Can a function always be represented as a product of two functions? [on hold]
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If the $g_i$ form a complete set in the given subspace of $f$ can than the function be represented such that
$$f(x, y) = sum_i g_i(x;y)h_i(y)
$$
where $g_i(x;y)$ depend explicitly on $x$ but only parametically on $y$?
functional-analysis numerical-linear-algebra
asked Nov 16 at 17:33
Aylin
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put on hold as unclear what you're asking by Hagen von Eitzen, Lord Shark the Unknown, Paul Frost, Scientifica, Rebellos Nov 17 at 13:58
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
1
down vote
favorite
If the $g_i$ form a complete set in the given subspace of $f$ can than the function be represented such that
$$f(x, y) = sum_i g_i(x;y)h_i(y)
$$
where $g_i(x;y)$ depend explicitly on $x$ but only parametically on $y$?
functional-analysis numerical-linear-algebra
asked Nov 16 at 17:33
Aylin
61
New contributor
Aylin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as unclear what you're asking by Hagen von Eitzen, Lord Shark the Unknown, Paul Frost, Scientifica, Rebellos Nov 17 at 13:58
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
It's really unclear to me what You are asking. What does "depend only parametrically" mean, what are $h_i$ for functions and through what set do the $i$'s run?
– Peter Melech
Nov 16 at 17:38
1
What does "depend explicitly on $x$ but only parametically on $y$" mean? Do you mean something like "for each value of $y$ the set ${g_i(x;y)}$ is a complete set of functions"?
– Winther
Nov 16 at 17:38
I am sorry if the question is unclear, I am no mathematician. @Winther: so yes. for each value of $y$ the $g_i(x;y)$ form a complete set such that if you fix $y$ for a given value $g_i(x;y)$ still depends explicitly on x. The $h_i$ would then be the coefficients in a sense. The $i$'s run through the natural numbers till infinity.
– Aylin
Nov 16 at 17:50
or even: is $f(x,y) = g(x;y)h(y)$ true? where $g(x;y)$ implies that the function is defined for each $y$ from a fixed a set ${y}$ such that it still depends on $x$?
– Aylin
Nov 16 at 18:16
add a comment |
up vote
1
down vote
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up vote
1
down vote
favorite
If the $g_i$ form a complete set in the given subspace of $f$ can than the function be represented such that
$$f(x, y) = sum_i g_i(x;y)h_i(y)
$$
where $g_i(x;y)$ depend explicitly on $x$ but only parametically on $y$?
functional-analysis numerical-linear-algebra
asked Nov 16 at 17:33
Aylin
61
New contributor
Aylin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
If the $g_i$ form a complete set in the given subspace of $f$ can than the function be represented such that
$$f(x, y) = sum_i g_i(x;y)h_i(y)
$$
where $g_i(x;y)$ depend explicitly on $x$ but only parametically on $y$?
functional-analysis numerical-linear-algebra
functional-analysis numerical-linear-algebra
asked Nov 16 at 17:33
Aylin
61
New contributor
Aylin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 16 at 17:33
Aylin
61
New contributor
Aylin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 16 at 17:54
asked Nov 16 at 17:33
Aylin
61
New contributor
Aylin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 16 at 17:33
Aylin
61
asked Nov 16 at 17:33
Aylin
61
61
New contributor
Aylin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Aylin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Aylin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as unclear what you're asking by Hagen von Eitzen, Lord Shark the Unknown, Paul Frost, Scientifica, Rebellos Nov 17 at 13:58
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as unclear what you're asking by Hagen von Eitzen, Lord Shark the Unknown, Paul Frost, Scientifica, Rebellos Nov 17 at 13:58
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
It's really unclear to me what You are asking. What does "depend only parametrically" mean, what are $h_i$ for functions and through what set do the $i$'s run?
– Peter Melech
Nov 16 at 17:38
1
What does "depend explicitly on $x$ but only parametically on $y$" mean? Do you mean something like "for each value of $y$ the set ${g_i(x;y)}$ is a complete set of functions"?
– Winther
Nov 16 at 17:38
I am sorry if the question is unclear, I am no mathematician. @Winther: so yes. for each value of $y$ the $g_i(x;y)$ form a complete set such that if you fix $y$ for a given value $g_i(x;y)$ still depends explicitly on x. The $h_i$ would then be the coefficients in a sense. The $i$'s run through the natural numbers till infinity.
– Aylin
Nov 16 at 17:50
or even: is $f(x,y) = g(x;y)h(y)$ true? where $g(x;y)$ implies that the function is defined for each $y$ from a fixed a set ${y}$ such that it still depends on $x$?
– Aylin
Nov 16 at 18:16
add a comment |
1
It's really unclear to me what You are asking. What does "depend only parametrically" mean, what are $h_i$ for functions and through what set do the $i$'s run?
– Peter Melech
Nov 16 at 17:38
1
What does "depend explicitly on $x$ but only parametically on $y$" mean? Do you mean something like "for each value of $y$ the set ${g_i(x;y)}$ is a complete set of functions"?
– Winther
Nov 16 at 17:38
I am sorry if the question is unclear, I am no mathematician. @Winther: so yes. for each value of $y$ the $g_i(x;y)$ form a complete set such that if you fix $y$ for a given value $g_i(x;y)$ still depends explicitly on x. The $h_i$ would then be the coefficients in a sense. The $i$'s run through the natural numbers till infinity.
