Shanks transformation

For a sequence ${\displaystyle \left\{a_{m}\right\}_{m\in \mathbb {N} }}$ the series

${\displaystyle A=\sum _{m=0}^{\infty }a_{m}\,}$

is to be determined. First, the partial sum ${\displaystyle A_{n}}$ is defined as:

${\displaystyle A_{n}=\sum _{m=0}^{n}a_{m}\,}$

and forms a new sequence ${\displaystyle \left\{A_{n}\right\}_{n\in \mathbb {N} }}$. Provided the series converges, ${\displaystyle A_{n}}$ will also approach the limit ${\displaystyle A}$ as ${\displaystyle n\to \infty .}$ The Shanks transformation ${\displaystyle S(A_{n})}$ of the sequence ${\displaystyle A_{n}}$ is the new sequence defined by[2][3]

${\displaystyle S(A_{n})={\frac {A_{n+1}\,A_{n-1}\,-\,A_{n}^{2}}{A_{n+1}-2A_{n}+A_{n-1}}}=A_{n+1}-{\frac {(A_{n+1}-A_{n})^{2}}{(A_{n+1}-A_{n})-(A_{n}-A_{n-1})}}}$

where this sequence ${\displaystyle S(A_{n})}$ often converges more rapidly than the sequence ${\displaystyle A_{n}.}$ Further speed-up may be obtained by repeated use of the Shanks transformation, by computing ${\displaystyle S^{2}(A_{n})=S(S(A_{n})),}$ ${\displaystyle S^{3}(A_{n})=S(S(S(A_{n}))),}$ etc.

Note that the non-linear transformation as used in the Shanks transformation is essentially the same as used in Aitken's delta-squared process so that as with Aitken's method, the right-most expression in ${\displaystyle S(A_{n})}$'s definition (i.e. ${\displaystyle S(A_{n})=A_{n+1}-{\frac {(A_{n+1}-A_{n})^{2}}{(A_{n+1}-A_{n})-(A_{n}-A_{n-1})}}}$) is more numerically stable than the expression to its left (i.e. ${\displaystyle S(A_{n})={\frac {A_{n+1}\,A_{n-1}\,-\,A_{n}^{2}}{A_{n+1}-2A_{n}+A_{n-1}}}}$). Both Aitken's method and the Shanks transformation operate on a sequence, but the sequence the Shanks transformation operates on is usually thought of as being a sequence of partial sums, although any sequence may be viewed as a sequence of partial sums.

Absolute error as a function of ${\displaystyle n}$ in the partial sums ${\displaystyle A_{n}}$ and after applying the Shanks transformation once or several times: ${\displaystyle S(A_{n}),}$ ${\displaystyle S^{2}(A_{n})}$ and ${\displaystyle S^{3}(A_{n}).}$ The series used is ${\displaystyle \scriptstyle 4\left(1-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}-\cdots \right),}$ which has the exact sum ${\displaystyle \pi .}$

As an example, consider the slowly convergent series[3]

${\displaystyle 4\sum _{k=0}^{\infty }(-1)^{k}{\frac {1}{2k+1}}=4\left(1-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+\cdots \right)}$

which has the exact sum π ≈ 3.14159265. The partial sum ${\displaystyle A_{6}}$ has only one digit accuracy, while six-figure accuracy requires summing about 400,000 terms.

In the table below, the partial sums ${\displaystyle A_{n}}$, the Shanks transformation ${\displaystyle S(A_{n})}$ on them, as well as the repeated Shanks transformations ${\displaystyle S^{2}(A_{n})}$ and ${\displaystyle S^{3}(A_{n})}$ are given for ${\displaystyle n}$ up to 12. The figure to the right shows the absolute error for the partial sums and Shanks transformation results, clearly showing the improved accuracy and convergence rate.

${\displaystyle n}$	${\displaystyle A_{n}}$	${\displaystyle S(A_{n})}$	${\displaystyle S^{2}(A_{n})}$	${\displaystyle S^{3}(A_{n})}$
0	4.00000000	—	—	—
1	2.66666667	3.16666667	—	—
2	3.46666667	3.13333333	3.14210526	—
3	2.89523810	3.14523810	3.14145022	3.14159936
4	3.33968254	3.13968254	3.14164332	3.14159086
5	2.97604618	3.14271284	3.14157129	3.14159323
6	3.28373848	3.14088134	3.14160284	3.14159244
7	3.01707182	3.14207182	3.14158732	3.14159274
8	3.25236593	3.14125482	3.14159566	3.14159261
9	3.04183962	3.14183962	3.14159086	3.14159267
10	3.23231581	3.14140672	3.14159377	3.14159264
11	3.05840277	3.14173610	3.14159192	3.14159266
12	3.21840277	3.14147969	3.14159314	3.14159265

The Shanks transformation ${\displaystyle S(A_{1})}$ already has two-digit accuracy, while the original partial sums only establish the same accuracy at ${\displaystyle A_{24}.}$ Remarkably, ${\displaystyle S^{3}(A_{3})}$ has six digits accuracy, obtained from repeated Shank transformations applied to the first seven terms ${\displaystyle A_{0},\ldots ,A_{6}.}$ As said before, ${\displaystyle A_{n}}$ only obtains 6-digit accuracy after about summing 400,000 terms.

