

          Tanh b = ( exp(2b) - 1 ) ( exp(2b) + 1 ) ^-1

    •  "ASHB"  arc sinh b = Ln [ b + ( b² + 1 )^1/2 ]

    •  "ACHB"  arc cosh b = Ln [ b + ( b² - 1 )^1/2 ]

    •  "ATHB"   arc tanh b = (1/2) [ Ln ( 1 + b ) - Ln ( 1 - b ) ]

    •  "SINB"   sine

          Sin b = ( exp(i.b) - exp(-i.b) ) / 2.i

    •  "COSB"  cosine

          Cos b = ( exp(i.b) + exp(-i.b) ) / 2

    •  "TANB"    Tan b = - i Tanh ( i.b )

    •  "ASINB"    arc sin b

          Arc Sin b     = - i  Arc Sinh ( i.b )

    •  "ACOSB"  arc cos b

          Arc Cos b = PI/2 - Arc Sin b
 

    •  "ATANB"  arc tan b

          Arc Tan b     = - i  Arc Tanh ( i.b )

Remark:

-I've used     arc cosh b = Ln [ b + ( b² - 1 )^1/2 ]

  instead of the standard definition    Arc Cosh b  = Ln [ b + ( b + 1 )^1/2 ( b - 1 )^1/2 ]

-So, "ACHB" may give a result that requires a sign change to get the principal value of  Arc Cosh b.
-See the notes below if you prefer the standard definition.
 
 

Data Registers:           •  R00 = 2n ( 4 , 8 , .... , 128 or 256 )    ( Registers R00 thru R2n are to be initialized before executing these programs )

                                      •  R01   ......  •  R2n = the components of the bi-on

   ( In several cases,             R2n+1 ..........  R4n  are used for temporary data storage )

  >>>  When the program stops:     R01   ......  R2n = the components of the result
 

Flag:  F01  ( sometimes )
Subroutines:   ST*A  ST/A  AMOVE  ASWAP  A+A  A-A  B*B1  Z+Z  Z*Z  Z/Z  SQRTZ

-Line 340 is a three-byte  GTO 03
 
 

 
          ( 728 bytes / SIZE 2n+1 or 4n+1 )
 
 

   where   1.2n   is the control number of the result.

Exceptions:

-For "B^X" or "X^B" , place x in register X
-For "B^Z"  or "Z*B" , place x in register X and y in register Y  ( z = x + i y )

Examples:   With the biquaternion  b = ( 1 + 0.9 i ) + ( 0.8 + 0.7 i ) e₁ + ( 0.6 + 0.5 i ) e₂ + ( 0.4 + 0.3 i ) e₃

>>>  Biquaternion, so   2n = 8  STO 00

>>>  1  STO 01   0.9  STO 02   0.8  STO 03   0.7  STO 04   0.6  STO 05   0.5  STO 06   0.4  STO 07   0.3  STO 08

-Register R00 is unchanged, but you'll have to store again b in R01 thru R08 after each example.
 

    •  XEQ "1/B"  >>>  1.008    and we find in registers  R01 thru  R08:

       1/b = ( 0.2710 - 0.2285 i ) + ( -0.2115 + 0.1835 i ) e₁ + ( -0.1521 + 0.1385 i ) e₂ + ( -0.0927 + 0.0936 i ) e₃             rounded to 4 decimals

    •  XEQ "E^B"   gives likewise

       exp(b) = ( 3.1095 - 0.0905 i ) + ( 0.7810 + 2.6836 i ) e₁ + ( 0.6211 + 1.9573 i ) e₂ + ( 0.4613 + 1.2310 i ) e₃             rounded to 4 D

   •   XEQ "LNB"

       Ln (b) = ( 0.6669 + 0.7167 i ) + ( 0.6100 - 0.0013 i ) e₁ + ( 0.4480 - 0.0118 i ) e₂ + ( 0.2860 - 0.0222 i ) e₃              rounded to 4 D

