By Dieudonne J. A.
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Additional resources for Treatise on Analysis, Vol. 2 (1970)(en)(422s)
R ) (2-75) where k = n - m. We discard the root j> = 0 because it corresponds to the first s roots. 72) as ny - x e n 1 0 = my - x e' v m 1 +a / 0 m . y l m - + •• • 1 Hence, equating the coefficients of e on both sides yields v ny ~ x n l =my - x +a - y ~ m l 0 Q m l m x Hence, FT _ m-\ a ~y~ 0 n n \ y r^ 1 — -my" ' x a _ m-l m-l ny ' 1 1 N RT _ ,~ a -m"n-m n m 7 ^x ' K } Therefore, the last n - m roots are given by x = ( ± e + JlLLL . . 75). 4. Transcendental Equations We consider the roots of Bessel's function Jo(x) for large x , that is, we consider the roots of the transcendental equation /<>(*) = 0 when x is large.
50) (3x1 ~ 12x 0 + l l ) x , = x2> - 2x Q whose solution is xl - 2x X, = — : 3X 1 When x 0 2 ,- , . 0 n (2-51) 12*0 + 11 . 51) that X , = - ± . 51) that X , = 0. 51) that x , = |. Hence, the third root is given by x =3 + | e + • • • Thus, in this case all roots go in powers of e. 52) gives (x 0 + « ! 56) that x , = 8. 56) that x form of the expansion is wrong. 58) v>0 and determine v in the course of analysis. 59) balance each other, 2 v must be equal to 1 or v - \, and x , = ±y/Ji. 58) that the second and third roots are given by x = 1±e 1 / 2 VJi + - •• This example illustrates the fact that difficulties arise whenever the assumed form of the expansion is not correct.
T T ( 1 . , are nonuniform) reduced equation tend to °°. In fact, r need not be exactly equal to 1 for the above expansions to break down. The expansions break down whenever the first-order term, second-order term, and so on are the order of the zeroth-order term, because the corrections to the zeroth-order term will not be small, contrary to the assumption underlying the method. , region of nonuniformity),'we determine the conditions under which successive terms are the same order. 23) that the zeroth- and first-order terms are the same order when =0(1) or l - r = 0(e) 1- r whereas the first- and second-order terms are the same order when ^2 1 - r O (1 - rf or (1 - r ) 2 = 0(e) or 1 - T= 0(€ ' ) 1 2 Since for small e, e is bigger than e, the region of nonuniformity is 1 - T = 0 ( e ^ ) , the larger of the above two regions.
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