本文介绍了exp(-x) - sinx = 0; _ MATLAB的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧! 问题描述 29岁程序员,3月因学历无情被辞! 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); ------------------------------ 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 stepiterator = 0;for i=0:eps:x0 x=x+exp(-x) - sin(x); iterator = iteratio + 1;enddisp('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 -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;enddisp('root ::: ');disp(x);disp('iterate count ::: ');disp(iterator); 感谢所有人。thanks to all. 这篇关于exp(-x) - sinx = 0; _ MATLAB的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持! 上岸,阿里云! 08-06 05:11