    •  XEQ "SHB"

      Sinh b = ( 1.6085 + 0.1583 i ) + ( 0.5553 + 1.2491 i ) e₁ + ( 0.4300 + 0.9076 i ) e₂ + ( 0.3047 + 0.5662 i ) e₃             rounded to 4 D

    •  XEQ "CHB"

      Cosh b = ( 1.5010 - 0.2488 i ) + ( 0.2256 + 1.4345 i ) e₁ + ( 0.1911 + 1.0497 i ) e₂ + ( 0.1566 + 0.6648 i ) e₃             rounded to 4 D

    •  XEQ "THB"

      Tanh b = ( 0.7548 - 1.1142 i ) + ( -0.8316 + 0.1404 i ) e₁ + ( -0.6083 + 0.1179 i ) e₂ + ( -0.3851 + 0.0953 i ) e₃          rounded to 4 D

    •  XEQ "ASHB"

      Asinh b = ( 1.3553 + 0.7183 i ) + ( 0.6025 + 0.0482 i ) e₁ + ( 0.4434 + 0.0247 i ) e₂ + ( 0.2843 + 0.0013 i ) e₃          rounded to 4 D

    •  XEQ "ACHB"

      Acosh b = ( 1.3521 + 0.7113 i ) + ( 0.6185 - 0.0512 i ) e₁ + ( 0.4534 - 0.0485 i ) e₂ + ( 0.2883 - 0.0459 i ) e₃          rounded to 4 D

    •  XEQ "ATHB"

      Atanh b = ( 0.2955 - 0.2011 i ) + ( 0.9797 + 0.2256 i ) e₁ + ( 0.7236 + 0.1484 i ) e₂ + ( 0.4675 + 0.0711 i ) e₃          rounded to 4 D

    •  XEQ "SINB"

      Sin b = ( 0.6434 + 1.7921 i ) + ( 1.3025 + 0.2589 i ) e₁ + ( 0.9613 + 0.1671 i ) e₂ + ( 0.6201 + 0.0753 i ) e₃             rounded to 4 D

    •  XEQ "COSB"

      Cos b = ( 1.6624 - 0.0748 i ) + ( -0.1571 - 1.5114 i ) e₁ + ( -0.1421 - 1.1074 i ) e₂ + ( -0.1272 - 0.7033 i ) e₃          rounded to 4 D

    •  XEQ "TANB"

      Tan b = ( -1.0993 - 0.1510 i ) + ( 0.8538 - 0.8196 i ) e₁ + ( 0.6127 - 0.6171 i ) e₂ + ( 0.3715 - 0.4146 i ) e₃             rounded to 4 D

    •  XEQ "ASINB"

      Asin b = ( 0.7517 + 0.0817 i ) + ( 1.0103 + 0.5534 i ) e₁ + ( 0.7519 + 0.3886 i ) e₂ + ( 0.4935 + 0.2238 i ) e₃          rounded to 4 D

    •  XEQ "ACOSB"

      Acos b = ( 0.8191 - 0.0817 i ) + ( -1.0103 - 0.5534 i ) e₁ + ( -0.7519 - 0.3886 i ) e₂ + ( -0.4935 - 0.2238 i ) e₃          rounded to 4 D

    •  XEQ "ATANB"

      Atan b = ( -0.2551 + 0.2588 i ) + ( 0.2065 + 1.0188 i ) e₁ + ( 0.1697 + 0.7447 i ) e₂ + ( 0.1329 + 0.4705 i ) e₃          rounded to 4 D

    •  "B^X"  with the same b and x = 3.14          3.14  XEQ "B^X"  yields

      b^x = ( 4.0377 - 5.3510 i ) + ( -2.4135 + 2.5059 i ) e₁ + ( -1.7284 + 1.8833 i ) e₂ + ( -1.0433 + 1.2608 i ) e₃          rounded to 4D

