Independent and Dependent Variables

... unspliced pgRNA and the SP1 splicing variant of HBV, primers SP1 (tgcccctatcctatcaacac), SP2 (actcccataggaattttccgaaa) and U2 (ttccaatgaggattaaagacag) were used Quantification of the splicing ratio by ... stems being marked by cyan bars and the RNase-accessible joint and loop regions by red bars (D) RNase footprinting analysis of products from the ribonuclease cleavage of the M3 mu...

Ngày tải lên: 28/03/2014, 23:20

... >O and f (2r - 1) ( Tk ) = v(2r- 1) (Zk) Then ~ 00 x 2(r -1 ) sin (2T'kx)dF(x )- f(2r-2) (0)-f(2r-2)(,tk)=2 ( -1 )r- - 00 = V (2r-2)( ) - v(2r-2)('Lk) - Arguing as before, 00 x rdF(x) - Co < cc and ... we use the fact that 1- k (t) = for I t1 < 2T, so that 11 h ( 1- k)ll =11 (f-g)u ( 1- k )11 , for any function u c V with the property that u(t) =11 -it for jtj...

Ngày tải lên: 02/07/2014, 20:20

... a(0)=(a- 1-1 ) +2( 1-a) 02+ b(0), where 00 ( b(~) _ - n )n n-3 {r(1-a)} In (2 35) substitute -z for to give -1 -1 )} Re )_(1-a) -2 r exp{-r(a Y'0 i rz{r(1 - a)} x xexp { -2 4 ' 2- rb[{r(1-a)}-Z~]} e -ir - (2 4.36) ... (2. 4 .29 ) p(x ;a, -1 )=0, and p(x ; a, 1)' {2n (1 -a)}-gal /2( 1-a)x- 1-1 /2( 1 -a) exp {-( 1-a)(a/x)"/( 1- )} x z (1+ a -n /2( 1-a) anx an...

Ngày tải lên: 02/07/2014, 20:20

... that Ih1(t)I SBN , so that (4. 4.15) implies that 00 lI < I BN exp{-c(2ma-2)} - ~Ih (t)I dt=o(1) l (4. 4.16) Combining (4. 4.13), (4 4. 14) , (4 4.16) we obtain (4. 4.11) Since S ... ()a m b n- m = m-na

Ngày tải lên: 02/07/2014, 20:20

... a, and therefore, as x-+ oo, F„(-x) = o(x -s) , G(-x) = o(x -a) , 1- F„ (x) = o (x - a) , 1- G (x) = o (x - a) Taking so that p > S -1 > a -1 , we have IF„(x)-G(x)IP= o(Ixl -Pa) so that (Fn - ... 1

Ngày tải lên: 02/07/2014, 20:20

... formula to (6. 1 5), we obtain the following local limit theorem if xm =o(n) as n-+oo, then b (m, n , p) - (2rcnpq) ' exp { - ixm - ( xm) } , (6 6) where ~(x) = nPq v=3 ~ xv Pv- 1-( -q)v-1 v(v-1) (npq) ... DEVIATIONS Chap Theorem 6. 1 If x>,0 and x = o (n ) as n + oo, then P(Z„>x) - exp x3 - 7x n2 \n ) 1-G(x) 1+O C (6. 1.9) x+l~ n2 and P (Zn < - x) G(-x) x3 = e...

Ngày tải lên: 02/07/2014, 20:20

... (- cn2 z) = exp (- cn2 Re z) < , (7. 2.15) I and therefore = O(exp(-nr7 (E ))) + (7. 2.16) JL on L2 and L4 Moreover, lexp (- un- z) I = exp (- cn - Re z) < , 1 (7 2.15) and therefore = O(exp(-ni ... (27r/nK"(z o))2 (1 +Bn -1 log8 n) Thus the first term on the right-hand side of (7. 2.25) is equal to a( 2itK"(zo)) -I- exp In (K(zo) - aTzo)}(1+Bn -1 log8 n) ,...

Ngày tải lên: 02/07/2014, 20:20

... R-1 f V(x-z)e-hwdWn(z) _ §~ n +1 x z J - 00 dWn(z)e-hz J - 00 Making the substitution ~ =1 1- z, R "+1 x f- Go Jf dW" (z) e -hz J 00 x =Rn+1 (8 2 _8) e-h(''-z)dn e -hsdV(~) (( 8) becomes V (h-Z) ... exp(-hdn-Iy)dFP(y)=(2n )- exp(-hin#y-iy )dyJ0 -Q n (0)+h6n* Jo e xp(-hdnly)Qn (y)dy (8 11) The last integral in (8. 3 11) is o Bn - hQnz exp (-hin y) dy = Bn - , (8. 3 12)...

Ngày tải lên: 02/07/2014, 20:20

... suitable so and ltl < so , (9 9) implies that 10(t)1 < 1- 4t2 (9 3.11) If we write so + < a (9. 3 12) and °1 - 2-a1 (9 3.13) - s0+3' then (9 11) shows that, for n - "`' < ~(t)1" < (1-n-2"`')"= B ... m and writing ,/, tr = r=so+3 Yr r -, , E r Kso+3 we have, from (9 4.5), p© < n - 41, 185 (9 14) (x) = n J -2 7r (9 2) and (9. 4.18), exp(-Znt +nK s0+3 (t)-...

Ngày tải lên: 02/07/2014, 20:20

... have (cf (9 5.2)) pn (x) n-u n2 _ - 7r exp (-4 nt +nKm (t)-n+itx)dt+B exp(-E2 n2 a) f-n-P (10 10) „ Application of the method of steepest descents The integrand of (10 10) is an entire function, ... ( 10. 4.11) = Bn a -2 Moreover, since z= ±in Re [n(?z 2- •rz)] =2n(~2-n-2 -2 •t ~) < in' u= - n2« , (10. 4.12) since ~ < n - '`/ p (n) and i < n - °/ p (n) (10...

Ngày tải lên: 02/07/2014, 20:20

... '"2du=Bx-1 eXp ( - x 2) 11 - ES (11 11 11) x In view of this, (11 11 9) gives P(n - ZS n >x) >, (I +o(1))(2rc )-2 e -2 " du (11 11 12) fx If in (11 11 7) we replace n a -2 by -n 2" -2 and reverse ... under (11 11 2), P(n S©+n -2 00 Y©>x) e - -j "z du 2m -l ( )z fx (11 11 8) Then, from (11 11 6) and (11 11 7), P(n -2 S©>x) (1+0(1))~ ~ e -2 "...

Ngày tải lên: 02/07/2014, 20:20

... jIzI>n~' -e Fn (x- z)g l (z) = B exp(-Znl-2E2) If x l

Ngày tải lên: 02/07/2014, 20:20

... give - J0 e -hvn 1/zv dFn (v) = (27r )-' x- (1 + 9C 11 h) (13 4.2) Since r=n -2 x, the substitution of (13 4.2) into (13 4.1) gives x3 1-Fn (x) _ (2~) -2 x -1 e -2 x exp ~,[2KI x x n2 n2 x (1+Bn -2 )(1+Bh)+6p ... (z, t) d- , l Izn-ina - '/2/P (n) zo+ina - '/2/P1(n) En (Z, t) dZ 12 = - 2761 in" - '/z/p j (n) zo - in a- '/2/P1(n) E n (z, t) dz , I3...

Ngày tải lên: 02/07/2014, 20:20