Why is...












1












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Problem A5 in the 1985 Putnam Competition: Let $I_m=int_0^{2pi}cos(x)cos(2x)cdotscos(mx)dx$. For which integers $m$, $1leq mleq10$, do we have $I_mneq0$?



The solution rewrites $cos(x)=frac{e^{ikx}+e^{-ikx}}{2}$. It then says that $$I_m=int_0^{2pi}prod_{k=1}^m{biggl(frac{e^{ikx}+e^{-ikx}}{2}biggr)}=2^{-m}sum_{epsilon_{k}=pm1}int_0^{2pi}{e^{i(epsilon_1+2epsilon_2+cdots+mepsilon_m)x}}$$
Where the sum ranges over the $2^m$ $m$-tuples $(epsilon_1,ldots,epsilon_m)$ with $epsilon_k=pm1$ for every $k$.



My question is, how do you make sense of the last step? Why is this true:
$$prod_{k=1}^m{(e^{ikx}+e^{-ikx})}=sum_{epsilon_{k}=pm1}{e^{i(epsilon_1+2epsilon_2+cdots+mepsilon_m)x}}$$






products







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  • $begingroup$
    mathworld.wolfram.com/WernerFormulas.html
    $endgroup$
    – lab bhattacharjee
    Dec 4 '18 at 11:45
















1












$begingroup$


Problem A5 in the 1985 Putnam Competition: Let $I_m=int_0^{2pi}cos(x)cos(2x)cdotscos(mx)dx$. For which integers $m$, $1leq mleq10$, do we have $I_mneq0$?



The solution rewrites $cos(x)=frac{e^{ikx}+e^{-ikx}}{2}$. It then says that $$I_m=int_0^{2pi}prod_{k=1}^m{biggl(frac{e^{ikx}+e^{-ikx}}{2}biggr)}=2^{-m}sum_{epsilon_{k}=pm1}int_0^{2pi}{e^{i(epsilon_1+2epsilon_2+cdots+mepsilon_m)x}}$$
Where the sum ranges over the $2^m$ $m$-tuples $(epsilon_1,ldots,epsilon_m)$ with $epsilon_k=pm1$ for every $k$.



My question is, how do you make sense of the last step? Why is this true:
$$prod_{k=1}^m{(e^{ikx}+e^{-ikx})}=sum_{epsilon_{k}=pm1}{e^{i(epsilon_1+2epsilon_2+cdots+mepsilon_m)x}}$$






products







share|cite|improve this question









$endgroup$












  • $begingroup$
    mathworld.wolfram.com/WernerFormulas.html
    $endgroup$
    – lab bhattacharjee
    Dec 4 '18 at 11:45














1












1








1


1



$begingroup$


Problem A5 in the 1985 Putnam Competition: Let $I_m=int_0^{2pi}cos(x)cos(2x)cdotscos(mx)dx$. For which integers $m$, $1leq mleq10$, do we have $I_mneq0$?



The solution rewrites $cos(x)=frac{e^{ikx}+e^{-ikx}}{2}$. It then says that $$I_m=int_0^{2pi}prod_{k=1}^m{biggl(frac{e^{ikx}+e^{-ikx}}{2}biggr)}=2^{-m}sum_{epsilon_{k}=pm1}int_0^{2pi}{e^{i(epsilon_1+2epsilon_2+cdots+mepsilon_m)x}}$$
Where the sum ranges over the $2^m$ $m$-tuples $(epsilon_1,ldots,epsilon_m)$ with $epsilon_k=pm1$ for every $k$.



My question is, how do you make sense of the last step? Why is this true:
$$prod_{k=1}^m{(e^{ikx}+e^{-ikx})}=sum_{epsilon_{k}=pm1}{e^{i(epsilon_1+2epsilon_2+cdots+mepsilon_m)x}}$$






products







share|cite|improve this question









$endgroup$




Problem A5 in the 1985 Putnam Competition: Let $I_m=int_0^{2pi}cos(x)cos(2x)cdotscos(mx)dx$. For which integers $m$, $1leq mleq10$, do we have $I_mneq0$?



The solution rewrites $cos(x)=frac{e^{ikx}+e^{-ikx}}{2}$. It then says that $$I_m=int_0^{2pi}prod_{k=1}^m{biggl(frac{e^{ikx}+e^{-ikx}}{2}biggr)}=2^{-m}sum_{epsilon_{k}=pm1}int_0^{2pi}{e^{i(epsilon_1+2epsilon_2+cdots+mepsilon_m)x}}$$
Where the sum ranges over the $2^m$ $m$-tuples $(epsilon_1,ldots,epsilon_m)$ with $epsilon_k=pm1$ for every $k$.



