Analytic “Lagrange” interpolation for a countably infinite set of points?Interpolation with entire functionSpecifying a holomorphic function by a sequence of valuesDoes for any sequence $(a_n)$ exist a power series $f$ such that $f(i)=a_i$?lagrange interpolation, polynomial of degree $2n-1$Lagrange interpolation not working?Any differences between Lagrange polynomial on Chebyshev points and Chebyshev polynomial?Lagrange interpolation with zeros at $pm infty$Polynomial interpolation with data points from derivative of original polynomialLagrangian interpolation: degree of the interpolation polynomialLagrangian interpolation at Chebyshev points - estimate on coefficients in monomic basisLagrange interpolation with multiplicitiesAbel Ruffini theorem for Lagrangian interpolation?Who knows this formula for polynomial interpolation?
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Analytic “Lagrange” interpolation for a countably infinite set of points?
Interpolation with entire functionSpecifying a holomorphic function by a sequence of valuesDoes for any sequence $(a_n)$ exist a power series $f$ such that $f(i)=a_i$?lagrange interpolation, polynomial of degree $2n-1$Lagrange interpolation not working?Any differences between Lagrange polynomial on Chebyshev points and Chebyshev polynomial?Lagrange interpolation with zeros at $pm infty$Polynomial interpolation with data points from derivative of original polynomialLagrangian interpolation: degree of the interpolation polynomialLagrangian interpolation at Chebyshev points - estimate on coefficients in monomic basisLagrange interpolation with multiplicitiesAbel Ruffini theorem for Lagrangian interpolation?Who knows this formula for polynomial interpolation?
$begingroup$
Suppose I have a finite set of points on the real plane, and I want to find the univariate polynomial interpolating all of them. Lagrange interpolation gives me the least-degree polynomial going through all of those.
Is there an analogous construct for a countably infinite, sparse set of points on the real plane, instead using analytic functions and power series?
There is obviously some difficulty in forming a perfect analogy, as Lagrange interpolation yields the "lowest degree" polynomial interpolating the points, whereas there is no such thing as a "lowest degree" power series. However, perhaps there is some generalized measure of the complexity of a power series that is decently workable, and which restricts to the lowest-degree polynomial in the finite case.
If so, how does this work? Is there an easy way to obtain the nth coefficient of the power series from the points?
real-analysis analytic-functions lagrange-interpolation interpolation-theory
asked Jul 20 '17 at 16:13
Mike BattagliaMike Battaglia
1,5421128
$endgroup$
add a comment |
$begingroup$
Suppose I have a finite set of points on the real plane, and I want to find the univariate polynomial interpolating all of them. Lagrange interpolation gives me the least-degree polynomial going through all of those.
Is there an analogous construct for a countably infinite, sparse set of points on the real plane, instead using analytic functions and power series?
There is obviously some difficulty in forming a perfect analogy, as Lagrange interpolation yields the "lowest degree" polynomial interpolating the points, whereas there is no such thing as a "lowest degree" power series. However, perhaps there is some generalized measure of the complexity of a power series that is decently workable, and which restricts to the lowest-degree polynomial in the finite case.
If so, how does this work? Is there an easy way to obtain the nth coefficient of the power series from the points?
real-analysis analytic-functions lagrange-interpolation interpolation-theory
asked Jul 20 '17 at 16:13
Mike BattagliaMike Battaglia
1,5421128
$endgroup$
2
$begingroup$
Your countably infinite sparse set $A$ of points needs to be so sparse that no sequence of distinct points from $A$ has a finite limit. In that case, there is an entire (i.e., analytic in the whole plane) interpolant. That follows from theorems of Weierstrass and Mittag-Leffler (the former gives an entire function that vanishes at exactly the points in $A$, and the latter gives a function with poles at exactly the points of $A$ and with prescribed principal parts at the poles).
$endgroup$
– Andreas Blass
Jul 20 '17 at 17:17
$begingroup$
do you know about Runge's phenomenon ?
$endgroup$
– G Cab
Jul 20 '17 at 17:24
$begingroup$
So suppose you have a function which is 0 at every nonzero integer, and 1 at x=0. Is there no way to interpolate this and get something like the sinc function?
$endgroup$
– Mike Battaglia
Jul 20 '17 at 23:00
$begingroup$
Never mind, misinterpreted - thanks
$endgroup$
– Mike Battaglia
Jul 21 '17 at 16:26
add a comment |
$begingroup$
Suppose I have a finite set of points on the real plane, and I want to find the univariate polynomial interpolating all of them. Lagrange interpolation gives me the least-degree polynomial going through all of those.
