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A novel recursive method to reconstruct multivariate functions on the unit cube

  • Zhihua Zhang EMAIL logo
Published/Copyright: December 29, 2017
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Open Mathematics
From the journal Open Mathematics Volume 15 Issue 1

Abstract

Due to discontinuity on the boundary, traditional Fourier approximation does not work efficiently for d−variate functions on [0, 1]d. In this paper, we will give a recursive method to reconstruct/approximate functions on [0, 1]d well. The main process is as follows: We reconstruct a d−variate function by using all of its (d−1)–variate boundary functions and few d–variate Fourier coefficients. We reconstruct each (d−1)–variate boundary function given in the preceding reconstruction by using all of its (d−2)–variate boundary functions and few (d−1)–variate Fourier coefficients. Continuing this procedure, we finally reconstruct each univariate boundary function in the preceding reconstruction by using values of the function at two ends and few univariate Fourier coefficients. Our recursive method can reconstruct multivariate functions on the unit cube with much smaller error than traditional Fourier methods.

Keywords: Recursive method; Hyperbolic cross truncation; Multivariate functions; Fourier series
MSC 2010: 42B05; 41A63

1 Introduction

Due to discontinuity on the boundary, traditional Fourier approximation does not work efficiently for d–variate functions on [0, 1]d [1, 2, 3, 4, 5, 6]. In order to solve it, we will give a recursive method to reconstruct/approximate functions on [0, 1]d well. We start from a smooth triple–variate function f on the unit cube [0, 1]3. There are six faces (or six 2-dimensional boundaries) on the unit cube:

(ν,y,z)(0≤y,z≤1,ν=0or1),(x,ν,z)(0≤x.z≤1,ν=0or1),(x,y,ν)(0≤x,y≤1,ν=0or1).

The corresponding six bivariate boundary functions on [0, 1]2 are

f(ν,y,z),f(x,ν,z),f(x,y,ν)(ν=0or1). (1)

There are twelve edges (or twelve 1-dimensional boundaries) on the unit cube:

(ν,μ,z)(0≤z≤1,ν,μ=0or1),(x,ν,μ)(0≤x≤1,ν,μ=0or1),(ν,y,μ)(0≤y≤1,ν,μ=0or1).

The corresponding twelve univariate boundary functions on [0, 1] are

f(ν,μ,z),f(x,ν,μ),f(ν,y,μ)(ν,μ=0or1).

The method of reconstructing a triple–variate smooth function f(x,y,z) on [0, 1]3 is as follows.

We construct g3(x,y,z) which is a simple combination of six bivariate boundary functions and six factors x, y, z 1−x, 1−y, 1−z such that g3(x,y,z) = f(x,y,z) on the boundary ∂ ([0, 1]3). Let

h3(x,y,z)=f(x,y,z)−g3(x,y,z).

Expand h3(x,y,z) into Fourier series:

h3(x,y,z)=∑n1,n2,n3cn1n2n3(h3)e2πi(n1x+n2y+n3z),

where ∑n1,n2,n3=∑n1=−∞∞∑n2=−∞∞∑n3=−∞∞ and cn1n2n3(h3) are Fourier coefficients of h3, and then use its hyperbolic truncation:

sN(hc)(h3;x,y,z)=∑|n1|,|n2|,|n3|<N|n1n2n3|<Ncn1n2n3(h3)e2πi(n1x+n2y+n3z)

to reconstruct h3(x, y, z) with small square errors. From this, the problem of reconstruction of the triple–variate function f is reduced to that of each bivariate function on [0, 1]2 which is given in (1).

