HSC 2014 MX2 Marathon ADVANCED (archive) (7 Viewers)

seanieg89 · May 12, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
An attempt at a more formal solution:

$\\ f \ $is differentiable on$ \ (a,b) \ $using this fact, prove that$ \ f' \ $is continuous on$ \ (a,b) \\ \\ $For$ \ f' \ $to be continuous on$ \ (a,b) \ $then$ \\ \\ \lim_{x \to c} f'(x) = f'( c) \ $for all$ \ c \in (a,b) \\ \\ \lim_{x \to c} \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \ $the inner limit exists as$ \ f \ $is differentiable on$ \ (a,b) \\ \\ h = x-c \\ \\ \Rightarrow \ \lim_{h \to 0} \frac{f(h + (h+c)) - f(h+c)}{h} = \lim_{h \to 0} f'(h+c) = f'(c)$

$\therefore \ \lim_{x \to c} f'(x) = f'(c) \ $for all$ \ c \in (a,b) \\ \\ $Due to the Extreme Value Theorem, if a maximum value of the function$ \ f' \ $on$ \ (a,b) \ $exists, then it exists inside the interval$ \ (a,b) \ $i.e. there is some$ \ f'(c) \geq f'(x) \ $true for some$ \ c \in (a,b) \\ \\ $Implying$ \ \int_a^b f'(c) \ dx \geq \int_a^b f'(x) \ dx \\ \\ $However, if a maximum does not exist in the interval$ \ (a,b) \ $for$ \ f' \ $then either$ \\ \\ L_b = \lim_{x \to b} f'(x) \geq f'(x) \ $or$ \ L_a = \lim_{x \to a} f'(x) \geq f'(x) \\ \\ \Rightarrow \ \int_a^bL_b \ dx > \int_a^bf'(x) \ dx \ $since$ \ f' \ $is non-constant$ \\ \\ \therefore \ L_b (b-a) > f(b) - f(a) \$

$$Since the inequality is strict, by bringing$ \ x \ $arbitrarily close to$ \ b \ $we can still choose some$ \ c \ $so that the strict inequality still holds, meaning further that$ \ f'(c) (b-a) > f(b) - f(a)$

Differentiability doesn't imply continuous differentiability (which is why I brought up the point). (Consider f(x)=(x^2)sin(1/x) at 0 as a counterexample).

So something is going wrong in your proof of this first assertion. (For starters, I don't follow your h=x-c line...it seems dodgy. More importantly though, the last equality before your "therefore" seems to use the assumption of continuity, the very thing you are trying to prove.)

(There would also be sign issues in the end of your argument, taking absolute values will reverse the inequality if the sides are both negative for example.)

mathing · May 16, 2014

Re: HSC 2014 4U Marathon - Advanced Level

I believe what we can do is either assume that f is continuous on [a, b] and apply the mean value theorem to know that there exists a a <= c <= b such that f'(c) = (f(b) - f(a)/(b-a) and as f' is non constant we take take that point to prove the inequality.

If we don't want to assume that f is continuous on [a, b] differentiability implies continuity on (a,b) so we just take a smaller closed interval that contains the non constant point of f' and use the same reasoning.

Sy123 · May 16, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\\ $Suppose the function$ \ f \ $which is differentiable on$ \ \mathbb{R} \ $is such that$ \\ \\ \int_{y}^{x+y} f(t) \ dt = \int_0^x f(y) \cos t + f(t) \cos y \ dt$

$\\ $Find$ \ f(x+y) \ $in terms of$ \ f(x) \ $and$ \ f(y) \\ \\ $Hence find$ \ f(x)$

seanieg89 · May 17, 2014

Re: HSC 2014 4U Marathon - Advanced Level

mathing said:
I believe what we can do is either assume that f is continuous on [a, b] and apply the mean value theorem to know that there exists a a <= c <= b such that f'(c) = (f(b) - f(a)/(b-a) and as f' is non constant we take take that point to prove the inequality.

