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  1. fan96

    Challenging (?) Proof Question

    If one wanted to actually find examples of M , here is an approach that works for M \in \mathbb R . I'll use the hyperbolic function \cosh and its inverse, which aren't in the MX2 syllabus but their definitions are quite simple to understand: \cosh x = \frac 12 \left(e^x+e^{-x}\right)...
  2. fan96

    Challenging (?) Proof Question

    Is the condition M \notin \mathbb Z really necessary? The only case of this I can see is 1 + 1/1 = 2 \in \mathbb Z, so it would be simpler to leave it unmentioned.
  3. fan96

    Perms & Coms

    If the queues are distinct, then we should have 17280 = \underbrace{\left( \binom{8}{4} \times 4! \times 4!\right)}_{\text{no restrictions}} - \underbrace{\left(\binom{6}{3} \times 4!\ \times 4!\right)\times 2}_{\text{Sean and Liam in diff. queues}}, or 17280 = \left( \binom{6}{4} \times 4...
  4. fan96

    Tricky projectile motion question, Thanks

    For both balls we have \begin{cases} y_1(t) &= -\frac{9.8}{2}t^2+100\sin(\pi/6)t+h \\ x_1(t) &=100\cos(\pi/6)t\end{cases} \begin{cases} y_2(t) &= -\frac{9.8}{2}t^2+h \\ x_2(t) &=100t\end{cases} Solve y_1(t) = 0 and y_2(t) = 0 to get the time of flight for each ball (call these t_1 and...
  5. fan96

    What do I transfer to?

    Computer Engineering requires only standard Phys 1B and it's not a prerequisite for anything else. There are quite a few ELEC courses required though.
  6. fan96


    Recall that for two vectors \bold x, \bold y we have \cos \theta = \frac{\bold x \cdot \bold y}{|\bold x||\bold y|}, where \theta is the angle between \bold x and \bold y .
  7. fan96

    4 types of relations

    Unfortunately it's not that simple. You can't "break up" the absolute value function like that. If y = |x-3| , then y = \pm(x - 3). You can try graphing these to visualise the effect of the absolute value function.
  8. fan96

    4 types of relations

    That's called the absolute value function, and it's defined by |x| = \begin{cases} x, &\text{if }x \ge 0, \\-x, &\text{if }x < 0.\end{cases} Basically, you only want the magnitude of the number, not its sign. For example, |-1| = 1, |-2| = 2, |1| = 1, |0| = 0. If x = |y| , then...
  9. fan96

    Help with complex

    Say you have a statement you're trying to prove by induction. A normal induction proof would go like "if this statement is true for k then it must also be true for k + 1." A strong induction proof would be more like "if this statement is true for all numbers less than or equal to k then it...
  10. fan96

    Common Mistakes in differentiation and trigonometry

    This is one of the more common differentiation mistakes. \frac{d}{dx} x^x \overset{?}=x \cdot x^{x-1} For trigonometry in general, people make a lot of mistakes. \cot x \overset{?}= \frac{1}{\tan x} \sin^{-1}( \sin x ) \overset{?}= x , or equivalently, \sin x = y \overset{?} \implies x...
  11. fan96

    Hardest geometry question in history answered by student trivially............ How?

    The quantity \frac{k^2-RX^2}{2\cdot PX \cdot QX} can probably be simplified so as to remove k . The answer I gave holds numerically for all such equilateral and isosceles triangles formed and most likely for the rest of them too. The other solutions are definitely much nicer though. I...
  12. fan96

    Hardest geometry question in history answered by student trivially............ How?

    Must the angles be expressed only in terms of s and t ? The best I could do is s' = \arccos\left( \cos s + \frac{k^2-RX^2}{2\cdot PX \cdot QX}\right), t' = \arccos\left( \cos t + \frac{k^2-PX^2}{2\cdot QX \cdot RX}\right), where k is the side length of the equilateral triangle and...
  13. fan96

    COMP1511 vs COMP1911

    HSC SDD has very little in common with the introductory programming courses. The latter focuses solely on being able to code properly. There's nothing about ethics, dev approaches or that sort of stuff. 1911 is a cut down version of 1511. 1811 is significantly different to both of those. If...
  14. fan96

    Complex roots questions

    The first factorisation, (z+1)^8 - z^8 = ((z+1)^4 + z^4)((z+1)^4 - z^4), is straightforward. Realising that you can turn a sum of squares into a difference of squares is not so trivial - that's the part I showed. (actually, this method generalises to any positive even power...)
  15. fan96

    Complex roots questions

    This is not ideal but one method is to use the difference of squares identity to continuously factorise the LHS of the equation into a product of seven polynomial factors of degree one. You can rewrite a sum of squares as a difference of squares by using i^2 = -1 . e.g. (z+1)^4+z^4 =...
  16. fan96

    help with simplification

    Let x = \sqrt{a+\sqrt b} - \sqrt{a-\sqrt b}. Now, \begin{aligned} x^2 &= (a + \sqrt b) - 2\sqrt{ (a + \sqrt b) (a - \sqrt b)} + (a - \sqrt b) \\ &= 2(a - \sqrt{a^2-b}).\end{aligned} Clearly x \ge 0 , so x = \sqrt{2(a - \sqrt{a^2-b})}. Set a = 11, \, b = 2 and you're done.
  17. fan96

    Correlation between Complex Locus and Circle Geometry (former harder 3u)

    I'm not familiar with the new syllabus but I can tell you how it was in the older syllabus. In the older syllabus, circle geometry problems were pretty limited in scope - some diagram involving circles is given and you have to bash out those circle geo-specific theorems you memorised to prove a...
  18. fan96

    HSC 2018-2019 MX2 Integration Marathon

    How about something a little bit different? \text{Let } I = \int_0^{\pi/2} e^{\sin x}\, \mathrm dx. Without numerically evaluating I, show that \text{a) } \frac{\pi}2 + 1 \leq I \leq \frac{\pi e}2, \text{b) } \mathrm{(\bold {Harder}.) \,\,} e \leq I \leq \frac{\pi e}2 - \frac{5\pi}8 +1.
  19. fan96

    Perms and Combs HSC question

    Choose the one car that gets the manually operated gate, then choose which gate. \binom{3}{1}\cdot \binom 21 = 6. The remaining two cars must go through the auto gates (of which there are three). \binom 31 \cdot \binom 31 = 9. Finally, 6 \cdot 9 = 54 . A gate can be used by more than one...
  20. fan96

    UNSW Trimesters

    Trimesters are fine. Especially if you haven't experienced semesters.