Is there an analogous construct for a countably infinite, sparse set of points on the real plane, instead using analytic functions and power series?
There is obviously some difficulty in forming a perfect analogy, as Lagrange interpolation yields the "lowest degree" polynomial interpolating the points, whereas there is no such thing as a "lowest degree" power series. However, perhaps there is some generalized measure of the complexity of a power series that is decently workable, and which restricts to the lowest-degree polynomial in the finite case.
If so, how does this work? Is there an easy way to obtain the nth coefficient of the power series from the points?
real-analysis analytic-functions lagrange-interpolation interpolation-theory
asked Jul 20 '17 at 16:13
Mike BattagliaMike Battaglia
1,5421128
$endgroup$
Suppose I have a finite set of points on the real plane, and I want to find the univariate polynomial interpolating all of them. Lagrange interpolation gives me the least-degree polynomial going through all of those.
Is there an analogous construct for a countably infinite, sparse set of points on the real plane, instead using analytic functions and power series?
There is obviously some difficulty in forming a perfect analogy, as Lagrange interpolation yields the "lowest degree" polynomial interpolating the points, whereas there is no such thing as a "lowest degree" power series. However, perhaps there is some generalized measure of the complexity of a power series that is decently workable, and which restricts to the lowest-degree polynomial in the finite case.
If so, how does this work? Is there an easy way to obtain the nth coefficient of the power series from the points?
real-analysis analytic-functions lagrange-interpolation interpolation-theory
real-analysis analytic-functions lagrange-interpolation interpolation-theory
asked Jul 20 '17 at 16:13
Mike BattagliaMike Battaglia
1,5421128
asked Jul 20 '17 at 16:13
Mike BattagliaMike Battaglia
1,5421128
edited Jul 20 '17 at 17:20
Mike Battaglia
asked Jul 20 '17 at 16:13
Mike BattagliaMike Battaglia
1,5421128
asked Jul 20 '17 at 16:13
Mike BattagliaMike Battaglia
1,5421128
asked Jul 20 '17 at 16:13
Mike BattagliaMike Battaglia
1,5421128
1,5421128
2
$begingroup$
Your countably infinite sparse set $A$ of points needs to be so sparse that no sequence of distinct points from $A$ has a finite limit. In that case, there is an entire (i.e., analytic in the whole plane) interpolant. That follows from theorems of Weierstrass and Mittag-Leffler (the former gives an entire function that vanishes at exactly the points in $A$, and the latter gives a function with poles at exactly the points of $A$ and with prescribed principal parts at the poles).
$endgroup$
– Andreas Blass
Jul 20 '17 at 17:17
$begingroup$
do you know about Runge's phenomenon ?
$endgroup$
– G Cab
Jul 20 '17 at 17:24
$begingroup$
So suppose you have a function which is 0 at every nonzero integer, and 1 at x=0. Is there no way to interpolate this and get something like the sinc function?
$endgroup$
– Mike Battaglia
Jul 20 '17 at 23:00
$begingroup$
Never mind, misinterpreted - thanks
$endgroup$
– Mike Battaglia
Jul 21 '17 at 16:26
add a comment |
2
$begingroup$
Your countably infinite sparse set $A$ of points needs to be so sparse that no sequence of distinct points from $A$ has a finite limit. In that case, there is an entire (i.e., analytic in the whole plane) interpolant. That follows from theorems of Weierstrass and Mittag-Leffler (the former gives an entire function that vanishes at exactly the points in $A$, and the latter gives a function with poles at exactly the points of $A$ and with prescribed principal parts at the poles).
$endgroup$
– Andreas Blass
Jul 20 '17 at 17:17
$begingroup$
do you know about Runge's phenomenon ?
$endgroup$
– G Cab
Jul 20 '17 at 17:24
$begingroup$
So suppose you have a function which is 0 at every nonzero integer, and 1 at x=0. Is there no way to interpolate this and get something like the sinc function?
