[Math] Newton-Horner Method Example

numerical methods

I'm struggling to understand Newton-Horner's method and to find information about it so pardon me if I seem a bit lost in expressing my question, my goal is to implement it in racket as a learning exercise.

Given a polynomial $P(x)$ of $n$ degree and that has $\bar x_n$ roots it is used to find said roots.

$$ x_{i + 1} = x_i – \frac{\frac{f(x_i)}{f'(x_i)}}{1 – \frac{f(x_i)}{f'(x_i)} \sum_{j=1}^{n}(\frac{1}{x_i – \bar x_j})} $$

I can identify Newton's Method, however I'm a bit lost in how exactly the whole method is applied.

I know that for the first iteration it is the result if using Newton's method with a supplied first guess value (because the Sum part is $0$). I've evaluated a sample polynomial $P(x) = 4x^5 – 3x^4 – 2x^2 +12x + 4$ for $x_0 = 0$ but on subsequent iterations the sum part is always zero.

I've searched quite extensively but all I find are the descriptions of Newton's Method and Horner's Method used in an independent manner I've also found lots of old homework exercises implemented in Mathematica or matlab, but none that describes the method thoroughly. If anyone has links to resources or can explain it I'd appreciate it a lot.

Best Answer

Methods of Horner and Newton-Raphson combined

Let $x = r$ be an approximation to a root of a polynomial $f(x)$, Dividing $f(x)$ by the linear factor $x-r$ using Horner method we get $$ f(x) = g(x)(x-r) + a $$ from which $f(r)=a$. On differentiating follows $$ f'(x) = g'(x)(x-r) + g(x) $$ so that $f'(r) = g(r)$. Now $g(r)$ can be found as before by dividing it by the linear factor $x-r$: $$ g(x) = h(x)(x-r) + b $$ giving $g(r) = b$.

The new Newton-Raphson approximation $r'$ is now found as $$ r' = r - f(r)/f'(r) = r - a/b. $$

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