HSC 2015 MX2 Marathon ADVANCED (archive) (5 Viewers)

integral95 · Feb 26, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Well I dunno if this is too hard for the integration thread, but i'll post it here anyway.

From UNSW integration bee.

$\int \frac{1}{x(x+1)(x+2)...(x+n)}\ \ dx$

seanieg89 · Feb 26, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Sy123 said:
$f(xy) = f(x) f(y) - f(x+y) + 1$

$\\ $Let$ \ x=y=0 \ \Rightarrow \ f(0) = f(0)^2 - f(0) + 1 \Rightarrow \ f(0) = 1$

$\\ $Let$ \ x = 1, y = -1 \ \Rightarrow \ f(-1)(f(1) - 1) = 0 \\ \\ \therefore \ f(-1) = 0 \ $or$ \ f(1) = 1$

$\\ $If$ \ f(1) = 1 \ $then let$ \ y =1 \ \Rightarrow \ f(x) + f(x+1) = f(x) + 1 \Rightarrow \ f(x) = 1 \ $is a solution$$

$\\ $If$ \ f(-1) = 0 \ $then let$ \ y = -1 \ \Rightarrow \ f(-x) + f(x-1) = 1$

$x \mapsto x + \frac{1}{2} \ \Rightarrow \ f\left(-x - \frac{1}{2} \right) + f\left(x - \frac{1}{2} \right) = 1$

$\\ $Let$ \ g(x) = f\left(x - \frac{1}{2} \right)$

$\\ \therefore \ g(x) + g(-x) = 1$

$\\ \therefore \ g(x) = \frac{1}{2} + O(x) \ $for some odd function$ \ O: \mathbb{Q} \rightarrow \mathbb{Q}$

$\\ \therefore \ f(x) = \frac{1}{2} + O\left(x + \frac{1}{2} \right) \ $for any odd function$ \ O: \mathbb{Q} \rightarrow \mathbb{Q}$

$\\ \therefore \ $The functions$ \ f: \mathbb{Q} \rightarrow \mathbb{Q} \ $are$ \\ \\ f(x) = 1 \ $(if f(1) = 1)$ \\ \\ f(x) = \frac{1}{2} + O\left(x+ \frac{1}{2} \right) \ $(otherwise)$$

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I don't think I made any mistakes, is that the solution you're looking for?

Your logic is only 1-directional when you introduce the arbitrary odd function O.

Indeed any f must be of such a form (apart from the constant solution 1), but a generic odd O will not do.

Eg O(x)=0 does not work.

Sy123 · Feb 26, 2015

Re: HSC 2015 4U Marathon - Advanced Level

integral95 said:
Well I dunno if this is too hard for the integration thread, but i'll post it here anyway.

From UNSW integration bee.

$\int \frac{1}{x(x+1)(x+2)...(x+n)}\ \ dx$

$\\ \frac{A_0}{x} + \frac{A_1}{x+1} + \dots + \frac{A_n}{x+n} = \frac{1}{x(x+1) \dots (x+n)} \\ \\ \Rightarrow \ \sum_{k=0}^n A_k \frac{x(x+1)\dots(x+n)}{x+k} = 1$

$\\ x = -k \Rightarrow \ A_k (n-k)(n-k-1)\dots(1)(-1)(-2)\dots(-k) = (-1)^k(n-k)! k!$

$\\ \therefore \ \int \frac{dx}{x(x+1)\dots(x+n)} = \sum_{k=0}^n \int \frac{A_k}{x+k} \ dx = \ln \prod_{k=0}^n (x+k)^{(-1)^k (n-k)!k!} + c$

SilentWaters · Feb 26, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Sy123 said:
$\\ \frac{A_0}{x} + \frac{A_1}{x+1} + \dots + \frac{A_n}{x+n} = \frac{1}{x(x+1) \dots (x+n)} \\ \\ \Rightarrow \ \sum_{k=0}^n A_k \frac{x(x+1)\dots(x+n)}{x+k} = 1$

$\\ x = -k \Rightarrow \ A_k (n-k)(n-k-1)\dots(1)(-1)(-2)\dots(-k) = (-1)^k(n-k)! k!$

$\\ \therefore \ \int \frac{dx}{x(x+1)\dots(x+n)} = \sum_{k=0}^n \int \frac{A_k}{x+k} \ dx = \ln \prod_{k=0}^n (x+k)^{(-1)^k (n-k)!k!} + c$

Damn, difficult question. I agree with you up to about the final line.

