HSC 2015 MX2 Integration Marathon (archive) (7 Viewers)

seanieg89 · Apr 9, 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: \\ \\ $C_1||z|-1|^{-\alpha}<\int_0^{2\pi} |z-\textrm{cis}(\theta)|^{-\alpha} \, d\theta < C_2||z|-1|^{-\alpha}$ for $|z|\neq 1$.$

(This integral clearly blows up as $z$ approaches the unit circle. The point of the question is to quantify how quickly this happens, which is generally a useful thing to know.)

I should clarify that you need to prove the existence of constants such that these inequalities hold for ALL z not on the unit circle. Your constants cannot depend on z themselves! (Otherwise these inequalities don't really tell you how fast the integral grows without knowing how C1 and C2 grow).

If you find it too hard to spot the clever idea that I think is necessary to prove this result, then you can try proving a weaker upper and lower bound.

Ie. Find an upper bound and a lower bound for this integral of the forms $C_1f(|z|),C_2g(|z|)$ and improve these as much as you can. You don't need to find explicit values for the constants, the functions are the important part.

(It should be clear that the integral only depends on |z|.)

InteGrand · Apr 10, 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: \\ \\ $C_1||z|-1|^{-\alpha}<\int_0^{2\pi} |z-\textrm{cis}(\theta)|^{-\alpha} \, d\theta < C_2||z|-1|^{-\alpha}$ for $|z|\neq 1$.$

(This integral clearly blows up as $z$ approaches the unit circle. The point of the question is to quantify how quickly this happens, which is generally a useful thing to know.)

Note sure if this makes any progress.

We want to show that there exist constants $C_1,C_2>0$ independent of z such that

$C_1\bigl \lvert |z|-1 \bigl \rvert ^{-\alpha}<\int_0^{2\pi} |z-\textrm{cis}(\theta)|^{-\alpha} \text{ d}\theta < C_2\bigl \lvert|z|-1\bigr\rvert^{-\alpha}\Leftrightarrow C_1<\int_0^{2\pi} \left(\frac{\bigl\lvert|z|-1\bigr\rvert}{|z-\textrm{cis}(\theta)|} \right )^{\alpha} \text{ d}\theta < C_2$

Let $I = \int_0^{2\pi} \left(\frac{\bigl\lvert|z|-1\bigr\rvert}{|z-\textrm{cis}(\theta)|} \right )^{\alpha} \text{ d}\theta$

By the reverse triangle inequality ( $|\omega_1 - \omega_2|\geq \bigl \lvert |\omega_1| - |\omega_2| \bigr \rvert$ ), we have

$\left(\frac{\bigl\lvert|z|-1\bigr\rvert}{|z-\textrm{cis}(\theta)|} \right )^{\alpha}\leq \left(\frac{\bigl\lvert|z|-1\bigr\rvert}{\bigl\lvert|z|-|\textrm{cis}(\theta)|\bigr\rvert} \right )^{\alpha}=\left(\frac{\bigl\lvert|z|-1\bigr\rvert}{\bigl\lvert|z|-1\bigr\rvert} \right )^{\alpha}=1$

which implies

$I<\int _0 ^{2\pi}1\text{ d}\theta = 2\pi$

so $C_2 = 2\pi$ is suitable.

seanieg89 · Apr 10, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
Note sure if this makes any progress.

We want to show that there exist constants $C_1,C_2>0$ independent of z such that

$C_1\bigl \lvert |z|-1 \bigl \rvert ^{-\alpha}<\int_0^{2\pi} |z-\textrm{cis}(\theta)|^{-\alpha} \text{ d}\theta < C_2\bigl \lvert|z|-1\bigr\rvert^{-\alpha}\Leftrightarrow C_1<\int_0^{2\pi} \left(\frac{\bigl\lvert|z|-1\bigr\rvert}{|z-\textrm{cis}(\theta)|} \right )^{\alpha} \text{ d}\theta < C_2$

Let $I = \int_0^{2\pi} \left(\frac{\bigl\lvert|z|-1\bigr\rvert}{|z-\textrm{cis}(\theta)|} \right )^{\alpha} \text{ d}\theta$

By the reverse triangle inequality ( $|\omega_1 - \omega_2|\geq \bigl \lvert |\omega_1| - |\omega_2| \bigr \rvert$ ), we have

$\left(\frac{\bigl\lvert|z|-1\bigr\rvert}{|z-\textrm{cis}(\theta)|} \right )^{\alpha}\leq \left(\frac{\bigl\lvert|z|-1\bigr\rvert}{\bigl\lvert|z|-|\textrm{cis}(\theta)|\bigr\rvert} \right )^{\alpha}=\left(\frac{\bigl\lvert|z|-1\bigr\rvert}{\bigl\lvert|z|-1\bigr\rvert} \right )^{\alpha}=1$

which implies

$I<\int _0 ^{2\pi}1\text{ d}\theta = 2\pi$

so $C_2 = 2\pi$ is suitable.

