A Numerical Analytic Continuation and Its Application to Fourier Transform

Numerical analytic continuation
Numerical Fourier transform
Numerical examples
Summary
A Numerical Analytic Continuation and
Its Application to Fourier Transform
Hidenori Ogata
Dept. Computer and Network Engineering,
The Graduate School of Informatics and Engineering,
The Univerisity of Electro-Communications, Tokyo, Japan
18 September, 2018
Hidenori Ogata A Numerical Analytic Continuation and Its Application to Fourier

Numerical examples
Summary
Contents
1 Proposition of a numerical method of
the analytic continuation of analytic functions
using continued fractions.
2 Application of the proposed method of analytic continuation
to the computation of Fourier transforms.
3 Numerical examples
which show the eﬀectiveness of the proposed method
4 Summary

Numerical examples
Summary
1. Analytic continuation using continued fractions
Let’s consider an analytic function f (z) given in a Taylor series
f (z) =
∞
n=0
cnzn
and its analytic continuation.

Numerical examples
Summary
We get an analytic continuation of the analytic function f (z)
by transforming it into a continued fraction.
f (z) =
∞
n=0
cnzn
⇒ f (z) =
a1
1 +
a2z
1 +
a3z
1 +
...
.

Numerical examples
Summary
We get an analytic continuation of the analytic function f (z)
f (z) =
∞
n=0
cnzn
⇒ f (z) =
a1
1 +
a2z
1 +
a3z
1 +
...
.
Why continued fractions?
The reasons are as follows.
1 It converges in a wide region.
2 It is easy to compute.

Numerical examples
Summary
1 A continued fraction converges in a wide region.

Numerical examples
Summary
Taylor series
f (z) =
∞
n=0
cnzn
.
R
converges
in a disk |z| < R

Numerical examples
Summary
The Taylor series can be transformed into a continued fraction
under some condition.
continued fraction
f (z) =
a1
1 +
a2z
1 +
a3z
1 +
...
R
converges∗
in a plane with a cut
∗
There is a possibility that f (z) = ∞.

Numerical examples
Summary
continued fraction
f (z) =
a1
1 +
a2z
1 +
a3z
1 +
...
R
analytic continuation
disk → plane with a cut∗
∗
There is a possibility that f (z) = ∞.
We can get an analytic continuation of f (z)

Numerical examples
Summary
1. Analytic continuation using a continued fraction
2 A continued fraction is easy to compute.
We can easily compute the continued fraction
by the recurrence formula for the approximants
fn(z) =
pn(z)
qn(z)
≡
a1
1
+
a2z
1
+
a3z
1
+ · · · +
anz
1
,
p0 = 0, q0 = 1, p1 = a1, q1 = 1,
pn(z) = anz pn−2(z) + pn−1(z)
qn(z) = anz qn−2(z) + qn−1(z),
( n = 2, 3, . . .).
The approximant converges fast, i.e.,
fn(z) → f (z) as n → ∞, exponentially.

Numerical examples
Summary
1. Analyticity and convergence of the continued fraction
Once we obtain the continued fraction,
it satisﬁes the following theorem
on its analyticity and convergence.
f (z) =
a1
1
+
a2z
1
+
a3z
1
+ · · · ,
a = lim
n→∞
an = 0,
S : a plane with a cut as in the ﬁgure
S
a
−1/(4a)
O

Numerical examples
Summary
1. Analyticity and convergence of the continued fraction
f (z) =
a1
1
+
a2z
1
+
a3z
1
+ · · · ,
a = lim
n→∞
an = 0,
S
a
−1/(4a)
O
1 The continued fraction is meromorphic in S.

