REWARD!!! The first person that solves this gets 10k platinums in game:

X^2 + X^2/(X+1)^2 = 1

There are 4 solutions: 2 real and 2 complex. I want complete and detail answers to this. Also, I am not looking for computer solutions or the quadtic formula to solve this, since it will take all the fun out of it. Elementary solutions preferred, since this question was given at a high school level.

I have 40k sitting in bank atm so I guess I can afford to pay 10k per different solution(in terms of approach).
--------------------------
UPDATE(11/29) : Congradulations to Palarran and Laeyakk for 2 winning solutions!!! Great job guys.

I still have 40kpp in the bank so I can afford to pay for 4 other solutions if anyone is still interested. Also the following if you really enjoy math(upper lvl college):

Given p(X) = X^4 + 2X^3+X^2-2X-1, irreducible over the rationals. Compute the splitting field of p(X). The answer for this will get 20kpp


Enjoy it!
Foreverlive.

I think the most elementary approach = "When in doubt, choose C"

a - 2
b - 4
c - 1
d - -1

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


x = -1/2 + 1/(radical)2 - 1/2(radical)-1+2(radical)2

x ~ -.469

I remember this problem from Sophomore math class :(

Anyone know how to make the radical sign?

Talid,

Please clarify your answer some more. Kind of hard to comprehend. use sqrt(asdasdkbasd) for the radical of (asdasdkbasd).


Thanks,
Foreverlive.

X is equal to the negative value of one divided by two plus one divided by the squareroot of two plus the negative value of one divided by two multiplied by the squareroot of negative one plus two times the square root out two.

When solving for X, which is why the question asks for, X is found to be approximately the negaitve value of 0.46899...

I rounded negative .46899... to -.469

I can't get any more basic unless you want me to make a jpg image of the equation I formed.

you forgot the other root talid, .883

Yes, that would be awesome if you can make a scan of your work.


Thanks,
Foreverlive.

I'm going to die.

I don't have a real scanner, I had a small photo scanner. I went to plug it in and left it on my computer desk...it fell on the floor and smashed..

Good thing it was cheap.

I'll try doing something in like...paint or something

i bet this guy wont pay up, i have it solved with the complex answers but i have a feeling he needs it for school and is just asking for the work so he can turn his take him quiz and get an A because hes a (insert slur) too busy with eq to actually learn.

Tierfin,

Please read the first post completely before you reply.

-----
Talid,

Sorry about your scanner.


P.S. Already received 2 different approaches in solving this from my colleagues. Owe them dinners already. Purpose of this is for self-enrichment only. Got this question back in 1995 or 1996...what I was still in high school.


P.S.S. Still have 40kpp in bank waiting for 4 different ways to solve this.

Thanks,
Foreverlive.

Wait, so you're saying my answer wasn't correct?

Because I'm 100% sure that it is. Why am I not one of the four winners?

because he wants all of the roots

as in yours, the other real root and the 2 i roots

I think i'd rather buy 10kpp off of Yansuck.

is the question A)

X^2 + X^2
----------- = 1
(X+1)^2

or B)

X^2
-------- + X^2= 1
(X+1)^2

B

-1, if we're counting 1/0 as 0. But that's not usually the case.

Lets play the gemoetry game!

X^2 + X^2/(X+1)^2 = 1

X^2*(1 + 1/(X+1)^2) = 1

1+1/(X+1)^2 = 1/X^2

I've now reduced this to a geometrical problem.

You have two different hyperbolic curves (squared), and we want to find the point of intersection.

The left curve is translated one unit left, and one unit up.

Now, what does a squared hyperbola do in the complex numbers?

Going polar:
1/(r * e^(it))^2 = 1/(r^2 * e^(2it))

So, it is basically a hyperbola that rotates 360 degrees in the complex numbers every half turn in the domain.

Unfortunetally, the +1 does wierd things to the phase of the answer, which means solving it seperately in phase and magnatude is probably not reasonable. Oh well, still interesting. =)

What year of HS? Newtons method might be a way to go about finding the solutions, doable at the high school level.

Aye, Bowler. It's B.

Laeyakk, yes, you can view it as intersection of curves. Your approach is to divide both sides by x^2. I really am not very good in looking at these kind of questions this way. This question was on a math competition in CA back when I was in high school in mid 90's. This question is very interesting in the sense that it looks easy but can be extremely hard. By all means, people can solve this using more sophiscated math such as Algebraic Geometry, Groebner Basis, etc. But this question was targeted to high school students so I am really looking for solutions they can cook up. For the solutions I have seen so far, they are all very clever and not overwhelmingly long.


Thanks,
Foreverlive.

P.S. I hope to be able to give out some of the rewards by Saturday before raid starts at 1pm EST. :-)

If you don't mind me asking, what HS did you go to?

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

i get rid of paranthesis by squaring each figure in them to coem up with:

x^2
------- +x^2=1
x^2+1


then my x^2 in the fractions cancel each other out and i get

1+x^2=1

i subtract 1 from both sides:
x^2=0
final answer:
x=0







--Chooch--

i get rid of paranthesis by squaring each figure in them to coem up with:

Gotta factor them out.

(X+1)^2

Becomes
(X+1)(X+1)

Which is
X^2+2X+1

Aside from that, 0 doesn't work plugging it back into the original problem anyways.

oops oh well :/

I simplified this down out of boredom and came to the astounding conclusion that 1=1 /sigh

X^2 + X^2/(X+1)^2 = 1

X^2 * (X+1)^2 + X^2 = (X+1)^2

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

X^4 + 2x^3 + 2X^2 = X^2 + 2x + 1

F(x) = X^4 + 2X^3 + X^2 - 2X - 1 = 0

Since I'm too lazy to factor this crap, I'll derive and graph.

F'(x) = 4X^3 + 6X^2 + 2x - 2

So much for an easy time with critical numbers.

Anyhow, for the largest real root I grab an arbitrary large number, say 1000

By Newton's method, 1000-F(1000)/F`(1000)= 749.875
Repeating gives: 562, 421, 316... and eventually jumps around among various numbers between -2 and 2, ad nauseum. Newton's Method tends to zero in on the nearest real root to the approximation.

Other stuff:

F(-1) = 1 - 2 + 1 + 2 - 1 = 1

F(0) = -1, obviously

F(1) = 1 + 2 + 1 - 2 - 1 = 1

The intermediate value theorem says that there should be roots on both (-1, 0) and (0, 1) because there is a sign change in the continuous (polynomial) function.

F'(-1) = -4 + 6 - 2 - 2 = -2

F'(0) = - 2

F' (1) = 4 + 6 + 2 -2 = 10

Rolle's theorem says there is a local extremum on (-1, 0), as the function has a point with slope zero somewhere there.

I can't think of anything else till I sober up. Good luck with that problem.

Hey Forever, ill let you have the next rogue poker that drops for 10k! :p

I unfortunately am too stupid to understand the math that other people on this post did. Here's my "elementary" start:


To Start:
x^2 + x^2/(x+1)^2 = 1

Take the Square Root of both sides:
sqrt(x^2 + x^2/(x+1)^2) = sqrt(1)

Leaves you with:
x + x/(x+1) = 1

Times both sides by x+1:
x+1 (x + x/(x+1)) = x+1(1)

Leaves you with:
x^2 + x = 1

Or:
x^2 + x - 1 = 0

Ok, so this is supposed to be elementary, but from here, I don't understand how to solve it without using the Quadratic formula because you can't factor it... can you? Your original post said no Quadratic formula's, so I'm really stumped.