    •  Likewise   3.14  XEQ "X^B"  returns:

     x^b = ( 4.0060 - 0.4653 i ) + ( 0.9349 + 3.5690 i ) e₁ + ( 0.7498 + 2.6050 i ) e₂ + ( 0.5648 + 1.6409 i ) e₃                rounded to 4 D

    •  "Z*B"  with the same b and  z = 2 + 3 i     3  ENTER^  2   XEQ "Z*B"

     z.b =  ( -0.7 + 4.8 i ) + ( -0.5 + 3.8 i ) e₁ + ( -0.3 + 2.8 i ) e₂ + ( -0.1 + 1.8 i ) e₃

    •  Likewise   3  ENTER^  2   XEQ "B^Z"   returns:

     b^z = ( -0.4735 + 2.3593 i ) + ( -1.8006 - 0.3988 i ) e₁ + ( -1.3296 - 0.2611 i ) e₂ + ( -0.8587 - 0.1233 i ) e₃          rounded to 4 D

Notes:

-The standard definition of  Arc Cosh b =  Ln [ b + ( b + 1 )^1/2 ( b - 1 )^1/2 ]  is applied by the following routine:
 
 

 
   ( 60 bytes / SIZE 6n+1 )
 

-However, since the SIZE = 6n+1,  this variant cannot compute the Arc Cosh  of a bi-64-on !
-This routine may of course be inserted in the programs above to save bytes.

Other elementary functions

-The "self-power function" is defined unambiguously by   b^b = exp ( ( Ln b ) b )
 
 

 
( 31 bytes / SIZE 4n+1 )
 

-The more general function  b^b'  may be defined in several ways.
-If you define it by  b^b' = exp ( ( Ln b ) b' ) ,

>>>  simply replace line 05 above by  XEQ "B*B"  and delete lines 02-03
 

Gudermannian Function
 

Formulae:    Gd(b) = 2 ArcTan [ Tanh (b/2) ]  and its inverse:   Agd(b) = 2 ArcTanh [ Tan(b/2) ]
 
 

 
   ( SIZE 4n+1 )
 
 

 
Example:           b = ( 1 + 0.9 i ) + ( 0.8 + 0.7 i ) e₁ + ( 0.6 + 0.5 i ) e₂ + ( 0.4 + 0.3 i ) e₃                       ( biquaternion  ->  8  STO 00 )

 1  STO 01   0.9  STO 02   0.8  STO 03   0.7  STO 04   0.6  STO 05   0.5  STO 06   0.4  STO 07   0.3  STO 08

  XEQ "GDB"   >>>>   1.008       and you find in registers R01 thru R08:

  Gd( b ) = ( -1.0420 + 0.7537 i ) + ( 0.5771 + 1.7617 i ) e₁ + ( 0.4551 + 1.2838 i ) e₂ + ( 0.3331 + 0.8058 i ) e₃         rounded to 4D

-Likewise, you'll find:

  Agd( b ) = ( 0.8355 - 0.6779 i ) + ( 1.4782 + 0.6366 i ) e₁ + ( 1.0970 + 0.4414 i ) e₂ + ( 0.7158 + 0.2463 i ) e₃         rounded to 4D

Notes:

-Of course, this list of functions is not exhaustive...

-Here are also 2 M-Code routines:  Z*B & Z*IMB
-The 1st one does the same job as the focal routine "Z*B" above
-The 2nd one multiplies by z = x+i.y the "imaginary part" of a bion i-e the coefficients of  e₁ , e₂ , e₃ , ............
-It's equivalent to  .002  XEQ 14  in the long program above.
 

