text stringlengths 30 4k | source stringlengths 60 201 |
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lli’s law:
2
H = + = constant.
p
ρ
u
2
(439)
The constancy of H in irrotational flow is a famous result, and has many simple qualitative
consequences. It requires that the pressure in a fluid is smaller when the velocity is larger,
and is a statement of conservation of energy when viscous dissipation can safely be ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
the time evolution of Γ? We know that
DΓ
Dt
=
(cid:90)
d
dt C(t)
(cid:90)
u · dl =
Du
Dt
· dl + u ·
D(dl)
Dt
.
(443b)
(444)
The first term concerns changes in the velocity field and is identically zero since
(cid:90)
Du
Dt
· dl =
(cid:90)
1
ρ
−\p · dl =
(cid:90)
1
ρ
(\ × \p) · ndA = 0.
(445)
The seco... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
the flow is irrotational, we just need to find solutions to Laplace’s equation
to obtain solutions to the Euler equations. Let’s write down a couple of these to gain some
intuition: our aim being to acquire techniques to begin to think about airplane flight.
18.1 Point source
We know from electrostatics that a solutio... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
θ2
(cid:18)
r
d
dr
(cid:19)
r
dg
dr
= gk2 .
For k = 0,
(cid:54)=
f = C sin(kθ) + D cos(kθ),
(452)
(453)
(454a)
(454b)
(455a)
with continuity of u requiring k to be an integer. Turning to the radial part we guess that
g = r . The radial equation then requires that α = ±k, giving
α
However, if k = 0 then ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
This is called a point vortex solution. If
we consider the circulation about a loop containing the origin
(cid:90)
u · dl =
(cid:90)
2π
0
uθrdθ = 2πD = Γ
(461)
Thus D = Γ/2π, where Γ is the circulation about the point vortex.
18.4 Flow around a cylindrical wing
Okay, so this isn’t the true shape of an airplane... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
(cid:90)
2π
p(R, θ) R sin θdθ,
0
(cid:90)
2π
p(R, θ) R cos θdθ.
0
We can determine the pressure distribution from Bernoulli’s Law
ρ
p = p0 − (\φ)2
2
r=R
= p0 −
(cid:18)
ρ
2
2u0 sin θ −
(cid:19)
2
Γ
2πR
.
Putting this into the above relations we find that
FD = 0.
(465a)
(465b)
(466)
(467)
This result i... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
:1, it is acceptable to consider 2D flow. In the next part, we will study how things
change if we alter the shape of the wing. To do so will require conformal mapping.
19 Stream functions and conformal maps
There is a useful device for thinking about two dimensional flows, called the stream function
of the flow. The s... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
an imaginary part v(x, y), where u and v are real functions, i.e.:
f (z) = f (x + iy) = u(x, y) + iv(x, y).
(472)
For example, if f (z) = z then u = x2 − y and v = 2xy. What then is df /dz? Since we
are now in two-dimensions we can approach a particular point z from the x-direction or the
y-direction (or any other... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
analytic function of the complex variable z = x + iy. We call w the complex potential.
Another important property of 2D incompressible flow is that both φ and ψ satisfy Laplace’s
equation. For example, using the Cauchy-Riemann equations we see that
∂ψ
∂x2
+
∂ψ
∂y2
= −
∂2ψ
∂x∂y
+
∂2ψ
∂y∂x
= 0.
(474)
The same... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
Also
w(z) = −i ln z is the complex potential for a point vortex since
Re (w(z)) = Re −i ln(re
iθ)
= θ,
and we know that φ = θ is the real potential for a point vortex. Thus
(cid:18)
w(z) = u0 z +
(cid:19)
R2
z
−
iΓ
2π
ln z
(478)
(479)
is the complex potential for flow past a cylinder with circulation Γ.
... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
with the key
question: Given flow past a circular cylinder in the z-plane can we choose a mapping so
as to obtain in the Z-plane uniform flow past a more wing-like shape? (Note that we have
brushed passed some technical details here, such as the requirement that dF/dz = 0 at any
point, as this would cause a blow-up o... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
c
a
2
+
Y
2
a − c
a
2
= 1,
which is the equation of an ellipse, provided c < a.
20 Classical aerofoil theory
(483)
(484)
(485)
We now know that through conformal mapping it is possible to transform a circular wing into
a more realistic shape, with the bonus of also getting the corresponding inviscid, irrot... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
489)
(490)
U − iV =
dw/dz
dZ/dz
=
(cid:16)
u0 e
−iα − e
(cid:17)
iαR2
iΓ
z2 − 2πz
1 − R2/z2
On the surface of the body we have z = Reiθ, so the velocities become
�
u0 e
U − iV =
�
−iα − e−2iθ iα − iΓe
e
−2iθ
−iθ
2πR
1 − e
.
(491)
At θ = 0 and θ = π we are in trouble because the velocities are infinite. Nota... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
we
can do by considering a shifted circle, that passes through z = R but encloses z = −R. In
this case we obtain an aerofoil with a rounded nose but a sharp trailing edge. The boundary
of the appropriate circle is prescribed by
iθ
z = −λ + (a + λ)e ,
(495)
where θ is a parameter. First we must modify the complex... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
oukowski theorem.