My question is, how do you make sense of the last step? Why is this true:
$$prod_{k=1}^m{(e^{ikx}+e^{-ikx})}=sum_{epsilon_{k}=pm1}{e^{i(epsilon_1+2epsilon_2+cdots+mepsilon_m)x}}$$






products




products






share|cite|improve this question













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asked Dec 4 '18 at 11:27









David KDavid K

1007




1007












  • $begingroup$
    mathworld.wolfram.com/WernerFormulas.html
    $endgroup$
    – lab bhattacharjee
    Dec 4 '18 at 11:45


















  • $begingroup$
    mathworld.wolfram.com/WernerFormulas.html
    $endgroup$
    – lab bhattacharjee
    Dec 4 '18 at 11:45
















$begingroup$
mathworld.wolfram.com/WernerFormulas.html
$endgroup$
– lab bhattacharjee
Dec 4 '18 at 11:45




$begingroup$
mathworld.wolfram.com/WernerFormulas.html
$endgroup$
– lab bhattacharjee
Dec 4 '18 at 11:45










1 Answer
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$begingroup$


We obtain
begin{align*}
color{blue}{prod_{k=1}^m}&color{blue}{left(e^{ikx}+e^{-ikx}right)}tag{1}\
&=prod_{k=1}^mleft(sum_{varepsilon_{k}=pm 1}e^{ivarepsilon_{k}kx}right)\
&=left(sum_{varepsilon_{1}=pm 1}e^{ivarepsilon_1 x}right)left(sum_{varepsilon_{2}=pm 1}e^{ivarepsilon_22 x}right)
cdots left(sum_{varepsilon_{m}=pm 1}e^{ivarepsilon_mm x}right)\
&=sum_{varepsilon_{1}=pm 1}sum_{varepsilon_{2}=pm 1}cdots
sum_{varepsilon_{m}=pm 1}e^{ivarepsilon_1 x}e^{ivarepsilon_22 x}cdots e^{ivarepsilon_mm x}\
&=sum_{{1leq k leq m}atop{varepsilon_{k}=pm 1}}e^{ivarepsilon_1 x}e^{ivarepsilon_22 x}cdots e^{ivarepsilon_mm x}\
&,,color{blue}{=sum_{{1leq k leq m}atop{varepsilon_{k}=pm 1}}e^{ileft(varepsilon_1 +2varepsilon_2+cdots +m varepsilon_{m}right)x}}tag{2}
end{align*}
and the claim follows.




The product (1) consists of $m$ factors $e^{ikx}+e^{-ikx}$ where $1leq kleq m$. From each factor we select either $e^{ikx}$ or $e^{-ikx}$ giving a total of $2^m$ summands. These $2^m$ summands are explicitly stated in (2).






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    1 Answer
    1






    active

    oldest

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    active

    oldest

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    active

    oldest

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    1












    $begingroup$


    We obtain
    begin{align*}
    color{blue}{prod_{k=1}^m}&color{blue}{left(e^{ikx}+e^{-ikx}right)}tag{1}\
    &=prod_{k=1}^mleft(sum_{varepsilon_{k}=pm 1}e^{ivarepsilon_{k}kx}right)\
    &=left(sum_{varepsilon_{1}=pm 1}e^{ivarepsilon_1 x}right)left(sum_{varepsilon_{2}=pm 1}e^{ivarepsilon_22 x}right)
    cdots left(sum_{varepsilon_{m}=pm 1}e^{ivarepsilon_mm x}right)\
    &=sum_{varepsilon_{1}=pm 1}sum_{varepsilon_{2}=pm 1}cdots
    sum_{varepsilon_{m}=pm 1}e^{ivarepsilon_1 x}e^{ivarepsilon_22 x}cdots e^{ivarepsilon_mm x}\
    &=sum_{{1leq k leq m}atop{varepsilon_{k}=pm 1}}e^{ivarepsilon_1 x}e^{ivarepsilon_22 x}cdots e^{ivarepsilon_mm x}\
    &,,color{blue}{=sum_{{1leq k leq m}atop{varepsilon_{k}=pm 1}}e^{ileft(varepsilon_1 +2varepsilon_2+cdots +m varepsilon_{m}right)x}}tag{2}
    end{align*}
    and the claim follows.




    The product (1) consists of $m$ factors $e^{ikx}+e^{-ikx}$ where $1leq kleq m$. From each factor we select either $e^{ikx}$ or $e^{-ikx}$ giving a total of $2^m$ summands. These $2^m$ summands are explicitly stated in (2).






