HSC 2016 MX2 Marathon ADVANCED (archive) (1 Viewer)

lita1000 · Oct 29, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Sy123 said:
You're thinking of the factorisation of (t+h)^q + t^q

$\\ 1 + x + \dots + x^{q-1} = \frac{x^q -1}{x-1}$

$\\ $Let$ \ x = \frac{t+h}{t} \ $and multiply both sides by$ \ t^{q-1}$

Yep yep I screwed up hahaha

Sy123 · Oct 29, 2015

Re: HSC 2016 4U Marathon - Advanced Level

$\\ $Find the minimum value of$ \ f(x) \ $for$ \ 0 <x < 1 \ $where$ \ f(x) = \int_0^{\frac{\pi}{4}} |\tan t - x| \ $d$t$

Sy123 · Oct 29, 2015

Re: HSC 2016 4U Marathon - Advanced Level

$\\ $Show that there is no polynomial$ \ p(x) \ $that has integer coefficients and that there are three distinct integers$ \ a,b,c \ $such that$ \\\\ p(a) = b, \ p(b) = c, \ p(c) =a$

Paradoxica · Oct 29, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Sy123 said:
$\\ $Show that there is no polynomial$ \ p(x) \ $that has integer coefficients and that there are three distinct integers$ \ a,b,c \ $such that$ \\\\ p(a) = b, \ p(b) = c, \ p(c) =a$

$$Fact: $x-y|P(x)-P(y)$ if P: $\mathbb Z \to \mathbb Z.$

$$From this, we establish three facts.$

$\frac{P(a)-P(b)}{a-b} = k_1$

$\frac{P(b)-P(c)}{b-c} = k_2$

$\frac{P(c)-P(a)}{c-a} = k_3$

$$Where $\{k_1, k_2, k_3\} \in \mathbb Z$

$The product of these three integers is equal to one.$

$This means all three are integers and hence fractions are either 1 or -1.$

$WLOG, assume $k_1 = 1.$

$Then the other two can simultaneously be considered as either -1 or 1.$

$Both cases result in a=b=c. Contradiction. Hence, no such P exists.$

RealiseNothing · Oct 29, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Suppose that it is Easter weekend coming up, so that Friday and Monday are both public holidays (i.e. 4 day weekend).

John wants to go to the beach on exactly one day of the long weekend (so Friday, Saturday, Sunday, Monday).

However he wants to go on the hottest day of the long weekend.

Assuming that John is rational, and that he only finds out the temperature a particular day will be at 9am of the day itself (and so can only decide if he wants to go to the beach at 9am on that day), what is the probability that John goes to the beach on the hottest day as he wants to?

(Assume the temperature on any of the 4 days is completely random and not dependent on other factors such as the previous temperature etc. Also assume the temperature can not be the same on any of the 4 days to avoid ties).

Paradoxica · Oct 29, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Sy123 said:
$\\ $Find the minimum value of$ \ f(x) \ $for$ \ 0 <x < 1 \ $where$ \ f(x) = \int_0^{\frac{\pi}{4}} |\tan t - x| \ $d$t$

$It is assumed $t$ and $x$ are independent variables,$

$and that it is valid to treat $x$ as constant.$

$$Solving for $ \tan{t} - x = 0,$ we have $t = \tan^{-1}{x}.$

$This is the point where the integrated function abruptly turns.$

$Use this fact to split the integrand into two parts.$

$The left segment is negative before the absolute value, so we negate it.$

$The right segment remains as is, being positive already.$

$f(x) = \int_{0}^{\tan^{-1}{x}}{(x - \tan{t})}dt + \int_{\tan^{-1}{x}}^{\frac{\pi}{4}}{(\tan{t} - x)}dt$

$f(x) = \left \[ xt + \ln(\cos{t}) \right \]_0^{\tan^{-1}{x}} + \left \[ \ln(\sec{t}) - xt \right \]_{\tan^{-1}{x}}^{\frac{\pi}{4}}$

