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August 31st, 2012, 09:32 PM   #21
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Re: Solvable Quintics

[attachment=0:3juv0j7f]Quintic function - Wikipedia, the free encyclopedia.png[/attachment:3juv0j7f]

I found the skipjack problem!

the excerpt above is from this wiki link

http://en.m.wikipedia.org/wiki/Quintic_equation

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September 1st, 2012, 01:51 AM   #22
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Re: Solvable Quintics

Quote:
 Originally Posted by agentredlum I found the skipjack problem!
Yes! Nice!

But as you see, it's actually an enormous solution and the derivation of the solution is not so elegant

@skipjack : How many terms there are inside the fifth-root symbol ?

Balarka

.

 September 1st, 2012, 03:33 AM #23 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Solvable Quintics Consider about the general Bring-Jerrard quintic $x^5 + mx + n= 0$ Set $x= z m^{\frac{1}{5}}$ The equation becomes $z^5 + z + a= 0$ dependent on a single constant a. This is the Bring-Jerrard normal form. Bring introduces a function, namely $\text{BR}(x)$ which is the root of the above equation. (the root is $x= \text{BR}(a)$) This is called the Bring-radical or the hyper root. This is a transcendental function. We can easily get an expansion of it by the Taylor series of the reverse function of $f(x)= -x - x^5$ : $x= \text{BR}(a) = -a + a^5 - 5a^9 + 35a^{13} - \, . \, . \, .$ Feel free to ask anything, Balarka .
September 1st, 2012, 04:16 AM   #24
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Re: Solvable Quintics

Quote:
 Originally Posted by agentredlum I found the skipjack problem!
I think a simpler method is possible using Spearman-Williams parametrization of the given Bring-quintic.

September 1st, 2012, 04:28 AM   #25
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Quote:
 Originally Posted by mathbalarka I could just send you the paper by Dummit.
It's available online (search for "Solving Solvable Quintics") for free.

September 1st, 2012, 04:35 AM   #26
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Re: Solvable Quintics

Quote:
 Originally Posted by skipjack It's available online (search for "Solving Solvable Quintics") for free.
Well, guess I didn't knew that!

 September 1st, 2012, 04:53 AM #27 Global Moderator   Joined: Dec 2006 Posts: 19,865 Thanks: 1833 The 1994 paper by Speaman and Williams (and a later paper by the same authors) are also available online for free.
September 2nd, 2012, 08:14 PM   #28
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Re: Solvable Quintics

Quote:
 Originally Posted by mathbalarka Consider about the general Bring-Jerrard quintic $x^5 + mx + n= 0$ Set $x= z m^{\frac{1}{5}}$ The equation becomes $z^5 + z + a= 0$.
How did you get this?

September 2nd, 2012, 08:22 PM   #29
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Re: Solvable Quintics

Quote:
 Originally Posted by agentredlum How did you get this?
Sorry, I made a mistake while writing the transformation. I meant $x= z \cdot m^{\frac{1}{4}}$.

September 2nd, 2012, 08:32 PM   #30
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Re: Solvable Quintics

Quote:
 Originally Posted by mathbalarka Consider about the general Bring-Jerrard quintic $x^5 + mx + n= 0$ Set $x= z m^{\frac{1}{4}}$ The equation becomes $z^5 + z + a= 0$ dependent on a single constant a. This is the Bring-Jerrard normal form. Feel free to ask anything, Balarka .
All right, suppose a = 1, please show me the calculation of the Bring hyper radical.

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