TABLES OF SOME INDEFINITE INTEGRALS OF BESSEL FUNCTIONS
|
|
- Ἑνώχ Γλυκύς
- 7 χρόνια πριν
- Προβολές:
Transcript
1 Werner Rosenheinrich Ernst - Abbe - Hochschule Jena First variant: University of Applied Sciences Germany TABLES OF SOME INDEFINITE INTEGRALS OF BESSEL FUNCTIONS Integrals of the type J 0 d or J 0 aj 0 b d are well-known Most of the following integrals are not found in the widely used tables of Gradstein/Ryshik, Bateman/Erdélyi, Abramowitz/ Stegun, Prudnikov/Brychkov/Marichev or Jahnke/Emde/Lösch The goal of this table was to get tables for practicians So the integrals should be epressed by Bessel and Struve functions Indeed, there occured some eceptions Generally, integrals of the type µ J ν d may be written with Lommel functions, see 8], 10-74, or ], III In many cases reccurence relations define more integrals in a simple way Partially the integrals may be found by MAPLE as well In some cases MAPLE gives results with hypergeometric functions, see also ], 96, or 4] Some known integrals are included for completeness Here Z ν denotes some Bessel function or modified Bessel function of the first or second kind Partially the functions Y ν sometimes called Neumann s functions or Weber s functions and denoted by N ν ] and the Hankel functions H ν 1 and H ν are also considered The same holds for the modified Bessel function of the second kind K ν When a formula is continued in the net line, then the last sign + or - is repeated in the beginning of the new line On page 456 the used special functions and defined functions are described *E* - This sign marks formulas, that were incorrect in previous editions The pages with corrected errors are listed in the errata in the end I wish to epress my thanks to B Eckstein, S O Zafra, Yao Sun, F Nouguier, M Carbonell and R Oliver for their remarks 1
2 References: 1] M Abramowitz, I Stegun: Handbook of Mathematical Functions, Dover Publications, NY, 1970 ] Y L Luke: Mathematical Functions and their Approimations, Academic Press, NY, 1975 ] Y L Luke: Integrals of Bessel Functions, McGraw-Hill, NY, 196 4] A P Prudnikov, A Bryqkov, O I Mariqev: Integraly i r dy, t : Specialьnye funkcii, Nauka, Moskva, 00; FIZMATLIT, 00 5] E Jahnke, F Emde, F Lösch: Tafeln höherer Funktionen, 6 Auflage, B G Teubner, Stuttgart, ] I S Gradstein, I M Ryshik: Summen-, Produkt- und Integraltafeln / Tables of Series, Products, and Integrals, Band 1 / Volume 1, Verlag Harri Deutsch, Thun Frankfurt/M, ] I S Gradstein, I M Ryshik: Summen-, Produkt- und Integraltafeln / Tables of Series, Products, and Integrals, Band / Volume, Verlag Harri Deutsch, Thun Frankfurt/M, ] G N Watson: A Treatise on the Theory of Bessel Functions, Cambridge, University Press, 19 / ] P Humbert: Bessel-integral functions, Proceedings of the Edinburgh Mathematical Society Series, 19, : ] B A Peavy, Indefinite Integrals Involving Bessel Functions JOURNAL OF RESEARCH of the National Bureau of Standards - B, vol 718, Nos and, April - September 1967, pp ] B G Korenev: Vvedenie v teori besselevyh funkci i, Nauka, Moskva, ] S K H Auluck: Some integral identities involving products of general solutions of Bessel s equation of integral order, arivorg/abs/ ] H Bateman, A Erdélyi: Tables of Integral Transforms, vol I, McGraw-Hill Book Company, Inc, New York, Toronto, London, ] H Bateman, A Erdélyi: Tables of Integral Transforms, vol II, McGraw-Hill Book Company, Inc, New York, Toronto, London, 1954
3 1 1 n Z ν with integer values of n Contents 1 Integrals with one Bessel function 111 n Z n+1 Z n 1 Z n Z n Z n+1 Z n Z ν, ν > Second Antiderivatives of n+1 ν Z ν 44 a n+1 Z 0 44 b n Z Higher Antiderivatives Some Integrals of the Type n+1 Z 1 + α/ + α 50 1 Elementary Function and Bessel Function 11 n+1/ Z ν 57 a Z ν d 57 b Integrals 6 c Recurrence { Formulas } 67 1 n e ± Iν 69 K ν a Integrals with e 69 b Integrals with e 71 { } { } sinh 1 n Iν 75 cosh K ν 14 n { sin cos } J ν n e a Z ν 81 a General facts 81 b The Case a > 0 8 c The Case a < 0 90 d Integrals 95 e Special Cases 100 { } sin 16 n 1/ cos 17 n 1/ e ± J ν 10 } 106 { Iν K ν a n 1/ e Z ν 106 b n 1/ e Z ν 107 c General formulas 109 { } { } sinh 18 n 1/ Iν 114 cosh K ν 19 n+1 ln Z n ln Z n+ν ln Z ν 1 a The Functions Λ k and Λ k 1 b Basic Integrals 17 c Integrals of n ln Z 0 10 c Integrals of n+1 ln Z 1 1
4 11 n e ± ln Z ν n e J ν α 150 a The Case α = 1, Basic Integrals 150 b Integrals α = c General Case α 1, Basic Integrals 159 d Integrals α 1 16 e Special Cases: ν = f Special Cases: ν = Special Function and Bessel Function 11 Orthogonal Polynomials 169 a Legendre Polynomials P n 169 b Chebyshev Polynomials T n 174 c Chebyshev Polynomials U n 179 d Laguerre Polynomials L n 18 e Hermite Polynomials H n Eponential Integral Sine and Cosine Integral Error function: n erf J ν α 196 a The Case α = b General Case 197 Products of two Bessel Functions 1 Bessel Functions with the the same Argument : 11 n+1 Zν 00 1 n Zν 05 1 n Zν 08 a The Functions Θ and Ω 08 b Integrals n Z 0 Z n+1 Z 0 Z n+1 Z 0 Z n+1 J ν I ν and n+1 J ν K ν 7 a ν = 0 7 b ν = 1 8 c Recurrence Relations 9 18 n J ν I 1 ν and n J ν K 1 ν 0 a ν = 0 0 b ν = 1 1 c Recurrence Relations 19 n+1] J µ Y ν a n+1 J 0 Y 0 b n J 0 Y 0 4 c n J 0 Y 1 5 d n 1 J 0 Y 1 6 e n J 1 Y 0 7 f n 1 J 0 Y 0 8 g n+1 J 1 Y 1 9 h n J 1 Y 1 40 Bessel Functions with different Arguments α and β : 1 n+1 Z ν αz ν β 41 4
5 a ν = 0 41 b ν = 1 51 n Z 0 αz 1 β 6 n Z ν αz ν β 77 a Basic Integrals 77 b Integrals 87 4 n+1 Z 0 αz 1 β 94 5 n+1 J 0 αi 0 β 98 6 n J 0 αi 1 β 00 7 n J 1 αi 0 β 0 8 n+1 J 1 αi 1 β 04 9 n+1 J ν αy ν β 06 Bessel Functions with different Arguments and + α 1 1 Z ν Z 1 + α and + α] 1 Z 1 + αz 1 d 07 4 Elementary Function and two Bessel Functions 41 n+1 ln Zν d and n ln Z 0 Z 1 d 08 4 n ln Z ν Zν d 16 a Integrals with 4n+ ln J 0 Z 0 16 a Integrals with 4n+1 ln J 1 Z 1 16 c Integrals with n+1 ln I ν K ν 17 d Integrals with n+ ln I ν K 1 ν 19 4 Some Cases of n ln Z ν Zν α d 44 1 ep/ sin / cos Z ν Z 1 d 1 45 Some Cases of n e α Z ν Z 1 d 46 Some Cases of { } sin / cos n α Z sinh / cosh µ Zν β d 5 a { } sin n α Z µ Z ν β d 5 cos b { sinh n cosh } α Z µ Z ν β d 41 Products of three Bessel Functions 1 n Z0 m Z1 m 45 a Basic Integral Z0 45 b Basic Integral Z 0 Z1 49 c Basic Integral Z1 5 d Basic Integral 1 Z0 56 e n Z0 60 f n Z0 Z 1 6 g n Z 0 Z1 66 h n Z1 68 i Recurrence Relations 7 j Z1 Z0 7 n Z κ α Z µ β Z ν γ 74 a n Z κ Z µ Z ν 74 b n Z κ α Z µ β Z ν α + β 85 c n Z κ α Z µ β Z ν α ± β Products of four Bessel Functions 5
6 41 m Z0 n Z1 4 n 414 a Eplicit Integrals 414 b Basic Integral Z c Basic Integral Z0 Z1 414 d Basic Integral Z e Integrals of m Z f Integrals of m Z0 Z 1 49 g Integrals of m Z0 Z1 4 h Integrals of m Z 0 Z1 46 i Integrals of m Z j Recurrence relations 49 5 Quotients Denominator p Z 0 + q Z a Typ f Z µ /p Z 0 + q Z 1 ] Denominator p Z 0 + q Z 1 ] 445 a Typ f Z µ /p Z 0 + q Z 1 ] 445 b Typ f Z0 n Z1 n /p Z 0 + q Z 1 ], n = 0, 1, Denominator p Z 0 + q Z 1 ] 449 a Typ f Z µ /p Z 0 + q Z 1 ] Denominator p Z 0 + q Z 1 ] a Typ f Z µ /p Z 0 + q Z 1 ] Denominator p Z0 + q Z1 45 a Typ f Z0 n Z1 n /p Z0 + q Z1 ], n = 0, 1, 45 b Typ f Z0 n Z1 n /p Z0 + q Z 0 Z 1 + rz1 ], n = 0, 1, Denominator a Z 0 + b Z 1 + p Z0 0 Z 1 + r Z 45 6 Miscellaneous Used special functions and defined functions Errata
7 1 Integrals with one Bessel Function: See also 10], 11 n Z ν with integer values of n 111 Integrals of the type n Z 0 d Let Φ = π J 1 H 0 J 0 H 1 ], where H ν denotes the Struve function, see 1], chapter 1117, 1118 and 1 And let Ψ = π I 0 L 1 I 1 L 0 ] be defined with the modified Struve function L ν Furthermore, let Φ Y = π Y 1 H 0 Y 0 H 1 ], Φ 1 π ] H = H 1 1 H 0 H 1 0 H 1, Φ π ] H = H 1 H 0 H 0 H 1 and Ψ K = π K 0 L 1 + K 1 L 0 ] In the following formulas J ν may be substituted by Y ν and simultaneously Φ by Φ Y or H ν p, p = 1, and Φ p H Well-known integrals: J 0 d = J 0 + Φ = Λ 0 I 0 d = I 0 + Ψ = Λ 0 K 0 d = K 0 + Ψ K The new-defined function Λ 0 is discussed in 111 a on page 1 and so is Λ 0 on page 15 See also 1], 111 H p 0 Y 0 d = Y 0 + Φ Y d = Hp 0 + Φp H, p = 1, J 0 d = J 1 Φ I 0 d = I 1 + Ψ K 0 d = K 1 + Ψ K 4 J 0 d = 4 9 J 1 + J 0 + 9Φ 4 I 0 d = I 1 I 0 + 9Ψ 4 K 0 d = K 1 K 0 + 9Ψ K E 7
8 6 J 0 d = J J 0 5Φ 6 I 0 d = I I 0 + 5Ψ 6 K 0 d = K K 0 + 5Ψ K and so on 8 J 0 d = J J Φ 8 I 0 d = I I Ψ 10 J 0 d = J J Φ 10 I 0 d = I I Ψ 1 J 0 d = J 0 + Let J Φ 1 I 0 d = I I Ψ n!! = and n!! = 1 in the case n 0 General formulas: and + + n J 0 d = n 1 k=0 n { 4 n n, n = m 1 5 n n, n = m + 1 n 1 k n 1!!] n k 1 J 0 + n 1 k!!] n k!!] k=0 ] 1 k n 1!! n k J n n 1!!] Φ = n 1 k!! = n 1 k=0 k=0 1 k n!] n k! n k 1! n k 1 k+1 n! J 0 + n k! n k! ] 1 k n! n k! k n k n! n k! n k=0 n I 0 d = n 1 n 1!!] n k 1 n 1 k!!] n k!!] ] n! J n n Φ n! ] n 1!! n k I 1 n 1 k!! k=0 I 0 + n 1!!] Ψ = 8
9 n k=0 Recurrence formulas: Ascending: Descending: = n 1 ] n! n k! k n k I 1 n! n k! k=0 n!] n k! n k 1! n k 1 k+1 n! n k! n k! ] n! I 0 + n Ψ n! n+ J 0 d = n + 1 n+1 J 0 + n+ J 1 n + 1 n+ I 0 d = n + 1 n+1 I 0 + n+ I 1 + n + 1 n+ K 0 d = n + 1 n+1 K 0 n+ K 1 + n + 1 n J 0 d = 1 n 1 n 1 I 0 d = n 1 n K 0 d = 1 n 1 n 1 n 1 J 0 + n J 1 n 1 n 1 I 0 n I 1 + n 1 n 1 K 0 + n K 1 + In the case n < 0 the previous formulas give J0 d = J J 0 Φ I0 d = 1 I 0 I 1 + Ψ K0 d = 1 K 0 + K 1 + Ψ K K0 6 d = 1 5 J0 8 d = J0 4 d = ] 9 J 0 1 J 1 + Φ I0 4 d = 1 4 ] 9 I I 1 + Ψ K0 4 d = 1 4 ] 9 K K 1 + Ψ K J0 6 d = J I0 6 d = I K ] K 1 + Ψ K n J 0 d n I 0 d n K 0 d ] n J 0 d ] n I 0 d ] n K 0 d ] J 0 Φ ] I 1 + Ψ and so on J ] 5 6 J 1 + Φ E E 9
10 I0 8 d = I ] I 1 + Ψ J0 10 d = J I0 10 d = ] J 0 Φ I I 1 + Ψ 9 ] J0 1 d = I0 1 d = J ] J 1 + Φ I I 1 + Ψ General formula: With n!! as defined on page 8 holds J0 d 1 n n 1 n = n 1!!] + 1 k k + 1!! k 1!! k 1 J 0 k=0 ] n 1 1 k k + 1!!] k J 1 + Φ = k=0 = 1n n n! n! 1 n k=0 { n k k +! k! k+1 k + 1! k! k+1 J 0 1 k k+ k=0 ] } k +! k+1 J 1 + Φ k + 1! With obviously modifications one gets the the formulas for the integrals n I 0 d and n K 0 d ] 10
11 11 Integrals of the type n+1 Z 0 d In the following formulas J ν may be substituted by Y ν or H p ν, p = 1, J 0 d = J 1 I 0 d = I 1 K 0 d = K 1 J 0 d = J J 1 ] I 0 d = + 4 I 1 I 0 ] K 0 d = + 4 K 1 + K 0 ] 5 J 0 d = 4 J J 1 ] 5 I 0 d = I I 0 ] 5 K 0 d = K K 0 ] 7 J 0 d = J J 1 ] 7 I 0 d = I I 0 ] 7 K 0 d = K K 0 ] 9 J 0 d = = J J 1 ] 9 I 0 d = = I I 0 ] 9 K 0 d = = K K 0 ] Let m J 0 d = P m J 0 + Q m J 1 ] and m I 0 d = Q mi 1 P mi 0 ], m K 0 d = Q mk 1 + P mk 0 ], then holds P 11 = Q 11 = P11 = Q 11 = *E* 11
12 P 1 = Q 1 = P1 = Q 1 = P 15 = Q 15 = P 15 = = *E* Q 15 = Recurrence formulas: n+1 J 0 d = n n J 0 + n+1 J 1 4n n+1 I 0 d = n n I 0 + n+1 I 1 + 4n n+1 K 0 d = n n K 0 n+1 K 1 + 4n k=0 n 1 J 0 d General formula: With n!! as defined on page 8 holds n 1 n+1 J 0 d = 1 k n!!] n k n k!!] n k!!] = n 1 1 k k=0 n ] + 1 k n!! n+1 k J 1 = n k!! k=0 n J 0 + k+1 n! n k n k! n k 1! k=0 n 1 I 0 d n 1 K 0 d J 0 + ] 1 k k n! n+1 k J 1 n k! With obviously modifications one gets the the formulas for the integrals n+1 I 0 d and n+1 K 0 d *E* 1
13 11 Integrals of the type n 1 Z 0 d The basic integral J0 d can be epressed by 0 1 J 0 t t J 0 t dt dt or = Ji 0, t see 1], equation and the following formulas There are given asymptotic epansions and polynomial approimations as well Tables of these functions may be found by 1], 111] or 11] The function Ji 0 is introduced and discussed in 9] For fast computations of this integrals one should use approimations with Chebyshev polynomials, see ], tables 9 I got the information from F Nouguier, that there is an error in a formula in 9], p 78 The true formula is The power series in Ji 0 ln = sin π π I0 d can be used without numerical problems γ ln + sin π π = ln + k=1 s=1 1 s 1 s Ji 0s ln s] 1 k k k! E In the following formulas J ν may be substituted by Y ν or H ν p, p = 1, J0 d = J 0 + J J0 d 4 I0 d = I 0 I I0 d 4 J0 d 1 5 = J J J0 d 64 I0 d 1 5 = I I I0 d 64 J0 d 7 = J J 1 1 J0 d 04 I0 d 7 = I I I0 d 04 J0 d 9 = = J J 1 + I0 d 9 = = I I 1 + J0 d 11 = E J J0 d J I0 d 11 = I I J0 d I0 d I0 d 1
14 Descending recurrence formulas: n 1 J 0 d = 1 4n n 1 I 0 d = 1 4n n+1 J 1 n n J 0 n+1 I 1 n n I 0 + General formula: With n!! as defined on page 8 holds J0 d n+1 = ] n+1 J 0 d ] n+1 I 0 d { n 1 n 1 = 1n k k +!! k!! } n!!] 1 k+ J 0 1 k k!!] J0 d k+1 J 1 + = E k=0 k=0 { n 1 n 1 = 1n n n! 1 k k+1 k + 1! k! k+ J 0 1 k k k! } J0 d k+1 J 1 + E k=0 With obviously modifications one gets the the formula for the integral n 1 I 0 d k=0 14
15 114 Integrals of the type n Z 1 d In the following formulas J ν may be substituted by Y ν or H p ν, p = 1, J 1 d = J 0 I 1 d = I 0 K 1 d = K 0 J 1 d = J 1 J 0 ] I 1 d = I 0 I 1 ] K 1 d = K 0 + K 1 ] 4 J 1 d = 4 16 J 1 8 J 0 ] 4 I 1 d = + 8 I I 1 ] 4 K 1 d = + 8 K K 1 ] 6 J 1 d = J J 0 ] 6 I 1 d = I I 1 ] 6 K 1 d = K K 1 ] 8 J 1 d = = J J 0 ] 8 I 1 d = = I I 1 ] 8 K 1 d = = K K 1 ] 10 J 1 d = J J 0 ] 10 I 1 d = I I 1 ] 10 K 1 d = K
16 K 1 ] Let m J 1 d = Q m J 1 P m J 0 ] and m I 1 d = P mi 0 Q mi 1 ], m K 1 d = P mi 0 + Q mi 1 ], then holds P 1 = Q 1 = P 1 = Q 1 = *E* P 14 = Q 14 = P14 = Q 14 = Recurrence formulas: n+ J 1 d = n+ J 0 + n + n+1 J 1 4nn + 1 n J 1 d n+ I 1 d = n+ I 0 n + n+1 I 1 + 4nn + 1 n+ K 1 d = n+ K 0 n + n+1 K 1 + 4nn + 1 n I 1 d n K 1 d General formula: With n!! as defined on page 8 holds n 1 n k n!!] n!!] n 1 k J 1 d = 1 n k!!] = n 1 k=0 1 k k=0 n 1 k=0 n 1 n!! n!! n k n k!!] n k!!] J 0 = 1 k k+1 n! n 1! n 1 k n 1 k!] J 1 k=0 1 k k n! n 1!! n k J 0 n k! n 1 k! J 1 With obviously modifications one gets the the formulas for the integrals n I 1 d and n K 1 d 16
