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# Multi-Digit Multiplication

## Objective

Solve multi-step word problems involving multiplication, addition, and subtraction and assess the reasonableness of answers.

## Common Core Standards

### Core Standards

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• 4.OA.A.3 — Solve multistep word problems posed with whole numbers and having whole-number answers using the four operations, including problems in which remainders must be interpreted. Represent these problems using equations with a letter standing for the unknown quantity. Assess the reasonableness of answers using mental computation and estimation strategies including rounding.

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• 3.MD.C.7.B

• 3.MD.D.8

• 4.NBT.B.4

• 3.OA.D.8

## Criteria for Success

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1. Make sense of a three-act task and persevere in solving it (MP.1).
2. Solve multi-step word problems involving multiplication, addition, and subtraction (MP.4).
3. Assess the reasonableness of an answer (MP.1).

## Tips for Teachers

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• This lesson provides an opportunity to connect two domains in the grade, 4.OA and 4.NBT, since students will be solving multi-step word problems involving addition, subtraction, and multiplication and assessing the reasonableness of their answers using rounding, thus connecting standards 4.OA.3, 4.NBT.3, 4.NBT.4, and 4.NBT.5.
• Let students work on each problem in the Problem Set independently and circulate to see whether students are solving correctly. If not, come back together to discuss how/what to draw on a tape diagram, then allow them to try again on their own.
• Encourage students to use a strategy of their choice to assess the reasonableness of their answer.

#### Remote Learning Guidance

This lesson does not have any identified priority Anchor Tasks, but students should complete the Problem Set independently rather than skipping the lesson entirely. Find more guidance on adapting our math curriculum for remote learning here.

#### Fishtank Plus

• Problem Set
• Student Handout Editor
• Vocabulary Package

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### Problem 1

Act 1: Look at the following images.

Which batch of chocolate chip cookies has the most chocolate chips in it? How many more?

#### References

Cookie Problem by Gaelen is made available on 101Questions under the CC BY 3.0 license. Accessed Jan. 8, 2018, 2:47 p.m..

Modified by The Match Foundation, Inc.

Chocolate Chip Cookie Problem by Gaelen is made available on 101Questions under the CC BY 3.0 license. Accessed Jan. 8, 2018, 2:46 p.m..

Modified by The Match Foundation, Inc.

### Problem 2

Act 2: Use the following information to solve.

 Batch 1 Batch 2 One cookie: One cookie: I counted three times as many chocolate chips in this cookie than in the cookie from Batch 1. The whole batch: The whole batch:

#### References

Chocolate Chip Cookie Problem by Gaelen is made available on 101Questions under the CC BY 3.0 license. Accessed Jan. 8, 2018, 2:46 p.m..

Modified by The Match Foundation, Inc.

Cookie Problem by Gaelen is made available on 101Questions under the CC BY 3.0 license. Accessed Jan. 8, 2018, 2:47 p.m..

Modified by The Match Foundation, Inc.

### Problem 3

Act 3:

 Batch 1 Batch 2

#### References

Chocolate Chip Cookie Problem by Gaelen is made available on 101Questions under the CC BY 3.0 license. Accessed Jan. 8, 2018, 2:46 p.m..

Modified by The Match Foundation, Inc.

Cookie Problem by Gaelen is made available on 101Questions under the CC BY 3.0 license. Accessed Jan. 8, 2018, 2:47 p.m..

Modified by The Match Foundation, Inc.

### Problem 4

Act 4: There are about 300 chocolate chips in a cup. About how many cups of chocolate chips are in both batches?

#### References

Cookie Problem by Gaelen is made available on 101Questions under the CC BY 3.0 license. Accessed Jan. 8, 2018, 2:47 p.m..

Modified by The Match Foundation, Inc.

Chocolate Chip Cookie Problem by Gaelen is made available on 101Questions under the CC BY 3.0 license. Accessed Jan. 8, 2018, 2:46 p.m..

Modified by The Match Foundation, Inc.

## Problem Set & Homework

#### Discussion of Problem Set

• How was #1 similar to the other problems we did today? How was it different?
• How was setting up #2 similar to setting up #3? At what point did the two problems become quite different?
• What piece of information did you need to know to solve #2 that was not given to you in the problem?

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Solve. Show or explain your work.

Michael earns $9 per hour. He works 28 hours each week. David earns$8 per hour. He works 40 hours each week. After 6 weeks, who earns more money? How much more money?

#### References

EngageNY Mathematics Grade 4 Mathematics > Module 3 > Topic D > Lesson 13Exit Ticket, Questions #1-3

Grade 4 Mathematics > Module 3 > Topic D > Lesson 13 of the New York State Common Core Mathematics Curriculum from EngageNY and Great Minds. © 2015 Great Minds. Licensed by EngageNY of the New York State Education Department under the CC BY-NC-SA 3.0 US license. Accessed Dec. 2, 2016, 5:15 p.m..

Modified by The Match Foundation, Inc.

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# FDNY to the Rescue!

Time Limit: Java: 2000 ms / Others: 2000 ms

Memory Limit: Java: 65536 KB / Others: 65536 KB

## Description

The Fire Department of New York (FDNY) has always been proud of their response time to fires in New York City, but they want to make their response time even better. To help them with their response time, they want to make sure that the dispatchers know the closest firehouse to any address in the city. You have been hired to write this software and are entrusted with maintaining the proud tradition of FDNY. Conceptually, the software will be given the address of the fire, the locations of the firehouses, street intersections, and the time it takes to cover the distance between each intersection. It will then use this information to calculate how long it takes to reach an address from each firehouse.

Given a specific fire location in the city, the software will calculate the time taken from all the fire stations located in the city to reach the fire location. The list of fire stations will be sorted from shortest time to longest time. The dispatcher can then pick the closest firestation with available firefighters and equipment to dispatch to the fire.

## Input

Line 1:
# of intersections in the city, a single integer (henceforth referred to as N) N<20

Lines 2 to N+1:
A table (square matrix of integer values separated by one or more spaces) representing the time taken in minutes between every pair of intersections in the city. In the sample input shown below the value ��3�� on the 1st row and the 2nd column represents the time taken from intersection #1 to reach intersection #2.

Similarly the value ��9�� on the 4th row and the 2nd column represents the time taken from intersection #4 to reach intersection #2.

A value of -1 for time means that it is not possible to go directly from the origin intersection (row #) to the destination intersection (column #). All other values in the table are non-negative.

Line N+2:
An integer value n (<= N) indicating the intersection closest to the fire location followed by one or more integer values for the intersections closest to the fire stations (all on one line, separated by one or more spaces) will follow the input matrix.

Notes on input format:

1. The rows and columns are numbered from 1 to N.
2. All input values are integers
3. All fire locations are guaranteed reachable from all firehouses.
4. All distance calculations are made from the intersection closest to each firehouse to the intersection closest to the fire.

## Output

Line 1:
A label line with the headings for each column, exactly as shown in the example.

Line 2 onwards (one line for each fire station):
A sorted list (based on time) showing the fire station (origin), the destination site, time taken and a complete shortest path of nodes from the originating fire station to the fire location.

Notes on output format:
1. Columns are tab separated.
2. If two or more firehouses are tied in time they can be printed in any order.
3. If more than one path exists that has the same minimal time for a given location & firehouse, either one can be printed on the output.
4. If the fire location and the fire station locations happen to be the same intersection, the output will indicate that the origin and destination have the same intersection number, the time will be ��0�� and the nodes in the shortest path will show just one number, the fire location.

This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

## Sample Input

1
6
0 3 4 -1 -1 -1
-1 0 4 5 -1 -1
2 3 0 -1 -1 2
8 9 5 0 1 -1
7 2 1 -1 0 -1
5 -1 4 5 4 0
2 4 5 6

In the above input the last line indicates that ��2�� is the location of the
fire and ��4��, ��5�� and ��6�� are the intersections where fire stations are located.

## Sample Output

Org     Dest    Time    Path

5       2       2       5       2
4       2       3       4       5       2
6       2       6       6       5       2

None

## Source

Mid-Atlantic USA 2001

format: plain


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# Calculate pressure drop in dense phase pneumatic conveying

I've came across many references for pressure drop calculations for dilute phase pneumatic conveying, but i've yet to encounter any for dense phase. Anyone has any idea how do i calculate that? Thank you


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# International Mathematical Olympiad '59, Second day

The goal of this set of notes is to improve our problem solving and proof writing skills. You are encouraged to submit a solution to any of these problems, and join in the discussion in #imo-discussion on Saturday at 9:00 pm IST ,8 30 PDT. For more details, see IMO Problems Discussion Group.

Here are the problems from the second day of the 1959 International mathematical Olympiad. They range from easy to harder Geometry and Combinatorics. Try your hand at them. Don't be discouraged if you cant completely solve them. Do post your inspirations and ideas towards the problems. The discussion for theses questions will be held soon. Happy Problem Solving!

Q4. Construct a right-angled triangle whose hypotenuse $$c$$ is given if it is known that the median from the right angle equals the geometric mean of the remaining two sides of the triangle. (HUN)

Q5. A segment $$AB$$ is given and on it a point $$M$$. On the same side of $$AB$$ squares $$AMCD$$ and $$BMFE$$ are constructed. The circumcircles of the two squares, whose centers are $$P$$ and $$Q$$, intersect in $$M$$ and another point $$N$$.

(a) Prove that lines $$FA$$ and $$BC$$ intersect at $$N$$.

(b) Prove that all such constructed lines $$MN$$ pass through the same point $$S$$, regardless of the selection of $$M$$.

(c) Find the locus of the midpoints of all segments $$PQ$$, as $$M$$ varies along the segment $$AB$$.(ROM)

Q6. Let $$\alpha$$ and $$\beta$$ be two planes intersecting at a line $$p$$. In $$\alpha$$ a point $$A$$ is given and in $$\beta$$ a point $$C$$ is given, neither of which lies on $$p$$. Construct $$B$$ in $$\alpha$$ and $$D$$ in $$\beta$$ such that $$ABCD$$ is an equilateral trapezoid, $$AB || CD$$, in which a circle can be inscribed. (CZS)

###### This is part of the set International Mathematical Olympiad

Note by Sualeh Asif
1 year, 8 months ago

Sort by:

$$Q1.$$ Let the legs of the right triangle $$\Delta ABC$$ (right-angled at $$C$$) be $$a$$ & $$b$$.

The median from $$C$$ is defined to have a length of $$\sqrt{ab}$$.

$$\:$$

$$Lemma:$$ The length of the median of a right triangle, from the right angle to the hypotenuse, is half the length of the hypotenuse.

$$Proof:$$ Suppose we have a right triangle $$\Delta ABC$$ (right-angled at $$C$$). Let the mid-point of $$AB$$ be $$D$$.

