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B Basic Challenge of the Week #3 04/02/2017

  1. Mar 3, 2017 #1
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  3. Mar 31, 2017 #2

    QuantumQuest

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    Find the limit ##\lim_{x\to \frac{\pi}{2}} \frac{(1 - \sin x)(1 - \sin^2 x)\cdots (1 - \sin^{n} x)}{\cos^{2n} x}## , ##n \in \mathbb{N}##
     
    Last edited: Mar 31, 2017
  4. Mar 31, 2017 #3

    mfb

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    $$\begin{align} =& \lim_{x\to 0} \frac{(1 - \cos x)(1 - \cos^2 x)\cdots (1 - \cos^{n} x)}{\sin^{2n} x} \\
    =& \lim_{x\to 0} \frac{1 - \cos x}{\sin^2 x} \lim_{x\to 0} \frac{1 - \cos^2 x}{\sin^2 x} \cdots \lim_{x\to 0} \frac{1 - \cos^n x}{\sin^2 x}\end{align}$$
    We have to show that all those limits exist as real numbers. They are all of the type 0/0, use l'Hospital (is that [B] level?):
    $$\begin{align}=&\lim_{x\to 0} \frac{\sin x}{2\sin x \cos x} \lim_{x\to 0} \frac{2\sin x \cos x}{2\sin x \cos x} \cdots \lim_{x\to 0} \frac{n \sin x \cos^{n-1} x}{2\sin x \cos x}\\
    =& \frac 1 2 \cdot \frac 2 2 \cdots \frac n 2 \\
    =& \frac{n!}{2^n}\end{align}$$This is more like a homework problem. I think challenges can be a bit more complicated.

    Interesting MathJax bug: The formula numbers increase every time you use the preview or edit function. They only reset back if I reload the page.
     
  5. Mar 31, 2017 #4

    QuantumQuest

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    Thanks for the feedback @mfb. This is more at the introductory undergrad level but it has a small trick to solve it, so I regard it as an easy challenge. Now, I didn't know about this way you present for the solution - a very good one. My take was to leverage ##\frac{1 - \sin ^k x}{\cos^2 x} = \frac{(1 - \sin x)(1 + \sin x + \sin^2 x + \cdots +\sin^{k -1} x)}{1 - \sin^2 x} = \cdots = \frac{1 + \sin x + \sin^2 x + \cdots +\sin^{k -1} x}{1 + sin x}##. So, ##\lim_{x\to\frac{\pi}{2}}\frac{1 - \sin ^k x}{\cos^2 x} = \lim_{x\to\frac{\pi}{2}}\frac{1 + \sin x + \sin^2 x + \cdots +\sin^{k -1} x}{1 + sin x} = \frac{k}{2}##, for every ##k\in \mathbb{N}##. So

    ##\lim_{x\to \frac{\pi}{2}} \frac{(1 - \sin x)(1 - \sin^2 x)\cdots (1 - \sin^{n} x)}{\cos^{2n} x} = \lim_{x\to \frac{\pi}{2}}\frac{1 - \sin x}{\cos^2 x}\frac{1 - \sin^2 x}{cos^2 x}\cdots\frac{1 - \sin^n x}{cos^2 x} = \frac{1}{2}\cdot\frac{2}{2}\cdot\frac{3}{2}\cdots\frac{n}{2} = \frac{n!}{2^n}##.
     
  6. Mar 31, 2017 #5

    Charles Link

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    @QuantumQuest I like your solution to it. As @mfb said, it's not a real difficult challenge, but some of micromass' at the high school/first year college level were also somewhat easy. We're trying to generate interest in math and physics and I think some of the students would find this of interest.
     
  7. Mar 31, 2017 #6

    mfb

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    We could replace 1/cos^n by tan^n, just to hide the factors a bit and to add some additional terms that you have to take care of (although these are easy to split out). It doesn't make it a hard challenge, but it is a little bit more to think about.
     
  8. Mar 31, 2017 #7

    QuantumQuest

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    I think that @mfb is right. It would make a better challenge to tweak it a little bit and elevate its level but not go to "I", as it is good to post some challenges for people at the college / introductory undergrad level.
     
  9. Apr 5, 2017 #8

    QuantumQuest

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    Find the smallest integer ##n## ,##n > 1## for which the number ##[\frac{1^2 + 2^2 +\cdots+n^2}{n}]^\frac{1}{2}## is integer.
     
  10. Apr 6, 2017 #9

    mfb

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    Is there some nice mathematical trick, or do you just calculate the first 3xx elements in a spreadsheet?

    There is no other number up to n=2000, and a heuristic argument suggests that there is no larger number either (probability of 1/n to have n as factor, probability 1/n^1.5 to be square, the sum converges).
     
  11. Apr 6, 2017 #10

    QuantumQuest

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    With a little bit of math you find a Diophantine equation namely ##(n + 1)(2n +1) = 6m^2##. Then modulo divisibility must be checked. There are two possible cases to be examined and it involves some thought in order to find the value of n (##n = 337##). I think that it has some work for an undergrad in the B level and it has been given many years before as a challenge (I have it in my notes).
     
