text stringlengths 16 3.88k | source stringlengths 60 201 |
|---|---|
mean bit-score: ∑
n
pk log (
2
k =1
pk
)
qk
If qk =
1
w
4
then mean RelEnt = 2w - Hmotif = Imotif
RelEnt is a measure of information, not entropy/uncertainty.
In general RelEnt is different from Hbefore - Hafter and is a better
measure when background is non-random
Example: qA = qT = 3/8, qC = qG = 1/8
Sup... | https://ocw.mit.edu/courses/7-91j-foundations-of-computational-and-systems-biology-spring-2014/5b2e28907ea0396609fd2cbccd2dd462_MIT7_91JS14_Lecture9.pdf |
Chapter 4
Nonlinear equations
4.1 Root finding
Consider the problem of solving any nonlinear relation g(x) = h(x) in the
real variable x. We rephrase this problem as one of finding the zero (root)
of a function, here f (x) = g(x) − h(x). The minimal assumption we need on
f, g, h is that they’re continuous.
We have... | https://ocw.mit.edu/courses/18-330-introduction-to-numerical-analysis-spring-2012/5b325bfa56a599794c7196de926844b0_MIT18_330S12_Chapter4.pdf |
Let
x ∗ be the unknown root. The error obeys
|x ∗ − mk| ≤ |bk − ak| = 2−k|b0 − a0|.
Every step of the bisection discovers a new correct digit in the binary
expansion of x ∗ .
The advantage of the bisection method is that it is guaranteed to con
verge to a root, by construction. On the other hand, convergence is
... | https://ocw.mit.edu/courses/18-330-introduction-to-numerical-analysis-spring-2012/5b325bfa56a599794c7196de926844b0_MIT18_330S12_Chapter4.pdf |
= 1.4167...,
= 1.4142157... The true value of 2 is 1.4142135...
x3 =
Convergence is very fast, when it occurs. Assume that f '' is continuous,
and that f ' (x) = 0 in some neighborhood of the root x ∗ (large enough
so that all our iterates stay in this neighborhood.) Put En = xn − x ∗ .
2 √
12
2
4.1. ROOT FINDI... | https://ocw.mit.edu/courses/18-330-introduction-to-numerical-analysis-spring-2012/5b325bfa56a599794c7196de926844b0_MIT18_330S12_Chapter4.pdf |
not on n.) It
follows that
1
C
We say the method “converges quadratically” because the exponent of
En is 2. The number of correct digits is squared at each iteration. In
contrast, the bisection method only converges linearly. We also some
times refer to “linear convergence” as first-order convergence, although
t... | https://ocw.mit.edu/courses/18-330-introduction-to-numerical-analysis-spring-2012/5b325bfa56a599794c7196de926844b0_MIT18_330S12_Chapter4.pdf |
xn −
fn
.
f [xn−1, xn]
The geometrical idea is to replace the tangent line at xn by the secant
line supported by xn−1 and xn. The secant method requires two points
x0 and x1 as starting guesses.
Notice that at each step, only one evaluation of f is necessary, because
f (xn−1) is already known from the previous ... | https://ocw.mit.edu/courses/18-330-introduction-to-numerical-analysis-spring-2012/5b325bfa56a599794c7196de926844b0_MIT18_330S12_Chapter4.pdf |
equations we get
En+1 =
'' (ξ)
1
f
2 f [xn−1, xn]
EnEn−1.
Again, thanks to the same assumptions on f as in Newton’s method,
f rr(ξ)
the ratio
has a finite limit as n → ∞, hence is bounded by
f [xn−1,xn]
some number C > 0. We get
|En+1| ≤ C|En||En−1|.
The decay of En is somewhere between first (linear) and sec... | https://ocw.mit.edu/courses/18-330-introduction-to-numerical-analysis-spring-2012/5b325bfa56a599794c7196de926844b0_MIT18_330S12_Chapter4.pdf |
. . . , n.
We take the same number of equations and unknowns, so that we may be in
a situation where there is one solution (rather than a continuum of solutions
or no solution at all.) Whether the system has zero, one or several solutions
is still a question that needs to be addressed separately. The shorthand
not... | https://ocw.mit.edu/courses/18-330-introduction-to-numerical-analysis-spring-2012/5b325bfa56a599794c7196de926844b0_MIT18_330S12_Chapter4.pdf |
as a root-finding problem: f1 = f2 = 0 with
x2 = sin(x1).
2 = 1,
f1(x1, x2) = x 2
1 + x 2
2 − 1,
f2(x1, x2) = x2 − sin(x1).
6
4.2. OPTIMIZATION PROBLEMS
The Jacobian matrix is
J = Vf (x) =
�
∂f1
∂x1
∂f2
∂x1
�
∂f1
∂x2
∂f2 =
∂x2
2x1
− cos(x1)
2x2
1
.
Use the formula for the inverse of a 2-by-2 matrix:
−... | https://ocw.mit.edu/courses/18-330-introduction-to-numerical-analysis-spring-2012/5b325bfa56a599794c7196de926844b0_MIT18_330S12_Chapter4.pdf |
(x), so it suffices to
talk about minimization.
When F (x) is smooth, and x is allowed to run over all real numbers (not
restricted to an interval or other set), then it suffices to solve F ' (x) = 0 (and
check that F '' (x) > 0) in order to find a local minimum. Hence it suffices
to apply Newton’s method or any other ro... | https://ocw.mit.edu/courses/18-330-introduction-to-numerical-analysis-spring-2012/5b325bfa56a599794c7196de926844b0_MIT18_330S12_Chapter4.pdf |
This function has a unique minimum for x1 ∈ R and x2 > 0. We compute
∇F (x1, x2) =
(cid:19)
(cid:18) 2x1
2 log x2
x2
and
(cid:32)
∇∇F (x1, x2) =
Newton’s iteration is therefore
0
2
0 2−2 log x2
x2
2
(cid:33)
.
(cid:19)
(cid:18)x1,n+1
x2,n+1
=
(cid:19)
(cid:18)x1,n
x2,n
−
(cid:32) 1
2
0
0
x2
2,n
2−2 log x2,n
(cid:33) (... | https://ocw.mit.edu/courses/18-330-introduction-to-numerical-analysis-spring-2012/5b325bfa56a599794c7196de926844b0_MIT18_330S12_Chapter4.pdf |
Lecture # 17
Solar Thermal Energy
Ahmed Ghoniem
April 6, 2020
Renewables: Some characteristics and specifics.
