Show that x^n = (a_i) where a_i = { 1 if i = n, 0 if otherwise| for all n >= 0}

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- Oct 28th 2009, 10:35 PMCoda202Polynomial rings
Show that x^n = (a_i) where a_i = { 1 if i = n, 0 if otherwise| for all n >= 0}

- Oct 29th 2009, 10:13 AMCoda202Progress thus far
So now I have:

x = (ai) where ai = { 1 if i = 1, 0 if i <> 1

x(bi) = (ai)(bi) = (di)

where di = sum from j=0 to i of aj(bi-j) = a1(bi-1) = 1(bi-1) = bi-1, since aj=0 if j<>1

but I don't know where to go from here..