17 115 Integrals of the type n Z 1 d About the integrals see 11, page 1 J0 d and I0 d In the following formulas J 0 may be substituted by Y 0 and simultaneously J 1 by Y 1 J1 d = 1 J J0 d I1 d = 1 I I0 d J1 d 4 = 1 8 J J 1 1 J0 d 16 I1 d 4 = 1 8 I I I0 d 16 J1 d 6 = = J J J0 d 84 I1 d 6 = I I I0 d 84 J1 d 8 = = J J 1 1 J0 d 184 I1 d 8 = I I I0 d 184 J1 d 10 = Recurrence formulas: = J J J0 d I1 d 10 = = I I I0 d E E J1 d J 0 n+ = 4nn + 1 n I1 d I 0 n+ = 4nn + 1 n J 1 n + n+1 1 4nn + 1 I 1 n + n nn + 1 J1 d n I1 d n 17
18 General formula: With n!! as defined on page 8 holds J1 d 1 n+1 n = n!! n!! { n n 1 k k +!! k!! } 1 k+ J 0 1 k k!!] J0 d k+1 J 1 + = k=0 k=0 1 n+1 = n 1 n! n 1! n 1 1 k k+1 k + 1! k! k+ J 0 1 k k k! ] J0 d k+1 J 1 + n k=0 With obviously modifications one gets the the formula for the integral n I 1 d k=0 18
19 116 Integrals of the type n+1 Z 1 d Φ, Φ Y, Ψ and Ψ K are the same as in 111, page 7 In the following formulas J ν may be substituted by Y ν and simultaneously Φ by Φ Y or H ν p, p = 1, and Φ p H J 1 d = Φ I 1 d = Ψ K 1 d = Ψ K J 1 d = J 1 J 0 Φ I 1 d = I 1 + I 0 Ψ K 1 d = K 1 K 0 + Ψ K 5 J 1 d = J J Φ 5 I 1 d = I I 0 45Ψ 5 K 1 d = K K Ψ K 7 J 1 d = J J Φ E 7 I 1 d = I I Ψ 7 K 1 d = K K Ψ K 9 J 1 d = = J J Φ 9 I 1 d = = I I Ψ 9 K 1 d = = K K Ψ General formula: With n!! as defined on page 8 holds n 1 n+1 k n + 1!! n 1!! n k J 1 d = 1 n 1 k!!] J 1 n 1 k=0 k=0 k n + 1!! n 1!! n+1 k 1 J n n + 1!! n 1!! Φ = n + 1 k!! n 1 k!! 19
20 n 1 k=0 n 1 = 1 k n +! n! n k!] n k k+1 n + 1! n! n k!] J 1 k=0 k n +! n! n + 1 k! n k! n+1 k 1 k J 0 + n + 1! n! n + k! n k! + 1 n n +! n! n+1 n + 1! n! Φ With obviously modifications one gets the the formulas for the integrals n+1 I 1 d and n+1 K 1 d Recurrence formulas: n+1 J 1 d = n+1 J 0 + n + 1 n J 1 n 1n + 1 n 1 J 1 d n+1 I 1 d = n+1 I 0 n + 1 n I 1 + n 1n + 1 n+1 K 1 d = n+1 K 0 n + 1 n K 1 + n 1n + 1 n 1 I 1 d n 1 K 1 d Descending: J1 d J 0 n+1 = 4n 1 n 1 J 1 n + 1 n 1 J1 d 4n 1 n 1 I1 d I 0 n+1 = 4n 1 n 1 I 1 n + 1 n + 1 I1 d 4n 1 n 1 K1 d K 0 n+1 = 4n 1 n 1 K 1 n + 1 n + 1 K1 d 4n 1 n 1 J1 d = J 0 J 1 + Φ I1 d = I 0 I 1 + Ψ K1 d = K 0 K 1 Ψ K J1 d = 1 ] 1 J J 0 Φ I1 d = 1 ] + 1 I I 0 + Ψ K1 d = 1 ] + 1 K 1 1 K 0 Ψ K J1 5 d = J 0 4 ] J 1 + Φ I1 5 d = I ] I 1 + Ψ K1 5 d = K ] K 1 Ψ K J1 7 d = J ] J 0 Φ 0
21 I1 7 d = K1 7 d = = = = J1 11 d = 6 I K J1 9 d = 7 J I1 9 d = 7 I K1 9 d = ] I 0 + Ψ ] K 0 Ψ k ] J 1 + Φ ] I 1 + Ψ 7 K J I1 11 d = K1 11 d = 9 ] J 0 Φ I ] I 0 + Ψ K 0 + Ψ K General formula: With n!! as defined on page 8 holds { J1 d 1 n n+1 = + n + 1!! n 1!! n 1 k=0 ] ] I 1 Ψ K K 1 1 k k + 1!! k 1!! k+1 J 0 } n 1 k k + 1!!] 1 1 J 1 + Φ = k=0 k+ { = n+1 n 1 n + 1! n! 1 k k +! k! n +! n! k+1 k + 1! k! k+1 J 0 k=0 } n k k +!] k+ k + 1!] k+ J 1 + Φ k=0 With obviously modifications one gets the the formulas for the integrals n 1 I 1 d and n 1 K 1 d 1
22 117 Integrals of the type n Z ν d, ν > 1 : From the well-known recurrence relations one gets immadiately J ν+1 d = J ν + J ν 1 d and I ν+1 d = I ν I ν 1 d With this formulas follows J ν t dt = Λ 0 0 κ=1 n J κ 1, J ν+1 t dt = 1 J 0 0 κ=1 n J κ 0 I ν t dt = 1 n Λ 0+ n 1 n+κ I κ 1, κ=1 0 I ν+1 t dt = 1 n I 0 1]+ n 1 n+κ I κ The integrals Λ 0 and Λ 0 are defined on page 7 and discussed on page 1 and 15 Holds n n Y ν d = Y 0 + Φ Y Y κ 1, Y ν+1 d = Y 0 Y κ 1 H 1 ν H ν d = H1 0 + Φ1 d = H 0 + Φ κ=1 H n κ=1 H n κ=1 H 1 κ 1, H κ 1, H 1 ν+1 d = H1 H ν+1 d = H { K ν d = 1 n K 0 + π } K 0L 1 + K 1 L 0 ] + K ν+1 d = 1 n+1 K 0 + About the functions Φ Y, Φ 1 H, Φ H see page 7 Further on, holds t J ν+1 t dt = ν + 1Λ 0 J 0 + ] ν 1 t J ν t dt = J 1 + J κ+1 κ=1 t I ν+1 t dt = 1 ν+1 ν + 1Λ 0 I 0 κ=1 κ=1 κ=1 0 n κ=1 0 n κ=1 H 1 κ 1 H κ 1 n 1 n+κ K κ 1, κ=1 n 1 n+κ+1 K κ κ=1 ] ν J κ κ=1 ν 1 4 κ=0 ν 1 + ν1 J 0 ] 4 κ=1 ν κj κ+1 ν κj κ ] ν ν 1 1 κ I κ 4 1 κ ν κi κ+1 ] ν 1 ν 1 t I ν t dt = 1 I ν κ I κ+1 + ν1 I 0 ] 4 1 κ ν κi κ Some of the previous sums may cause numerical problems, if is located near 0 For instance, the sum 0 gives with = 0 t I 6 t dt = J 1 J + J J 0 + 8J 4J = = , κ=1 κ=0 κ=1
23 which means the loss of 10 decimal digits For that reason the value of such integrals should be computed by the power series or other formulas See also the following remark In the following the integrals are epressed by Z 0 and Z 1 Integrals with n 4 are written eplicitely: at first n = 0, 1,,, 4, after them n = 1, In the other cases the functions P ν n, Q n ν and the coefficients R n ν, S ν n describe the integral n J ν d = P ν n J 0 + Q n ν J 1 + R n ν Λ 0 + S ν n J0 d Furthermore, let n I ν d = P n, ν I 0 + Q n, ν I 1 + R n, ν Λ 0 + S ν n, I0 d Concerning 1 Z 0 d see 11, page 1 Simple recurrence formula: n J ν+1 d = ν n 1 J ν d n J ν 1 d n I ν+1 d = ν n 1 J ν d + n J ν 1 d The integrals of n Z 0 and n Z 1 to start this recurrences are already described Remark: Let F ν m denote the antiderivative of m Z ν as given in the following tables They do not eist in the point = 0 in the case ν + m < 0 However, even if ν + m 0 the value of F ν m 0 sometimes turns out to be a limit of the type For instance, holds J d = J 0 J 1 = F with lim F = With L ν,m = lim 0 F ν m for the Bessel functions J ν and L ν,m for the modified Bessel functions I ν one has the following limits in the tables of integrals The values L ν,m = 0 are omitted: L, 1 = 1/, L, 1 = 1/ L,0 = 1, L, = 1/8; L,0 = 1, L, = 1/8 L 4,1 = 4, L 4, 1 = 1/4, L 4, = 1/48; L 4,1 = 4, L 4, 1 = 1/4, L 4, = 1/48 L 5, = 4, L 5,0 = 1, L 5, = 1/4, L 5, 4 = 1/84; L 5, = 4, L 5,0 = 1, L 5, = 1/4, L 5, 4 = 1/84 L 6, = 19, L 6,1 = 6, L 6, 1 = 1/6, L 6, = 1/19, L 6, 5 = 1/840; L 6, = 19, L 6,1 = 6, L 6, 1 = 1/6, L 6, = 1/19, L 6, 5 = 1/840 L 7,4 = 190, L 7, = 48, L 7,0 = 1, L 7, = 1/48, L 7, 4 = 1/190, L 7, 6 = 1/46080; L 7,4 = 190, L 7, = 48, L 7,0 = 1, L 7, = 1/48, L 7, 4 = 1/190, L 7, 6 = 1/46080 L 8,5 = 040, L 8, = 480, L 8,1 = 8, L 8, 1 = 1/8, L 8, = 1/480, L 8, 5 = 1/040; L 8,5 = 040, L 8, = 480, L 8,1 = 8, L 8, 1 = 1/8, L 8, = 1/480, L 8, 5 = 1/040 L 9,6 = 560, L 9,4 = 5760, L 9, = 80, L 9,0 = 1, L 9, = 1/80, L 9, 4 = 1/5760, L 9, 6 = 1/560; L 9,6 = 560, L 9,4 = 5760, L 9, = 80, L 9,0 = 1, L 9, = 1/80, L 9, 4 = 1/5760, L 9, 6 = 1/560 L 10,7 = , L 10,5 = 80640, L 10, = 960, L 10,1 = 10, L 10, 1 = 1/10, L 10, = 1/960, L 10, 5 = 1/80640; L 10,7 = , L 10,5 = 80640, L 10, = 960, L 10,1 = 10, L 10, 1 = 1/10, L 10, = 1/960, L 10, 5 = 1/80640
