DiffEq, Binomial Expansion and limits

[QuietMind]

Homework Statement

Use algebra to show that U(x) = −√x − 1 and L(x) = −√x satisfy the ’funnel condition’ U(x) − L(x) → 0 as x → ∞

Homework Equations

Funnel condition: The two fences come together asymptotically, i.e. U(x) − L(x) is small for large x.

The Attempt at a Solution

I think that the binomial expansion is appropriate here? But I am unclear if it can be used for fractions.

$U(x)-L(x) = -\sqrt{x-1} +\sqrt{x}$
the first term can be expressed as: (hopefully did the binomial right)
$x^\frac{1}{2} - \frac{1}{2}x^\frac{-1}{2}- \frac{1}{8}x^\frac{-3}{2}...$

so then
$-\sqrt{x-1} +\sqrt{x} = x^\frac{1}{2} -x^\frac{1}{2} + \frac{1}{2}x^\frac{-1}{2} + \frac{1}{8}x^\frac{-3}{2}...$
the first 2 terms cancel, then the rest go to 0 as x goes to infinity. Does that look right?

 

[Ray Vickson]

Homework Statement

Use algebra to show that U(x) = −√x − 1 and L(x) = −√x satisfy the ’funnel condition’ U(x) − L(x) → 0 as x → ∞

Homework Equations

Funnel condition: The two fences come together asymptotically, i.e. U(x) − L(x) is small for large x.

The Attempt at a Solution

I think that the binomial expansion is appropriate here? But I am unclear if it can be used for fractions.

$U(x)-L(x) = -\sqrt{x-1} +\sqrt{x}$
the first term can be expressed as: (hopefully did the binomial right)
$x^\frac{1}{2} - \frac{1}{2}x^\frac{-1}{2}- \frac{1}{8}x^\frac{-3}{2}...$

so then
$-\sqrt{x-1} +\sqrt{x} = x^\frac{1}{2} -x^\frac{1}{2} + \frac{1}{2}x^\frac{-1}{2} + \frac{1}{8}x^\frac{-3}{2}...$
the first 2 terms cancel, then the rest go to 0 as x goes to infinity. Does that look right?

Yes.

 

[haruspex]

8x−32...−x−1+x=x12−x12+12x−12+18x−32...-\sqrt{x-1} +\sqrt{x} = x^\frac{1}{2} -x^\frac{1}{2} + \frac{1}{2}x^\frac{-1}{2} + \frac{1}{8}x^\frac{-3}{2}...
the first 2 terms cancel, then the rest go to 0 as x goes to infinity. Does that look right?

That's a little shaky. You have an infinite sum of terms each of which tends to zero. That's not quite the same as the sum tending to zero.
Also, are you sure you are permitted to lean on the binomial expansion here?
It is quite easy, and perhaps more instructive, to get the result from a simple epsilon-delta proof. I.e. given ε > 0, find x_ε such that x > x_ε implies etc.

 

[QuietMind]

I have no idea what's allowed. The question just says "use algebra" so I took binomial expansion as being part of the bag of tricks. Any other ideas what "use algebra" entails?

I have very little experience with epsilon delta proofs but I looked them up and will give it a try:

Want to prove this
Expression1: $\lim_{x\rightarrow +\infty}{\sqrt{x}-\sqrt{x-1}} = 0$

given $\epsilon > 0$

I looked up the procedure for when the limit approaches infinity, and they say the first step is to work backwards from this:
Expression2: $\epsilon > |-\sqrt{x-1} + \sqrt{x}-0|$

, and try to express x in terms of epsilon. Is that a good first step? Here's my work:

multiplying both sides by the sum of the squareroots, and getting rid of abs values, yields
$(\sqrt{x} + \sqrt{x-1})\epsilon > 1$
$(\sqrt{x} + \sqrt{x-1}) > \frac{1}{\epsilon}$
Adding this expression to the original Expression 2 (adding the greater than side to the greater than side to preserve inequality) yields:
$\epsilon + \sqrt{x} + \sqrt{x-1} > \frac{1}{\epsilon} + \sqrt{x} - \sqrt{x-1}$

Chug through algebra to get:
Expression 3: $x > \frac{1}{4}*(\frac{1}{\epsilon}-\epsilon)^2 + 1$

And then it tells me to start from expression 3, and re-derive expression 2 to formally constitute the proof, which I have trouble doing (the step where I add 2 inequalities together seems irreversible). Am I way off base?

 

[Ray Vickson]

I have no idea what's allowed. The question just says "use algebra" so I took binomial expansion as being part of the bag of tricks. Any other ideas what "use algebra" entails?

I have very little experience with epsilon delta proofs but I looked them up and will give it a try:

Want to prove this
Expression1: $\lim_{x\rightarrow +\infty}{\sqrt{x}-\sqrt{x-1}} = 0$

given $\epsilon > 0$

I looked up the procedure for when the limit approaches infinity, and they say the first step is to work backwards from this:
Expression2: $\epsilon > |-\sqrt{x-1} + \sqrt{x}-0|$

, and try to express x in terms of epsilon. Is that a good first step? Here's my work:

multiplying both sides by the sum of the squareroots, and getting rid of abs values, yields
$(\sqrt{x} + \sqrt{x-1})\epsilon > 1$
$(\sqrt{x} + \sqrt{x-1}) > \frac{1}{\epsilon}$
Adding this expression to the original Expression 2 (adding the greater than side to the greater than side to preserve inequality) yields:
$\epsilon + \sqrt{x} + \sqrt{x-1} > \frac{1}{\epsilon} + \sqrt{x} - \sqrt{x-1}$

Chug through algebra to get:
Expression 3: $x > \frac{1}{4}*(\frac{1}{\epsilon}-\epsilon)^2 + 1$

And then it tells me to start from expression 3, and re-derive expression 2 to formally constitute the proof, which I have trouble doing (the step where I add 2 inequalities together seems irreversible). Am I way off base?

You could also use the fact that $F(x) = \sqrt{x-1} - \sqrt{x}$ can be re-written as
F(x) = \frac{( \sqrt{x-1} - \sqrt{x}) ( \sqrt{x-1} + \sqrt{x})}{ \sqrt{x-1} + \sqrt{x}} = \frac{-1}{ \sqrt{x-1} + \sqrt{x}},
and that really is "algebra". In this last form you do not have "cancellation", so it is easier to deal with the small differences between $\sqrt{x-1}$ and $\sqrt{x}$.

 

[haruspex]

$(\sqrt{x} + \sqrt{x-1}) > \frac{1}{\epsilon}$

Down to there was fine (indeed, very good). The next step is to find a condition like "x> some function of epsilon" which ensures that statement is true. There's no need to be precious about finding the finest such condition. E.g. if √(x-1)>a, what can you say about √x in relation to a?