Historical Trends …
Solar Thermals: Concentrators and Plants
Renewable Sources and Their Utilization
Biomass
Geothermal
Solar
Wind/Wave
Chemical
Thermal
photo
Kinetic
Combustion
Gasification... | https://ocw.mit.edu/courses/2-60j-fundamentals-of-advanced-energy-conversion-spring-2020/5b41ce8e948ad3b93015bb1827897026_MIT2_60s20_lec17.pdf |
. The receiver is a forced circulation radiant boiler
receiving ~ 55 MWt of concentrated solar radiation. Storage capacity is 20 MWht, sufficient to
operate the turbines for 50 minutes at 50% capacity.
© Solucar. All rights reserved. This content is excluded from our Creative Commons
license. For more information, s... | https://ocw.mit.edu/courses/2-60j-fundamentals-of-advanced-energy-conversion-spring-2020/5b41ce8e948ad3b93015bb1827897026_MIT2_60s20_lec17.pdf |
of the civilized world.” Robert H. Thurston - 1901, the
Smithsonian Institution annual report.
“… the human race must finally utilize direct sun power or revert to barbarism because
eventually all coal and oil will be used up. I would recommend all far-sighted engineers and
inventors to work in this direction to th... | https://ocw.mit.edu/courses/2-60j-fundamentals-of-advanced-energy-conversion-spring-2020/5b41ce8e948ad3b93015bb1827897026_MIT2_60s20_lec17.pdf |
, 64 MWe, built over 250 acres (1.3 sq km), using 760 troughs.
Expected power 130 million kWh/y, capacity factor ~ 25%). Cost $250M (~$110M for IGCC and ~$35M for NGCC).
Ivanpah solar plant (2014), Dry Lake, CA, world largest CSP, 392 MW, capacity factor 28.72% . 4000 acres, 173,500
heliostats, $2.2 B ($1.6 B loan guar... | https://ocw.mit.edu/courses/2-60j-fundamentals-of-advanced-energy-conversion-spring-2020/5b41ce8e948ad3b93015bb1827897026_MIT2_60s20_lec17.pdf |
line is at 5.75 kWh/m2day, showing both direct and diffuse
radiation. Location affects number of hours/day of sun, solar angle, weather conditions, ..
14
Intermittency is tricky! Role of storage,
backup and multiple sources/technologies
How Much? On ... | https://ocw.mit.edu/courses/2-60j-fundamentals-of-advanced-energy-conversion-spring-2020/5b41ce8e948ad3b93015bb1827897026_MIT2_60s20_lec17.pdf |
Flat collectors:
limit heat transfer fluid tempertaure.
Typical values of β, is 80%,
Collection efficiency at Tc ~ 60 C, ~ 50%.
goes down linearly with temperature! must limit heat loss
© Ahmed F. Ghoniem
16
Concentrating Collectors:
1. Trough
2. Tower
3. Cone
© Source unknown. All rights reserved... | https://ocw.mit.edu/courses/2-60j-fundamentals-of-advanced-energy-conversion-spring-2020/5b41ce8e948ad3b93015bb1827897026_MIT2_60s20_lec17.pdf |
collectors deliver higher T than
cylindrical (2) collectors.
T requirements for
different engines and the
corresponding
concentration ratio
col (Tc − Ta )
net absorbed flux: qAcol = β A conc I − hAˆ
define: CR = Aconc / Acol
h Tc
then:
β depends on reflective and transmissive properies
of glass cover and absorba... | https://ocw.mit.edu/courses/2-60j-fundamentals-of-advanced-energy-conversion-spring-2020/5b41ce8e948ad3b93015bb1827897026_MIT2_60s20_lec17.pdf |
rights reserved. This content is excluded from our Creative Commons license. For more information, see https://ocw.mit.edu/fairuse.
22
Heat-Collection Element (HCE)
Space between absorber pipe
and glass shield is evacuated
• Reduces convective ... | https://ocw.mit.edu/courses/2-60j-fundamentals-of-advanced-energy-conversion-spring-2020/5b41ce8e948ad3b93015bb1827897026_MIT2_60s20_lec17.pdf |
courtesy of NREL.
• Total reflective area > 2.3 M. m2
• More than 117,000 Heat Collecting Elements
• 30 MW increment based on regulated power
block size
27
Hybridized Parabolic-Trough System
Image courtesy of DOE.
© Ahmed F. Ghoniem
Source: US DOE
28... | https://ocw.mit.edu/courses/2-60j-fundamentals-of-advanced-energy-conversion-spring-2020/5b41ce8e948ad3b93015bb1827897026_MIT2_60s20_lec17.pdf |
content is excluded from our Creative Commons license. For more information, see https://ocw.mit.edu/fairuse.
35
The solar potential in the MENA Region
AQUA-CSP: Concentrating solar power for seawater desalination
German Aerospace Center (DLR) http://www.dlr.de/tt/aqua-csp
© German Aerospace Center. All rights re... | https://ocw.mit.edu/courses/2-60j-fundamentals-of-advanced-energy-conversion-spring-2020/5b41ce8e948ad3b93015bb1827897026_MIT2_60s20_lec17.pdf |
an equivalent to planting 1.5 million trees, or taking 15,000 cars off the road.
40
Solar Chimney
the Hydroelectric Power for the desert”
r the desert”
Vch =
ΔT
T
2gH ch
Courtesy Elsevier, Inc., http://www... | https://ocw.mit.edu/courses/2-60j-fundamentals-of-advanced-energy-conversion-spring-2020/5b41ce8e948ad3b93015bb1827897026_MIT2_60s20_lec17.pdf |
Mokheimer, and A. F. Ghoniem, Journal of Hydrogen Energy, 40(7): 2939-2949, 2015
45
Solar Fuels?