$f(x) = x \tan^{-1}{x} - \frac{1}{2} \ln(x^2+1) - \frac{\pi}{4} x +\frac{1}{2} \ln{2} + x \tan^{-1}{x} - \frac{1}{2} \ln(x^2+1)$

$f(x) = 2x \tan^{-1}{x} - \ln(x^2+1) - \frac{\pi}{4} x + \frac{1}{2} \ln{2}$

$f'(x) = 2\tan^{-1}{x} + \frac{2x}{x^2+1} - \frac{2x}{x^2+1} - \frac{\pi}{4}$

$f'(x) = 2\tan^{-1}{x} - \frac{\pi}{4} = 0$

$x= \tan{\frac{\pi}{8}} = \frac{\sin{\frac{\pi}{8}}}{\cos{\frac{\pi}{8}}} = \frac{2 \sin{\frac{\pi}{8}}\cos{\frac{\pi}{8}}}{2 \cos^2{\frac{\pi}{8}}} = \frac{\sin{\frac{\pi}{4}}}{1+\cos{\frac{\pi}{4}}} = \frac{\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}}= \frac{1}{\sqrt{2}+1}$

$= \sqrt{2}-1 \approx 0.414$

$So a stationary point exists in the domain of the function.$

$Taking the second derivative, we have:$

$f''(x) = \frac{2}{x^2+1}$

$Which is positive for $ x \in \mathbb R,$ in particular, positive for the domain where $f$ is defined.$

$Hence, $f(x)$ is an everywhere convex function, and the stationary point is a global minimum, in particular, a local minimum.$

$$Therefore, the minimum value of $f(x)$ is $f(\sqrt{2}-1) = f(\tan{\frac{\pi}{8}})$

$=2\left ( \tan{\frac{\pi}{8}}\right ) \frac{\pi}{8} - \frac{\pi}{4}\tan{\frac{\pi}{8}} + \frac{1}{2} \ln{2} - \ln{\left (1+\tan^2{\frac{\pi}{8}}\right )} = \frac{1}{2} \ln{2} - 2 \ln{\sec{\frac{\pi}{8}}}$

$\approx 0.1882264065$

InteGrand · Oct 29, 2015

Re: HSC 2016 4U Marathon - Advanced Level

RealiseNothing said:
Suppose that it is Easter weekend coming up, so that Friday and Monday are both public holidays (i.e. 4 day weekend).

John wants to go to the beach on exactly one day of the long weekend (so Friday, Saturday, Sunday, Monday).

However he wants to go on the hottest day of the long weekend.

Assuming that John is rational, and that he only finds out the temperature a particular day will be at 9am of the day itself (and so can only decide if he wants to go to the beach at 9am on that day), what is the probability that John goes to the beach on the hottest day as he wants to?

(Assume the temperature on any of the 4 days is completely random and not dependent on other factors such as the previous temperature etc. Also assume the temperature can not be the same on any of the 4 days to avoid ties).

$Can we assume that $T_1,T_2,T_3,T_4$ are identically distributed (so they are i.i.d., since we're told they're independent), where $T_i$ is the temperature of day $i$ in Celsius, for $i=1,2,3,4$? Because otherwise, the answer would depend on the actual distributions I think. For example if days 2, 3 and 4 had distributions that constrained $T_k$ to be less than say $20$, and on day 1 the temperature was greater than 20, then he would be guaranteed to get the hottest day (we are assuming he knows the distributions for each day too). Even if all the days had temperature probability distributions with support $[0,\infty)$ or $[0,T_\text{max.}]$ say, the actual distributions of each day would influence the answer.$

$(Did you intend for each $T_i$ to have a uniform distribution on $[0,T_\text{max.}]$ for a common positive number $T_\text{max.}$?)$

RealiseNothing · Oct 30, 2015

Re: HSC 2016 4U Marathon - Advanced Level

InteGrand said:
$(Did you intend for each $T_i$ to have a uniform distribution on $[0,T_\text{max.}]$ for a common positive number $T_\text{max.}$?)$

Pretty much this.

