Imprmis, notice that f(x)=−f(−x). This gives f(0)=0. Let x=0. Define, a0=x, an=2an−12−1. If for some n, an=0, it follows that f(a0)=0. The sequence bn=cos2ny satisfies bn=2bn−12−1. The set {2nπ(2k+21)∣n,k∈Z+} is dense in [−1,1]. From Heine - Borel we conclude that [−1,1] is a compact Hausdorff space. Now since f and ≡0 agree on a dense subspace of [−1,1], it follows that they agree on [−1,1].
Another one?
Problem 117**
Find all continuous functions defined for each x∈R, which satisfy f(xp+yq)=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 Another good reference which comes to my mind is Waerden's Algebra - it is good but a bit out of date.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.
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; g1,g2,..gn, and m boys; b1,b2,...bm.
Now, for every pair of boys (bi,bj) there exist 2 girls that danced with both of them and let these be (gi,gj).
If m>n, then as there are nC2 pairs of girls and mC2 pairs of boys, and mC2>nC2, 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.
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 980122n=0∑∞(n!)43964n(4n)!(1103+26390n)=π1.
i) Find the approximation to π 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 π, those using inverse trigonometric functions.
These series are typically written in the form π=af(x)+bg(y), where a, b, x and y are constants (not necessarily rational or positive) that do not contain π 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=5, find a series in the given form and, hence, find a series expansion for π, giving your answer in sigma notation and calculating the number of decimal places of π given by the first three terms.
iii) Using what you have now learnt, find a series expansion for π, whereby the first three terms give an approximation to a greater degree of accuracy that the series found in part ii.
Let a,b,c be positive real numbers. Then, cyc∑a+ba2b2(b−c)≥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...
cyc∑a+ba2b2(b−c)=cyc∑(a+b)(b−c)a2b2(b−c)2[br]≥∑cyc(a+b)(b−c)(∑cycab(b−c))2=(a2+b2+c2)−(ab+bc+ca)(ab2+bc2+ca2−3abc)2[br]=(a−b)2+(b−c)2+(c−a)22(ab2+bc2+ca2−3abc)2≥0 as required.
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
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.
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 S1,S2,...,Sn be subsets of {1,2,...,n} such that for any 1≤k≤n the union of any k of the subsets contains at least k elements. Prove that there is a permutation (a1,a2,...,an) of (1,2,...,n) such that ai∈Si.
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.
If we are to have integer roots then the discriminant must be positive, so we have:
(4pq+1)2−4(p2+q2)2≥0⇒(1+2(p+q)2)(1−2(p−q)2)≥0
The left bracket is positive so the right bracket must be positive too, giving:
(1−2(p+q)2)≥0⇒(p−q)2≤21
Now since p and q are integers, their difference must be 0 so we have p=q giving us the quadratic:
2p2(x−1)2=x
The trivial solution is p=q=x=0, but consider p=0. Then it is clear that the left hand side is greater than the right hand side for all x≥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=21. Therefore the pair (0,0) is the only solution.
I claim that all solutions are represented by (1,1), (F2k−1,F2k+1) - up to permutation.
Suppose (m0,n0) be a solution such that m0+n0 is minimal. Assume, without loss of generality, m0≤n0. The case m0=n0 leads to m0=n0=1 which is a contradiction. Thus, we suppose m0<n0. Notice that l=n0m02+1<m0<n0. Moreover, l2+1≡0(modm0), and m02+1≡0(modl). Hence (l,m0) is a solution for which l+m0<m0+n0; thus, we have either (l,m0)=(1,1) or (l,m0)=(F2k−1,F2k+1); these possibilities imply (m0,n0)=(1,2) or (m0,n0)=(F2k+1,F2k+3) - contradiction. Therefore all solutions belong to the Fibonacci sequence.