restart:

# STEFFENSEN'S ALGORITHM 2.6

# To find a solution to g(x) = x

# given an initial approximation p0:

# INPUT: initial approximation p0: tolerance TOL:

# maximum number of iterations N0.

# OUTPUT: approximate solution p or

# a message that the method fails.

print(`This is Steffensens Method.\134n`):

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

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

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

G := unapply(G,x):

OK := FALSE:

print(`Input initial approximation\134n`):

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

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, `STEFFENSEN'S METHOD\134n`):

if FLAG = 2 then

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

fi:

# Step 1

I1 := 1:

OK := TRUE:

# Step 2

while I1 <= N0 and OK = TRUE do

# Step 3

# Compute P(1) with superscript (I-1)

P1 := G(P0):

# Compute P(2) with superscript (I-1)

P2 := G(P1):

if abs(P2-2*P1+P0) < 1.0e-20 then

FLAG := 1:

D1 := 10:

fprintf(OUP,`Denominator = 0, method fails\134n`):

fprintf(OUP,`best possible is P2(%2d) = %15.8f\134n`,I1,P2):

OK := FALSE:

else

D1 := (P1-P0)*(P1-P0)/(P2-2*P1+P0):

fi:

# Compute P(0) with superscript (I)

P := P0-D1:

if FLAG = 2 then

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

fi:

# Step 4

if abs(D1) < TOL then

# Procedure completed successfully

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

fprintf(OUP, `Number of iterations := %3d`, I1):

fprintf(OUP, ` Tolerance := %15.8e\134n`,TOL):

OK := FALSE:

else

# Step 5

I1 := I1+1:

# Step 6

# Update P0

P0 := P:

fi:

od:

if OK = TRUE then

# Step 7

# Procedure completed unsuccessfully

fprintf(OUP, `\134nIteration number %3d`, N0):

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

fprintf(OUP, `not within tolerance %14e\134n`,TOL):

fi:

if OUP <> default then

fclose(OUP):

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

fi:

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IiIj

STEFFENSEN'S METHOD I P 1 1.36526522e+00 2 1.36523001e+00 Approximate solution := 1.36523001 Number of iterations := 2 Tolerance := 5.00000000e-04

JSFH