A Summer of Maths

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  • View Poll Results: Have you studied any Group Theory already?
    Yes, I did some Group Theory during/at my A-level.
    		24 34.78%
    No, but I plan to study some before Uni.
    		15 21.74%
    No, I haven't.
    		30 43.48%
    Voters: 69. You may not vote on this poll

  1. jack.hadamard's Avatar
    161 07-07-2012  00:19
    • Benevolent Member
    • Posts: 696
    Re: A Summer of Maths
    (Original post by David_Skiller)
    your mistaken.
    Either way, my focus is on first-year material.
  2. Lord of the Flies's Avatar
    162 07-07-2012  00:24
    • The foul fiend Flibbertigibbet
    • Location: Paris, France
    • Warning points: 2
    Re: A Summer of Maths
    (Original post by jack.hadamard)
    The difficult part was to note the induction step.

    Spoiler:
    Show

    Notice that \displaystyle \lim_{x\to\infty} \left(1 + \frac{y}{f(x)}\right)^{f(x)} = e^y for probably most of the unbounded functions f(x).
    Ah yes of course! Why didn't I think of this! Also, why in the world did I disregard the "!" in my first post... Oh well, instead of being lazy I'll show it:

    Spoiler:
    Show

    \forall n>0:

    a_n(n)=\displaystyle \left( \frac{b}{2^n}+1 \right)^{2^n}-1

    \quad\quad\;\; =\displaystyle \sum_{k=0}^{2^n} \binom{2^n}{k} \left( \frac{b}{2^n} \right)^k -1


    \forall k we have:

    \; \displaystyle \binom{2^n}{k} \left( \frac{b}{2^n} \right)^k = \frac{\displaystyle \prod_{m=0}^{k-1} (2^n-m)}{k!} \cdot \left( \frac{b}{2^n} \right)^k

    \quad\quad\quad\quad\quad\quad\; = \displaystyle \left( \frac{2^{kn}}{2^{kn}}\right) \frac{b^k}{k!} \displaystyle \prod_{m=0}^{k-1} \left( 1-\dfrac{m}{2^n} \right)}{k!}

    \quad\quad\quad\quad\quad\quad\; = \displaystyle \frac{b^k}{k!} \prod_{m=0}^{k-1} \left( 1-\dfrac{m}{2^n} \right)}{k!}

    Therefore \displaystyle \lim_{n\to\infty} \binom{2^n}{k} \left( \frac{b}{2^n} \right) = \frac{b^k}{k!}

    So \displaystyle \lim_{n\to\infty}a_n(n)=\sum_{i= 0}^{\infty} \frac{b^k}{k!} -1 = e^b -1


    Yes?
    Last edited by Lord of the Flies; 07-07-2012 at 00:26.
  3. David_Skiller's Avatar
    163 07-07-2012  00:28
    • Full Member
    • Posts: 126
    Re: A Summer of Maths
    (Original post by ben-smith)
    you cannot be for real.
    oryt ben?

    David- Eton College
  4. Lord of the Flies's Avatar
    164 07-07-2012  00:28
    • The foul fiend Flibbertigibbet
    • Location: Paris, France
    • Warning points: 2
    Re: A Summer of Maths
    (Original post by TheMagicMan)
    One of the most universally disliked IMO questions around...


    This was posted from The Student Room's iPhone/iPad App
    Wasn't aware... But ""
  5. David_Skiller's Avatar
    165 07-07-2012  00:28
    • Full Member
    • Posts: 126
    Re: A Summer of Maths
    (Original post by jack.hadamard)
    Either way, my focus is on first-year material.
    aint you got girls to smash