Quadratic Formula:

(-b +/- Sqrt(b^2 - 4ac))/2a = x

Which will come out in this case to:

.5 + sqrt(5)/2
and
.5 - sqrt(5)/2

Dying for an answer,
Helan

Hi Helan,

My original post says no quadtic (4th degree formula). The argument you have falls apart when you take the square root of both sides, while the right side is 1, the left side does not reduce. i.e. sqrt(a^2 + b^2) is not a + b.

P.S. I really am very appreciative that everyone is trying to solve this. This question is very meaningful for me.

P.S.S. To be honest,
10kpp
____________ << plats you make in Rathe Mountain
avg. time spent

but solving this will be a great mental exercise!


Thanks,
Foreverlive.

I tied my shoes all by myself today.

Ah, ok, well now I at least understand half of Willgatus's math. :)

The other half... whoosh!

Helan

Attempt #2, with the booze slept off. Going back to the formula with my starting points on [-1,0] and [0,1], which the mean value theorem a few posts above shows as intervals where the function has roots.

F(x) = X^4 + 2X^3 + X^2 - 2X - 1 = 0

Newton's method starting from 0.5 zeroes in on roughly 0.8832
From -0.5, the method gets me -0.4689 or so.

Plugging in the numbers to confirm: X = -0.4689, F(X) = -0.00018. The root is there, that's just rounding error.

The other number can be plugged in as well to get a similar result. I have no clue as to how to even get started on the complex roots.

In short, Talid got it right to begin with. I probably need the plat more, though =P

Edit: First paragraph should read "intermediate value theorem" instead of "mean value theorem." Iamsuck.

The other real number is .883 or something like that. I'm horribly with Significant digits. 8/

hmm.. i tried solving this today but couldn't get further than X^4 + 2X^3 + X^2 - 2X - 1 = 0, well i got further but i got the feeling i was going in the wrong direction...
also i was gonna divide it but couldn't pick what to divide it with.. a calculator might have helped hehe

I think the only one do deserves a win is

Willgatus Airslasher
Dragon bait of Autonomous Collective

I can pop the formula into my ti89 and get the answers too.. the whole idea is to show your work and how you got the answer..

also just to answer one other guys question.. the proof for the quadratic formula is done by completing the square.. If you want to avoid the formula just complete the square :)

Cleveland! The answer is Cleveland!

No no, it's 42.

Thank god for Quicken.. Otherwise my checkbook would be fucked.

Hi Willgatus,

Let me see if I can paraphrase your work:
1.) multiply everything by (x+1)^2 to get rid of fractions.
2.) move all terms to one side so we have some p(x) = 0.
3.) Intermediate Value Theorem shows the existence of a 0.
4.) apply Newton's Approximation Method.

Your approach is good to show the existence of solution but that is pretty much a given since step 2 gives a 4th degree polynomial and we know it has 4 roots(in real and complex). Newton's Method gives an arbitrarily close solution to a zero of the polynomial but it is not exact nonetheless. My point: to get a solution of x^2 - 2 = 0, approximation methods will yield x = 1.21... but the exact solutions that we all like are sqrt(2) and -sqrt(2). Good try nonetheless.

Many people would simply plug this into a TI 83 or some computer program and get the answer. Engineers do that. :-)


40kpp still available in bank.
Foreverlive.

Before I waste any more time on this, let me share my thoughts... since the distinct possibility that I am barking up the wrong tree has occurred to me.

Let's start at the 4th degree polynomial generated by expanding the fractions (you can see this done above):

x^4 + 2x^3 + x^2 - 2x - 1 = 0

Given a 4th degree polynomial, there are two 2nd degree polynomials that it is the product of. If I can determine those two polynomials, I can then either factor them, or use the quadratic formula to solve them. Those two binomials are represented as

(ax^2 + bx + c) = 0
(dx^2 + ex + f) = 0

Based on our knowledge of how two binomials are multiplied, we get the following system of equations:

ad = 1
ae + bd = 2
af + dc + be = 1
bf + ec = -2
cf = -1

So now the task (which is long and arduous and I have not completed :) ) is to solve these equations in terms of each other and eliminate variables until I can make one into a number. For example,

d = 1/a

and thus

ae + bd = 2 ==> ae + b/a = 2 ==> ae = 2 - b/a ==>
e = 2/a - b/a^2

and so on. I finally got to the point where I had gotten down to two variables (trust me on intermediate steps, as it doesn't matter because I have not finished...) and had the following equation:

-a^2(a-c)^2 + c^2(a-c)^2 + c(2ac)(2a-4c) - ac(a-c)^2 = 0

and it occurred to me that in solving this for c, unless something cancels out, I am going to end up with a result where c is equal to some 3rd degree polynomial of a. In which case, plugging it back in to one of the base equations will likely land me with a polynomial of degree 4 again, in terms of a. My question to myself was
1) is this going to be the same freakin' polynomial I started with? Or,
2) is it going to be another 4th degree polynomial that I have no hope of factoring?

So anyways, if someone with a lot of time, interest, and attention to details wants to pursue that system of equations to the end, I'd sure be interested in seeing how this movie turns out...

*edit* - Interestingly enough, I think this approach is simpler than using the quartic formula... of course, all for naught if it does not work...

based off
mathworld.wolfram.com/Qua...ation.html (http://mathworld.wolfram.com/QuarticEquation.html)

Start with:
X^2(X+1)^2 + X^2 = (X+1)^2
Substitute using X = (Z - 1/2), giving
(Z-1/2)^2 * (Z+1/2)^2 + (Z-1/2)^2 = (Z+1/2)^2
(Z-1/2)^2 (Z+1/2)^2 + (Z^2 - Z + 1/4) = (Z^2 + Z + 1/4)
(Z-1/2)^2 (Z+1/2)^2 = 2 Z
(Z^2 - 1/4)^2 = 2 Z
or
Z^4 - Z^2 / 2 - 2 Z + 1/16 = 0

For every u, adding and subtracting Z^2 u + u^2 / 4 will do nothing. Or:

(Z^4 + Z^2 u + u^2 /4) + (-Z^2 u - u^2 / 4 - Z^2 / 2 - 2 Z + 1/16) = 0
grouping into two piles we get:
(Z^2 + 1/2 u)^2 - [(u+1/2)Z^2 + 2 Z + (u^2 /4 - 1/16)] = 0

Now, what we want is this to be
P^2 - Q^2 = 0
or
(P-Q)(P+Q) = 0
from some P and Q, with both P and Q at most quadradic.

We are close. If you take another look:
(Z^2 + 1/2 u)^2 - [(u+1/2)Z^2 + 2 Z + (u^2 /4 - 1/16)] = 0
The first term is already a perfect square, for any choice of u.

Our goal is to find a u that makes the second term a perfect square.
Q^2 = (u+1/2)Z^2 + 2Z + (u^2 /4 - 1/16)
Tidying up
Q^2 = (u+1/2) [Z^2 + 2/(u+1/2) Z + (u^2 /4 - 1/16)/(u+1/2)] (tidy)
we want:
Q^2 = (u+1/2) [ Z - rt ( (u^2 /4 - 1/16)/(u + 1/2) ) ]^2 (want)

Now, for this to be true, the Z coefficent has to work out.