24 In the described cases of limits of the type the numerical computation of F ν m causes difficulties, if 0 < << 1 Then it is preferable to use the power series, which has a fast convergengence for such values of With m + ν 0 holds 0 t m J ν t dt = m+ν+1 ν 1 k k k! ν + k! 4 k m + ν k k=0 and From this one has For instance, 0 t m I ν t dt = m+ν+1 ν F m ν = L ν,m k=0 k k! ν + k! 4 k m + ν k t m J ν t dt and F,m ν = L ν,m + J d = J 0 J = 0 t m I ν t dt = = = = It was a loss of seven decimal digits at = 000 This value may be found without problems by the power series: F 000 = = ] = = = In the previous value, signed by *, the last digit should be instead of 4 and the result had to finish with 8 The integrals with I ν may be computed in the same way This method can be used even if ν + m < 0 For instance, J 4 1 J 4 J 4 7 d = d d and the second integral is given in the following tables For the first one holds with the power series of the function J 4 1 J d = = d = 1 1 = d = = ln = = = = Here are no differences of nearly the same values = 4
25 Z : J d = J 1 + Λ 0 I d = I 1 Λ 0 J d = J 0 J 1 I d = I 0 + I 1 J d = J 0 J 1 + Λ 0 I d = I 0 + I 1 + Λ 0 J d = 4 J 0 8 J 1 I d = 4 I I 1 4 J d = 5 J 0 15 J 1 15Λ 0 4 I d = 5 + I I Λ 0 J d = J 1 I d = I 1 J d = 1 J 0 + J Λ 0 I d = 1 I 0 I Λ 0 P 5 = 6 8, Q 5 = , R 5 = 0, S 5 = 0 P 5, = 6 + 8, Q 5, = , R 5, = 0, S 5, = 0 P 6 = , Q 6 = , R 6 = 15, S 6 = 0 P 6, = , Q 6, = , R 6, = 15, S 6, = 0 P 7 = , Q 7 = , R 7 = 0, S 7 = 0 P 7, = , Q 7, = , R 7, = 0, S 7, = 0 P 8 = , Q 8 = , R 8 = 14175, S 8 = 0 P 8, = , Q 8, = , R 8, = 14175, S 8, = 0 P 9 = , Q 9 = , R 9 = 0, S 9 = 0 5
26 P 9, = , Q 9, = , R 9, = 0, S 9, = 0 P 10 = , Q 10 = , R 10 = , S 10 = 0 P 10, = , Q 10, = , R 10, = , S 10, = 0 P = 1 4, Q = + 4 8, R = 0, S = 1 8 P, = 1 4, Q, = 4 8, R, = 0, S, = 1 8 P 4 = 15, Q 4 = , R 4 = 1 15, S 4 = 0 P 4, = + 15, Q 4, = , R 4, = 1 15, S 4, = 0 P 5 = , Q 5 = , R 5 = 0, S 5 = 1 96 P 5, = , Q 5, = , R 5, = 0, S 5, = 1 96 P 6 = , Q 6 = , R 6 = 1 15, S 6 = 0 P 6, = , Q 6, = , R 6, = 1 15, S 6, = 0 Z : J d = J 0 4 J 1 I d = I 0 4 I 1 J d = J 0 8J 1 + Λ 0 I d = I 0 8I 1 + Λ 0 J d = 8J 0 6J 1 I d = + 8I 0 6I 1 J d = 15 J 0 7 J Λ 0 I d = + 15 I 0 7 I 1 15Λ 0 4 J d = 4 J J 1 4 I d = + 4 I I 1 J d = 4 J J Λ 0 6
27 I d = 4 I I 1 1 Λ 0 J d = J 0 J 1 I d = I 0 I 1 P 5 = , Q 5 = , R 5 = 105, S 5 = 0 P 5, = , Q 5, = , R 5, = 105, S 5, = 0 P 6 = , Q 6 = , R 6 = 0, S 6 = 0 P 6, = , Q 6, = , R 6, = 0, S 6, = 0 P 7 = , Q 7 = , R 7 = 85, S 7 = 0 P 7, = , Q 7, = , R 7, = 85, S 7, = 0 P 8 = , Q 8 = , R 8 = 0, S 8 = 0 P 8, = , Q 8, = , R 8, = 0, S 8, = 0 P 9 = , Q 9 = , R 9 = 15595, S 9 = 0 P 9, = , Q 9, = , R 9, = 15595, S 9, = 0 P 10 = , Q 10 = , R 10 = 0, S 10 = 0 P 10, = , Q 10, = , R 10, = 0, S 10, = 0 P = , Q = , R = 1 15, S = 0 P, = 1 15, Q, = , R, = 1 15, S, = 0 P 4 = , Q 4 = , R 4 = 0, S 4 = 1 48 P 4, = , Q 4, = , R 4, = 0, S 4, = 1 48 P 5 = , Q 5 = , R 5 = 1 105, S 5 = 0 P 5, = , Q 5, = , R 5, = 1 105, S 5, = 0 P 6 = , Q 6 = , R 6 = 0, S 6 = P 6, = , Q 6, = , R 6, = 0, S 6, =
28 Z 4 : J 4 d = 8J 0 16J 1 + Λ 0 I 4 d = 8I I 1 + Λ 0 J 4 d = 8J J 1 I 4 d = 8I I 1 J 4 d = 9J J Λ 0 I 4 d = 9I I 1 15Λ 0 J 4 d = 10 48J J 1 I 4 d = I I 1 4 J 4 d = J J Λ 0 4 I 4 d = I I Λ 0 J4 d = 6J J 1 I4 d = 6J J 1 J4 d = J J Λ 0 I4 d = 7 15 I I Λ 0 P 5 4 = , Q 5 4 = , R 5 4 = 0, S 5 4 = 0 P 5, 4 = , Q 5, 4 = , R 5, 4 = 0, S 5, 4 = 0 P 6 4 = , Q 6 4 = , R 6 4 = 945, S 6 4 = 0 P 6, 4 = , Q 6, 4 = , R 6, 4 = 945, S 6, 4 = 0 P 7 4 = , Q 7 4 = , R 7 4 = 0, S 7 4 = 0 P 7, 4 = , Q 7, 4 = , R 7, 4 = 0, S 7, 4 = 0 P 8 4 = , Q 8 4 = , R 8 4 = 1185, S 8 4 = 0 8
29 P 8, 4 = , Q 8, 4 = , R 8, 4 = 1185, S 8, 4 = 0 P 9 4 = , Q 9 4 = , R 9 4 = 0, S 9 4 = 0 P 9, 4 = , Q 9, 4 = , R 9, 4 = 0, S 9, 4 = 0 P 10 4 = , Q 10 4 = , R 10 4 = 0705, S 10 4 = 0 P 10, 4 = , Q 10, 4 = , R 10, 4 = 0705, S 10, 4 = 0 P 4 = 4 4, Q 4 = 8 5, R 4 = 0, S 4 = 0 P, 4 = 4 4, Q, 4 = + 8 5, R, 4 = 0, S, 4 = 0 P 4 4 = , Q 4 4 = , R 4 4 = 1 105, S 4 4 = 0 P 4, 4 = , Q 4, 4 = , R 4, 4 = 1 105, S 4, 4 = 0 P 5 4 = , Q 5 4 = , R 5 4 = 0, S 5 4 = 1 84 P 5, 4 = , Q 5, 4 = , R 5, 4 = 0, S 5, 4 = 1 84 P 6 4 = , Q 6 4 = , R 6 4 = 1 945, S 6 4 = 0 P 6, 4 = , Q 6, 4 = , R 6, 4 = 1 945, S 6, 4 = 0 Z 5 : J 5 d = 48 J J 1 9
30 I 5 d = + 48 I I 1 J 5 d = 64 J J 1 + 5Λ 0 I 5 d = + 64 I I 1 5Λ 0 J 5 d = 7J J 1 I 5 d = + 7I I 1 J 5 d = 87 J J Λ 0 I 5 d = + 87 I I Λ 0 4 J 5 d = J J 1 4 I 5 d = I I 1 J5 d = J J Λ 0 I5 d = I I Λ 0 J5 d = 4 J J 1 I5 d = + 4 I I 1 P 5 5 = , Q 5 5 = , R 5 5 = 945, S 5 5 = 0 P 5, 5 = , Q 5, 5 = , R 5, 5 = 945, S 5, 5 = 0 P 6 5 = , Q 6 5 = , R 6 5 = 0, S 6 5 = 0 P 6, 5 = , Q 6, 5 = , R 6, 5 = 0, S 6, 5 = 0 P 7 5 = , Q 7 5 = , R 7 5 = 1095, S 7 5 = 0 P 7, 5 = , Q 7, 5 = , R 7, 5 = 1095, S 7, 5 = 0 P 8 5 = , Q 8 5 = , R 8 5 = 0, S 8 5 = 0 P 8, 5 = , Q 8, 5 = , R 8, 5 = 0, S 8, 5 = 0 P 9 5 = , Q 9 5 = , R 9 5 = , S 9 5 = 0 0
31 P 9, 5 = , Q 9, 5 = , R 9, 5 = , S 9, 5 = 0 P 10 5 = , Q 10 5 = , R 10 5 = 0, S 10 5 = 0 P 10, 5 = , Q 10, 5 = , R 10, 5 = 0, S 10, 5 = 0 P 5 = , Q 5 = , R 5 = 1 105, S 5 = 0 P, 5 = , Q, 5 = , R, 5 = 1 105, S, 5 = 0 P 4 5 = 4 6, Q 4 5 = , R 4 5 = 0, S 4 5 = 0 P 4, 5 = + 4 6, Q 4, 5 = , R 4, 5 = 0, S 4, 5 = 0 P 5 5 = , Q 5 5 = , R 5 5 = 1 945, S 5 5 = 0 P 5, 5 = , Q 5, 5 = , R 5, 5 = 1 945, S 5, 5 = 0 P 6 5 = , Q 6 5 = , R 6 5 = 0, S 6 5 = P 6, 5 = , Q 6, 5 = , R 6, 5 = 0, S 6, 5 = Z 6 : J 6 d = J J 1 + Λ 0 I 6 d = I I 1 Λ 0 J 6 d = J J 1 I 6 d = I I 1 J 6 d = J J 1 + 5Λ 0 I 6 d = I I 1 + 5Λ 0 J 6 d = 0 768J J 1 1
32 I 6 d = I I 1 4 J 6 d = J J Λ 0 4 I 6 d = I I 1 945Λ 0 J6 d = J J 1 I6 d = I I 1 J6 d = J J Λ 0 I6 d = I I Λ 0 P 5 6 = , Q 5 6 = , R 5 6 = 0, S 5 6 = 0 P 5, 6 = , Q 5, 6 = , R 5, 6 = 0, S 5, 6 = 0 P 6 6 = , Q 6 6 = , R 6 6 = 1095, S 6 6 = 0 P 6, 6 = , Q 6, 6 = , R 6, 6 = 1095, S 6, 6 = 0 P 7 6 = , Q 7 6 = , R 7 6 = 0, S 7 6 = 0 P 7, 6 = , Q 7, 6 = , R 7, 6 = 0, S 7, 6 = 0 P 8 6 = , Q 8 6 = , R 8 6 = 1515, S 8 6 = 0 P 8, 6 = , Q 8, 6 = , R 8, 6 = 1515, S 8, 6 = 0 P 9 6 = , Q 9 6 = , R 9 6 = 0, S 9 6 = 0 P 9, 6 = , Q 9, 6 = , R 9, 6 = 0, S 9, 6 = 0 P 10 6 = , Q 10 6 = , R 10 6 = , S 10 6 = 0 P 10, 6 = , Q 10, 6 = , R 10, 6 = , S 10, 6 = 0
33 P 6 = , Q 6 = , R 6 = 0, S 6 = 0 P, 6 = , Q, 6 = , R, 6 = 0, S, 6 = 0 P 4 6 = , Q 4 6 = , R 4 6 = 1 945, S 4 6 = 0 P 4, 6 = , Q 4, 6 = , R 4, 6 = 1 945, S 4, 6 = 0 P 5 6 = , Q 5 6 = , R 5 6 = 0, S 5 6 = 0 P 5, 6 = , Q 5, 6 = , R 5, 6 = 0, S 5, 6 = 0 P 6 6 = , Q 6 6 = , R 6 6 = , S 6 6 = 0 P 6, 6 = , Q 6, 6 = , R 6, 6 = , S 6, 6 = 0 Z 7 : J 7 d = J J 1 I 7 d = I I 1 J 7 d = J J 1 + 7Λ 0 I 7 d = I I 1 + 7Λ 0 J 7 d = J J 1 I 7 d = I I 1 J 7 d = J J Λ 0 I 7 d = I I 1 15Λ 0 4 J 7 d = J J 1