If we don't want to assume that f is continuous on [a, b] differentiability implies continuity on (a,b) so we just take a smaller closed interval that contains the non constant point of f' and use the same reasoning.

We definitely need to assume continuity on [a,b] for the claim to be true. Otherwise take a continuous function g on [a,b] that is differentiable with bounded derivative in (a,b) and then choose f to agree with g on [a,b) and make f(b) really large to get a counterexample.

I also don't quite follow what you mean by: "and as f' is non constant we take take that point to prove the inequality". The claim is that an interior point exists such that strict inequality is obtained, it is not quite as simple as just choosing one of the mean value theorem points, since at these points we only have equality.

The essential point is that since f' is nonconstant, if the claim wasn't true then all interior points c would satisfy the (weak) reversed inequality, and at lease one point p would satisfy the strict reversed inequality. If we look at the definition of the derivative this means that "f will not change fast enough" near p to give you the full difference |f(b)-f(a)| given that we have a weak upper bound on how quickly f can change elsewhere. All of this can be formalised in a few lines using just the triangle inequality and the mean value theorem.

(Use of the Mean Value Theorem here is essential. Without further assumptions on the regularity of f (absolute continuity is necessary and sufficient), it is impossible to integrate f').

Davo_01 · May 17, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $Suppose the function$ \ f \ $which is differentiable on$ \ \mathbb{R} \ $is such that$ \\ \\ \int_{y}^{x+y} f(t) \ dt = \int_0^x f(y) \cos t + f(t) \cos y \ dt$

$\\ $Find$ \ f(x+y) \ $in terms of$ \ f(x) \ $and$ \ f(y) \\ \\ $Hence find$ \ f(x)$

$Here is my attempt and again nice question$

$I'm assuming that x,y and t are independent variables$

$\int_{0}^{x+y} f(t) \ dt = \int_0^x f(y) \cos t + f(t) \cos y \ dt+\int_0^y f(t)dt$

$\dfrac{dF}{d(x+y)}=\dfrac{dF}{dx}\dfrac{dx}{d(x+y)}=\dfrac{dF}{dx}\dfrac{1}{(1+\dfrac{dy}{dx})}$

$=\dfrac{dF}{dx}\;\; $x and y are independent variables so$ \dfrac{dy}{dx}=0$

$\therefore \dfrac{d}{dx}\int_{0}^{x+y} f(t) \ dt=f(x+y) \;\; \\ \\ $By the fundamental theorem of calculus$$

$f(x+y)=\dfrac{d}{dx} \int_0^x f(y) \cos t + f(t) \cos y \ dt+\dfrac{d}{dx}\int_0^y f(t)dt$

$f(x+y)=f(y)cosx+f(x)cosy$

$f(\dfrac{\pi}{2}-x)=f(\dfrac{\pi}{2})\cos{x}$

$\therefore \; $functions that satisfy this property are in the form$$ $\;\;f(x)=c sinx$

seanieg89 · May 17, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy probably just wanted the sin(x) solution, but usually when you are asked to solve a functional equation like this, you need to find ALL solutions to the given equation.

You should try to prove that all solutions to this problem are of the form c*sin(x). (It comes quite quickly from the functional equation and the assumption of differentiability.)

Sy123 · May 17, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Davo_01 said:
$Here is my attempt and again nice question$

$I'm assuming that x,y and t are independent variables$

$\int_{0}^{x+y} f(t) \ dt = \int_0^x f(y) \cos t + f(t) \cos y \ dt+\int_0^y f(t)dt$

$\dfrac{dF}{d(x+y)}=\dfrac{dF}{dx}\dfrac{dx}{d(x+y)}=\dfrac{dF}{dx}\dfrac{1}{(1+\dfrac{dy}{dx})}$

$=\dfrac{dF}{dx}\;\; $x and y are independent variables so$ \dfrac{dy}{dx}=0$

$\therefore \dfrac{d}{dx}\int_{0}^{x+y} f(t) \ dt=f(x+y) \;\; \\ \\ $By the fundamental theorem of calculus$$

