Determinant of this $N$ x $N$ matrix
$begingroup$
Let $nu$ be the minimum integer which satisfies $2sin(frac{pi nu}{2(N+1)})>tau$, for $N$ an integer, and $tau$ an arbitrary positive number. Since the LHS is bounded and RHS is not, there might be cases where no such value of $nu$ satisfying the inequality does exist. In that case, we will take $nu=0$.
Now consider the polynomial:($a=tau^2-2)$
$$sum_{i=1}^{nu-1}ac_{i}^2-sum_{i=nu}^{N}ac_{i}^2+2sum_{i=1}^{nu-2}c_{i}c_{i+1}+2ic_{nu}c_{nu-1}-2sum_{i=nu}^{N}c_{i}c_{i+1}$$
Suppose I want to express the above as $c^{T}Ac$ with $c^{T}=begin{matrix}
(c_{1} & c_{2} ... &c_{N})
end{matrix}$
I want to calculate the NxN determinant of A:
$$ begin{matrix}
a & 1 & 0 & 0 .....&0 \
1 & a & 1 & 0.......&0 \
0 & 1 & a &1....... &0 \
0 & 0 & 1 &a.....\
.\
.\
0.\
end{matrix}
$$
So we have repeating blocks of $begin{matrix}
a & 1\
1 & a\
end{matrix}$ until the $(nu-2)^{th}$ row and similar repeating blocks of $begin{matrix}
-a & -1\
-1 & -a\
end{matrix}$ from the $(nu+1)^{th}$ row and a block $begin{matrix}
a & i\
i & -a\
end{matrix}$ in the $(nu-1)^{th}$ and $(nu)^{th}$ rows. I could not write down the matrix here, so I would be obliged greatly if you check it yourself. As you can guess I am doing to this calculate $int_{-infty}^{infty}dc e^{-c^{T}Ac}$
determinant
edited Jan 7 at 3:21
Gnumbertester
1675
asked Jan 7 at 2:16
Mani JhaMani Jha
93
$endgroup$
add a comment |
$begingroup$
Let $nu$ be the minimum integer which satisfies $2sin(frac{pi nu}{2(N+1)})>tau$, for $N$ an integer, and $tau$ an arbitrary positive number. Since the LHS is bounded and RHS is not, there might be cases where no such value of $nu$ satisfying the inequality does exist. In that case, we will take $nu=0$.
Now consider the polynomial:($a=tau^2-2)$
$$sum_{i=1}^{nu-1}ac_{i}^2-sum_{i=nu}^{N}ac_{i}^2+2sum_{i=1}^{nu-2}c_{i}c_{i+1}+2ic_{nu}c_{nu-1}-2sum_{i=nu}^{N}c_{i}c_{i+1}$$
Suppose I want to express the above as $c^{T}Ac$ with $c^{T}=begin{matrix}
(c_{1} & c_{2} ... &c_{N})
end{matrix}$
I want to calculate the NxN determinant of A:
$$ begin{matrix}
a & 1 & 0 & 0 .....&0 \
1 & a & 1 & 0.......&0 \
0 & 1 & a &1....... &0 \
0 & 0 & 1 &a.....\
.\
.\
0.\
end{matrix}
$$
So we have repeating blocks of $begin{matrix}
a & 1\
1 & a\
end{matrix}$ until the $(nu-2)^{th}$ row and similar repeating blocks of $begin{matrix}
-a & -1\
-1 & -a\
end{matrix}$ from the $(nu+1)^{th}$ row and a block $begin{matrix}
a & i\
i & -a\
end{matrix}$ in the $(nu-1)^{th}$ and $(nu)^{th}$ rows. I could not write down the matrix here, so I would be obliged greatly if you check it yourself. As you can guess I am doing to this calculate $int_{-infty}^{infty}dc e^{-c^{T}Ac}$
determinant
edited Jan 7 at 3:21
Gnumbertester
1675
asked Jan 7 at 2:16
Mani JhaMani Jha
93
$endgroup$
add a comment |
$begingroup$
Let $nu$ be the minimum integer which satisfies $2sin(frac{pi nu}{2(N+1)})>tau$, for $N$ an integer, and $tau$ an arbitrary positive number. Since the LHS is bounded and RHS is not, there might be cases where no such value of $nu$ satisfying the inequality does exist. In that case, we will take $nu=0$.
