text stringlengths 16 3.88k | source stringlengths 60 201 |
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.22 illustrates
downsampling with and without aliasing.
M
2π
1
2
Sampling Rate Expansion by an Integer Factor
A typical system for increasing the sampling rate of a discrete sequence by an integer factor is
illustrated in OSB Figure 4.24. Expressed in terms of Fourier transforms, the expander output
is:
Xe(ejω... | https://ocw.mit.edu/courses/6-341-discrete-time-signal-processing-fall-2005/1b110cd38e3cbd3efd56f43c9ddf8b2a_lec05.pdf |
Topic 12 Notes
Jeremy Orloff
12 Laplace transform
12.1 Introduction
The Laplace transform takes a function of time and transforms it to a function of a complex variable
. Because the transform is invertible, no information is lost and it is reasonable to think of a function
() and its Laplace transform () as two v... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
+ 8 ′ + 7 = ()
This equation models a damped harmonic oscillator, say a mass on a spring with a damper, where
() is the force on the mass and () is its displacement from equilibrium. If we consider to be
the input and the output, then this is a linear time invariant (LTI) system.
Example 12.2. There are many varia... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
)
− (cid:243)
0
=
1
−
divergent otherwise
if Re() > Re()
The last formula comes from plugging ∞ into the exponential. This is 0 if Re( − ) < 0 and
undefined otherwise.
Example 12.4. Let () = . Compute () = ( ; ) directly. Give the region in the complex
-plane where the integral converges.
∞
(; ) = ∫
0
{ ... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
function ().
̂() = ∫
If we assume () = 0 for < 0, this becomes
∞
−∞
()e− .
̂() = ∫
Now if = then the Laplace transform is
0
∞
()e− .
( ; ) = ( ; ) = ∫
0
∞
()e−
(1)
(2)
Comparing these two equations we see that ̂() = ( ; ). We see the transforms are basically
the same things using different notatio... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
12.7. Here is a list of some functions of exponential type.
(cid:240)
< 2eRe()
() = e ∶
() = 1 ∶ () < 2 = 2e0⋅ (exponential type 0)
(exponential type 0)
()
(cid:240)
(cid:240)
(cid:240)
(cid:240) ()(cid:240) ≤ 1
(exponential type Re())
() = cos() ∶
In the above, all of the inequalities are for ≥ 0.
For ()... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
Proof. We prove this using integration by parts.
∞
( ′; ) = ∫
0
′()e− = ()e− ∞
∞
()e− = − (0) + ().
(cid:243)
(cid:243)0 + ∫
0
∞, ()e−
In the last step we used the fact that at =
about exponential type.
= 0, which follows from the assumption
12 LAPLACE TRANSFORM
Equation 4 gives us formulas for all der... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
integration and taking the derivative.
′() =
∫
0
∞
∞
()e− = ∫
0
− ()e− = (− (); ).
This proves Equation 7.
Equation 7 is called the -derivative rule. We can extend it to more derivatives in : Suppose
( ; ) = (). Then,
( (); ) = − ′()
( (); ) = (−1) ()()
(8)
(9)
Equation 8 is the same as Equation 7 a... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
− = e− ().
The properties in Equations 3-10 will be used in examples below. They are also in the table at the
end of these notes.
12.6 Differential equations
Coverup method. We are going to use partial fractions and the coverup method. We will assume
you have seen partial fractions. If you don’t remember them well o... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
1
, so =
1
(2 − 1)
=
1
( − 1)( + 1)
=
+
− 1
+
+ 1
The coverup method gives = −1, = 1∕2, = 1∕2. So,
1
2
= + e + e− = −1 +
e +
e−.
1
2
12 LAPLACE TRANSFORM
7
12.7 System functions and the Laplace transform
When we introduced the Nyquist criterion for stability we stated without any justi... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
) + 7(3 + 2 + 5) = 73 + 242 + 20 + 51.
() = e2
() = 2e2
2() = 4e2
Therefore: (2 + 8 + 7) = 4e2 + 8(2)e2 + 7e2 = (4 + 16 + 7)e2 = (2)e2.
The substitution rule is a straightforward statement about the derivatives of exponentials.
Theorem 12.13. (Substitution rule) For a polynomial differential operator () we have
... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
=
2
.
Let () = (; ). Applying the Laplace transform to the equation we get
(2 () − (0) − ′(0)) + 8( () − (0)) + 7 () = 0
Algebra:
(2 + 8 + 7) () −
1 −
2 − 8
1 = 0 ⇔ =
1 + 8
1 +
2
2 + 8 + 7
Factoring the denominator and using partial fractions, we get
() =
1 + 8
1 +
2
2 + 8 + 7
=
1 + 8
1 +
2
( + 1)(... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
to the exponential modes of the system, i.e. the poles of
() correspond to the exponential modes.
The system is called stable if the modes all decay to 0 as goes to infinity. That is, if all the poles
have negative real part.
Example 12.17. This example is to emphasize that not all system functions are of the form 1... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
an inverse.
We start with the bad news: Unfortunately this is not strictly true. There are many functions with
the same Laplace transform. We list some of the ways this can happen.
12 LAPLACE TRANSFORM
10
1. If () = () for ≥ 0, then clearly () = (). Since the Laplace transform only concerns
≥ 0, the functions ca... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
path integral along the vertical line = .
Solution: Proving Equation 13 is straightforward: It is clear that
has only one pole which is at = . Since,
e
−
(
∑
Res
)
,
= e
e
−
we have proved Equation 13.
Proving Equation 14 is more involved. We should first check the convergence of the integral. In this
ca... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
2, this proves Equation 14.
To verify step 2 we look at one side at a time.
2:
2 is parametrized by = () = + , with − ≤ ≤ . So,
(cid:243)
(cid:243)
(cid:243)
(cid:243)
(cid:243)
∫
2
e
−
(cid:243)
(cid:243)
(cid:243)
(cid:243)
(cid:243)
(cid:243)
(cid:243)
(cid:243)
(cid:243)
e(+)
(cid:243)
(cid:243)
− (cid:24... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
12 LAPLACE TRANSFORM
12
This is similar to the previous example. Since decays like 1∕2 we can actually allow ≥ 0
Theorem 12.20. Laplace inversion 1. Assume is continuous and of exponential type . Then for
> we have
()e .
