HSC 2014 MX2 Marathon ADVANCED (archive) (1 Viewer)

FrankXie · Nov 25, 2014

Re: HSC 2014 4U Marathon - Advanced Level

ok here are the details. My previous calculation is for the case X, Y and theire wives all in one boat.

$5\times{8 \choose 2}^2\times{6 \choose 2}^2\times{4 \choose 2}^2=31752000$

They can be in any of the five boates, which is why 5 times; 8 C 2 squared is for two men and two women chosen for the first boat of the remaining four boats; similarly for all other boats.

Now calculation for X and his wife in one boat and Y and his wife in the other boat (explanation quite similar so omitted)

$5\times4\times{8 \choose 1}^2\times{7 \choose 1}^2\times{6 \choose 2}^2\times{4 \choose 2}^2=508032000$

Therefore, total number of ways =31752000+508032000=539784000

abecina · Nov 25, 2014

Re: HSC 2014 4U Marathon - Advanced Level

I think what's going on here is that you are assuming that different boats means different arrangement. I think the question is asking how many different ways can they be separated. Sort of life "how many ways can 10 people be divided into two teams" doesn't assume which team is Team A and which team is Team B

abecina · Nov 25, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$$Case I: \textbf{X and his wife and Y and his wife are all in the same boat}. \\\\This means that the remaining eight men can be distributed amongst the other boats in $ \frac{8!}{4!(2!)^4} $ ways. Here I am assuming the boats themselves are indistinguishable. Once the men have taken their seats, the women can be seated in$ \frac{8!}{(2!)^4} $ways, because once the men are seated the order the women are distributed matters. Multiplying these two together we get $\frac{(8!)^2}{4!(2!)^8}\\\\ $Case II: \textbf{X and his wife in one boat, Y and his wife in another boat}\\\\ Person X and his wife take up two spots on one boat. Person Y and his wife take up another two spots on another boat. Between these two boats, two men can fill up the remaining spots in $ 8.7 $ ways, and the women likewise. The remaining six pairs can are distributed in a similar fashion to Case I: i.e.$ \frac{(8.7)^2(6!)^2}{3!(2!)^6 }. $\\\\Adding Case I and Case II together we get\\\\$ \frac{17.(8!)^2}{4!(2!)^8}$

FrankXie · Nov 25, 2014

Re: HSC 2014 4U Marathon - Advanced Level

abecina said:
I think what's going on here is that you are assuming that different boats means different arrangement. I think the question is asking how many different ways can they be separated. Sort of life "how many ways can 10 people be divided into two teams" doesn't assume which team is Team A and which team is Team B

but boats are different, aren't they?

Anyway, my answer divided by 5! will be the same as yours

FrankXie · Nov 26, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
(can someone post a good question please)

This one should be a good one:

$$ Let $ P(x)=\frac{d^n}{dx^n}(x^2-1)^n. $ Evaluate $ \int_{-1}^1P^2(x)dx.$

glittergal96 · Nov 26, 2014

Re: HSC 2014 4U Marathon - Advanced Level

FrankXie said:
This one should be a good one:

$$ Let $ P(x)=\frac{d^n}{dx^n}(x^2-1)^n. $ Evaluate $ \int_{-1}^1P^2(x)dx.$

Nice question

.

Let $u_m:=\frac{d^m}{dx^m}((x^2-1)^n).$

Then by repeated integration by parts:

$\int_{-1}^1 u_n^2\, dx=(-1)^n\int_{-1}^1 u_{2n}u_0 \, dx+\sum_{k=0}^{n-1}(-1)^k[u_{n+k}u_{n-k-1}]^1_{-1}.$

Each of the summands in the second term vanish, because the polynomials $u_{n+k}u_{n-k-1}$ are all odd, so we are left with

$(-1)^n\int_{-1}^1 u_{2n}u_0 \, dx=(-1)^n(2n)!\int_{-1}^1 (x^2-1)^n\, dx=(-1)^n(2n)!J_n.$

But for $n > 1$ , we can integrate by parts to obtain a recurrence for $J_n$ .

$J_n=\int_{-1}^1 \frac{d}{dx}(x)(x^2-1)^n\, dx$

$=-\int_{-1}^1 2nx^2(x^2-1)^{n-1}\, dx$

$=-2n(J_n+J_{n-1})=-\frac{2n}{2n+1}J_{n-1}.$

Hence the expression we need to evaluate is equal to

$(-1)^n(2n)!\cdot \prod_{k=1}^n (-\frac{2k}{2k+1})\cdot J_0$

$=(2n)!\frac{2\cdot 4 \cdots 2n}{3\cdot 5 \cdots (2n+1)}\cdot 2$

$=\frac{2^{2n+1}(n!)^2}{2n+1}.$

FrankXie · Nov 26, 2014

Re: HSC 2014 4U Marathon - Advanced Level

glittergal96 said:
Nice question .

