How to solve this complicated expression to get real roots?

The following is a complicated expression for which I need to find a real value for such that the expression equals to zero:

clear all; clc;

c0 =340; 
M=41;
dm = [0.721	0.680	0.640	0.602	0.565	0.531	0.5000	0.471	0.447	0.427	0.412...
    0.403	0.400	0.403	0.412	0.427	0.447	0.471	0.500	0.531	0.565	0.602...
    0.640	0.680	0.721	0.763	0.806	0.850	0.894	0.939	0.984	1.030	1.077...
    1.123	1.170	1.217	1.265	1.312	1.360	1.408	1.456]';
tm = dm/c0;
Hm = 1./dm;

% A
A1 = sum( abs(Hm(:)).^2 .* [-ones(M,1)  +3*tm(:) -3*tm(:).^2 tm(:).^3]);
A =0.25* sum( abs(Hm(:)).^2)*conv(A1,A1);

% B
B1 = sum( abs(Hm(:)).^2 .* [ones(M,1)  -4*tm(:) +6*tm(:).^2 -4*tm(:).^3 tm(:).^4]);
B2 = sum( abs(Hm(:)).^2 .* [-ones(M,1) tm(:)]);
B = 0.25*conv(conv(B2,B2),B1);

   
P = A+B;
t0_roots = roots(P); % roots

When I compute for by above code resulted in complex roots:

t0_roots =

   0.0013 + 0.0015i
   0.0013 - 0.0015i
   0.0020 + 0.0002i
   0.0020 - 0.0002i
   0.0017 + 0.0006i
   0.0017 - 0.0006i
   

Now I have taken one of the roots,

 t0 = t0_roots(1);
    
 E = .25*(sum(abs(Hm.').^2)*sum(abs(Hm.').^2 .* (tm(:).' - t0).^3, 2).^2 ...
        + sum(abs(Hm.').^2 .* (tm(:).' - t0).^4, 2) .* sum(abs(Hm.').^2 .* (tm(:).' - t0), 2).^2); % expression value

The resulted E

 E = 1.2177e-25 + 2.5445e-26i

Lets assume the root as a real number and calculate E,

 E = 2.7709e-13

In this regard, could someone help to find real value for t0 for the expression to be zero.

[Matt J]

How close to zero do you aspire to get? I assume you know that, because of floating point imprecision, you cannot achieve zero exactly.

[David Goodmanson]

< I had an incorrect argument here, as pointed out by Matt J >

[Matt J]

The two terms are added together so it appears thatfor real t this result cannot possibly be zero unless [1] some of the tm are complex, or [2] the trivial case where all the Hm are zero.

No, you could have, for example,

Hm=[1,1];
tm=[1,-1];
t0=0;

[David Goodmanson]

Certainly in this particular case Hm has many nonzero values, there are many different values of tm with corresponding nonzero values of Hm. So both the first sum and B1 are positive. Both A1^2 and B2^2 are positive or zero, so for the whole works to be zero, both A1 and B2 must be zero. For B2, the solution is

t0 = Sum(tm.*Hm.^2) / Sum(Hm.^2)
   t0 = 0.001642868116609

and when plugged into A1, the result has to be zero. The result, A1 = 5.7296e-08 is surprisingly small but I beleive it is not consistent with zero. Evidently there is not a solution for real t0.

Answer 1

[Matt J]

Based on the plot, the minimum value of the expression would lie somewhere between -0.005 and 0.008, so using fminbnd(),

[x,fval]=fminbnd(@expression,-0.002,0.006)

x = 0.0020

fval = 3.0272e-14

warning off; 
fplot(@expression,[-0.005,0.008]);

function E = expression(t0)

    c0 =340; 
    M=41;
    dm = [0.721	0.680	0.640	0.602	0.565	0.531	0.5000	0.471	0.447	0.427	0.412...
        0.403	0.400	0.403	0.412	0.427	0.447	0.471	0.500	0.531	0.565	0.602...
        0.640	0.680	0.721	0.763	0.806	0.850	0.894	0.939	0.984	1.030	1.077...
        1.123	1.170	1.217	1.265	1.312	1.360	1.408	1.456]';
    tm = dm/c0;
    Hm = 1./dm;

 E = .25*(sum(abs(Hm.').^2)*sum(abs(Hm.').^2 .* (tm(:).' - t0).^3, 2).^2 ...
        + sum(abs(Hm.').^2 .* (tm(:).' - t0).^4, 2) .* sum(abs(Hm.').^2 .* (tm(:).' - t0), 2).^2); % expression value

end