(15)
As usual, this formula holds for > 0.
() =
+∞
1
2 ∫−∞
Proof. The proof uses the... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
feedback factor e−.
The system function for the closed loop system is
() =
1 + e−
Note even if you start with a rational function the system function of the closed loop with delay is
not rational. Usually it has an infinite number of poles.
Example 12.22. Suppose () = 1, = 1 and = 1 find the poles of ().
Solution: ... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
3 −3 + …
e
e
() = () () = () − e
− () + 2e−2 () − 3e−3 () + …
Using the shift formula Equation 10, we have
() = () − ( − ) + 2 ( − 2) − 3 ( − 3) + …
(This is not really an infinite series because () = 0 for < 0.) If the input is bounded and < 1
then even for large the series is bounded. So bounded input produces... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
)
1∕2
!∕+1
∕(2 + 2)
∕(2 + 2)
( − )∕(( − )2 + 2)
∕(( − )2 + 2)
1
−
e
∕(2 − 2)
∕(2 − 2)
1
(2 + 2)2
(2 + 2)2
2
(2 + 2)2
!∕( − )+1
1ø
Γ( + 1)
+1
Region of convergence
Re() > 0
Re() > Re()
Re() > 0
Re() > 0
Re() > 0
Re() > 0
Re() > Re()
Re() > Re()
all
all
Re() >
Re() >
Re() > 0
Re() > 0
... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1b1cbc6540b338b58b7bc1b1cc1f280b_MIT18_04S18_topic12.pdf |
MATH 18.152 COURSE NOTES - CLASS MEETING # 7
18.152 Introduction to PDEs, Fall 2011
Professor: Jared Speck
Class Meeting # 7: The Fundamental Solution and Green Functions
1. The Fundamental Solution for ∆ in Rn
Here is a situation that often arises in physics. We are given a function f x on Rn representing
( )
the spat... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
.0.3)
def
Φ x
( )
1
2π ln ∣x∣
2,
1
ωn∣x∣n−2 n 3,
i=1(xi)2 and ωn is the surface area of a unit ball in Rn (e.g. ω3
= { −
n =
≥
n
√
def
=
where as usual ∣x∣
Remark 1.0.1. Some people prefer to define their Φ to be the negative of our Φ.
∑
4π).
=
that this holds for now. We then claim that the solution to (1.0.1) is u x
(... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
∆ ∂2
r
r Φ +
r
4
def
n = 3. Note that Φ(x) = Φ(r) (r = ∣x∣) is spherically symmetric.
0 for spherically symmetric functions, we have that
(cid:3)
>
version of the aforementioned heuristic results.
Theorem 1.1 (Solution to Poisson’s equation in Rn
∞
C0
). Let f x
on Rn). Then for
3, the Laplace e
smooth, compactly suppo... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
n
Remark 1.0.2. As we alluded to above, Theorem 1.1 shows that ∆Φ x
“delta distribution.” For on the one hand, as we have previously discussed, we have that f
On the other hand, our proof of Theorem 1.1 below will show that f ∆
(
Thus, for any f, we have δ
δ x , where δ is the
(
δ f.
= ∗
∆Φ f.
) ∗
f ∆Φ f, and so ∆Φ δ.
... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
y
∣
We first show that I goes to 0 as (cid:15) →
∆xu(x) = −
∆
yf (x − y) d3
y
−
1
∫
4π Bc 0
(cid:15)
( ) ∣
1
y
∣
+
0
. To this end, let
∆
yf (x − y) d y = I
+ I.
I
3
def
(1.0.8)
def
M = sup
y R3
∈
∣f (y)∣ + ∣∇f (y)∣ + ∣∆yf (y)∣.
Then using spherical coordinates r, ω for the y variable, and recalling that d3y
ω ∈ ∂B
that... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
w
)
(
f x as desired.
To show (1.0.10), we will use integration by parts via Green’s identity and simple
estimates to
control the boundary terms. Recall that Green’s identity for two functions u, v is
(1.0.7),
(1.0.9), and (1.0.10) and let (cid:15)
+
0 to deduce
→
(1.0.11)
( ) −
Ω
∫ v x ∆u x
( )
1.0.1,
Lemma
Using (1.0... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
c
c
(cid:15)
=
1
∂B
(cid:15)
0 . This corresponds to the
( )
(cid:15) 0 ,
)
(
on the right-hand side of
,
0
( )
Bc
in the standard
form
ulation of Green’s identity for
(1.0.13)
− ∫
1
(cid:15) 0 y
Bc
( ) ∣
∣
∆yf (x − y) d3y = − ∫
( )
∂B1 0
(cid:15)ω ⋅ (∇f )(x − (cid:15)ω) dω + ∫
( )
∂B1 0
f (x (cid:15)ω dω.
−
)
Using (1... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
.14)
∣u(x)∣ =
1
4π
∣ ∫
1
f (
y d3y
)
∣ ≤
M
∣ ∫ ( )
2π x BR 0
∣
1 d3y
=
2R3M
3 x
∣
∣
,
and we have shown (1.0.5) in the case n
BR 0 x y
( ) ∣ − ∣
= 3
.
4
MATH 18.152 COURSE NOTES - CLASS MEETING # 7
To prove uniqueness, we will make use of Corollary 4.0.4,
are two solutions with the assumed decay conditions at
that w
u... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
has a unique solution u
theorem
Then
.
a bounde
C 2
Ω ∩
( )
d Lipschitz domain, and let
C(Ω).
∈
g
∈
(cid:3)
Definition 2.0.2. Let Ω ⊂
x, y Ω Ω verifying the following conditions for each fixed x Ω
(
Rn be a domain. A Green function in Ω is defined to be a function of
∈
) ∈ ×
∶
(2.0.17)
(2.0.18)
∆yG(
G x, σ
(
x, y) = δ(
0,... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
− (
φ x, y also verifies the boundary condition (2.0.18).
