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    Is applied mathematics allowed on here?
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    (Original post by rayquaza17)
    Problem 494 ***

    Evaluate:
    \displaystyle \int_{0}^{\infty }\frac{cos(ax)}{1+x^2}dx for a\geq 0.

    Sorry if it's been asked before.
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    \\Let\ g(a)= e^{-|a|}\ \forall a \in \mathbb{R}.\\\





Then\ \int_{-\infty}^{\infty}e^{-|a|}e^{iax}da = \int_{0}^{\infty}e^{-a}e^{iax}da\ + \int_{-\infty}^{0}e^{a}e^{iax}da = \int_{0}^{\infty}e^{-a(-1 + ix)}da\ + \int_{-\infty}^{0}e^{a(1+ix)}da = \frac{-1}{-1 + ix}\ + \frac{1}{1 + ix} = \frac{2}{1 + x^{2}}\\



So\ \widehat{g}(x) = \frac{1}{2}f(x)\ \forall x \in \mathbb{R},\ where\ f(x) = \frac{1}{1 + x^{2}}\\\



Let\ I(a) = \int_{0}^{\infty}\frac{cos(ax))}  {1 + x^{2}}dx = \frac{1}{2}\int_{0}^{\infty} \frac{e^{iax}}{1 + x^2}dx\ +\ \frac{1}{2}\int_{0}^{\infty} \frac{e^{-iax}}{1 + x^2}dx = \int_{-\infty}^{\infty}\frac{e^{iax}}{1 + x^2}dx = \int_{-\infty}^{\infty}\frac{e^{-iax}}{1 + x^2}dx = \pi g(a)

    The last bit of working is via two changes of variables; the first one on the second integral in the sum, via  x\mapsto -y and the second change being on the second to last equality  x\mapsto -y again, and then invoking Fourier Inversion, which is valid because both g and \widehat{g} are in L^{1}.
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    (Original post by SParm)
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    \\Let\ g(a)= e^{-|a|}\ \forall a \in \mathbb{R}.\\\





Then\ \int_{-\infty}^{\infty}e^{-|a|}e^{iax}da = \int_{0}^{\infty}e^{-a}e^{iax}da\ + \int_{-\infty}^{0}e^{a}e^{iax}da = \int_{0}^{\infty}e^{-a(-1 + ix)}da\ + \int_{-\infty}^{0}e^{a(1+ix)}da = \frac{-1}{-1 + ix}\ + \frac{1}{1 + ix} = \frac{2}{1 + x^{2}}\\



So\ \widehat{g}(x) = \frac{1}{2}f(x)\ \forall x \in \mathbb{R},\ where\ f(x) = \frac{1}{1 + x^{2}}\\\



Let\ I(a) = \int_{0}^{\infty}\frac{cos(ax))}  {1 + x^{2}}dx = \frac{1}{2}\int_{0}^{\infty} \frac{e^{iax}}{1 + x^2}dx\ +\ \frac{1}{2}\int_{0}^{\infty} \frac{e^{-iax}}{1 + x^2}dx = \int_{-\infty}^{\infty}\frac{e^{iax}}{1 + x^2}dx = \int_{-\infty}^{\infty}\frac{e^{-iax}}{1 + x^2}dx = \pi g(a)

    The last bit of working is via two changes of variables; the first one on the second integral in the sum, via  x\mapsto -y and the second change being on the second to last equality  x\mapsto -y again, and then invoking Fourier Inversion, which is valid because both g and \widehat{g} are in L^{1}.
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    The answer is \frac{\pi }{2}e^{-a}.
    I think your final answer is just missing a half at the end?
    PS: Jeff Winger. <3



    (Original post by Arieisit)
    Is applied mathematics allowed on here?
    Yeah. Maybe if it's some easy-ish applied maths I might be able to actually answer a question.
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    (Original post by rayquaza17)
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    The answer is \frac{\pi }{2}e^{-a}.
    I think your final answer is just missing a half at the end?
    PS: Jeff Winger. <3




    Yeah. Maybe if it's some easy-ish applied maths I might be able to actually answer a question.
    Yeah sorry very possible I messed up by a factor of a half, won't be the first or last time.
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    Me looking through this thread
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    (Original post by ThatPerson)
    From IMO 2013 Q1.