Novel, looping based reformer
Parabolic
Solar Collector
Solar Radiation
Solar Window
Receiver Reactor
Solar Radiation
Courtesy Elsevier, Inc., http://www.sciencedirect.co... | https://ocw.mit.edu/courses/2-60j-fundamentals-of-advanced-energy-conversion-spring-2020/5b41ce8e948ad3b93015bb1827897026_MIT2_60s20_lec17.pdf |
Lecture 4: Stochastic
Thinking and Random
Walks
(cid:1010)(cid:856)(cid:1004)(cid:1004)(cid:1004)(cid:1006)(cid:3)(cid:62)(cid:286)(cid:272)(cid:410)(cid:437)(cid:396)(cid:286)(cid:3)(cid:1008)
1
Relevant Reading
Pages 235-238
Chapter 14
6.0002 LECTURE 4
2
The World is Hard to Under... | https://ocw.mit.edu/courses/6-0002-introduction-to-computational-thinking-and-data-science-fall-2016/5b4ec0ef29910b74e6a4bbd5ccf4dda5_MIT6_0002F16_lec4.pdf |
random element
def rollDie():
""" returns an int between 1 and 6"""
def rollDie():
""" returns a randomly chosen int
between 1 and 6"""
6.0002 LECTURE 4
8
Implementing a Random Process
import random
def rollDie():
"""returns a random int between 1 and 6"""
return random.... | https://ocw.mit.edu/courses/6-0002-introduction-to-computational-thinking-and-data-science-fall-2016/5b4ec0ef29910b74e6a4bbd5ccf4dda5_MIT6_0002F16_lec4.pdf |
Independence should not be taken for granted
6.0002 LECTURE 4
13
Will One of the Patriots and Broncos Lose?
Patriots have winning percentage of 7/8, Broncos of
6/8
Probability of both winning next Sunday is 7/8 * 6/8 =
42/64
Probability of at least one losing is 1 – 42/64 = 22/64
What a... | https://ocw.mit.edu/courses/6-0002-introduction-to-computational-thinking-and-data-science-fall-2016/5b4ec0ef29910b74e6a4bbd5ccf4dda5_MIT6_0002F16_lec4.pdf |
We’ll
talk lots more in later lectures about how to know
when we have enough trials.
Moral 2: One should not confuse the sample
probability with the actual probability
Moral 3: There was really no need to do this by
simulation, since there is a perfectly good closed form
answer. We will see many examples where... | https://ocw.mit.edu/courses/6-0002-introduction-to-computational-thinking-and-data-science-fall-2016/5b4ec0ef29910b74e6a4bbd5ccf4dda5_MIT6_0002F16_lec4.pdf |
',
1 - numerator/denom)
Suppose we want the probability of 3 people sharing
6.0002 LECTURE 4
20
Why 3 Is Much Harder Mathematically
For 2 the complementary problem is “all birthdays
distinct”
For 3 people, the complementary probl... | https://ocw.mit.edu/courses/6-0002-introduction-to-computational-thinking-and-data-science-fall-2016/5b4ec0ef29910b74e6a4bbd5ccf4dda5_MIT6_0002F16_lec4.pdf |
Lot
To model systems that are mathematically intractable
To extract useful intermediate results
Lend themselves to development by successive
refinement and “what if” questions
Start by simulating random walks
6.0002 LECTURE 4
25
Why Random Walks?
Random walks are important in man... | https://ocw.mit.edu/courses/6-0002-introduction-to-computational-thinking-and-data-science-fall-2016/5b4ec0ef29910b74e6a4bbd5ccf4dda5_MIT6_0002F16_lec4.pdf |
not until the next lecture
6.0002 LECTURE 4
34
MIT OpenCourseWare
https://ocw.mit.edu
6.0002 Introduction to Computational Thinking and Data Science
Fall 2016
For information about citing these materials or our Terms of Use, visit: https://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/6-0002-introduction-to-computational-thinking-and-data-science-fall-2016/5b4ec0ef29910b74e6a4bbd5ccf4dda5_MIT6_0002F16_lec4.pdf |
6.045: Automata, Computability, and
Complexity
Or, Great Ideas in Theoretical
Computer Science
Spring, 2010
Class 4
Nancy Lynch
Today
• Two more models of computation:
– Nondeterministic Finite Automata (NFAs)
• Add a guessing capability to FAs.
• But provably equivalent to FAs.
– Regular expres... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b6d125bc94344e53285a1cdbdb33371_MIT6_045JS11_lec04.pdf |
state, and
– F ⊆ Q is the set of accepting, or final states.
Formal Definition of an NFA
• An NFA is a 5-tuple ( Q, Σ, δ, q0, F ), where:
– Q is a finite set of states,
– Σ is a finite set (alphabet) of input symbols,
– δ: Q × Σε → P(Q) is the transition function,
– q0 ∈ Q, is the start state, and
– F ⊆ ... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b6d125bc94344e53285a1cdbdb33371_MIT6_045JS11_lec04.pdf |
( Q, Σ, δ, q0, F ),
where:
– Q is a finite set of states,
– Σ is a finite set (alphabet) of input symbols,
– δ: Q × Σε → P(Q) is the transition function,
Σε means Σ ∪ {ε }.
– q0 ∈ Q, is the start state, and
– F ⊆ Q is the set of accepting, or final states.
NFA Examples
Example 1:
0,1
Example 2:
0
a
b... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b6d125bc94344e53285a1cdbdb33371_MIT6_045JS11_lec04.pdf |
a } Æ { a, b } Æ { a, b } Æ { a, c } Æ { a, b }
1
0,1
ε
ε
a
Example 2
0
1
b
e
c
f
1
0
d
g
• L(M) = { w | w ends with 01 or 10 }
• Computations for w = 0010:
– Possible states after no input: { a, b, e }
– After 0: { a, b, e, c }
– After 0: { a, b, e, c }
– After 1: { a, b, e, d, f }
– After 0: { a... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b6d125bc94344e53285a1cdbdb33371_MIT6_045JS11_lec04.pdf |
by the entire input
string, possibly
interspersed with εs,
leading to an
accepting state.
Here, leads to accepting
state d.
a
ε
b
a
1
ε
e
ε
b
ε
e
f
1
Stuck: No
moves on ε
or 1
ε
ε
b
0
c
1
d
e
Stuck: No
moves on ε
or 0
Done,
accept
Formal definition of computation
• Define E(q) = set of ... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b6d125bc94344e53285a1cdbdb33371_MIT6_045JS11_lec04.pdf |
!
