Discrete inequality plotting

I have an equation like this: (20k)⋅k∑i=0(ki)⋅i≤1+f{20 \choose k}\cdot \sum_{i=0}^k {k \choose i}\cdot i \leq 1+f.

I want to plot this inequality as the dependency between k and f.
I found a commend RegionPlot but I don’t know how to correctly use it in such equation.

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You have an inequality, not an equation. 🙂
– Svend Tveskæg
Dec 9 ’13 at 13:58

  

 

@SvendTveskæg Yes, you are right ^^
– Ziva
Dec 9 ’13 at 15:32

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3 Answers
3

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ListPlot[Table[{k, Binomial[20, k] Sum[i Binomial[k, i] – 1, {i, 0, k}]}, {k, 1, 20}], Filling->Top]

Or the same using a Log scale

If you want only integer values you can put k in Round:

RegionPlot[Binomial[20, Round[k]] Sum[
i Binomial[Round[k], i], {i, 0, Round[k]}] <= 1 + f, {k, 1, 20}, {f, 0, 5 10^9}] Your function is explicit, so it will be better to use ListPlot as in belisarius's answer: ListPlot[Table[{k - 0.5, Binomial[20, k] Sum[i Binomial[k, i], {i, 0, k}] - 1}, {k, 20}], Filling -> Top, Joined -> True, InterpolationOrder -> 0]

Also one can note that

Sum[i Binomial[k, i], {i, 0, k}]

2^(-1 + k) k

The inequality you are considering is simplified as follow.

Consequently, we can use RegionPlot easily.

2^(-1+k) k Binomial[20,k]<=1+f RegionPlot[2^(-1+k) k Binomial[20,k]<=1+f, {k,-4,10},{f,-20,40}]