Acceleration of Fourier Seriesbugs.unica.it/SC2011/slides/slides/moore.pdf · Charles Moore...

Acceleration of Fourier Series

Charles Moore

Kansas State University, U.S.A.

International Conference on Scientific ComputingOctober 13, 2011

IntroductionFunctions with multiple jumps

Comments, further results

Outline

IntroductionFourier seriesAccelerationPrevious work

Functions with multiple jumpsNotation and lemmasMain resultsSome examples

Charles Moore Acceleration of Fourier Series

Fourier seriesAccelerationPrevious work

Fourier Series

For a function f integrable on [−π, π] and n ∈ Z, we define thenth Fourier coefficient by

f̂ (n) :=1

π∫−π

f (x)e−inxdx .

For n = 0,1,2, . . . and x ∈ [−π, π], define the nth partial sum ofthe Fourier series:

Snf (x) :=n∑

k=−n

f̂ (k)eikx .

Fourier Series

For a function f integrable on [−π, π] and n ∈ Z, we define thenth Fourier coefficient by

f̂ (n) :=1

π∫−π

f (x)e−inxdx .

For n = 0,1,2, . . . and x ∈ [−π, π], define the nth partial sum ofthe Fourier series:

Snf (x) :=n∑

k=−n

f̂ (k)eikx .

Accelerating Convergence

The convergence of Fourier series can be quite slow and oftendepends on subtle cancellation.

This is especially true for functions with jump discontinuities.

For a sequence {sn} that converges to s, we say that atransformation {tn} accelerates the convergence of {sn} ifthere exists a positive integer k such that each tn depends onlyon s1, s2, ..., sn+k and {tn} converges to s faster than {sn}, thatis

limn→∞

tn − ssn − s

limn→∞

tn − ssn − s

limn→∞

tn − ssn − s

The δ2 Transformation

Given a numerical sequence sn, the δ2 process transforms it tothe sequence

tn := sn −(sn+1 − sn)(sn − sn−1)

(sn+1 − sn)− (sn − sn−1),

where we set tn = sn if the denominator of the fraction is zero.

If a complex series and its δ2 transform converge, then theirsums are equal (Tucker, 1967).

It is well known that the δ2 process behaves badly for manysequences; it can turn a convergent sequence into one withmultiple convergent subsequences. It can also turn divergentsequences into convergent sequences.

tn := sn −(sn+1 − sn)(sn − sn−1)

(sn+1 − sn)− (sn − sn−1),

tn := sn −(sn+1 − sn)(sn − sn−1)

(sn+1 − sn)− (sn − sn−1),

Applying this to Fourier SeriesWe consider the possibility of applying the δ2 transformpointwise to the partial sums Snf (x) of an integrable function fon [−π, π]. This results in the sequence of functions Tnf (x)given by:

Tnf (x) := Snf (x)− (Sn+1f (x)− Snf (x))(Snf (x)− Sn−1f (x))(Sn+1f (x)− Snf (x))− (Snf (x)− Sn−1f (x))

Various authors have done numerical experiments: Smith andFord, Drummond.

If f (x) = −1 on [−π,0) and f (x) = 1 on [0, π], then Snf (π2 ) givesthe partial sums for the Leibnitz series1 = f (π2 ) =

4π (1−

13 + 1

5 −17 + . . . ).

The convergence of this series is extremely slow but isdramatically accelerated by applying the δ2 process.

4π (1−

13 + 1

5 −17 + . . . ).

4π (1−

13 + 1

5 −17 + . . . ).

The Transform for Functions With One Jump

Theorem (joint with E. Abebe and P. Graber, 2007)(a) Suppose f ∈ C2([−π, π]) and that f (−π) 6= f (π). Then thetransformed sequence of partial sums Tnf (x) diverges at everyx of the form x = 2aπ, where a ∈ [−.5, .5] is irrational.

(b) Suppose f is as above and let x := 2jπk where j

k is in lowestterms and k is odd. Then Tnf (x) has three limit points, f (x) andf (x)± α2 sin2 (x/2)

α sin (x/2)+2β cos (x/2) , where α = [f (π)− f (−π)]/π andβ = [f ′(π)− f ′(−π)]/π.

