Question

R2240 Computation Methods Name: The function: y = 5 x" has a root between 1 and 3 Calculate the root using the secant method ex/2 (x +2) r 2 has a root between 1 and 3. a. Use 1.0000 as the initial guess. b. c. Use a perturbation (8) of o.020 for secant calculation Make three iterations d. Sho w all work (each section/step) below for full credit. 2 1.0000 1.747868 1.0100 0.9900

Answer 1

Dear friend,

Assuming the modified secant method updates the x_{t+1} accoding to following formula
x_{t+1} = x_{t} - Y(x_{t})*(delta/( Y(x_{t}+delta) - Y(x_{t})))

I have written follwoing python code for you.

===============

import math

def Y(x):    
    V = (5 * math.exp(x/2))/(x+2) -x*x
    return V

delta=0.02
x=1

print("i x      Y(x) (x+delta/2) (x-delta/2) Y(x+delta/2) Y(x-delta/2)");
for i in range(0,5):
   print(str(i)+" "+str("{0:.4f}".format(x))+" "+str("{0:.4f}".format(Y(x)))+" "+str("{0:.4f}".format(x+delta/2))+"      "+str("{0:.4f}".format(x-delta/2))+"      "+str("{0:.4f}".format(Y(x+delta/2)))+"     "+str("{0:.4f}".format(Y(x-delta/2))));
   x = x - delta*(Y(x)/( Y(x+delta) - Y(x))) #Modify here if required

=============================

On execution I get following output

----------------------------------------------

i x      Y(x) (x+delta/2) (x-delta/2) Y(x+delta/2) Y(x-delta/2)
0 1.0000 1.7479 1.0100      0.9900      1.7324     1.7632
1 2.1217 -0.9973 2.1317      2.1117      -1.0308     -0.9639
2 1.8247 -0.0740 1.8347      1.8147      -0.1028     -0.0454
3 1.7990 -0.0009 1.8090      1.7890      -0.0293     0.0273
4 1.7987 -0.0000 1.8087      1.7887      -0.0284     0.0282

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Please fill the values in your table.

In case a fifferent function to update the x_{t+1} is explained in your class you can easity modify the code. I have written " #Modify here if required" to let you know where to modify. Re run the code and fill in the table.

Cheers!