An Interesting Problem

Discussion in 'Physics & Math' started by §outh§tar, Jul 1, 2008.

  1. Guest254 Valued Senior Member

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    But you can - you only need to know how to expand log(1+x) for small x...
    That's pretty much the definition of a power series... (or, more formally, an asymptotic series). What I've written is perfectly rigorous.

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  3. Guest254 Valued Senior Member

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    Excellent, well done. This is actually something I've given in interviews, as it's easier to construct an argument by conversing, than it is to actually write down. I.e, one says "well, there can't be a 5 because 5=2+3, and 2x3=6>5 etc...
     
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  5. §outh§tar is feeling caustic Registered Senior Member

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    4,832
    If a, b > 0, find \(S_p (a,b) = \left( {\frac{{a^p + b^p }}{2}} \right)^{\frac{1}{p}} \)
    as p tends to +infinity and as p tends to -infinity
     
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  7. CptBork Valued Senior Member

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    If \(a\geq b\), then the limit is \(a\) as \(p\mapsto\infty\) and the limit is \(b\) as \(p\mapsto -\infty\).
     
  8. §outh§tar is feeling caustic Registered Senior Member

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    4,832
    Correct, CptBork. We'd like a proof though.

    ---

    What is the largest number of coins from which one can detect a counterfeit in three weighings with a pan balance, if it known only that the counterfeit coin differs in weight from the others?
     
  9. CptBork Valued Senior Member

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    Well I'm too lazy to type out the whole entire argument, I did write it out with pen and paper though. Start with \(a\geq b\). Then we write:

    \(\left(\frac{a^p+b^p}{2}\right)^{1/p}=a\left(\frac{1+\left(\frac{b}{a}\right)^p}{2} \right)^{1/p}\)

    Next we write

    \(\left(\frac{1}{2}\right)^{1/p}\leq\left(\frac{1+\left(\frac{b}{a}\right)^p}{2}\right)^{1/p}\leq1\)

    and apply the squeeze theorem as \(p\mapsto\infty\). That takes care of the first part. Second part, \(p\mapsto-\infty\), is similar and just requires a little extra manipulation.
     
  10. §outh§tar is feeling caustic Registered Senior Member

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    4,832
    Show that if \(\frac{{b_n }}{{b_{n + 1} }} = 1 + \beta _n \), n=1,2,..., and the series \(\sum\limits_{n = 1}^\infty {\beta _n } \) converges absolutely, then the limit \( {\lim }\limits_{n \to \infty } b_n \) exists.
     
  11. §outh§tar is feeling caustic Registered Senior Member

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    4,832
    Show that if p is a real number then the series with nth term \(\frac{{(n + p)n^p - (n + 1)^p n}}{{(n + 1)^p n}}\) converges.
     
  12. §outh§tar is feeling caustic Registered Senior Member

    Messages:
    4,832
    Show that if \(a_n > 0\) and \( {\lim }\limits_{n \to \infty } \frac{{a_{n + 1} }}{{a_n }}\) exists then \( {\lim }\limits_{n \to \infty } \frac{{a_{n + 1} }}{{a_n }} = {\lim }\limits_{n \to \infty } \sqrt[n]{{a_n }}\)


    In particular, if the left hand limit exists in the following then the right hand limit exists and equality holds: \( {\lim }\limits_{n \to \infty } a_{n + 1} - a_n = {\lim }\limits_{n \to \infty } \frac{{a_n }}{n}\)
     
  13. CptBork Valued Senior Member

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    6,355
    I'm really glad you asked this question. These are some of the very results I learned in analysis class and were then used to provide clever proofs for other results. I've been doing some thinking on the pennies problem so I'll get back to you later.
     
  14. CptBork Valued Senior Member

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    6,355
    Mind you, I should have said analysis classes, in plural, since I've taken 4 of them to date. But I always found the first 2 I took to be the most interesting and direct, after that it was all metric spaces and measure theory and abstract stuff.

    Anyhow, I did some thinking on the coins problem and I believe the answer is that if you're allowed up to 3 weighings, you can guarantee finding the counterfeit coin in 3 or less measurements if you have 8 or less coins to start with. If you start with 9 or more coins, then there's always a chance that you will fail to detect the counterfeit in 3 measurements or less no matter what algorithm you use. I can post my reasoning if you want, but first I want to check if I have the right answer. My method was to treat it on a case by case basis, considering the worst possible cases for 3, 4, 5 coins etc. until you get up to 9 or more coins.
     
  15. §outh§tar is feeling caustic Registered Senior Member

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    4,832
    CptBork, I'll have you know that I don't know what "the" answer is to the coins problem. The book I got the problem from doesn't come with answers, not to mention some of its questions are known to be completely wrong..
     
  16. §outh§tar is feeling caustic Registered Senior Member

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    4,832
    What is the smallest number of questions to be answered "yes" or "no" that one must pose in order to be sure of determining a 7-digit telephone number?
     
  17. Amadeusboy Registered Member

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    22
    I don't mean to sidetrack anyone from the current problem, but I have a question regarding SouthStar's second problem. The question was:

    Evaluate the limits

    \(\lim_{x\to0}\frac{\ln\cos x}{(\sin x)^2}\).


    Additionally, on July 3, SouthStar wrote:


    I know I haven't looked at problems similar to this in a while, but I don't quite understand from where this came. Did you apply l'Hopital's Rule?
     
  18. Vkothii Banned Banned

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    That looks like a standard rearrangement of trig functions, to me. You know what cos(x), or cos(2x) can be substituted with?
     
  19. §outh§tar is feeling caustic Registered Senior Member

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    4,832
    Amadeusboy, the second equality comes from standard properties of a*ln x = ln (x^a) and (cos x)^2 = 1- (sin x)^2. The third comes from a use of power series. Is that what you have trouble with?

    (The point of my solution was that it involved (almost) no calculus (depending on how you arrive at sin x ~ x as x -> 0))
     
  20. Reiku Banned Banned

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    11,238
    In other words, you proved his point.
     
  21. Amadeusboy Registered Member

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    I don't believe any trigonometric substitutions were made. Even if they were, what happened to the \(\ln \) term in the third equality?

    After more work and refreshing my memory of l'Hopital's rule, my approach was as follows:

    Let \(f(x)=\ln(\cos x) \) and \(g(x)=\sin^2 x=(\sin x)^2\).

    Then

    \(f^\prime (x)=-\frac {\sin x}{\cos x} \),

    and

    \(g^\prime (x)=2 \sin x \cos x\).

    Accordingly,

    \(\Large\frac {f^\prime (x)}{g^\prime (x)}=\frac{-\frac{\sin x}{\cos x}}{2\sin x\cos x}=-\frac{1}{2\cos^2x}\).

    Thus

    \(\Large\lim_{x\to0}\frac{f^\prime (x)}{g^\prime (x)}=\lim_{x\to0}\frac{-1}{2\cos x}=-\frac{1}{2}\).

    If my line of reasoning is incorrect or I have made a mathematical error, please let me know.

    (SouthStar, I didn't see your last post before I finished writing and posting mine. I looked at the problem again using your hints and I understand your approach. Thanks for your response. Amadeusboy)
     
    Last edited: Jul 8, 2008
  22. Reiku Banned Banned

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    Looks good to me...
     
  23. CptBork Valued Senior Member

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    And what exactly would that point be? I never got around to answering the question yet.
     

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