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$


      We obtain
      begin{align*}
      color{blue}{prod_{k=1}^m}&color{blue}{left(e^{ikx}+e^{-ikx}right)}tag{1}\
      &=prod_{k=1}^mleft(sum_{varepsilon_{k}=pm 1}e^{ivarepsilon_{k}kx}right)\
      &=left(sum_{varepsilon_{1}=pm 1}e^{ivarepsilon_1 x}right)left(sum_{varepsilon_{2}=pm 1}e^{ivarepsilon_22 x}right)
      cdots left(sum_{varepsilon_{m}=pm 1}e^{ivarepsilon_mm x}right)\
      &=sum_{varepsilon_{1}=pm 1}sum_{varepsilon_{2}=pm 1}cdots
      sum_{varepsilon_{m}=pm 1}e^{ivarepsilon_1 x}e^{ivarepsilon_22 x}cdots e^{ivarepsilon_mm x}\
      &=sum_{{1leq k leq m}atop{varepsilon_{k}=pm 1}}e^{ivarepsilon_1 x}e^{ivarepsilon_22 x}cdots e^{ivarepsilon_mm x}\
      &,,color{blue}{=sum_{{1leq k leq m}atop{varepsilon_{k}=pm 1}}e^{ileft(varepsilon_1 +2varepsilon_2+cdots +m varepsilon_{m}right)x}}tag{2}
      end{align*}
      and the claim follows.




      The product (1) consists of $m$ factors $e^{ikx}+e^{-ikx}$ where $1leq kleq m$. From each factor we select either $e^{ikx}$ or $e^{-ikx}$ giving a total of $2^m$ summands. These $2^m$ summands are explicitly stated in (2).






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$


        We obtain
        begin{align*}
        color{blue}{prod_{k=1}^m}&color{blue}{left(e^{ikx}+e^{-ikx}right)}tag{1}\
        &=prod_{k=1}^mleft(sum_{varepsilon_{k}=pm 1}e^{ivarepsilon_{k}kx}right)\
        &=left(sum_{varepsilon_{1}=pm 1}e^{ivarepsilon_1 x}right)left(sum_{varepsilon_{2}=pm 1}e^{ivarepsilon_22 x}right)
        cdots left(sum_{varepsilon_{m}=pm 1}e^{ivarepsilon_mm x}right)\
        &=sum_{varepsilon_{1}=pm 1}sum_{varepsilon_{2}=pm 1}cdots
        sum_{varepsilon_{m}=pm 1}e^{ivarepsilon_1 x}e^{ivarepsilon_22 x}cdots e^{ivarepsilon_mm x}\
        &=sum_{{1leq k leq m}atop{varepsilon_{k}=pm 1}}e^{ivarepsilon_1 x}e^{ivarepsilon_22 x}cdots e^{ivarepsilon_mm x}\
        &,,color{blue}{=sum_{{1leq k leq m}atop{varepsilon_{k}=pm 1}}e^{ileft(varepsilon_1 +2varepsilon_2+cdots +m varepsilon_{m}right)x}}tag{2}
        end{align*}
        and the claim follows.




        The product (1) consists of $m$ factors $e^{ikx}+e^{-ikx}$ where $1leq kleq m$. From each factor we select either $e^{ikx}$ or $e^{-ikx}$ giving a total of $2^m$ summands. These $2^m$ summands are explicitly stated in (2).






        share|cite|improve this answer











        $endgroup$




        We obtain
        begin{align*}
        color{blue}{prod_{k=1}^m}&color{blue}{left(e^{ikx}+e^{-ikx}right)}tag{1}\
        &=prod_{k=1}^mleft(sum_{varepsilon_{k}=pm 1}e^{ivarepsilon_{k}kx}right)\
        &=left(sum_{varepsilon_{1}=pm 1}e^{ivarepsilon_1 x}right)left(sum_{varepsilon_{2}=pm 1}e^{ivarepsilon_22 x}right)
        cdots left(sum_{varepsilon_{m}=pm 1}e^{ivarepsilon_mm x}right)\
        &=sum_{varepsilon_{1}=pm 1}sum_{varepsilon_{2}=pm 1}cdots
        sum_{varepsilon_{m}=pm 1}e^{ivarepsilon_1 x}e^{ivarepsilon_22 x}cdots e^{ivarepsilon_mm x}\
        &=sum_{{1leq k leq m}atop{varepsilon_{k}=pm 1}}e^{ivarepsilon_1 x}e^{ivarepsilon_22 x}cdots e^{ivarepsilon_mm x}\
        &,,color{blue}{=sum_{{1leq k leq m}atop{varepsilon_{k}=pm 1}}e^{ileft(varepsilon_1 +2varepsilon_2+cdots +m varepsilon_{m}right)x}}tag{2}
        end{align*}
        and the claim follows.




        The product (1) consists of $m$ factors $e^{ikx}+e^{-ikx}$ where $1leq kleq m$. From each factor we select either $e^{ikx}$ or $e^{-ikx}$ giving a total of $2^m$ summands. These $2^m$ summands are explicitly stated in (2).







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 9 '18 at 12:05

























        answered Dec 6 '18 at 23:07









        Markus ScheuerMarkus Scheuer

        61.2k456145




        61.2k456145






























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