Asymptotic anylasis of two tricky functions

[pleasehelpmeno]

Homework Statement

Hi I have two questions related to the above topic.
1)I am trying to practice this technique and am stuck on a couple of problems, one is in taking


a)\sqrt{10-\left(\frac{y}{x\sqrt{1+\frac{y^2}{x^2}}}\right)} asymptotically as y\rightarrow -\infty

b)\exp\left(\pm i \int_{x_0}^{x} \sqrt{f(x)}dx\right), \mbox {as } \rightarrow \pm\infty f(x)=\sqrt{a^2 + a^{2}\frac{y^2}{x^2} }

I know the answers are just a)-\frac{y}{x \sqrt{2}} and b) (\frac{2y}{x})^{\pm ip^{2}/2}e^{\pm \frac{iy^{2}}{2}}e^{\pm \frac{ip^{2}}{4}}but when trying to find it I just get stuck.

If this is in the wrong section I apologise, and can someone please redirect it thanks.

The Attempt at a Solution

I have attempted to simplify expression a) but only end up with the same answer that i get for +infinity which is:+\frac{y}{x \sqrt{2}} and isn't correct.
As to b) I don't really know what to do because of the integral being present.

 

[mfb]

Your first limit is wrong. It cannot depend on y, and the denominator is wrong as well.
There are some convenient ways to simplify limits:
If $f(y) \to A$, then $\sqrt{f(y)} \ to \sqrt{A}$, and similar with any other continuous function. This should be sufficient to get the first limit. Hint: It does not depend on x.

The second limit is x -> +- infinity? And why does it have x both in the integrand and as limit of the integral? And where is the point in those two square roots (instead of one 4th root)?

 

[pleasehelpmeno]

sry my mistake

I see your point for the first one.
As to the integral it should read:

e^{\int^{n}_{n_0}\pm i p \sqrt{1+\frac{x^2}{p^2}}d x} as x \rightarrow \infty

I( just don't know how to apply asymptotic anylasis to integrals etc

 

[mfb]

That is a completely different expression now. And it is meaningless - the expression does not depend on x, so taking the limit of x to anything is pointless.
Please make sure that you copy the expression exactly.

 

[pleasehelpmeno]

That is the correct expression, and it does depend on x?

 

[mfb]

No, you integrate over x.

$\int_a^b f(x) dx = F(b)-F(a)$ where F is an antiderivative of f. There is no x appearing on the right side, which is equal to the left side. In addition, x is just a dummy variable, you can replace it by anything else if you like: $\int_a^b f(x) dx = \int_a^b f(w) dw$.

 

[Ray Vickson]

Your first limit is wrong. It cannot depend on y, and the denominator is wrong as well.
There are some convenient ways to simplify limits:
If $f(y) \to A$, then $\sqrt{f(y)} \ to \sqrt{A}$, and similar with any other continuous function. This should be sufficient to get the first limit. Hint: It does not depend on x.

The second limit is x -> +- infinity? And why does it have x both in the integrand and as limit of the integral? And where is the point in those two square roots (instead of one 4th root)?

The asymptotic form for y → ∞ CAN in principle depend on y, although in this case (i.e., for (a)) it happens to NOT depend on y. That is, the "claimed" answer is incorrect.

 

[mfb]

The asymptotic form for y → ∞ CAN in principle depend on y

How? Can you give an example?

 

[pleasehelpmeno]

ive done it now incidently

 

[Ray Vickson]

How? Can you give an example?

"Asymptotic" does NOT always mean "approach an asymptote". By definition,
f(x,y) \sim g(x,y) \;\text{ as } y \to \infty
if
\frac{f(x,y)}{g(x,y)} \to 1 \; \text{ as } y \to \infty
The symbol '~' means "is asymptotic to".

An important example in probability is the tail behaviour of the standard normal distribution: if $Z$ has density
f(z) = \frac{1}{\sqrt{2 \pi}} e^{-z^2/2},
then
\Pr \{ Z > y \} \sim \frac{1}{\sqrt{2 \pi}}\frac{1}{y} e^{-y^2/2},
for large y > 0. Use of the asymptotic form allows compution of tail probabilities far out in the distribution. Of course, the tail probability → 0 as y → ∞, but using the previous asymptotic form allows much better estimates: rather than saying "the probability is really, really small", we can say it is about 2.357e-08, for example.

 

[mfb]

Ah, well, okay. But there is no y-dependent limit then.