text:
# Trouble with this integral and only an hour to solve it

1. Aug 9, 2004

### Brianjw

Well first here is the question:

A deep-sea diver is suspended beneath the surface of Loch Ness by a cable of length h that is attached to a boat on the surface . The diver and his suit have a total mass of m and a volume of V. The cable has a diameter of d and a linear mass density of mu. The diver thinks he sees something moving in the murky depths and jerks the end of the cable back and forth to send transverse waves up the cable as a signal to his companions in the boat.

Calculate the tension in the cable a distance above the diver. The buoyant force on the cable must be included in your calculation. Take the free fall acceleration to be g.

I solve this one, I got:

$$F(x) = (\mu*x+m)*g-\rho_{water}*g*((d/2)^2*pi*x+V)$$

For the next one I can't solve it, the integral is nuts for me, I must be overlooking something

The speed of transverse waves on the cable is given by v = sqrt(F/mu). The speed therefore varies along the cable, since the tension is not constant. (This expression neglects the damping force that the water exerts on the moving cable.) Integrate to find the time required for the first signal to reach the surface. Take the free fall acceleration to be g.

From what I understand I have to plug in my F(x) from part b and integrate, but I'm not sure how to integrate it since its just so huge. Any ideas?

Thanks!

Brian

2. Aug 10, 2004

### ehild

Write $$F/\mu$$ in the form $$F/\mu=ax+b$$ .

Remember that

$$v=x'=\frac{dx}{dt}$$ .

So you have got the differential equation

$$\frac{dx}{dt}=\sqrt{ax+b}$$

with the starting condition x=0 at t=0.
You can integrate it by separating the variables:

$$\int_0^h\frac{dx}{\sqrt{ax+b}}=\int_0^T{dt}=T$$

where T is the time needed for the signal to reach the surface and h is the lenght of the cable. The integrand at the left side is

$$(ax+b)^{-1/2}$$ .

I think you can do it...

ehild


text:
# Legendre Transform of Entropy

1. Jan 10, 2013

### kd215

Trying an example from a textbook but I don't understand Legendre transform at all. "construct Legendre transforms of the entropy that are natural functions of (1/T, V, n) and (1/T, V, μ/T)". I don't really understand where to start. An example prior to this exercise just states: A = E - TS = A(T,V,n) ; and G = E - TS - (-PV) = E - TS +PV = G(T,P,N).

If someone explains how the examples used Legendre transforms that should be enough for me to figure out the entropy problem. Thanks

2. Jan 11, 2013

### vanhees71

You always start from the first and second law of thermodynamics, i.e.,
$$\mathrm{d}U=T \mathrm{d} S-p \mathrm{d} V+\mu \mathrm{d} N.$$
Here, $U$ is the internal energy, $T$ the temperature, $S$ the entropy, $p$ the pressure, $V$ the volume, $\mu$ the chemical potential, and $N$ a (conserved!) particle-number like quantity. In nonrelativistic physics often you can take $N$ as the particle number itself, while in non-relativistic physics usually there are only conserved charges like elecctric charge, net-baryon number, etc. Here, we just consider the case of one such particle-number like quantities and thus only one chemical potential. The generalization to more variables is straight forward.

The above equation shows that the "natural" independent variables for the internal energy are $S$, $V$, and $N$. Since by construction $\mathrm{d} U$ is a total differential, you read off relations like
$$\left (\frac{\partial U}{\partial S} \right)_{V,N}=T.$$
Here, the variables listed in the subscript indicate which variables have to be kept constant when differentiating. It's pretty important in this business to keep track of what's left constant when taking partial derivatives of the various quantities. Since from $U$ you can derive other quantities by taking partial derivatives one calls $U$ a thermodynamical potential, and which other quantities you get by taking derivatives in the most straight-forward way is determined by what are the "natural" independent quantities to use as explained above.

Let's now look at the entropy. From the equation above we get
$$\mathrm{d} S=\frac{1}{T} \mathrm{d} U+\frac{p}{T} \mathrm{d} V-\frac{\mu}{T} \mathrm{d}N.$$
The "natural" independent variables to be used together with the entropy are thus $U$, $V$, and $N$, and the corresponding partial derivatives wrt. to one of these three variables (keeping precisely the other two fixed) are the coefficients in front of the differentials. Again we use the argument that $S$ is a state variable and thus $\mathrm{d} S$ is a total differential. For the following it's convenient to introduce new variables, namely
$$\beta=1/T, \quad \alpha=\mu/T.$$
Then we have
$$\mathrm{d} S=\beta \mathrm{d} U+\frac{p}{T} \mathrm{d} V -\alpha \mathrm{d} N.$$
With a Legendre transformation you define new quantities such that other variables become the "natural" independent variables. Let's start with the first example. You are supposed to find a Legendre transformation of $S$ such that the natural quantities become $(\beta,V,N)$. Now, for the enropy $V$ and $N$ are already natural variables, but not $\beta$, but we have the relation
$$\beta=\left (\frac{\partial S}{\partial U} \right )_{V,N}.$$
Now to make $\beta$ a natural variable of another quantitiy, you simply have to define the new quantity as
$$X=S-\beta U$$
Then from the above given differential you have
$$\mathrm{d} X=\mathrm{d} S - \mathrm{d}(\beta U)=\mathrm{d} S - U \mathrm{d} \beta-\beta \mathrm{d}U = \frac{p}{T} \mathrm{d} V-\alpha \mathrm{d} N-U \mathrm{d} \beta.$$
Now obviously the natural variables for $X$ are indeed $(\beta,V,N)$.

As next step you want still a new quantity such that the natural variables are $\beta,V,N$. Now you read off the last equation
$$\left (\frac{\partial X}{\partial N} \right)_{\beta,V}=-\alpha.$$
Thus, by another Legendre transform you see that the searched quanity must be
$$\Omega=X+\alpha N=S-\beta U+\alpha N$$
$$\mathrm{d} \Omega=\frac{p}{T} \mathrm{d} V+N \mathrm{d} \alpha-U \mathrm{d} \beta.$$
That's the grand-canonical potential with the natural independent variables $V, \quad \beta, \quad \alpha$.

3. Jan 12, 2013

### jfy4

The goal is to make a new function, $\tilde{f}(p)$, of a new variable, $p$, and, in doing so maximize
$$xp-f(x)$$
for a given value of $x$. To maximize anything we have drilled in us to take the derivative and set it equal to zero
$$xp-f(x)\implies p=\frac{df(x)}{dx}\equiv g(x)$$
where I renamed the derivative as a new function, $g(x)$. Now we are interested in the $x$ that maximizes the first expression, so we solve for $x$
$$x=g^{-1}(p)$$
then
$$\tilde{f}(p)=p \, g^{-1}(p) - f \left( g^{-1}(p) \right)$$
this is the new function. Now let's start with the expression for $p$.
$$p=\frac{df}{dx}\implies p \, dx = df \implies d(px) - x \, dp = df \implies d(xp-f) = x \, dp$$
which is true by looking back to the first part. So you can revert back and forth between $x$ and $p$ with this little "reverse product rule" move. You can try it on your thermodynamic pair of choice. Like $TdS = d(TS) - S \, dT$ and you can reshuffle the total differentials to make it look nice :)

format: plain


text:
Science

# 30% reduction in deaths from bowel cancer

30% reduction in deaths from bowel cancerThe rate of new cases of bowel cancer in Austria has fallen by around 20% in the last ten years, while the associated mortality rate has fallen by nearly 30%. This trend is primarily due to improvements in preventive screening colonoscopy, in which precancerous stages are removed before the disease can take hold.The rate of new cases of bowel cancer in Austria has fallen by around 20% in the last ten years, while the associated mortality rate has fallen by nearly 30%. This trend is primarily due to improvements in preventive screening colonoscopy, in which precancerous stages are removed before the disease can take hold.

Source: https://www.sciencedaily.com/releases/2017/04/170419093149.htm


text:
Santino Bautista

2022-06-21

Reading about random sets, I've come across the phrase "$\mathcal{B}\left(\mathcal{F}\right)$ is the Borel $\sigma$-algebra generated by the topology of closed convergence." where $\mathcal{F}$ is the family of closed sets in $\mathbb{R}.$
I don't know a lot about topology, so I'm not sure how to understand/interpret this. In general, a Borel $\sigma$-algebra on a set $S$ is the $\sigma$-algebra generated by the open sets of $S$, and I suppose a topology gives a notion of "openness", but I would appreciate if someone could provide me some intuition behind this particular σ-algebra and the "topology of closed convergence".

### Answer & Explanation

Patricia Curry

For any family $\mathcal{F}$ of subsets of a set $X$ we can talk about the $\sigma$-algebra generated by $\mathcal{F}$ on $X$, which is the defined to be the intersection of all $\sigma$-algebras on $X$ that contain $\mathcal{F}$ as a subset. This is standard in measure theory. It's well-defined as the intersection of $\sigma$-algebras is a $\sigma$-algebra again and we always have at least one σ-algebra containing $\mathcal{F}$, namely the power set of $X$.
A topology on $X$ is just another such family $\mathcal{F}$ so the definition applies. The $\sigma$-algebra will thus contain all open sets, all closed sets (their complements), all ${G}_{\delta }$ sets and ${F}_{\sigma }$ sets etc. (all in that topology). It's also called the Borel $\sigma$-algebra of a topological space.

Do you have a similar question?

Recalculate according to your conditions!


text:
Dayan & Abbott Week 2

Spike Trains and Firing Rates (cont.)

1. The brain is big complicated and messy.
2. We can only measure/simulate very simplified versions.
3. Almost always treat action potentials as identical even though they vary in duration/amplitude/shape.
4. Different ways of talking about spike sequences and firing rates:
1. $\rho(t)$ = neural response function = list of times of action potentials.
2. $r$ = spike-count rate = frequency of firing over extended period calculated from one trial.
3. $r(t)$ = time-dependent firing rate = average over many trials for the same stimulus chance to fire at time $t$. In other words, rate that changes over small periods of time. Must be calculated over multiple trials.
4. $\langle r \rangle$ = average firing rate = frequency of firing over extended period averaged over multiple trials.
5. Figure 1.4 shows different methods for calculating r(t).
6. Would be really good to generate a map labelling known stimulus with the region in the brain that responds to them.
7. From Figure 1.5 interesting that curve shows Gaussian response to $s_{max}$. What is the benefit over binary?
8. Error bars in Fig 1.9 seem small compared to interpretation of Figures on left.
9. Are the following two different interpretations of Fig 1.10c significantly different and equally plausible? a) Neuron is passing different information when it fires in rapid succession. b) Conditions for the neuron firing twice withing 5ms are arbitrary and only F1.10a is important.

Homogeneous Poisson Processes

Standard Questions/Notes

• To simplify things we can assume that the history is less important
• Point process: Firing could depend on all history.
• Renewal Process: Firing depends on history since the last firing.
• Poisson Process: No dependence at all on previous events.
• Homogeneous Poisson Process: firing rate is independent of time. Just fitting random spikes over the time period with probability dependent on the known firing rate.

Technical Questions/Notes

Python script for generating and plotting homogenous Poisson distribution with given length (s) and firing rate (Hz):
import sys
from numpy import *
import matplotlib.pyplot as plt

#Initialise
T = 100  #length of trial
r = 10 #rate of firing in Hz
if (len(sys.argv) == 3):
T = int(sys.argv[1])
r = int(sys.argv[2])
print "Using given values. Length = %is, Rate = %iHz" % (T, r)
else:
print "Using default values. Length = %is, Rate = %iHz" % (T, r)

#Generate Data
intervals = []
while (sum(intervals) < T): #add random intervals using Poisson distribution until over length of trial
intervals.append(-log(random.rand())/r)
intervals.pop() #remove last one as it's over the length of the trial

#Output Results
print "Number of spikes: %i (expected approx. %i)" % (len(intervals), r*T)
#Coefficient of variation = deviation of interval / average interval
CV = std(intervals) / mean(intervals)
print "Coefficient of Variation (should be approx 1) = %f" % CV

sp_times = cumsum(intervals) #ordered array of each spike time.