Yep, so this establishes the upper bound.

Geometrically this is the same as noting that we are really integrating around the unit circle, and our integrand is maximised at the point on the unit circle closest to z, then replacing the whole integrand by this upper bound.

Now try the lower bound (this is the more delicate half as you might have gathered.)

Ekman · Apr 10, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
NEW Q:

$If $I_n = \int_{0}^{1} x(1-x^3)^{n}\text{ d}x$ for $n\,\geq 0$, show that $I_n = \frac{3n}{3n+2}I_{n-1}$ for $n$\,\geq 1.$

Bump. Can a 2015er solve this?

studynerd · Apr 10, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Bump. Can a 2015er solve this?

Should be able to... its an extension of IBP

VBN2470 · Apr 10, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Bump. Can a 2015er solve this?

Yes, it is a standard recurrence question that may require a trick or two to solve, but is probably one of the more easier recurrence questions I have seen. There are definitely much harder ones out there.

VBN2470 · Apr 10, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
NEW Q:

$If $I_n = \int_{0}^{1} x(1-x^3)^{n}\text{ d}x$ for $n\,\geq 0$, show that $I_n = \frac{3n}{3n+2}I_{n-1}$ for $n$\,\geq 1.$

SOLUTION

$$I_n = \int_{0}^{1} x(1-x^3)^{n}\text{ d}x$ \\ = $[\frac{x^2}{2}(1-x^3)^n]_{0}^{1}+\frac{3n}{2}\int_{0}^{1} x^4(1-x^3)^{n-1}\text{ d}x$ \\ = $\frac{3n}{2}\int_{0}^{1} x(1-(1-x^3))(1-x^3)^{n-1}\text{ d}x$ \\ = $\frac{3n}{2}(I_{n-1}-I_n)$ \\ So $I_n = \frac{3n}{2}(I_{n-1}-I_n)$ \Rightarrow\, $I_n = \frac{3n}{3n+2}I_{n-1}$$

Drsoccerball · Apr 10, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
SOLUTION

$$I_n = \int_{0}^{1} x(1-x^3)^{n}\text{ d}x$ \\ = $[\frac{x^2}{2}(1-x^3)^n]_{0}^{1}+\frac{3n}{2}\int_{0}^{1} x^4(1-x^3)^{n-1}\text{ d}x$ \\ = $-\frac{3n}{2}\int_{0}^{1} x(1-x^3-1)(1-x^3)^{n-1}\text{ d}x$ \\ = $\frac{3n}{2}(-I_n+I_{n-1})$ \\ So $I_n = \frac{3n}{2}(-I_n+I_{n-1})$ \Rightarrow\, $I_n = \frac{3n}{3n+2}I_{n-1}$$

More recurence D:

VBN2470 · Apr 10, 2015

Re: MX2 2015 Integration Marathon

EASIER RECURRENCE Q:

$Let $I_n = \int_{0}^{1} (1-x^2)^{\frac{n}{2}}\text{ d}x$, where $n\,\geq0$ is an integer.$ \\ $(i)$ \, $Show that $I_n=\frac{n}{n+1}I_{n-2}$ for every integer $n$\,\geq2.$ \\ $(ii)$ \, $Evaluate $I_5$$

Ekman · Apr 10, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
Yes, it is a standard recurrence question that may require a trick or two to solve, but is probably one of the more easier recurrence questions I have seen. There are definitely much harder ones out there.

I know a 2015er can solve this, I just wanted to see more solutions by 2015ers on this marathon. Btw an alternate method to your question is to split the (1-x^3)^n into (1-x^3)^(n-1) * (1-x^3). Then you expand and use IBP once to get your final solution.

