Arc Length of a Parametric Curve for the HP-41
Overview
1°) 3-Dimensional Problem
a) With Newton-Cotes Integration
b) Discrete Data
c) With Romberg's Method
2°) N-Dimensional Problem
a) With Newton-Cotes Integration
b) With Romberg's Method ( N <
10 )
Latest Update:
-The program in paragraph 1-a) has been improved to take into account
the cylindrical & spherical coordinates.
-In paragraph 1-b) "CRVL" may be used if we only know a finite number
of data points, equally spaced.
1°) 3-Dimensional Problem
a) With Newton-Cotes Integration
"CRVL" evaluates the integral §ab
[ ( dx/dt )2 + ( dy/dt )2 + ( dz/dt )2
] 1/2 dt with cartesian
coordinates ( CF 02-CF03 )
or §ab [ ( dr/dt )2 +
r2 ( dL/dt )2 + ( dz/dt )2 ] 1/2
dt
with cylindrical coordinates ( SF 02-CF 03 )
or §ab [ ( dr/dt )2 +
r2 ( Cos b )2 ( dL/dt )2 + r2
( db/dt )2 ] 1/2 dt
with spherical coordinates ( CF 02-SF 03 )
-The derivatives are estimated by the following formula of order 10
df/dx ~ (1/2520.h).[ 2100.( f1 - f-1 ) - 600.( f2 - f-2 ) + 150.( f3 - f-3 ) - 25.( f4 - f-4 ) + 2.( f5 - f-5 ) ]
( f(x+k.h) is denoted fk to simplify these expressions )
-And the integral is approximated by Newton-Cotes 7-point formula:
§x1x7 f(x).dx ~ h [ 41f(x1)+216f(x2)+27f(x3)+272f(x4)+27f(x5)+216f(x6)+41f(x7) ] / 140 with xi+1 - xi = h = Cste
-The intervall [a,b] may be divided into n parts to get more and more
accurate results.
Data Registers: • R00 = function name ( Registers R00 thru R03 are to be initialized before executing "CRVL" )
• R01 = a R04 = L
R06 to R19: temp
• R02 = b R05 = h ( to
estimate the derivatives )
• R03 = n
Flags: F02-F03
CF 02 CF 03 -> Cartesian coordinates
SF 02 CF 03 -> Cylindrical coordinates
CF 02 SF 03 -> Spherical coordinates
Subroutine: 1 program that takes t in X-register
and returns x(t) , y(t) , z(t) in registers X , Y , Z
respectively
or r(t) , L(t) , z(t) ---------------------------------
or r(t) , L(t) , b(t) ---------------------------------
-Line 28 is a three-byte GTO 01
01 LBL "CRVL"
02 STO 05 03 RCL 02 04 RCL 01 05 - 06 RCL 03 07 6 08 * 09 STO 09 10 / 11 STO 06 12 2 13 STO 10 14 STO 19 15 CLX 16 STO 04 17 600 18 CHS 19 STO 16 20 150 21 STO 17 22 14 23 * 24 STO 15 25 25 26 CHS 27 STO 18 28 GTO 01 29 LBL 00 30 RCL 06 31 RCL 09 32 * 33 RCL 01 |
34 +
35 STO 11 36 XEQ 02 37 CLX 38 SIGN 39 FC? 02 40 FS? 03 41 X=0? 42 GTO 00 43 D-R 44 FC? 43 45 ST* 13 46 SIGN 47 FS? 03 48 D-R 49 FC? 43 50 ST* 14 51 RCL 11 52 XEQ IND 00 53 ST* 13 54 FS? 03 55 ST* 14 56 X<> Z 57 COS 58 FS? 03 59 ST* 13 60 LBL 00 61 RCL 12 62 X^2 63 RCL 13 64 X^2 65 RCL 14 66 X^2 |
67 +
