Hey there! Sign in to join this conversationNew here? Join for free
    Offline

    1
    ReputationRep:
    Solution 104

    Imprmis, notice that f(x)=-f(-x). This gives f(0)=0.
    Let x \not= 0. Define, a_{0}=x, a_{n}=2a_{n-1}^{2}-1. If for some n, a_{n}=0, it follows that f(a_{0})=0.
    The sequence b_{n}= \cos 2^{n}y satisfies b_{n}=2b_{n-1}^{2}-1.
    The set \{\frac{\pi(2k+\frac{1}{2})}{2^{  n}}| n,k \in \mathbb{Z^{+}} \} is dense in \left[-1,1 \right].
    From Heine - Borel we conclude that \left[-1,1 \right] is a compact Hausdorff space. Now since f and  \equiv 0 agree on a dense subspace of \left[-1,1 \right], it follows that they agree on \left[-1,1 \right].

    Another one?

    Problem 117**

    Find all continuous functions defined for each x \in \mathbb{R}, which satisfy f(x^{p}+y^{q})=f(x)^{p}+f(y)^{q}, where p and q are positive integers at least one of which is greater than 1.

    By the way, very elegant solution to problem 112. I solved it using my beloved complex analysis.

    I noticed discussion regarding mathematical books. I do not know whether or not Spivak's book is a must, but Rudin's Principles of Mathematical Analysis definitely is. For algebra - look at Lang's book - his approach is rigorous and modern, yet he does not give many examples, not to mention his proofs. In other words, his books will teach you how to run if you know how to walk. Another good reference which comes to my mind is Waerden's Algebra - it is good but a bit out of date. For topology - Munkres is good, but I prefer Kelley's approach.

    (Original post by Jkn)
    Come do some STEP on my thread! It's really empty
    I shall check it out!
    Offline

    2
    ReputationRep:
    (Original post by james22)
    Does this require spotting some complicated foruier series or is there an elementary solution?
    It may or may not be...
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show

    ...unproven :pierre:

    (Original post by Mladenov)
    I shall check it out!
    Lad!
    Offline

    0
    ReputationRep:
    (Original post by Mladenov)
    Nice. I will give you something different.

    Check your solutions to problem 86. You have missed solutions and (1,1) is not a solution, for example.

    Problem 87*

    A group of boys and girls went to a dance party. It is known that for every pair of boys, there are exactly two girls who danced with both of them; and for every pair of girls there are exactly two boys who danced with both of them. Prove that the numbers of girls and boys are equal.
    Okay, I don't know if this has already been solved but,

    Solution 87

    Let there be n girls; g_1, g_2,.. g_n, and m boys; b_1, b_2,... b_m.

    Now, for every pair of boys (b_i,b_j) there exist 2 girls that danced with both of them and let these be (g_i,g_j).

    If m>n, then as there are ^n\mathrm{C}_2 pairs of girls and ^m\mathrm{C}_2 pairs of boys, and ^m\mathrm{C}_2>^n\mathrm{C}_2, this would imply that more than 2 girls would have danced with atleast one pair of boys, by the pigeonhole principle.

    So, m is less than or equal to n.
    And so, by symmetry, we have n is less than or equal to m.

    And so, we can assert that m = n.
    • Thread Starter
    Offline

    16
    ReputationRep:
    (Original post by Jkn)
    It may or may not be...
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show
    Spoiler:
    Show

    ...unproven :pierre:


    Lad!
    Ahem... Please read the terms set out in the OP.
    Offline

    2
    ReputationRep:
    (Original post by und)
    Ahem... Please read the terms set out in the OP.
    That's before Mladenov showed up, game changer

    But I managed to find a derivation online :pierre:. Izzzaaaaaall good!
    Offline

    2
    ReputationRep:
    (Original post by und)
    Ahem... Please read the terms set out in the OP.
    I just got warning points for that post! :eek: what's the world coming too :lol:
    Offline

    0
    ReputationRep:
    (Original post by Jkn)
    I just got warning points for that post! :eek: what's the world coming too :lol:
    I smell Llewellyn
    Offline

    2
    ReputationRep:
    (Original post by bananarama2)
    I smell Llewellyn
    Really? :eek: They docked more when I emailed them explaining that they were wrong and that it did not constitute spam.

    Instead of getting rid of the points.... or even replying to my email.... they docked two more!!!!! :lol: :lol: fml

    Doubt it! He seems nice. I spoke to him on the STEP thread
    Offline

    2
    ReputationRep:
    I've deleted my other question. This is quite possibly the funnest thing of all time

    Problem 116*


    In the early 20th century, Srinivasa Ramanujan proved that \displaystyle\frac{2\sqrt{2}}{98  01}\sum_ {n=0}^{\infty}\frac{(4n)!(1103+2  6390n)}{(n!)^4396^{4n}}=\frac{1}  { \pi}.

    i) Find the approximation to \pi given by the first term of this series.

