Optimizing f(p,q) with Close Proximity of p and q in (0,1): A Formal Analysis

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Homework Statement



I have the function f(p,q)=p(1-p)/[q(1-q)] where p and q are in (0,1). I want to say that if p is close to q, f(p,q) is 'close' to 1. What is a formal way of saying how close to p q should be?

The Attempt at a Solution



Basically I want to say f(p,q)= 1 + \epsilon(p,q) where \epsilon(p,q) is small if some condition "X" is true. Is there an obvious way of saying what "X" is? Maybe I can substitute q=p+u... but then what? Taylor series or L'hopital's or something? My calc is pretty rusty, so I'd appreciate any reminders...
 
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Substituting p = q+u sounds like a good idea. Then use
<br /> (q+u)(1-q-u) = q-q^2-2qu+u-u^2 = q(1-q)-2qu+u-u^2<br />
 
If you define a function

<br /> g(x) = x (1 - x)<br />

can you translate the question you asked in terms of it?
 
Why not just write that the limit[f(p,q)] = 1 as p \rightarrow q ?
 
Raskolnikov said:
Why not just write that the limit[f(p,q)] = 1 as p \rightarrow q ?

I agree.
The problem statement is nearly the very definition of a Limit.
You just did not use the letters epsilon and delta x.
 
\frac{p(1-p)}{q(1- q)}= \left(\frac{p}{q}\right)\left(\frac{1-p}{1-q}\right)
Does that help?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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