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Topic: solve (Read 1869 times) |
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perash
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$(x^4+1)=2x(x^2 +1)$
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jollytall
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Re: solve
« Reply #1 on: Jan 3rd, 2007, 12:26pm » |
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What does "$" mean in this eq.?
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towr
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Re: solve
« Reply #2 on: Jan 3rd, 2007, 12:29pm » |
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He was probably hoping it would be formatted as a latex formula..
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| « Last Edit: Jan 3rd, 2007, 12:31pm by towr » |
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ThudnBlunder
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Re: solve
« Reply #3 on: Jan 3rd, 2007, 2:49pm » |
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on Jan 3rd, 2007, 11:56am, perash wrote:
Applying this method will give you
2x = 1 + sqrt(3) +/- [sqrt(2sqrt(3)) or sqrt(2sqrt(3))i]
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| « Last Edit: Sep 14th, 2007, 9:20am by ThudnBlunder » |
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towr
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Re: solve
« Reply #4 on: Jan 3rd, 2007, 3:06pm » |
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I solved it rather more adhoc.. I first did a plot, which showed there to be two real solutions. Then a numeric solve, from which I spotted/guessed that the imaginary solutions were the roots of
x^2 + (sqrt(3) - 1)·x + 1
multiplying that by (x-a)(x-b), expanding and then solving for a and b to the known values from the original equations gives the same answer..
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| « Last Edit: Jan 3rd, 2007, 3:07pm by towr » |
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ThudnBlunder
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Re: solve
« Reply #5 on: Jan 3rd, 2007, 3:38pm » |
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on Jan 3rd, 2007, 3:06pm, towr wrote:| I solved it rather more adhoc.. |
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OK, sorry if I stole your thunder.
While the general theory is rather interesting, once you have done one...
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| « Last Edit: Jan 3rd, 2007, 3:39pm by ThudnBlunder » |
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towr
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Re: solve
« Reply #6 on: Jan 3rd, 2007, 3:51pm » |
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on Jan 3rd, 2007, 3:38pm, THUDandBLUNDER wrote:| OK, sorry if I stole your thunder. |
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Nah, it's nice to know there is a sound way to get an answer 
It would be hard to guess the right numbers in general, after all.
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| « Last Edit: Jan 3rd, 2007, 3:52pm by towr » |
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Eigenray
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Re: solve
« Reply #7 on: Jan 3rd, 2007, 4:50pm » |
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Note that the polynomial
p(x) = x4 - 2x3 - 2x + 1
is symmetric, hence if r is a root, then 1/r is too. So we may write
p(x) = (x-r)(x-1/r)(x-s)(x-1/s) = (x2 - ax + 1)(x2 - bx + 1),
for some a=r+1/r and b=s+1/s. Comparing coefficients gives a+b=2 and ab=-2, so a and b themselves are roots of y2-2y-2=0, i.e., a,b = 1 +/- sqrt(3). Thus
p(x) = (x2 - [1+sqrt(3)]x + 1)(x2 - [1-sqrt(3)]x + 1),
so
2x = [1 +/- sqrt(3)] +/- sqrt([1 +/- sqrt(3)]2 - 4)
= 1 +/- sqrt(3) +/- sqrt[+/- 2sqrt(3)],
where the first and third +/- have the same sign.
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ThudnBlunder
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Re: solve
« Reply #8 on: Jan 4th, 2007, 6:52am » |
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Well spotted, Eigenray. 
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balakrishnan
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Re: solve
« Reply #9 on: Jan 5th, 2007, 9:23pm » |
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alternatively
the eq can be written as
(x^2+1/x^2)=2(x+1/x)
putting x+1/x as t,we get
t^2-2=2t
or
t^2-2t-2=0==>
t=1+/-sqrt(3)=x+1/x
=>
x=0.5*[t+sqrt(t^2-4)] where t=1+/-sqrt(3)
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srn437
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Re: solve
« Reply #10 on: Sep 13th, 2007, 9:39pm » |
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on Jan 3rd, 2007, 12:26pm, jollytall wrote:| What does "$" mean in this eq.? |
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n$=n!n![sup]n![/sup]...(going on for n!).
Hopefully you understand the !, if not:n!=n(n-1)(n-2)...(2)(1).
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pex
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Re: solve
« Reply #11 on: Sep 13th, 2007, 11:14pm » |
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on Sep 13th, 2007, 9:39pm, srn347 wrote:n$=n!n![sup]n![/sup]...(going on for n!).
Hopefully you understand the !, if not:n!=n(n-1)(n-2)...(2)(1). |
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In this puzzle, $ has nothing to do with superfactorials. (In fact, the standard notation seems to be something that looks like a dollar sign, but is actually an exclamation mark with superimposed capital S.) I think that towr's first response to jollytall is correct - and that it would be hard to find someone on this forum not familiar with the n! notation for factorials.
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towr
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Re: solve
« Reply #12 on: Sep 13th, 2007, 11:36pm » |
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on Sep 13th, 2007, 9:39pm, srn347 wrote:
n$=n!n![sup]n![/sup]...(going on for n!).
Hopefully you understand the !, if not:n!=n(n-1)(n-2)...(2)(1). |
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And what, pray tell, would $n be then?
Regardless, $ isn't an operator in this case, it's a typesetting artifact; it's exactly how you would typeset of formula in LaTeX.
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