The Z coefficient in the equation (tidy) is 2/(u+1/2).
The Z coefficient in the equation (want) is - 2 rt ((u^2 /4 - 1/16)/(u+1/2))
so, our goal is:
2/(u+1/2) = - 2 rt ((u^2 /4 - 1/16)/(u+1/2))

Square it, and rearrange
4 = 4 (u^2 / 4 - 1/16) (u+1/2)
4 = (u^2 - 1/4) (u + 1/2)
0 = u^3 + 1/2 u^2 - 1/4 u - 4

Oh fun, a cubic! We can solve those. I hope.

But I'm bored now. See
mathworld.wolfram.com/CubicEquation.html (http://mathworld.wolfram.com/CubicEquation.html)

Santerre,

Very nice progress so far. Your approach is very similar to one of the solution that my colleague gave. He did add a trick to it so I am not certain if your work will yield a solution.


Laeyakk,

I have not had time to check your work yet. You might be able to reduce it to a cubic equation (I believe that's how they deal with the 4th degree equation: by reduction to 3rd degrees) but again, I am not sure if one can solve a cubic easily or not. Quadratic, on the other hand, is godly since we have that infamous formula.



Keep it up guys!
Foreverlive.

Cubic->Quadradic is detailed in the 2nd link. It is easier than Quardic->Quadradic. =)

Lets go play with Lagrange resolvents.

Let w = e^(2 pi i / 3)
(ie, the complex number corresponding to a 1/3 rotation in phase space)

let u1 and u2 be arbitrary complex numbers

Now consider:
[x - (u1 + u2)] [x - (w u1 + w^2 u2)] [x - (w^2 u1 + w u2)] = 0
The roots are quite clearly
u1 + u1
w u1 + w^2 u2
w^2 u1 + w u2

And, if you multiply it out, you get
x^3 - 3 u1 u2 x - (u1^3 + u2^3) = 0

So, if you have an equation of the form
x^3 + a x + b = 0
(which, using the X = Z-C trick, you can turn any 3rd degree polynomial into)
Let
m = u1^3
n = u2^3
Then, just solve for
m n = - (a/3)^3
m + n = b

and you have a finite set of possible roots.

There are more ways to solve cubics over in:
mathworld.wolfram.com/CubicEquation.html (http://mathworld.wolfram.com/CubicEquation.html)

Personally, I'm not sure if putting anyone through the fun of manually solving a quartic, or even cubic, equation is allowed under the UN charter of rights and freedoms. Maybe it is allowed on high school students?

Yet another step

we did want to solve
0 = u^3 + 1/2 u^2 - 1/4 u - 4 (hope)
but there is that annoying 1/2 term in front of the u^2. Bad term!

So, set
u = v - 1/6
and examine the equation (hope) in terms of v.

I believe you end up with (cheating, didn't do the math here)

0 = v^3 + (-3/4 - 1/2)/3 v + (9/8 - 27*4 - 2/8) /27
or
0 = v^3 - 5/12 v - 857 / 216

So, now we need to solve for n and m.

The answer to any hard math problem is usually 0 or infinity because those are the only two solutions that never make sense and baffle the math geeks.

m n = - (a/3)^3
m + n = b
where
a = -5/12
b = -857/216

n m = - 125 / (6^3)^2
n + m = -857 / (6^3)

let n` = n * 6^3, m` = m * 6^3
n` m` = -125 (eqn 1)
n + m = -857 (eqn 2)

n` = d + ei
m` = f + gi

g = -e from (1) and (2)
f = d - 857 from (2)

n` = d + ei
m` = d - 867 - ei

n` m` = d^2 - e^2 - e 867 i
proving that e = 0, n = m, and showing that I made a mistake somewhere above. =)

Like I said, cruel and unusual punishment!

mambo dogface to the banana patch.

Laeyakk,

Another hint I will point out is that none of the different approaches that led so a solution so far involves using the 4th degree or the cubic. They do all involve the quadratic. But again, there might be other approaches as well.

P.S. The progress you have so far is nearly half way. I really hope you can look at it carefully and pick up at least 1 of the 10k prize I have up.


Good Lucks,
Foreverlive.

Once 2 real roots are found it should be trivial to find the complex roots. I remember synthetic division from high school as a good mechanical way to factor out known roots.

As far as finding those two initial roots...my initial thoughts were along the lines of letting y = x/(x+1), resulting in the system x^2 + y^2 = 1 and y = x/(x+1). I don't remember any good ways to find the intersection of the two lines though. Maybe converting to polar coordinates would help...
y = r cos theta
x = r sin theta

So the system would become...
r=1
r cos theta = (r sin theta)/((r sin theta) + 1)

resulting in:
cos theta = (sin theta)/((sin theta) + 1)

Solve for theta, then let x = sin theta.

I'm not sure if that goes anywhere, it's been a long time since I've tried to manipulate trig expressions. :P

However, thinking about what the two lines look like suggests that the two real roots listed above (somewhere around 0.8 and -0.4) are likely correct.

Foreverlive, I gave the problem to my math prof. He got about as far as I =/ Going to give it another shot later tonight, maybe with Santerre's method.

Warning, lots of math ahead. :P

Let y = x/(x+1), resulting in the system:
x^2 + y^2 = 1
y = x/(x+1)

Let's switch to polar coordinates...
y = r cos theta
x = r sin theta

So the system becomes:
r=1
r cos theta = (r sin theta)/((r sin theta) + 1)

resulting in:
cos theta = (sin theta)/((sin theta) + 1)

Multiply both sides by ((sin theta) + 1):
cos theta((sin theta) + 1) = sin theta
(cos theta)(sin theta) + cos theta = sin theta
(cos theta)(sin theta) = (sin theta - cos theta)

Use identity sin (2*theta) = 2(sin theta)(cos theta):
(1/2)sin(2*theta) = (sin theta - cos theta)

Square both sides:
(1/2)^2(sin(2*theta))^2 = (sin theta - cos theta)^2
(1/4)(sin(2*theta))^2 = (sin theta)^2 - 2(sin theta)(cos theta) + (cos theta)^2

Use identity (sin theta)^2 + (cos theta)^2 = 1:
(1/4)(sin(2*theta))^2 = 1 - 2(sin theta)(cos theta)

Use identity sin (2*theta) = 2(sin theta)(cos theta):
(1/4)(sin(2*theta))^2 = 1 - (sin(2*theta))

Let A=sin(2*theta).
(1/4)A^2 = 1-A

(1/4)A^2 + A - 1 = 0
A^2 + 4A - 4 = 0

Yay, we have a quadratic; we can solve for A:

A = (-4 +/- sqrt(4^2 - 4(1)(-4)))/2(1)
A = -2 +/- 2*sqrt(2)

so:
(I): sin(2*theta) = -2 + 2*sqrt(2)
or
(II): sin(2*theta) = -2 - 2*sqrt(2)

Clearly (II) is impossible, so (I) must be the case.
sin(2*theta) = -2 + 2*sqrt(2)

Use identity sin (2*theta) = 2(sin theta)(cos theta):
2(sin theta)(cos theta) = -2 + 2*sqrt(2)

Using x = r cos theta, y = r sin theta (and r=1):
2xy = -2 + 2*sqrt(2)
xy = -1 + sqrt(2)

Using y = x/(x+1):
(x^2) / (x+1) = -1 + sqrt(2)
x^2 = (-1 + sqrt(2)) (x+1)
x^2 + (1 - sqrt(2))(x+1) = 0
x^2 + (1 - sqrt(2))x + (1 - sqrt(2)) = 0

Let K = (1 - sqrt(2)). K^2 = 3 - 2*sqrt(2)
x^2 + Kx + K = 0

x = (-K +/- sqrt(K^2 - 4(1)(K))) / 2(1)
x = -K/2 +/- sqrt(K^2 - 4K)/2
x = -1/2 + (1/2)sqrt(2) +/- (1/2)sqrt(-1 + 2*sqrt(2))

Let s = (1/2) (sqrt(2) - 1 + sqrt(2*sqrt(2) - 1))
Let t = (1/2) (sqrt(2) - 1 - sqrt(2*sqrt(2) - 1))
x = s or t

So, s and t are the two real solutions.

