HSC 2013 MX2 Marathon (archive) (1 Viewer)

Sy123 · Jul 3, 2013

Re: HSC 2013 4U Marathon

$$i) Show that$ \ \ \cos(n\theta) \cos \theta = \frac{1}{2}(\cos(n+1)\theta + \cos(n-1)\theta) \ \ \ \fbox{1}$

$ii) A sequence of polynomials$ \ \ C_n \ \ $is defined by$

$C_n(\cos \theta) = \cos(n\theta)$

$Prove the recurrence formulae$

$C_0(x) = 1$

$C_1(x) = x$

$C_{n+1}(x) = 2xC_{n} (x) - C_{n-1} (x) \ \ \ \ \fbox{3}$

$iii) Similarly show that, for some sequence of polynomials$ \ \ S_n \ \ $defined by$

$S_n(\cos \theta) = \frac{\sin (n+1)\theta}{\sin \theta}$

$Prove the recurrence relations$

$S_0(x) = 1$

$S_1(x) = 2x$

$S_{n+1}(x) = 2xS_{n}(x) - S_{n-1}(x) \ \ \ \fbox{2}$

$iv) Prove that$

$\frac{d^2}{dx^2} C_n(x) = n \frac{nC_n(x) - xS_{n-1}(x)}{x^2-1} \ \ \ \ \fbox{5}$

braintic · Jul 3, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Solve the equation$

$a^2+b^2+c^2+d^2 - ab - bc -cd - d + \frac{2}{5} = 0$

I assume that is a typo: da instead of d?

No real solution I believe?
I hope you weren't after complex solutions. But then you would get dependence on a parameter.

Sy123 · Jul 4, 2013

Re: HSC 2013 4U Marathon

braintic said:
I assume that is a typo: da instead of d?

No real solution I believe?
I hope you weren't after complex solutions. But then you would get dependence on a parameter.

Nope, its supposed to be d, there is no real significance in the number and the apparent symmetry though.

seanieg89 · Jul 4, 2013

Re: HSC 2013 4U Marathon

But are you looking for real or complex solutions?

Sy123 · Jul 4, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
But are you looking for real or complex solutions?

Oh yes real solutions definitely.

braintic · Jul 4, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Oh yes real solutions definitely.

I let 2/5 = k, then treated the equation as a quadratic in a, with b, c,d and k as constants.
I then treated the discriminant as a quadratic in b, with c, d and k as constants.
And so on, taking 4 discriminants.
It is unlikely that I did the calculation error-free, but I convinced myself that provided there were no errors that the original equation would have an infinite number of solutions if k > 1/15. And I think that a numerical error would simply change that boundary value for k.

In fact, I can't see how such an equation could have a finite number of solutions, unless there was something special about the choice of 2/5.

But I'm waiting to be proved wrong.

seanieg89 · Jul 4, 2013

Re: HSC 2013 4U Marathon

braintic said:
I let 2/5 = k, then treated the equation as a quadratic in a, with b, c,d and k as constants.
I then treated the discriminant as a quadratic in b, with c, d and k as constants.
And so on, taking 4 discriminants.
It is unlikely that I did the calculation error-free, but I convinced myself that provided there were no errors that the original equation would have an infinite number of solutions if k > 1/15. And I think that a numerical error would simply change that boundary value for k.

In fact, I can't see how such an equation could have a finite number of solutions, unless there was something special about the choice of 2/5.

But I'm waiting to be proved wrong.

Such an equation can have a finite number of solutions for the same reason that the equation x^2+y^2+z^2=0 has a finite number of solutions. The function on the LHS has a finite (in this case consisting of only one point) set of minima.

This is the case here. Viewed as a function f(a,b,c,d), the unique stationary point of f is at (1/5,2/5,3/5,4/5) which is a global minimum at which f=0. (set all partials = 0 to find stat point, compute Hessian at this point, check that f gets big away from the origin). As this is multivariable calculus and certainly not an MX2 level technique I am not going to use it to justify my answer, but it tells us what the answer should be. Now it is a matter of MX2 level inequalities to prove this is a solution and it is unique. I will post a solution here later today, off to uni now.

