Re: Re: Write an expression in a specific form
- To: mathgroup at smc.vnet.net
- Subject: [mg108111] Re: [mg108025] Re: Write an expression in a specific form
- From: DrMajorBob <btreat1 at austin.rr.com>
- Date: Mon, 8 Mar 2010 06:16:46 -0500 (EST)
- References: <hmo293$qbh$1@smc.vnet.net> <201003050935.EAA29559@smc.vnet.net>
- Reply-to: drmajorbob at yahoo.com
Why shouldn't I integrate the product, if it gives me the unknown multiplier? Also see my other post, where Integrate gives us c, m, AND s with no (visible) algebra. Bobby On Sun, 07 Mar 2010 23:50:09 -0600, dh <dh at metrohm.com> wrote: > Hi Bob, > you were not supposed to integrate the product of two Gaussian. But the > product of two Gaussian is another Gaussian whose parameters are asked. > Daniel > > On 05.03.2010 13:14, DrMajorBob wrote: >> I missed where c4 is defined in that... but starting out, we know it has >> to be equal to >> >> G[m_,s_,x_]:=(1/(s*Sqrt[2*Pi]))*Exp[-(x-m)^2/(2*s^2)]; >> Integrate[G[m1,s1,x] >> G[m2,s2,x],{x,-Infinity,Infinity},Assumptions->{s1>0,s2>0}] >> >> E^(-((m1-m2)^2/(2 (s1^2+s2^2))))/(Sqrt[2 \[Pi]] Sqrt[s1^2+s2^2]) >> >> No? >> >> I think c4 was supposed to be c3, and sure enough, retrieving the RHS of >> the rule for c3, I have: >> >> (E^(-((m1-m2)^2/(2 (s1^2+s2^2)))) Sqrt[(s1^2 s2^2)/(s1^2+s2^2)])/(Sqrt[2 >> \[Pi]] s1 s2)//PowerExpand >> >> E^(-((m1-m2)^2/(2 (s1^2+s2^2))))/(Sqrt[2 \[Pi]] Sqrt[s1^2+s2^2]) >> >> Bobby >> >> On Fri, 05 Mar 2010 03:35:13 -0600, dh <dh at metrohm.com> wrote: >> >>> Hi Ares, >>> Mathematica really seems to have difficulties (that is I do not have >>> the nerves >>> to wait so long) with a calculation that can done by hand. It look like >>> it needs some human help. >>> We may get an equation for the exponents with only 2 unknowns. >>> We denot by m1,s1 and m2,s2 the parameters of the given Gaussian. m3,s3 >>> and c3 denote the searched Gaussian: c4 G[m3,s3]. tmp is an >>> intermediate >>> result we are not interested in: >>> sol1 = Reduce[{ForAll[ >>> x, (x - m1)^2/(2 s1^2) + (x - m2)^2/(2 s2^2) == >>> tmp + (x - m3)^2/(2 s3^2)], s1 > 0, s2 > 0, s3 > 0}, {m3, s3, tmp}, >>> Reals] >>> this gives us m3,s3 and an intermediate result tmp. With this we may >>> solve the equation for the pre factor. Toward this aim we change sol1 >>> to >>> rules and use it in the equation: >>> G[m1, s1, x] G[m2, s2, x] == c3 G[m3, s3, x] /. sol1 /. sol1 >>> Note that in sol1 tmp is given in term of m3,s3. We must therefore >>> apply >>> /.sol1 twice. >>> I also delete some superfluous info from the result. Here is the whole >>> code: >>> ================ >>> G[m_, s_, x_] := (1/(s*Sqrt[2*Pi]))*Exp[-(x - m)^2/(2*s^2)]; >>> sol1 = Reduce[{ForAll[ >>> x, -((x - m1)^2/(2 s1^2)) - (x - m2)^2/(2 s2^2) == >>> tmp - (x - m3)^2/(2 s3^2)], s1 > 0, s2 > 0, s3 > 0}, {m3, s3, >>> tmp}, Reals]; >>> sol1 = Drop[sol1, 2] // ToRules >>> sol2 = Reduce[{G[m1, s1, x] G[m2, s2, x] == c3 G[m3, s3, x] /. >>> sol1 /. sol1, s1 > 0, s2 > 0, s3 > 0, c3 > 0}, {c3}, >>> Reals][[4]] // ToRules >>> =============== >>> >>> Finally we may test if the calculation is correct: >>> G[m1, s1, x] G[m2, s2, x] == c4 G[m3, s3, x] /. sol2 /. sol1 /. >>> sol1 // Simplify >>> >>> Daniel >>> >>> On 04.03.2010 11:33, Ares Lagae wrote: >>>> Hi all, >>>> >>>> I am a beginner in Mathematica, and I have the following "problem": >>>> How can >>>> I write an expression in a specific form? >>>> >>>> For example: >>>> >>>> - Define a Gaussian: >>>> >>>> G[m_, s_, x_] := (1/(s*Sqrt[2*Pi]))*Exp[-(x - m)^2/(2*s^2)]; >>>> >>>> - Product of two Gaussians: >>>> >>>> G[m1, s1, x] * G[m2, s2, x] >>>> >>>> - How can I get Mathematica to write the result in terms of c * G[m_, >>>> s_, >>>> x_]? I.e., get the values for c, m and s. >>>> >>>> Thanks, >>>> >>>> Ares Lagae >>>> >>>> >>> >>> >> >> > > -- DrMajorBob at yahoo.com
- References:
- Re: Write an expression in a specific form
- From: dh <dh@metrohm.com>
- Re: Write an expression in a specific form