φ x, y verifies equation (2.0.17).
ve
e ha
that
Φ x
y
)
)
(
x, σ
) =
0
whenev
er σ ∈
∂Ω. Thus, Φ x
(
−
(cid:3)
MATH 18.152 COURSE NOTES - CLASS MEETING # 7
5
The following technical proposition will play later in this section when we derive representation
formula... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
„„„‚„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„¶
σ
‚„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„
( )∇
+ ∫
u σ
∂Ω
∂Ω
·„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„
single layer potential
double layer potential
Proof. We’ll do the proof for n 3, in whic
h case Φ x
=
(
) = −
identity (1.0.1... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
− ∫ u σ
∂Ω
( ∇ (
)
ˆN
1
x σ
∣ −
∣
) dσ
1
∣x − σ
In the last two integrals above, N σ denotes the radially outward unit
normal to the oundary of
b
( )
the ball B(cid:15) x . This corresponds to the “opposite” choice of normal that appears in the standard
(
formulation
identity, but we have compensated by adjusting the s... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
∣ ≤ ∫B(cid:15)(x) ∣x
M
≤
∫ ( )
B(cid:15) x
∣
1
− y
∣
1
x y
−
∣
∆u y d3y
)∣
(
d
3y
→
0 as (cid:15) 0.
↓
∣
This shows that L converges to
Ω x y ∆u y d3y as (cid:15) 0.
∫ ∣ − ∣
The limits for R1 and R2 are obvious since these terms do not dep
We now address R3. To this end,
end
u y . We then estimate R3 by
( )
on
(cid:15)... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
∂
x, we ha
estimate
(2.0.28)
1
∣ R4 − [ − u(x)]∣ = ∣u(x) +
4π
1
π
4
∫ ( )
B(cid:15) x
∂
u σ
( )∇ ˆ
1
4π
1
=
∣
∫∂ (cid:15)( )
B
x
u
( ( ) − ( ))
(
u
σ
x
≤ 4π ∫∂B(cid:15)(x)
u x
∣ ( ) − ( )∣(
u σ
1
) dσ
∣
N (σ)( x σ
− ∣
∣
1
∣x − σ
1
∣x − σ
dσ
∣2
)
)
2
∣
dσ
∣
1
4π
max
(
∈
σ ∂B x
(cid:15)
max u x
∣ ( )
)
∈
σ ∂B x
u
) ∣ (
)... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
„„„„„„„„‚„„„„„„„„„„„„„„„„„„„„„„„
kernel
Poisson
(2.0.30) u(x) = − ∫ Φ
Recall also that
Ω
(x − y)f y
( ) dny + ∫ Φ
∂Ω
(x − σ)∇ ˆN (σ)u(σ) dσ − ∫ g
∂Ω
dσ.
(σ)∇ ˆN (σ)Φ(x − σ) dσ.
(2.0.31)
(2.0.32)
Applying the Green
G(x, y) = Φ(x − y
) − φ(x, y
)
G x, σ
(
) =
0 when
σ ∂Ω.
∈
identity (1.0.11) to the functions u y and φ x,... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
in the case that Ω BR 0
( )
called the method of images that works for special domains.
G x, y and Poisson kernel P x, σ
)
the
(
)
of radius R centered at
(
ball
function
R3 is a
⊂
def
= −∇
origin.
(
ˆN G x, σ from (2.0.29)
We’ll use a technique
)
Warning 3.0.1. Brace yourself for a bunch of tedious computations that a... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
y∣
= q2
(∣x∗
∣
2
− 2x∗
⋅ y + R2
).
Then
performing simple algebra, we ha
= ∣x y
−
ve
(3.0.38)
2
∣x∗
∣
+ R2
− q2
(R2
+ ∣x∣
2
) = 2y ⋅ (x∗
− q2x).
Now since the left-hand side of (3.0.38) does not
term on the right-hand side vanishes. This implies that x
depend on
=
∗
y, it must be the case that the second
q2x, and also ... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
,
∣x
∣∣
x ≠ 0,
1
4π x y∣
∣
−
1
1
R
4π
−
y
∣
∣
.
yG x, y
∇ (
) =
x
y
−
x y
4π
∣ −
∗
1 R x
y
∣3 − 4π ∣x∣ ∣x∗ − y∣3
−
Now when σ ∂B 0 , (3.0.36) and (3.0.40)
imply
that
∈
R
( )
(3.0.47)
∣
Therefore, using (3.0.46) and (3.0.47), w
x∗ σ
− ∣ =
∣
e compute
R
x
∣
that
x σ .
∣ − ∣
(3.0.48)
σG
∇ (
x, σ
) =
∣
Using (3.0.48) and t... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
cen
(3.0.49)
would be replaced with
tered
at
the p
t
oin
p ∈ R
R
σ) =
(
1
2
− ∣x∣2
4πR ∣x
of
3 instead
.
origin, then the formula
σ 3
∣
−
the
(3.0.50)
ˆ G x, σ
(
N (σ)
∇
)
def
= ∇ (
σG x, σ N σ
) ⋅
ˆ
( ) =
Theorem
3.1 (Poisson’s formula). Let B
R p
( )
3 , and let x = (x1, x2, x3) denote a point in
1, p2
, p
p
(
)
C BR... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
the surface area of the unit ball in
n
f σ
)
(
− σ
∣
dσ,
n
∂BR p x
( )
∣
Rn.
Proof. The identity (3.0.52) follows immediately from Theorem 2.2 and (3.0.50).
4. Harnack’s inequality
Theorem 4.1 (Harnac
Rn. Then for any x BR 0 , we have that
∈
k’s
( )
inequality). Let u be harmonic and non-negative in the ball BR(0
(cid:... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
�2 4πR ∂BR 0
( )
∣
−
e that
, we hav
y
ert
∫
mean value
f σ dσ.