$f(x+y)=\dfrac{d}{dx} \int_0^x f(y) \cos t + f(t) \cos y \ dt+\dfrac{d}{dx}\int_0^y f(t)dt$

$f(x+y)=f(y)cosx+f(x)cosy$

$$A function that satisfies this property is$ \;\;f(x)=sinx$

Yes well done, my solution was a little more direct:

$\\ $If$ \ F \ $is the primitive of$ \ f \ $which exists since$ \ f \ $is differentiable and hence continuous$ \\ \\ \therefore \ F(x+y) - F(y) = f(y) \sin x + F(x) \cos y$

$\\ \frac{d}{dx} (F(x+y) - F(y)) = \frac{d}{dx} (f(y) \sin x + F(x) \cos y) \\ \\ f(x+y) = f(y) \cos x + f(x) \cos y$

To show that c*sin(x) is the only solution:

$\ $Let$ \ y = \frac{\pi}{2} \\ \\ f\left(x + \frac{\pi}{2} \right) = f \left(\frac{\pi}{2} \right) \cos x \\ \\ \Rightarrow \ f(x) = c \sin x$

Davo_01 · May 17, 2014

Re: HSC 2014 4U Marathon - Advanced Level

seanieg89 said:
Sy probably just wanted the sin(x) solution, but usually when you are asked to solve a functional equation like this, you need to find ALL solutions to the given equation.

You should try to prove that all solutions to this problem are of the form c*sin(x). (It comes quite quickly from the functional equation and the assumption of differentiability.)

Oh right, I see it! Thanks for letting me know

Il update it now

Davo_01 · May 23, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$$Prove$\sum_{k=0}^{n} \binom{4n}{2k}^2=\dfrac{1}{4}\binom{8n}{4n}+\dfrac{1}{4}\binom{4n}{2n}+\dfrac{1}{2}\binom{4n}{2n}^2$

seanieg89 · May 23, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Davo_01 said:
$$Prove$\sum_{k=0}^{n} \binom{4n}{2k}=\dfrac{1}{4}\binom{8n}{4n}+\dfrac{1}{4}\binom{4n}{2n}+\dfrac{1}{2}\binom{4n}{2n}^2$

I really don't think this is true...the last RHS term on its own grows WAY faster than the LHS does. Also, a quick check shows this is not true for n=1.

Davo_01 · May 23, 2014

Re: HSC 2014 4U Marathon - Advanced Level

seanieg89 said:
I really don't think this is true...the last RHS term on its own grows WAY faster than the LHS does. Also, a quick check shows this is not true for n=1.

Right i made a typo my bad, i fixed it the LHS was meant to be squared, thanks for pointing out the error

Sy123 · May 23, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\\ $Consider the constant of$ \\ \\ (1+x)^{4n}\left(1 + \frac{1}{x} \right)^{4n} + (1+x)^{4n} \left(1 - \frac{1}{x} \right)^{4n} = \frac{(1+x)^{8n}}{x^{4n}} + \frac{(x^2-1)^{4n}}{x^{4n}} \\ \\ \sum_{k=0}^{4n} \binom{4n}{k}^2 + \sum_{k=0}^{4n}(-1)^k\binom{4n}{k}^2 = \binom{8n}{4n} + \binom{4n}{2n} \\ \\ \Rightarrow \ 2\sum_{k=0}^{2n} \binom{4n}{2k}^2 = \binom{8n}{4n} + \binom{4n}{2n} \\ \\ $As$ \ \binom{4n}{2k} = \binom{4n}{4n-2k} \\ \\ \therefore \ 2\left(2\sum_{k=0}^{n} \binom{4n}{2k}^2 - \binom{4n}{2n}^2 \right) = \binom{8n}{4n} + \binom{4n}{2n} \\ \\ \Rightarrow \ \sum_{k=0}^n \binom{4n}{2k}^2 = \frac{1}{4} \binom{8n}{4n} + \frac{1}{4} \binom{4n}{2n} + \frac{1}{2} \binom{4n}{2n}^2$

