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\(\Delta\!\!\!|\,\) is a divided difference operator and \(\Delta\) is a forward difference operator, f(x) and g(x) are two functions and a and b are constants, then :

  • A.\(\Delta\!\!\!|\,[af(x) + bg(x)]\) \(= a \Delta\!\!\!|\,f(x) + b \Delta\!\!\!|\,g(x)\)
  • B.\(\Delta\!\!\!|\,[af(x) + bg(x)]\) \(= a \Delta\!\!\!|\,(f(x)\, g(x)) + b \Delta\!\!\!|\,(f(x) . g(x))\)
  • C.\(\Delta\!\!\!|\,[af(x) + bg(x)]\) \(= a \Delta f(x) + b \Delta g(x)\)
  • D.\(\Delta\!\!\!|\,[af(x) + bg(x)]\) \(= a\Delta(f(x)\, g(x)) + b \Delta (f(x)\, g(x))\)

Answer: A

The correct property is linearity of the divided difference: the operator applied to \(af(x) + bg(x)\) equals a times its value on f plus b times its value on g.

A first divided difference is

\[ [x_{0}, x_{1}]f = \frac{f(x_{1}) - f(x_{0})}{x_{1} - x_{0}} \]

and replacing f by \(af + bg\) splits the numerator into a times the f part plus b times the g part. Higher divided differences are built recursively from the same quotients, so they are linear too.

Option C mixes in the forward difference \(\Delta\), a different operator. Options B and D replace f and g by their product, which has nothing to do with a linear combination.

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