Is there a name for this modified newton's method?

Question

I know that Newton's method has the following formula:

$$f_{t+1} (x)= f_{t}(x)-f'(x)/f''(x)$$

The source code at the end of the post seems to use the following construction instead: $$f_{t+1} (x)= f_{t}(x)-f'(x)/(f''(x)* c +f'(x))$$

I wanted to figure out the property of this optimization, but I cannot look up what this is because I do not know if this thing has a name for it. What is this modified newton's method called?

The below is the source code and I marked where the newton method occurs.

double alhood(double a, double ss, int D, int K)
{ return(D * (lgamma(K * a) - K * lgamma(a)) + (a - 1) * ss); }

double d_alhood(double a, double ss, int D, int K)
{ return(D * (K * digamma(K * a) - K * digamma(a)) + ss); }

double d2_alhood(double a, int D, int K)
{ return(D * (K * K * trigamma(K * a) - K * trigamma(a))); }

double opt_alpha(double ss, int D, int K)
{
    double a, log_a, init_a = 100;
    double f, df, d2f;
    int iter = 0;

    log_a = log(init_a);
    do
    {
        iter++;
        a = exp(log_a);
        if (isnan(a))
        {
            init_a = init_a * 10;
            printf("warning : alpha is nan; new init = %5.5f\n", init_a);
            a = init_a;
            log_a = log(a);
        }
        f = alhood(a, ss, D, K);
        df = d_alhood(a, ss, D, K);
        d2f = d2_alhood(a, D, K);
        log_a = log_a - df/(d2f * a + df); ####<-------HERE
        printf("alpha maximization : %5.5f   %5.5f\n", f, df);
    }
    while ((fabs(df) > NEWTON_THRESH) && (iter < MAX_ALPHA_ITER));
    return(exp(log_a));
}

Answer 1

Note that your iteration reads $$ \ln(a_+)=\ln(a)-\frac{f'(a)}{f''(a)·a+f'(a)} $$ when ultimately you want to find solutions to $f'(a)=0$.

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Set $x=\ln(a)$, and consider the function $$g(x)=f'(e^x)·e^{α·x}.$$ The derivative of $g$ is $$g'(x)=f''(e^x)·e^{(1+α)·x}+α·f'(e^x)·e^{α·x}$$ and the corresponding Newton method $$ x_+=x-\frac{f'(e^x)}{f''(e^x)·e^x+α·f'(e^x)}\ \text{ or }\ \ln(a_+)=\ln(a)-\frac{f'(a)}{f''(a)·a+α·f'(a)} $$ Since the additional factor is always positive the root set does not change, $0=g(\ln a)$ has the same solutions as $0=f'(a)$. The formula from the algorithm results for $α=1$.

Answer 2

Given the nature of the question, I assume you meant... $$p_{n+1}=p_n-{{f(p_n) \cdot f^{'}(p_n)} \over {[f^{'}(p_n)^2]-f(p_n) \cdot f^{''}(p_n)}}$$

This is the result of applying newton's method to $$u(x)={{f(x)} \over {f^{'}(x)}}$$

and looking for a p such that $u(x)$ is $0$. Since f(x) should be $0$ if $u(x)$ is $0$, this is a fair thing to do. This method is so much better because it allows quadratic convergence to a root of $f(x)=0$, assuming the conditions are met. Of course, its also harder to set up.