    Problem 492**

    Prove that for any pair of positive integers k and  n, there exist  k positive integers  m_1,m_2,...,m_k (not necessarily different) such that

     1 + \dfrac{2^k - 1}{n} = \left(1 + \dfrac{1}{m_1}\right)\left(1 + \dfrac{1}{m_2}\right)...\left(1 + \dfrac{1}{m_k}\right)
    After 15 hours, I have finally solved it... Finally...
    Now I can do some revision haha
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    (Original post by Renzhi10122)
    After 15 hours, I have finally solved it... Finally...
    Now I can do some revision haha
    Nice. Although I'm not sure if I should feel somewhat guilty for distracting you .
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    (Original post by ThatPerson)
    Nice. Although I'm not sure if I should feel somewhat guilty for distracting you .
    Haha, around 13 of those 15 hours were spent on the question a few weeks ago, so no worries.
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    (Original post by Renzhi10122)
    After 15 hours, I have finally solved it... Finally...
    Now I can do some revision haha
    Proof or gtfo
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    (Original post by HeavisideDelts)
    Proof or gtfo
    Fine then (warning, slightly long and inelegant)

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    First note that we may write any  n in binary and also that:

     1+ \dfrac{2^k-1}{n}=\dfrac{2^k+n-1}{n}=(\dfrac{m_1+1}{m_1}) (\dfrac{m_2+1}{m_2})...(\dfrac{m  _k+1}{m_k})

    Let  A denote the number of 1s when  n is written in binary,  B denote the number of 0s to the left of the furthest right 1 and  C denote the number of 0s to the right of the furthest right 1. Let us first take the case of  k&gt;n-1 .

    We have:

     n has  A+B+C digits and let  D be the number of 0s after the second leftmost 1 in  2^k+n-1 e.g. if  n=101_2 and  k=5 then  2^k+n-1=100100_2 and  A=2, B=1, C=0, D=2 . Realise that  k=A+B+C+D

    Let us first take the denominator of  \dfrac{1}{n} and the aim is to write a series of fractions that give  \dfrac{2^a}{n} where  a is an integer. This is achieved by the following algorithm:

    Take all factors of 2 out of  n and call this number  n_0 which is now odd. Let the first fraction on the RHS be  \dfrac{n_0+1}{n_0} and realise that  n_0+1 is even. Repeat the algorithm for  n_0+1

    From this, we see that the number of fractions needed until we reach  \dfrac{2^a}{n} is  B+1 and  a=-A-B-C

    This is illustrated by use of an example:

     \dfrac{1}{100101101100_2}=\dfrac  {1001011100_2}{1001011011_2} \dfrac{10011000_2}{10010111_2} \dfrac{10100_2}{10011_2} \dfrac{110_2}{101_2} \dfrac{100_2}{11_2}

    After each fraction, the rightmost 0 becomes a 1 and then an extra fraction is needed to convert the last string of 1s into a power of 2 so  B+1 fractions are needed (I hope this doesn't need too much justification). It should now be clear that the 'extra' powers of two generated by this series of fractions is equal to  2^{A+B+C}

    We now deal with the numerator. Let  2^k+n-1=m and take out all factors of two resulting in  m_0 which is odd. Now let the first fraction dealing with the numerator be
     \dfrac{m_0}{m_0-1}
    We now take all factors of two out of  m_0-1 calling the new number  m_1 which is odd and let the second fraction be  \dfrac{m_1}{m_1-1}

    The number of fractions needed is simply the number of 1s in  m apart from the leftmost 1 (since that gives us a power of two) because each fraction gets rid of a 1 (from the  -1 on the denominator). The number of 1s in m (apart from the leftmost 1) is  A+C-1 since we take off 1 from  n and the rightmost one in  n becomes a 0 and all 0s to the right of that become 1s. The 'extra' powers of two generated is simply the number of digits in m minus the leftmost one, but this time on the denominator so  2^{-A-B-C-D} .

    Putting the two together, we use  A+B+C fractions and get  2^{-D} 'extra', but  k=A+B+C+D so we let the other  D fractions be  \dfrac{2}{1} getting rid of the  2^{-D} .

