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        riddles >> hard >> solve
        
(Message started by: perash on Jan 3^rd, 2007, 11:56am)

Title: solve
Post by perash on Jan 3^rd, 2007, 11:56am

$(x^4+1)=2x(x^2 +1)$

Title: Re: solve
Post by jollytall on Jan 3^rd, 2007, 12:26pm

What does "$" mean in this eq.?

Title: Re: solve
Post by towr on Jan 3^rd, 2007, 12:29pm

He was probably hoping it would be formatted as a latex formula..

Title: Re: solve
Post by THUDandBLUNDER on Jan 3^rd, 2007, 2:49pm

on 01/03/07 at 11:56:58, perash wrote:

(x^4+1)=2x(x^2 +1)

Applying this method (http://mathworld.wolfram.com/QuarticEquation.html) will give you
2x = 1 + sqrt(3) +/- [sqrt(2sqrt(3)) or sqrt(2sqrt(3))i]

Title: Re: solve
Post by towr on Jan 3^rd, 2007, 3:06pm

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..

Title: Re: solve
Post by THUDandBLUNDER on Jan 3^rd, 2007, 3:38pm

on 01/03/07 at 15:06:37, towr wrote:

I solved it rather more adhoc..

OK, sorry if I stole your thunder.
While the general theory is rather interesting, once you have done one...

Title: Re: solve
Post by towr on Jan 3^rd, 2007, 3:51pm

on 01/03/07 at 15:38:56, THUDandBLUNDER wrote:

OK, sorry if I stole your thunder.

Nah, it's nice to know there is a sound way to get an answer :P
It would be hard to guess the right numbers in general, after all.

Title: Re: solve
Post by Eigenray on Jan 3^rd, 2007, 4:50pm

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.

Title: Re: solve
Post by THUDandBLUNDER on Jan 4^th, 2007, 6:52am

Well spotted, Eigenray. :)

Title: Re: solve
Post by balakrishnan on Jan 5^th, 2007, 9:23pm

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)

Title: Re: solve
Post by srn347 on Sep 13^th, 2007, 9:39pm

on 01/03/07 at 12:26:57, jollytall wrote:

What does "$" mean in this eq.?

n$=n!^n![sup]n![/sup]...(going on for n!).
Hopefully you understand the !, if not:n!=n(n-1)(n-2)...(2)(1).

Title: Re: solve
Post by pex on Sep 13^th, 2007, 11:14pm

on 09/13/07 at 21:39:19, 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).

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.

Title: Re: solve
Post by towr on Sep 13^th, 2007, 11:36pm

on 09/13/07 at 21:39:19, 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).

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 (http://en.wikipedia.org/wiki/LaTeX).