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Concerned about my intelligence / math ability

  1. Aug 6, 2014 #1
    Hello everyone.

    So as the title suggests, I am really concerned about my intelligence and/or math(s) potential. I'll explain in just a minute, but first allow me to say that any response or advice from you guys will be greatly appreciated, as basically I started to question my whole life and I just feel hopeless about it.

    So ever since primary school, I quite enjoyed math classes. They were great deal of fun and usually paying attention at them was enough for me to understand the topic and get a good grade afterwards. I didn't do any competitions though - I just didn't find them interesting at the time. Things continued to be the same on high school, but there was an issue - the high school itself. The quality and amount of math classes was ridiculous, but honestly, I wasn't aware at the time.

    It wasn't until my senior year (I am from Europe, so equivalent) before it struck me - I had to do something about my life and my future career. I somehow decided that I actually want to go to a university, but not just that - abroad, in the UK. So I did a little research and at first, I was looking at like top30. After some time, I shifted my focus on top10 and I ended up applying even to Oxford. But once I saw their admission test - 2 months before the real thing - I realized just how terribly behind I was with my math knowledge. So I went to our high school principal and asked for a break. I bought myself British textbooks and I just studied everything from scratch. I spent 8 hours a day for 2 months doing that, and let me tell you, I loved it. I ended up getting rejected by Oxford before the interview stage, but I got offers from Warwick, St. Andrews, Bristol and Edinburgh and I eventually fulfilled them by graduating with straight As.

    Thing is, I still kinda suck at math. I understand each topic I studied thoroughly, but I just lack the real "talent" to solve a bit more challenging problems, like those on Oxford's MAT (Link to a paper from their website). I see some of those questions and I just feel like crying. I think the issue is that I learned everything by myself, kind of "mechanically", and now I feel desperate because well, I can't just go back. I can now also clearly see how much it hurts me not to have any experience with competitive math, because I guess my mind wasn't guided to "think outside of the box". Or even worse, I just don't have the brain power to be able to cope with these math challenges...

    Is there anything I can do about this? Is there still a way how to develop this "6th sense"?

    Thank you for any answer.
     
  2. jcsd
  3. Aug 6, 2014 #2

    jbunniii

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    Which questions in particular did you have trouble with? Most of the multiple choice ones aren't too bad:

    A: simple application of the quadratic formula, be sure to subtract 1 from both sides first
    B: easy if you do it graphically
    C: introduce a new function ##a(x) = f'(x)##, differentiate it to find a formula for ##a'(x)## in terms of ##h##, then substitute ##2x## for ##x##
    D: consider what happens when ##x,y## are large to rule out (c), then plug in ##x=0## and ##y=0## to determine which of the remaining ones it must be
    E: work out the leading term for each polynomial and carry out the differentiation to notice that they cancel
    F: I got bored with this one and skipped it :tongue:
    G: Use the formula ##\sum_{k=1}^{n} k = n(n+1)/2##, and write ##p_n(x) = q(x)p_{n-1}(x) + r(x)##, you'll see that ##q(x)## must be ##(n+1)/n##; solve for ##r(x)##
    H: Looks like annoying calculus, I skipped it.
    I: Tricky. I played with it for a while but didn't find a solution yet.
    J: Ugh, I'll come back to it later.

    Of course given the 2.5 hour time limit, one would have to do the above pretty quickly to leave time for the 5 "real" questions. Of those, I only tried 1. It's actually pretty cool if you formulate it as a linear algebra problem:
    $$\begin{pmatrix}1 & -k \\ -k & 1 \end{pmatrix} \begin{pmatrix}f(t) \\ f(1-t) \end{pmatrix} = \begin{pmatrix}t \\ 1-t \end{pmatrix}$$
    The rest of problem 1 is pretty easy if you approach it this way.

    I'm not sure how much of this I would have been able to do as a high school student, though. :tongue:

    I'll play around with problems 2-5 now. I'm guessing they get harder.