Alt text

We construct a line $$\parallel$$ to $$AB$$ from $$D$$. Let it intersect $$BC$$ at $$E$$.

Since $$DE \parallel AB \Rightarrow \angle DEC = \angle DEB = 90^\circ$$

Also, $$EB=EC$$ (Mid-point Theorem) and $$DE$$ is common to both $$\Delta DEB$$ & $$\Delta DEC$$

$\therefore \Delta DEB \cong \Delta DEC$

$\Rightarrow DA = DB = DC$

Hence, proved.

$$\:$$

We now use the above statement to proceed,

$\therefore \sqrt{ab} = \frac{c}{2} \Rightarrow ab = \frac{c^2}{4}$

$\therefore 2 \cdot \frac{a}{c} \cdot \frac{b}{c} = \frac{1}{2}$

$\therefore 2\sin A \cos A = \frac{1}{2} \Rightarrow \sin 2A = \frac{1}{2}$

Since, $$0 \leq 2A \leq \pi$$, $\Rightarrow 2A = \frac{\pi}{6}, \frac{5\pi}{6}$

$\therefore A = \frac{\pi}{12}, \frac{5\pi}{12} \Leftrightarrow A = 15^\circ, 75^\circ$

$$\:$$

In order to construct such a triangle, we will need to construct $$60^\circ$$ angles. To do so, we construct a segment (say, $$QR$$). We take a radius less than the length of $$QR$$ and construct an arc, with center $$Q$$, intersecting $$QR$$ at $$R'$$. Using the same radius and with center $$R'$$, we construct an arc intersecting the first arc at $$P$$. $$\angle PQR$$ is the required angle. (We can construct a $$120^\circ$$ angle by intersecting the first arc again, with center $$P$$ and the same radius, as before)

We also need to bisect angles. To do so, we construct an arbitrary angle (say, $$\angle PQR$$). We take a radius less than the length of $$QR$$ (or $$QP$$, if $$QP<QR$$)and construct an arc, with center $$Q$$, intersecting $$QR$$ and $$QP$$ at $$R'$$ and $$P'$$, respectively. Using $$R'$$ and $$P'$$ as centers, using the same radius, we construct an arc from both of them. These arcs intersect at two points, which we join and extend to $$Q$$ to construct the angle bisector of $$\angle PQR$$.

Using the above two methods, we can now construct such a right triangle $$\Delta ABC$$ with these angles.

We construct a segment $$AB$$ with the given length $$c$$. On $$A$$, we construct a $$60^\circ$$ angle and bisect it twice, towards $$AB$$ giving us $$15^\circ$$. On $$B$$, we construct a $$120^\circ$$ angle and bisect the angle between the $$60^\circ$$ and the $$120^\circ$$ mark twice, towards the $$60^\circ$$ mark giving us $$75^\circ$$. We extend the new segments and label their intersection point $$C$$, which is right-angled. Thus, $$\Delta ABC$$ is the required triangle. · 1 year, 8 months ago

Good clear presentation. Keep it up!

At the IMO, this will be worth 7-.

A slightly easier way to construct a $$15 ^ \circ$$ angle is to construct $$60^\circ$$ through an equilateral triangle and then $$45^\circ$$ through a right isosceles triangle. Staff · 1 year, 7 months ago

Good work in finding the angles of such a triangle! Now ponder on this: In Euclidean Geometry when you are asked to construct a figure you are expected to use a straight edge and a compass only. @Shamay Samuel your job is to find a way to construct this triangle. This solution wont be valid at the IMO. Since it does not give a way to construct this triangle. So find a way to complete this solution!

Secondly, In the third line you say " We know that...." As Sir @Calvin Lin pointed out this is not acceptable at proof-writing. Please write why you know that, and how the others can understand it!

However great start! try and finish it. · 1 year, 8 months ago

Thanks, @Sualeh Asif for your advice! I have made changes to my solution according to it. Please check it to see if it can be improved upon. · 1 year, 8 months ago

@Shamay Samuel can you define how to make an angle of $$60^{\circ}$$. The solution is still incomplete.That would be another problem. Great proving the firs fact. · 1 year, 8 months ago

2) This is my attempt at Question 2

(a)

The triangles $$AFM$$ and $$CBM$$ are congruent by the $$SAS$$ criterion ($$AM = CM$$, $$FM=BM$$ and $$\angle AMF = \angle BMC$$). This implies that $$\Delta CBM$$ is similar to $$\Delta CN'F$$ by the $$AA$$ criterion ($$\angle CBM = \angle CFN'$$ and $$\angle BCM = \angle FCN'$$). This means that $$\angle CMB = \angle CN'F = \angle CN'A = 90^{\circ}$$. Since $$\angle FN'B + \angle FEB = 180^{\circ}$$ (because they are both $$90^{\circ}$$, $$FN'BE$$ is a cyclic quadrilateral. Furthermore, since $$\angle CN'A + \angle CMA = 180^{\circ}$$ (because they are both $$90^{\circ}$$, $$CN'MA$$ is a cyclic quadrilateral. Therefore, $$N'$$ lies on both circumcircles; hence it is the same point as $$N$$.

(b)

Note that $$\dfrac {AM}{MB} = \dfrac {CM}{MB} = \dfrac {AN}{NB}$$. Thus, by the converse of angle bisector theorem, $$NM$$ is the angle bisector of $$\angle ANB$$

Consider a circle with diameter $$AB$$. It follows that point $$N$$ is on the circumference since $$\angle ANB = 90^{\circ}$$ (Angle in semicircle). The angle $$\angle ANB$$ subtends arc $$AB$$ (This arc has not got point $$N$$ on it). Since $$MN$$ bisects $$\angle ANB$$, $$MN$$ must also bisect the arc $$AB$$. Since the bisector of an arc is a constant point, the line $$MN$$ passes through a constant point.

(c)

The average of the distances of $$P$$ and $$Q$$ from $$AB$$ gives the distance of the midpoint of $$PQ$$ from $$AB$$. Since $$P$$ is a distance of $$\dfrac {1}{2} AM$$ and $$Q$$ is distance of $$\dfrac {1}{2} MB$$, the midpoint is a distance of $$\dfrac {1}{4} AB$$ from the line $$AB$$, which is a constant. Therefore, the locus of this midpoint is a line segment. · 1 year, 8 months ago

In a), what is $$N'$$? Unfortunately, as I am not a mind-reader, your solution doesn't make sense.

In b), where did that first equation come from? You should also prove why "the bisector of an arc is a constant point", and it helps to state what that point is.

In c), you have to specify what is the exact locus, and not just what the shape of the locus is. IE Give the exact endpoints, and prove that they can be achieved, and that no other points can be achieved.

At the IMO, this solution is worth 0+. It is a classic case of where you as the solver knows enough to completely solve the problem, but when writing up the solution you miss out crucial bits of information and explanation which completely destroys the building of your proof. Staff · 1 year, 7 months ago

2nd one is quite easy using coordinates. Fix A as origin. a) First one is simple, just solve some simple equations and we are done. b) For 2nd one use the idea that equation of radical axis of two circles is S1 - S2 . c) For 3rd one writing the coordinates of mid-point we see that it's y coordinate is always fixed. But I have not yet figured out what the locus of x coordinate means geometrically.

I will post my solution later as I have my exam tomorrow. · 1 year, 8 months ago

Good work. For c, note that you're just asked to find the locus of the points, and do not need to give a geometric interpretation of it. Defining the locus is the crucial part here. Staff · 1 year, 7 months ago


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# Automation: steady state error calculation block diagram

Hello I have a problem determining the value of K in this block diagram, to calculate this value, I applied this method, but the value at the end of K I is different from 216, you could explain to me where I'm wrong

So, for this problem, you first need to find the transfer function from U to E. An easy way to do this to use the forward gain over one plus the loop gain. This simply mean $\frac{E(s)}{U(s)} = \frac{forward~gain}{1+loop~gain}$.
The forward gain in this problem is 1 (U to E has no gains between it) while the loop gain is $\frac{K}{(s+2)(s+3)}*\frac{1}{s+4}$ (all gains around loop from E back to U).
The steady state error is $e_{ss} = \lim_{s\to0} s\frac{E(s)}{U(s)} U(s)$. This simplifies to what you had above. If you substitute everything is, you have $0.1 = \lim_{s\to0}\frac{1}{1+\frac{K}{(s+2)(s+3)(s+4)}}$.
Multiply both top and bottom by $(s+2)(s+3)(s+4)$ and set s equal to 0. Multiply it all out and you should get 216.


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# How to prove two propositions equivalent?

I was given the question:

Determine if $\exists x (P(x) \implies Q(x))$ and $\exists xP(x) \implies \exists x Q(x)$ are equivalent, i.e., always have the same truth value. If they are not equivalent figure out if one implies the other. Prove that you have correctly determined the most general logical relationship between the two expressions. Be sure to include examples if the two expressions are not equivalent.

How do I go about doing this? Would I simplify each proposition and see if they simplify to the same thing? And if they are not equivalent, how would I see if one implies the other? Thanks.

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"Equal" is not the same as "equivalent". –  Andres Caicedo Nov 2 '12 at 0:44
Have you tried out any concrete "examples"? "There exists something such that (if it is P, then it is Q)" vs. "if there exists something such that it is P, then there exists something such that it is Q"...I've translated generically; perhaps it's enough for you to better understand the differences (hint) between the two propositions? –  amWhy Nov 2 '12 at 0:51
To prove equivalence, you need to be able to derive one from the other, or "simplify" as you mention. Equivalence can also be proven by establishing that each statement implies the other. Equivalence would be the most general relationship then. If the expressions are not equivalent, you can prove this by providing a valid counterexample. Then you'd need to determine whether one of the statements implies the other (whether if one is true, the other cannot be false); if so you'd need to prove it. Implication would be the most general relationship then. –  amWhy Nov 2 '12 at 0:57
@AndresCaicedo I edited accordingly. –  amWhy Nov 2 '12 at 1:22

Let the universe be all the people on Earth, $\,P(x):=x\,$ is a mother , $\,Q(x):=x\,$ is a man, so

$$\exists\,x\left(P(x)\rightarrow Q(x)\right)=\,\text{exists a person s.t. if it is a mother then it is a man}$$

$$(\exists\,x\,P(x))\rightarrow (\exists\,x\,Q(x))=\,\text{if there exists a mother then there exists a man}$$

Do you think the above are equivalent?

