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  1. Sy123

    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon This was on the Wikipedia page for the Divisor function: Note how it says: 'the Riemann Hypothesis is equivalent to the statement that' Doesn't that mean that if we prove this that we prove the Riemann hypothesis, and as far as I know, the Riemann hypothesis is...
  2. Sy123

    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon I just made this now: $A polynomial is said to be palindromic if the co-efficients are palindromic, that is$ P(x)= a_0 x^n + a_1 x^{n-1}+...+a_{1}x+a_0 \ \ \ \ \rightarrow \ \ a_i=a_{n-i} $Take the palindromic polynomial$ P(x)= c_0 x^{2n+1} + c_1 x^{2n} + c_2...
  3. Sy123

    Oo nice, I will check the forum out then and look at the papers. Thanks bro

    Oo nice, I will check the forum out then and look at the papers. Thanks bro
  4. Sy123

    Hi again, I didn't want to ask you this in NS, but do you know any other good math resources...

    Hi again, I didn't want to ask you this in NS, but do you know any other good math resources other than STEP, HSC and school papers? School papers are all pretty easy and I'm using HSC questions as homework and stuff, and I am looking for questions that are like in STEP papers, since they have...
  5. Sy123

    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon Your first product is the product of n+1 consecutive integers, then you go on and prove that the product of n+1 consecutive integers is divisible by n!. Though the idea is correct
  6. Sy123

    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon Yeah I know that about the Basel Problem, someone helped me prove it as an extension (but I didn't post about it) - And I would appreciate it if you didn't post the solution to the question straight away, maybe after 4-5 hours or something.
  7. Sy123

    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon Not necessarily starting from 1, 2, 3 Generally from r, r+1 etc. But its not worth answering lol its very simple, I just thought it was a cool result. I was able to arrive at: \sum_{k=2}^{n} 1 - \frac{2}{k+1} = \frac{n}{2}(n-1) - 2\sum_{k=2}^{n} \frac{1}{k+1} And...
  8. Sy123

    HSC 2013-14 MX1 Marathon (archive)

    Re: HSC 2013 3U Marathon Thread Correct - this question was more about finding the quickest solution, and you had the solution that I had in mind.
  9. Sy123

    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon Haha nice ============= $Prove that the product of n consecutive integers is divisible by $ \ \ n!
  10. Sy123

    HSC 2013-14 MX1 Marathon (archive)

    Re: HSC 2013 3U Marathon Thread P(2ap, ap^2) \ \ \ $lies on the parabola with the equation$ \ \ \ x^2=4ay $A tangent to the parabola is constucted at P, and the tangent intersects the x-axis at R, and the y-axis at S $Prove that $ \ \ \ SR = RP
  11. Sy123

    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon My way is pretty weird but it works, probably not the intended way but here it goes: The diagram is needed in order to understand solution (or not) Ok, first sketch the graphs: f(x)=\sqrt{x}+\frac{1}{x} g(x)=x h(x)=1 Via calculus, I can show that x >...
  12. Sy123

    In a dilemma - please help

    +1 As a person who has experienced both the public and private high school system, I can definitely say that they are a lot different. A private school has much more discipline in terms of rules, it is more competitive, naturally it will have better facilities than a local comprehensive simply...
  13. Sy123

    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon I have seen the joke before and that is my response most of the time with the reasoning x \neq $x$ , just to be a smartass :P
  14. Sy123

    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon $x$ \neq x $Anyway, if I were to expand it, I will first assume that there are n brackets$ (x-a)(x-b)(x-c)...(x-z)=\sum_{k=0}^{n} x^{k} (-1)^{k} \sum \prod_{i=0}^{n-k}a_i
  15. Sy123

    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon Precisely. $Find the values of $ \ \ a, b, x, y \ \ $that satisfy the equations$ \\ \\ a+b=1 \\ ax+by=\frac{1}{3} \\ ax^2+by^2=\frac{1}{5} \\ ax^3+by^3=\frac{1}{7} $You may wish to start by multiplying the second equation by$ \ \ x+y
  16. Sy123

    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon I originally thought of using a^x > log_a x somehow but this came to me first. Do you mind posting your method? I'm curious
  17. Sy123

    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon $Prove that$ \ \ \ \frac{1}{n!} < \frac{1}{2^ {n-1}} \ \ \ $for $ \ \ \ n \geq 3 $WITHOUT using a proof by induction$
  18. Sy123

    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon Well we can define the actual definition of e^x to do something similar: e^x = \lim_{n \to \infty}(1+\frac{x}{n})^n Now looking back at the new thing you gave us: 9999^{10001} = 9999^{9999} \cdot 9999^2 = a 10001^{9999}= (10001/9999)^ {9999} \cdot 9999^{9999} = b...
  19. Sy123

    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon a=9999^ {10 000} = 9999 \cdot 9999^{9999} b= 10 000^ {9999} = (9999 \cdot \frac{10 000}{9999})^{9999} = 9999^{9999} \cdot (10000/9999)^{9999} a > b I can justify this, since I know the limit: \lim_{n \to \infty} (1+\frac{1}{n})^{n} = e Hence the product...
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