The Transform for Functions With One Jump

Theorem (joint with E. Abebe and P. Graber, 2007)(a) Suppose f ∈ C2([−π, π]) and that f (−π) 6= f (π). Then thetransformed sequence of partial sums Tnf (x) diverges at everyx of the form x = 2aπ, where a ∈ [−.5, .5] is irrational.

(b) Suppose f is as above and let x := 2jπk where j

k is in lowestterms and k is odd. Then Tnf (x) has three limit points, f (x) andf (x)± α2 sin2 (x/2)

Fourier series and transformed Fourier series forf (x) = x

0 0.5 1 1.5 2 2.5 30

Figure: S100f for f (x) = x .

0 0.5 1 1.5 2 2.5 30

Figure: T100f

The δ2 process is same as ε(n)2 in family of transforms ε(n)k .

Brezinski (see also Wynn): To Snf add conjugate function S̃nf ;create analytic function Gnf (z). Apply epsilon algorithm toGnf (eiθ); take real part.

Brezinski gave numerical experiments to demonstrate that thisprocedure reduces the Gibbs phenomenon.

Beckermann, Matos, and Wielonsky show that this methodaccelerates convergence for functions of the form f = f1 + f2,where f1 has prescribed discontinuities but is smoothelsewhere, f2 has quickly decaying Fourier coefficients, andG(f1) = limn→∞Gn(f1) is a certain type of hypergeometricfunction.

Consider the analytic function∞∑

eick log k eikx

k12+α

α ∈ R and c > 0. This was studied by Hardy and Littlewood.

If α > 0 then the partial sums are uniformly convergent;consequently, it is the Fourier series of a continuous function.

TheoremSuppose c > 0, 0 < α ≤ 1

2 . Consider the partial sums

Sn(x) =n∑

eick log k eikx

k12+α

and let Tn(x) be the sequence of functions which results fromapplying the δ2 process to the sequence Sn(x). Then at everyx, Tn(x) fails to converge to the same limit as Sn(x). In fact, if0 < α < 1

2 , {Tn(x)} has subsequences which becomeunbounded.

eick log k eikx

k12+α

α ∈ R and c > 0. This was studied by Hardy and Littlewood.If α > 0 then the partial sums are uniformly convergent;consequently, it is the Fourier series of a continuous function.

Sn(x) =n∑

eick log k eikx

k12+α

eick log k eikx

k12+α

α ∈ R and c > 0. This was studied by Hardy and Littlewood.If α > 0 then the partial sums are uniformly convergent;consequently, it is the Fourier series of a continuous function.

Sn(x) =n∑

eick log k eikx

k12+α

Notation and lemmasMain resultsSome examples

Some Notation

Throughout this section, f is a piecewise smooth functionintegrable on [−π, π] having a finite number of jumpdiscontinuities at a1,a2, ...,am ∈ (−π, π).

Suppose that for each j , f (aj±) = limx→aj

± f (x) andf ′(aj

±) = limx→aj± f ′(x) exist and are finite.

Set dj = f (a+j )− f (a−j ) and d∗j = f ′(a−j )− f ′(a+

j ) for allj ∈ {1, ...,m}.

Some Notation

j ) for allj ∈ {1, ...,m}.

Some Notation

j ) for allj ∈ {1, ...,m}.

Partial Sums of the Fourier Series

After some computation, the N th partial sum of the Fourierseries is

SN f (x) = f̂ (0) +N∑

m∑j=1

[dj sin k(x − aj)

]+ εk

,where

εk =1

m∑j=1

[d∗j cos k(x − aj)

]− f̂ ′′(k)eikx + f̂ ′′(−k)e−ikx

We note that εN = O( 1N2 ).

