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Shanks transformation

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In numerical analysis, the Shanks transformation is a non-linear series acceleration method to increase the rate of convergence of a sequence. This method is named after Daniel Shanks, who rediscovered this sequence transformation in 1955. It was first derived and published by R. Schmidt in 1941.

One can calculate only a few terms of a perturbation expansion, usually no more than two or three, and almost never more than seven. The resulting series is often slowly convergent, or even divergent. Yet those few terms contain a remarkable amount of information, which the investigator should do his best to extract.
This viewpoint has been persuasively set forth in a delightful paper by Shanks (1955), who displays a number of amazing examples, including several from fluid mechanics.

Milton D. Van Dyke (1975) Perturbation methods in fluid mechanics, p. 202.

Formulation

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

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

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

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

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

S ( A n ) = A n + 1 A n 1 A n 2 A n + 1 2 A n + A n 1 = A n + 1 ( A n + 1 A n ) 2 ( A n + 1 A n ) ( A n A n 1 ) {\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 S ( A n ) {\displaystyle S(A_{n})} often converges more rapidly than the sequence A n . {\displaystyle A_{n}.} Further speed-up may be obtained by repeated use of the Shanks transformation, by computing S 2 ( A n ) = S ( S ( A n ) ) , {\displaystyle S^{2}(A_{n})=S(S(A_{n})),} S 3 ( A n ) = S ( 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 S ( A n ) {\displaystyle S(A_{n})} 's definition (i.e. S ( A n ) = A n + 1 ( A n + 1 A n ) 2 ( A n + 1 A n ) ( A n A n 1 ) {\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. S ( A n ) = A n + 1 A n 1 A n 2 A n + 1 2 A n + A n 1 {\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.

Example

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

As an example, consider the slowly convergent series

4 k = 0 ( 1 ) k 1 2 k + 1 = 4 ( 1 1 3 + 1 5 1 7 + ) {\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 A 6 {\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 A n {\displaystyle A_{n}} , the Shanks transformation S ( A n ) {\displaystyle S(A_{n})} on them, as well as the repeated Shanks transformations S 2 ( A n ) {\displaystyle S^{2}(A_{n})} and S 3 ( A n ) {\displaystyle S^{3}(A_{n})} are given for n {\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.

n {\displaystyle n} A n {\displaystyle A_{n}} S ( A n ) {\displaystyle S(A_{n})} S 2 ( A n ) {\displaystyle S^{2}(A_{n})} S 3 ( 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 S ( A 1 ) {\displaystyle S(A_{1})} already has two-digit accuracy, while the original partial sums only establish the same accuracy at A 24 . {\displaystyle A_{24}.} Remarkably, S 3 ( A 3 ) {\displaystyle S^{3}(A_{3})} has six digits accuracy, obtained from repeated Shank transformations applied to the first seven terms A 0 , , A 6 . {\displaystyle A_{0},\ldots ,A_{6}.} As mentioned before, A n {\displaystyle A_{n}} only obtains 6-digit accuracy after summing about 400,000 terms.

Motivation

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

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

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

A n 1 = A + α q n 1 , A n = A + α q n and A n + 1 = A + α q n + 1 . {\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: A , {\displaystyle A,} α {\displaystyle \alpha } and q . {\displaystyle q.} Solving for A {\displaystyle A} gives

A = A n + 1 A n 1 A n 2 A n + 1 2 A n + A n 1 . {\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 A n = A {\displaystyle A_{n}=A} for all n . {\displaystyle n.}

Generalized Shanks transformation

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

S k ( A n ) = | A n k A n 1 A n Δ A n k Δ A n 1 Δ A n Δ A n k + 1 Δ A n Δ A n + 1 Δ A n 1 Δ A n + k 2 Δ A n + k 1 | | 1 1 1 Δ A n k Δ A n 1 Δ A n Δ A n k + 1 Δ A n Δ A n + 1 Δ A n 1 Δ A n + k 2 Δ A n + k 1 | , {\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 Δ A p = A p + 1 A p . {\displaystyle \Delta A_{p}=A_{p+1}-A_{p}.} It is the solution of a model for the convergence behaviour of the partial sums A n {\displaystyle A_{n}} with k {\displaystyle k} distinct transients:

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

This model for the convergence behaviour contains 2 k + 1 {\displaystyle 2k+1} unknowns. By evaluating the above equation at the elements A n k , A n k + 1 , , A n + k {\displaystyle A_{n-k},A_{n-k+1},\ldots ,A_{n+k}} and solving for A , {\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: S 1 ( A n ) = S ( A n ) . {\displaystyle S_{1}(A_{n})=S(A_{n}).}

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

Note: The calculation of determinants requires many arithmetic operations to make, however Peter Wynn discovered a recursive evaluation procedure called epsilon-algorithm which avoids calculating the determinants.

See also

Notes

  1. Weniger (2003).
  2. ^ Bender & Orszag (1999), pp. 368–375.
  3. ^ Van Dyke (1975), pp. 202–205.
  4. ^ Bender & Orszag (1999), pp. 389–392.
  5. Wynn (1956)
  6. Wynn (1962)

References

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