    [edit] Oh wait, problem 1 was the A-J questions. The f(t) / f(1-t) problem was 2. No bonus points for attention to detail for me.
     
    Last edited: Aug 6, 2014
  4. Aug 6, 2014 #3
    My advice is don't worry about it. In order to think outside the box you have to have a thorough understanding of it first, i.e. lots of mechanical learning.
    Take Picasso for example; he is very famous for his crazy 'outside the box style', however he did not even begin to create this style until he had a thorough knowledge on the elements of art. (To the point where he surpassed his artist father, who consequently quit art)

    Art is not maths and I am no mathematician but I think this can apply to any skill-based thing... I would imagine having mastery over mathematics would take a very long time, but boy would it be satisfying.

    Plus - I don't really believe there is such thing as talent. Sure there are people who grasp concepts faster than others, but that can only take you so far. The most important quality you could ever have is unrelenting persistence, as my bio teacher oft repeats to me. :p
     
  5. Aug 6, 2014 #4

    mathwonk

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    don't worry about it, just work. i.e. there is nothing you can do about your IQ, there is a lot you can do about how much you work.
     
  6. Aug 7, 2014 #5

    jbunniii

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    OK, 1.J turned out to be horribly grungy. I typed out the solution but I don't want to spam the thread with it - send me a PM if you want to see it. :yuck:

    An important part of this sort of exam (the Math subject GRE is in the same category), where timing is critical, is to be able to anticipate which questions are going to offer little reward (points) for lots of time-consuming work, and skip them. I personally never mastered this skill, and I'm sure my exam results were suboptimal as a result.

    If you're lucky, you may spot a trick which will make the computation much easier. But you can't always know in advance if this is going to happen, until you commit some time to the problem. Be ruthless about cutting your losses and moving on to the next one if you get bogged down.

    Or for a multiple choice question, perhaps you can carry out enough of the calculation to see which of the choices the answer must be. By around halfway through my solution to 1.J, it was pretty clear which was the answer, but getting to that point took longer than this problem was worth in terms of points.
     
  7. Aug 7, 2014 #6

    pasmith

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    Congratulations. Which offer did you accept?

    You have been accepted to study mathematics (I assume, although you don't actually say what courses you received offers for) at four universities of excellent repute. You demonstrably do not "suck at math".

    Going into exam without having done adequate preparation will cause problems, and that is what you would have done.

    Reading A-level textbooks and doing A-level exercises and A-level past papers is not adequate preparation for MAT or STEP. You need to do MAT or STEP past papers and to start with you need to look at model answers while doing so until you get a sense of what you should be doing, and you need to be doing this for far longer before the test than two months. For STEP there is also the booklet http://www.admissionstestingservice.org/images/121908-advanced-problems-in-core-mathematics.pdf [Broken], which is essentially a large collection of past STEP questions annotated by a former STEP examiner. There seems to be no similar resource for MAT, but I expect Advanced Problems will still be useful preparation in how to approach MAT questions.

    Practice. The way to get better at doing mathematics is to do mathematics. Constructive feedback on your attempts is necessary, of course, but at university that is exactly what you will get. Also I offer these words of Tim Gowers:

