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Question 11 of 37

Q.Δ2y0=\Delta^2 y_0 =

(a) y2+y1+2y0y_2 + y_1 + 2y_0
(b) y2−2y1+y0y_2 - 2y_1 + y_0
(c) y2+2y1−y0y_2 + 2y_1 - y_0
(d) y2+2y1+y0y_2 + 2y_1 + y_0
Tamil Nadu DgeTamil Nadu HSC (DGE) Commerce Board 2020MCQ· 1mImportance★★★★★
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Apply the forward-difference operator twice: Δy0=y1−y0\Delta y_0 = y_1 - y_0, then Δ2y0=Δy1−Δy0=y2−2y1+y0\Delta^2 y_0 = \Delta y_1 - \Delta y_0 = y_2 - 2y_1 + y_0.

Step 1 — First forward differences.

Δy0=y1−y0,Δy1=y2−y1.\Delta y_0 = y_1 - y_0, \qquad \Delta y_1 = y_2 - y_1.

Step 2 — Second forward difference. …

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