# Homework Help: Asymptotic anylasis of two tricky functions

1. Mar 7, 2013

1. The problem statement, all variables and given/known data
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.

3. 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.

2. Mar 7, 2013

### Staff: Mentor

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)?

3. Mar 7, 2013

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

4. Mar 8, 2013

### Staff: Mentor

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.

5. Mar 9, 2013

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

6. Mar 9, 2013

### Staff: Mentor

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: $\int_a^b f(x) dx = \int_a^b f(w) dw$.

7. Mar 9, 2013

### Ray Vickson

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.

8. Mar 10, 2013

### Staff: Mentor

How? Can you give an example?

9. Mar 10, 2013

ive done it now incidently

10. Mar 10, 2013

### Ray Vickson

"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.

11. Mar 11, 2013

### Staff: Mentor

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