Quiz  

 1)  What  is  the  formula  to  calculate  the  

percentage  rela7ve  of  numerical  error?  2)  Con7nue  the  following  Taylor  series  formula:      f(x+h)  =  f(x)  +  …    

 

 

Roots  of  Polynomials  

Revision  on  Basic  Laws  of  Indices  

Examples:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

The general form of polynomial equations

•  Linear equation (or polynomial of order 1) •  Quadratic equation (or polynomial of order 2) •  Cubic equation (or polynomial of order 3)

(2.1)

Simple  Linear  Equa<ons  

(1) Solve

(i) By cross-multiplying

(ii) Removing brackets

(iii) Rearranging

(2) Solve

Quadra<c  Equa<ons  

two unequal real roots

The expression is called the discriminant.

Roots of a general quadratic eqn.

two unequal complex roots the roots are real and equal

Quadratic Formula 1

(2.2)

are (2.3) x1,2 =

�b±pb

2 � 4ac

2a

Quadra<c  Equa<ons  

How do we get these roots? Assume that the equation factors as

Roots of a general quadratic eqn.

Quadratic Formula 2 (from Lagrange resolvents)

are

(2.4)

(2.5)

x

2 + px+ q = (x� ↵)(x� �)

Continue!

Quadra<c  Equa<ons  

Then, either

Factorization

or

(1)

(2)

MATLAB  Implementa<ons  

function x = quadraticSolver(a,b,c) x = zeros(2,1); d = sqrt(b^2 - 4*a*c);x(1) = ( -b + d ) / (2*a);x(2) = ( -b - d ) / (2*a);

Create a solver by writing functions, for example:

This solver accepts 3 input numbers to find roots of a quadratic function

and returns a vector output (answers).

MATLAB  Implementa<ons  Call the function from the command line (or Command Window): This solver can be downloaded at: http://people.bu.edu/andasari/courses/Fall2015/quadraticSolver.m

>> quadraticSolver(3,-4,2)

ans

0.6667 + 0.4714i 0.6667 - 0.4714i

Tschirnhaus  Transforma<on  

be a polynomial of order n Let

where

Then the substitution

converts Pn into a depressed polynomial

Cubic  Equa<on  (Cardano’s  Formula)  

is by the substitution

The cubic equation

reduced to

Letting

(2.6)

(2.7) we obtain

Cubic  Equa<on  (Cardano’s  Formula)  

and substituting into the depressed cubic eqn. (2.7):

Now let and

Cubic  Equa<on  (Cardano’s  Formula)  

Solving using the quadratic formula, we get

The 6th order polynomial can be solved by using the quadratic formula, by taking and the equation reduces to

Cubic  Equa<on  (Cardano’s  Formula)  

Substituting k back to u

or

If we take the positive root

and substitute to

we get the expression for

It goes the same way if we take the negative root for u3.

Cubic  Equa<on  (Cardano’s  Formula)  

Putting

the roots can be interpreted as follows: (i) if D > 0, then one root is real and two are complex conjugates (ii) if D = 0, then all roots are real, and at least two are equal (iii) if D < 0, then all roots are real and unequal

and

is called the discriminant.

Cubic  Equa<on  (Cardano’s  Formula)  

then

are roots of the equation If

The roots of are (prove it???)

MATLAB  Implementa<on  Write a Matlab solver to calculate roots of cubic equations.

MATLAB  Implementa<on  function [x,y] = CardanoFormula(a,b,c,d)% Returns roots and discriminant of cubic equations % ax^3 + bx^2 + cx + d = 0 x = zeros(3,1); Q = ((3*a*c)-(b^2))/(9*a^2);R = ((9*a*b*c)-(27*a^2*d)-(2*b^3))/(54*a^3); y = sqrt(Q^3 + R^2); u = (R+y)^(1/3);v = (R-y)^(1/3); Bb = u + v;zz1 = -(1/2)*Bb + (1/2)*sqrt(-1)*sqrt(3)*sqrt(Bb^2+4*Q);zz2 = -(1/2)*Bb - (1/2)*sqrt(-1)*sqrt(3)*sqrt(Bb^2+4*Q);

% Roots of the cubic equationx(1) = BB - b/(3*a);x(2) = -b/(3*a) + zz1;x(3) = -b/(3*a) + zz2;

MATLAB  Implementa<on  The solver can be downloaded at: http://people.bu.edu/andasari/courses/Fall2015/CardanoFormula.m

Polynomial  Division  

Divide by

. .

The  Factor  Theorem  If is a root of the equation then

is a factor of

Example: Factorize and use it to solve the cubic equation

Let If

Since then is a factor.

Next, we divide by the factor

Hence

. .

