Using Newtons method to find multiple polynomial roots

Question

How is Newtons method used to find multiple polynomial roots?

I’m reading here on Wikipedia that “When one root r has been found, one may use Euclidean division for removing the factor x – r from the polynomial.” and then “To reduce this error, one may, for each root that is found, restart Newton's method with the original polynomial, and this approximate root as starting value”.

I think restarting the method using approximate root as a starting value would only give a better approximation of the first root found and not the second.

If an example can be provided that would be helpful.

I’m working off the following pictures:

Ref: link

Answer 1

The challenge of computing multiple roots with Netwon's method is to find good starting approximations of each of them (good enough that the iterations do converge to the respective roots).

The method explained in Wikipedia to obtain these initial approximations is to solve the deflated equations, i.e. what you obtain by dividing the polynomial by the binomials $x-r_k$, so that the reduced polynomial does not have $r_k$ as a root (but still has the next ones).

When you have a good starting value for all roots, Newton's iterations with the full polynomial will perform equally well for all roots. You need the refinement steps because deflation introduces inaccuracies.

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Note that nothing is said about which starting value to choose to solve for the first root nor for those of the deflated polynomials.

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Simplistic example: $x^2-2=0$.

• Iterations for the first root: $1, 1.5, 1.41666$
• Deflated polynomial: $x+1.41666$
• Iterations for the second root: $1,-1.41666,-1.41666$
• Final iterations for the first root: $1.41666,1.4142156747561,1.4142135623747$
• Final iterations for the second root: $-1.41666,-1.4142156747561,-1.4142135623747$

Answer 2

Taking as example a simple polynomial $(x-a)(x-b)(x-z)\tag{1}$ with $z$ the first root that Newton found by chance. Then you may, after defining $x-z\ne0$, divide $(1)$ by $(x-z)$ and attack the remaining roots by solving $(x-a)(x-b)\tag{2}$ -- so far the theory.

But because Newton's method is an approximation it returns $z\pm r$ with $r$ a little error, simplified in the following as $r$. Now dividing $(1)$ by $(x-(z+r))$ will not reduce it, since $\displaystyle\frac{x-z}{x-z-r}\tag{3}$ is not $1$. Taking only its first three terms of Taylor at $z$ it's $\approx\displaystyle -\frac{1}{r}\left(x-z\right)-\frac{1}{r^2}\left(x-z\right)^2-\frac{1}{r^3}\left(x-z\right)^3\tag{4}$what is for sure close to $1$, alas not exactly. Thus, after Euclidean division of $(1)$ with approximated $z$ you'll try to solve $\displaystyle(x-a)(x-b)\left(-\frac{1}{r}\left(x-z\right)-\frac{1}{r^2}\left(x-z\right)^2-\frac{1}{r^3}\left(x-z\right)^3\right)\tag{5}$ -- no chance to get $a$ or $b$ of original polynomial $(1)$.

That is why it's a good idea to have Newton run $(1)$ again with a "near hit" found for $(5)$.

After you supplemented your query with details about Newton's method, some additional remarks.
Within the process of finding several or (if possible) all roots of a polynomial Newton's method is only one component to introduce errors, the second is polynomial division. If you "work off Newtons method" only, you ignore a source of errors.

Newton's method is an approximation. Thus if your pocket calculator (or system you use) returns an integer as root found by Newton, the error is smaller than your calculette with 10..12 mantissa digits may show -- in fact, the result is rounded and may by chance be an integer. If the following polynomial division, an Euclidean division, produces no rest (whew!), you may continue your quest for roots with the reduced polynomial.
If not, and you ignore/drop/neglect the remnant of the Euclidean division (as described by Yves Daoust and called "deflation"), the "deflated polynomial" will not have the same roots than the original polynomial.

Still not convinced? Here an example with numbers: The polynomial $(x-2/7)(x-9)^2=\displaystyle\frac{7x^3-128x^2+603x-162}{7}$ has deliberately one root not representable as decimal. This is to demonstrate the error of Newton running on a pocket calculator (or similar system).
The first root found starting with $x_0=1$ is $0.2857142857$, but since polynomial division is daunting if you do have to do it manually, you round this result to $0.2857143$. The polynomial division yields
$x^2-17.9999999857x+80.9999997469$ with rest $0.0000010848396137$
This rest shows, the reduced polynomial is only close to the original, its roots are not the same as those of the original, they must be verified using the original polynomial.
As the reduced polynomial is a quadratic I take the shortcut w/o Newton and get the roots $x_1=8.99964716175,x_2=9.00035282397$ while it should be $x_{1/2}=9$, an effect of dropping the remnant of the Euclidean division.

The situation may even be worse. To demonstrate this I simulate a slightly bigger error in the result of the first root approximation by cutting off few digits to $0.28571$. Then the polynomial division yields
$x^2-18.0000042857x+81.0000759184$ with a remnant of $-0.000325452215165$
This quadratic has no real roots while the original polynomial has.