Approximaths

Two properties of Padé Approximants

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Property 1

Let g(x) = \frac{1}{f(x)}, with f(0) \neq 0, and assume f is at least C^{m+n} at x = 0. If P(m,n)_f = \frac{P(x)}{Q(x)}, then the [n/m] Padé approximant of g(x):

\displaystyle P(n,m)_g = \frac{Q(x)}{P(x)},

provided P(0) \neq 0.

Proof: Given P(m,n)_f = \frac{P(x)}{Q(x)}, we have:

\displaystyle f(x) Q(x) - P(x) = \epsilon(x) x^{m+n+1}.

Since g(x) = \frac{1}{f(x)}, consider:

\displaystyle g(x) P(x) - Q(x) = \frac{1}{f(x)} P(x) - Q(x) = \frac{P(x) - f(x) Q(x)}{f(x)} = -\frac{\epsilon(x) x^{m+n+1}}{f(x)}.

Since f(0) \neq 0, \frac{1}{f(x)} is bounded near x = 0, and:

\displaystyle g(x) P(x) - Q(x) = O(x^{m+n+1}),

indicating that \frac{Q(x)}{P(x)} is the [n/m] Padé approximant of g(x), as it matches the Taylor series of g(x) up to x^{m+n}. For example, for f(x) = \sqrt{1+x}, the [2/2] Padé approximant can be computed, and g(x) = \frac{1}{\sqrt{1+x}} yields a consistent [2/2] approximant by taking the reciprocal.

Property 2

If f is even (f(x) = f(-x)) and at least C^{m+n}, and the [m/n] Padé approximant exists and is unique (guaranteed if the Hankel determinant is non-zero), then P(m,n)_f = \frac{P(x)}{Q(x)} is even, i.e., P(x) = P(-x) and Q(x) = Q(-x).

Proof: Since f(x) = f(-x), the Taylor series of f contains only even powers. For P(m,n)_f = \frac{P(x)}{Q(x)}, we have:

\displaystyle f(x) Q(x) - P(x) = O(x^{m+n+1}).

Evaluate at -x:

\displaystyle f(-x) Q(-x) - P(-x) = f(x) Q(-x) - P(-x) = O(x^{m+n+1}),

since f(-x) = f(x), and the error term remains of order x^{m+n+1}. Thus, \frac{P(-x)}{Q(-x)} satisfies the same Padé condition as \frac{P(x)}{Q(x)}. By uniqueness of the [m/n] approximant (assuming non-zero Hankel determinant), we conclude:

\displaystyle P(x) = P(-x), \quad Q(x) = Q(-x).

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