Iteration method or successive approximation method.

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(x) = x³+x²-100 = 0

f(1) = -98

f(2) = -88

f(3) = -64

f(4) = 64+16-100 = -20 (-ve)

f(5) = 125+25-100 = 50 (+ve)

f(4) and f(5) are of opposite sign.

So the root of equation lie between 4 and 5.

The given equation can be written as,

x²(x+1) = 100

or x = 10/√(x+1)

or x = 𝜙(x), where 𝜙(x) = 10/√(x+1)

Therefore 𝜙'(x) = 10×(-1/2)× 1/(x+1)³⁄ ² = -5/(x+1)³⁄ ²

or |𝜙'(x)| = |-5/(x+1)³⁄ ²| < 1 in the interval (4,5).

So, the iteration method can be applied.

Taking x₀ = 4.2, then approximations are

x₁ = 𝜙(x₀) = 10/√(4.2+1) = 4.3852

x₂ = 𝜙(x₁) = 10/√ (4.3852+1) = 4.3092

x₃ = 𝜙(x₂) = 10/√ (4.3092+1) = 4.33995

x₄ = 𝜙(x₃) = 10/√ (4.33995+1) = 4.32744

x₅ = 𝜙(x₄) = 10/√ (4.32744+1) = 4.33252

x₆= 𝜙(x₅) = 10/√ (4.33252+1) = 4.33045

x₇ = 𝜙(x₆) = 10/√ (4.33045+1) = 4.33129

x₈= 𝜙(x₇) = 10/√ (4.33129+1) = 4.33095

x₉ = 𝜙(x₈) = 10/√ (4.33095+1) = 4.33109

x₁₀ = 𝜙(x₉) = 10/√ (4.33109+1) = 4.33103

Since x₉ = x₁₀ = 4.3310 correct up to four decimal places.

Therefore, root is x = 4.3310 Ans.

Comments

Questions (Forward difference operator, Backward difference operator and Shifting operator)

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...

Gauss's central difference formula for equal intervals.

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+...

Different operators

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) ...

Numerical analysis

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