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I'm trying to simplify the following expression $$\frac{n.r \ p.q - (n.p + n.q) (p.r + q.r)}{(n.p + n.q)\ p.q}\to \frac{n.r}{n.p+p.q}-\frac{p.r+q.r}{p.q}$$

When I put it into Mathematica and use FullSimplify nothing happens though. For me it would be "simpler" to express this as a sum of "small" fractions.

I've tried using Apart but that also does nothing. Could anyone recommend a way for me to force Mathematica to simplify an expression like this?

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    $\begingroup$ you have some mismatched parenthesis.. also it may help if you show what the expected simple form should be. (for the record I didnt't downvote..) $\endgroup$
    – george2079
    May 23, 2014 at 20:23
  • $\begingroup$ I've added a simpler and clearer example, together with the expected simplification. Although this one is almost trivial to spot, it gets annoying in longer formulae! $\endgroup$
    – Edward Hughes
    May 23, 2014 at 21:05
  • $\begingroup$ Also could the downvoter explain how I can improve my post? I'm fairly active on other .SE sites, and as far as I'm aware it's poor etiquette to downvote without saying why! $\endgroup$
    – Edward Hughes
    May 23, 2014 at 21:06
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    $\begingroup$ Could you please post the actual MMA code? Also, not sure if my browser rendering is off, but your TeX looks strange. $\endgroup$
    – Yves Klett
    May 23, 2014 at 21:15
  • $\begingroup$ Always post actual Mathematica code/expressions in the form: "this is my input", "this is my expected output". It's not clear what all the dots mean in your expression and how you typed this into Mathematica. $\endgroup$
    – Szabolcs
    May 23, 2014 at 21:36

1 Answer 1

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let n=a[1] , r=a[2] , p=a[3] , q=a[4]:

term = (a[1]*a[2]*a[3]*
a[4] - (a[1]*a[3] + a[1]*a[4])*(a[3]*a[2] + a[4]*a[2]))/((a[1]*
a[3] + a[1]*a[4])*a[3]*a[4]);
Expand[term]
Apart[term]

Result is (Expand): $$-\frac{a(1) a(2)}{a(1) a(3)+a(1) a(4)}-\frac{a(1) a(4) a(2)}{a(3) (a(1) a(3)+a(1) a(4))}-\frac{a(1) a(3) a(2)}{a(4) (a(1) a(3)+a(1) a(4))}$$

Apart:

$$-\frac{a(2)}{a(3)}-\frac{a(2)}{a(4)}+\frac{a(2)}{a(3)+a(4)}$$

you can use Rules to replace your own parameters.

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  • $\begingroup$ Thank you - do you know why it doesn't just work directly with my variables? $\endgroup$
    – Edward Hughes
    May 23, 2014 at 21:40
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    $\begingroup$ i just don't!sometimes it's better to use these patterns if you have so many parameters.just a suggestion! $\endgroup$
    – user3127040
    May 23, 2014 at 21:44

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