// Workers AI · dad joke modeWhat did "Perturbation problem beyond all orders" say to its date? You disrupt my order.

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 $e^{-1/\epsilon }$ in a Taylor series in $\epsilon >0$ about 0. All terms in a naïve Taylor expansion are identically zero. This is because the function $e^{-1/z}$ possesses an essential singularity at $z=0$ in the complex $z$-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 $e^{-1/\epsilon }$ are considered to be "beyond all orders" of the standard perturbative power series.

See also

Asymptotic expansion

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 Archived 2017-01-09 at the Wayback Machine