    David- Eton College
    Post rating:
    2
    2
  6. jack.hadamard's Avatar
    166 07-07-2012  00:29
    • Benevolent Member
    • Posts: 696
    Re: A Summer of Maths
    (Original post by Lord of the Flies)
    Yes?
    From a quick scan, the derivation is correct and yes, that's the answer. Nice?
  7. ben-smith's Avatar
    167 07-07-2012  00:30
    • Overlord in Training
    • Location: Hilbert Space
    • Posts: 2,366
    Re: A Summer of Maths
    (Original post by David_Skiller)
    aint you got girls to smash

    David- Eton College
    bitches don't put out unless you can do representation theory on lie groups.
    Well known fact.
    Post rating:
    1
    1
  8. Lord of the Flies's Avatar
    168 07-07-2012  00:31
    • The foul fiend Flibbertigibbet
    • Location: Paris, France
    • Warning points: 2
    Re: A Summer of Maths
    (Original post by jack.hadamard)
    From a quick scan, the derivation is correct and yes, that's the answer. Nice?
    Enjoyed it yes! Thanks!
  9. David_Skiller's Avatar
    169 07-07-2012  00:34
    • Full Member
    • Posts: 126
    Re: A Summer of Maths
    (Original post by ben-smith)
    bitches don't put out unless you can do representation theory on lie groups.
    Well known fact.
    easy don

    David- Eton College
  10. Rahul.S's Avatar
    170 07-07-2012  00:51
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    Re: A Summer of Maths
    (Original post by David_Skiller)
    oryt ben?

    David- Eton College
    whos this idiot? Ima rob you if I see you around PUSSSSSSSSY!!!

    Rahul- I dont go Eton College I go LOX
    Post rating:
    2
  11. f1mad's Avatar
    171 07-07-2012  00:56
    • TSR Demigod
    • Posts: 5,423
    Re: A Summer of Maths
    (Original post by Rahul.S)
    whos this idiot? Ima rob you if I see you around PUSSSSSSSSY!!!

    Rahul- I dont go Eton College I go LOX
    Easy don .
  12. jack.hadamard's Avatar
    172 07-07-2012  01:31
    • Benevolent Member
    • Posts: 696
    Re: A Summer of Maths
    The above posts ^^.

    I don't see the relevance of this college; is it going to change to one of the Cambridge/Oxford colleges?

    ***

    How do we show this

    3n^2 + 3n + 7 = k^3 has no solutions for n,k \in \mathbb{N}

    with elementary tools. (I stole it from another thread )


    Lets give it a go.

    Spoiler:
    Show

    3n^2 + 3n + 7 = k^3 \iff 3(n^2 + n + 2) = k^3 - 1 = (k - 1)(k^2 + k + 1)

    Notice that n^2 + n is always even. Therefore, n^2 + n + 2 is always even, and k^2 + k + 1 is always odd.


    Now, we conclude (n^2 + n + 2) \nmid (k^2 + k + 1) \implies (n^2 + n + 2) \mid (k - 1) \implies k \geq 5.

    However, since \min(k^2 + k + 1) = 3, we conclude that the equation has no solutions for n,k \in \mathbb{N}.
  13. Blutooth's Avatar
    173 07-07-2012  01:50
    • Peer Of The TSR Realm
    • Location: London
    • Posts: 1,943
    Re: A Summer of Maths
    (Original post by jack.hadamard)
    The above posts ^^.

    I don't see the relevance of this college; is it going to change to one of the Cambridge/Oxford colleges?

    ***

    How do we show this

    3n^2 + 3n + 7 = k^3 has no solutions for n,k \in \mathbb{N}

    with elementary tools. (I stole it from another thread )


    Lets give it a go.

    Spoiler:
    Show

    3n^2 + 3n + 7 = k^3 \iff 3(n^2 + n + 2) = k^3 - 1 = (k - 1)(k^2 + k + 1)

    Notice that n^2 + n is always even. Therefore, n^2 + n + 2 is always even, and k^2 + k + 1 is always odd.


    Now, we conclude (n^2 + n + 2) \nmid (k^2 + k + 1) \implies (n^2 + n + 2) \mid (k - 1) \implies k \geq 5.