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That makes it much more simple to understand. I would say no, they are not equivalent. But I also don't see how one would imply the either using this universe/example? –  Jakemmarsh Nov 2 '12 at 2:59

HINTS:

1. What if $Q(x)$ is $x\ne x$, and $P$ is such that $\exists xP(x)$ and $\exists x\big(\lnot P(x)\big)$ are both true?

2. One of the propositions does imply the other.

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I like to develop intuition with a specific universe. In this case, I take $x$ as having two values, say $0$ and $1$.

Then the first statement becomes $(\lnot P(0) \lor Q(0)) \lor (\lnot P(1) \lor Q(1))$.

The second statement becomes $(\lnot(P(0) \lor P(1)) \lor (Q(0) \lor Q(1))$.

It is straightforward to see that the assignment $P(0) = F, P(1)=T, Q(0) = F, Q(1) = F$ results in the first formula having the value $T$, while the second is $F$. Hence the formulas are not equivalent.

Rewriting the formulas as $\lnot P(0) \lor \lnot P(1) \lor Q(0) \lor Q(1)$ and $(\lnot P(0) \land \lnot P(1)) \lor Q(0) \lor Q(1)$ suggests that the second formula implies the first.

To prove this (loosely) we need only show that if the second formula is true, then so is the first.

The second formula can be true in two ways:

(1) $\exists x P(x)$ is false, in which case $\forall x \lnot P(x)$ is true, from which is follows that $\forall x (\lnot P(x) \lor Q(x))$ is true, from which we have $\forall x (P(x) \Rightarrow Q(x))$ and finally $\exists x (P(x) \Rightarrow Q(x))$ is true.

(2) $\exists x P(x)$ is true and $\exists x Q(x)$ is true. Since $\exists x Q(x)$ is true, it follows that $\exists x (\lnot P(x) \lor Q(x))$ is true, which is equivalent to $\exists x ( P(x) \Rightarrow Q(x))$.

Formalizing the proof needs more input from you...

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What are the rules for converting a proposition like this to ands/ors/negations? –  Jakemmarsh Nov 2 '12 at 19:57
Depends on your rules... –  copper.hat Nov 2 '12 at 20:10


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• # question_answer An NPN transistor is used in common emitter configuration as an amplifier with $1k\Omega$load resistance. Signal voltage of 10 mV is applied across the base-emitter. This produces a 3 mA change in the collector current and $15\mu A$ change in the base current of the amplifier. The input resistance and voltage gain are: [JEE Main 9-4-2019 Morning] A) $0.33k\Omega ,1.5$              B) $0.67k\Omega ,200$C) $0.33k\Omega ,300$             D) $0.67k\Omega ,300$

input current $=15\times {{10}^{6}}$ output current $=3\times {{10}^{3}}$ resistance output = 1000 ${{V}_{input}}=10\times {{10}^{3}}$ Now ${{V}_{input}}={{r}_{input}}\times {{i}_{input}}$ $10\times {{10}^{-3}}={{r}_{input}}\times 15\times {{10}^{-6}}$ ${{r}_{input}}=\frac{2000}{3}=0.67K\Omega .$ voltage gain =$=\frac{{{V}_{output}}}{{{V}_{input}}}=\frac{1000\times 3\times {{10}^{-3}}}{10\times {{10}^{-3}}}=300$ Option [d]


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# NetBIOS and Samba Ports

Service Name Port Protocol
Name Resolution Service 137 UDP
Datagram Services (browsing) 138 UDP
Session Service 139 TCP
SMB 445 TCP Input, UDP Output

## Observations

These are the contents of /etc/services:

netbios-ns 137/udp   # NetBIOS Name Service
netbios-dgm 138/udp  # NetBIOS Datagram Service
netbios-ssn 139/tcp  # NetBIOS Session Service
microsoft-ds 445/tcp # Microsoft Directory Service

These are the ports that the Samba server listens on. A few things here:

1. The first three are ports ports used by Windows to ‘find’ other computers (a la “My Network”)
2. The last is for directory services
3. UDP 137 and 138 are serviced by nmbd
4. TCP 139 and 445 are serviced by smbd


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1.6k views

Which of the following tuple relational calculus expression(s) is/are equivalent to $\forall t \in r \left(P\left(t\right)\right)$?

1. $\neg \exists t \in r \left(P\left(t\right)\right)$
2. $\exists t \notin r \left(P\left(t\right)\right)$
3. $\neg \exists t \in r \left(\neg P\left(t\right)\right)$
4. $\exists t \notin r \left(\neg P\left(t\right)\right)$
1. I only
2. II only
3. III only
4. III and IV only
Change the category of this question from Mathematical Logic to Databases.
Thanks :)
I think it belongs to both Mathematical Logic & Databases. Keep Database tag, add Mathematical logic too !
relational calculus implies that..

I understand the meaning of the individual symbols used in this question, but I am not able to clearly interprete their meaning in the way they are used.

1. For example in tuple relational calculus we have ∀t (predicates) or we may have single predicate ∀t P(t). But I am first time seeing "belongs to ∈" symbol following quantifier. Somehow I feel it should be ∀t (r(P(t)))

2. Next we use r(t) in dbms to say tuple t belongs to relation r. But I never saw predicate used inside r(). Should it be "r(t) ^ P(t)" to make it more sensible or better to say to follow more standard usage?

So the whole expression should be ∀t (r(t) ^ P(t))

Am I correct on these? Or I am just plain wrong in understanding the usage of these symbols?

what is the exact meaning of third statement??

Let predicate $P(t)$ means "student is intelligent"

't' is some person and as for our predicate domain is 'student' so we'll talk when this person belongs to domain 'student'.

$\forall t\in r (P(t))$ will then mean "Every person who is a student is intelligent or simply every student is intelligent"

in other words we can also say it like "there exists no student ($\sim \exists t$) who is dull ($\sim P(t)$)"

Mathematical way

$\forall t\in r (P(t))$

$\equiv \sim \sim$( $\forall t\in r (P(t))$ )  //Use one negation to apply Demorgan law

$\equiv \sim \exists t\in r (\sim P(t))$

Only III is correct.
The given statement means for all tuples from r, P is true. III means there does not exist a tuple in r where P is not true. Both are equivalent.

IV is not correct as it as saying that there exist a tuple, not in r for which P is not true, which is not what the given expression means.
selected by
(1)  gives all tuples that dont satisfy P. So its equivalent to all tuples that satisfy P, right?

But it would describe the complement of set that contains all tuples that satisfy P.
@Arjun sir I got the answer

but in (1) if we give -ve inside what happens?

will it not be $\forall t \in r \left(P\left(t\right)\right)$
That is option 3 rt?
ok u mean

$\neg \exists t \in r \left(P\left(t\right)\right)$

and $\neg (\exists t \in r \left(P\left(t\right)\right))$

are same?
Yes, if we say in English they should mean the same.
just curious if we can reduce $∀t∈r(P(t))$ to $¬∃t∈r(¬P(t))$ by using propositional calculus manipulations, instead of interpreting English meaning of those expressions in order to find their equivalence? can we do this?
sir t is tuple variable, r is table and P ? P is attribute?
@Aspirant yes, you can do.

@Resilient P is a predicate- some condition working on a tuple.
okay sir got it....thank you.
doubt : negation of belongs to is belongs to how?
Is r a tuple variabkle or a relation?

(IV) is incorrect as Arjun sir have explained.Above is simplification of given expression to (III).

Note that, ∀x∈t (P(x))=∀x (x∊t-->P(x))

∃x∈t (P(x))= ∃x (x∊t^P(x))

Concept:- A statement is equivalent to its double negation.

i.e. x=~(~x)

Similarly compare below expression with above expression.

$\forall t \in r \left(P\left(t\right)\right)$ = $\neg$ ( $\exists t \in r \left(\neg P\left(t\right)\right)$ )

Hence C is Ans.

∀t∈r(P(t)):  All boys in the class are in a relationship.

I) ¬∃t∈r(P(t)) : There is not a single boy in the class who is in a relationship.  (X)
II) ∃t∉r(P(t)) : There is at least one boy in the class who does not belong those boys who are in a relationship. (X)
III) ¬∃t∈r(¬P(t)) : There is not a single boy in the class who is not in a relationship. (Matches the qsn)
IV) ∃t∉r(¬P(t)) : There is at least one boy in the class who does not belong to those boys who are not in a relationship. (X)

Only IV is little bit tricky. Please read again carefully.  We need "all boys" instead of "at least one" to match our original qsn.
So only option III is true.

+1 vote

∃t∈r(⌝ P(t))--means that there are some tuples in relation r which do not satisfy the constraint P(t)

and if we put not before this statement we get there are no tuples in relation r which do not satisfy the constraint P(t) which is equivalent to saying all tuples in relation r satify predicate condition P(t)

I think even (1) is same as yours.


text:
GOE. The most well known model of random matrices is probably the Gaussian Orthogonal Ensemble (GOE) of symmetric $n\times n$ random matrices, with Gaussian density proportional to $$X\mapsto\mathrm{e}^{-\tfrac{1}{2}\mathrm{Trace}(X^2)}.$$ The eigen-decomposition $X=ODO^\top$ and the change of variable $S\mapsto (O,D)$ give that the eigenvalues of GOE have density in $\mathbb{R}^n$ proportional to $$\mathrm{e}^{-\tfrac{1}{2}\sum_{i=1}^n\lambda_i^2}\prod_{i<j}|\lambda_i-\lambda_j|=\mathrm{e}^{-\Bigr(\tfrac{1}{2}\sum_{i=1}^n\lambda_i^2+\sum_{i<j}\log\frac{1}{|\lambda_i-\lambda_j|}\Bigr)}.$$ The term $\prod_{i<j}|\lambda_i-\lambda_j|$ is the absolute value of the determinant of the Jacobian of the change of variable. It expresses the way the Lebesgue measure on symmetric matrices is deformed by the change of variable. It vanishes when two eigenvalues are the same, revealing an eigenvalues repulsion. What is the simplest way to understand this repulsion phenomenon?

2×2 fact. It is actually a basic non-linear fact checkable on $2×2$ fixed matrices : $$X=\begin{pmatrix} x_1 & x_3\\x_3 & x_2\end{pmatrix},\quad x_1,x_2,x_3\in\mathbb{R}.$$ Namely, the eigenvalues $\lambda_1$ and $\lambda_2$, the roots of $\lambda^2-(x_1+x_2)\lambda+x_1x_2-x_3^2=0$, are $$\frac{x_1+x_2\pm\sqrt{(x_1+x_2)^2-4(x_1x_2-x_3^2)}}{2}$$ hence the eigenvalues spacing is given by $$|\lambda_1-\lambda_2|=\sqrt{(x_1-x_2)^2+4x_3^2}.$$ Now if $x_3=0$ then the matrix is diagonal and $|\lambda_1-\lambda_2|=|x_1-x_2|,$ while if $x_3\neq0$ then $|\lambda_1-\lambda_2|>2|x_3|>0$ which is precisely an eigenvalues repulsion phenomenon.