68 + 69 SQRT 70 RTN 71 LBL 02 72 5 73 STO 07 74 19 75 STO 08 76 CLX 77 STO 12 78 STO 13 79 STO 14 80 LBL 03 81 RCL 11 82 RCL 05 83 RCL 07 84 * 85 + 86 XEQ IND 00 87 RCL IND 08 88 ST* T 89 ST* Z 90 * 91 ST+ 12 92 RDN 93 ST+ 13 94 X<>Y 95 ST+ 14 96 RCL 11 97 RCL 05 98 RCL 07 99 * |
100 -
101 XEQ IND 00 102 RCL IND 08 103 ST* T 104 ST* Z 105 * 106 ST- 12 107 RDN 108 ST- 13 109 X<>Y 110 ST- 14 111 DSE 08 112 DSE 07 113 GTO 03 114 RTN 115 LBL 01 116 XEQ 00 117 82 118 RCL 10 119 / 120 * 121 ST+ 04 122 CLX 123 SIGN 124 ST- 09 125 STO 10 126 XEQ 00 127 216 128 * 129 ST+ 04 130 DSE 09 131 XEQ 00 132 27 |
133 *
134 ST+ 04 135 DSE 09 136 XEQ 00 137 272 138 * 139 ST+ 04 140 DSE 09 141 XEQ 00 142 27 143 * 144 ST+ 04 145 DSE 09 146 XEQ 00 147 216 148 * 149 ST+ 04 150 DSE 09 151 GTO 01 152 XEQ 00 153 41 154 * 155 RCL 04 156 + 157 RCL 06 158 * 159 RCL 05 160 / 161 352800 162 / 163 STO 04 164 END |
( 262 bytes / SIZE
020 )
STACK | INPUT | OUTPUT |
X | h | L |
Example1: x(t) = t2 ,
y(t) = exp(t) , z(t) = Ln t ; a = 1 ,
b = 2
01 LBL "T"
02 X^2 03 LASTX 04 E^X 05 LASTX 06 LN 07 RTN |
"T" ASTO 00 1 STO 01 2 STO
02 CF 02 CF 03
-If you choose n = 1 & h = 0.1 1 STO 03
0.1 XEQ "CRVL" >>>> L = 5.609133827 ---Execution time = 95s---
-With n = 2 - and the same h-value - it yields L = 5.609132375
-The exact result is L = 5.609132359...
Example2: r(t) = exp(t) , L(t)
= t , b(t) = t2 ; a = 0 ,
b = 1 ( all angles in radians )
01 LBL "T"
02 X^2 03 LASTX 04 ENTER^ 05 E^X 06 RTN |
XEQ "RAD" SF 02 CF 03
0 STO 01 1 STO 02
-With n = 1 & h = 0.1 1 STO 03
0.1 XEQ "CRVL" >>>> L = 2.645584611 ---Execution time = 84s---
-With n = 2 & h = 0.1 we get L = 2.645588402
-The exact value is 2.645588439...
Example3: r(t) = exp(t) , L(t)
= t2 , b(t) = t3 ;
a = 0 , b = 1 ( all angles in radians )
01 LBL "T"
02 ENTER^ 03 X^2 04 ST* Y 05 LASTX 06 E^X 07 RTN |
XEQ "RAD" CF 02 SF 03
0 STO 01 1 STO 02
-With n = 1 & h = 0.1 1 STO 03
0.1 XEQ "CRVL" >>>> L = 3.449957472 ---Execution time = 90s---
-With n = 2 & h = 0.1 we get L = 3.449386896
-The exact result is 3.449386684...
Notes:
-"CRVL" works in RAD mode & in DEG mode but not in GRAD mode
-The spherical coordinates that are used here are the radius "vector"
r , the longitude L and the latitude b
-If you prefer the co-latitude, replace line 57 COS by
SIN
b) Discrete Data
-The following variant calculates the same integrals with the Simpson's rule and the derivatives with approximate formulas of order 4.
-The coordinates of the N points are to be stored in Rbb thru Ree
( bb > 15 )
Warning: the number of points N must be odd and at least
5
Data Registers: R00: temp ( Registers Rbb thru Ree are to be initialized before executing "CRVL" )
R02 to R15: temp
• Rbb = x1
r1 r1
.................................. • Re-2 = xn
rn rn
• Rb+1 = y1 or L1
or L1 ..................................