#Plotting the generated spike chain.
y_axis = [1] * len(sp_times)
plt.bar(sp_times, y_axis, width=(1/(float(T*r))))
plt.show()

The results show the problems with not modelling the refractory period with many spikes very close when rate is set high.

Python script for checking generation leads to Gaussian distribution for number of spikes over the time period:

import sys
from numpy import *
import matplotlib.pyplot as plt

#Initialise
T = 10  #length of trial
r = 10 #rate of firing in Hz
trials = 10000
if (len(sys.argv) >= 3):
T = int(sys.argv[1])
r = int(sys.argv[2])
if (len(sys.argv) >= 4):
trials = int(sys.argv[3])
print "Using given values. Length = %is, Rate = %iHz for %i trials" % (T, r, trials)
else:
print "Using default values. Length = %is, Rate = %iHz for %i trials" % (T, r, trials)

#Generate Data
no_spikes = []
while (len(no_spikes) < trials):
if(len(no_spikes) % 10 == 0):
print '{0}{1}{2}\r'.format(len(no_spikes)/float(trials)*100, "%", " completed..."),
intervals = []
while (sum(intervals) < T): #add random intervals using Poisson distribution until over length of trial
intervals.append(-log(random.rand())/r)
no_spikes.append(len(intervals)-1)

#Output Results
print "Average number of spikes: %f (expected approx. %i)" % (mean(no_spikes), r*T)

density = []
for i in range(0, max(no_spikes)+1):
density.append(no_spikes.count(i))
print density

plt.plot(density, 'o')
plt.show()

Sample output:


text:
English

Problem: Repository setup is unsuccessful

Description

During the repository set up, while running the post installation setup, the following error message occurs:

"Repository setup is unsuccessful".

The wise_err.log contains a line similar to:

"illegal option -- Files\arcgis\arcsde\sqlexe\".

Cause

ArcSDE is being installed to a directory with spaces in the name. For example: C:\Program Files\ArcGIS.

Solution or Workaround

1. Open the wise_err.log file found at <ArcSDE installation location>ArcSDE\<database>exe\etc.

2. Look for spaces in any incomplete directory name in the first line after the 'illegal option'. If the incomplete path is caused by a space in the installation directory, select one of the two options given below:

• Run the repository set up from the command line.
Navigate to %SDEHOME%\bin and manually launch sdesetup<database>.exe
• Uninstall ArcSDE. Re-install it to a path without spaces; for example: C:\arcgis. Run the Post installation setup.


text:
Contiki-NG
psock.h
Go to the documentation of this file.
1 /*
2  * Copyright (c) 2004, Swedish Institute of Computer Science.
4  *
5  * Redistribution and use in source and binary forms, with or without
6  * modification, are permitted provided that the following conditions
7  * are met:
8  * 1. Redistributions of source code must retain the above copyright
9  * notice, this list of conditions and the following disclaimer.
10  * 2. Redistributions in binary form must reproduce the above copyright
11  * notice, this list of conditions and the following disclaimer in the
12  * documentation and/or other materials provided with the distribution.
13  * 3. Neither the name of the Institute nor the names of its contributors
14  * may be used to endorse or promote products derived from this software
15  * without specific prior written permission.
16  *
17  * THIS SOFTWARE IS PROVIDED BY THE INSTITUTE AND CONTRIBUTORS AS IS'' AND
18  * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
19  * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
20  * ARE DISCLAIMED. IN NO EVENT SHALL THE INSTITUTE OR CONTRIBUTORS BE LIABLE
21  * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
22  * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
23  * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
24  * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
25  * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
26  * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
27  * SUCH DAMAGE.
28  *
29  * This file is part of the Contiki operating system.
30  *
32  *
33  */
34
35 /**
37  * @{
38  */
39
40 /**
41  * \defgroup psock Protosockets library
42  * @{
43  *
44  * The protosocket library provides an interface to the uIP stack that is
45  * similar to the traditional BSD socket interface. Unlike programs
46  * written for the ordinary uIP event-driven interface, programs
47  * written with the protosocket library are executed in a sequential
48  * fashion and does not have to be implemented as explicit state
49  * machines.
50  *
51  * Protosockets only work with TCP connections.
52  *
53  * The protosocket library uses \ref pt protothreads to provide
54  * sequential control flow. This makes the protosockets lightweight in
55  * terms of memory, but also means that protosockets inherits the
56  * functional limitations of protothreads. Each protosocket lives only
57  * within a single function block. Automatic variables (stack
58  * variables) are not necessarily retained across a protosocket
59  * library function call.
60  *
61  * \note Because the protosocket library uses protothreads, local variables
62  * will not always be saved across a call to a protosocket library
63  * function. It is therefore advised that local variables are used
64  * with extreme care.
65  *
66  * The protosocket library provides functions for sending data without
67  * having to deal with retransmissions and acknowledgements, as well
68  * as functions for reading data without having to deal with data
69  * being split across more than one TCP segment.
70  *
71  * Because each protosocket runs as a protothread, the protosocket has to be
72  * started with a call to PSOCK_BEGIN() at the start of the function
73  * in which the protosocket is used. Similarly, the protosocket protothread can
74  * be terminated by a call to PSOCK_EXIT().
75  *
76  */
77
78 /**
79  * \file
80  * Protosocket library header file
81  * \author
83  *
84  */
85
86 #ifndef PSOCK_H_
87 #define PSOCK_H_
88
89 #include "contiki.h"
90 #include "contiki-lib.h"
91 #include "contiki-net.h"
92
93  /*
94  * The structure that holds the state of a buffer.
95  *
96  * This structure holds the state of a uIP buffer. The structure has
97  * no user-visible elements, but is used through the functions
98  * provided by the library.
99  *
100  */
101 struct psock_buf {
102  uint8_t *ptr;
103  unsigned short left;
104 };
105
106 /**
107  * The representation of a protosocket.
108  *
109  * The protosocket structure is an opaque structure with no user-visible
110  * elements.
111  */
112 struct psock {
113  struct pt pt, psockpt; /* Protothreads - one that's using the psock
114  functions, and one that runs inside the
115  psock functions. */
116  const uint8_t *sendptr; /* Pointer to the next data to be sent. */
117  uint8_t *readptr; /* Pointer to the next data to be read. */
118
119  uint8_t *bufptr; /* Pointer to the buffer used for buffering
120  incoming data. */
121
122  uint16_t sendlen; /* The number of bytes left to be sent. */
123  uint16_t readlen; /* The number of bytes left to be read. */
124
125  struct psock_buf buf; /* The structure holding the state of the
126  input buffer. */
127  unsigned int bufsize; /* The size of the input buffer. */
128
129  unsigned char state; /* The state of the protosocket. */
130 };
131
132 void psock_init(struct psock *psock, uint8_t *buffer, unsigned int buffersize);
133 /**
134  * Initialize a protosocket.
135  *
136  * This macro initializes a protosocket and must be called before the
137  * protosocket is used. The initialization also specifies the input buffer
138  * for the protosocket.
139  *
140  * \param psock (struct psock *) A pointer to the protosocket to be
141  * initialized
142  *
143  * \param buffer (uint8_t *) A pointer to the input buffer for the
144  * protosocket.
145  *
146  * \param buffersize (unsigned int) The size of the input buffer.
147  *
148  * \hideinitializer
149  */
150 #define PSOCK_INIT(psock, buffer, buffersize) \
151  psock_init(psock, buffer, buffersize)
152
153 /**
154  * Start the protosocket protothread in a function.
155  *
156  * This macro starts the protothread associated with the protosocket and
157  * must come before other protosocket calls in the function it is used.
158  *
159  * \param psock (struct psock *) A pointer to the protosocket to be
160  * started.
161  *
162  * \hideinitializer
163  */
164 #define PSOCK_BEGIN(psock) PT_BEGIN(&((psock)->pt))
165
166 PT_THREAD(psock_send(struct psock *psock, const uint8_t *buf, unsigned int len));
167 /**
168  * Send data.
169  *
170  * This macro sends data over a protosocket. The protosocket protothread blocks
171  * until all data has been sent and is known to have been received by
172  * the remote end of the TCP connection.
173  *
174  * \param psock (struct psock *) A pointer to the protosocket over which
175  * data is to be sent.
176  *
177  * \param data (uint8_t *) A pointer to the data that is to be sent.
178  *
179  * \param datalen (unsigned int) The length of the data that is to be
180  * sent.
181  *
182  * \hideinitializer
183  */
184 #define PSOCK_SEND(psock, data, datalen) \
186
187 /**
188  * \brief Send a null-terminated string.
189  * \param psock Pointer to the protosocket.
190  * \param str The string to be sent.
191  *
192  * This function sends a null-terminated string over the
193  * protosocket.
194  *
195  * \hideinitializer
196  */
197 #define PSOCK_SEND_STR(psock, str) \
198  PT_WAIT_THREAD(&((psock)->pt), psock_send(psock, (uint8_t *)str, strlen(str)))
199
201  unsigned short (*f)(void *), void *arg));
202
203 /**
204  * \brief Generate data with a function and send it
205  * \param psock Pointer to the protosocket.
206  * \param generator Pointer to the generator function
207  * \param arg Argument to the generator function
208  *
209  * This function generates data and sends it over the
210  * protosocket. This can be used to dynamically generate
211  * data for a transmission, instead of generating the data
212  * in a buffer beforehand. This function reduces the need for
213  * buffer memory. The generator function is implemented by
214  * the application, and a pointer to the function is given
215  * as an argument with the call to PSOCK_GENERATOR_SEND().
216  *
217  * The generator function should place the generated data
218  * directly in the uip_appdata buffer, and return the
219  * length of the generated data. The generator function is
220  * called by the protosocket layer when the data first is
221  * sent, and once for every retransmission that is needed.
222  *
223  * \hideinitializer
224  */
225 #define PSOCK_GENERATOR_SEND(psock, generator, arg) \
227  psock_generator_send(psock, generator, arg))
228
229
230 /**
231  * Close a protosocket.
232  *
233  * This macro closes a protosocket and can only be called from within the
234  * protothread in which the protosocket lives.
235  *
236  * \param psock (struct psock *) A pointer to the protosocket that is to
237  * be closed.
238  *
239  * \hideinitializer
240  */
241 #define PSOCK_CLOSE(psock) uip_close()
242
244 /**
245  * Read data until the buffer is full.
246  *
247  * This macro will block waiting for data and read the data into the
248  * input buffer specified with the call to PSOCK_INIT(). Data is read
249  * until the buffer is full..
250  *
251  * \param psock (struct psock *) A pointer to the protosocket from which
252  * data should be read.
253  *
254  * \hideinitializer
255  */
258
259
260 /**
261  * Read data until at least len bytes have been read.
262  *
263  * This macro will block waiting for data and read the data into the
264  * input buffer specified with the call to PSOCK_INIT(). Data is read
265  * until the buffer is full or len bytes have been read.
266  *
267  * \param psock (struct psock *) A pointer to the protosocket from which
268  * data should be read.
269  * \param len (uint16_t) The minimum number of bytes to read.
270  *
271  * \hideinitializer
272  */
275
277 /**
278  * Read data up to a specified character.
279  *
280  * This macro will block waiting for data and read the data into the
281  * input buffer specified with the call to PSOCK_INIT(). Data is only
282  * read until the specified character appears in the data stream.
283  *
284  * \param psock (struct psock *) A pointer to the protosocket from which
285  * data should be read.
286  *
287  * \param c (char) The character at which to stop reading.
288  *
289  * \hideinitializer
290  */
293
294 /**
295  * The length of the data that was previously read.
296  *
297  * This macro returns the length of the data that was previously read
299  *
300  * \param psock (struct psock *) A pointer to the protosocket holding the data.
301  *
302  * \hideinitializer
303  */
304 #define PSOCK_DATALEN(psock) psock_datalen(psock)
305
306 uint16_t psock_datalen(struct psock *psock);
307
308 /**
309  * Exit the protosocket's protothread.
310  *
311  * This macro terminates the protothread of the protosocket and should
312  * almost always be used in conjunction with PSOCK_CLOSE().
313  *
314  * \sa PSOCK_CLOSE_EXIT()
315  *
316  * \param psock (struct psock *) A pointer to the protosocket.
317  *
318  * \hideinitializer
319  */
320 #define PSOCK_EXIT(psock) PT_EXIT(&((psock)->pt))
321
322 /**
323  * Close a protosocket and exit the protosocket's protothread.
324  *
325  * This macro closes a protosocket and exits the protosocket's protothread.
326  *
327  * \param psock (struct psock *) A pointer to the protosocket.
328  *
329  * \hideinitializer
330  */
331 #define PSOCK_CLOSE_EXIT(psock) \
332  do { \
333  PSOCK_CLOSE(psock); \
334  PSOCK_EXIT(psock); \
335  } while(0)
336
337 /**
338  * Declare the end of a protosocket's protothread.
339  *
340  * This macro is used for declaring that the protosocket's protothread
341  * ends. It must always be used together with a matching PSOCK_BEGIN()
342  * macro.
343  *
344  * \param psock (struct psock *) A pointer to the protosocket.
345  *
346  * \hideinitializer
347  */
348 #define PSOCK_END(psock) PT_END(&((psock)->pt))
349
350 char psock_newdata(struct psock *s);
351
352 /**
353  * Check if new data has arrived on a protosocket.
354  *
355  * This macro is used in conjunction with the PSOCK_WAIT_UNTIL()
356  * macro to check if data has arrived on a protosocket.
357  *
358  * \param psock (struct psock *) A pointer to the protosocket.
359  *
360  * \hideinitializer
361  */
362 #define PSOCK_NEWDATA(psock) psock_newdata(psock)
363
364 /**
365  * Wait until a condition is true.
366  *
367  * This macro blocks the protothread until the specified condition is
368  * true. The macro PSOCK_NEWDATA() can be used to check if new data
369  * arrives when the protosocket is waiting.
370  *
371  * Typically, this macro is used as follows:
372  *
373  \code
375  {
376  PSOCK_BEGIN(s);
377
378  PSOCK_WAIT_UNTIL(s, PSOCK_NEWDATA(s) || timer_expired(t));
379
380  if(PSOCK_NEWDATA(s)) {
382  } else {
383  handle_timed_out(s);
384  }
385
386  PSOCK_END(s);
387  }
388  \endcode
389  *
390  * \param psock (struct psock *) A pointer to the protosocket.
391  * \param condition The condition to wait for.
392  *
393  * \hideinitializer
394  */
395 #define PSOCK_WAIT_UNTIL(psock, condition) \
396  PT_WAIT_UNTIL(&((psock)->pt), (condition));
397