By substituting x = -k, we obtain the constants:

$A_k = \frac{1}{(-1)^{k}(n-k)!k!}$

So the final answer should be:

$\ln\prod_{k=0}^n (x+k)^{\frac{1}{(-1)^{k}(n-k)!k!}} \, + C$

EDIT: Note the fractional exponent.

@InteGrand: 0! = 1

FrankXie · Feb 26, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Chlee1998 said:
Let ABCD be a trapezium with AB||CD such that

(i) its vertices A,B,C,D lie in a circle with centre O ,
(ii) its diagonals AC and BD intersect at a point M and ∠AMD=60 ∘ ,
(iii) MO=10 .

Find the difference between the lengths of AB and CD .

$10\sqrt{3}$

Drsoccerball · Feb 26, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Chlee1998 said:
Working?

Draw it i geogebra the test wouldnt have it in words

FrankXie · Feb 27, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Chlee1998 said:
Working?

it is clear that the trapezium is isosceles and angle ACD=angle BCD=30 degree. So angle BOC=60 degree and triangle BOC is equilateral. Note that the centre O must lie inside the trapezium. Let the radius=R and angle OCD=\alpha.
$\angle{MCO}=30^\circ-\alpha, \angle{BCM}=60^\circ-(30^\circ-\alpha)=30^\circ+\alpha, \angle{AOB}=60^\circ+2\alpha.$
$$ Therefore, $ MO=R\cos(30^\circ+\alpha)-R\sin(30^\circ+\alpha)\tan30^\circ$
$\therefore 10=R\left(\frac{\sqrt3}{3}\cos\alpha-\sin\alpha\right) $ (1) $$
$the desired difference $ CD-AB=2R[\cos\alpha-\sin(30^\circ+\alpha)]=2R\left(\frac{1}{2}\cos\alpha-\frac{\sqrt3}{2}\sin\alpha\right)=10\sqrt3 $ using (1)$

InteGrand · Feb 27, 2015

Re: HSC 2015 4U Marathon - Advanced Level

$\text{Let } w(x) = \frac{\alpha}{\lambda}\left ( \frac{x}{\lambda} \right )^{\alpha - 1}\text{e}^{-\left ( \frac{x}{\lambda} \right )^\alpha},\text{ for real non-negative } x, \text{ where }\alpha \text{ and }\lambda \text{ are positive parameters}. \\ \text{ For what values of }\alpha \text{ is } w(x) \text{ a one-to-one function?}$

seanieg89 · Feb 27, 2015

Re: HSC 2015 4U Marathon - Advanced Level

InteGrand said:
$\text{Let } w(x) = \frac{\alpha}{\lambda}\left ( \frac{x}{\lambda} \right )^{\alpha - 1}\text{e}^{-\left ( \frac{x}{\lambda} \right )^\alpha},\text{ for real non-negative } x, \text{ where }\alpha \text{ and }\lambda \text{ are positive parameters}. \\ \text{ For what values of }\alpha \text{ is } w(x) \text{ a one-to-one function?}$

Do you insist that the domain contains 0? Because that forces $\alpha \geq 1$ otherwise we blow up at 0.

And for $\alpha >1$ then we tend to 0 from above at both 0 and infinity, which makes injectivity impossible.

For $\alpha=1$ this is of course just a decreasing exponential function that is certainly injective because it is monotone.

If you only require the function to be defined on the positive reals, then $\alpha\in (0,1)$ words as well. In this case w is the product of two decreasing positive functions and is hence decreasing (hence injective). In fact in this case w is a bijection from the positive reals to themselves because it blows up at 0 and exponentially decays to 0 in the large x limit.

InteGrand · Feb 27, 2015

Re: HSC 2015 4U Marathon - Advanced Level

seanieg89 said:
Do you insist that the domain contains 0? Because that forces $\alpha \geq 1$ otherwise we blow up at 0.