• Re-1 = yn or Ln
or Ln bbb >
015
• Rb+2 = z1
z1 b1
................................. • Ree =
zn zn
bn
Flags: F02-F03
CF 02 CF 03 -> Cartesian coordinates
SF 02 CF 03 -> Cylindrical coordinates
CF 02 SF 03 -> Spherical coordinates
Subroutines: /
01 LBL "CRVL"
02 STO 02 03 STO 03 04 INT 05 DSE X 06 CHS 07 LASTX 08 FRC 09 E3 10 * 11 STO 01 12 STO 10 13 + 14 3 15 STO 15 16 / 17 4 18 STO 13 19 - 20 STO 00 21 9 22 STO 14 23 CLX 24 STO 04 25 STO 05 26 STO 06 27 STO 07 28 STO 08 29 STO 09 30 SIGN 31 STO 11 32 STO 12 33 FC? 02 34 FS? 03 35 D-R |
36 FC? 43
37 STO 11 38 SIGN 39 FS? 03 40 D-R 41 FC? 43 42 STO 12 43 25 44 XEQ 02 45 48 46 CHS 47 XEQ 02 48 36 49 XEQ 02 50 16 51 CHS 52 XEQ 02 53 XEQ 02 54 13 55 XEQ 03 56 X<> 01 57 STO 10 58 RCL 02 59 STO 03 60 CLX 61 STO 04 62 STO 05 63 STO 06 64 STO 07 65 STO 08 66 STO 09 67 3 68 XEQ 02 69 10 70 XEQ 02 |
71 18
72 CHS 73 XEQ 02 74 6 75 XEQ 02 76 SIGN 77 CHS 78 XEQ 02 79 10 80 XEQ 03 81 2 82 STO 09 83 X^2 84 STO 08 85 * 86 ST+ 01 87 GTO 01 88 LBL 02 89 SIGN 90 LBL 04 91 RCL IND 03 92 LASTX 93 * 94 ST- IND 13 95 RCL IND 10 96 LASTX 97 * 98 ST+ IND 14 99 1 100 ST+ 03 101 ST- 10 102 ST+ 13 103 ST- 14 104 DSE 15 105 GTO 04 |
106 3
107 ST- 13 108 ST+ 14 109 STO 15 110 RTN 111 LBL 03 112 ENTER 113 X<> 10 114 + 115 2 116 X<>Y 117 ST+ Y 118 RCL IND Y 119 COS 120 RCL IND Y 121 FS? 02 122 ST* 08 123 FS? 03 124 ST* 09 125 * 126 FS? 03 127 ST* 08 128 2 129 RCL 03 130 RCL 10 131 2 132 + 133 - 134 ST+ Y 135 RCL IND Y 136 COS 137 RCL IND Y 138 FS? 02 139 ST* 05 140 FS? 03 |
141 ST* 06
142 * 143 FS? 03 144 ST* 05 145 RCL 04 146 X^2 147 RCL 05 148 RCL 11 149 * 150 X^2 151 RCL 06 152 RCL 12 153 * 154 X^2 155 + 156 + 157 SQRT 158 RCL 07 159 X^2 160 RCL 08 161 RCL 11 162 * 163 X^2 164 + 165 RCL 09 166 RCL 12 167 * 168 X^2 169 + 170 SQRT 171 + 172 RTN 173 LBL 05 174 SIGN 175 LBL 06 |
176 RCL IND 02
177 LASTX 178 * 179 ST+ IND 13 180 1 181 ST+ 02 182 ST+ 13 183 DSE 15 184 GTO 06 185 3 186 ST- 13 187 STO 15 188 RTN 189 LBL 01 190 CLX 191 STO 04 192 STO 05 193 STO 06 194 SIGN 195 XEQ 05 196 8 197 CHS 198 XEQ 05 199 ST+ 02 200 8 201 XEQ 05 202 SIGN 203 CHS 204 XEQ 05 205 2 206 RCL 02 207 9 208 - 209 ST+ Y 210 RCL IND Y |
211 COS
212 RCL IND Y 213 FS? 02 214 ST* 05 215 FS? 03 216 ST* 06 217 * 218 FS? 03 219 ST* 05 220 RCL 04 221 X^2 222 RCL 05 223 RCL 11 224 * 225 X^2 226 RCL 06 227 RCL 12 228 * 229 X^2 230 + 231 + 232 SQRT 233 RCL 08 234 X<> 09 235 STO 08 236 * 237 ST+ 01 238 12 239 ST- 02 240 DSE 00 241 GTO 01 242 36 243 ST/ 01 244 RCL 01 245 END |
( 361 bytes / SIZE
var )
STACK | INPUT | OUTPUT |
X | bbb.eee | L |
With bbb > 015
Example1: x(t) = t2 , y(t) = exp(t) , z(t) = Ln t ; a = 1 , b = 2 but you only have the coordinates of 9 points corresponding to t = 1 , 9/8 , ... , 15/8 , 2
( you can store these coordinates with the following routine, for example
in R21 to R47 )
LBL 10
9 STO 00 47 STO 01 LBL 01 RCL 00 1 - 8 / 1 + LN STO IND 01 DSE 01 LASTX E^X STO IND 01 E^X DSE 01 LASTX X^2 STO IND 01 DSE 01 DSE 00 GTO 01 |
CF 02 CF 03 21.047 XEQ
"CRVL" >>>> L = 5.609125853
---Execution time = 66s---
-The exact result is L = 5.609132359...
