HSC 2015 MX2 Integration Marathon (archive) (1 Viewer)

Ekman · Jul 4, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
i got $\frac{\pi}{2n}ln|\frac{n+1}{n-1}|$

Yeah I got that as well.

glittergal96 · Jul 4, 2015

Re: MX2 2015 Integration Marathon

seanieg89 said:
Here is a question that concerns inequalities arising from

integration. (It is harder conceptually than most questions in this thread, but easier than

a lot of them in terms of how technically demanding the required manipulations are.)

$Let $\alpha>0$ be a positive parameter.\\ Show that there exist positive constants $C_1,C_2$ such that: \\ \\$C_1||z|-1|^{1-\alpha}<\int_0^{2\pi} |z-\textrm{cis}(\theta)|^{-\alpha} \, d\theta < C_2||z|-1|^{1-\alpha}$ for $|z|\neq 1$.$

First a couple of simplifying observations:

Note: The constant C in the following proof means an arbitrary positive constant. In
particular, C will typically change between lines as we absorb other constants into it. Eg
$(C+(x^2+1)C) \geq C$ is true in this sense, because the RHS is just shorthand for $K=2C$ .

1. By rotational symmetry we may assume $z = x$ is a non-negative real number.

2. If we can find such pairs of constants for $x\in F_j$ where $\{F_j\}$
is a finite collection of subsets of the non-negative reals, with union the non-negative
reals, then we can find such a pair of constants for x in the full set on non-negative
reals. (We simply take the lowest constant for the lower bound and the largest constant for
the upper bound.)

3. For a similar reason, and symmetry, it suffices to only consider integration in $[0,\pi]$ .

4. If we take $F_1=[0,1/2),F_2=[1/2,3/2),F_3=[3/2,\infty)$ , then the upper bound is
trivial for the first and third sets. The integrand is bounded by $Cd^{-\alpha}\leq C\cdot \frac{1}{2}\cdot{d^{-\alpha}}=Cd^{1-\alpha}$ .

5. The integrand is positive and decreasing in theta, whatever x is.

The key to the rest of the work is choosing a nice way cutting up the domain of integration into intervals of varying width and using upper/lower rectangle type approximations. But of course, we first need to understand how $d(\theta)$ varies in theta.

$d(\theta)=|\cis\theta-x|^{-\alpha}\\ = ((\cos\theta-x)^2+\sin^2\theta)^{1/2}\\=(|1-x|^2+2x(1-\cos\theta))^{1/2}.$

We also have $C(a+b)\leq\sqrt{a^2+b^2}\leq C(a+b)$ for non-negative a and b, by using the AM-QM for the first and adding in a 2ab under the square root to prove the upper bound.

Cobbling this together with the standard polynomial bounds for cos we get from repeated integration of the trig functions, we get

$d(\theta)\leq C(|1-x|+\sqrt{2x}\sqrt{1-\cos\theta})\leq C(d+\sqrt{2x}\sqrt{\theta^2/2})\leq C(d+\theta\sqrt{x})$

and

$d(\theta)\geq C(|1-x|+\sqrt{2x}\sqrt{1-\cos\theta})\geq C(d+\sqrt{2x}\sqrt{\theta^2/2-\theta^4/4!})\geq C(d+\theta\sqrt{x}).$

We don't need such delicate analysis to prove the lower bounds, as we have

$I\geq \int_0^d d(\theta)^{-\alpha}\, d\theta\geq d\cdot(2-d)^{-\alpha} \geq d\cdot (3d)^{-\alpha}=Cd^{1-\alpha}$

on $F_1$ and

$I\geq \int_0^d d(\theta)^{-\alpha}\, d\theta\ged d\cdot d(d)^{-\alpha}\geq Cd(d+d\sqrt{x})^{-\alpha}\geq Cd^{1-\alpha}$

on $F_2$

$I\geq d\cdot(d+2)^{-\alpha}\geq d\cdot (5d)^{-\alpha}=Cd^{1-\alpha}$

on $F_3$ .

So the only real work to do is on $F_2$ and this is a bit tricky. (It is unsurprising this is the hard bit, as this is where the integrand starts to get big.)

Because x is now a small bounded distance from 1 (indeed this was the purpose of my partitioning choice), our earlier distance estimates simplify to

$C_1(r+\theta)\leq d(\theta)\leq C_2(r+\theta).$

We want to partition the interval of integration into small sets and use the above estimates. The form of the upper and lower bounds make it convenient to consider scaled intervals of the form $\theta\in[da,db)$ .

