How to Change the Interval of Interpolation from [-1,1] to [a,b] for Chebyshev Nodes
(According to this website:http://fac-staff.seattleu.edu/difranco/web/Math_371_W11/Files/Chebyshevnodes.pdf)
Between [-1,1], the Chebyshev Nodes are given as:
$x_k = cosBig((2k-1)pi/2n)Big), k=1,......,n$
and over [a,b] it is given as:
$x_k= 0.5(a+b) +0.5(b-a)cosBig((2k-1)(3.14159)/2n)Big)$
What is the logic behind this transformation?
Similarly, the maximum error over [-1,1] is given as : $1/2^{n-1}$
Over [a,b], why is the error: $(b-a)^{n+1}/2^{2n+1}$?
chebyshev-polynomials
edited Aug 7 '15 at 7:41
gammatester
16.6k21632
asked Aug 7 '15 at 0:08
stats555
114
add a comment |
(According to this website:http://fac-staff.seattleu.edu/difranco/web/Math_371_W11/Files/Chebyshevnodes.pdf)
Between [-1,1], the Chebyshev Nodes are given as:
$x_k = cosBig((2k-1)pi/2n)Big), k=1,......,n$
and over [a,b] it is given as:
$x_k= 0.5(a+b) +0.5(b-a)cosBig((2k-1)(3.14159)/2n)Big)$
What is the logic behind this transformation?
Similarly, the maximum error over [-1,1] is given as : $1/2^{n-1}$
Over [a,b], why is the error: $(b-a)^{n+1}/2^{2n+1}$?
chebyshev-polynomials
edited Aug 7 '15 at 7:41
gammatester
16.6k21632
asked Aug 7 '15 at 0:08
stats555
114
I replace $3.14159$ by $pi$, note that you erroneously used $pi$ instead on $n$ at two places.
– gammatester
Aug 7 '15 at 7:45
add a comment |
(According to this website:http://fac-staff.seattleu.edu/difranco/web/Math_371_W11/Files/Chebyshevnodes.pdf)
Between [-1,1], the Chebyshev Nodes are given as:
$x_k = cosBig((2k-1)pi/2n)Big), k=1,......,n$
and over [a,b] it is given as:
$x_k= 0.5(a+b) +0.5(b-a)cosBig((2k-1)(3.14159)/2n)Big)$
What is the logic behind this transformation?
Similarly, the maximum error over [-1,1] is given as : $1/2^{n-1}$
Over [a,b], why is the error: $(b-a)^{n+1}/2^{2n+1}$?
chebyshev-polynomials
edited Aug 7 '15 at 7:41
gammatester
16.6k21632
asked Aug 7 '15 at 0:08
stats555
114
(According to this website:http://fac-staff.seattleu.edu/difranco/web/Math_371_W11/Files/Chebyshevnodes.pdf)
Between [-1,1], the Chebyshev Nodes are given as:
$x_k = cosBig((2k-1)pi/2n)Big), k=1,......,n$
and over [a,b] it is given as:
$x_k= 0.5(a+b) +0.5(b-a)cosBig((2k-1)(3.14159)/2n)Big)$
What is the logic behind this transformation?
Similarly, the maximum error over [-1,1] is given as : $1/2^{n-1}$
Over [a,b], why is the error: $(b-a)^{n+1}/2^{2n+1}$?
chebyshev-polynomials
chebyshev-polynomials
edited Aug 7 '15 at 7:41
gammatester
16.6k21632
asked Aug 7 '15 at 0:08
stats555
114
edited Aug 7 '15 at 7:41
gammatester
16.6k21632
asked Aug 7 '15 at 0:08
stats555
114
edited Aug 7 '15 at 7:41
gammatester
16.6k21632
edited Aug 7 '15 at 7:41
gammatester
16.6k21632
edited Aug 7 '15 at 7:41
gammatester
16.6k21632
16.6k21632
asked Aug 7 '15 at 0:08
stats555
114
asked Aug 7 '15 at 0:08
stats555
114
asked Aug 7 '15 at 0:08
stats555
114
114
I replace $3.14159$ by $pi$, note that you erroneously used $pi$ instead on $n$ at two places.
– gammatester
Aug 7 '15 at 7:45
add a comment |
I replace $3.14159$ by $pi$, note that you erroneously used $pi$ instead on $n$ at two places.
– gammatester
Aug 7 '15 at 7:45
I replace $3.14159$ by $pi$, note that you erroneously used $pi$ instead on $n$ at two places.
– gammatester
Aug 7 '15 at 7:45
I replace $3.14159$ by $pi$, note that you erroneously used $pi$ instead on $n$ at two places.
