(((0.091*28.82)-x)*92.736)*100 = ((7.264*100)*(41-(28.82-x)))

[Ales Dvorak]

<BLOCKQUOTE>quote:</font><HR>Originally posted by Murph:

tell us how this equation applied to CM...<HR></BLOCKQUOTE>

Maybe this equation is more related with his

occupation.

[Juardis]

<BLOCKQUOTE>quote:</font><HR>Originally posted by Baker350:

I am sorry if my algebra skills are not up to your high standards Juardis. I haven't had to mess with it since college. You could of just helped a CM brother out instead of chastising my meager abilities.<HR></BLOCKQUOTE>

It's not your algebra skills I'm chastising, it's the fact that you expect us to beleive that this is related to CM. So you could have come clean instead of feeding us a bull****e line about this being related to CM. But you got your answer so all is well in the world again.

In the off chance that I am wrong and this IS somehow related to CM, I'd sure like to know how.

[MrSpkr]

<BLOCKQUOTE>quote:</font><HR>Originally posted by Juardis:

Yeah, I can do that problem. Took me about 3 minutes. Extremely easy if you've taken algegra already. You should try it yourself sometime.<HR></BLOCKQUOTE>

And you say you aren't sure if you are ready to wander back into the muck of the Cess right now.

[Pvt. Ryan]

<BLOCKQUOTE>quote:</font><HR>Originally posted by Ales Dvorak:

Maybe this equation is more related with his

occupation.<HR></BLOCKQUOTE>

I don't think Algebra has any real life applications.

[ElGuapo]

Show your work? I'll bet you a copy of CB:BB that this has nothing to do with CM. This is homework of some type and he just suckered you all into doing it for him. The way that you arrived at it would not matter at all if he was just trying to plug it into some equation.

Gimme a break, man. Do your own damn homework.

Guap, shaking his head

[Joe Shaw]

<BLOCKQUOTE>quote:</font><HR>Show your work? I'll bet you a copy of CB:BB that this has nothing to do with CM. This is homework of some type and he just suckered you all into doing it for him. <HR></BLOCKQUOTE> Actually it probably WAS CM related ... his Mom probably said he couldn't play CM 'till he got his algebra homework done.

Joe

[ElGuapo]

<BLOCKQUOTE>quote:</font><HR>Originally posted by Joe Shaw:

Actually it probably WAS CM related ... his Mom probably said he couldn't play CM 'till he got his algebra homework done.

Joe<HR></BLOCKQUOTE>

*grin* Point taken, but he's not playing me, so I don't care. :mad:

[Marco Bergman]

Yes I'd considered the homework angle. But I thought that in the case of genuine algebraic difficulty a solution might lead to insight.( I used to be a part-time maths tutor).

It cost me nothing to take him at face value; I enjoyed the equation.

If he's just a lazy slacker it'll cost HIM far more when he fails the next test...

[lcm1947]

Good for you Marco. My hat off to you sir.

[Pascal DI FOLCO]

No, its really CM related : x is penetration value for APDS shell arriving at 28.82 degree angle over a composite armour plate ... that has African sparrow sh*t spread over :eek: :eek: :eek:

[Ales Dvorak]

<BLOCKQUOTE>quote:</font><HR>Originally posted by Pvt. Ryan:

I don't think Algebra has any real life applications. <HR></BLOCKQUOTE>

Look at his occupation! ( loan )!

[WineCape]

... and I thought wine were a simple business... tsk, tsk

----------------

WINE, noun: Fermented grape-juice known to the Women's Christian Union as "liquor," sometimes as "rum." --> Wine, Madam, is God's next best gift to man.

-- Ambrose Bierce (1842-1914), "The Devil's Dictionary" (1911)

[Nabla]

What really amazes me is that this topic is two pages long...

[Kanonier Reichmann]

<BLOCKQUOTE>quote:</font><HR>Originally posted by Pascal DI FOLCO:

No, its really CM related : x is penetration value for APDS shell arriving at 28.82 degree angle over a composite armour plate ... that has African sparrow sh*t spread over :eek: :eek: :eek:<HR></BLOCKQUOTE>

Where's rexford when you need him?

Regards

Jim R.

[Wesreidau]

Reading grog-porn I bet.

