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Prove that E=1+Δ=ehD
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By definition

E.g

f(x)=f(x+h)(1)

Δf(x)=f(x+h)f(x)

Δf(x)=Ef(x)f(x)

Δf(x)=(E1)f(x)

Δ=E1

E=1+Δ(2)

Also

E.g

f(x)=f(x+h)

f(x)+hf1(x)h22!f1(x)+h33!f1(x)+............ Taylor series

$f(x) +hDf(x)\dfrac …

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