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Math Expressions Grade 3 Student Activity Book Unit 1 Lesson 6 Answer Key Building Fluency with 2s and 5s

This handy Math Expressions Grade 3 Student Activity Book Answer Key Unit 1 Lesson 6 Building Fluency with 2s and 5s provides detailed solutions for the textbook questions.

Math Expressions Grade 3 Student Activity Book Unit 1 Lesson 6 Building Fluency with 2s and 5s Answer Key

Use the Target

Question 1.
Discuss how you can use the Target to find the product for 8 × 5.
8 × 5 = 40,

Explanation:
Select number 8 in the column and 5 in the row of the target present in the green line, check the column of 5 and row of 8 meet, they meet at 40, hence 8 × 5 = 40.

Question 2.
Discuss how you can use the Target to practice division.
Yes we can use the target for even division,

Explanation:
For example select a random number 63, to give a division equation for 63, select the number in the green line given in the column which consists of 63 the number is 7, if 63 ÷ 7 we get the number which is in the green line of the row in which 63 is present then 63 ÷ 7 = 9.

Question 3.
Practice using the Target.
Let’s take an example and practice the target,
Question: 5 × 4 = 20,
Solution:
Select number 5 in the column and 4 in the row of the target present in the green line, check the column of 4 and row of 5 meet, they meet at 20, hence 5 × 4 = 20.

Question 4.
When using the Target, how are multiplication and division alike? How are they different?
When dealing with multiplication with the help of target if we select a number from row and a number from column and see where they meet and that come out as the product of the two numbers,
Example:
Question: 6 × 7 = 72,
Solution:
Select number 6 in the column and 7 in the row of the target present in the green line, check the column of number 7 and row of number 6 and find where they meet, they meet at 42, hence 6 × 7 = 42,
For division also we follow the same method of checking the columns and rows with help of the number where the column and row meet,
Example:
Question: 42 ÷ 7 = 6 or 42 ÷ 6 = 7,
Solution:
Select the number 42 from the target and when you see in the green line of this column we will have 7, when 42 ÷ 7 we get the answer as the number 6 which is in the green line of the row 42 or Select the number 42 from the target and when you see in the green line of this column we will have 6, when 42 ÷ 6 we get the answer as the number 7 which is in the green line of the row 42.

Make Sense of Problems

Write an equation and solve the problem.

Question 5.
Mrs. Cheng bought 8 pairs of mittens. How many individual mittens did she buy?
Mrs. Cheng bought 16 individual mittens,

Explanation:
Number of pairs of mittens = 8, Number of items in a pair = 2, Number of individual mittens Mrs. Cheng bought = Number of pairs of mittens × Number of items in a pair = 8 × 2 = 16.

Question 6.
Brian divided 10 crayons equally between his two sisters. How many crayons did each girl get?
Each sister of Brian got 5 crayons,

Explanation:
Number of crayons = 10, Number of sisters Brian has = 2, Number of crayons each sister get = $$\frac{Number of crayons}{Number of sisters Brian has}$$ = $$\frac{10}{2}$$ = 5.

Question 7.
Maria has 5 piles of flash cards. There are 9 cards in each pile. How many flash cards does Maria have?
Maria has 45 flash cards,

Explanation:
Number of piles = 5, Number of flash cards in each pile = 9, Total number of flash cards = Number of piles × Number of flash cards in each pile = 5 × 9 = 45.

Question 8.
A parking lot has 5 rows of parking spaces with the same number of spaces in each row. There are 35 parking spaces in the lot. How many spaces are in each row?
There are 7 spaces in each row,

Explanation:
Total number of spaces in the parking lot = 35, Number of rows = 5, Number of spaces in each row = $$\frac{Total number of spaces in the parking lot}{Number of rows}$$ = $$\frac{35}{5}$$ = 7.

Question 9.
Ari arranged his bottle cap collection into 9 rows with 2 bottle caps in each row. How many bottle caps are in his collection?
Ari collected a total of 18 bottle caps,

Explanation:
Number of rows = 9, Number of bottle caps in each row = 2, Total number of bottle caps = Number of rows × Number of bottle caps in each row = 9 × 2 = 18.

Write a Word Problem

Question 10.
Write a word problem that can be solved using the equation 45 ÷ 5 = where 5 is the number of groups.
Total number of apples Maria has = 45, Number of apples in each group = 5, Number of groups = 45 ÷ 5 = $$\frac{Total number of apples Maria has}{Number of apples in each group}$$ = $$\frac{45}{5}$$ = 9, Therefore there are 9 groups of apples.


