The Best Approximation of the Sinc Function by a Polynomial of Degree n with the Square Norm

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Volume 2010, Article ID 307892,12pages doi:10.1155/2010/307892

Research Article

The Best Approximation of the Sinc Function by a Polynomial of Degree n with the Square Norm

Yuyang Qiu and Ling Zhu

College of Statistics and Mathematics, Zhejiang Gongshang University, Hangzhou 310018, China

Correspondence should be addressed to Yuyang Qiu,[email protected] Received 9 April 2010; Accepted 31 August 2010

Academic Editor: Wing-Sum Cheung

Copyrightq2010 Y. Qiu and L. Zhu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

The polynomial of degree n which is the best approximation of the sinc function on the interval0, π/2with the square norm is considered. By using Lagrange’s method of multipliers, we construct the polynomial explicitly. This method is also generalized to the continuous function on the closed intervala, b. Numerical examples are given to show the eﬀectiveness.

1. Introduction

Let sincx sinx/x be the sinc function; the following result is known as Jordan inequality1:

2

π ≤sincx<1,0< x≤ π

2, 1.1

where the left-handed equality holds if and only ifxπ/2. This inequality has been further refined by many scholars in the past few years2–30. ¨Ozban12presented a new lower bound for the sinc function and obtained the following inequality:

2 π 1

π³

π²−4x²

4π−3 π³

x−π

2 2

≤sin cx. 1.2

The above inequality was generalized to an upper bound by Zhu26:

sin cx≤ 2 π 1

π³

π²−4x²

12−π² π³

x− π

2 2

. 1.3

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Later, Agarwal and his collaborators2proposed a more refined two-sided inequality:

1−4

−6643π−7π²

π² x−4

124−83π14π²

π³ x²−412−4π π⁴ x³

≤sin cx≤1−4

−7549π−8π²

π² x4

−14295π−16π²

π³ x²−412−4π π⁴ x³,

1.4 where the two-sided equalities hold ifxtends to zero orxπ/2.

Note that the bounds of the sinc function sincx listed above are estimated by the given polynomials with the boundary constraints; the smaller the residual between the polynomial and the sinc function is, the more refined the estimation will be. Hence, our aim is to seek a polynomial of degreen,p_nx, which is the best approximation of the sinc function with the square norm. In view of that, the sinc function is defined on 0, π/2 and two boundary constrained conditions are imposed. So we want to solve the following minimum problem:

pnminx∈Pn

_π/2

0

sin cx−p_nx2

dx _1/2

s.t. lim

x→0p_nx lim

x→0sin cx, lim

x→π/2p_nx lim

x→π/2sincx,

1.5

wherePnis the set of the polynomial of degreenand it is denoted by Pn pn|pnx a0a1x· · ·anxⁿ, ai∈R, i1,2, . . . , n

1.6 In this paper, an explicit representation for the approximating polynomial of sincx is presented by using Lagrange’s method of multipliers, and numerical examples are given to show the eﬀectiveness. Moreover, this method can be generalized to the continuous function gxon the closed interval a, b. However, the residual error between the approximating polynomialpnxandgxis concussive, that is, it cannot keep positive or negative always.

The rest of paper is organized as follows. In Section2, we solve the problem5by Lagrange’s method of multipliers and this method is generalized to a continuous function on a, bin Section3. Numerical examples are given in Section4to display the eﬀectiveness of our estimations.

2. The Best Approximation of the Sinc Function by a Polynomial of Degree n on 0, π/2

Obviously, the constraints of1.5imply

a01, pn

π 2

2

π. 2.1

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Note that _π/2

0

sin cx−pnx₂ dx

_π/2

0

sin²cx 1−2sin cx−2 n

i1

aixⁱ⁻¹sinx2 n

i1

aixⁱ

2 ⁿ

1≤i<j≤n

a_ia_jx^ijⁿ

i1

a²_ix²ⁱ

⎞

⎠dx

_π/2

0

sin²cx 1−2sin cx−2 n

i1

aixⁱ⁻¹sinx

dx

ⁿ

i1

2a_i i1

π 2

_i1

1≤i<j≤n

2a_ia_j ij1

π 2

_ij1

ⁿ

i1

a²_i 2i1

π 2

_2i1 .