34 4 I 7 d = I I 1 J7 d = J J Λ 0 I7 d = I I Λ 0 J7 d = J J 1 I7 d = I I 1 P 5 7 = , Q 5 7 = , R 5 7 = 1095, S 5 7 = 0 P 5, 7 = , Q 5, 7 = , R 5, 7 = 1095, S 5, 7 = 0 P 6 7 = , Q 6 7 = , R 6 7 = 0, S 6 7 = 0 P 6, 7 = , Q 6, 7 = , R 6, 7 = 0, S 6, 7 = 0 P 7 7 = , Q 7 7 = , R 7 7 = 1515, S 7 7 = 0 P 7, 7 = , Q 7, 7 = , R 7, 7 = 1515, S 7, 7 = 0 P 8 7 = , Q 8 7 = , R 8 7 = 0, S 8 7 = 0 P 8, 7 = , Q 8, 7 = , R 8, 7 = 0, S 8, 7 = 0 P 9 7 = , Q 9 7 = , R 9 7 = 0705, S 9 7 = 0 P 9, 7 = , Q 9, 7 = , R 9, 7 = 0705, S 9, 7 = 0 P 10 7 = , Q 10 7 = , R 10 7 = 0, S 10 7 = 0 P 10, 7 = , Q 10, 7 = , R 10, 7 = 0, S 10, 7 = 0 P 7 = , Q 7 = , R 7 = 1 15, S 7 = 0 4
35 P, 7 = , Q, 7 = , R, 7 = 1 15, S, 7 = 0 P 4 7 = , Q 4 7 = , R 4 7 = 0, S 4 7 = 0 P 4, 7 = , Q 4, 7 = , R 4, 7 = 0, S 4, 7 = 0 P 5 7 = , Q 5 7 = , R 5 7 = , S 5 7 = 0 P 5, 7 = , Q 5, 7 = , R 5, 7 = , S 5, 7 = 0 P 6 7 = , Q 6 7 = , R 6 7 = 0, S 6 7 = 0 P 6, 7 = , Q 6, 7 = , R 6, 7 = 0, S 6, 7 = 0 Z 8 : J 8 d = J J 1 + Λ 0 I 8 d = I I 1 + Λ 0 J 8 d = J J 1 I 8 d = I I 1 J 8 d = J J 1 + 6Λ 0 I 8 d = I I 1 6Λ 0 J 8 d = J J 1 I 8 d = I I 1 4 J 8 d = J J Λ 0 4 I 8 d = I I Λ 0 J8 d = J J 1 I8 d = I I 1 J8 d = J 0 5
36 J Λ 0 I8 d = I I Λ 0 P 5 8 = , Q 5 8 = , R 5 8 = 0, S 5 8 = 0 P 5, 8 = , Q 5, 8 = , R 5, 8 = 0, S 5, 8 = 0 P 6 8 = , Q 6 8 = , R 6 8 = 1515, S 6 8 = 0 P 6, 8 = , Q 6, 8 = , R 6, 8 = 1515, S 6, 8 = 0 P 7 8 = , Q 7 8 = , R 7 8 = 0, S 7 8 = 0 P 7, 8 = , Q 7, 8 = , R 7, 8 = 0, S 7, 8 = 0 P 8 8 = , Q 8 8 = , R 8 8 = 0705, S 8 8 = 0 P 8, 8 = , Q 8, 8 = , R 8, 8 = 0705, S 8, 8 = 0 P 9 8 = , Q 9 8 = , R 9 8 = 0, S 9 8 = 0 P 9, 8 = , Q 9, 8 = , R 9, 8 = 0, S 9, 8 = 0 P 10 8 = , Q 10 8 = , R 10 8 = , S 10 8 = 0 P 10, 8 = , Q 10, 8 = , R 10, 8 = , S 10, 8 = 0 P 8 = , Q 8 = , R 8 = 0, S 8 = 0 P, 8 = , Q, 8 = , R, 8 = 0, S, 8 = 0 P 4 8 = , 6
37 Q 4 8 = , R 4 8 = 1 465, S 4 8 = 0 P 4, 8 = , Q 4, 8 = , R 4, 8 = 1 465, S 4, 8 = 0 P 5 8 = , Q 5 8 = , R 5 8 = 0, S 5 8 = 0 P 5, 8 = , Q 5, 8 = , R 5, 8 = 0, S 5, 8 = 0 P 6 8 = , Q 6 8 = , R = 1515, S 6 8 = 0 P 6, 8 = , Q 6, 8 = , R 6, 8 = , S 6, 8 = 0 Z 9 : J 9 d = J J 1 I 9 d = I I 1 J 9 d = = J J 1 + 9Λ 0 I 9 d = = I I 1 9Λ 0 J 9 d = J J 1 I 9 d = I I 1 J 9 d = J 0 + 7
38 J Λ 0 4 I 9 d = I I Λ 0 4 J 9 d = J J 1 4 I 9 d = I I 1 J9 d = J J Λ 0 I9 d = I I Λ 0 J9 d = J J 1 I9 d = I I 1 P 5 9 = , Q 5 9 = , R 5 9 = 45045, S 5 9 = 0 P 5, 9 = , Q 5, 9 = , R 5, 9 = 45045, S 5, 9 = 0 P 6 9 = , Q 6 9 = , R 6 9 = 0, S 6 9 = 0 P 6, 9 = , Q 6, 9 = , R 6, 9 = 0, S 6, 9 = 0 P 7 9 = , Q 7 9 = , R 7 9 = 0705, S 7 9 = 0 P 7, 9 = , Q 7, 9 = , R 7, 9 = 0705, S 7, 9 = 0 P 8 9 = , Q 8 9 = , R 8 9 = 0, S 8 9 = 0 P 8, 9 = , Q 8, 9 = , R 8, 9 = 0, S 8, 9 = 0 P 9 9 = , 8
39 Q 9 9 = , R 9 9 = , S 9 9 = 0 P 9, 9 = , Q 9, 9 = , R 9, 9 = , S 9, 9 = 0 P 10 9 = , Q 10 9 = , R 10 9 = 0, S 10 9 = 0 P 10, 9 = , Q 10, 9 = , R 10, 9 = 0, S 10, 9 = 0 P 9 = , Q 9 = , R 9 = 1 69, S 9 = 0 P, 9 = , Q, 9 = , R, 9 = 1 69, S, 9 = 0 P 4 9 = , Q 4 9 = , R 4 9 = 0, S 4 9 = 0 P 4, 9 = , Q 4, 9 = , R 4, 9 = 0, S 4, 9 = 0 P 5 9 = , Q 5 9 = , R 5 9 = , S 5 9 = 0 P 5, 9 = , Q 5, 9 = , R 5, 9 = , S 5, 9 = 0 P 6 9 = , Q 6 9 = , R 6 9 = 0, S 6 9 = 0 9
40 P 6, 9 = , Q 6, 9 = , R 6, 9 = 0, S 6, 9 = 0 Z 10 : J 10 d = J J 1 + Λ 0 8 I 10 d = I I 1 Λ 0 8 J 10 d = J J 1 I 10 d = I I 1 J 10 d = J J Λ 0 6 I 10 d = I I Λ 0 6 J 10 d = J J 1 I 10 d = I I 1 4 J 10 d = J J Λ I 10 d = I I Λ 0 J10 d 4 = J 0 40
41 J 1 I10 d 9 = I I 1 9 J10 d = J J Λ 0 I10 d = I I Λ 0 P 5 10 = , Q 5 10 = , R 5 10 = 0, S5 10 = 0 P 5, 10 = , Q 5, 10 = , R 5, 10 = 0, S 5, 10 = 0 P 6 10 = , Q 6 10 = , R 6 10 = , S6 10 = 0 P 6, 10 = , Q 6, 10 = , R 6, 10 = , S 6, 10 = 0 P 7 10 = , Q 7 10 = , R 7 10 = 0, S7 10 = 0 P 7, 10 = , Q 7, 10 = , R 7, 10 = 0, S 7, 10 = 0 P 8 10 = , Q 8 10 = , R 8 10 = , S8 10 = 0 P 8, 10 = , Q 8, 10 = , R 8, 10 = , S 8, 10 = 0 41
42 P 9 10 = , Q 9 10 = , R 9 10 = 0, S9 10 = 0 P 9, 10 = , Q 9, 10 = , R 9, 10 = 0, S 9, 10 = 0 P = , Q = , R = , S = 0 P 10, 10 = , Q 10, 10 = , R 10, 10 = , S 10, 10 = 0 P 10 = , Q 10 = , R 10 = 0, S 10 = 0 P, 10 = , Q, 10 = , R, 10 = 0, S, 10 = 0 P 4 10 = , Q 4 10 = , R 4 10 = , S 4 10 = 0 P 4, 10 = , Q 4, 10 = , R 4, 10 = , S 4, 10 = 0 P 5 10 = , Q 5 10 = , R 5 10 = 0, S 5 10 = 0 P 5, 10 = , Q 5, 10 = , R 5, 10 = 0, S 5, 10 = 0 P 6 10 = , Q 6 10 = 4
43 , R = , S 6 10 = 0 P 6, 10 = , Q 6, 10 = , R 6, 1 10 = , S 6, 10 = 0 4
44 118 Higher antiderivatives: Φ, Ψ and Ψ K are the same as in I, page 7 a n+1 Z 0 : With the functions Φ, Ψ and Ψ K as defined on page??? holds: J 0 = d Φ d, I 0 = d Ψ d, K 0 = d Ψ K d J 0 = d d J J 1 9 Φ ] I 0 = d d I 0 5 I Ψ ] K 0 = d d K K 1 9 Ψ K ] 5 J 0 = d d J J 1 + 5Φ ] 5 I 0 = d d I I Ψ ] 5 K 0 = d d K K 1 5 Ψ K ] 7 J 0 = d d J J Φ ] 7 I 0 = d d I I Ψ ] 7 K 0 = d d K K Ψ K ] The formulas for I 1 and K 1 vary in two signs 9 J 0 = d d J J Φ ] 9 I 0 = d d I I Ψ ] 11 J 0 = d d J J Φ ] 11 I 0 = d 11 d I I Ψ ] 1 J 0 = d d J J Φ ] 1 I 0 = d d I I Ψ ] 15 J 0 = d d
TABLES OF SOME INDEFINITE INTEGRALS OF BESSEL FUNCTIONS
Werner Rosenheinrich 1604015 Ernst - Abbe - Hochschule Jena First variant: 40900 University of Applied Sciences Germany TABLES OF SOME INDEFINITE INTEGRALS OF BESSEL FUNCTIONS Integrals of the type J 0
Διαβάστε περισσότεραSPECIAL FUNCTIONS and POLYNOMIALS
SPECIAL FUNCTIONS and POLYNOMIALS Gerard t Hooft Stefan Nobbenhuis Institute for Theoretical Physics Utrecht University, Leuvenlaan 4 3584 CC Utrecht, the Netherlands and Spinoza Institute Postbox 8.195
Διαβάστε περισσότεραCHAPTER 25 SOLVING EQUATIONS BY ITERATIVE METHODS
CHAPTER 5 SOLVING EQUATIONS BY ITERATIVE METHODS EXERCISE 104 Page 8 1. Find the positive root of the equation x + 3x 5 = 0, correct to 3 significant figures, using the method of bisection. Let f(x) =
Διαβάστε περισσότερα3.4 SUM AND DIFFERENCE FORMULAS. NOTE: cos(α+β) cos α + cos β cos(α-β) cos α -cos β
3.4 SUM AND DIFFERENCE FORMULAS Page Theorem cos(αβ cos α cos β -sin α cos(α-β cos α cos β sin α NOTE: cos(αβ cos α cos β cos(α-β cos α -cos β Proof of cos(α-β cos α cos β sin α Let s use a unit circle
Διαβάστε περισσότερα2 Composition. Invertible Mappings
Arkansas Tech University MATH 4033: Elementary Modern Algebra Dr. Marcel B. Finan Composition. Invertible Mappings In this section we discuss two procedures for creating new mappings from old ones, namely,
Διαβάστε περισσότεραSection 8.3 Trigonometric Equations
99 Section 8. Trigonometric Equations Objective 1: Solve Equations Involving One Trigonometric Function. In this section and the next, we will exple how to solving equations involving trigonometric functions.