082  "B"
02A "*"
01A "Z"
284  CLRF 7
03B  JNC+07
082  "B"
00D "M"
009  "I"
02A "*"
01A "Z"
288  SETF 7
248  SETF 9
321   ?NCXQ                   executes the initialization
3BC  EFC8                       routine listed in paragraph 1°)b)
378  C=c
03C  RCR 3
226   C=C+1 S&X
226   C=C+1 S&X
226   C=C+1 S&X
106   A=C S&X
0B0  C=N ALL
306   ?A<C S&X
0B5   ?NCGO
0A2   DATAERROR         There will be a DATA ERROR message if  R00 = 2n < 4
28C  ?FSET 7
023   JNC+04
23A  C=C+1 M
23A  C=C+1 M
070  N=C ALL
0F8  C=X
068  Z=C
0B8  C=Y
028  T=C
0B0  C=N ALL                  LOOP
268  Q=C
10E  A=C ALL
270  RAMSLCT
038  READATA
0AE  A<>C ALL
266  C=C-1 S&X
270  RAMSLCT
038  READATA
0EE  B<>C ALL
046  C=0 S&X
270  RAMSLCT
0EE  B<>C ALL
0E8  X=C
0AE  A<>C ALL
0A8  Y=C
259   ?NCXQ            The 1st executable word of my Z*Z routine is @F696 in my ROM
3D8   Z*Z                  Change these 2 words according to your own ROM
0B8  C=Y
10E  A=C ALL
0F8  C=X
0EE  B<>C ALL
278  C=Q
070  N=C ALL
270  RAMSLCT
0AE  A<>C ALL
2F0  WRITDATA
0B0  C=N ALL
260  SETHEX
266  C=C-1 S&X
070  N=C ALL
270  RAMSLCT
0EE  B<>C ALL
2F0  WRITDATA
046  C=0 S&X
270  RAMSLCT
0B0  C=N ALL
266  C=C-1 S&X
070  N=C ALL
106  A=C S&X
03C  RCR 3
306  ?A<C S&X
2B3  JNC-42d           GOTO LOOP
046  C
270  =
038  T
0A8  Y=C
078  C=Z
0E8  X=C
3E0  RTN

( 83 words )
 
 

 
-See above for a numerical example.
 

3°)  Polynomials with Complex Coefficients
 

BPVAL  evaluates  p(b) = c_m b^m + c_m-1 b^m-1 + ................ + c₁ b + c₀    for  a given bion b  and  (m+1) complex numbers  c_m , .............. , c₁ , c₀
 

-Store the bion b in registers  R01 thru R2n  as usual, and the coefficients of the polynomial in Rbb thru Ree  with  bb > 4n
 

Data Registers:           •  R00 = 2n > 3                    ( Registers R00 thru R2n & Rbb thru Ree are to be initialized before executing "BPVAL" )

                                      •  R01   ......  •  R2n = b
                                         Rn+1 ........   R4n = b   ( b is saved in R2n+1 thru R4n )

                                      •  Rbb = Re(c_m) ,  •  Rb+1 = Im(c_m) , ................. , •  Ree-1 = Re(c₀)   •  Ree = Im(c₀)

  >>>  When the program stops:     R01   ......  R2n = the n components of p(b)

Flags: /
Subroutine:    ST*A   "B*B1"   STO W   RCL W  ( where W is an extra register cf"A few M-Code Routines for the HP-41" )

-A variant that does not use register W is also listed below.
 
 

 
  ( 54 bytes )
 
 

   where  bbb.eee  is the control number of the polynomial    ( bbb > 4n )

Example:     b = ( 1 + 2 i ) + ( 3 + 4 i ) e₁ + ( 5 + 6 i ) e₂ + ( 7 + 8 i ) e₃                        ( biquaternion  ->  8  STO 00 )

  and     p(b) = ( 2 + 3 i ) b⁴ + ( 4 - 7 i ) b³ + ( 3 - 5 i ) a² + ( 1 + 4 i ) a + ( 6 + 2 i )

