Question

Design a Root Finding solver in Matlab that satisfies the arguments below;
• Inputs will be coeficients of the equation.
• Outputs will be roots
• It should work with any polynomial
?=?0+?1?1+?2?2+?3?3+⋯+????

Answer 1

The below function gives you the roots of the any polynomial equation which is in the form of  p1xn+...+pnx+pn+1=0.

Polynomial equations contain a single variable with nonnegative exponents.

r = roots(p) ------------- Predefined matlab function.

Where r is the root of the equation.

p is the column vector, which is the list of coeficients.

Case 1:

For example, create a vector to represent the polynomial x2−x−6, then calculate the roots.

p = [1 -1 -6];
r = roots(p)

r =

     3
    -2

By convention, matlab returns the roots in a column vector.

The poly function converts the roots back to polynomial coefficients.

p2 = poly(r)

p2 =

     1    -1    -6

Case 2:

for a quadratic polynomial x4−1=0.

p = [1 0 0 0 -1];
r = roots(p)

r = 4×1 complex

  -1.0000 + 0.0000i
   0.0000 + 1.0000i
   0.0000 - 1.0000i
   1.0000 + 0.0000i

Case 3 :

For example, find the values of θ that solve the equation.

3cos2(θ)−sin(θ)+3=0.

Use the fact that cos2(θ)=1−sin2(θ) to express the equation entirely in terms of sine functions:

−3sin2(θ)−sin(θ)+6=0.

Use the substitution x=sin(θ) to express the equation as a simple polynomial equation:

−3x2−x+6=0.

p = [-3 -1 6];

r = roots(p)

r = 

    -1.5907
    1.2573

To resolve the substitution, use θ=sin−1(x). The asin function calculates the inverse sine.

theta = asin(r)

theta = 

  -1.5708 + 1.0395i
   1.5708 - 0.7028i