Numerical examples
Summary
1.Analyticity and convergence of the continued fraction
2 Let T(⊂ S) be an arbitrary compact set.
For arbitrary ǫ > 0, there exists m ∈ N s.t.
|f ∗
m,n(z) − f ∗
m(z)| ≦ C(θT + ǫ)n
( ∀n > m, ∀z ∈ T ),
S
a
−1/(4a)
O
T
where
f ∗
m(z) =
amz
1
+
am+1z
1
+ · · ·
a tail of
the continued fraction
,
f ∗
m,n(z) =
amz
1
+
am+1z
1
+ · · · +
anz
1
,
θT ( 0 < θT < 1 ) : const. depending on T only.

Numerical examples
Summary
1.Analyticity and convergence of the continued fraction
2 Let T(⊂ S) be an arbitrary compact set.
For arbitrary ǫ > 0, there exists m ∈ N s.t.
|f ∗
m,n(z) − f ∗
m(z)| ≦ C(θT + ǫ)n
( ∀n > m, ∀z ∈ T ),
exponential convergence
S
a
−1/(4a)
O
T
where
f ∗
m(z) =
amz
1
+
am+1z
1
+ · · ·
a tail of
the continued fraction
,
f ∗
m,n(z) =
amz
1
+
am+1z
1
+ · · · +
anz
1
,
θT ( 0 < θT < 1 ) : const. depending on T only.

Numerical examples
Summary
1. Numerical analytic continuation
As seen above, once we obtain the continued fraction
corresponding to the analytic function f (z),
it gives an analytic continuation of f (z)
which we can easily compute.

Numerical examples
Summary
1. Numerical analytic continuation
As seen above, once we obtain the continued fraction
corresponding to the analytic function f (z),
it gives an analytic continuation of f (z)
which we can easily compute.
How can we obtain the continued fraction?

Numerical examples
Summary
1. Numerical analytic continuation: QD algorithm
How can we get the continued fraction corresponding to f (z) = cnzn

Numerical examples
Summary
Quotient-diﬀerence (QD) algorithm
We generate the sequences e
(n)
k and q
(n)
k by
e
(n)
0 = 0, q
(n)
1 = cn+1/cn ( n = 0, 1, 2, . . . ),
e
(n)
k = q
(n+1)
k − q
(n)
k + e
(n+1)
k−1 , q
(n)
k+1 = (e
(n+1)
k /e
(n)
k )q
(n+1)
k
( n = 0, 1, . . .; k = 1, 2, . . . ).
f (z) =
c0
1
−
q
(0)
1 z
1
−
e
(0)
1 z
1
−
q
(0)
2 z
1
−
e
(0)
2 z
1
− · · · .

Numerical examples
Summary
Quotient-diﬀerence (QD) algorithm
We generate the sequences e
(n)
k and q
(n)
k by
e
(n)
0 = 0, q
(n)
1 = cn+1/cn ( n = 0, 1, 2, . . . ),
e
(n)
k = q
(n+1)
k − q
(n)
k + e
(n+1)
k−1 , q
(n)
k+1 = (e
(n+1)
k /e
(n)
k )q
(n+1)
k
( n = 0, 1, . . .; k = 1, 2, . . . ).
f (z) =
c0
1
−
q
(0)
1 z
1
−
e
(0)
1 z
1
−
q
(0)
2 z
1
−
e
(0)
2 z
1
− · · · .
The QD algorithm is numerically unstable.
⇒ multiple precision arithmetics

Numerical examples
Summary
2. Numerical Fourier transform
We consider a Fourier transform
F[f ](ξ) =
∞
−∞
f (x)e−2πiξx
dx.
Familiar and important in science and engineering.
It is diﬃcult to compute one
by conventional numerical quadrature rules
especially for slowly decaying functions f (x).
We can compute F[f ](ξ) by the proposed method of
analytic continuation.

