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The Proof is Trivial!

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Reply 800
Solution 104

Imprmis, notice that f(x)=f(x)f(x)=-f(-x). This gives f(0)=0f(0)=0.
Let x0x \not= 0. Define, a0=xa_{0}=x, an=2an121a_{n}=2a_{n-1}^{2}-1. If for some nn, an=0a_{n}=0, it follows that f(a0)=0f(a_{0})=0.
The sequence bn=cos2nyb_{n}= \cos 2^{n}y satisfies bn=2bn121b_{n}=2b_{n-1}^{2}-1.
The set {π(2k+12)2nn,kZ+}\{\frac{\pi(2k+\frac{1}{2})}{2^{n}}| n,k \in \mathbb{Z^{+}} \} is dense in [1,1]\left[-1,1 \right].
From Heine - Borel we conclude that [1,1]\left[-1,1 \right] is a compact Hausdorff space. Now since ff and 0 \equiv 0 agree on a dense subspace of [1,1]\left[-1,1 \right], it follows that they agree on [1,1]\left[-1,1 \right].

Another one?

Problem 117**

Find all continuous functions defined for each xRx \in \mathbb{R}, which satisfy f(xp+yq)=f(x)p+f(y)qf(x^{p}+y^{q})=f(x)^{p}+f(y)^{q}, where pp and qq are positive integers at least one of which is greater than 11.

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

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 Another good reference which comes to my mind is Waerden's Algebra - it is good but a bit out of date.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 :wink:


I shall check it out!
Reply 801
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


Original post by Mladenov

I shall check it out!

Lad!
Reply 802
Original post by Mladenov
Nice. I will give you something different.

Check your solutions to problem 86. You have missed solutions and (1,1)(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; g1,g2,..gng_1, g_2,.. g_n, and m boys; b1,b2,...bmb_1, b_2,... b_m.

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

If m>n, then as there are nC2^n\mathrm{C}_2 pairs of girls and mC2^m\mathrm{C}_2 pairs of boys, and mC2>nC2^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.
(edited 10 years ago)
Reply 803
Original post by Jkn
It may or may not be...

Spoiler



Lad!

Ahem... Please read the terms set out in the OP. :biggrin:
Reply 804
Original post by und
Ahem... Please read the terms set out in the OP. :biggrin:

That's before Mladenov showed up, game changer :wink:

But I managed to find a derivation online :pierre:. Izzzaaaaaall good!
Reply 805
Original post by und
Ahem... Please read the terms set out in the OP. :biggrin:

I just got warning points for that post! :eek: what's the world coming too :lol:
Original post by Jkn
I just got warning points for that post! :eek: what's the world coming too :lol:


I smell Llewellyn :tongue:
Reply 807
Original post by bananarama2
I smell Llewellyn :tongue:

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 :tongue:
Reply 808
I've deleted my other question. This is quite possibly the funnest thing of all time :biggrin:

Problem 116*


In the early 20th century, Srinivasa Ramanujan proved that 229801n=0(4n)!(1103+26390n)(n!)43964n=1π\displaystyle\frac{2\sqrt{2}}{9801}\sum_ {n=0}^{\infty}\frac{(4n)!(1103+26390n)}{(n!)^4396^{4n}}=\frac{1}{ \pi}.

i) Find the approximation to π\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 :colone:) and so we will now deal with another type of series that approximates π\pi, those using inverse trigonometric functions.

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

ii) By considering the equation 4cosθ+23sinθ=54cos\theta+2\sqrt{3}sin\theta=5, find a series in the given form and, hence, find a series expansion for π\pi, giving your answer in sigma notation and calculating the number of decimal places of π\pi given by the first three terms.

iii) Using what you have now learnt, find a series expansion for π\pi, whereby the first three terms give an approximation to a greater degree of accuracy that the series found in part ii.
(edited 10 years ago)
Reply 809
Original post by Mladenov

Problem 95** warm-up

Let a,b,ca,b,c be positive real numbers. Then, cyca2b2(bc)a+b0\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 :colone:

Solution 95

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

cyca2b2(bc)a+b=cyca2b2(bc)2(a+b)(bc)[br](cycab(bc))2cyc(a+b)(bc)=(ab2+bc2+ca23abc)2(a2+b2+c2)(ab+bc+ca)[br]=2(ab2+bc2+ca23abc)2(ab)2+(bc)2+(ca)20\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)}[br]\displaystyle\geq\frac{(\sum_{cyc} 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)}[br]\displaystyle=\frac{2(ab^2+bc^2+ca^2-3abc)^2}{(a-b)^2+(b-c)^2+(c-a)^2} \geq 0
as required.
(edited 10 years ago)
Reply 810
Original post by MW24595

Spoiler



Good job.


