restart:

# BISECTION ALGORITHM 2.1

# To find a solution to f(x) = 0 given the continuous function

# f on the interval [a,b], where f(a) and f(b) have

# opposite signs:

# INPUT: endpoints a,b: tolerance TOL:

# maximum number of iterations NO.

# OUTPUT: approximate solution p or

# a message that the algorithm fails.

print(`This is the Bisection Method.`):

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

print(`For example: cos(x)`):

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

F := unapply(F,x):

OK := FALSE:

while OK = FALSE do

print(`Input endpoints A < B separated by blank`):

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

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

if A > B then

X := A:

A := B:

B := X:

fi:

if A = B then

print(`A cannot equal B`):

else

FA := F(A):

FB := F(B):

if FA*FB > 0 then

print(`F(A) and F(B) have same sign`):

else

OK := TRUE:

fi:

od:

OK := FALSE:

while OK = FALSE do

print(`Input tolerance`):

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

if TOL <= 0 then

print(`Tolerance must be positive`):

else

OK := TRUE:

fi:

od:

OK := FALSE:

while OK = FALSE do

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

NO := scanf(`%d`)[1]: print(`Maximum number of iterations = `): print(NO):

if NO <= 0 then

print(`Must be positive integer`):

else

OK := TRUE:

fi:

od:

if OK = TRUE then

print(`Select output destination`):

print(`1. Screen`):

print(`2. Text file`):

print(`Enter 1 or 2`):

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

if FLAG = 2 then

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

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

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

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

else

OUP := default:

fi:

print(`Select amount of output`):

print(`1. Answer only`):

print(`2. All intermediate approximations`):

print(`Enter 1 or 2`):

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

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

if FLAG = 2 then

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

fi:

# Step 1

K := 1:

# Step 2

OK := TRUE:

while K <= NO and OK = TRUE do

# Step 3

# Compute P(I)

C := (B - A) / 2.0:

P := A + C:

# Step 4

FP := F(P):

if FLAG = 2 then

fprintf(OUP,`%3d %15.8e %15.7e \134n`,K,P,FP):

fi:

if abs(FP) < 1.0e-20 or C < TOL then

# Procedure completed successfully

fprintf(OUP,`\134nApproximate solution P = %11.8f \134n`,P):

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

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

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

OK := FALSE:

else

# Step 5

K := K+1:

# Step 6

# Compute A(I) and B(I)

if FA*FP > 0 then

A := P:

FA := FP:

else

B := P:

FB := FP:

fi:

od:

if OK = TRUE then

# Step 7

# Procedure completed unsuccessfully

fprintf(OUP,`\134nIteration number %3d`,NO):

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

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

fi:

if OUP <> default then

fclose(OUP):

printf(`Output file %s created successfully`,NAME):

fi:

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SUZJbnB1dH50aGV+ZnVuY3Rpb25+Rih4KX5pbn50ZXJtc35vZn54RzYi

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SUlJbnB1dH5lbmRwb2ludHN+QX48fkJ+c2VwYXJhdGVkfmJ5fmJsYW5rRzYi

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SVZJbnB1dH5tYXhpbXVtfm51bWJlcn5vZn5pdGVyYXRpb25zfi1+bm9+ZGVjaW1hbH5wb2ludEc2Ig==

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IiIj

Bisection Method I P F(P) 1 1.50000000e+00 2.3750000e+00 2 1.25000000e+00 -1.7968750e+00 3 1.37500000e+00 1.6210938e-01 4 1.31250000e+00 -8.4838867e-01 5 1.34375000e+00 -3.5098267e-01 6 1.35937500e+00 -9.6408842e-02 7 1.36718750e+00 3.2355780e-02 8 1.36328125e+00 -3.2149969e-02 9 1.36523438e+00 7.2030000e-05 10 1.36425781e+00 -1.6046697e-02 11 1.36474609e+00 -7.9892590e-03 Approximate solution P = 1.36474609 with F(P) = -0.00798926 Number of iterations = 11 Tolerance = 5.00000000e-04

JSFH