Note that:
s+t = sqrt(2) - 1
s*t = ((sqrt(2) - 1)^2 - 2*sqrt(2) + 1)/4
s*t = 1 - sqrt(2)

That'll make synthetic division much cleaner than I initially expected.

<TABLE BORDER="1">
<TR><TD>s<TD>1<TD>2<TD>1<TD>-2<TD>-1</tr>
<TR><TD><TD><TD>s<TD>s^2+2s<TD>s^3+2s^2+s<TD>s^4+2s^3+s^2-2s</tr>
<TR><TD>t<TD>1<TD>s+2<TD>s^2+2s+1<TD>s^3+2s^2+s-2<TD>s^4+2s^3+s^2-2s-1=0</tr>
<TR><TD><TD><TD>t<TD>t+t*sqrt(2)<TD>ts^2+ts+sqrt(2)*t^2 + sqrt(2)*t=s-sqrt(2)*s+sqrt(2)*t^2+sqrt(2)*t
<TR><TD><TD>1<TD>s+t+2=sqrt(2)+1<TD>s^2+2s+t+t*sqrt(2)+1=s(s+1)+sqrt(2)(t+1)<TD>...=0<TD></tr>
</table>

So we want to solve for x:
x^2 + (sqrt(2) + 1)x + s(s+1) + sqrt(2)(t+1) = 0

Hmm, let me try completing the square...

x^2 + (sqrt(2) + 1)x = (x + (1/2)sqrt(2) + (1/2))^2 - (3/4) - (1/2)sqrt(2)

(x + (1/2)sqrt(2) + (1/2))^2 = (3/4) - (1/2)sqrt(2) - s(s+1) - sqrt(2)(t+1)
x + (1/2)sqrt(2) + (1/2) = +/- sqrt((3/4) - (1/2)sqrt(2) - s(s+1) - sqrt(2)(t+1))
x = -1/2 - (1/2)sqrt(2) +/- sqrt((3/4) - (1/2)sqrt(2) - s(s+1) - sqrt(2)(t+1))
x = -1/2 - (1/2)sqrt(2) +/- i*sqrt((-3/4) + (1/2)sqrt(2) + s(s+1) + sqrt(2)(t+1))

s+1 = (1/2)(sqrt(2) + 1 + sqrt(2*sqrt(2) - 1))
s(s+1) = (1/2)sqrt(2) + (sqrt(2)/2)sqrt(2*sqrt(2) - 1)
t+1 = (1/2)(sqrt(2) + 1 - sqrt(2*sqrt(2) - 1))
sqrt(2)(t+1) = 1 + (1/2)sqrt(2) - (sqrt(2)/2)sqrt(2*sqrt(2) - 1)

x = -1/2 - (1/2)sqrt(2) +/- i*sqrt((-3/4) - sqrt(2)/2 + sqrt(2)/2 + (sqrt(2)/2)sqrt(2*sqrt(2) - 1) + 1 + sqrt(2)/2 - (sqrt(2)/2)sqrt(2*sqrt(2)-1))
x = -1/2 - (1/2)sqrt(2) +/- i*sqrt((1/4)+(sqrt(2)/2))
x = (1/2)(-sqrt(2) - 1 +/- i*sqrt(1 + 2*sqrt(2)))

So, our two complex roots should be
x = (1/2)(-sqrt(2) - 1 + i*sqrt(1 + 2*sqrt(2)))
and
x = (1/2)(-sqrt(2) - 1 - i*sqrt(1 + 2*sqrt(2)))
to go with the real roots
x = (1/2)(sqrt(2) - 1 + sqrt(-1 + 2*sqrt(2)))
and
x = (1/2)(sqrt(2) - 1 - sqrt(-1 + 2*sqrt(2)))

Hmm, some of the signs may be wrong...

Palarran, where do you get the trig functions from?

The ones I used are pretty fundamental (well, the sin (2*theta) can be derived from sin(a+b)).
I did cheat a bit though--I had to look up what they were by doing a google search for "trig identities". :)
http://aleph0.clarku.edu/~djoyce/java/trig/identities.html

I started working on the synthetic division to factor out the two real solutions, resulting in a quadratic that can be easily solved to find the two complex solutions. While it's guaranteed to work, it's painfully long...maybe I can find a more elegant method. (Edit: Added what I think are the complex roots above...)

As far as finding those two initial roots...my initial thoughts were along the lines of letting y = x/(x+1), resulting in the system x^2 + y^2 = 1 and y = x/(x+1). I don't remember any good ways to find the intersection of the two lines though. Maybe converting to polar coordinates would help...
y = r cos theta
x = r sin theta

How do you know x and y can be represented thus?

[quoe]So the system would become...
r=1
r cos theta = (r sin theta)/((r sin theta) + 1)[/quote]

Where did r=1 come from, left field?

I mean, if you knew r=1, then you could make some of the above claims. Possibly you just lost me.

1 the answer is one

Saberius

I defined y = x/(x+1); it's just a substitution. Plug it back in and it works. y^2 = x^2/(x+1)^2

x^2 + y^2 = 1 is, of course, the equation for a circle of radius 1 centered about the origin. The equivalent polar equation is r=1. (Any point on the circle is a fixed distance of 1 from the origin regardless of angle theta.)

Palarran,

I don't think I can digest all that you have typed. I think I will make an attempt tomorrow night to read thru it. That's some serious work you put into it! /bow. The correct answer does involve sqrt of sqrts. I will get back to you tomorrow night hopefully.


Thanks,
Foreverlive.

Heh, yeah, being restricted to text only makes it look pretty ugly. It didn't look so bad on the whiteboard I used to work things out before typing. :)

The neat thing is, all of the concepts used came from high school math: substitution, polar coordinates, intersection of lines and solving systems of equations, the quadratic formula, synthetic division, trig identities...

A program like Maple or Mathematica might help to interpret the longer bits, printing the expression in a more readable form. It would also make it easier to verify the complex roots. :) (Heh, I should probably just throw a couple functions into my calculator for dealing with complex numbers so I can check it myself. Either that or I could continue digging through my stuff looking for my Maple cd. :P )

Gah, where's Palimax when you need him.

x^2 + y^2 = 1 is, of course, the equation for a circle of radius 1 centered about the origin. The equivalent polar equation is r=1. (Any point on the circle is a fixed distance of 1 from the origin regardless of angle theta.)

I agree, if you assume r=1, then you are talking about only points on the unit circle in the complex plane, and are no longer talking about the general case. Right?

