Maximize $~~|sum_{n=1}^N a_n|^2-maxlimits_{1leq kleq N}|sum_{n=1}^N(-1)^{delta_{k,n}}a_n|^2~~$ subject to...
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I wanted to solve the following optimization problem, where $delta_{k,n}$ is a Kronecker delta:
begin{eqnarray*}
~\
~textrm{Maximize}~~~left|sumlimits_{n=1}^N a_nright|^2-maxlimits_{1leq kleq N}left|sumlimits_{n=1}^N(-1)^{delta_{k,n}}a_nright|^2\
textrm{subject to}~~sumlimits_{n=1}^{N}{{{left|a_nright|}^{2}}}=1,~~a_ninmathbb{C}, n=1,ldots N.\
end{eqnarray*}
I suspect that the maximum is reached for a uniform distribution of $a_ninmathbb{C}$, but I do not know where to start to prove it. For example, if ${{a}_{n}}=frac{1}{sqrt{N}}~forall n$, then $left|sum_{n=1}^N a_nright|^2=N$, $maxlimits_{1leq kleq N}left|sum_{n=1}^N(-1)^{delta_{k,n}}a_nright|^2=frac{(N-2)^2}{N}$, and $left|sum_{n=1}^N a_nright|^2-maxlimits_{1leq kleq N}left|sum_{n=1}^N(-1)^{delta_{k,n}}a_nright|^2=frac{4(N-1)}{N}$. Is this really a maximum if $a_ninmathbb{C}$? Any help is much appreciated. Any idea where I should start? Thank you.
calculus sequences-and-series analysis optimization
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I wanted to solve the following optimization problem, where $delta_{k,n}$ is a Kronecker delta:
begin{eqnarray*}
~\
~textrm{Maximize}~~~left|sumlimits_{n=1}^N a_nright|^2-maxlimits_{1leq kleq N}left|sumlimits_{n=1}^N(-1)^{delta_{k,n}}a_nright|^2\
textrm{subject to}~~sumlimits_{n=1}^{N}{{{left|a_nright|}^{2}}}=1,~~a_ninmathbb{C}, n=1,ldots N.\
end{eqnarray*}
I suspect that the maximum is reached for a uniform distribution of $a_ninmathbb{C}$, but I do not know where to start to prove it. For example, if ${{a}_{n}}=frac{1}{sqrt{N}}~forall n$, then $left|sum_{n=1}^N a_nright|^2=N$, $maxlimits_{1leq kleq N}left|sum_{n=1}^N(-1)^{delta_{k,n}}a_nright|^2=frac{(N-2)^2}{N}$, and $left|sum_{n=1}^N a_nright|^2-maxlimits_{1leq kleq N}left|sum_{n=1}^N(-1)^{delta_{k,n}}a_nright|^2=frac{4(N-1)}{N}$. Is this really a maximum if $a_ninmathbb{C}$? Any help is much appreciated. Any idea where I should start? Thank you.
calculus sequences-and-series analysis optimization
$endgroup$
add a comment |
$begingroup$
I wanted to solve the following optimization problem, where $delta_{k,n}$ is a Kronecker delta:
begin{eqnarray*}
~\
~textrm{Maximize}~~~left|sumlimits_{n=1}^N a_nright|^2-maxlimits_{1leq kleq N}left|sumlimits_{n=1}^N(-1)^{delta_{k,n}}a_nright|^2\
textrm{subject to}~~sumlimits_{n=1}^{N}{{{left|a_nright|}^{2}}}=1,~~a_ninmathbb{C}, n=1,ldots N.\
end{eqnarray*}
I suspect that the maximum is reached for a uniform distribution of $a_ninmathbb{C}$, but I do not know where to start to prove it. For example, if ${{a}_{n}}=frac{1}{sqrt{N}}~forall n$, then $left|sum_{n=1}^N a_nright|^2=N$, $maxlimits_{1leq kleq N}left|sum_{n=1}^N(-1)^{delta_{k,n}}a_nright|^2=frac{(N-2)^2}{N}$, and $left|sum_{n=1}^N a_nright|^2-maxlimits_{1leq kleq N}left|sum_{n=1}^N(-1)^{delta_{k,n}}a_nright|^2=frac{4(N-1)}{N}$. Is this really a maximum if $a_ninmathbb{C}$? Any help is much appreciated. Any idea where I should start? Thank you.
calculus sequences-and-series analysis optimization
$endgroup$
I wanted to solve the following optimization problem, where $delta_{k,n}$ is a Kronecker delta:
begin{eqnarray*}
~\
~textrm{Maximize}~~~left|sumlimits_{n=1}^N a_nright|^2-maxlimits_{1leq kleq N}left|sumlimits_{n=1}^N(-1)^{delta_{k,n}}a_nright|^2\
textrm{subject to}~~sumlimits_{n=1}^{N}{{{left|a_nright|}^{2}}}=1,~~a_ninmathbb{C}, n=1,ldots N.\
end{eqnarray*}
I suspect that the maximum is reached for a uniform distribution of $a_ninmathbb{C}$, but I do not know where to start to prove it. For example, if ${{a}_{n}}=frac{1}{sqrt{N}}~forall n$, then $left|sum_{n=1}^N a_nright|^2=N$, $maxlimits_{1leq kleq N}left|sum_{n=1}^N(-1)^{delta_{k,n}}a_nright|^2=frac{(N-2)^2}{N}$, and $left|sum_{n=1}^N a_nright|^2-maxlimits_{1leq kleq N}left|sum_{n=1}^N(-1)^{delta_{k,n}}a_nright|^2=frac{4(N-1)}{N}$. Is this really a maximum if $a_ninmathbb{C}$? Any help is much appreciated. Any idea where I should start? Thank you.
calculus sequences-and-series analysis optimization
calculus sequences-and-series analysis optimization
asked Dec 14 '18 at 5:36
HS TQHS TQ
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