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# Can "Infinity" ever be more than a mathematical abstraction?

Discussion in 'Physics & Math' started by Seattle, Jun 24, 2018.

1. ### someguy1Registered Senior Member

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686
Mathematically, it's not a function, it's a distribution. There are no infinitesimal intervals in the real numbers.

This is merely a heuristic characterization. The Dirac delta is not a function in the traditional sense as no function defined on the real numbers has these properties.[17] The Dirac delta function can be rigorously defined either as a distribution or as a measure.

https://en.wikipedia.org/wiki/Dirac_delta_function#Definitions

Surprised you didn't check on Wiki. It's perfectly obvious that it's not a mathematical function. The only question is the mathematical formalisms used to make it work out. I thought this was fairly well known. Then again, that's because I learned what I know about QM from the math side.

Now I can see you put "function" in quotes so you're indicating you know this. But it doesn't take an infinite value on an infinitesimal interval. That's the physicist's intuition, but it's not the mathematical formalism.

Last edited: Aug 13, 2018

3. ### arfa branecall me arfValued Senior Member

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5,711
I know that.
Then calculus is in big trouble. Can you prove there are no infinitesimal intervals in the real numbers?
Ah. And yet
If it doesn't take an infinite value on an infinitesimal interval, that's something you could demonstrate mathematically then?

5. ### someguy1Registered Senior Member

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686
I don't think so.

Of course. If $[x, y]$ is an interval of reals with $x \leq y$ then if $x < y$ the length of the interval is $y - x > 0$. And if $x = y$ then the length of the interval is zero. So the length of any interval of reals is either a positive real number or zero. That makes sense since there are no infinitesimals in the real numbers.

Yes, the Wiki page is pretty clear. The Dirac delta is not a function. But you say you know that. The physicists are free to think of it any way they want.

Last edited: Aug 13, 2018

7. ### arfa branecall me arfValued Senior Member

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5,711
There are plenty of online sources from universities which all define the Dirac delta function thus:

$\delta(x) = \begin{cases} \infty\; when\; x = 0 \\ 0\; when\; x \ne 0 \end{cases}; \int_{-\infty}^{+\infty} \delta(x) dx = 1$
It seems to unequivocally define it as being infinite when x = 0, and 0 otherwise. What's going on?

8. ### someguy1Registered Senior Member

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686
Hard to say since you didn't link your source. But the Wiki article is pretty clear. This isn't something I'm going to argue with you about.

9. ### someguy1Registered Senior Member

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ps -- How exactly do you evaluate that integral? Can you show me? On the contrary, since the value at a single point never changes the value of either a Riemann or Lebesgue integral, by your definition it must be zero. So what you wrote isn't logically consistent with delta being defined as a function. The only way to make the integral work out is by properly defining delta as a distribution, as is shown in the Wiki article.

Last edited: Aug 13, 2018
10. ### arfa branecall me arfValued Senior Member

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5,711
Almost. Apparently 0 is the only real infinitesimal.
And from nLab:
--https://ncatlab.org/nlab/show/infinitesimal number

I can agree with what that quote says, because in calculus you consider dx as an infinitesimal change in the value of x. Plenty of textbooks do this.
What they also do is consider dx to be indistinguishable from 0, although heuristically it's nonzero.

11. ### someguy1Registered Senior Member

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686
arfa brane You quoted me and generated an alert! Thank you!!

That article is crap. 0 is not an infinitesimal. An infinitesimal is a number that is greater than 0 and less than 1/n for every positive integer n. By that definition there are clearly no infinitesimals in the real numbers. The def you quoted is simply wrong. The bit about "infintieth" is garbage.

The site you quoted seems reputable. I can't account for their strange claims. Why are you aiming this stuff at me? I can't do anything about everything on the Internet. It's a category theory based site. Maybe they have a different definition.

12. ### someguy1Registered Senior Member

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686
ps -- I read farther into the article you quoted. They're not idiots, they are very aware of their choice and they explicitly justify it:

Traditionally, one adds to the definition the requirement that i≠0, but this leads to a less useful notion of the space of all infinitesimals. We allow 0 to be infinitesimal for some of the same reasons that we allow it to be a natural number
.

Ok, fine. They acknowledge that their definition is different from everyone else's; and they justify themselves. That's cool. What of it?

