HSC 2013 MX2 Marathon (archive) (1 Viewer)

seanieg89 · May 9, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Can you please criticise this proof? I am starting to have doubts about it.

===

Start with the polynomial,

$x=p+\sqrt{q}$

We need to make the lowest reducible rational polynomial that has that as a root, so we simply:

$(x-p)^2 - q = 0$

The above polynomial has the root $p - \sqrt{q}$
It is the simplest most reducible rational polynomial that has p + root(q) as a root.

Now consider:

$P(x) = a_1(x) a_2(x) \dots a_n(x)$

Where a_1, a_2 ... are all rational polynomial functions of lowest reducibility.

If $p+\sqrt{q}$ is a root then among the series a_1, a_2... there must exist a_k for some k such that $p+\sqrt{q}$ is a root.
Since a_k is irreducible and rational, then that polynomial must be:

$a_k(x) = (x-p)^2 - q$

And that polynomial also has the root $p-\sqrt{q}$

Therefore that is also a root of P(x).
Is there a gap or something in this? Or is it all good?

Its pretty good, certainly everything you have said is true and this method extends nicely to more complicated situations.

The main question is why a given algebraic number must be the root of a unique minimal (with respect to degree) monic irreducible polynomial with rational coeffs? (note we need uniqueness as well as existence here, think about why.)

If you resolve this question, it is clear that the minimal polynomial of this particular root must have degree at least 2 (else it would be rational). And therefore the polynomial you have constructed is the "minimal polynomial" of p+sqrt(q). The rest of the argument is fine.

My argument is less general but works well here. (And is what I hoped a student would find by thinking of the complex conjugate root theorem and trying to mimic it.)

$For real numbers of the form $x=a+b\sqrt{q}$ with $a,b\in\mathbb{Q}$, define $\overline{a+b\sqrt{q}}=a-b\sqrt{q}.$ Note that the sum or product of two real numbers $x$ and $y$ of this surdic form is also of this form. Moreover, by simple computation we have: $\overline{x+y}=\overline{x}+\overline{y}$ and $\overline{xy}=\overline{x}\cdot\overline{y}.$ Hence for a polynomial with rational coefficients and root $\alpha$ of this surdic form we have:\\ \\$P(\overline{\alpha})=\sum_{k=0}^n a_k \overline{\alpha}^k=\sum_{k=0}^n a_k \overline{\alpha^k}=\sum_{k=0}^n \overline{a_k \alpha^k}=\overline {\sum_{k=0}^n a_k \alpha^k}=\overline{P(\alpha)}=\overline{0}=0.$$

At the beginning I should probably have proven that two numbers of that form are equal iff their rational and surdic parts are equal, which is easy to do but am tired and lazy now. Other than that it is a pretty rigorous argument.

Sy123 · May 9, 2013

Re: HSC 2013 4U Marathon

Registered User said:
In this proof I will assume that $x -\frac{x^2}{2}+\frac{x^3}{3}\cdots$ converges to $\ln (1+x)$ (Taylor Series).
$Let,$

$S_1 = \sum_{n=1}^{\infty}\frac{\cos n\theta}{n}$

$S_2 = \sum_{n=1}^{\infty}\frac{\sin n\theta}{n}$

$Then$ $S_1 + iS_2 = \sum_{n=1}^{\infty}\frac{\cos(n\theta)+i\sin(n \theta)}{n}=\sum_{n=1}^{\infty}\frac{e^{in\theta}}{n}$

$Now, from the Taylor expansion,$ $\ln (1+x) = x -\frac{x^2}{2}+\frac{x^3}{3}\cdots$

$\implies -\ln(1-x) = x+ \frac{x^2}{2}+\frac{x^3}{3} ... = \sum_{n=1}^{\infty}\frac{x^n}{n}$

$\therefore S_1+iS_2 = -\ln(1-e^{i\theta})$

$=-\ln(1-\cos\theta-i\sin \theta)$

$=-\ln(2\sin^2\theta/2 - 2i\sin(\theta/2)\cos(\theta/2))$

$=-\ln(2\sin\theta/2)-\ln(\sin\theta/2-i\cos\theta/2)$

$=-\ln(2\sin\theta/2)+\ln(\sin\theta/2+i\cos\theta/2)$

$=-\ln(2\sin\theta/2)+\ln(e^{i(\pi/2-\theta/2)})$

$=-\ln(2\sin\theta/2)+i(\pi/2-\theta/2)$

$Taking the imaginary part of both sides,$

$S_2 = \frac{\pi}{2} - \frac{\theta}{2}$

==================================================

$Prove that$

$\sum_{k=-\infty}^{\infty}\left(\dfrac{\sin(k)}{k}\right)^{2}=\pi$

Nice work, it can also be done by considering the geometric series:

$z+z^2+ \dots +z^n = \frac{z(z^n-1)}{z-1}$

Substituting cis theta, equating real parts for a series in cosine. Then integrating with bounds alpha and pi yields the LHS and the RHS plus a limit to infinity of an integral which one must prove converges to zero.

seanieg89 said:
Its pretty good, certainly everything you have said is true and this method extends nicely to more complicated situations.

