The value of f(4)-f(3) is | vedclass

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(A) Using the definition of the forward difference operator $\Delta f(x) = f(x+1) - f(x)$,we have $f(4) - f(3) = \Delta f(3)$.Since $\Delta f(x) = f(x

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$\Delta f(2)+\Delta^{2} f(1)+\Delta^{3} f(1)$

Vedclass · Answer

(A) Using the definition of the forward difference operator $\Delta f(x) = f(x+1) - f(x)$,we have $f(4) - f(3) = \Delta f(3)$.Since $\Delta f(x) = f(x+1) - f(x)$,we can write $f(3) = f(2) + \Delta f(2)$.Thus,$\Delta f(3) = \Delta [f(2) + \Delta f(2)] = \Delta f(2) + \Delta^2 f(2)$.Further,since $\Delta^2 f(x) = \Delta^2 f(x-1) + \Delta^3 f(x-1)$,we expand $\Delta^2 f(2)$ as $\Delta^2 [f(1) + \Delta f(1)] = \Delta^2 f(1) + \Delta^3 f(1)$.Substituting this back,we get $f(4) - f(3) = \Delta f(2) + \Delta^2 f(1) + \Delta^3 f(1)$.

The value of $f(4)-f(3)$ is

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