How do you know the solution is in the unit circle of the complex plane? How can you justify this assumption? Or, is there some trick I missed that makes what I think you assumed not what you assumed?

We were told there were two real solutions. So, initially I was trying to find those.

The unit circle I'm using is not in the complex plane; it's in the plane formed by real values for variables x and y, rather than real and imaginary components for x. Any real solutions to x^2 + x^2/(x+1)^2 = 1 must lie on that circle, since we defined y = x/(x+1), resulting in x^2 + y^2 = 1.

Edit: Oh, and we can get to r=1 by simple substitution too.
Given the standard conversion from rectangular to polar coordinates:
y = r cos theta
x = r sin theta
...and the identity (sin x)^2 + (cos x)^2 = 1
we substitute into x^2 + y^2 = 1:
(r sin theta)^2 + (r cos theta)^2 = 1
r^2 (sin theta)^2 + r^2 (cos theta)^2 = 1
r^2 ((sin theta)^2 + (cos theta)^2) = 1
r^2 (1) = 1
r^2 = 1
r = 1, -1

r=1 and r=-1 are equivalent; we choose r=1 for convenience.

(Ugh, I always end up using the royal "we" when talking about math, heh...)

palarran, are you a math professor?

I still think that r is fishy. So I'm gonna avoid using it.

So, we know
x^2 + y^2 = 1

We have an equation constraining what y can be as a function of x.

We know x is between -1 and 1. So, we define theta as:
theta = sin_inverse x

Which gives us
sin^2 theta + y^2 = 1
or
y^2 = 1 - sin^2 theta
but
cos^2 theta = 1 - sin^2 theta

thus giving us
cos^2 theta = y^2

or
cos^2 theta =
sin^2 theta
--------------
(sin theta + 1)^2

Ok, I've found a justification that satisfies me. =)

Well...it would have to be theta = (sin_inverse(x)) / r initially. Otherwise you're already assuming r=1, since x = r sin theta.

Palarran, I recovered equations similar to yours without ever using an r parameter.

I used your definition of y. I then restricted x to be between -1 and 1 (this restriction is easy to justify). I then defined theta to simply be
theta = sin_inverse x
there is no r involved in this definition.

Possibly I should have used a different variable name for my theta to avoid confusion, but because I ended up with the same conclusions I reused your variable name.

In the end, the relations you illustrated:
x^2 = sin^2 theta
x^2/(x+1)^2 = cos^2 theta
still hold, but I reached them through a different route.

Ohh...so rather than using the standard polar substitutions, you used your own which led to the same results (and with a bit less work), after which the remaining steps were the same?

Ohh...so rather than using the standard polar substitutions, you used your own which led to the same results (and with a bit less work), after which the remaining steps were the same?

/shrug, possibly. When I ran into parts of your math that didn't contain the justification for why it was justified (r=1, some of the early polar stuff), I balked, so I haven't even looked at the rest of your work. The main situations I've used r & angle values is when working on complex numbers.

Now I'll go peer at the rest of the work. =)

Hmm, I just thought of something.

Sine and cosine are defined on the complex numbers, not just the reals, and when given complex numbers the range is not confined to [-1, 1].

At one point I broke into cases (I) and (II), and immediately dismissed (II) because it had sine returning something outside of [-1, 1]. (I) led to the two real solutions for the problem. I wonder if (II) leads to the two complex solutions directly, rather than finding the two real solutions, factoring them out of the 4th degree polynomial and using the remaining quadratic formula to find the other two roots.

The use of sins and cosines turns out to be fluff.

X^2 + X^2 / (X+1)^2 = 1 (1)
Define B = X/(X+1) (2)
X^2 + B^2 = 1 (3) from (1) and (2)

from (3), we know
-1 <= X <= 1 (4)
-1 <= B <= 1 (5)

Multiply (2) by (X+1)
B(X+1) = X
XB + B = X
XB = X - B
Sqaure it
X^2B^2 = X^2 - 2 XB + B^2
using (3), we eliminate X^2 and B^2
X^2B^2 = 1 - 2 XB

let Z = XB, then
Z^2 = 1 - 2 Z
Z^2 + 2 Z - 1 = 0
The roots are Z_1 and Z_2:
Z_1 = -1 + sqrt( 2 )
Z_2 = -1 - sqrt( 2 )

However, Z=XB, and from (4) and (5)
-1 <= XB <= 1
so, the Z_2 root is eliminated, as Z_2 < -1.

So:
XB = sqrt(2)-1
is a solution. Substituting using (2) we get

X^2/(X+1) = sqrt(2)-1
X^2 = (sqrt(2)-1)X + (sqrt(2)-1)
X^2 - (sqrt(2)-1)X - (sqrt(2)-1) = 0

Let A = sqrt(2)-1 (6)
note A^2 = 2 - 2 sqrt 2 + 1
A^2 = 3 - 2 sqrt 2
A^2 = 1 - 2 (sqrt(2) - 1)
A^2 = 1 - 2 A (7)

So we have:
X^2 - A X - A = 0
Quadradic equation time! Solutions X_1 and X_2

X_1 = (A + sqrt(A^2 + 4 A))/2
X_2 = (A - sqrt(A^2 + 4 A))/2

using (7)

X_1 = (A + sqrt((1-2A) + 4 A))/2
X_2 = (A - sqrt((1-2A) + 4 A))/2

X_1 = (A + sqrt(1+2A)/2
X_2 = (A - sqrt(1+2A)/2

using (6)

X_1 = (1-sqrt2 + sqrt(2 sqrt(2) - 1)/2
X_2 = (1-sqrt2 - sqrt(2 sqrt(2) - 1)/2

Those are both real roots, and the same ones you generateI believe (you called them s and t).

What discipline btw?

Err...what do you mean what discipline? :)

Edit C^2 = 2C +1 I think, oops.

At one point I broke into cases (I) and (II), and immediately dismissed (II) because it had sine returning something outside of [-1, 1]. (I) led to the two real solutions for the problem. I wonder if (II) leads to the two complex solutions directly, rather than finding the two real solutions, factoring them out of the 4th degree polynomial and using the remaining quadratic formula to find the other two roots.

Point. That has an analogy in my version.

(4) and (5) no longer hold, which means XB can be beyond -1 and +1.

So, going at it

Z = Z_2
Z = XB, Z_2 = -1 - sqrt( 2 ), so:
XB = -1 -sqrt(2)
using (2) we get:
X^2/(X+1) = -1 -sqrt(2)
Let C = (1 + sqrt(2))
Note that C^2 = 1 + 2 sqrt2 + 2 = 2C + 1 (8) Edit here and below
X^2/(X+1) = -C
X^2 + CX + C = 0

Quadradic formula reviels 2 roots, X_3 and X_4:
X_3 = (-C+sqrt(C^2 - 4 C))/2
X_4 = (-C-sqrt(C^2 - 4 C))/2

Using (8)
X_3 = (-C+sqrt(2C + 1 - 4 C))/2
X_4 = (-C-sqrt(2C + 1 - 4 C))/2

X_3 = (-C+sqrt(1 - 2 C))/2
X_4 = (-C-sqrt(1 - 2 C))/2

replacing C with 1 + sqrt2 we get:

X_3 = (-1 - sqrt(2) +sqrt(-1 - 2 sqrt(2)))/2
X_4 = (-1 - sqrt(2) -sqrt(-1 - 2 sqrt(2)))/2

X_3 = -1/2 - 1/sqrt(2) + i * sqrt(1/4 + 1/sqrt(2))
X_4 = -1/2 - 1/sqrt(2) - i * sqrt(1/4 + 1/sqrt(2))

Err...what do you mean what discipline?