Let $u_m:=\frac{d^m}{dx^m}((x^2-1)^n).$

Then by repeated integration by parts:

$\int_{-1}^1 u_n^2\, dx=(-1)^n\int_{-1}^1 u_{2n}u_0 \, dx+\sum_{k=0}^{n-1}(-1)^k[u_{n+k}u_{n-k-1}]^1_{-1}.$

Each of the summands in the second term vanish, because the polynomials $u_{n+k}u_{n-k-1}$ are all odd, so we are left with

$(-1)^n\int_{-1}^1 u_{2n}u_0 \, dx=(-1)^n(2n)!\int_{-1}^1 (x^2-1)^n\, dx=(-1)^n(2n)!J_n.$

But for $n > 1$ , we can integrate by parts to obtain a recurrence for $J_n$ .

$J_n=\int_{-1}^1 \frac{d}{dx}(x)(x^2-1)^n\, dx$

$=-\int_{-1}^1 2nx^2(x^2-1)^{n-1}\, dx$

$=-2n(J_n+J_{n-1})=-\frac{2n}{2n+1}J_{n-1}.$

Hence the expression we need to evaluate is equal to

$(-1)^n(2n)!\cdot \prod_{k=1}^n (-\frac{2k}{2k+1})\cdot J_0$

$=(2n)!\frac{2\cdot 4 \cdots 2n}{3\cdot 5 \cdots (2n+1)}\cdot 2$

$=\frac{2^{2n+1}(n!)^2}{2n+1}.$

well done. nice question and nicer solution. lol

glittergal96 · Nov 26, 2014

Re: HSC 2014 4U Marathon - Advanced Level

glittergal96 said:
Nice question .

Let $u_m:=\frac{d^m}{dx^m}((x^2-1)^n).$

Then by repeated integration by parts:

$\int_{-1}^1 u_n^2\, dx=(-1)^n\int_{-1}^1 u_{2n}u_0 \, dx+\sum_{k=0}^{n-1}(-1)^k[u_{n+k}u_{n-k-1}]^1_{-1}.$

Each of the summands in the second term vanish, because the polynomials $u_{n+k}u_{n-k-1}$ are all odd, so we are left with

$(-1)^n\int_{-1}^1 u_{2n}u_0 \, dx=(-1)^n(2n)!\int_{-1}^1 (x^2-1)^n\, dx=(-1)^n(2n)!J_n.$

But for $n > 1$ , we can integrate by parts to obtain a recurrence for $J_n$ .

$J_n=\int_{-1}^1 \frac{d}{dx}(x)(x^2-1)^n\, dx$

$=-\int_{-1}^1 2nx^2(x^2-1)^{n-1}\, dx$

$=-2n(J_n+J_{n-1})=-\frac{2n}{2n+1}J_{n-1}.$

Hence the expression we need to evaluate is equal to

$(-1)^n(2n)!\cdot \prod_{k=1}^n (-\frac{2k}{2k+1})\cdot J_0$

$=(2n)!\frac{2\cdot 4 \cdots 2n}{3\cdot 5 \cdots (2n+1)}\cdot 2$

$=\frac{2^{2n+1}(n!)^2}{2n+1}.$

Actually, thats not a correct explanation of why the sum vanishes, I will post again after lunch fixing that part.

FrankXie · Nov 26, 2014

Re: HSC 2014 4U Marathon - Advanced Level

maybe it is good, one odd and the other even, always like that

glittergal96 · Nov 26, 2014

Re: HSC 2014 4U Marathon - Advanced Level

glittergal96 said:
Actually, thats not a correct explanation of why the sum vanishes, I will post again after lunch fixing that part.

Ahh simple, because each of these terms has a factor that comes from hitting an (x^2-1)^n with less than n derivatives and evaluating at the roots of this polynomial. As this polynomials roots are of multiplicity n, these factors all vanish, and hence the whole sum does.

glittergal96 · Nov 26, 2014

Re: HSC 2014 4U Marathon - Advanced Level

FrankXie said:
maybe it is good, one odd and the other even, always like that

No its not, because odd is not what we actually need there. But my above post tidies up this inaccuracy.