SN f (x) = f̂ (0) +N∑

m∑j=1

[dj sin k(x − aj)

]+ εk

,where

εk =1

m∑j=1

]− f̂ ′′(k)eikx + f̂ ′′(−k)e−ikx

SN f (x) = f̂ (0) +N∑

m∑j=1

[dj sin k(x − aj)

]+ εk

,where

εk =1

m∑j=1

]− f̂ ′′(k)eikx + f̂ ′′(−k)e−ikx

Transformed Series

Applying the δ2 process and simplifying gives

SN f (x)− TN f (x) ≈(m∑

[dj sin (N + 1)(x − aj)

])( m∑j=1

[dj sin N(x − aj)

Nπm∑

j=1[dj sin (N + 1)(x − aj)]− (N + 1)π

m∑j=1

[dj sin N(x − aj)]

Investigate the behavior of both the numerator and denominatorof the large fraction on the right.

Denominator estimatesLemmaFor any fixed real number x0, on a subinterval of

[x0, x0 +

g(x) =m∑

j=1dj [sin (N + 1)(x − aj)− sin N(x − aj)] is bounded

above by CN .

proof: Roughly g(x0) and g(x0 +π

N+1) have opposite sign.

Thus, on 2(N + 1) subintervals of [−π, π] we estimate∣∣∣∣Nπ m∑j=1

[dj sin (N + 1)(x − aj)]− (N + 1)πm∑

∣∣∣∣≤ Nπ|g(x)|+ π

m∑j=1

|dj | ≤ C

[x0, x0 +

g(x) =m∑

above by CN .

[dj sin (N + 1)(x − aj)]− (N + 1)πm∑

∣∣∣∣≤ Nπ|g(x)|+ π

m∑j=1

|dj | ≤ C

[x0, x0 +

g(x) =m∑

above by CN .

[dj sin (N + 1)(x − aj)]− (N + 1)πm∑

∣∣∣∣≤ Nπ|g(x)|+ π

m∑j=1

|dj | ≤ C

Set αN =m∑

j=1dj cos Naj , βN =

m∑j=1

dj sin Naj , AN =

√αN

2 + βN2,

φN = arctan(βN

), so that with this notation

m∑j=1

dj sin N(x − aj) =m∑

dj(sin Nx cos Naj − cos Nx sin Naj)

= αN sin Nx − βN cos Nx = AN sin(Nx − φN).

Note that if aj =pjπqj

in lowest terms, AN is periodic of periodL = 2 lcm{qj}.

Set αN =m∑

m∑j=1

dj sin Naj , AN =

√αN

2 + βN2,

φN = arctan(βN

m∑j=1

Set αN =m∑

m∑j=1

dj sin Naj , AN =

√αN

2 + βN2,

φN = arctan(βN

m∑j=1

Another denominator estimatem∑

dj [ sin (N + 1)(x − aj)− sin N(x − aj)]

= AN+1 sin ((N + 1)x − φN+1)− AN sin (Nx − φN)

LemmaSuppose each aj is a rational multiple of π, aj =

pjqjπ, in lowest

terms. Suppose x = 2aπ, where a ∈ [−.5, .5] is irrational. FixJ ∈ {1, ...,L}, L = 2 lcm{qj}, which has AJAJ+1 6= 0. Then thereare an infinite number of integers N = kL + J such that

|AN+1 sin ((N + 1)x − φN+1)− AN sin (Nx − φN)|

≤ max{AJ ,AJ+1}24L2π

proof: Chebyshev’s theorem: Given a irrational, θ real,|na−m − θ| < 3/n, infinite # of m,n

pjqjπ, in lowest

Numerator estimatesWe want to estimate m∑

[dj sin (N + 1)(x − aj)

] m∑j=1

[dj sin N(x − aj)

]= AN+1 sin((N + 1)x − φN+1)AN sin(Nx − φN)

Suppose x is a point at which

|AN+1 sin((N + 1)x − φN+1)− AN sin(Nx − φN)| ≤CN

Squaring each side and rearranging gives

ANAN+1 sin ((N + 1)x − φN+1) sin (Nx − φN)

≥ 12

min{AN ,AN+1}2[sin2 ((N + 1)x − φN+1) + sin2 (Nx − φN)

Except for a small set, the term in brackets is bounded below.

