0
Bisection Method:
Begin with the interval [0, pi/2]. The midpoint of the interval is x1 = pi/4. Calculate the value of the function at x1: f(x1) = pi/4 - cos(pi/4). Since f(x1) > 0, the solution must be in the interval [0, pi/4]. Now consider the midpoint of this interval, x2 = pi/8. Calculate the value of the function at x2: f(x2) = pi/8 - cos(pi/8). Since f(x2) < 0, the solution must be in the interval [pi/8, pi/4]. Now consider the midpoint of this interval, x3 = 3pi/16. Calculate the value of the function at x3: f(x3) = 3pi/16 - cos(3pi/16). Since f(x3) > 0, the solution must be in the interval [pi/8, 3pi/16]. Continue this process, calculating the midpoint of the interval and the value of the function at the midpoint, until the difference between the lower and upper bounds of the interval is less than or equal to the error of 0.005.
Newton’s Method:
Try an initial guess of x0 = 1. Calculate the value of the function at x0: f(x0) = 1 - cos(1). Calculate the derivative of the function at x0: f'(x0) = 1 + sin(1). Calculate the next x-value using the Newton’s method formula: x1 = x0 - f(x0)/f'(x0) = 1 - (1 - cos(1))/(1 + sin(1)) = 0.6247. Calculate the value of the function at x1: f(x1) = 0.6247 - cos(0.6247). Calculate the derivative of the function at x1: f'(x1) = 1 + sin(0.6247). Calculate the next x-value using the Newton’s method formula: x2 = x1 - f(x1)/f'(x1) = 0.6247 - (0.6247 - cos(0.6247))/(1 + sin(0.6247)) = 0.739. Continue this process until the difference between two successive x-values is less than or equal to the error of 0.005.
Secant Method:
Start with two initial x-values, x0 = 0 and x1 = 1. Calculate the value of the function at x0 and x1: f(x0) = 0 - cos(0) = 0, f(x1) = 1 - cos(1). Calculate the next x-value using the Secant method formula: x2 = x1 - f(x1)(x1 - x0)/(f(x1) - f(x0)) = 1 - (1 - cos(1))(1 - 0)/(1 - cos(1) - 0) = 0.6247. Calculate the value of the function at x2: f(x2) = 0.6247 - cos(0.6247). Calculate the next x-value using the Secant method formula: x3 = x2 - f(x2)(x2 - x1)/(f(x2) - f(x1)) = 0.6247 - (0.6247 - cos(0.6247))(0.6247 - 1)/(0.6247 - cos(0.6247) - 1) = 0.7396. Continue this process until the difference between two successive x-values is less than or equal to the error of 0.005.
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FranWe start with the interval
[
�
,
�
]
=
[
0
,
�
/
2
]
[a,b]=[0,π/2] since
�
(
0
)
=
−
1
f(0)=−1 and
�
(
�
/
2
)
=
�
/
2
−
1
>
0
f(π/2)=π/2−1>0, indicating a root between
0
0 and
�
/
2
π/2.
Iteration 1:
�
=
(
0
�
/
2
)
/
2
=
�
/
4
c=(0+π/2)/2=π/4,
�
(
�
/
4
)
=
�
/
4
−
cos
(
�
/
4
)
≈
0.553
f(π/4)=π/4−cos(π/4)≈0.553, which is positive. So, we set
[
�
,
�
]
=
[
0
,
�
/
4
]
[a,b]=[0,π/4].
Iteration 2:
�
=
(
�
/
4
0
)
/
2
=
�
/
8
c=(π/4+0)/2=π/8,
�
(
�
/
8
)
=
�
/
8
−
cos
(
�
/
8
)
≈
0.132
f(π/8)=π/8−cos(π/8)≈0.132, still positive. So,
[
�
,
�
]
=
[
0
,
�
/
8
]
[a,b]=[0,π/8].
Iteration 3:
�
=
(
�
/
8
0
)
/
2
=
�
/
16
c=(π/8+0)/2=π/16,
�
(
�
/
16
)
=
�
/
16
−
cos
(
�
/
16
)
≈
−
0.191
f(π/16)=π/16−cos(π/16)≈−0.191, which is negative. So,
[
�
,
�
]
=
[
�
/
16
,
�
/
8
]
[a,b]=[π/16,π/8].
We continue the bisection iterations until the width of the interval is less than
0.005
0.005. The final result is approximately
�
≈
0.209
x≈0.209.
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