And remember these people are doing category theory. This is an entirely different way of thinking about mathematics and mathematical objects than people are used to. There is literally no accounting for how category theorists will express basic notions in order to better fit their framework. There are a lot of oddities. https://en.wikipedia.org/wiki/Category_theory

Last edited: Aug 14, 2018
13. ### arfa branecall me arfValued Senior Member

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5,711
But "they" are not the only source I've seen that claims 0 is the only infinitesimal real number. Another example is here.
Of course anyone could google "zero and infinitesimal", say.

You seem to be implying that the notion of an infinitesimal depends on how it's defined. Or perhaps the best conclusion a student of calculus can make is that an infinitesimal is only a heuristic? Unless they happen on Robinson's extensions to the reals--nonstandard numbers?

14. ### someguy1Registered Senior Member

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686
Um ... yeah. Doesn't everything in math? The notion of the number 3 depends on how it's defined. Are you some kind of Platonist about infinitesimals? More power to you. Was the quote I bolded above unclear? The authors acknowledge that they changed the usual definition of infinitesimal for their own reasons. Changing the definition changes the notion, or at least provides a variation on the notion.

Student of calculus? Pull down the exponent and subtract 1. Chain rule. Product rule. Integration by parts. Related rate problems. And underneath it all, limits. Not infinitesimals. Limits. Epsilons and deltas.

The material you're discussing is not part of the subject of calculus. Philosophy and history of calculus, maybe, but even then you're pretty far afield.

I'm glad you read a Wiki page on those. What do you want to know about them?

Last edited: Aug 14, 2018

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17. ### arfa branecall me arfValued Senior Member

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Maybe. The reals form a ring under addition and multiplication, right? It's a commutative ring. Addition has inverses (negative reals).
$\mathbb R$ also has well-defined relations: <, >, =, ...

Such that (this is pretty obvious) if a and b are real, then if a < b, a - b < 0; if a > b, a - b > 0; if a = b, a - b = 0 (and -b is the additive inverse of a).
Using the relations you can prove 0 is the only solution where -a < x < a for any real a.
The real number line is also called Archimedean, when it's a commutative ring.
So, is it? and why would it make a difference?

18. ### someguy1Registered Senior Member

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686
Making a mountain out of a molehill. It's like arguing whether zero is a natural number. It makes no difference. If you like you could just RENAME 0 to 1 and 1 to 2 and 2 to 3 and so forth and nothing would change. I agree that having a zero is the best thing since sliced bread when it comes to algebra, but it's not nearly good enough. Nobody ever goes, "Oh we can't do algebra in the natural numbers unless we include 0." Because even with 0, the arithmetic of the natural numbers doesn't have all the nice group, ring, and module properties. For that you need the integers, the positive and negative numbers and zero. So the argument over whether to include 0 is TRIVIAL and so is the argument about adding 0 to the infinitesimals.

arfa brane You did show me something I didn't know, which is that SOME authors include 0 with the infinitesimals. And Keisler's name I know well, he wrote the first calculus text based on the hyperreals. So I learned something.

But you are making much more of it than is warranted. [Of course that's what I WOULD say, having just lost the point! LOL]

No model of the real numbers that contains infinitesimals can be Archimedean. I don't have a link to the proof at hand. Maybe I'll find it later. Being a commutative ring has nothing to do with an ordered set being Archimedean. Commutative rings are essentially generalizations of the integers. They're about algebraic properties, not continuous properties or order properties.

ps -- Ah here is the proof that the hyperreals are not Archimedean. https://en.wikipedia.org/wiki/Hyperreal_number

In the hyperreals there is an element $\omega$ such that $1<\omega, \ \ 1 + 1 < \omega, \ \ 1 + 1 + 1 < \omega$, etc. That violates the definition of an Archimedean order.

Last edited: Aug 14, 2018
19. ### arfa branecall me arfValued Senior Member

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5,711
Introduce a relation: ≈, "infinitely close to".

So that if a ≈ b, then a - b ≈ 0. You also have if a ≈ 0 and b ≈ 0, then a + b ≈ 0.
Multiplication means if at least one of a,b ≈ 0, then ab ≈ 0. Also if b ≈ 0, then 1/b is almost infinite. 1/0 is (still) not defined.

This "new" relation changes the structure of the ring somewhat, it redefines addition, multiplication and their inverses.

Just sayin'

Write4U likes this.
20. ### someguy1Registered Senior Member

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686
That's what they do in expositions of the hyperreals, but so what? You're just sayin', but what are you just sayin'?

