advanced28 min studyLesson 316 in Mathematics

Rational approximation and Padé approximants

Match a local power series with a quotient while distinguishing local accuracy, genuine singularities, and unstable pole-zero cancellations.

On this page

What you’ll understand

  • Set up normalized Padé coefficient-matching equations.
  • Derive a small approximant and identify its poles.
  • Explain why local Taylor matching does not guarantee global accuracy or stable poles.

Polynomials cannot contain finite poles, so they may use many coefficients to mimic a function whose nearby singularities dominate its shape. Rational functions add a denominator and can represent some such behavior compactly. Padé approximation chooses that quotient from local Taylor information. Its extra flexibility is valuable, but a pole in the approximant is not automatically a singularity of the target, and local matching is not a global error certificate.

Define the matching problem precisely

Let f be analytic near zero and choose nonnegative integers m and n. A normalized type-(m,n) Padé representation seeks polynomials p and q of degrees at most m and n, with q(0)=1, for which qf−p vanishes through degree m+n. Because q is nonzero in some neighborhood of zero, the quotient then matches the Taylor expansion of f to that order. The type gives degree upper bounds, so a successful representation can have smaller effective degrees.

The normalization is convenient but must not be assumed feasible for every index and every Taylor series. Some matching systems are defective or require a denominator vanishing at the expansion point before cancellation; we only call the displayed normalized construction successful when its equations have a solution. Common factors can make coefficient representations nonunique, even when the resulting rational function is unique. Cross-multiplying two successful normalized quotients shows their difference numerator has degree at most m+n but vanishes to higher order, so it is identically zero.

deg⁡p≤m,deg⁡q≤n,q(0)=1,q(z)f(z)−p(z)=O ⁣(zm+n+1)\deg p\le m,\quad\deg q\le n,\quad q(0)=1,\qquad q(z)f(z)-p(z)=O\!\left(z^{m+n+1}\right)
These are the normalized Padé matching conditions when a solution exists. They imply a local quotient error of the same order near zero, where q stays nonzero.↔ Scroll wide equations horizontally.

Solve for the denominator first

Write the Taylor coefficients of f and the unknown coefficients of q. In the product qf, coefficients above degree m through degree m+n must vanish because p cannot contain them. These equations determine possible denominator coefficients; the remaining low-degree coefficients define p. The system is linear in the unknown coefficients even though the quotient is nonlinear as a function of its denominator. This is a useful reduction, not a guarantee that the linear system is well conditioned.

Exact coefficient matching can be sensitive when the data nearly admit a lower-degree rational representation. A nearly canceling numerator zero and denominator zero may appear as a spurious pole-zero pair, often called a Froissart doublet. Rounding or noisy Taylor coefficients can move such features dramatically. Rank-revealing or regularized computations may prefer a lower effective degree, but that is an additional algorithmic choice. A tiny series residual alone does not certify a meaningful pole or accurate values near that pole.

ez−1+z/21−z/2=−z312+O(z4)(z→0)e^z-\frac{1+z/2}{1-z/2}=-\frac{z^3}{12}+O(z^4)\qquad(z\to0)
The normalized [1/1] approximant matches through degree two. Its denominator vanishes at z=2 although the exponential is entire.↔ Scroll wide equations horizontally.

Local information has a limited reach

A high-order zero of the local error describes behavior as the argument approaches the expansion point. It does not bound the quotient on a distant interval, especially near denominator zeros. A rational approximation optimized uniformly over a prescribed interval solves a different problem, with restrictions that exclude poles on that interval. Rational interpolation at distinct sample points is another different construction. Sharing a numerator/denominator representation does not make their objectives or convergence theorems interchangeable.

Before using a rational approximant, state the region of interest, inspect the denominator there, and evaluate an actual error or residual bound suited to the application. If the target has known singularities, they can guide the approximation space, but fitted poles should still be interpreted cautiously. Padé approximants can exactly recover a sufficiently low-degree rational target from its series; that favorable case does not establish convergence for every analytic function or every sequence of degree pairs.

