Re: simultaneous equation
- To: mathgroup at smc.vnet.net
- Subject: [mg128766] Re: simultaneous equation
- From: Alexei Boulbitch <Alexei.Boulbitch at iee.lu>
- Date: Sat, 24 Nov 2012 02:27:26 -0500 (EST)
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Hi: When I solve a simultaneous equation, the output looks as follows. How do I remove those ?1.? and the annoying wiggles in the superscript position in so many places. With regards trichy Here is the whole thing. It's a small program... When I ran it anew, the superscript-wiggles are gone but the "1." persists. Thanks for the help. trichy In[1]: x1 = 0.5 + (1/(4 t1)) (rsm - rsk); x2 = 0.5 + (1/(4 t2)) (rLm - rLk); x3 = 0.5 + (1/4) (rLk - n rsk); x4 = 0.5 + (1/4) (rLm - n rsm); ssk = a b x1 + (1 - a) g x3; sLk = a (1 - b) x2 + (1 - a) g (1 - x3); ssm = a b (1 - x1) + (1 - a) (1 - g) x4; sLm = a (1 - b) (1 - x2) + (1 - a) (1 - g) (1 - x4); Pik = ssk rsk + sLk rLk; Pim = ssm rsm + sLm rLm; Pidsm = D[Pim, rsm]; PidLm = D[Pim, rLm]; Solve[Pidsm == 0 && PidLm == 0, {rsm, rLm}] Out[1]: {{rsm -> -((0.5 a b + 0.5 (1. - 1. a) (1. - 1. g) + (0.25 a b rsk)/ t1) (-0.5 (1. - 1. a) (1. - 1. g) - (0.5 a (1. - 1. b))/ t2) - 1. (0.25 (1. - 1. a) (1. - 1. g) + 0.25 (1. - 1. a) (1. - 1. g) n) (0.5 a (1. - 1. b) + 0.5 (1. - 1. a) (1. - 1. g) + (0.25 a (1. - 1. b) rLk)/ t2))/(-1. (0.25 (1. - 1. a) (1. - 1. g) + 0.25 (1. - 1. a) (1. - 1. g) n)^2 + (-0.5 (1. - 1. a) (1. - 1. g) n - (0.5 a b)/t1) (-0.5 (1. - 1. a) (1. - 1. g) - ( 0.5 a (1. - 1. b))/t2)), rLm -> (1. (0.25 (1. - 1. a) (1. - 1. g) + 0.25 (1. - 1. a) (1. - 1. g) n) ((0.5 a b + 0.5 (1. - 1. a) (1. - 1. g) + (0.25 a b rsk)/ t1) (-0.5 (1. - 1. a) (1. - 1. g) - (0.5 a (1. - 1. b))/ t2) - 1. (0.25 (1. - 1. a) (1. - 1. g) + 0.25 (1. - 1. a) (1. - 1. g) n) (0.5 a (1. - 1. b) + 0.5 (1. - 1. a) (1. - 1. g) + (0.25 a (1. - 1. b) rLk)/ t2)))/((-1. (0.25 (1. - 1. a) (1. - 1. g) + 0.25 (1. - 1. a) (1. - 1. g) n)^2 + (-0.5 (1. - 1. a) (1. - 1. g) n - (0.5 a b)/ t1) (-0.5 (1. - 1. a) (1. - 1. g) - (0.5 a (1. - 1. b))/ t2)) (-0.5 (1. - 1. a) (1. - 1. g) - (0.5 a (1. - 1. b))/ t2)) - (0.5 a (1. - 1. b) + 0.5 (1. - 1. a) (1. - 1. g) + ( 0.25 a (1. - 1. b) rLk)/t2)/(-0.5 (1. - 1. a) (1. - 1. g) - ( 0.5 a (1. - 1. b))/t2)}} Hi, is this better? x1 = 1/2 + (1/(4 t1)) (rsm - rsk); x2 = 1/2 + (1/(4 t2)) (rLm - rLk); x3 = 1/2 + (1/4) (rLk - n rsk); x4 = 1/2 + (1/4) (rLm - n rsm); ssk = a b x1 + (1 - a) g x3; sLk = a (1 - b) x2 + (1 - a) g (1 - x3); ssm = a b (1 - x1) + (1 - a) (1 - g) x4; sLm = a (1 - b) (1 - x2) + (1 - a) (1 - g) (1 - x4); Pik = ssk rsk + sLk rLk; Pim = ssm rsm + sLm rLm; Pidsm = D[Pim, rsm]; PidLm = D[Pim, rLm]; Solve[Pidsm == 0 && PidLm == 0, {rsm, rLm}] {{rsm -> -(((a b)/2 + 1/2 (1 - a) (1 - g) + (a b rsk)/( 4 t1)) (-(1/2) (1 - a) (1 - g) - (a (1 - b))/( 2 t2)) - (1/4 (1 - a) (1 - g) + 1/4 (1 - a) (1 - g) n) (1/2 a (1 - b) + 1/2 (1 - a) (1 - g) + (a (1 - b) rLk)/( 4 t2)))/(-(1/4 (1 - a) (1 - g) + 1/4 (1 - a) (1 - g) n)^2 + (-(1/2) (1 - a) (1 - g) n - ( a b)/(2 t1)) (-(1/2) (1 - a) (1 - g) - (a (1 - b))/(2 t2))), rLm -> ((1/4 (1 - a) (1 - g) + 1/4 (1 - a) (1 - g) n) (((a b)/2 + 1/2 (1 - a) (1 - g) + ( a b rsk)/(4 t1)) (-(1/2) (1 - a) (1 - g) - (a (1 - b))/( 2 t2)) - (1/4 (1 - a) (1 - g) + 1/4 (1 - a) (1 - g) n) (1/2 a (1 - b) + 1/2 (1 - a) (1 - g) + (a (1 - b) rLk)/( 4 t2))))/((-(1/4 (1 - a) (1 - g) + 1/4 (1 - a) (1 - g) n)^2 + (-(1/2) (1 - a) (1 - g) n - ( a b)/(2 t1)) (-(1/2) (1 - a) (1 - g) - (a (1 - b))/( 2 t2))) (-(1/2) (1 - a) (1 - g) - (a (1 - b))/( 2 t2))) + (-a rLk + a b rLk - 2 t2 + 2 a b t2 + 2 g t2 - 2 a g t2)/(2 (-a + a b - t2 + a t2 + g t2 - a g t2))}} Have fun, Alexei Alexei BOULBITCH, Dr., habil. IEE S.A. ZAE Weiergewan, 11, rue Edmond Reuter, L-5326 Contern, LUXEMBOURG Office phone : +352-2454-2566 Office fax: +352-2454-3566 mobile phone: +49 151 52 40 66 44 e-mail: alexei.boulbitch at iee.lu