$endgroup$
– Mike Battaglia
Jul 20 '17 at 23:00
$begingroup$
Never mind, misinterpreted - thanks
$endgroup$
– Mike Battaglia
Jul 21 '17 at 16:26
2
2
$begingroup$
Your countably infinite sparse set $A$ of points needs to be so sparse that no sequence of distinct points from $A$ has a finite limit. In that case, there is an entire (i.e., analytic in the whole plane) interpolant. That follows from theorems of Weierstrass and Mittag-Leffler (the former gives an entire function that vanishes at exactly the points in $A$, and the latter gives a function with poles at exactly the points of $A$ and with prescribed principal parts at the poles).
$endgroup$
– Andreas Blass
Jul 20 '17 at 17:17
$begingroup$
Your countably infinite sparse set $A$ of points needs to be so sparse that no sequence of distinct points from $A$ has a finite limit. In that case, there is an entire (i.e., analytic in the whole plane) interpolant. That follows from theorems of Weierstrass and Mittag-Leffler (the former gives an entire function that vanishes at exactly the points in $A$, and the latter gives a function with poles at exactly the points of $A$ and with prescribed principal parts at the poles).
$endgroup$
– Andreas Blass
Jul 20 '17 at 17:17
$begingroup$
do you know about Runge's phenomenon ?
$endgroup$
– G Cab
Jul 20 '17 at 17:24
$begingroup$
do you know about Runge's phenomenon ?
$endgroup$
– G Cab
Jul 20 '17 at 17:24
$begingroup$
So suppose you have a function which is 0 at every nonzero integer, and 1 at x=0. Is there no way to interpolate this and get something like the sinc function?
$endgroup$
– Mike Battaglia
Jul 20 '17 at 23:00
$begingroup$
So suppose you have a function which is 0 at every nonzero integer, and 1 at x=0. Is there no way to interpolate this and get something like the sinc function?
$endgroup$
– Mike Battaglia
Jul 20 '17 at 23:00
$begingroup$
Never mind, misinterpreted - thanks
$endgroup$
– Mike Battaglia
Jul 21 '17 at 16:26
$begingroup$
Never mind, misinterpreted - thanks
$endgroup$
– Mike Battaglia
Jul 21 '17 at 16:26
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
There is this theorem:
Given two sequences $z_n$ and $w_n$ of complex numbers such that $|z_n| to infty$, there exists a holomorphic function $f$ such that $f(z_n) = w_n$ for all $n$.
It is a consequence of the Weierstrass factorization theorem and the Mittag-Leffler theorem.
See this question.
answered Jul 20 '17 at 17:20
lhflhf
167k11172404
$endgroup$
1
$begingroup$
See also math.stackexchange.com/a/1529860/589.
$endgroup$
– lhf
Jul 20 '17 at 17:21
2
$begingroup$
So it doesn't seem the interpolating function must be unique. For example, if we sample at the integers, we can add sin(2*pi*x) to the result to obtain another valid interpolant. Is there some condition we can place on the interpolant to obtain a simplest one, in some sense?
$endgroup$
– Mike Battaglia
Jul 21 '17 at 17:11
1
$begingroup$
@MikeBattaglia, I don't know. Ask a separate question.
$endgroup$
– lhf
Jul 21 '17 at 17:12
add a comment |
$begingroup$
I was trying something like this myself. Not 100% sure if this is what you mean.
Let $ a_i $ be the infinite sequence you want to interpolate by polynomials in $t$.
We construct series of $n$-th degree polynomials $X^n(t)$ such that : $forall i le n : X^n(i) = a_i$ like this:
$X^0(t)= fraca_00!0!$
$X^1(t)= fraca_00!0! - (t-0)fraca_00!1! -fraca_11!0! $
$X^2(t) = fraca_00!0! - (t-0)fraca_00!1! -fraca_11!0! - (t-1) fraca_00!2! - fraca_11!1! + fraca_22!0! $
$X^3(t) = fraca_00!0! - (t-0)fraca_00!1! -fraca_11!0! - (t-1) fraca_00!2! - fraca_11!1! + fraca_22!0! - (t-2) fraca_00!3! -fraca_11!2! +fraca_22!1! - fraca_33!0! $
The idea of course is that every $X^n(t)$ is 'cut off' at some point when we fill in an integer $p < n$, resulting in the polynomial $X^p(p)$ for which we know the relation holds.
This would lead to the general formula :
$beginarrayl
X^n(t) = \ fraca_00!0! \- (t-0)fraca_00!1! -fraca_11!0! \- (t-1) fraca_00!2! - fraca_11!1! + fraca_22!0! \- (t-2) fraca_00!3! -fraca_11!2! +fraca_22!1! - fraca_33!0!\
- (t-3) fraca_00!4! -fraca_11!3! +fraca_22!2! - fraca_33!1! +fraca_44!0!
