solve not returning two answers

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Syeda Amberin
Syeda Amberin 2019-7-7
评论： Syeda Amberin 2019-7-8
Hi,
I have a nonlinear equation that I need to sovle for 'x'. Following is my code:
syms x q N
q=.95;
N=500;
eqn = 2*log((((N-x)/(q*N))^(N-x))*((x/((1-q)*N))^x)) == 3.841;
solx = solve(eqn, x);
I am expecting two values for 'x' but am getting just one (i.e. ~35). I am not getting the other value of 16. Also I get the following warning:
Warning: Cannot solve symbolically. Returning a numeric approximation instead.
Please tell me what I am doing wrong.
Thanks,

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采纳的回答

infinity
infinity 2019-7-7
Hello,
Maybe you could try with another approach, for example,
eq = @(y)2*log((((N-y)./(q*N)).^(N-y)).*((y./((1-q)*N)).^y)) - 3.841;
fsolve(eq,[0 100])
It will give two solutions.
ans =
16.0510 35.1056
But in this case, you should be aware that if you chose different interval of solutions, for example [-100, 100] in fsolve, you may get wrong results.

更多回答（1 个）

Walter Roberson
Walter Roberson 2019-7-8
When an equation is sufficiently complicated, solve cannot find all the exact solutions. Often it does not even know how to find one exact solution. In such cases, solve() uses vpasolve() to find one numeric solution.
  2 个评论
Walter Roberson
Walter Roberson 2019-7-8
It is always a mistake to use solve() for equations with floating point coefficients such as 3.841. solve() is for finding exact solutions, which is something that you cannot do for equations involving inexact quantities such as 3.841. 3.841 represents a range of numbers, everything from 38405/10000 to 38149/10000. 3.841 does *not* mean 3841/1000 exactly.

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