Shanks transformation explained

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.[1]

Formulation

For a sequence

\left\{am\right\}m\inN

the series

A=

infty
\sum
m=0

am

is to be determined. First, the partial sum

An

is defined as:

An=

n
\sum
m=0

am

and forms a new sequence

\left\{An\right\}n\inN

. Provided the series converges,

An

will also approach the limit

A

as

n\toinfty.

The Shanks transformation

S(An)

of the sequence

An

is the new sequence defined by[2] [3]

S(An)=

AAn-1-
2
A
n
n+1
An+1-2An+An-1

=An+1-

(An+1-An)2
(An+1-An)-(An-An-1)

where this sequence

S(An)

often converges more rapidly than the sequence

An.

Further speed-up may be obtained by repeated use of the Shanks transformation, by computing
2(A
S
n)=S(S(A

n)),

3(A
S
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(An)

's definition (i.e.

S(An)=An+1-

(An+1-An)2
(An+1-An)-(An-An-1)
) is more numerically stable than the expression to its left (i.e.

S(An)=

AAn-1-
2
A
n
n+1
An+1-2An+An-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

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

4

infty
\sum
k=0

(-1)k

1
2k+1

=4\left(1-

1
3

+

1
5

-

1
7

+\right)

which has the exact sum π ≈ 3.14159265. The partial sum

A6

has only one digit accuracy, while six-figure accuracy requires summing about 400,000 terms.

In the table below, the partial sums

An

, the Shanks transformation

S(An)

on them, as well as the repeated Shanks transformations
2(A
S
n)
and
3(A
S
n)
are given for

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

An

S(An)

2(A
S
n)
3(A
S
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(A1)

already has two-digit accuracy, while the original partial sums only establish the same accuracy at

A24.

Remarkably,
3(A
S
3)
has six digits accuracy, obtained from repeated Shank transformations applied to the first seven terms

A0,\ldots,A6.

As mentioned before,

An

only obtains 6-digit accuracy after summing about 400,000 terms.

Motivation

The Shanks transformation is motivated by the observation that — for larger

n

— the partial sum

An

quite often behaves approximately as[2]

An=A+\alphaqn,

with

|q|<1

so that the sequence converges transiently to the series result

A

for

n\toinfty.

So for

n-1,

n

and

n+1

the respective partial sums are:

An-1=A+\alphaqn-1,    An=A+\alphaqn    and    An+1=A+\alphaqn+1.

These three equations contain three unknowns:

A,

\alpha

and

q.

Solving for

A

gives[2]

A=

AAn-1-
2
A
n
n+1
An+1-2An+An-1

.

In the (exceptional) case that the denominator is equal to zero: then

An=A

for all

n.

Generalized Shanks transformation

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

Sk(An)=

\begin{vmatrix
A

n-k&&An-1&An\\ \DeltaAn-k&&\DeltaAn-1&\DeltaAn\\ \DeltaAn-k+1&&\DeltaAn&\DeltaAn+1\\ \vdots&&\vdots&\vdots\\ \DeltaAn-1&&\DeltaAn+k-2&\DeltaAn+k-1\\ \end{vmatrix} }{ \begin{vmatrix} 1&&1&1\\ \DeltaAn-k&&\DeltaAn-1&\DeltaAn\\ \DeltaAn-k+1&&\DeltaAn&\DeltaAn+1\\ \vdots&&\vdots&\vdots\\ \DeltaAn-1&&\DeltaAn+k-2&\DeltaAn+k-1\\ \end{vmatrix} },

with

\DeltaAp=Ap+1-Ap.

It is the solution of a model for the convergence behaviour of the partial sums

An

with

k

distinct transients:

An=A+

k
\sum
p=1

\alphap

n.
q
p

This model for the convergence behaviour contains

2k+1

unknowns. By evaluating the above equation at the elements

An-k,An-k+1,\ldots,An+k

and solving for

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:

S1(An)=S(An).

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

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.[5] [6]

See also

References

Notes and References

  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)