Without loss of generalization, we reconstruct the bivariate function f(x,y,0) on [0, 1]2. First, we construct g2(x,y) which is a combination of four univariate boundary functions on [0, 1]:

f(x,0,0),f(x,1,0),f(0,y,0),f(1,y,0)

and four factors x, y, 1−x, 1−y such that g2(x,y) is equal to f(x,y,0) on the boundary ∂ ([0, 1]2). Secondly, we expand h2(x,y) = f(x,y,0)−g2(x,y) into bivariate Fourier series. Finally, we can reconstruct h2(x,y) using the corresponding hyperbolic cross truncations of the Fourier series of h2(x,y). So the problem is reduced to that of reconstruction of univariate functions on [0, 1]

f(x,0,0),f(x,1,0),f(0,y,0),f(1,y,0),

using f(0,0,0), f(0,1,0), f(1,0,0), f(1,1,0) and few univariate Fourier coefficients.

The triple–variate function f(x,y,z) is reconstructed eventually by the values of this triple–variate function at vertexes of the cube [0, 1]3 and few univariate, bivariate & triple–variate Fourier coefficients.

In general, the reconstruction of a d–variate function on [0, 1]d can be reduced to that of (d−1)–variate functions on [0, 1]d−1, up to the reconstruction of univariate functions on [0, 1]. If ∂2df∂x12⋯∂xd2 is continuous on [0, 1]d, then we will prove that the square errors e~N(d) of this reconstruction scheme satisfy

e~N(d)2∼log4d−4⁡N~dN~d3,

where N͠d is the number of the used Fourier coefficients and N͠d∼ N logd−1N. Here An∼ Bn means that there are two constants C > 0 and D > 0 such that CBn ≤ An ≤ DBn for all n.

On the other hand, if we use directly the partial sums of Fourier series of f to reconstruct the function f, then the square errors e~N(d) satisfy (e~N(d))2∼(N~d)−1d, and if we directly reconstruct f by hyperbolic cross truncations of the Fourier series of f, then the square errors e~N(d) satisfy (e~N(d))2∼log2⁡N~dN~d. Thus, our recursive method can reconstruct multivariate functions on the unit cube with much smaller error than traditional Fourier methods.

2 Univariate functions on [0, 1]

Let f(x) be a function on [0, 1] and f″(x) be continuous on [0, 1] and φ(x) = f(x)−f(0)(1−x)−f(1)x. Expand φ(x) into Fourier series φ(x) = ∑ncn(φ) e2πinx, where Fourier coefficients cn(φ) satisfy

cn(φ)=∫01φ(x)e−2πinxdx=12πin∫01φ′(x)e−2πinxdx=O1n2.

Here An = O(Bn) means that there is a constant C > 0 such that |An| < C|Bn| for all n.

Take the partial sum of the Fourier series of φ [7, 8, 9, 10]:

sN(φ;x)=∑|n|≤Ncn(φ)e2πinx.

Then the square approximation errors eN(1)=∫01|φ(x)−sN(φ;x)|2dx12 satisfy

eN(1)2=∑|n|>N|cn(φ)|2=O∑|n|>N1n4=O1N3.

From this, we see that f(x) is reconstructed by f(0), f(1), and few Fourier coefficients {cn(φ)}|n| ≤ N as follows.

Step 1. Take φ(x) = f(x)−f(0)(1−x)−f(1) x (0 ≤ x ≤ 1).

Step 2. Find the Fourier coefficients of φ:cn(φ)=∫01φ(x)e−2πinxdx.

Step 3. Reconstruct f by the formula: f(x)≈f(0)(1−x)+f(1)x+∑n=−NNcn(φ)e2πinx(0≤x≤1) with square errors eN(1) satisfying (eN(1))2=O1N3.

3 Bivariate functions on [0, 1]2

Let f(x,y) be a function on [0, 1]2 and ∂i+jf∂xi∂yj (i,j = 0,1,2) be a continuous function on [0, 1]2. We reconstructed four boundary functions f(x,0), f(x,1), f(0,y), f(1,y) using the method shown in Section 2, and then reconstructed f by four boundary function and few bivariate Fourier coefficients as follows.

