Calculate Pade Approximation for f(x)=1-\frac{1}{2}x+\frac{1}{3}x^2-...

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SUMMARY

The discussion focuses on calculating the Pade Approximation for the function f(x) = 1 - (1/2)x + (1/3)x² using the formula [N/D] = (a₀ + a₁x + ... + aₙxⁿ) / (1 + b₁x + ... + bₖxᵏ). The specific approximation sought is [1/1], which simplifies to (1 + (1/6)x) / (1 + (2/3)x). Key steps include determining coefficients a₀, a₁, and b₁ through manipulation of the series expansion and matching terms. The final expression reveals the relationship between the coefficients and the behavior of the series as x approaches zero.

PREREQUISITES
  • Understanding of Pade Approximation and its application in approximating functions.
  • Familiarity with Maclaurin series and Taylor series expansions.
  • Basic algebraic manipulation of polynomials and series.
  • Knowledge of coefficient matching techniques in series expansions.
NEXT STEPS
  • Study the derivation of Pade Approximations for various functions.
  • Learn about the convergence properties of Pade Approximations compared to Taylor series.
  • Explore the application of Pade Approximations in numerical analysis and computational methods.
  • Investigate the role of higher-order terms in series expansions and their impact on approximation accuracy.
USEFUL FOR

Students and researchers in mathematics, particularly those focusing on numerical analysis, approximation theory, and series expansions. This discussion is beneficial for anyone looking to deepen their understanding of Pade Approximations and their practical applications.

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Homework Statement


Pade approximation
[tex][N/D]=\frac{a_0+a_1x+...+a_Nx^N}{1+b_1x+...+b_Dx^D}[/tex]
With this approximation we approximate Maclaurin series
[tex]f(x)=\sum^{\infty}_{i=0}c_ix^i=[N/D]+O(x^{N+D+1})[/tex]
How to calculate [tex][1/1][/tex] for [tex]f(x)=1-\frac{1}{2}x+\frac{1}{3}x^2-...[/tex] ?


Homework Equations


[tex][N/D]=\frac{a_0+a_1x+...+a_Nx^N}{1+b_1x+...+b_Dx^D}[/tex]
[tex]\sum^{\infty}_{i=0}c_ix^i=[N/D]+O(x^{N+D+1})[/tex]

The Attempt at a Solution


[tex](1+b_1x)(1-\frac{1}{2}x)=a_0+a_1x[/tex]
[tex]a_0=1[/tex]
[tex]b_1-\frac{1}{2}=a_1[/tex]

How to calculate [tex]a_1,b_1[/tex]

In solution
[tex][1/1]=\frac{1+\frac{1}{6}x}{1+\frac{2}{3}x}[/tex]
 
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Write more terms on both sides
[tex](1+b_1 x)(1-x/2+x^2/3) = a_0 + a_1 x + \mathcal{O}(x^3)[/tex]
So what happens to the term proportional to [itex]x^2[/itex]?
 

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