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MathGroup Archive 2000

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Re: Re: ArcCos[]

  • To: mathgroup at smc.vnet.net
  • Subject: [mg24798] Re: [mg24793] Re: [mg24730] ArcCos[]
  • From: Andrzej Kozlowski <andrzej at tuins.ac.jp>
  • Date: Sun, 13 Aug 2000 03:16:38 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

I have had a further exchange of messages with Gianluca about this topic
which lead to conclude that the problem is simpler than I assumed. Below is
an extract from a message I sent to Gianluca:


I think now that actually it would be possible to implement this, (although
it may not be worth the effort). What one would really need to to is to find
a mathematical characterization of the minimal polynomials of algebraic
numbers of the form Cos[r*Pi] for rational r. This should not be too hard.
Then given a radical expression like Sqrt[2 + Sqrt[2]]/2 one simply applies:

In[51]:=
RootReduce[Sqrt[2 + Sqrt[2]]/2]
Out[51]=
             2       4
Root[1 - 8 #1  + 8 #1  & , 4]

and then one "looks" at the minimal polynomial. For example, look at my
other example (Cos[Pi/15]):

In[64]:=
RootReduce[((-1 + Sqrt[5]))/8 + (Sqrt[(3*(5 + Sqrt[5]))/2])/4]
Out[64]=
                     2       3        4
Root[1 - 8 #1 - 16 #1  + 8 #1  + 16 #1  & , 4]

One can see some very clear regularity, suggesting certain conjectures. By
comparison, taking a radical "at random":

In[57]:=
RootReduce[Sqrt[(1 + Sqrt[7])/4]]
Out[57]=
              2       4
Root[-3 - 4 #1  + 8 #1  & , 2]

This clearly looks quite different. So I now believe a rigorous mathematical
solution is possible and very likely well known to number theorists, but
perhaps no sufficiently important for Mathematica developers to bother
about.

Andrzej


on 8/10/00 6:33 AM, Andrzej Kozlowski at andrzej at tuins.ac.jp wrote:

> on 8/9/00 8:31 AM, Gianluca Gorni at gorni at dimi.uniud.it wrote:
> 
>> 
>> Hello!
>> 
>> I have just come across one more example that shows how there is
>> room for improving Mathematica's trig functions.
>> 
>> With Mathematica 4:
>> 
>> v = Cos[ Pi/8 ] // FunctionExpand   gives   Sqrt[2 + Sqrt[2]]/2
>> 
>> Still, neither
>> 
>> ArcCos[v] // FunctionExpand   nor    ArcCos[v] // FullSimplify
>> 
>> give  Pi/8, as I would expect, but just
>> 
>> ArcSec[2/Sqrt[2 + Sqrt[2]]]
>> 
>> %%%%%%%%%%%%%%%
>> 
>> An unrelated problem: the following instructions consistently crash
>> my Mac Mathematica 4 kernel:
>> 
>> a = Root[-t + 2*#1 + 2*t^2*#1 + #1^3 & , 1];
>> b = D[a, t];
>> Solve[b == 0, t]
>> 
>> %%%%%%%%%%%%%%%
>> 
>> Best regards,
>> 
>> Gianluca Gorni
> 
> I am not an expert on this sort of thing,  but "mathematical common sense"
> suggest to me that this may not be easy. Let's consider carefully the
> problem of finding a "radical" expression for Cos[Pi/8] (or indeed any
> Tr[Pi*m] where Tr is a trigonometric function and m a rational). Although
> they do not look like it at first sight these expressions are in fact
> algebraic numbers.  The point is that Cos[Pi/8] is just the real part of the
> cyclotomic number Cos[2*Pi/16]+I*Sin[2*Pi/16] which is just Root[#^16 - 1 &,
> 16]  as you can see from:
> 
> In[54]:=
> (Root[#^16 - 1 &, 16] - Cos[2*Pi/16] - I*Sin[2*Pi/16]) // FullSimplify
> Out[54]=
> 0
> 
> So we can write Cos[Pi/8] as 1/2(Root[#^16 - 1 &, 16] + 1/Root[#^16 - 1 &,
> 16]), which is, of course,  an algebraic number. Using RootReduce we can get
> 
> In[58]:=
> RootReduce[1/2(Root[#^16 - 1 &, 16] + 1/Root[#^16 - 1 &, 16])]
> Out[58]=
> 2       4
> Root[1 - 8 #1  + 8 #1  & , 4]
> 
> and then 
> 
> In[60]:=
> ToRadicals[%] // FullSimplify
> Out[60]=
> Sqrt[2 + Sqrt[2]]
> -----------------
> 2
> 
> Of course in this case we could have much easier got this expression
> starting from the known values of the trigonometric functions of Pi/4 and
> then using half angle formulas. But my point is that there is a procedure
> that can be tried in general in such cases.
> 
> However, conversely:  given a radical expresion  e.g. v=((-1 + Sqrt[5]))/8 +
> (Sqrt[(3*(5 + Sqrt[5]))/2])/4 how do you go about deciding if it is the real
> or complex part of a cyclotomic number?   In fact,  v= Cos[Pi/15] but I
> doubt that there is any algorithm which has a reasonable chcance of
> determining if an arbitrary radical expression (with value between -1 and 1,
> say)  is the value of a trigonometric function of some rational multiple of
> Pi.
> (If I am wrong I would like to hear about this).
> 
> 



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