Define

g(x,y)=f(x,0)(1−y)+f(x,1)y+f(0,y)(1−x)+f(1,y)x−f(0,0)(1−x)(1−y)−f(0,1)(1−x)y−f(1,0)x(1−y)−f(1,1)xy (2)

and h(x,y) = f(x,y)−g(x,y). It is easy to check that

h(x,y)=0(x,y)∈∂([0,1]2). (3)

To reconstruct f, now we only need to reconstruct h. Expand h(x,y) into Fourier series:

h(x,y)=∑m,ncmn(h)e2πi(mx+ny),

where Fourier coefficients

cmn(h)=∫[0,1]2h(x,y)e−2πi(mx+ny)dxdy.

By (3),

cmn(h)=1(2πm)2∫01∂h∂x(1,y)−∂h∂x(0,y)e−2πinydy−1(2πm)2∫[0,1]2∂2∂x2(x,y)e−2πi(mx+ny)dxdy.

Again, by (3), ∂h∂x(1,0)=∂h∂x(1,1)=∂h∂x(0,1)=∂h∂x(0,0)=0 and ∂2h∂x2(x,0)=∂2h∂x2(x,1)=0. So

∫01∂h∂x(1,y)−∂h∂x(0,y)e−2πinydy=O1n2,∫01∂2h∂x2(x,y)e−2πinydy=O1n2.

Finally, we have

cmn(h)=O1m2n2(m≠0,n≠0). (4)

For m = 0 or n = 0,

cm0(h)=O1m2,c0n(h)=O1n2. (5)

We take Nth hyperbolic cross truncation of the Fourier series [11, 12]:

sN(hc)(h;x,y)=∑|m|,|n|≤N|mn|≤Ncmn(h)e2πi(mx+ny). (6)

By the Parseval identity and estimates (4), (5) of Fourier coefficients, we deduce that the square errors eN(2) satisfy

llllleN(2)2=∫[0,1]2h(x,y)−sN(hc)(h;x,y)2dxdy=∑|n|>N|cm0(h)|2+∑|n|>N|c0n(h)|2+∑|n|≥N∑m≠0|cmn(h)|2+∑0<|n|<N∑|m|≥N|n||cmn(h)|2=O1N3+O(1)∑0<|n|<N1n4∑|m|≥N|n|1m4=Olog⁡NN3.

Hence f(x,y) is reconstructed by four boundary functions f(x,0), f(x,1), f(0,y), f(1,y) and Fourier coefficients cmn(h) (|mn| ≤ N, |m|,|n| ≤ N) as follows.

Step 1. Reconstruct four boundary functions f(x,0), f(x,1), f(0,y), f(1,y) using the scheme proposal in Section 3.

Step 2. Construct a function g(x,y) which is stated in (2) based on these four boundary functions.

Step 3. Expand h(x,y) = f(x,y)−g(x,y) into Fourier series:

h(x,y)=∑mncmn(h)e2πi(mx+ny)((x,y)∈[0,1]2).

Step 4. Take the hyperbolic cross truncations sN(hc) (h;x,y) which are stated in (6).

Step 5. Reconstruct f by the formula: f(x,y)≈ g(x,y)+ sN(hc) (x,y) ((x,y) ∈ [0, 1]2) with the square errors eN(2) satisfying (eN(2))2=O(log⁡NN3).

4 Triple–variate function on [0, 1]3

Let f(x,y,z) be a function on [0, 1]3 and ∂i+j+kf∂xi∂yj∂zk (i,j,k = 0,1,2) be a continuous function on [0, 1]3. We first reconstruct six boundary functions f(ν, y, z), f(x,ν,z), f(x,y,ν) (ν = 0,1) using the method shown in in Section 3. Secondly, we construct a combination g(x,y,z) of six boundary functions and six factors x, y, z, 1−x, 1−y, 1−z such that g(x,y,z) = f(x,y,z) on the boundary ∂ ([0, 1]3) as follows.

