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

Sy123 · Apr 24, 2014

Re: HSC 2014 4U Marathon - Advanced Level

seanieg89 said:
. Right idea and pretty close to being done, but we can have a pair of the k's being -1 instead of all of them having to be 1.

Ah yes my bad

$k_1, k_2, k_3 \in \{-1,1 \}$

$\\ $Case 1:$ \ \ k_1 = k_2 = k_3 = 1 \ \Rightarrow \ $(dealt with above)$ \\ \\ $Case 2:$ \ \ $2 of$ \ k_1, k_2, k_3 \ $is$ \ -1 \ $w.l.o.g$ \ k_1=k_2 = -1 \ , \ k_3 = 1 \\ \\ (2) \Rightarrow \ (c-a) = -(b-c) \Rightarrow \ a=b \ \ $but$ \ a,b \ $are distinct, a contradiction!$

$\\ \therefore \square$

Sy123 · Apr 24, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\sum_{n=1}^{\infty} (-1)^{\frac{n}{2}(n-1)}\frac{1}{n}$

Davo_01 · Apr 24, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $Find$ \ \ \lim_{n \to \infty} \sum_{k=1}^n \frac{1}{n + ik} \\ \\ $Where$ \ i = \sqrt{-1}$

Nice question heres my attempt

$\lim_{n \to \infty} \sum_{k=1}^n \dfrac{1}{n + ik}$ $= \lim_{n \to \infty} \sum_{k=1}^n \dfrac{n-ik}{n^2 + k^2}=$ $\lim_{n \to \infty} \sum_{k=1}^n \dfrac{n}{n^2 + k^2}- i\lim_{n \to \infty} \sum_{k=1}^n \dfrac{k}{n^2 + k^2}$

$=\lim_{n \to \infty} \sum_{k=1}^n \dfrac{1}{n}\cdot\dfrac{1}{1 + (\frac{k}{n})^2}- i\lim_{n \to \infty} \sum_{k=1}^n \dfrac{1}{n}\cdot \frac{\frac{k}{n}}{1 + (\frac{k}{n})^2}=$ $\int_0^1\dfrac{1}{1+x^2}\cdot dx-i\int_0^1 \dfrac{x}{1+x^2}\cdot dx$

$=\dfrac{\pi-2iln(2)}{4}$

ayecee · Apr 25, 2014

Re: HSC 2014 4U Marathon - Advanced Level

I had this idea the other day how the spokes on a wheel have no net moment on the wheel, I and wanted to prove it using maths. I liked it so much, I made a question out of it.
If you want more of a challenge, skip straight to the last part ( b ii ), otherwise, the questions should lead you through the problem.
(Let me know if I made some mistakes... I've never made a question like this before, and I'm not perfect)

Enjoy!View attachment wheel.pdf

Davo_01 · Apr 25, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Prove that the average value of a function between the closed interval $[a,b]$ is given by

$\dfrac{\int_a^b f(x)\cdot dx}{b-a}$

RealiseNothing · Apr 25, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Davo_01 said:
Prove that the average value of a function between the closed interval $[a,b]$ is given by

$\dfrac{\int_a^b f(x)\cdot dx}{b-a}$

The average value of a function is just the sum of all values of $f(x)$ in $[a,b]$ , divided by the number of values in that interval.

For simplicity's sake, I'm going to shift the interval to some new interval $[0,k]$ , this will still preserve everything.

Now we add up all value of $f(x)$ in our new interval of $[0,k]$

$\lim_{n \to \infty} \sum_{m=1}^{kn} f(\frac{m}{n})$

However, we must now divide by the amount of values within the interval, i.e.:

$\lim_{n \to \infty} \frac{1}{kn} \sum_{m=1}^{kn} f(\frac{m}{n})$

Noting the Riemann sum we now have gives:

$\frac{1}{k} \int_{0}^{k} f(x)~dx$

Shifting the interval from $[0,k]$ back to $[a,b]$ gives the result:

$\frac{1}{b-a} \int_{a}^{b} f(x)~dx$

Sy123 · Apr 25, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\sum_{n=1}^{\infty} (-1)^{\frac{n}{2}(n-1)}\frac{1}{n}$

Bump.

Davo_01 · Apr 25, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
Bump.

Okay so this is what i came up with, not fully confident but here it is:

$\dfrac{n(n-1)}{2}=0+1+2+...+(n-1)$

Using the fact that an $odd+even=odd$ and $odd+odd=even$
The sum gives o, then an odd number, then another odd, then an even twice and an odd twice and so on...