text:
$$\require{cancel}$$

# 5.22: Capacitance

When separate regions of positive and negative charge exist in proximity, Coulomb forces (Section 5.1) will attempt to decrease the separation between the charges. As noted in Section 5.8, this can be interpreted as a tendency of a system to reduce its potential energy. If the charges are fixed in place, then the potential energy remains constant. This potential energy is proportional to the Coulomb force. Referring back to Section 5.1, the Coulomb force is:

• Proportional to quantity of positive charge squared
• Inversely proportional to the separation between the charges squared
• Inversely proportional to the permittivity of the material separating the charges

Therefore, the potential energy of the system is likewise dependent on charge, separation, and permittivity. Furthermore, we see that the ability of a system to store energy in this manner depends on the geometry of the charge distribution and the permittivity of the intervening material.

Now recall that the electric field intensity $${\bf E}$$ is essentially defined in terms of the Coulomb force; i.e., $${\bf F}=q{\bf E}$$ (Section 2.2). So, rather than thinking of the potential energy of the system as being associated with the Coulomb force, it is equally valid to think of the potential energy as being stored in the electric field associated with the charge distribution. It follows from the previous paragraph that the energy stored in the electric field depends on the geometry of the charge distribution and the permittivity of the intervening media. This relationship is what we mean by capacitance. We summarize as follows:

Capacitance is the ability of a structure to store energy in an electric field.

and

The capacitance of a structure depends on its geometry and the permittivity of the medium separating regions of positive and negative charge.

Note that capacitance does not depend on charge, which we view as either a stimulus or response from this point of view. The corresponding response or stimulus, respectively, is the potential associated with this charge. This leads to the following definition: $\boxed{ C \triangleq \frac{Q_+}{V} } \label{m0112_Cdef}$ where $$Q_+$$ (units of C) is the total positive charge, $$V$$ (units of V) is the potential associated with this charge (defined such that it is positive), and $$C$$ (units of F) is the associated capacitance. So:

In practice, capacitance is defined as the ratio of charge present on one conductor of a two-conductor system to the potential difference between the conductors (Equation \ref{m0112_Cdef}).

In other words, a structure is said to have greater capacitance if it stores more charge – and therefore stores more energy – in response to a given potential difference.

Figure $$\PageIndex{1}$$ shows the relevant features of this definition.

Figure $$\PageIndex{1}$$: Electrostatic interpretation of capacitance.

Here, a battery imposes the potential difference $$V$$ between two regions of perfectly-conducting material. $$Q_+$$ is the total charge on the surface of the PEC region attached to the positive terminal of the battery. An equal amount of negative charge appears on the surface of the PEC region attached to the negative terminal of the battery (Section 5.19). This charge distribution gives rise to an electric field. Assuming the two PEC regions are fixed in place, $$Q_+$$ will increase linearly with increasing $$V$$, at a rate determined by the capacitance $$C$$ of the structure.

A capacitor is a device that is designed to exhibit a specified capacitance. We can now make the connection to the concept of the capacitor as it appears in elementary circuit theory. In circuit theory, the behavior of devices is characterized in terms of terminal voltage $$V_T$$ in response to terminal current $$I_T$$, and vice versa. First, note that current does not normally flow through a capacitor,1 so when we refer to “terminal current” for a capacitor, what we really mean is the flow of charge arriving or departing from one of the conductors via the circuit, which is equal to the flow of charge departing or arriving (respectively) at the other conductor. This gives the appearance of current flow through the capacitor when the current is examined from outside the capacitor. With that settled, we proceed as follows. Using Equation \ref{m0112_Cdef}, we express the voltage $$V_T$$ across the terminals of a capacitor having capacitance $$C$$:

$V_T = \frac{Q_+}{C}$

We seek a relationship between $$V_T$$ and $$I_T$$. Current is charge per unit time, so the charge on either conductor is the integral of $$I_T$$ over time; i.e.:

$Q_+(t) = \int_{t_0}^{t} { I_T(\tau)~d\tau } + Q_+(t_0)$

where $$t_0$$ is an arbitrarily-selected start time. In other words, amps integrated over time is charge. If we define $$I_T$$ as being positive in the direction of the flow of positive charge as is the usual convention, then we have:

$V_T(t) = \frac{1}{C}\int_{t_0}^{t} { I_T(\tau)~d\tau } + \frac{1}{C}Q_+(t_0)$

Again applying Equation \ref{m0112_Cdef}, we see that the second term is simply $$V_T(t_0)$$. This is the expected relationship from elementary circuit theory.

Finally, solving for $$I_T$$ we obtain the differential form of this relationship: $I_T(t) = C\frac{d}{dt}V_T(t)$


text:
## EXOEXOEXO one year ago if the point (3,-7) is the center of a circle and the point (8,5) is on the circle, what is the circumference of the circle?

1. jdoe0001

|dw:1439943435490:dw| find the radius keeping in mind that $$\bf \textit{circumference}=2\pi r\qquad r=radius$$

2. jdoe0001

so the radius is really the distance between, the center point, and point on the circle thus $$\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({\color{red}{ 3}}\quad ,&{\color{blue}{ -7}})\quad % (c,d) &({\color{red}{ 8}}\quad ,&{\color{blue}{ 5}}) \end{array}\qquad % distance value d = \sqrt{({\color{red}{ x_2}}-{\color{red}{ x_1}})^2 + ({\color{blue}{ y_2}}-{\color{blue}{ y_1}})^2}\to radius$$

Find more explanations on OpenStudy


text:
## Requirements🔗

You can only add support for plugins that have a public API. This means Maven must be able to resolve the dependency from an online repository. Adding dependencies from you local hard-drive is NOT allowed as this stops everyone from building the plugin.

Open up the pom.xml file located in the project's folder. Check if the new dependencies' repository already exists in our list of repositories. If that's the case, search for the dependency block related to that repository - there are comments above these blocks indicating that.

If there is no such repository tag, add it. New repository tags need to be added in this format: betonquest-<repoName>-repo. Then add a new dependency block for that repository. There needs to be a comment above that dependency block that indicates which repository holds this dependency. Take a look at the other blocks for guidance.

## Setting up the mirror for the new repo🔗

We have configured our Maven project to speed our builds up using mirror repositories. When adding new repositories we need to add them as mirrors to our Maven repository. As long as they are not added, your local Maven build will fail. We will add the repository as a mirror when you open your pull request on GitHub.

However, to test your changes you need to compile. Therefore, you need to bypass the mirrors so your dependency will be downloaded from the original repository. There are three ways do this:

The first option is to temporarily add the following to the mirrorOf entry in the file .mvn/settings.xml. This allows Maven to find the new repo as it is excluded from our mirror repo.

      <mirrorOf>*,!betonquest-repoName-repo</mirrorOf>

Second option, you add the following to the command line when executing Maven to disable our mirrors.

-DskipProjectSettings=true

Third option, you ask us on GitHub (Issue or PullRequest) to add the new repository to our mirrors.


text:
# 3.10 Frequency response

 Page 1 / 1

Recall from [link] that the convolution integral

$y\left(t\right)={\int }_{-\infty }^{\infty }x\left(\tau \right)h\left(t-\tau \right)d\tau$

has the Fourier Transform:

$Y\left(j\Omega \right)=H\left(j\Omega \right)X\left(j\Omega \right)$

where $H\left(j\Omega \right)$ and $X\left(j\Omega \right)$ are the Fourier Transforms of $h\left(t\right)$ and $x\left(t\right)$ , respectively. Solving for $H\left(j\Omega \right)$ gives the frequency response :

$H\left(j\Omega \right)=\frac{Y\left(j\Omega \right)}{X\left(j\Omega \right)}$

The frequency response, the Fourier Transform of the impulse response of a filter, is useful since it gives a highly descriptive representation of the properties of the filter. The frequency response can be considered to be the gain of the filter, expressed as a function of frequency. The magnitude of the frequency response evaluated at $\Omega ={\Omega }_{0}$ , $|H\left(j{\Omega }_{0}\right)|$ gives the factor the frequency component of $x\left(t\right)$ at $\Omega ={\Omega }_{0}$ would be scaled by. The phase of the frequency response at $\Omega ={\Omega }_{0}$ , $\angle H\left(j{\Omega }_{0}\right)$ gives the phase shift the component of $x\left(t\right)$ at $\Omega ={\Omega }_{0}$ would undergo. This idea will be discussed in greater detail in [link] . A lowpass filter is a filter which only passes low frequencies, while attenuating or filtering out higher frequencies. A highpass filter would do just the opposite, it would filter out low frequencies and allow high frequencies to pass. [link] shows examples of these various filter types.

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text:
Chapter P - P.2 - Linear Models and Rates of Change - Exercises - Page 18: 83

Distance $=\dfrac{5\sqrt{2}}{2}$

Work Step by Step

Point: $(-2,1)$ $;$ Line: $x-y-2=0$ The distance between the point $(x_{1},y_{1})$ and the line $Ax+By+C=0$ is given by $\dfrac{|Ax_{1}+By_{1}+C|}{\sqrt{A^{2}+B^{2}}}$. In this case, $(x_{1},y_{1})=(-2,1)$ and $A=1$, $B=-1$ and $C=-2$ Substitute the known values into the formula and evaluate: Distance $=\dfrac{|(1)(-2)+(-1)(1)-2|}{\sqrt{1^{2}+(-1)^{2}}}=...$ $...=\dfrac{|-2-1-2|}{\sqrt{1+1}}=\dfrac{|-5|}{\sqrt{2}}=\dfrac{5}{\sqrt{2}}=...$ Rationalize the denominator: $...=\dfrac{5}{\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}}=\dfrac{5\sqrt{2}}{(\sqrt{2})^{2}}=\dfrac{5\sqrt{2}}{2}$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


text:
# Using CAPM to find correlation of two assets with each other

I stumpled upon an exercise in an investments book:

The data below describe a three-stock financial market that satisfies the single-index model.