20.3 Blasius’ lemma
To derive Blasius’ lemma, we consider the force acting per unit length on the wing, f =
F /(cid:96) = (fx, fy), which is obtained by integrating the pressure over the (now arbitrary)
surface contour ∂S
f = −
(cid:73)
∂S
p n ds
(499)
where n is the outward surface normal vector and ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
)
x − v2
y − 2ivxvy)(dx + idy)
2
= (v2
ydx − 2ivxvydx + vxidy − v2
= v2dx − v2
x
xdy + v2
ydx − 2iv2
xdx − v2
= v2
xidy − v2
ydx − v2
xdx + v2
= v2
x + v2
= (v2
x + v2
y)dx − (v2
y)idy
= (v2
2
x + vy)(dx − idy)
= |v|2 dz¯
xidy − v2
yidy
yidy + 2vxvydy
yidy + 2v2
ydx
¯f = i
(cid:73)
ρ
2
v2 dz = i
(cid:73) (cid:18) dw (c... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
510)
26Let’s assume some otherwise analytic
a−1 of the Laurent series f (z) = ∞
simple closed curve γ around z = 0
(cid:80)
function f (z) has a pole at z = 0. The residue is the coefficient
k=−∞ akz . The residue theorem states that for a positively oriented
k
(cid:73)
γ
f (z)dz = 2πi a−1.
97
Computing the integral on ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
a manifestation of d’Alembert’s paradox (now for arbitrarily shaped
wings), which can be traced back to the fact that we neglected the viscosity terms in the
Navier-Stokes equations.
21 Rotating flows
In our above discussion of airfoils, we have neglected viscosity which led to d’Alembert’s
paradox. To illustrate f... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
517)
For the rotating earth, the effect of this force is to simply distort the shape of the object
from a sphere into an oblate ellipsoid. The second term is the Coriolis acceleration which
is velocity dependent. Hopefully you have heard about it in classical mechanics.
We are going to be interested in flows which are mu... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
equations
say that uI and vI are independent of z. Then the last equation says that wI is independent
of z. Thus, the entire fluid velocity is independent of z ! This result, which is remarkable,
is called the Taylor-Proudman theorem. Proudman discovered the theorem, but Taylor
discovered what is perhaps its most remark... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
2u − 2Ω × u,
(520)
(521a)
(521b)
We will use these ideas to revisit the famous problem of the spin-down of a coffee cup that
we discussed at the very beginning of class. You might recall that the problem we had
was that our simple theory of how the spin-down occurred turned out to be entirely false.
We shall now constru... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
with u, we get
u · \p = −ρu · (2Ω × u) = 0.
(525)
This formula states that the velocity field moves perpendicular to the pressure gradient,
which is somewhat against one’s intuition. Hence, the fluid actually moves along lines
of constant pressure. Pressure work is not performed either on the fluid or by the fluid.
G... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
would now like to return to our coffee cup problem, to get the right answer. To do
so, we shall consider the effect of walls on the inviscid flow we calculated in the previous
lecture. For starters, lets consider a jar with the top moving at angular velocity ΩT and
the bottom moving at angular velocity ΩB . Clearly, if... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
p
I ,
ρ ∂y
0 =
1 ∂pI
ρ ∂z
.
(526a)
(526b)
(526c)
In the same way as before, we expect the pressure gradient of the outer flow to force the
boundary layer at the rotating wall. Let’s consider the structure of the boundary layer at
the bottom wall, z = 0. There the equations are
−2Ωv = −
2Ωu = −
1 ∂p
I
ρ ∂x
1 ∂pI
ρ ∂y
0 =... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
ν.
102
(529a)
(529b)
(530)
We require that as z∗ → ∞, f → 0. This implies that A = 0. We are in the frame
of reference moving with the bottom plate, so the no slip boundary condition at z = 0
requires that f (z = 0) = −uI − ivI . Splitting f into its real an imaginary parts implies
u = uI − e−z∗
v = vI − e−z∗
(uI cos(... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
the bottom. What we just assumed was
that the boundary is moving at frequency Ω. If it is not, but instead moving at an angular
frequency ΩB relative to the rotating frame, then we need to change the boundary conditions
a little in the rotating frame. In this case
w
=
(cid:16) ν (cid:17)1 2
/
Ω B
(cid:18) ωˆI − ΩB
2
(c... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
assume the coffee
cup to be a cylinder with a top and a bottom both rotating with angular velocity Ω + (cid:15).
At t = 0 the angular velocity of the boundaries is reduced to Ω. How long does it take to
reach a steady state?
We use the time-dependent formula
∂uI
∂t
∂vI
∂
t
−
2ΩvI =
−
+ 2ΩuI =
−
,
1 ∂pI
ρ ∂x
1 ∂p
I .
ρ ∂... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
end up at the centre of your cup.
