In numerical analysis, Aitken's delta-squared process or Aitken extrapolation is a series acceleration method used for accelerating the rate of convergence of a sequence. It is named after Alexander Aitken, who introduced this method in 1926. [1] It is most useful for accelerating the convergence of a sequence that is converging linearly. A precursor form was known to Seki Kōwa (1642 – 1708) and applied to the rectification of the circle, i.e., to the calculation of π.
Given a sequence with Aitken's delta-squared process associates to this sequence the new sequence
which can also be written as
with and Both are the same sequence algebraically but the latter has improved numerical stability in computational implementation.
is ill-defined if the sequence contains a zero element, which occurs if the sequence of forward differences, has any repeated term. From a theoretical point of view, if that occurs only for a finite number of indices, one could apply the Aitken process to only the part of the sequence with indices such that is the last index for which the sequence repeats. In practice, the first few terms of the sequence usually provide desired precision; also, when numerically computing the sequence, one has to take care to stop the computation before rounding errors in the denominator become too large, as the sequence transformation may cancel significant digits.
Aitken's delta-squared process is an acceleration of convergence method and a particular case of a nonlinear sequence transformation.
A sequence that converges to a limiting value is said to converge linearly, or more technically Q-linearly, if there is some number for which
This means that asymptotically, the distance between the sequence and its limit shrinks by nearly the same proportion, on every step and the ratio of reduction becomes closer and closer to that proportion. This is also sometimes called "geometric convergence," since it is a characteristic property for geometric series, or "exponential convergence," since it is convergence like
Aitken's method will accelerate the convergence of a sequence if with terms defined above, satisfies
is not a linear operator on sequences, but it is linear with respect to addition of constant sequences: if is any constant sequence , constant for all This is clear from the expression of in terms of the finite difference operator
The new process does not in general converge quadratically, but for an iterated function sequence satisfying for some function converging to a fixed point, the accelerated sequence's convergence is quadratic. In this case, the technique is known as Steffensen's method.
Empirically, the A-operation eliminates the "most important error term". One can check this by considering a sequence of the form , where : The sequence will then go to the limit like goes to zero.
Geometrically, the graph of an exponential function that satisfies , and has an horizontal asymptote at (if ).
One can also show that if a sequence converges to its limit at a rate strictly greater than 1, does not have a better rate of convergence. (In practice, one rarely has e.g. quadratic convergence which would mean over 30 (respectively 100) correct decimal places after 5 (respectively 7) iterations (starting with 1 correct digit); usually no acceleration is needed in that case.)
In practice, often converges much faster to the limit than does, as demonstrated by the example calculations below. Usually, it is much cheaper to calculate (involving only calculation of differences, one multiplication and one division) than to calculate many more terms of the sequence . Care must be taken, however, to avoid introducing errors due to insufficient precision when calculating the differences in the numerator and denominator of the expression.
Example 1: The value of can be approximated by assuming an initial value for and iterating the following sequence, called Heron's method: Starting with
n | X | A[X] |
---|---|---|
0 | 1 | 1.4285714 |
1 | 1.5 | 1.4141414 |
2 | 1.4166667 | 1.4142136 |
3 | 1.4142157 | -- |
4 | 1.4142136 | -- |
It is worth noting here that Aitken's method does not save the cost of calculating two iterations here; computation of the first three values required the first five values. Also, the second value is less accurate than the 4th value, which is not surprising due to the fact that Aitken's process is best suited for sequences that converge linearly, rather than quadratically, and Heron's method for calculating square roots converges quadratically.[ citation needed ]
Example 2: The value of may be calculated as an infinite sum via the Leibniz formula for π:
n | Series Terms | X = Partial Sums | A[X] |
---|---|---|---|
0 | 1 | 1 | 0.79166667 |
1 | −0.33333333 | 0.66666667 | 0.78333333 |
2 | 0.2 | 0.86666667 | 0.78630952 |
3 | −0.14285714 | 0.72380952 | 0.78492063 |
4 | 0.11111111 | 0.83492063 | 0.78567821 |
5 | −9.0909091×10−2 | 0.74401154 | 0.78522034 |
6 | 7.6923077×10−2 | 0.82093462 | 0.78551795 |
7 | -6.6666667×10−2 | 0.75426795 | -- |
8 | 5.8823529×10−2 | 0.81309148 | -- |
In this example, Aitken's method is applied to a sublinearly converging series and accelerates convergence considerably. The convergence is still sublinear, but much faster than the original convergence: the first value, whose computation required the first three values, is closer to the limit than the eighth value.
The following is an example of using the Aitken extrapolation to help find the limit of the sequence when given some initial where the limit of this sequence is assumed to be a fixed point (say ). For instance, if the sequence is given by with starting point then the function will be which has as a fixed point (see Methods of computing square roots); it is this fixed point whose value will be approximated.
This pseudo code also computes the Aitken approximation to . The Aitken extrapolates will be denoted by aitkenX
. During the computation of the extrapolate, it is important to check if the denominator becomes too small, which could happen if we already have a large amount of accuracy; without this check, a large amount of error could be introduced by the division. This small number will be denoted by epsilon
. Because the binary representation of the fixed point could be infinite (or at least too large to fit in the available memory), the calculation will stop once the approximation is within tolerance
of the true value.
%These choices depend on the problem being solvedx0=1%The initial valuef(x)=(1/2)*(x+2/x)%The function that finds the next element in the sequencetolerance=10^-10%10 digit accuracy is desiredepsilon=10^-16%Do not divide by a number smaller than thismaxIterations=20%Do not allow the iterations to continue indefinitelyhaveWeFoundSolution=false%Were we able to find the solution to within the desired tolerance? not yetfori=1:maxIterationsx1=f(x0)x2=f(x1)if(x1~=x0)lambda=absoluteValue((x2-x1)/(x1-x0))%OPTIONAL: Computes an approximation of |f'(fixedPoint)|, which is denoted by lambdaenddenominator=(x2-x1)-(x1-x0);if(absoluteValue(denominator)<epsilon)%To avoid greatly increasing error, do not divide by too small of a numberprint('WARNING: denominator is too small')break%Leave the loopendaitkenX=x2-((x2-x1)^2)/denominatorif(absoluteValue(aitkenX-x2)<tolerance)%If the value is within toleranceprint("The fixed point is ",aitkenX))%Display the result of the Aitken extrapolationhaveWeFoundSolution=truebreak%Done, so leave the loopendx0=aitkenX%Update x0 to start again endif(haveWeFoundSolution==false)%If we were not able to find a solution to within the desired toleranceprint("Warning: Not able to find solution to within the desired tolerance of ",tolerance)print("The last computed extrapolate was ",aitkenX)end
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