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Mathematics Calculus Wolfram Language

Make a function that can find the 3 rod of a number?

Martin Friis

Posted 6 years ago

Here you can see my code, as it is for now. Must admit, I'm kinda lost why it doesn't work. I think there is something about the syntax I'm not understanding. Best regards, kobikrod2[x0_] := Module[{xny, xgl}, xs = If[Positive[x0], xs = N[x0], Print["Tallet er ikke positivt"]] xgl = xs // 2 ; xny++ 1 = 1/3 (2xgl + (xs/xgl^2)); While[xny != xgl, xgl = xny; xny = xgl + 1 = 1/3 (2xgl + (xs/xgl^2))]; xny]

Here you can see my code, as it is for now. Must admit, I'm kinda lost why it doesn't work. I think there is something about the syntax I'm not understanding.

Best regards,

kobikrod2[x0_] := Module[{xny, xgl},
  xs = If[Positive[x0], xs = N[x0], Print["Tallet er ikke positivt"]]
     xgl = xs // 2 ;
  xny++ 1 = 1/3 (2*xgl + (xs/xgl^2));

  While[xny != xgl, xgl = xny; 
   xny = xgl + 1 = 1/3 (2*xgl + (xs/xgl^2))];
  xny]

POSTED BY: Martin Friis

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Murray Eisenberg

Murray Eisenberg, University of Massachusetts Amherst

Posted 6 years ago

The stopping criterion presented in that answer (`Abs[func@#] > precision &`) is a dangerous! If the function is "flat" near a zero, you could stop when the magnitude of the current approximation is close to 0 yet the approximation is far from 0. It is much better to base the stopping criterion primarily on how far apart the x-values are from each other (e.g., compare the current and preceding approximations). Of course it doesn't hurt to add also a condition about the y-value of the function; but usually it's better, then, to use some kind of relative error rather than absolute error.

POSTED BY: Murray Eisenberg

Rohit Namjoshi

Posted 6 years ago

There are several problems with the code. e.g. `xs // 2`. In Mathematica, `//` is the postfix form of function application so `xs // 2` evaluates to `2[xs]`. Take a look at this and this to get more familiar with Mathematica. From the code it looks like you are trying to implement cube root finding using the Newton-Raphson method. Here is one way to do that. Need to find the value of x for which the function f is zero. f[x_, n_] := x^3 - n Generic Newton-Raphson method newtonRaphson[func_, guess_, precision_: 0.0001] := NestWhile[# - func[#]/func'[#] &, guess, Abs[func@#] > precision &] Cube root of 3 newtonRaphson[Curry[f][3], 1.0] (* 1.44225 ) Cube root of 10 newtonRaphson[Curry[f][10], 1.0] ( 2.15443 *)

POSTED BY: Rohit Namjoshi

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