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Re: Trig functions expressed as Radicals
*To*: mathgroup at smc.vnet.net
*Subject*: [mg60080] Re: Trig functions expressed as Radicals
*From*: "Scout" <not at nothing.net>
*Date*: Thu, 1 Sep 2005 02:13:14 -0400 (EDT)
*References*: <df3bfp$kmm$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Hi Mike,
I think that Mathematica treats unvaluated expressions
in order to save memory space, according with the "depth" of
representation of that particular expression.
If you try the Depth[expr] command you will receive how many
nested lists you need to represent the information.
e.g.
In[1]:= FunctionExpand[Cos[Pi/7]];
In[2]:= Depth[%]
Out[2]= 15
In[3]:= FunctionExpand[Cos[Pi/2]];
In[4]:= Depth[%]
Out[4]= 1
Compare your own these results:
In[5]:= Cos[Pi/8]
Out[5]= Cos[Pi/8]
In[6]:= Depth[%]
Out[6]= 3
In[7]:= FunctionExpand[Cos[Pi/8]]
Out[7]= Sqrt[2+Sqrt[2]]/2
In[8]:= Depth[%]
Out[8]= 5
Bye,
~Scout~
<mike_in_england2000 at yahoo.co.uk>
> Hi
>
> When I ask mathematica what the value of Cos[Pi/4] is, it returns the
> well-known value 1/Sqrt[2]. Of course there are other values that it
> will evaluate in this way..for example Cos[Pi/6] returns Sqrt[3]/2
>
> Now if I put Cos[Pi/8] in then Mathematica returns it unevaluated. I
> can get what I want using FunctionExpand[Cos[Pi/8]] but what I am
> wondering is why does Mathematica give the radical form of trig
> expressions like Cos[Pi/6] automatically but rquires the use of
> FunctionExpand[] for values like Cos[Pi/8].
>
> Is it something to do with the complexity of the result? I know that
> FunctionExpand[Cos[Pi/23]] gives an extremely complicated result if you
> have the patience to wait for it (don't try and evaluate this unless
> you have a while) and so clearly it makes sense for mathematica to
> leave expressions like Cos[Pi/23] unevaluated unless the user explicity
> asks for it with FunctionExpand[]. If this is the case then what rules
> does Mathematica use in order to decide whether to leave the expression
> unevaluated or to give the radical form?
>
> Thanks
>
> Mike
>
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