• Theorem: If M is an NFA then L(M) is DFA-recognizable.
•
Proof:
– Given NFA M1 = ( Q1, Σ, δ1, q01, F1 ), produce an equivalent DFA M2
= ( Q2, Σ, δ2, q02, F2 ).
• Equivalent means they recognize the same language, L(M2) =
L(M1).
– Each state of M2 represents a set of states of M1: Q2 = P(Q1).
– Start state of ... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b6d125bc94344e53285a1cdbdb33371_MIT6_045JS11_lec04.pdf |
:
1
0
0
{a}
{a,b}
F2
{a,c}
1
0
1
• Other 5 subsets aren’t reachable from start state,
don’t bother drawing them.
NFAs vs. DFAs
• NFAs and DFAs have the same power.
• But sometimes NFAs are simpler than equivalent DFAs.
• Example: L = strings ending in 01 or 10
– Simple NFA, harder DFA (LTTR)
• Examp... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b6d125bc94344e53285a1cdbdb33371_MIT6_045JS11_lec04.pdf |
(M3) = L(M1) ∪ L(M2).
– Idea: Run M1 and M2 “in parallel” on the same input. If
either reaches an accepting state, accept.
Closure under union
• Example:
M1: Substring 01
1
a
0
0,1
0
1
b
c
M2: Odd number of 1s
0
d
0
e
1
1
M3:
(cid:31)
0
0
ad
1
1
ae
0
0
bd
be
1
1
0
0
cd
ce
1
1
Closure under union, general rule
•... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b6d125bc94344e53285a1cdbdb33371_MIT6_045JS11_lec04.pdf |
.
• Concatenation, star:
– Now try NFA-based constructions.
Closure under concatenation
• L1 ◦ L2 = { x y | x ∈ L1 and y ∈ L2 }
• Theorem: FA-recognizable languages are closed
under concatenation.
• Proof:
– Start with NFAs M1 and M2.
– Get another NFA, M3, with L(M3) = L(M1) ◦ L(M2).
M1
ε
ε
M2
These are ... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b6d125bc94344e53285a1cdbdb33371_MIT6_045JS11_lec04.pdf |
1
0
1
0
ε
ε
Closure, summary
• FA-recognizable (regular) languages are
closed under set operations, concatenation,
and star.
• Regular operations: Union, concatenation,
and star.
• Can be used to build regular expressions,
which denote languages.
• E.g., regular expression ( 0 ∪ 1 )* 0 0*
denotes the l... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b6d125bc94344e53285a1cdbdb33371_MIT6_045JS11_lec04.pdf |
Sometimes omit parens, use precedence of operations: * highest,
then °, then ∪ .
• Example: Abbreviate above as ( ( 0 ∪ 1 ) ε )* ∪ 0
• Example: ( 0 ∪ 1 )* 111 ( 0 ∪ 1 )*
How regular expressions
denote languages
• Define the languages recursively, based on the
expression structure:
• Definition:
– L(a) = { ... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b6d125bc94344e53285a1cdbdb33371_MIT6_045JS11_lec04.pdf |
1 0* 1 0* )*
• Example: L = strings with substring 01 or 10.
( 0 ∪ 1 )* 01 ( 0 ∪ 1 )* ∪ ( 0 ∪ 1 )* 10 ( 0 ∪ 1 )*
Abbreviate (writing Σ for ( 0 ∪ 1 )):
Σ* 01 Σ* ∪ Σ* 10 Σ*
More examples
• Example: L = strings with substring 01 or 10.
( 0 ∪ 1 )* 01 ( 0 ∪ 1 )* ∪ ( 0 ∪ 1 )* 10 ( 0 ∪ 1 )*
Abbreviate:
Σ* 01 Σ*... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b6d125bc94344e53285a1cdbdb33371_MIT6_045JS11_lec04.pdf |
regular expression, then L(R)
is a regular language (recognized by a FA).
Easy.
• Proof:
• Theorem 2: If L is a regular language, then there
is a regular expression R with L = L(R).
Harder, more technical.
• Proof:
Theorem 1
• Theorem 1: If R is a regular expression, then L(R)
is a regular language (recogni... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b6d125bc94344e53285a1cdbdb33371_MIT6_045JS11_lec04.pdf |
• Theorem 1: If R is a regular expression, then L(R)
is a regular language (recognized by a FA).
• Proof:
– Case 6: R = (R1)*
• M1 recognizes L(R1),
• Same construction we used to show regular languages are
closed under star.
ε
M1
ε
ε
Example for Theorem 1
• L = ab ∪ a*
• Construct machines recursively:
• ... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b6d125bc94344e53285a1cdbdb33371_MIT6_045JS11_lec04.pdf |
remove x:
a ∪ bb* a
b*a
q0
b a*
y
qf
qf
• New label b*a describes all strings that can move the
machine from q0 to y, visiting (just) x any number of times.
• New label a ∪ bb* a describes all strings that can move the
machine from y to y, visiting (just) x any number of times.
Theorem 2
a ∪ bb* a
b*a
... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b6d125bc94344e53285a1cdbdb33371_MIT6_045JS11_lec04.pdf |
R ∪ SU*T
z
R
• If y = z:
y
S
U
x
R ∪ SU*T
y
T
we get:
Next time…
• Existence of non-regular languages
• Showing specific languages aren’t regular
• The Pumping Lemma
• Algorithms that answer questions about
FAs.
• Reading: Sipser, Section 1.4; some pieces
from 4.1
MIT OpenCourseWare
http://ocw.mit.e... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b6d125bc94344e53285a1cdbdb33371_MIT6_045JS11_lec04.pdf |
6.045: Automata, Computability, and
Complexity
Or, Great Ideas in Theoretical
Computer Science
Spring, 2010
Class 7
Nancy Lynch
Today
• Basic computability theory
• Topics:
– Decidable and recognizable languages
– Recursively enumerable languages
– Turing Machines that solve problems involving FAs
– Undecidability ... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b9f68b5081985b135f6ad521a93d81d_MIT6_045JS11_lec07.pdf |
-decidable
if there is some TM that recognizes L.
if there is some TM that decides L.
• The classes of Turing-recognizable and Turing-decidable
languages are different.