  1. Take the sum of linear interpolation of f between the boundary functions

    g1(x,y,z)=(f(x,y,0)(1−z)+f(x,y,1)z)+(f(x,0,z)(1−y)+f(x,1,z)y)+(f(0,y,z)(1−x)+f(1,y,z)x)=:g11+g12+g13. (7)

  2. Take the sum of bivariate interpolation of f at boundary points:

    g2(x,y,z)=(f(x,0,0)(1−y)(1−z)+f(x,0,1)(1−y)z+f(x,1,0)y(1−z)+f(x,1,1)yz)+(f(0,y,0)(1−x)(1−z)+f(0,y,1)(1−x)z+f(1,y,0)x(1−z)+f(1,y,1)xz)+(f(0,0,z)(1−x)(1−y)+f(0,1,z)(1−x)z+f(1,0,z)x(1−y)+f(1,1,z)xy)=:g21+g22+g23 (8)

  3. Take triple–variate interpolation of f at vertexes of [0, 1]3:

    g3(x,y,z)=f(0,0,0)(1−x)(1−y)(1−z)+f(1,0,0)x(1−y)(1−z)+f(0,1,0)(1−x)y(1−z)+f(0,0,1)(1−x)(1−y)z+f(1,1,0)xy(1−z)+f(1,0,1)x(1−y)z+f(0,1,1)(1−x)yz+f(1,1,1)xyz. (9)

In detail, each term in g1, such as g12, is the linear interpolation of F1(t) = f(x,t,z) (0 ≤ t ≤ 1) at the each-point t = 0 or t = 1; each term in g2, such as g21, is the bivariate interpolation polynomial of F2(t,s) = f(x,t,s) (0 ≤ s,t ≤ 1) at the vertexes of [0, 1]2. Define

g(x,y,z)=g1(x,y,z)−g2(x,y,z)+g3(x,y,z). (10)

Then g is a combination of the boundary values of f and six factors x, (1−x), y, (1−y), z, (1−z).

Theorem 4.1

Let g(x,y,z) be constructed as above. Then g(x,y,z) and f(x,y,z) have same boundary values, i.e., g(x,y,z) = f(x,y,z) on ∂ ([0, 1]3).

Proof

By similarity, we only prove g(0,y,z) = f(0,y,z) (0 ≤ y,z ≤ 1).

Note that

g1(0,y,z)=f(0,y,0)(1−z)+f(0,y,1)z+f(0,0,z)(1−y)+f(0,y,z)−f(0,1,z)y,g2(0,y,z)=f(0,0,0)(1−y)(1−z)+f(0,0,1)(1−y)z+f(0,1,0)y(1−z)+f(0,1,1)yz+f(0,0,z)(1−y)+f(0,1,z)y+f(0,y,0)(1−z)+f(0,y,1)z,g3(0,y,z)=f(0,0,0)(1−y)(1−z)+f(0,1,0)y(1−z)+f(0,0,1)(1−y)z+f(0,1,1)yz.

Then g(0,y,z) = f(0,y,z). □

Corollary 4.2

Let h(x,y,z) = f(x,y,z)−g(x,y,z). Then h(x,y,z) satisfies:

  1. h(x,y,z) = 0 on ∂ ([0, 1]3);

  2. ∂h∂x(x,y,z)=∂h∂y(x,y,z)=∂h∂z(x,y,z)=0 at the vertexes {0,1}3.

Proof

By Theorem 4.1, we obtain immediately (i).

For vertexes {0,1}3, by similarity, we only prove ∂h∂x (1,0,1) = 0.

Note that (x,y,1) ∈ ∂ ([0, 1]3) for (x,y) ∈ [0, 1]2. By (i), h(x,y,1) = 0 for (x,y) ∈ [0, 1]2. So ∂h∂x (x,y,1) = 0 ((x,y) ∈ [0, 1]2). Especially, ∂h∂x (1,0,1) = 0. □

Now we only need to reconstruct h(x,y,z) by few Fourier coefficients.

