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Re: simplify a trig expression
*To*: mathgroup at smc.vnet.net
*Subject*: [mg65438] Re: [mg65415] simplify a trig expression
*From*: Adam Strzebonski <adams at wolfram.com>
*Date*: Sat, 1 Apr 2006 05:38:57 -0500 (EST)
*References*: <200603311109.GAA15029@smc.vnet.net> <E2726B55-498F-412D-871F-5B1CEE7D1175@mimuw.edu.pl>
*Reply-to*: adams at wolfram.com
*Sender*: owner-wri-mathgroup at wolfram.com
Andrzej Kozlowski wrote:
> This is one of those cases where FullSimplify will not work because it
> lacks a suitable transformation function. In this particular case the
> transformation function is of the form:
>
> f[n_*Log[a_]] := Log[a^n]
>
> Of course this is only valid with various assumptions on n and a, but I
> won't bother with this here.
Indeed, the two expressions are not equal for all values of x.
In[1]:= {2 Log[Cos[x/2] + Sin[x/2]], Log[Sin[x] + 1]} /. x->5Pi/2
Log[2]
Out[1]= {2 (I Pi + ------), Log[2]}
2
If we assume that 2 Log[Cos[x/2] + Sin[x/2]] is real we can use
the following transformation to simplify it.
In[2]:= g[x_]:=Log[TrigReduce[E^x]]
In[3]:= g[2 Log[Cos[x/2] + Sin[x/2]]]
Out[3]= Log[1 + Sin[x]]
> Anyway, observe that:
>
>
> FullSimplify[Integrate[Cos[x]/(Sin[x] + 1), x],
> TransformationFunctions -> {Automatic, f}]
>
>
> Log[Sin[x] + 1]
>
> Note also that Simplify will not work even when you add f.
This is because f itself does not make the expression simpler.
Simplify does not use TrigReduce, FullSimplify does.
Best Regards,
Adam Strzebonski
Wolfram Research
>
> I am not sure if there are good reasons for adding a version of f
> (taking account of suitable assumptions) to the default transformation
> functions of FullSimplify. It may however be a good idea to have
> another possible value for the option TransformationFunctions besides
> only Automatic and user defined ones. In fact I have suggested in the
> past one or two other useful TransformationFunctions; perhaps it might
> be a good idea to define more and collect them into a single option
> value or maybe several.
>
> Andrzej Kozlowski
>
>
>
>
>
>
> On 31 Mar 2006, at 13:09, Murray Eisenberg wrote:
>
>> A direct substitution (with paper and pencil) gives that the integral of
>> Cos[x]/(Sin[x] + 1) is Log[Sin[x] + 1]. This is valid provided Sin[x]
>> is not -1.
>>
>> Mathematica gives:
>>
>> Integrate[Cos[x]/(Sin[x] + 1), x]
>> 2 Log[Cos[x/2] + Sin[x/2]]
>>
>> Is there some simple way to coerce the latter Mathematica-supplied
>> result into the paper-and-pencil answer?
>>
>> The closest I could get is:
>>
>> Log[TrigExpand[Expand[(Cos[x/2] + Sin[x/2])^2]]] /.
>> {Sin[x/2] -> Sqrt[(1 - Cos[x])/2],
>> Cos[x/2] -> Sqrt[(1 + Cos[x])/2]}
>> Log[1 + Sqrt[1 - Cos[x]]*Sqrt[1 + Cos[x]]]
>>
>> Am I not seeing some easier TrigExpand or TrigReduce method?
>>
>> --
>> Murray Eisenberg murray at math.umass.edu
>> Mathematics & Statistics Dept.
>> Lederle Graduate Research Tower phone 413 549-1020 (H)
>> University of Massachusetts 413 545-2859 (W)
>> 710 North Pleasant Street fax 413 545-1801
>> Amherst, MA 01003-9305
>>
>
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