MTH6P2C: ITERATIVE METHODS

Iterative Method

• In computational mathematics, an iterative method is a mathematical procedure that uses an initial guess to generate a sequence of improving approximate solutions for a class of problems, in which the n-th approximation is derived from the previous ones.
• A specific implementation of an iterative method, including the termination criteria, is an algorithm of the iterative method.
• An iterative method is called convergent if the corresponding sequence converges for given initial approximations.
• A mathematically rigorous convergence analysis of an iterative method is usually performed; however, heuristic-based iterative methods are also common.
• Iteration is a way of solving equations. You would usually use iteration when you cannot solve the equation any other way.An iteration formula might look like the following:x_n+1 = 2  + 1 
  x_n .You are usually given a starting value, which is called x₀. If x₀ = 3, for example, you would substitute 3 into the original equation where it says x_n. This will give you x₁. (This is because if n = 0, x₁ = 2 + 1/x₀ and x₀ = 3).
  x₁ = 2 + 1/3 = 2.333 333 (by substituting in 3).To find x₂, substitute the value you found for x₁.
  x₂ = 2 + 1/(2.333 333) = 2.428 571Repeat this until you get an answer to a suitable degree of accuracy. This may be about the 5th value for an answer correct to 3s.f. In this example, x₅ = 2.414…

In contrast, direct methods attempt to solve the problem by a finite sequence of operations. In the absence of rounding errors, direct methods would deliver an exact solutio.

Iterative methods are often the only choice for nonlinear equations. However, iterative methods are often useful even for linear problems involving a large number of variables (sometimes of the order of millions), where direct methods would be prohibitively expensive (and in some cases impossible) even with the best available computing power.

Let the given equation be f(x) = 0 and the value of x to be determined. By using the Iteration method you can find the roots of the equation. To find the root of the equation first we have to write equation like below
x = pi(x)
Let x=x₀ be an initial approximation of the required root α then the first approximation x₁ is given by x₁ = pi(x0).

Similarly for second, thrid and so on. approximation
x₂ = pi(x₁)
x₃ = pi(x₂)
x₄ = pi(x₃)
x_n = pi(x_n-1)

Iteration Method Example:
Find the real root of the equation x³ + x ² = 1 by iteration method.
Solution:
We can rewrite the above equation by
x³ + x ² – 1 = 0;
Let f(x) = x³ + x ² – 1
f(0) = -1 (positive)
f(1) = 1 (negative)
Hence the root value lie between 0 to 1

x³ + x ² – 1 = 0
x² (x + 1) = 1
x² = 1/ (x + 1)
x = 1/ √(x + 1)
pi(x) = 1/ √(x + 1)

Let the initial approximation be x₀ = 0.5

x₁ = pi(x₀) = 1/√1+ 0.5 = 0.81649

x₂ = pi(x₁) = 1/√1+ 0.81649 = 0.74196

x₃ = pi(x₂) = 1/√1+ 0.74196 = 0.75767

x₄ = pi(x₃) = 1/√1+ 0.75767 = 0.75427

x₅ = pi(x₄) = 1/√1+ 0.75427 = 0.75500

x₆ = pi(x₅) = 1/√1+ 0.75500 = 0.75485

x₇ = pi(x₆) = 1/√1+ 0.75485 = 0.75488

Since the difference between x₆ and x₇ are very small, so the root is 0.75488.

Iteration method Practice problem:
1. Solve by iteration method 2x – logx – 7 = 0
2. Find the root of the equation x log x = 1.2 by iteration method
3. Compute the real root of 3x – cosx – 1 = 0 by iteration method
4. Find the root of the equation sin x = 1 + x3 between ( -2,-1) to 3 decimal places by Iteration method

ITERATIVE METHODS TUTORIAL