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solve the equation:
where n is a given positive integer
Last edited by mathkeep (2009-08-02 01:04:12)
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Hi mathkeep
Check Juriguen's post below.
Last edited by bobbym (2009-08-02 01:32:58)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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hi bobbym..yeah..you r right but what are the other solutions..i was able to make it out..that
is always a solution..n am looking for the method of solving this question tooLast edited by mathkeep (2009-08-02 02:05:17)
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Hi mathkeep!
Lets see, I would do the following:
Then, checking the equation, we realize that for the first set of solutions not all k are valid, since sin^n(x) should be equal to -1 only, and then:
Jose
Last edited by juriguen (2009-08-02 02:00:21)
Make everything as simple as possible, but not simpler. -- Albert Einstein
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Hi mathkeep;
A general solution will probably be incomplete as there are imaginary solutions too. Nice work, Juriguen!
Last edited by bobbym (2009-08-02 02:04:13)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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hi juriguen!
yeah...ur solution seems to be correct but i could not understand this step:
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Thanks bobbym!
However, mathkeep you're right, jaja. I think I mixed things up a little, and I was lucky, since I wrote the first equation down using:
But, the expression that you give is all the time with the power of n. The solutions are correct, though, but just skip the substitution using z...
I am going to think about the problem in general again!
Indeed, it would be something like:
But this is not so easy any more!
Jose
Last edited by juriguen (2009-08-02 02:04:23)
Make everything as simple as possible, but not simpler. -- Albert Einstein
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hi juriguen, also m not able to get it..why are u taking n to be a even when the solution is
....cuzat x= ... and ... i mean it doesnt matter wheher n is even or odd...Offline
Hi Juriguen and mathkeep;
5π is not a solution when n=3
Also mathkeep a general solution as I have indicated above will be incomplete.
Last edited by bobbym (2009-08-02 01:56:36)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi again
Completely agree with you mathkeep! I think I answered too fast...
Sorry for that!
Jose
Last edited by juriguen (2009-08-02 01:56:33)
Make everything as simple as possible, but not simpler. -- Albert Einstein
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I agree again bobbym
Corrected
Make everything as simple as possible, but not simpler. -- Albert Einstein
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Hi Juriguen;
Your idea was a lot better than mine so no apology necessary.
Hi mathkeep;
Also for n>3 and odd, x takes imaginary values. For n=3 we have x = 2 π m+(2.35619449... + .881373587... i) m ∈ Z. The complex coefficients do not have a simple form.
Last edited by bobbym (2009-08-02 02:13:26)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Indeed, it would be something like:
But this is not so easy any more!
Jose
yeas u r right now...but as u said this not so easy anymore....
n i have another problem:
when n is odd x={2k(pi)+1}....
will always be =-1 n will always be =0...i mean that can be the solution only if n is even.Offline
Hi Juriguen;
Your idea was a lot better than mine so no apology necessary.
Hi mathkeep;
Also for n>3 and odd, x takes imaginary values. For n=3 we have x = 2 π m+(2.35619449... + .881373587... i) m ∈ Z. The complex coefficients do not have a simple form.
nice work but i want real solutions only...
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Hi mathkeep;
Yes, only when n is even. Juriguen has that indicated in post #4.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi mathkeep;
Yes, only when n is even. Juriguen has that indicated in post #4.
ok..i did not see that ...thanks for telling ..i think we have got all the solutions now..from post #4...
what do u people say??
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Hi mathkeep;
We might be done for all real solutions, maybe...
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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hmm..any more replies???
i also think we are done with all the solutions...
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Yeah!
I find no other way to solve it...
By the way bobbym, how did you get any of those complex solutions, I feel curious!
Make everything as simple as possible, but not simpler. -- Albert Einstein
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Hi Juriguen;
A friend runs them for me. She uses newtons iteration in the complex plane.
Sorry for that above explanation, after being called stupid for 1/2 an hour, I have been reminded that no good numerical analyst would blindly allow newtons to frolic in the complex plane. Root location methods are brought to bear and robust root finding methods used until one gets close to the roots. Only then is newton used.
Last edited by bobbym (2009-08-05 12:08:02)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline