A comparison on binomial expansion

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Discussion Overview

The discussion revolves around the inequality involving the binomial expansion, specifically whether the expression \(\sum_{k=n+1}^{2n}\left(\begin{array}{c} 2n\\k\end{array}\right)x^{k}\left(1-x\right)^{2n-k}\leq2x\) holds for any \(x\in(0,1)\) and any positive integer \(n\). The scope includes mathematical reasoning and exploration of the properties of binomial coefficients.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions the validity of the inequality and seeks assistance in proving it.
  • Another participant notes that the left-hand side (LHS) represents the "second half" of the binomial expansion and attempts to relate it to the total sum, suggesting that the LHS approaches 1 as \(x\) approaches 0.
  • A different participant agrees that as \(x\) approaches 0, the LHS tends to 0, raising concerns about the inequality holding in that limit.
  • One participant expresses confusion, suggesting that their reasoning implies the LHS could be negative, indicating a potential flaw in their understanding.
  • Another participant asserts that the LHS is always positive for \(x\in(0,1)\), countering the previous concerns.
  • One participant mentions empirical checks for values of \(n\) from 1 to 1000, noting that the inequality holds in those cases but still seeks a general proof.

Areas of Agreement / Disagreement

Participants express differing views on the validity of the inequality, with some questioning its correctness under certain conditions while others maintain that it holds true. The discussion remains unresolved regarding a general proof.

Contextual Notes

There are limitations in the reasoning presented, particularly concerning the behavior of the LHS as \(x\) approaches 0 and the implications of the inequality in that context. The discussion does not resolve these mathematical uncertainties.

forumfann
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Could anyone help me on this question? Is it true that
[tex]\sum_{k=n+1}^{2n}\left(\begin{array}{c}<br /> 2n\\k\end{array}\right)x^{k}\left(1-x\right)^{2n-k}\leq2x[/tex]
for any [tex]x\in(0,1)[/tex] and any positive integer [tex]n[/tex]?

Any help on that will be greatly appreciated!
 
Last edited:
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Well, the LHS seems to be the "second half" of the terms of the expansion of ( x+ (1-x) )^(2n) = 1.

This gives that the LHS is [tex]1 - \sum_{k=0}^n \left(\begin{array}{c}<br /> 2n\\k\end{array}\right)x^{k}\left(1-x\right)^{2n-k}[/tex].

I can't see how that's incorrect, but then I note the inequality doesn't seem to hold anymore for when x is small, for the LHS would seem to tend to 1 whilst the RHS goes to zero. Maybe I'm going crazy..
 
Gib Z said:
the LHS would seem to tend to 1

When k=0, [tex]x^k(1-x)^{2n-k}[/tex] goes to 1 as x goes to 0, so [tex]1-\sum_{k=0}^{n}\left(\begin{array}{c}<br /> 2n\\<br /> k\end{array}\right)x^{k}\left(1-x\right)^{2n-k}[/tex] tends to 0 as x goes to 0 because the other terms with k>0 has a factor x. But thanks anyway.
 
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Well if that is true, haven't you shown that the LHS is negative? I must have done something completely wrong as it seems my idea implies the original LHS must be negative..

Today obviously isn't my day. Sorry for wasting your time !
 
The LHS is always positive for [tex]x\in(0,1)[/tex]. That is fine.
 
In fact, I have checked the cases of n=1 through 1000, and it holds for all the cases. But I still don't know how to show it in general. Any other help?
 

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