• Theorem 2: If L is Turing-decidable then L is Turing-
recognizable.
• Obviously.
• But the other direction does not hold---there are languages
tha... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b9f68b5081985b135f6ad521a93d81d_MIT6_045JS11_lec07.pdf |
2 one after the other because
the first one might never halt.
– Run them in parallel, until one accepts?
– How? We don’t have a parallel Turing Machine model.
Decidable and recognizable
languages
• Theorem 4: L is Turing decidable if and only if L and Lc are
both Turing-recognizable.
• Proof: ⇐
– M1 recognizes L, a... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b9f68b5081985b135f6ad521a93d81d_MIT6_045JS11_lec07.pdf |
rejects.
Examples
• Example: Let X = be the set of binary
representations of natural numbers for
which the following procedure halts:
while x ≠ 1 do
if x is odd then x := 3x + 1
if x is even then x := x/2
halt
– Obviously, X is Turing-recognizable: just
simulate this procedure and accept if/when it
halts.
– Is it ... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b9f68b5081985b135f6ad521a93d81d_MIT6_045JS11_lec07.pdf |
recognizability:
• Theorem 8: L is recursively enumerable if and
only if L is Turing-recognizable.
• Proof: ⇒
– Given E, an enumerator for L, construct Turing machine
M to recognize L.
– M: On input x:
• M simulates E (on no input, as usual).
• Whenever E prints, M checks to see if the new output is x.
• If it ever s... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b9f68b5081985b135f6ad521a93d81d_MIT6_045JS11_lec07.pdf |
, print out the associated input.
– Dovetail all computations of M.
– Complicated bookkeeping, messy to work out in detail.
– But can do algorithmically, hence on a Turing machine.
Turing Machines that solve problems
for other domains besides strings
Turing Machines that solve problems
for other domains
• [Sipser S... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b9f68b5081985b135f6ad521a93d81d_MIT6_045JS11_lec07.pdf |
also consider computability for domains
that are sets of machines:
• DFAs:
– Encode DFAs using bit strings, by defining standard
naming schemes for states and alphabet symbols.
– Then a DFA tuple is again a list.
– Example:
1
0,1
1
0
2
Encode as:
( (1, 2), (0, 1), ( (1, 1, 1), (1, 0, 2), (2, 0, 2), (2, 1, 2) ), (1)... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b9f68b5081985b135f6ad521a93d81d_MIT6_045JS11_lec07.pdf |
a Turing machine.
• Proves that L2 is Turing-decidable.
Turing Machines that solve DFA problems
• Example: Acceptance for DFAs
L3 = { < M, w > | w ∈ L(M) } is Turing-decidable.
• Domain is (DFA, input) pairs.
• Algorithm simply runs M on w.
• Formalize as a Turing machine.
• Proves that L3 is Turing-decidable.
Moving... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b9f68b5081985b135f6ad521a93d81d_MIT6_045JS11_lec07.pdf |
• AccTM = { < M, w > | M is a (basic) Turing machine, w is a
word in M’s alphabet, and M accepts w }.
• HaltTM = { < M, w > | M is a Turing machine, w is a word in
M’s alphabet, and M halts (either accepts or rejects) on w }.
• EmptyTM = { < M > | M is a Turing machine and L(M) = ∅ }
– Recall: L(M) refers to the set ... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b9f68b5081985b135f6ad521a93d81d_MIT6_045JS11_lec07.pdf |
M accepts w }.
• U: On input < M, w >:
– Simulate M on input w.
– If M accepts, accept.
– If M rejects, reject.
– Otherwise, U loops forever.
• U recognizes AccTM.
• Does U decide AccTM ?
• No.
– If M loops forever on w, U loops forever on <M,w>,
never accepts or rejects.
– To decide, U would have to detect when M is... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b9f68b5081985b135f6ad521a93d81d_MIT6_045JS11_lec07.pdf |
or if M loops on <M>.
– If H′ exists, then so does D: D runs H′ and outputs the
opposite.
Undecidability of Acceptance
• Theorem 2: AccTM is not Turing-decidable.
• Proof, cont’d:
– D(<M>):
• rejects if M accepts <M>,
• accepts if M rejects <M> or if M loops on <M>.
– Now, what happens if we run D on <D>?
– Plug in... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b9f68b5081985b135f6ad521a93d81d_MIT6_045JS11_lec07.pdf |
ither accepts or rejects) w }.
• Compare with AccTM = { < M, w > | M is a Turing machine
and M accepts w }.
• Terminology caution: Sipser calls AccTM the “halting
problem”, and calls HaltTM just HaltTM.
• Theorem: HaltTM is not Turing-decidable.
• Proof:
– Let’s not use diagonalization.
– Rather, take advantage of d... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b9f68b5081985b135f6ad521a93d81d_MIT6_045JS11_lec07.pdf |
If M
accepts, accept. If rejects, reject.
– If R rejects, then reject.
– Claim S decides AccTM: 3 cases:
• If M accepts w, then R accepts <M,w>, and the simulation leads
S to accept.
• If M rejects w, then R accepts <M,w>, and the simulation leads
S to reject.
• If M loops on w, then R rejects <M,w>, and S rejects.
... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b9f68b5081985b135f6ad521a93d81d_MIT6_045JS11_lec07.pdf |
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
6.436J/15.085J
Lecture 3
Fall 2018
CONDITIONING AND INDEPENDENCE
Most of the material in this lecture is covered in [Bertsekas & Tsitsiklis] Sec-
tions 1.3-1.5 and Problem 48 (or problem 43, in the 1st edition), available at
http://athenasc.com/Prob-2nd-Ch1.pdf. Solutions to ... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/5ba99e3803701c145f1d06c26dcf61b8_MIT6_436JF18_lec03.pdf |
X
i=1
P(Ai | B).
As a result, suppose PB : F → [0, 1] is defined by PB(A) = P(A | B).
Then, PB is a probability measure on (
, F).
(b) Let A be an event. If the events Bi, i ∈ N, form a partition of
P(Bi) > 0 for every i, then
, and
P(A) =
∞
X
i=1
P(A | Bi)P(Bi).