2.2

Denote

Ga1, . . . , an h

1≤i<j≤n

2aiaj

ij1 π

2 _ij1

ⁿ

i1

a²_i 2i1

π 2

_2i1

2.3

with

h _π/2

0

−2ⁿ

i1

a_ixⁱ⁻¹sinx

dxⁿ

i1

2a_i i1

π 2

_i1

, 2.4

wherea_i∈ R, i1,2, . . . , n. So1.5is equivalent to solving the following minimum problem:

minai∈RGa1, . . . , a_n s.t. a₁π

2 · · ·a_nπ 2

n

2 π −1.

2.5

This can be solved by using Lagrange’s method of multipliers. We construct the Lagrange function by

La1, a₂, . . . , a_n, λ Ga1, . . . , a_n λ

a₁π

2 · · ·a_nπ 2

n

− 2 π 1

2.6

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with Lagragian multiplierλ. Thus we need to equate to zero the partial derivatives ofLwith respect to eacha_jj1,2, . . . , nandλ, that is,

∂L

∂aj 0, j1, . . . , n, a₁π

2 · · ·a_nπ 2

n

− 2

π 10.

2.7

It gives a system of linear equations

Auf, 2.8

where

A

⎛

⎜⎜

⎝ 2 3

π 2

3 2 4

π 2

4

. . . 2 n2

π 2

_n2 π 2 2

4 π

2

4 2 5

π 2

5

. . . 2 n3

π 2

_n3 π 2

2

2 5

π 2

₅ 2 6

π 2

₆

. . . 2 n4

π 2

_n4 π 2

₃

... ... . . . ... ...

2 n2

π 2

_n2 2 n3

π 2

_n3

. . . 2 2n1

π 2

_2n1 π 2

n

π 2

₂

. . . π 2

_n

0

⎞

⎟⎟

⎠

, 2.9

u

⎛

⎜⎜

⎜⎝ a₁ a2

... a_n

λ

⎞

⎟⎟

⎟⎠

, f −

⎛

⎜⎜

⎝

∂h

∂a₁

∂h

∂a2

...

∂h

∂an

1− 2 π

⎞

⎟⎟

⎠

. 2.10

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To consider the consistence of the equations2.8, we introduce the following lemma for the square matrixAof ordern1.

Lemma 2.1. The square matrixAof ordern1 defined by2.9is nonsingular.

Proof. We want to prove that detA/0. Note that

detA π 2

_34···n21

det

⎛

⎜⎜

⎝ 2 3

2 4

π 2

1

. . . 2 n2

π 2

_n−1 π 2

₋₂

2 4

2 5

π 2

1

. . . 2 n3

π 2

_n−1 π 2

₋₂

2 5

2 6

π 2

1

. . . 2 n4

π 2

_n−1 π 2

₋₂ ... ... . . . ... ...

2 n2

2 n3

π 2

1

. . . 2 2n1

π 2

_n−1 π 2

₋₂

1 π

2 1

. . . π 2

_n−1

0

⎞

⎟⎟

⎠

π 2

n5n/2n−1n/2−1

det

⎛

⎜⎜

⎜⎝ 2 3

2

4 . . . 2 n2 1 2

4 2

5 . . . 2 n3 1 2

5 2

6 . . . 2 n4 1 ... ... . . . ... ... 2

n2 2

n3 . . . 2 2n1 1

1 1 . . . 1 0

⎞

⎟⎟

⎟⎠

π 2

_nn2−1

det

2H_n e e^T 0

,

2.11

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where

H_n

⎛

⎜⎜

⎝ 1 3

1

4 . . . 1 n2 1

4 1

5 . . . 1 n3 1

5 1

6 . . . 1 n4 ... ... . . . ... 1

n2 1

n3 . . . 1 2n1

⎞

⎟⎟

⎠

, e

⎛

⎜⎜

⎝ 1 1 1 ... 1

⎞

⎟⎟

⎠

. 2.12

SinceHn is onen-order principal square submatrix ofn2-order Hilbert matrix, together with Hilbert matrix being positive definite31, volume 1, page 401, thenH_nis also positive definite. Hence,H_n⁻¹exists and it is positive definite, which impliese^TH_n⁻¹e /0. Moreover,

det

2H_n e e^T 0

det

⎛

⎝2H_n e 0 −1

2e^TH_n⁻¹e

⎞

⎠. 2.13

So, det

2Hne e^T 0

/0, that is,Ais nonsingular.