$\therefore S=1-\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}-\dfrac{1}{7}+...$

$= [x-\dfrac{x^2}{2}-\dfrac{x^3}{3}+\dfrac{x^4}{4}+\dfrac{x^5}{5}-\dfrac{x^6}{6}-\dfrac{x^7}{7}+... ]_0^1$

$=1-\int_0^1 (x+x^2-x^3-x^4+x^5+x^6-x^7-x^8+...)\cdot dx$

$=1-\int_0^1 \dfrac{x+x^2}{1+x^2}\cdot dx$

$=\dfrac{\pi-2ln(2)}{4}$

Sy123 · Apr 26, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Davo_01 said:
Okay so this is what i came up with, not fully confident but here it is:

$\dfrac{n(n-1)}{2}=0+1+2+...+(n-1)$

Using the fact that an $odd+even=odd$ and $odd+odd=even$
The sum gives o, then an odd number, then another odd, then an even twice and an odd twice and so on...

$\therefore S=1-\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}-\dfrac{1}{7}+...$

$= [x-\dfrac{x^2}{2}-\dfrac{x^3}{3}+\dfrac{x^4}{4}+\dfrac{x^5}{5}-\dfrac{x^6}{6}-\dfrac{x^7}{7}+... ]_0^1$

$=1-\int_0^1 (x+x^2-x^3-x^4+x^5+x^6-x^7-x^8+...)\cdot dx$

$=1-\int_0^1(x+x^2+x^5+x^6+x^9+x^{10}+...)-(x^3+x^4+x^7+x^8+x^{11}+x^{12}+...)\cdot dx$

$=1-\int_0^1 \dfrac{x+x^2}{1-x^4}-\dfrac{x^3+x^4}{1-x^4}\cdot dx$

$=1-\int_0^1 \dfrac{x^2+x}{1+x^2}\cdot dx$

$=\dfrac{\pi-2ln(2)}{4}$

Yes well done

seanieg89 · Apr 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

1. Do there exist functions f and g defined on the real numbers such that:

$f(g(x))=x^2\textrm{ and } g(f(x))=x^3\textrm{ for all }x.$

2. Do there exist functions f and g defined on the real numbers such that:

$f(g(x))=x^2\textrm{ and } g(f(x))=x^4\textrm{ for all }x.$

Justify your answers.

RealiseNothing · Apr 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

seanieg89 said:
1. Do there exist functions f and g defined on the real numbers such that:

$f(g(x))=x^2\textrm{ and } g(f(x))=x^3\textrm{ for all }x.$

Here's my attempt.

Assume that there does exists real functions defined as above. Then:

$f(g(f(x))) = [f(x)]^2$

Since:

$g(f(x)) = x^3$

We can deduce:

$f(x^3) = [f(x)]^2$

Let $x=-1,~0,~1$

$f(-1) = [f(-1)]^2$

$f(0) = [f(0)]^2$

$f(1) = [f(1)]^2$

These are true IFF:

$f(-1) = 0~or~1$

$f(0) = 0~or~1$

$f(1) = 0~or~1$

Now if we consider our original definition of our function:

$g(f(x)) = x^3$

Then we arrive at:

$g(f(-1)) = -1$

So either:

$g(0) = -1 ~or~ g(1) = -1$

Similarly:

$g(f(0)) = 0$

So either:

$g(0) = 0 ~or~ g(1) = 0$

Also:

$g(f(1)) = 1$

So either:

$g(0) = 1 ~or~ g(1) = 1$

Arbitrarily set $g(0) = -1$ and $g(1) = 0$ .

Now from the above possibilities, either $g(0)=1 ~or~ g(1)=1$ .

In these cases, we will reach either:

$g(0) = -1 ~and~ g(0) = 1$ or $g(1) = 0 ~and~ g(1) = 1$

In both cases there is a contradiction as this says that $g(x)$ is NOT a function as for some x-value there exists two y-values.

Thus there does not exist real valued functions such that the given definitions hold.

seanieg89 · Apr 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Correct.

You should have a crack at the second

.

RealiseNothing · Apr 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

seanieg89 said:
Correct.

You should have a crack at the second .

Soon (if not later tonight then tomorrow).

Enough procrastination for now.

Sy123 · May 3, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\\ $i) Show that for a convex function$ \ f \ $then$ \\ \\ w_1 f(x_1) + w_2f(x_2) \geq f(w_1x_1 + w_2x_2) \\ \\ $Where$ \ \ w_1 + w_2 = 1 \ \ $for non-negative$ \ w_1, w_2$

$\\ $ii) Hence prove that for some positive integer$ \ n \\ \\ \sum_{j=1}^n w_j f(x_j) \geq f \left(\sum_{j=1}^n w_j x_j \right) \\ \\ $For$ \ \ \sum_{j=1}^n w_j = 1 \ \ \ $for non-negative$ \ \ w_j$

$\\ $iii) Hence prove for positive$ \ \ a_j \\ \\ \frac{a_1 + a_2 + \dots + a_{n-1} + a_n}{n} \geq \sqrt[n]{a_1 \cdot a_2 \dots \cdot a_{n-1} \cdot a_n}$

seanieg89 · May 3, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Assuming a,b,c are positive, then b^2(a^2+c^2), just by expanding and comparing terms.