[dj sin (N + 1)(x − aj)

] m∑j=1

[dj sin N(x − aj)

≥ 12

[dj sin (N + 1)(x − aj)

] m∑j=1

[dj sin N(x − aj)

≥ 12

Theorems

Theorem (joint with E. Jennings, D. Muñiz, A. Toth)Let f be a piecewise C2 function on [−π, π] having a finitenumber of jump discontinuities at a1 < a2 < ... < am ∈ (−π, π),and suppose f (aj

±) and f ′(aj±) exist and are finite.

(a) Suppose N is a sufficiently large positive integer withANAN+1 6= 0. Then there exists intervals on which|TN f (x)− SN f (x)| ≥ m min{AN ,AN+1}2 − C

N , where m and Care constants which do not depend on N or x .Proof: For (a) there are 2(N + 1) intervals of length C

N2 ,uniformly spaced, where the denominator of TN f (x)− SN f (x) isbounded. The numerator is estimated using the previouslemma.

Theorems

N , where m and Care constants which do not depend on N or x .

Proof: For (a) there are 2(N + 1) intervals of length CN2 ,

uniformly spaced, where the denominator of TN f (x)− SN f (x) isbounded. The numerator is estimated using the previouslemma.

Theorems

N , where m and Care constants which do not depend on N or x .Proof: For (a) there are 2(N + 1) intervals of length C

N2 ,uniformly spaced, where the denominator of TN f (x)− SN f (x) isbounded. The numerator is estimated using the previouslemma.

Theorem (continued)Let f be a piecewise C2 function on [−π, π] having a finitenumber of jump discontinuities at a1 < a2 < ... < am ∈ (−π, π),and suppose f (aj

(b) Suppose that the aj are rational multiples of π, aj =pjqjπ (in

lowest terms), so that the sequence AN has periodL = 2 lcm{qj}. Suppose that there exists a J ∈ {1, . . . ,L} suchthat AJAJ+1 6= 0. Then TN f does not converge to SN f uniformly.

For (b), the AN are in this case periodic, say of period L. Thenfor N = kL + J, there are x where

|TN f (x)−SN f (x)| ≥ m min{AN ,AN+1}2−CN

= m min{AJ ,AJ+1}2−CN

TheoremSuppose x = 2aπ, where a is an irrational number, andsuppose that there exists J ∈ {1, ...,L} such that AJAJ+1 6= 0.Then {TN f (x)} fails to converge.

Proof.For x = 2πa, a irrational, second denominator Lemmaimmediately gives an infinite number of k (which depend on x)for which the denominator with N = kL + J is bounded. Thenprevious lemma gives a lower bound for the numerator at thesepoints.

TheoremSuppose x = 2aπ, where a is an irrational number, andsuppose that there exists J ∈ {1, ...,L} such that AJAJ+1 6= 0.Then {TN f (x)} fails to converge.

Proof.For x = 2πa, a irrational, second denominator Lemmaimmediately gives an infinite number of k (which depend on x)for which the denominator with N = kL + J is bounded. Thenprevious lemma gives a lower bound for the numerator at thesepoints.

Indicator function of the interval [−1,1].

−3 −2 −1 0 1 2 3−0.5

Figure: T20f

−3 −2 −1 0 1 2 3−0.5

Figure: T30f

A situation where the hypotheses are not satisfiedDefine f1 by

f1(x) =

3πx + 1 if x ∈ [−π, −2π

2πx + 1 if x ∈ (−2π3 ,0)

0 if x ∈ (0, 2π3 )

3πx − 1 if ∈ (2π

3 , π].

which is a function with jumps of 1 evenly spaced. The graphsof S20f1 and T20f1 are shown in figure 2. Here, ANAN+1 = 0 forall N and our theorems do not apply; indeed it is difficult todistinguish the graphs.

Recall: αN =m∑

m∑j=1

dj sin Naj ,

√αN

2 + βN2, φN = arctan

)Charles Moore Acceleration of Fourier Series

A situation where the hypotheses are not satisfied

−3 −2 −1 0 1 2 3−1.5

−0.5

Figure: S20f

−3 −2 −1 0 1 2 3−1.5

−0.5

Figure: T20f

A small change in the last example; hypotheses nowhold

Define f2 by

f2(x) =

3πx + 1 if x ∈ [−π, −2π

4πx + 12 if x ∈ (−2π

3 ,0)0 if x ∈ (0, 2π

3 )3πx − 1 if ∈ (2π

3 , π].

