integration of chebyshev polynomials of first kind with an exponential funcionIntegration of a Chebyshev series multiplied by an exponential functionIntegrating Chebyshev polynomial of the first kindWhy might one be inclined to think that polynomials of the form $cos(narccosx)$ would minimize error in Lagrange interpolation?Prove that $T_n$ satisfy $ sum_k=0^N-1T_i(x_k)T_j(x_k) = begincases 0 &: ine j \ lneq 0 &: i=j endcases ,! $Integral with Chebyshev polynomialsChebyshev Polynomials of 1/nrelation between first kind Chebyshev poly and second kind Chebyshev polyInner Product of Chebyshev polynomials of the second kind with $x$ as weightingRecurrence relation for integralHow to bound the summation by an integralIntegration of a Chebyshev series multiplied by an exponential function
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integration of chebyshev polynomials of first kind with an exponential funcion
Integration of a Chebyshev series multiplied by an exponential functionIntegrating Chebyshev polynomial of the first kindWhy might one be inclined to think that polynomials of the form $cos(narccosx)$ would minimize error in Lagrange interpolation?Prove that $T_n$ satisfy $ sum_k=0^N-1T_i(x_k)T_j(x_k) = begincases 0 &: ine j \ lneq 0 &: i=j endcases ,! $Integral with Chebyshev polynomialsChebyshev Polynomials of 1/nrelation between first kind Chebyshev poly and second kind Chebyshev polyInner Product of Chebyshev polynomials of the second kind with $x$ as weightingRecurrence relation for integralHow to bound the summation by an integralIntegration of a Chebyshev series multiplied by an exponential function
$begingroup$
In my class, my tutor raised a question of the following integral:
$int T_n(x)*exp(a*x)dx,$ where $T_n(x)$ is an n power of Chebyshev polynomials of first kind and a is a constant. The hint he gave us is using the recurrence relation of Chebyshev polynomials. I have tried a few times but I didn't get lucky. I think the question is wrong, isn't it.
integration special-functions chebyshev-polynomials
edited Mar 29 at 14:16
J. M. is not a mathematician
61.2k5152290
asked Jun 15 '17 at 19:33
RayRay
767
$endgroup$
add a comment |
$begingroup$
In my class, my tutor raised a question of the following integral:
$int T_n(x)*exp(a*x)dx,$ where $T_n(x)$ is an n power of Chebyshev polynomials of first kind and a is a constant. The hint he gave us is using the recurrence relation of Chebyshev polynomials. I have tried a few times but I didn't get lucky. I think the question is wrong, isn't it.
integration special-functions chebyshev-polynomials
edited Mar 29 at 14:16
J. M. is not a mathematician
61.2k5152290
asked Jun 15 '17 at 19:33
RayRay
767
$endgroup$
$begingroup$
First forget that they are Chebyshev's. If they were just polynomials what would you do? Integration by parts $n$ times, no? And of course the part in which one uses that they are Tchebyshev's.
$endgroup$
– OR.
Jun 15 '17 at 19:40
add a comment |
$begingroup$
In my class, my tutor raised a question of the following integral:
$int T_n(x)*exp(a*x)dx,$ where $T_n(x)$ is an n power of Chebyshev polynomials of first kind and a is a constant. The hint he gave us is using the recurrence relation of Chebyshev polynomials. I have tried a few times but I didn't get lucky. I think the question is wrong, isn't it.
integration special-functions chebyshev-polynomials
edited Mar 29 at 14:16
J. M. is not a mathematician
61.2k5152290
asked Jun 15 '17 at 19:33
RayRay
767
$endgroup$
In my class, my tutor raised a question of the following integral:
$int T_n(x)*exp(a*x)dx,$ where $T_n(x)$ is an n power of Chebyshev polynomials of first kind and a is a constant. The hint he gave us is using the recurrence relation of Chebyshev polynomials. I have tried a few times but I didn't get lucky. I think the question is wrong, isn't it.
