Determinant of this $N$ x $N$ matrix












0












$begingroup$


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







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    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







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited Jan 7 at 3:21









      Gnumbertester

      1675




      1675










      asked Jan 7 at 2:16









      Mani JhaMani Jha

      93




      93






















          1 Answer
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          $begingroup$

          $$D_n=begin{vmatrix}
          a & 1 & 0 & cdots & 0 & 0 & 0\
          1 & a & 1 & cdots & 0 & 0 & 0\
          0 & 1 & a & cdots & 0 & 0 & 0\
          cdots & cdots & cdots & cdots &cdots & cdots & cdots\
          0 & 0 & 0 & cdots & a & 1 & 0\
          0 & 0 & 0 & cdots & 1 & a & 1\
          0 & 0 & 0 &cdots & 0 & 1 & aend{vmatrix}$$



          $$D_1=a,\D_2={{a}^{2}}-1,\D_3={{a}^{3}}-2 a,\D_4={{a}^{4}}-3 {{a}^{2}}+1,\D_5={{a}^{5}}-4 {{a}^{3}}+3a,\ dots\ D_n=aD_{n-1}-D_{n-2}$$
          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






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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











          Your Answer





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          1 Answer
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          1 Answer
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          0












          $begingroup$

          $$D_n=begin{vmatrix}
          a & 1 & 0 & cdots & 0 & 0 & 0\
          1 & a & 1 & cdots & 0 & 0 & 0\
          0 & 1 & a & cdots & 0 & 0 & 0\
          cdots & cdots & cdots & cdots &cdots & cdots & cdots\
          0 & 0 & 0 & cdots & a & 1 & 0\
          0 & 0 & 0 & cdots & 1 & a & 1\
          0 & 0 & 0 &cdots & 0 & 1 & aend{vmatrix}$$



          $$D_1=a,\D_2={{a}^{2}}-1,\D_3={{a}^{3}}-2 a,\D_4={{a}^{4}}-3 {{a}^{2}}+1,\D_5={{a}^{5}}-4 {{a}^{3}}+3a,\ dots\ D_n=aD_{n-1}-D_{n-2}$$
          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






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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
















          0












          $begingroup$

          $$D_n=begin{vmatrix}
          a & 1 & 0 & cdots & 0 & 0 & 0\
          1 & a & 1 & cdots & 0 & 0 & 0\
          0 & 1 & a & cdots & 0 & 0 & 0\
          cdots & cdots & cdots & cdots &cdots & cdots & cdots\
          0 & 0 & 0 & cdots & a & 1 & 0\
          0 & 0 & 0 & cdots & 1 & a & 1\
          0 & 0 & 0 &cdots & 0 & 1 & aend{vmatrix}$$



          $$D_1=a,\D_2={{a}^{2}}-1,\D_3={{a}^{3}}-2 a,\D_4={{a}^{4}}-3 {{a}^{2}}+1,\D_5={{a}^{5}}-4 {{a}^{3}}+3a,\ dots\ D_n=aD_{n-1}-D_{n-2}$$
          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






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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














          0












          0








          0





          $begingroup$

          $$D_n=begin{vmatrix}
          a & 1 & 0 & cdots & 0 & 0 & 0\
          1 & a & 1 & cdots & 0 & 0 & 0\
          0 & 1 & a & cdots & 0 & 0 & 0\
          cdots & cdots & cdots & cdots &cdots & cdots & cdots\
          0 & 0 & 0 & cdots & a & 1 & 0\
          0 & 0 & 0 & cdots & 1 & a & 1\
          0 & 0 & 0 &cdots & 0 & 1 & aend{vmatrix}$$



          $$D_1=a,\D_2={{a}^{2}}-1,\D_3={{a}^{3}}-2 a,\D_4={{a}^{4}}-3 {{a}^{2}}+1,\D_5={{a}^{5}}-4 {{a}^{3}}+3a,\ dots\ D_n=aD_{n-1}-D_{n-2}$$
          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






          share|cite|improve this answer









          $endgroup$



          $$D_n=begin{vmatrix}
          a & 1 & 0 & cdots & 0 & 0 & 0\
          1 & a & 1 & cdots & 0 & 0 & 0\
          0 & 1 & a & cdots & 0 & 0 & 0\
          cdots & cdots & cdots & cdots &cdots & cdots & cdots\
          0 & 0 & 0 & cdots & a & 1 & 0\
          0 & 0 & 0 & cdots & 1 & a & 1\
          0 & 0 & 0 &cdots & 0 & 1 & aend{vmatrix}$$



          $$D_1=a,\D_2={{a}^{2}}-1,\D_3={{a}^{3}}-2 a,\D_4={{a}^{4}}-3 {{a}^{2}}+1,\D_5={{a}^{5}}-4 {{a}^{3}}+3a,\ dots\ D_n=aD_{n-1}-D_{n-2}$$
          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







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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


















          • $begingroup$
            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


















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