intro Link to heading

If the derivative of function, whose root is to be found, does not exist or hard to find, secant method can be used instead of Newton raphson method, since it does not required the derivative but it requires two initial guesses for the roots (Mohan, 2021). Since this method retains only the most recent estimate, the root does not necessary bracketed (Weisstein). Because of that secant method may not converge (Han, 2017). There is a report explaining origin and evolution of secont method in 1-d (Papakonstantinou & Tapia, 2013).

formula Link to heading

  • The Newton-Raphson method has a iterative formula in finding root of $f(x)$ as follow $$\tag{1} x_{n+2} = x_{n+1} - \frac{(x_{n+1} - x_n) f(x_{n+1})}{f(x_{n+1}) - f(x_n)} $$ with $n = 0, 1, 2, 3, \dots$, $x_0$ and $x_1$ are the initial guesses for the root.
  • As $n$ increases, the value of $f(x_n)$ is nearer to $0$.

flowchart Link to heading

  • The Newton-Raphson method can be represented in following flowchart.
    flowchart TD B --> I --> P1 --> O1a O1b --> P2 --> C --"N"--> P3 --> O1c C --"Y"--> O --> E O[/"x[n+2]"/] C{"|f(x[n+2])| < ϵ"} P2["x[n+2] = x[n+1]
    — ( x[n+1] — x[n] ) f(x[n])
    / ( f(x[n+1]) — f(x[n]) )"] P1["n = 0"] P3["n = n + 1"] I[/"f(x), ε, x[0], x[1]"/] O1a(("1")); O1b(("1")); O1c(("1")); B(["Begin"]) E(["End"]) classDef style1 fill:none, color:inherit classDef style2 fill:#888, color:#fff class P1,P2,P3,P4 style1 class I,C,O,B,E style1 class O1a,O1b,O1c, style2 linkStyle default stroke:#8a8
  • Parameter $\varepsilon$ is small value that is acceptable representing $0$, since it might require a lot of iterations to find $f(x) = 0$, where $f(x) \approx 0$ is easier.

algorithm Link to heading

  1. Start.
  2. Input $f(x)$, $\varepsilon$, $x_0$, $x_1$.
  3. Set $n = 0$.
  4. Calculate $x_{n+2} = x_{n+1} - (x_{n+1} - x_n) f(x_{n+1}) / [ f(x_{n+1}) - f(x_n) ]$.
  5. If $f(x_{n+2}) < \varepsilon$, go to Step 8.
  6. Increase $n$ using $n = n + 1$.
  7. Go to Step 4.
  8. Print $x_{n+2}$.
  9. End.

problem Link to heading

  • Equation to be solved is $$\tag{2} f(x) = x^2 - 102.345x + 234.5. $$

code Link to heading

  • An example of code in JS is as follow.
    function f0(x) {
  let y = x**2 - 102.345*x + 234.5;
  return y;
}

let eps = 1E-5;
let n = 0
let x = [];
x[n] = 5;
x[n+1] = 10;
let fx = f0(x[n]);

console.log("n   x[n]           f(x[n])     eps");
  console.log(
    n, ' ',
    x[n].toExponential(7), ' ',
    fx.toExponential(4), ' ',
    eps
  );
  console.log(
    n+1, ' ',
    x[n+1].toExponential(7), ' ',
    fx.toExponential(4), ' ',
    eps
  );

while(Math.abs(fx) > eps) {
  xx = x[n+1] - f0(x[n+1]) * (x[n+1] - x[n]) / (f0(x[n+1]) - f0(x[n])); 
  x.push(xx);

  fx = f0(x[n+2]);

  console.log(
    n+2, ' ',
    x[n+2].toExponential(7), ' ',
    fx.toExponential(4), ' ',
    eps
  );

  n = n + 1;
}

console.log();
console.log("root = ", x[n+1].toFixed(5));
    url https://onecompiler.com/javascript/3zytsa2wj.
  • Previous code produces following results.
    n   x[n]           f(x[n])     eps
0   5.0000000e+0   -2.5223e+2   0.00001
1   1.0000000e+1   -2.5223e+2   0.00001
2   2.1123132e+0   2.2777e+1   0.00001
3   2.3647403e+0   -1.9273e+0   0.00001
4   2.3450469e+0   -4.5833e-3   0.00001
5   2.3450000e+0   9.2667e-7   0.00001

root =  2.34500

graphics Link to heading

  • Previous data can be visualized for $r_1$, the root, as follow.
  • It can also visualized for $f(r_1)$ as follow.
  • You may hover your mouse on a point to see its pair values.

challenges Link to heading

  • Explain in brief the background of Newton-Raphson method and how it is formulated until the iterative formula in Equation (1) is obtained. If necessary draw also some graphs to make the exlplanation clearer.
  • Find resources explaining that convergence of Newton-Raphson method is quadratic, while convergence of secant method is linear. And read it.
  • Try to find the root of $$\tag{3} f(x) = x - 3.45, $$ and show which part of the given code should be modified.
  • Compare the flowchart for scanning method in 000s, bisection method in 0004 and Newton-Raphson method in 0014. Which one is easier to understand? Why?
  • Find roots of $$\tag{4} f(x) = (x - 1.23456)(x - 9.87654)(x - 23.45678), $$ with different initial guesses, e.g. 0, 5, 11, 40. Create table to show the relation between initial guess and root found.
  • Compare the result obtained previously using Newton-Raphson method in 0014. Which one is more efficient?
  • What will happen if $x_0 = 5 \pm \delta$ in finding root of $$\tag{5} f(x) = (x - 2)(x - 8) $$ by varying value of $\delta$ to assure that both roots are found.
  • Until now all examples are in the form of polynomial, will this method also work for non-polynomial function? Test it for at least such functions. Report the results.
  • What would be the advantages and the disadvantages from this method compared to Newton-Raphson method?
  • Can this method apply to a constant function, e.g. $f(x) = c$, where $c$ is a constant value. Explain in brief.

refs Link to heading