( )
(4.0.57)
u(0) =
1
4πR2 ∫
∂BR
)
(0
f σ dσ.
( )
Th
us, combining (4.0.56) and (4.0.57), we have that
(4.0.58)
u x
( ) ≤
Rn−2(R x
∣
+
n
R x
∣)
( − ∣
−
∣) ,
1
which implies one of the inequalities in (4.0.54). The other
remaining triangle inequality.
one c... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
)
(
v 0 . Th
( ) = ( )
R
x
−2
n
1
u
0
.
( )
us, v is a constant-valued function
(
v x , and we argue as ab
o handle the case u x M, we simply consider the function w x
T
)
≤
)
ve.
o
(
def
( ) = − ( ) + ∣
u x M in place of
∣
(cid:3)
MIT OpenCourseWare
http://ocw.mit.edu
18.152 Introduction to Partial Differential Equat... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/1b2023e54a5d0c4023418a27274787ef_MIT18_152F11_lec_07.pdf |
18.405J/6.841J: Advanced Complexity Theory
Spring 2016
Prof. Dana Moshkovitz
(cid:47)(cid:72)(cid:70)(cid:87)(cid:88)(cid:85)(cid:72)(cid:3)(cid:20)(cid:23)(cid:29)(cid:3)(cid:44)(cid:81)(cid:87)(cid:85)(cid:82)(cid:3)(cid:87)(cid:82)(cid:3)(cid:51)(cid:38)(cid:51)
Scribe: Dana Moshkovitz
Scribe Date: Spring 2015
1 Ch... | https://ocw.mit.edu/courses/18-405j-advanced-complexity-theory-spring-2016/1b2c1d36df1e2a9cfa79d6531ffb8d29_MIT18_405JS16_IntroPCP.pdf |
[your favorite math conjecture here] have a proof of length n?"
2. Max-3Sat: Given clauses C1; : : : ; Cm over variables x1; : : : ; xn, where each clause is of the
form (x _ y _ z) where x; y; z are either variables or their negations, (cid:12)nd an assignment to the
variables that satis(cid:12)es as many clauses as p... | https://ocw.mit.edu/courses/18-405j-advanced-complexity-theory-spring-2016/1b2c1d36df1e2a9cfa79d6531ffb8d29_MIT18_405JS16_IntroPCP.pdf |
designing approximation algorithms, devising
faster exponential-time algorithms: time (1:1)n is much better than 2n and may even be practical
for small enough n’s, or considering heuristics that do reasonably well in practice.
2 Approximation Algorithms
We will focus on approximations:
De(cid:12)nition 1 (Approximation... | https://ocw.mit.edu/courses/18-405j-advanced-complexity-theory-spring-2016/1b2c1d36df1e2a9cfa79d6531ffb8d29_MIT18_405JS16_IntroPCP.pdf |
are also used.
A few prominent examples of approximation algorithms appear in Table 1.
2.1 Example: Approximation Algorithm for Max-3Sat
Next we show the 7 -approximation algorithm of Max-3Sat. This algorithm demonstrates an
important technique: the random assignment method.
8
Input: Clauses C1; : : : ; Cm over Boolean... | https://ocw.mit.edu/courses/18-405j-advanced-complexity-theory-spring-2016/1b2c1d36df1e2a9cfa79d6531ffb8d29_MIT18_405JS16_IntroPCP.pdf |
of the clauses is at least
Proof. Denote the fraction of assignments that satisfy strictly less than 7 fraction of the clauses by
8 m (cid:0) 1 )8
p. Since these assignments satisfy a natural number of clauses, this number is at most ( 7
clauses. By Lemma 2.1,
8
(
)
So, p (cid:20) m
m+1 .
p (cid:1)
7
8
(cid:0) 1
8m
+ (... | https://ocw.mit.edu/courses/18-405j-advanced-complexity-theory-spring-2016/1b2c1d36df1e2a9cfa79d6531ffb8d29_MIT18_405JS16_IntroPCP.pdf |
).
3.2 Probabilistic Checking of Proofs
Perhaps surprisingly, in 1991 it was discovered that the problem of proving hardness of approxima-
tion is intimately related to a deep question about checking of proofs [FGL+96]:
\Can any mathematical proof be written in a form that can be checked probabilistically
by making onl... | https://ocw.mit.edu/courses/18-405j-advanced-complexity-theory-spring-2016/1b2c1d36df1e2a9cfa79d6531ffb8d29_MIT18_405JS16_IntroPCP.pdf |
their symbols. Such proofs are necessarily
also probabilistic, since deterministically locally checkable proofs are simply constant-sized proofs.
Let us make the appropriate de(cid:12)nitions:
De(cid:12)nition 2 (Probabilistic veri(cid:12)er). A probabilistic veri(cid:12)er is a probabilistic polynomial time
Turing mac... | https://ocw.mit.edu/courses/18-405j-advanced-complexity-theory-spring-2016/1b2c1d36df1e2a9cfa79d6531ffb8d29_MIT18_405JS16_IntroPCP.pdf |
es:
(cid:15) Completeness: For all x 2 L, there exists a proof (cid:25), such that
Pr
w2f0;1
r
g
[V (cid:25)(x; w) = 1]
(cid:21) c:
(cid:15) Soundness: For all x 2= L, for any proof (cid:25),
Pr
2f0;1gr
w
[V (cid:25)(x; w) = 1]
(cid:20) s:
If V 2 V[r; q](cid:6), we say that L 2 P CPc;s[r; q](cid:6) (\PCP" is \Probabili... | https://ocw.mit.edu/courses/18-405j-advanced-complexity-theory-spring-2016/1b2c1d36df1e2a9cfa79d6531ffb8d29_MIT18_405JS16_IntroPCP.pdf |
fact, any
veri(cid:12)er can be efficiently transformed to a veri(cid:12)er that makes only two queries. The new veri(cid:12)er
may need a non-binary alphabet, and, more importantly, its soundness error may be much larger
than the soundness error of the original veri(cid:12)er.