A hint for anyone who wants to try this in the mean time:

It collapses nicely if you make the substitutions

x=a-1/5
y=b-2/5
z=c-3/5
w=d-4/5

as many terms disappear, and you just have to show than some quadratic expression is non-negative, and zero only at the origin.

braintic · Jul 4, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
Such an equation can have a finite number of solutions for the same reason that the equation x^2+y^2+z^2=0 has a finite number of solutions. The function on the LHS has a finite (in this case consisting of only one point) set of minima.

This is the case here. Viewed as a function f(a,b,c,d), the unique stationary point of f is at (1/5,2/5,3/5,4/5) which is a global minimum at which f=0. (set all partials = 0 to find stat point, compute Hessian at this point, check that f gets big away from the origin). As this is multivariable calculus and certainly not an MX2 level technique I am not going to use it to justify my answer, but it tells us what the answer should be. Now it is a matter of MX2 level inequalities to prove this is a solution and it is unique. I will post a solution here later today, off to uni now.

A hint for anyone who wants to try this in the mean time:

It collapses nicely if you make the substitutions

x=a-1/5
y=b-2/5
z=c-3/5
w=d-4/5

as many terms disappear, and you just have to show than some quadratic expression is non-negative, and zero only at the origin.

But x^2+y^2+z^2=c generally doesn't have a finite number of solutions.
I made my comment based on Sy's comment that "there is no real significance in the number".
I DID say "I can't see how such an equation could have a finite number of solutions, unless there was something special about the choice of 2/5."
Multivariable calculus is too far back for me to remember. But I want to ask, are there a finite number of solutions for values of the constant other than 2/5? Or IS 2/5 special?
And by extension, for what values of this constant does the equation have no solutions/finitely many solutions/infinitely many solutions?
I'm asking so that I can check the logic/correctness of what I did.

seanieg89 · Jul 4, 2013

Re: HSC 2013 4U Marathon

braintic said:
But x^2+y^2+z^2=c generally doesn't have a finite number of solutions.
I made my comment based on Sy's comment that "there is no real significance in the number".
I DID say "I can't see how such an equation could have a finite number of solutions, unless there was something special about the choice of 2/5."
Multivariable calculus is too far back for me to remember. But I want to ask, are there a finite number of solutions for values of the constant other than 2/5? Or IS 2/5 special?
And by extension, for what values of this constant does the equation have no solutions/finitely many solutions/infinitely many solutions?
I'm asking so that I can check the logic/correctness of what I did.

Yeah I am not sure what Sy meant by that statement.

Replacing -2/5 with a greater number removes the possibility of any real solutions existing, and replacing it with a smaller number leads to an infinitude of solutions as you said.

This is (for example) a straightforward consequence of the intermediate value theorem after we observe that f is large for large |(a,b,c,d)|.

It is not in general an easy matter to figure out whether or not there are infinitely many solutions to f(x)=c in R^n.

(But I suppose for polynomials one can prove that this equation has finitely many solutions if and only if c is a global extreme value of f. This result would not need very sophisticated tools, the intermediate value theorem is the key.)

Out of interest, what did you do?

Sy123 · Jul 4, 2013

Re: HSC 2013 4U Marathon

$a^2+b^2+c^2+d^2 - ab-bc-cd-d+ 2/5 = 0$

$\Rightarrow \ \ \left(a - \frac{b}{2} \right)^2 + \frac{3}{4} \left(b - \frac{2c}{3}\right)^2 + \frac{2}{3} \left( c- \frac{3d}{4} \right)^2 + \frac{5}{8} \left( d - \frac{4}{5} \right)^2 = 0$

So just completing the square gives you this, since this is a sum of squares which equal to zero, (and because a, b, c, d are real), therefore, each square must be zero if the sum is zero, i.e:

$x^2+y^2+z^2 \geq 0$

With equality when: $x=y=z=0$

So the equality happens when each bracket = 0, this gives 4 equations, given d=4/5, then we can find c, then b then a.