Stock Capitalization Beta Mean Excess Return Standard Deviation
A     $3,000 1.0 10% 40% B$1,940         0.2  2%                 30%


text:
# ECSE 506: Stochastic Control and Decision Theory

## Aditya Mahajan Winter 2022

Theory: Monotone value functions and policies

TL;DR General conditions are presented under which the optimal policy is monotone. Such a structural property is useful because it makes it easy to search and implement the optimal policy.

Consider the matrix formulation of MDPs and suppose the state space $$\ALPHABET S$$ is totally ordered. In many applications, it is useful to know if the value function is increasing (or decreasing) in state.

Theorem 1

Consider an MDP where the state space $$\ALPHABET S$$ is totally ordered. Suppose the following conditions are satisfied.

C1. For every $$a \in \ALPHABET A$$, the per-step cost $$c_t(s,a)$$ is weakly inceasing in $$s$$.

C2. For every $$a \in \ALPHABET A$$, the transition matrix $$P(a)$$ is stochastically monotone.

Then, the value function $$V_t(s)$$ is weakly increasing in $$s$$.

Note

The result above also applies to models with continuous (and totally ordered) state space provided the measurable selection conditions hold so that the arg min at each step of the dynamic program is attained.

#### Proof

We proceed by backward induction. By definition, $$V_{T+1}(s) = 0$$, which is weakly increasing. This forms the basis of induction. Assume that $$V_{t+1}(s)$$ is weakly increasing. Now consider, $Q_t(s,a) = c_t(s,a) + \EXP[V_{t+1}(S_{t+1}) | S_t = s, A_t = a].$ For any $$a \in \ALPHABET A$$, $$Q_t(s,a)$$ is a sum of two weakly increasing functions in $$s$$; hence $$Q_t(s,a)$$ is weakly increasing in $$s$$.

Now consider $$s_1, s_2 \in \ALPHABET S$$ such that $$s_1 > s_2$$. Suppose $$a_1^*$$ is the optimal action at state $$s_1$$. Then $V_t(s^1) = Q_t(s^1, a_1^*) \stackrel{(a)}\ge Q_t(s^2,a_1^*) \stackrel{(b)}\ge V_t(s_2),$ where $$(a)$$ follows because $$Q_t(\cdot, u^*)$$ is weakly increasing and $$(b)$$ follows from the definition of the value function.

# 1 Submodularity

Definition

Let $$\ALPHABET X$$ and $$\ALPHABET Y$$ be partially ordered sets. A function $$f \colon \ALPHABET X \times \ALPHABET Y \to \reals$$ is called submodular if for any $$x^+ \ge x^-$$ and $$y^+ \ge y^-$$, we have $$$\label{eq:submodular} f(x^+, y^+) + f(x^-, y^-) \le f(x^+, y^-) + f(x^-, y^+).$$$

The function is called supermodular if the inequality in \eqref{eq:submodular} is reversed.

A continuous and differentiable function on $$\reals^2$$ is submodular iff $\frac{ \partial^2 f(x,y) }{ \partial x \partial y } \le 0, \quad \forall x,y.$ If the inequality is reversed, then the function is supermodular.

Submodularity is a useful property because it implies monotonicity of the arg min.

Theorem 2

Let $$\ALPHABET X$$ be a partially ordered set, $$\ALPHABET Y$$ be a totally ordered set, and $$f \colon \ALPHABET X \times \ALPHABET Y \to \reals$$ be a submodular function. Suppose that for all $$x$$, $$\arg \min_{y \in \ALPHABET Y} f(x,y)$$ exists. Then, $π(x) := \max \{ y^* \in \arg \min_{y \in \ALPHABET Y} f(x,y) \}$ is weakly increasing in $$x$$.

#### Proof

Consider $$x^+, x^- \in \ALPHABET X$$ such that $$x^+ \ge x^-$$. Since $$f$$ is submodular, for any $$y \le π(x^-)$$, we have $$$\label{eq:1} f(x^+, π(x^-)) - f(x^+, y) \le f(x^-, π(x^-)) - f(x^-, y) \le 0,$$$ where the last inequality follows because $$π(x^-)$$ is the arg min of $$f(x^-, y)$$. Eq. \eqref{eq:1} implies that for all $$y \le π(x^-)$$, $f(x^+, π(x^-)) \le f(x^+, y).$ Thus, $$π(x^+) \ge π(x^-)$$.

The analogue of Theorem 2 for supermodular functions is as follows.

Theorem 3

Let $$\ALPHABET X$$ be a partially ordered set, $$\ALPHABET Y$$ be a totally ordered set, and $$f \colon \ALPHABET X \times \ALPHABET Y \to \reals$$ be a supermodular function. Suppose that for all $$x$$, $$\arg \min_{y \in \ALPHABET Y} f(x,y)$$ exists. Then, $π(x) := \min \{ y^* \in \arg \min_{y \in \ALPHABET Y} f(x,y) \}$ is weakly decreasing in $$x$$.

#### Proof

The proof is similar to Theorem 2.

Consider $$x^+, x^- \in \ALPHABET X$$ such that $$x^+ \ge x^-$$. Since $$f$$ is supermodular, for any $$y \ge π(x^-)$$, we have $$$\label{eq:2} f(x^+, y) - f(x^+, π(x^-)) \ge f(x^-, y) - f(x^-, π(x^-)) \ge 0,$$$ where the last inequality follows because $$π(x^-)$$ is the arg min of $$f(x^-, y)$$. Eq. \eqref{eq:2} implies that for all $$y \ge π(x^-)$$, $f(x^+, y) \ge f(x^+, π(x^-)).$ Thus, $$π(x^+) \le π(x^-)$$.

# 2 Monotonicity of optimal policy

Theorem 4

Consider an MDP where the state space $$\ALPHABET S$$ and the action space $$\ALPHABET A$$ are totally ordered. Suppose that, in addition to (C1) and (C2), the following condition is satisfied.

C3. For any weakly increasing function $$v$$, $c_t(s,a) + \EXP[ v(S_{t+1}) | S_t = s, A_t = a]$ is submodular in $$(s,a)$$.

Let $$π^*_t(s) = \max\{ a^* \in \arg \min_{a \in \ALPHABET A} Q_t(s,a) \}$$. Then, $$π^*(s)$$ is weakly increasing in $$s$$.

#### Proof

Conditions (C1) and (C2) imply that the value function $$V_{t+1}(s)$$ is weakly increasing. Therefore, condition (C3) implies that $$Q_t(s,a)$$ is submodular in $$(s,a)$$. Therefore, the arg min is weakly increasing in $$x$$

It is difficult to verify condition (C3). The following conditions are sufficient for (C3).

Lemma 1

Consider an MDP with totally ordered state and action spaces. Suppose

1. $$c_t(s,a)$$ is submodular in $$(s,a)$$.
2. For all $$s' \in \ALPHABET S$$, $$H(s' | s,a) = 1 - \sum_{z \le s'} P_{sz}(a)$$ is submodular in $$(s,a)$$.

The condition (C3) of the previous theorem holds.

#### Proof

Consider $$s^+, s^- \in \ALPHABET S$$ and $$a^+, a^- \in \ALPHABET A$$ such that $$s^+ > s^-$$ and $$a^+ > a^-$$. Define

\begin{align*} μ_1(s) &= \tfrac 12 P_{s^- s}(a^-) + \tfrac 12 P_{s^+ s}(a^+), \\ μ_2(s) &= \tfrac 12 P_{s^- s}(a^+) + \tfrac 12 P_{s^+ s}(a^-). \end{align*} Since $$H(s' | s,a)$$ is submodular, we have $H(s' | s^+, a^+) + H(s' | s^-, a^-) \le H(s' | s^+, a^-) + H(s' | s^-, a^+)$ or equivalently, $\sum_{z \le s'} \big[ P_{s^+ z}(a^+) + P_{s^- z}(a^-) \big] \ge \sum_{z \le s'} \big[ P_{s^+ z}(a^-) + P_{s^- z}(a^+) \big].$ which implies $M_1(s') \ge M_2(s')$ where $$M_1$$ and $$M_2$$ are the CDFs of $$μ_1$$ and $$μ_2$$. Thus, $$μ_1 \preceq_s μ_2$$.

Hence, for any weakly increasing function $$v \colon \ALPHABET S \to \reals$$, $\sum_{s' \in \ALPHABET S} μ_1(s') v(s') \le \sum_{s' \in \ALPHABET S} μ_2(s') v(s').$ Or, equivalently, $H(s^+, a^+) + H(s^-, a^-) \le H(s^-, a^+) + H(s^+, a^-)$ where $$H(s,a) = \EXP[ v(X_{t+1}) | X_t = s, U_t = a]$$.

Therefore, $$c_t(s,a) + H_t(s,a)$$ is submodular in $$(s,a)$$.

The analogue of Theorem 4 for supermodular functions is as follows.

Theorem 5

Consider an MDP where the state space $$\ALPHABET S$$ and the action space $$\ALPHABET A$$ are totally ordered. Suppose that, in addition to (C1) and (C2), the following condition is satisfied.

C4. For any weakly increasing function $$v$$, $c_t(s,a) + \EXP[ v(S_{t+1}) | S_t = s, A_t = a]$ is supermodular in $$(s,a)$$.

Let $$π^*_t(s) = \min\{ a^* \in \arg \min_{a \in \ALPHABET S} Q_t(s,a) \}$$. Then, $$π^*(s)$$ is weakly decreasing in $$s$$.

#### Proof

Conditions (C1) and (C2) imply that the value function $$V_{t+1}(s)$$ is weakly increasing. Therefore, condition (C4) implies that $$Q_t(s,a)$$ is supermodular in $$(s,a)$$. Therefore, the arg min is decreasing in $$s$$

It is difficult to verify condition (C4). The following conditions are sufficient for (C4).

Lemma 2

Consider an MDP with totally ordered state and action spaces. Suppose

1. $$c_t(s,a)$$ is supermodular in $$(s,a)$$.
2. For all $$s' \in \ALPHABET S$$, $$H(s' | s,a) = 1 - \sum_{z \le s'} P_{sz}(a)$$ is supermodular in $$(s,a)$$.

The condition (C4) of the previous theorem holds.

#### Proof

Consider $$s^+, s^- \in \ALPHABET S$$ and $$a^+, a^- \in \ALPHABET A$$ such that $$s^+ > s^-$$ and $$a^+ > a^-$$. Define

\begin{align*} μ_1(s) &= \tfrac 12 P_{s^- s}(a^-) + \tfrac 12 P_{s^+ s}(a^+), \\ μ_2(s) &= \tfrac 12 P_{s^- s}(a^+) + \tfrac 12 P_{s^+ s}(a^-). \end{align*} Since $$H(s' | s,a)$$ is supermodular, we have $H(s' | s^+, a^+) + H(s' | s^-, a^-) \ge H(s' | s^+, a^-) + H(s' | s^-, a^+)$ or equivalently, $\sum_{s' \le s'} \big[ P_{s^+ s'}(a^+) + P_{s^- s'}(a^-) \big] \le \sum_{s' \le s'} \big[ P_{s^+ s'}(a^-) + P_{s^- s'}(a^+) \big].$ which implies $M_1(s') \le M_2(s')$ where $$M_1$$ and $$M_2$$ are the CDFs of $$μ_1$$ and $$μ_2$$. Thus, $$μ_1 \succeq_s μ_2$$.

Hence, for any weakly increasing function $$v \colon \ALPHABET S \to \reals$$, $\sum_{s' \in \ALPHABET S} μ_1(s') v(s') \ge \sum_{s' \in \ALPHABET S} μ_2(s') v(s').$ Or, equivalently, $H(s^+, a^+) + H(s^-, a^-) \ge H(s^-, a^+) + H(s^+, a^-)$ where $$H(s,a) = \EXP[ v(X_{t+1}) | X_t = s, U_t = a]$$.

Therefore, $$c_t(s,a) + H_t(s,a)$$ is supermodular in $$(s,a)$$.

