Numerical analysis

Newton- Raphson method Let x₀ be an approximation value of a root of the equation f(x) = 0. Let x₁ be the exact root closer to  x₀. So that x₁ =  x₀ + h. Where h is small. Since x₁ is the exact root of f(x) = 0, we have f(x₁) =0, i.e f( x₀ + h) =0 i.e f( x₀)+h/1![f'( x₀)]+h²/2![f"( x₀)]+....=0 by Taylor's theorem. Since h is small, h² and higher powers of h may be omitted. Hence  f( x₀)+h[f'( x₀)] = 0 => h = -f( x₀)/f'( x₀) .....(1) ∴ x₁ = x₀ + h = x₀ -f( x₀)/f'( x₀) [using(1)] Taking x₁ as an approximation value of the root, a still better approximation x₂ can be obtained by x₂ =  x₁ -f( x₁)/f'( x₁) The iterative process is continued until we get the required accuracy. The iterative formula is  xₙ₊₁ = xₙ - f(xₙ)/f'(xₙ) (for n= 0,1,2,.....) is called the Newton- Raphson method or Newton's iteration formula. Remarks 1) When the derivative of f(x) can be easily found and is a simple expression, then the real root of the equation f(x)=0 can be compu

 Accelerating convergence A technique called Aitken's ∆² method is used to accelerate the convergence of a sequence that is linearly convergent, regardless of its origin of application. Aitken's ∆² method: Suppose {pₙ}ₙ᪲₌₀ is a linearly convergent sequence with limit p. To motivate the construction of a sequence  {p̂ₙ}ₙ᪲₌₀ that converges more rapidly to p than {pₙ}ₙ᪲₌₀. Let us first assume that  (pₙ₊₁ - p)/(pₙ - p) ≈ (pₙ₊₂ - p)/(pₙ₊₁ - p) => (pₙ₊₁ - p)² ≈ (pₙ₊₂ - p)×(pₙ - p) => (pₙ₊₁)² -2pₙ₊₁p + p² ≈ pₙ₊₂pₙ -(pₙ+ pₙ₊₂)p+p² =>  (pₙ₊₁)² -2pₙ₊₁p ≈ pₙ₊₂pₙ -(pₙ+ pₙ₊₂)p => (pₙ₊₂ + pₙ - 2pₙ₊₁)p ≈ pₙ₊₂pₙ - (pₙ₊₁)² => p ≈  [pₙ₊₂pₙ - (pₙ₊₁)²] /  (pₙ₊₂ + pₙ - 2pₙ₊₁) Adding and subtracting the terms (pₙ)² and 2pₙ₊₁pₙ in the numerator and grouping terms appropriately gives p ≈ [pₙ₊₂pₙ -2pₙ₊₁pₙ +(pₙ)² - (pₙ₊₁)² +2pₙ₊₁pₙ -(pₙ)² ] / (pₙ₊₂ + pₙ - 2pₙ₊₁) => p ≈ [(pₙ₊₂ -2pₙ₊₁+pₙ)pₙ - {(pₙ₊₁)² -2pₙ₊₁pₙ +(pₙ)² }] / (pₙ₊₂ + pₙ - 2pₙ₊₁) => p ≈ pₙ - {(pₙ₊₁ -pₙ)² / (pₙ₊₂ + pₙ - 2

 Iteration method To find the real root of an equation f (x)=0 ...(1) Which can be expressed in the form x=𝜙(x) ......(2) First we find an initial approximate value x₀ for equation (1). The better approximation x₁ for the root is obtained by replacing x by x₀  in R.H.S of equation (2) i.e x₁ = 𝜙(x₀) A still better approximation x₂ for the root is obtained by putting x=x₁ in the R.H.S of equation (2). Thus x₂ = 𝜙(x₁). This procedure is continued and we get x₃ =  𝜙(x₂) x₄ =  𝜙(x₃) ................. xₙ =  𝜙(xₙ₋₁) If the sequence x₀,x₁,x₂,.......xₙ of approximate roots converges to a limit 𝛂, then 𝛂 is taken as the root of the equation f(x)=0. Remarks The sufficient condition for convergence of iterations: 1) If I is the interval in which the root  𝛂 of the equation x=𝜙(x) lies, then |𝜙'(x)|< 1 for all x in the interval I. 2) The initial approximation x₀ for the root lies in I. Example: Find the real root of the equation x³+x²-100=0, by iteration method. Solution. Given f

 Algebraic and transcendental equations; An equation of the form f(x)=0 is called algebraic or transcendental equations depending upon whether f(x) is purely a polynomial of x or contains exponential, trigonometric or logarithmic functions. For example: 2x³+3x²+17=0 is an algebraic equation while 5x³+4logx+2sinx=0 is transcendental equation.Examples of transcendental functions include the exponential function, the logarithm, and the trigonometric functions. The process of finding the roots of an equation (or zeroes of an equation) is called the solution of f(x)=0. Bisection or Bolzano method; Let f(x)=0 be an algebraic or a transcendental equation and let f(a) and f(b) are of opposite sign i.e f(a)f(b)<0 then root of equation f(x)=0 lie between a and b. Then the first approximation to the root is x₁ = (a+b)/2. Now if f(x₁) = 0, then x₁ is the root of f(x)=0, otherwise the root lie between a and x₁ or x₁ and b according as f(x) is positive or negative. Then we bisect the interval as

  Rounding off If a number has large number of digits then it is necessary to cut them to a usable number of figures. This process of omitting some digits is called rounding off. Rule for rounding off The rules for rounding off numbers to the appropriate significant figures are as follows: a) If the digit to be dropped is less than 5, then the preceding digit is left unchanged and if it is more than 5 then the preceding digit is raised by 1. b) If the digit to be dropped is 5, then the preceding digit is retained unchanged if it is even, but increased by 1 if it is odd.  Example: The number below are round off to four significant figures. 1.6583 to 1.658 30.0567 to 30.06 0.859378 to 0.8594 3.14159 to 3.142 Errors Numerical errors arise from the use of approximation to represent exact mathematical operation and quantities.  Truncation errors Truncation errors result when approximation are used to represent exact mathematical procedures. They are caused, when an infinite process is repla

  Kinds of Number  There are two kinds of number, exact and approximate. Approximate numbers Approximate numbers are those that represent the number to a certain degree of accuracy. Example; An approximate value of π is 3.1416 or if we wish to have a better approximation, it is 3.14169265. But we cannot write the exact value of π. Exact numbers The numbers which are accurately expressed are called exact numbers. For example; 1,2,3…., 1/2,3/2,…. √2, π, e etc written in this manner. Significant digits The digits that are used to express a number are called significant digits or significant figures. Example; The numbers 3.1416, 0.66667, 4.0687 contains five significant digits each. The number 0.00025 has however, only two significant digits viz 2 and 5 since zeroes serve only to fix the position of decimal points. Similarly, the numbers 0.00145, 0.000145, 0.0000145 all have three significant digits. Rule for finding significant digits Rule 1:   Every non zero digit is significant. Example