Original post by Jkn

Spoiler



Problem 74 remains unsolved. :tongue:
(edited 10 years ago)
For Problem 90 my solution is now complete, so can you remove the 'first part' bit please?
Reply 812
Original post by Mladenov

Problem 94 remains unsolved. :tongue:

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

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



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. :tongue:
Reply 814
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)(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 nn be a positive integer. Let S1,S2,...,SnS_{1}, S_{2},..., S_{n} be subsets of {1,2,...,n}\{1, 2, ..., n\} such that for any 1kn1 \le k \le n the union of any kk of the subsets contains at least kk elements. Prove that there is a permutation (a1,a2,...,an)(a_{1}, a_{2},..., a_{n}) of (1,2,...,n)(1, 2, ..., n) such that aiSia_{i} \in S_{i}.
Reply 815
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:
Reply 816
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. :tongue:

Hahaha :lol: Tell that to "Pierre" :wink:

Spoiler

He takes no ****
Reply 817
Solution 108

(pxq)2+(qxp)2=x(p2+q2)x2(4pq+1)x+p2+q2=0(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)24(p2+q2)20(1+2(p+q)2)(12(pq)2)0(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:

(12(p+q)2)0(pq)212(1-2(p+q)^2) \geq 0 \Rightarrow (p-q)^2 \leq \frac{1}{2}

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

2p2(x1)2=x2p^{2}(x-1)^2=x

The trivial solution is p=q=x=0p=q=x=0, but consider p0p \neq 0. Then it is clear that the left hand side is greater than the right hand side for all x3x\geq 3, restricting us to the possibilities x=1x=1 and x=2x=2. However, the first case leads to x=0x=0 which is a contradiction, and in the second case we have an additional root x=12x=\frac{1}{2}. Therefore the pair (0,0)(0,0) is the only solution.
Reply 818
Original post by und

Spoiler


Dayuuuuum puts my solution to shame :colondollar:

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

:lol: Awesome bro!
Reply 819
Original post by Jkn
Defend. :pierre:


I claim that all solutions are represented by (1,1)(1,1), (F2k1,F2k+1)(F_{2k-1},F_{2k+1}) - up to permutation.

Suppose (m0,n0)(m_{0},n_{0}) be a solution such that m0+n0m_{0}+n_{0} is minimal. Assume, without loss of generality, m0n0m_{0} \le n_{0}. The case m0=n0m_{0}=n_{0} leads to m0=n0=1m_{0}=n_{0}=1 which is a contradiction.
Thus, we suppose m0<n0m_{0} < n_{0}. Notice that l=m02+1n0<m0<n0\displaystyle l = \frac{m_{0}^{2}+1}{n_{0}} < m_{0}<n_{0}. Moreover, l2+10(modm0)l^{2}+1 \equiv 0 \pmod {m_{0}}, and m02+10(modl)m_{0}^{2}+1 \equiv 0 \pmod l. Hence (l,m0)(l,m_{0}) is a solution for which l+m0<m0+n0l+m_{0} < m_{0}+n_{0}; thus, we have either (l,m0)=(1,1)(l,m_{0}) = (1,1) or (l,m0)=(F2k1,F2k+1)(l,m_{0}) = (F_{2k-1},F_{2k+1}); these possibilities imply (m0,n0)=(1,2)(m_{0},n_{0})=(1,2) or (m0,n0)=(F2k+1,F2k+3)(m_{0},n_{0})=(F_{2k+1},F_{2k+3}) - contradiction.
Therefore all solutions belong to the Fibonacci sequence.

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