    Ramanujan's mathematics is a bit too high level for this thread (according to some people ) and so we will now deal with another type of series that approximates \pi, those using inverse trigonometric functions.

    These series are typically written in the form \displaystyle\pi=af(x)+bg(y), where a, b, x and y are constants (not necessarily rational or positive) that do not contain \pi and f(x) and g(y) are inverse trigonometric functions (they may or may not be the same).

    ii) By considering the equation 4cos\theta+2\sqrt{3}sin\theta=5, find a series in the given form and, hence, find a series expansion for \pi, giving your answer in sigma notation and calculating the number of decimal places of \pi given by the first three terms.

    iii) Using what you have now learnt, find a series expansion for \pi, whereby the first three terms give an approximation to a greater degree of accuracy that the series found in part ii.
    Offline

    2
    ReputationRep:
    (Original post by Mladenov)
    Problem 95** warm-up

    Let a,b,c be positive real numbers. Then, \displaystyle \sum_{cyc} \frac{a^{2}b^{2}(b-c)}{a+b} \ge 0.
    Don't think I gave this much thought when I tried it before. Looking at it now, it turns out the proof really does live up the the name of this thread

    Solution 95

    Using Cauchy-Schwartz, in engel form, we get...

    \displaystyle\sum_{cyc} \frac{a^{2}b^{2}(b-c)}{a+b} = \sum_{cyc} \frac{a^{2}b^{2}(b-c)^2}{(a+b)(b-c)}

\displaystyle\geq\frac{(\sum_{cy  c} ab(b-c))^2}{\sum_{cyc} (a+b)(b-c)} = \frac{(ab^2+bc^2+ca^2-3abc)^2}{(a^2+b^2+c^2)-(ab+bc+ca)}

\displaystyle=\frac{2(ab^2+bc^2+  ca^2-3abc)^2}{(a-b)^2+(b-c)^2+(c-a)^2} \geq 0
    as required.
    Offline

    1
    ReputationRep:
    (Original post by MW24595)
    Spoiler:
    Show
    Okay, I don't know if this has already been solved but,

    Solution 87

    Let there be n girls; g_1, g_2,.. g_n, and m boys; b_1, b_2,... b_m.

    Now, for every pair of boys (b_i,b_j) there exist 2 girls that danced with both of them and let these be (g_i,g_j).

    If m>n, then as there are ^n\mathrm{C}_2 pairs of girls and ^m\mathrm{C}_2 pairs of boys, and ^m\mathrm{C}_2>^n\mathrm{C}_2, this would imply that more than 2 girls would have danced with atleast one pair of boys, by the pigeonhole principle.

    So, m is less than or equal to n.
    And so, by symmetry, we have n is less than or equal to m.

    And so, we can assert that m = n.
    Good job.


    (Original post by Jkn)
    Spoiler:
    Show
    Don't think I gave this much thought when I tried it before. Looking at it now, it turns out the proof really does live up the the name of this thread

    Solution 95

    Using Cauchy-Schwartz, in engel form, we get...

    \displaystyle\sum_{cyc} \frac{a^{2}b^{2}(b-c)}{a+b} = \sum_{cyc} \frac{a^{2}b^{2}(b-c)^2}{(a+b)(b-c)}

\displaystyle\geq\frac{(\sum_{cy  c} ab(b-c))^2}{\sum_{cyc} (a+b)(b-c)} = \frac{(ab^2+bc^2+ca^2-3abc)^2}{(a^2+b^2+c^2)-(ab+bc+ca)}

\displaystyle=\frac{2(ab^2+bc^2+  ca^2-3abc)^2}{(a-b)^2+(b-c)^2+(c-a)^2} \geq 0
    as required.
    Problem 74 remains unsolved.
    Offline

    10
    ReputationRep:
    For Problem 90 my solution is now complete, so can you remove the 'first part' bit please?
    Offline

    2
    ReputationRep:
    (Original post by Mladenov)
    Problem 94 remains unsolved.
    74 I think you mean! I tried it earlier today! I'll have another look tonight. I'll get it one day :')
    (Original post by metaltron)
    For Problem 90 my solution is now complete, so can you remove the 'first part' bit please?
    Been looking on brilliant have we :pierre:
    Spoiler:
    Show
    lol jk post away :lol:
    Offline

    10
    ReputationRep:
    (Original post by Jkn)
    74 I think you mean! I tried it earlier today! I'll have another look tonight. I'll get it one day :')

    Been looking on brilliant have we :pierre:
    Spoiler:
    Show
    lol jk post away :lol:
    Just want to make clear my solution has been unedited for a few days, so no I didn't look on brilliant! And stop rubbing in the fact I'm not yet on Lvl4.
    Offline

    1
    ReputationRep:
    (Original post by Jkn)
    74 I think you mean! I tried it earlier today! I'll have another look tonight. I'll get it one day :')
    Ops, you are right!

    It is just slightly harder than 95.