The main question is why a given algebraic number must be the root of a unique minimal (with respect to degree) monic irreducible polynomial with rational coeffs? (note we need uniqueness as well as existence here, think about why.)

If you resolve this question, it is clear that the minimal polynomial of this particular root must have degree at least 2 (else it would be rational). And therefore the polynomial you have constructed is the "minimal polynomial" of p+sqrt(q). The rest of the argument is fine.

My argument is less general but works well here. (And is what I hoped a student would find by thinking of the complex conjugate root theorem and trying to mimic it.)

$For real numbers of the form $x=a+b\sqrt{q}$ with $a,b\in\mathbb{Q}$, define $\overline{a+b\sqrt{q}}=a-b\sqrt{q}.$ Note that the sum or product of two real numbers $x$ and $y$ of this surdic form is also of this form. Moreover, by simple computation we have: $\overline{x+y}=\overline{x}+\overline{y}$ and $\overline{xy}=\overline{x}\cdot\overline{y}.$ Hence for a polynomial with rational coefficients and root $\alpha$ of this surdic form we have:\\ \\$P(\overline{\alpha})=\sum_{k=0}^n a_k \overline{\alpha}^k=\sum_{k=0}^n a_k \overline{\alpha^k}=\sum_{k=0}^n \overline{a_k \alpha^k}=\overline {\sum_{k=0}^n a_k \alpha^k}=\overline{P(\alpha)}=\overline{0}=0.$$

At the beginning I should probably have proven that two numbers of that form are equal iff their rational and surdic parts are equal, which is easy to do but am tired and lazy now. Other than that it is a pretty rigorous argument.

Ah alright, is that not a definition of an algebraic number? To be a root of a rational polynomial?

Also your method is very clever.

Fus Ro Dah · May 9, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Also your method is very clever.

It is a good method, but why do you say this as if it is alien to you? You can imagine that the proof for this would be very similar to the proof of the existence of complex conjugate pairs in real polynomials, since the complex conjugate is analogous to the irrational conjugate p-sqrt(q).

Sy123 · May 9, 2013

Re: HSC 2013 4U Marathon

Fus Ro Dah said:
It is a good method, but why do you say this as if it is alien to you? You can imagine that the proof for this would be very similar to the proof of the existence of complex conjugate pairs in real polynomials, since the complex conjugate is analogous to the irrational conjugate p-sqrt(q).

Yes that is true.

===

Also Registered User, are you sure your series is correct?

$\sum_{k=-\infty}^{\infty} \frac{\sin^2 k}{k^2}$ Doesn't look very nice because of the sines of integers, is it a typo?

asianese · May 9, 2013

Re: HSC 2013 4U Marathon

Carrotsticks said:
Not necessarily. A lot of reports involving statistics (such as graduate Biology papers) are also compiled in TeX.

Well, yes, scientific papers in general.

Registered User · May 9, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Yes that is true.

===

Also Registered User, are you sure your series is correct?

$\sum_{k=-\infty}^{\infty} \frac{\sin^2 k}{k^2}$ Doesn't look very nice because of the sines of integers, is it a typo?

It is definitely correct.

Registered User · May 9, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
It can also be done by considering the geometric series:

$z+z^2+ \dots +z^n = \frac{z(z^n-1)}{z-1}$

Substituting cis theta, equating real parts for a series in cosine. Then integrating with bounds alpha and pi yields the LHS and the RHS plus a limit to infinity of an integral which one must prove converges to zero.

Didn't think of that. Seems like an easier method.
Did you make up this question?

Sy123 · May 9, 2013

Re: HSC 2013 4U Marathon

Registered User said:
Didn't think of that. Seems like an easier method.
Did you make up this question?

I made up the solution (with help from sean in its rigour), the problem is a discarded one from STEP.

hayabusaboston · May 9, 2013

Re: HSC 2013 4U Marathon

I found fun question. I wonder how u guys will solve it.
"Show that there is a multiple of 2013 whose last four digits are 2012"

I dont know how to find a nice elegant solution. Mines messy as hell. U guys have something elegant?