Example2: r(t) = exp(t) , L(t) = t , b(t) = t2 ; a = 0 , b = 1 ( all angles in radians )
but you only have the coordinates of 9 points corresponding to t = 0 , 1/8 , ... , 7/8 , 1
( you can store these coordinates with the following routine, for example
in R21 to R47 )
LBL 10
9 STO 00 47 STO 01 LBL 01 RCL 00 1 - 8 / X^2 STO IND 01 DSE 01 LASTX STO IND 01 DSE 01 E^X STO IND 01 DSE 01 DSE 00 GTO 01 |
XEQ "RAD" SF 02 CF 03
21.047 XEQ "CRVL" >>>> L = 2.645598006 ---Execution time = 66s---
-The exact value is 2.645588439...
Example3: r(t) = exp(t) , L(t) = t2 , b(t) = t3 ; a = 0 , b = 1 ( all angles in radians )
but you only have the coordinates of 9 points corresponding to t = 0 , 1/8 , ... , 7/8 , 1
( you can store these coordinates with the following routine, for example
in R21 to R47 )
LBL 10
9 STO 00 47 STO 01 LBL 01 RCL 00 1 - 8 / X^2 LASTX * STO IND 01 DSE 01 LASTX X^2 STO IND 01 DSE 01 LASTX E^X STO IND 01 DSE 01 DSE 00 GTO 01 |
XEQ "RAD" CF 02 SF 03
21.047 XEQ "CRVL" >>>> L = 3.449756039 ---Execution time = 67s---
-The exact result is 3.449386684...
Notes:
-"CRVL" works in RAD mode & in DEG mode but not in
GRAD mode
-This routine does not check if the control number is correct !
-The spherical coordinates that are used here are the radius "vector"
r , the longitude L and the latitude b
-If you prefer the co-latitude, replace lines 119-136-211
by SIN
c) With Romberg Method
-This variant uses Pythagoreas theorem and Romberg method to evaluate
the integral
Data Registers: • R00 = function name ( Register R00 is to be initialized before executing "CRVL" )
R01 = a R03 = L R04
to R16......: temp
R02 = b
Flags: /
Subroutine: 1 program that takes t
in X-register and R07 and returns X(t) , Y(t) , Z(t) in registers
X , Y , Z respectively
-Line 96 is a three-byte GTO 01
01 LBL "CRVL"
02 STO 02 03 X<>Y 04 STO 01 05 1 06 STO 04 07 LBL 01 08 RCL 02 09 RCL 01 10 - 11 RCL 04 12 STO 05 13 / 14 STO 11 15 CLX 16 STO 06 17 RCL 02 18 STO 07 19 XEQ IND 00 20 STO 08 |
21 RDN
22 STO 09 23 X<>Y 24 STO 10 25 LBL 02 26 RCL 05 27 RCL 11 28 ST* Y 29 - 30 RCL 01 31 + 32 STO 07 33 XEQ IND 00 34 ENTER 35 X<> 08 36 - 37 X^2 38 X<>Y 39 ENTER 40 X<> 09 |
41 -
42 X^2 43 + 44 X<>Y 45 ENTER 46 X<> 10 47 - 48 X^2 49 + 50 SQRT 51 ST+ 06 52 DSE 05 53 GTO 02 54 RCL 06 55 SIGN 56 ST* X 57 RCL 04 58 ST+ 04 59 X#Y? 60 GTO 00 |
61 STO 14
62 LASTX 63 STO 16 64 GTO 01 65 LBL 00 66 4 67 STO 12 68 16 69 STO 13 70 RCL 14 71 STO 15 72 LASTX 73 ISG 14 74 LBL 03 75 ENTER 76 ENTER 77 X<> IND 13 78 - 79 RCL 12 80 4 |
81 ST* 12
82 SIGN 83 ST+ 13 84 - 85 / 86 + 87 DSE 15 88 GTO 03 89 STO 03 90 STO IND 13 91 VIEW X 92 RND 93 X<>Y 94 RND 95 X#Y? 96 GTO 01 97 RCL 03 98 END |
( 132 bytes / SIZE
016+??? )
STACK | INPUTS | OUTPUTS |
Y | a | / |
X | b | L(a,b) |
Example: X(t) = t4 ,
Y(t) = t2 , Z(t) = exp t ; a =
0 , b = 1
01 LBL "T"
02 E^X 03 LASTX 04 X^2 05 ENTER^ 06 X^2 07 RTN |
-If you try 9 decimals
T ASTO 00
FIX 9
0 ENTER^
1 XEQ "CRVL"
>>>> L = 2.342116456
---Execution time = 97s---
Notes:
-The HP41 displays the successive approximations.
