exp(-x) - sinx = 0

hi, i'm planning to reach the root of this function:

exp(-x) - sinx = 0;

here is my code:

------------------------------

x0 = input('please enter x0 ::: ');

eps = input('please enter steps ::: ');

% eps is the step

iterator = 0;

for i=0:eps:x0

    x=x+exp(-x) - sin(x);

    iterator = iteratio + 1;

end

disp('root ::: '); disp(x);

disp('iterate count ::: '); disp(iterator);

------------------------------

well i can't make it work. can you help me with that?

Answer 1

[Mischa Kim]

Try the fzero function. Unless, of course, you would like to do it on your own.

Answer 2

[reza]

Now it's working,

x0 = input('please enter x0 ::: '); eps = input('please enter steps ::: '); iterator = 0; x=0; iterator = 0; for i=0:eps:x0 x=x+exp(-x) - sin(x); iterator = iterator + 1; end disp('root ::: '); disp(x); disp('iterate count ::: '); disp(iterator);

thanks to all.

[Mischa Kim]

Just want to make sure you are getting the results you are looking for. What root-finding method are you using? For those kind of problems I'd recommend Newton-Raphson (if you prefer coding yourself). Also note that this function has an infinte number of roots. Here are two of them:

r1 = fzero(@(x) exp(-x) - sin(x), 0)
r1 =
    0.5885
r2 = fzero(@(x) exp(-x) - sin(x), 2)
r2 =
    3.0964

[Youssef Khmou]

that is good remark, so how, for example, we find the period of the solutions using only fzero as you illustrated above?

[Mischa Kim]

Not sure I understand what you are asking. If it's about finding more than just a handful of roots, you could use a for -loop. For positive x-values the function (and therefore the roots) is dominated by the sine-term, which allows you to get pretty accurate starting values for the searches.

[Youssef Khmou]

yes, here is the result :

ct=1;
for n=0:100
F(ct)=fzero(@(x) exp(-x)-sin(x),n);
ct=ct+1;
end
stem(F)

can we conclude that there re 4 solutions?

[Mischa Kim]

By solutions you mean roots? Here is the function plot:

[Youssef Khmou]

alright, infinite number of solutions with period of ~3.1

thanks