PeterNZ

[Diceman]

<BLOCKQUOTE>quote:</font><HR>Originally posted by PeterNZer:

Reading grog-porn I bet.

PeterNZ<HR></BLOCKQUOTE>

Mmmm, grog-porn - in the imortal words of Hakko Ichiu aka Ethan:

Swinging his manly bulk over the shot-trap, SS-Oberspankführer Großeier mounted his Pzkw V Ausf. D with tight-fitting Zimmerit coating and anti-rocket corsets. Lovingly he stroked the long, hardened steel barrel of the 75L48 cannon. His piercing gaze fell upon the naked torso of his driver, Gefreiter Schaffevögeln, rhythmically swabbing out the barrel. Although he could feel in the core of his being that his commander was giving him the eye, Schaffevögeln suddenly started to slack.

"Ram harder, damn you!" screamed Großeier as he saw his driver's long tool begin to droop. "Give that barrel a damn good reaming out!"

Schaffevögeln redoubled his efforts; sweat began to drip down his chest onto his shiny leather boots. The force of it made the Panzer's barrel quiver. "That's more like it!" screamed Großeier, "Harder, harder, harder! Yes, that's it!"

Completely satisfied with his driver's reaming, Großeier turned his attention to the ammo cart that was being pulled by a team of strapping, sweaty horses...

Well, I hope that's cleared the air a bit. Sure made me feel manly.

------

------------------

Ethan

Link: http://www.battlefront.com/cgi-bin/bbs/ultimatebb.cgi?ubb=get_topic&f=1&t=011566&p=20

Who sais the search function doesn't work?

[ 10-06-2001: Message edited by: Diceman ]

[Ales Dvorak]

<BLOCKQUOTE>quote:</font><HR>Originally posted by Nabla:

What really amazes me is that this topic is two pages long... <HR></BLOCKQUOTE>

Almost three.

[Mikael]

Make that three.

[Mikael]

Oops...wrong timing.

[Ales Dvorak]

Almost.

[StellarRat]

While all you math wizards are at it, could somebody supply a proof that this equation has no solutions?

x^n + y^n = z^n where n is bigger than 2

[russellmz]

<BLOCKQUOTE>quote:</font><HR>Originally posted by StellarRat:

While all you math wizards are at it, could somebody supply a proof that this equation has no solutions?

x^n + y^n = z^n where n is bigger than 2

<HR></BLOCKQUOTE>

i feel smart looking this up: if you want the 150 page version of the proof you're on your own. http://www.pbs.org/wgbh/nova/proof/wiles.html

onto the Riemann hypothesis math boyz!

.

.

. http://www.best.com/~cgd/home/flt/flt08.htm

The Proof of Fermat's Last Theorem

For a quick definition of many of the terms used here, you may refer to the Glossary.

External references for this section: [Gou], [Rib], [Rih]

Contents:

Frey curves

Proof of Fermat's Last Theorem from the Taniyama-Shimura conjecture

Proof of the semistable case of the Taniyama-Shimura conjecture

Frey curves

Suppose there were a nontrivial solution of the Fermat equation for some number n, i. e. nonzero integers a, b, c, n such that

Then we recall that around 1982 Frey called attention to the elliptic curve

Call this curve E. Frey noted it had some very unusual properties, and guessed it might be so unusual it could not actually exist.

To begin with, various routine calculations enable us to make some useful simplifying assumptions, without loss of generality. For instance, n may be supposed to be prime and 5. b can be assumed to be even, a 3 (mod 4), and c 1 mod 4. a, b, and c can be assumed relatively prime.

The "minimal discriminant" of E, can be computed to be - a power of 2 times a perfect prime power. One unusual thing about E is how large the discriminant is.

The conductor is a product of primes at which E has bad reduction, which is the same as the set of primes that divide the minimal discriminant. However, the exact power of each prime occurring in the conductor depends on what type of singularity the curve possesses modulo the primes of bad reduction. The definition of the conductor provides that p divides the conductor only to the first power if x(x-a)(x+ has only a double root rather than a triple root mod p. Now, any prime can divide only a or b but not both, since otherwise it would also divide c, and we have assumed a, b, and c are relatively prime. Hence the the polynomial will have the form x(x+d) mod p, where (p,d) = 1. Hence there is only at most a double root modulo any prime, and therefore the conductor is square-free. In other words, E is semistable.