High School math, or university in one discipline or another, or self taught? =)

bah, need to remove a 8) from my post.

Ahh...I was a math/computer science dual major as an undergraduate in college. I mostly studied discrete math (number theory, graph theory, etc.)

(Edit: Argh, I should have read Laeyakk's edited posts first. Nothing new here then I guess. :P Laeyakk's start is simpler than mine too, though very quickly both our solutions merge.)

Yup, you're right. sin(x), sin(2x) and cos(x) really just ended up being variables that could have been named anything, and in that light the substitutions were unnecessary to begin with (aside from the initial y = x/(x+1) that is, and even that is just for convenience).

Given:
[A] x^2 + x^2/(x+1)^2 = 1

Starting with an equation that is always true:
x = x
x^2 = x^2
x^2 = x(x+1) - x
(x^2)/(x+1) = (x(x+1) - x)/(x+1)
(x^2)/(x+1) = x - x/(x+1)

Square both sides:
((x^2)/(x+1))^2 = x^2 - 2((x^2)/(x+1)) + x^2/(x+1)^2

Substitute [A]:
[b] ((x^2)/(x+1))^2 = 1 - 2((x^2)/(x+1))
[C] Let Y = (x^2)/(x+1).
Substitute [C] into [b]:
Y^2 = 1 - 2Y
Y^2 + 2Y = 1
(Y+1)^2 - 1 = 1
(Y+1)^2 = 2
Y+1 = +/- sqrt(2)
[D] Y = -1 +/- sqrt(2)
[D1] Y = -1 + sqrt(2), or
[D2] Y = -1 - sqrt(2)

Using [C]:
Y = (x^2)/(x+1)
x^2 = Y(x+1)
***edited from here down***
x^2 - Yx = Y
(x - Y/2)^2 - (1/4)Y^2 = Y
(x - Y/2)^2 = Y + (1/4)Y^2
x - Y/2 = +/- sqrt(Y + (1/4)Y^2)
[E] x = Y/2 +/- sqrt(Y + (1/4)Y^2)

Case [D1]:
Y = -1 + sqrt(2)
Y^2 = 3 - 2*sqrt(2)
Substitute [D1] into [E]:
x = (-1/2) + (1/2)sqrt(2) +/- sqrt(-1 + sqrt(2) + (1/4)(3 - 2*sqrt(2))
[F] x = (-1/2) + (1/2)sqrt(2) +/- sqrt(-1/4 + (1/2)sqrt(2))
[F1] x = (-1/2) + (1/2)sqrt(2) + sqrt(-1/4 + (1/2)sqrt(2)), or
[F2] x = (-1/2) + (1/2)sqrt(2) - sqrt(-1/4 + (1/2)sqrt(2))

Case [D2]:
Y = -1 - sqrt(2)
Y^2 = 3 + 2*sqrt(2)
Substitute [D1] into [E]:
x = (-1/2) - (1/2)sqrt(2) +/- sqrt(-1 - sqrt(2) + (1/4)(3 + 2*sqrt(2))
x = (-1/2) - (1/2)sqrt(2) +/- sqrt((-1/4) - (1/2)sqrt(2))
[G] x = (-1/2) - (1/2)sqrt(2) +/- i*sqrt((1/4) + (1/2)sqrt(2))
[G1] x = (-1/2) - (1/2)sqrt(2) + i*sqrt((1/4) + (1/2)sqrt(2)), or
[G2] x = (-1/2) - (1/2)sqrt(2) - i*sqrt((1/4) + (1/2)sqrt(2))


So we get all four roots immediately: [F1], [F2], [G1], [G2].

(Edit: And this time I checked the results. :)

Hi Palarran,

I have not go thru the posts with you and Laeyakk yet but your solution is very close as of now. I think you made some arithmatic/substitution error on your latest response:

------------------
Given:
[A] x^2 + x^2/(x+1)^2 = 1

Starting with an equation that is always true:
x = x
x^2 = x^2
x^2 = x(x+1) - x
(x^2)/(x+1) = (x(x+1) - x)/(x+1)
(x^2)/(x+1) = x - x/(x+1)

Square both sides:
((x^2)/(x+1))^2 = x^2 - 2((x^2)/(x+1)) + x^2/(x+1)^2

Substitute [A]:
((x^2)/(x+1))^2 = 1 - 2((x^2)/(x+1))
[C] Let Y = (x^2)/(x+1).
Substitute [C] into <!--EZCODE BOLD START--><strong>:
Y^2 = 1 - 2Y
Y^2 + 2Y = 1
(Y+1)^2 - 1 = 1
(Y+1)^2 = 2
Y+1 = +/- sqrt(2)
[D] Y = -1 +/- sqrt(2)
[D1] Y = -1 + sqrt(2), or
[D2] Y = -1 - sqrt(2)
--------------
*Everything upto here is very slick/brilliant, whichever you prefer. =)
--------------

Using [C]:
Y = (x^2)/(x+1)
x^2 = Y(x+1)
x^2 - Yx = -1
--------------
*Things fall apart here. It should be x^2 -Yx = -Y,
i.e., x^2 - (x^2/(x+1))x = -x^2/(x+1).

I think you have gotten about 75% of the correct answer at this point. play with your work for a little more and I think will will have the correct solution. Fix it up and you can claim your prize. /cheer
------------
(x - Y/2)^2 - (1/4)Y^2 = -1
(x - Y/2)^2 = -1 + (1/4)Y^2
x - Y/2 = +/- sqrt(-1 + (1/4)Y^2)
[E] x = Y/2 +/- sqrt(-1 + (1/4)Y^2)

Case [D1]:
Y = -1 + sqrt(2)
Y^2 = 3 - 2*sqrt(2)
Substitute [D1] into [E]:
x = -1/2 + (1/2)sqrt(2) +/- sqrt(-1 + (1/4)(3 - 2*sqrt(2))
x = -1/2 + (1/2)sqrt(2) +/- sqrt(-1/4 - sqrt(2)/2)
[F] x = -1/2 + (1/2)sqrt(2) +/- i*sqrt(1/4 + sqrt(2)/2)
[F1] x = -1/2 + (1/2)sqrt(2) + i*sqrt(1/4 + sqrt(2)/2), or
[F2] x = -1/2 + (1/2)sqrt(2) - i*sqrt(1/4 + sqrt(2)/2)

Case [D2]:
Y = -1 - sqrt(2)
Y^2 = 3 + 2*sqrt(2)
Substitute [D1] into [E]:
x = -1/2 - (1/2)sqrt(2) +/- sqrt(-1 + (1/4)(3 + 2*sqrt(2))
[G] x = -1/2 - (1/2)sqrt(2) +/- sqrt(-1/4 + sqrt(2)/2)
[G1] x = -1/2 - (1/2)sqrt(2) + sqrt(-1/4 + sqrt(2)/2), or
[G2] x = -1/2 - (1/2)sqrt(2) - sqrt(-1/4 + sqrt(2)/2)

So we get all four roots immediately: [F1], [F2], [G1][/b]

Laeyakk,

Can you also write up your solution in a 1 response format? It will be much easier for me to check your work. I am hoping to see both you and Palarran to get the prize at the point.


