Looking for set of functions with special properties wrt integration.
$begingroup$
Good day,
I look for set of analytic (if possible elementary) functions with special properties.
Let $(a,b)$ be an interval $a<b$ and $x_n$ a sequence with
1) $x_{n+1}<x_n$ and
2) $x_n to a$ as $ntoinfty$
The sequence can be chosen any way one wants, some special way is OK for me. The set of functions $f_n(x)$ should satisfy:
$$int_a^{x_m}f_n(x) dx = delta_{m,n}$$
It is easy to do with non-analytic functions, e.g.
$f_n(x) = 1/(x_n-x_{n-1})$ for $x in(x_{n-1},x_n)$,
$f_n(x) = -1/(x_{n+1}-x_{n})$ for $x in(x_{n},x_{n+1})$
and zero elsewhere.
Thank you.
EDIT: Motivation
One often approximates a function $f$ by approximation $A_f$ by imposing an infinite number of constraints. Examples:
1) Taylor/Neumann series and Pade match derivatives:
$$frac{d^nf}{dx^n}=frac{d^n A_f}{dx^n},~n=1,2,3,dots$$
2) Fourier (like) series match integrals
$$ int sin(nx)f(x)dx= int sin(nx) A_f(x)dx,$$
$$ int cos(nx)f(x)dx= int cos(nx) A_f(x)dx$$
$$n=1,2,3,dots$$
and more examples could be found. Often, in combination with further restriction (smoothness, analyticity) the approximation is valid.
I try to combine integral and differential approach hoping for expansion
$$A_g(x)= sum c_n f_n(x)$$
$$c_n = int_a^{x_n} g(x)dx$$
I have no proof this expansion has to work, well it certainly works in some trivial cases ($g(x)=0$ or $g(x)=sum_k c_k f_k(x)$). I hope the class of function which can be approximated is much larger...
integration sequences-and-series functions
$endgroup$
add a comment |
$begingroup$
Good day,
I look for set of analytic (if possible elementary) functions with special properties.
Let $(a,b)$ be an interval $a<b$ and $x_n$ a sequence with
1) $x_{n+1}<x_n$ and
2) $x_n to a$ as $ntoinfty$
The sequence can be chosen any way one wants, some special way is OK for me. The set of functions $f_n(x)$ should satisfy:
$$int_a^{x_m}f_n(x) dx = delta_{m,n}$$
It is easy to do with non-analytic functions, e.g.
$f_n(x) = 1/(x_n-x_{n-1})$ for $x in(x_{n-1},x_n)$,
$f_n(x) = -1/(x_{n+1}-x_{n})$ for $x in(x_{n},x_{n+1})$
and zero elsewhere.
Thank you.
EDIT: Motivation
One often approximates a function $f$ by approximation $A_f$ by imposing an infinite number of constraints. Examples:
1) Taylor/Neumann series and Pade match derivatives:
$$frac{d^nf}{dx^n}=frac{d^n A_f}{dx^n},~n=1,2,3,dots$$
2) Fourier (like) series match integrals
$$ int sin(nx)f(x)dx= int sin(nx) A_f(x)dx,$$
$$ int cos(nx)f(x)dx= int cos(nx) A_f(x)dx$$
$$n=1,2,3,dots$$
and more examples could be found. Often, in combination with further restriction (smoothness, analyticity) the approximation is valid.
I try to combine integral and differential approach hoping for expansion
$$A_g(x)= sum c_n f_n(x)$$
$$c_n = int_a^{x_n} g(x)dx$$
I have no proof this expansion has to work, well it certainly works in some trivial cases ($g(x)=0$ or $g(x)=sum_k c_k f_k(x)$). I hope the class of function which can be approximated is much larger...
integration sequences-and-series functions
$endgroup$
$begingroup$
It could be useful to add some context. Where did this problem originate from? What application are the solutions going to have?
$endgroup$
– Yuriy S
Feb 21 '18 at 12:02
$begingroup$
I have in mind new kind of expansion of a function $g$. Let $c_n=int_a^{x_n} g(x) dx$. Then, presumably, this function could be approximated by $g(x) approx sum_n c_n f_n(x)$.
