Question

Find the missing entries from the following.

x	0	1	2	3	4	5
y = f(x)	0	-	8	15	-	35

Answer 1

Since four values of f(x) are given we assume the polynomial of degree three.

Hence fourth-order differences are zeros.

i.e. Δ⁴y₀ = 0

i.e. (E – 1)⁴y_k = 0

(E⁴ – 4E³ + 6E² – 4E + 1)y_k = 0  ........(1)

Put k = 0 in (1)

(E⁴ – 4E³ + 6E² – 4E + 1)y₀ = 0

E⁴y₀ – 4E³y₀ + 6E³y₀ – 4Ey₀ + y₀ = 0

y₄ – 4y₃ + 6y₂ – 4y₁ + y₀ = 0

y₄ – 4(15) + 6(8) – 4y₁ + 0 = 0

y₄ – 4y₁ = 12  .......(2)

Put k = 1 in eqn (1)

(E⁴ – 4E³ + 6E² – 4E + 1) y₁ = 0

y₅ – 4y₄ + 6y₃ – 4y₂ + y₁ = 0

35 – 4(y₄) + 6(15) – 4(8) + y₁ = 0

35 – 4y₄ + 90 – 32 + y₁ = 0

– 4y₄ + y₁ + 125 – 32 = 0

– 4y₄ + y₁ = – 93  ........(3)

Solving equation (2) and (3)

Equation (2) × 4 ⇒   4y₄  – 16y₁ =   48

Equation (3)       ⇒ – 4y₄ +     y₁ = – 93
                                       – 15y₁ = – 45 

y₁ = `(-45)/(-15) = 45/15` = 3

⇒ y₁ = 3

Substitute y = 3 in equation (2)

y₄ – 4(3) = 12

y₄ – 12 = 12

y₄ = 12 + 12

∴ y₄ = 24

The required two missing entries are 3 and 24.