Surmise. Let us put now a Gaussian law on this $2×2$ matrix, making it GOE, with density $$\mathrm{e}^{-\tfrac{1}{2}\mathrm{Trace}\begin{pmatrix}x_1 & x_3\\x_3 & x_2\end{pmatrix}^2}=\mathrm{e}^{-\tfrac{1}{2}(x_1^2+x_2^2+2x_3^2)}.$$ Then the law of the spacing $S:=|\lambda_1-\lambda_2|$ can be computed as follows : $$S\overset{d}{=}\sqrt{(X_1-X_2)^2+4X_3^2},\quad\text{with } X_1,X_2,\sqrt{2}X_3\text{ iid }\mathcal{N}(0,1),$$ therefore
$$S^2\overset{d}{=}(X_1-X_2)^2+4X_3^2\sim\Gamma(\tfrac{1}{2},\tfrac{1}{4})*\Gamma(\tfrac{1}{2},\tfrac{1}{4})=\Gamma(1,\tfrac{1}{4})=\mathrm{Exp}(\tfrac{1}{4}),$$and thus the spacing $S$ has density proportional to $$s\mathrm{e}^{-\tfrac{1}{4}s^2}=\mathrm{e}^{-(\tfrac{1}{4}s^2-\log(s))}.$$ This density vanishes at the origin $s=0$, which corresponds to a logarithmic repulsion in the energy. This simple computation, made by Wigner, led him to his famous surmise (or conjecture) on the distribution of the eigenvalues spacings for high dimensional random matrices.

Non-linear effect and removable singularity. The squared spacing $S^2$ follows the exponential law of density proportional to $t\mapsto\mathrm{e^{-t/4}}$, which does not vanish at $t=0$, and this shows that the singularity at the origin of the density of the spacing $S$ can be removed by a non-linear change of variable. More generally, if for instance $t\mapsto f(t)$ is a density which may not vanish at $t=0$, then the change of variable $s=\sqrt{t}$ leads to a density proportional to $s\mapsto sf(s^2)$. In the case of a diagonal matrix, $x_3=0$, then $S^2=(X_1-X_2)^2\sim\Gamma(\tfrac{1}{2},\tfrac{1}{4})$, which has density proportional to $t\mapsto\mathrm{e}^{-t/4}/\sqrt{t}$, while $S$ has density proportional to $s\mapsto\mathrm{e}^{-s/4}$, which is not singular at $s=0$. What is important is the singularity or the non-singularity of the spacing $S$ because the squared spacing $S^2$ is not proportional to the coordinates of interest (the $\lambda$’s).

Jacobian. Let us check on the $2×2$ example that the absolute value of determinant of the Jacobian of the change of variable $X=ODO^\top\to (O,D)$ equals $|\lambda_1-\lambda_2|=\lambda_+-\lambda_-$ (here $n=2$). The matrix $X$ has $3=\frac{n^2-n}{2}+n$ degrees of freedom : $x_1,x_2,x_3$, while for $(O,D)$ we have two degrees of freedom $\lambda_\pm$, defining the diagonal matrix $D$, and another degree of freedom, say $\theta$, defining the $2×2$ orthogonal matrix $O$, which can be coded as $$O=\begin{pmatrix}\cos(\theta) & -\sin(\theta)\\\sin(\theta) & \cos(\theta)\end{pmatrix}.$$ Now $X=ODO^\top$ gives \begin{align*}x_1&=\lambda_+\cos(\theta)^2+\lambda_-\sin(\theta)^2\\x_2&=\lambda_+\sin(\theta)^2+\lambda_-\cos(\theta)^2\\x_3&=(\lambda_+-\lambda_-)\cos(\theta)\sin(\theta)\end{align*} and the determinant of the Jacobian is then \begin{align*}\det J&=\begin{vmatrix}\partial_{\lambda_+}x_1 & \partial_{\lambda_-}x_1 & \partial_{\theta}x_1\\\partial_{\lambda_+}x_2 & \partial_{\lambda_-}x_2 & \partial_{\theta}x_3\\\partial_{\lambda_+}x_3 & \partial_{\lambda_-}x_3 & \partial_{\theta}x_3\end{vmatrix}\\&=\begin{vmatrix}\cos(\theta)^2&\sin(\theta)^2&-2(\lambda_+-\lambda_-)\cos(\theta)\sin(\theta)\\\sin(\theta)^2&\cos(\theta)^2&2(\lambda_+-\lambda_-)\cos(\theta)\sin(\theta)\\\cos(\theta)\sin(\theta)&-\cos(\theta)\sin(\theta)&(\lambda_+-\lambda_-)(\cos(\theta)^2-\sin(\theta)^2)\end{vmatrix}\\&=\lambda_+-\lambda_-.\end{align*}

2×2 fact, non-Hermitian version. Let us consider now the fixed matrix $$X=\begin{pmatrix} x_1 & x_3\\x_4 & x_2\end{pmatrix},\quad x_1,x_2,x_3,x_4\in\mathbb{C}.$$ The eigenvalues $\lambda_1$ and $\lambda_2$, the roots in $\mathbb{C}$ of $\lambda^2-(x_1+x_2)\lambda+x_1x_2-x_3x_4=0$, are $$\frac{x_1+x_2\pm\sqrt{(x_1+x_2)^2-4(x_1x_2-x_3x_4)}}{2}$$ hence the eigenvalues spacing is given by $$|\lambda_1-\lambda_2|=\sqrt{|(x_1-x_2)^2+4x_3x_4|}.$$ Now if $x_3x_4=0$ then the matrix is tridiagonal and $|\lambda_1-\lambda_2|=|x_1-x_2|,$ while if $x_3x_4\neq0$ then the situation is more subtle than in the case of symmetric or Hermitian matrices.


text:
# Here's a question on light ray optics and focal length.

1. Aug 2, 2007

### Rainbow

Here's a question on ray optics.

We say that the focal length of a lens is half times the radius of curvature, right. Now, consider any lens(say convex), made of any material(say glass). Now a beam of light parallel to the principle axis, passes through the lens. All the rays in the beam will converge at a point, which we call the focus. For now, let us assume that the focal length of this particular lens is half time the radius of curvature of the lens. Now, take another lens whose structure is identical to that of the previous lens, i.e., it has the same radius of curvature and everything, the only difference being that this time the lens is made of a material that is denser than that of the previous lens. This time, if we pass a beam of light parallel to the principle axis through the lens, all the rays in the beam will converge at a different focus, with focal length shorter than that in case of the previous lens. So, we can see that, this time the focal length is not equal to half times the radius of curvature. This shows that saying that the focal length of a lens is half times the radius of curvature is not always true.

2. Aug 2, 2007

### Hootenanny

Staff Emeritus
Actually, the focal length of a spherical mirror is half the radius of curvature of the mirror. The focal length of a lens of non-negligible thickness is indeed dependent on the refractive index of the material among other factors.

3. Aug 2, 2007

### Rainbow

It doesn't matter. The situation that I have mentioned, is equally applicable to spherical lens.

4. Aug 2, 2007

### mgb_phys

Hootenanny said the equation you quoted is only for MIRRORS - you are correct that the focal length of any lens must depend on it's refractive index, but the focal lenght of a mirror doesn't.

5. Aug 2, 2007

### Staff: Mentor

The focal length of a thin spherical lens is given by the Lensmaker Formula; it's not simply R/2 as for a spherical mirror--it depends on both radii and the indices of refraction.

6. Aug 3, 2007

### Rainbow

But then why in most places it is written that, f=R/2? As, you already mentioned that the f is given by the lens-maker's formula, and that it depends on both radii and the indices of refraction.

7. Aug 3, 2007

### Staff: Mentor

Where have you seen this?

Exactly. Which is certainly not f = R/2.

8. Aug 4, 2007

### Rainbow

I have been taught this in my physics class. And I also found it many books like 'Fundamentals of Physics- by Halliday, Resnick and Walker', and my school text book, and many other references.

9. Aug 4, 2007

### Staff: Mentor

I don't think so. Why don't you scan in the page in Halliday and Resnick where it says that, so we know what it is you're reading.

10. Aug 4, 2007

### Staff: Mentor

Or give us a short quotation from a specific book (authors and edition), along with the chapter and page number. If we're lucky, one of us will have the book and can look it up.

The introductory textbooks that I've seen all discuss spherical mirrors first, and give their focal length as $f = - R / 2$. Then they move on to thin lenses, and introduce or derive the lensmaker's formula:

$$\frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$

(for a lens in vacuum, or to a good approximation, in air)

In my experience, it's common for students to glom onto the first formula they see for a certain quantity, and try to use it everwhere thereafter, even in situations where it doesn't apply.

11. Aug 5, 2007

### Rainbow

You're correct. And I just realised that I am one of those students. The formula f=R/2 is for a spherical mirror and not a lens.

I'm sorry to waste your time for this stupid question. But thank you for having enough patience and reminding me of the fact.

12. Aug 5, 2007

### Staff: Mentor

I don't think it was a waste of time. True, as we suspected, you mixed up mirrors with lenses. But you showed excellent reasoning in your first post to prove that that formula didn't make sense for lenses. And now you've gotten it straight.


text:
Tag Info

Code that is written to the 2011 version of the C++ standard, sometimes known by its pre-publication name of "C++0x". Use in conjunction with the 'c++' tag.

Use this tag for code that is to be compiled as C++11, published as ISO/IEC 14882:2011 in September 2011.

• Predecessor: C++03
• Successor: C++14

Some of the main improvements of C++11, compared to C++03, include:

• an expanded standard library
• type inference using the auto keyword
• lambda expressions, as in [](int x) { return x * x; }
• a range-based for loop, as in for (auto &x : my_array) ...
• move constructors
• constructor chaining
• a more natural syntax for the initialization of objects


text:
# Obtaining $\beta = \frac{1}{k_B T}$ from first principles derivation

I'm nearly at the end of this derivation but totally stuck so I'd appreciate a nudge in the right direction

Consider a set of N identical but distinguishable particles in a system of energy E. These particles are to be placed in energy levels $E_i$ for $i = 1, 2 .. r$. Assume that we have $n_i$ particles in each energy level. The two constraints we impose are that $\sum_{i}^{r}n_i = N$ and $\sum_{i}^{r}E_i n_i = E$.