$endgroup$
– F. Jatpil
Feb 21 '18 at 12:55
$begingroup$
I don't really get how that would work, but maybe you could add this to the post, as it is relevant to the question and might attract more attention
$endgroup$
– Yuriy S
Feb 21 '18 at 15:10
$begingroup$
I added and edit... :]
$endgroup$
– F. Jatpil
Feb 22 '18 at 7:56
add a comment |
$begingroup$
Good day,
I look for set of analytic (if possible elementary) functions with special properties.
Let $(a,b)$ be an interval $a<b$ and $x_n$ a sequence with
1) $x_{n+1}<x_n$ and
2) $x_n to a$ as $ntoinfty$
The sequence can be chosen any way one wants, some special way is OK for me. The set of functions $f_n(x)$ should satisfy:
$$int_a^{x_m}f_n(x) dx = delta_{m,n}$$
It is easy to do with non-analytic functions, e.g.
$f_n(x) = 1/(x_n-x_{n-1})$ for $x in(x_{n-1},x_n)$,
$f_n(x) = -1/(x_{n+1}-x_{n})$ for $x in(x_{n},x_{n+1})$
and zero elsewhere.
Thank you.
EDIT: Motivation
One often approximates a function $f$ by approximation $A_f$ by imposing an infinite number of constraints. Examples:
1) Taylor/Neumann series and Pade match derivatives:
$$frac{d^nf}{dx^n}=frac{d^n A_f}{dx^n},~n=1,2,3,dots$$
2) Fourier (like) series match integrals
$$ int sin(nx)f(x)dx= int sin(nx) A_f(x)dx,$$
$$ int cos(nx)f(x)dx= int cos(nx) A_f(x)dx$$
$$n=1,2,3,dots$$
and more examples could be found. Often, in combination with further restriction (smoothness, analyticity) the approximation is valid.
I try to combine integral and differential approach hoping for expansion
$$A_g(x)= sum c_n f_n(x)$$
$$c_n = int_a^{x_n} g(x)dx$$
I have no proof this expansion has to work, well it certainly works in some trivial cases ($g(x)=0$ or $g(x)=sum_k c_k f_k(x)$). I hope the class of function which can be approximated is much larger...
integration sequences-and-series functions
$endgroup$
Good day,
I look for set of analytic (if possible elementary) functions with special properties.
Let $(a,b)$ be an interval $a<b$ and $x_n$ a sequence with
1) $x_{n+1}<x_n$ and
2) $x_n to a$ as $ntoinfty$
The sequence can be chosen any way one wants, some special way is OK for me. The set of functions $f_n(x)$ should satisfy:
$$int_a^{x_m}f_n(x) dx = delta_{m,n}$$
It is easy to do with non-analytic functions, e.g.
$f_n(x) = 1/(x_n-x_{n-1})$ for $x in(x_{n-1},x_n)$,
$f_n(x) = -1/(x_{n+1}-x_{n})$ for $x in(x_{n},x_{n+1})$
and zero elsewhere.
Thank you.
EDIT: Motivation
One often approximates a function $f$ by approximation $A_f$ by imposing an infinite number of constraints. Examples:
1) Taylor/Neumann series and Pade match derivatives:
$$frac{d^nf}{dx^n}=frac{d^n A_f}{dx^n},~n=1,2,3,dots$$
2) Fourier (like) series match integrals
$$ int sin(nx)f(x)dx= int sin(nx) A_f(x)dx,$$
$$ int cos(nx)f(x)dx= int cos(nx) A_f(x)dx$$
$$n=1,2,3,dots$$
and more examples could be found. Often, in combination with further restriction (smoothness, analyticity) the approximation is valid.
