Numerical Fourier transform based on hyperfunction theory

Hyperfunction theory
Fourier transform in hyperfunction theory
Numerical Fourier transform
Numerical examples
Summary
Numerical Fourier Transform Based on
Hyperfunction Theory
Hidenori Ogata
Dept. Computer and Network Engineering,
The Graduate School of Informatics and Engineering,
The Univerisity of Electro-Communications, Tokyo, Japan
21 June, 2018
Hidenori Ogata Numerical Fourier Transform Based on Hyperfunction Theory

Numerical examples
Summary
Aim of the study
Aim of the study
Numerical Fourier transform based on hyperfunction theory

Numerical examples
Summary
Aim of the study
Aim of the study
Fourier transform
F[f ](ξ) =
∞
−∞
f (x) exp(−2πiξx) dx.
It is very familiar in science and engineering.
But, it is diﬃcult to compute it by conventional methods,
especially, if f (x) decays slowly as x → ±∞.

Numerical examples
Summary
Fourier transforms are diﬃcult to compute!
Numerical integration by the DE rule.
(1)
∞
0
dx
1 + x2
, (2)
∞
0
cos x
1 + x2
dx.
-16
-14
-12
-10
-8
-6
-4
-2
0
0 10 20 30 40 50 60 70
log10(error)
N
(1)
(2)
vertical axis: log10(error)
horizontal axis: number
of sampling points N

Numerical examples
Summary
Aim of the study
Aim of the study
Fourier transform
F[f ](ξ) =
∞
−∞
f (x) exp(−2πiξx) dx.
It is very familiar in science and engineering.
But, it is diﬃcult to compute it by conventional methods,
especially, if f (x) decays slowly as x → ±∞.
It is easy to compute it using hyperfunction theory.

Numerical examples
Summary
Contents
Contents
1 Basis of hyperfunction theory
2 Fourier transform in hyperfunction theory
3 Numerical Fourier transform
4 Numerical examples
5 Summary

Numerical examples
Summary
1. Hyperfunction theory
M. Sato, 1958
a theory of generalized functions
based on complex function theory
Hyperfunction
the diﬀerence between the boundary values
of an analytic function F(z)
f (x) = [F(z)] ≡ F(x + i0) − F(x − i0).
F(z) : deﬁning function of the hyperfunction f (x)

Numerical examples
Summary
1. Hyperfunction: f (x) = F(x + i0) − F(x − i0)
Examples
δ(x) = −
1
2πi
1
x + i0
−
1
x − i0
= lim
ǫ↓0
1
π
ǫ
x2 + ǫ2
,
χ(−1,1)(x) =
1 x ∈ (−1, 1)
0 x ∈ [−1, 1]
= −
1
2πi
log
(x + i0) + 1
(x + i0) − 1
− log
(x − i0) + 1
(x − i0) − 1
.
log z: the principal value, i.e., log x ∈ R for x > 0.
-1 -0.5 0 0.5 1Re z -1
-0.5
0
0.5
1
Im z
-6
-4
-2
0
2
4
6
Re F(z)
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2Re z -2-1.5-1-0.500.511.52
Im z
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
Re F(z)
deﬁning function F(z) of δ(x) deﬁning function F(z) of χ(−1,1)(x)

Numerical examples
Summary
2. Fourier transform in hyperfunction theory
How to treat Fourier transforms in hyperfunction theory.
F[f ](ξ) =
∞
−∞
f (x)e−2πiξx
dx.

Numerical examples
Summary
F[f ](ξ) =
0
−∞
f (x)e−2πi(ξ+i0)x
dx +
+∞
0
f (x)e−2πi(ξ−i0)x
dx .
We can deﬁne F[f ](ξ) thanks to e−2πǫ|x|
even if the integral is not convergent in the conventional sense.

Numerical examples
Summary
F[f ](ξ) =
0
−∞
dx +
+∞
0
dx .
(Example)
F[1](ξ) =
0
−∞
e−2πi(ξ+i0)x
dx +
∞
0
e−2πi(ξ−i0)x
dx
= −
1
2πi
1
ξ + i0
−
1
ξ − i0
= δ(ξ).