Numerical examples
Summary
Rewrite F[f ](ξ) =
∞
−∞
f (x)e−2πiξx
dx (1)
⇓
F[f ](ξ) = lim
ǫ→0+
0
−∞
f (x)e−2πi(ξ+iǫ)x
dx +
+∞
0
f (x)e−2πi(ξ−iǫ)x
dx .
e−2πǫ|x| : convergence factor
We can deﬁne F[f ](ξ) even if the integral (1) does not exist
in the conventional sense.
∗ deﬁnition of a Fourier transform in hyperfunction theory.

Numerical examples
Summary
By rewriting F[f ](ξ), the problem of the Fourier transform is
reduced to that of an analytic continuation.
F[f ](ξ) =
∞
−∞
f (x)e−2πiξx
dx
= lim
ǫ→0+
{F+(ξ + iǫ) − F−(ξ − iǫ)} ( ξ ∈ R ),
where
F+(ζ) =
0
−∞
f (x)e−2πiζx
dx analytic in Im ζ > 0,
F−(ζ) = −
∞
0
f (x)e−2πiζx
dx analytic in Im ζ < 0.
We can get F[f ](ξ)
by the analytic continuation of F±(ζ) onto R.

Numerical examples
Summary
Actually, we get F[f ](ξ) by the following algorithm.
1 We compute F±(ζ) in ± Im ζ > 0.
First, we choose ζ
(±)
0 s.t. ± Im ζ
(±)
0 > 0.
F±(ζ) =
∞
n=0
c(±)
n (ζ − ζ
(±)
0 )n
( ± Im ζ
(±)
0 > 0 ),
c(±)
n =
1
n!
F
(n)
± (ζ
(±)
0 ) = ±
1
n!
∞
0
(±2πix)n
f (∓x)e±2πiζ
(±)
0 x
exponential decay
dx.
We can compute c
(±)
n by conventional numerical quadrature rules.
2 We get F[f ](ξ) by the analytic continuation of F±(ζ) onto R.
F[f ](ξ) = lim
ǫ→0+
{F+(ξ + iǫ) − F−(ξ − iǫ)} ( ξ ∈ R ).

Numerical examples
Summary
3. Numerical examples: Analytic continuation
The analytic continuation of
f (z) = 1 −
z
2
+
z2
3
− · · · =
log(1 + z)
z
( |z| < 1 ).
Throughout this study, we carried out all the computations
using C++ programs
in 100 decimal digit precision (using exﬂib)

Numerical examples
Summary
f (z) = 1 −
z
2
+
z2
3
− · · · =
log(1 + z)
z
( |z| < 1 ).
f (z) =
a1
1
+
a2z
1
+
a3z
1
+ · · · ,
n an n an
1 1.0000 . . . 8 0.28571 42857 14285 . . .
2 0.5000 . . . 9 0.22222 . . .
3 0.16666 . . . 10 0.27777 . . .
4 0.33333 . . . 11 0.22727 27272 72727 . . .
5 0.20000 . . . 12 0.27272 72727 27272 . . .
6 0.30000 . . . 13 0.23076 92307 69230 . . .
7 0.21428 57142 85714 . . . 14 0.26923 07692 30769 . . .

Numerical examples
Summary
f (z) = 1 −
z
2
+
z2
3
− · · · =
log(1 + z)
z
( |z| < 1 ).
The error of the analytic continuation of f (z).
-3 -2 -1 0 1 2 3Re(z)
-3
-2
-1
0
1
2
3
Im(z)
1.0e-50
1.0e-40
1.0e-30
1.0e-20
1.0e-10
1.0e+00
|error|
1.0e-60
1.0e-50
1.0e-40
1.0e-30
1.0e-20
1.0e-10
1.0e+00
-3 -2 -1 0 1 2 3
|error|
Re(z)
Im(z)=0
Im(z)=0.5
Im(z)=1
We can compute the analytic continuation in S = C {x ≦ −1}.