FrankXie · Nov 26, 2014

Re: HSC 2014 4U Marathon - Advanced Level

yep well done

Sy123 · Nov 29, 2014

Re: HSC 2014 4U Marathon - Advanced Level

(HSC Methods only)

$\\ $Let$ \ H_n = \sum_{k=1}^n \frac{1}{k} \\ \\ $Show that$ \ \lim_{n \to \infty} (H_n - \ln n) \ $is finite$$

---

$\\ $If$ \ S_n = \sum_{k=1}^n \frac{(-1)^k}{k} \\ \\ $Show that$ \ S_{2n} = H_n - H_{2n} \\ \\ $Hence show that$ \ - \frac{1}{1} + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \dots = \ln 2$

FrankXie · Nov 29, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
(HSC Methods only)

$\\ $Let$ \ H_n = \sum_{k=1}^n \frac{1}{k} \\ \\ $Show that$ \ \lim_{n \to \infty} (H_n - \ln n) \ $is finite$$

---

$\\ $If$ \ S_n = \sum_{k=1}^n \frac{(-1)^k}{k} \\ \\ $Show that$ \ S_{2n} = H_n - H_{2n} \\ \\ $Hence show that$ \ - \frac{1}{1} + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \dots = \ln 2$

Part (a) Consider the area under the curve y=1/x between 1 and n. Divide the interval into (n-1) equal parts. If we use upper rectangles to approximate the area, we have

$1+\frac12+\frac13+\cdots+\frac{1}{n-1}>\ln{n}$

While we use lower rectangles to approximate the area, we obtain

$\frac12+\frac13+\cdots+\frac{1}{n}<\ln{n}$

Conbine these two inequalities, we arrive at

$\frac{1}{n}<H_n<1$

Therefore, if the limit exists, it must be finite.

NOTE: We assume the limit exists. To prove this, it seems to me we need monotone convergence which is beyond HSC scope.

FrankXie · Nov 29, 2014

Re: HSC 2014 4U Marathon - Advanced Level

For part (b) (i):

$H_{2n}+S_{2n}=\sum_{k=1}^{2n}\left(\frac1k+\frac{(-1)^k}{k}\right)=\sum_{m=1}^n\frac{2}{2m}=H_n$

(since odd numbers go away, even numbers k=2m get doubled)

Part (b)(ii):

$$ We again need the assumption that $ \lim_{n\rightarrow\infty}(H_n-\ln{n})=A$

From which it follows

$H_n-H_{2n}=(H_n-\ln{n})-(H_{2n}-\ln{2n})-(\ln{2n}-\ln{n})\rightarrow A-A-\ln2$

$$ Therefore, $ 1-\frac12+\frac13-\frac14+\cdots=\ln2$

Sy123 · Nov 29, 2014

Re: HSC 2014 4U Marathon - Advanced Level

FrankXie said:
NOTE: We assume the limit exists. To prove this, it seems to me we need monotone convergence which is beyond HSC scope.

Right. My bad.

SilentWaters · Nov 29, 2014

Re: HSC 2014 4U Marathon - Advanced Level

FrankXie said:
NOTE: We assume the limit exists. To prove this, it seems to me we need monotone convergence which is beyond HSC scope.

Not really. Investigating monotonic behaviour in functions and looking at boundaries is well within the scope of Extension 2.

Consider $a_n = H_n - \ln n$ . So:

$a_n - a_{n-1} = \frac{1}{n} - \ln n + \ln (n-1) \\= \frac{1}{n}+\ln \left(1-\frac{1}{n}\right) (*).$

We know by its positive square second derivative that $\ln (1-x)$ is concave down, and that by its first derivative, $y=-x$ is tangential at $x=0$ (thus exceeding the former function for all non-zero $x$ ). Popping $1/n$ in its place, where $n\geq1,$ we can conclude that (*) is negative. Hence our sequence is monotonic decreasing.

Additionally, looking at the relationship between the area beneath the curve $y=1/x$ to the right of $x=1$ , and the sums of lower-bound rectangles of unit width, we can say $\sum\limits_{k=1}^n {\frac{1}{k}} > \int_1^{n+1} \frac{1}{x} \,dx$ or $\ln (n+1).$ But $\ln (n+1) - \ln n > 0$ due to the monotonic increasing nature of the logarithmic curve. Hence $H_n -\ln n >0.$

In summary, we have a decreasing sequence bounded below by zero. So as $n\to\infty,$ it converges to a finite value.

SilentWaters · Nov 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

FrankXie said:
Part (a) Consider the area under the curve y=1/x between 1 and n. Divide the interval into (n-1) equal parts. If we use upper rectangles to approximate the area, we have

$1+\frac12+\frac13+\cdots+\frac{1}{n-1}>\ln{n}$

While we use lower rectangles to approximate the area, we obtain

$\frac12+\frac13+\cdots+\frac{1}{n}<\ln{n}$

Conbine these two inequalities, we arrive at

$\frac{1}{n}<H_n<1$

Therefore, if the limit exists, it must be finite.

Also I think the third line should read $H_n - \ln n.$

RealiseNothing · Nov 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

SilentWaters said:
Last line - RHS should read $-\ln 2.$ Fine work anyhow

no it shouldn't, it's defs positive

seanieg89 · Nov 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

SilentWaters said:
In summary, we have a decreasing sequence bounded below by zero. So as $n\to\infty,$ it converges to a finite value.

Sure, but this fact cannot be proven within the scope of MX2.

One would need to rigorously define the reals to prove such a statement as it is reliant on completeness or an equivalent axiom.

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