# 3 Constraints on actions

In the results above, we have assumed that the action set $$\ALPHABET A$$ is the same for all states. The results also extend to the case when the action at state $$s$$ must belong to some set $$\ALPHABET A(s)$$ provided the following conditions are satisfied:

1. For any $$s \ge s'$$, $$\ALPHABET A(s) \supseteq \ALPHABET A(s')$$
2. For any $$s \in \ALPHABET S$$ and $$a \in \ALPHABET A(s)$$, $$a' < a$$ implies that $$a' \in \ALPHABET A(s)$$.

# 4 Monotone dynamic programming

If we can establish that the optimal policy is monontone, then we can use this structure to implement the dynamic program more efficient. Suppose $$\ALPHABET S = \{1, \dots, n\}$$ and $$\ALPHABET A = \{1, \dots. m\}$$. The main idea is as follows. Suppose $$V_{t+1}(\cdot)$$ has been caclulated. Insead of computing $$Q_t(s,a)$$ and $$V_t(s)$$, proceed as follows:

1. Set $$s = 1$$ and $$α = 1$$.

2. For all $$u \in \{α, \dots, m\}$$, compute $$Q_t(s,a)$$ as usual.

3. Compute

$V_t(s) = \min_{ α \le a \le m } Q_t(s,a)$

and set

$π_t^*(s) = \max \{ a \in \{α, \dots, m\} : V_t(s) = Q_t(s,a) \}.$

4. If $$s = n$$, then stop. Otherwise, set $$α = π_t^*(s)$$ and $$s = s+1$$ and go to step 2.

# 5 Example: A machine replacement model

In this section, we use Theorem 4 to establish the structure of the optimal policy in a model of machine replacement.

Consider a manufacturing process, where the machine used for manufacturing deteriorates over time. Let $$\ALPHABET S = \{0, 1, \dots \}$$ represent the condition of the machine. The higher the value of $$s$$, the worse the condition of the equipment. Note that, for simplicity, we have assumed that the state space is countable.

At each decision epoch, the decision maker can choose actions from the set $$\ALPHABET A = \{0,1\}$$. Action $$0$$ corresponds to operating the equipment as is for an additional period, while action 1 corresponds to scrapping the equipment and replacing it with a new and identical piece of equipment. We assume that there is a PMF $$μ$$ on $$\ALPHABET S$$ such that $P_{sz}(0) = \begin{cases} 0, & z < s \\ μ_{z - s}, & z \ge s \end{cases} \quad\text{and}\quad P_sz(1) = μ_z.$ The transition matrix $$P(0)$$ states that the equipment detiorates from $$s$$ to $$s + k$$, with probability $$μ_k$$ for all states $$s$$.

We assume that running the machine in state $$s$$ costs $$h(s)$$, where $$h(\cdot)$$ is a weakly increasing function. Replacing the machine costs a constant amount~$$K$$.

We now verify the conditions (C1)–(C4) for the model.

C1. For $$a = 0$$, $$c(s,0) = h(s)$$, which is weakly increasing by assumption. For $$a = 1$$, $$c(s,1) = K$$, which is trivially weakly increasing.

C2. For $$a = 0$$, $$P(0)$$ is stochastically monotone (because the CDF of $$P(\cdot | s, 0)$$ lies above the CDF of $$P(\cdot | s+1, 0)$$). For $$a = 1$$, all rows of $$P(1)$$ are the same; therefore $$P(1)$$ is stochastically monotone.

Since (C1) and (C2) are satisfied, by Theorem 1, we can assert that the value function is weakly increasing.

C3. $$c(s,1) - c(s,0) = K - h(s)$$, which is weakly decreasing in $$s$$. Therefore, $$c(s,a)$$ is submodular in $$(s,a)$$.

C4. Recall that $$H(s'|s,a) = 1 - \sum_{z \le s'} P_{sz}(a).$$ Therefore,

$H(s'|s,0) = 1 - \sum_{z = s}^{s'} μ_{z -s} = 1 - \sum_{k = 0}^{s' - s} μ_k = 1 - M_{s' - s},$ where $$M$$ is the CMF of $$μ$$, and $H(s'|s,1) = 1 - \sum_{z \le s'} μ_z = 1 - M_{s'},$

Therefore, $$H(s'|s,1) - H(s'|s,0) = M_{s'-s} - M_{s'}$$. For any fixed $$s'$$, $$H(s'|s,1) - H(s'|s,0)$$ is weakly decreasing in $$s$$. There $$H(s'|s,a)$$ is submodular in $$(s,a)$$.

Since (C1)–(C4) are satisfied, the optimal policy is weakly increasing in~$$s$$. Since there are only two actions, it means that for every time, there exists a state $$s^*_t$$ with the property that if $$s$$ exceeds $$s^*_t$$, the optimal decision is to replace the machine; and if $$s \le s^*_t$$, then the optimal decision is to operate the machine for another period.

# Exercises

1. Consider the example of machine repair presented in notes on matrix formulation of MDPs. Prove that the optimal policy for that model is weakly increasing.

2. Suppose the state space $$\ALPHABET S$$ is a symmetric subset of integers of the form $$\{-L, -L + 1, \dots, L-1, L\}$$ and the action space $$\ALPHABET A$$ is discrete. Let $$\ALPHABET X_{\ge 0}$$ denote the set $$\{0, \dots, L\}$$.

Let $$P(a)$$ denote the controlled transition matrix and $$c_t(s,a)$$ denote the per-step cost. To avoid ambiguity, we define the optimal policy as $π^*_t(s) = \begin{cases} \max\bigl\{ a' \in \arg\min_{a \in \ALPHABET A} Q_t(s,a) \bigr\}, & \text{if } s \ge 0 \\ \min\bigl\{ a' \in \arg\min_{a \in \ALPHABET A} Q_t(s,a) \bigr\}, & \text{if } s < 0 \end{cases}$ The purpose of this exercise is to identify conditions under which the value function and the optimal policy are even and quasi-convex. We do so using the following steps.

1. We say that the transition probability matrix $$P(a)$$ is even if for all $$s, s' \in \ALPHABET S$$, $$P(s'|s,a) = P(-s'|-s,a)$$. Prove the following result.

Suppose the MDP satisfies the following properties:

(A1) For every $$t$$ and $$a \in \ALPHABET A$$, $$c_t(s,a)$$ is even function of $$s$$.

(A2) For every $$a \in \ALPHABET A$$, $$P(a)$$ is even.

Then, for all $$t$$, $$V_t$$ and $$π_t$$ are even functions.

2. Given any probability mass function $$μ$$ on $$\ALPHABET S$$, define the folded probability mass function $$\tilde μ$$ on $$\ALPHABET X_{\ge 0}$$ as follows: $\tilde μ(s) = \begin{cases} μ(0), & \text{if } s = 0 \\ μ(s) + μ(-s), & \text{if } s > 0. \end{cases}$

For ease of notation, we use $$\tilde μ = \mathcal F μ$$ to denote this folding operation. Note that an immediate consequence of the definition is the following (you don’t have to prove this).

If $$f \colon \ALPHABET S \to \reals$$ is even, then for any probability mass function $$μ$$ on $$\ALPHABET S$$ and $$\tilde μ = \mathcal F μ$$, we have $\sum_{s \in \ALPHABET S} f(s) μ(s) = \sum_{s \in \ALPHABET X_{\ge 0}} f(s) \tilde μ(s).$

Thus, the expectation of the function $$f \colon \ALPHABET S \to \reals$$ with respect to the PMF $$μ$$ is equal to the expectation of the function $$f \colon \ALPHABET X_{\ge 0} \to \reals$$ with respect to the PMF $$\tilde μ = \mathcal F μ$$.

Now given any probability transition matrix $$P$$ on $$\ALPHABET S$$, we can define a probability transition matrix $$\tilde P$$ on $$\ALPHABET X_{\ge 0}$$ as follows: for any $$s \in \ALPHABET S$$, $$\tilde P_s = \mathcal F P_s$$, where $$P_s$$ denotes the $$s$$-th row of $$P$$. For ease of notation, we use $$\tilde P = \mathcal F P$$ to denote this relationship.

Now prove the following:

Given the MDP $$(\ALPHABET S, \ALPHABET A, P, \{c_t\})$$, define the folded MDP as $$(\ALPHABET S_{\ge 0}, \ALPHABET A, \tilde P, \{c_t\})$$, where $$\tilde P(a) = \mathcal F P(a)$$ for all $$a \in \ALPHABET A$$. Let $$\tilde Q_t \colon \ALPHABET S_{\ge 0} \times \ALPHABET A \to \reals$$, $$\tilde V_t \colon \ALPHABET S_{\ge 0} \to \reals$$ and $$\tilde π_t^* \colon \ALPHABET S_{\ge 0} \to \ALPHABET A$$ denote the action-value function, value function and the policy of the folded MDP. Then, if the original MDP satisfies conditions (A1) and (A2) then, for any $$s \in \ALPHABET S$$ and $$a \in \ALPHABET A$$, $Q_t(s,a) = \tilde Q_t(|s|, a), \quad V_t(s) = \tilde V_t(|s|), \quad π_t^*(s) = \tilde π_t^*(|s|).$

3. The result of the previous part implies that if the value function $$\tilde V_t$$ and the policy $$\tilde π^*_t$$ are monotone increasing, then the value function $$V_t$$ and the policy $$π^*_t$$ are even and quasi-convex. This gives us a method to verify if the value function and optimal policy are even and quasi-convex.

Now, recall the model of the Internet of Things presented in Q2 of Assignment 3. The numerical experiments that you did in Assignment 3 suggest that the value function and the optimal policy are even and quasi-convex. Prove that this is indeed the case.

4. Now suppose the distribution of $$W_t$$ is not Gaussian but is some general probability density $$\varphi(\cdot)$$ and the cost function is $c(e,a) = \lambda a + (1 - a) d(e).$ Find conditions on $$\varphi$$ and $$d$$ such that the value function and optimal policy are even and quasi-convex.

# References

Ross (1974) has an early treatment of monotonicity of optimal policies. The general theory was developed by Topkis (1998). The presentation here follows Puterman (2014). Exercise 2 is from Chakravorty and Mahajan (2018).

Chakravorty, J. and Mahajan, A. 2018. Sufficient conditions for the value function and optimal strategy to be even and quasi-convex. IEEE Transactions on Automatic Control 63, 11, 3858–3864. DOI: 10.1109/TAC.2018.2800796.
Puterman, M.L. 2014. Markov decision processes: Discrete stochastic dynamic programming. John Wiley & Sons. DOI: 10.1002/9780470316887.
Ross, S.M. 1974. Dynamic programming and gambling models. Advances in Applied Probability 6, 3, 593–606. DOI: 10.2307/1426236.
Topkis, D.M. 1998. Supermodularity and complementarity. Princeton University Press.

This entry was last updated on 25 Aug 2022 and posted in MDP and tagged stochastic monotonicity, submodularity, monotone policies.


text:
# Degree of boolean functions

How can we compute the degree of boolean function? I encountered with this,while solving a problem given in my assignment module which is,

How many different boolean functions of degree 1 and 2 are there?

The suggest answer in my module is $4$ and $16$,but I don't understand how and what exactly meant by the degree of boolean functions,could somebody explain?

EDIT: I encountered this link today,but I don't understand is it $2^{2^p}$ or $2^{2p}$ ?

-
In your book how is the degree of a function defined? Even better give some more info about the context. – Arjang May 1 '11 at 12:38
@Arjang:The degree is not defined,that is the problem. – Quixotic May 1 '11 at 16:10
StackOverflow link. – Quixotic May 1 '11 at 16:17
I think it is 2^(2^p). On the page you linked to, they found the truth table for the $2^{2^2}$ possible boolean functions of two variables. – please delete me May 4 '11 at 11:20

It would seem that degree refers to the arity of the function, the number of input variables. (But you should check the definition, and the question is not really mathematically sensible without a precise definition.) Since the number of rows of a truth table with $p$ Boolean variables is $2^p$, and each row can get one of two values (true or false), the number of such truth tables is $2^{2^p}$. This is $2$ to the number of rows, since there are two possibilities for each of the $2^p$ rows.
Every such truth table determines a Boolean function in $p$ variables, and every Boolean function in $p$ variables is determined by its corresponding truth table. Thus, the total number of Boolean functions in $p$-variables is $2^{2^p}$. (And this is not the same as $2^{2p}$.)
Thus, when $p$ is $1$, we have $2^{2^1}=4$ Boolean unary functions, and when $p$ is $2$, we have $2^{2^2}=16$ Boolean binary functions.


text:
Integer type:  int32  int64  nag_int  show int32  show int32  show int64  show int64  show nag_int  show nag_int

Chapter Contents
Chapter Introduction
NAG Toolbox

# NAG Toolbox: nag_ode_bvp_fd_lin_gen (d02gb)

## Purpose

nag_ode_bvp_fd_lin_gen (d02gb) solves a general linear two-point boundary value problem for a system of ordinary differential equations, using a deferred correction technique.