Forget about any physical restrictions on max/min temperatures though. Think of it as say a God just choosing the temperature to be whatever the first number that comes to his mind is (so I guess it would be uniformally distributed).

Sy123 · Oct 30, 2015

Re: HSC 2016 4U Marathon - Advanced Level

edit; nvm

InteGrand · Oct 30, 2015

Re: HSC 2016 4U Marathon - Advanced Level

RealiseNothing said:
Pretty much this.

Forget about any physical restrictions on max/min temperatures though. Think of it as say a God just choosing the temperature to be whatever the first number that comes to his mind is (so I guess it would be uniformally distributed).

Yeah but wouldn't it need to have a finite max. (and min.) temp.? A uniform distribution on an infinite support doesn't make sense.

I guess you essentially mean that whatever the temperature is on a day, the next day's one (or any other day's one) is equally likely to be above it as below it (and can't equal it).

RealiseNothing · Oct 30, 2015

Re: HSC 2016 4U Marathon - Advanced Level

InteGrand said:
Yeah but wouldn't it need to have a finite max. (and min.) temp.? A uniform distribution on an infinite support doesn't make sense.

I guess you essentially mean that whatever the temperature is on a day, the next day's one is equally likely to be above it as below it (and can't equal it).

Yep.

I guess you can have a max/min temperature and it won't make a difference I think. I'm not too experienced with distributions, it's more so just from an equally as likely to be higher/lower than the previous day perspective I'm looking for.

Sy123 · Oct 30, 2015

Re: HSC 2016 4U Marathon - Advanced Level

This may or may not be completely wrong and/or based on a misinterpretation of the question

------

Pr(John picks the right day) =
Pr(John picks the first day, and it is the hottest the second day) +
Pr(John picks the second day, and it is the hottest the second day) +
Pr(John picks the third day, and it is the hottest the third day) +
Pr(John picks the fourth day, and it is the hottest the fourth day)

= 1/4*1/4 + 1/4*1/2 + 1/4*3/4 + 1/4 * 1 = 5/8

At the start of the first day, to him, every day is equally possible for it to be the hottest day, so the probability he picks it is 1/4, and the probability the first is the hottest is 1/4 (non-epistemic)

At the start of the second day, depending on how the first day went, it might be rational for John to not go the second day at all (i.e. if T2 < T1)
Given that John does pick the second day, the probability that it is the hottest is 1/2. This is because:

There are 24 different orderings of temperature (which I will denote by the numbers 1, 2, 3, 4, the lower the number the lower the temperature)

(e.g. if the order is: 1, 4, 3, 2, it means that 2nd day is hottest followed by day 3, then day 4 then day 1)

There are 24 = 4! orderings, 12 of these orderings are such that 2nd day is greater than 1, and of these 6 of them are such that 2nd day = 4

Thus 6/12 = 1/2

Similarly for the 3rd day, of the 24 orderings, there are 8 orderings so that the 3rd is the hottest out of the first 3 days
Of these 8, there are 6 that are such that 3rd day is indeed the hottest day,

Thus Pr(3rd is hottest) given John picks 3rd day is 6/8 = 3/4

Similarly for day 4 it is 1

-------

InteGrand · Oct 30, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Sy123 said:
This may or may not be completely wrong and/or based on a misinterpretation of the question

------

Pr(John picks the right day) =
Pr(John picks the first day, and it is the hottest the second day) +
Pr(John picks the second day, and it is the hottest the second day) +
Pr(John picks the third day, and it is the hottest the third day) +
Pr(John picks the fourth day, and it is the hottest the fourth day)

= 1/4*1/4 + 1/4*1/2 + 1/4*3/4 + 1/4 * 1 = 5/8

At the start of the first day, to him, every day is equally possible for it to be the hottest day, so the probability he picks it is 1/4, and the probability the first is the hottest is 1/4 (non-epistemic)