integration special-functions chebyshev-polynomials
integration special-functions chebyshev-polynomials
edited Mar 29 at 14:16
J. M. is not a mathematician
61.2k5152290
asked Jun 15 '17 at 19:33
RayRay
767
edited Mar 29 at 14:16
J. M. is not a mathematician
61.2k5152290
asked Jun 15 '17 at 19:33
RayRay
767
edited Mar 29 at 14:16
J. M. is not a mathematician
61.2k5152290
edited Mar 29 at 14:16
J. M. is not a mathematician
61.2k5152290
edited Mar 29 at 14:16
J. M. is not a mathematician
61.2k5152290
61.2k5152290
asked Jun 15 '17 at 19:33
RayRay
767
asked Jun 15 '17 at 19:33
RayRay
767
asked Jun 15 '17 at 19:33
RayRay
767
767
$begingroup$
First forget that they are Chebyshev's. If they were just polynomials what would you do? Integration by parts $n$ times, no? And of course the part in which one uses that they are Tchebyshev's.
$endgroup$
– OR.
Jun 15 '17 at 19:40
add a comment |
$begingroup$
First forget that they are Chebyshev's. If they were just polynomials what would you do? Integration by parts $n$ times, no? And of course the part in which one uses that they are Tchebyshev's.
$endgroup$
– OR.
Jun 15 '17 at 19:40
$begingroup$
First forget that they are Chebyshev's. If they were just polynomials what would you do? Integration by parts $n$ times, no? And of course the part in which one uses that they are Tchebyshev's.
$endgroup$
– OR.
Jun 15 '17 at 19:40
$begingroup$
First forget that they are Chebyshev's. If they were just polynomials what would you do? Integration by parts $n$ times, no? And of course the part in which one uses that they are Tchebyshev's.
$endgroup$
– OR.
Jun 15 '17 at 19:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The question isn't wrong, the hint is... a bit misleading, I'd say. You might use
$$T_n(x)=frac12(U_n(x)-U_n-2(x)),$$ for $nge2$ together with
$$T'_n(x)=n,U_n-1(x)$$ and partial integration, to get a recurrence relation for your integrals (the $U_n$ are Chebyshev polynomials of second kind).
Here are more details: Let's define $$I_n(x)=int T_n(x),e^ax,dx.$$ Since $$T_n(x)=frac12left(fracT'_n+1(x)n+1-fracT'_n-1(x)n-1right),$$ multiplying by $e^ax$ and integrating (by parts) gives
$$I_n(x)=frac12left(fracT_n+1(x)n+1-fracT_n-1(x)n-1right),e^ax-fraca2left(fracI_n+1(x)n+1-fracI_n-1(x)n-1right).$$ I'll omit integration constants, because they cancel out when calculating definite integrals. If people insist in them, feel free to add any constant you like best. Now you can solve for $I_n+1(x)$, and calculate it
starting from $n=2$ recursively from $I_n-1(x)$ and $I_n(x).$ To prime the pump, you'll have to calculate $I_0(x)$, $I_1(x)$ and $I_2(x)$ manually, but with $T_0(x)=1$, $T_1(x)=x$ and $T_2(x)=2x^2-1$ that's a very simple task I won't explain here.
answered Jun 15 '17 at 20:04
Professor VectorProfessor Vector
11.2k11333
$endgroup$
$begingroup$
Hey professor Vector, would you mind giving me more steps such that I can follow your answer? Many thanks in advance.
$endgroup$
– Ray
Jun 15 '17 at 20:44
$begingroup$
In principle, the above derivation is valid for $n=1,$ too, but there are caveats: it's valid if you assume $U_-1(x)=0.$ That's in agreement with the definition $$U_n(cos t)=fracsin(n+1)tsin t,$$ and even $T'_0(x)=0cdot U_-1(x)$ is true, you just can't divide both sides by $0,$ you have to omit the terms originating from $U_-1$. So the above recurrence for $n=1$ becomes $$I_1(x)=frac14T_2(x),e^ax-fraca4I_2(x).$$
$endgroup$
– Professor Vector
Jun 16 '17 at 7:16
$begingroup$
Thanks Professor Vector.
