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

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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,... ?

### Accepted Answer

Ameer Hamza
on 2 Jun 2020

Edited: Ameer Hamza
on 2 Jun 2020

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

##### 5 Comments

Ameer Hamza
on 3 Jun 2020

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.

### More Answers (1)

Steven Lord
on 2 Jun 2020

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.

##### 2 Comments

Steven Lord
on 2 Jun 2020

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

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