I try to combine integral and differential approach hoping for expansion
$$A_g(x)= sum c_n f_n(x)$$
$$c_n = int_a^{x_n} g(x)dx$$
I have no proof this expansion has to work, well it certainly works in some trivial cases ($g(x)=0$ or $g(x)=sum_k c_k f_k(x)$). I hope the class of function which can be approximated is much larger...
integration sequences-and-series functions
integration sequences-and-series functions
edited Dec 1 '18 at 19:19
Mostafa Ayaz
15.6k3939
15.6k3939
asked Feb 21 '18 at 11:50
F. JatpilF. Jatpil
1347
1347
$begingroup$
It could be useful to add some context. Where did this problem originate from? What application are the solutions going to have?
$endgroup$
– Yuriy S
Feb 21 '18 at 12:02
$begingroup$
I have in mind new kind of expansion of a function $g$. Let $c_n=int_a^{x_n} g(x) dx$. Then, presumably, this function could be approximated by $g(x) approx sum_n c_n f_n(x)$.
$endgroup$
– F. Jatpil
Feb 21 '18 at 12:55
$begingroup$
I don't really get how that would work, but maybe you could add this to the post, as it is relevant to the question and might attract more attention
$endgroup$
– Yuriy S
Feb 21 '18 at 15:10
$begingroup$
I added and edit... :]
$endgroup$
– F. Jatpil
Feb 22 '18 at 7:56
add a comment |
$begingroup$
It could be useful to add some context. Where did this problem originate from? What application are the solutions going to have?
$endgroup$
– Yuriy S
Feb 21 '18 at 12:02
$begingroup$
I have in mind new kind of expansion of a function $g$. Let $c_n=int_a^{x_n} g(x) dx$. Then, presumably, this function could be approximated by $g(x) approx sum_n c_n f_n(x)$.
$endgroup$
– F. Jatpil
Feb 21 '18 at 12:55
$begingroup$
I don't really get how that would work, but maybe you could add this to the post, as it is relevant to the question and might attract more attention
$endgroup$
– Yuriy S
Feb 21 '18 at 15:10
$begingroup$
I added and edit... :]
$endgroup$
– F. Jatpil
Feb 22 '18 at 7:56
$begingroup$
It could be useful to add some context. Where did this problem originate from? What application are the solutions going to have?
$endgroup$
– Yuriy S
Feb 21 '18 at 12:02
$begingroup$
It could be useful to add some context. Where did this problem originate from? What application are the solutions going to have?
$endgroup$
– Yuriy S
Feb 21 '18 at 12:02
$begingroup$
I have in mind new kind of expansion of a function $g$. Let $c_n=int_a^{x_n} g(x) dx$. Then, presumably, this function could be approximated by $g(x) approx sum_n c_n f_n(x)$.
$endgroup$
– F. Jatpil
Feb 21 '18 at 12:55
$begingroup$
I have in mind new kind of expansion of a function $g$. Let $c_n=int_a^{x_n} g(x) dx$. Then, presumably, this function could be approximated by $g(x) approx sum_n c_n f_n(x)$.
$endgroup$
– F. Jatpil
Feb 21 '18 at 12:55
$begingroup$
I don't really get how that would work, but maybe you could add this to the post, as it is relevant to the question and might attract more attention
$endgroup$
– Yuriy S
Feb 21 '18 at 15:10
$begingroup$
I don't really get how that would work, but maybe you could add this to the post, as it is relevant to the question and might attract more attention
$endgroup$
– Yuriy S
Feb 21 '18 at 15:10
$begingroup$
I added and edit... :]
$endgroup$
– F. Jatpil
Feb 22 '18 at 7:56
$begingroup$
I added and edit... :]
$endgroup$
– F. Jatpil
Feb 22 '18 at 7:56
add a comment |
1 Answer
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Well, it seems I have found such functions, but approximation is not working. Here is logical method of construction.
