Numerical Methods Formulas

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1. Iterative method of Solving Equations

(i) Successive Bisection method
If f(x) is a continuous function in the interval [a, b] and f(a), f(b) are of opposite sign, then there exist at least one value of x say a ∈ (a, b) such that f(α) = 0 and a < α < b.
Working Rule

• Find f(a) and f(b).
• Let f(a) be negative and f(b) be positive then take \(\alpha=\frac{a+b}{2}\).
• If f(α) = 0, then a is a required root. If f(α) > 0 then roots lie between a and α. If f(α) < 0 then roots lies between α and b.
• Repeat the process until we get the root correct up to desired level of accuracy.

(ii) False Position method (Regula- falsi method)
If the root of f(x) = 0 belongs to the interval (x₀, x₁) and f(x₀), f(x₁) are of opposite sign (say f(x₀) < 0, f(x₁) > 0), then
x₂ = x₀ – \(\frac{\left(x_{1}-x_{0}\right) f\left(x_{0}\right)}{f\left(x_{1}\right)-f\left(x_{0}\right)}\)
Working rule

• Calculate f(x₀) and f(x₁), if these are of opposite sign then the root lies between x₀ and x₁.
• Calculate x₂ by the above formula.
• Now if f(x₂) = 0, then x₂ is the required root.
• If f(x₂) is negative, then the root lies in (x₂, x₁)
• If f(x₂) is positive, then the root lies in (x₀, x₂).

(iii) Newton- Raphson method
If f(x) = 0 and f(a) and f(b) are of opposite sign, then to find the root in interval (a, b).

• Find |f (a)|, |f (b)|. If |f(a)| < |f (b)| then assume a = x₀ otherwise b = x₀
  
• Find x₁ = x₀ – \(\frac{f\left(x_{0}\right)}{f^{\prime}\left(x_{0}\right)}\)
• If f(x₁) = 0, then x₁ is the required root, otherwise by taking x₁ as starting point find x₂

⇒ x₂ = x₁ – \(\frac{f\left(x_{1}\right)}{f^{\prime}\left(x_{1}\right)}\)
This process is continued till we get a value for the root up to the desired level of accuracy.

2. Trapezoidal rule

Let y = f(x) be a function defined on [a, b] which is subdivided into n equal sub intervals each of width h so that b – a = nh.
\(\int_{a}^{b}\)f(x)dx = \(\frac{h}{2}\)[(y₀ + y_n) + 2(y₁ + y₂ + ……. + y_n-1)]
where n is any positive integer and yr is the value of f(x) for x = a + rh.

3. Simpson’s one third rule

Let y = f(x) be a function defined in the interval [a, b] which is divided into n (an even number) equal parts of width h so that b – a = nh and yr, is the value of f(x) for x = a + rh, Then
\(\int_{a}^{b}\)f(x)dx = \(\frac{h}{3}\)[(y₀ + y_n) + 4(y₁ + y₃ + ……. + y_n-1) + 2(y₂ + y₄ + ……. + y_n-2)]

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