At the start of the second day, depending on how the first day went, it might be rational for John to not go the second day at all (i.e. if T2 < T1)
Given that John does pick the second day, the probability that it is the hottest is 1/2. This is because:

There are 24 different orderings of temperature (which I will denote by the numbers 1, 2, 3, 4, the lower the number the lower the temperature)

(e.g. if the order is: 1, 4, 3, 2, it means that 2nd day is hottest followed by day 3, then day 4 then day 1)

There are 24 = 4! orderings, 12 of these orderings are such that 2nd day is greater than 1, and of these 6 of them are such that 2nd day = 4

Thus 6/12 = 1/2

Similarly for the 3rd day, of the 24 orderings, there are 8 orderings so that the 3rd is the hottest out of the first 3 days
Of these 8, there are 6 that are such that 3rd day is indeed the hottest day,

Thus Pr(3rd is hottest) given John picks 3rd day is 6/8 = 3/4

Similarly for day 4 it is 1

-------

$But couldn't he also say that given the first day's temperature, he knows the probability that one of the next three days is hotter than Day 1 is $1-\left(\frac{1}{2} \right)^3 = 0.875>0.5$, so he should always \textsl{not} pick Day 1? (Haven't thought about it fully though; I think my reasoning here may be wrong.)$

RealiseNothing · Oct 30, 2015

Re: HSC 2016 4U Marathon - Advanced Level

InteGrand said:
$But couldn't he also say that given the first day's temperature, he knows the probability that one of the next three days is hotter than Day 1 is $1-\left(\frac{1}{2} \right)^3 = 0.875>0.5$, so he should always \textsl{not} pick Day 1? (Haven't thought about it fully though; I think my reasoning here may be wrong.)$

Your logic is right, but the probability is wrong I think.

InteGrand · Oct 30, 2015

Re: HSC 2016 4U Marathon - Advanced Level

RealiseNothing said:
Your logic is right, but the probability is wrong I think.

I did 1 minus the probability that all 3 of the next days are colder than Day 1 (given that it is Day 1 currently). Is this the wrong way to calculate that probability?

RealiseNothing · Oct 30, 2015

Re: HSC 2016 4U Marathon - Advanced Level

InteGrand said:
I did 1 minus the probability that all 3 of the next days are colder than Day 1 (given that it is Day 1 currently). Is this the wrong way to calculate that probability?

Wouldn't it just be 1/4 chance that day 1 is the hottest (we have no information on any other days yet), so 3/4 chance that it isn't?

I understand how you got that probability though and now I'm kinda mind-fucked.

InteGrand · Oct 30, 2015

Re: HSC 2016 4U Marathon - Advanced Level

RealiseNothing said:
Wouldn't it just be 1/4 chance that day 1 is the hottest (we have no information on any other days yet), so 3/4 chance that it isn't?

I understand how you got that probability though and now I'm kinda mind-fucked.

Lol idk, maybe this is the new Monty Hall haha.

I have a feeling that before Day 1, the chance that Day 1 is hottest is 1/4, but once you know Day 1's temperature on Day 1, it changes to 1/8?? Or maybe this is a problem with having the infinite uniform distribution. Or maybe it's like the Envelope paradox thing. Actually, remembering this paradox, I think the following is true: if the four days' temperatures are fixed beforehand (by an omniscient genie say), then the probability is indeed 3/4. But if they're not fixed and truly can be higher or lower than Day 1 on each later day with 50-50 chance, then the probability should be the one calculated above (0.875).

I think the answer to this problem will be different if we consider the temperature as NOT fixed beforehand (which is how I interpreted it) compared to if they were fixed; haven't thought of it in much detail, this is kinda based on my instinct and the relation to the envelope paradox.

RealiseNothing · Oct 30, 2015

Re: HSC 2016 4U Marathon - Advanced Level

InteGrand said:
Lol idk, maybe this is the new Monty Hall haha.

