Why is...
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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
asked Dec 4 at 11:27
David K
1007
add a comment |
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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
asked Dec 4 at 11:27
David K
1007
mathworld.wolfram.com/WernerFormulas.html
– lab bhattacharjee
Dec 4 at 11:45
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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
asked Dec 4 at 11:27
David K
1007
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
asked Dec 4 at 11:27
David K
1007
asked Dec 4 at 11:27
David K
1007
asked Dec 4 at 11:27
David K
1007
asked Dec 4 at 11:27
David K
1007
asked Dec 4 at 11:27
David K
1007
1007
mathworld.wolfram.com/WernerFormulas.html
– lab bhattacharjee
Dec 4 at 11:45
add a comment |
mathworld.wolfram.com/WernerFormulas.html
– lab bhattacharjee
Dec 4 at 11:45
mathworld.wolfram.com/WernerFormulas.html
– lab bhattacharjee
Dec 4 at 11:45
mathworld.wolfram.com/WernerFormulas.html
– lab bhattacharjee
Dec 4 at 11:45
add a comment |
1 Answer
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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).
answered Dec 6 at 23:07
Markus Scheuer
59.7k455142
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
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).
answered Dec 6 at 23:07
Markus Scheuer
59.7k455142
add a comment |
up vote
1
down vote
accepted
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).
answered Dec 6 at 23:07
Markus Scheuer
59.7k455142
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
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).
answered Dec 6 at 23:07
Markus Scheuer
59.7k455142
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).
answered Dec 6 at 23:07
Markus Scheuer
59.7k455142
edited Dec 9 at 12:05
answered Dec 6 at 23:07
Markus Scheuer
59.7k455142
answered Dec 6 at 23:07
Markus Scheuer
59.7k455142
answered Dec 6 at 23:07
Markus Scheuer
59.7k455142
59.7k455142
add a comment |
add a comment |
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mathworld.wolfram.com/WernerFormulas.html
– lab bhattacharjee
Dec 4 at 11:45