Now consider the polynomial:($a=tau^2-2)$
$$sum_{i=1}^{nu-1}ac_{i}^2-sum_{i=nu}^{N}ac_{i}^2+2sum_{i=1}^{nu-2}c_{i}c_{i+1}+2ic_{nu}c_{nu-1}-2sum_{i=nu}^{N}c_{i}c_{i+1}$$
Suppose I want to express the above as $c^{T}Ac$ with $c^{T}=begin{matrix}
(c_{1} & c_{2} ... &c_{N})
end{matrix}$
I want to calculate the NxN determinant of A:
$$ begin{matrix}
a & 1 & 0 & 0 .....&0 \
1 & a & 1 & 0.......&0 \
0 & 1 & a &1....... &0 \
0 & 0 & 1 &a.....\
.\
.\
0.\
end{matrix}
$$
So we have repeating blocks of $begin{matrix}
a & 1\
1 & a\
end{matrix}$ until the $(nu-2)^{th}$ row and similar repeating blocks of $begin{matrix}
-a & -1\
-1 & -a\
end{matrix}$ from the $(nu+1)^{th}$ row and a block $begin{matrix}
a & i\
i & -a\
end{matrix}$ in the $(nu-1)^{th}$ and $(nu)^{th}$ rows. I could not write down the matrix here, so I would be obliged greatly if you check it yourself. As you can guess I am doing to this calculate $int_{-infty}^{infty}dc e^{-c^{T}Ac}$
determinant
edited Jan 7 at 3:21
Gnumbertester
1675
asked Jan 7 at 2:16
Mani JhaMani Jha
93
$endgroup$
Let $nu$ be the minimum integer which satisfies $2sin(frac{pi nu}{2(N+1)})>tau$, for $N$ an integer, and $tau$ an arbitrary positive number. Since the LHS is bounded and RHS is not, there might be cases where no such value of $nu$ satisfying the inequality does exist. In that case, we will take $nu=0$.
Now consider the polynomial:($a=tau^2-2)$
$$sum_{i=1}^{nu-1}ac_{i}^2-sum_{i=nu}^{N}ac_{i}^2+2sum_{i=1}^{nu-2}c_{i}c_{i+1}+2ic_{nu}c_{nu-1}-2sum_{i=nu}^{N}c_{i}c_{i+1}$$
Suppose I want to express the above as $c^{T}Ac$ with $c^{T}=begin{matrix}
(c_{1} & c_{2} ... &c_{N})
end{matrix}$
I want to calculate the NxN determinant of A:
$$ begin{matrix}
a & 1 & 0 & 0 .....&0 \
1 & a & 1 & 0.......&0 \
0 & 1 & a &1....... &0 \
0 & 0 & 1 &a.....\
.\
.\
0.\
end{matrix}
$$
So we have repeating blocks of $begin{matrix}
a & 1\
1 & a\
end{matrix}$ until the $(nu-2)^{th}$ row and similar repeating blocks of $begin{matrix}
-a & -1\
-1 & -a\
end{matrix}$ from the $(nu+1)^{th}$ row and a block $begin{matrix}
a & i\
i & -a\
end{matrix}$ in the $(nu-1)^{th}$ and $(nu)^{th}$ rows. I could not write down the matrix here, so I would be obliged greatly if you check it yourself. As you can guess I am doing to this calculate $int_{-infty}^{infty}dc e^{-c^{T}Ac}$
determinant
determinant
edited Jan 7 at 3:21
Gnumbertester
1675
asked Jan 7 at 2:16
Mani JhaMani Jha
93
edited Jan 7 at 3:21
Gnumbertester
1675
asked Jan 7 at 2:16
Mani JhaMani Jha
93
edited Jan 7 at 3:21
Gnumbertester
1675
edited Jan 7 at 3:21
Gnumbertester
1675
edited Jan 7 at 3:21
Gnumbertester
1675
1675
asked Jan 7 at 2:16
Mani JhaMani Jha
93
asked Jan 7 at 2:16
Mani JhaMani Jha
93
asked Jan 7 at 2:16
Mani JhaMani Jha
93
93
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$D_n=begin{vmatrix}
a & 1 & 0 & cdots & 0 & 0 & 0\
1 & a & 1 & cdots & 0 & 0 & 0\
0 & 1 & a & cdots & 0 & 0 & 0\
cdots & cdots & cdots & cdots &cdots & cdots & cdots\
0 & 0 & 0 & cdots & a & 1 & 0\
0 & 0 & 0 & cdots & 1 & a & 1\
0 & 0 & 0 &cdots & 0 & 1 & aend{vmatrix}$$
$$D_1=a,\D_2={{a}^{2}}-1,\D_3={{a}^{3}}-2 a,\D_4={{a}^{4}}-3 {{a}^{2}}+1,\D_5={{a}^{5}}-4 {{a}^{3}}+3a,\ dots\ D_n=aD_{n-1}-D_{n-2}$$
Determinants $D_n$ is Chebyshev polynomials of the second kind:
$$D_n=U_nleft(frac{a}2right)$$
https://en.wikipedia.org/wiki/Chebyshev_polynomials
http://mathworld.wolfram.com/ChebyshevPolynomialoftheSecondKind.html
answered Jan 7 at 8:26
Aleksas DomarkasAleksas Domarkas
8546
$endgroup$
$begingroup$
Thank you for your answer but after there are blocks of $begin{matrix} -a & -1\ -1 & -a\ end{matrix}$ after the $(nu)^{th}$ row
$endgroup$
– Mani Jha
Jan 7 at 11:43
$begingroup$
Can you find the inverse of the matrix you calculated the determinant of? The adjoint calculation gets messy..is there a shorter way?