    (Original post by Jkn)
    Hmm well in that case I suppose the interesting question here is weather or not there are an infinite number of pairs (m,n) such that m and n are not both Fibonacci numbers?
    The answer is no. All the solutions belong to the Fibonacci sequence.

    Problem 118**

    Let n be a positive integer. Let S_{1}, S_{2},..., S_{n} be subsets of \{1, 2, ..., n\} such that for any 1 \le k \le n the union of any k of the subsets contains at least k elements. Prove that there is a permutation (a_{1}, a_{2},..., a_{n}) of (1, 2, ..., n) such that a_{i} \in S_{i}.
    Offline

    2
    ReputationRep:
    (Original post by Mladenov)
    Ops, you are right!

    It is just slightly harder than 95.
    :eek:
    The answer is no. All the solutions belong to the Fibonacci sequence.
    Defend. :pierre:
    Offline

    2
    ReputationRep:
    (Original post by metaltron)
    Just want to make clear my solution has been unedited for a few days, so no I didn't look on brilliant! And stop rubbing in the fact I'm not yet on Lvl4.
    Hahaha :lol: Tell that to "Pierre"
    Spoiler:
    Show
    :pierre:
    He takes no ****
    • Thread Starter
    Offline

    16
    ReputationRep:
    Solution 108

    (px-q)^2+(qx-p)^2=x \Rightarrow (p^2+q^2)x^2-(4pq+1)x+p^2+q^2=0

    If we are to have integer roots then the discriminant must be positive, so we have:

    (4pq+1)^2-4(p^2+q^2)^2 \geq 0 \Rightarrow (1+2(p+q)^2)(1-2(p-q)^2) \geq 0

    The left bracket is positive so the right bracket must be positive too, giving:

    (1-2(p+q)^2) \geq 0 \Rightarrow (p-q)^2 \leq \frac{1}{2}

    Now since p and q are integers, their difference must be  0 so we have p=q giving us the quadratic:

    2p^{2}(x-1)^2=x

    The trivial solution is p=q=x=0, but consider p \neq 0. Then it is clear that the left hand side is greater than the right hand side for all x\geq 3, restricting us to the possibilities x=1 and x=2. However, the first case leads to x=0 which is a contradiction, and in the second case we have an additional root x=\frac{1}{2}. Therefore the pair (0,0) is the only solution.
    Offline

    2
    ReputationRep:
    (Original post by und)
    Spoiler:
    Show
    Solution 108

    (px-q)^2+(qx-p)^2=x \Rightarrow (p^2+q^2)x^2-(4pq+1)x+p^2+q^2=0

    If we are to have integer roots then the discriminant must be positive, so we have:

    (4pq+1)^2-4(p^2+q^2)^2 \geq 0 \Rightarrow (1+2(p+q)^2)(1-2(p-q)^2) \geq 0

    The left bracket is positive so the right bracket must be positive too, giving:

    (1-2(p+q)^2) \geq 0 \Rightarrow (p-q)^2 \leq \frac{1}{2}

    Now since p and q are integers, their difference must be  0 so we have p=q giving us the quadratic:

    2p^{2}(x-1)^2=x

    The trivial solution is p=q=x=0, but consider p \neq 0. Then it is clear that the left hand side is greater than the right hand side for all x\geq 3, restricting us to the possibilities x=1 and x=2. However, the first case leads to x=0 which is a contradiction, and in the second case we have an additional root x=\frac{1}{2}. Therefore the pair (0,0) is the only solution.
    Dayuuuuum puts my solution to shame

    Loved the part with trapping the difference! Like the twist at the end of a brilliant film

    :lol: Awesome bro!
    Offline

    1
    ReputationRep:
    (Original post by Jkn)
    Defend. :pierre:
    I claim that all solutions are represented by (1,1), (F_{2k-1},F_{2k+1}) - up to permutation.

    Suppose (m_{0},n_{0}) be a solution such that m_{0}+n_{0} is minimal. Assume, without loss of generality, m_{0} \le n_{0}. The case m_{0}=n_{0} leads to m_{0}=n_{0}=1 which is a contradiction.
    Thus, we suppose m_{0} < n_{0}. Notice that \displaystyle l = \frac{m_{0}^{2}+1}{n_{0}} < m_{0}<n_{0}. Moreover, l^{2}+1 \equiv 0 \pmod {m_{0}}, and m_{0}^{2}+1 \equiv 0 \pmod l. Hence (l,m_{0}) is a solution for which l+m_{0} < m_{0}+n_{0}; thus, we have either (l,m_{0}) = (1,1) or (l,m_{0}) = (F_{2k-1},F_{2k+1}); these possibilities imply (m_{0},n_{0})=(1,2) or (m_{0},n_{0})=(F_{2k+1},F_{2k+3}  ) - contradiction.
    Therefore all solutions belong to the Fibonacci sequence.
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Has a teacher ever helped you cheat?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Write a reply...
    Reply
    Hide
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.