5
3.3 The Connection Between Hardness of A... | https://ocw.mit.edu/courses/18-405j-advanced-complexity-theory-spring-2016/1b2c1d36df1e2a9cfa79d6531ffb8d29_MIT18_405JS16_IntroPCP.pdf |
for some s < 1. Let us show a reduction from
3Sat to Max-3Sat1;s′ for some s′ < 1:
1. Let V be the probabilistic veri(cid:12)er for 3Sat. Note that V can be converted to a new veri(cid:12)er
V ′ whose tests are of the form (x _ y _ z) where x, y, z are either variables or their negations.
The completeness of V ′ will r... | https://ocw.mit.edu/courses/18-405j-advanced-complexity-theory-spring-2016/1b2c1d36df1e2a9cfa79d6531ffb8d29_MIT18_405JS16_IntroPCP.pdf |
, and by 1992, the question was already answered positively:
Theorem 4 (The PCP Theorem [AS98, ALM+98]). N P (cid:18) P CP1; 1 [O(log n); O(1)].
2
Under plausible complexity assumptions, the randomness must be Ω(log n). An interesting question
is what is the constant in the Ω. This corresponds to the exponent of n in t... | https://ocw.mit.edu/courses/18-405j-advanced-complexity-theory-spring-2016/1b2c1d36df1e2a9cfa79d6531ffb8d29_MIT18_405JS16_IntroPCP.pdf |
AS92]
N. Alon and J. H. Spencer. The Probabilistic Method. Wiley, 1992.
[AS98]
S. Arora and S. Safra. Probabilistic checking of proofs: a new characterization of NP.
Journal of the ACM, 45(1):70{122, 1998.
[FGL+96] U. Feige, S. Goldwasser, L. Lovasz, S. Safra, and M. Szegedy. Interactive proofs and
the hardness of appr... | https://ocw.mit.edu/courses/18-405j-advanced-complexity-theory-spring-2016/1b2c1d36df1e2a9cfa79d6531ffb8d29_MIT18_405JS16_IntroPCP.pdf |
3.15
Optical Fibers and Photonic Deviccs
C.A. Ross, DMSE, MIT
References:
Braithwnite and Weaver chapter 6.4 (fibers)
Sanger. How fiber optics works, The Industrial Physicist p l 8 . FebIMar 2002
Savage. Linking lvith Light, IEEE Spectrum p32. Aug. 2002
Saleh and Teich, Furldamentals of' Photonics. Wiley 199 I .... | https://ocw.mit.edu/courses/3-15-electrical-optical-magnetic-materials-and-devices-fall-2006/1b318e5a5a7d1ba5a21d9ddf6edc5c63_lecture15_16.pdf |
. e.g.
core n = 1.53, cladding n = 1.50 gives $, = 78.6'
Step index fibers can have rrtohl dispersion: different modes of light traveling at
different angles traverse different path lengths so get out of phase. Leads to spreading of
pulses. A graded index fiber can cure this proble~n, or use a single-mode fiber. (c... | https://ocw.mit.edu/courses/3-15-electrical-optical-magnetic-materials-and-devices-fall-2006/1b318e5a5a7d1ba5a21d9ddf6edc5c63_lecture15_16.pdf |
laser, external ~notlulation. amplifiers every 100 km, dispersion compensation. 40
\vnvelengths simultnneously around 1550 nm. 10 Gbit/sec x 40 channels, spaced 0.8 nm
apart ( I 0 0 GHz).
Laser -> Attenwitor -> Modulator -> Multiplexer ->fiber with amplifiers -> Add/drop
niulriplexer -> Demultiplexer -> Attenuator ... | https://ocw.mit.edu/courses/3-15-electrical-optical-magnetic-materials-and-devices-fall-2006/1b318e5a5a7d1ba5a21d9ddf6edc5c63_lecture15_16.pdf |
except for one frequency which it transmirs). These can be used in a demultiplexer.
Handout 8
, .. | https://ocw.mit.edu/courses/3-15-electrical-optical-magnetic-materials-and-devices-fall-2006/1b318e5a5a7d1ba5a21d9ddf6edc5c63_lecture15_16.pdf |
6.867 Machine learning, lecture 7 (Jaakkola)
1
Lecture topics:
• Kernel form of linear regression
• Kernels, examples, construction, properties
Linear regression and kernels
Consider a slightly simpler model where we omit the offset parameter θ0, reducing the
model to y = θT φ(x) + � where φ(x) is a particular fe... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/1b55295adbca700abfa7356a4f832e33_lec7.pdf |
feature vectors:
θ =
n1 �
λ
t=1
αtφ(xt)
(3)
The implication is that the optimal θ (however high dimensional) will lie in the span of the
feature vectors corresponding to the training examples. This is due to the regularization
Cite as: Tommi Jaakkola, course materials for 6.867 Machine Learning, Fall 2006. MIT... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/1b55295adbca700abfa7356a4f832e33_lec7.pdf |
n × n matrix. This is the cost of
dealing with inner products as opposed to handing feature vectors directly. In some cases,
the benefit is substantial since the feature vectors in the inner products may be infinite
dimensional but never needed explicitly.
As a result of finding ˆαt we can cast the predictions for new... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/1b55295adbca700abfa7356a4f832e33_lec7.pdf |
corresponding to the polynomial kernel. The
components of these feature vectors were polynomial terms up to degree p with specifically
chosen coefficients. The restricted choice of coefficients was necessary in order to collapse
the inner product calculations.
