Re: Forcing Trig Identities
- To: mathgroup at smc.vnet.net
- Subject: [mg94640] Re: Forcing Trig Identities
- From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
- Date: Thu, 18 Dec 2008 07:22:57 -0500 (EST)
- Organization: Uni Leipzig
- References: <giao6o$8qt$1@smc.vnet.net>
- Reply-to: kuska at informatik.uni-leipzig.de
Hi,
Mathematica 7. gives
pmn[\[Rho]_, x_, z_] :=
PDF[MultinormalDistribution[{0, 0}, {{1, \[Rho]}, {\[Rho], 1}}], {x,
z}]
res = Integrate[
pmn[\[Rho], x, z], {x, -Infinity, 0}, {z, -Infinity, 0},
Assumptions -> {\[Rho]^2 < 1, \[Rho] > 0}];
FunctionExpand[res] //
FullSimplify[#, Assumptions -> {\[Rho]^2 < 1, \[Rho] > 0}] &
(Pi + 2*ArcSin[\[Rho]])/(4*Pi)
Regards
Jens
Gary McClelland wrote:
> In earlier versions of Mathematica (I think 5 and before), the following code:
>
> pmn[\[Rho]_ , x_, z_] :=
> PDF[MultinormalDistribution[{0, 0}, {{1, \[Rho]}, {\[Rho], 1}}], {x,
> z}]
> Integrate[pmn[\[Rho], x, z], {x, -Infinity, 0}, {z, -Infinity, 0},
> Assumptions -> {\[Rho] ^2 < 1, \[Rho] > 0}]
>
> yielded:
> (\[Pi] + 2 ArcSin[\[Rho]])/(4 \[Pi])
>
> now (Mathematica 7), I get the mathematically equivalent but less desirable:
> (\[Pi] - ArcTan[Sqrt[-1 + 1/\[Rho]^2]])/(2 \[Pi])
>
> TrigReduce and FullSimplify fail to simplify what Mathematica 7 gives to the simpler expression Mathematica 5 yielded. any suggestions on how to force trig identities?
>