The number of microstates in a given macrostate is given by $$\Omega = \frac{N!}{\sum_{i}^r n_{i}!}$$

We want to maximize this and for ease of notation, we work with $\ln\Omega$ and we use Stirling's approximation ($\ln x! = x\ln x - x$) to obtain $$\ln\Omega = N\ln N - N - \sum_{i}^{r}n_i\ln n_i - n_i$$

Maximizing this function subject to the constraints $\sum_{i}^{r}n_i = N$ and $\sum_{i}^{r}E_i n_i = E$ is a classic Lagrange multiplier problem. We represent the undetermined multipliers to be $\alpha$ and $\beta$ for the two constraints and obtain \begin{align} \frac{\partial\ln\Omega}{\partial n_i} &= \alpha\frac{\partial n_i}{\partial n_i} + \beta\frac{\partial E_i n_i}{\partial n_i} \\ \nonumber \ln n_i &= \alpha + \beta E_i \\ \nonumber \therefore n_i &= e^{\alpha}e^{\beta E_i} \end{align}

Now, we use the first constraint equation to determine $\alpha$. We get \begin{align} \sum_i^r n_i &= N \\ \nonumber \sum_i^r e^{\alpha}e^{\beta E_i} &= N \\ \nonumber e^\alpha &= \frac{N}{\sum_i^re^{\beta E_i}} \\ \nonumber e^\alpha &= \frac{N}{Z} \end{align}

We have introduced the partition function, $Z=\sum_i^re^{\beta E_i}$ in the last line. Next, we have the second constraint equation that determines $\beta$ \begin{align} \sum_i^r E_i n_i &= E \\ \nonumber \frac{\sum_{i}^{r} E_i e^{\beta E_i}}{\sum_i^r e^{\beta E_i}} &= E \\ \nonumber \end{align}

I'm assuming I should somehow connect $E$ with $T$ so let's say $E=Nk_B T$. Then we have \begin{align} \frac{N}{Z}\frac{\partial Z}{\partial\beta} &= E \\ \frac{\partial\ln Z}{\partial\beta} &= k_B T \end{align}

How do I get to $\beta = -\frac{1}{k_B T}$ here? Notice that this derivation requires an extra minus sign compared to the usual definition of $\beta$ and this should come out naturally too, shouldn't it?

• Hint: what actually is $T$? (Think experiment...) – lemon Aug 21 '16 at 15:47
• @lemon Oh dear, I'm really blanking! Did you mean that I shouldn't be relating E and T as I have done? – user1936752 Aug 21 '16 at 17:39
• Indeed. Think back before the time of statistical mechanics; temperature $T$ was simply the thing that thermometers measured, and can be defined in terms of the relevant equation of state. In fact, the usual way to connect $\beta$ with $T$ is to simply derive the ideal gas law from your above formalism... – lemon Aug 21 '16 at 19:01
• @lemon sorry, I can't see how to bring about P or V into this derivation since I have not said anything about either. Integrating my last equation, we simply get $\ln Z = \beta k_B T + constant$ but that doesn't illuminate the problem for me either. Could you elaborate? – user1936752 Aug 22 '16 at 6:19


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# COMPLETE MOMENT CONVERGENCE OF MOVING AVERAGE PROCESSES WITH DEPENDENT INNOVATIONS

• Kim, Tae-Sung ;
• Ko, Mi-Hwa ;
• Choi, Yong-Kab
• Published : 2008.03.31
• 67 11

#### Abstract

Let ${Y_i;-\infty<i<\infty}$ be a doubly infinite sequence of identically distributed and $\phi$-mixing random variables with zero means and finite variances and ${a_i;-\infty<i<\infty}$ an absolutely summable sequence of real numbers. In this paper, we prove the complete moment convergence of ${{\sum}_{k=1}^{n}\;{\sum}_{i=-\infty}^{\infty}\;a_{i+k}Y_i/n^{1/p};n\geq1}$ under some suitable conditions.

#### Keywords

moving average process;complete moment convergence$\phi$-mixing

#### References

1. J. I. Baek, T. S. Kim, and H. Y. Liang, On the convergence of moving average processes under dependent conditions, Aust. N. Z. J. Stat. 45 (2003), no. 3, 331-342 https://doi.org/10.1111/1467-842X.00287
2. R. M. Burton and H. Dehling, Large deviations for some weakly dependent random processes, Statist. Probab. Lett. 9 (1990), no. 5, 397-401 https://doi.org/10.1016/0167-7152(90)90031-2
3. Y. S. Chow, On the rate of moment convergence of sample sums and extremes, Bull. Inst. Math. Acad. Sinica 16 (1988), no. 3, 177-201
4. D. L. Li, M. B. Rao, and X. C. Wang, Complete convergence of moving average processes, Statist. Probab. Lett. 14 (1992), no. 2, 111-114 https://doi.org/10.1016/0167-7152(92)90073-E
5. Y. X. Li and L. X. Zhang, Complete moment convergence of moving-average processes under dependence assumptions, Statist. Probab. Lett. 70 (2004), no. 3, 191-197 https://doi.org/10.1016/j.spl.2004.10.003
6. H. Y. Liang, Complete convergence for weighted sums of negatively associated random variables, Statist. Probab. Lett. 48 (2000), no. 4, 317-325 https://doi.org/10.1016/S0167-7152(00)00002-X
7. Q. M. Shao, Almost sure invariance principles for mixing sequences of random variables, Stochastic Process. Appl. 48 (1993), no. 2, 319-334 https://doi.org/10.1016/0304-4149(93)90051-5
8. L. X. Zhang, Complete convergence of moving average processes under dependence assumptions, Statist. Probab. Lett. 30 (1996), no. 2, 165-170 https://doi.org/10.1016/0167-7152(95)00215-4
9. I. A. Ibragimov, Some limit theorems for stationary processes, Teor. Verojatnost. i Primenen. 7 (1962), 361-392

#### Cited by

1. Toeplitz lemma, complete convergence, and complete moment convergence vol.46, pp.4, 2017, https://doi.org/10.1080/03610926.2015.1026996
2. On Complete Convergence of Moving Average Process for AANA Sequence vol.2012, 2012, https://doi.org/10.1155/2012/863931
3. Complete moment convergence of widely orthant dependent random variables vol.46, pp.14, 2017, https://doi.org/10.1080/03610926.2016.1148728
4. Convergence of Moving Average Processes for Dependent Random Variables vol.40, pp.13, 2011, https://doi.org/10.1080/03610921003797761
5. Complete moment convergence of moving average processes under ρ-mixing assumption vol.61, pp.6, 2011, https://doi.org/10.2478/s12175-011-0063-9
6. Complete moment convergence for moving average process generated by ρ − $\rho^{-}$ -mixing random variables vol.2015, pp.1, 2015, https://doi.org/10.1186/s13660-015-0766-5
7. Complete Convergence for Moving Average Process of Martingale Differences vol.2012, 2012, https://doi.org/10.1155/2012/128492


text:
## augmentation ideal and $\rho$-isotopic spaces

These are two questions regarding Rubins paper – Global units and Ideal class groups in Inventiones 1987. The two questions are from section 3.

Let $$p$$ be a rational prime and $$K \subset E \subset F$$, where $$F$$ is an abelian extension of $$K$$ containing the Hilbert Class Field of $$K$$, with $$G=Gal(F/K)$$. Let $$\rho$$ be an irreducible $$\mathbb{Z}_p$$-representation of $$\Delta = Gal(F/E)$$. For any $$\mathbb{Z}[\Delta]$$-module $$M$$, write $$M^{\rho}$$ for $$\rho$$-eigenspace of $$\hat{M}= \lim M/p^n M$$. Explicitly, $$M^{\rho}= e_{\rho}\hat{M}$$, where $$e_{\rho}$$ is the $$\rho$$-idempotent. Let $$W$$ be a submodule of $$(O_F^*)^{\rho}$$ such that $$(O_F^*)^{\rho}$$ has no $$\mathbb{Z}_p$$ torsion. Let $$N=p^n$$.

Since $$\rho \ne 1$$, I have two questions:

1) Why is $$(\mathbb{Z}/N\mathbb{Z})[G]^{\rho}$$ contained in the augmentation ideal of $$(\mathbb{Z}/N\mathbb{Z})[G]$$? I couldnt understand the action of $$\rho$$ on a $$(\mathbb{Z}/N\mathbb{Z})[G]$$-module $$M$$.

2) Let $$V=W/W^N$$. Why is $$O_K^* \cap V = 0$$?


text:
# september 30

$$\def\Real{\mathbb{R}}\def\Comp{\mathbb{C}}\def\Rat{\mathbb{Q}} \def\Field{\mathbb{F}} \def\Fun{\mathbf{Fun}} \def\e{\mathbf{e}} \def\f{\mathbf{f}}$$

#### Lyapunov’s direct method, cont’d

For complex spaces quadratic forms are not really suitable (if one wants just a real number as a result, the signatures are all $$(n,n)$$

Sylvester criterion for positive definiteness.
(Bonus: Rayleigh characterization of eigenvalues.)

#### Interlude: complex eigenvalues and eigenvectors.

Jordan normal forms in real vector spaces.

#### Quadratic forms as Lyapunov functions

If the operator defines an asymptotically stable system, it is Hurwitz. For a Hurwitz operator, a quadratic form exists which is a (strict) Lyapunov function. A strict Lyapunov function implies AS.

Lyapunov equation.

Discrete time systems.

#### Bounded Input Bounded Output

GES implies BIBO; not vice versa.


text:
# State Transition Matrix

by M11
Tags: matrix, state, transition
 P: 16 1. The problem statement, all variables and given/known data I am trying to find the state transition matrix of a 5x5 matrix. Is there a way to use MATLAB in this problem ? 2. Relevant equations $$\Phi$$ = (T^-1) x e^(Dt) x T Where D is 5x5 matrix containing the eigenvalues in the main diagonal, rest of elements are zeros. T is also 5x5 matrix contain the eigen vectors corresponding to the eign vectors in its columns. T^-1 is the inverse of T 3. The attempt at a solution I found the matrices using matlab. The problem is with the variable t . Is there a way to solve problem using matlab or any other software ?


text:
Holzmann Krauth 1999

Jump to: navigation, search

M. Holzmann, W. Krauth Transition temperature of the homogeneous, weakly interacting Bose gas Physical Review Letters 83 2687 (1999)

Abstract: We present a Monte Carlo calculation for up to $N \sim 20 000$ bosons in 3 D to determine the shift of the transition temperature due to small interactions a. We generate independent configurations of the ideal gas. At finite N, the superfluid density changes by a certain correlation function in the limit $a \to 0$ ; the $N \to \infty$ limit is taken afterwards. We argue that our result is independent of the order of limits. Detailed knowledge of the non-interacting system for finite $N$ allows us to avoid finite-size scaling assumptions.

