Numerical analysis

Stirling's formula for equal interval: Let u = (x-x₀)/h Then Gauss's forward interpolation formula is f(x) = y₀+u/1!∆y₀+{u(u-1)/2!}∆²y₋₁+{(u+1)(u)(u-1)/3!}∆³y₋₁+{(u+1)u(u-1)(u-2)/4!}∆⁴y₋₂ +.......    .....(1) and Guass backward interpolation formula is f(x) = y₀+u/1!∆y₋₁+{u(u+1)/2!}∆²y₋₁+{(u+1)(u)(u-1)/3!}∆³y₋₂+{(u+2)(u+1)u(u-1)/4!}∆⁴y₋₂ +......     .......(2) Taking the mean of (1) and (2), we get f(x) = = y₀+u/1![(∆y₀+∆y₋₁)/2] + (u²/2!) ∆²y₋₁ + u(u-1)(u+1)/3![(∆³y₋₁+∆³y₋₂)/2] + [u²(u²-1)/4!]∆⁴y₋₂ +..... .....(3) Which is called Stirling formula. It is used when u lies between -1/4 and 1/4. Example: Employ Stirling formula to compute y₁₂.₂ from the following table (yₓ = 1+log₁₀sinx):       x°:             10         11          12       13           14 10⁵yₓ:     23967  28060   31788...

 Gauss's central difference formula for equal intervals: We shall develop central difference formulae which are best suitable for interpolation near the middle of the tabulated set (table). x:             x₋₂  x₋₁  x₀  x₁  x₂ y=f(x):    y₋₂  y₋₁  y₀  y₁  y₂ Difference table. Gauss's forward interpolation formula for equal intervals: f(x) = y₀+[u/1!]∆y₀+{[u(u-1)]/2!}∆²y₋₁+{[(u+1)(u)(u-1)]/3!}∆³y₋₁+{[(u+1)u(u-1)(u-2)]/4!}∆⁴y₋₂ +......., Where u= (x-x₀)/h Remark: This formula is applicable when u lies between 0 and 1 i.e (0<u<1). Example: Using Gauss's forward formula to evaluate y₃₀ given that y₂₁=18.4708, y₂₅=17.8144, y₂₉=17.1070, y₃₃=16.3432 and y₃₇=15.5154. Solution. The difference table is Forward difference table. To find y=f(x) at x=30, i.e f(30): Taking x₀ = 29, h=4, x=30, then u= (x-x₀)/h = (30-29)/4 = 0.25 Using Gauss's forward difference formula f(x) = y₀+[u/1!]∆y₀+{[u(u-1)]/2!}∆²y₋₁+...

Questions Example: Given that y₅=4, y₆=3, y₇=4, y₈=10, y₉=24, find the value of ∆⁴y₅: (i) By using the difference table and (ii) Without using the difference table. Solution: (i) Forward difference table. From the difference table ∆⁴y₅=0. Ans. (ii) We have ∆⁴y₅= (E-1)⁴y₅ [∵ ∆=E-1]                               = (E⁴-4E³+6E²-4E+1)y₅                               = E⁴y₅-4E³y₅+6E²y₅-4Ey₅+1y₅                               = y₉ -4y₈+6y₇-4y₆+y₅ [Eⁿyₓ=yₓ₊ₙ]                               = 24-4×10+6×4-4×3+4                               = 24-40-24-12+4 =0Ans. Example: Given x:  1     2     3...

 Relation between the operators: We can express each of Δ,∇,𝛿,μ and D in terms of E.  1) Δ = E - 1 Proof: By definition of  Δ  Δf(x) = f(x+h) - f(x)           =Ef(x) - f(x)         [∵Ef(x) = f(x+h)] Hence Δ = E -1 2) ∇ = 1 - E⁻¹ Proof: By definition of ∇ ∇f(x) = f(x) - f(x-h)          = f(x) - E⁻¹f(x)   [∵E⁻¹f(x) = f(x-h)] => ∇f(x) =  (1 - E⁻¹)f(x) => ∇ = 1 - E⁻¹ 3) 𝛿 =E ¹/² - E⁻ ¹/² Proof: By definition of 𝛿 𝛿f(x) = f[x+(h/2)] - f[x-(h/2)]           = E ¹/²f(x) -  E⁻ ¹/²f(x) Hence  𝛿 =E ¹/² - E⁻ ¹/² 4) 𝝁 = 1/2 (E ¹/² + E⁻ ¹/²) Proof: By definition of  𝝁     𝝁f(x) = (1/2){f[x+(h/2)] + f[x-(h/2)]}              = (1/2){E ¹/²f(x) +  E⁻ ¹/²f(x)}              =(1/2){E ¹/² + E⁻ ¹/²}f(x) Hence        𝝁 = 1/2 (E ¹/² +...

 Different operators We will study the following operators: 1) The Shifting Operator (E): Ef(x) = f(x+h) i.e Ey₀= y₁ E²f(x) = f(x+2h) i.e E²y₀ = y₂ Eⁿf(x) = f(x+nh) i.e Eⁿy₀ = yₙ Here n takes integral or fractional, positive or negative values. For example: E⁻¹f(x) = f(x-h) i.e E⁻¹ y₂ = y₁ E¹/²f(x) = f[x+(1/2)h] i.e E¹/² y₁ = y₃ₗ₂ Properties of Operator E 1) Operator E is distributive. 2) Operator E is commutative with respect to constant. 3) Operator E obeys laws of indices. 2) Forward difference operator (∆): If x₀, x₁,x₂,......., xₙ are equally spaced with interval of differencing h and if y = f(x), then  ∆f(x) = f(x+h) - f(x) i.e ∆yᵢ = yᵢ₊₁ - yᵢ for i = 0,1,2,3,..... n-1, The symbol ∆ is called forward difference operator and  ∆yᵢ is called first forward difference. Similarly, the second forward differences are ∆²yᵢ =∆yᵢ₊₁ - ∆yᵢ For example: ∆²y₀ =∆y₁ - ∆y₀ = (y₂ - y₁)-(y₁-y₀)                           ...

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...