I want the domain to contain 0 if it doesn't blow up ( $\alpha \geq 1$ ), but not to contain 0 if it does blow up ( $0<\alpha <1$ ).

But it doesn't seem to matter to the question if 0 is excluded.

InteGrand · Feb 27, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Determine whether $\sum_{n=1}^{\infty}\tan^{-1}\frac{1}{n}$ converges.

Sy123 · Feb 27, 2015

Re: HSC 2015 4U Marathon - Advanced Level

InteGrand said:
Determine whether $\sum_{n=1}^{\infty}\tan^{-1}\frac{1}{n}$ converges.

$\\ $By drawing the graph of$ \ f(x) = \tan^{-1} \left( \frac{1}{x} \right) \ $and drawing the upper rectangles from$ \ x = 1 \ $to$ \ x = N \ $for a positive integer$ \ N \\ $It becomes clear that$ \\ \\ \int_1^{N+1}\tan^{-1} \frac{1}{x} \ dx < \sum_{n=1}^N \tan^{-1} \frac{1}{n}$

$\\ \int \tan^{-1} \frac{1}{x} \ dx = x\tan^{-1}x + \int \frac{x}{1+x^2} \ dx \ \ $(using IBP)$ \\ \\ \therefore \ I_N = \int_1^{N+1} \tan^{-1} \frac{1}{x} \ dx = \frac{1}{2} \ln (N^2+2N+2) + (N+1) \tan^{-1} \frac{1}{N+1} - \ln \sqrt{2} + \frac{\pi}{4}$

$\\ $Since$ \ (N+1)\tan^{-1} \frac{1}{N+1} > 0 \ $and$ \ \ln (N^2+2N+2) \to \infty \ $as$ \ N \to \infty \\ \\ \therefore \ I_N \to \infty \ $as$ \ N \to \infty \\ \\ $Since$ \ S_N > I_N \\ \\ \lim_{N \to \infty}S_N > \lim_{N \to \infty} I_N = \infty \\ \\ \therefore \ \sum_{n=1}^{\infty} \tan^{-1} \frac{1}{n} \ $diverges$$

SilentWaters · Feb 27, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Alternatively, observing the standard limit:

$\lim_{\frac{1}{x}\to0}\frac{\tan ^{-1}\left( \frac{1}{x} \right)}{\left( \frac{1}{x} \right)}$

we can say that as x tends to infinity this summation tends to the harmonic series. By looking at the upper bound rectangle sum for y = 1/x from x = 1 onwards, we establish the relationship:

$\sum_{n=1}^k \frac{1}{x} > \int_1^{k+1} \frac{1}{x}\,dx$

where k is positive integral. Now note the improper integral

$\int_1^\infty \frac{1}{x} = \ln\infty = \infty$

So since this latter summation diverges, the former summation of the arc-tangent must do the same.

InteGrand · Feb 27, 2015

Re: HSC 2015 4U Marathon - Advanced Level

SilentWaters said:
Hence if we consider the sum of upper-bound rectangles of unit width from x = 1 onwards, noting that the corresponding exact area beneath the graph is infinite, we can say the summation diverges.

How did you know the area under the arctan graph was infinite (without doing the integration)?

(Or did you just mean by doing an integration, you see the area is infinite? Because without doing an integration, it doesn't seem obvious to me.)

SilentWaters · Feb 27, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Woops. Edited my original post

FrankXie · Feb 27, 2015

Re: HSC 2015 4U Marathon - Advanced Level

new question

$$ Two sequences $ \{a_n\} $ and $ \{b_n\} $ are such that (i) $ a_1=1 $ (ii) $ a_n $ and $ a_{n+1} $ are roots of the equation $ x^2-b_nx+2^n=0 $ for $ n=1, 2, 3, \cdots. $ Find the $n$th partial sum $ S_n $ of the sequence $ \{b_n\}.$

Sy123 · Feb 27, 2015

Re: HSC 2015 4U Marathon - Advanced Level

FrankXie said:
new question

$$ Two sequences $ \{a_n\} $ and $ \{b_n\} $ are such that (i) $ a_1=1 $ (ii) $ a_n $ and $ a_{n+1} $ are roots of the equation $ x^2-b_nx+2^n=0 $ for $ n=1, 2, 3, \cdots. $ Find the $n$th partial sum $ S_n $ of the sequence $ \{b_n\}.$