\
vdots\
-(t-n+1) fraca_00!n! -fraca_11!(n-1)! + cdots cdots cdots cdots\ +(-1)^n-2 fraca_n-2(n-2)!2! +(-1)^n-1 fraca_n-1(n-1)!1! +(-1)^n fraca_nn!0! \ cdots \
endarray$
Which can be verified with the help of the formula $sum_k=0^n (-1)^kbinomnk=0$
Now I assumed that the data points were equally spaced (say every $y=a_i$ can be found at $x=i$). If this is not the case we could regard the parameter $t$ as a parameter on a 2-dimensional curve $left( X^n(t), Y^n(t)right)$.
I hope if someone reads this they can verify the above result. Does this procedure have a name?
answered Mar 21 at 12:32
Rutger MoodyRutger Moody
1,51911019
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There is this theorem:
Given two sequences $z_n$ and $w_n$ of complex numbers such that $|z_n| to infty$, there exists a holomorphic function $f$ such that $f(z_n) = w_n$ for all $n$.
It is a consequence of the Weierstrass factorization theorem and the Mittag-Leffler theorem.
See this question.
answered Jul 20 '17 at 17:20
lhflhf
167k11172404
$endgroup$
1
$begingroup$
See also math.stackexchange.com/a/1529860/589.
$endgroup$
– lhf
Jul 20 '17 at 17:21
2
$begingroup$
So it doesn't seem the interpolating function must be unique. For example, if we sample at the integers, we can add sin(2*pi*x) to the result to obtain another valid interpolant. Is there some condition we can place on the interpolant to obtain a simplest one, in some sense?
$endgroup$
– Mike Battaglia
Jul 21 '17 at 17:11
1
$begingroup$
@MikeBattaglia, I don't know. Ask a separate question.
$endgroup$
– lhf
Jul 21 '17 at 17:12
add a comment |
$begingroup$
There is this theorem:
Given two sequences $z_n$ and $w_n$ of complex numbers such that $|z_n| to infty$, there exists a holomorphic function $f$ such that $f(z_n) = w_n$ for all $n$.
It is a consequence of the Weierstrass factorization theorem and the Mittag-Leffler theorem.
See this question.
answered Jul 20 '17 at 17:20
lhflhf
167k11172404
$endgroup$
1
$begingroup$
See also math.stackexchange.com/a/1529860/589.
$endgroup$
– lhf
Jul 20 '17 at 17:21
2
$begingroup$
So it doesn't seem the interpolating function must be unique. For example, if we sample at the integers, we can add sin(2*pi*x) to the result to obtain another valid interpolant. Is there some condition we can place on the interpolant to obtain a simplest one, in some sense?
$endgroup$
– Mike Battaglia
Jul 21 '17 at 17:11
1
$begingroup$
@MikeBattaglia, I don't know. Ask a separate question.
$endgroup$
– lhf
Jul 21 '17 at 17:12
add a comment |
$begingroup$
There is this theorem:
Given two sequences $z_n$ and $w_n$ of complex numbers such that $|z_n| to infty$, there exists a holomorphic function $f$ such that $f(z_n) = w_n$ for all $n$.
It is a consequence of the Weierstrass factorization theorem and the Mittag-Leffler theorem.
See this question.
answered Jul 20 '17 at 17:20
lhflhf
167k11172404
$endgroup$
There is this theorem:
Given two sequences $z_n$ and $w_n$ of complex numbers such that $|z_n| to infty$, there exists a holomorphic function $f$ such that $f(z_n) = w_n$ for all $n$.
It is a consequence of the Weierstrass factorization theorem and the Mittag-Leffler theorem.
See this question.
answered Jul 20 '17 at 17:20
lhflhf
167k11172404
answered Jul 20 '17 at 17:20
lhflhf
167k11172404
answered Jul 20 '17 at 17:20
lhflhf
167k11172404
answered Jul 20 '17 at 17:20
lhflhf
167k11172404
167k11172404
1
$begingroup$
See also math.stackexchange.com/a/1529860/589.
$endgroup$
– lhf
Jul 20 '17 at 17:21
2
$begingroup$
So it doesn't seem the interpolating function must be unique. For example, if we sample at the integers, we can add sin(2*pi*x) to the result to obtain another valid interpolant. Is there some condition we can place on the interpolant to obtain a simplest one, in some sense?