Sy123 · May 3, 2014

Re: HSC 2014 4U Marathon - Advanced Level

seanieg89 said:
Assuming a,b,c are positive, then b^2(a^2+c^2), just by expanding and comparing terms.

Yea it was supposed to go in the normal level haha

Davo_01 · May 3, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $i) Show that for a convex function$ \ f \ $then$ \\ \\ w_1 f(x_1) + w_2f(x_2) \geq f(w_1x_1 + w_2x_2) \\ \\ $Where$ \ \ w_1 + w_2 = 1 \ \ $for non-negative$ \ w_1, w_2$

$\\ $ii) Hence prove that for some positive integer$ \ n \\ \\ \sum_{j=1}^n w_j f(x_j) \geq f \left(\sum_{j=1}^n w_j x_j \right) \\ \\ $For$ \ \ \sum_{j=1}^n w_j = 1 \ \ \ $for non-negative$ \ \ w_j$

$\\ $iii) Hence prove for positive$ \ \ a_j \\ \\ \frac{a_1 + a_2 + \dots + a_{n-1} + a_n}{n} \geq \sqrt[n]{a_1 \cdot a_2 \dots \cdot a_{n-1} \cdot a_n}$

A property of convex functions is that the line passing through the points $(x_1,f(x_1))$ and $(x_2,f(x_2))$ is greater than or equal to the function at every point within the closed interval $[x_1,x_2]$ .

$\therefore y=f(x_1)+\dfrac{f(x_2)-f(x_1)}{x_2-x_1} (x-x_1) \geq f(x)$ in the closed interval $[x_1,x_2]$ .

Substitute $x=\omega_1 x_1+\omega_2 x_2$ where $\omega_1+\omega_2=1$

Simplification gives:

$w_1 f(x_1) + w_2f(x_2) \geq f(w_1x_1 + w_2x_2)$

ii) Using the inequality for convex functions

$f(a)+f(b) \geq 2f(\frac{a+b}{2})$ and showing that:

$w_1 f(x_1) + w_2 f(x_2)+f(x_3) + 2f(x_4)+4f(x_5)+...+2^{n-3}f(x_n)$

$\geq f(w_1 x_1+w_2 x_2)+ f(x_3) +2f(x_4)+4f(x_5)+...+2^{n-3}f(x_n)$

$\geq 2f(\frac{w_1 x_1+w_2 x_2+x_3}{2})+2f(x_4)+4f(x_5)+...+2^{n-3}f(x_n)$

$\geq 4f(\frac{w_1 x_1+w_2 x_2+x_3+2x_4}{4})+4f(x_5)+...+2^{n-3}f(x_n)$

$...\geq 2^{n-2}f( \frac{w_1 x_1+w_2 x_2+x_3+2x_4+...+2^{n-3}x_n}{2^{n-2}})$

Replacing coefficients of $x_j$ with $\omega_j$

$\therefore \sum_{j=1}^n \omega_j =\dfrac{w_1+w_2+1+2+4+...+2^{n-3}}{2^{n-2}}=1$

$\therefore \sum_{j=1}^n \omega_j f(x_j) \geq f \left(\sum_{j=1}^n \omega_j x_j \right)$

iii) let $f(x)=e^x$ (A convex function)
Define $a_j=e^{x_j}$
Also let $w_j=\dfrac{1}{n}$

Applying part ii

$\therefore \frac{a_1 + a_2 + \dots + a_{n-1} + a_n}{n}$ $\geq e^{\frac{x_1 + x_2 + \dots + x_{n-1} + x_n}{n}}$ $=\sqrt[n]{e^{x_1}e^{x_2}e^{x_3}...e^{x_n}} = \sqrt[n]{a_1 a_2 a_3 ... a_n}$

edit: Ok i noticed a problem with part ii, what i did only verifies the inequality but does not prove it. Another problem is I let w=1/n in part 3 which isnt consistent with part ii. any suggestions?

Sy123 · May 4, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Davo_01 said:
A property of convex functions is that the line passing through the points $(x_1,f(x_1))$ and $(x_2,f(x_2))$ is greater than or equal to the function at every point within the closed interval $[x_1,x_2]$ .

$\therefore y=f(x_1)+\dfrac{f(x_2)-f(x_1)}{x_2-x_1} (x-x_1) \geq f(x)$ in the closed interval $[x_1,x_2]$ .

