CARDENAS, RICO JAY B.
ID No. 19-22711
BSCE-3A
Using shift solve function of my calculator and putting the nearest value of x equals to -1,
I assume that my expected value for the equation is -1.067.
2. Let y= cos (2x)-x/2
1st iteration: x1 = -2,
y1 = 0.346
x2 = -1,
y2 = 0.84
𝑥𝑟 = −1 −
xr = -0.681
y1 (yr) =
(Focus on root in between -2.0 and -1.0)
0.84(−2 − (−1))
0.346 − (0.84)
Thus, yr = 0.548
0.190> 0
; the root lies in between x2 and xr
CARDENAS, RICO JAY B.
ID No. 19-22711
BSCE-3A
2nd iteration: x1 = -0.681,
x2 = -1,
𝑥𝑟 = −1 −
y2 = 0.84
0.84(−0.681 − (−1))
0.548 − (0.84)
xr = -1.058
y1 (yr) =
y1 = 0.548
Thus, yr =0.011
0.006> 0
; the root lies in between x2 and xr
3rd iteration: x1 = -1.058,
x2 = -1,
𝑥𝑟 = −1 −
y2 = 0.84
0.84(−1.058 − (−1))
0.011 − (0.84)
xr = -1.066
y1 (yr) =
y1 = 0.011
Thus, yr =0.001
0.00001> 0 ; the root lies in between x2 and xr
4th iteration: x1 = -1.066,
x2 = -1,
𝑥𝑟 = −1 −
y2 = 0.84
0.84(−1.066 − (−1))
0.001 − (0.84)
xr = -1.067
y1 (yr) =
y1 = 0.001
Thus, yr =0.000
0.00001> 0 ; the root lies in between x2 and xr
SUMMARY OF FPM ON ROOT BETWEEN -2 AND -1
Iteration
𝑥1
𝑦1
𝑥2
𝑦2
𝑥𝑟
𝑦𝑟
𝑦1 𝑦𝑟
1.000
-2.000
0.346
-1.000
0.084
-0.681
0.548
0.190
2.000
-0.681
0.548
-1.000
0.084
-1.058
0.011
0.006
3.000
-1.058
0.011
-1.000
0.084
-1.066
0.001
0.00001
4.000
-1.066
0.001
-1.000
0.084
-1.067
0.000
0.000
ROOT 2= -1.067
ROOT = -1.067
CARDENAS, RICO JAY B.
ID No. 19-22711
BSCE-3A
Again, using shift solve function of my calculator and putting the nearest value of x equals to
-2, I assume that my expected value for the equation is -1.798.
Let y= cos (2x)-x/2
1st iteration: x1 = -3,
x2 = -2,
𝑥𝑟 = −2 −
xr = -1.836
y1 (yr) =
y1 = 2.460
y2 = 0.346
0.346(−3 − (−2))
2.460 − (0.346)
Thus, yr = 0.056
0.137> 0
; the root lies in between x2 and xr
2nd iteration: x1 = -1.836,
x2 = -2,
𝑥𝑟 = −2 −
xr = -1.805
y1 (yr) =
0.346(−1.836 − (−2))
0.056 − (0.346)
Thus, yr = 0.010
0.137> 0
x2 = -2,
𝑥𝑟 = −2 −
y1 (yr) =
y1 = 0.056
y2 = 0.346
; the root lies in between x2 and xr
3rd iteration: x1 = -1.805,
xr = -1.799
(Focus on root in between -3.0 and -2.0)
y1 =0.010
y2 = 0.346
0.346(−1.805 − (−2))
0.010 − (0.346)
Thus, yr = 0.0020
0.00002> 0 ; the root lies in between x2 and xr
CARDENAS, RICO JAY B.
ID No. 19-22711
BSCE-3A
4th iteration: x1 = -1.799,
x2 = -2,
𝑥𝑟 = −2 −
y2 = 0.346
0.346(−1.799, −(−2))
0.002 − (0.346)
xr = -1.798
y1 (yr) =
y1 = yr = 0.0020
Thus, yr = 0.00
0.0000> 0
; the root lies in between x2 and xr
SUMMARY OF FPM ON ROOT BETWEEN -3 AND -2
𝑥1
Iteration
𝑦1
𝑥2
𝑦2
𝑥𝑟
𝑦𝑟
𝑦1 𝑦𝑟
1.000
-3.000
2.460
-2.000
0.346
-1.836
0.056
0.137
2.000
-1.836
0.056
-2.000
0.346
-1.805
0.010
0.001
3.000
-1.805
0.010
-2.000
0.346
-1.799
0.002
0.00002
4.000
-1.799
0.002
-2.000
0.346
-1.798
0.000
0.000
ROOT 2= -1.798
ROOT = -1.798
Again, using shift solve function of my calculator and putting the nearest value of x equals to
1, I assume that my expected value for the equation is 0.626.
Let y= cos (2x)-x/2
1st iteration: x1 = 0,
x2 = 1,
𝑥𝑟 = −1 −
xr = 0.522
y1 (yr) =
y1 = 1
(Focus on root in between 0 and 1.0)
y2 = -0.916
−0.916(−0 − (1))
1 − (−0.916 )
Thus, yr = 0.242
0.242> 0
; the root lies in between x2 and xr
CARDENAS, RICO JAY B.
