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Integration of a Chebyshev series multiplied by an exponential function

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1

$begingroup$

I would like to evaluate the following integral:

$$I(x_0)=int_-1^1 left(sum_k=0^n a_k , T_k(x)right) , mathrme^b(x-x_0),mathrmdx$$

with $a_k$ some constant coefficients comming from an Chebyshev interpolation of a function $f(x)$, $T_k(x)$ the Chebyshev polynomials, $binmathbbC$, $ninmathbbN$ and $x_0>1$.

I assume, I can rewrite the integral as

$$I(x_0)=mathrme^-b,x_0sum_k=0^n a_k left(int_-1^1 T_k(x), mathrme^b,x,mathrmdxright)$$

In this question this integral is solved by a recurrence relation. I used this recurrence relation to evaluate the sum, but sadly the recurrence formular is not stable for large $n$. I also tryed to adopt the Clenshaw algorithm (Link) to overcome this problem, but still the evaluation diverges.

Is there a possibility to write this integral in explicit form?

Edit:
I am wondering if replacing

$$sum_k=0^n a_k , T_k(x)$$
by the barycentric interpolation formula
$$sum_j=0^N a_j , T_j(x)
= fracdisplaystyle sum_j=0^N fracw_jx-x_jf_jdisplaystyle sum_j=0^N fracw_jx-x_j,
$$
where
$$
w_j = left{
beginarraycc
(-1)^j/2, & j=0text or j=N,\
(-1)^j, & textotherwise
endarray
right.
$$
is any better!

integration definite-integrals exponential-function chebyshev-polynomials

share|cite|improve this question

edited Mar 29 at 17:14

Michael_K

asked Mar 29 at 9:59

Michael_KMichael_K

400212

$endgroup$

add a comment | 

1

$begingroup$

I would like to evaluate the following integral:

$$I(x_0)=int_-1^1 left(sum_k=0^n a_k , T_k(x)right) , mathrme^b(x-x_0),mathrmdx$$

with $a_k$ some constant coefficients comming from an Chebyshev interpolation of a function $f(x)$, $T_k(x)$ the Chebyshev polynomials, $binmathbbC$, $ninmathbbN$ and $x_0>1$.

I assume, I can rewrite the integral as

$$I(x_0)=mathrme^-b,x_0sum_k=0^n a_k left(int_-1^1 T_k(x), mathrme^b,x,mathrmdxright)$$

In this question this integral is solved by a recurrence relation. I used this recurrence relation to evaluate the sum, but sadly the recurrence formular is not stable for large $n$. I also tryed to adopt the Clenshaw algorithm (Link) to overcome this problem, but still the evaluation diverges.

Is there a possibility to write this integral in explicit form?

Edit:
I am wondering if replacing

$$sum_k=0^n a_k , T_k(x)$$
by the barycentric interpolation formula
$$sum_j=0^N a_j , T_j(x)
= fracdisplaystyle sum_j=0^N fracw_jx-x_jf_jdisplaystyle sum_j=0^N fracw_jx-x_j,
$$
where
$$
w_j = left{
beginarraycc
(-1)^j/2, & j=0text or j=N,\
(-1)^j, & textotherwise
endarray
right.
$$
is any better!

integration definite-integrals exponential-function chebyshev-polynomials

share|cite|improve this question

edited Mar 29 at 17:14

Michael_K

asked Mar 29 at 9:59

Michael_KMichael_K

400212

$endgroup$

add a comment | 

$begingroup$

I would like to evaluate the following integral:

$$I(x_0)=int_-1^1 left(sum_k=0^n a_k , T_k(x)right) , mathrme^b(x-x_0),mathrmdx$$

with $a_k$ some constant coefficients comming from an Chebyshev interpolation of a function $f(x)$, $T_k(x)$ the Chebyshev polynomials, $binmathbbC$, $ninmathbbN$ and $x_0>1$.

I assume, I can rewrite the integral as

$$I(x_0)=mathrme^-b,x_0sum_k=0^n a_k left(int_-1^1 T_k(x), mathrme^b,x,mathrmdxright)$$

In this question this integral is solved by a recurrence relation. I used this recurrence relation to evaluate the sum, but sadly the recurrence formular is not stable for large $n$. I also tryed to adopt the Clenshaw algorithm (Link) to overcome this problem, but still the evaluation diverges.

Is there a possibility to write this integral in explicit form?

Edit:
I am wondering if replacing

$$sum_k=0^n a_k , T_k(x)$$
by the barycentric interpolation formula
$$sum_j=0^N a_j , T_j(x)
= fracdisplaystyle sum_j=0^N fracw_jx-x_jf_jdisplaystyle sum_j=0^N fracw_jx-x_j,
$$
where
$$
w_j = left{
beginarraycc
(-1)^j/2, & j=0text or j=N,\
(-1)^j, & textotherwise
endarray
right.
$$
is any better!

integration definite-integrals exponential-function chebyshev-polynomials

share|cite|improve this question

edited Mar 29 at 17:14

Michael_K

asked Mar 29 at 9:59

Michael_KMichael_K

400212

$endgroup$

share|cite|improve this question

edited Mar 29 at 17:14

Michael_K

asked Mar 29 at 9:59

Michael_KMichael_K

400212

share|cite|improve this question

edited Mar 29 at 17:14

Michael_K

asked Mar 29 at 9:59

Michael_KMichael_K

400212

asked Mar 29 at 9:59

Michael_KMichael_K

400212

asked Mar 29 at 9:59

Michael_KMichael_K

400212

1 Answer
1

active

oldest

votes

1

$begingroup$

In "The Fourier Transforms of the Chebyshev and Legendre Polynomials" (Fokas&Smitheman https://arxiv.org/abs/1211.4943 ) Eq. (2.4) gives an expression for $hat T_m(lambda) = int_-1^1 exp(-ilambda x) T_m(x) dx$.

