We obtain a numerical result that the Radau rules are better in accuracy than the Gauss–Legendre (GL) formula for the integral of even functions on the interval <InlineEquation ID="IEq1"> <EquationSource Format="TEX">$[-1,1]$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mo stretchy="false">[</mo> <mo>-</mo> <mn>1</mn> <mo>,</mo> <mn>1</mn> <mo stretchy="false">]</mo> </mrow> </math></EquationSource> </InlineEquation>. There exist two types of the Radau rules having a node at either 1 or <InlineEquation ID="IEq2"> <EquationSource Format="TEX">$-1$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </math></EquationSource> </InlineEquation>, as well as interior nodes. The two Radau rules are the same for even functions, so the numerical result suggests that their averaged rule (a-R) is superior to the GL rule, regardless of odd or even. We investigate this phenomenon for the integral with the Jacobi weight function <InlineEquation ID="IEq3"> <EquationSource Format="TEX">$(1-x)^\alpha (1+x)^\beta $</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msup> <mrow> <mo stretchy="false">(</mo> <mn>1</mn> <mo>-</mo> <mi>x</mi> <mo stretchy="false">)</mo> </mrow> <mi>α</mi> </msup> <msup> <mrow> <mo stretchy="false">(</mo> <mn>1</mn> <mo>+</mo> <mi>x</mi> <mo stretchy="false">)</mo> </mrow> <mi>β</mi> </msup> </mrow> </math></EquationSource> </InlineEquation> (<InlineEquation ID="IEq4"> <EquationSource Format="TEX">$\alpha >-1$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>α</mi> <mo>></mo> <mo>-</mo> <mn>1</mn> </mrow> </math></EquationSource> </InlineEquation>, <InlineEquation ID="IEq5"> <EquationSource Format="TEX">$\beta >-1$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>β</mi> <mo>></mo> <mo>-</mo> <mn>1</mn> </mrow> </math></EquationSource> </InlineEquation>). The a-R with the ratio <InlineEquation ID="IEq6"> <EquationSource Format="TEX">$(n+\alpha ):(n+\beta )$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mo stretchy="false">(</mo> <mi>n</mi> <mo>+</mo> <mi>α</mi> <mo stretchy="false">)</mo> <mo>:</mo> <mo stretchy="false">(</mo> <mi>n</mi> <mo>+</mo> <mi>β</mi> <mo stretchy="false">)</mo> </mrow> </math></EquationSource> </InlineEquation> of the n-point Radau rules with degree of exactness <InlineEquation ID="IEq7"> <EquationSource Format="TEX">$2n-2$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mn>2</mn> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </math></EquationSource> </InlineEquation> is of degree <InlineEquation ID="IEq8"> <EquationSource Format="TEX">$2n-1$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mn>2</mn> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </math></EquationSource> </InlineEquation>. The degree rises to 2n when <InlineEquation ID="IEq9"> <EquationSource Format="TEX">$\alpha +\beta =-1$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>α</mi> <mo>+</mo> <mi>β</mi> <mo>=</mo> <mo>-</mo> <mn>1</mn> </mrow> </math></EquationSource> </InlineEquation>, and to <InlineEquation ID="IEq10"> <EquationSource Format="TEX">$4n-3$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mn>4</mn> <mi>n</mi> <mo>-</mo> <mn>3</mn> </mrow> </math></EquationSource> </InlineEquation> when <InlineEquation ID="IEq11"> <EquationSource Format="TEX">$\alpha =\beta =-\tfrac{1}{2}$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>α</mi> <mo>=</mo> <mi>β</mi> <mo>=</mo> <mo>-</mo> <mstyle displaystyle="false" scriptlevel="0"> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> </mstyle> </mrow> </math></EquationSource> </InlineEquation>. This indicates that the two Radau rules have errors of opposite sign, so each rule is the anti-rule to the other one. When <InlineEquation ID="IEq12"> <EquationSource Format="TEX">$\alpha =\beta =0$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>α</mi> <mo>=</mo> <mi>β</mi> <mo>=</mo> <mn>0</mn> </mrow> </math></EquationSource> </InlineEquation> (the Legendre weight), the a-R has the same degree of exactness as that of the n-point GL rule, the dominant error term of the a-R being smaller in magnitude than that of the GL rule. The propositions above are also numerically verified. Further numerical results show that the a-R can be effectively used to estimate the errors of the Radau rules.

On two Gauss–Radau quadrature formulae and their weighted averaged-rule: Is Gauss–Legendre always better than Radau?

期刊论文
发表日期：

2025年5月15日

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Hiroshi Sugiura,
Takemitsu Hasegawa

摘要

We obtain a numerical result that the Radau rules are better in accuracy than the Gauss–Legendre (GL) formula for the integral of even functions on the interval $[-1,1]$ $[- 1, 1]$ . There exist two types of the Radau rules having a node at either 1 or $-1$ $- 1$ , as well as interior nodes. The two Radau rules are the same for even functions, so the numerical result suggests that their averaged rule (a-R) is superior to the GL rule, regardless of odd or even. We investigate this phenomenon for the integral with the Jacobi weight function $(1-x)^\alpha (1+x)^\beta $ ${(1 - x)}^{α} {(1 + x)}^{β}$ ( $\alpha >-1$ $α > - 1$ , $\beta >-1$ $β > - 1$ ). The a-R with the ratio $(n+\alpha ):(n+\beta )$ $(n + α) : (n + β)$ of the n-point Radau rules with degree of exactness $2n-2$ $2 n - 2$ is of degree $2n-1$ $2 n - 1$ . The degree rises to 2n when $\alpha +\beta =-1$ $α + β = - 1$ , and to $4n-3$ $4 n - 3$ when $\alpha =\beta =-\tfrac{1}{2}$ $α = β = - \frac{1}{2}$ . This indicates that the two Radau rules have errors of opposite sign, so each rule is the anti-rule to the other one. When $\alpha =\beta =0$ $α = β = 0$ (the Legendre weight), the a-R has the same degree of exactness as that of the n-point GL rule, the dominant error term of the a-R being smaller in magnitude than that of the GL rule. The propositions above are also numerically verified. Further numerical results show that the a-R can be effectively used to estimate the errors of the Radau rules.