I have a feeling that before Day 1, the chance that Day 1 is hottest is 1/4, but once you know Day 1's temperature on Day 1, it changes to 1/8?? Or maybe this is a problem with having the infinite uniform distribution. Or maybe it's like the Envelope paradox thing. Actually, remembering this paradox, I think the following is true: if the four days' temperatures are fixed beforehand (by an omniscient genie say), then the probability is indeed 3/4. But if they're not fixed and truly can be higher or lower than Day 1 on each later day with 50-50 chance, then the probability should be the one calculated above (0.875).

I think the answer to this problem will be different if we consider the temperature as NOT fixed beforehand (which is how I interpreted it) compared to if they were fixed, haven't thought of it in much detail, this is kinda based on my instinct and the relation to the envelope paradox.

That's the same instinct I had whilst thinking about why we get two different probabilities. Might look into it a bit more since it seems way more counter-intuitive than even the Monty Hall problem.

When I did this question I was assuming the temperatures were all fixed. Feel free to do it however you want though, I only have the answer (confirmed by other sources too so all good) using the assumption that the temperatures are fixed beforehand however.

glittergal96 · Oct 30, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Sy123 said:
Revamping solution (and making some of the inferences more rigorous), had fun writing this

$\\ p(z(\cos \alpha + i \sin \alpha)) = p(z) \\\\ \Rightarrow \ p(z(\cos k\alpha + i \sin k\alpha)) = p(z) \ $for all integers$ \ k \geq 0$

$\\ \alpha \ $is either a rational multiple of$ \ 2\pi \ $or not, if it is not, then$ \ \cos k \alpha + i \sin k \alpha \ $is distinct for all values of$ \ k \ $(L1)$$

$\\ $Since$ \ p(1) = p(\cos k \alpha + i \sin k \alpha) \ $for all values of$ \ k \ $then$ \ p(z) - p(1) \ $has infinitely many roots, meaning$ \ p(z) = C \ $for some constant$ \ C$

$\\ $Suppose on the other hand that$ \ \alpha = 2\pi\frac{p}{q} \ $then$ \\\\ p(t) = p\left(t \left( \cos \frac{2\pi p}{q} i \sin \frac{2\pi p}{q} \right) \right) = \dots = p \left(t \left( \cos \frac{2\pi p (q-1)}{q} i \sin \frac{2\pi p (q-1)}{q} \right)\right)$

$\\ $Then$ \ p(z) - p(t) = A(z-t)\left(z - t \left( \cos \frac{2\pi p}{q} i \sin \frac{2\pi p}{q} \right) \right) \dots \left(z - t \left( \cos \frac{2\pi p (q-1)}{q} i \sin \frac{2\pi p (q-1)}{q} \right) \right) M(z)$

$\\ $Where$ \ M(z) \ $is a non-zero polynomial$$

$\\ (z-t)\left(z - t \left( \cos \frac{2\pi p}{q} i \sin \frac{2\pi p}{q} \right) \right) \dots \left(z - t \left( \cos \frac{2\pi p (q-1)}{q} i \sin \frac{2\pi p (q-1)}{q} \right) \right) \\\\ = z^q - t^q \ $(L2)$$

$\\ $Then$ \ p(z) - p(t) = A(z^q - t^q) M(z)$

$\\ $Let$ \ z = t +h \ $for some small$ \ h > 0$

$p(t+h) - p(t) = A((t+h)^q - t^q) M(t+h)$

$\\ \Rightarrow p(t+h) - p(t) = Ah( (t+h)^{q-1} + t(t+h)^{q-2} + \dots + t^{q-1}) M(t+h)$

$\\ \Rightarrow p'(t) = A \lim_{h \to 0}( (t+h)^{q-1} + t(t+h)^{q-2} + \dots + t^{q-1}) M(t+h) = A \cdot M(t) q t^{q-1} \ \ (*)$