VBN2470 · Apr 10, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
I know a 2015er can solve this, I just wanted to see more solutions by 2015ers on this marathon. Btw an alternate method to your question is to split the (1-x^3)^n into (1-x^3)^(n-1) * (1-x^3). Then you expand and use IBP once to get your final solution.

Lol my bad, I misinterpreted your question.

Drsoccerball · Apr 10, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
EASIER RECURRENCE Q:

$Let $I_n = \int_{0}^{1} (1-x^2)^{\frac{n}{2}}\text{ d}x$, where $n\,\geq0$ is an integer.$ \\ $(i)$ \, $Show that $I_n=\frac{n}{n+2}I_{n-2}$ for every integer $n$\,\geq2.$ \\ $(ii)$ \, $Evaluate $I_5$$

I always get stuck wheb doing recurrence i got to :
$I_n=n \int{x^2(1-x^2)^{\frac{n-2}{2}}$

InteGrand · Apr 10, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
I always get stuck wheb doing recurrence i got to :
$I_n=n \int{x^2(1-x^2)^{\frac{n-2}{2}}$

The common trick to use for these recurrence questions is to now expand the integrand like this:

$x^2 (1-x^2)^{\frac{n-2}{2}}=(1-(1-x^2))(1-x^2)^{\frac{n-2}{2}}$ (writing the $x^2$ as $1-(1-x^2)$ – this is the key step)

$=1\cdot(1-x^2)^{\frac{n-2}{2}} - (1-x^2)(1-x^2)^{\frac{n-2}{2}}$ (expanding)

$=(1-x^2)^{\frac{n-2}{2}} - (1-x^2)^{\frac{n}{2}}$ (using index laws)

$\to I_{n-2}-I_n$

VBN2470 · Apr 10, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
I always get stuck wheb doing recurrence i got to :
$I_n=n \int{x^2(1-x^2)^{\frac{n-2}{2}}$

Change the $x^2$ into $-(1-x^2-1)$ and see how you can go from there...

Drsoccerball · Apr 10, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
Change the $x^2$ into $-(1-x^2-1)$ and see how you can go from there...

I end up with $I_n = \frac{nI_{n-2}}{n+1}$

VBN2470 · Apr 10, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
I end up with $I_n = \frac{nI_{n-2}}{n+1}$

InteGrand said:
The common trick to use for these recurrence questions is to now expand the integrand like this:

$x^2 (1-x^2)^{\frac{n-2}{2}}=(1-(1-x^2))(1-x^2)^{\frac{n-2}{2}}$ (writing the $x^2$ as $1-(1-x^2)$ – this is the key step)

$=1\cdot(1-x^2)^{\frac{n-2}{2}} - (1-x^2)(1-x^2)^{\frac{n-2}{2}}$ (expanding)

$=(1-x^2)^{\frac{n-2}{2}} - (1-x^2)^{\frac{n}{2}}$ (using index laws)

$\to I_{n-2}-I_n$

.

Drsoccerball · Apr 10, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
EASIER RECURRENCE Q:

$Let $I_n = \int_{0}^{1} (1-x^2)^{\frac{n}{2}}\text{ d}x$, where $n\,\geq0$ is an integer.$ \\ $(i)$ \, $Show that $I_n=\frac{n}{n+2}I_{n-2}$ for every integer $n$\,\geq2.$ \\ $(ii)$ \, $Evaluate $I_5$$

Yeah it should definitely be $I_n=\frac{n}{n+1}I_{n-2}$
spent so much time D: eventually did a random value and subbed it into wolfram so it has to be n+1 not n+2

VBN2470 · Apr 10, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
Yeah it should definitely be $I_n=\frac{n}{n+1}I_{n-2}$
spent so much time D: eventually did a random value and subbed it into wolfram so it has to be n+1 not n+2

I don't know why, I always make some stupid typo when typing up these questions

Drsoccerball · Apr 10, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
I don't know why, I always make some stupid typo when typing up these questions

All goods

2010 hsc question 8 (kills self)

Ekman · Apr 10, 2015

Re: MX2 2015 Integration Marathon

Here's another recurrence question:

$$ Let $ I_{n} = \int_{0}^{\frac{\pi}{2}} \sin^{n}x \cos^{2} x dx$

$$ Show that $ I_{n} = \frac{n-1}{n+2} \Cdot I_{n-2}$

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