BecauseAis nonsingular, the solution of the equations2.8exists and is unique, as well as the best approximation of sin cxby a polynomial of degreen. Therefore, we obtain the following theorem.

Theorem 2.2. Let 0< x≤ π/2; suppose the matrixAand vectorf are denoted by2.9. Then the best approximation of sin cx by a polynomial of degreenon interval0, π/2with the square norm is given by

pnx 1a1x· · ·a_n−1xⁿ⁻¹anxⁿ, 2.14 wherea1, . . . , anis the 1,2,. . . n-th components of the vectorA⁻¹f.

3. The Best Approximation of the Continuous Function gx by a Polynomial of Degree n on a, b

In this section, we generalize the above conclusion to the continuous functiongxon interval a, b, that is, we want to consider the following minimum problem:

pnminx∈Pn

b a

gx−pnx₂ dx

1/2

3.1

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with the constraints

p_na ga, p_nb gb, 3.2

where the polynomialp_nxis rewritten as

p_nx a₀a₁x−a · · ·a_nx−aⁿ 3.3 andPnis defined by1.6. If we settx−a, problem3.1is equivalent to

min

pnt∈Pn

_b−a

0

gta−pnt₂ dt

_1/2

3.4

with

a0ga, pnb−a gb, 3.5

where

p_nt a₀a₁t· · ·a_ntⁿ. 3.6 If we replacea₀ 1,π/2, sincx,p_nxin Section2bya₀ ga,b−a,gx, andp_nt, respectively, then2.4is rewritten as

h _b

a

−2ⁿ

i1

a_igxx−aⁱ

dxⁿ

i1

2a_iga

i1 b−aⁱ¹, 3.7

A

⎛

⎜⎜

⎜⎝

2b−a³ 3

2b−a⁴

4 . . . 2b−aⁿ²

n2 b−a 2b−a⁴

4

2b−a⁵

5 . . . 2b−aⁿ³

n3 b−a² 2b−a⁵

5

2b−a⁶

6 . . . 2b−aⁿ⁴

n4 b−a³ ... ... . . . ... ... 2b−aⁿ²

n2

2b−aⁿ³

n3 . . . 2b−a²ⁿ¹

2n1 b−aⁿ b−a b−a² . . . b−aⁿ 0

⎞

⎟⎟

⎟⎠

, f −

⎛

⎜⎜

⎝

∂h

∂a₁

∂h

∂a₂ ...

∂h

∂a_n ga−gb

⎞

⎟⎟

⎠ .

3.8 So we have the following theorem.

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Theorem 3.1. Letgxbe continuous ona, b, and we denote the matrixAandf by3.8. Then the best approximation ofgxby the polynomial of degreenona, bwith the square norm is given by

p_nx ga a₁x−a · · ·a_n−1x−aⁿ⁻¹a_nx−aⁿ, 3.9

wherea₁, . . . , a_nis the 1,2,. . . n-th components of the vectorA⁻¹f.

Remark 3.2. The intervala, bin Theorem3.1can be generalized toa, b, where

xlim→agx, lim

x→b⁻gx both exist. 3.10

4. Numerical Examples

In this section, we present some numerical examples to illustrate the eﬀectiveness of our methods based on Theorems2.2and3.1. For any functiongx, two kinds of errors are used as measures of accuracy. One is the residual error

_gx−p_n gx−pnx. 4.1

The other is the integration error

^int_gx−p

n

_b

a

gx−pnx₂ dx

_1/2

. 4.2

Example 4.1. Let a 0, b π/2, and gx sin cx; we compare the approximation eﬀectiveness between the approximating polynomial of degree 3 and sin cxby Theorem2.2 and that in2. Denote the left-handed polynomial in inequality1.4byp^l₃x, and the right- handed one byp^r₃x, that is,

p^l₃x 1−4

−6643π−7π²

π² x− 4

124−83π14π²

π³ x²−412−4π π⁴ x³, p^r₃x 1−4

−7549π−8π²

π² x 4

−14295π−16π²

π³ x²−48−16π π⁴ x³.