In particular, if B is an event with P(B) > ... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/5ba99e3803701c145f1d06c26dcf61b8_MIT6_436JF18_lec03.pdf |
(B) = 1. Also
Ω
Ω
∪∞P
i=1
�
P B ∩ (∪∞ Ai)
�
P(B)
Since the sets B ∩ Ai, i ∈ N are disjoint, countable additivity, applied to the
P(∪∞ (B ∩ Ai))
P(B)
Ai | B =
i=1
i=1
=
.
2
right-hand side, yields
∪∞
i=1
P
�
Ai | B =
P
∞
i=1
P(B ∩ Ai)
P(B)
=
∞
X
i=1
P(Ai | B),
as claimed.
(b) We have
... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/5ba99e3803701c145f1d06c26dcf61b8_MIT6_436JF18_lec03.pdf |
to
∩∞ Ai. By the continuity property of probability measures, we have P ∩∞
i=1
i=1
Ai = limn→∞ P ∩i
n
=1 Ai . Note that
i=1Ai
�
�
∩n
i=1
P
�
Ai = P(A1) ·
P(A1)
·
P(A1 ∩ A2)
P(A1 ∩ A2) P(A1 ∩ A2 ∩ A3)
· · ·
P(A1 ∩ · · · ∩ An)
P(A1 ∩ · · · ∩ An−1)
= P(A1)
n
Y
i=2
P(Ai | A1 ∩ · · · ∩ Ai−1).
T... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/5ba99e3803701c145f1d06c26dcf61b8_MIT6_436JF18_lec03.pdf |
As |
s ∈ S} be a family (set) of events. The events in this family are said to be
independent if for every finite subset S0 of S, we have
P ∩s∈S0 As =
�
P(As).
Y
s∈S0
(c) Let F1 ⊂ F and F2 ⊂ F be two ˙-fields. We say that F1 and F2 are
independent (write F1 ⊥ F2) if any two events A1 ∈ F1 and A2 ∈ F2 are
indep... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/5ba99e3803701c145f1d06c26dcf61b8_MIT6_436JF18_lec03.pdf |
| i ∈ N} are independent. This statement
captures the intuitive idea of “independent” coin tosses.
(c) Let Fn be the collection of all events whose occurrence can be decided by looking
at the results of tosses 2n and 2n + 1. (Note that each Fn is a ˙-field comprised of
finitely many events.) Then, the families Fn, n ... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/5ba99e3803701c145f1d06c26dcf61b8_MIT6_436JF18_lec03.pdf |
2.1 How to check independence of ˙-algebras? p-systems.
How can one establish that two complicated ˙-fields (e.g., as in the last example
above) are independent?
It turns out that one only needs to check indepen-
dence for smaller collections of sets – see the theorem below. This is similar to
the question of unique... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/5ba99e3803701c145f1d06c26dcf61b8_MIT6_436JF18_lec03.pdf |
Theorem 2. Let 1 and 2 be p-systems and Fi = ˙(i), i = 1, 2. If
P(A ∩ B) = P(A)P(B)
(1)
for every A ∈ 1, B ∈ 2, then F1 and F2 are independent.
Proof. Fix an arbitrary B ∈ 2 and define a collection of sets
LB
, {E ∈ F1 : P(E ∩ B) = P(E)P(B)}.
By assumption 1 ⊆ LB. We also have:
1. Clearly ∈ LB .
Ω
2. If ... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/5ba99e3803701c145f1d06c26dcf61b8_MIT6_436JF18_lec03.pdf |
Thus (1) holds for all
A ∈ F1 and B ∈ 2. By symmetry it also holds for all A ∈ 1 and B ∈ F2.
And applying the above argument again (with 2 replaced by F2) for all of F1
and F2.
α
Proposition 1. Let be a p-system on
satisfying the following:
Ω
. Let D be a collection containing
1.
Ω
∈ D
2. For all A, B ∈ ... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/5ba99e3803701c145f1d06c26dcf61b8_MIT6_436JF18_lec03.pdf |
∩ C = (B ∩ C) \ (A ∩ C). Thus
LC = D0 by minimality of D0. Hence D0 is closed under intersections with
elements of .
Ω
Next take an arbitrary D ∈ D0. We have
LD = {A ∈ D0 : A ∩ D ∈ D0}
containing (and
(same reasoning). Thus LD = D0 and D0 is closed under intersections.
) by the previous argument and closed un... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/5ba99e3803701c145f1d06c26dcf61b8_MIT6_436JF18_lec03.pdf |
antelli lemma is a tool that is often used to establish that a certain
event has probability zero or one. Given a sequence of events An, n ∈ N,
the event {An i.o.} (read as “An occurs infinitely often”) is defined to be the
that belong to infinitely many An. Show that
event consisting of all ! ∈
equivalently
Ω
{An i... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/5ba99e3803701c145f1d06c26dcf61b8_MIT6_436JF18_lec03.pdf |
1. Suppose that 0 ≤ pi ≤ 1 for every i ∈ N. Then:
∞
X
i=1
∞
X
i=1
pi = ∞ ⇒
∞
(1 − pi) = 0
Y
i=1
∞
pi = ∞ ⇐
(1 − pi) = 0, pi < 1
Y
i=1
(2)
(3)
Proof. Note that log(1 − x) is a concave function of its argument, and its deriva-
tive at x = 0 is −1. It follows that log(1 − x) ≤ −x, for x ∈ [0, 1]. We then
7
... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/5ba99e3803701c145f1d06c26dcf61b8_MIT6_436JF18_lec03.pdf |
result is automatic. Hence, we may also assume pi → 0. Then
taking n so large that pi ≤ 1 − e−1 for all i ≥ n we may apply the lower bound
log(1 − x) ≥ −
e
e − 1
x
∀0 ≤ x ≤ 1 − e
−1
.
Then, for arbitrary large C we have for all sufficiently large n
−C ≥ log
(1 − pi) ≥ −
n
Y
i=1
e
e − 1
n
X
i=1
pi ,
∞
i... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/5ba99e3803701c145f1d06c26dcf61b8_MIT6_436JF18_lec03.pdf |
will imply the desired result because
P(Ac) = P ∪∞ Bc ≤
�
n=1 n
∞
X
n=1
P(Bc ) = 0.
n
Let us fix some n and some m ≥ n. We have, using independence (show
that independence of {An} implies independence of {Ac }) n
P(∩m Ac
i=n i ) =
m
Y
i=n
P(Ac
i ) =
m
Y �
i=n
1 − P(Ai)
.