For Fourier coefficients:

cn1n2n3(h)=∫[0,1]3h(x,y,z)e−2πi(n1x+n2y+n3z)dxdydz, (11)

we first consider the inner integral. By Corollary 4.2,

∫[0,1]2h(x,y,z)e−2πi(n1x+n2y+n3z)dxdy=1(2πn1)2∫01∂h∂x(1,y,z)−∂h∂x(0,y,z)e−2πin2ydy−1(2πn1)2∫[0,1]2∂2h∂x2(x,y,z)e−2πi(n1x+n2y)dxdy=:J1(z)+J2(z) (12)

Since h(x,1,z) = h(x,0,z) = 0 (0 ≤ x,z ≤ 1),

∂h∂x(x,1,z)=∂h∂x(x,0,z)=0(0≤x,z≤1),

and so

J1(z)=−i(2πn1)2(2πn2)∫01∂2h∂x∂y(1,y,z)−∂2h∂x∂y(0,y,z)e−2πin2ydy=1(2πn1)2(2πn2)2∂2h∂x∂y(1,y,z)−∂2h∂x∂y(0,y,z)−1(2πn1)2(2πn2)2∫01∂3h∂x∂y2(1,y,z)−∂3h∂x∂y2(0,y,z)e−2πin2ydy (13)

By ∂2h∂x2(x,ν,z)=0(ν=0,1),

J2(z)=−1(2πn1)2(2πin2)∫[0,1]2∂3h∂x2∂y(x,y,z)e−2πi(n1x+n2y)dxdy=−1(2πn1)2(2πn2)2∫01∂3h∂x2∂y(x,1,z)−∂3h∂x2∂y(x,0,z)e−2πin1xdx+1(2πn1)2(2πn2)2∫[0,1]2∂4h∂x2∂y2(x,y,z)e−2πi(n1x+n2y)dxdy.

Noticing that h(x,y,0) = h(x,y,1) = 0 (0 ≤ x,y ≤ 1), we get

J1(0)=J1(1)=J2(0)=J2(1)=0.

From this and (11)-(13), it follows that

cn1n2n3=∫01J1(z)e−2πin3zdz+∫01J2(z)e−2πin3zdz=O1n12n22n32.

Theorem 4.3

  1. cn1n2n3(h)=O1n12n22n32(n1≠0,n2≠0,n3≠0).

  2. If one of these three numbers n1,n2,n3 vanishes, for example, n2 = 0, then

    cn1,0,n3(h)=O1n12n32(n2≠0,n3≠0).

  3. If two numbers in these three numbers n1,n2,n3 vanish, for example, n1 = n2 = 0, then

    c0,0,n3(h)=O1n32.

To reconstruct h(x,y,z), take the hyperbolic cross truncation of Fourier series of h:

sN(hc)(h;x,y,z)=∑|n1|,|n2|,|n3|<N|n1n2n3|<Ncn1n2n3(h)e−2πi(n1x+n2y+n3z). (14)

By the Parseval identity and Theorem 4.3, we deduce that the square errors eN(3) satisfy

eN(3)2=∫[0,1]3h(x,y,z)−sN(hc)(h;x,y,z)2dxdydz=O(1)∑0<|n1|<N1n14∑1<|n3|<N1n34∑|n2|>N|n1n3|1n24+O1N3=Olog2⁡NN3

The number Nd of Fourier coefficients in sN(hc) (h) is equivalent to Nlog2 N. So we get the following theorem.

Theorem 4.4

Let h be stated as above. Then

eN(3)2=∫[0,1]3h(x,y,z)−sN(hc)(x,y,z)2dxdydz=Olog8⁡NdNd3.

For a function f on [0,1]l, the number of Fourier coefficients in Nth hyperbolic cross truncation of its Fourier series is denoted by Nl. To construct f on [0, 1]3, we need one Nth hyperbolic cross truncation of the triple–variate Fourier series, six Nth hyperbolic cross truncations of the bivariate Fourier series, and twelve partial sums of the univariate Fourier series. Hence, the total number of Fourier coefficients are

N~3=N3+6N2+12N1.