-The result depends on the display format ( lines 92-94 )
-The exact result is L = 2.342116459....
2°) N-Dimensional Problem
a) With Newton-Cotes Integration
"CRVLN" evaluates the integral §ab
[ ( dX1/dt )2 + ( dX2/dt )2
+ ............ + ( dXN/dt )2 ] 1/2
dt
-The derivatives and the integral are estimated by the same formulas of order 10.
-The inervall [a,b] may be divided in m parts to get more and more accurate
results.
Data Registers: R00: temp
R01 = a R04 = h
R06 to R15: temp
R02 = b R05 = L
R03 = m
Flags: /
Subroutine: N programs named "X1"
"X2" ............. "XN" that take t in X-register and return
X1(t) , X2(t) , .......... , XN(t) respectively
in X-register
01 LBL "CRVLN"
02 "DIM=?" 03 PROMPT 04 STO 14 05 "H^M^A^B" 06 PROMPT 07 STO 02 08 X<>Y 09 STO 01 10 - 11 X<>Y 12 STO 03 13 6 14 * 15 STO 07 16 / 17 STO 06 18 X<>Y 19 STO 04 20 6 21 * 22 STO 09 23 FIX 0 24 CF 29 |
25 XEQ 00
26 41 27 * 28 STO 05 29 DSE 07 30 GTO 01 31 LBL 00 32 RCL 06 33 RCL 07 34 * 35 RCL 01 36 + 37 STO 11 38 CLX 39 STO 13 40 RCL 14 41 STO 15 42 LBL 04 43 "X" 44 ARCL 15 45 XEQ 02 46 X^2 47 ST+ 13 48 DSE 15 |
49 GTO 04
50 RCL 13 51 SQRT 52 RTN 53 LBL 02 54 ASTO 00 55 RCL 09 56 STO 10 57 XEQ 03 58 ST+ X 59 STO 08 60 XEQ 03 61 25 62 * 63 ST- 08 64 XEQ 03 65 150 66 * 67 ST+ 08 68 XEQ 03 69 600 70 * 71 ST- 08 72 XEQ 03 |
73 2100
74 * 75 RCL 08 76 + 77 RCL 04 78 2520 79 * 80 / 81 RTN 82 LBL 03 83 RCL 04 84 ST- 10 85 RCL 11 86 RCL 10 87 - 88 XEQ IND 00 89 STO 12 90 RCL 11 91 RCL 10 92 + 93 XEQ IND 00 94 RCL 12 95 - 96 RTN |
97 LBL 01
98 XEQ 00 99 216 100 * 101 ST+ 05 102 DSE 07 103 XEQ 00 104 27 105 * 106 ST+ 05 107 DSE 07 108 XEQ 00 109 272 110 * 111 ST+ 05 112 DSE 07 113 XEQ 00 114 27 115 * 116 ST+ 05 117 DSE 07 118 XEQ 00 119 216 120 * |
121 ST+ 05
122 DSE 07 123 CLX 124 XEQ 00 125 STO M 126 82 127 * 128 ST+ 05 129 DSE 07 130 GTO 01 131 CLX 132 X<> M 133 41 134 * 135 ST- 05 136 RCL 05 137 RCL 06 138 * 139 140 140 / 141 STO 05 142 FIX 9 143 SF 29 144 END |
( 251 bytes / SIZE 016 )
STACK | INPUT | OUTPUT |
X | / | L(a,b) |
Example: X1(t) = t4
, X2(t) = t2 , X3(t)
= t3 , X4(t) = exp t ;
a = 0 , b = 1
01 LBL "X1"
02 X^2 03 X^2 04 RTN 05 LBL "X2" 06 X^2 07 RTN 08 LBL "X3" 09 ENTER^ 10 X^2 11 * 12 RTN 13 LBL "X4" 14 E^X 15 RTN |
-With h = 0.1 and m = 1
XEQ "CRVLN" >>>> "DIM=?"