1) Start by finding the anti-derivative (integral) $F$ of $f$ (i.e. $F'=f$) on the interval $(0,1)$. A good starting point is
$$F(x)=sin(pi/x)$$
and take its crossing with $y=0$ zero as $x_n$, i.e.
$$x_n = 1/n.$$
2) Because this is anti-derivative, you want the value to be defined at $x=0$, the lower limit of the interval. Thus you make the modification
$$F(x)=x ~sin(pi/x).$$
This does not change its crossing points with zero, and can be evaluated at upper ($x=x_n$) and lower ($x=0$) boundary of each integral (at least as limit)
$$ int_0^{x_n} f(x)dx = [F(x)]^{x_n}_0 = 0-0 = 0$$
3) So far so good. Now you need $F$ to get at one crossing not the value 0 but 1! To do this you define
$$F_n(x)=frac{x~sin(pi/x)}{(-1)^{(n+1)}npi(x-1/n)}.$$
With this construction all zeros are kept except the zero at $x=1/n=x_n$ where the value (well, the limit) is 1.
The set of function I looked for then can be written as:
$$f_n(x)= frac{d}{dx}left[ frac{x~sin(pi/x)}{(-1)^{(n+1)}npi(x-1/n)} right]$$
which, I do not write explicitly because of length but can easily be done.
Conclusion: Trying the approximation I see it fails. It gets the integral correct thanks to wild oscillations, but even a large number of terms does not force the approximant to approximate the function.
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Well, it seems I have found such functions, but approximation is not working. Here is logical method of construction.
1) Start by finding the anti-derivative (integral) $F$ of $f$ (i.e. $F'=f$) on the interval $(0,1)$. A good starting point is
$$F(x)=sin(pi/x)$$
and take its crossing with $y=0$ zero as $x_n$, i.e.
$$x_n = 1/n.$$
2) Because this is anti-derivative, you want the value to be defined at $x=0$, the lower limit of the interval. Thus you make the modification
$$F(x)=x ~sin(pi/x).$$
This does not change its crossing points with zero, and can be evaluated at upper ($x=x_n$) and lower ($x=0$) boundary of each integral (at least as limit)
$$ int_0^{x_n} f(x)dx = [F(x)]^{x_n}_0 = 0-0 = 0$$
3) So far so good. Now you need $F$ to get at one crossing not the value 0 but 1! To do this you define
$$F_n(x)=frac{x~sin(pi/x)}{(-1)^{(n+1)}npi(x-1/n)}.$$
With this construction all zeros are kept except the zero at $x=1/n=x_n$ where the value (well, the limit) is 1.
The set of function I looked for then can be written as:
$$f_n(x)= frac{d}{dx}left[ frac{x~sin(pi/x)}{(-1)^{(n+1)}npi(x-1/n)} right]$$
which, I do not write explicitly because of length but can easily be done.
Conclusion: Trying the approximation I see it fails. It gets the integral correct thanks to wild oscillations, but even a large number of terms does not force the approximant to approximate the function.
$endgroup$
add a comment |
$begingroup$
Well, it seems I have found such functions, but approximation is not working. Here is logical method of construction.
1) Start by finding the anti-derivative (integral) $F$ of $f$ (i.e. $F'=f$) on the interval $(0,1)$. A good starting point is
$$F(x)=sin(pi/x)$$
and take its crossing with $y=0$ zero as $x_n$, i.e.
$$x_n = 1/n.$$
2) Because this is anti-derivative, you want the value to be defined at $x=0$, the lower limit of the interval. Thus you make the modification
$$F(x)=x ~sin(pi/x).$$
This does not change its crossing points with zero, and can be evaluated at upper ($x=x_n$) and lower ($x=0$) boundary of each integral (at least as limit)
$$ int_0^{x_n} f(x)dx = [F(x)]^{x_n}_0 = 0-0 = 0$$
3) So far so good. Now you need $F$ to get at one crossing not the value 0 but 1! To do this you define
$$F_n(x)=frac{x~sin(pi/x)}{(-1)^{(n+1)}npi(x-1/n)}.$$
With this construction all zeros are kept except the zero at $x=1/n=x_n$ where the value (well, the limit) is 1.
The set of function I looked for then can be written as:
$$f_n(x)= frac{d}{dx}left[ frac{x~sin(pi/x)}{(-1)^{(n+1)}npi(x-1/n)} right]$$
which, I do not write explicitly because of length but can easily be done.
Conclusion: Trying the approximation I see it fails. It gets the integral correct thanks to wild oscillations, but even a large number of terms does not force the approximant to approximate the function.
