newton's method to solve nonlinear equation

hw prompt is below my work. please check my code. i am new to using newton's method and need some recommendations about what to do or fix. thank you.

what i did first:

x=[-3:0.01:3];
f = sin(x)-x.*cos(x); 
plot(x,f);

what i did next:

x=0;
Tol = 0.0000001;
count = 0;
dx=0; 
f=sin(0)-0*cos(0);
fprintf('step x dx f(x)\n')
fprintf('---- ----------- --------- ----------\n')
fprintf('%3i %12.8f %12.8f %12.8f\n',count,x,dx,f)
while (dx > Tol) 
    count = count + 1;
    fprime = x.*sin(x);
    xnew = x - (f./fprime); 
    dx=abs(x-xnew); 
    x = xnew;
    f = sin(x)-x.*cos(x); 
    fprintf('%3i %12.8f %12.8f %12.8f\n',count,x,dx,f) 
end

on the other homework problems, i received several answers, yet on this problem i received only one. did i make a mistake?

Answer 1

[John D'Errico]

Is there a good reason why you initialized dx to be zero?

Do you see that immediately after that, you start a while loop. What is the codition?

while (dx > Tol) 

Now, ask yourself. What was dx again when the loop started?

What is the value of Tol?

Is this true? (dx > Tol)?

Do you expect the while loop to ever allow entrance into the code inside that loop? What would happen if you wrote this code fragment?

while false
    disp('The sky is falling! (sincerely, Chicken Little)')
end

How many times would you expect to see anything written to the command window?

[N/A]

so i should make the initial guess to be 1? okay, i will do that and post my updated code momentarily. please be paitent with me. this is my first programming class and i am taught online.

[N/A]

I changed it to 0.5. do you like this better?

x=0.5;

Tol = 0.0000001;

count = 0;

dx=0.5;

f=sin(0.5)-0.5*cos(0.5);

fprintf('step x dx f(x)\n')

fprintf('---- ----------- --------- ----------\n')

fprintf('%3i %12.8f %12.8f %12.8f\n',count,x,dx,f)

while (dx > Tol)

count = count + 1;

fprime = x.*sin(x);

xnew = x - (f./fprime);

dx=abs(x-xnew);

x = xnew;

f = sin(x)-x.*cos(x);

fprintf('%3i %12.8f %12.8f %12.8f\n',count,x,dx,f)

end

[N/A]

1 0.33048772 0.16951228 0.01190128

2 0.21951457 0.11097316 0.00350893

3 0.14610690 0.07340767 0.00103744

4 0.09733515 0.04877175 0.00030710

5 0.06486959 0.03246556 0.00009095

6 0.04324032 0.02162926 0.00002694

7 0.02882509 0.01441524 0.00000798

8 0.01921619 0.00960889 0.00000237

9 0.01281064 0.00640555 0.00000070

10 0.00854038 0.00427026 0.00000021

11 0.00569357 0.00284681 0.00000006

12 0.00379571 0.00189786 0.00000002

13 0.00253047 0.00126524 0.00000001

14 0.00168698 0.00084349 0.00000000

15 0.00112465 0.00056233 0.00000000

16 0.00074977 0.00037488 0.00000000

17 0.00049985 0.00024992 0.00000000

18 0.00033323 0.00016662 0.00000000

19 0.00022215 0.00011108 0.00000000

20 0.00014810 0.00007405 0.00000000

21 0.00009874 0.00004937 0.00000000

22 0.00006582 0.00003291 0.00000000

23 0.00004388 0.00002194 0.00000000

24 0.00002925 0.00001463 0.00000000

25 0.00001950 0.00000975 0.00000000

26 0.00001300 0.00000650 0.00000000

27 0.00000867 0.00000433 0.00000000

28 0.00000578 0.00000289 0.00000000

29 0.00000385 0.00000193 0.00000000

30 0.00000257 0.00000128 0.00000000

31 0.00000171 0.00000086 0.00000000

32 0.00000114 0.00000057 0.00000000

33 0.00000076 0.00000038 0.00000000

34 0.00000051 0.00000025 0.00000000

35 0.00000034 0.00000017 0.00000000

36 0.00000023 0.00000011 0.00000000

37 0.00000015 0.00000008 0.00000000

[N/A]

This answer seems more right. Thank you for your recommendations. Please let me know if you have another.