Davo_01 · May 24, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $Consider the constant of$ \\ \\ (1+x)^{4n}\left(1 + \frac{1}{x} \right)^{4n} + (1+x)^{4n} \left(1 - \frac{1}{x} \right)^{4n} = \frac{(1+x)^{8n}}{x^{4n}} + \frac{(x^2-1)^{4n}}{x^{4n}} \\ \\ \sum_{k=0}^{4n} \binom{4n}{k}^2 + \sum_{k=0}^{4n}(-1)^k\binom{4n}{k}^2 = \binom{8n}{4n} + \binom{4n}{2n} \\ \\ \Rightarrow \ 2\sum_{k=0}^{2n} \binom{4n}{2k}^2 = \binom{8n}{4n} + \binom{4n}{2n} \\ \\ $As$ \ \binom{4n}{2k} = \binom{4n}{4n-2k} \\ \\ \therefore \ 2\left(2\sum_{k=0}^{n} \binom{4n}{2k}^2 - \binom{4n}{2n}^2 \right) = \binom{8n}{4n} + \binom{4n}{2n} \\ \\ \Rightarrow \ \sum_{k=0}^n \binom{4n}{2k}^2 = \frac{1}{4} \binom{8n}{4n} + \frac{1}{4} \binom{4n}{2n} + \frac{1}{2} \binom{4n}{2n}^2$

Nice solution bro

RealiseNothing · May 26, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Find the amount of times 0 appears between 1 and $10^n$

Sy123 · May 26, 2014

Re: HSC 2014 4U Marathon - Advanced Level

RealiseNothing said:
Find the amount of times 0 appears between 1 and $10^n$

$\\ $Every number between$ \ 1 \ $and$ \ 10^n \ $not inclusive can be represented by the sum$ \\ \\ 10^k a_k + 10^{k-1}a_{k-1} + \dots + 10a_1 + a_0 \\ \\ $for$ \ k=0,1,2,\dots,n-1 \ \ $and$ \ \ a_k = \{0,1,2,3,4,5,6,7,8,9\} \\ (a_k \neq 0 \ $and if$ \ k=0 \Rightarrow \ a_0 \neq 1) \\ \\ $To find the number of times$ \ 0 \ $appears, simply look at the numbers$ \ a_0, a_1, \dots , a_{k-1} \\ $and see that each number can be any single digit$ \\ \\ $Consider the case where there are$ \ m \ $times, that$ \ 0 \ $appears in the set$ \{a_k\} \\ \\ $Firstly, there are$ \ \binom{k}{m} \ $ways to pick the position of the$ \ m \ $zeroes$ \\ $then there are$ \ 9^{k-m} \ $ways to choose the remaining digits in the other slots$ \\ \\ \therefore \ $The number of ways$ \ m \ $zeroes appear in the set of$ \ k \ $is$ \\ \binom{k}{m} 9^{k-m} 1^m$

$\\$Hence the number of times a zero appears if there is$ \ m \ $zeroes is clearly$ \ \ m\binom{k}{m}9^{k-m} \\ \\ $Since$ \ m = 0,1,2,\dots,k \\ \\ $The number of zeroes that appear in the set of$ \{ a_k \} \ $would be$ \\ \\ \sum_{m=0}^k m \binom{k}{m} 9^{k-m}$

$\\ (1+x)^k = \sum_{m=0}^k \binom{k}{m} x^{k-m} \\ \\ k(1+x)^{k-1} = \sum_{m=0}^k (k-m) \binom{k}{m} x^{k-m-1} \\ \\ \sum_{k=0}^m m\binom{k}{m}x^{k-m} = k \sum_{k=0}^m \binom{k}{m} x^{k-m} - kx(1+x)^{k-1} = k(1+x)^{k-1}$