$\\ $Let instead$ \ z = 0 \ $then$ \ p(t) = A\cdot M(t) t^q + B \ \ (**)$

$\\ $Differentiating$ \ (**) \ $we get$ \ p'(t) = A( M'(t) t^q + M(t) q t^{q-1})$

$\\ $Equating$ \ (**) \ $and$ \ (*) \ $we get$ \ M'(t) = 0 \Rightarrow M(t) = C$

$\\ $Therefore$ \ p(z) - p(t) = A(z^q - t^q)$

$\\ \Rightarrow \ p(z) = a z^q + b \ $for constants$ \ a,b$

$\\ $Thus, the final solution is$ \ p(z) = \begin{cases}c , \ \alpha \ $is not a rational multiple of$ \ 2\pi \\ a z^q + b , \ \alpha \ $is a rational multiple of$ \ 2\pi \ $with denominator$ \ q \end{cases}$

-----------

$\\ $(L1) proof as follows:$ \ $Suppose there exists$ \ m < n \ $so that$ \ cis m \alpha = cis n \alpha \ $with$ \ m,n \ $natural$ \\ $Then$ \ m\alpha + 2\pi k = n\alpha \ $for some integer$ \ k \ $as$ \ cis \ $is periodic in$ \ 2\pi \\ \\ $So,$ \ \alpha = 2\pi \frac{k}{m-n} \ $which is a rational multiple of$ \ 2\pi \\\\ $Contradiction, therefore$ \ $all distinct if not rational multiple$$

-----------

$\\ $(L2) proof as follows: It is sufficient to prove that$ \\\\ \prod_{i=0}^{q-1} \left(z - t\left(\cos \frac{2\pi i p}{q} + i\sin \frac{2 \pi i p}{q} \right) \right) = \prod_{i=0}^{q-1} \left(z - t\left(\cos \frac{2\pi i}{q} + i\sin \frac{2 \pi i }{q} \right) \right)$

$\\ $This can be done by showing that$ \\\\ $(1):$ \ \cos \frac{2\pi i p}{q} + i\sin \frac{2 \pi i p}{q} = \cos \frac{2\pi j}{q} + i\sin \frac{2 \pi j}{q} \ $for some integers$ \ i,j \\\\ $(2): From the above equation, for each different$ \ i \ $there is a unique$ \ j$

$\\ $(1) and (2) then imply that the set of: $ \\\\ \left(z - t\left(\cos \frac{2\pi i p}{q} + i\sin \frac{2\pi i p}{q} \right) \right) \ $is a permutation of$ \\\\\ \left(z - t\left(\cos \frac{2\pi i}{q} + i\sin \frac{2 \pi i }{q} \right) \right)$

$\\ $Meaning their products are equal$$

$\\ $Proof of (1) is that for each$ \ ip = r_i + k_iq \ $for some integers$ \ 0 \leq r_i \leq q-1 \ $and integer$ \ k_i$

$\frac{2\pi ip}{q} = \frac{2\pi r_i}{q} + 2\pi k_i$

$\\ $Since$ \ \cos \theta + i \sin \theta \ $is periodic in$ \ 2\pi \ $then$$

$\cos \frac{2\pi i p}{q} + i\sin \frac{2 \pi i p}{q} = \cos \frac{2\pi j}{q} + i\sin \frac{2 \pi j}{q} \ $where$ \ j = r_i$

$\\ $Proof of (2) is that for any two distinct integers$ \ m,n \ $then if$ \ mp = r_m + k_mq \ $and$ \ np = r_n + k_nq \ $then$ \ r_n \neq r_m \ $as it is the remainders that get mapped onto set of$ \ j$

$\\ $Suppose to the contrary that$ \ r_m = r_n \ $then$ \ \frac{(m-n)p}{q} = K \ $for some integer$ \ K$

$\\ $Given the definition of$ \ p,q \ $then$ \ p \ $and$ \ q \ $have no common factors, moreover, since$ \ 0 < m-n < q-1 \ $then$ \ (m-n) \ $is not divisible by$ \ q \ $then in order for$ \ \frac{(m-n)p}{q} = K \ $all the factors of$ \ q \ $must divide into either$ \ (m-n) \ $or$ \ p \ $but since it can't be$ \ p \ $then it must be$ \ (m-n)$