Comment: This articles contains the direct-sampling algorithm for ideal bosons which is the work horse in the Markov-chain algorithm used in a number of papers.


text:
# Thread: trig functions: verifying identities

1. ## trig functions: verifying identities

how would you verify SIN(X)/ 1+COS(X) + SIN(X)/1-COS(X) = 2CSC(X)

2. Originally Posted by loganmeyers
how would you verify SIN(X)/ 1+COS(X) + SIN(X)/1-COS(X) = 2CSC(X)
start with the left hand side. you are going to manipulate it until you get the right hand side.

begin by combining the fractions, then simplify

3. Originally Posted by loganmeyers
SIN(X)/ 1+COS(X) + SIN(X)/1-COS(X) = 2CSC(X)
$\frac{{\sin x}}
{{1 + \cos x}} + \frac{{\sin x}}
{{1 - \cos x}} = \sin x\left( {\frac{1}
{{1 + \cos x}} + \frac{1}
{{1 - \cos x}}} \right) = \frac{{2\sin x}}
{{\sin ^2 x}}$

Can you take it from there?


text:
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Ellipsoidal coordinates are a three-dimensional orthogonal coordinate system $(\lambda, \mu, \nu)$ that generalizes the two-dimensional elliptic coordinate system. Unlike most three-dimensional orthogonal coordinate systems that feature quadratic coordinate surfaces, the ellipsoidal coordinate system is not produced by rotating or projecting any two-dimensional orthogonal coordinate system.

## Basic formulae

The Cartesian coordinates $(x, y, z)$ can be produced from the ellipsoidal coordinates $( \lambda, \mu, \nu )$ by the equations

$x^{2} = \frac{\left( a^{2} + \lambda \right) \left( a^{2} + \mu \right) \left( a^{2} + \nu \right)}{\left( a^{2} - b^{2} \right) \left( a^{2} - c^{2} \right)}$
$y^{2} = \frac{\left( b^{2} + \lambda \right) \left( b^{2} + \mu \right) \left( b^{2} + \nu \right)}{\left( b^{2} - a^{2} \right) \left( b^{2} - c^{2} \right)}$
$z^{2} = \frac{\left( c^{2} + \lambda \right) \left( c^{2} + \mu \right) \left( c^{2} + \nu \right)}{\left( c^{2} - b^{2} \right) \left( c^{2} - a^{2} \right)}$

where the following limits apply to the coordinates

$- \lambda < c^{2} < - \mu < b^{2} < -\nu < a^{2}.$

Consequently, surfaces of constant $\lambda$ are ellipsoids

$\frac{x^{2}}{a^{2} + \lambda} + \frac{y^{2}}{b^{2} + \lambda} + \frac{z^{2}}{c^{2} + \lambda} = 1,$

whereas surfaces of constant $\mu$ are hyperboloids of one sheet

$\frac{x^{2}}{a^{2} + \mu} + \frac{y^{2}}{b^{2} + \mu} + \frac{z^{2}}{c^{2} + \mu} = 1,$

because the last term in the lhs is negative, and surfaces of constant $\nu$ are hyperboloids of two sheets

$\frac{x^{2}}{a^{2} + \nu} + \frac{y^{2}}{b^{2} + \nu} + \frac{z^{2}}{c^{2} + \nu} = 1$

because the last two terms in the lhs are negative.

## Scale factors and differential operators

For brevity in the equations below, we introduce a function

$S(\sigma) \ \stackrel{\mathrm{def}}{=}\ \left( a^{2} + \sigma \right) \left( b^{2} + \sigma \right) \left( c^{2} + \sigma \right)$

where $\sigma$ can represent any of the three variables $(\lambda, \mu, \nu )$. Using this function, the scale factors can be written

$h_{\lambda} = \frac{1}{2} \sqrt{\frac{\left( \lambda - \mu \right) \left( \lambda - \nu\right)}{S(\lambda)}}$
$h_{\mu} = \frac{1}{2} \sqrt{\frac{\left( \mu - \lambda\right) \left( \mu - \nu\right)}{S(\mu)}}$
$h_{\nu} = \frac{1}{2} \sqrt{\frac{\left( \nu - \lambda\right) \left( \nu - \mu\right)}{S(\nu)}}$

Hence, the infinitesimal volume element equals

$dV = \frac{\left( \lambda - \mu \right) \left( \lambda - \nu \right) \left( \mu - \nu\right)}{8\sqrt{-S(\lambda) S(\mu) S(\nu)}} \ d\lambda d\mu d\nu$

and the Laplacian is defined by

$\nabla^{2} \Phi = \frac{4\sqrt{S(\lambda)}}{\left( \lambda - \mu \right) \left( \lambda - \nu\right)} \frac{\partial}{\partial \lambda} \left[ \sqrt{S(\lambda)} \frac{\partial \Phi}{\partial \lambda} \right] \ + \$
$\frac{4\sqrt{S(\mu)}}{\left( \mu - \lambda \right) \left( \mu - \nu\right)} \frac{\partial}{\partial \mu} \left[ \sqrt{S(\mu)} \frac{\partial \Phi}{\partial \mu} \right] \ + \ \frac{4\sqrt{S(\nu)}}{\left( \nu - \lambda \right) \left( \nu - \mu\right)} \frac{\partial}{\partial \nu} \left[ \sqrt{S(\nu)} \frac{\partial \Phi}{\partial \nu} \right]$

Other differential operators such as $\nabla \cdot \mathbf{F}$ and $\nabla \times \mathbf{F}$ can be expressed in the coordinates $(\lambda, \mu, \nu)$ by substituting the scale factors into the general formulae found in orthogonal coordinates.

• Focaloid (shell given by two coordinate surfaces)

## Bibliography

• Morse PM, Feshbach H (1953). Methods of Theoretical Physics, Part I. New York: McGraw-Hill. p. 663.
• Zwillinger D (1992). Handbook of Integration. Boston, MA: Jones and Bartlett. p. 114. ISBN 0-86720-293-9.
• Sauer R, Szabó I (1967). Mathematische Hilfsmittel des Ingenieurs. New York: Springer Verlag. pp. 101–102. LCCN 6725285 Check |lccn= value (help).
• Korn GA, Korn TM (1961). Mathematical Handbook for Scientists and Engineers. New York: McGraw-Hill. p. 176. LCCN 5914456 Check |lccn= value (help).
• Margenau H, Murphy GM (1956). The Mathematics of Physics and Chemistry. New York: D. van Nostrand. pp. 178–180. LCCN 5510911 Check |lccn= value (help).
• Moon PH, Spencer DE (1988). "Ellipsoidal Coordinates (η, θ, λ)". Field Theory Handbook, Including Coordinate Systems, Differential Equations, and Their Solutions (corrected 2nd ed., 3rd print ed.). New York: Springer Verlag. pp. 40–44 (Table 1.10). ISBN 0-387-02732-7.

### Unusual convention

• Landau LD, Lifshitz EM, Pitaevskii LP (1984). Electrodynamics of Continuous Media (Volume 8 of the Course of Theoretical Physics) (2nd ed.). New York: Pergamon Press. pp. 19–29. ISBN 978-0-7506-2634-7. Uses (ξ, η, ζ) coordinates that have the units of distance squared.
 266 videos foundNext >
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## Circuit Lab B/C

azboy1910
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### Re: Circuit Lab B/C

Umaroth wrote:
October 10th, 2020, 11:05 pm
azboy1910 wrote:
October 10th, 2020, 10:33 pm
1. 50 mA through 120-ohm resistor, 0 mA through 330-ohm resistor
2. 13 mA through both resistors
1. Your mistake here is that you're assuming that the full 6V is applied across the 120 ohm resistor.
Start by treating the circuit as if the diode weren't there, and then add the diode and see what changes you have to make.
Ok, so here's my new answer:

1. A zener diode acts as a regular diode would in forward bias. A geranium diode usually has a forward voltage of about 0.3 V. The 330-ohm resistor in parallel would also have a voltage drop of 0.3 V across it due to beeing in parallel with the zener diode, where the current here would be about 0.00091 A or 0.91 mA from $\frac{0.3}{330}$. The current across the 120-ohm reisistor would be the difference of the voltage of the voltage source and the zener diode voltage, which is 5.7 V.

$\frac{5.7}{120}$ = 0.0475 A or 47.5 mA

Answer: The current flowing through the 120-ohm resistor is 47.5 mA and the current flowing through the 330-ohm resistor is 0.91 mA mA.

2. $6(\frac{330}{120+330}) = 4.4$

4.4 V > 2.4 V -> Since the voltage across the diode is greater than its breakdown voltage, it will experience avalanche breakdown and current will flow through the diode. This current is used to regulate the voltage across the diode so that the voltage across is close to its breakdown voltage.

The current across the 120-ohm resistor can be represented by the expression $\frac{6 - 2.4}{120}$, since the node where current first flows through the zener diode reads 2.4 volts when there is a ground node set at the node on the other side. This current has a value of about 0.03 A, or 30 mA. Since the 330-ohm resistor is in parallel with the zener diode, the two both have the same voltage. From here, we just do $\frac{2.4}{330}$, giving us a current value of about 0.0073 A or 7.3 mA.

Answer: The current flowing through the 120-ohm resistor is 30 mA and the current flowing through the 330-ohm resistor is 7.3 mA.
Last edited by azboy1910 on October 11th, 2020, 9:18 am, edited 1 time in total.
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### Re: Circuit Lab B/C

azboy1910 wrote:
October 11th, 2020, 9:17 am
Umaroth wrote:
October 10th, 2020, 11:05 pm
azboy1910 wrote:
October 10th, 2020, 10:33 pm
1. 50 mA through 120-ohm resistor, 0 mA through 330-ohm resistor
2. 13 mA through both resistors
1. Your mistake here is that you're assuming that the full 6V is applied across the 120 ohm resistor.
Start by treating the circuit as if the diode weren't there, and then add the diode and see what changes you have to make.
Ok, so here's my new answer:

1. A zener diode acts as a regular diode would in forward bias. A geranium diode usually has a forward voltage of about 0.3 V. The 330-ohm resistor in parallel would also have a voltage drop of 0.3 V across it due to beeing in parallel with the zener diode, where the current here would be about 0.00091 A or 0.91 mA from $\frac{0.3}{330}$. The current across the 120-ohm reisistor would be the difference of the voltage of the voltage source and the zener diode voltage, which is 5.7 V.