You can play around with this and try obvious partitionings like $[0,d), [d,2d),\ldots$ , etc, but you will find you have convergence issues generally when you sum up the upper/lower bounds. The problem is that we need to have very sharp control of what happens at the "big" part, where d is kind of small. This motivates the choice of a dyadic style decomposition.

Now, to prove the upper bound on $F_2$ :

Let $I_{-1}=[0,d),I_k=[2^k d,2^{k+1}d)$ for $k=0,2,...,\lfloor\log_2(d^{-1})\rfloor$

Then

$I\leq \sum_{-1\leq k\leq -\log_2(d)}\int_{I_k} d(\theta)\, d\theta \leq d\cdot d(0)+\sum_{k=0}^{\lfloor\log_2(d^{-1})\rfloor}\int_{I_k}d(\theta)\, d\theta \\ \leq d^{1-\alpha}+ \sum_{k=0}^{\lfloor\log_2(d^{-1})\rfloor} 2^kd\cdot d(2^k d)^{-\alpha}\\ \leq d^{1-\alpha}+ \sum_{k=0}^{\lfloor\log_2(d^{-1})\rfloor} 2^kd\cdot d(2^k d)^{-\alpha}\\ \leq Cd^{1-\alpha}\left(\sum_{k=0}^{\lfloor\log_2(d^{-1})\rfloor}(1+2^k)^{-\alpha}+1\right)\\ \leq Cd^{1-\alpha}\left(\sum_{k=0}^{\lfloor\log_2(d^{-1})\rfloor}2^{-\alpha k}+1\right).$

But the geometric series has common ratio < 1 and so converges to something finite, and all partial sums are bounded above by this constant! This (finally) completes the proof.

Jesus christ that was exhausting...

Apologies for my abuse of notation having two d's (to clarify, $d(\theta)$ means the distance from 1 to the unit circle point with argument $\theta$ , and $d = ||x|-1|$ is the distance of x from the unit circle.), there is no way I am going to go back and fix them lol. Feel free to ask me if anything is unclear, it is entirely possible I made little algebra mistakes here or there.

glittergal96 · Jul 4, 2015

Re: MX2 2015 Integration Marathon

In case anyone wants to try to prove the similar question I posted based on the factorial before Sy posts his, it is MUCH MUCH shorter and easier than the above.

Sy123 · Jul 4, 2015

Re: MX2 2015 Integration Marathon

glittergal96 said:
In case anyone wants to try to prove the similar question I posted based on the factorial before Sy posts his, it is MUCH MUCH shorter and easier than the above.

btw is there a non-strict lower bound for your question? It seems to me (at least on the positive integers) that the expression is monotone decreasing tending towards a limit, but then there is no non-strict lower bound

glittergal96 · Jul 4, 2015

Re: MX2 2015 Integration Marathon

Sy123 said:
btw is there a non-strict lower bound for your question? It seems to me (at least on the positive integers) that the expression is monotone decreasing tending towards a limit, but then there is no non-strict lower bound

Well if it tends to a limit from above, wouldn't any real number less than that limit be a non-strict lower bound?

There are probably lots of ways of finding one, but I had a particular method in mind.

glittergal96 · Jul 4, 2015

Re: MX2 2015 Integration Marathon

(Ps, if you are actually able to prove that it is monotonically decreasing and bounded below, then you would have filled in the blank in my question which then completes a proof of Stirling's formula. I'm not sure that's true though...will check.)

Update: A quick check with small numbers seems to indicate this is true, can you prove it though?

Update 2: Ahh it's actually not that hard to prove monotonicity at all. That's cool. That means:

(Proof of montonicity)+(Proof of lower bound)+(Evaluation of limit, given that one exists) = (Proof of Stirling's formula.)

porcupinetree · Jul 6, 2015

Re: MX2 2015 Integration Marathon

Next Q:

$\int_{0}^{\frac{\pi }{2}}\frac{dx}{6-sin^{2}x}$

Kaido · Jul 6, 2015

Re: MX2 2015 Integration Marathon

Hmm not sure if I was supposed to use something like F(a-x)

I just multiplied by sec^2x/sec^2x and used u=tanx. Then standard arctan integral

Drsoccerball · Jul 6, 2015

Re: MX2 2015 Integration Marathon

porcupinetree said:
Next Q:

$I= \int_{0}^{\frac{\pi }{2}}\frac{dx}{6-sin^{2}x}$

$I= \int_0^{\frac{\pi}{2}}{\frac{dx}{6cos^2x +5sin^2x}} = \int{\frac{sec^2x}{6+5tan^2x}}$

$let u=tanx$

$\frac{1}{5}\int_0^{\infty}{\frac{1}{\frac{6}{5}+u^2}}$

$\frac{1}{6\sqrt{5}} \cdot \frac{\pi}{2}$

porcupinetree · Jul 6, 2015

Re: MX2 2015 Integration Marathon

Kaido said:
Hmm not sure if I was supposed to use something like F(a-x)

I just multiplied by sec^2x and used u=tanx. Then standard arctan integral

Yep - that's the way that I first did it to get an answer of $\frac{\pi}{2\sqrt{30}}$

But then I started fiddling around with partial fractions and t method and solved it a different way - but I must have made a mistake somewhere because it yielded the wrong answer.

