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The binomial expansion is of the form

$(x+y)^n = \sum\limits_{i=0}^n {n \choose i} x^i y^{n-i}$.

I'd like to generate the appropriate Mathematica output (so I can then convert it to $\LaTeX$ source through TeXForm[]) showing each term. For instance, for $(x+y)^5$ I would like (after conversion and typesetting):

$1 x^0 y^5 + 5 x^1 y^4 + 10 x^2 y^3 + 10 x^3 y^2 + 5 x^4 y^1 + 1 x^5 y^0$.

The first step in my first attempt is:

myelements = Table[{ToString[Binomial[5, i]], StringForm["\!\(\*SuperscriptBox[\(x\), \(``\)]\)", i], StringForm["\!\(\*SuperscriptBox[\(y\), \(``\)]\)", 5 - i]}, {i, 0, 5}]

giving

{{"1", StringForm["\!\(\*SuperscriptBox[\(x\), \(``\)]\)", 0], StringForm["\!\(\*SuperscriptBox[\(y\), \(``\)]\)", 5]}, {"5", StringForm["\!\(\*SuperscriptBox[\(x\), \(``\)]\)", 1], StringForm[ "\!\(\*SuperscriptBox[\(y\), \(``\)]\)", 4]}, {"10", StringForm[ "\!\(\*SuperscriptBox[\(x\), \(``\)]\)", 2], StringForm[ "\!\(\*SuperscriptBox[\(y\), \(``\)]\)", 3]}, {"10", StringForm[ "\!\(\*SuperscriptBox[\(x\), \(``\)]\)", 3], StringForm[ "\!\(\*SuperscriptBox[\(y\), \(``\)]\)", 2]}, {"5", StringForm[ "\!\(\*SuperscriptBox[\(x\), \(``\)]\)", 4], StringForm[ "\!\(\*SuperscriptBox[\(y\), \(``\)]\)", 1]}, {"1", StringForm[ "\!\(\*SuperscriptBox[\(x\), \(``\)]\)", 5], StringForm[ "\!\(\*SuperscriptBox[\(y\), \(``\)]\)", 0]}}

to get the individual terms, but unfortunately these elements are not easily joined (since they are not all Strings or Text), nor is the separating "+" sign appearing. Then:

Flatten@Riffle[myelements, Table["+", {5}]]

There must be a way using Text or Boxes or whatever that yields a form that can be converted to $\LaTeX$, but I'm seeking the most elegant method.

Any ideas?

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1 Answer 1

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The order is reversed, but you could try:

Sum[With[{b = Binomial[5, i], xexp = i, yexp = 5 - i}, 
  HoldForm[b] HoldForm[x^xexp] HoldForm[ y^yexp]], {i, 0, 5}];
% // TeXForm

$1 x^5 y^0+5 x^4 y^1+10 x^3 y^2+10 x^2 y^3+5 x^1 y^4+1 x^0 y^5$

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  • $\begingroup$ With the trivial fix of xexp = 5-i and yexp = i, this is perfect. Thanks! $\endgroup$ May 5, 2017 at 22:45

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