$\\ $The definitions of the sequences are$ \ a_{n} + a_{n+1} = b_n \ (1) \ $and$ \ a_n a_{n+1} = 2^n \ (2) \\ a_1 = 1 \ $and$ \ b_1 = 3$

$\\ \frac{2^{n+1}}{2^n} = \frac{a_{n+2} a_{n+1}}{a_{n+1}a_n} \Rightarrow \ a_{n+2} = 2a_n$

$\\ $So, given$ \ m=1,2,3,\dots \\ \ x_m = a_{2m} \\ y_m = a_{2m-1}$

$\\ $So,$ \ x_{m+1} = 2x_m \ $and$ \ y_{m+1} = 2y_m \ $and$ \ x_1 = 2, y_1 = 1$

$\\ S_n = b_1 + b_2 + \dots + b_n = (a_1+ a_2) + (a_2+a_3) + \dots + (a_n + a_{n+1}) \\ = (a_{n+1} + a_1) + 2(a_2 + a_3 + \dots + a_{n-1} + a_n) \\ \\ $If$ \ n = 2m \Rightarrow \ S_{2m} = y_{m+1} + y_1 + 2((x_1 + x_2 + \dots + x_m) + (y_2 + \dots + y_m))$

$\\ $If$ \ n = 2m -1 \Rightarrow \ S_{2m-1} = x_m + y_1 + 2((x_1 + \dots + x_{m-1}) + (y_2 + \dots + y_m))$

$\\ \therefore \ S_{2m} = 7\cdot 2^m - 7$

$\\ \therefore \ S_{2m-1} = 7\cdot 2^{m} - 7 \\\\ \therefore \ S_n = 7 \cdot 2^{\lceil \frac{n}{2} \rceil } - 7$

FrankXie · Feb 27, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Sy123 said:
$\\ $The definitions of the sequences are$ \ a_{n} + a_{n+1} = b_n \ (1) \ $and$ \ a_n a_{n+1} = 2^n \ (2) \\ a_1 = 1 \ $and$ \ b_1 = 3$

$\\ \frac{2^{n+1}}{2^n} = \frac{a_{n+2} a_{n+1}}{a_{n+1}a_n} \Rightarrow \ a_{n+2} = 2a_n$

$\\ $So, given$ \ m=1,2,3,\dots \\ \ x_m = a_{2m} \\ y_m = a_{2m-1}$

$\\ $So,$ \ x_{m+1} = 2x_m \ $and$ \ y_{m+1} = 2y_m \ $and$ \ x_1 = 2, y_1 = 1$

$\\ S_n = b_1 + b_2 + \dots + b_n = (a_1+ a_2) + (a_2+a_3) + \dots + (a_n + a_{n+1}) \\ = (a_{n+1} + a_1) + 2(a_2 + a_3 + \dots + a_{n-1} + a_n) \\ \\ $If$ \ n = 2m \Rightarrow \ S_{2m} = y_{m+1} + y_1 + 2((x_1 + x_2 + \dots + x_m) + (y_2 + \dots + y_m))$

$\\ $If$ \ n = 2m -1 \Rightarrow \ S_{2m-1} = x_m + y_1 + 2((x_1 + \dots + x_{m-1}) + (y_2 + \dots + y_m))$

$\\ \therefore \ S_{2m} = 7\cdot 2^m - 7$

$\\ \therefore \ S_{2m-1} = 7\cdot 2^{m} - 7 \\\\ \therefore \ S_n = 7 \cdot 2^{\lceil \frac{n}{2} \rceil } - 7$

s1 is not equal to b1 ?

Sy123 · Feb 27, 2015

Re: HSC 2015 4U Marathon - Advanced Level

FrankXie said:
s1 is not equal to b1 ?

Isn't the nth partial sum of the sequence S(n) = b1 + b2 + .. + bn?

Therefore, S(1) = b1?

InteGrand · Feb 28, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Sy123 said:
Isn't the nth partial sum of the sequence S(n) = b1 + b2 + .. + bn?

Therefore, S(1) = b1?

He's saying that your current formula for $S_n$ doesn't agree with $b_1 = 3$ .

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