ID No. 19-22711
BSCE-3A
2nd iteration: x1 = 0.522,
x2 = 1,
𝑥𝑟 = −1 −
y2 = -0.916
−0.916(0.522 − (1))
0.242 − (−0.916)
xr = 0.622
y1 (yr) =
y1 =0.242
Thus, yr = 0.010
0.003> 0
; the root lies in between x2 and xr
3rd iteration: x1 = 0.622,
x2 = 1,
𝑥𝑟 = −1 −
y2 = -0.916
−0.916(0.622 − (1))
0.010 − (−0.916)
xr = 0.626
y1 (yr) =
y1 =0.010
Thus, yr = 0.0002
0.000002> 0 ; the root lies in between x2 and xr
SUMMARY OF FPM ON ROOT BETWEEN 0 AND 1
Iteration
𝑥1
𝑦1
𝑥2
𝑦2
𝑥𝑟
𝑦𝑟
𝑦1 𝑦𝑟
1.000
0.000
1.000
1.000
-0.916
0.522
0.242
0.242
2.000
0.522
0.242
1.000
-0.916
0.622
0.010
0.003
3.000
0.622
0.010
1.000
-0.916
0.626
0.0002
0.000002
ROOT 2= 0.626
ROOT = 0.626
CARDENAS, RICO JAY B.
ID No. 19-22711
2. Using Newton-Raphson Method (NRM)
For my initial guess I will use -2 to be the value of x to find the real root
Solution:
Plot the function f(x) = Cos(2x)-x/2
2. Determine the first derivative the function
y = Cos(2x)-x/2
y’= -2 Sin (2x)-1/2
Perform NRM
1st iteration: try 𝑥1 = -2.0
y = Cos(2x)-x/2
y’ =-2 Sin (2x)-1/2
For 𝑥2
𝑥2 = −2 −
. 346
−2.014
Therefore 𝒙𝟐 = -1.828
BSCE-3A
CARDENAS, RICO JAY B.
ID No. 19-22711
BSCE-3A
2nd iteration: try 𝑥1 = -1.828
y = Cos(2x)-x/2
y’ =-2 Sin (2x)-1/2
For 𝑥2
𝑥2 = −1.828 −
0.043
−1.484
Therefore 𝒙𝟐 = -1.799
3rd iteration: try 𝑥1 = -1.799
y = Cos(2x)-x/2
y’ =-2 Sin (2x)-1/2
For 𝑥2
𝑥2 = −1.799 −
0.002
−1.381
Therefore 𝒙𝟐 = -1.798
SUMMARY OF NRM
iteration
X1
f(x1)
f'(x)
X2
%Error
1
-2.000
0.346
-2.014
-1.828
1.309
2
-1.828
0.043
-1.484
-1.799
0.032
3
-1.799
0.002
-1.381
-1.798
-0.015
Therefore, the real roots for the equation
f(x)= Cos(2x)-x/2
Using NRM method is -1.789
CARDENAS, RICO JAY B.
ID No. 19-22711
SOLVING FOR THE 2ND ROOT
For my initial guess I will use -1.2 to be the value of x to find the real root
1. Determine the first derivative the function
y = Cos(2x)-x/2
y’= -2 Sin (2x)-1/2
Perform NRM
1st iteration: try 𝑥1 = -1.20
y = Cos(2x)-x/2
y’ =-2 Sin (2x)-1/2
For 𝑥2
𝑥2 = −1.2 −
−.137
. 851
Therefore 𝒙𝟐 = -1.039
2nd iteration: try 𝑥1 = -1.039
y = Cos(2x)-x/2
y’ =-2 Sin (2x)-1/2
For 𝑥2
𝑥2 = −1.039 −
−.034
1.249
Therefore 𝒙𝟐 = -1.066
BSCE-3A
CARDENAS, RICO JAY B.
ID No. 19-22711
BSCE-3A
3rd iteration: try 𝑥1 = -1.066
y = Cos(2x)-x/2
y’ =-2 Sin (2x)-1/2
For 𝑥2
𝑥2 = −1.066 −
−.001
1.192
Therefore 𝒙𝟐 = -1.067
SUMMARY OF NRM
iteration
X1
f(x1)
f'(x)
X2
%Error
1
-1.200
-0.137
0.851
-1.039
2.668
2
-1.039
0.034
1.249
-1.066
0.091
3
-1.066
0.001
1.193
-1.067
0.031
Therefore, the real roots for the equation
f(x)= Cos(2x)-x/2
Using NRM method is -1.067
CARDENAS, RICO JAY B.
ID No. 19-22711
SOLVING FOR THE 3rD ROOT
For my initial guess I will use 0.3 to be the value of x to find the real root
1. Determine the first derivative the function
y = Cos(2x)-x/2
y’= -2 Sin (2x)-1/2
Perform NRM
1st iteration: try 𝑥1 = .30
y = Cos(2x)-x/2
y’ =-2 Sin (2x)-1/2
For 𝑥2
𝑥2 = .30 −
. 675
−1.629
Therefore 𝒙𝟐 = .714
2nd iteration: try 𝑥1 = .714
y = Cos(2x)-x/2
y’ =-2 Sin (2x)-1/2
For 𝑥2
𝑥2 = .714 −
−.216
−2.48
Therefore 𝒙𝟐 = .627
BSCE-3A
CARDENAS, RICO JAY B.
ID No. 19-22711
BSCE-3A
3rd iteration: try 𝑥1 = .627
y = Cos(2x)-x/2
y’ =-2 Sin (2x)-1/2
For 𝑥2
𝑥2 = .627 −
−.003
−2.401
Therefore 𝒙𝟐 = .626
SUMMARY OF NRM
iteration
X1
f(x1)
f'(x)
X2
%Error
1
0.300
0.675
-1.629
0.714
14.137
2
0.714
-0.216
-2.480
0.627
0.228
3
0.627
-0.003
-2.401
0.626
0.028
Therefore, the real roots for the equation
f(x)= Cos(2x)-x/2
Using NRM method is 0.626