$endgroup$
– Ray
Jun 17 '17 at 0:55
add a comment |
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$begingroup$
The question isn't wrong, the hint is... a bit misleading, I'd say. You might use
$$T_n(x)=frac12(U_n(x)-U_n-2(x)),$$ for $nge2$ together with
$$T'_n(x)=n,U_n-1(x)$$ and partial integration, to get a recurrence relation for your integrals (the $U_n$ are Chebyshev polynomials of second kind).
Here are more details: Let's define $$I_n(x)=int T_n(x),e^ax,dx.$$ Since $$T_n(x)=frac12left(fracT'_n+1(x)n+1-fracT'_n-1(x)n-1right),$$ multiplying by $e^ax$ and integrating (by parts) gives
$$I_n(x)=frac12left(fracT_n+1(x)n+1-fracT_n-1(x)n-1right),e^ax-fraca2left(fracI_n+1(x)n+1-fracI_n-1(x)n-1right).$$ I'll omit integration constants, because they cancel out when calculating definite integrals. If people insist in them, feel free to add any constant you like best. Now you can solve for $I_n+1(x)$, and calculate it
starting from $n=2$ recursively from $I_n-1(x)$ and $I_n(x).$ To prime the pump, you'll have to calculate $I_0(x)$, $I_1(x)$ and $I_2(x)$ manually, but with $T_0(x)=1$, $T_1(x)=x$ and $T_2(x)=2x^2-1$ that's a very simple task I won't explain here.
answered Jun 15 '17 at 20:04
Professor VectorProfessor Vector
11.2k11333
$endgroup$
$begingroup$
Hey professor Vector, would you mind giving me more steps such that I can follow your answer? Many thanks in advance.
$endgroup$
– Ray
Jun 15 '17 at 20:44
$begingroup$
In principle, the above derivation is valid for $n=1,$ too, but there are caveats: it's valid if you assume $U_-1(x)=0.$ That's in agreement with the definition $$U_n(cos t)=fracsin(n+1)tsin t,$$ and even $T'_0(x)=0cdot U_-1(x)$ is true, you just can't divide both sides by $0,$ you have to omit the terms originating from $U_-1$. So the above recurrence for $n=1$ becomes $$I_1(x)=frac14T_2(x),e^ax-fraca4I_2(x).$$
$endgroup$
– Professor Vector
Jun 16 '17 at 7:16
$begingroup$
Thanks Professor Vector.
$endgroup$
– Ray
Jun 17 '17 at 0:55
add a comment |
$begingroup$
The question isn't wrong, the hint is... a bit misleading, I'd say. You might use
$$T_n(x)=frac12(U_n(x)-U_n-2(x)),$$ for $nge2$ together with
$$T'_n(x)=n,U_n-1(x)$$ and partial integration, to get a recurrence relation for your integrals (the $U_n$ are Chebyshev polynomials of second kind).
Here are more details: Let's define $$I_n(x)=int T_n(x),e^ax,dx.$$ Since $$T_n(x)=frac12left(fracT'_n+1(x)n+1-fracT'_n-1(x)n-1right),$$ multiplying by $e^ax$ and integrating (by parts) gives
$$I_n(x)=frac12left(fracT_n+1(x)n+1-fracT_n-1(x)n-1right),e^ax-fraca2left(fracI_n+1(x)n+1-fracI_n-1(x)n-1right).$$ I'll omit integration constants, because they cancel out when calculating definite integrals. If people insist in them, feel free to add any constant you like best. Now you can solve for $I_n+1(x)$, and calculate it
starting from $n=2$ recursively from $I_n-1(x)$ and $I_n(x).$ To prime the pump, you'll have to calculate $I_0(x)$, $I_1(x)$ and $I_2(x)$ manually, but with $T_0(x)=1$, $T_1(x)=x$ and $T_2(x)=2x^2-1$ that's a very simple task I won't explain here.
answered Jun 15 '17 at 20:04
Professor VectorProfessor Vector
11.2k11333
$endgroup$
$begingroup$
Hey professor Vector, would you mind giving me more steps such that I can follow your answer? Many thanks in advance.