$\\ $Hence the number of times a zero appears in$ \ \{a_k \} \ $is$ \ \ k10^{k-1} \\ \\ $We now need to consider the leading co-efficient, if k=0$ \\ \\ $then we multiply by 8, otherwise multiply by 9 (to consider all 9 possibilities), then add all together$ \\ \\ \therefore \ $The final answer is$ \\ \\ S = 8 \times 0 \times 10^{-1} + 9\sum_{k=1}^{n-1} k10^{k-1} = 9 \sum_{k=1}^{n-1} k10^{k-1}$

$1 + x + x^2 + \dots + x^{n-1} = \frac{x^n-1}{1-x} \\ \\ 1x^0 + 2x + 3x^2 + \dots+ (n-1)x^{n-2} = \frac{nx^{n-1} - (n-1)x^n -1}{(x-1)^2} \\ \\ \therefore \ S = \frac{n 10^{n-1} - (n-1)10^{n} - 1}{81}$

Great question, I hope this is correct.

Sy123 · May 26, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\\ $Find$ \ \lim_{n \to \infty} \left(\frac{(2n)!}{n^n n!} \right)^{1/n}$

seanieg89 · May 26, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $Find$ \ \lim_{n \to \infty} \left(\frac{(2n)!}{n^n n!} \right)^{1/n}$

$\log(T_n)=\frac{1}{n}\sum_{k=1}^n \log(1+k/n)\rightarrow \int_1^2 \log(t)\, dt = 2\log(2)-1\\ \\ \Rightarrow T_n \rightarrow 4e^{-1}.$

Sy123 · May 26, 2014

Re: HSC 2014 4U Marathon - Advanced Level

seanieg89 said:
$\log(T_n)=\frac{1}{n}\sum_{k=1}^n \log(1+k/n)\rightarrow \int_1^2 \log(t)\, dt = 2\log(2)-1\\ \\ \Rightarrow T_n \rightarrow 4e^{-1}.$

Yep I got that as well.

What I did:

$u_n = \left(\frac{(2n)!}{n^nn!} \right)^{1/n}$

$2u_n^n \frac{2n+1}{n+1} \frac{1}{(1+1/n)^n} = u_{n+1}^{n+1}$

$\lim_{n \to \infty} u_n = L$

$L = 4e^{-1}$

seanieg89 · May 28, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Here is a bit of general calculus (for the uni students here, this idea is important in extending elements of Sobolev spaces beyond the boundary of a domain in a way that's pretty smooth).

(Recall that "f is bounded" means there is a positive M such that |f(x)| =< M for all x in the domain.)

Let f be a real valued function defined on the non-negative reals which is bounded and continuous, and has continuous second derivative over the positive reals. (And this continuity extends to the second order one-sided derivative at 0, which we assume exists).

Prove that there exists a function g satisfying the same properties that is defined on the whole real line and agrees with f at non-negative reals.

mathing · May 28, 2014

Re: HSC 2014 4U Marathon - Advanced Level

seanieg89 said:
Here is a bit of general calculus (for the uni students here, this idea is important in extending elements of Sobolev spaces beyond the boundary of a domain in a way that's pretty smooth).

(Recall that "f is bounded" means there is a positive M such that |f(x)| =< M for all x in the domain.)

Let f be a real valued function defined on the non-negative reals which is bounded and continuous, and has continuous second derivative over the positive reals.

Prove that there exists a function g satisfying the same properties that is defined on the whole real line and agrees with f at non-negative reals.

Can we define g(x) = f(x) for x >=0 and g(x) = f(-x) for x < 0?

HSC 2014 MX2 Marathon ADVANCED (archive) (7 Viewers)

Well-Known Member

New Member

This too shall pass

Well-Known Member

Active Member

Well-Known Member

This too shall pass

Active Member

Active Member

Well-Known Member

Active Member

This too shall pass

Active Member

what is that?It is Cowpea

This too shall pass

This too shall pass

Well-Known Member

This too shall pass

Well-Known Member

New Member

Users Who Are Viewing This Thread (Users: 0, Guests: 7)