Root of exponential function not complete

12 次查看（过去 30 天）
Needless Needless Also
评论： madhan ravi 2018-10-28
I am trying to find the roots of the equation
exp(-0.5*x)-0.4/(1+0.5*x)
and with the following code, only one of the 2 roots is found:
syms x
eq1=exp(-0.5*x)-0.4/(1+0.5*x)==0;
sol=solve(eq1,x);
vpa(sol,6)
ans =
-1.64929
But there is another root of this equation which is found using the fzero command
eq1=@(x) exp(-0.5*x)-0.4/(1+0.5*x);
sol=fzero(eq1,[-10 10]);
vpa(sol,6)
ans =4.04463
Again the given solution is only one of the 2. How can I get both of them at any time with one command? Thanks

请先登录，再回答此问题。

回答（1 个）

madhan ravi
madhan ravi 2018-10-28
编辑：madhan ravi 2018-10-28
syms x
eq1=exp(-0.5*x)-0.4/(1+0.5*x)==0;
fplot(exp(-0.5*x)-0.4/(1+0.5*x)) %always compare the result with the graph
solution1=vpasolve(eq1,x,[-2 0])
solution2=vpasolve(eq1,x,[0 5])
grid on
  7 个评论
显示 5更早的评论
Needless Needless Also
Needless Needless Also 2018-10-28
This works but, what about any other equation? Shouldn't it be a pretty simple task for MatLab to find the roots just with vpasolve(eq1,x)? Of course I can always check the graph but I think you get what I mean.
madhan ravi
madhan ravi 2018-10-28
Yes I know even I have thought about it ;) but yet no discernible idea in my mind though

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Loops and Conditional Statements 的更多信息

标签

产品


版本

R2018a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by