– Aylin
Nov 16 at 17:50
or even: is $f(x,y) = g(x;y)h(y)$ true? where $g(x;y)$ implies that the function is defined for each $y$ from a fixed a set ${y}$ such that it still depends on $x$?
– Aylin
Nov 16 at 18:16
1
1
It's really unclear to me what You are asking. What does "depend only parametrically" mean, what are $h_i$ for functions and through what set do the $i$'s run?
– Peter Melech
Nov 16 at 17:38
It's really unclear to me what You are asking. What does "depend only parametrically" mean, what are $h_i$ for functions and through what set do the $i$'s run?
– Peter Melech
Nov 16 at 17:38
1
1
What does "depend explicitly on $x$ but only parametically on $y$" mean? Do you mean something like "for each value of $y$ the set ${g_i(x;y)}$ is a complete set of functions"?
– Winther
Nov 16 at 17:38
What does "depend explicitly on $x$ but only parametically on $y$" mean? Do you mean something like "for each value of $y$ the set ${g_i(x;y)}$ is a complete set of functions"?
– Winther
Nov 16 at 17:38
I am sorry if the question is unclear, I am no mathematician. @Winther: so yes. for each value of $y$ the $g_i(x;y)$ form a complete set such that if you fix $y$ for a given value $g_i(x;y)$ still depends explicitly on x. The $h_i$ would then be the coefficients in a sense. The $i$'s run through the natural numbers till infinity.
– Aylin
Nov 16 at 17:50
I am sorry if the question is unclear, I am no mathematician. @Winther: so yes. for each value of $y$ the $g_i(x;y)$ form a complete set such that if you fix $y$ for a given value $g_i(x;y)$ still depends explicitly on x. The $h_i$ would then be the coefficients in a sense. The $i$'s run through the natural numbers till infinity.
– Aylin
Nov 16 at 17:50
or even: is $f(x,y) = g(x;y)h(y)$ true? where $g(x;y)$ implies that the function is defined for each $y$ from a fixed a set ${y}$ such that it still depends on $x$?
– Aylin
Nov 16 at 18:16
or even: is $f(x,y) = g(x;y)h(y)$ true? where $g(x;y)$ implies that the function is defined for each $y$ from a fixed a set ${y}$ such that it still depends on $x$?
– Aylin
Nov 16 at 18:16
add a comment |
1 Answer
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If you have a function $f(x,y)$ that is square integrable on a rectangle $a le x le b$ and $c le y le d$, then you can expand $f$ in a double Fourier series, which gives an infinite sum of functions $sum a_j(x)b_k(y)$.
answered Nov 16 at 19:18
DisintegratingByParts
57.6k42376
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
votes
up vote
0
down vote
If you have a function $f(x,y)$ that is square integrable on a rectangle $a le x le b$ and $c le y le d$, then you can expand $f$ in a double Fourier series, which gives an infinite sum of functions $sum a_j(x)b_k(y)$.
answered Nov 16 at 19:18
DisintegratingByParts
57.6k42376
add a comment |
up vote
0
down vote
If you have a function $f(x,y)$ that is square integrable on a rectangle $a le x le b$ and $c le y le d$, then you can expand $f$ in a double Fourier series, which gives an infinite sum of functions $sum a_j(x)b_k(y)$.
answered Nov 16 at 19:18
DisintegratingByParts
57.6k42376
add a comment |
up vote
0
down vote
up vote
0
down vote
If you have a function $f(x,y)$ that is square integrable on a rectangle $a le x le b$ and $c le y le d$, then you can expand $f$ in a double Fourier series, which gives an infinite sum of functions $sum a_j(x)b_k(y)$.
answered Nov 16 at 19:18
DisintegratingByParts
57.6k42376
If you have a function $f(x,y)$ that is square integrable on a rectangle $a le x le b$ and $c le y le d$, then you can expand $f$ in a double Fourier series, which gives an infinite sum of functions $sum a_j(x)b_k(y)$.
answered Nov 16 at 19:18
DisintegratingByParts
57.6k42376
answered Nov 16 at 19:18
DisintegratingByParts
57.6k42376
answered Nov 16 at 19:18
DisintegratingByParts
57.6k42376
answered Nov 16 at 19:18
DisintegratingByParts
57.6k42376
57.6k42376
add a comment |
add a comment |
1
It's really unclear to me what You are asking. What does "depend only parametrically" mean, what are $h_i$ for functions and through what set do the $i$'s run?
– Peter Melech
Nov 16 at 17:38
1
What does "depend explicitly on $x$ but only parametically on $y$" mean? Do you mean something like "for each value of $y$ the set ${g_i(x;y)}$ is a complete set of functions"?
– Winther
Nov 16 at 17:38
I am sorry if the question is unclear, I am no mathematician. @Winther: so yes. for each value of $y$ the $g_i(x;y)$ form a complete set such that if you fix $y$ for a given value $g_i(x;y)$ still depends explicitly on x. The $h_i$ would then be the coefficients in a sense. The $i$'s run through the natural numbers till infinity.
– Aylin
Nov 16 at 17:50
or even: is $f(x,y) = g(x;y)h(y)$ true? where $g(x;y)$ implies that the function is defined for each $y$ from a fixed a set ${y}$ such that it still depends on $x$?
– Aylin
Nov 16 at 18:16