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We know that $n\ge0$ and $m>0$ and both are integers. We also know that $a,b>0$. I think for any $0\le x\le b$, we should have

$(a(x-b))^{m+n}(-a(x-b))^{-m-n}=(-1)^{m+n}$

But when I try to simplify it in Mathematica, I get

FullSimplify[(a(x-b))^(m+n)*(-a(x-b))^(-m-n)]=((x-b)a)^(m+n)*((-x+b)a)^(-m-n)

which is the exact same input expression. Why is this happening?

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  • $\begingroup$ You need to tell FullSimplify about the ancillary conditions given in the question by means of Assumptions. Check the documentation. $\endgroup$
    – bbgodfrey
    Jan 13 at 14:03

1 Answer 1

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Well, you haven't told FullSimplify about those assumptions, so it cannot make the simplification that would result from them because it would not be correct in general. For instance, your relationship would not hold for complex values of the parameters, which is the default assumption in Mathematica.

If you include them, however, you do get the simplification that you seek:

FullSimplify[
  (a(x - b))^(m + n)*(-a(x - b))^(-m - n),
  Assumptions -> {
     n >= 0, m > 0, 
     a > 0, b > 0, 
     0 <= x <= b
  }
]

(* Out: (-1)^(m + n) *)
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