21. ### Write4UValued Senior Member

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7,597
I love the logic.
Would that be a 2 dimensional linear measurement? If so, is there a vertical version of that symbol indicating "infinitely close" in a 3 dimensional space?

22. ### arfa branecall me arfValued Senior Member

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5,711
What am I just sayin'? Maybe nothing so much. But, here we are in a thread about the mathematical abstraction: infinity. And we're talking about physical infinity (whatever that is).

Both of these things have occupied minds for thousands of years. Do we now have all the mathematical tools we need? Is physics done and dusted? Who really knows? I don't think sciforums is going to decide that one . . .

Even in that definition of (what today is called) Dirac's delta function, which isn't a function; look at what might be called the restriction: $\int_{-\infty}^{+\infty} \delta(x) dx = 1$

The limits of integration are infinite. That is, the integral is over the whole of $\mathbb R$.
I can still remember a question some student asked in 1st year calc: What do the infinite limits mean? The lecturer said something like, in this case you can consider $-\infty$ and $+\infty$ to be equivalent (??). That is, you consider the real line to be extended indefinitely to the left and to the right, such that both ends "go to infinity", i.e. the same place. (like the real line meets itself at infinity, or some such).

In fact, the real line plus a point at infinity is a well-known construction.

23. ### someguy1Registered Senior Member

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686
You never seem to have a point or a thesis. You communicate like a bot.

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Henrik Ljungstrand
Posts: 1
Joined: Mon Mar 08, 2021 7:00 am

Hello everyone, i am pretty new to Sagittal, less new to xenharmonic alliance or the regular tuning paradigm. Some musical experience, mostly from church choir and amateur playing, nothing professional. Would say i have a good musical ear and a keen mathematical mind though.

After having noticed that standard Sagittal notation for 13-limit JI uses the symbols , , , and for 81/80, 64/63, 33/32 and 27/26 (down, down, up, down) as standards, and noticing that 33/32 is almost exactly a fourth root of 9/8 and 27/26 is almost exactly a cube root of 9/8, thus being likely to be perceived as a quarter tone and a third tone respectively, i wondered about what temperament we would get by setting 27/26 = 26/25 = 25/24 (thus (27/26)^3 = 9/8) and setting 36/35 = 33/32 and 36/35 * 33/32 = 35/33 (thus (33/32)^4 = 9/8).
The answer is Quadritikleismic temperament, a rank 2 (linear) 13-limit temperament, supported by 72edo, 68edo and 140edo.
Having commas 676/675, 625/624, 385/384 and 9801/9800 tempered to unison, among many others.

I now wonder which accidentals would be best for using in this temperament, besides and ? It should preferrably be a good compromise between 72edo and 68edo, and possibly 140edo.
Are and still good alternatives for 5 and 7 commas, or are there better ones? Should the 11 and 13 commas also be changed perhaps?

The idea is for this temperament to possibly be a good microtemperament for 13-limit music performance, especially real time performance, if there is any interest for such in 13-limit almost JI based on diatonicism plus pretty simple extensions.
Just an idea, i haven't really had the opportunity to train myself very much in any instrument except voice yet, although planning to learn recorder, piano/synth and guitar better, and hopefully be able to retune such instruments to e.g. Quadritikleismic, and other tunings/temperaments.

Dave Keenan
Posts: 2092
Joined: Tue Sep 01, 2015 2:59 pm
Location: Brisbane, Queensland, Australia
Contact:

Hi Henrik. Welcome to the Sagittal forum.

the 5-comma 81/80 as 1/9-tone,
the 7-comma 64/63 as 1/8-tone,
the 11-M-diesis 33/32 as 1/4-tone,
the 13-L-diesis 27/26 as 1/3-tone,
where 1 tone is 9/8.

422ef-ET certainly covers it (having 72 steps to the 9/8) but of course it begs the question: Why not use the best approximations of 33/32 and 27/26 at 19 and 23 steps (rather than the exact 1/4-tone and 1/3-tone at 18 and 24 steps)?

So understandably you'd like a higher-rank temperament that would effectively let you choose a subset of 422ef-ET that still gives say a 13-limit diamond, or some useful number of 13-limit consonances, without too many notes.

I'm not convinced that Quadritikleismic is what you want here. Can you explain in what sense it satisfies your above specification? I see that it satifies the last two criteria (re 11 and 13). But I don't see that it satisfies the first two (re 5 and 7).