4 R/S "H^M^A^B"
0.1 ENTER^
1 ENTER^
0 ENTER^
1 R/S
>>>> L = 2.579075091
---Execution time = 180s---
-With m = 2 - and the same h-value - it yields L = 2.579079247
-The exact result is L = 2.579079231....
b) With Romberg Method
( n < 10 )
-Like the 2nd version of "CRVL", this variant of "CRVLN" uses an extrapolation
to the limit.
Data Registers: R00 = n
R01 = a R03 = L R04
to R26......: temp
R02 = b
Flags: /
Subroutine: N programs named "X1"
"X2" ....... "XN" that take t in X-register & R07 and return
X1(t) , X2(t) , ....... , XN(t) respectively
in X-register
-Line 115 is a three-byte GTO 01
01 LBL "CRVLN"
02 STO 02 03 RDN 04 STO 01 05 X<>Y 06 STO 00 07 1 08 STO 04 09 RCLFLAG 10 STO 20 11 LBL 01 12 FIX 00 13 CF 29 14 RCL 00 15 STO 18 16 17 17 STO 19 18 LBL 02 19 "X" 20 ARCL 18 21 ASTO L 22 RCL 02 23 STO 07 24 XEQ IND L |
25 STO IND 19
26 DSE 19 27 DSE 18 28 GTO 02 29 CLX 30 STO 06 31 RCL 02 32 RCL 01 33 - 34 RCL 04 35 STO 05 36 / 37 STO 08 38 LBL 03 39 RCL 05 40 RCL 08 41 ST* Y 42 - 43 RCL 01 44 + 45 STO 07 46 RCL 00 47 STO 18 48 17 |
49 STO 19
50 CLX 51 STO 21 52 LBL 04 53 "X" 54 ARCL 18 55 ASTO L 56 RCL 07 57 XEQ IND L 58 ENTER 59 X<> IND 19 60 - 61 X^2 62 ST+ 21 63 DSE 19 64 DSE 18 65 GTO 04 66 RCL 21 67 SQRT 68 ST+ 06 69 DSE 05 70 GTO 03 71 RCL 20 72 STOFLAG |
73 RCL 06
74 SIGN 75 ST* X 76 RCL 04 77 ST+ 04 78 X#Y? 79 GTO 00 80 STO 24 81 LASTX 82 STO 26 83 GTO 01 84 LBL 00 85 4 86 STO 22 87 26 88 STO 23 89 RCL 24 90 STO 25 91 LASTX 92 ISG 24 93 LBL 05 94 ENTER 95 ENTER 96 X<> IND 23 |
97 -
98 RCL 22 99 4 100 ST* 22 101 SIGN 102 ST+ 23 103 - 104 / 105 + 106 DSE 25 107 GTO 05 108 STO 03 109 STO IND 23 110 VIEW X 111 RND 112 X<>Y 113 RND 114 X#Y? 115 GTO 01 116 RCL 03 117 END |
( 184 bytes / SIZE
026+??? )
STACK | INPUTS | OUTPUTS |
Z | n < 10 | / |
Y | a | / |
X | b | L(a,b) |
Example: X1(t) = t4
, X2(t) = t2 , X3(t)
= t3 , X4(t) = exp t ;
a = 0 , b = 1
01 LBL "X1"
02 X^2 03 X^2 04 RTN 05 LBL "X2" 06 X^2 07 RTN 08 LBL "X3" 09 ENTER^ 10 X^2 11 * 12 RTN 13 LBL "X4" 14 E^X 15 RTN |
-If you try to find 9 decimals
FIX 9
4 ENTER^
0 ENTER^
1 XEQ "CRVLN" >>>>
L = 2.579079231
---Execution time = 4m16s---
Notes:
-The HP41 displays the successive approximations.
-The result depends on the display format ( lines 111-113 )
-The exact result is L = 2.579079231....
-So, there is no roundoff error in this example ( this is not always
the case... ).