Sorry for being misleading when saying that 2/5 is not significant, it seems like it is important in order for the equation to dissolve into a sum of squares.

seanieg89 · Jul 4, 2013

Re: HSC 2013 4U Marathon

braintic said:
(Besides an incomplete Electrical Engineering degree) BSC (Maths/Computing), DipEd (Maths). But we are talking mid 80s. Other than the maths I use in my teaching, and some things I've decided to relearn, I only remember some broad concepts. (I'd probably remember more if I didn't spend most of my Uni days in the cricket nets, playing Ultimate Frisbee, or lying by the pool).

Haha actually meant what was your approach to the problem? (Ignoring the calculation error).

Cool though. The idea of teaching mathematics is very attractive to me if I decide at some stage down the road I am not cut out for academia.

asianese · Jul 4, 2013

Does a PhD. Not cut out for academia. W0t

seanieg89 · Jul 4, 2013

Re: HSC 2013 4U Marathon

asianese said:
Does a PhD. Not cut out for academia. W0t

I don't see how this is that surprising, PhD dropout rates are pretty significant...and many people who complete them don't always end up going into academia (the job market at the moment is a deterrent for one).

"Cut out" in this context mostly refers to how strong ones work ethic is and how dedicated they are to research. I love what I am doing at the moment but there are ups and downs, and I might change my mind in the future.

asianese · Jul 4, 2013

Haha DW about me I was just stirring. I know what you mean - a PhD doesn't mean necessarily research or academia for long term.

Sy123 · Jul 5, 2013

Re: HSC 2013 4U Marathon

$Prove that$

$\frac{a^{m} + a^{m-1} + \dots + 1}{m+1} > \frac{a^{m-1} + a^{m-2} + \dots + 1}{m}$

$For some real$

$a> 1$

HeroicPandas · Jul 5, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Prove that$

$\frac{a^{m} + a^{m-1} + \dots + 1}{m+1} > \frac{a^{m-1} + a^{m-2} + \dots + 1}{m}$

$For some real$

$a> 1$

$a^{m-1} + \dots +1 = a^{m-1} + \dots + 1$

$\therefore \underbrace{a^m + a^{m-1}+ \dots +1}_{m+1 terms} > \underbrace{a^{m-1} + \dots +1}_{ mterms}$

$\therefore $The average of the LHS $> RHS \\ \\ \therefore \frac{a^m + a^{m-1} + \dots +1}{m+1} > \frac{a^{m-1}+\dots +1}{m}$

Sy123 · Jul 5, 2013

Re: HSC 2013 4U Marathon

HeroicPandas said:
$a^{m-1} + \dots +1 = a^{m-1} + \dots + 1$

$\therefore \underbrace{a^m + a^{m-1}+ \dots +1}_{m+1 terms} > \underbrace{a^{m-1} + \dots +1}_{ mterms}$

$\therefore $The average of the LHS $> RHS \\ \\ \therefore \frac{a^m + a^{m-1} + \dots +1}{m+1} > \frac{a^{m-1}+\dots +1}{m}$

Hmm, I'm not sure if your third line is always true.

seanieg89 · Jul 5, 2013

Re: HSC 2013 4U Marathon

It isn't.

seanieg89 · Jul 5, 2013

Re: HSC 2013 4U Marathon

Proof that the statement is true for a=2, for all positive integers m.

$LHS-RHS=\frac{2^{m+1}-1}{m+1}-\frac{2^m-1}{m}=\frac{(m-1)2^m+1}{m(m+1)}>0.$

Sy123 · Jul 5, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Prove that$

$\frac{a^{m} + a^{m-1} + \dots + 1}{m+1} > \frac{a^{m-1} + a^{m-2} + \dots + 1}{m}$

$For some real$

$a> 1$

$$Let$ \ LHS= P$

$RHS = Q$

$\frac{P(m+1) - a^m}{m} = Q$

$m(P-Q) = a^m - P$

$P-Q = \frac{a^m - P}{m}$

$$Since$ \ \ a > 1$

$P = \frac{a^m + a^{m-1} + \dots + 1}{m+1} < \frac{a^m + a^m + \dots + a^m}{m+1} = a^m$

$\therefore P < a^m$

$\therefore P - Q > 0$

$\therefore P > Q$

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