√
23 Water waves
If you look out onto the River Charles, the waves that are immediately apparent are surface
waves on the water. However, there are many different types of waves in the rivers and
oceans, which have profound effects on our surroundings. The most dramatic example is a
Tsun... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
relation we get
∂φ
∂t
1
2
+
(u2 + v2) + gh = 0
(544)
on h(x, t), where we have chosen the constant C(t) appropriately to simplify things.
The equations we have derived so far take account of the effect of gravity on the free
surface. We have ignored one important factor, however, which is surface tension. It costs
energ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
knowing that these satisfy Laplace’s equation (we have ignored terms of the form e−ky, as
the surface is at y = 0 and we need all terms to disappear as y → −∞). Putting these into
the surface boundary conditions (8) and (9) gives
ω(cid:15) = Ak,
ωA = g(cid:15) +
γk2(cid:15)
ρ
.
Eliminating A we get the dispersion relat... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
−ω(k)t]
k
dk.
(553)
Now the dominant wavelength in the disturbance corresponds to the diameter of the rock d.
We shall call the corresponding wavenumber k0 = 2π/d. Other wavenumbers will also be
excited but we argue that hk will be very small except when k is very near to k0. Near k0
we have that
ˆ
where ω(cid:48) = ∂ω... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
would be a decay in amplitude due to power
conservation).
This is not so for water waves, which have a different dispersion relation, and we can
highlight the difference between these two cases by considering the wakes behind an airplane
and a boat.
107
(557a)
(557b)
(558)
23.3 The wake of an airplane
The equation gover... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
θ = √
1
M 2 − 1
.
(563)
(564)
Only a narrow region behind the plane knows it exists, and the air ahead doesn’t know
what’s coming!
23.4 Flow created by a 1D ‘boat’
We now consider a boat moving at constant speed across the surface of water. The motion
of the boat generates a disturbance at point (x(cid:48), t(cid:48)).... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
]−Γkt
(cid:90) U t
−∞
dx h e{−i[(k−ω(k)/U ]+(Γ/U )}x(cid:48)
(cid:48) ˆ
k
To simplify things a bit, let’s focus on spatial perturbations at time t = 0
h(x, 0) ∝
(cid:90) ∞
ˆdk h eikx
k
(cid:90) 0
−∞
dx(cid:48)
e{−i[(k − ω(k)/U ] + (Γ/U )}x
(cid:48)
−∞
(cid:90)
= U
∞
dk
−∞
eikx
−i[kU − ω(k)] + Γk
.
(566)
(567)
Now this ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
the Fourier component, and the contour C includes the real k-axis with other contributions
vanishing. If we are considering gravity waves, then U0 is a positive quantity. The integral
has to be evaluated around a contour C in the complex plane. For x > 0 there is no pole
inside the semicircle and the integral is zero. ... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
extended to account for the V-shaped wake
behind a boat, also known as the Kelvin wedge. In 2D, the disturbance generated by the
boat is
h(x, t|
x , t ) = dk h(k)eik(x−x')e iω(k)(t−t') Γk(t−t'
'
e
ˆ
'
).
(570)
tribute significantly are those whose phase velocity in
As before, the only waves that con
the direction... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
suddenly leaving it behind, rolled forward with great velocity, assuming the
form of a large solitary elevation, a rounded, smooth and well-defined heap of water, which
continued its course along the channel apparently without change of form or diminution of
speed. I followed it on horseback, and overtook it still ro... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
KdV) equation
A continuum model for solitary waves in water was first introduced by Boussinesq in 1871.
The theory was developed further by Lord Rayleigh in 1876 and by Korteweg and de Vries
(KdV) in 1895. According to their theory, the spatio-temporal evolution of weakly nonlinear
shallow water waves is described by
∂t... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
2 + f 3.
c
2
This equation possesses an explicit solution with f (X) = 0 as X → ±∞, given by
φ(t, x) =
sech2
c
2
(cid:20) √
c
2
(cid:21)
(x − ct − a) .
(579b)
(580)
This solution describes a right-moving solitary wave-front of speed c.
In 1965, Zabesky and Kruskal showed how one can derive this equation in the continuu... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/6163837c5efa5d6ed969e535ec8f890c_MIT18_354JS15_lectureNotes.pdf |
6.801/6.866: Machine Vision, Lecture 10
Professor Berthold Horn, Ryan Sander, Tadayuki Yoshitake
MIT Department of Electrical Engineering and Computer Science
Fall 2020
These lecture summaries are designed to be a review of the lecture. Though I do my best to include all main topics from the
lecture, the lectures ... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/617445f0e31836831b40d42cb2f11a10_MIT6_801F20_lec10.pdf |
Hapke surfaces, we were able to solve the SfS problem in a particular direction, which
we could then determine object surface orientation along. We can integrate along this direction (the profile direction) to find
height z, but recall that we gain no information from the other direction.
For the Hapke example, we hav... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/617445f0e31836831b40d42cb2f11a10_MIT6_801F20_lec10.pdf |
solve the 3 ODEs above using a forward Euler method with a given step size. For small steps, this approach will be
accurate enough, and accuracy here is not too important anyway since we will have noise in our brightness measurements.