$endgroup$
– Ray
Jun 15 '17 at 20:44
$begingroup$
In principle, the above derivation is valid for $n=1,$ too, but there are caveats: it's valid if you assume $U_-1(x)=0.$ That's in agreement with the definition $$U_n(cos t)=fracsin(n+1)tsin t,$$ and even $T'_0(x)=0cdot U_-1(x)$ is true, you just can't divide both sides by $0,$ you have to omit the terms originating from $U_-1$. So the above recurrence for $n=1$ becomes $$I_1(x)=frac14T_2(x),e^ax-fraca4I_2(x).$$
$endgroup$
– Professor Vector
Jun 16 '17 at 7:16
$begingroup$
Thanks Professor Vector.
$endgroup$
– Ray
Jun 17 '17 at 0:55
add a comment |
$begingroup$
The question isn't wrong, the hint is... a bit misleading, I'd say. You might use
$$T_n(x)=frac12(U_n(x)-U_n-2(x)),$$ for $nge2$ together with
$$T'_n(x)=n,U_n-1(x)$$ and partial integration, to get a recurrence relation for your integrals (the $U_n$ are Chebyshev polynomials of second kind).
Here are more details: Let's define $$I_n(x)=int T_n(x),e^ax,dx.$$ Since $$T_n(x)=frac12left(fracT'_n+1(x)n+1-fracT'_n-1(x)n-1right),$$ multiplying by $e^ax$ and integrating (by parts) gives
$$I_n(x)=frac12left(fracT_n+1(x)n+1-fracT_n-1(x)n-1right),e^ax-fraca2left(fracI_n+1(x)n+1-fracI_n-1(x)n-1right).$$ I'll omit integration constants, because they cancel out when calculating definite integrals. If people insist in them, feel free to add any constant you like best. Now you can solve for $I_n+1(x)$, and calculate it
starting from $n=2$ recursively from $I_n-1(x)$ and $I_n(x).$ To prime the pump, you'll have to calculate $I_0(x)$, $I_1(x)$ and $I_2(x)$ manually, but with $T_0(x)=1$, $T_1(x)=x$ and $T_2(x)=2x^2-1$ that's a very simple task I won't explain here.
answered Jun 15 '17 at 20:04
Professor VectorProfessor Vector
11.2k11333
$endgroup$
The question isn't wrong, the hint is... a bit misleading, I'd say. You might use
$$T_n(x)=frac12(U_n(x)-U_n-2(x)),$$ for $nge2$ together with
$$T'_n(x)=n,U_n-1(x)$$ and partial integration, to get a recurrence relation for your integrals (the $U_n$ are Chebyshev polynomials of second kind).
Here are more details: Let's define $$I_n(x)=int T_n(x),e^ax,dx.$$ Since $$T_n(x)=frac12left(fracT'_n+1(x)n+1-fracT'_n-1(x)n-1right),$$ multiplying by $e^ax$ and integrating (by parts) gives
$$I_n(x)=frac12left(fracT_n+1(x)n+1-fracT_n-1(x)n-1right),e^ax-fraca2left(fracI_n+1(x)n+1-fracI_n-1(x)n-1right).$$ I'll omit integration constants, because they cancel out when calculating definite integrals. If people insist in them, feel free to add any constant you like best. Now you can solve for $I_n+1(x)$, and calculate it
starting from $n=2$ recursively from $I_n-1(x)$ and $I_n(x).$ To prime the pump, you'll have to calculate $I_0(x)$, $I_1(x)$ and $I_2(x)$ manually, but with $T_0(x)=1$, $T_1(x)=x$ and $T_2(x)=2x^2-1$ that's a very simple task I won't explain here.
answered Jun 15 '17 at 20:04
Professor VectorProfessor Vector
11.2k11333
edited Jun 16 '17 at 4:25
answered Jun 15 '17 at 20:04
Professor VectorProfessor Vector
11.2k11333
answered Jun 15 '17 at 20:04
Professor VectorProfessor Vector
11.2k11333
answered Jun 15 '17 at 20:04
Professor VectorProfessor Vector
11.2k11333
11.2k11333
$begingroup$
Hey professor Vector, would you mind giving me more steps such that I can follow your answer? Many thanks in advance.