Summarizing up the preceding results, we can reconstruct the function f(x,y,z) as follows.

Step 1. Reconstruct six boundary functions:

f(0,y,z),f(1,y,z),f(x,0,z),f(x,1,z),f(x,y,0),f(x,y,1)

using the scheme of reconstructions of bivariate functions on [0, 1]2 proposal in Section 3.

Step 2. Construct a function g(x,y,z) which is stated in (7)-(10) based on these six boundary functions.

Step 3. Expand h(x,y,z) = f(x,y,z)−g(x,y,z) into a triple-variate Fourier series:

h(x,y,z)=∑n1,n2,n3cn1n2n3(h)e2πi(n1x+n2y+n3z).

Step 4. Take Nth hyperbolic cross truncations sN(hc) (h;x,y,z) which is stated in (14).

Step 5. Reconstruct f by the formula f(x,y,z)≈ g(x,y,z)+ sN(hc) (h;x,y,z) with the square error eN(3) satisfying (eN(3))2=O(log2⁡NN3). The total number of Fourier coefficients are equivalent to Nlog2 N in the reconstruction scheme.

5 The d–variate functions on [0, 1]d

The problem of high-dimension approximation has attracted a wide attention [13,14,15, 16, 17, 18]. Here we will present a novel recursive method to approximate d-variate functions. Let f(x1,…,xd) be a function on [0, 1]d and ∂i1+⋯+idf∂x1i1⋯∂xdid∈C([0,1]d)(i1,...,id=0,1,2). The 2d boundary functions are the following:

f(ν,x2,...,xd),f(x1,ν,...,xd),...,f(x1,...,xd−1,ν)(ν=0,1) (15)

which are (d−1)–variate functions on [0, 1]d−1. We construct these boundary functions by the method reconstructing d−1 variate functions, and then construct a combination g(x1,…,xd) of 2d boundary functions and 2d factors x1,x2,…,xd, 1−x1, 1−x2,…,1−xd such that g(x1,…,xd) = f(x1,…,xd) on ∂ ([0, 1]d).

Let h(x1,…,xd) = f(x1,…,xd)−g(x1,…,xd). Expand h(x1,…,xd) into the d–variate Fourier series:

h(x1,...,xd)=∑n1,n2,...,ndcn1n2⋯nde2πi(n1x1+⋯ndxd),

where cn1n2⋯nd=O(1n12n22⋯nd2). Take the hyperbolic cross truncations of the Fourier series:

sN(hc)(h;x1,...,xd)=∑|n1|,|n2|,...,|nd|<N|n1⋯nd|<Ncn1n2⋯nd(h)e2πi(n1x1+⋯+ndxd).

By the Parseval identity and Fourier coefficients estimate, we deduce that the square error eN(d) satisfies

eN(d)2=∫[0,1]dh(x1,...,xd)−sN(hc)(h;x1,...,xd)2dx1⋯dxd=Ologd−1⁡NN3.

The number of Fourier coefficients in sN(hc) is

Nd=∑|n1|,...,|nd|<N|n1⋯nd|<N1∼∫1Ndx1∫1Nx1dx2⋯∫1Nx1⋯xddxd∼Nlogd−1⁡N.

So

eN(d)2=Olog4d−4⁡NdNd3.

Note that the reconstruction of each boundary function in (5.1) needs Nlogd−2 N Fourier coefficients of one function on [0, 1]d−1 and 2d−2 functions on [0, 1]d−2 with the error eN(d−1) satisfying (eN(d−1))2=Ologd−2⁡NN3.

Continuing this procedure, we deduce that the reconstruction of functions on [0, 1]d needs N͠d Fourier coefficients and the values of f(x1,…,xd) at vertexes on [0, 1]d, where

Adlogd−1⁡N≤N~d≤Bdlogd−1⁡N,

where Ad and Bd are constants depending on d. The total square errors e~N(d) satisfy

e~N(d)2=Ologd−1⁡NN3=Olog4d−4⁡N~dN~d3.