$\\ $But not all the factors of$ \ q \ $can divide into$ \ (m-n) \ $as that would mean$ \ (m-n) \ $is divisible by$ \ q \ $which is already shown to be impossible$$

$\\ $Therefore$ \ \frac{(m-n)p}{q} \neq K \ $thus$ \ r_m \neq r_n \ $thus$ \ (2) \ $is proven$$

$\\ $Thus the proposition in question is proven$$

-----------------

$\\ $NOTE: (*) means that at HSC level at least, this at most finds all$ \ \textbf{real} \ $polynomials with the aforementioned properties$$

Still not quite right...in your (**) line, substituting z=0 would give p(t)= p(0) + At^qM(0).

Also, there is a deeper problem going wrong in your assumption that p(z)-p(t)=(z^n-t^n)M(z).

Your reasoning (presumably) was that for nonzero t, we obtain n distinct points z (by rotating t according to the assumption of the question) at which p(z)=p(t) and so q(z):=p(z)-p(t) has these n roots. Multiplying them together we get (z^n-t^n) and so (z^n-t^n) divides the polynomial q(z) for each t.
HOWEVER, q(z) depends on the parameter t! This means that the coefficients of the polynomial M obtained by dividing q(z) by the factor (z^n-t^n) will generally depend on t as well as z.

A quick way to see why this answer is incorrect is that taking a=pi in the original question implies that p(z)=p(-z) for all z on the unit circle. As a nonzero polynomial can only have finitely many roots, this is equivalent to the seemingly stronger assertion that p(z)=p(-z) identically on the complex plane...that is p is even.

If your answer was correct, it would imply that the only even polynomials are of the form ax^2 + b.

My method for when a is a rational multiple of pi:

$Let $a=m\pi/n$, with $m,n$ coprime. Let $w=\textrm{cis}(a)$. Our transformation law in the original question tells us $p(wz)=p(z)$ for all $z$ on the unit circle. As nonzero polynomials have finitely many roots, we have that $p(wz)=p(z)$ for ALL $z$ in the complex plane. We now formally differentiate this expression k times to obtain\\ \\ $w^kp^{(k)}(wz)=p^{(k)}(z)$. By taking $z=0$, we see that $p^{(k)}(0)(w^k-1)=0$. Since $w^k=\textrm{cis}(mk\pi/n)$ and $m,n$ are coprime, $w^k=1$ iff $2n$ divides $k$. For all other $k$, we have $p^{(k)}(0)=0$, which implies that the $c_k$ coefficient of $p$ is zero for all $k$ not divisible by $2n$.\\ \\ Hence $p(z)=q(z^{2n})$ for some polynomial $q$. Indeed any $q$ will do, as $p(wz)=q(w^{2n}z^{2n})=q(z^{2n})=p(z)$.$

$To sum up: For $a$ an irrational multiple of $\pi$, $p(z)=C$.\\ \\ For $a=m\pi/n$, with $m,n$ coprime positive integers, we have $p(z)=q(z^{2n})$ for arbitrary $q$. Another way of saying this is that $p$ is any polynomial whose nonzero terms all have degree divisible by $2n$. Eg: any polynomial that is an even function is simply one with all nonzero terms having even degree.$

glittergal96 · Oct 30, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Also note that it is not a significant leap of faith to differentiate polynomials with complex coefficients/domain.

For such objects differentiation can be viewed as a purely formal linear map that sends the monomial

z^n to nz^(n-1).

One can study the relationship of this operator to the properties of a polynomial without having the slightest notion of what calculus was. Eg the proof that we can get the multiplicity of a root of a polynomial by evaluating derivatives at that root.

The cool thing is this extends to polynomials defined in much more abstract settings, where we can't do actual calculus.

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