– gammatester
Aug 7 '15 at 7:45
add a comment |
1 Answer
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This is the simple linear transformation $T: [-1,1] rightarrow [a,b],quad T(x)=alpha x + beta,$ and $T(-1)=a,; T(+1)=b.$ Now use
$$a=T(-1)=-alpha + beta,quad b=T(+1)=+alpha + beta$$
to get the values
$$alpha = frac{1}{2}(b-a),quad beta = frac{1}{2}(b+a)$$
It is not quite clear what $n$ is in your example (you have non standard indices $1..n,$ but the error scales with $alpha^{n+1}$ for a polynomial with degree $n+1$ and you get
$$mathrm{err} = frac{1}{2^n} alpha^{n+1}=frac{1}{2^n} left(frac{b-a}{2}right)^{n+1}=frac{1}{2^{n+1}} left(b-aright)^{n+1}$$
answered Aug 7 '15 at 7:15
gammatester
16.6k21632
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
This is the simple linear transformation $T: [-1,1] rightarrow [a,b],quad T(x)=alpha x + beta,$ and $T(-1)=a,; T(+1)=b.$ Now use
$$a=T(-1)=-alpha + beta,quad b=T(+1)=+alpha + beta$$
to get the values
$$alpha = frac{1}{2}(b-a),quad beta = frac{1}{2}(b+a)$$
It is not quite clear what $n$ is in your example (you have non standard indices $1..n,$ but the error scales with $alpha^{n+1}$ for a polynomial with degree $n+1$ and you get
$$mathrm{err} = frac{1}{2^n} alpha^{n+1}=frac{1}{2^n} left(frac{b-a}{2}right)^{n+1}=frac{1}{2^{n+1}} left(b-aright)^{n+1}$$
answered Aug 7 '15 at 7:15
gammatester
16.6k21632
add a comment |
This is the simple linear transformation $T: [-1,1] rightarrow [a,b],quad T(x)=alpha x + beta,$ and $T(-1)=a,; T(+1)=b.$ Now use
$$a=T(-1)=-alpha + beta,quad b=T(+1)=+alpha + beta$$
to get the values
$$alpha = frac{1}{2}(b-a),quad beta = frac{1}{2}(b+a)$$
It is not quite clear what $n$ is in your example (you have non standard indices $1..n,$ but the error scales with $alpha^{n+1}$ for a polynomial with degree $n+1$ and you get
$$mathrm{err} = frac{1}{2^n} alpha^{n+1}=frac{1}{2^n} left(frac{b-a}{2}right)^{n+1}=frac{1}{2^{n+1}} left(b-aright)^{n+1}$$
answered Aug 7 '15 at 7:15
gammatester
16.6k21632
add a comment |
This is the simple linear transformation $T: [-1,1] rightarrow [a,b],quad T(x)=alpha x + beta,$ and $T(-1)=a,; T(+1)=b.$ Now use
$$a=T(-1)=-alpha + beta,quad b=T(+1)=+alpha + beta$$
to get the values
$$alpha = frac{1}{2}(b-a),quad beta = frac{1}{2}(b+a)$$
It is not quite clear what $n$ is in your example (you have non standard indices $1..n,$ but the error scales with $alpha^{n+1}$ for a polynomial with degree $n+1$ and you get
$$mathrm{err} = frac{1}{2^n} alpha^{n+1}=frac{1}{2^n} left(frac{b-a}{2}right)^{n+1}=frac{1}{2^{n+1}} left(b-aright)^{n+1}$$
answered Aug 7 '15 at 7:15
gammatester
16.6k21632
This is the simple linear transformation $T: [-1,1] rightarrow [a,b],quad T(x)=alpha x + beta,$ and $T(-1)=a,; T(+1)=b.$ Now use
$$a=T(-1)=-alpha + beta,quad b=T(+1)=+alpha + beta$$
to get the values
$$alpha = frac{1}{2}(b-a),quad beta = frac{1}{2}(b+a)$$
It is not quite clear what $n$ is in your example (you have non standard indices $1..n,$ but the error scales with $alpha^{n+1}$ for a polynomial with degree $n+1$ and you get
$$mathrm{err} = frac{1}{2^n} alpha^{n+1}=frac{1}{2^n} left(frac{b-a}{2}right)^{n+1}=frac{1}{2^{n+1}} left(b-aright)^{n+1}$$
answered Aug 7 '15 at 7:15
gammatester
16.6k21632
edited Aug 7 '15 at 7:35
answered Aug 7 '15 at 7:15
gammatester
16.6k21632
answered Aug 7 '15 at 7:15
gammatester
16.6k21632
answered Aug 7 '15 at 7:15
gammatester
16.6k21632
16.6k21632
add a comment |
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I replace $3.14159$ by $pi$, note that you erroneously used $pi$ instead on $n$ at two places.
– gammatester
Aug 7 '15 at 7:45