From which we get

Use  factoriza7on  or      quadra7c  formula  

The  Remainder  Theorem  For quadratic equations, the remainder theorem states

For cubic equations, the remainder theorem states

is divided by If the remainder will be

is divided by If the remainder will be

Bairstow’s  Method  Bairstow’s method is an algorithm used to find the roots of a polynomial of arbitrary degree (usually order 3 and higher). The method divides a polynomial

are guessed. The division gives us a new polynomial by a quadratic function

and the remainder

, where r and s

(2.8)

(2.9)

(2.10)

Bairstow’s  Method  

recurrence relationship

obtained by the standard polynomial division. The quotient and the remainder are

The coefficients can be calculated by the following

(2.11c)

(2.11b)

(2.11a)

Bairstow’s  Method  

to determining the value of r and s such that ,

remainder If is an exact factor of then the

are the roots of . is zero and the roots of

hence, Because b0 and b1 are functions of

The Bairstow’s method reduces

r and s, they can be expanded using Taylor series, as:

(2.12a)

(2.12b)

Bairstow’s  Method  By setting both equations equal to zero

To solve the system of equations above, we need partial derivatives of b0 and b1 with respect to r and s. Bairstow showed that these partial derivatives can be obtained by a synthetic division of the b’s in a fashion similar to the

(2.13b)

(2.13a)

Bairstow’s  Method  way in which the b’s themselves were derived, that is by replacing ai’s with bi’s, and bi’s with ci’s, such that,

where

(2.14a)

(2.14b)

(2.14c)

Bairstow’s  Method  Then substituting into equations (2.13a) and (2.13b) gives

These equations can be solved for Δr and Δs, which can be used to improved the initial guess of r and s.

At each step, an approximate error in r and s estimated as

and (2.15)

Bairstow’s  Method  If and criterion, the values of the roots can be determined by

At this point, there exist three possibilities (1) If the quotient polynomial fn-2 is a third (or higher)

where is a stopping

(2.16)

order polynomial, the Bairstow’s method can be

applied to the quotient to evaluate new values for

r and s. The previous values of r and s can serve as

the starting guesses.

Bairstow’s  Method  (2) If the quotient polynomial fn-2 is a quadratic function,

then use eqn. (2.16) to obtain the remaining two

roots of fn(x). (3) If the quotient polynomial fn-2 is a linear function,

then the single remaining root is given by

(2.17)

Newton-­‐Raphson  Method  The Newton-Raphson, or simply Newton’s method is one of the most useful and best known algorithms that relies on the continuity of derivatives of a function. The method is usually used to to find the solution of nonlinear equations f(x) = 0 whose derivatives, f′(x) and f′′(x), are continuous near a root. We can derive the formula for Newton’s method from Taylor series of 1st order:

(2.18)

Newton-­‐Raphson  Method  Setting the left hand side, f(xi+1) to zero and solving for xi+1, we get which is the Newton’s iteration function for finding root of the equation f(x) = 0.

(2.19)

Newton-­‐Raphson  Method  Order of a Root Assume that f(x) and its derivatives f′(x), f′′(x), … f(M)(x) are defined and continuous on an interval about x = p. We say that f(x) = 0 has a root of order M at x = p if and only if

f(p) = 0, f′(p) = 0, …, f(M-1)(p) = 0 and f(M)(p) ≠ 0

(2.20)

Newton-­‐Raphson  Method  For:

- root of order M = 1, it is called a simple root - M > 1 it is called a multiple root - M = 2, it is called a double root - etc

Newton-­‐Raphson  Method  If p is a root of a function then the function f(x) has a simple root at p = −2 and a double root at p = 1. This can be verified by considering the derivatives and . At the value p = −2, we have:

f(−2) = 0 and f′(-2) = 9, so M = 1, hence p = −2 is a simple root. At the value p = 1, we have:

f(1) = 0, f′(1) = 0, and f′′(1) = 6, so M = 2 and p = 1 is a double root.

MATLAB  Implementa<on  function x0 = newtonRaphson() x0 = 1.2;iter = 0; while abs(f(x0)) > 1e-4

iter = iter + 1;x1 = x0 - f(x0)/fprime(x0) x0 = x1;

end disp(['roots found after ', num2str(iter), ' iterations']); function y = f(x)y = x^3 - 3*x + 2; function y = fprime(x)y = 3*x^2 – 3;

MATLAB  Implementa<on  

>> p = [1 0 -7 -6];>> polyval(p, [1 2 3])

ans

-12 -12 0

at

The polynomial is evaluated

Using MATLAB’s built-in functions

MATLAB  Implementa<on  

>> polyder(p)

ans

3 0 -7

by Evaluate the derivative

Using MATLAB’s built-in functions

MATLAB  Implementa<on  

>> a = [1 -3.5 2.75 2.125 -3.875 1.25];>> r = roots(a)

r =

2.0000 -1.0000

1.0000 + 0.5000i1.0000 – 0.5000i0.5000

To find roots of Using MATLAB’s built-in functions

MATLAB  Implementa<on  

>> format long>> r = roots(a)

r =

2.000000000000005 -1.000000000000000 0.999999999999998 + 0.500000000000001i 0.999999999999998 - 0.500000000000001i 0.500000000000000

All computations in MATLAB are done in double precision. To switch between different output, use the command “format” (type “help format” for more info).

•  Numerical  Methods  for  Engineers,  S.C.  Chapra  and  R.P.  Canale.  

•  Numerical  Methods  Using  MATLAB,  John  H.  Mathews  and  Kur7s  D.  Fink.  

References