The feature “vectors” corresponding to the radial basis kern... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/1b55295adbca700abfa7356a4f832e33_lec7.pdf |
(x, x�)
Cite as: Tommi Jaakkola, course materials for 6.867 Machine Learning, Fall 2006. MIT OpenCourseWare
(http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].(cid:13)(cid:10)
6.867 Machine learning, lecture 7 (Jaakkola)
4
are all valid kernels. While simple, these rules are... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/1b55295adbca700abfa7356a4f832e33_lec7.pdf |
φ(x�)
φi,j (x)φi,j(x�)
�
i,j
�
i,j
�
[
=
=
=
i (x)φ(2)
φ(1)
j (x)φ(1)
j (x�)
i (x�)φ(2)
�
i (x)φ(1)
φ(1)
i (x�)][
j (x)φ(2)
φ(2)
j (x�)]
i
j
[φ(1)(x)T φ(1)(x�)][φ(2)(x)T φ(2)(x�)]
=
= K1(x, x�)K2(x, x�)
(17)
(18)
(19)
(20)
(21)
(22)
(23)
These construction rules can also be used to verif... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/1b55295adbca700abfa7356a4f832e33_lec7.pdf |
of simple products xT x� and is therefore a kernel based on the
second and third rules; the first rule allows us to incorporate f (x) and f (x�).
String kernels. It is often necessary to make predictions (classify, assess risk, determine
user ratings) on the basis of more complex objects such as variable length seque... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/1b55295adbca700abfa7356a4f832e33_lec7.pdf |
subsequence. For example,
φon(the common construct) = 2
(28)
The number of components in such feature vectors is very large (exponential in k). Yet,
the inner product
�
u∈Ak
φu(x)φu(x�)
(29)
can be computed efficiently (there are only a limited number of possible contiguous subse
quences in x and x�). The reaso... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/1b55295adbca700abfa7356a4f832e33_lec7.pdf |
any immediate effect from the sequence length:
K˜ (x, x�) = �
K(x
, x�)
�
K(x, x) K(x�, x�)
Appendix (optional): Kernel linear regression with offset
Given a feature expansion specified by φ(x) we try to minimize
J(θ, θ0) =
n
� �
yt − θT φ(xt) − θ0 + λ�θ�2
�2
(32)
(33)
t=1
where we have chosen not to regulari... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/1b55295adbca700abfa7356a4f832e33_lec7.pdf |
6.867 Machine learning, lecture 7 (Jaakkola)
7
We can therefore construct the optimal θ in terms of prediction differences αt and the
feature vectors as before:
θ =
n1 �
λ
t=1
αtφ(xt)
(36)
Using this form of the solution for θ and Eq.(34) we can also express the optimal θ0 as a
function of the prediction diffe... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/1b55295adbca700abfa7356a4f832e33_lec7.pdf |
same matrix notation as before, and letting 1 = [1, . . . , 1]T , we can rewrite the
above condition as
a = (I − 11T /n) y − (I − 11T /n)Ka
(42)
�
C
��
�
1
λ
where C = I − 11T /n is a centering matrix. Any solution to the above equation has
to satisfy 1T a = 0 (just left multiply the equation with 1T ). Note ... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/1b55295adbca700abfa7356a4f832e33_lec7.pdf |
Global Illumination and Monte Carlo
MIT EECS 6.837 Computer Graphics
Wojciech Matusik
with many slides from Fredo Durand and Jaakko Lehtinen
© ACM. All rights reserved. This content is excluded from our Creative Commons
license. For more information, see http://ocw.mit.edu/help/faq-fair-use/.
1
Today
• Lots of r... | https://ocw.mit.edu/courses/6-837-computer-graphics-fall-2012/1b5985f78c68379e15543fe27d41b72c_MIT6_837F12_Lec18.pdf |
from?
– It is the light reflected towards x from the surface point in
direction l ==> must compute similar integral there!
• Recursive!
– AND if x happens
to be a light source,
we add its contribution
directly
x
11
The Rendering Equation
• The rendering equation describes the appearance of
the scen... | https://ocw.mit.edu/courses/6-837-computer-graphics-fall-2012/1b5985f78c68379e15543fe27d41b72c_MIT6_837F12_Lec18.pdf |
• Systematically sample light sources at each hit
– Don’t just wait the rays will hit it by chance
21
Results
n
e
s
n
e
J
n
n
a
W
k
i
r
n
e
H
Courtesy of Henrik Wann Jensen. Used with permission.
22
Monte Carlo Path Tracing
• Trace only one secondary ray per recursion
– Otherwise number of rays explodes!... | https://ocw.mit.edu/courses/6-837-computer-graphics-fall-2012/1b5985f78c68379e15543fe27d41b72c_MIT6_837F12_Lec18.pdf |
✔
29
Why Use Random Numbers?
• Fixed random sequence
• We see the structure in the error
n
e
s
n
e
J
n
n
a
W
k
i
r
n
e
H
Courtesy of Henrik Wann Jensen. Used with permission.
30
Demo
• http://madebyevan.com/webgl-path-tracing/
Image removed due to copyright restrictions. Please see the above link for fu... | https://ocw.mit.edu/courses/6-837-computer-graphics-fall-2012/1b5985f78c68379e15543fe27d41b72c_MIT6_837F12_Lec18.pdf |
designed by H.J. Wegner has been removed due to copyright restrictions.
Please see http://tora_2097.cgsociety.org/portfolio/project-detail/786738/ for further details.
Image: Pure
43
Photon Mapping
• Preprocess: cast rays from light sources, let them
bounce around randomly in the scene
• Store “photons”
44
P... | https://ocw.mit.edu/courses/6-837-computer-graphics-fall-2012/1b5985f78c68379e15543fe27d41b72c_MIT6_837F12_Lec18.pdf |
osity (with hierarchies & wavelets)
– Precomputed Radiance Transfer
• This would warrant a class of its own!
51
What Else Can We Integrate?
• Pixel: antialiasing
• Light sources: Soft shadows
• Lens: Depth of field
• Time: Motion blur
• BRDF: glossy reflection
• (Hemisphere: indirect lighting)
Courtesy of... | https://ocw.mit.edu/courses/6-837-computer-graphics-fall-2012/1b5985f78c68379e15543fe27d41b72c_MIT6_837F12_Lec18.pdf |
get different image
• What is the noise/variance/standard deviation?
– And what’s really going on anyway?
© source unknown. All rights reserved. This content is excluded from our Creative
Commons license. For more information, see http://ocw.mit.edu/help/faq-fair-use/.
