HSC 2016 MX2 Marathon (archive) (1 Viewer)
- Thread starter Trebla
- Start date
- Status
Re: HSC 2016 4U Marathon
\right)^n$ is a palindromic polynomial for any palindromic polynomial $P(x)$ (where $n$ is a positive integer). Since $1+x+x^2$ is a palindromic polynomial, the result follows.$)
$ is a palindromic polynomial, and $a$ is a root of $p$, then $a^{-1}$ is a root of $p$ with same multiplicity. (Note $0$ can't be a root of a palindromic polynomial $p(x)$. This is because $p(x)$ has constant term same as its leading coefficient, which is non-zero. This means $a^{-1}$ is well-defined for any root $a$ of $p$.) Also, conversely, if $a$ and $a^{-1}$ occur as roots of same multiplicity of a polynomial $p(x)$ for all roots $a$ of $p$, then $p$ is either palindromic or \textsl{anti-palindromic}. (Excluding the trivial case of the zero polynomial.)$)
$ and $Q(x)$ be palindromic polynomials. We want to show that $p(x):= P(x)Q(x)$ is palindromic. It is clear that the roots of $p(x)$ are just those of $P(x)$ together with those of $Q(x)$. Since $P(x)$ and $Q(x)$ were palindromic, their roots had the property mentioned above (for any root $a$, $a^{-1}$ is a root of same multiplicity). Since $p(x)$ has the roots of $P(x)$ and $Q(x)$ as its roots, the roots of $p(x)$ also satisfy this property. So $p(x)$ is either anti-palindromic or palindromic. But it is easy to see by expansion that $P(x)Q(x)$ has equal leading coefficient and constant term (since $P(x) = Ax^m + \cdots + A, Q(x) = Bx^n + \cdots + B\right),$ where $A,B \neq 0$), so $p(x)$ can't be anti-palindromic. So $p(x)$ is palindromic, as desired.$)
$'s roots has this property, recall that the roots of the \textsl{reciprocal polynomial} $q^{*}(x) = x^n q\left(x^{-1}\right)$ are precisely the reciprocals of the roots of $q$ (this fact crops up in the HSC a lot, when finding polynomials with reciprocal roots and multiplying through by $x^n$). Since $q= q^{*}$ (definition of palindromic polynomial), the result follows.$)
 is either anti-palindromic or palindromic, let $p(x)$ be a polynomial with such roots and let $p^{*}(x)$ be the corresponding reciprocal polynomial. As said above, the roots of $p^{*}(x)$ are just precisely the reciprocals of the roots of $p(x)$ (used in the HSC a lot, when finding a polynomial with reciprocal roots). But the assumed property of $p(x)$'s roots together with the fact that reciprocating is an involution means that in fact $p^{*} (x)$ has precisely the same roots as $p(x)$. Since the product of roots of $p(x)$ is $\pm 1$ (could be a minus, since $-1$ roots could be present), the constant term in $p(x)$ is $(-1)^n$ times its leading coefficient, where $n$ is the degree of $p$. So the constant term is either plus or minus the leading term of $p$.$)
 = \pm p^{*}(x)$, i.e. $p(x)$ is palindromic or anti-palindromic.$)
Further reading: https://en.wikipedia.org/wiki/Reciprocal_polynomial#Palindromic_polynomial .
Further reading: https://en.wikipedia.org/wiki/Reciprocal_polynomial#Palindromic_polynomial .
Last edited:
Paradoxica
-insert title here-
Re: HSC 2016 4U Marathon
Or maybe not, if someone can manage the cases correctly to obtain a valid expansion that doesn't use the multinomial theorem.
The fastest way would probably be to use induction on n, Binomial expansions look tedious from this perspective.
Or maybe not, if someone can manage the cases correctly to obtain a valid expansion that doesn't use the multinomial theorem.
Re: HSC 2016 4U Marathon
Yeah I probably wouldn't do it in the above way for that Q in a HSC exam (I was just showing it as an interesting and more general result). And yeah, you can (and are probably expected to) do it via something like the binomial theorem (or you could try trinomial theorem).
RealiseNothing
what is that?It is Cowpea
Paradoxica
-insert title here-
Paradoxica
-insert title here-
Re: HSC 2016 4U Marathon
You could obtain the answer to this new one from the answer to the old one given by Paradoxica. Just take conjugates, since the new equation is satisfied precisely by the conjugates of the old one.
KingOfActing
lukewarm mess
Paradoxica
-insert title here-
seanieg89
Well-Known Member
- Joined
- Aug 8, 2006
- Messages
- 2,662
- Gender
- Male
- HSC
- 2007
Re: HSC 2016 4U Marathon
(*) By this I mean that the questions as stated would never be on an official MX2 exam because of the lack of guidance, but if one was to break down a proof of them into multiple guided steps, these questions are of the level that the guided steps could be fashioned into parts of a Q16. Deeper facts have definitely been proven in official exams.
Higher really*, but some four unit students would definitely be able to do them without assuming anything wildly out of syllabus.
(*) By this I mean that the questions as stated would never be on an official MX2 exam because of the lack of guidance, but if one was to break down a proof of them into multiple guided steps, these questions are of the level that the guided steps could be fashioned into parts of a Q16. Deeper facts have definitely been proven in official exams.
Last edited:
seanieg89
Well-Known Member
- Joined
- Aug 8, 2006
- Messages
- 2,662
- Gender
- Male
- HSC
- 2007
Re: HSC 2016 4U Marathon
Well inequalities are usually considered part of Harder 3U, so I would classify them as such. As for what tools from elsewhere in the syllabus are useful to attack these questions, calculus is always good to keep in mind, and the presence of symmetric sums in the latter inequality (2) indicates that some polynomial knowledge can be helpful. I am sure there are several ways of proving these other than the ways I have envisaged though.
- Status