Recall brightness of Hapke surfaces is given by:
r
E =
p
f rac1 + psp + qsq 1 + p... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/617445f0e31836831b40d42cb2f11a10_MIT6_801F20_lec10.pdf |
Case
Let us now generalize our Hapke case above to be any surface. We begin with our Image Irradiance Equation:
E(x, y) = R(p, q)
Let us now consider taking a small step δx, δy in the image, and our goal is to determine how z changes from this small step.
We can do so using differentials and surface orientation:
δz... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/617445f0e31836831b40d42cb2f11a10_MIT6_801F20_lec10.pdf |
z
∂y∂x
∂2
z
∂x∂y
∂2
z
∂y2
⎤
⎥ δx
⎦
δy
⎡
px py
⎤
=
⎣
⎦
qy
qx
δx
δy
= H
δx
δy
Where H is the Hessian of second spatial derivatives (x and y) of z. Note that from Fubuni’s theorem and from our Taylor
expansion from last lecture, py = qx.
Intuition:
2
• First spatial derivatives
∂z
∂x
... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/617445f0e31836831b40d42cb2f11a10_MIT6_801F20_lec10.pdf |
∂y
∂R
∂p
=
r +
Ey =
=
=
s +
+
∂R ∂q
∂q ∂x
+
∂R ∂q
∂q ∂y
s
∂R ∂p
∂p ∂x
∂R
∂q
∂R ∂p
∂p ∂y
∂R
t
∂q
In matrix-vector form, we have:
Ex
Ey
r
s
=
⎡ ⎤
⎡ ⎤
∂R
∂R
∂p
∂p
s ⎢ ⎥
⎢ ⎥
⎣ ⎦ = H ⎣ ⎦
t
∂R
∂q
∂R
∂q
Notice that we have the same Hessian matrix that we had derived for our su... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/617445f0e31836831b40d42cb2f11a10_MIT6_801F20_lec10.pdf |
and side is our gradient of the reflectance
map in gradient space (p, q), and ξ is the step size. Intuitively, this is the direction where we can “make progress”. Substituting
this equality into our update equation for p and q, we have:
δp
δq
= H
ξ
Rp
Rq
Ex
Ey
δξ
=
Therefore, we can formally wr... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/617445f0e31836831b40d42cb2f11a10_MIT6_801F20_lec10.pdf |
ODE quantities of interest.
• The step in the image (x, y) depends on the gradient of the reflectance map in (p, q).
• Analogously, the step in the reflectance map (p, q) depends on the gradient of the image in (x, y).
• This system of equations necessitates that we cannot simply optimize with block gradient ascent or... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/617445f0e31836831b40d42cb2f11a10_MIT6_801F20_lec10.pdf |
factor. Therefore, we can simply omit this factor when we update these variables after taking a step. With this, our updates
become:
1. δx ← ps
2. δy ← qs
3. δz ← (psp + qsq) = rsE2 − 1
Which are consistent with our prior results using our Hapke surfaces. Next, we will apply this generalized approach to Scanning
... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/617445f0e31836831b40d42cb2f11a10_MIT6_801F20_lec10.pdf |
will be taking steps along the brightness gradient. Our solution generates characteristic strips that contain
information about the surface orientation. To continue our analysis of this SfS problem, in addition to defining characteristic
strips, it will also be necessary to define the base characteristic.
1.3 Base Cha... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/617445f0e31836831b40d42cb2f11a10_MIT6_801F20_lec10.pdf |
of one another - this allows computation over them to be parallelizable. Intuition: These sets
of base characteristics can be thought of like a wavefront propagating outward from the initial curve. We expect the
solutions corresponding to these base characteristics to move at similar “speeds”.
1.4 Analyzing “Speed” ... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/617445f0e31836831b40d42cb2f11a10_MIT6_801F20_lec10.pdf |
the brightness gradient and the gradient of the reflectance map in (p, q) space.
Achieved by: Dividing by ( ∂E +
)δξ = ((Ex, Ey ) · (Rp, Rq))δξ =⇒ δE = 1.
∂R
∂x ∂p
∂E ∂R
∂y ∂q
1.5 Generating an Initial Curve
While only having to measure a single initial curve for profiling is far better than having to measure an en... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/617445f0e31836831b40d42cb2f11a10_MIT6_801F20_lec10.pdf |
Autogenerate an Initial Curve
An even more sweeping question: do we even need an initial curve? Are there any special points on the object where we already
know the orientation without a measurement? These points are along the edge, or occluding boundary, of our objects of
interest. Here, we know the surface normal ... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/617445f0e31836831b40d42cb2f11a10_MIT6_801F20_lec10.pdf |
derivatives of x and y are zero:
dx
dξ
dy
dξ
= Rp = 0
= Rq = 0
Additionally, if we have stationary brightness points in image space, we encounter the “dual” of this problem. By definition
stationary points in the image space imply that:
∂E
∂x
=
∂E
∂y
= 0
6
This in turn implies th... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/617445f0e31836831b40d42cb2f11a10_MIT6_801F20_lec10.pdf |
enough to the stationary point, we can approximate that these neighboring points have nearly the same surface
orientation.