Διαβάστε περισσότεραMath221: HW# 1 solutions
Math: HW# solutions Andy Royston October, 5 7.5.7, 3 rd Ed. We have a n = b n = a = fxdx = xdx =, x cos nxdx = x sin nx n sin nxdx n = cos nx n = n n, x sin nxdx = x cos nx n + cos nxdx n cos n = + sin
Διαβάστε περισσότεραDiracDelta. Notations. Primary definition. Specific values. General characteristics. Traditional name. Traditional notation
DiracDelta Notations Traditional name Dirac delta function Traditional notation x Mathematica StandardForm notation DiracDeltax Primary definition 4.03.02.000.0 x Π lim ε ; x ε0 x 2 2 ε Specific values
Διαβάστε περισσότεραCRASH COURSE IN PRECALCULUS
CRASH COURSE IN PRECALCULUS Shiah-Sen Wang The graphs are prepared by Chien-Lun Lai Based on : Precalculus: Mathematics for Calculus by J. Stuwart, L. Redin & S. Watson, 6th edition, 01, Brooks/Cole Chapter
Διαβάστε περισσότεραANSWERSHEET (TOPIC = DIFFERENTIAL CALCULUS) COLLECTION #2. h 0 h h 0 h h 0 ( ) g k = g 0 + g 1 + g g 2009 =?
Teko Classes IITJEE/AIEEE Maths by SUHAAG SIR, Bhopal, Ph (0755) 3 00 000 www.tekoclasses.com ANSWERSHEET (TOPIC DIFFERENTIAL CALCULUS) COLLECTION # Question Type A.Single Correct Type Q. (A) Sol least
Διαβάστε περισσότεραSCITECH Volume 13, Issue 2 RESEARCH ORGANISATION Published online: March 29, 2018
Journal of rogressive Research in Mathematics(JRM) ISSN: 2395-028 SCITECH Volume 3, Issue 2 RESEARCH ORGANISATION ublished online: March 29, 208 Journal of rogressive Research in Mathematics www.scitecresearch.com/journals
Διαβάστε περισσότεραApproximation of distance between locations on earth given by latitude and longitude
Approximation of distance between locations on earth given by latitude and longitude Jan Behrens 2012-12-31 In this paper we shall provide a method to approximate distances between two points on earth
Διαβάστε περισσότεραFourier Series. MATH 211, Calculus II. J. Robert Buchanan. Spring Department of Mathematics
Fourier Series MATH 211, Calculus II J. Robert Buchanan Department of Mathematics Spring 2018 Introduction Not all functions can be represented by Taylor series. f (k) (c) A Taylor series f (x) = (x c)
Διαβάστε περισσότεραEE512: Error Control Coding
EE512: Error Control Coding Solution for Assignment on Finite Fields February 16, 2007 1. (a) Addition and Multiplication tables for GF (5) and GF (7) are shown in Tables 1 and 2. + 0 1 2 3 4 0 0 1 2 3
Διαβάστε περισσότεραCHAPTER 101 FOURIER SERIES FOR PERIODIC FUNCTIONS OF PERIOD
CHAPTER FOURIER SERIES FOR PERIODIC FUNCTIONS OF PERIOD EXERCISE 36 Page 66. Determine the Fourier series for the periodic function: f(x), when x +, when x which is periodic outside this rge of period.
Διαβάστε περισσότεραderivation of the Laplacian from rectangular to spherical coordinates
derivation of the Laplacian from rectangular to spherical coordinates swapnizzle 03-03- :5:43 We begin by recognizing the familiar conversion from rectangular to spherical coordinates (note that φ is used
Διαβάστε περισσότεραConcrete Mathematics Exercises from 30 September 2016
Concrete Mathematics Exercises from 30 September 2016 Silvio Capobianco Exercise 1.7 Let H(n) = J(n + 1) J(n). Equation (1.8) tells us that H(2n) = 2, and H(2n+1) = J(2n+2) J(2n+1) = (2J(n+1) 1) (2J(n)+1)
Διαβάστε περισσότερα4.6 Autoregressive Moving Average Model ARMA(1,1)
84 CHAPTER 4. STATIONARY TS MODELS 4.6 Autoregressive Moving Average Model ARMA(,) This section is an introduction to a wide class of models ARMA(p,q) which we will consider in more detail later in this
Διαβάστε περισσότεραSection 7.6 Double and Half Angle Formulas
09 Section 7. Double and Half Angle Fmulas To derive the double-angles fmulas, we will use the sum of two angles fmulas that we developed in the last section. We will let α θ and β θ: cos(θ) cos(θ + θ)
Διαβάστε περισσότεραOther Test Constructions: Likelihood Ratio & Bayes Tests
Other Test Constructions: Likelihood Ratio & Bayes Tests Side-Note: So far we have seen a few approaches for creating tests such as Neyman-Pearson Lemma ( most powerful tests of H 0 : θ = θ 0 vs H 1 :
Διαβάστε περισσότεραA summation formula ramified with hypergeometric function and involving recurrence relation
South Asian Journal of Mathematics 017, Vol. 7 ( 1): 1 4 www.sajm-online.com ISSN 51-151 RESEARCH ARTICLE A summation formula ramified with hypergeometric function and involving recurrence relation Salahuddin
Διαβάστε περισσότεραIf we restrict the domain of y = sin x to [ π, π ], the restrict function. y = sin x, π 2 x π 2
Chapter 3. Analytic Trigonometry 3.1 The inverse sine, cosine, and tangent functions 1. Review: Inverse function (1) f 1 (f(x)) = x for every x in the domain of f and f(f 1 (x)) = x for every x in the
Διαβάστε περισσότεραHomework 3 Solutions
Homework 3 Solutions Igor Yanovsky (Math 151A TA) Problem 1: Compute the absolute error and relative error in approximations of p by p. (Use calculator!) a) p π, p 22/7; b) p π, p 3.141. Solution: For
Διαβάστε περισσότεραExpIntegralE. Notations. Primary definition. Specific values. Traditional name. Traditional notation. Mathematica StandardForm notation
ExpIntegralE Notations Traditional name Exponential integral E Traditional notation E Mathematica StandardForm notation ExpIntegralE, Primary definition 06.34.0.000.0 E t t t ; Re 0 Specific values Specialied
Διαβάστε περισσότεραIf we restrict the domain of y = sin x to [ π 2, π 2
Chapter 3. Analytic Trigonometry 3.1 The inverse sine, cosine, and tangent functions 1. Review: Inverse function (1) f 1 (f(x)) = x for every x in the domain of f and f(f 1 (x)) = x for every x in the
Διαβάστε περισσότεραBessel functions. ν + 1 ; 1 = 0 for k = 0, 1, 2,..., n 1. Γ( n + k + 1) = ( 1) n J n (z). Γ(n + k + 1) k!
Bessel functions The Bessel function J ν (z of the first kind of order ν is defined by J ν (z ( (z/ν ν Γ(ν + F ν + ; z 4 ( k k ( Γ(ν + k + k! For ν this is a solution of the Bessel differential equation
Διαβάστε περισσότεραPARTIAL NOTES for 6.1 Trigonometric Identities
PARTIAL NOTES for 6.1 Trigonometric Identities tanθ = sinθ cosθ cotθ = cosθ sinθ BASIC IDENTITIES cscθ = 1 sinθ secθ = 1 cosθ cotθ = 1 tanθ PYTHAGOREAN IDENTITIES sin θ + cos θ =1 tan θ +1= sec θ 1 + cot
Διαβάστε περισσότεραFinite Field Problems: Solutions
Finite Field Problems: Solutions 1. Let f = x 2 +1 Z 11 [x] and let F = Z 11 [x]/(f), a field. Let Solution: F =11 2 = 121, so F = 121 1 = 120. The possible orders are the divisors of 120. Solution: The
Διαβάστε περισσότεραMatrices and Determinants
Matrices and Determinants SUBJECTIVE PROBLEMS: Q 1. For what value of k do the following system of equations possess a non-trivial (i.e., not all zero) solution over the set of rationals Q? x + ky + 3z
Διαβάστε περισσότεραNotations. Primary definition. Specific values. General characteristics. Traditional name. Traditional notation. Mathematica StandardForm notation
KelvinKei Notations Traditional name Kelvin function of the second kind Traditional notation kei Mathematica StandardForm notation KelvinKei Primary definition 03.5.0.000.0 kei kei 0 Specific values Values
Διαβάστε περισσότεραHOMEWORK 4 = G. In order to plot the stress versus the stretch we define a normalized stretch:
HOMEWORK 4 Problem a For the fast loading case, we want to derive the relationship between P zz and λ z. We know that the nominal stress is expressed as: P zz = ψ λ z where λ z = λ λ z. Therefore, applying
Διαβάστε περισσότεραSolutions to Exercise Sheet 5
Solutions to Eercise Sheet 5 jacques@ucsd.edu. Let X and Y be random variables with joint pdf f(, y) = 3y( + y) where and y. Determine each of the following probabilities. Solutions. a. P (X ). b. P (X
Διαβάστε περισσότεραEvery set of first-order formulas is equivalent to an independent set
Every set of first-order formulas is equivalent to an independent set May 6, 2008 Abstract A set of first-order formulas, whatever the cardinality of the set of symbols, is equivalent to an independent
Διαβάστε περισσότερα9.09. # 1. Area inside the oval limaçon r = cos θ. To graph, start with θ = 0 so r = 6. Compute dr
9.9 #. Area inside the oval limaçon r = + cos. To graph, start with = so r =. Compute d = sin. Interesting points are where d vanishes, or at =,,, etc. For these values of we compute r:,,, and the values
Διαβάστε περισσότεραSecond Order Partial Differential Equations
Chapter 7 Second Order Partial Differential Equations 7.1 Introduction A second order linear PDE in two independent variables (x, y Ω can be written as A(x, y u x + B(x, y u xy + C(x, y u u u + D(x, y
Διαβάστε περισσότεραThe Simply Typed Lambda Calculus
Type Inference Instead of writing type annotations, can we use an algorithm to infer what the type annotations should be? That depends on the type system. For simple type systems the answer is yes, and
Διαβάστε περισσότεραProblem Set 9 Solutions. θ + 1. θ 2 + cotθ ( ) sinθ e iφ is an eigenfunction of the ˆ L 2 operator. / θ 2. φ 2. sin 2 θ φ 2. ( ) = e iφ. = e iφ cosθ.