  12. Apr 6, 2017 #11

    mfb

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    Well, checking the spreadsheet took a minute, and it is a mathematically sound solution, just not very elegant. But the division by n always works, good point. Then there could be larger solutions. Nothing up to n=30,000, however.
     
  13. Apr 6, 2017 #12

    QuantumQuest

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    I think that in any case if we put it as a challenge, anyone wishing to participate will have to write down and justify the steps. There is often some software helping to do this in various cases but fair use of software is in the rules anyway, like for instance the case for an integral to be solved. Any participant can use Wolfram Alpha to see what the solution is but not copy - paste the steps that Wolfram Alpha does. So, I don't think that we will have any issues regarding unfair use of software. On the other hand, what I wonder is if it is a good enough challenge. I find it good for B level but team's opinions matter!
     
  14. Apr 6, 2017 #13

    mfb

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    n has to be odd, write n as 2k-1
    (2k)(4k-1) = 6 m2
    k(4k-1) = 3 m2

    If k=3h, then
    h(12h-1) = m2
    h and 12h-1 don't have common divisors apart from 1, therefore both h and 12h-1 have to be perfect squares. Unfortunately 12h-1 = 2 mod 3, it cannot be a perfect square. This case is impossible.

    If k=3h+1, then
    (3h+1)(4h+1) = m2
    Again (3h+1) and (4h+1) don't have common divisors, they both have to be perfect squares.

    Let 3h+1=p2 and 4h+1=(p+q)2, then h=2pq+q2.
    p2-(6q)p-3q2-1=0
    p=3q + sqrt(12q2+1)

    The first solution (apart from q=0) is
    q=2, p=13 => n=337
    The next solutions:
    q=28, p=181 => n=65,521
    q=390, n=12710881
    q=5432, n=2465845537
    Sequence for q at oeis - a(n) = 14*a(n-1)-a(n-2) with a(0) = 0, a(1) = 2.

    There is an infinite number of solutions. n increases by a factor of about 142=196 each time.

    Edit: Now with oeis sequence.

    I guess I went a bit beyond the [b] level.
     
    Last edited: Apr 6, 2017
  15. Apr 7, 2017 #14

    QuantumQuest

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    Yes indeed, this is a complete solution. I think - as I remember from this challenge, it suffices to find ##n = 337##. My solution was in a similar fashion

    We substitute ##1^2 + 2^2 +\cdots+n^2## for ##\frac{n(n + 1)(2n + 1)}{6}##, so with a little bit of math we find the Diophantine ##(n + 1)(2n +1) = 6m^2##.
    Now ##6## divides ##(n + 1)(2n +1)## if and only if ##n \equiv 5## or ##1 \pmod 6##

    Case 1

    ##n = 6k + 5##. Substituting in the Diophantine we have ##m^2 = (k + 1)(12k + 11)##. Due to primality between the factors, each must be a perfect square so ##k + 1 = x^2## and ##12k + 11 = y^2##. So ##12x^2 = y^2 + 1##. Now, this last equation has no solution ##\pmod 4##.

    Case 2

    ##n = 6k + 1##. Substituting in the Diophantine we have ##m^2 = (3k + 1)(4k + 1)##. Again as in case 1 due to primality between the factors we have ##3k + 1 = x^2## and ##4k + 1 = y^2##. Then, ##(2x - 1)(2x + 1) = 3y^2##. Every possible prime factor of ##3y^2## excluding ##3## has to be in an even power. Now, we just have to try putting values for x such that both ##2x - 1## and ##2x + 1## not be prime (besides ##3##). If ##x = 1## we find ##n = 1## that violates the given of the problem. Doing some testing it turns out that ##x = 13## does the trick, so after some math we find ##n = 337## as the solution.
     
  16. Apr 7, 2017 #15

    Charles Link

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    @QuantumQuest I've been busy with some other items so I didn't get the chance to respond in detail. The problem uses the sum of the first n-squares in an interesting manner, and from that standpoint is quite educational. (I remember seeing that in a solid state physics calculation as a first year graduate student, and it wasn't until a couple years later that I worked through a systematic way to compute the formula for the sum of the first n cubes, or the sum of the first n to the fourth, etc.) I didn't get the chance to try to solve the part where n=337, but I would say it is a very good challenge problem. :) :)
     
  17. Apr 7, 2017 #16

    QuantumQuest

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    @Charles Link I also think that it's a good B level challenge. Now, if @mfb agrees I'll ask Greg Bernhardt to post it next week.
     
  18. Apr 7, 2017 #17

    mfb

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    We can ask for all solutions as I-level challenge.
     
  19. Apr 7, 2017 #18

    QuantumQuest

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    I agree. So, it will be

    Find all integers ##n, n > 1## for which the number ##[\frac{1^2 + 2^2 +\cdots+n^2}{n}]^\frac{1}{2}## is integer.
     
  20. Apr 7, 2017 #19

    QuantumQuest

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    @mfb I'll ask Greg to post it with both our (nick)names as posters and judges because with the whole work you already did it's fair to be so
     
  21. Apr 7, 2017 #20

    mfb

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    We can drop the n>1 requirement if we ask for all integers anyway.
     
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