Derive the [1/1] approximant to the exponential

Take p(z)=a+bzp(z)=a+bz and q(z)=1+czq(z)=1+cz. Match ez=1+z+z2/2+⋯e^z=1+z+z^2/2+\cdots through degree two, then evaluate the quotient at z=1 and locate its pole.

  1. Multiplying q by the series gives 1+(1+c)z+(1/2+c)z2+⋯1+(1+c)z+(1/2+c)z^2+\cdots. Matching the constant and linear terms gives a=1 and b=1+c.

  2. The quadratic coefficient must vanish because p is only linear. Thus c=−1/2 and b=1/2, producing (1+z/2)/(1−z/2). Its series has cubic coefficient one quarter rather than the exponential's one sixth.

  3. At z=1 the quotient is three, whereas the exponential is approximately 2.71828. Its denominator vanishes at two and its numerator does not, giving a genuine pole of the approximant at a point where the target remains analytic.

The result

Local matching through degree two yields a useful quotient near zero, but neither its value at one nor its pole at two is guaranteed by the matching order alone.

Explore the model

Interactive lab

Interactive lab · root-finding

Locate a denominator hazard separately

Pause and Reset, then set c = 4 and U = 3. Predict bisection approaches x = 2, the zero of the exponential [1/1] approximant's denominator 1−x/2; Step and compare the bracket. Reset and set both c = 1 and U = 2 to locate the pole of the [0/1] approximant 1/(1−x) after one exact midpoint. The plotted residual is x²−c, used only to locate these positive coordinates; it is not a Padé error function.

Locate a denominator hazard separatelyBisection of x² − c on [0,U], capped at 50 iterations for floating-point resolution. The error bound is the maximum distance from the displayed midpoint to either bracket endpoint (half the width in exact arithmetic). Compensated square evaluation preserves the bracket when ordinary multiplication rounds a residual to zero. A truly exact endpoint or midpoint collapses the bracket. This is one monotone quadratic, not a general root solver. Current readings: Midpoint after 0 refinements: 1.5; Bracket error bound: 1.5; Computed residual · x² − c: 0.25. All parameters are adjustable with the labeled controls below.Bisection on x squared minus 2: bracket from 0 to 3 after 0 refinements.00.751.52.253−2.360.192.745.297.84x · bracket stays nonnegativef(x) = x² − cn = 0 · [ 0, 3 ]
Quadratic / midpointCurrent bracketPositive root
Midpoint after 0 refinements
1.5
Bracket error bound
1.5
Computed residual · x² − c
0.25
t = 0 s

Bisection of x² − c on [0,U], capped at 50 iterations for floating-point resolution. The error bound is the maximum distance from the displayed midpoint to either bracket endpoint (half the width in exact arithmetic). Compensated square evaluation preserves the bracket when ordinary multiplication rounds a residual to zero. A truly exact endpoint or midpoint collapses the bracket. This is one monotone quadratic, not a general root solver. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forFinding a denominator zero and certifying target approximation are separate tasks; an accurately located fitted pole can still be spurious for the target.

Check your understanding

A normalized [m/n] Padé approximant matches an analytic target through Taylor degree m+n. Which conclusion is justified?

Try it yourself

Derive the normalized [0/1] approximant to eze^z by matching its constant and linear terms. Identify its pole and its first unmatched coefficient, and explain why this does not contradict analyticity of the target.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use a constant numerator a and denominator 1+bz. Force the linear coefficient of (1+bz)eᶻ to vanish, then expand the resulting quotient geometrically.

Compare with the solution

The product has constant term one and linear coefficient 1+b, so a=1 and b=−1. The approximant is 1/(1−z)=1+z+z2+⋯1/(1-z)=1+z+z^2+\cdots near zero. Its quadratic coefficient is one rather than the exponential's one half, so ez−1/(1−z)=−z2/2+O(z3)e^z-1/(1-z)=-z^2/2+O(z^3). The quotient has a pole at one, although the target is entire. This is consistent because the matching conditions concern only local coefficients through degree one and impose no global analyticity requirement on the rational approximant.

Take it with you

  • Normalized Padé matching is a local series condition and can be defective.
  • A fitted pole need not be a target singularity; near-canceling pairs may be highly sensitive.
  • Local matching, uniform rational approximation, and stable numerical computation require separate checks.

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