This has evenly spaced jumps, but the jumps are not equal,and it is easy to check that Theorem 2 does apply. The nextframe gives the graphs of S20f2 and T20f2.

A small change in the last example; hypotheses nowhold

−3 −2 −1 0 1 2 3−1.5

−0.5

Figure: S20f

−3 −2 −1 0 1 2 3−1.5

−0.5

Figure: T20f

Iteration of δ2

Consider the iterated transform, T 2n . Suppose f is C2 except for

one jump, and x is of the form x = 2πjk , k odd, where j and k

have no common factors.

Claim: T 2n f (x) has subsequences which do not converge to

f (x).

Recall: Tnf (x) has three limit points, f (x) andf (x)± α2 sin2 (x/2)

Iteration of δ2

f (x).

Iteration of δ2

f (x).

Iteration of δ2

The claim follows from:

LemmaSuppose an is a sequence and l , k are fixed positive integerswith k ≥ 2, such that an → a if n 6= l + mk and m = 1,2, . . . andan → a + b for n = l + mk , m = 1,2, . . . . Then T (an)→ a + b

2for n = k + ml as m→∞.Proof. Notice that for n = l + mk ,

T (an) =an −(an+1 − an)(an − an−1)

(an+1 − an)− (an − an−1)

→ (a + b)− [a− (a + b)][(a + b)− a][a− (a + b)]− [(a + b)− a]

= a +b2.

Iteration of δ2

The claim follows from:

LemmaSuppose an is a sequence and l , k are fixed positive integerswith k ≥ 2, such that an → a if n 6= l + mk and m = 1,2, . . . andan → a + b for n = l + mk , m = 1,2, . . . . Then T (an)→ a + b

2for n = k + ml as m→∞.Proof. Notice that for n = l + mk ,

T (an) =an −(an+1 − an)(an − an−1)

(an+1 − an)− (an − an−1)

→ (a + b)− [a− (a + b)][(a + b)− a][a− (a + b)]− [(a + b)− a]

= a +b2.

Shanks transformations – ε algorithm

Analyzing ε4(SN f (x)) seems to be much more difficult.

ε−1(n) := 0 for every n, ε0(n) := Snf (x)εk+1(n) = εk−1(n + 1) + 1

εk (n+1)−εk (n).

ε4(n) = Tn+2 +1

[ 1Tn+2−Tn+2

] + [ 1Tn+3−Tn+2

]− [ 1Tn+2−Tn+1

Suppose f is C2 with one jump, x = 2πjk , where j and k have no

common factors and k is odd. There are subsequences alongwhich Tn(x)→ f (x) + α2 sin2(x/2)

α sin(x/2)+2β cos(x/2) .

Direct computation shows that curiously, the straysubsequences of Tnf (x) which diverged from f (x) aretransformed back to the correct limit.

ε−1(n) := 0 for every n, ε0(n) := Snf (x)

εk+1(n) = εk−1(n + 1) + 1εk (n+1)−εk (n)

ε4(n) = Tn+2 +1

[ 1Tn+2−Tn+2

] + [ 1Tn+3−Tn+2

]− [ 1Tn+2−Tn+1

εk (n+1)−εk (n).

ε4(n) = Tn+2 +1

[ 1Tn+2−Tn+2

] + [ 1Tn+3−Tn+2

]− [ 1Tn+2−Tn+1

εk (n+1)−εk (n).

ε4(n) = Tn+2 +1

[ 1Tn+2−Tn+2

] + [ 1Tn+3−Tn+2

]− [ 1Tn+2−Tn+1

εk (n+1)−εk (n).

ε4(n) = Tn+2 +1

[ 1Tn+2−Tn+2

] + [ 1Tn+3−Tn+2

]− [ 1Tn+2−Tn+1

The Lubkin transform

Let ρN f (x) = SN+1f (x)−SN f (x)SN f (x)−SN−1f (x) . The Lubkin transform of SN f (x) is

TN f (x) := SN f (x) +(SN+1f (x)− SN f (x))(1− ρN+1f (x))

1− 2ρN+1f (x) + ρN f (x)ρN+1f (x).