• Approach 1: Construct a local tangent plane by extruding a circle in the plane centered at the stationary point with
radius - this means all points in this plane will have the same surface o... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/617445f0e31836831b40d42cb2f11a10_MIT6_801F20_lec10.pdf |
∂x ∂y
) = (8x, 32y) = (0, 0) at the origin x = 0, y = 0.
Can we use the brightness gradient to estimate local shape? It turns out the answer is no, again because of stationary points.
But if we look at the second derivatives of brightness:
=
(8x) = 8
=
(32y) = 32
Exx =
Eyy =
Exy =
∂2E
∂x2
∂2E
∂y2
∂2E
∂x∂y ... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/617445f0e31836831b40d42cb2f11a10_MIT6_801F20_lec10.pdf |
6.866 Machine Vision
Fall 2020
For information about citing these materials or our Terms of Use, visit: https://ocw.mit.edu/terms | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/617445f0e31836831b40d42cb2f11a10_MIT6_801F20_lec10.pdf |
Chapter 5
Coupled Fluids with Heat and Mass
Transfer
5.1 November 26, 2003: Coupled Fluids, Heat and Mass Transfer!
Mechanics:
• Congrats to Jenny and David for winning the contest, prize: $5 Tosci’s.
• PS8 on Stellar, due Fri 12/5.
• Evaluations next Wednesday 12/3.
Muddy from last time:
• Time smoothing: what... | https://ocw.mit.edu/courses/3-185-transport-phenomena-in-materials-engineering-fall-2003/6185b0ff143062ed97aea16a782c8603_chap5.pdf |
U∞
Thermal if flow uniform, same criterion:
Dimensionless:
δT
x
δT = 3.6
�
αx
U∞
3.6
= �
U∞ x
α
=
√
3.6
RexPr
When is flow uniform? In a solid, or for much larger thermal boundary layer than fluid, so α >> ν, Pr<< 1.
66
Another way to look at it:
Large Prandtl number (>.5) means
δT = 0.72Pr−1/2
δu
δT
... | https://ocw.mit.edu/courses/3-185-transport-phenomena-in-materials-engineering-fall-2003/6185b0ff143062ed97aea16a782c8603_chap5.pdf |
T
∂y
= −k
dT dθ ∂βT
dθ dβT ∂y
qy = k(Ts − T∞)
�
�
U
1
∞ = hx(Ts − T∞)
√
π αx
�
k
hx = √
π αx
U∞ =
∞
kρcpU
πx
Likewise for mass transfer, ρcp is effectively one, so:
hDx =
�
DU
∞
πx
Next time: average, dimensional analysis, δT < δu case.
67
5.2 December 1, 2003: Nusselt Number, Heat and Mass Transf... | https://ocw.mit.edu/courses/3-185-transport-phenomena-in-materials-engineering-fall-2003/6185b0ff143062ed97aea16a782c8603_chap5.pdf |
/δu or δT /δu = 0.72Pr−1/2 .
Physical meaning: grows as sqrt of diffusivity, so ratio is ratio of square roots of diffusivity, which is
inverse sqrt(Pr).
• Case 2: smaller thermal(/concentration) boundary layer (Pr>5 or so): consider T/C BL to have linear
velocity, smaller velocity means thicker T/C BL. Here:
• Movin... | https://ocw.mit.edu/courses/3-185-transport-phenomena-in-materials-engineering-fall-2003/6185b0ff143062ed97aea16a782c8603_chap5.pdf |
D =
hDx
Df l
68
Looks like the Biot number, right? But it’s not, it’s actually quite different.
Bi =
hD L
Dsolid
=
L/Dsolid =
1/h
Resistance to conduction in solid
Resistance due to BL in liquid
Uses L=solid thickness, Dsolid. Heat transfer note: you get one extra dimensionless number, due to heating
by v... | https://ocw.mit.edu/courses/3-185-transport-phenomena-in-materials-engineering-fall-2003/6185b0ff143062ed97aea16a782c8603_chap5.pdf |
gives Nu, gives h (maybe
Bi).
69
5.3 December 3: Natural Convection
Mechanics:
• Course evals today!
Muddy from last time:
• What’s the relationship between hx or hL and the friction factor? Hmm... Meaning: heat transfer
coefficient, kinetic energy transfer coefficient. Types: local, global/average. Laminar flow va... | https://ocw.mit.edu/courses/3-185-transport-phenomena-in-materials-engineering-fall-2003/6185b0ff143062ed97aea16a782c8603_chap5.pdf |
d(1/V ) = −dV /V 2 , dV = d(L3) = 3L2dL
3L2dL
V dT
Word explanation: heat a solid cube, length increases 1% in each direction, volume increases 3%. Both have
units 1/K. Ideal gases:
dV
V 2dT
3 dL
L dT
β = V
= 3α.
=
=
ρ =
P
RT
, β = −
1 dρ
ρ dT
= −
RT
P
−
P
RT 2
= 1/T.