$endgroup$
– Mike Battaglia
Jul 21 '17 at 17:11
1
$begingroup$
@MikeBattaglia, I don't know. Ask a separate question.
$endgroup$
– lhf
Jul 21 '17 at 17:12
add a comment |
1
$begingroup$
See also math.stackexchange.com/a/1529860/589.
$endgroup$
– lhf
Jul 20 '17 at 17:21
2
$begingroup$
So it doesn't seem the interpolating function must be unique. For example, if we sample at the integers, we can add sin(2*pi*x) to the result to obtain another valid interpolant. Is there some condition we can place on the interpolant to obtain a simplest one, in some sense?
$endgroup$
– Mike Battaglia
Jul 21 '17 at 17:11
1
$begingroup$
@MikeBattaglia, I don't know. Ask a separate question.
$endgroup$
– lhf
Jul 21 '17 at 17:12
1
1
$begingroup$
See also math.stackexchange.com/a/1529860/589.
$endgroup$
– lhf
Jul 20 '17 at 17:21
$begingroup$
See also math.stackexchange.com/a/1529860/589.
$endgroup$
– lhf
Jul 20 '17 at 17:21
2
2
$begingroup$
So it doesn't seem the interpolating function must be unique. For example, if we sample at the integers, we can add sin(2*pi*x) to the result to obtain another valid interpolant. Is there some condition we can place on the interpolant to obtain a simplest one, in some sense?
$endgroup$
– Mike Battaglia
Jul 21 '17 at 17:11
$begingroup$
So it doesn't seem the interpolating function must be unique. For example, if we sample at the integers, we can add sin(2*pi*x) to the result to obtain another valid interpolant. Is there some condition we can place on the interpolant to obtain a simplest one, in some sense?
$endgroup$
– Mike Battaglia
Jul 21 '17 at 17:11
1
1
$begingroup$
@MikeBattaglia, I don't know. Ask a separate question.
$endgroup$
– lhf
Jul 21 '17 at 17:12
$begingroup$
@MikeBattaglia, I don't know. Ask a separate question.
$endgroup$
– lhf
Jul 21 '17 at 17:12
add a comment |
$begingroup$
I was trying something like this myself. Not 100% sure if this is what you mean.
Let $ a_i $ be the infinite sequence you want to interpolate by polynomials in $t$.
We construct series of $n$-th degree polynomials $X^n(t)$ such that : $forall i le n : X^n(i) = a_i$ like this:
$X^0(t)= fraca_00!0!$
$X^1(t)= fraca_00!0! - (t-0)fraca_00!1! -fraca_11!0! $
$X^2(t) = fraca_00!0! - (t-0)fraca_00!1! -fraca_11!0! - (t-1) fraca_00!2! - fraca_11!1! + fraca_22!0! $
$X^3(t) = fraca_00!0! - (t-0)fraca_00!1! -fraca_11!0! - (t-1) fraca_00!2! - fraca_11!1! + fraca_22!0! - (t-2) fraca_00!3! -fraca_11!2! +fraca_22!1! - fraca_33!0! $
The idea of course is that every $X^n(t)$ is 'cut off' at some point when we fill in an integer $p < n$, resulting in the polynomial $X^p(p)$ for which we know the relation holds.
This would lead to the general formula :
$beginarrayl
X^n(t) = \ fraca_00!0! \- (t-0)fraca_00!1! -fraca_11!0! \- (t-1) fraca_00!2! - fraca_11!1! + fraca_22!0! \- (t-2) fraca_00!3! -fraca_11!2! +fraca_22!1! - fraca_33!0!\
- (t-3) fraca_00!4! -fraca_11!3! +fraca_22!2! - fraca_33!1! +fraca_44!0!
\
vdots\
-(t-n+1) fraca_00!n! -fraca_11!(n-1)! + cdots cdots cdots cdots\ +(-1)^n-2 fraca_n-2(n-2)!2! +(-1)^n-1 fraca_n-1(n-1)!1! +(-1)^n fraca_nn!0! \ cdots \
endarray$
Which can be verified with the help of the formula $sum_k=0^n (-1)^kbinomnk=0$
Now I assumed that the data points were equally spaced (say every $y=a_i$ can be found at $x=i$). If this is not the case we could regard the parameter $t$ as a parameter on a 2-dimensional curve $left( X^n(t), Y^n(t)right)$.