## Syntax

[c, d, gam, x, y, np, ifail] = d02gb(a, b, tol, fcnf, fcng, c, d, gam, x, np, 'n', n, 'mnp', mnp)
[c, d, gam, x, y, np, ifail] = nag_ode_bvp_fd_lin_gen(a, b, tol, fcnf, fcng, c, d, gam, x, np, 'n', n, 'mnp', mnp)

## Description

nag_ode_bvp_fd_lin_gen (d02gb) solves a linear two-point boundary value problem for a system of n$\mathit{n}$ ordinary differential equations in the interval [a,b$a,b$]. The system is written in the form
 y′ = F(x)y + g(x) $y′=F(x)y+g(x)$ (1)
and the boundary conditions are written in the form
 Cy(a) + Dy(b) = γ . $Cy(a)+Dy(b)=γ .$ (2)
Here F(x)$F\left(x\right)$, C$C$ and D$D$ are n$\mathit{n}$ by n$\mathit{n}$ matrices, and g(x)$g\left(x\right)$ and γ$\gamma$ are n$\mathit{n}$-component vectors. The approximate solution to (1) and (2) is found using a finite difference method with deferred correction. The algorithm is a specialization of that used in function nag_ode_bvp_fd_nonlin_gen (d02ra) which solves a nonlinear version of (1) and (2). The nonlinear version of the algorithm is described fully in Pereyra (1979).
You supply an absolute error tolerance and may also supply an initial mesh for the construction of the finite difference equations (alternatively a default mesh is used). The algorithm constructs a solution on a mesh defined by adding points to the initial mesh. This solution is chosen so that the error is everywhere less than your tolerance and so that the error is approximately equidistributed on the final mesh. The solution is returned on this final mesh.
If the solution is required at a few specific points then these should be included in the initial mesh. If, on the other hand, the solution is required at several specific points, then you should use the interpolation functions provided in Chapter E01 if these points do not themselves form a convenient mesh.

## References

Pereyra V (1979) PASVA3: An adaptive finite-difference Fortran program for first order nonlinear, ordinary boundary problems Codes for Boundary Value Problems in Ordinary Differential Equations. Lecture Notes in Computer Science (eds B Childs, M Scott, J W Daniel, E Denman and P Nelson) 76 Springer–Verlag

## Parameters

### Compulsory Input Parameters

1:     a – double scalar
a$a$, the left-hand boundary point.
2:     b – double scalar
b$b$, the right-hand boundary point.
Constraint: b > a${\mathbf{b}}>{\mathbf{a}}$.
3:     tol – double scalar
A positive absolute error tolerance. If
 a = x1 < x2 < ⋯ < xnp = b $a=x1
is the final mesh, z(x)$z\left(x\right)$ is the approximate solution from nag_ode_bvp_fd_lin_gen (d02gb) and y(x)$y\left(x\right)$ is the true solution of equations (1) and (2) then, except in extreme cases, it is expected that
 ‖z − y‖ ≤ tol $‖z-y‖≤tol$ (3)
where
 ‖u‖ = max max |ui(xj)|. 1 ≤ i ≤ n 1 ≤ j ≤ np
$‖u‖ = max 1≤i≤n max 1≤j≤np |ui(xj)| .$
Constraint: tol > 0.0${\mathbf{tol}}>0.0$.
4:     fcnf – function handle or string containing name of m-file
fcnf must evaluate the matrix F(x)$F\left(x\right)$ in (1) at a general point x$x$.
[f] = fcnf(x)

Input Parameters

1:     x – double scalar
x$x$, the value of the independent variable.

Output Parameters

1:     f( : $:$) – double array
f(j,i)${\mathbf{f}}\left(\mathit{j},\mathit{i}\right)$ must contain the (i,j)$\left(\mathit{i},\mathit{j}\right)$th element of the matrix F(x)$F\left(x\right)$, for i = 1,2,,n$\mathit{i}=1,2,\dots ,\mathit{n}$ and j = 1,2,,n$\mathit{j}=1,2,\dots ,\mathit{n}$. (See Section [Example] for an example.)
5:     fcng – function handle or string containing name of m-file
fcng must evaluate the vector g(x)$g\left(x\right)$ in (1) at a general point x$x$.
[g] = fcng(x)

Input Parameters

1:     x – double scalar
x$x$, the value of the independent variable.

Output Parameters

1:     g( : $:$) – double array
The i$\mathit{i}$th element of the vector g(x)$g\left(x\right)$, for i = 1,2,,n$\mathit{i}=1,2,\dots ,\mathit{n}$. (See Section [Example] for an example.)
6:     c(n,n) – double array
7:     d(n,n) – double array
8:     gam(n) – double array
n, the first dimension of the array, must satisfy the constraint n2${\mathbf{n}}\ge 2$.
The arrays c and d must be set to the matrices C$C$ and D$D$ in (2)). gam must be set to the vector γ$\gamma$ in (2).
9:     x(mnp) – double array
mnp, the dimension of the array, must satisfy the constraint mnp32${\mathbf{mnp}}\ge 32$.
If np4${\mathbf{np}}\ge 4$ (see np), the first np elements must define an initial mesh. Otherwise the elements of x$x$ need not be set.
Constraint:
 a = x(1) < x(2) < ⋯ < x(np) = b,   np ≥ 4 . $a=x1 (4)
10:   np – int64int32nag_int scalar
Determines whether a default mesh or user-supplied mesh is used.
np = 0${\mathbf{np}}=0$
A default value of 4$4$ for np and a corresponding equispaced mesh x(1),x(2),,x(np)${\mathbf{x}}\left(1\right),{\mathbf{x}}\left(2\right),\dots ,{\mathbf{x}}\left({\mathbf{np}}\right)$ are used.
np4${\mathbf{np}}\ge 4$
You must define an initial mesh x as in (4).

### Optional Input Parameters

1:     n – int64int32nag_int scalar
Default: The dimension of the array gam and the first dimension of the arrays c, d and the second dimension of the arrays c, d. (An error is raised if these dimensions are not equal.)
The number of equations; that is n$\mathit{n}$ is the order of system (1).
Constraint: n2${\mathbf{n}}\ge 2$.
2:     mnp – int64int32nag_int scalar
Default: The dimension of the array x.
The maximum permitted number of mesh points.
Constraint: mnp32${\mathbf{mnp}}\ge 32$.

w lw iw liw

### Output Parameters

1:     c(n,n) – double array
2:     d(n,n) – double array
3:     gam(n) – double array
The rows of c and d and the components of gam are reordered so that the boundary conditions are in the order:
 (i) conditions on y(a)$y\left(a\right)$ only; (ii) condition involving y(a)$y\left(a\right)$ and y(b)$y\left(b\right)$; and (iii) conditions on y(b)$y\left(b\right)$ only.
The function will be slightly more efficient if the arrays c, d and gam are ordered in this way before entry, and in this event they will be unchanged on exit.
Note that the problems (1) and (2) must be of boundary value type, that is neither C$C$ nor D$D$ may be identically zero. Note also that the rank of the matrix [C,D]$\left[C,D\right]$ must be n$\mathit{n}$ for the problem to be properly posed. Any violation of these conditions will lead to an error exit.
4:     x(mnp) – double array
x(1),x(2),,x(np)${\mathbf{x}}\left(1\right),{\mathbf{x}}\left(2\right),\dots ,{\mathbf{x}}\left({\mathbf{np}}\right)$ define the final mesh (with the returned value of np) satisfying the relation (4).
5:     y(n,mnp) – double array
The approximate solution z(x)$z\left(x\right)$ satisfying (3), on the final mesh, that is
 y(j,i) = zj(xi),  i = 1,2, … ,np and j = 1,2, … ,n $yji=zj(xi), i=1,2,…,np and j=1,2,…,n$
where np is the number of points in the final mesh.
The remaining columns of y are not used.
6:     np – int64int32nag_int scalar
The number of points in the final (returned) mesh.
7:     ifail – int64int32nag_int scalar
${\mathrm{ifail}}={\mathbf{0}}$ unless the function detects an error (see [Error Indicators and Warnings]).

## Error Indicators and Warnings

Errors or warnings detected by the function:

Cases prefixed with W are classified as warnings and do not generate an error of type NAG:error_n. See nag_issue_warnings.

ifail = 1${\mathbf{ifail}}=1$
One or more of the parameters n, tol, np, mnp, lw or liw is incorrectly set, ba${\mathbf{b}}\le {\mathbf{a}}$ or the condition (4) on x is not satisfied.
ifail = 2${\mathbf{ifail}}=2$
There are three possible reasons for this error exit to be taken:
 (i) one of the matrices C$C$ or D$D$ is identically zero (that is, the problem is of initial value and not boundary value type). In this case, iw(1) = 0$\mathit{iw}\left(1\right)=0$ on exit; (ii) a row of C$C$ and the corresponding row of D$D$ are identically zero (that is, the boundary conditions are rank deficient). In this case, on exit iw(1)$\mathit{iw}\left(1\right)$ contains the index of the first such row encountered; and (iii) more than n$\mathit{n}$ of the columns of the n$\mathit{n}$ by 2n$2\mathit{n}$ matrix [C,D]$\left[C,D\right]$ are identically zero (that is, the boundary conditions are rank deficient). In this case, on exit iw(1)$\mathit{iw}\left(1\right)$ contains minus the number of non-identically zero columns.
W ifail = 3${\mathbf{ifail}}=3$
The function has failed to find a solution to the specified accuracy. There are a variety of possible reasons including:
 (i) the boundary conditions are rank deficient, which may be indicated by the message that the Jacobian is singular. However this is an unlikely explanation for the error exit as all rank deficient boundary conditions should lead instead to error exits with either ${\mathbf{ifail}}={\mathbf{2}}$ or 5${\mathbf{5}}$; see also (iv); (ii) not enough mesh points are permitted in order to attain the required accuracy. This is indicated by ${\mathbf{np}}={\mathbf{mnp}}$ on return from a call to nag_ode_bvp_fd_lin_gen (d02gb). This difficulty may be aggravated by a poor initial choice of mesh points; (iii) the accuracy requested cannot be attained on the computer being used; and (iv) an unlikely combination of values of Fx$Fx$ has led to a singular Jacobian. The error should not persist if more mesh points are allowed.
ifail = 4${\mathbf{ifail}}=4$
A serious error has occurred in a call to nag_ode_bvp_fd_lin_gen (d02gb). Check all array subscripts and function parameter lists in calls to nag_ode_bvp_fd_lin_gen (d02gb). Seek expert help.
ifail = 5${\mathbf{ifail}}=5$
There are two possible reasons for this error exit which occurs when checking the rank of the boundary conditions by reduction to a row echelon form:
 (i) at least one row of the n$\mathit{n}$ by 2n$2\mathit{n}$ matrix [C,D]$\left[C,D\right]$ is a linear combination of the other rows and hence the boundary conditions are rank deficient. The index of the first such row encountered is given by iw(1)$\mathit{iw}\left(1\right)$ on exit; and (ii) as (i) but the rank deficiency implied by this error exit has only been determined up to a numerical tolerance. Minus the index of the first such row encountered is given by iw(1)$\mathit{iw}\left(1\right)$ on exit.
In case (ii) there is some doubt as to the rank deficiency of the boundary conditions. However even if the boundary conditions are not rank deficient they are not posed in a suitable form for use with this function.
For example, if
C =
 ( 1 0 ) 1 ε
,  D =
 ( 1 0 ) 1 0
,  γ =
 ( γ1 ) γ2
$C= 1 0 1 ε , D= 1 0 1 0 , γ= γ1 γ2$
and ε$\epsilon$ is small enough, this error exit is likely to be taken. A better form for the boundary conditions in this case would be
C =
 ( 1 0 ) 0 1
,  D =
 ( 1 0 ) 0 0
,  γ =
 ( γ1 ) ε − 1(γ2 − γ1)
.
$C= 1 0 0 1 , D= 1 0 0 0 , γ= γ1 ε-1(γ2-γ1) .$

## Accuracy

The solution returned by the function will be accurate to your tolerance as defined by the relation (3) except in extreme circumstances. If too many points are specified in the initial mesh, the solution may be more accurate than requested and the error may not be approximately equidistributed.

