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The shift operator $E$ is defined as $E [f(x_i)] = f (x_i+h)$ and $E'[f(x_i)]=f (x_i -h)$ then $\triangle$ (forward difference) in terms of $E$ is

  1. $E-1$
  2. $E$
  3. $1-E^{-1}$
  4. $1-E$
in Numerical Methods by Veteran (105k points) | 1.4k views

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Forward difference operator(delta (D))

D(f(x)) = f(x + h) - f(x)

Shift operator , Ef(x) = f(x+h) =  f(x+h) - f(x) + f(x) = (1 + D)f(x)

so E= 1 + D

gives D = E - 1

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