POTE generator: 316.989
Map: [ 4 0 4 7 17 0], 0 6 5 4 -3 14] ]
EDOs: 68, 72, 140, 212

But I decided to have a look at a notation for 140-edo anyway.

I assume you've seen the standard Sagittal notations for 68 and 72, e.g. on the Periodic Table.

140edo has the same best fifth as 70edo and has CD=24, EF=10, #=14. The following proposal has valid flag arithmetic. I was forced to use primes 19 and 17 to get symbols for 1 and 2 steps.

140edo
Steps   Symbol  Comma  Other valid symbols and their commas
-----------------------------------------------------------
1		19-schisma
2		1/17-kleisma
3		1/5-comma		 1/7C	 55C
4		19/5-comma		 19/7C
5		7/11-comma
6		11-medium-diesis	 1/35M, 13M
7		5/49-medium-diesis
8		1/11-large-dieisis	 35L, 1/13L

But that's not a notation for Quadritikleismic and I don't think it really helps you get to one.

herman.miller
Posts: 33
Joined: Sun Sep 06, 2015 8:27 am

One thing to note if you're wanting to notate quadritikleismic is that (64/63) and (55/54) are equivalent in this temperament: both work out to be [+1, -16>. This is also the case with the combinations (36/35) and (33/32) = [+0, +3>, as well as (729/704) vs. (8505/8192) = [-2, +39>. So you might want to pick one or the other flag in each of these three pairs to make the addition easier to remember. But note that and map to different intervals.

I haven't looked into quadritikleismic, but with a generator of 17 cents, you're going to need a lot of accidentals for the small intervals. Here's some options:

[+0, +1> (17.09): 99/98
[+0, +2> (34.17): 49/48
[+0, +3> (51.26): 36/35 or 33/32
[+0, +4> (68.35): 25/24
[+0, +5> (85.43): 21/20
[+0, +6> (102.52): 35/33 (approx.)

Note that at least with the TOP tuning of 17.086 cent generators with a 300.00 cent period, the schisma accent is not negligible, but it's still a relatively small interval at 5.584 cents. You might be able to leave out the schisma accents if you don't stray too far from your reference pitch.

Dave Keenan
Posts: 2092
Joined: Tue Sep 01, 2015 2:59 pm
Location: Brisbane, Queensland, Australia
Contact:


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# What is the derivative of the given function?

Which of these are correct form for $\large \frac{d}{dx} (4.7^{x+2.4})$? There may be more than one correct answer

    1. 4.7^(2.4) * 4.7^(x)ln*(4.7)
2. (x+2.4)(4.7^(x+1.4))
3. 2.4(4.7^(x+1.4))
4. (4.7^(x+2.4))ln(4.7)
5. 2.4(4.7^(x+1.4))ln(4.7)
6. All are correct
7. None are correct

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Welcome to MSE! It really helps readability to format thing using MathJax. I did part of it to show you, but it would help to format questions this way! Regards –  Amzoti Feb 1 '13 at 18:53

Since this is in the form of $y=a^{f(x)}$ to derive this its just $y'= (a^x)\cdot\ln x\cdot f'(x)$

so in your case we have $y=(4.7)^{(x+2.4)}$ and now if we apply that rule we will get

$$y' = [(4.7)^{(x+2.4)}]\cdot\ln(4.7)\cdot(1) = \left[(4.7)^{(x+2.4)}\right]\cdot\ln(4.7)$$

so the answer is $4$ but if you notice closely we can break up the exponent by the exponent laws

to obtain $\left(4.7^{2.4}\right)\cdot\left(4.7^x\right)\cdot(\ln 4.7)$ and this is answer 1 :)

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thank you! I had gotten 1 as an answer and wasnt really sure if any of the other ones were the same. :) –  user56852 Feb 1 '13 at 20:11
thanks brian for editing it, i wanted to know whats the language you guys use on here? latex? –  MathGeek Feb 2 '13 at 4:41

Hint: What is $4.7^{x + 2.4}$ written in terms of $\exp$ and $\ln$?


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# Correlated equilibrium

Correlated equilibrium
A solution concept in game theory
Relationship
Superset ofNash equilibrium
Significance
Proposed byRobert Aumann
ExampleChicken

In game theory, a correlated equilibrium is a solution concept that is more general than the well known Nash equilibrium. It was first discussed by mathematician Robert Aumann in 1974.[1][2] The idea is that each player chooses their action according to their observation of the value of the same public signal. A strategy assigns an action to every possible observation a player can make. If no player would want to deviate from the recommended strategy (assuming the others don't deviate), the distribution is called a correlated equilibrium.

