solve return imaginary answer for a problem should have a real solution

Hi fellows,

I have written a code to solve a expotinal function, matlab returns an answer which seems like contain imaginary part. But I tried to plot the function and I think it indicates the function should has a real solution between 5 and 6. Could you help me find out what is the problem here? Really appreciated! The code is below (yes the loop seems unnecessary, because I just show a example so there is not need to let it do the whole loop).

%this is the maxmin when c2=0

sub=[0.3,0.3,0.3,0.08];               
d1=[0.200000000000000,-0.400000000000000,0.600000000000000]
rn=size(d1,1)
lb1=sub(1);
lb2=sub(2);
lb3=sub(3);
r=sub(4);
for i=1:1

     if (d1(i,1)==min(d1(i,:)))
              p1=1-lb2-lb3;
              p2=lb2;
              p3=lb3;
     elseif (d1(i,2)==min(d1(i,:)))
              p1=lb1;
              p2=1-lb1-lb3;
              p3=lb3;
     elseif (d1(i,2)==min(d1(i,:)))
              p1=lb1;
              p2=lb2;
              p3=1-lb1-lb2;
     end
     syms x;
  f=simplify(p1*d1(i,1)*exp(-r*x*d1(i,1))+p2*d1(i,2)*exp(-r*x*d1(i,2))+p3*d1(i,3)*exp(-r*x*d1(i,3)))
  y=solve(f,x)
  %x=-100:1:100;
  %solve(0.06*exp(-0.016*x)-0.16*exp(0.032*x)+0.18*exp(-0.048*x),x)
  %y=0.06*exp(-0.016*x)-0.16*exp(0.032*x)+0.18*exp(-0.048*x);
  %plot(x,y)

end

[Walter Roberson]

Please show us what f comes out as before the solve() call, and also what y contains after the solve()

[xueqi]

f =

(3*exp((4*x)/125) - 8*exp((2*x)/25) + 9)/(50*exp((6*x)/125))

y =

(125*log(z1))/2 + 125*pi*k*i

Answer 1

[Walter Roberson]

It appears you are not being shown the complete solution. The general solution is of the form

(125/2)*ln(RootOf(-3*z^2+8*z^5-9, z))

Here RootOf(f(z),z) stands for the set of values of z such that f(z) becomes 0. Which is to say, the roots of a quintic. There is no general analytic solution for quintics.

The z1 that you are being shown in the solution is MuPAD's way of making the formula shorter, along the lines of saying,

"(125/2)*ln(z1) WHERE z1 = RootOf(-3*z^2+8*z^5-9, z)"

except that in R2010a, the interface between MuPAD and MATLAB does not know to show you the "WHERE z1 =" part.

What you should probably do is use

syms x real

and then in theory you should get the real root of the solution.

[xueqi]

yes add an real works! Thanks!

Answer 2

[Matt J]

Well, it does have a real solution doesn't it? It looks like for k=0, the solution returned to you is real. So what's the problem?

[xueqi]

but according to to plot, it should has solution between 5 and 6. I prefer to an explicit anwers instead of this. To be honest I dont really understand the structure of the answers matlab returns...

Answer 3

[Shashank Prasanna]

What version of MATLAB are you using? On MATLAB R2012a:

>> syms x
>> f = (3*exp((4*x)/125) - 8*exp((2*x)/25) + 9)/(50*exp((6*x)/125));
>> y=solve(f,x)
y =

5.6756938340216981622204109791808

[xueqi]

2010a. Is there anyway to do it in 2010a?

Answer 4

[Ankit Kumar Karmakar]

I am facing the same problem on 2015a version, my pc is running on 32bit Win10 with 2Gb ram.

for testing purpose I made a simple coding

syms x;
a=2;
b=a+3;
c=a*b;
eq='4*x-((a-b)*(x-c)^4)+(c-x)';
f=solve(eq);
disp(f)

it is not even evaluating the values of a, b, c variables. Then the output is horrible.

RootOf(b*z^4 - a*z^4 - 4*b*c*z^3 + 4*a*c*z^3 + 6*b*c^2*z^2 - 6*a*c^2*z^2 - 4*b*c^3*z + 4*a*c^3*z + 3*z + b*c^4 + c - a*c^4, z)[1]

RootOf(b*z^4 - a*z^4 - 4*b*c*z^3 + 4*a*c*z^3 + 6*b*c^2*z^2 - 6*a*c^2*z^2 - 4*b*c^3*z + 4*a*c^3*z + 3*z + b*c^4 + c - a*c^4, z)[2]

RootOf(b*z^4 - a*z^4 - 4*b*c*z^3 + 4*a*c*z^3 + 6*b*c^2*z^2 - 6*a*c^2*z^2 - 4*b*c^3*z + 4*a*c^3*z + 3*z + b*c^4 + c - a*c^4, z)[3]

RootOf(b*z^4 - a*z^4 - 4*b*c*z^3 + 4*a*c*z^3 + 6*b*c^2*z^2 - 6*a*c^2*z^2 - 4*b*c^3*z + 4*a*c^3*z + 3*z + b*c^4 + c - a*c^4, z)[4]

Please tell us what to do

[Ankit Kumar Karmakar]

then I tried real command on it, the output shows different value for f and x, but they should be equal.

output::::

Warning: The solutions are valid under the following conditions: in(RootOf(b*z^4 - a*z^4 - 4*b*c*z^3 + 4*a*c*z^3 +

6*b*c^2*z^2 - 6*a*c^2*z^2 - 4*b*c^3*z + 4*a*c^3*z + 3*z + b*c^4 + c - a*c^4, z)[1], 'real'); in(RootOf(b*z^4 - a*z^4 -

4*b*c*z^3 + 4*a*c*z^3 + 6*b*c^2*z^2 - 6*a*c^2*z^2 - 4*b*c^3*z + 4*a*c^3*z + 3*z + b*c^4 + c - a*c^4, z)[2], 'real');

in(RootOf(b*z^4 - a*z^4 - 4*b*c*z^3 + 4*a*c*z^3 + 6*b*c^2*z^2 - 6*a*c^2*z^2 - 4*b*c^3*z + 4*a*c^3*z + 3*z + b*c^4 + c -

a*c^4, z)[3], 'real'); in(RootOf(b*z^4 - a*z^4 - 4*b*c*z^3 + 4*a*c*z^3 + 6*b*c^2*z^2 - 6*a*c^2*z^2 - 4*b*c^3*z +

4*a*c^3*z + 3*z + b*c^4 + c - a*c^4, z)[4], 'real'). To include parameters and conditions in the solution, specify the

'ReturnConditions' option.

> In solve>warnIfParams (line 517)

In solve (line 360)

In Untitled2 (line 6)

value of x

ans =

120

value of f

ans =

102

Code is as follows:

syms x real;
a=2;
b=a+3;
c=a*b;
eq='4*x-((a-b)*(x-c)^4)+(c-x)';
f=solve(eq,x);
disp ('value of x');
real 'x'
disp ('value of f');
real 'f'