Chemistry 362 Dr Jean M Standard Problem Set 9 Solutions The ˆ L 2 operator is defined as Verify that the angular wavefunction Y θ,φ) Also verify that the eigenvalue is given by 2! 2 & L ˆ 2! 2 2 θ 2 +
Διαβάστε περισσότεραUniform Convergence of Fourier Series Michael Taylor
Uniform Convergence of Fourier Series Michael Taylor Given f L 1 T 1 ), we consider the partial sums of the Fourier series of f: N 1) S N fθ) = ˆfk)e ikθ. k= N A calculation gives the Dirichlet formula
Διαβάστε περισσότεραAreas and Lengths in Polar Coordinates
Kiryl Tsishchanka Areas and Lengths in Polar Coordinates In this section we develop the formula for the area of a region whose boundary is given by a polar equation. We need to use the formula for the
Διαβάστε περισσότεραOrdinal Arithmetic: Addition, Multiplication, Exponentiation and Limit
Ordinal Arithmetic: Addition, Multiplication, Exponentiation and Limit Ting Zhang Stanford May 11, 2001 Stanford, 5/11/2001 1 Outline Ordinal Classification Ordinal Addition Ordinal Multiplication Ordinal
Διαβάστε περισσότεραSCHOOL OF MATHEMATICAL SCIENCES G11LMA Linear Mathematics Examination Solutions
SCHOOL OF MATHEMATICAL SCIENCES GLMA Linear Mathematics 00- Examination Solutions. (a) i. ( + 5i)( i) = (6 + 5) + (5 )i = + i. Real part is, imaginary part is. (b) ii. + 5i i ( + 5i)( + i) = ( i)( + i)
Διαβάστε περισσότεραInverse trigonometric functions & General Solution of Trigonometric Equations. ------------------ ----------------------------- -----------------
Inverse trigonometric functions & General Solution of Trigonometric Equations. 1. Sin ( ) = a) b) c) d) Ans b. Solution : Method 1. Ans a: 17 > 1 a) is rejected. w.k.t Sin ( sin ) = d is rejected. If sin
Διαβάστε περισσότεραPartial Differential Equations in Biology The boundary element method. March 26, 2013
The boundary element method March 26, 203 Introduction and notation The problem: u = f in D R d u = ϕ in Γ D u n = g on Γ N, where D = Γ D Γ N, Γ D Γ N = (possibly, Γ D = [Neumann problem] or Γ N = [Dirichlet
Διαβάστε περισσότεραChapter 6: Systems of Linear Differential. be continuous functions on the interval
Chapter 6: Systems of Linear Differential Equations Let a (t), a 2 (t),..., a nn (t), b (t), b 2 (t),..., b n (t) be continuous functions on the interval I. The system of n first-order differential equations
Διαβάστε περισσότεραAppendix to On the stability of a compressible axisymmetric rotating flow in a pipe. By Z. Rusak & J. H. Lee
Appendi to On the stability of a compressible aisymmetric rotating flow in a pipe By Z. Rusak & J. H. Lee Journal of Fluid Mechanics, vol. 5 4, pp. 5 4 This material has not been copy-edited or typeset
Διαβάστε περισσότεραAreas and Lengths in Polar Coordinates
Kiryl Tsishchanka Areas and Lengths in Polar Coordinates In this section we develop the formula for the area of a region whose boundary is given by a polar equation. We need to use the formula for the
Διαβάστε περισσότεραC.S. 430 Assignment 6, Sample Solutions
C.S. 430 Assignment 6, Sample Solutions Paul Liu November 15, 2007 Note that these are sample solutions only; in many cases there were many acceptable answers. 1 Reynolds Problem 10.1 1.1 Normal-order
Διαβάστε περισσότεραExample Sheet 3 Solutions
Example Sheet 3 Solutions. i Regular Sturm-Liouville. ii Singular Sturm-Liouville mixed boundary conditions. iii Not Sturm-Liouville ODE is not in Sturm-Liouville form. iv Regular Sturm-Liouville note
Διαβάστε περισσότεραDESIGN OF MACHINERY SOLUTION MANUAL h in h 4 0.
DESIGN OF MACHINERY SOLUTION MANUAL -7-1! PROBLEM -7 Statement: Design a double-dwell cam to move a follower from to 25 6, dwell for 12, fall 25 and dwell for the remader The total cycle must take 4 sec
Διαβάστε περισσότεραCongruence Classes of Invertible Matrices of Order 3 over F 2
International Journal of Algebra, Vol. 8, 24, no. 5, 239-246 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/.2988/ija.24.422 Congruence Classes of Invertible Matrices of Order 3 over F 2 Ligong An and
Διαβάστε περισσότεραSrednicki Chapter 55
Srednicki Chapter 55 QFT Problems & Solutions A. George August 3, 03 Srednicki 55.. Use equations 55.3-55.0 and A i, A j ] = Π i, Π j ] = 0 (at equal times) to verify equations 55.-55.3. This is our third
Διαβάστε περισσότεραTridiagonal matrices. Gérard MEURANT. October, 2008
Tridiagonal matrices Gérard MEURANT October, 2008 1 Similarity 2 Cholesy factorizations 3 Eigenvalues 4 Inverse Similarity Let α 1 ω 1 β 1 α 2 ω 2 T =......... β 2 α 1 ω 1 β 1 α and β i ω i, i = 1,...,
Διαβάστε περισσότεραAn Inventory of Continuous Distributions
Appendi A An Inventory of Continuous Distributions A.1 Introduction The incomplete gamma function is given by Also, define Γ(α; ) = 1 with = G(α; ) = Z 0 Z 0 Z t α 1 e t dt, α > 0, >0 t α 1 e t dt, α >
Διαβάστε περισσότερα6.1. Dirac Equation. Hamiltonian. Dirac Eq.
6.1. Dirac Equation Ref: M.Kaku, Quantum Field Theory, Oxford Univ Press (1993) η μν = η μν = diag(1, -1, -1, -1) p 0 = p 0 p = p i = -p i p μ p μ = p 0 p 0 + p i p i = E c 2 - p 2 = (m c) 2 H = c p 2
Διαβάστε περισσότεραSecond Order RLC Filters
ECEN 60 Circuits/Electronics Spring 007-0-07 P. Mathys Second Order RLC Filters RLC Lowpass Filter A passive RLC lowpass filter (LPF) circuit is shown in the following schematic. R L C v O (t) Using phasor
Διαβάστε περισσότεραArithmetical applications of lagrangian interpolation. Tanguy Rivoal. Institut Fourier CNRS and Université de Grenoble 1
Arithmetical applications of lagrangian interpolation Tanguy Rivoal Institut Fourier CNRS and Université de Grenoble Conference Diophantine and Analytic Problems in Number Theory, The 00th anniversary
Διαβάστε περισσότεραPhys460.nb Solution for the t-dependent Schrodinger s equation How did we find the solution? (not required)
Phys460.nb 81 ψ n (t) is still the (same) eigenstate of H But for tdependent H. The answer is NO. 5.5.5. Solution for the tdependent Schrodinger s equation If we assume that at time t 0, the electron starts
Διαβάστε περισσότεραEvaluation of some non-elementary integrals of sine, cosine and exponential integrals type
Noname manuscript No. will be inserted by the editor Evaluation of some non-elementary integrals of sine, cosine and exponential integrals type Victor Nijimbere Received: date / Accepted: date Abstract
Διαβάστε περισσότεραforms This gives Remark 1. How to remember the above formulas: Substituting these into the equation we obtain with
Week 03: C lassification of S econd- Order L inear Equations In last week s lectures we have illustrated how to obtain the general solutions of first order PDEs using the method of characteristics. We
Διαβάστε περισσότεραLecture 2. Soundness and completeness of propositional logic
Lecture 2 Soundness and completeness of propositional logic February 9, 2004 1 Overview Review of natural deduction. Soundness and completeness. Semantics of propositional formulas. Soundness proof. Completeness
Διαβάστε περισσότεραSimilarly, we may define hyperbolic functions cosh α and sinh α from the unit hyperbola
Universit of Hperbolic Functions The trigonometric functions cos α an cos α are efine using the unit circle + b measuring the istance α in the counter-clockwise irection along the circumference of the
Διαβάστε περισσότεραInstruction Execution Times
1 C Execution Times InThisAppendix... Introduction DL330 Execution Times DL330P Execution Times DL340 Execution Times C-2 Execution Times Introduction Data Registers This appendix contains several tables
Διαβάστε περισσότεραST5224: Advanced Statistical Theory II
ST5224: Advanced Statistical Theory II 2014/2015: Semester II Tutorial 7 1. Let X be a sample from a population P and consider testing hypotheses H 0 : P = P 0 versus H 1 : P = P 1, where P j is a known
Διαβάστε περισσότεραMathCity.org Merging man and maths
MathCity.org Merging man and maths Exercise 10. (s) Page Textbook of Algebra and Trigonometry for Class XI Available online @, Version:.0 Question # 1 Find the values of sin, and tan when: 1 π (i) (ii)
Διαβάστε περισσότεραNew bounds for spherical two-distance sets and equiangular lines
New bounds for spherical two-distance sets and equiangular lines Michigan State University Oct 8-31, 016 Anhui University Definition If X = {x 1, x,, x N } S n 1 (unit sphere in R n ) and x i, x j = a
Διαβάστε περισσότεραStrain gauge and rosettes
Strain gauge and rosettes Introduction A strain gauge is a device which is used to measure strain (deformation) on an object subjected to forces. Strain can be measured using various types of devices classified
Διαβάστε περισσότεραSection 9.2 Polar Equations and Graphs
180 Section 9. Polar Equations and Graphs In this section, we will be graphing polar equations on a polar grid. In the first few examples, we will write the polar equation in rectangular form to help identify
Διαβάστε περισσότερα6.3 Forecasting ARMA processes
122 CHAPTER 6. ARMA MODELS 6.3 Forecasting ARMA processes The purpose of forecasting is to predict future values of a TS based on the data collected to the present. In this section we will discuss a linear
Διαβάστε περισσότεραChapter 6: Systems of Linear Differential. be continuous functions on the interval
Chapter 6: Systems of Linear Differential Equations Let a (t), a 2 (t),..., a nn (t), b (t), b 2 (t),..., b n (t) be continuous functions on the interval I. The system of n first-order differential equations
Διαβάστε περισσότεραEcon 2110: Fall 2008 Suggested Solutions to Problem Set 8 questions or comments to Dan Fetter 1
Eon : Fall 8 Suggested Solutions to Problem Set 8 Email questions or omments to Dan Fetter Problem. Let X be a salar with density f(x, θ) (θx + θ) [ x ] with θ. (a) Find the most powerful level α test
Διαβάστε περισσότεραGAUSS-LAGUERRE AND GAUSS-HERMITE QUADRATURE ON 64, 96 AND 128 NODES
GAUSS-LAGUERRE AND GAUSS-HERMITE QUADRATURE ON 64, 96 AND 128 NODES RICHARD J. MATHAR Abstract. The manuscript provides tables of abscissae and weights for Gauss- Laguerre integration on 64, 96 and 128
Διαβάστε περισσότεραLanczos and biorthogonalization methods for eigenvalues and eigenvectors of matrices
Lanzos and iorthogonalization methods for eigenvalues and eigenvetors of matries rolem formulation Many prolems are redued to solving the following system: x x where is an unknown numer А a matrix n n
Διαβάστε περισσότεραω ω ω ω ω ω+2 ω ω+2 + ω ω ω ω+2 + ω ω+1 ω ω+2 2 ω ω ω ω ω ω ω ω+1 ω ω2 ω ω2 + ω ω ω2 + ω ω ω ω2 + ω ω+1 ω ω2 + ω ω+1 + ω ω ω ω2 + ω
0 1 2 3 4 5 6 ω ω + 1 ω + 2 ω + 3 ω + 4 ω2 ω2 + 1 ω2 + 2 ω2 + 3 ω3 ω3 + 1 ω3 + 2 ω4 ω4 + 1 ω5 ω 2 ω 2 + 1 ω 2 + 2 ω 2 + ω ω 2 + ω + 1 ω 2 + ω2 ω 2 2 ω 2 2 + 1 ω 2 2 + ω ω 2 3 ω 3 ω 3 + 1 ω 3 + ω ω 3 +
Διαβάστε περισσότεραComputing the Macdonald function for complex orders
Macdonald p. 1/1 Computing the Macdonald function for complex orders Walter Gautschi wxg@cs.purdue.edu Purdue University Macdonald p. 2/1 Integral representation K ν (x) = complex order ν = α + iβ e x
Διαβάστε περισσότεραΚΥΠΡΙΑΚΗ ΕΤΑΙΡΕΙΑ ΠΛΗΡΟΦΟΡΙΚΗΣ CYPRUS COMPUTER SOCIETY ΠΑΓΚΥΠΡΙΟΣ ΜΑΘΗΤΙΚΟΣ ΔΙΑΓΩΝΙΣΜΟΣ ΠΛΗΡΟΦΟΡΙΚΗΣ 24/3/2007
Οδηγίες: Να απαντηθούν όλες οι ερωτήσεις. Όλοι οι αριθμοί που αναφέρονται σε όλα τα ερωτήματα μικρότεροι του 10000 εκτός αν ορίζεται διαφορετικά στη διατύπωση του προβλήματος. Αν κάπου κάνετε κάποιες υποθέσεις
Διαβάστε περισσότεραPractice Exam 2. Conceptual Questions. 1. State a Basic identity and then verify it. (a) Identity: Solution: One identity is csc(θ) = 1
Conceptual Questions. State a Basic identity and then verify it. a) Identity: Solution: One identity is cscθ) = sinθ) Practice Exam b) Verification: Solution: Given the point of intersection x, y) of the
Διαβάστε περισσότεραSolution Series 9. i=1 x i and i=1 x i.