P.S. I am not sure if you guys are still in high school or in college/beyond. But for one thing I am certain is that, if you can solve this question in high school and especially in different approaches, you are fit to be in some of the best colleges available.

Best Wishes,
Foreverlive.

Heh, yeah, I sort of rushed that response since it was late at night, and I didn't work things out before typing them. :)

My original solution did produce the correct answers (verified all 4) but I got there in a roundabout way. I'll fix up the streamlined version when I get home from snowboarding. In the meantime, I think Laeyakk's solution was almost identical except for how he got started (building the initial equation of the form xy = x-y.

I'm a college graduate, but very little of the math I learned in college applied to this problem (or at least to my approach to it). I think about all I used was my experience in writing informal proofs, some of which really should be taught at the high school level, but in my case wasn't.

I did not learn any of this math in High School and I was in the best math classes they offered.

Gotta love rural town education

I'm with you on this bowler.. I went to a college prepetry highschool. I satified my math credits junior year and quit math(I hate it). Junior year was Trig and I have no idea what any of those number mean..LOL O fourse this was over 10 years ago and I tested out for College the next fall so in otherwords i feel dumb..HEHEHE

There, fixed it up. :)

I'm curious about what other approaches there are...nothing has come to mind so far.

We start with
X^2 + X^2 / (X+1)^2 = 1 (1)
Define B = X/(X+1) (2)
So,
X^2 + B^2 = 1 (3)

Multipy (2) by (X+1), giving us
B(X+1) = X
XB + B = X
XB = X-B
Square it
X^2B^2 = X^2 - 2XB + B^2
by (3)
X^2B^2 = 1 - 2XB

Define Z = XB
Z^2 = 1 - 2Z
Z^2 + 2Z -1 = 0
Using the quadradic equation, we find the roots Z_1 and Z_2
Z_1 = -1 + sqrt(2)
Z_2 = -1 - sqrt(2)

As it happens, Z_1 corresponds to the real roots, and Z_2 to the imaginary roots. |Z_2|>1, so it cannot correspond to a real root. ((3) -> |X| < 1 and |B| < 1 on R, thus |Z| < 1 for real valued X).

Getting back to the solution:

Ok, now we know
Z = XB = X^2/(X+1)
by the defintion of Z and B
Let S be one of Z_1 or Z_2. Then
X^2/(X+1) = S
X^2 = SX + S
X^2 - SX - S = 0

Using the quadradic equation once again we get 2 solutions for X in terms of S:

(S + sqrt(S^2 + 4S)) / 2
(S - sqrt(S^2 + 4S)) / 2

There are two possible values for S, giving us 4 solutions.

Let
P = Z_1 = sqrt(2) - 1
Q = Z_2 = -sqrt(2) - 1
note
P^2 = 1 - 2 P
and
Q^2 = 1 - 2 Q

The four solutions X_1 to X_4 are:
X_1 = (P + sqrt(P^2 + 4P)) / 2
X_2 = (P - sqrt(P^2 + 4P)) / 2
X_3 = (Q + sqrt(Q^2 + 4Q)) / 2
X_4 = (Q - sqrt(Q^2 + 4Q)) / 2

X_1 = (P + sqrt(1-2P + 4P)) / 2
X_2 = (P - sqrt(1-2P + 4P)) / 2
X_3 = (Q + sqrt(1-2Q + 4Q)) / 2
X_4 = (Q - sqrt(1-2Q + 4Q)) / 2

X_1 = P/2 + sqrt(1/4 + P/2)
X_2 = P/2 - sqrt(1/4 + P/2)
X_3 = Q/2 + sqrt(1/4 + Q/2)
X_4 = Q/2 - sqrt(1/4 + Q/2)

Plugging in P and Q, we get:
X_1 = -1/2 + 1/sqrt(2) + sqrt(-1/4 + 1/sqrt(2))
X_2 = -1/2 + 1/sqrt(2) - sqrt(-1/4 + 1/sqrt(2))
X_3 = -1/2 - 1/sqrt(2) + i sqrt(1/4 + 1/sqrt(2))
X_4 = -1/2 - 1/sqrt(2) - i sqrt(1/4 + 1/sqrt(2))

I did not learn any of this math in High School and I was in the best math classes they offered.

Well, the solution above requires nothing except the quadradic formula, and lots of symbol manipulations.

I'm curious about what other approaches there are...nothing has come to mind so far.

Well, I did try to manually do a 4th power polynomial general solution. =) But, that is quite honestly better done by computer.

I did find it interesting that all the roots start with -1/2. Which means the substitution

W := X+1/2
or
X = W - 1/2

might be somehow "easier". (and it does generate a 4th degree polynomial with no 3rd degree term).

Hi Palarran and Laeyakk,

Sorry that I have not gotten back to you guys for the whole week. Been extremely busy and holiday. Congradulations for your solutions!!! Please send me a tell tomorrow before 11am or after 4pm EST to claim your prizes. I am not sure what my play schedule will be for the remainder of Dec because of Final exams and holiday; so find me tomorrow if you need the pp for xmas shopping =). Both solutions are very similar and use the completing the square trick. Very good.

P.S. I have another 40kpp in my bank still so I can afford to pay for 4 other different solutions to the same problem.

P.S.S. If you are a mathematician and enjoy algebra, try to answer the following:

Given p(X) = X^4 + 2X^3+X^2-2X-1, irreducible over the rationals. Compute the splitting field of p(X). The answer for this will get 20kpp



Best,
Foreverlive.

my brain seriously hurts now and I'm going to sue.

You didn't specify how it was to be described.

edit rewrite.
The roots of that polynomial are:
X_1 = -1/2 + 1/sqrt(2) + sqrt(-1/4 + 1/sqrt(2))
X_2 = -1/2 + 1/sqrt(2) - sqrt(-1/4 + 1/sqrt(2))
X_3 = -1/2 - 1/sqrt(2) + i sqrt(1/4 + 1/sqrt(2))
X_4 = -1/2 - 1/sqrt(2) - i sqrt(1/4 + 1/sqrt(2))

As such, one description of the splitting field is
Q[X_1, X_2, X_3, X_4]

This contains all the roots of the polynomial, and is generated by the roots.

Where you looking for a concise description of the splitting field? Or do I win the "I know the defintion of a splitting field" points?

Hmm...
(whee, more edits)

Q[ sqrt(-1+2*sqrt(2)), i*sqrt(1+2*sqrt(2)) ] ?

Very fast replies, guys. =) I am sory that I did not provide enough information on what I was looking for. Here's what I had in mind:

Compute the splitting field of p(X). I want you to characterize the extension of Q (denotes the rationals now onward) where p(X) factors completely, and by splitting field, we want the smallest extension. A very intuitive and straight forward guess is definitely Q(x1, x2, x3, x4) where the x's are the roots of p(X). p(X) definitely factors completely in this field. Does it factor in any subfield? If so, find it. If not, prove it doesn't. We also would want to view this as a simple extension of the rationals, ie, Q(w). Also need the degree of the extension, [Q(w):Q].

This is definitely getting more and more interesting if you like algebra and/or number theory.


Best,
Forevelive.

Given...