4.3

With Theorem2.2, it is easy to compute that

p₃x 1−2

13440−1440π−960π²−4π³7π⁴

π⁵ x

4

40320−4800π−2640π²−16π³13π⁴

π⁶ x²

−56

3840−480π−240π²−2π³π⁴

π⁷ x³.

4.4

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1.5 1

0.5

x

−0.0025

−0.002

−0.0015

−0.001

−0.0005 0 0.0005

Figure 1: The residual errors between the approximating polynomial of degree 3 and sincxwith the square norm, where we denote the yellow dotted line by_{sin cx−p}l

3, green dash line by_{sin cx−p}₃^r_x, and red line by_{sin cx−p}₃.

Table 1: The residual error sin cx−pn and integration error ^int_sin_cx−p_n between the approximating polynomial of degreenand sin cxwith the square norm on interval0, π/2, wheren2,3,4,5.

n maximal_{sin cx−p}_n minimal_{sin cx−p}_n _{sin cx−p}^int

n

2 4.12∗10⁻³ −4.73∗10⁻³ 3.97∗10⁻³

3 3.51∗10⁻⁴ −4.68∗10⁻⁴ 4.59∗10⁻⁴

4 3.28∗10⁻⁵ −2.16∗10⁻⁵ 5.08∗10⁻⁴

5 1.72∗10⁻⁶ −1.16∗10⁻⁶ 5.12∗10⁻⁴

We plot the residual error forp^l₃x,p^r₃x, andp3x, respectively. In Figure1, we will find that the total error of p3xis smaller than that of p^l₃xandp^r₃x. However, the curve of _sin_cx−p₃is concussive aty0.

Example 4.2. In this example, we consider the residual error _gx−p_n and integration error ^int_gx−p

n forn 2,3,4,5 withgx sincxand interval0, π/2. In Table1, we will find that the order of the residual errors_sin_cx−p_n will decrease with increasingn. However, the precision of integration error^int_sin_cx−p

ncan remain 10⁻⁴whenn3,4,5.

Example 4.3. In this example, let gx cosx and the interval be0, π; we consider its approximating polynomial of degree 3:p₃x. By Theorem3.1, we have

p3x 1−3

140π²3π⁴−1680

π⁵ x−21

60π²π⁴−720

π⁶ x²

14

60π²π⁴−720 π⁷ x³,

4.5

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3 2

1 x

−0.006

−0.004

−0.002 0 0.002 0.004 0.006

Figure 2: The residual errorcosx−p3xbetween cosxandp3xon0, π.

and the residual error cosx−p3 can be represented by Figure 2 . Obviously, the curve is concussive; however, the residual error can reach 10⁻³.

Example 4.4. Letgx sinxand the interval be π/2, π; we consider its approximating polynomial of degree 4p4xby Theorem3.1. It is easy to verify

p4x 1−23π⁵8400π³−127680π²−1532160π5806080 π⁶

x−π

2

14

11π⁵6000π³−110400π²−1209600π4700160 π⁷

x−π

2 2

− 56

7π⁵4560π³−95040π²−979200π3870720 π⁸

x−π 2

3

336

π⁵720π³−16320π²−161280π645120 π⁹

x−π

2 4

.

4.6

We plot the residual errorsinx−p4xin Figure3, where we find it can reach 10⁻⁴.

Acknowledgment

The work of the first author was supported in part by National Science Foundation for Young ScholarsGrant no. 60803076,61003186, the Natural Science Foundation of Zhejiang Province Grant no. Y6090211, and Foundation of Education Department of Zhejiang ProvinceGrant no. Y201017555, Y201017322.

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3 2.8 2.6 2.4 2.2 2 1.8

x

−0.0001

−0.00005 0 0.00005 0.0001 0.00015

Figure 3: The residual errorsinx−p4xbetween sinxandp4xonπ/2, π.

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