�
�
Also βC =
−
1 dρ
ρ dC .
... | https://ocw.mit.edu/courses/3-185-transport-phenomena-in-materials-engineering-fall-2003/6185b0ff143062ed97aea16a782c8603_chap5.pdf |
ν
∞�
2 ux +
−ρg + ρ∞g
ρ∞
Now assumptions 5 and 6, xmomentum becomes:
ux
∂ux + uy
∂x
∂ux
∂y
= ν
∞�
2 ux + gβ(T − T∞)
With Ts > T∞ and gx = −g, this gives driving force in the positivex direction, which is up, like it’s supposed
to. Okay, that’s all for today, more next time.
71
5.4 December 5: Wrapup Natu... | https://ocw.mit.edu/courses/3-185-transport-phenomena-in-materials-engineering-fall-2003/6185b0ff143062ed97aea16a782c8603_chap5.pdf |
Gr, Pr).
Detour: recall falling film
ux =
uav =
g sin θ(2Lz − z2)
2ν
g sin θL2
3ν
g cos βδ3
3ν2
uavδ
ν
=
Re =
So Gr is a natural convection Reynolds number, determines the rate of growth of the BL.
Graphs of dimensionless T = (T − T∞)/(Ts − T∞), dimensionless ux = Rex/2
Grx on P&G
p. 232 corresponding to d... | https://ocw.mit.edu/courses/3-185-transport-phenomena-in-materials-engineering-fall-2003/6185b0ff143062ed97aea16a782c8603_chap5.pdf |
�
4
NuL =
GrL/4
0.902Pr1/2
(0.861 + Pr)1/4
72
Special for 0.6 < Pr < 10, laminar:
NuL = 0.56(GrLPr)1/4
Turbulence, Ra between 109 and 1012 (p. 259):
NuL =
0.0246Gr2/5P r7/15
L
(1 + 0.494Pr2/3)2/5
Again, velocity0.8 in a way, sorta like turbulent forced convection boundary layers.
73
5.5 December 8: Wra... | https://ocw.mit.edu/courses/3-185-transport-phenomena-in-materials-engineering-fall-2003/6185b0ff143062ed97aea16a782c8603_chap5.pdf |
BL). For large Pr, ν > α, so the momentum diffusion happens faster,
thin thermal and thick velocity.
• Dimensionless curves: crazy nonintuitive axis value x
Grx/4! Well, not much worse than Blassius:
Rex.
But I’ll give you that the dimensionless velocity is a bit odd.
ux/U∞ vs. β = y U∞/νx = y
x
�
√
�
4y
• Wh... | https://ocw.mit.edu/courses/3-185-transport-phenomena-in-materials-engineering-fall-2003/6185b0ff143062ed97aea16a782c8603_chap5.pdf |
(estimate) heat/mass transfer coefficients for forced and natural convection, laminar
or turbulent. (D’oh! Forgot this closing part after the muddy stuff.)
Stream Function and Vorticity Vorticity introduced in turbulence video, measure of local rotation,
definition:
ω = � × �u
2D scalar, 3D vector. Some formulations... | https://ocw.mit.edu/courses/3-185-transport-phenomena-in-materials-engineering-fall-2003/6185b0ff143062ed97aea16a782c8603_chap5.pdf |
athing through nose. (D’oh! Forgot to
mention breathing through the nose.)
Decisions... Finish the term with the Bernoulli equation, or continuous flow reactors? Bernoulli wins the
vote.
75
5.6 December 10, 2003: Bernoulli, Semester Wrapup
TODO: get rooms for review sessions!
Mechanics:
• Review sessions: me Fr... | https://ocw.mit.edu/courses/3-185-transport-phenomena-in-materials-engineering-fall-2003/6185b0ff143062ed97aea16a782c8603_chap5.pdf |
− ρgz = 0
ds
1
2
ρV 2 + p + ρgz = constant
KE + P + P E = constant
Example 1: draining tub with a hole in the bottom. Set z = 0 at the bottom: PE=ρgh at top, P
at bottom corner is that plus atmospheric pressure, 2 ρV 2 beyond outlet (further accelerating). Potential
energy becomes pressure ΔP = ρgh, then become... | https://ocw.mit.edu/courses/3-185-transport-phenomena-in-materials-engineering-fall-2003/6185b0ff143062ed97aea16a782c8603_chap5.pdf |
! Mentioned linear to multiple nonlinear PDEs, un
derstanding of solution. More generally, learned to start with a simple conservation relation: accum = in
out + gen, turn into really powerful results, on macro or micro scale, for diffusion, thermal energy, mass,
momentum, even kinetic energy.
Covered all topics i... | https://ocw.mit.edu/courses/3-185-transport-phenomena-in-materials-engineering-fall-2003/6185b0ff143062ed97aea16a782c8603_chap5.pdf |
+ h2)
patm
PE
ρg(h1 + h2)
ρgh2
ρgh2
0
Batch and Continuous Flow Reactors For those interested.