There's another expression for $hat T_m(lambda)$ in "The finite Fourier transform of classical polynomials" (Dixit,Jiu,Moll,Vignat https://arxiv.org/abs/1402.5544) in Thm 4.1 which looks slightly more inviting:

$$ hat T_n(lambda) = sum_k=0^n (-1)^k+1 fracn 2^k (n+k)! k!(2k+1)!(n-k)! frac(-1)^n-k e^-ilambda - e^ilambda (-2ilambda)^k+1 $$

(If you can work in the Legendre basis, the transform of a Legendre polynomial is a half-integer Bessel, which would at least look nicer)

share|cite|improve this answer

answered Mar 29 at 18:37

JCKJCK

111

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for this information. I am wondering if it is possible to reduce the double sum to something simpler from a computational point of view.
  $endgroup$
  – Michael_K
  Mar 29 at 21:31
  
  

  
  
  
  

add a comment | 

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1

$begingroup$

In "The Fourier Transforms of the Chebyshev and Legendre Polynomials" (Fokas&Smitheman https://arxiv.org/abs/1211.4943 ) Eq. (2.4) gives an expression for $hat T_m(lambda) = int_-1^1 exp(-ilambda x) T_m(x) dx$.

There's another expression for $hat T_m(lambda)$ in "The finite Fourier transform of classical polynomials" (Dixit,Jiu,Moll,Vignat https://arxiv.org/abs/1402.5544) in Thm 4.1 which looks slightly more inviting:

$$ hat T_n(lambda) = sum_k=0^n (-1)^k+1 fracn 2^k (n+k)! k!(2k+1)!(n-k)! frac(-1)^n-k e^-ilambda - e^ilambda (-2ilambda)^k+1 $$

(If you can work in the Legendre basis, the transform of a Legendre polynomial is a half-integer Bessel, which would at least look nicer)

share|cite|improve this answer

answered Mar 29 at 18:37

JCKJCK

111

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for this information. I am wondering if it is possible to reduce the double sum to something simpler from a computational point of view.
  $endgroup$
  – Michael_K
  Mar 29 at 21:31
  
  

  
  
  
  

add a comment | 

1

$begingroup$

In "The Fourier Transforms of the Chebyshev and Legendre Polynomials" (Fokas&Smitheman https://arxiv.org/abs/1211.4943 ) Eq. (2.4) gives an expression for $hat T_m(lambda) = int_-1^1 exp(-ilambda x) T_m(x) dx$.

There's another expression for $hat T_m(lambda)$ in "The finite Fourier transform of classical polynomials" (Dixit,Jiu,Moll,Vignat https://arxiv.org/abs/1402.5544) in Thm 4.1 which looks slightly more inviting:

$$ hat T_n(lambda) = sum_k=0^n (-1)^k+1 fracn 2^k (n+k)! k!(2k+1)!(n-k)! frac(-1)^n-k e^-ilambda - e^ilambda (-2ilambda)^k+1 $$

(If you can work in the Legendre basis, the transform of a Legendre polynomial is a half-integer Bessel, which would at least look nicer)

share|cite|improve this answer

answered Mar 29 at 18:37

JCKJCK

111

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for this information. I am wondering if it is possible to reduce the double sum to something simpler from a computational point of view.
  $endgroup$
  – Michael_K
  Mar 29 at 21:31
  
  

  
  
  
  

add a comment | 

$begingroup$

In "The Fourier Transforms of the Chebyshev and Legendre Polynomials" (Fokas&Smitheman https://arxiv.org/abs/1211.4943 ) Eq. (2.4) gives an expression for $hat T_m(lambda) = int_-1^1 exp(-ilambda x) T_m(x) dx$.

There's another expression for $hat T_m(lambda)$ in "The finite Fourier transform of classical polynomials" (Dixit,Jiu,Moll,Vignat https://arxiv.org/abs/1402.5544) in Thm 4.1 which looks slightly more inviting:

$$ hat T_n(lambda) = sum_k=0^n (-1)^k+1 fracn 2^k (n+k)! k!(2k+1)!(n-k)! frac(-1)^n-k e^-ilambda - e^ilambda (-2ilambda)^k+1 $$

(If you can work in the Legendre basis, the transform of a Legendre polynomial is a half-integer Bessel, which would at least look nicer)

share|cite|improve this answer

answered Mar 29 at 18:37

JCKJCK

111

$endgroup$

share|cite|improve this answer

answered Mar 29 at 18:37

JCKJCK

111

answered Mar 29 at 18:37

JCKJCK

111

answered Mar 29 at 18:37

JCKJCK

111