Numerical examples
Summary
(1) f (z) = 1 − z/2 + z2
/3 − · · · (= z−1
log(1 + z)) ( |z| < 1 ),
(2) f (z) = z − z3
/3 + z5
/5 − · · · (= arctan z) ( |z| < 1 ).
-35
-30
-25
-20
-15
-10
-5
0
0 5 10 15 20 25 30 35 40
log10(error)
n
z=1
z=2
z=1+i
-35
-30
-25
-20
-15
-10
-5
0
0 5 10 15 20 25 30 35 40
log10(error)
n
z=1
z=2
z=1+i
(1) (2)
Errors with the continued fractions truncated at the n-th term.
( vertical axis: log10(error), horizontal axis: n )
Our method gives exponential convergence.

Numerical examples
Summary
3. Numerical examples: Zeta functions
We can also apply the numerical analytic continuation
to the compution of ζ(s).
ζ(s) =
∞
n=1
1
ns
=
fs (1)
1 − 21−s
, where fs (z) =
∞
n=0
(−1)n
(n + 1)s
zn
(|z| < 1 ).
We compute fs(1) by the numerical analytic continuation.

Numerical examples
Summary
3. Numerical examples: Zeta functions
We can also apply the numerical analytic continuation
to the compution of ζ(s).
ζ(s) =
∞
n=1
1
ns
=
fs (1)
1 − 21−s
, where fs (z) =
∞
n=0
(−1)n
(n + 1)s
zn
(|z| < 1 ).
We compute fs(1) by the numerical analytic continuation.
Results with the continued fractions truncated at the n-th term.
n ζ(3) error
5 1.2020 46303 07249 1.1e-05
10 1.2020 56904 61243 1.5e-09
15 1.2020 56903 15940 2.0e-13
20 1.2020 56903 15959 2.9e-17
· · ·
exact value 1.2020 56903 15959 . . .

Numerical examples
Summary
3. Numerical examples: Fourier transforms
The Fourier transforms
(1) F[tanh(πx)](ξ), (2) F[(1+x2
)−2
](ξ), (3) F[log |x|](ξ).
1.0e-45
1.0e-40
1.0e-35
1.0e-30
1.0e-25
1.0e-20
1.0e-15
1.0e-10
1.0e-05
1.0e+00
-4 -2 0 2 4
|error|
xi
(1)
(2)
(3)
vertical axis: error
horizontal axis: ξ
Our method works well (especially for (1)).

Numerical examples
Summary
3. Numerical examples
Comparison with the previous methods
1 Sugihara’s method
using the DE rule & the Richardson extrapolation (1987)
2 DE-type rule for oscillatory integrals by T. Ooura & Mori (1991)

Numerical examples
Summary
3. Numerical examples
Comparison with the previous methods
1 Sugihara’s method
using the DE rule & the Richardson extrapolation (1987)
2 DE-type rule for oscillatory integrals by T. Ooura & Mori (1991)
F[tanh(πx)](ξ = 1) = −i cosechπ.
number of the evaluations
of f (x) = tanh(πx) error
our method 666 7.4e-43
(1) DE & Richardson 17156 7.8e-21
(2) DE for
oscillatory integrals 1892 1.5e-46
Our method is superior to the previous methods for this example.

Numerical examples
Summary
Summary
1 We proposed a numerical method of analytic continuation
using continued fractions
2 We applied the method to Fourier transforms
3 Numerical examples show the eﬀectiveness of our method.
Problem for future study
1 reduce the cost of transforming a given analytic function
into a continued fraction
(The present method needs multiple precision arithmetics
due to the instability of the QD algorithm).

Numerical examples
Summary
Summary
1 We proposed a numerical method of analytic continuation
using continued fractions
2 We applied the method to Fourier transforms
3 Numerical examples show the eﬀectiveness of our method.
Problem for future study
1 reduce the cost of transforming a given analytic function
into a continued fraction
(The present method needs multiple precision arithmetics
due to the instability of the QD algorithm).
Thank you very much!

A Numerical Analytic Continuation and Its Application to Fourier Transform

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