P(Ai) = ∞. Using
The assum... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/5ba99e3803701c145f1d06c26dcf61b8_MIT6_436JF18_lec03.pdf |
6.581 / 20.482 J
Foundations of Algorithms and Computational Techniques in Systems Biology
Adjoint Sensitivity Analysis for Optimization
From: Cao Y, Li ST, Petzold L, Serban R, Adjoint sensitivity
analysis or differential-algebraic equations: The adjoint DAE
system and its numerical solution, SIAM Journal on Scie... | https://ocw.mit.edu/courses/20-482j-foundations-of-algorithms-and-computational-techniques-in-systems-biology-spring-2006/5bb25cdfb8333d9ef2d0631fcffc8e01_adjoint.pdf |
2
6.581J / 20.482JFoundations of Algorithms and Computational Techniques in Systems BiologyProfessor Bruce TidorProfessor Jacob K. White
Trick 2: Integration By Parts
dG
dp
=
T
∫0
g
dx
x
dp
{
Hard
dx
+ g p +λ*(F
+ F
dt
)
x&
p
dp {
{ { Easy
+ F
x
{
dx&
dp
Hard
Hard
Easy
T
∫
0
*
λ
&
dx
F
x&... | https://ocw.mit.edu/courses/20-482j-foundations-of-algorithms-and-computational-techniques-in-systems-biology-spring-2006/5bb25cdfb8333d9ef2d0631fcffc8e01_adjoint.pdf |
T
x
x
x
T
T
⎧ d (λ Fx& ) = λ F + g
⎪ dt
⎪
dG⎨
⎪
⎪ dp
⎩
( g p +λ F
p )
∫0 14243
Easy
=
T
T
dt
+
⎛
⎜λT F
x&
⎝
dx
dp
t T=
t =0
3
6.581J / 20.482JFoundations of Algorithms and Computational Techniques in Systems BiologyProfessor Bruce TidorProfessor Jacob K. White
Trick 4: For Index 0 and 1 DAEs... | https://ocw.mit.edu/courses/20-482j-foundations-of-algorithms-and-computational-techniques-in-systems-biology-spring-2006/5bb25cdfb8333d9ef2d0631fcffc8e01_adjoint.pdf |
t
x
f p x t
( ,
+
p A x A x x B u B2 x
) = 1 + 2 ⊗ +
1
) =
⊗ x
p A A I
,
+ ⊗ ) +
+ 2 (
x
I
⊗
1
( p) ⊗ + 1
p A x A x x B u B u (
) = 1
+ 1
B I
2
+ 1
u
⊗
( p)
,
p
(
)
u
For Your Objective Function:
y )
− )* (
g x t
( ,
−
,
) = (
p Cx
y Cx
) = 2 * (
p C Cx − y)
p) = 0
,
,
( ,
g x t
x
( ,
g x t
p ... | https://ocw.mit.edu/courses/20-482j-foundations-of-algorithms-and-computational-techniques-in-systems-biology-spring-2006/5bb25cdfb8333d9ef2d0631fcffc8e01_adjoint.pdf |
Testing
6.170 Lecture 5
Fall 2005
Reading: Chapter 10 of Program Development in Java by Barbara Liskov
1
Program verification techniques and input space partitioning
The goal of testing — like that of all other program verification techniques — is to ensure that
a program functions correctly. Testing cannot prove ... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/5bb92f87c17d1cb32a055db3af9cb4cc_lec5.pdf |
we are guaranteed that the program is correct:
it operates correctly no matter what input it is given. Checking the output is usually done
by recording the expected answer and comparing the actual answer against that, though the
checking may also be done procedurally. The desired outputs can be generated by hand (fo... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/5bb92f87c17d1cb32a055db3af9cb4cc_lec5.pdf |
members. Testing
one member is equivalent, in terms of finding errors, to testing every member of the group.
Thus, a test suite that includes one (arbitrary) input from each equivalence group provides
complete testing of the implementation.
The difficulty with this idea is that it is just as hard to find the equivalenc... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/5bb92f87c17d1cb32a055db3af9cb4cc_lec5.pdf |
is the key heuristic used in testing. Coverage criteria aims to make some
input partition (some test case) exercise each code or specification construct. Since different
construct are likely to have different behavior, a test suite that uses each one is more likely
to be complete than one that doesn’t.
Complete covera... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/5bb92f87c17d1cb32a055db3af9cb4cc_lec5.pdf |
an integer output, it would be wise to supply
negative, positive, and zero inputs.
• Boundary cases When drawing equivalence class boundaries, even if the general idea is
correct, the boundaries may be slightly incorrect: perhaps they are shifted a bit, and/or
there are small equivalence classes at the boundaries o... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/5bb92f87c17d1cb32a055db3af9cb4cc_lec5.pdf |
one for inputs less than 10, one for the input 10, and
one for inputs greater than 10. (Maybe the input 10 should have behaved like inputs greater
than 10, but your code contains an error.) Again, your test suite fails to cover one of the
three true partitions.
Tester's guess at partition�
4� 5� 6�
7� 8�
9� 10� ... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/5bb92f87c17d1cb32a055db3af9cb4cc_lec5.pdf |
reverse order
//
// throws: NullPointerException if v1 is null or v2 is null
static void appendVector (Vector v1, Vector v2) {
to the end of v1
while (v2.size() > 0) {
Object elt = v2.remove(v2.size()-1);
v1.add(elt);
}
}
It works correctly unless its two arguments are the same Vector, in which case it loops fo... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/5bb92f87c17d1cb32a055db3af9cb4cc_lec5.pdf |
— and are just as valid and complete — even if a new implementation
is substituted for the old one.
Given the following specification,
// returns the absolute value of its input
int abs(int x);
the three heuristics produce the following test cases:
• coverage
– input: include some test. The input domain is monoli... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/5bb92f87c17d1cb32a055db3af9cb4cc_lec5.pdf |
perhaps by using either -1 and 2 or 1
and -2.)
– non-explicit values: again, none are obvious
• duplicates There is only a single input, a single output, and no non-explicit values, so the
only tests suggested are one in which the input and output differ and one in which the input
and output are the same.
