       Re: Re: Re: Simplifying algebraic expressions

```On 6 Jun 2006, at 11:46, Carl K. Woll wrote:

> Andrzej Kozlowski wrote:
>> Here is, I think, an optimized version of the "simplification" I
>> sent  earlier:
>> rule1 = {2x -> u, 3y -> v}; rule2 = Map[Reverse, rule1];
>> Simplify[TrigFactor[Simplify[ExpToTrig[
>>       Simplify[ExpToTrig[(-1)^(2*x + 3*y) /. rule1],
>>         Mod[u, 2] == 0] /. rule2], y â?? Integers]],
>>    y â?? Integers]
>> (-1)^y
>> "Optimized" means that I can't see any obvious way to make this
>> simpler.
>> Unlike other answers to the original post, this works in
>> Mathematica  5.1 and 5.2, and involves only reversible operations.
>> In other words,  it constitutes a proof. On the other hand,
>> obviously, it would be  ridiculous to use this approach in
>> practice: there must be a simple  transformation rule (or rules)
>> the development version.
>
> Andrzej,
>
> It seems to me that your simplification approach isn't as general
> as the replacement method I advocated (a variant of which is given
> below), as rule1 and rule2 are specific to the input. Also, my
> replacement method is based on the fact that:
>
> In:= Simplify[((-1)^a)^b == (-1)^(a b), Element[{a, b}, Integers]]
>
> Out= True
>
> So, I think my approach is generally valid.

You are right, I apologise. I did not look carefully at your code.
But now that I see some challenge ;-) I also see  that I can
completley remove rule1 and rule 2 form my code and make it equally
general:

Simplify[TrigFactor[Simplify[ExpToTrig[
Simplify[ExpToTrig[(-1)^PolynomialMod[2*x + 3*y,
2]], Mod[u, 2] == 0]], y â?? Integers]],
y â?? Integers]

(-1)^y

What's more, I can actually produce an even shorter code that will
produce the required simplification:

f[(-1)^(n_)] := (-1)^PolynomialMod[n, 2]

Simplify[f[(-1)^(2*x + 3*y)], TransformationFunctions ->
{Automatic, f}]

(-1)^y

Of course this ought to be modified so as to be performed only under
the assumption that x and y are integers (not very hard to do).  I
suspect that this is close to what has been done in the development
version.

Andrzej Kozlowski

>
>> Note also the following problem, which I suspect is related:
>> Simplify[(-1)^(u + v), (u | v) â?? Integers &&
>>     Mod[u, 2] == 0 && Mod[v, 2] == 1]
>> -1
>> Simplify[(-1)^(u + v), (u | v) â?? Integers &&
>>     Mod[u, 2] == 0 && Mod[v, 2] == 0]
>> 1
>> But
>> Simplify[(-1)^(u + v), (u | v) â?? Integers &&
>>     Mod[u, 2] == 0]
>> (-1)^(u + v)
>> I am sure this also gives (-1)^v in the development version.
>
> To obtain the desired simplification in 5.2, we can modify my
> previous approach as follows:
>
> power[e_, a_Plus] := Times @@ (Simplify /@ (power[e, #1]&) /@ List
> @@ a)
> power[e_, a_Integer b_?(Refine[Element[#,Integers]]&)] := (e^a)^b
> power[e_, a_] := e^a
>
> simp[expr_, assum_] := Block[{\$Assumptions = assum},
>   Simplify[expr /. Power -> power]
> ]
>
> In the above I rely on the following two equivalences, verified by
> Mathematica:
>
> In:=
> Simplify[e^a e^b == e^(a+b)]
>
> Out=
> True
>
> In:=
> Simplify[(e^a)^b == e^(a b), Element[{a,b},Integers]]
>
> Out=
> True
>
> The tricky part is that Mathematica automatically converts e^a e^b
> --> e^(a+b), and as you point out above Simplify[(-1)^(u+v),...]
> doesn't work. However, Simplify[(-1)^u, ...] does work, which is
> why the Simplify appears in the definition of power[e_,a_Plus]
> before Times is applied.
>
> Here are some examples:
>
> In:=
> simp[(-1)^(u+v),Mod[u,2]==0]
>
> Out=
>     v
> (-1)
>
> In:=
> simp[(-2)^(u+v),Mod[u,2]==0]
>
> Out=
>     v  u + v
> (-1)  2
>
> In:=
> simp[(-1)^(2x+3y),Element[{x,y},Integers]]
>
> Out=
>     y
> (-1)
>
> In:=
> simp[(-2)^(2x+3y),Element[{x,y},Integers]]
>
> Out=
>     y  2 x + 3 y
> (-1)  2
>
> Carl Woll
> Wolfram Research
>
>> Andrzej Kozlowski
>> On 2 Jun 2006, at 17:09, Andrzej Kozlowski wrote:
>>> On 1 Jun 2006, at 19:54, Amitabha Roy wrote:
>>>
>>>
>>>> Hello:
>>>>
>>>> I would like Mathematica to be able to take an expression, say,
>>>>
>>>> (-1)^{2 x  + 3 y} and be able to simplify to (-1)^y.
>>>>
>>>> Is there a way one can do this ?
>>>>
>>>> Thanks
>>>>
>>>
>>> Note that you are using {} instead of (). Different kind of brackets
>>> have completely different meaning in Mathematica.
>>> I have found it amazingly hard to force Mathematica to perform this
>>> simplification. The best I could do is this. We need two rules.
>>> rule1
>>> will replace 2x by u and 3y by v. rule 2 does the opposite: it
>>> replaces u by 2x and v by 2y.
>>>
>>> rule1 = {2x -> u, 3y -> v};rule2 = Map[Reverse, rule1];
>>>
>>> Now:
>>>
>>>
>>> Simplify[TrigFactor[
>>>    FullSimplify[ExpToTrig[
>>>      FullSimplify[
>>>        ComplexExpand[
>>>         (-1)^(2*x + 3*y) /.
>>>          rule1], Mod[u, 2] ==
>>>         0] /. rule2],
>>>     y â?? Integers]],
>>>   y â?? Integers]
>>>
>>>
>>> (-1)^y
>>>
>>> Uff... Surely, this ought to be easier...
>>>
>>> Andrzej Kozlowski
>>> Tokyo, Japan
>>>
>

```

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