The time taken by nag_ode_bvp_fd_lin_gen (d02gb) depends on the difficulty of the problem, the number of mesh points (and meshes) used and the number of deferred corrections.
You are strongly recommended to set ifail to obtain self-explanatory error messages, and also monitoring information about the course of the computation. You may select the channel numbers on which this output is to appear by calls of nag_file_set_unit_error (x04aa) (for error messages) or nag_file_set_unit_advisory (x04ab) (for monitoring information) – see Section [Example] for an example. Otherwise the default channel numbers will be used.
In the case where you wish to solve a sequence of similar problems, the final mesh from solving one case is strongly recommended as the initial mesh for the next.

## Example

This example solves the problem (written as a first-order system)
 ε y ′ ′ + y′ = 0 $ε y′′+y′=0$
with boundary conditions
 y(0) = 0,  y(1) = 1 $y(0)=0, y(1)=1$
for the cases ε = 101$\epsilon ={10}^{-1}$ and ε = 102$\epsilon ={10}^{-2}$ using the default initial mesh in the first case, and the final mesh of the first case as initial mesh for the second (more difficult) case. We give the solution and the error at each mesh point to illustrate the accuracy of the method given the accuracy request tol = 1.0e−3${\mathbf{tol}}=\text{1.0e−3}$.
Note the call to nag_file_set_unit_advisory (x04ab) prior to the call to nag_ode_bvp_fd_lin_gen (d02gb).
function nag_ode_bvp_fd_lin_gen_example
% For communication with f.
global eps;

% Set up initial values before calling NAG routine for the first time.
mnp = 70;
a = 0;
b = 1;
tol = 0.001;
c = [1, 0; 0, 0];
d = [0, 0; 1, 0];
gam = [0; 1];
x = zeros(1,mnp);
np = 0;
n = 2;
xkeep = zeros(1,2);
npkeep = zeros(2);
ykeep = zeros(1,2);
espkeep = zeros(2);

fprintf('nag_ode_bvp_fd_lin_gen example program results\n');

for i = 1:2
eps = double(10)^(-i);

[c, d, gam, x, y, np, ifail] = ...
nag_ode_bvp_fd_lin_gen(a, b, tol, @f, @g, c, d, gam, x,...
int64(np), 'n', int64(n), 'mnp', int64(mnp));
if (ifail ~= 0)
% Problems in integration, or incorrect parameters.  Print
% message and exit.
error('Warning: nag_ode_bvp_fd_lin_gen returned with ifail = %d ',ifail);
else
% Output results.
fprintf('\nProblem with epsilon = %1.2e\n\n',eps);
fprintf('Approximate solution on final mesh of %1.0f points \n',np)
disp('      X(I)       Y(1,I)');
for j = 1:int32(np) % can't use int64 in loop range
fprintf('     %1.4f    %8.4f\n',x(j),y(1,j));
xkeep(j,i) = x(j);
ykeep(j,i) = y(1,j);
end
end
% Store the number of points and epsilon for use in plot.
npkeep(i) = np;
espkeep(i) = eps;
end
% Plot results.
fig = figure('Number', 'off');
display_plot(xkeep, ykeep, npkeep, espkeep)

function f = f(x)
% For communication with main routine.
global eps;

% Evaluate matrix.
f=zeros(2,2);
f(1,1) = 0;
f(1,2) = 1;
f(2,1) = 0;
f(2,2) = -1/eps;
function g = g(x)
% Evaluate vector.
g=zeros(2,1);
function display_plot(xkeep, ykeep, npkeep, espkeep)
% Formatting for title and axis labels.
titleFmt = {'FontName', 'Helvetica', 'FontWeight', 'Bold', 'FontSize', 14};
labFmt = {'FontName', 'Helvetica', 'FontWeight', 'Bold', 'FontSize', 13};
set(gca, 'FontSize', 13); % for legend, axis tick labels, etc.
% Plot the two curves.
plot(xkeep(1:npkeep(1),1), ykeep(1:npkeep(1),1), '-+', ...
xkeep(1:npkeep(2),2), ykeep(1:npkeep(2),2), ':x');
set(gca, 'YLim', [0 1.1]);
title('Solution to Simple Second-order Problem', titleFmt{:});
% Label the axes.
xlabel('x', labFmt{:});
ylabel('Solution', labFmt{:});
% Label the curves.
legend(['\epsilon = ', num2str(espkeep(1))], ...
['\epsilon = ', num2str(espkeep(2))], ...
'Location', 'Best');

nag_ode_bvp_fd_lin_gen example program results

Problem with epsilon = 1.00e-01

Approximate solution on final mesh of 15 points
X(I)       Y(1,I)
0.0000      0.0000
0.0278      0.2425
0.0556      0.4263
0.1111      0.6708
0.1667      0.8112
0.2222      0.8917
0.2778      0.9379
0.3333      0.9644
0.4444      0.9883
0.5556      0.9962
0.6667      0.9988
0.7500      0.9995
0.8333      0.9998
0.9167      0.9999
1.0000      1.0000

Problem with epsilon = 1.00e-02

Approximate solution on final mesh of 49 points
X(I)       Y(1,I)
0.0000      0.0000
0.0009      0.0884
0.0019      0.1690
0.0028      0.2425
0.0037      0.3095
0.0046      0.3706
0.0056      0.4262
0.0065      0.4770
0.0074      0.5232
0.0083      0.5654
0.0093      0.6038
0.0111      0.6708
0.0130      0.7265
0.0148      0.7727
0.0167      0.8111
0.0185      0.8431
0.0204      0.8696
0.0222      0.8916
0.0241      0.9100
0.0259      0.9252
0.0278      0.9378
0.0306      0.9529
0.0333      0.9643
0.0361      0.9730
0.0389      0.9795
0.0417      0.9845
0.0444      0.9883
0.0472      0.9911
0.0500      0.9933
0.0528      0.9949
0.0556      0.9961
0.0648      0.9985
0.0741      0.9994
0.0833      0.9998
0.0926      0.9999
0.1019      1.0000
0.1111      1.0000
0.1389      1.0000
0.1667      1.0000
0.2222      1.0000
0.2778      1.0000
0.3333      1.0000
0.4444      1.0000
0.5556      1.0000
0.6667      1.0000
0.7500      1.0000
0.8333      1.0000
0.9167      1.0000
1.0000      1.0000

function d02gb_example
% For communication with f.
global eps;

% Set up initial values before calling NAG routine for the first time.
mnp = 70;
a = 0;
b = 1;
tol = 0.001;
c = [1, 0; 0, 0];
d = [0, 0; 1, 0];
gam = [0; 1];
x = zeros(1,mnp);
np = 0;
n = 2;
xkeep = zeros(1,2);
npkeep = zeros(2);
ykeep = zeros(1,2);
espkeep = zeros(2);

fprintf('d02gb example program results\n');

for i = 1:2
eps = double(10)^(-i);

[c, d, gam, x, y, np, ifail] = ...
d02gb(a, b, tol, @f, @g, c, d, gam, x,...
int64(np), 'n', int64(n), 'mnp', int64(mnp));
if (ifail ~= 0)
% Problems in integration, or incorrect parameters.  Print
% message and exit.
error('Warning: d02gb returned with ifail = %d',ifail);
else
% Output results.
fprintf('\nProblem with epsilon = %1.2e\n\n',eps);
fprintf('Approximate solution on final mesh of %1.0f points \n',np)
disp('      X(I)       Y(1,I)');
for j = 1:int32(np) % can't use int64 in loop range
fprintf('     %1.4f    %8.4f\n',x(j),y(1,j));
xkeep(j,i) = x(j);
ykeep(j,i) = y(1,j);
end
end
% Store the number of points and epsilon for use in plot.
npkeep(i) = np;
espkeep(i) = eps;
end
% Plot results.
fig = figure('Number', 'off');
display_plot(xkeep, ykeep, npkeep, espkeep)

function f = f(x)
% For communication with main routine.
global eps;

% Evaluate matrix.
f=zeros(2,2);
f(1,1) = 0;
f(1,2) = 1;
f(2,1) = 0;
f(2,2) = -1/eps;
function g = g(x)
% Evaluate vector.
g=zeros(2,1);
function display_plot(xkeep, ykeep, npkeep, espkeep)
% Formatting for title and axis labels.
titleFmt = {'FontName', 'Helvetica', 'FontWeight', 'Bold', 'FontSize', 14};
labFmt = {'FontName', 'Helvetica', 'FontWeight', 'Bold', 'FontSize', 13};
set(gca, 'FontSize', 13); % for legend, axis tick labels, etc.
% Plot the two curves.
plot(xkeep(1:npkeep(1),1), ykeep(1:npkeep(1),1), '-+', ...
xkeep(1:npkeep(2),2), ykeep(1:npkeep(2),2), ':x');
set(gca, 'YLim', [0 1.1]);
title('Solution to Simple Second-order Problem', titleFmt{:});
% Label the axes.
xlabel('x', labFmt{:});
ylabel('Solution', labFmt{:});
% Label the curves.
legend(['\epsilon = ', num2str(espkeep(1))], ...
['\epsilon = ', num2str(espkeep(2))], ...
'Location', 'Best');

d02gb example program results

Problem with epsilon = 1.00e-01

Approximate solution on final mesh of 15 points
X(I)       Y(1,I)
0.0000      0.0000
0.0278      0.2425
0.0556      0.4263
0.1111      0.6708
0.1667      0.8112
0.2222      0.8917
0.2778      0.9379
0.3333      0.9644
0.4444      0.9883
0.5556      0.9962
0.6667      0.9988
0.7500      0.9995
0.8333      0.9998
0.9167      0.9999
1.0000      1.0000

Problem with epsilon = 1.00e-02

Approximate solution on final mesh of 49 points
X(I)       Y(1,I)
0.0000      0.0000
0.0009      0.0884
0.0019      0.1690
0.0028      0.2425
0.0037      0.3095
0.0046      0.3706
0.0056      0.4262
0.0065      0.4770
0.0074      0.5232
0.0083      0.5654
0.0093      0.6038
0.0111      0.6708
0.0130      0.7265
0.0148      0.7727
0.0167      0.8111
0.0185      0.8431
0.0204      0.8696
0.0222      0.8916
0.0241      0.9100
0.0259      0.9252
0.0278      0.9378
0.0306      0.9529
0.0333      0.9643
0.0361      0.9730
0.0389      0.9795
0.0417      0.9845
0.0444      0.9883
0.0472      0.9911
0.0500      0.9933
0.0528      0.9949
0.0556      0.9961
0.0648      0.9985
0.0741      0.9994
0.0833      0.9998
0.0926      0.9999
0.1019      1.0000
0.1111      1.0000
0.1389      1.0000
0.1667      1.0000
0.2222      1.0000
0.2778      1.0000
0.3333      1.0000
0.4444      1.0000
0.5556      1.0000
0.6667      1.0000
0.7500      1.0000
0.8333      1.0000
0.9167      1.0000
1.0000      1.0000


text:
If I were a month, I’d be October.
If I were a day of the week, I’d be
Sunday.
If I were a time of day, I’d be
2:30 in the A.M.
If I were a direction, I’d be
North.
If I were a liquid, I’d be Absinthe.

If I were a gemstone, I’d be a
Tigers Eye.
If I were a tree, I’d be a
Weeping Willow.
If I were a tool, I’d be a
Bone Saw.
If I were a flower, I’d be a Lily of the Valley.

If I were a kind of weather, I’d be
Dark and warm.
If I were a color, I’d be Scarlet Red.

