exp(-x) - sinx = 0; _ MATLAB [英] exp(-x) - sinx = 0; _ MATLAB
问题描述
我打算到达这个功能的根源:
exp(-x) - sinx = 0 ;
这是我的代码:
------------ ------------------
x0 =输入('请输入x0 :::');
eps =输入('请输入步骤:::');
%eps是步骤
iterator = 0;
for i = 0:eps:x0
x = x + exp(-x) - sin(x);
iterator = iteratio + 1;
结束
disp('root :::');
disp(x) ;
disp('iterate count :::');
disp(迭代器);
< br $>
------------------------------
我不能使它工作。你可以帮我吗?
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?
推荐答案
我强烈建议首先得到图表的感觉。
1)sin( x)周期性地在2⋅π并且每个n⋅π具有根(∀n∈N 0 )
2)sin(x)≈x(x远小于π/ 2)
3)e -x 将monotoinc变为零(∀x∈R,x≥0)
4)e < sup> -x 的样本值为:e 0 = 1,e -π≈0.0432,e -2⋅π ≈0.0019,e -3⋅π≈0.00008,e -4⋅π≈0.0000034,...
从这里,你可以假设e -x - sin(x)的根是
- ≈n⋅π(∀n∈N)(因为e -x 对于值x =π及以上可以忽略)
- ≈0.59(对于0和π之间的根) )(通过设置e -x ≈x并估计并粗略计算π/ 6和π/ 4之间的某个值)
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)
现在它正在工作,
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);
感谢所有人。
thanks to all.
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