An upperbound for an algebraic expression

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Discussion Overview

The discussion revolves around finding a tight upper bound for the algebraic expression \(\frac{x}{1-x(1-y)}\) under the constraints \(0 \leq x, y \leq 1\). The scope includes mathematical reasoning and exploration of bounds for the function.

Discussion Character

  • Exploratory, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant, Bincy, requests a tight upper bound for the expression given the constraints on \(x\) and \(y\).
  • Another participant argues that there is no upper or lower bound for the function, noting that on the critical curve \(x=1/(1-y)\), the function approaches negative infinity, while for \(y\to 1/2\) and \(x\to 2\), it approaches positive infinity.
  • A later reply reiterates the previous point about the absence of bounds, emphasizing that both \(x\) and \(y\) must remain between 0 and 1.
  • Another participant acknowledges a mistake and points out that as \(y\to 0\) and \(x\to 1\), the function also diverges, confirming that there is no upper bound.

Areas of Agreement / Disagreement

Participants generally agree that there is no upper or lower bound for the function, but the discussion includes multiple perspectives on the behavior of the function under different conditions.

Contextual Notes

The discussion highlights the critical points where the function diverges, but does not resolve the implications of these behaviors fully. The dependence on the specific values of \(x\) and \(y\) is also noted.

bincy
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Hii Everyone,
Plz tell me a tight upperbound for [math]\frac{x}{1-x\left(1-y\right)}
[/math] where o<=x,y<=1

regards,
Bincy
 
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I don't think there is an upper or lower bound for that function. On the critical curve $x=1/(1-y)$, the function goes to negative infinity, and as, say, $y\to 1/2$ and $x\to 2$, the function goes to positive infinity.
 
Ackbach said:
I don't think there is an upper or lower bound for that function. On the critical curve $x=1/(1-y)$, the function goes to negative infinity, and as, say, $y\to 1/2$ and $x\to 2$, the function goes to positive infinity.
Plz note that BOTH x and y are in between 0 and 1
 
Ah, my mistake. You still have a problem when $y\to 0$ and $x\to 1$. The function blows up there, and there is no upper bound on it.
 

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