How to know results

I have a polynomial and parameters in it,

a=2, c=1, d=2 , g=0.5,  k=2 , f=2
e=[2,8,5,4]
b=[1,2,3,4]
p=[ -c.*k+ (f.^2)*(e.^2)/a+2*e.*g, d.*e-(f*b.*e)/a] 
roots(p)

I calculate roots of polynomial given parameter values.

But e an b parameters have vector values.

I want to know if it obtains roots for every combination of the e and b values or

it simply gets roots when e=2, b=1 , then b=8,e=2,... ?

[Ani Asoyan]

I want to know if it takes into account cross-combinations of e and b ?

Answer 1

[Ameer Hamza]

Your current code does not solve the problem, as you described. You need to use for-loop

a=2, c=1, d=2 , g=0.5,  k=2 , f=2
e=[2,8,5,4]
b=[1,2,3,4]

r = zeros(numel(e), 1) % polynomial is 1st order so there will be only one root
for i=1:numel(e)
    p = [ -c.*k+(f.^2)*(e(i).^2)/a+2*e(i).*g, d.*e(i)-(f*b(i).*e(i))/a] 
    r(i) = roots(p)
end

[Ani Asoyan]

does this calculate the roots in the case of every combinations of e and b ?

[Ameer Hamza]

Yes, it calculate 4 roots corresponding to the 4 elements in e and b.

[Ani Asoyan]

can I make this for all combinations ?

[Ani Asoyan]

like in cross combinations, for example e=8,b=1 or e=8,b=3 ,ect...

[Ameer Hamza]

Following code use cross-combination

a=2, c=1, d=2 , g=0.5,  k=2 , f=2
e=[2,8,5,4]
b=[1,2,3,4]

[E, B] = ndgrid(e, b);
e = E(:);
b = B(:);

r = zeros(numel(e), 1) % polynomial is 1st order so there will be only one root
for i=1:numel(e)
    p = [ -c.*k+(f.^2)*(e(i).^2)/a+2*e(i).*g, d.*e(i)-(f*b(i).*e(i))/a] 
    r(i) = roots(p)
end

sol = [e, b, r]

1st column of 'sol' is 'e' value, 2nd is 'b' values, and 3rd is the corresponding root.

Answer 2

[Steven Lord]

As written your code solves for the roots of the 7th order polynomial whose coefficients are:

p =
     8   134    53    34     2     0    -5    -8

This is:

8*x^7 + 134*x^6 + 53*x^5 + 34*x^4 + 2*x^3 - 5*x - 8

If instead you want to find the roots of the four first order polynomials created by specifying each element of e and b in the expression for p, use a for loop instead.

>> for q = 1:numel(e)
p=[ -c.*k+ (f.^2)*(e(q).^2)/a+2*e(q).*g, d.*e(q)-(f*b(q).*e(q))/a]
end

p =
     8     2

p =
   134     0

p =
    53    -5

p =
    34    -8

If you want to find the roots of the sixteen first order polynomials with elements taken from e and from b (and not necessarily the same element of each vector) the easiest way is to use a double for loop.

[Ani Asoyan]

sorry I didn't understand... my polynomial is 1st order

[Steven Lord]

It is not first order as you've written it. Use one or two for loops to construct and solve the four or sixteen first order polynomials or use the explicit formula for the solution of a linear equation.

% if q*x = w then x = w./q
a=2, c=1, d=2 , g=0.5,  k=2 , f=2
e=[2,8,5,4]
b=[1,2,3,4]

q = -c.*k+ (f.^2)*(e.^2)/a+2*e.*g;
w = d.*e-(f*b.*e)/a;
x = w./q % Using ./ instead of / because w and q are vectors

check = q.*x - w % Should be all 0's or close to it