I hope if someone reads this they can verify the above result. Does this procedure have a name?
answered Mar 21 at 12:32
Rutger MoodyRutger Moody
1,51911019
$endgroup$
add a comment |
$begingroup$
I was trying something like this myself. Not 100% sure if this is what you mean.
Let $ a_i $ be the infinite sequence you want to interpolate by polynomials in $t$.
We construct series of $n$-th degree polynomials $X^n(t)$ such that : $forall i le n : X^n(i) = a_i$ like this:
$X^0(t)= fraca_00!0!$
$X^1(t)= fraca_00!0! - (t-0)fraca_00!1! -fraca_11!0! $
$X^2(t) = fraca_00!0! - (t-0)fraca_00!1! -fraca_11!0! - (t-1) fraca_00!2! - fraca_11!1! + fraca_22!0! $
$X^3(t) = fraca_00!0! - (t-0)fraca_00!1! -fraca_11!0! - (t-1) fraca_00!2! - fraca_11!1! + fraca_22!0! - (t-2) fraca_00!3! -fraca_11!2! +fraca_22!1! - fraca_33!0! $
The idea of course is that every $X^n(t)$ is 'cut off' at some point when we fill in an integer $p < n$, resulting in the polynomial $X^p(p)$ for which we know the relation holds.
This would lead to the general formula :
$beginarrayl
X^n(t) = \ fraca_00!0! \- (t-0)fraca_00!1! -fraca_11!0! \- (t-1) fraca_00!2! - fraca_11!1! + fraca_22!0! \- (t-2) fraca_00!3! -fraca_11!2! +fraca_22!1! - fraca_33!0!\
- (t-3) fraca_00!4! -fraca_11!3! +fraca_22!2! - fraca_33!1! +fraca_44!0!
\
vdots\
-(t-n+1) fraca_00!n! -fraca_11!(n-1)! + cdots cdots cdots cdots\ +(-1)^n-2 fraca_n-2(n-2)!2! +(-1)^n-1 fraca_n-1(n-1)!1! +(-1)^n fraca_nn!0! \ cdots \
endarray$
Which can be verified with the help of the formula $sum_k=0^n (-1)^kbinomnk=0$
Now I assumed that the data points were equally spaced (say every $y=a_i$ can be found at $x=i$). If this is not the case we could regard the parameter $t$ as a parameter on a 2-dimensional curve $left( X^n(t), Y^n(t)right)$.
I hope if someone reads this they can verify the above result. Does this procedure have a name?
answered Mar 21 at 12:32
Rutger MoodyRutger Moody
1,51911019
$endgroup$
add a comment |
$begingroup$
I was trying something like this myself. Not 100% sure if this is what you mean.
Let $ a_i $ be the infinite sequence you want to interpolate by polynomials in $t$.
We construct series of $n$-th degree polynomials $X^n(t)$ such that : $forall i le n : X^n(i) = a_i$ like this:
$X^0(t)= fraca_00!0!$
$X^1(t)= fraca_00!0! - (t-0)fraca_00!1! -fraca_11!0! $
$X^2(t) = fraca_00!0! - (t-0)fraca_00!1! -fraca_11!0! - (t-1) fraca_00!2! - fraca_11!1! + fraca_22!0! $
$X^3(t) = fraca_00!0! - (t-0)fraca_00!1! -fraca_11!0! - (t-1) fraca_00!2! - fraca_11!1! + fraca_22!0! - (t-2) fraca_00!3! -fraca_11!2! +fraca_22!1! - fraca_33!0! $
The idea of course is that every $X^n(t)$ is 'cut off' at some point when we fill in an integer $p < n$, resulting in the polynomial $X^p(p)$ for which we know the relation holds.
This would lead to the general formula :
$beginarrayl
X^n(t) = \ fraca_00!0! \- (t-0)fraca_00!1! -fraca_11!0! \- (t-1) fraca_00!2! - fraca_11!1! + fraca_22!0! \- (t-2) fraca_00!3! -fraca_11!2! +fraca_22!1! - fraca_33!0!\
- (t-3) fraca_00!4! -fraca_11!3! +fraca_22!2! - fraca_33!1! +fraca_44!0!