Our method can not only approximate smooth multivariate functions on the cube well, but also approximate piecewise-smooth multivariate functions with on a general region. Under that case, based on the smoothness of functions, we divide the general region into the union of some cubes with difference size which contain smooth parts and a very small region which contains non-smooth parts. Because of piecewise smoothness, we can make the size of the region containing non-smooth parts as small as possible. Finally, we will apply our novel recursive method on each cube.

6 Numerical results

In this section, we will give some numerical results in order to compare our method with traditional Fourier methods. Due to the similarity, we only discuss one-dimensional and two-dimensional cases.

First we consider a univariate function f(x) = ex (x ∈ [0, 1]). Using the method in Section 2, we will construct a new function φ(x) such that

φ(x)=f(x)−f(0)(1−x)−f(1)x=ex+(1−e)x−1.

Then we will expend φ(x) into Fourier series and numerically compute the Fourier coefficients by using discrete Fourier transform. After that, we will reconstruct f(x) by using f(0), f(1) and the Fourier coefficients of φ(x). The approximation error will be measured as

∫01(f(x)−frecon(x))2dx∫01f2(x)dx

Figure 1 shows approximation errors by using traditional Fourier method and our method. It is clear that our method can reconstruct f(x) with much smaller error than traditional Fourier methods.

Figure 1 Approximation of f(x) = ex by traditional Fourier method and our method
Figure 1

Approximation of f(x) = ex by traditional Fourier method and our method

Next we consider a bivariate function f(x,y) = ex+y on [0, 1]2. Using the method in Section 3, we will construct g(x,y):

g(x,y)=ex(1−y)+ex+1y+ey(1−x)+ey+1x−(1−x)(1−y)−e(1−x)y−ex(1−y)−e2xy.

and h(x,y) = f(x,y)−g(x,y). Since g(x,y) is completely determined by four boundary functions f(x,0), f(x,1), f(0,y) and f(1,y). In order to reconstruct g(x,y) well, with the help of our method in Section 2, we need to construct the following function:

φ1(x)=f(x,0)−f(0,0)(1−x)−f(1,0)x=ex−(1−x)−exφ2(x)=f(x,1)−f(0,1)(1−x)−f(1,1)x=ex+1−e(1−x)−e2xφ3(y)=f(0,y)−f(0,0)(1−y)−f(0,1)y=ey−(1−y)−eyφ4(y)=f(1,y)−f(1,0)(1−y)−f(1,1)y=e1+y−e(1−y)−e2y

Then we will expand h(x,y), φ1(x), φ2(x), φ3(y), and φ4(y) into Fourier series. Finally, we will reconstruct f(x,y) by f(0,0), f(0,1), f(1,0), f(1,1), partial sums of Fourier series of four univariate functions φ1(x), φ2(x), φ3(y) and φ4(y), as well as the hyperbolic cross truncations of bivariate Fourier series of h(x,y). Figure 2 shows approximation errors by using traditional Fourier method based on hyperbolic cross truncations and our method. It is clear that our method can reconstruct f(x,y) with much smaller error than traditional Fourier method.

Figure 2 Approximation of f(x,y) = ex+y by traditional Fourier method and our method
Figure 2

Approximation of f(x,y) = ex+y by traditional Fourier method and our method

Acknowledgement

The author is grateful to the reviewers for their valuable comments. This research is partially supported by National Key Science Programme for Global Change Research 2015CB953602, Beijing Higher Education Young Elite Teacher Project, and Scientific Research Foundation for the Returned Overseas Chinese Scholars, State Education Ministry.

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Received: 2016-11-3
Accepted: 2017-11-16
Published Online: 2017-12-29

© 2017 Zhang

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 License.

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Keywords for this article
Recursive method; Hyperbolic cross truncation; Multivariate functions; Fourier series
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