X_1 = -1/2 + 1/sqrt(2) + sqrt(-1/4 + 1/sqrt(2))
X_2 = -1/2 + 1/sqrt(2) - sqrt(-1/4 + 1/sqrt(2))
X_3 = -1/2 - 1/sqrt(2) + i sqrt(1/4 + 1/sqrt(2))
X_4 = -1/2 - 1/sqrt(2) - i sqrt(1/4 + 1/sqrt(2))


Let a=sqrt(-1+2*sqrt(2))
Let b=i*sqrt(1+2*sqrt(2))
Let S = Q[ a, b ].

X_1 = (-1/4) + (1/4)a^2 + (1/2)a
X_2 = (-1/4) + (1/4)a^2 - (1/2)a
X_3 = (-3/4) - (1/4)a^2 + (1/2)b
X_4 = (-3/4) - (1/4)a^2 - (1/2)b

So p(x) factors completely in S.

Now to prove that p(x) won't factor completely in any subfield of S...umm...

...I think I'll leave that until tomorrow. :P It's really getting into unfamiliar territory for me.

a and b are roots of 4th degree polynomials, so I guess [S:Q] = 4*4 = 16?

One thing I noticed was that x^4 + 2x^3 + x^2 - 2x - 1 can be factored into...
(x^2 + cx + c) (x^2 + dx + d)
where c = 1-sqrt(2) and d = 1+sqrt(2).

Any field generated by a polynomials roots is the smallest field containing those roots.

By definition almost, and the proof is pretty easy if I recall correctly edit from here on:, and it is one of those fundamental results like "dx^n/dx = nx^(n-1)".



note that
F_1 := Q[X_1, X_2, X_3, X_4] is the SAME field as F_2 := Q[a,b] above. Because each contains the other. (:= shorthand for "defined to be equals").

F_2 is a more concise way of describing F_1. =p~

You didn't prove that "a" was in every field that the polynomial factors in. It's pretty easy, but it should be done.

a = (X_1-X_2)/2
b = (X_3-X_4)/2

Thus, Q[a,b] is a subfield of Q[X_1,..,X_4].

You already proved that
Q[X_1,...,X_4] was a subfield of Q[a,b], hence
Q[a,b] = Q[X_1,...,X_4]

Hard part, I think, is showing
Q[ b ] != Q[a,b]

Q[a] is clearly a strict subfield of Q[a,b], because Q[a] contains no imaginary elements.

I will assume for now that Q[ b ] is a strict subfield of Q[a,b]. (I can think of two ways to prove it. First, that Q[ b ] contains no real terms of degree 4 over the rationals, or second showing that b/i and a are Q-linear independant.)

Let a=sqrt(-1+2*sqrt(2))
Let b=i*sqrt(1+2*sqrt(2))

a^2 = -1 + 2 sqrt(2)
b^2 = 1 + 2 sqrt(2)

notice a^2 and b^2 are Q-linearly dependant.

So, let c = sqrt(2)

Then, I believe the following are all Q-linear independant:
a, b, c (aka a^2 or b^2), ac (aka a^3), bc (aka b^3)

That means our field is of degree 5 over Q. Proof will need more work.

Heh, I've got some reading to do just to understand all of that. My one abstract algebra course only covered groups (with brief mention of fields).

Yup, I was aware that Q[a,b] was the same field as Q[X_1,X_2,X_3,X_4]. I was hoping a cleaner definition would make it easy to find the degree of the extension, but I guess there's more to it than that. :)

Heh Palarran, I'm just using mathworld mostly.

I don't remember much of my galois theory course. It never really sunk in. =/ I think my problem is I never used it in a later course.

The definition of a splitting field is pretty simple: it is the smallest field containing Q in which a particular polynomial factors linearly. (Q is the rational numbers, otherwise known as fractions).

So, if you take the field Q, then add in the roots of your polynomial, then extend it to a field (add in powers and inverses of the roots), you end up with the smallest field that contains Q and the roots of the polynomial.

He originally asked for the splitting field, so I gave the smart-ass answer of "Q with the roots of the polynomial". ;)

Reducing it to Q[a,b] is a good step. (Q[a,b] is shorthand for "Q extended by a and b".

Now, ideally we'd find a number w for which Q[w] is the splitting field. This is called a simple extension. If you have a simple extension, then the dimension of Q[w] over Q (aka, Q[w]:Q) is simply the degree of w.

The degree of w is the degree of the smallest polynomial which has w as a root.

However, I think that for a non-simple extension, things aren't as simple. However, my certainty has plummetted at this point: from here on out, I'm basically doing bullshit that I think might be right. =)

Because a^2 and b^2 are linearly dependant over Q, we end up with less dimensions than we'd expect.

In fact, we want to look at a number of elements, and see which are Q-linearly dependant.

a^0, a, a^2, a^3, b, b^2, b^3, ab, ab^2, ab^3, a^2b, a^3b, a^2b^3, a^3b^2, a^3b^3.

We know a^2 and b^2 can be replaced with sqrt(2), up to multiplication and addition by Q.

So, after replacing a^2 and b^2 with sqrt(2), and throwing out collisions, we get:
1, a, sqrt(2), a sqrt(2), b, b sqrt(2), a b, ab sqrt(2)

Now, recall:
a = sqrt( 2sqrt(2) - 1 )
b = i sqrt( 2sqrt(2) + 1 )

then a b = i sqrt( 8 - 1 ) = i sqrt( 7 )

changing things to:
1, a, sqrt(2), a sqrt(2), b, b sqrt(2), i sqrt(7), i sqrt(14)

Which seems to be saying the degree of Q[a,b]:Q is 8.

Which leaves the questions: is this right, and how do I prove it? =)

thank you...
















































This thread, made my eyes bleed. :(

God I've been meaning to post this...
Q: What do you call a Polak in a F15?

A: A simple pole in a complex plane (http://35.8.247.219/home/modules/pdf_modules/m59.pdf).

Rofl!

And after a long delay... Some proof.

The polynomial in question is irreduceable and of degree 4. That means that the degree of each root over Q is 4.

My previous post showed that the splitting field is of at most degree 8.

We know that the degree of the splitting field of an irreduceable polynomial of degree 4 is divisable by 4. So, the degree is either 4 or 8.

One of the roots of the polynomial is a real root. The smallest field containing the the real root (aka Q[a], "a" defined in my previous post) is of degree 4.

The complex root is a strict extension to the smallest field containing the real root. This means the dimension of Q[a][b] over Q[a] is at least 2.

Edit: damn stupid ezcodes.

The degree of Q[a,b]:Q = Q[a][b]:Q = Q[a][b]:Q[a] * Q[a]:Q >= 2 * 4

But, Q[a,b] is also <= 8, so Q[a,b] = 8.

OMG let this thread die already rofl, jesus someone shoot these math nerds =)

So....with all that whats been basically proven is 8 divided by 4 equals 2.

We also know Q[a,b]:Q <= 8.

Thus Q[a

...at which case he passed out from overload from thinking so hard that late at night.

All mathematical proofs are one large conditional tautology. =p

Well, all valid mathematical proofs that is. =)

I just wanted to finish this.

I am not completely sure that Q[a,b]:Q = Q[a,b]:Q[b] * Q[b]:Q, I am certain Q[a,b]:Q <= Q[a,b]:Q[b] * Q[b]:Q. This vaguely worries me, but I am pretty sure what I need is true. And I'm slack... =)