$endgroup$
– Ray
Jun 15 '17 at 20:44
$begingroup$
In principle, the above derivation is valid for $n=1,$ too, but there are caveats: it's valid if you assume $U_-1(x)=0.$ That's in agreement with the definition $$U_n(cos t)=fracsin(n+1)tsin t,$$ and even $T'_0(x)=0cdot U_-1(x)$ is true, you just can't divide both sides by $0,$ you have to omit the terms originating from $U_-1$. So the above recurrence for $n=1$ becomes $$I_1(x)=frac14T_2(x),e^ax-fraca4I_2(x).$$
$endgroup$
– Professor Vector
Jun 16 '17 at 7:16
$begingroup$
Thanks Professor Vector.
$endgroup$
– Ray
Jun 17 '17 at 0:55
add a comment |
$begingroup$
Hey professor Vector, would you mind giving me more steps such that I can follow your answer? Many thanks in advance.
$endgroup$
– Ray
Jun 15 '17 at 20:44
$begingroup$
In principle, the above derivation is valid for $n=1,$ too, but there are caveats: it's valid if you assume $U_-1(x)=0.$ That's in agreement with the definition $$U_n(cos t)=fracsin(n+1)tsin t,$$ and even $T'_0(x)=0cdot U_-1(x)$ is true, you just can't divide both sides by $0,$ you have to omit the terms originating from $U_-1$. So the above recurrence for $n=1$ becomes $$I_1(x)=frac14T_2(x),e^ax-fraca4I_2(x).$$
$endgroup$
– Professor Vector
Jun 16 '17 at 7:16
$begingroup$
Thanks Professor Vector.
$endgroup$
– Ray
Jun 17 '17 at 0:55
$begingroup$
Hey professor Vector, would you mind giving me more steps such that I can follow your answer? Many thanks in advance.
$endgroup$
– Ray
Jun 15 '17 at 20:44
$begingroup$
Hey professor Vector, would you mind giving me more steps such that I can follow your answer? Many thanks in advance.
$endgroup$
– Ray
Jun 15 '17 at 20:44
$begingroup$
In principle, the above derivation is valid for $n=1,$ too, but there are caveats: it's valid if you assume $U_-1(x)=0.$ That's in agreement with the definition $$U_n(cos t)=fracsin(n+1)tsin t,$$ and even $T'_0(x)=0cdot U_-1(x)$ is true, you just can't divide both sides by $0,$ you have to omit the terms originating from $U_-1$. So the above recurrence for $n=1$ becomes $$I_1(x)=frac14T_2(x),e^ax-fraca4I_2(x).$$
$endgroup$
– Professor Vector
Jun 16 '17 at 7:16
$begingroup$
In principle, the above derivation is valid for $n=1,$ too, but there are caveats: it's valid if you assume $U_-1(x)=0.$ That's in agreement with the definition $$U_n(cos t)=fracsin(n+1)tsin t,$$ and even $T'_0(x)=0cdot U_-1(x)$ is true, you just can't divide both sides by $0,$ you have to omit the terms originating from $U_-1$. So the above recurrence for $n=1$ becomes $$I_1(x)=frac14T_2(x),e^ax-fraca4I_2(x).$$
$endgroup$
– Professor Vector
Jun 16 '17 at 7:16
$begingroup$
Thanks Professor Vector.
$endgroup$
– Ray
Jun 17 '17 at 0:55
$begingroup$
Thanks Professor Vector.
$endgroup$
– Ray
Jun 17 '17 at 0:55
add a comment |
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$begingroup$
First forget that they are Chebyshev's. If they were just polynomials what would you do? Integration by parts $n$ times, no? And of course the part in which one uses that they are Tchebyshev's.
$endgroup$
– OR.
Jun 15 '17 at 19:40