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I am trying to solve this problem here but have to understand basics. Suppose I have a multiplication such as $a*(b+c+d+...+x)$. How can I multiply it to the result $a*b+a*c+...+a*x$?


[Not working] Reduce[(a + b)*c]

[Not working] Simplify[(a + b)*c] or FullSimplify[(a + b)*c]

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Try Distribute[a * (x + y + z)]. –  b.gatessucks Dec 9 '12 at 12:51
@b.gatessucks plesae, make it into an answer. It worked? It is giving wrong result with "Distribute[(a + b)*(a + b)]" , why? –  hhh Dec 9 '12 at 12:52
Or Expand[a*(b + c)]. –  Mark McClure Dec 9 '12 at 12:56
Your new example is explicitly covered in the documentation. You should read it. –  b.gatessucks Dec 9 '12 at 13:08
There are many ways besides Expand, e.g. PolynomialReduce[a (b + c), 1, c][[1, 1]]. Take a look here : mathematica.stackexchange.com/questions/9111/… –  Artes Dec 9 '12 at 13:35

2 Answers 2

up vote 4 down vote accepted

I know this has been answered in the comments, but just for the sake of having an actual answer..



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While covered by the comments and another answer, I would like to expand on this a bit. The word "simplify" has very different connotations between what you intend and what Mathematica intends. For Mathematica, it means simply the least complex solution by some measure. This defaults to the number of subexpressions and integer digits, but it can be set to whatever you wish in an attempt to coerce Simplify to generate a specific form (or avoid others), such as this example from the docs

f[e_] := 100 Count[e, _ChebyshevT, {0, Infinity}] + LeafCount[e]
FullSimplify[ChebyshevT[n, x], ComplexityFunction -> f]
(* Cos[n ArcCos[x]] *)

where ChebyshevT is being made more expensive than its alternative forms. I would also look at the example for Abs below that. Using this definition, then, the suggestions of Expand make sense as it is not simplest (least complex) form.

Your use of the word simplify is very different. In assigning problems to students, we use the word simplify imprecisely, and what we mean is transform it into a specific form which is not necessarily the simplest in form. I think this is where the confusion lies; this is not what Mathematica means, but it may be convinced that it is correct.

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