Theorem (with Boggess, Bunch, 2008)Suppose that f ∈ C2([−π, π]) and that f (−π) 6= f (π). Considerthe sequence TN f (x) formed by applying the Lubkin transformto the sequence SN f (x). Then TN f (x) fails to converge to f (x)at every x of the form x = 2πa, where a ∈ (−1

4 ,14) is irrational.

The Lubkin transform

Let ρN f (x) = SN+1f (x)−SN f (x)SN f (x)−SN−1f (x) . The Lubkin transform of SN f (x) is

TN f (x) := SN f (x) +(SN+1f (x)− SN f (x))(1− ρN+1f (x))

1− 2ρN+1f (x) + ρN f (x)ρN+1f (x).

Theorem (with Boggess, Bunch, 2008)Suppose that f ∈ C2([−π, π]) and that f (−π) 6= f (π). Considerthe sequence TN f (x) formed by applying the Lubkin transformto the sequence SN f (x). Then TN f (x) fails to converge to f (x)at every x of the form x = 2πa, where a ∈ (−1

4 ,14) is irrational.

f (x) = x and the Lubkin transform

0.0 0.5 1.0 1.5 2.0 2.5 3.0

Figure: Fifty terms of the Fourierseries

0.0 0.5 1.0 1.5 2.0 2.5 3.0

Figure: Lubkin transform– 50thterm

Thanks again to the organizers for the invitationand thank you for listening.

Appendix References

References I

A.C. Aitken, On Bernoulli’s numerical solution of algebraicequations, Proc. Roy. Soc. Edinburgh 46 (1926), 289-305.

E. Abebe, J. Graber, and C. N. Moore, Fourier Series andthe δ2 Process, J. Comput. Appl. Math. 224 (2009), no. 1,146-151.

J. Boggess, E. Bunch, C. N. Moore, Fourier series and theLubkin W-transform, Numer. Algorithms 47 (2008),133-142.

C. Brezinski, Accélération de la Convergence en AnalyseNumérique, Springer-Verlag, Berlin, Heidelberg, New York.1977.

Appendix References

References II

C. Brezinski, Extrapolation algorithms for filtering series offunctions, and treating the Gibbs phenomenon, Numer.Algorithms 36 (2004), 309-329.

C. Brezinski. and M. Redivo-Zaglia, Extrapolation Methods.Theory and Practice. North-Holland, Amsterdam, 1991.

J-.P Delahaye, Sequence Transformations, Springer Seriesin Computational Mathematics 11, Springer Verlag, Berlin,

J. E. Drummond, Convergence speeding, convergence andsummability, J. Comput. Appl. Math. 11 (1984), 145-159.

A. Ya. Khinchin, Continued Fractions, The University ofChicago Press, Chicago, 1964,

Appendix References

References III

C. N. Moore, Acceleration of Fourier Series, J. Analysis, 17(2009), 1-20.

D. Shanks, Non-linear transformations of divergent andslowly convergent sequences, J. Math. Phys. 34 (1955),1-42.

A. Sidi, Practical Extrapolation Methods: Theory andApplications. Cambridge University Press, 1st Edition,2003.

D. A. Smith, and W. F. Ford, Numerical Comparisons ofNonlinear Convergence Accelerators, Math. Comp. 38, no.158 (1982), 481-499.

Appendix References

References IV

R. Tucker, The δ2-Process and related topics, Pacific J.Math. 22 , no. 2 (1967), 349-359.

R. Tucker, The δ2-Process and related topics II, Pacific J.Math. 28 , no. 2 (1969), 455-463.

J. Wimp, Sequence Transformations and TheirApplications, Academic Press, New York. 1981.

P. Wynn, Transformations to accelerate the convergence ofFourier series, in: Gertrude Blanch Anniversary Volume(Wright Patterson Air Force Base, Aerospace ResearchLaboratories, Office of Aerospace Research, United StatesAir Force, 1967), 339-379.

Appendix References

References V

A. Zygmund, Trigonometric Series, Second Edition.Cambridge University Press, Cambridge, 1959.

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