\
vdots\
-(t-n+1) fraca_00!n! -fraca_11!(n-1)! + cdots cdots cdots cdots\ +(-1)^n-2 fraca_n-2(n-2)!2! +(-1)^n-1 fraca_n-1(n-1)!1! +(-1)^n fraca_nn!0! \ cdots \
endarray$
Which can be verified with the help of the formula $sum_k=0^n (-1)^kbinomnk=0$
Now I assumed that the data points were equally spaced (say every $y=a_i$ can be found at $x=i$). If this is not the case we could regard the parameter $t$ as a parameter on a 2-dimensional curve $left( X^n(t), Y^n(t)right)$.
I hope if someone reads this they can verify the above result. Does this procedure have a name?
answered Mar 21 at 12:32
Rutger MoodyRutger Moody
1,51911019
$endgroup$
I was trying something like this myself. Not 100% sure if this is what you mean.
Let $ a_i $ be the infinite sequence you want to interpolate by polynomials in $t$.
We construct series of $n$-th degree polynomials $X^n(t)$ such that : $forall i le n : X^n(i) = a_i$ like this:
$X^0(t)= fraca_00!0!$
$X^1(t)= fraca_00!0! - (t-0)fraca_00!1! -fraca_11!0! $
$X^2(t) = fraca_00!0! - (t-0)fraca_00!1! -fraca_11!0! - (t-1) fraca_00!2! - fraca_11!1! + fraca_22!0! $
$X^3(t) = fraca_00!0! - (t-0)fraca_00!1! -fraca_11!0! - (t-1) fraca_00!2! - fraca_11!1! + fraca_22!0! - (t-2) fraca_00!3! -fraca_11!2! +fraca_22!1! - fraca_33!0! $
The idea of course is that every $X^n(t)$ is 'cut off' at some point when we fill in an integer $p < n$, resulting in the polynomial $X^p(p)$ for which we know the relation holds.
This would lead to the general formula :
$beginarrayl
X^n(t) = \ fraca_00!0! \- (t-0)fraca_00!1! -fraca_11!0! \- (t-1) fraca_00!2! - fraca_11!1! + fraca_22!0! \- (t-2) fraca_00!3! -fraca_11!2! +fraca_22!1! - fraca_33!0!\
- (t-3) fraca_00!4! -fraca_11!3! +fraca_22!2! - fraca_33!1! +fraca_44!0!
\
vdots\
-(t-n+1) fraca_00!n! -fraca_11!(n-1)! + cdots cdots cdots cdots\ +(-1)^n-2 fraca_n-2(n-2)!2! +(-1)^n-1 fraca_n-1(n-1)!1! +(-1)^n fraca_nn!0! \ cdots \
endarray$
Which can be verified with the help of the formula $sum_k=0^n (-1)^kbinomnk=0$
Now I assumed that the data points were equally spaced (say every $y=a_i$ can be found at $x=i$). If this is not the case we could regard the parameter $t$ as a parameter on a 2-dimensional curve $left( X^n(t), Y^n(t)right)$.
I hope if someone reads this they can verify the above result. Does this procedure have a name?
answered Mar 21 at 12:32
Rutger MoodyRutger Moody
1,51911019
edited Mar 21 at 16:53
answered Mar 21 at 12:32
Rutger MoodyRutger Moody
1,51911019
answered Mar 21 at 12:32
Rutger MoodyRutger Moody
1,51911019
answered Mar 21 at 12:32
Rutger MoodyRutger Moody
1,51911019
1,51911019
add a comment |
add a comment |
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Your countably infinite sparse set $A$ of points needs to be so sparse that no sequence of distinct points from $A$ has a finite limit. In that case, there is an entire (i.e., analytic in the whole plane) interpolant. That follows from theorems of Weierstrass and Mittag-Leffler (the former gives an entire function that vanishes at exactly the points in $A$, and the latter gives a function with poles at exactly the points of $A$ and with prescribed principal parts at the poles).
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– Andreas Blass
Jul 20 '17 at 17:17
$begingroup$
do you know about Runge's phenomenon ?
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– G Cab
Jul 20 '17 at 17:24
$begingroup$
So suppose you have a function which is 0 at every nonzero integer, and 1 at x=0. Is there no way to interpolate this and get something like the sinc function?
$endgroup$
– Mike Battaglia
Jul 20 '17 at 23:00
$begingroup$
Never mind, misinterpreted - thanks
$endgroup$
– Mike Battaglia
Jul 21 '17 at 16:26