Basic definitions, motivating examples. Economics: batch better for flexibility, continuous for quality
and no setup time (always on).
Two types: volumetric and surface reactors. Volume V, generation due to chemical r... | https://ocw.mit.edu/courses/3-185-transport-phenomena-in-materials-engineering-fall-2003/6185b0ff143062ed97aea16a782c8603_chap5.pdf |
Batch:
prodection rate is
• Plug:
tR =
1
k
ln(CA,in/CA,out) =
4.6
k
V
4.6
k + tchange
=
kV
4.6 + ktchange
Q =
kV
ln(CA,in/CA,out)
=
kV
4.6
Better than batch, likely better quality too, less flexible.
• Perfect mixing:
Q =
kV
CA,in/CA,out − 1
=
kV
99
Much smaller than either of the others!
Dead z... | https://ocw.mit.edu/courses/3-185-transport-phenomena-in-materials-engineering-fall-2003/6185b0ff143062ed97aea16a782c8603_chap5.pdf |
18.S997: High Dimensional Statistics
Lecture Notes
(This version: July 14, 2015)
Philippe Rigollet
Spring 2015
Preface
These lecture notes were written for the course 18.S997: High Dimensional
Statistics at MIT. They build on a set of notes that was prepared at Princeton
University in 2013-14.
Over the past decade, st... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
to understand regression with an infinite number of parameters. Much
insight from this work can be gained to understand high-dimensional or sparse
regression and it comes as no surprise that Donoho and Johnstone have made
the first contributions on this topic in the early nineties.
i
Preface
ii
Acknowledgements. These n... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
xf (x)
∇xf (x)
Distributions
Gradient of f at x
|x=x0 Gradient of f at x0
iii
Preface
iv
(µ, σ2)
N
Nd(µ, Σ)
subG(σ2)
subGd(σ2)
subE(σ2)
Ber(p)
Bin(n, p)
Lap(λ)
PX
∈
IR and variance σ2 > 0
IRd
d
×
Univariate Gaussian distribution with mean µ
IRd and covariance matrix Σ
d-variate distribution with mean µ
Univariate sub-... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
.2 Least squares estimators . . . . . . . . . . . . . . . . . . . . . .
2.3 The Gaussian Sequence Model
. . . . . . . . . . . . . . . . . .
2.4 High-dimensional linear regression . . . . . . . . . . . . . . . .
2.5 Problem set . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3 Misspecified Linear Models
. . . .... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
. . . . . . . . . . . . . 102
v
Contents
vi
. . . . . . . . . . . . . . 103
5.2 Reduction to finite hypothesis testing
. . . . . . . . . . . . . 105
5.3 Lower bounds based on two hypotheses
5.4 Lower bounds based on many hypotheses
. . . . . . . . . . . . 110
5.5 Application to the Gaussian sequence model . . . . . . .... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
need Y to depend nontrivially on X. Our task would be
done if we had access to the conditional distribution of Y given X. This is the
world of the probabilist. The statistician does not have access to this valuable
information but rather, has to estimate it, at least partially. The regression
function gives a simpl... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
1
Introduction
2
variable Z small if IE[Z 2] = [IEZ]2 + var[Z] is small. Indeed in this case, the
expectation of Z is small and the fluctuations of Z around this value are also
g(X)]2 is called the L2 risk of g defined for
small. The function R(g) = IE[Y
IEY 2 <
−
.
∞
For any measurable function g :
posed as
... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
We have proved that the regression function f (x) = IE[Y
|
enjoys the best prediction property, that is
X = x], x
,
∈ X
IE[Y
−
f (X)]2 = inf IE[Y
g
−
g(X)]2 ,
where the infimum is taken over all measurable functions g :
IR.
X →
Prediction and estimation
As we said before, in a statistical problem, we do not ... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
(
)2 :=
h
h
I
I
Introduction
3
It follows from the proof of the best prediction property above that
R(fˆ n) = IE[Y
−
= inf IE[Y
g
−
f (X)]2 +
fˆ n
I
g(X)]2 +
f
−
fˆ n
2
2
I
f
2
2
I
−
I
−
In particular, the prediction risk will always be at least equal to the positive
f (X)]2 . Since we tend to pref... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
nonparametric
(1/2, 1) for example.
statistics where typically φn = O(n−
Note that such bounds do not characterize the size of the deviation of the
2
random variable
2 around its expectation. As a result, it may be
therefore appropriate to accompany such a bound with the second option
below.
α) for some α
fˆ n ... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
reading of [BLM13] to the interested reader. This book presents many
aspects of concentration that are particularly well suited to the applications
covered in these notes.
]
Introduction
4
Other measures of error
IE
We have chosen the L2 risk somewhat arbitrarily. Why not the Lp risk defined
by g
1? The main rea... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
error is more customary
in statistics whereas, risk bounds are preferred in learning theory.
−
I
Here is a list of choices for the pseudo-distance employed in the estimation
error.
•
Pointwise error. Given a point x0, the pointwise error measures only
the error at this point. It uses the pseudo-distance:
d0(fˆ n... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
1
}
R(h) = IP(Y
=
h(X))
.