Numerical examples
Summary
F[f ](ξ) =
0
−∞
dx +
+∞
0
dx .
Fourier transform F[f ](ξ)
Hyperfunction with the deﬁning functions F±(ζ)
F[f ](ξ) = F+(ξ + i0) − F−(ξ − i0),
F+(ζ) =
0
−∞
f (x)e−2πiζx
dx ( Im ζ > 0 ),
F−(ζ) = −
∞
0
f (x)e−2πiζx
dx ( Im ζ < 0 ).

Numerical examples
Summary
3. Numerical Fourier transform: strategy
We use the deﬁnition of Fourier transforms in hyperfunction theory
for numerical Fourier transforms.
F[f ](ξ) =
∞
−∞
f (x)e−2πiξx
dx = F+(ξ + i0) − F−(ξ − i0),
where F+(ζ) =
0
−∞
f (x)e−2πiζx
dx ( Im ζ > 0 ),
F−(ζ) = −
∞
0
f (x)e−2πiζx
dx ( Im ζ < 0 ).
We can compute F±(ζ) in C because the integrands include
the exponentially decaying factor exp(−2π| Im ζ||x|).

Numerical examples
Summary
3. Numerical Fourier transform: strategy
We use the deﬁnition of Fourier transforms in hyperfunction theory
for numerical Fourier transforms.
F[f ](ξ) =
∞
−∞
f (x)e−2πiξx
dx = F+(ξ + i0) − F−(ξ − i0),
where F+(ζ) =
0
−∞
f (x)e−2πiζx
dx ( Im ζ > 0 ),
F−(ζ) = −
∞
0
f (x)e−2πiζx
dx ( Im ζ < 0 ).
Strategy
1 Get F±(ζ) in { ζ ∈ C | ± Im ζ > 0 }.
2 Get F±(ξ ± i0) on R by analytic continuation.

Numerical examples
Summary
3. Numerical Fourier transform: strategy 1/2
1. Get F±(ζ) in { ζ ∈ C | ± Im ζ > 0 }.
F+(ζ) =
0
−∞
f (x)e−2πiζx
dx ( Im ζ > 0 ),
F−(ζ) = −
∞
0
f (x)e−2πiζx
dx ( Im ζ < 0 ).

Numerical examples
Summary
1. Get F±(ζ) in { ζ ∈ C | ± Im ζ > 0 }.
F±(ζ) = ±
∞
0
f (∓x)e±2πiζx
dx ( ± Im ζ > 0 ).

Numerical examples
Summary
1. Get F±(ζ) in { ζ ∈ C | ± Im ζ > 0 }.
F±(ζ) = ±
∞
0
f (∓x)e±2πiζx
dx ( ± Im ζ > 0 ).
We get F±(ζ) in Taylor series.
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.

Numerical examples
Summary
2. Get F±(ξ ± i0) on R by analytic continuation.
Taylor series → continued fraction
F±(ζ) =
∞
n=0
cn(ζ − ζ
(±)
0 )n
=
a
(±)
0
1 +
a
(±)
1 (ζ − ζ
(±)
0 )
1 +
a
(±)
2 (ζ − ζ
(±)
0 )
1 +
...
R
R
ζ
(±)
0
convergence region
c
(±)
n → a
(±)
n by quotient diﬀerence (QD) algorithm
The QD algorithm is unstable. → multiple precision arithmetic

Numerical examples
Summary
2. Get F±(ξ ± i0) on R by analytic continuation.
Taylor series → continued fraction
F±(ζ) =
∞
n=0
cn(ζ − ζ
(±)
0 )n
=
a
(±)
0
1 +
a
(±)
1 (ζ − ζ
(±)
0 )
1 +
a
(±)
2 (ζ − ζ
(±)
0 )
1 +
...
R
R
ζ
(±)
0
convergence region
F[f ](ξ) = F+(ξ + i0) − F−(ξ − i0).

