129 256 513 Eone n 1.6 10-4 2.9 10-5 4.1 10-6 four.six 10-7 four.0 10-8 four.3 10-9 3.3 10-10 three.eight 10-11 condone

129 256 513 Eone n 1.six 10-4 two.9 10-5 four.1 10-6 4.6 10-7 4.0 10-8 four.3 10-9 3.3 10-10 three.eight 10-11 Methyl jasmonate supplier condone 1.99 2.07 two.13 2.13 2.13 two.14 two.14 2.14 Size m.l.s. Emix n eight.7 10-5 1.7 10-7 1.eight 10-9 1.five 10-11 condmix two.60 three.03 three.ten 3.one mix(four, 5) (16, 17) (64, 65) (256, 257)Table 3. Instance 2: Ordinary and Mixed Collocation approaches [4]. Size o.l.s. 5 9 17 33 65 129 257 513 Enonecondone 1.99 2.07 two.13 two.13 two.13 2.14 two.14 two.Size m.l.s.Enmixcondmix 1.31 1.45 1.52 1.1.1 10-2 3.0 10-3 7.two 10-4 1.5 10-4 3.0 10-5 5.7 10-6 9.9 10-7 1.7 10-(5, 4) (17, 16) (65, 64) (257, 256)1.two 10-2 1.5 10-3 2.2 10-5 eight.8 10-Example three. Let us think about the following equation: f (y) – 1-f ( x ) cos(250x ) u = v0.1,0.1 ,1 – x2 dx = (1 – y) two cos yw = = v0.5,0.In this test, the kernel k ( x, y) = cos(250x ) presents a quick oscillating behaviour. Its graphic is reported in Figure 2. Therefore, the solution formula makes it possible for us to overcome the drawbacks deriving in the use of your Gauss acobi rule. Regarding the price of convergence, due to the fact g W5 (u), we Scaffold Library supplier expect that the errors are O m-5 . We’ve reported the values attained by ONM and MNM in three unique points of your interval (-1, 1) (Table four) along with the maximum errors on the whole interval (Table 5). In all the circumstances the theoretical estimates are attained. Additionally, in both procedures the situation numbers of the linear systems are comparable.Mathematics 2021, 9,12 ofFigure 2. Instance three: graphic of k( x ) = cos(250x ). Table four. Example three: numerical values in the weighted answer attained by ONM and MNM. x = -0.eight m four 9 16 33 64 129 256 513 m four 9 16 33 64 129 256( fn u)( x )two.733837532248313 two.733857147494405 2.733857151078503 two.733857151493599 two.733857151490818 two.733857151490921 2.733857151490914 2.733857151490910 x=( f u)( x )two.733837532248313 two.733857116533352 2.733857151078503 2.733857151494889 2.733857151490818 2.733857151490937 two.733857151490914 2.( fn u)( x )9.973699234411486 10-1 9.973916652404244 10-1 9.973916692130872 10-1 9.973916696731837 10-1 9.973916696702131 10-1 9.973916696702052 10-1 9.973916696702078 10-1 9.973916696702031 10-1 x = 0.( f u)( x )9.973699234411486 10-1 9.973916609227826 10-1 9.973916692130872 10-1 9.973916696740153 10-1 9.973916696702131 10-1 9.973916696702041 10-1 9.973916696702078 10-1 9.973916696702035 10-m 4 9 16 33 64 129 256( fn u)( x )1.502019304836581 10-1 1.502230544539280 10-1 1.502230583137010 10-1 1.502230587607230 10-1 1.502230587578369 10-1 1.502230587578292 10-1 1.502230587578317 10-1 1.502230587578272 10-( f u)( x )1.502019304836581 10-1 1.502230511114784 10-1 1.502230583137010 10-1 1.502230587564985 10-1 1.502230587578369 10-1 1.502230587578573 10-1 1.502230587578317 10-1 1.502230587578276 10-Mathematics 2021, 9,13 ofTable 5. Example three. Size o.l.s. four 9 16 33 64 129 256 513 Eone n 2.2 10-5 four.4 10-9 4.six 10-10 3.0 10-12 9.8 10-14 1.8 10-14 4.4 10-15 4.five 10-16 condone 1.00 1.01 1.02 1.02 1.03 1.04 1.35 1.36 Size m.l.s. Emix n three.9 10-8 two.1 10-12 three.0 10-14 eps condmix 1.00 1.01 1.02 1.(four, five) (16, 17) (64, 65) (256, 257)Example 4. Let us think about the following equation: 1 f (y) –(1 – x2 )3/10 1 f (x) 2 dx = y – two ( x + 5-2 )5/w = = v0.three,0.9u = v0.two,0.2 ,In this case, and u satisfy the assumptions for the convergence of the Nystr strategies, while they do not satisfy those in the collocation methods [4]. We recall that for the convergence of both the collocation procedures smoother kernels and much more restrictive assumptions on the weights are necessary. Regarding the rate of convergence, considering that g W4 (u),.

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