Question

   Assignment ID is p1

NOTE: Algebraic expressions follow FORTRAN conventions.
Use full calculator precision for intermediate values.


Use the bisection method with the function defined by:

A=(X + 1.62) / 5.39
B=(X - 3.77) / 5.39
P1=((A-1)**2)*(1+2*A)
P2=((B+1)**2)*(1-2*B)
F(X) = (P1 * (-1.24) ) + (P2 * 4.13)

Start with interval (Xleft,Xright) = (-1.62, 3.77)
The function values at these end points are
F(Xleft) = -1.24
F(Xright) = 4.13

The new approximation interval bracketing the root
after ONE bisection is (______1______,______2______).
Function values at these end points are: ______3______, and 1.445

The new approximation interval bracketing the root
after ONE MORE bisection is (______4______,______5______).
Function values at these end points are: -0.400937 and ______6______.

The new approximation interval bracketing the root
after ONE MORE bisection is (______7______,______8______).

If Xmid satisfies the convergence criterior |f(Xmid)|<=0.00001,
then the root Xmid is ______9______.

ANSWER SECTION:
SELECTIONS FOR BLANK NUMBER 1
(a) 0.73
(b) 1.075
(c) -1.62
(d) -0.21
(e) -0.68
SELECTIONS FOR BLANK NUMBER 2
(a) 1.075
(b) 3.77
(c) -0.235
(d) -0.895
(e) -1.62
SELECTIONS FOR BLANK NUMBER 3
(a) 1.445
(b) -3.23
(c) -1.24
(d) -3.89
(e) 2.07
SELECTIONS FOR BLANK NUMBER 4
(a) 1.075
(b) -0.2725
(c) -2.3525
(d) 1.3875
(e) -1.1025
SELECTIONS FOR BLANK NUMBER 5
(a) 3.525
(b) 2.4225
(c) -0.2725
(d) -0.145
(e) 1.075
SELECTIONS FOR BLANK NUMBER 6
(a) 1.445
(b) 3.29094
(c) 0.575
(d) -0.745
(e) -0.400937
SELECTIONS FOR BLANK NUMBER 7
(a) 1.075
(b) 0.40125
(c) 1.2775
(d) 0.6575
(e) -0.2725
SELECTIONS FOR BLANK NUMBER 8
(a) -1.41875
(b) 1.74875
(c) 0.40125
(d) 2.68125
(e) -0.2725
SELECTIONS FOR BLANK NUMBER 9
(a) 1.8501
(b) 1.4001
(c) 1.74875
(d) -0.829902
(e) 0.0600983

Answer 1

Given,

A=(X + 1.62) / 5.39
B=(X - 3.77) / 5.39
P1=((A-1)**2)*(1+2*A)
P2=((B+1)**2)*(1-2*B)
F(X) = (P1 * (-1.24) ) + (P2 * 4.13)

As, we know in FORTRAN notation ' ** ' implies exponentiation operator, Thus we can re-write the given equations
as,
A=(X + 1.62) / 5.39
B=(X - 3.77) / 5.39
P1=((A-1)^2)*(1+2*A)
P2=((B+1)^2)*(1-2*B)
F(X) = (P1 * (-1.24) ) + (P2 * 4.13)

By re-writing the equations P1 and P2 , i.e., by substituting A,B in P1 and P2 respectively, we obtain,

Now by substituting these P1 and P2 in F(X) we obtain the equation:

Thus Start with interval (Xleft,Xright) = (-1.62, 3.77)
The function values at these end points are
F(Xleft) = -1.24 (Given), which can be checked by substituting X=-1.62 in F(X), i.e, F(-1.62)= -1.24
F(Xright) = 4.13 (Given), which can be checked by substituting X=3.77 in F(X), i.e, F(3.77)= 4.13
Thus we can observe that F(X) is an increasing function, and the solution for F(X)=0 lies in the given starting interval (-1.62, 3.77) for sure.

If we bisect the given initial interval (Xleft,Xright)=(-1.62, 3.77), into two bisections, First part is (Xleft,X1)=(-1.62,1.075) and the other part is (X1,Xright)=(1.075,3.77).
If we find the F(X) value at X1=1.075, i.e., the bisecting point,
F(X1)=F(1.075)=1.445, thus making clear that solution for F(X)=0 lies in the first bisection (Xleft,X1)=(-1.62,1.075).
Thus,The new approximation interval bracketing the root after ONE bisection is (-1.62,1.075)
Function values at these end points are: -1.24 and 1.445

Now again if we bisect our new interval (Xleft,Xright)=(-1.62,1.075) into two bisections, we get First part as (Xleft,X2)=(-1.62,-0.2725) and the other part is (X2,X1)=(-0.2725,1.075).
If we find the F(X) value at X2=-0.2725, i.e., the bisecting point,
F(X2)=F(-0.2725)=-0.4009375, thus making clear that solution for F(X)=0 lies in the second bisection of this iteration (X2,X1)=(-0.2725,1.075).
Thus,The new approximation interval bracketing the root after ONE MORE bisection is (-0.2725,1.075)
Function values at these end points are: -0.4009375 and 1.445

Now again if we bisect our new interval (X2,X1)=(-0.2725,1.075) into two bisections, we get First part as (X2,X3)=(-0.2725,0.40125) and the other part is (X3,X1)=(0.40125,1.075).
If we find the F(X) value at X3=0.40125, i.e., the bisecting point,
F(X3)=F(0.40125)=0.4591015625, thus making clear that solution for F(X)=0 lies in the first bisection of this iteration (X2,X3)=(-0.2725,0.40125).
Thus,The new approximation interval bracketing the root after ONE MORE bisection is (-0.2725,0.40125)
Function values at these end points are: -0.4009375 and 0.4591015625

Thus if we go on bisecting the intervals on iterations we get as follows,
Next bisection: (-0.2725,0.064375)
Next bisection: (-0.1040625,0.064375)
Next bisection: (-0.01984375,0.064375)
Next bisection:(0.0222656225,0.064375)
Next bisection:(0.0433203125,0.064375)
Thus keeping on iterating increases the accuracy, and if we observe the last bisection we can understand that the solution for F(X)=0 lies in (0.0433203125,0.064375). Precise MATLAB code gave the answer of X=0.060102443099022 such that F(X)=0, From options for question.9 we can understand that only 0.0600983 lies in our interval and is closest to the derived answer.

SELECTIONS FOR BLANK NUMBER 1: (c) -1.62
SELECTIONS FOR BLANK NUMBER 2: (a) 1.075
SELECTIONS FOR BLANK NUMBER 3: (c) -1.24
SELECTIONS FOR BLANK NUMBER 4: (b) -0.2725
SELECTIONS FOR BLANK NUMBER 5: (e) 1.075
SELECTIONS FOR BLANK NUMBER 6: (a) 1.445
SELECTIONS FOR BLANK NUMBER 7: (e) -0.2725
SELECTIONS FOR BLANK NUMBER 8: (c) 0.40125
SELECTIONS FOR BLANK NUMBER 9: (e) 0.0600983