Introduction
5
We will not cover this problem in this course.
Finally, we will devote a large part of these notes to the study of linear
= IRd and f is linear (or affine), i.e., f (x) = x⊤θ
models. For such models,
for some unknown θ
In this case, it is traditional to measure error
... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
candidate
empirical mean ( 1
i=1
n
to estimate the expected value IEX of a random variable X from a sequence of
i.i.d copies X1, . . . , Xn of X, is their empirical average
�
¯X =
n1
n
i=1
n
Xi .
In many instances, it corresponds the maximum likelihood estimator of IEX.
IE(X))2 is estimated by
Another example ... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
a certain class
linear functions) or it is smooth (e.g., the L2-norm of its second derivative is
≥
G
)
2ERM may also mean Empirical Risk Minimizer
Introduction
6
5
.
1
0
.
1
5
.
0
0
.
0
5
.
0
−
0
.
1
−
5
.
1
−
0
.
2
−
y
5
.
1
0
.
1
5
.
0
0
.
0
5
.
0
−
0
.
1
−
5
.
1
−
0
.
2
−
y
5
.
1
0
.
1
5
.
0
0
.
0
5
.
0
−... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
course.
Unlike traditional (low dimensional) statistics, computation plays a key role
in high-dimensional statistics. Indeed, what is the point of describing an esti
mator with good prediction properties if it takes years to compute it on large
datasets? As a result of this observation, much of the modern estimator... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
∈
⊤θ∗
θ, θ∗) for some specific pseudo-distance d(
·
i=1
n
Note that θˆ may not be unique. In the case of a linear model, where we assume
that the regression function is of the form f (x) = x
for some unknown
IRd, we will need assumptions to ensure identifiability if we want to prove
θ∗
bounds on d(ˆ
). Nevertheles... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
size that tends to infinity.
Rather, we will pay a systematic approximation error. When
is a linear
subspace as above, and the pseudo distance is given by the squared L2 norm
d(fˆ n, f ) =
2
2, it follows from the Pythagorean theorem that
fˆ n
G
f
I
−
I
fˆ n
f
2 =
2
fˆ n
2
2 +
f¯
I
f¯
f
2
2 ,
I
I
−
−
I
wher... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
when we restrict our attentions to estimator in the
class
. Going back to the gap in knowledge between a probabilist who knows
the whole joint distribution of (X, Y ) and a statistician who only see the data,
it can only see the whole distribution
the oracle sits somewhere in-between:
through the lens provided by... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
techniques employed for this goal will essentially be the same
as the ones employed to minimize the prediction risk. The extra assumptions
on the Xis will then translate in interesting properties on θˆ itself, including
uniqueness on top of the prediction properties of the function fˆ n(x) = x⊤θˆ.
High dimension an... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
C > 0 is a constant and in Chapter 5, we will show that this cannot
be improved apart perhaps for a smaller multiplicative constant. Clearly such
n and actually, in view of its optimality,
a bound is uninformative if d
we can even conclude that the problem is too difficult statistically. However,
the situation is no... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
d
θ2 ≥ · · · ≥
i
j
C
α for some α > 0
|
|
−
d
1
j=1 1I(θj+1 = θj )
−
=
k
≤
Structured in another basis: θ = Ψµ, for some orthogonal matrix and µ
is in one of the structured classes described above.
�
P
Sparsity plays a significant role in statistics because, often, structure translate
into sparsity in a certain ... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
j=1
n
q
θj
|
|
≤
1
}
3
|
θ
where
vectors in the unit ℓq ball can be approximated by sparse vectors.
k
Note that the set of k-sparse vectors of IRd is a union of
j=0
|q is often called ℓq-norm . As we will see, the smaller q is, the better
linear
subspaces with dimension at most k and that are spanned by at mos... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
1.
6
Introduction
10
θj
θj
Monotone
Smooth
θj
j
j
Piecewise constant
Smooth in a different basis
θj
j
j
Figure 2. Examples of structures vectors θ
∈ IR50
Nonparametric regression
Nonparametric does not mean that there is no parameter to estimate (the
regression function is a parameter) but rather that the para... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
essentially)
known whenever this is needed.
|
|
k
Even if infinity is countable, we still have to estimate an infinite number
of coefficients using a finite number of observations. It does not require much
statistical intuition to realize that this task is impossible in general. What if
we know something about the sequ... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
n
|≤
|
Note that it depends on the unknown αk and define the estimator
fˆ n =
αˆkϕk ,
k0
k
n
|≤
where ˆαk are some data-driven coefficients (obtained by least-squares for ex
ample). Then by the Pythagorean theorem and Parseval’s identity, we have
|
fˆ n
2 =
2
f
I
−
¯
f
I
−
I
=
f
2 +
I2
2 +
αk
fˆ n
I
¯
f
I
... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
make assump
tions on the regression function (here in terms of smoothness) in order to
4Here we illustrate a convenient notational convention that we will be using through
out these notes: a constant C may be different from line to line. This will not affect the
interpretation of our results since we are interested i... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
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