56
Integration
• Compute integral of arbitra... | https://ocw.mit.edu/courses/6-837-computer-graphics-fall-2012/1b5985f78c68379e15543fe27d41b72c_MIT6_837F12_Lec18.pdf |
.random()
y=random.random()
if x*x+y*y<1: success = success+1 return
4.0*float(success)/float(n)
63
Why Not Use Simpson Integration?
• You’re right, Monte Carlo is not very efficient for
computing
• When is it useful?
– High dimensions: Convergence is independent of
dimension!
– For d dimensions, Simp... | https://ocw.mit.edu/courses/6-837-computer-graphics-fall-2012/1b5985f78c68379e15543fe27d41b72c_MIT6_837F12_Lec18.pdf |
available at http://www.cs.virginia.edu/~jdl/.
71
Sampling a BRDF
Slide courtesy of Jason Lawrence
Image removed due to copyright restrictions – please see Jason Lawrence’s slide 9-12 in the talk slides on “Efficient BRDF
Importance Sampling Using a Factored Representation,” available at http://www.cs.virginia.edu/... | https://ocw.mit.edu/courses/6-837-computer-graphics-fall-2012/1b5985f78c68379e15543fe27d41b72c_MIT6_837F12_Lec18.pdf |
• But mostly for low-dimensional domains
– Again, subdivision of N-D needs Nd domains like
trapezoid, Simpson’s, etc.!
• With very high dimensions, Monte Carlo is pretty
much the only choice
79
Questions?
•
Image from the ARNOLD Renderer by Marcos Fajardo
Images removed due to copyright restrictions -- Pleas... | https://ocw.mit.edu/courses/6-837-computer-graphics-fall-2012/1b5985f78c68379e15543fe27d41b72c_MIT6_837F12_Lec18.pdf |
6.096 Lecture 3:
Functions
How to reuse code
Geza Kovacs
#include <iostream>
using namespace std;
int main() {
int threeExpFour = 1;
for (int i = 0; i < 4; i = i + 1) {
threeExpFour = threeExpFour * 3;
}
cout << "3^4 is " << threeExpFour << endl;
return 0;
}
Copy-paste
coding
#include <i... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/1b954ce3fc96fd1168f8cd123c5748e4_MIT6_096IAP11_lec03.pdf |
}
cout << "12^10 is " << twelveExpTen << endl;
return 0;
}
With a
function
#include <iostream>
using namespace std;
// some code which raises an arbitrary integer
// to an arbitrary power
int main() {
int threeExpFour = raiseToPower(3, 4);
cout << "3^4 is " << threeExpFour << endl;
return 0;
}... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/1b954ce3fc96fd1168f8cd123c5748e4_MIT6_096IAP11_lec03.pdf |
= i + 1) {
result = result * base;
}
return result;
}
Function Declaration Syntax
Return type
int raiseToPower(int base, int exponent)
{
int result = 1;
for (int i = 0; i < exponent; i = i + 1) {
result = result * base;
}
return result;
}
Function Declaration Syntax
Argument 1
int raiseToPowe... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/1b954ce3fc96fd1168f8cd123c5748e4_MIT6_096IAP11_lec03.pdf |
int i = 0; i < exponent; i = i + 1) {
result = result * base;
}
return result;
}
Return statement
Function
declaration
#include <iostream>
using namespace std;
int raiseToPower(int base, int exponent) {
int result = 1;
for (int i = 0; i < exponent; i = i + 1) {
result = result * base;
}
return r... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/1b954ce3fc96fd1168f8cd123c5748e4_MIT6_096IAP11_lec03.pdf |
}
int main() {
int x = 4;
printNumberIfEven(x);
// even number; number is 3
int y = 5;
printNumberIfEven(y);
// odd number
}
Argument Type Matters
void printOnNewLine(int x)
{
cout << x << endl;
}
• printOnNewLine(3) works
• printOnNewLine("hello") will not compile
Argument Type Matters
vo... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/1b954ce3fc96fd1168f8cd123c5748e4_MIT6_096IAP11_lec03.pdf |
1 Integer: 3”
• printOnNewLine(2, 3) prints “2 Integers: 2 and 3”
• Function declarations need to occur before invocations
int foo()
{
return bar()*2; // ERROR - bar hasn’t been declared yet
}
int bar()
{
return 3;
}
• Function declarations need to occur before invocations
– Solution 1: reorder fu... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/1b954ce3fc96fd1168f8cd123c5748e4_MIT6_096IAP11_lec03.pdf |
);
}
int square(int x)
{
return x*x;
}
Recursion
• Functions can call themselves.