Lecturer: Prof. Dr. Mete SONER Coordinator: Yilin WANG Solution Series 9 Q1. Let α, β >, the p.d.f. of a beta distribution with parameters α and β is { Γ(α+β) Γ(α)Γ(β) f(x α, β) xα 1 (1 x) β 1 for < x
Διαβάστε περισσότεραStatistical Inference I Locally most powerful tests
Statistical Inference I Locally most powerful tests Shirsendu Mukherjee Department of Statistics, Asutosh College, Kolkata, India. shirsendu st@yahoo.co.in So far we have treated the testing of one-sided
Διαβάστε περισσότεραCHAPTER 48 APPLICATIONS OF MATRICES AND DETERMINANTS
CHAPTER 48 APPLICATIONS OF MATRICES AND DETERMINANTS EXERCISE 01 Page 545 1. Use matrices to solve: 3x + 4y x + 5y + 7 3x + 4y x + 5y 7 Hence, 3 4 x 0 5 y 7 The inverse of 3 4 5 is: 1 5 4 1 5 4 15 8 3
Διαβάστε περισσότεραRight Rear Door. Let's now finish the door hinge saga with the right rear door
Right Rear Door Let's now finish the door hinge saga with the right rear door You may have been already guessed my steps, so there is not much to describe in detail. Old upper one file:///c /Documents
Διαβάστε περισσότεραΑπόκριση σε Μοναδιαία Ωστική Δύναμη (Unit Impulse) Απόκριση σε Δυνάμεις Αυθαίρετα Μεταβαλλόμενες με το Χρόνο. Απόστολος Σ.
Απόκριση σε Δυνάμεις Αυθαίρετα Μεταβαλλόμενες με το Χρόνο The time integral of a force is referred to as impulse, is determined by and is obtained from: Newton s 2 nd Law of motion states that the action
Διαβάστε περισσότεραk A = [k, k]( )[a 1, a 2 ] = [ka 1,ka 2 ] 4For the division of two intervals of confidence in R +
Chapter 3. Fuzzy Arithmetic 3- Fuzzy arithmetic: ~Addition(+) and subtraction (-): Let A = [a and B = [b, b in R If x [a and y [b, b than x+y [a +b +b Symbolically,we write A(+)B = [a (+)[b, b = [a +b
Διαβάστε περισσότεραNotations. Primary definition. Specific values. General characteristics. Series representations. Traditional name. Traditional notation
Pi Notations Traditional name Π Traditional notation Π Mathematica StandardForm notation Pi Primary definition.3... Π Specific values.3.3.. Π 3.5965358979338663383795889769399375589795937866868998683853
Διαβάστε περισσότεραLecture 34 Bootstrap confidence intervals
Lecture 34 Bootstrap confidence intervals Confidence Intervals θ: an unknown parameter of interest We want to find limits θ and θ such that Gt = P nˆθ θ t If G 1 1 α is known, then P θ θ = P θ θ = 1 α
Διαβάστε περισσότεραF19MC2 Solutions 9 Complex Analysis
F9MC Solutions 9 Complex Analysis. (i) Let f(z) = eaz +z. Then f is ifferentiable except at z = ±i an so by Cauchy s Resiue Theorem e az z = πi[res(f,i)+res(f, i)]. +z C(,) Since + has zeros of orer at
Διαβάστε περισσότερα2. THEORY OF EQUATIONS. PREVIOUS EAMCET Bits.
EAMCET-. THEORY OF EQUATIONS PREVIOUS EAMCET Bits. Each of the roots of the equation x 6x + 6x 5= are increased by k so that the new transformed equation does not contain term. Then k =... - 4. - Sol.
Διαβάστε περισσότεραAquinas College. Edexcel Mathematical formulae and statistics tables DO NOT WRITE ON THIS BOOKLET
Aquinas College Edexcel Mathematical formulae and statistics tables DO NOT WRITE ON THIS BOOKLET Pearson Edexcel Level 3 Advanced Subsidiary and Advanced GCE in Mathematics and Further Mathematics Mathematical
Διαβάστε περισσότεραJesse Maassen and Mark Lundstrom Purdue University November 25, 2013
Notes on Average Scattering imes and Hall Factors Jesse Maassen and Mar Lundstrom Purdue University November 5, 13 I. Introduction 1 II. Solution of the BE 1 III. Exercises: Woring out average scattering
Διαβάστε περισσότεραParametrized Surfaces
Parametrized Surfaces Recall from our unit on vector-valued functions at the beginning of the semester that an R 3 -valued function c(t) in one parameter is a mapping of the form c : I R 3 where I is some
Διαβάστε περισσότεραExercises 10. Find a fundamental matrix of the given system of equations. Also find the fundamental matrix Φ(t) satisfying Φ(0) = I. 1.
Exercises 0 More exercises are available in Elementary Differential Equations. If you have a problem to solve any of them, feel free to come to office hour. Problem Find a fundamental matrix of the given
Διαβάστε περισσότεραEQUATIONS OF DEGREE 3 AND 4.
EQUATIONS OF DEGREE AND 4. IAN KIMING Consider the equation. Equations of degree. x + ax 2 + bx + c = 0, with a, b, c R. Substituting y := x + a, we find for y an equation of the form: ( ) y + py + 2q
Διαβάστε περισσότεραMock Exam 7. 1 Hong Kong Educational Publishing Company. Section A 1. Reference: HKDSE Math M Q2 (a) (1 + kx) n 1M + 1A = (1) =
Mock Eam 7 Mock Eam 7 Section A. Reference: HKDSE Math M 0 Q (a) ( + k) n nn ( )( k) + nk ( ) + + nn ( ) k + nk + + + A nk... () nn ( ) k... () From (), k...() n Substituting () into (), nn ( ) n 76n 76n
Διαβάστε περισσότεραΚΥΠΡΙΑΚΗ ΕΤΑΙΡΕΙΑ ΠΛΗΡΟΦΟΡΙΚΗΣ CYPRUS COMPUTER SOCIETY ΠΑΓΚΥΠΡΙΟΣ ΜΑΘΗΤΙΚΟΣ ΔΙΑΓΩΝΙΣΜΟΣ ΠΛΗΡΟΦΟΡΙΚΗΣ 19/5/2007
Οδηγίες: Να απαντηθούν όλες οι ερωτήσεις. Αν κάπου κάνετε κάποιες υποθέσεις να αναφερθούν στη σχετική ερώτηση. Όλα τα αρχεία που αναφέρονται στα προβλήματα βρίσκονται στον ίδιο φάκελο με το εκτελέσιμο
Διαβάστε περισσότεραProblem Set 3: Solutions
CMPSCI 69GG Applied Information Theory Fall 006 Problem Set 3: Solutions. [Cover and Thomas 7.] a Define the following notation, C I p xx; Y max X; Y C I p xx; Ỹ max I X; Ỹ We would like to show that C
Διαβάστε περισσότεραHomework 8 Model Solution Section
MATH 004 Homework Solution Homework 8 Model Solution Section 14.5 14.6. 14.5. Use the Chain Rule to find dz where z cosx + 4y), x 5t 4, y 1 t. dz dx + dy y sinx + 4y)0t + 4) sinx + 4y) 1t ) 0t + 4t ) sinx
Διαβάστε περισσότεραFractional Colorings and Zykov Products of graphs
Fractional Colorings and Zykov Products of graphs Who? Nichole Schimanski When? July 27, 2011 Graphs A graph, G, consists of a vertex set, V (G), and an edge set, E(G). V (G) is any finite set E(G) is
Διαβάστε περισσότεραF-TF Sum and Difference angle
F-TF Sum and Difference angle formulas Alignments to Content Standards: F-TF.C.9 Task In this task, you will show how all of the sum and difference angle formulas can be derived from a single formula when
Διαβάστε περισσότεραFigure A.2: MPC and MPCP Age Profiles (estimating ρ, ρ = 2, φ = 0.03)..
Supplemental Material (not for publication) Persistent vs. Permanent Income Shocks in the Buffer-Stock Model Jeppe Druedahl Thomas H. Jørgensen May, A Additional Figures and Tables Figure A.: Wealth and
Διαβάστε περισσότεραSolutions to the Schrodinger equation atomic orbitals. Ψ 1 s Ψ 2 s Ψ 2 px Ψ 2 py Ψ 2 pz
Solutions to the Schrodinger equation atomic orbitals Ψ 1 s Ψ 2 s Ψ 2 px Ψ 2 py Ψ 2 pz ybridization Valence Bond Approach to bonding sp 3 (Ψ 2 s + Ψ 2 px + Ψ 2 py + Ψ 2 pz) sp 2 (Ψ 2 s + Ψ 2 px + Ψ 2 py)
Διαβάστε περισσότερα