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I need to divide a Gaussian Mixture by it's widest component. When I do this, the exponents of the output end up a mess of terms in need of simplification, but Simplify[] doesn't do it. How can I make this work?

gauMix[x_, means_, vars_] := 
  (1/Length[vars])*Total[(E^-(((x -means)^2)/(2*vars)))/Sqrt[2*Pi*vars]];
means = {-2, 2, 5};
vars = {1, 2, 2};
widest = Flatten[Position[vars, _?(# == Max[vars] &)]];
h[x_, v_] := 
  gauMix[x, means, vars + v]/gauMix[x, {Mean[means[[widest]]]},{vars[[Min[widest]]] + v}];
Expand[h[x, v]]

$$\frac{\sqrt{v+2} \exp \left(\frac{\left(x-\frac{7}{2}\right)^2}{2 (v+2)}-\frac{(x+2)^2}{2 (v+1)}\right)}{3 \sqrt{v+1}}+\frac{1}{3} \exp \left(\frac{\left(x-\frac{7}{2}\right)^2}{2 (v+2)}-\frac{(x-5)^2}{2 (v+2)}\right)+\frac{1}{3} \exp \left(\frac{\left(x-\frac{7}{2}\right)^2}{2 (v+2)}-\frac{(x-2)^2}{2 (v+2)}\right)$$

I would like to see, the exponents individually Simplify[]'d and Together[]'d into something like this:

$$\frac{\sqrt{v+2} \exp \left(\frac{-44 v x+33 v-4 x^2-60 x+17}{8 (v+1) (v+2)}\right)}{3 \sqrt{v+1}}+\frac{1}{3} e^{\frac{33-12 x}{8 v+16}}+\frac{1}{3} e^{\frac{3 (4 x-17)}{8 (v+2)}}$$

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  • $\begingroup$ h[x, v] // Simplify ? $\endgroup$
    – Dr. belisarius
    Commented Feb 21, 2015 at 0:34
  • $\begingroup$ $\frac{1}{3} e^{\frac{(7-2 x)^2}{8 (v+2)}} \left(e^{-\frac{(x-5)^2}{2 (v+2)}}+e^{-\frac{(x-2)^2}{2 (v+2)}}+\frac{e^{-\frac{(x+2)^2}{2 (v+1)}}}{\sqrt{\frac{v+1}{v+2}}}\right)$ $\endgroup$
    – Dr. belisarius
    Commented Feb 21, 2015 at 0:36
  • $\begingroup$ @belisarius No, that's just dividing back out the wide gaussian that I just multiplied in. What I want is for each of those exponents to get individually simplified. For example, in the 2nd and 3rd terms, the x^2 term of the exponent polynomials will cancel out. $\endgroup$
    – Jerry Guern
    Commented Feb 21, 2015 at 0:41
  • $\begingroup$ perhaps you should write down the expected result $\endgroup$
    – Dr. belisarius
    Commented Feb 21, 2015 at 0:46
  • $\begingroup$ @belisarius Okay, I did that. $\endgroup$
    – Jerry Guern
    Commented Feb 21, 2015 at 1:16

2 Answers 2

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Explicitly replacing the exponents with simplifications:

expr /. {Exp[x_] :> Exp[Together@FullSimplify[x]]}

This results in:

$$ \frac{\sqrt{v+2} \exp \left(\frac{-44 v x+33 v-4 x^2-60 x+17}{8 (v+1) (v+2)}\right)}{3 \sqrt{v+1}}+\frac{1}{3} e^{\frac{3 (4 x-17)}{8 (v+2)}}+\frac{1}{3} e^{-\frac{3 (4 x-11)}{8 (v+2)}} $$

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  • $\begingroup$ P.S. A simpler way to find the parameters of the widest gaussian: Mean@MaximalBy[Transpose[{means, vars}], Last]. You can apply Sequence to put it into your function without extracting the parts separately. $\endgroup$
    – 2012rcampion
    Commented Feb 21, 2015 at 2:28
  • $\begingroup$ I didn't really understand you PS comment, but thank you for this answer. $\endgroup$
    – Jerry Guern
    Commented Apr 10, 2016 at 7:38
  • $\begingroup$ Genius and simple trick! How didn't I think of that? Kudos !! $\endgroup$
    – Massimo
    Commented Sep 14, 2021 at 10:48
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$\begingroup$ 
gauMix[x_, means_, vars_] := (1/Length[vars])*
   Total[(E^-(((x - means)^2)/(2*vars)))/Sqrt[2*Pi*vars]];
means = {-2, 2, 5};
vars = {1, 2, 2};
widest = Flatten[Position[vars, _?(# == Max[vars] &)]];
h[x_, v_] := 
  gauMix[x, means, vars + v]/
   gauMix[x, {Mean[means[[widest]]]}, {vars[[Min[widest]]] + v}];
Simplify[Expand[h[x, v]] /. E^(a__ /b__ + c__ /b__) -> E^((a + c)/b)]

$ \frac{1}{3} e^{\frac{8 x^2-44 x+65}{4 v+8}}+\frac{1}{3} e^{\frac{(x-5)^2+\left(x-\frac{7}{2}\right)^2}{v+2}}+\frac{e^{\frac{1}{2} \left(\frac{\left(x-\frac{7}{2}\right)^2}{v+2}-\frac{(x+2)^2}{v+1}\right)}}{3 \sqrt{\frac{v+1}{v+2}}} $

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  • 1
    $\begingroup$ This gives wrong results. The x^2 terms should be cancelling out of the exponents of two of the terms. $\endgroup$
    – Jerry Guern
    Commented Feb 21, 2015 at 1:15

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