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AP space and difference operator

Authors
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    Name
    Vu Hung
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Problem Statement

Let A(a,d)\mathcal{A}(a,d) denote an AP. Define

I=A(1,0),J=A(0,1).\mathcal{I}=\mathcal{A}(1,0),\qquad \mathcal{J}=\mathcal{A}(0,1).
  • Show any AP is uniquely aI+dJa\mathcal{I}+d\mathcal{J}.
  • Let Δxn=xn+1xn\Delta x_n=x_{n+1}-x_n. Show ΔI=0\Delta\mathcal{I}=0 and ΔJ=I\Delta\mathcal{J}=\mathcal{I}.
  • Deduce Δ(aI+dJ)=dI\Delta(a\mathcal{I}+d\mathcal{J})=d\mathcal{I}.

Hints

Compare first term and common difference.


Solutions

An AP has nnth term a+(n1)d=a1+d(n1)a+(n-1)d=a\cdot1+d(n-1), so X=aI+dJX=a\mathcal{I}+d\mathcal{J}, unique by matching (a,d)(a,d). Also I=(1,1,1,)\mathcal{I}=(1,1,1,\dots) so ΔI=0\Delta\mathcal{I}=0, and J=(0,1,2,)\mathcal{J}=(0,1,2,\dots) so ΔJ=(1,1,1,)=I\Delta\mathcal{J}=(1,1,1,\dots)=\mathcal{I}. Hence

Δ(aI+dJ)=aΔI+dΔJ=dI.\Delta(a\mathcal{I}+d\mathcal{J})=a\Delta\mathcal{I}+d\Delta\mathcal{J}=d\mathcal{I}.

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