Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN: 9780133594140
Author: James Kurose, Keith Ross
Publisher: PEARSON
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Question

I'm a girl so use arc cos. The program should run just like the example in the picture. Example of Newton's method is given below. Thank you.

Instructions:
1. Use Newton's method to solve for arccot 0.2345Use initial estimate xº = 0.67, with
dk = 0.01
2.
Use arc cot if you are a boy.
3. Use arc cos if you are a girl.
CALiloy Files\DevC program MOSS.cpp - [Executing] - Dev-C++ 5.11
File Edit Search View Project Execute Tools AStyle Window Help
相电|吕
CE
JU (globals)
Project Classes
Line: 1
4.
Use DevC to create a program that will solve the equation in step 1.
5. The program will ask for the initial estimate and the required error.
6.
The program will output the answer showing all the needed iterations. Please see figure below.
Compiler
Abort Compil
f
Shorten compil
CALiloy Files\DevC program MOSS.exe
Solution for f(x) = x^4-1.5x^3-x+1.2
Relative error is 0.01 and initial estimate 0.6
Enter intial estimate: 0.6
Enter target relative error: 0.01
Iter k
12
17
18
19
xk+1 0.81964
Col: 1
75°F
Rain to stop
Sel: 0
xk
0.60000
1.00560
8.69725
0.92789
0.74294
0.88955
0.77031
8.86647
0.78787
0.85172
0.79945
0.84206
0.88716
0.83565
0.81232
0.83139
0.81577
0.82854
0.81809
8.82664
Lines: 89
xk+1
1.00560
0.69725
8.92789
0.74294
0.88955
0.77031
0.86647
8.78787
0.85172
8.79945
0.84206
0.80716
8.83565
0.81232
0.83139
0.81577
0.82854
0.81809
8.82664
0.81964
Length: 1402
dk+1
0.40334
0.44224
0.24857
0.24894
0.16481
0.15480
0.11099
0.09976
0.07496
0.06538
0.05059
0.04323
0.03409
0.02872
0.02294
0.01914
0.01541
0.01278
0.01034
0.00854
Insert
0
Remarks
>0.01000
>0.01000
>0.01000
>0.01000
>0.01000
>0.01000
>0.01000
>0.01000
>0.01000
>0.01000
>0.01000
>0.01000
>0.01000
>0.01000
>0.01000
>0.01000
>0.01000
>0.01000
>0.01000
<0.01000
TOM-GCC 4.9.2 64-bit Release
Done parsing in 0.734 seconds
DEV
X
4)
X
10:40 PM
9/28/2022
expand button
Transcribed Image Text:Instructions: 1. Use Newton's method to solve for arccot 0.2345Use initial estimate xº = 0.67, with dk = 0.01 2. Use arc cot if you are a boy. 3. Use arc cos if you are a girl. CALiloy Files\DevC program MOSS.cpp - [Executing] - Dev-C++ 5.11 File Edit Search View Project Execute Tools AStyle Window Help 相电|吕 CE JU (globals) Project Classes Line: 1 4. Use DevC to create a program that will solve the equation in step 1. 5. The program will ask for the initial estimate and the required error. 6. The program will output the answer showing all the needed iterations. Please see figure below. Compiler Abort Compil f Shorten compil CALiloy Files\DevC program MOSS.exe Solution for f(x) = x^4-1.5x^3-x+1.2 Relative error is 0.01 and initial estimate 0.6 Enter intial estimate: 0.6 Enter target relative error: 0.01 Iter k 12 17 18 19 xk+1 0.81964 Col: 1 75°F Rain to stop Sel: 0 xk 0.60000 1.00560 8.69725 0.92789 0.74294 0.88955 0.77031 8.86647 0.78787 0.85172 0.79945 0.84206 0.88716 0.83565 0.81232 0.83139 0.81577 0.82854 0.81809 8.82664 Lines: 89 xk+1 1.00560 0.69725 8.92789 0.74294 0.88955 0.77031 0.86647 8.78787 0.85172 8.79945 0.84206 0.80716 8.83565 0.81232 0.83139 0.81577 0.82854 0.81809 8.82664 0.81964 Length: 1402 dk+1 0.40334 0.44224 0.24857 0.24894 0.16481 0.15480 0.11099 0.09976 0.07496 0.06538 0.05059 0.04323 0.03409 0.02872 0.02294 0.01914 0.01541 0.01278 0.01034 0.00854 Insert 0 Remarks >0.01000 >0.01000 >0.01000 >0.01000 >0.01000 >0.01000 >0.01000 >0.01000 >0.01000 >0.01000 >0.01000 >0.01000 >0.01000 >0.01000 >0.01000 >0.01000 >0.01000 >0.01000 >0.01000 <0.01000 TOM-GCC 4.9.2 64-bit Release Done parsing in 0.734 seconds DEV X 4) X 10:40 PM 9/28/2022
NEWTON'S METHOD
EXAMPLE: USING NEWTON'S METHOD WITH d<0.001 ESTIMATE
VIN" (0.52) USING INITIAL ESTIMATE Xº = I
vir [x=cm (0.52)] vin
rinx = 0-52
vin x-0.52=0
•EQUATE f(x)=0
f(x)=vinx -0.52
• DERIVE
f'(x) = cos x
TER K
xkxkx²x²= f(x)
F'(x)
0
TT/6
059669
1 0.54669 0.54685
X0.59685
0.04223
0.00029
REMARKS
70.001
<0.0⁰1
expand button
Transcribed Image Text:NEWTON'S METHOD EXAMPLE: USING NEWTON'S METHOD WITH d<0.001 ESTIMATE VIN" (0.52) USING INITIAL ESTIMATE Xº = I vir [x=cm (0.52)] vin rinx = 0-52 vin x-0.52=0 •EQUATE f(x)=0 f(x)=vinx -0.52 • DERIVE f'(x) = cos x TER K xkxkx²x²= f(x) F'(x) 0 TT/6 059669 1 0.54669 0.54685 X0.59685 0.04223 0.00029 REMARKS 70.001 <0.0⁰1
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