Perturbation problem beyond all orders
In mathematics, perturbation theory works typically by expanding unknown quantity in a power series in a small parameter. However, in a perturbation problem beyond all orders, all coefficients of the perturbation expansion vanish and the difference between the function and the constant function 0 cannot be detected by a power series.
A simple example is understood by an attempt at trying to expand [math]\displaystyle{ e^{-1/\epsilon} }[/math] in a Taylor series in [math]\displaystyle{ \epsilon \gt 0 }[/math] about 0. All terms in a naïve Taylor expansion are identically zero. This is because the function [math]\displaystyle{ e^{-1/z} }[/math] possesses an essential singularity at [math]\displaystyle{ z = 0 }[/math] in the complex [math]\displaystyle{ z }[/math]-plane, and therefore the function is most appropriately modeled by a Laurent series -- a Taylor series has a zero radius of convergence. Thus, if a physical problem possesses a solution of this nature, possibly in addition to an analytic part that may be modeled by a power series, the perturbative analysis fails to recover the singular part. Terms of nature similar to [math]\displaystyle{ e^{-1/\epsilon} }[/math] are considered to be "beyond all orders" of the standard perturbative power series.
See also
References
- J P Boyd, "The Devil's Invention: Asymptotic, Superasymptotic and Hyperasymptotic Series", https://link.springer.com/article/10.1023/A:1006145903624
- C. M. Bender and S. A. Orszag, "Advanced Mathematical Methods for Scientists and Engineers", https://link.springer.com/book/10.1007%2F978-1-4757-3069-2
- C. M. Bender, Lectures on Mathematical Physics, https://www.perimeterinstitute.ca/video-library/collection/11/12-psi-mathematical-physics
 |  |
