Minimize $int_{-pi}^{pi}|x-a-be^{3ix}-c^{5ix}|^2dx$ for $a,b,c in mathbb{R}$
Minimize the term $int_{-pi}^{pi}|x-a-be^{3ix}-c^{5ix}|^2dx$ for $a,b,c in mathbb{R}$
Thoughts-
So it's a rather common problem, i know that if $ {1, e^{3ix}, e^{5ix} } $ span a vector space, then the minimum of the term is the minimum of $ || x -g|| $ for $g in span{1, e^{3ix}, e^{5ix} }$. therefore, the minimum will be the projection of $x$ onto $span{1, e^{3ix}, e^{5ix} }$. That is -
$a = dfrac{<x,1>}{||1||^2} , b = dfrac{<x,e^{3ix}>}{||e^{3ix}||^2} , c = dfrac{<x,e^{5ix}>}{||e^{5ix}||^2}$
Is it right? what do you think?
harmonic-analysis
asked Dec 26 '18 at 21:59
Yariv Levy
7572416
add a comment |
Minimize the term $int_{-pi}^{pi}|x-a-be^{3ix}-c^{5ix}|^2dx$ for $a,b,c in mathbb{R}$
Thoughts-
So it's a rather common problem, i know that if $ {1, e^{3ix}, e^{5ix} } $ span a vector space, then the minimum of the term is the minimum of $ || x -g|| $ for $g in span{1, e^{3ix}, e^{5ix} }$. therefore, the minimum will be the projection of $x$ onto $span{1, e^{3ix}, e^{5ix} }$. That is -
$a = dfrac{<x,1>}{||1||^2} , b = dfrac{<x,e^{3ix}>}{||e^{3ix}||^2} , c = dfrac{<x,e^{5ix}>}{||e^{5ix}||^2}$
Is it right? what do you think?
harmonic-analysis
asked Dec 26 '18 at 21:59
Yariv Levy
7572416
Your intuition is totally correct. Just check your calculations if that's the case.
– Rebellos
Dec 26 '18 at 22:00
Thanks. just one question, isn't $||e^{3ix}||^2$, $||e^{5ix}||^2$ $ = 0 $ ?
– Yariv Levy
Dec 26 '18 at 22:17
Norm of a nonzero function cannot be $0$. $int_{-pi}^{pi}|e^{3ix}|^{2}dx=2pi$.
– Kavi Rama Murthy
Dec 26 '18 at 23:41
add a comment |
Minimize the term $int_{-pi}^{pi}|x-a-be^{3ix}-c^{5ix}|^2dx$ for $a,b,c in mathbb{R}$
Thoughts-
So it's a rather common problem, i know that if $ {1, e^{3ix}, e^{5ix} } $ span a vector space, then the minimum of the term is the minimum of $ || x -g|| $ for $g in span{1, e^{3ix}, e^{5ix} }$. therefore, the minimum will be the projection of $x$ onto $span{1, e^{3ix}, e^{5ix} }$. That is -
$a = dfrac{<x,1>}{||1||^2} , b = dfrac{<x,e^{3ix}>}{||e^{3ix}||^2} , c = dfrac{<x,e^{5ix}>}{||e^{5ix}||^2}$
Is it right? what do you think?
harmonic-analysis
asked Dec 26 '18 at 21:59
Yariv Levy
7572416
Minimize the term $int_{-pi}^{pi}|x-a-be^{3ix}-c^{5ix}|^2dx$ for $a,b,c in mathbb{R}$
Thoughts-
So it's a rather common problem, i know that if $ {1, e^{3ix}, e^{5ix} } $ span a vector space, then the minimum of the term is the minimum of $ || x -g|| $ for $g in span{1, e^{3ix}, e^{5ix} }$. therefore, the minimum will be the projection of $x$ onto $span{1, e^{3ix}, e^{5ix} }$. That is -
$a = dfrac{<x,1>}{||1||^2} , b = dfrac{<x,e^{3ix}>}{||e^{3ix}||^2} , c = dfrac{<x,e^{5ix}>}{||e^{5ix}||^2}$
Is it right? what do you think?
harmonic-analysis
harmonic-analysis
asked Dec 26 '18 at 21:59
Yariv Levy
7572416
asked Dec 26 '18 at 21:59
Yariv Levy
7572416
asked Dec 26 '18 at 21:59
Yariv Levy
7572416
asked Dec 26 '18 at 21:59
Yariv Levy
7572416
asked Dec 26 '18 at 21:59
Yariv Levy
7572416
7572416
Your intuition is totally correct. Just check your calculations if that's the case.
– Rebellos
Dec 26 '18 at 22:00
Thanks. just one question, isn't $||e^{3ix}||^2$, $||e^{5ix}||^2$ $ = 0 $ ?
– Yariv Levy
Dec 26 '18 at 22:17
Norm of a nonzero function cannot be $0$. $int_{-pi}^{pi}|e^{3ix}|^{2}dx=2pi$.
– Kavi Rama Murthy
Dec 26 '18 at 23:41
add a comment |
Your intuition is totally correct. Just check your calculations if that's the case.
– Rebellos
Dec 26 '18 at 22:00
Thanks. just one question, isn't $||e^{3ix}||^2$, $||e^{5ix}||^2$ $ = 0 $ ?
– Yariv Levy
Dec 26 '18 at 22:17
Norm of a nonzero function cannot be $0$. $int_{-pi}^{pi}|e^{3ix}|^{2}dx=2pi$.
– Kavi Rama Murthy
Dec 26 '18 at 23:41
Your intuition is totally correct. Just check your calculations if that's the case.
– Rebellos
Dec 26 '18 at 22:00
Your intuition is totally correct. Just check your calculations if that's the case.
– Rebellos
Dec 26 '18 at 22:00
Thanks. just one question, isn't $||e^{3ix}||^2$, $||e^{5ix}||^2$ $ = 0 $ ?
– Yariv Levy
Dec 26 '18 at 22:17
Thanks. just one question, isn't $||e^{3ix}||^2$, $||e^{5ix}||^2$ $ = 0 $ ?
– Yariv Levy
Dec 26 '18 at 22:17
Norm of a nonzero function cannot be $0$. $int_{-pi}^{pi}|e^{3ix}|^{2}dx=2pi$.
– Kavi Rama Murthy
Dec 26 '18 at 23:41
Norm of a nonzero function cannot be $0$. $int_{-pi}^{pi}|e^{3ix}|^{2}dx=2pi$.
– Kavi Rama Murthy
Dec 26 '18 at 23:41
add a comment |
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Your intuition is totally correct. Just check your calculations if that's the case.
– Rebellos
Dec 26 '18 at 22:00
Thanks. just one question, isn't $||e^{3ix}||^2$, $||e^{5ix}||^2$ $ = 0 $ ?
– Yariv Levy
Dec 26 '18 at 22:17
Norm of a nonzero function cannot be $0$. $int_{-pi}^{pi}|e^{3ix}|^{2}dx=2pi$.
– Kavi Rama Murthy
Dec 26 '18 at 23:41