Re: a^n*b^n != (a*b)^n
- To: mathgroup@smc.vnet.net
- Subject: [mg12290] Re: [mg12267] a^n*b^n != (a*b)^n
- From: Daniel Lichtblau <danl@wolfram.com>
- Date: Thu, 7 May 1998 18:51:26 -0400
- References: <199805050730.DAA17180@smc.vnet.net.>
Michael Milirud wrote: > > This is not so much about the Mathematica as a software as about > mathematica as a subject. Mathematica just confirmed it and I am REALLY > puzzled on this one. > > I always considered it trivial that a^n*b^n == (a*b)^n when a,b are > complex and n is real. However: > > 6 == Sqrt[36] == Sqrt[-4*-9] == Sqrt[-4]*Sqrt[-9] == 2i*3i == 6*i^2 == > -6 > > Hence 6 == -6 > > ARGHHH!!!! > > After quite some time, I found the problem to be in the step: > > Sqrt[-4*-9] == Sqrt[-4]*Sqrt[-9] > > which as Mathematica claims does NOT equal to each other!!! > > So generally that would mean: a^n*b^n != (a*b)^n > > I tried to go and search for the basic proof of this equality. Obviously > enough I couldn't find any :( > For a, b being real and n being positive integer the equality is > obvious. But for other cases - I don't know how to approach it. > > While playing around with different examples I noticed that the above > equality upholds for all the cases except when we have a and b being > negative REAL numbers and n being p/q with q=2k > > ANYTHING at all will be greatly appriciated, as I am completely stuck!!! > ;( > > Michael More generally it is easy to find examples that fail by taking a and b in the second quadrant (this is not strictly necessary, though). Here is a simple example. Note that I do not use an even root. In[16]:= a = -4+I; b = -3+I; N[(a*b)^(1/3) - a^(1/3)*b^(1/3)] Out[16]= 3.08489 - 2.66504 I We do not even need a rational power. In[17]:= x=Log[2] Out[17]= Log[2] In[18]:= N[(a*b)^x - a^x*b^x] Out[18]= 9.51975 + 2.06854 I The identity to use is a^b == Exp[Log[a]*b]. This uses the principal branch of the logarithm (along the negative real axis) and it DEFINES exponentiation for other bases. Since E is positive you of course have Exp[x]*Exp[y]==Exp[x+y]. But high-school "logarithm laws" such as log(a^b)==b*log(a) etc. do not hold in the complex plane. So identities you obtain will depend on where log laws do indeed hold. Mathematica internally makes some use of (valid) simplifications in "canonicalizing" products of powers. Daniel Lichtblau Wolfram Research
- References:
- a^n*b^n != (a*b)^n
- From: "Michael Milirud" <mmichael@idirect.com>
- a^n*b^n != (a*b)^n