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I have a simple code for bisection method that isn't working correctly. Can someone please tell me what is wrong.

 f[x_] := x*Log[x] + x^2;
Plot[f[x], {x, 0, 1}]
i = 1;
Subscript[a, i] = 0.4;
Subscript[b, i] = 0.6;
Subscript[error, i] = Abs[f[Subscript[x, i]]];
Subscript[error, i] = 1;
While[
 Subscript[error, i] >= 0.0001,
 Subscript[x, i] = (Subscript[a, i] + Subscript[b, i])/2;
 If[
   N@f[Subscript[x, i]] > 0, 
   Subscript[a, i + 1] == Subscript[a, i] && 
    Subscript[b, i + 1] ==  Subscript[x, i], 
   Subscript[a, i + 1] == Subscript[x, i] && 
    Subscript[b, i + 1] == Subscript[b, i]
   ]
  Print["iteration ", i , ", output", ": ", Subscript[x, i], ", ", 
   Subscript[a, i], ", ", Subscript[b, i]]
  i++
 ]
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  • 2
    $\begingroup$ Avoid the subscripts, except for pretty-printing purposes. It just makes debugging harder to do. $\endgroup$ Apr 1, 2018 at 17:19

1 Answer 1

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Equal (a.k.a. ==) is a logical operation. You have to use Set (a.k.a. =) in order to assign values to variables. Moreover, And (a.k.a. &&) is a logical operation; use a semicolon ; instead. This is the cleaned code, incoporating J.M.'s remark on Subscript.

f[x_] := x Log[x] + x^2;
Plot[f[x], {x, 0, 1}]
i = 1;
imax = 100;
ClearAll[a, b, x, error, c];
a[i] = 0.4;
b[i] = 0.6;
x[i] = b[i];
error[i] = Abs[f[x[i]]];
While[error[i] >= 0.0001 && i < imax,
 x[i + 1] = (a[i] + b[i])/2.;
 c = f[x[i + 1]];
 If[c > 0.,
  a[i + 1] = a[i]; b[i + 1] = x[i + 1],
  a[i + 1] = x[i + 1]; b[i + 1] = b[i]
  ];
 error[i + 1] = Abs[c];
 Print["iteration ", i, ", output", ": ", x[i], ", ", a[i], ", ", b[i]];
 i++;
 ]

Remark

Performancewise, it would be better to forget old iterates and simply reuse the variables, a, b, x, and error:

i = 1;
imax = 100;
ClearAll[a, b, x, error];
a = 0.4;
b = 0.6;
x = b;
error = Abs[f[x]];
While[error >= 0.0001 && i < imax,
 x = (a + b)/2.;
 c = f[x];
 If[c > 0.,
  a = a; b = x,
  a = x; b = b
  ];
 error = Abs[c];
 i++;
 ]
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2
  • $\begingroup$ Thank you very much :) $\endgroup$
    – user57225
    Apr 1, 2018 at 18:20
  • $\begingroup$ You're welcome! $\endgroup$ Apr 1, 2018 at 18:21

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