With the... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/5bb92f87c17d1cb32a055db3af9cb4cc_lec5.pdf |
stutter, the location of the stutter (its distance from the beginning or the end), the
number of stutters in the string, and the number of maximal-length stutters. These are
all non-negative integers. It would be wise to include tests which make them zero; the
boundary cases will take care of that as well.
• bounda... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/5bb92f87c17d1cb32a055db3af9cb4cc_lec5.pdf |
by looking at the implementation. Such tests are sometimes
called “glass-box” or “white-box” (by contrast with “black-box”), or “structural” (by contrast
with “functional”), because they are chosen by considering the lexical or other structure of the
code.
Such test cases are especially effective at finding problems ... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/5bb92f87c17d1cb32a055db3af9cb4cc_lec5.pdf |
the lecture notes,
many different heuristics may produce the same test cases. That is less often the case with real
code.)
Achieving black box coverage does not guarantee clear-box coverage; four tests are needed. For
any three tests, at least one statement is not executed. That might well fail to locate a cut-and-
... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/5bb92f87c17d1cb32a055db3af9cb4cc_lec5.pdf |
such as local variables, to which the
heuristics can be applied (to obtain tests which cause those variables to contain a range of values).
As another example, a good test suite will try to execute each loop zero, one, two, and many times
on different executions of a procedure.
5
Test strategy and automation
An ad... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/5bb92f87c17d1cb32a055db3af9cb4cc_lec5.pdf |
before D; and C before E or F. Bottom-up testing
requires the writing of drivers which exercise the functionality of the class. Drivers simulate both its
environment (how it is called) in the system you are building, and all other environments in which
it may be called according to its specification. A driver is requ... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/5bb92f87c17d1cb32a055db3af9cb4cc_lec5.pdf |
a new bug when you fix an old one, but many people don’t bother to
re-validate their implementation after making corrections, extensions, or other modifications. If
tests are easy to run (for instance, you have written a test driver), then regression testing is very
easy to do.
Another effective testing strategy is to... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/5bb92f87c17d1cb32a055db3af9cb4cc_lec5.pdf |
never a use of a variable
before it is assigned to.
• think about the code. Even if no automatic proof is possible, a human may be able to prove
properties of code, either formally or by informal reasoning. Even the process of writing down
the representation invariant, abstraction function, and pre- and post-condit... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/5bb92f87c17d1cb32a055db3af9cb4cc_lec5.pdf |
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 18 - Solution of F.E. equations
Prof. K.J. Bathe
In structures,
F (u, p) = R.
In heat transfer,
F (θ) = Q
In fluid flow,
F (v, p, θ) = R
In structures/solids
�
F (m) � �
F =
=
m
m
0V (m)
Elastic materials
Example p. 590 textbook
MIT O... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/5bbd789c4c008f43a78ffc97d314ed5d_MIT2_094S11_lec18.pdf |
0L
tτ11 = tL
tτ11
∴
0
L tτ11 = E˜ ·
tL
1
2
��
t
u
1 + 0
L
�2
�
− 1
⇒ tτ11A =
tP =
��
1 +
E˜A
2
tu �2
0L
� �
− 1
1 +
tu �
0L
This is because of the material-law assumption (18.5) (okay for small strains . . . )
Hyperelasticity
tW = f (Green-Lagrange strains, material constants)
0
�
�
tSij =
... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/5bbd789c4c008f43a78ffc97d314ed5d_MIT2_094S11_lec18.pdf |
F
�
(i−1) +
U
= f
∂f
∂U
�
�
�
�
t+Δt
(i−1)
U
�
U ∗ − t+ΔtU (i−1) + H.O.T.
·
�
where t+ΔtU (i−1) is the value we just calculated and an approximation to U ∗.
Assume t+ΔtR is independent of the displacements.
�
t+ΔtR
0 =
−
t+ΔtF (i−1)
�
−
∂ t+ΔtF
∂U
�
�
�
�
t+Δt
(i−1)
U
· ΔU (i)
We obtain
t+ΔtK (i−... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/5bbd789c4c008f43a78ffc97d314ed5d_MIT2_094S11_lec18.pdf |
+ΔtR
−
t+ΔtF (1)
Convergence Use
�ΔU (i)�2 < �
�a�2 =
��
(ai)2
i
But, if incremental displacements are small in every iteration, need to also use
�t+ΔtR − t+ΔtF (i−1)�2 < �R
18.1 Slender structures
(beams, plates, shells)
t
Li
� 1
79
(18.23)
(18.24)
(18.25)
(18.26)
(18.27)
(18.28)
18. Solution of... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/5bbd789c4c008f43a78ffc97d314ed5d_MIT2_094S11_lec18.pdf |
Massachusetts Institute of Technology
Department of Electrical Engineering and Computer Science
6.245: MULTIVARIABLE CONTROL SYSTEMS
by A. Megretski �
Interpretations for Standard Optimization Setup1
This is the second lecture on standard feedback optimization setup. It describes a variety
of ways to come up with... | https://ocw.mit.edu/courses/6-245-multivariable-control-systems-spring-2004/5bca558808b8b652b14b2be6128d8e7d_lec2_6245_2004.pdf |
= c0 +
t
0
f (δ )dδ,
where c0 is an arbitrary constant. In general, a system’s output is not necessarily unique
because it may also depend on a set of auxiliary parameters (e.g. the initial states of
the system), as in Figure 2.1. In the case of the pure integrator system described above,
input�
initial state ... | https://ocw.mit.edu/courses/6-245-multivariable-control-systems-spring-2004/5bca558808b8b652b14b2be6128d8e7d_lec2_6245_2004.pdf |
(�t)
at rate T = 1/M for M ≥= 1).
An alternative way of representing a discrete time signal f = f (t) with sampling rate
T > 0 is by specifying T and the sequence of its values f [k] at time instances tk = kT ,
i.e.
f [k] = f (kT ), k = 0, 1, 2, . . . .
Thus, a DT signal f (t) is completely defined by its sampl... | https://ocw.mit.edu/courses/6-245-multivariable-control-systems-spring-2004/5bca558808b8b652b14b2be6128d8e7d_lec2_6245_2004.pdf |
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