HP 11C real root finder [Newton Method]

[Carlos CM (Mexico)]

LBL 0
RCL 1
GSB 1
RCL 1
GSB 2
/
CHS
RCL 1
+
STO 3
RCL 1
-
ABS
RCL 2
X>Y?
RTN
RCL 3
STO 1
GTO 0

LBL 1
f(x)=0 code

LBL 2
f'(x)=0 code

R1: old value (or initial value)
R2: Tolerance of error (0.001, 0.0001, 0.000000001 etc)

THE RESULT:
R3: new value (root)

EXAMPLE

Find the root of f(x)=x^3 - 3x^2-6x+8

rewritten as f(x)= [(x-3)x-6]x +8

f(x) code:

LBL1
ENTER
ENTER
ENTER
3
-
*
6
-
*
8
+
RTN


f'(x)= 3x(x-6)-6
f'(x) code

LBL 2
ENTER
ENTER
6
-
*
3
*
6
-
RTN

The roots are [-2, 1, 4]

give a initial value, for example: -3

-3 STO 1

give a tolerance of error, for example: 0.0001

0.0001 STO 2

BEGIN THE PROGRAM...

GSB 0

runnning...

when the error < TOL the program stop and display:

0.0001

you can find the root in the register 3

RCL 3

DISPLAY: -2.0001

try with initial value = 2

















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