The Shanks transformation is motivated by the observation that — for larger ${\displaystyle n}$ — the partial sum ${\displaystyle A_{n}}$ quite often behaves approximately as[2]

${\displaystyle A_{n}=A+\alpha q^{n},\,}$

with ${\displaystyle |q|<1}$ so that the sequence converges transiently to the series result ${\displaystyle A}$ for ${\displaystyle n\to \infty .}$ So for ${\displaystyle n-1,}$ ${\displaystyle n}$ and ${\displaystyle n+1}$ the respective partial sums are:

${\displaystyle A_{n-1}=A+\alpha q^{n-1}\quad ,\qquad A_{n}=A+\alpha q^{n}\qquad {\text{and}}\qquad A_{n+1}=A+\alpha q^{n+1}.}$

These three equations contain three unknowns: ${\displaystyle A,}$ ${\displaystyle \alpha }$ and ${\displaystyle q.}$ Solving for ${\displaystyle A}$ gives[2]

${\displaystyle A={\frac {A_{n+1}\,A_{n-1}\,-\,A_{n}^{2}}{A_{n+1}-2A_{n}+A_{n-1}}}.}$

In the (exceptional) case that the denominator is equal to zero: then ${\displaystyle A_{n}=A}$ for all ${\displaystyle n.}$

The generalized kth-order Shanks transformation is given as the ratio of the determinants:[4]

${\displaystyle S_{k}(A_{n})={\frac {\begin{vmatrix}A_{n-k}&\cdots &A_{n-1}&A_{n}\\\Delta A_{n-k}&\cdots &\Delta A_{n-1}&\Delta A_{n}\\\Delta A_{n-k+1}&\cdots &\Delta A_{n}&\Delta A_{n+1}\\\vdots &&\vdots &\vdots \\\Delta A_{n-1}&\cdots &\Delta A_{n+k-2}&\Delta A_{n+k-1}\\\end{vmatrix}}{\begin{vmatrix}1&\cdots &1&1\\\Delta A_{n-k}&\cdots &\Delta A_{n-1}&\Delta A_{n}\\\Delta A_{n-k+1}&\cdots &\Delta A_{n}&\Delta A_{n+1}\\\vdots &&\vdots &\vdots \\\Delta A_{n-1}&\cdots &\Delta A_{n+k-2}&\Delta A_{n+k-1}\\\end{vmatrix}}},}$

with ${\displaystyle \Delta A_{p}=A_{p+1}-A_{p}.}$ It is the solution of a model for the convergence behaviour of the partial sums ${\displaystyle A_{n}}$ with ${\displaystyle k}$ distinct transients:

${\displaystyle A_{n}=A+\sum _{p=1}^{k}\alpha _{p}q_{p}^{n}.}$

This model for the convergence behaviour contains ${\displaystyle 2k+1}$ unknowns. By evaluating the above equation at the elements ${\displaystyle A_{n-k},A_{n-k+1},\ldots ,A_{n+k}}$ and solving for ${\displaystyle A,}$ the above expression for the kth-order Shanks transformation is obtained. The first-order generalized Shanks transformation is equal to the ordinary Shanks transformation: ${\displaystyle S_{1}(A_{n})=S(A_{n}).}$

The generalized Shanks transformation is closely related to Padé approximants and Padé tables.[4]

• Aitken's delta-squared process
• Rate of convergence
• Richardson extrapolation
• Sequence transformation

• Shanks, D. (1955), "Non-linear transformation of divergent and slowly convergent sequences", Journal of Mathematics and Physics, 34: 1–42, doi:10.1002/sapm19553411
• Schmidt, R. (1941), "On the numerical solution of linear simultaneous equations by an iterative method", Philosophical Magazine, 32: 369–383
• Van Dyke, M.D. (1975), Perturbation methods in fluid mechanics (annotated ed.), Parabolic Press, ISBN 0-915760-01-0
• Bender, C.M.; Orszag, S.A. (1999), Advanced mathematical methods for scientists and engineers, Springer, ISBN 0-387-98931-5
• Weniger, E.J. (1989). "Nonlinear sequence transformations for the acceleration of convergence and the summation of divergent series". Computer Physics Reports. 10 (5–6): 189–371. arXiv:math.NA/0306302. Bibcode:1989CoPhR..10..189W. doi:10.1016/0167-7977(89)90011-7.