Sy123 · May 9, 2013

Re: HSC 2013 4U Marathon

Registered User said:
$Prove that$

$\sum_{k=-\infty}^{\infty}\left(\dfrac{\sin(k)}{k}\right)^{2}=\pi$

Using the result from the previous question:

$\sum_{k=1}^{n} \frac{\sin(k\theta)}{k} = 0.5(\pi -\theta) + \int_{\pi}^{\theta} \frac{\cos(n\alpha) + \cos(n+1)\alpha}{4\sin^2 \alpha/2} \ d\alpha$

Integrating both sides from pi to 2, and using the double angle rule, the result from the Basel problem and so on, we arrive at:

$2\sum_{k=1}^{\infty} \frac{\sin^2 k}{k^2} = \pi - 1+ \lim_{n \to \infty} \varepsilon(\theta, \alpha)$

Then noticing that $\frac{\sin^2 (-k)}{(-k)^2} = \frac{\sin^2 k}{k^2}$

We can notice that the LHS of what I have right now, is the LHS of the question - 1 (since at k=0 the function approaches 1). This then cancels out the term on the RHS, therefore leaving it with:

$\sum_{k = -\infty}^{\infty} \frac{\sin^2 k}{k^2} = \pi + \lim_{n \to \infty}\varepsilon(\theta, \alpha)$

$\varepsilon(\theta, \alpha) = \int_{\pi}^2 \int_{\pi}^{\theta} \frac{\cos(n\alpha) + \cos(n+1)\alpha}{4\sin^2 \alpha /2} \ d\alpha \ d\theta$

I know that it should evaluate to zero, however the same problem from when I encountered trying the Basel problem comes up where I cannot justify interchanging the theta integral and the limit to infinity as the alpha integral converges to zero due to by parts. However I cannot justify putting it in there unfortunately :/

Makematics · May 9, 2013

Re: HSC 2013 4U Marathon

hayabusaboston said:
I found fun question. I wonder how u guys will solve it.
"Show that there is a multiple of 2013 whose last four digits are 2012"

I dont know how to find a nice elegant solution. Mines messy as hell. U guys have something elegant?

I still can't get it

Got the first question though

The junior one. took about 90 minutes though -.- very satisfying though.

Sy123 · May 9, 2013

Re: HSC 2013 4U Marathon

hayabusaboston said:
I found fun question. I wonder how u guys will solve it.
"Show that there is a multiple of 2013 whose last four digits are 2012"

I dont know how to find a nice elegant solution. Mines messy as hell. U guys have something elegant?

IF there does exist a multiple of 2013 like that, then for some pair of integers (n,m) The following equality must hold:

$2013n = m$

And for some value k, if m has last four digits 2012 then k is integer.

$k= \frac{m - 2012}{10 000}$

On the k-m plane it is a linear function and hence there must exist some pair of integers (k,m) which satisfy the above equality as the domain of the function is all real k and m.

Hence there exists an integer value of m for which k is an integer, hence there must exist a value m with last four digits 2012.

EDIT: Done

Makematics · May 9, 2013

Re: HSC 2013 4U Marathon

EDIT: Find an example of a function f(x) such that its inverse is the same as its derivative.

asianese · May 9, 2013

Re: HSC 2013 4U Marathon

Makematics said:
Find the function f(x) such that its inverse is the same as its derivative.

lol unsw?

Carrotsticks · May 9, 2013

Re: HSC 2013 4U Marathon

Makematics said:
Find the function f(x) such that its inverse is the same as its derivative.

There are infinitely many functions satisfying this condition?

Makematics · May 9, 2013

Re: HSC 2013 4U Marathon

asianese said:
lol unsw?

yeh lol

Makematics · May 9, 2013

Re: HSC 2013 4U Marathon

Carrotsticks said:
There are infinitely many functions satisfying this condition?

My bad, edited.

hayabusaboston · May 9, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
IF there does exist a multiple of 2013 like that, then for some pair of integers (n,m) The following equality must hold:

$2013n = m$

And for some value k, if m has last four digits 2012 then k is integer.

$k= \frac{m - 2012}{10 000}$

On the k-m plane it is a linear function and hence there must exist some pair of integers (k,m) which satisfy the above equality as the domain of the function is all real k and m.

Hence there exists an integer value of m for which k is an integer, hence there must exist a value m with last four digits 2012.

EDIT: Done

I am inexpressibly mesmerised by your brilliance...

Makematics · May 9, 2013

Re: HSC 2013 4U Marathon

hayabusaboston said:
I am inexpressibly mesmerised by your brilliance...

+1

Sy123 · May 10, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
After completing Sy123's question on conjugate roots, it is a nice challenge exercise to consider the analogous problem for cube roots:

$If $P(x)$ is a polynomial with rational coefficients and $\alpha+\sqrt[3]{\beta}$ is a root of $P$ for some rational $\alpha$ and some rational $\beta$ that is not the cube of a rational number, formulate and prove an assertion similar to the conjugate root theorem. (ie find another root or several other roots that $P$ must possess).$

So is this how complex numbers are involved:

$(x-\alpha)^3 = \beta$

Is the minimal rational polynomial with roots alpha + cuberoot(beta).

Following the same template as the square root conjugate proof we can prove that the roots of this polynomial above can be the root of P(x).

Solving the cubic as roots of unity.

$x= \alpha + \sqrt[3]{\beta}$

$x = \alpha + \sqrt[3]{\beta} (cis(2\pi /3))$

$x= \alpha + \sqrt[3]{\beta} (cis(4\pi/3))$

And of course this can be extended to nth roots (right?)

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