If I were a sound, I’d be
Siren song.
If I were a food, I’d be Caramel.
If I were a place, I’d be
A Castle.
If I were a taste, I’d taste like
Strawberries.
If I were a scent, I’d be
Lust and Fire.
If I were an object, I’d be
Leather Restraints.
If I were a body part, I’d be
Legs.
If I were a facial expression, I’d be
Mischievous.
If I were a song, I’d be
Sunspots by Nine Inch Nails.
If I were a pair of shoes, I’d be
Stilettos.
Scroll Up


text:
# C# 版本和.NET 版本以及VS版本的对应关系

https://en.wikipedia.org/wiki/C_Sharp_(programming_language)#Versions

http://stackoverflow.com/questions/247621/what-are-the-correct-version-numbers-for-c

These are the versions of C# known about at the time of this writing:

• C# 1.0 released with .NET 1.0 and VS2002 (January 2002)
• C# 1.2 (bizarrely enough); released with .NET 1.1 and VS2003 (April 2003). First version to call Dispose on IEnumerators which implemented IDisposable. A few other small features.
• C# 2.0 released with .NET 2.0 and VS2005 (November 2005). Major new features: generics, anonymous methods, nullable types, iterator blocks
• C# 3.0 released with .NET 3.5 and VS2008 (November 2007). Major new features: lambda expressions, extension methods, expression trees, anonymous types, implicit typing (var), query expressions
• C# 4.0 released with .NET 4 and VS2010 (April 2010). Major new features: late binding (dynamic), delegate and interface generic variance, more COM support, named arguments, tuple data type and optional parameters
• C# 5.0 released with .NET 4.5 and VS2012 (August 2012). Major features: async programming, caller info attributes. Breaking change: loop variable closure.
• C# 6.0 released with .NET 4.6 and VS2015 (July 2015). Implemented by Roslyn. Features: initializers for automatically implemented properties, using directives to import static members, exception filters, indexed members and element initializers, await in catch and finally, extension Add methods in collection initializers.

There is no such thing as C# 3.5 - the cause of confusion here is that the C# 3.0 is present in .NET 3.5. The language and framework are versioned independently, however - as is the CLR, which is at version 2.0 for .NET 2.0 through 3.5, .NET 4 introducing CLR 4.0, service packs notwithstanding. The CLR in .NET 4.5 has various improvements, but the versioning is unclear: in some places it may be referred to as CLR 4.5 (this MSDN page used to refer to it that way, for example), but the Environment.Version property still reports 4.0.xxx.

More detailed information about the relationship between the language, runtime and framework versions is available on the C# in Depth site. This includes information about which features of C# 3.0 you can use when targeting .NET 2.0. (If anyone wants to bring all of the content into this wiki answer, they're welcome to.)

posted @ 2014-11-09 20:18  ChuckLu  阅读(4983)  评论(2编辑  收藏  举报


text:
# values of a $\displaystyle\lim_{x \to -\infty} x^a$

For which values of $a$ the following limit is true?

$$x,a\in \mathbb R$$ $$\displaystyle\lim_{x \to -\infty} x^a = -\infty$$

For whole even numbers that would be false, and for whole odd numbers, that would be true.

What about the rest?

• Your results so far are corretcd only for $a>0$. You can easily solve the remaining cases $a\in\mathbb Z\setminus\mathbb N$. How do you define $x^a$ for $x<0$ and $a\notin \mathbb Z$? Aug 6, 2014 at 17:06
• what do you mean how do i define, isnt $x^a$ well defined for every $a\in \mathbb R$? Aug 6, 2014 at 17:08
• What is $(-1)^{1/2}$? $(-1)^\pi$? Aug 6, 2014 at 17:09
• What is $(-\pi)^\pi$? Aug 6, 2014 at 17:09
• how about defining it as the limit of $x^(a_n)$ such that $a_n\in \mathbb Q$ , $a_n \to a$ Aug 6, 2014 at 17:12

Suppose $a=j/k$ is a fraction in lowest terms and all numbers under consideration are assumed to be real. If both $j$ and $k$ are odd then $\lim_{x \to -\infty} =-\infty$. If $j$ is even then $k$ is odd and so $\lim_{x \to -\infty} =\infty$. If $j$ is odd and $k$ is even then the limit doesn't exist. If $a$ is an irrational number then in any neighborhood of $a$ there are fractions whose denominators are even and so the limit does not exist. This last part assumes that we are defining an irrational number using limits of sequences of rational numbers. See http://en.wikipedia.org/wiki/Dedekind_cut


text:
# How do you simplify 4sqrt5-2sqrt45?

I found: $- 2 \sqrt{5}$
$4 \sqrt{5} - 2 \sqrt{5 \cdot 9} =$
$= 4 \sqrt{5} - 2 \cdot 3 \sqrt{5} =$
$= 4 \sqrt{5} - 6 \sqrt{5} = - 2 \sqrt{5}$


text:
+0

0
75
1

A central angle with measure 120 degrees is drawn in a circle with a circumference of 8 pi miles. What is the length of the intercepted arc, rounded to the tenth of a mile?

Guest May 24, 2017
Sort:

#1
+4777
+1

$$\frac{\text {arc length}}{\text{circumference of circle}}=\frac{120^{\circ}} {360^{\circ}} \\~\\ \frac{\text{arc length}}{8\pi}=\frac{120^{\circ}} {360^{\circ}} \\~\\ \begin{array} \ \text{arc length}&=&\frac{120^{\circ}} {360^{\circ}}*8\pi \\~\\ &=&\frac13*8\pi \\~\\ &=& \frac{8\pi}{3} \\~\\ &\approx& 8.4 \, \text{miles} \end{array}$$

hectictar  May 24, 2017

### 18 Online Users

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text:
# Polar Coordinates Worksheet

Polar Coordinates Worksheet
• Page 1
1.
Which of the following is correct?
 a. Polar coordinate system is a line b. Polar coordinate system is a Point c. Polar coordinate system is a plane d. Polar coordinate system is a space

#### Solution:

A polar coordinate system is a plane on which every point is associated with two polar coordinates.
[Definition.]

2.
What is the number of coordinates of a point in a polar coordinate system?
 a. Infinite b. 4 c. 3 d. 1

#### Solution:

Every point of a plane in a polar coordinate system has infinite number of polar coordinates.

3.
What are the polar coordinates of the pole?
 a. ($\theta$, 0) b. (0, 0) c. (1, 2) d. (0, $\theta$)

#### Solution:

Pole has polar coordinates (0, θ), where θ is any angle.

4.
Which of the following is correct?
 a. Every point of a polar coordinate system has unique pair of polar coordinates. b. Each pair of polar coordinates determine a unique point. c. Every point of a polar coordinate system has 2 pairs of polar coordinates. d. Each pair of polar coordinates does not determine a unique point.

#### Solution:

For each pair of polar coordinates, (r, θ), there exists a unique point in the plane whose directed distance from the pole is r and whose directed angle is θ.

However, the polar coordinates of a point with directed distance r from the pole, directed angle θ are many like (r, θ), (r, 2π + θ), (r, 4π + θ).

So, each pair of polar coordinates determine a unique point and for a point, the pairs of polar coordinates are not unique.

5.
From the polar coordinate system shown, the coordinates of P are

 a. (7, 60o) b. (5, 60o) c. (4, 60o) d. (6, 100o)

#### Solution:

The directed distance of the point P from the pole is r = 5.
[From the figure.]

The measure of the angle made by the line OP with polar axis in the counter clockwise direction is 60o.

So, the polar coordinates of P are (5, 60o).

6.
The polar coordinates of point P are ($\sqrt{2}$, $\frac{\pi }{4}$). Which of the following are the rectangular coordinates for point P?
 a. (- 1, - 1) b. (1, 1) c. (- 1, 1) d. (1, - 1)

#### Solution:

Polar coordinates of P are (r, θ) = (2, π4).

r = 2, θ = π4

Let (x, y) be the rectangular coordinates of P.

So, x = r cos θ = 2 cos π4 = 1 and y = r sin θ = 2 sin π4 = 1

(x, y) = (1, 1).

7.
Let P be a point in a polar coordinate system. If $r$ is the directed distance from the pole O to P and $\theta$ is the directed angle measured in counter clockwise direction with the initial side on the polar axis and the terminal side on the line OP, then write the polar coordinates of P.
 a. (- $r$, $\theta$) b. ($r$, $\theta$) c. (- $\theta$, $r$) d. ($\theta$, $r$)

#### Solution:

The first polar coordinate of a point in a polar coordinate system is its directed distance from the pole O.
[Use the definition of directed distance.]

Here for P, the directed distance from the pole O is r.

The second polar coordinate of a point in a polar coordinate system is its directed angle whose initial side is on the polar axis and the terminal side is on the line OP.
[Use the definition of the directed angle.]

Here for P, the directed angle is θ.

So, the polar coordinates of P are (r, θ).

8.
The polar coordinates of a point P are ($r$, $\theta$). Which of the following is the polar coordinate pair of P other than ($r$, $\theta$)?
 a. ($r$, $\theta$ + $\frac{3\pi }{2}$) b. ($r$, $\theta$ + 2$\pi$) c. ($r$, $\theta$ + $\frac{\pi }{2}$) d. ($r$, $\theta$ - $\frac{\pi }{2}$)

#### Solution:

If (r, θ) are the polar coordinates of a point, then any other polar coordinates of the point must be of the form (r, θ + 2nπ) or (- r, θ + (2n + 1)π), where n is any integer.

For n = 1, (r, θ + 2nπ) = (r, θ + 2π)

So, among the choices, (r, θ + 2π) are the polar coordinates of P other than (r, θ).

9.
Which of the following is correct?
 a. Both (2, $\frac{\pi }{6}$) and (2, $\frac{7\pi }{6}$) represent the same point. b. Both (2, $\frac{\pi }{6}$) and (2, $\frac{13\pi }{6}$) do not represent the same point. c. Both (2, $\frac{\pi }{6}$) and (2, $\frac{7\pi }{6}$) do not represent the same point. d. Both (2, $\frac{\pi }{6}$) and (2, $\frac{13\pi }{6}$) represent the same point.

#### Solution:

For integral values of n, both (r, θ + 2nπ) and (r, θ) represent the same point in a polar coordinate system.

Then (r, θ + 2π) and (r, θ) represent the same point.
[Substitute n = 1.]

(2, 13π6) = (2, π6 + 2π)
[Write 13π6 as π6 + 2π.]

So, both (2, π6) and (2, 13π6) represent the same point.

10.
Which of the following is correct?
 a. Both (3, $\frac{\pi }{3}$) and (- 3, $\frac{\pi }{3}$) represent the same point. b. Both (3, $\frac{\pi }{3}$) and (- 3, - $\frac{\pi }{3}$) represent the same point. c. Both (3, $\frac{\pi }{3}$) and (- 3, - $\frac{2\pi }{3}$) represent the same point. d. Both (3, $\frac{\pi }{3}$) and (3, - $\frac{\pi }{3}$) represent the same point.

#### Solution:

For integral values of n, both (r, θ), (- r, θ + (2n + 1) π) represent the same point in a polar coordinate system.

So, (- r, θ - π) and (r, θ) represent the same point.
[Substitute n = -1.]

(-3, - 2π3) = (-3, π3 - π)
[Write - 2π3 as π3 - π.]

So, both (3, π3) and (- 3, - 2π3) represent the same point.


text:
World Science Scholars

### 2.6 Modern Understanding of Consciousness

Discussions
Viewing 4 reply threads
• Dr. Koch lays out all of the mental processes that are not consciousness. How do you personally define consciousness? Does your definition include emotion, learning, or other mental functions that have been apparently ruled out as part of consciousness?

• I do agree with the given definition.I have often observed myself – upon retrospect – doing things without giving much thought to it. But it is quite hard to imagine being concious without actually thinking about something. For me, though conciousness can exist without emotion, learning or other functions I believe meaningful consciousness requires some amount of self-awareness .

• I disagree. ALL these things are rightly consciousness.

• La conciencia es parte de uno , con la misma naturaleza impredecible que es uno mismo , es tan nuestra a la ves tan autónoma puesto que nos damos cuenta que existe en nuestras acciones o comportamientos .

• Todos os processos tem um pouco de autoconsciência. A consciência pode existir sem ambas as ter mas requer autoconsciência

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