Can someone have a look at this working and see what's wrong with it? I'm sure it's something very simple https://imgur.com/6mA5WzM

Drsoccerball · Jul 6, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
$I= \int_0^{\frac{\pi}{2}}{\frac{dx}{6cos^2x +5sin^2x}} = \int{\frac{sec^2x}{6+5tan^2x}}$

$let u=tanx$

$\frac{1}{5}\int_0^{\infty}{\frac{1}{\frac{6}{5}+u^2}}$

$\frac{1}{6\sqrt{5}} \cdot \frac{\pi}{2}$

EDIT : should be root 6 my bad

Drsoccerball · Jul 6, 2015

Re: MX2 2015 Integration Marathon

Alright lets see if this works:

$$ Show that: $ \int_0^{\frac{\pi}{2}}{\frac{dx}{1+sinx \cdot cos\alpha}} =\frac{2}{sin\alpha}tan^{-1}{\frac{sin\alpha}{1+cos\alpha}$

porcupinetree · Jul 6, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
Alright lets see if this works:

$$ Show that: $ \int_0^{\frac{\pi}{2}}{\frac{dx}{1+sinx \cdot cos\alpha}} =\frac{2}{sin\alpha}tan^{-1}{\frac{sin\alpha}{1+cos\alpha}$

I got $\frac{2}{sin\alpha }[tan^{-1}\frac{1+cos\alpha }{sin\alpha } - \frac{\pi}{2} + \alpha ]$
Idk if it can be simplified though

Drsoccerball · Jul 6, 2015

Re: MX2 2015 Integration Marathon

porcupinetree said:
I got $\frac{2}{sin\alpha }[tan^{-1}\frac{1+cos\alpha }{sin\alpha } - \frac{\pi}{2} + \alpha ]$
Idk if it can be simplified though

Thats the answer but you simplified th previous steps making the answer unattainable

Kaido · Jul 6, 2015

Re: MX2 2015 Integration Marathon

lol, "let's see"

used f(a-x), then let cosa be any constant, used t formula and standard arctan integral

Drsoccerball · Jul 6, 2015

Re: MX2 2015 Integration Marathon

Kaido said:
lol, "let's see"

used f(a-x), then let cosa be any constant, used t formula and standard arctan integral

using f(a-x) you just change the sin to the cos and thus wasting a step (since youre using t formula)

Kaido · Jul 6, 2015

Re: MX2 2015 Integration Marathon

lol, i always try it first whenever i see a definite integral

soz boss

porcupinetree · Jul 6, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
Thats the answer but you simplified th previous steps making the answer unattainable

How was I meant to do it?

Sy123 · Jul 6, 2015

Re: MX2 2015 Integration Marathon

glittergal96 said:
Well if it tends to a limit from above, wouldn't any real number less than that limit be a non-strict lower bound?

There are probably lots of ways of finding one, but I had a particular method in mind.

glittergal96 said:
(Ps, if you are actually able to prove that it is monotonically decreasing and bounded below, then you would have filled in the blank in my question which then completes a proof of Stirling's formula. I'm not sure that's true though...will check.)

Update: A quick check with small numbers seems to indicate this is true, can you prove it though?

Update 2: Ahh it's actually not that hard to prove monotonicity at all. That's cool. That means:

(Proof of montonicity)+(Proof of lower bound)+(Evaluation of limit, given that one exists) = (Proof of Stirling's formula.)

I wasn't able to think of a way to approximate the lower bound as $\sqrt{2\pi}$ (though technically if you allow for a strict lower bound like 2pi as an answer to the question, then a strict lower bound of 1 would be sufficient too and easy to prove, though such a bound would not be useful).

How did you get it?

Drsoccerball · Jul 6, 2015

Re: MX2 2015 Integration Marathon

Sy123 said:
I wasn't able to think of a way to approximate the lower bound as $\sqrt{2\pi}$ (though technically if you allow for a strict lower bound like 2pi as an answer to the question, then a strict lower bound of 1 would be sufficient too and easy to prove, though such a bound would not be useful).

How did you get it?

integrand got the lower bound ages ago if you want to look for it by fowarding to when the question was posted

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