How do I solve a polynomial equation of the form: q^n + a(1)*q^(n-1) +...+ a(k)*q^(n-k) +...+ a(n) = 0 ? In these equation, a(k) are known quaternion constants (coefficients), and q - unknown quaternion variable. The order n is relatively small but not limited to 1 or 2 - i.e., when a simple closed form solution is not possible, is there any good iterative algorithm? Does it always have n roots? Can you point to some sources where these equations are discussed?

-- Oleg (opiany@lsuhsc.edu), January 11, 2003

Answers

Hello Oleg:

If a(n) is constrainted to real numbers, then the problem is the same as polynomial equations of complex numbers. This is the most useful class of quaternion polynomials, doing things like representing functions like sine and exp. If all a(n) happen to commute with q (perhaps they are powers of k), then the problem is again just like the complex polynomial.

If this is not the case, then I would break the problem in two. I would express each a(n) as a_parallel_to_q and a_perpendicular_to_q. The first case will again be exactly like the complex polynomical. The second only involves the curl. It is often fairly easy to characterized.

Because the problem of quaternion polynomials can often be simplified to complex polynomials, many similar statements can be made. I do think there are always n roots. My focus has not been on quaternion polynomials, so I do not have any good references to pass along.

doug

-- Douglas Sweetser (sweetser@theworld.com), January 18, 2003.

Oleg asked asked if a quaternionic polynomial equation of order n has n roots. Doug suggested that the answer is in the affirmative.

It is easy to show by counter-example that there can be an infinitude of roots for a quaternionic polynomial equation of order 2. Consider the quadratic equation q^2 + 1 = 0. And consider the quaternion q = i cos(v) + j sin(v) where v is any real number. Then the square of the above expression for q is:

q^2 = i^2 [cos(v)]^2 + ij cos(v)sin(v) + ji sin(v)cos(v) + j^2 [sin(v)]^2

= (-1)[cos(v)]^2 + (k)cos(v)sin(v) + (-k)sin(v)cos(v) + (-1)[sin(v)]^2

= (-1)[cos(v)^2 + k cos(v)sin(v) - k cos(v)sin(v) + (-1)[sin(v)]^2

= (-1)[cos(v)^2 + 0 + (-1)[sin(v)]^2

= (-1){ [cos(v)]^2 + [sin(v)]^2 }

= (-1) { 1 }

= -1

So the above choice for q = q(v) satisfies the equation q^2 + 1 = 0 for all real values of v. Of course, Hamilton discovered 3 roots (i,j and k) for the above equation of order 2. My example proves there are a continuum of roots.

Best regards,

Matthew

-- Matthew McCann PE (mmccann@franciscan.edu), March 06, 2004.

Hello Matthew:

I guess the message is that we cannot pin down where the 3-vector points to, so there is an infinite selection. I view the "i" of a complex number as the sum of i, j, and k of a quaternion. Working with co-directional quaternions is like fixing i, j, and k.

doug

-- Douglas Sweetser (sweetser@alum.mit.edu), March 07, 2004.

Hello, Doug,

The sum of the quaternions i,j,k cannot be equated with the imaginary unit of the complex field. The latter always commutes in a product with a quaternion, but the former cannot always commute in a product with any quaternion.

Best regards,

Matthew

-- Matthew McCann PE (mmccann@franciscan.edu), March 07, 2004.

Does that mean, if I have a 4-tap filter h[n]= [h0,h1,h2,h3] where all coeff are quaternions. I cant find the roots of the filter h[n], that is H(z).

If any one knows the answer pls let me know.

Cant we use the complex numbers to decompose quaternions. For SVD of quaternion , we do the similar trick.

cheers Pancham

-- Pancham Shukla (spancham@yahoo.com), March 15, 2005.