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Henri Padé Henri Padé (1863–1953) [4]

Taylor series approximate a function with a polynomial. Padé approximants use the same coefficients but form a ratio of two polynomials. This lets them capture poles and branch points, and converge where Taylor diverges.

The Padé approximant

Given $f(x) = c_0 + c_1 x + c_2 x^2 + \cdots$, the $[L/M]$ Padé approximant is

\[R_{[L/M]}(x) = \frac{a_0 + a_1 x + \cdots + a_L x^L}{1 + b_1 x + \cdots + b_M x^M} \tag{1}\]

whose Taylor expansion matches $f(x)$ through order $L + M$:

\[R_{[L/M]}(x) = c_0 + c_1 x + \cdots + c_{L+M} x^{L+M} + \mathcal{O}(x^{L+M+1}) \tag{2}\]

Equating coefficients in (1) gives a linear system for the $b_j$, then explicit formulas for the $a_k$. The total number of parameters is $L + M + 1$, the same as a degree-$(L+M)$ polynomial, but the denominator can vanish, letting the approximant represent singularities.

Interactive comparison

Compare Taylor and Padé with independent slider controls:

Taylor
Padé $[L/M]$

Practical examples

1. Integrating $\ln(1+x)$ beyond convergence

\[I = \int_0^3 \ln(1+x)\,dx = 4\ln 4 - 3 \approx 2.5452 \tag{3}\]

The series $\ln(1+x) = x - x^2/2 + x^3/3 - \cdots$ converges only for $\lvert x \rvert \le 1$. At $x = 3$ the polynomial blows up:

\[\text{Taylor } T_6: \quad \int_0^3 T_6\,dx \approx -33.17 \quad (1403\%\text{ error, wrong sign})\]

The $[3,3]$ Padé from the same 7 coefficients:

\[R_{[3,3]}(x) = \frac{x(11x^2 + 60x + 60)}{3(x^3 + 12x^2 + 30x + 20)}\] \[\int_0^3 R_{[3,3]}\,dx \approx 2.5435 \quad (0.07\%\text{ error})\]

The rational integral evaluates in closed form via partial fractions.

2. Differentiating $1/\sqrt{1+x}$

\[f(x) = \frac{1}{\sqrt{1+x}}, \qquad f'(x) = -\frac{1}{2}(1+x)^{-3/2} \tag{4}\]

Exact: $f’(2) \approx -0.0962$. The $[2,2]$ Padé from 5 coefficients:

\[R_{[2,2]}(x) = \frac{x^2 + 12x + 16}{5x^2 + 20x + 16}\] \[\text{Taylor } T_4': \quad f'_T(2) \approx 6.0 \quad (\text{diverged, wrong sign})\] \[\text{Padé } [2,2]': \quad f'_P(2) \approx -0.0942 \quad (2.1\%\text{ error})\]

The Padé captures the branch point at $x = -1$. Its derivative (via quotient rule) is again rational.

3. Cantilever with singular load

A cantilever of length $L$ carries a distributed load with a square-root singularity at the free end:

\[q(\xi) = \frac{q_0}{\sqrt{1 - \xi}}, \qquad \xi = x/L \tag{5}\] q(x) L x = 0 x = L

The Euler-Bernoulli deflection satisfies

\[EI\,w^{(4)}(\xi) = \frac{q_0}{\sqrt{1-\xi}} \tag{6}\]

The Taylor series of the load converges only for $\lvert\xi\rvert < 1$, so the polynomial diverges at the tip. Replacing the load by its Padé approximant before integrating keeps the deflection accurate across the full span:

Taylor
Padé $[L/M]$

Set Taylor to $T_5$ and Padé to $[3,2]$: same 6 coefficients, but the Padé deflection tracks the exact curve while Taylor underestimates the tip displacement.

Why rational functions are practical

Operation Polynomial Rational function
Evaluate Horner’s method Horner on numerator and denominator, one division
Differentiate Power rule Quotient rule, result is rational
Integrate Term by term Partial fractions, then log/arctan terms
Capture poles Cannot Denominator zeros model singularities

References

  1. Baker, G. A. & Graves-Morris, P. (1996). Padé Approximants, 2nd ed. Cambridge University Press.
  2. Bender, C. M. & Orszag, S. A. (1999). Advanced Mathematical Methods for Scientists and Engineers. Springer.
  3. Gonnet, P., Güttel, S. & Trefethen, L. N. (2013). Robust Padé approximation via SVD. SIAM Review, 55(1), 101–117.
  4. Portrait of Henri Padé. Wikimedia Commons, public domain. Source.