$endgroup$
– Mani Jha
Jan 8 at 10:44
add a comment |
Your Answer
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1 Answer
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1 Answer
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active
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votes
$begingroup$
$$D_n=begin{vmatrix}
a & 1 & 0 & cdots & 0 & 0 & 0\
1 & a & 1 & cdots & 0 & 0 & 0\
0 & 1 & a & cdots & 0 & 0 & 0\
cdots & cdots & cdots & cdots &cdots & cdots & cdots\
0 & 0 & 0 & cdots & a & 1 & 0\
0 & 0 & 0 & cdots & 1 & a & 1\
0 & 0 & 0 &cdots & 0 & 1 & aend{vmatrix}$$
$$D_1=a,\D_2={{a}^{2}}-1,\D_3={{a}^{3}}-2 a,\D_4={{a}^{4}}-3 {{a}^{2}}+1,\D_5={{a}^{5}}-4 {{a}^{3}}+3a,\ dots\ D_n=aD_{n-1}-D_{n-2}$$
Determinants $D_n$ is Chebyshev polynomials of the second kind:
$$D_n=U_nleft(frac{a}2right)$$
https://en.wikipedia.org/wiki/Chebyshev_polynomials
http://mathworld.wolfram.com/ChebyshevPolynomialoftheSecondKind.html
answered Jan 7 at 8:26
Aleksas DomarkasAleksas Domarkas
8546
$endgroup$
$begingroup$
Thank you for your answer but after there are blocks of $begin{matrix} -a & -1\ -1 & -a\ end{matrix}$ after the $(nu)^{th}$ row
$endgroup$
– Mani Jha
Jan 7 at 11:43
$begingroup$
Can you find the inverse of the matrix you calculated the determinant of? The adjoint calculation gets messy..is there a shorter way?
$endgroup$
– Mani Jha
Jan 8 at 10:44
add a comment |
$begingroup$
$$D_n=begin{vmatrix}
a & 1 & 0 & cdots & 0 & 0 & 0\
1 & a & 1 & cdots & 0 & 0 & 0\
0 & 1 & a & cdots & 0 & 0 & 0\
cdots & cdots & cdots & cdots &cdots & cdots & cdots\
0 & 0 & 0 & cdots & a & 1 & 0\
0 & 0 & 0 & cdots & 1 & a & 1\
0 & 0 & 0 &cdots & 0 & 1 & aend{vmatrix}$$
$$D_1=a,\D_2={{a}^{2}}-1,\D_3={{a}^{3}}-2 a,\D_4={{a}^{4}}-3 {{a}^{2}}+1,\D_5={{a}^{5}}-4 {{a}^{3}}+3a,\ dots\ D_n=aD_{n-1}-D_{n-2}$$
Determinants $D_n$ is Chebyshev polynomials of the second kind:
$$D_n=U_nleft(frac{a}2right)$$
https://en.wikipedia.org/wiki/Chebyshev_polynomials
http://mathworld.wolfram.com/ChebyshevPolynomialoftheSecondKind.html
answered Jan 7 at 8:26
Aleksas DomarkasAleksas Domarkas
8546
$endgroup$
$begingroup$
Thank you for your answer but after there are blocks of $begin{matrix} -a & -1\ -1 & -a\ end{matrix}$ after the $(nu)^{th}$ row
$endgroup$
– Mani Jha
Jan 7 at 11:43
$begingroup$
Can you find the inverse of the matrix you calculated the determinant of? The adjoint calculation gets messy..is there a shorter way?
