Re: Question:
- To: mathgroup at smc.vnet.net
- Subject: [mg34186] Re: [mg34165] Question:
- From: Andrzej Kozlowski <andrzej at bekkoame.ne.jp>
- Date: Wed, 8 May 2002 01:57:58 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
On Wednesday, May 8, 2002, at 06:13 AM, Per Lundgren wrote:
> Please, if you have the time, show me in detail, Andrzej, how you think
> in this last step
> Again thank you
> Per Lundgren, Sweden
I was a bit in a hurry so I did not copy and paste the Mathematica code
(which was more messy than what I am sending now), but , actually all I
did was equivalent to:
In[8]:=
Select[Flatten[Table[{s, t} /.
Solve[{s + t == (2*i + 1)*(Pi/7), s - t == Pi*(j/6)},
{s, t}], {i, 7}, {j, 0, 6}], 2],
First[#1] != Last[#1] && 0 <= First[#1] <= (3/7)*Pi &&
0 <= Last[#1] <= (3/7)*Pi & ]
Out[8]=
{{(25*Pi)/84, (11*Pi)/84}, {(8*Pi)/21, Pi/21}}
In[9]:=
Select[Flatten[Table[{s, t} /. Solve[{s - t == (2*i)*(Pi/7),
s + t == Pi*(j/6)}, {s, t}], {i, 7}, {j, 6}], 2],
First[#1] != Last[#1] && 0 <= First[#1] <= (3/7)*Pi &&
0 <= Last[#1] <= (3/7)*Pi & ]
Out[9]=
{{(13*Pi)/42, Pi/42}, {(11*Pi)/28, (3*Pi)/28}}
Andrzej Kozlowski
Toyama International University
JAPAN
http://platon.c.u-tokyo.ac.jp/andrzej/
On Wednesday, May 8, 2002, at 06:13 AM, Per Lundgren wrote:
> Thank you very much, Andrzej
> I am sorry I wrote 2Pi/7 instead of 3Pi/7. Glad you could correct it so
> easy.
> I am impressed by your solution. I am far below your level. I can
> follow in the beginning but in the end I do not get how you get the
> four points of solution from:
>
> (s + t) == k Pi/6
>> or
>> (s-t) == k Pi/6 (k integer) while the second if
>>
>> s - t == 2Pi k/7 (k integer) or
>> (s + t) == (2k + 1)Pi/7 k integer
>
> Please, if you have the time, show me in detail, Andrzej, how you think
> in this last step
> Again thank you
> Per Lundgren, Sweden
>
>
> ----- Original Message -----
> From: "Andrzej Kozlowski" <andrzej at platon.c.u-tokyo.ac.jp>
To: mathgroup at smc.vnet.net
> To: "Per Lundgren" <yesgoyes at ebox.tninet.se>
> Cc: <mathgroup at smc.vnet.net>
> Sent: Tuesday, May 07, 2002 5:02 PM
> Subject: [mg34186] Re: [mg34165] Question:
>
>
>> First of all, looking at the graph of
>>
>> ParametricPlot[{(Cos[6 t])^2, 2Sin[7 t]}, {t, 0, 2Pi/7}]
>>
>> I can see no points of self-intersection at all! Did you not mean
>> 3Pi/7?
>> Assuming the letter:
>> this is the sort of thing that is easier done exactly and by hand (with
>> just a bit of help form Mathematica, maybe) than with accuracy
>> 1x10^-7 !
>>
>> Basically, you want to find values s and t satisfying
>>
>> Cos[6s] == +Cos[6t] or -Cos[6t]
>> Sin[7s]==Sin[7t]
>>
>> the first condition will be satisfied if either
>>
>> (s + t) == k Pi/6
>> or
>> (s-t) == k Pi/6 (k integer) while the second if
>>
>> s - t == 2Pi k/7 (k integer) or
>> (s + t) == (2k + 1)Pi/7 k integer
>>
>> it is now not hard to find the following four pairs of solutions (with
>> t<=3Pi/7)
>>
>> In[5]:=
>> sols = {{{t -> Pi/21}, {t -> (8*Pi)/21}},
>> {{t -> (3*Pi)/28}, {t -> (11*Pi)/28}},
>> {{t -> Pi/42}, {t -> (13*Pi)/42}}, {{t -> (11*Pi)/84},
>> {t -> (25*Pi)/84}}};
>>
>> Checking
>>
>> In[6]:=
>> N[{(Cos[6 t])^2,2Sin[7 t]}/.sols]
>>
>> Out[6]=
>> {{{0.38874,1.73205},{0.38874,1.73205}},{{0.188255,1.41421},{0.188255,
>>
>> 1.41421}},{{0.811745,1.},{0.811745,1.}},{{0.61126,0.517638},{0.61126,
>> 0.517638}}}
>>
>>
>> On Tuesday, May 7, 2002, at 04:54 PM, Per Lundgren wrote:
>>
>>>
>>> Hi,
>>>
>>> Here is my question: How do I calculate the t-values for the four
>>> points (x,y) where the curve below intersects itself with an accuracy
>>> of 1x10^-7
>>>
>>> x==(Cos[6 t])^2
>>>
>>> y==2Sin[7 t]
>>>
>>> in the intervall: [0,2Pi/7]
>>>
>>> Thank you in advance
>>>
>>> Per Lundgren, Sweden
>>>
>>> (Plot the parametric curve and you will understand what I am asking
>>> for)
>>>
>>>
>>>
>>>
>>
>>
>