restart:

# SECANT ALGORITHM 2.4

# To find a solution to the equation f(x) = 0

# given initial approximations p0 and p1:

# INPUT: initial approximation p0, p1: tolerance TOL:

# maximum number of iterations N0.

# OUTPUT: approximate solution p or

# a message that the algorithm fails.

:print(`This is the Secant Method\134n`):

:print(`Input the function F(x) in terms of x\134n`):

:print(`For example: cos(x)\134n`):

F := scanf(`%a`)[1]:print(F):

F := unapply(F,x):

OK := FALSE:

while OK = FALSE do

:print(`Input initial approximations P0 and P1 separated by a blank\134n`):

P0 := scanf(`%f`)[1]:print(`P0 = `):print(P0):

P1 := scanf(`%f`)[1]:print(`P1 = `):print(P1):

if P0 = P1 then

:print(`P0 cannot equal P1\134n`):

else

OK := TRUE:

fi:

od:

OK := FALSE:

while OK = FALSE do

:print(`Input tolerance\134n`):

TOL := scanf(`%f`)[1]:print(`Tolerance = `):print(TOL):

if TOL <= 0 then

:print(`Tolerance must be positive\134n`):

else

OK := TRUE:

fi:

od:

OK := FALSE:

while OK = FALSE do

:print(`Input maximum number of iterations - no decimal point\134n`):

N0 := scanf(`%d`)[1]:print(`Maximum number of iterations = `):print(N0):

if N0 <= 0 then

:print(`Must be positive integer\134n`):

else

OK := TRUE:

fi:

od:

if OK = TRUE then

:print(`Select output destination\134n`):

:print(`1. Screen\134n`):

:print(`2. Text file\134n`):

:print(`Enter 1 or 2\134n`):

FLAG := scanf(`%d`)[1]:print(`Entry = `):print(FLAG):

if FLAG = 2 then

:print(`Input the file name in the form - drive:\134\134name.ext\134n`):

:print(`For example: A:\134\134OUTPUT.DTA\134n`):

NAME := scanf(`%s`)[1]:

OUP := fopen(NAME,WRITE,TEXT):

else

OUP := default:

fi:

:print(`Select amount of output\134n`):

:print(`1. Answer only\134n`):

:print(`2. All intermediate approximations\134n`):

:print(`Enter 1 or 2\134n`):

FLAG := scanf(`%d`)[1]:print(`Entry = `):print(FLAG):

fprintf(OUP, `Secant Method\134n`):

if FLAG = 2 then

fprintf(OUP, ` I P F(P)\134n`):

fi:

# Step 1

I1 := 2:

F0 := F(P0):

F1 := F(P1):

OK := TRUE:

# Step 2

while I1 <= N0 and OK = TRUE do

# Step 3

# Compute P(I)

P := P1-F1*(P1-P0)/(F1-F0):

FP := F(P):

if FLAG = 2 then

fprintf(OUP,`%3d %15.8e %15.8e\134n`,I1,P,FP):

fi:

# Step 4

if abs(P-P1) < TOL then

# Procedure completed successfully

fprintf(OUP,`\134nApproximate solution P = %12.8f\134n`,P):

fprintf(OUP,`with F(P) = %12.8f\134n`,FP):

fprintf(OUP,`Number of iterations = %d\134n`,I1):

fprintf(OUP,`Tolerance = %14.8e\134n`,TOL):

OK := FALSE:

# Step 5

else

I1 := I1+1:

# Step 6

# Update P0, F0, P1, F1

P0 := P1:

F0 := F1:

P1 := P:

F1 := FP:

fi:

od:

if OK = TRUE then

# Step 7

# Procedure completed unsuccessfully

fprintf(OUP,`\134nIteration number %d`,N0):

fprintf(OUP,` gave approximation %12.8f\134n`,P):

fprintf(OUP,`with F(P) = %12.8f not within tolerance %15.8e\134n`,FP,TOL):

fi:

if OUP <> default then

fclose(OUP):

:print(`Output file `,NAME,` created successfully`):

fi:

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IiIj

Secant Method I P F(P) 2 7.36384139e-01 4.51771860e-03 3 7.39058139e-01 4.51772000e-05 4 7.39085149e-01 -2.69000000e-08 Approximate solution P = 0.73908515 with F(P) = -0.00000003 Number of iterations = 4 Tolerance = 5.00000000e-04

JSFH