How to solve the following type of equation that includes summation in matlab?

How to find an expression for using matlab, so that the below equation equals to zero.

Answer 1

[Bruno Luong]

Assuming

H_m and t_m are stored in (1 x M )arrays

% Dummy test data
M = 10;
H = rand(1,M);
t = rand(1,M);

P = sum( abs(H(:)).^2 .* [-ones(M,1) +3*t(:) -3*t(:).^2 +t(:).^3]);
t0 = roots(P)

t0 = 
   0.2914 + 0.3069i
   0.2914 - 0.3069i
   0.3144 + 0.0000i

sum(abs(H).^2 .* (t - t0).^3, 2)

ans = 
   1.0e-15 *

  -0.2845 + 0.1978i
  -0.2845 - 0.1978i
   0.1162 + 0.0000i

[Torsten]

To apply the "roots" command to determine the roots of a polynomial, this polynomial is written as the vector of its coefficients.

E.g. p(x) = 1*x^2 +3*x + 5 is written as p = [1 3 5].

Now the coefficients of the polynomial (tm-t0)^3 (as a polynomial in t0) are

[-1 3*tm -3*tm^2 tm^3]. 

This polynomial is multiplied by abs(Hm)^2

abs(Hm)^2 * [-1 3*tm -3*tm^2 tm^3]

and summed over m

sum (abs(Hm)^2 * [-1 3*tm -3*tm^2 tm^3])

[Kalasagarreddi Kottakota]

Hi @Torsten,

Thank you, here the polynomial is (tm-t0)^3 and its expansion is simple, but if there is (tm-t0)^7. Is there any way to get the coefficients in matlab?

[Torsten]

(tm-t0)^n = sum_{k=0}^{n} nchoosek(n,k) * (-1)^(n-k) * tm^k * t0^(n-k)

so the coefficients are

k = 0:n;
p = nchoosek(n,k).*(-1).^(n-k).*tm.^k

[Kalasagarreddi Kottakota]

Hi @Bruno Luong and @Torsten,

I have a much more complex equation to be solved for ,

In this case, the equation becomes very complex by expanding it and arranging it for coefficents to solve for . Do you have any suggestions to solve this kind of problem in matlab?

[Bruno Luong]

In the form of polynomial as array coefficients we show earlier, the product of two polynomial P and Q is

% conv(P,Q)

Use the fact to compute the polynomial equations .

For example

instead of usi,g binomial formula

t = rand(1);
P3 = [1 +3*t -3*t.^2 +t.^3] % P(t0) = (t-t0)^3

P3 = 1×4

1.0000 0.7917 -0.2089 0.0184

% you can do
P = [-1 t]; % (t-t0)
P3 = conv(conv(P,P),P)

P3 = 1×4

-1.0000 0.7917 -0.2089 0.0184

I let you figure out how to perform the sum P+Q

[Torsten]

@Kalasagarreddi Kottakota

And next time you have something even more complicated than the complicated before ?

I suggest you present your final problem first for that it's not necessary for us to invent a new solution each time.

Here is the command how to multiply polynomials given in coefficient vector form:

https://de.mathworks.com/help/matlab/ref/conv.html

[Kalasagarreddi Kottakota]

Hi @Bruno Luong, I have a situation for which I need to find a real value for such that the expression equals to zero, but when I compute the roots of this expression, I get complex values. So that I followed a trail and error method and found a real value for t0 for the expression to be zero. Is there any way where I can get a real root with out junk methods.

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.

[Bruno Luong]

That means your equation does not have a real solution.

If you run

ezplot(@(t) polyval(P,t),0.1*[-1 1])

And zoom in you see it never reach 0, but getting very close.

To find out the closest point to 0 you can find a real stationary point.

>> Q=polyder(P)

Q =

1.0e+06 *

4.5696   -0.0375    0.0001   -0.0000    0.0000   -0.0000

>> t0 = roots(Q)

ans =

0.0014 + 0.0012i
0.0014 - 0.0012i
0.0019 + 0.0000i
0.0018 + 0.0004i
0.0018 - 0.0004i

There is only one real roots

>> t0real = t0(3)

t0real =

    0.0019

>> polyval(P,t0real)

ans =

   3.0150e-14

>> 

BTW your eqt doesn't look right to me unit-wise. You keep multiply abs(H)^2 with different power of t. I just think there is something wrong about it.

[Kalasagarreddi Kottakota]

@Bruno Luong Thanks,

I dont see any mistake in the matlab expression. Could you please elaborate your commment.

[Bruno Luong]

Elaborate what? Your orginal equation you post seems fishy to me, not the MATLAB implementation of your equation.

Another fischy fact : the expression only have sum of strictly positive quantities, you cannot expect there is a real root of it. Something is wrong with the maths/physics where you derive the expression/equation.