   1  STO 01   2  STO 02   3  STO 03  ...............................  8  STO 08

-If you choose  bb = 21

   2  STO 21  3  STO 22   4  STO 23   -7  STO 24   3  STO 25   -5  STO 26   1  STO 27   4  STO 28   6  STO 29   2  STO 30

-Control number =  21.030

    21.030   XEQ "BPVAL"  >>>>   1.008                                    ---Execution time = 11s---

    and you get in registers R01 thru R08

  p(b) = ( -39087 - 128086 i ) + ( -863 + 23966 i ) e₁ + ( 571 + 37456 i ) e₂ + ( 2005 + 50946 i ) e₃

Note:

-Here is another version without register W ( in this case, choose  bb > 4n+1 )

   X+1 may be replaced by  1  +
   X/E3 -------------------   E3  /
 
 

 
     ( 63 bytes )
 
 

   where  bbb.eee  is the control number of the polynomial    ( bbb > 4n+1 )

-Same example >>> same result.
 

4°)  Bionic Equations F(b) = b
 

-The following program again uses the same iterative method that is used for anionic equations:

-The equation must be rewritten in the form:    f ( b ) = b
  and if  f  satisfies a Lipschitz condition   | f(b) - f(b') | < h | b - b' |   with  h < 1  ,   provided b and b' are close to a solution,
  then the sequence  b_n+1 = f ( b_n )  converges to a root.
 

Data Registers:           •  R00 =  2n  ( 4 , 8 , .... )     ( Registers R00 thru R2n are to be initialized before executing "BEQ" )

                                      •  R01 to R2n = the n components of the bion  b          R4n+1 to R6n: temp  ,  R6n+1 = fname

                                          R2n+1 to R4n  may be used by the subroutine.

     >>>>     When the program stops,        R01 to R2n = the n components of  a solution  b

Flags: /
Subroutine:  A program that takes the bion b   in R01 thru R2n , calculates and stores  f ( b )  in R01 thru R2n
                       without disturbing R4n+1 to R6n+1

-Line 43  X/E3  may be replaced with   E3  /
 
 

 
        ( 70 bytes / SIZE 6n+2 )
 
 

    where  1.2n  = control number of the solution

Example:     Find a solution of the biquaternionic equation

   b²  - Ln b  + ( 0.1 + 0.2 i ) + ( 0.3 + 0.4 i ) e₁ + ( 0.5 + 0.6 i ) e₂ + ( 0.7 + 0.8 i ) e₃ = 0      near   b = 1

-First, we re-write this equation:    b  =  [ Ln b  - {  ( 0.1 + 0.2 i ) + ( 0.3 + 0.4 i ) e₁ + ( 0.5 + 0.6 i ) e₂ + ( 0.7 + 0.8 i ) e₃ } ] ^1/2  =  f ( b )

-The short routine "T" hereunder computes  f ( b )
 
 

 
-Then,  2n = 8    STO 00

-Initial approximation:     1  STO 01  CLX  STO 02  STO 03  STO 04  STO 05  STO 06  STO 07  STO 08

-Place the subroutine name in the alpha register:  alpha  "T"  alpha

    XEQ "BEQ"   >>>>    the successive approximations of the real part of the 1st component of the solution are displayed, and after several minutes, we get:

    X =  1.008 = control number of the solution.

-We find in registers R01 thru R08

  b = ( 1.014983106 + 0.292115635 i )  +  ( -0.307551610 - 0.118387777 i )  e₁
                                                               + ( -0.489251534 - 0.160080803 i )  e₂ + ( -0.670951458 - 0.201773829 i )  e₃

Notes:

-To avoid an infinite loop, replace line 33 by  E-8   X<Y?

-The convergence is very slow.
-If  f doesn't satisfy the required Lipschitz condition or if we choose a bad initial guess, the algorithm may be divergent.
-As usual, rewriting the equation in a proper form is often very difficult.

-If you need more registers for your function, replace 3 by a larger number like 4 ( lines 03-14-19-36 ) and replace line 10 by 14 or more.