Approximaths

Hankel determinant

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In the previous posts, we have seen that in order to compute the Padé coefficients P(m,n) corresponding to a given geometric series we have to be invert the following matrix:

\displaystyle \begin{pmatrix} C_{m-n+1} & C_{m-n+2} & \ldots & C_{m} \\ C_{m-n+2} & C_{m-n+3} & \ldots & C_{m+1} \\ \vdots & & & \vdots \\ C_{m} & C_{m+1} & \ldots & C_{m+n-1} \end{pmatrix}

Where C_l are the coefficients of the Taylor series. The matrix above is called a “Hankel matrix”. A ‘Hankel Matrix’ is a symmetric square matrix in which each ascending skew-diagonal from left to right is constant. For Example a Hankel matrix of size 5 can be written like this:

\displaystyle \begin{pmatrix} a & b & c & d & e \\ b & c & d & e & f \\ c & d & e & f & g \\ d & e & f & g & h \\ e & f & g & h & i \end{pmatrix}

Let’s make some observations on the Hankel determinant H_{n,m}(f):

\displaystyle H_{n,m}(f) := \begin{vmatrix} C_{m-n+1} & C_{m-n+2} & \ldots & C_{m} \\ C_{m-n+2} & C_{m-n+3} & \ldots & C_{m+1} \\ \vdots & & & \vdots \\ C_{m} & C_{m+1} & \ldots & C_{m+n-1} \end{vmatrix}

This determinant has n colons and n rows. We can also numerate the terms of the determinant following the notation:

\displaystyle H_{n,m}(f) := \begin{vmatrix} d(1,1) & d(1,2) & \ldots & d(1,n) \\ d(2,1) & d(2,2) & \ldots & d(2,n) \\ \vdots & & & \vdots \\ d(n,1) & d(n,2) & \ldots & d(n,n) \end{vmatrix}

Where d(1,1) := C_{m-n+1} etc. This notation of the terms of the determinants implies that:

\displaystyle d(i,j) = C_{m-n+i+j-1}

So that:

\displaystyle d(1,1) = C_{m-n+1+1-1} = C_{m-n+1}
\displaystyle d(1,2) = C_{m-n+1+2-1} = C_{m-n+2}
\displaystyle d(2,1) = C_{m-n+2+1-1} = C_{m-n+2}
\displaystyle \ldots
\displaystyle d(n,n) = C_{m-n+n+n-1} = C_{m+n-1}

If f(x) is a even function of class C_{\infty}, we see that odd coefficients C_{2p+1} = \frac{f^{(2p+1)}(0)}{(2p+1)!} = 0. In this case, every second term of in the Hankel matrix is zero. If f(x) is even, we can establish that if m and n are odd then the Hankel determinant is zero:

The term with index d(i,j) of the Hankel determinant is C_{m-n+i+j-1}. As stated before, this term is zero for an even function if m-n+i+j-1 is odd. Now, if m and n are odd this means that n-m is even and m-n+i+j-1 is odd when i + j is even. It follows that, in the case of an even function, the Hankel determinant is of the form:

\displaystyle \begin{vmatrix} 0 & b & 0 & d & \hdots \\ b & 0 & d & 0 & \hdots \\ 0 & d & 0 & f & \hdots \\ \vdots & \vdots & \vdots & & \ddots \end{vmatrix}

We observe that the odd rows of this determinant are linear combinations of:

\displaystyle \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ \vdots \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ \vdots \end{pmatrix} \text{, etc.}

This implies that odd-numbered columns are linked and therefore the determinant is zero.

As example we will derive the P(3,3) of the cosine function. First we have to consider the geometric series of degrees up to 3 + 3 = 6 of cosine:

\displaystyle \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \mathcal{O}(x^8)

For m=3 and n=3 (m and n are both odd) the Hankel determinant is:

\displaystyle H_{3,3}(f) = \begin{vmatrix} C_{1} & C_{2} & C_{3} \\ C_{2} & C_{3} & C_{4} \\ C_{3} & C_{4} & C_{5} \end{vmatrix}

The corresponding Hankel determinant for calculating the coefficients of the P(3,3) Padé approximant of the geometric series of cosine is therefore:

\displaystyle H_{3,3}(cos) = \begin{vmatrix} 0 & -\frac{1}{2} & 0 \\ -\frac{1}{2} & 0 & \frac{1}{4!} \\ 0 & \frac{1}{4!} & 0 \end{vmatrix} = 0

This implies that we cannot calculate the P(3,3) approximant for the cosine function.

In conclusion, for an even function like cosine, when m and n are odd, the Hankel determinant Hn,m(f) is zero due to the linear dependence of the columns.

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