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05. Solution of Non Linear Equations (Method of Iteration)

Dated: 15-08-2025

Rewrite \(f(x) = 0\) into \(x = \phi(x)\) and take \(x = x_0\) as the initial approximation. Then keep iterating as such

  • \(x_1 = \phi(x_0)\)
  • \(x_2 = \phi(x_1)\)
  • \(x_3 = \phi(x_2)\)

If this sequence1 converges to \(\alpha\) then it is taken as the root.2

Theorem

Let \(f(x) = 0\) be written as \(x = \phi(x)\) and \(I\) be an interval3 containing \(\alpha\) (where \(\alpha\) is the root2) then, if \(|\phi^\prime(x)| < 1 \ \forall \ x \in I\) then the sequence1 \(\{x_0, x_1, \ldots, x_n\}\) converges to \(\alpha\).

\(\phi^\prime(x) = \frac d {dx} \phi(x)\)

Example

Find the real4 root2 of the equation \(x^3 + x^2 - 1 = 0\) by method of iterations.

Solution

\[\because f(x) = x^3 + x^2 - 1\]
\[\implies f(0) = 0^3 + 0^2 - 1 = -1 < 0\]
\[\implies f(1) = 1^3 + 1^2 - 1 = 1 > 0\]

Hence, a real4 root2 exists in \([0, 1]\).

\[x^3 + x^2 - 1 = 0\]
\[x^2(x + 1) = 1\]
\[x^2 = \frac 1 {x + 1} \implies x = \frac 1 {\sqrt {x + 1}} = \phi (x)\]
\[\phi^\prime(x) = - \frac 1 {2 (x + 1)^\frac 3 2}\]

Checking the extremes

\[\phi^\prime(0) = \frac 1 2 < 1\]
\[\phi^\prime(1) = \frac 1 {2^\frac 5 2} < 1\]

so, \(|\phi^\prime(x)| < 1 \ \forall \ x \in [0, 1]\).

Let \(x_0 = 0.65\)

\[ x_1 = \phi(x_0) = \frac{1}{\sqrt{x_0+1}} = \frac{1}{\sqrt{1.65}} = 0.7784989 \]
\[ x_2 = \phi(x_1) = \frac{1}{\sqrt{x_1+1}} = \frac{1}{\sqrt{1.7784989}} = 0.7498479 \]
\[ x_3 = \phi(x_2) = \frac{1}{\sqrt{x_2+1}} = \frac{1}{\sqrt{1.7498479}} = 0.7559617 \]
\[ x_4 = \phi(x_3) = \frac{1}{\sqrt{x_3+1}} = \frac{1}{\sqrt{1.7559617}} = 0.7546446 \]
\[ x_5 = \phi(x_4) = \frac{1}{\sqrt{x_4+1}} = \frac{1}{\sqrt{1.7546446}} = 0.7549278 \]
\[ x_6 = \phi(x_5) = \frac{1}{\sqrt{x_5+1}} = \frac{1}{\sqrt{1.7549278}} = 0.7548668 \]
\[ x_7 = \phi(x_6) = \frac{1}{\sqrt{x_6+1}} = \frac{1}{\sqrt{1.7548668}} = 0.7548799 \]
\[ x_8 = \phi(x_7) = \frac{1}{\sqrt{x_7+1}} = \frac{1}{\sqrt{1.7548799}} = 0.7548771 \]
\[ x_9 = \phi(x_8) = \frac{1}{\sqrt{x_8+1}} = \frac{1}{\sqrt{1.7548771}} = 0.7548777 \]
\[ x_{10} = \phi(x_9) = \frac{1}{\sqrt{x_9+1}} = \frac{1}{\sqrt{1.7548777}} = 0.7548776 \]
\[ x_{11} = \phi(x_{10}) = \frac{1}{\sqrt{x_{10}+1}} = \frac{1}{\sqrt{1.7548776}} = 0.7548776 \]

Therefore, root2 is \(0.7548776\).

In this question the accuracy up to \(7\) places is acquires or here the TOL is \(10^{-7}\).

References

Read more about notations and symbols.

  1. Read more about sequences. ↩↩

  2. Read more about roots of an equation. ↩↩↩↩↩

  3. Read more about intervals. ↩

  4. Read more about numbers. ↩↩