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Thread: Snead: Youthful Rams on the Rise

01112014 #1
Snead: Youthful Rams on the Rise
By Jonathan Webb
Approaching his third offseason as the Rams’ general manager, Les Snead spoke of an organization on a discernible positive progression in his first meeting with St. Louis media since the conclusion of the 2013 season.
During the team’s second consecutive sevenwin season, Snead oversaw the swift ascension of a bevy of firstyear players, including RB Zac Stacy—who was named the team’s rookie of the year after rushing for 973 yards, thirdmost in Rams history by a rookie—and LB Alec Ogletree, who established a new Rams rookie record with 155 tackles.
As encouraging as that start was, Snead anticipates it only represents the beginning for a rookie class that included team leaders in touchdowns (Stacy, seven), tackles (Ogletree), and allpurpose yardage (Tavon Austin).
“I’m harder on them than anybody, because there should never be a moment when you’re satisfied,” Snead said.
The 79 season was certainly not without its challenges, which were often related to the team’s considerable youth. For the second consecutive season, the Rams opened the year as the league’s youngest team, and Snead acknowledged the learning curve can at times be steep. The growing pains were evident during a 13 start to the season, a series of early setbacks punctuated by a rash of earlyseason penalties on special teams and a scuffling ground game. True to the vision of Snead and Head Coach Jeff Fisher, both areas proved to be strengths by season’s end behind the Rams’ youthful, yet talented personnel.
"If you do things this way, there’s definitely going to be some spilled milk,” Snead said. “But you learn the hard way, and I think this team is learning and getting experience. You’re building a foundation."
That foundation proved sturdy enough to win seven games in the NFL’s toughest division, despite having been dealt the loss of starting QB Sam Bradford for the season just seven weeks into the year. The Rams weathered much of that storm behind the performance of QB Kellen Clemens, who accounted for four of the Rams’ seven wins in his nine starts. Snead credited Clemens not only for his onfield performance in Bradford’s absence, but also for the leadership the eightyear veteran offered to the comparatively youthful Rams’ offensive weapons.
"One of the reasons we went to Kellen is that he knows these receivers,” Snead said. “He knows their strengths and weaknesses, he knows what they need to improve on. The guy could coach. He was a guy that would be after practice before he had to make starts, working with the guys. I like the way our receivers progressed."
The 2013 season could easily be defined by the word ‘progress,’ a description befitting an organization that remained a few steps from contention, but has demonstrated plenty reason for optimism in the years ahead.
"If you take the whole thing,” Snead said, “to be inconsistent is from a metrics standpoint," with the roster turnover and having the youngest team in the league, you can see this thing heading in the right direction."
Well Les .. I agree about the "right direction" but your explanation is over my head.
Introduction: basic definitions and statement of the results ... dense closed geodesics, at least from the C1generic metric standpoint:.
THE C1 GENERAL DENSITY THEOREM
FOR GEODESIC FLOWS
M´ ARIO BESSA AND MARIA JOANA TORRES
Abstract. Given a closed Riemannian manifold, we prove the C1general density theorem
for geodesic flows. More precisely, that C1generic metrics have dense closed
geodesics.
MSC 2000: primary 28D05; 54H20; secondary 37B20.
keywords: Periodic points, topological dynamics, closing lemma.
1. Introduction: basic definitions and statement of the results
A Riemannian manifold (M, g) is a C1manifold with an Euclidean inner product gx
in each TxM which varies smoothly with respect to x ∈ M. So a Riemannian metric is
a smooth section gM → Symm+
2 (TM), where Symm+
2 (TM) is the set of positive bilinear
and symmetric forms in TM. Pick a Riemannian metric on TM and denote by dTM(·, ·)
the geodesic distance associated to it on TM. Note that since all Riemannian metrics are
Lipschitz equivalent on compact subsets, the choice of the metric on TM is not important.
The geodesic flow of the metric g is the flow on TM defined by
φt
g : TM −→ TM
(x, v) 7−→ (γgx
,v(t), ˙ γgx
,v(t)),
where, γgx
,v : R −→ M denotes the geodesic starting at x with initial velocity v, x ∈ M,
v ∈ TxM. Since the speed of the geodesic is constant, we can consider the flow restricted
to UT(M) := {(x, v) ∈ TM : gx(v, v) = 1}.
Clearly, the orbit of a point (x, v) ∈ UT(M) consists of the tangent vectors to the
geodesic defined by (x, v) and periodic orbits for the geodesic flow correspond to closed
geodesics on M. We write Per(φt
g) ⊂ UT(M) for the set of closed geodesics. Recall
that (x, v) ∈ UT(M) belongs to the nonwandering set of φt
g, denoted by
(φt
g), if for
every neighborhood V of (x, v) there exists tn → ∞ such that φtn
g (V ) ∩ V 6= ∅. We
say that (x, v) ∈ UT(M) is a φt
grecurrent point, and we denote this set by R(φt
g), if
given any neighborhood V of (x, v), there exists tn such that φtn
g ((x, v)) ∈ V . We have
R(φt
g) ⊂
(φt
g). Along this paper we are going to consider that M is closed and with
dimension ≥ 2. We observe that the geodesic flow keeps the Liouville volume invariant.
Thus, Poincar´e’s recurrence theorem (see e.g. [8]) asserts that Lebesgue almost every point
is recurrent. Therefore, we conclude that Lebesgue almost every point is nonwandering
and so UT(M) =
(φt
g).
Let Rk(M) denote the set of Ck Riemannian metrics in M endowed with the Ck
topology, k ≥ 1. We denote by R3(M) the C1closure of R3(M) in R1(M), which is a
Baire space because R3(M) is clearly C1dense in R1(M) (i.e. R3(M) = R1(M)). A
property in R1(M) is said to be C1generic if it holds in a residual subset of R1(M),
that is, the subset of points in which the property is valid contains a C1dense Gδ. [/LEFT]
There are 4 more pages, but I'll stop here .. Hope you are correct Les!

01112014 #2
Re: Snead: Youthful Rams on the Rise
Metrics make my head hurt

01112014 #3
Re: Snead: Youthful Rams on the Rise
(VS 1) For all in V, (commutativity of addition).(VS 2) For all in V, (associativity of addition).
(VS 3) There exists an element in V denoted by such that for each in V.
(VS 4) For each element in V there exists an element in V such that .
(VS 5) For each element in V, .
(VS 6) For each pair of element in F and each element in V, .
(VS 7) For each element in F and each pair of elements in V, .
(VS 8) For each pair of elements in F and each pair of elements in V, .
Or am I wrong in finding a vectored solution?
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