    There are two key ideas in problem J: the first is that functions of the form [itex][f(x)][/itex] are piecewise constant, and change value where [itex]f(x)[/itex] attains integer values, which for [itex]f(x) = 2^x[/itex] on [itex]0 \leq x \leq n[/itex] is at [itex]\log_2(k)[/itex] for [itex]1 \leq k \leq 2^n[/itex]. Thus [tex]
    \int_0^n [2^x]\,dx = \sum_{k=1}^{2^{n} - 1} \int_{\log_2(k)}^{\log_2 (k+1)} k\,dx.[/tex] The form of the answers is then a hint that you should rearrange the right hand side into the form [itex]\log_2(\mbox{something})[/itex], and after some basic manipulation of logarithms you discover that [tex]\mbox{something} = \prod_{k=1}^{2^n-1} \left(\frac{k+1}{k}\right)^k
    = 2\left(\frac{3}{2}\right)^2\left(\frac{4}{3}\right)^3 \times \dots \times
    \left(\frac{2^n-1}{2^n - 2}\right)^{2^n-2}
    \left(\frac{2^n}{2^n - 1}\right)^{2^n-1}.[/tex] The second key idea is that there's always one more power of each integer between 2 and [itex]2^n - 1[/itex] in the denominator than there is in the numerator, and so there are a lot of cancellations: [tex]
    \prod_{k=1}^{2^n-1} \left(\frac{k+1}{k}\right)^k = \frac{2^{n(2^n-1)}}{(2^n-1)!} = \frac{2^{n2^n}}{(2^n)!}.[/tex]
     
    Last edited by a moderator: May 6, 2017
  8. Aug 7, 2014 #7

    jbunniii

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    That's how I did it too. It was pretty grungy considering it was only one-tenth of a problem (and more longwinded than the entirety of problem 2), hence my suspicion that there could be a trick that makes it much quicker. But if there is, I don't see it.
     
  9. Aug 7, 2014 #8

    AlephZero

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    The question doesn't ask you to evaluate the integral. It asks you which answer is correct.

    You can eliminate some of the possibilities by checking the results when ## n = 1##.

    You can eliminate the rest by seeing now the integral behaves as ##n \to \infty##.

    This a nice example of the OP's problem. These questions are meant to test whether you can "think like a mathematician", not whether you can grind out complicated algebra, calculus, trig, etc.
     
  10. Aug 7, 2014 #9

    jbunniii

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    For this problem, it's more like whether you can "think like someone taking a multiple choice math exam." :tongue:
     
  11. Aug 7, 2014 #10

    AlephZero

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    Well, they didn't have multiple choice exams when I was a school and university, so I never got to practice taking them :smile:

    But laziness, in the sense of not doing more work than you need to, is a good quality for mathematicians to have, IMO.

    The tiny amount of space on the exam paper to "show your rough working" should be a clue that brute force is not what they are looking for.

    But the OP shouldn't be too worried about not "getting" this type of question. The Cambridge equivalent of those Oxford tests are designed to rank the top 2% of all math A level students, which is a small subset of the students who get A and A* grades at A level.

    You don't have to be in that top 2% to make a success of a math degree course, but if everybody on the course except you IS in that top 2%, that's probably not the "best" university for you, whatever its world ranking.
     
  12. Aug 7, 2014 #11
    Life's far too short for you to be worrying about things over which you have no control.
     
  13. Aug 8, 2014 #12

    verty

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    For me, question 1.J was surely intended to be impossible under the time constraints. I can't think that the test setters realistically expected anyone to solve it in a suitably short amount of time. I may be wrong but for me it was a question that had to be skipped.
     
  14. Aug 8, 2014 #13

    jbunniii

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    Either skipped, or as AlephZero suggested, take advantage of the fact that it's a multiple choice question and try to quickly eliminate all possibilities but one.

    As he noted, if ##n=1## then the integral is ##1##, which rules out (a) and (c) immediately. At this point you could guess and have a 50% chance and apparently no penalty for guessing wrong.

    Or you could check ##n=2##: then the integral is ##1 + 2(\log_2(3) - \log_2(2)) + 3(\log_2(4) - \log_2(3)) = 5 - \log_2(3)## which matches (b) since
    $$n2^n - \log_2((2^n)!) = 8 - \log_2(24) = 8 - \log_2(8) - \log_2(3) = 5 - \log_2(3)$$
    Resorting to this sort of "test tactic" is annoying but a necessary evil. I seem to recall that the subject GRE had some similar questions where actually deriving the correct answer would take far too long.
     
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