$\frac{5.7}{120}$ = 0.0475 A or 47.5 mA

Answer: The current flowing through the 120-ohm resistor is 47.5 mA and the current flowing through the 330-ohm resistor is 0.91 mA mA.

2. $6(\frac{330}{120+330}) = 4.4$

4.4 V > 2.4 V -> Since the voltage across the diode is greater than its breakdown voltage, it will experience avalanche breakdown and current will flow through the diode. This current is used to regulate the voltage across the diode so that the voltage across is close to its breakdown voltage.

The current across the 120-ohm resistor can be represented by the expression $\frac{6 - 2.4}{120}$, since the node where current first flows through the zener diode reads 2.4 volts when there is a ground node set at the node on the other side. This current has a value of about 0.03 A, or 30 mA. Since the 330-ohm resistor is in parallel with the zener diode, the two both have the same voltage. From here, we just do $\frac{2.4}{330}$, giving us a current value of about 0.0073 A or 7.3 mA.

Answer: The current flowing through the 120-ohm resistor is 30 mA and the current flowing through the 330-ohm resistor is 7.3 mA.
 You got the 120 ohm resistor ones correct, but you're something important with the 330. What does a diode act like when a sufficient voltage is applied for charge to flow? In other words, what is the effective "resistance" of an ideal diode?
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### Re: Circuit Lab B/C

I believe the resistance of an ideal diode is 0 ohms, but what does this have to do with the current flowing through the 330-ohm resistor? I am confused here.
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### Re: Circuit Lab B/C

From the discussion on discord, azboy1910 is correct, though if you want to verify this in lab you'll need to either make sure your multimeter can measure low currents (less than 100uA for a good reading), or measure voltage across the resistors and compute the currents using ohm's law.

'If you're the smartest person in the room, you're in the wrong room' - Unknown

azboy1910
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### Re: Circuit Lab B/C

Alright, revive time!

Use correct significant figures and metric units for the following questions.

1. What is the energy stored in a 50.0 mF capacitor connected in series with a 12.0-volt battery and a 25.0-ohm resistor after 1.10 time constants?
2. Calculate the net force exerted on a proton when there is one electron in the east direction 300.0 cm away from it and another electron in the west direction 250.0 cm away from it.
3. A transformer has a ratio of 50:200 of the number of windings from the primary coil to the secondary coil. There is a voltage of 3.00 kV and a current of 20.0 mA across the primary windings. What is the current across the secondary coil in microamperes?
Last edited by azboy1910 on October 14th, 2020, 6:45 pm, edited 1 time in total.
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### Re: Circuit Lab B/C

azboy1910 wrote:
October 14th, 2020, 6:29 pm
Alright, revive time!

Use correct significant figures and metric units for the following questions.

1. What is the energy stored in a 50.0 mF capacitor connected in series with a 12.0-volt battery and a 25.0-ohm resistor after 1.10 time constants?
2. Calculate the net force exerted on a proton when there is one electron in the east direction 300.0 cm away from it and another electron in the west direction 250.0 cm away from it.
3. A transformer has a ratio of 50:200 of the number of windings from the primary coil to the secondary coil. There is a voltage of 3.00 kV and a current of 20.0 mA across the primary windings. What is the current across the secondary coil in microamperes?
Capacitors frustrate me
1. The voltage would be 12.0e^-1.10 = 3.99, the energy would be 0.5CV^2 = 0.399 J.
2. Using Coulomb's law twice, 1.536e-29 N west.
3. 20.0mA / 4 = 5.00 mA = 5000 microamperes.

azboy1910
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### Re: Circuit Lab B/C

UTF-8 U+6211 U+662F wrote:
October 14th, 2020, 7:00 pm
azboy1910 wrote:
October 14th, 2020, 6:29 pm
Alright, revive time!

Use correct significant figures and metric units for the following questions.

1. What is the energy stored in a 50.0 mF capacitor connected in series with a 12.0-volt battery and a 25.0-ohm resistor after 1.10 time constants?
2. Calculate the net force exerted on a proton when there is one electron in the east direction 300.0 cm away from it and another electron in the west direction 250.0 cm away from it.
3. A transformer has a ratio of 50:200 of the number of windings from the primary coil to the secondary coil. There is a voltage of 3.00 kV and a current of 20.0 mA across the primary windings. What is the current across the secondary coil in microamperes?
Capacitors frustrate me
1. The voltage would be 12.0e^-1.10 = 3.99, the energy would be 0.5CV^2 = 0.399 J.
2. Using Coulomb's law twice, 1.536e-29 N west.
3. 20.0mA / 4 = 5.00 mA = 5000 microamperes.

I agree with question 1, but I am not sure how you got your answers for 1 and 2. If you're willing, can you show how you got your answers? It may be possible that I am in the wrong, as I do sometimes get my own problems wrong, as you may have seen in other marathons.

Here's how I got mine.

Last edited by azboy1910 on October 15th, 2020, 9:22 am, edited 1 time in total.
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### Re: Circuit Lab B/C

azboy1910 wrote:
October 15th, 2020, 9:17 am
UTF-8 U+6211 U+662F wrote:
October 14th, 2020, 7:00 pm
azboy1910 wrote:
October 14th, 2020, 6:29 pm
Alright, revive time!

Use correct significant figures and metric units for the following questions.

1. What is the energy stored in a 50.0 mF capacitor connected in series with a 12.0-volt battery and a 25.0-ohm resistor after 1.10 time constants?
2. Calculate the net force exerted on a proton when there is one electron in the east direction 300.0 cm away from it and another electron in the west direction 250.0 cm away from it.
3. A transformer has a ratio of 50:200 of the number of windings from the primary coil to the secondary coil. There is a voltage of 3.00 kV and a current of 20.0 mA across the primary windings. What is the current across the secondary coil in microamperes?
Capacitors frustrate me
1. The voltage would be 12.0e^-1.10 = 3.99, the energy would be 0.5CV^2 = 0.399 J.
2. Using Coulomb's law twice, 1.536e-29 N west.
3. 20.0mA / 4 = 5.00 mA = 5000 microamperes.

I agree with question 1, but I am not sure how you got your answers for 1 and 2. If you're willing, can you show how you got your answers? It may be possible that I am in the wrong, as I do sometimes get my own problems wrong, as you may have seen in other marathons.

Here's how I got mine.

For 1, I did 12e^-1.10 instead of 12(1 - e^-1.10), whoops should've known better.
For 2, it looks like I forgot to square the denominator, which is a mistake I really should not be making.

1. Explain how you could make a 0-10 V voltmeter using a 0-5 mA galvanometer and resistors.
2. Explain how you could make a 0-10 A ammeter using a 0-5 mA galvanometer and resistors.
3. What is the name for a resistor that provides a very low resistance in order to the limit the amount of electric current passing around the component it is put in parallel with?
Last edited by UTF-8 U+6211 U+662F on October 15th, 2020, 6:04 pm, edited 1 time in total.

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# How to get properties of ColorDataFunction objects programmatically?

The display forms of ColorDataFunction objects (especially after one clicks on the white + sign on them) indicate that these objects have properties like Name, Class, etc. For example:

ColorData["StarryNightColors"]

ColorData[60]

I'd like to write a function that takes a ColorDataFunction object as an argument. How can this function access the properties of a ColorDataFunction object?

FWIW, expressions like the ones below won't work:

ColorData[60, "ColorFunction", "Properties"]

ColorData[60, "ColorFunction"]["Properties"]

• you can get somewhere inspecting FullForm and then addressing parts, eg ColorData[60, "ColorFunction"][[2]]->"Indexed". Not very satisfactory though. – george2079 May 4 '17 at 15:18
• Related: (76612) – Mr.Wizard May 23 '17 at 9:52

The ColorDataFunction is not atomic,

cdf = ColorData["StarryNightColors"];

AtomQ@cdf
(* False *)

so you have direct access to the arguments. Just replace the head with List and you have the four items listed in the box

List @@ cdf
(* {"StarryNightColors", "Gradients", {0, 1}, Blend["StarryNightColors", #1] &} *)

Strictly speaking, I am only talking about the most important property of ColorDataFunction, because you can use FullForm to get the other properties.

There is some additional information. ColorData[scheme] returns a ColorDataFunction. You can use the returned ColorDataFunction directly.

If you check the definition of ColorDataFunction, you will see only Function f really matters if you don't use illegal parameter as the input of ColorDataFunction/Function f.

In:

ColorData["Gradients"]

ColorData["AlpineColors"]
ColorData["AlpineColors"][x]
ColorData["AlpineColors"][1]

Manipulate[
ColorData["AlpineColors"][x],
{x, 0, 1, 0.01}]

Out:


text:
# Product of Ring Negatives

## Theorem

Let $\struct {R, +, \circ}$ be a ring.

Then:

$\forall x, y \in \struct {R, +, \circ}: \paren {-x} \circ \paren {-y} = x \circ y$

where $\paren {-x}$ denotes the negative of $x$.

## Proof

We have:

 $\displaystyle \paren {-x} \circ \paren {-y}$ $=$ $\displaystyle -\paren {x \circ \paren {-y} }$ Product with Ring Negative $\displaystyle$ $=$ $\displaystyle -\paren {-\paren {x \circ y} }$ Product with Ring Negative $\displaystyle$ $=$ $\displaystyle x \circ y$ Negative of Ring Negative

$\blacksquare$


text:
# A new prime sieve

by epsi00
Tags: primes sieve
P: 84
 Quote by baric Can't write code?!? aww, man.. that's the easy part! :)
well, if that's the easy part, you are welcome to experiment numerically with this sieve and take full credit for it. I am really curious to see how far we can go before we hit the rule of diminishing return. Obviously we can't keep improving the efficiency at little or no cost.
P: 12
 Quote by epsi00 well, if that's the easy part, you are welcome to experiment numerically with this sieve and take full credit for it. I am really curious to see how far we can go before we hit the rule of diminishing return. Obviously we can't keep improving the efficiency at little or no cost.
Like I said, it's totally dependent upon the computer's memory capacity although the size of the sieve does go up factorially with {p-1.. } where p is the series of primes, so you can't really get too far...

If you set the sieve limit to 1 billion, for example, your prime sieve stops at p(10)=29... i.e E!{p(1)-1*p(2)-1... *p(10)-1} = 1,021,870,080. At that point, you are filtering 84.2% of the integers. Adding 31 to the sieve increases the sieve size to 30 billion and raises the filtering to 84.7%, or an improvement of 3.33% (31/30). In addition, each entry in the sieve would occupy multiple bytes since we would be dealing with very large integers. (8 bytes each would allow representation of integers up to 1.8E19). So a sieve of 30 billion in size would require 240 gigabytes to maintain.