$endgroup$
add a comment |
$begingroup$
Well, it seems I have found such functions, but approximation is not working. Here is logical method of construction.
1) Start by finding the anti-derivative (integral) $F$ of $f$ (i.e. $F'=f$) on the interval $(0,1)$. A good starting point is
$$F(x)=sin(pi/x)$$
and take its crossing with $y=0$ zero as $x_n$, i.e.
$$x_n = 1/n.$$
2) Because this is anti-derivative, you want the value to be defined at $x=0$, the lower limit of the interval. Thus you make the modification
$$F(x)=x ~sin(pi/x).$$
This does not change its crossing points with zero, and can be evaluated at upper ($x=x_n$) and lower ($x=0$) boundary of each integral (at least as limit)
$$ int_0^{x_n} f(x)dx = [F(x)]^{x_n}_0 = 0-0 = 0$$
3) So far so good. Now you need $F$ to get at one crossing not the value 0 but 1! To do this you define
$$F_n(x)=frac{x~sin(pi/x)}{(-1)^{(n+1)}npi(x-1/n)}.$$
With this construction all zeros are kept except the zero at $x=1/n=x_n$ where the value (well, the limit) is 1.
The set of function I looked for then can be written as:
$$f_n(x)= frac{d}{dx}left[ frac{x~sin(pi/x)}{(-1)^{(n+1)}npi(x-1/n)} right]$$
which, I do not write explicitly because of length but can easily be done.
Conclusion: Trying the approximation I see it fails. It gets the integral correct thanks to wild oscillations, but even a large number of terms does not force the approximant to approximate the function.
$endgroup$
Well, it seems I have found such functions, but approximation is not working. Here is logical method of construction.
1) Start by finding the anti-derivative (integral) $F$ of $f$ (i.e. $F'=f$) on the interval $(0,1)$. A good starting point is
$$F(x)=sin(pi/x)$$
and take its crossing with $y=0$ zero as $x_n$, i.e.
$$x_n = 1/n.$$
2) Because this is anti-derivative, you want the value to be defined at $x=0$, the lower limit of the interval. Thus you make the modification
$$F(x)=x ~sin(pi/x).$$
This does not change its crossing points with zero, and can be evaluated at upper ($x=x_n$) and lower ($x=0$) boundary of each integral (at least as limit)
$$ int_0^{x_n} f(x)dx = [F(x)]^{x_n}_0 = 0-0 = 0$$
3) So far so good. Now you need $F$ to get at one crossing not the value 0 but 1! To do this you define
$$F_n(x)=frac{x~sin(pi/x)}{(-1)^{(n+1)}npi(x-1/n)}.$$
With this construction all zeros are kept except the zero at $x=1/n=x_n$ where the value (well, the limit) is 1.
The set of function I looked for then can be written as:
$$f_n(x)= frac{d}{dx}left[ frac{x~sin(pi/x)}{(-1)^{(n+1)}npi(x-1/n)} right]$$
which, I do not write explicitly because of length but can easily be done.
Conclusion: Trying the approximation I see it fails. It gets the integral correct thanks to wild oscillations, but even a large number of terms does not force the approximant to approximate the function.
answered Feb 22 '18 at 12:44
F. JatpilF. Jatpil
1347
1347
add a comment |
add a comment |
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$begingroup$
It could be useful to add some context. Where did this problem originate from? What application are the solutions going to have?
$endgroup$
– Yuriy S
Feb 21 '18 at 12:02
$begingroup$
I have in mind new kind of expansion of a function $g$. Let $c_n=int_a^{x_n} g(x) dx$. Then, presumably, this function could be approximated by $g(x) approx sum_n c_n f_n(x)$.
$endgroup$
– F. Jatpil
Feb 21 '18 at 12:55
$begingroup$
I don't really get how that would work, but maybe you could add this to the post, as it is relevant to the question and might attract more attention
$endgroup$
– Yuriy S
Feb 21 '18 at 15:10
$begingroup$
I added and edit... :]
$endgroup$
– F. Jatpil
Feb 22 '18 at 7:56