    Sorry, but I think this is faulty logic. N^2+n+2 could have odd factors that divide into k^2+k+1, then n^2+n+1 would not necessarily divide into (k-1)

    However, since \min(k^2 + k + 1) = 3, we conclude that the equation has no solutions for n,k \in \mathbb{N}.
    I think there's a mistake, posted my comments in the spoiler .


    Oh and, is considering something mod 2 that much more elementary than considering something mod 3 .
    Last edited by Blutooth; 07-07-2012 at 02:12.
  14. TheMagicMan's Avatar
    174 07-07-2012  08:04
    • Overlord in Training
    • Posts: 2,940
    (Original post by David_Skiller)
    your mistaken. you need SS to read that book.....I dont know whether you have got it. Magicman does ive already spoken to him in person. Also, for this summer, im looking to read erotic maths for the innocent- very passionate :yes:

    David-Eton College
    I don't know what you're talking about...but you don't go to Eton


    This was posted from The Student Room's iPhone/iPad App
  15. TheMagicMan's Avatar
    175 07-07-2012  08:05
    • Overlord in Training
    • Posts: 2,940
    (Original post by ben-smith)
    you cannot be for real.
    Don't worry he's not...


    This was posted from The Student Room's iPhone/iPad App
  16. TheMagicMan's Avatar
    176 07-07-2012  08:12
    • Overlord in Training
    • Posts: 2,940
    (Original post by jack.hadamard)
    This is Part II?

    Is the book by Beardon any good, and what are the prerequisites?
    It's a cool book... As with all of Beardon's work it's very clear and well presented. As for prerequisites, nothing more that some intermediate analysis and basic complex analysis are required...roughly what is covered across volumes 1&2 of zorich's analysis books


    This was posted from The Student Room's iPhone/iPad App
  17. TheMagicMan's Avatar
    177 07-07-2012  08:15
    • Overlord in Training
    • Posts: 2,940
    (Original post by Blutooth)
    That's interesting, why is that? Is it because the solution is messy?
    That, and the fact that there is a trap you can fall into: there is an 'obvious' route to a solution which doesn't actually yield anything useful and a lot of people ended up hitting total dead ends in the exam


    This was posted from The Student Room's iPhone/iPad App
  18. jack.hadamard's Avatar
    178 07-07-2012  09:50
    • Benevolent Member
    • Posts: 696
    Re: A Summer of Maths
    (Original post by Blutooth)
    I think there's a mistake, posted my comments in the spoiler.
    6 \nmid 9 means that 6 does not divide into 9, not that (6,9) = 1.

    Spoiler:
    Show

    Consider the \min(k^2 + k + 1) = 3.

    The other factor have to a positive integer, and have to be 1 regardless of how it cancels, because left-hand is the prime 3.
  19. sputum's Avatar
    179 07-07-2012  10:54
    • Adored and Respected Member
    • Posts: 425
    Re: A Summer of Maths
    (Original post by TheMagicMan)
    I have posted this question before, but I have yet to see anyone respond to it on here.

    Given that two parabolas with perpendicular directrices have 4 distinct points of intersection, show that these four intersection points form a cyclic quadrilateral.

    Required knowledge: there exist elementary solutions, however the two more elegant solutions I know require, for the first, some IMO geometry theorems and for the second, certain trig identities (probably not ones you usually use)
    it's a good question still working on it

    Spoiler:
    Show
    projective geometry should be all that's needed I think shame mine sucks
  20. Lord of the Flies's Avatar
    180 07-07-2012  15:28
    • The foul fiend Flibbertigibbet
    • Location: Paris, France
    • Warning points: 2
    Re: A Summer of Maths
    Question

    Find f:\mathbb{R}\to\mathbb{R} such that \forall (x,y)\in\mathbb{R}:

    \displaystyle f \left( x^2+f(y) \right)=x f(x)+y
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