The realistic memory limit has to be around there because I doubt that anyone would want to designate several terabytes of memory to maintain a sieve of {2...37}

I guess you could write the program to query the heap space and automatically limit the sieve size to the computer's memory capacity. It would not be a bad way to do it since a 3% improvement over the factorization of 1.8E19 numbers is going to save a lot more time than it takes to set up the sieve.

Then you could simply throw more hardware (memory) at the problem to speed things up.

I'll see if I can write up some code for this tonight.
 P: 84 Sometimes one may be interested in sieving between two numbers (N1,N2) instead of from (1,N). This sieve can probably be adapted to do that ( I haven't thought seriously about this case ). It's just a matter of figuring out how to calculate the max element of each column for the sieve of (1,N) then start from those max element and calculate the elements of columns up to N.
 P: 1 This is an interesting variation of the wheel sieve, but I don't have any idea how efficiently it could be implemented versus a classic wheel. Especially if extended to larger residue class systems (i.e., mod 30 or mod 210). My own sieve currently uses wheel sieves up to mod 2310 (2*3*5*7*11) for sieving large ranges of the number line, and is one of the fastest sieve of eratosthenes implementations I'm aware of. I've spent enough time optimizing it that I don't have any interest in coding this variation, but I encourage you to learn how to do it yourself. I'll note that returns diminish sufficiently that mod 2310 is as high as I could go before the sieve speed actually started to decrease (mod 30030 involves sieving over too many residue classes).
 P: 5 If you want be happy, you can arrest the algorithm at $$\sqrt{n}$$.
 P: 20 Hey this is correct but you can actually reduce to 1/5

 Related Discussions Linear & Abstract Algebra 2 General Math 6 Programming & Computer Science 6 Computing & Technology 14


text:
## SectionSupplementary Problems

###### 1

What is the generating function for the number of ways to pass out $k$ pieces of candy from an unlimited supply of identical candy to $n$ children (where $n$ is fixed) so that each child gets between three and six pieces of candy (inclusive)? Use the fact that

\begin{equation*} (1+x+x^2+x^3)(1-x) = 1-x^4 \end{equation*}

to find a formula for the number of ways to pass out the candy.

###### 2
1. In paying off a mortgage loan with initial amount $A\text{,}$ annual interest rate $p$% on a monthly basis with a monthly payment of $m\text{,}$ what recurrence describes the amount owed after $n$ months of payments in terms of the amount owed after $n-1$ months? Some technical details: You make the first payment after one month. The amount of interest included in your monthly payment is $.01p/12\text{.}$ This interest rate is applied to the amount you owed immediately after making your last monthly payment.

2. Find a formula for the amount owed after $n$ months.

3. Find a formula for the number of months needed to bring the amount owed to zero. Another technical point: If you were to make the standard monthly payment $m$ in the last month, you might actually end up owing a negative amount of money. Therefore it is ok if the result of your formula for the number of months needed gives a non-integer number of months. The bank would just round up to the next integer and adjust your payment so your balance comes out to zero.

4. What should the monthly payment be to pay off the loan over a period of 30 years?

###### 3

We have said that for nonnegative $i$ and positive $n$ we want to define $\binom{-n}{i}$ to be $\binom{n+i-1}{i}\text{.}$ If we want the Pascal recurrence to be valid, how should we define $\binom{-n}{-i}$ when $n$ and $i$ are both positive?

###### 4

Find a recurrence relation for the number of ways to divide a convex $n$-gon into triangles by means of non-intersecting diagonals. How do these numbers relate to the Catalan numbers?

###### 5

How does $\sum_{k=0}^n\binom{n-k}{k}$ relate to the Fibonacci Numbers?

###### 6

Let $m$ and $n$ be fixed. Express the generating function for the number of $k$-element multisets of an $n$-element set such that no element appears more than $m$ times as a quotient of two polynomials. Use this expression to get a formula for the number of $k$-element multisets of an $n$-element set such that no element appears more than $m$ times.

###### 7

One natural but oversimplified model for the growth of a tree is that all new wood grows from the previous year's growth and is proportional to it in amount. To be more precise, assume that the (total) length of the new growth in a given year is the constant $c$ times the (total) length of new growth in the previous year. Write down a recurrence for the total length $a_n$ of all the branches of the tree at the end of growing season $n\text{.}$ Find the general solution to your recurrence relation. Assume that we begin with a one meter cutting of new wood (from the previous year) which branches out and grows a total of two meters of new wood in the first year. What will the total length of all the branches of the tree be at the end of $n$ years?

###### 8

(Relevant to Appendix C) We have some chairs which we are going to paint with red, white, blue, green, yellow and purple paint. Suppose that we may paint any number of chairs red or white, that we may paint at most one chair blue, at most three chairs green, only an even number of chairs yellow, and only a multiple of four chairs purple. In how many ways may we paint $n$ chairs?

###### 9

What is the generating function for the number of partitions of an integer in which each part is used at most $m$ times? Why is this also the generating function for partitions in which consecutive parts (in a decreasing list representation) differ by at most $m$ and the smallest part is also at most $m\text{?}$


text:
### Home > CCG > Chapter 3 > Lesson 3.1.2 > Problem3-18

3-18.

Use the method from problem 3-2 to enlarge the shape below from the origin by a zoom factor of $4$.

Your enlarged shape should be $12$ units tall and $16$ units wide.

Explore enlarging the shape with the eTool below.
Click the link at right for the full version of the eTool: 3-18 HW eTool (Desmos)


text:
## Monday, January 3, 2022

### Semantic text clustering: testing homogeneity of text clusters using Shannon entropy.

Many natural language processing (NLP) projects are based on the task of semantic text clustering. In short, we have a set of statements (texts, phrases, or sentences), and our goal is to group them semantically. Nowadays, this is quite a classical problem with a rich literature on clustering methods.

In spite of the simple formulation of the task, there are many problems during its realization. To make things more difficult, let's assume we are dealing with unlabeled data, which means we must rely on unsupervised techniques.

#### The first problem:

as we know, we can find many categories according to which we can try to group the given set of texts. In order to do this task well, we need to know the business case better. In other words, we need to know the question we want to answer using clustering. This problem is not the purpose of this note, so we will skip it.

### The second problem:

suppose we have successfully performed text clustering. And the question arises: how do we know that all clusters contain the correct texts ? Answering this question is the purpose of my note .

Let us assume that a given method has generated $N$ groups/clusters , each of which contains a set of similar texts . It is known that every cluster of similar texts is more or less homogeneous. There are clusters that are completely wrong in the sense of similarity of texts. So to answer our original question
How do we know that all clusters contain correct texts ?
is to find those clusters that contain erroneous texts.

### How to find the worst clusters ?

In this note, I would like to propose the following method for determining the worst clusters:
1. checking intra-cluster similarity by computing the Shanon entropy of the set of texts belonging to a given cluster.
2. using dynamically determined threshold entropy to select the worst clusters (based on the method presented in my blog https://dataanalyticsforall.blogspot.com/2021/05/semantic-search-too-many-or-too-few.html).
As data to illustrate our task I use data known as reuters-21578 (https://paperswithcode.com/dataset/reuters-21578).
Since the clustering stage is not our goal, this part of the work was done using the fast clustering method based on Agglomerative Clustering ( https://www.sbert.net/examples/applications/clustering/README.html#fast-clustering, code: https://github.com/UKPLab/sentence-transformers/blob/master/examples/applications/clustering/fast_clustering.py).

Each set of texts is characterized by a different degree of mutual homogeneity. The general idea of the method is to calculate the Shanon entropy for each identified text cluster, and then using the elbow rule to determine the homogeneity threshold (maximum entropy value) below which the clusters satisfy the homogeneity condition (are accepted). All clusters with entropy above the threshold should be discarded, used to find a better clustering algorithm or reanalyzed.
In any clustering method, we will get multiple single element clusters. Since in our method we check for homogeneity between sentences in a cluster, therefore I calculate entropy only for multi-element clusters.

The only parameters that needs to be introduced in the proposed method are:
1. the 'approximation_order' which defines a number of of consecutive characters (known as ngrams) from which we create the probability distribution used later to calculate the Shanon entropy.
2. The sensitivity parameter S (to adjust the aggressiveness in kneed detection [ kneed]) used in the Python kneed library.
The complete code is available at
https://github.com/Lobodzinski/Semantic_text_clustering__testing_homogeneity_of_text_clusters_using_Shannon-entropy.

To illustrate the method I show two pictures with the final results for two values of the parameter used to determine the Shannon entropy (approximation_order: 2 and 3)

All the values of the Shannon entropy approximations are ordered from smallest to largest (y-axis). The kneed library, given a parameter S (in this case S = 1), determines the best inflection point and this entropy value defines our maximum entropy value for the accepted sentence clusters. For a given approximation_order parameter (=2), we obtain 14 clusters which are inappropriately homogeneous.
Questionable clusters are:
Cluster 13, #10 Elements
Cluster 23, #8 Elements
Cluster 26, #8 Elements
Cluster 28, #8 Elements
Cluster 49, #6 Elements
Cluster 56, #6 Elements
Cluster 59, #5 Elements
Cluster 62, #5 Elements
Cluster 127, #4 Elements
Cluster 137, #3 Elements
Cluster 145, #3 Elements
Cluster 152, #3 Elements
Cluster 247, #3 Elements
Cluster 275, #3 Elements

For comparison, the next picture is calculated for approximation_order=3. In this case, we get 15 incorrect clusters:
Rejected clusters:
Cluster 0, #396 Elements
Cluster 13, #10 Elements
Cluster 23, #8 Elements
Cluster 26, #8 Elements
Cluster 28, #8 Elements
Cluster 56, #6 Elements
Cluster 59, #5 Elements
Cluster 62, #5 Elements
Cluster 109, #4 Elements
Cluster 127, #4 Elements
Cluster 137, #3 Elements
Cluster 145, #3 Elements
Cluster 152, #3 Elements
Cluster 179, #3 Elements
Cluster 275, #3 Elements

Comparing the list of unacceptable clusters may give more information about the occurring heterogeneities in our text set.

I encourage you to test the method for yourself. In my case it turned out to be very useful.

The code:
https://github.com/Lobodzinski/Semantic_text_clustering__another_way_to_study_cluster_homogeneity. The outstanding advantage of the method is that it works fully autonomously without the need for human intervention.