• fib(n) = fib(n-1) + fib(n-2) can be easily
expressed via a recursive implementation
int fibonacci(int n) {
if (n == 0 || n == 1) {
return 1;
} else {
return fibonacci(n-2) + fibonacci(n-1);
}
}
Recursion... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/1b954ce3fc96fd1168f8cd123c5748e4_MIT6_096IAP11_lec03.pdf |
return result;
}
int max(int num1, int num2) {
numCalls = numCalls + 1;
int result;
if (num1 > num2) {
result = num1;
}
else {
result = num2;
}
return result;
}
int numCalls = 0;
Scope
• Scope: where a
variable was declared,
determines where it
can be accessed from
int raiseToPower(in... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/1b954ce3fc96fd1168f8cd123c5748e4_MIT6_096IAP11_lec03.pdf |
return result;
}
int numCalls = 0;
int raiseToPower(int base, int exponent) {
numCalls = numCalls + 1;
int result = 1;
for (int i = 0; i < exponent; i = i + 1) {
result = result * base;
}
// A
return result;
}
int max(int num1, int num2) {
numCalls = numCalls + 1;
int result;
if (num1 > ... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/1b954ce3fc96fd1168f8cd123c5748e4_MIT6_096IAP11_lec03.pdf |
}
// A
return result;
}
int max(int num1, int num2) {
numCalls = numCalls + 1;
int result;
if (num1 > num2) {
result = num1;
}
else {
result = num2;
}
// B
return result;
}
Global scoperaiseToPower function scopemax function scopeint baseint exponentint resultint num1int num2int resultint nu... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/1b954ce3fc96fd1168f8cd123c5748e4_MIT6_096IAP11_lec03.pdf |
double squareRoot(double num) {
double low = 1.0;
double high = num;
for (int i = 0; i < 30; i = i + 1) {
double estimate = (high + low) / 2;
if (estimate*estimate > num) {
double newHigh = estimate;
high = newHigh;
} else {
double newLow = estimate;
low = newLow;
}
if (i == 29)
return estimat... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/1b954ce3fc96fd1168f8cd123c5748e4_MIT6_096IAP11_lec03.pdf |
q in main " << q << endl;
}
Output
a in increment 4
q in main 3
Pass by value vs by reference
// pass-by-value
void increment(int a) {
a = a + 1;
cout << "a in increment " << a << endl;
}
int main() {
int q = 3; // HERE
increment(q); // does nothing
cout << "q in main " << q << endl;
}
Output ... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/1b954ce3fc96fd1168f8cd123c5748e4_MIT6_096IAP11_lec03.pdf |
main 4
Pass by value vs by reference
// pass-by-value
void increment(int &a) {
a = a + 1;
cout << "a in increment " << a << endl;
}
int main() {
int q = 3; // HERE
increment(q); // works
cout << "q in main " << q << endl;
}
Output
a in increment 4
q in main 4
main function scopeq=3
Pass by ... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/1b954ce3fc96fd1168f8cd123c5748e4_MIT6_096IAP11_lec03.pdf |
int r = 5; // HERE
swap(q, r);
cout << "q " << q << endl; // q 5
cout << "r " << r << endl; // r 3
}
main function scopeq=3r=5
Implementing Swap
void swap(int &a, int &b) { // HERE
int t = a;
a = b;
b = t;
}
int main() {
int q = 3;
int r = 5;
swap(q, r);
cout << "q " << q << endl; // q 5
co... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/1b954ce3fc96fd1168f8cd123c5748e4_MIT6_096IAP11_lec03.pdf |
r);
cout << "q " << q << endl; // q 5
cout << "r " << r << endl; // r 3
}
main function scopeswap function scopeq=5r=3abt=3
Returning multiple values
• The return statement only allows you to
return 1 value. Passing output variables by
reference overcomes this limitation.
int divide(int numerator, int denom... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/1b954ce3fc96fd1168f8cd123c5748e4_MIT6_096IAP11_lec03.pdf |
Massachusetts Institute of Technology
Department of Materials Science and Engineering
77 Massachusetts Avenue, Cambridge MA 02139-4307
3.205 Thermodynamics and Kinetics of Materials—Fall 2006
October 31, 2006
Kinetics Lecture 2: Mass Diffusion and Heat Conduction
Lecture References
1. Porter and Easterling, Phase ... | https://ocw.mit.edu/courses/3-205-thermodynamics-and-kinetics-of-materials-fall-2006/1baa318e2b9bd0996e61c1e5017c1dbf_lecture02_review.pdf |
to the crystal follows Fick’s first law, with a proportionality constant known as the
intrinsic diffusivity. The intrinsic diffusivities and the self-diffusivities are related by P&E Eq. 2.64
with 2.69 (also KoM Eq. 3.13) and the relation involves a thermodynamic factor. Nonideality can
either accelerate or retard in... | https://ocw.mit.edu/courses/3-205-thermodynamics-and-kinetics-of-materials-fall-2006/1baa318e2b9bd0996e61c1e5017c1dbf_lecture02_review.pdf |
Chapter 3
Grand Unified Theory
3.1 SU(5) Unification
Gauge bosons:
SU (3)
�
SU (2)
�
U (1): (commuting with SU (3)
SU (2))
×
Or Lie algebra:
2iλ
e
2iλ
e
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎝
2iλ
e
3iλ
e−
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎠
3iλ
e−
2
2
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎝
2
3
−
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎠
3
−
(3.1)
(3.2)
(3.3)
One gets breaking SU (5)
... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/1bb0ca3337af13703289df734a424e7c_chap3.pdf |
−
change conjugation
−→
(u
1
2
−
2
3
�
L
C
R)
−
(dC
R)
C(eR)1
1
3
all L – handed .
Multiplets? Clues:
15 =
10
antisymmetric tensor
����
+ 5
vector
Y = 0
����
�
Altogether?
6
1
× 6
+ 2
× −
1
2
+ 3
× −
2
3
+ 3
1
× 3
v
e
�
L
+ 1
=
0
make 5
1
2
−
(dC
R) +
1
3
�
Actually (using
αβ )
�
... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/1bb0ca3337af13703289df734a424e7c_chap3.pdf |
) : 2.
Y =
1
−6
(2
−
3) =
1
6 ⇒
�
u
d
�
L
Ψαβ , color: ¯3, SU (2) : singlet.
Y =
1
6
−
(2 + 2) =
2
3 ⇒
−
u C
R
Ψij, color: singlet, SU (2) : 2
Y =
1
6
(
3
−
−
−
3) = 1
e C
R
⇒
It clicks.
Normalizing ˜g:
(3.14)
(3.15)
(3.16)
SU (3) generators
:
gun.
SU (2) generators
:
gun.
fab
t... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/1bb0ca3337af13703289df734a424e7c_chap3.pdf |
⎟
⎟
⎟
⎟
⎠
3
−
3.1. SU(5) UNIFICATION
= g
1
−
3
⎛
1
3
−
�2
g
(3
(
1
3
·
)2 + 2
·
g
So “naive” prediction:
(
)2) =
1
2
5
�2( ) =
6
�2
g =
�
⎜
⎜
⎜
⎜
⎜
⎜
⎝
2
gun.
1
2
1
2
3
2
g
5 un.
2
gun.
1
3
−
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎠
1
2
1
2
g
S = g
2
2
ω
=
5
2
g
3 un.
sin 2 θω =
g
�2
g
�2 + g2
ω
=
3
5
1 +
3
... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/1bb0ca3337af13703289df734a424e7c_chap3.pdf |
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