# How to solve this complicated expression to get real roots?

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Kalasagarreddi Kottakota am 30 Mär. 2022
Bearbeitet: David Goodmanson am 31 Mär. 2022
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.
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David Goodmanson am 30 Mär. 2022
Bearbeitet: David Goodmanson am 31 Mär. 2022
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.

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### Akzeptierte Antwort

Matt J am 30 Mär. 2022
Bearbeitet: Matt J am 30 Mär. 2022
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
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