exp(-x) - sinx = 0; _ MATLAB

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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?

Posted 10-Jan-14 10:32am

m.r.m.40

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Comments

Ron Beyer 10-Jan-14 16:37pm

"I can't make it work" isn't a description of a problem. Please tell us what problem you are having, we don't have a lot of Matlab experts here so you'll have to help narrow your problem down.

[no name] 10-Jan-14 17:13pm

What is "sinx"...maybe I missed something, but I'm willing to learn.

And after this: If you have a minimum of the idea of the two functions in the equation you should find the answer.

If not please first try to learn about the pb and not simply ask about the solution!

m.r.m.40 10-Jan-14 17:58pm

actually that was:
sin(x)
i guess i could solve it myself(the solution below),
thank you.

Matt T Heffron 10-Jan-14 17:42pm

Assuming that x is correctly initialized to 0.0 before code given.
Remember that eps is used as the step size in the for loop, so it must be less than x0.
You seem to be attempting Newton's method, but it doesn't look correct.
I don't see the derivative of the exp(-x) - sin(x) represented anywhere.

m.r.m.40 10-Jan-14 17:58pm

i guess i could solve it myself(the solution below),
thank you.

2 solutions

Solution 1

Now it's working,

VB

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.

Posted 10-Jan-14 11:42am

m.r.m.40

Comments

Andreas Gieriet 10-Jan-14 18:36pm

This is not calculating the roots, or am I missing something?
You should get roots at approximately 0.59, π, 2π, 3π, 4π, 5π, 6π, 7π. etc.
Cheers
Andi

[no name] 10-Jan-14 20:16pm

my 1 :(
I recomend to check the solution #2

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Andreas Gieriet · Accepted Answer · 2014-01-10T12:34:00

I strongly suggest to first get a "feeling" for the graph.
1) sin(x) is periodically in 2⋅π and has roots every n⋅π (∀ n ∈ Ν₀)
2) sin(x) ≈ x (x much smaller than π/2)
3) e^-x goes monotoinc to zero (∀ x ∈ R, x ≥ 0)
4) e^-x has sample values at: e⁰ = 1, e^-π ≈ 0.0432, e^-2⋅π ≈ 0.0019, e^-3⋅π ≈ 0.00008, e^-4⋅π ≈ 0.0000034, ...

From this, you can postulate that the roots for e^-x - sin(x) are

≈ n⋅π (∀ n ∈ Ν) (since e^-x for values x=π and above can be neglected)
≈ 0.59 (for the root between 0 and π) (found by setting e^-x ≈ x and estimating and roughly calculating some value between π/6 and π/4)

Knowing that, you can compare your numeric calculation with the aproximately estimated roots from above.

Your immediate problem was that x was not set to 0 initially.

My understanding is that you don't calculate the roots with your solution #1. It's calculating "something" but not the roots.

Describe first in your own words how you want to calculate the roots (e.g. by Regula Falsi, Newton, ...) and then only code that choosen approach.

Cheers
Andi