    We now look to the case of  k \leq n . We deal with the denominator first. Instead of getting to a power of two in the last fraction, we get to whichever digit we add  2^k to on the denominator (call this digit  d ) so that on the numerator, we have the correct digits to the left of  d . We now take the numerator and put it on the denominator of a new fraction, after multiplying by some power of 2 so that the end digit coincides with the next 1 to the right of  d (hard to explain, I'll provide an example below). We then carry on until we reach  m

    E.g.

     k=6, n=10011010110_2 

\dfrac{10100010101_2}{1001101011  0_2}=\dfrac{1001101100_2}{100110  1011_2} \dfrac{10011100_2}{10011011_2} \dfrac{101000_2}{100111_2} \dfrac{1010001_2}{1010000_2} \dfrac{101000101_2}{101000100_2} \dfrac{10100010101_2}{1010001010  0_2}

    Let  E be the number of 1s to the right of  d in  n and  F be the number 0s to the right of  d but to the left of the rightmost 1 and let  G be the number of 0s to the right of the rightmost 1. We require  F+1 fractions for the first stage (getting the denominator) and then for the second stage, we need  E+G-1 fractions, i.e. a total of  E+F+G=k fractions. The powers of two generated from the two stages also cancel (from the first stage, we get  2^{E+F+G} and from the second, we get  2^{-E-F-G} ) which completes the proof.

    Bit of an inelegant solution, and also explained poorly at points.
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    (Original post by Renzhi10122)
    Fine then (warning, slightly long and inelegant)

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    First note that we may write any  n in binary…

     k=6, n=10011010110_2 

\dfrac{10100010101_2}{1001101011  0_2}=\dfrac{1001101100_2}{100110  1011_2} \dfrac{10011100_2}{10011011_2} \dfrac{101000_2}{100111_2} \dfrac{1010001_2}{1010000_2} \dfrac{101000101_2}{101000100_2} \dfrac{10100010101_2}{1010001010  0_2}

    Niiiiice. Basically by induction on the binary expansion, I suppose, but you've gone and done a constructive proof.
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    (Original post by Smaug123)
    Niiiiice. Basically by induction on the binary expansion, I suppose, but you've gone and done a constructive proof.
    Yeah, induction would probably have made it easier, and less time consuming
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    (Original post by Mladenov)


    There are hundreds of techniques when it comes to functional equations. It is annoying that there are no books on functional eqs; I could think of 1-2 really good for olympiad problems.


    .
    plz can you share them
    the methods and techniques
    im aweful
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    (Original post by demigawdz)
    plz can you share them
    the methods and techniques
    im aweful
    At and below BMO2 level, substitution is the main one. Sub in some values, see if you get any good relations, see if you can get any values for some f(x). At a higher level, check for injectivity and surjectivity (you may have to look on wikipedia if you don't know what these are). Functional equations also get more algebraic as they get harder, so then a set method doesn't really apply. Someone probably has better advice than this.
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    (Original post by Renzhi10122)
    At and below BMO2 level, substitution is the main one. Sub in some values, see if you get any good relations, see if you can get any values for some f(x). At a higher level, check for injectivity and surjectivity (you may have to look on wikipedia if you don't know what these are). Functional equations also get more algebraic as they get harder, so then a set method doesn't really apply. Someone probably has better advice than this.
    do you know whether for q 5 0 counts as being in the set of ''non negative integers'' as the positive integer sign means positive lol (which doesnt include 0)/
    http://www.bmoc.maths.org/home/bmo1-2002.pdf
    Thank you.
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    (Original post by demigawdz)
    do you know whether for q 5 0 counts as being in the set of ''non negative integers'' as the positive integer sign means positive lol (which doesnt include 0)/
    http://www.bmoc.maths.org/home/bmo1-2002.pdf
    Thank you.
    I would say 0 is in the set of non-negative integers.

    But Wolfram Alpha disagrees with what they define as the non-negative integers: http://mathworld.wolfram.com/NonnegativeInteger.html
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    (Original post by demigawdz)
    do you know whether for q 5 0 counts as being in the set of ''non negative integers'' as the positive integer sign means positive lol (which doesnt include 0)/
    http://www.bmoc.maths.org/home/bmo1-2002.pdf
    Thank you.
    0 is indeed in the set of non-negative integers, although I always thought that Z+ was the set of positive integers.
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    (Original post by Renzhi10122)
    0 is indeed in the set of non-negative integers, although I always thought that Z+ was the set of positive integers.
    precisely why i am confuzzled by this
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    (Original post by demigawdz)
    precisely why i am confuzzled by this
    In the question, I guess its the set of non-negative integers then.
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    (Original post by Renzhi10122)
    In the question, I guess its the set of non-negative integers then.
    im so aweful at maths i cannot even do this one.
    lemme guess its probably an easy one
 
 
 
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