Substitute $x=\omega_1 x_1+\omega_2 x_2$ where $\omega_1+\omega_2=1$

Simplification gives:

$w_1 f(x_1) + w_2f(x_2) \geq f(w_1x_1 + w_2x_2)$

ii) Using the inequality for convex functions

$f(a)+f(b) \geq 2f(\frac{a+b}{2})$ and showing that:

$w_1 f(x_1) + w_2 f(x_2)+f(x_3) + 2f(x_4)+4f(x_5)+...+2^{n-3}f(x_n)$

$\geq f(w_1 x_1+w_2 x_2)+ f(x_3) +2f(x_4)+4f(x_5)+...+2^{n-3}f(x_n)$

$\geq 2f(\frac{w_1 x_1+w_2 x_2+x_3}{2})+2f(x_4)+4f(x_5)+...+2^{n-3}f(x_n)$

$\geq 4f(\frac{w_1 x_1+w_2 x_2+x_3+2x_4}{4})+4f(x_5)+...+2^{n-3}f(x_n)$

$...\geq 2^{n-2}f( \frac{w_1 x_1+w_2 x_2+x_3+2x_4+...+2^{n-3}x_n}{2^{n-2}})$

Replacing coefficients of $x_j$ with $\omega_j$

$\therefore \sum_{j=1}^n \omega_j =\dfrac{w_1+w_2+1+2+4+...+2^{n-3}}{2^{n-2}}=1$

$\therefore \sum_{j=1}^n \omega_j f(x_j) \geq f \left(\sum_{j=1}^n \omega_j x_j \right)$

iii) let $f(x)=e^x$ (A convex function)
Define $a_j=e^{x_j}$
Also let $w_j=\dfrac{1}{n}$

Applying part ii

$\therefore \frac{a_1 + a_2 + \dots + a_{n-1} + a_n}{n}$ $\geq e^{\frac{x_1 + x_2 + \dots + x_{n-1} + x_n}{n}}$ $=\sqrt[n]{e^{x_1}e^{x_2}e^{x_3}...e^{x_n}} = \sqrt[n]{a_1 a_2 a_3 ... a_n}$

edit: Ok i noticed a problem with part ii, what i did only verifies the inequality but does not prove it. Another problem is I let w=1/n in part 3 which isnt consistent with part ii. any suggestions?

Yea I quickly wrote the question up, I thought with part(ii) it would be straightforward induction, but I don't think it is.

Here is one (I did this one)

$\\ $Let$ \ a,b,c \ $be non-negative reals such that$ \ a^2+b^2+c^2 = 8 \\ \\ $Prove that$ \ (4-a)(4-b)(4-c) + abc \geq 16$

RealiseNothing · May 4, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
Here is one (I did this one)

$\\ $Let$ \ a,b,c \ $be non-negative reals such that$ \ a^2+b^2+c^2 = 8 \\ \\ $Prove that$ \ (4-a)(4-b)(4-c) + abc \geq 16$

Taking the 16 to the LHS, then expanding and simplifying gives:

$(ab+bc+ca)-4(a+b+c) + 12$

However:

$a^2+b^2+c^2 = (a+b+c)^2-2(ab+bc+ca) = 8$

$(ab+bc+ca) = \frac{(a+b+c)^2-8}{2}$

Substituting this in and letting $a+b+c=x$ gives:

$\frac{x^2-8}{2} - 4x + 12$

$x^2-8x+16$

$(x-4)^2$

Which is a square number and is thus always positive, hence the required result.

An observation would be to see equality occurs when:

$x=a+b+c=4$

Trebla · May 4, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $i) Show that for a convex function$ \ f \ $then$ \\ \\ w_1 f(x_1) + w_2f(x_2) \geq f(w_1x_1 + w_2x_2) \\ \\ $Where$ \ \ w_1 + w_2 = 1 \ \ $for non-negative$ \ w_1, w_2$

$\\ $ii) Hence prove that for some positive integer$ \ n \\ \\ \sum_{j=1}^n w_j f(x_j) \geq f \left(\sum_{j=1}^n w_j x_j \right) \\ \\ $For$ \ \ \sum_{j=1}^n w_j = 1 \ \ \ $for non-negative$ \ \ w_j$

$\\ $iii) Hence prove for positive$ \ \ a_j \\ \\ \frac{a_1 + a_2 + \dots + a_{n-1} + a_n}{n} \geq \sqrt[n]{a_1 \cdot a_2 \dots \cdot a_{n-1} \cdot a_n}$

Sy123, I just wanna say SCREW YOU because this question (worded slightly differently) was a potential consideration for this year's BoS trials

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

This too shall pass

This too shall pass

Active Member

Member

Active Member

what is that?It is Cowpea

This too shall pass

Active Member

This too shall pass

Well-Known Member

what is that?It is Cowpea

Well-Known Member

what is that?It is Cowpea

This too shall pass

Well-Known Member

This too shall pass

Active Member

This too shall pass

what is that?It is Cowpea

Administrator

Users Who Are Viewing This Thread (Users: 0, Guests: 1)