Digit Sum Inequality

Inspired by [http://en.wikipedia.org/wiki/Digit_sum] "digit sums for binary being on average smaller than those for any other base." Here's my attempt at the proposition.

if R is the radix, & N is the given integer value, the number of digits n required on radix R to represent N = [ln(N*R)/lnR], yes [] is the greatest integer or the floor function....

Clearly, N = A0 + A1R + A2Rsq + A3Rcube + ...... An(R^n) = Sum(Ai*R^i) & the digit sum DS = A0 + A1 + A2 + A3 + ...... +An = Sum(Ai)

we know CS inequality..., check the special case #1 of this link for reference. [http://en.wikipedia.org/wiki/Cauchy–Schwarz_inequality]

put Ai = ai, & R^i = bi... we get, [Sum(AiR^i)]sq <= Sum(Aisq)Sum(R^i sq)

Look closely, its... Nsq <= Sum(Aisq)Sum(R^i sq)... & Sum(R^i sq) is a GP..., we get Nsq <= Sum(Aisq)[(Rsq^n -1)/(Rsq-1)]...

Now we also know PowerMean inequality [https://www.artofproblemsolving.com/Wiki/index.php/PowerMeanInequality] according to which... sqrt[Avg(Aisq)]>=Avg(Ai).... write avg(Ai) = (1/n)*sum(Ai)..., mind you sum(Ai) is DS. Now substitute this into the Nsq inequality... we will get....

Nsq <= [(DS)sq/n][(Rsq^n -1)/(Rsq-1)]... &..., DS >= Nsqrt[n*(Rsq-1)/(Rsq^n-1)]

You can see that for any N, if we increase R from 2, the curve agrees with the proposition. With equality holding only for n=1, which makes sense, if R>N, then N can be represented on R in a single digit & DS would = N.

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