Numerical examples
Summary
4. Numerical examples
(1) F[tanh(πx)](ξ) = −i cosech(πξ),
(2) F[(1 + x2
)−ν−1/2
] =
2πν+1/2
Γ(ν + 1/2)
|ξ|ν
Kν(2π|ξ|) ( ν = 1.5 ),
(3) F[log |x|](ξ) = −γδ(ξ) −
1
2|ξ|
.
The center of the Taylor series of F±(ζ): ζ
(±)
0 = ±i.
C++ programs,
multiple precision arithmetic (100 decimal digits).
exﬂib: multiple precision arithmetic library.

Numerical examples
Summary
4. Numerical examples: errors of our method
-60
-50
-40
-30
-20
-10
0
-4 -2 0 2 4
log10(error)
xi
(1)
(2)
(3)
vertical axis: log10(error)
horizontal axis: ξ
(1) F[tanh(πx)](ξ), (2) F[(1+x2
)−ν−1/2
](ξ), (3) F[log |x|].
The number of sampling points for numerical integration (the DE rule)
(1) 1330 (2) 1416 (3) 1430.

Numerical examples
Summary
4. Numerical examples: an interesting fact
We do not need to compute oscillatory integrals
for Fourier transforms.

Numerical examples
Summary
4. Numerical examples: an interesting fact
We do not need to compute oscillatory integrals
for Fourier transforms.
The deﬁning function ( F[f ](ξ) = F+(ξ + i0) − F−(ξ − i0) )
F±(ζ) = ±
∞
0
f (∓x)e±2πiζx
dx
=
∞
n=0
c
(±)
n (ζ ∓ i)n
( ± Im ζ > 0 ),
c
(±)
n = ±
1
n!
∞
0
(±2πix)n
f (∓x) exp(−2πx)dx.
including no oscillatory function.

Numerical examples
Summary
4. Numerical examples: comparisons
Comparison with the previous methods
1 DE rule & Richardson extrapolation
M. Sugihara, J. Comp. Appl. Math., 17 (1987) 47–68.
F[f ](ξ) = lim
n→∞
∞
−∞
f (x) exp(−2πiξx)exp(−2−n
x2
)dx.
2 DE-type rule for oscillatory integrals
T. Ooura & M. Mori,
J. Comp. Appl. Math., 38 (1991) 353–360.

Numerical examples
Summary
F[tanh(πx)](ξ) = −i cosech(πξ), ξ = 1.
multiple precision arithmetic (100 decimal digits, exﬂib)
the center of the Taylor series ζ
(±)
0 = 1 ± i.

Numerical examples
Summary
F[tanh(πx)](ξ) = −i cosech(πξ), ξ = 1.
multiple precision arithmetic (100 decimal digits, exﬂib)
the center of the Taylor series ζ
(±)
0 = 1 ± i.
number of sampling points
for integration error
our method 2610 2.8 × 10−28
DE & Richardson 10060 9.1 × 10−20
DE for
oscillatory integrals 948 5.0 × 10−25

Numerical examples
Summary
our method 2610 2.8 × 10−28
DE & Richardson 10060 9.1 × 10−20
DE for

Numerical examples
Summary
our method 2610 2.8 × 10−28
DE & Richardson 10060 9.1 × 10−20
DE for
Our method > DE & Richardson
Our method < DE for oscillatory integrals
Our method computes F[f ](ξ) as a function, i.e.,
once we obtain the coeﬃcients a
(±)
n of the continued fraction,
we can compute F[f ](ξ)’s for many ξ’s.
The conventional methods compute F[f ](ξ) as an integrals.

Numerical examples
Summary
Summary
1 Numerical Fourier transform based on hyperfunction theory
2 Fourier transform as a hyperfunction
F[f ](ξ) =
0
−∞
dx +
∞
0
dx.
deﬁning functions F±(ζ)
Get F±(ζ) in { ζ ∈ C | ± Im ζ > 0 }.
Get F±(ξ ± i0) on R by analytic continuation
(by the continued fraction).
3 Numerical examples shows the eﬀectiveness of our method.
Problems for future studies
1 Theoretical error estimate
2 Analytic continuation by continued fractions

Numerical examples
Summary
Thank you very much!

Numerical Fourier transform based on hyperfunction theory

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