$endgroup$
– Mani Jha
Jan 8 at 10:44
add a comment |
$begingroup$
$$D_n=begin{vmatrix}
a & 1 & 0 & cdots & 0 & 0 & 0\
1 & a & 1 & cdots & 0 & 0 & 0\
0 & 1 & a & cdots & 0 & 0 & 0\
cdots & cdots & cdots & cdots &cdots & cdots & cdots\
0 & 0 & 0 & cdots & a & 1 & 0\
0 & 0 & 0 & cdots & 1 & a & 1\
0 & 0 & 0 &cdots & 0 & 1 & aend{vmatrix}$$
$$D_1=a,\D_2={{a}^{2}}-1,\D_3={{a}^{3}}-2 a,\D_4={{a}^{4}}-3 {{a}^{2}}+1,\D_5={{a}^{5}}-4 {{a}^{3}}+3a,\ dots\ D_n=aD_{n-1}-D_{n-2}$$
Determinants $D_n$ is Chebyshev polynomials of the second kind:
$$D_n=U_nleft(frac{a}2right)$$
https://en.wikipedia.org/wiki/Chebyshev_polynomials
http://mathworld.wolfram.com/ChebyshevPolynomialoftheSecondKind.html
answered Jan 7 at 8:26
Aleksas DomarkasAleksas Domarkas
8546
$endgroup$
$$D_n=begin{vmatrix}
a & 1 & 0 & cdots & 0 & 0 & 0\
1 & a & 1 & cdots & 0 & 0 & 0\
0 & 1 & a & cdots & 0 & 0 & 0\
cdots & cdots & cdots & cdots &cdots & cdots & cdots\
0 & 0 & 0 & cdots & a & 1 & 0\
0 & 0 & 0 & cdots & 1 & a & 1\
0 & 0 & 0 &cdots & 0 & 1 & aend{vmatrix}$$
$$D_1=a,\D_2={{a}^{2}}-1,\D_3={{a}^{3}}-2 a,\D_4={{a}^{4}}-3 {{a}^{2}}+1,\D_5={{a}^{5}}-4 {{a}^{3}}+3a,\ dots\ D_n=aD_{n-1}-D_{n-2}$$
Determinants $D_n$ is Chebyshev polynomials of the second kind:
$$D_n=U_nleft(frac{a}2right)$$
https://en.wikipedia.org/wiki/Chebyshev_polynomials
http://mathworld.wolfram.com/ChebyshevPolynomialoftheSecondKind.html
answered Jan 7 at 8:26
Aleksas DomarkasAleksas Domarkas
8546
answered Jan 7 at 8:26
Aleksas DomarkasAleksas Domarkas
8546
answered Jan 7 at 8:26
Aleksas DomarkasAleksas Domarkas
8546
answered Jan 7 at 8:26
Aleksas DomarkasAleksas Domarkas
8546
8546
$begingroup$
Thank you for your answer but after there are blocks of $begin{matrix} -a & -1\ -1 & -a\ end{matrix}$ after the $(nu)^{th}$ row
$endgroup$
– Mani Jha
Jan 7 at 11:43
$begingroup$
Can you find the inverse of the matrix you calculated the determinant of? The adjoint calculation gets messy..is there a shorter way?
$endgroup$
– Mani Jha
Jan 8 at 10:44
add a comment |
$begingroup$
Thank you for your answer but after there are blocks of $begin{matrix} -a & -1\ -1 & -a\ end{matrix}$ after the $(nu)^{th}$ row
$endgroup$
– Mani Jha
Jan 7 at 11:43
$begingroup$
Can you find the inverse of the matrix you calculated the determinant of? The adjoint calculation gets messy..is there a shorter way?
$endgroup$
– Mani Jha
Jan 8 at 10:44
$begingroup$
Thank you for your answer but after there are blocks of $begin{matrix} -a & -1\ -1 & -a\ end{matrix}$ after the $(nu)^{th}$ row
$endgroup$
– Mani Jha
Jan 7 at 11:43
$begingroup$
Thank you for your answer but after there are blocks of $begin{matrix} -a & -1\ -1 & -a\ end{matrix}$ after the $(nu)^{th}$ row
$endgroup$
– Mani Jha
Jan 7 at 11:43
$begingroup$
Can you find the inverse of the matrix you calculated the determinant of? The adjoint calculation gets messy..is there a shorter way?
$endgroup$
– Mani Jha
Jan 8 at 10:44
$begingroup$
Can you find the inverse of the matrix you calculated the determinant of? The adjoint calculation gets messy..is there a shorter way?
$endgroup$
– Mani Jha
Jan 8 at 10:44
add a comment |
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