If u_{0}=8, u_{1}=3, u_{2}=12, u_{3}=51,then | vedclass

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(C) The forward difference operator $\Delta$ is defined as $\Delta u_{n} = u_{n+1} - u_{n}$.For the third-order forward difference,we use the formula

Vedclass · Accepted Answer

$16$

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(C) The forward difference operator $\Delta$ is defined as $\Delta u_{n} = u_{n+1} - u_{n}$.For the third-order forward difference,we use the formula $\Delta^{3} u_{0} = (E-1)^{3} u_{0}$,where $E$ is the shift operator such that $E u_{n} = u_{n+1}$.Expanding the operator,we get $\Delta^{3} u_{0} = (E^{3} - 3E^{2} + 3E - 1) u_{0}$.This simplifies to $\Delta^{3} u_{0} = u_{3} - 3u_{2} + 3u_{1} - u_{0}$.Substituting the given values $u_{0}=8, u_{1}=3, u_{2}=12, u_{3}=51$:$\Delta^{3} u_{0} = 51 - 3(12) + 3(3) - 8$.$\Delta^{3} u_{0} = 51 - 36 + 9 - 8$.$\Delta^{3} u_{0} = 15 + 9 - 8 = 24 - 8 = 16$.

If $u_{0}=8, u_{1}=3, u_{2}=12, u_{3}=51$,then the value of $\Delta^{3} u_{0}$ is

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