MATLAB: Why there is only one solution to the following equation

equationsolve

>> syms x;

>> x=solve(6*x-exp(x)==-1,x)

Best Answer

You did not constrain to real numbers, so there are an infinite number of solutions. The internal symbolic engine returns the entire set of them, but the user interface then selects a "representative" value out of the infinite set to return.

If you add the assumption of real to x, then two values will be returned.

If you need the entire set of them, then

>> evalin(symengine,'solve(6*x-exp(x)=1,x)')
 
ans =
 
{- lambertw(k, -exp(1/6)/6) + 1/6 | in(k, 'integer')}

and extracting the infinite number of them from that set would be up to you to figure out how to represent.

Related Solutions

MATLAB: Derivative of a Function

If you have the Symbolic Toolbox:

>> g = @(x) exp(x^2-2)
g = 
    @(x)exp(x^2-2)
>> syms x
>> diff(g(x))
ans =
2*x*exp(x^2 - 2)

MATLAB: Equation solution for large iteration number

Assuming you have the ‘lambertw’ function on your Matlab system:

d = a./b;
e1 = exp(1);
y = zeros(1,18858);
for ix = 1:1:18858
  y(ix) = lambertw(e1*d(ix))-1;
end

Best Answer

Related Solutions

MATLAB: Derivative of a Function

MATLAB: Equation solution for large iteration number

Related Question