There are other odd things about E, which have to do with specific properties of its Galois representations. Because of these, Ribet's results allow us to conclude that E cannot be modular.

Proof of Fermat's Last Theorem from the Taniyama-Shimura conjecture

After Frey drew attention to the unusual elliptic curve which would result if there were actually a nontrivial solution to the Fermat equation, Jean-Pierre Serre (who has made many contributions to modern number theory and algebraic geometry) formulated various conjectures which, sometimes alone and sometimes together with the Taniyama-Shimura conjecture, could be used to prove Fermat's Last Theorem.

Kenneth Ribet quickly found a way to prove one of these conjectures. The conjecture itself doesn't really talk about either Frey curves or FLT. Instead, it simply states that if the Galois representation associated with an elliptic curve E has certain properties, then E cannot be modular. Specifically, it cannot be modular in the sense that there exists a modular form which gives rise to the same Galois representation.

We need to introduce a little additional notation and terminology to explain this more precisely. Let S(N) be the (vector) space of cusp forms for (N) of weight 2. "Classical" theory of modular forms shows that S(N) can be identified with the space of "holomorphic differentials" on the Riemann surface X(N). Furthermore, the dimension of S(N) is finite and equal to the "genus" of X(N). "Genus" is a standard topological property of surfaces, which is intuitively the number of holes in the surface. (E. g. a torus, such as an elliptic curve, has genus 1.)

But there are relatively simple explicit formulas for the genus of X(N). These formulas, developed long ago by Hurwitz in the theory of Riemann surfaces, involve the index of (N) in G. A fact of crucial importance is that for N less than . 11, the genus of X(N), and hence the dimension of S(N), is zero. In other words, S(N) contains only the constant form 0 in that case. We shall use this fact about S(2) very soon.

There are certain operators called Hecke operators, after Erich Hecke, on spaces of modular forms, and for the subspace S(N) in particular, since they preserve the weight of a form. Hecke operators can be defined concretely in various ways. There is a Hecke operator T(n) for all n 1. There are formulas that relate T(n) for composite n to T(p) where is a prime dividing n, so T(p) for prime p determine all T(n).

All T(n) are linear operators on S(N). If there is an f in S(N) that is a simultaneous eigenvector of all T(n), i. e. T(n)(f) = (n)f, where (n) C, f is called an eigenform. (Nontrivial eigenforms need not exist, e. g. if S(N) has dimension 0.) f is said to be normalized if its leading Fourier series coefficient is 1. In that case, the eigenvalues (n) turn out to be the Fourier series coefficients in the expansion

It can be shown that if f(z) is a cusp form which is a normalized eigenfunction for all T(p), then there is an Euler product decomposition for the L-function L(f,s). This is obviously of great technical usefulness in relating L-functions of forms and those of elliptic curves (which are Euler products by definition).

If f S(N) is a normalized eigenform of all Hecke operators, it can in fact be shown that the coefficients in the Fourier expansion are all algebraic numbers and that they generate a finite extension K of Q.

Prime ideals of the ring of integers of K are the analogues of prime numbers of Q. In the case that f is a normalized eigenform it is possible to carry out the construction of a Galois representation (f,) of Gal(/Q) for any prime ideal of the ring of integers of K.

At last we can describe what Ribet proved. Suppose E is a semistable elliptic curve with conductor N and that its associated Galois representation (E,p) for some prime p has certain properties. Suppose 2 divides N (which is true for Frey curves). If E is modular, then there is a normalized eigenform f and a prime ideal lying over p (i. e. one of the prime factors of p in the extension field generated by the Fourier coefficients of f) such that the Galois representation (f,) is (E,p). Ribet showed that it is possible to find an odd prime q p which divides N such that there is another f' S(N/q), and a corresponding prime ideal ' of the ring of integers in the field generated by the coefficients of f' such that (f',') gives essentially the same Galois representation. This is known as the "level lowering" conjecture since it asserts that under the right conditions there is an eigenform of a lower level that gives essentially the same representation.

But this process can be repeated as long as N has any odd prime factors. It is important that the curve E is semistable so that N is square-free. This means all that all odd prime factors of N can been eliminated, so there must be a nontrivial eigenform of level 2, i. e. in S(2), that gives essentially the same Galois representation. And that is a contradiction, since S(2) has dimension 0, hence contains no non-trivial forms. The contradiction means that E can't be modular.

Now we invoke the "unusual" properties of the Frey curve resulting from a solution of FLT. These properties allow it to be shown that the associated Galois representation has the properties required to apply Ribet's result. Hence the Frey curve can't be modular.

But the Frey curve is semistable, so the semistable case of the Taniyama-Shimura conjecture, which Wiles proved, implies the curve is modular. This contradiction means that the assumption of the existence of a nontrivial solution of the Fermat equation must be wrong, and so FLT is proved.

Proof of the semistable case of the Taniyama-Shimura conjecture

Not very surprisingly (since it was such hard work), the proof is quite technical. However, the outline of it is relatively simple. In the following, we assume that E is a semistable elliptic curve with conductor N. We have to prove E is modular.

We know we can construct a Galois representation (E,p): G->GL(Z) for any prime p. To show that E is modular, we have to show this representation is modular in a suitable sense. The wonderful thing is, this needs to be done for only one prime p, and we can "shop around" for whatever prime is easiest to work with.

To show (E,p) is modular involves finding a normalized eigenform f in S(N) with appropiate properties. The properties required are that the eigenvalues of f, which are its Fourier series coefficients, should be congruent mod q to trace((E,p) ()) for all but a finite number of prime q. ( G is the "Frobenius element".) We know that the trace is, for q prime to pN, the coefficient a = q + 1 - #(E(F)) of the Dirichlet series of L(E,s).

The longest and hardest part of Wiles' work was to prove a general result which is roughly that if (E,p) is modular then so is (E,p). In other words, to show that E is modular, it is actually sufficient just to show that (E,p): G->GL(Z/pZ) is modular. This is called the "modular lifting problem".

The problem boils down to assuming that (E,p) is modular and attempting to "lift" the representation to (E,p). This is done mainly by working with the theory of representations as much as possible, without specific reference to the curve E. The proof uses a concept called "deformation", which suggests intuitively what goes on in the process of lifting.

The outcome of this part of Wiles work is:

Theorem: Suppose that E is a semistable elliptic curve over Q. Let p be an odd prime. Assume that the representation (E,p) is both irreducible and modular. Then E is a modular elliptic curve.

At this point, all we have to do is find a single prime p such that (E,p) is irreducible and modular. But Langlands and Tunnell had already proven in 1980-81 that (E,3) is modular.

Unfortunately, this isn't quite enough. If (E,3) is irreducible, we are done. But otherwise, one more step is required. So suppose (E,3) is reducible. Wiles then considered (E,5). That may be either reducible or irreducible as well. If it is reducible, Wiles proved directly that E is modular.

So the last case is if (E,5) is irreducible. Wiles showed that there is another semistable curve E' such that (E',3) is irreducible, and hence E' is modular by the above theorem. But Wiles could also arrange that the representations (E',5) and (E,5) are isomorphic. Hence (E,5) is irreducible and modular, so E is modular by the theorem.

--------------------------------------------------------------------------------

Back to Fermat's Last Theorem Home Page

Copyright © 1996 by Charles Daney, All Rights Reserved

Last updated: March 13, 1996

[ 10-06-2001: Message edited by: russellmz ]

[Kanonier Reichmann]

My head STILL hurts! (he says with a handkerchief on his head with knots tied in each corner).

Regards

Jim R.

[OGSF]

<BLOCKQUOTE>quote:</font><HR>Originally posted by russellmz:

So the last case is if (E,5) is irreducible. Wiles showed that there is another semistable curve E' such that (E',3) is irreducible, and hence E' is modular by the above theorem. But Wiles could also arrange that the representations (E',5) and (E,5) are isomorphic. Hence (E,5) is irreducible and modular, so E is modular by the theorem.

<HR></BLOCKQUOTE>

I'll get me coat....

[Porajkl]

Now we're really loooong - third page...

And by the way, this problem troubles me for two nights:

what is the square root of 1.3452?

My results are all almost around -4,55"

Is this right?

edit:

I was wrong, we aren't yet on Page 3. I guess we aren't pretty enough

[ 10-08-2001: Message edited by: Porajkl ]