## Formal definition

An ${\displaystyle N}$-player strategic game ${\displaystyle \displaystyle (N,A_{i},u_{i})}$ is characterized by an action set ${\displaystyle A_{i}}$ and utility function ${\displaystyle u_{i}}$ for each player ${\displaystyle i}$. When player ${\displaystyle i}$ chooses strategy ${\displaystyle a_{i}\in A_{i}}$ and the remaining players choose a strategy profile described by the ${\displaystyle N-1}$-tuple ${\displaystyle a_{-i}}$, then player ${\displaystyle i}$'s utility is ${\displaystyle \displaystyle u_{i}(a_{i},a_{-i})}$.

A strategy modification for player ${\displaystyle i}$ is a function ${\displaystyle \phi _{i}\colon A_{i}\to A_{i}}$. That is, ${\displaystyle \phi _{i}}$ tells player ${\displaystyle i}$ to modify his behavior by playing action ${\displaystyle \phi _{i}(a_{i})}$ when instructed to play ${\displaystyle a_{i}}$.

Let ${\displaystyle (\Omega ,\pi )}$ be a countable probability space. For each player ${\displaystyle i}$, let ${\displaystyle P_{i}}$ be his information partition, ${\displaystyle q_{i}}$ be ${\displaystyle i}$'s posterior and let ${\displaystyle s_{i}\colon \Omega \rightarrow A_{i}}$, assigning the same value to states in the same cell of ${\displaystyle i}$'s information partition. Then ${\displaystyle ((\Omega ,\pi ),P_{i},s_{i})}$ is a correlated equilibrium of the strategic game ${\displaystyle (N,A_{i},u_{i})}$ if for every player ${\displaystyle i}$ and for every strategy modification ${\displaystyle \phi _{i}}$:

${\displaystyle \sum _{\omega \in \Omega }q_{i}(\omega )u_{i}(s_{i}(\omega ),s_{-i}(\omega ))\geq \sum _{\omega \in \Omega }q_{i}(\omega )u_{i}\left(\phi _{i}\left(s_{i}(\omega )\right),s_{-i}(\omega )\right)}$

In other words, ${\displaystyle ((\Omega ,\pi ),P_{i})}$ is a correlated equilibrium if no player can improve his or her expected utility via a strategy modification.

## An example

 Dare Chicken out Dare 0, 0 7, 2 Chicken out 2, 7 6, 6 A game of Chicken

Consider the game of chicken pictured. In this game two individuals are challenging each other to a contest where each can either dare or chicken out. If one is going to Dare, it is better for the other to chicken out. But if one is going to chicken out it is better for the other to Dare. This leads to an interesting situation where each wants to dare, but only if the other might chicken out.

In this game, there are three Nash equilibria. The two pure strategy Nash equilibria are (D, C) and (C, D). There is also a mixed strategy equilibrium where each player Dares with probability 1/3.

Now consider a third party (or some natural event) that draws one of three cards labeled: (C, C), (D, C), and (C, D), with the same probability, i.e. probability 1/3 for each card. After drawing the card the third party informs the players of the strategy assigned to them on the card (but not the strategy assigned to their opponent). Suppose a player is assigned D, he would not want to deviate supposing the other player played their assigned strategy since he will get 7 (the highest payoff possible). Suppose a player is assigned C. Then the other player will play C with probability 1/2 and D with probability 1/2. The expected utility of Daring is 7(1/2) + 0(1/2) = 3.5 and the expected utility of chickening out is 2(1/2) + 6(1/2) = 4. So, the player would prefer chickening out.

Since neither player has an incentive to deviate, this is a correlated equilibrium. The expected payoff for this equilibrium is 7(1/3) + 2(1/3) + 6(1/3) = 5 which is higher than the expected payoff of the mixed strategy Nash equilibrium.

The following correlated equilibrium has an even higher payoff to both players: Recommend (C, C) with probability 1/2, and (D, C) and (C, D) with probability 1/4 each. Then when a player is recommended to play C, she knows that the other player will play D with (conditional) probability 1/3 and C with probability 2/3, and gets expected payoff 14/3, which is equal to (not less than) the expected payoff when she plays D. In this correlated equilibrium, both players get 5.25 in expectation. It can be shown that this is the correlated equilibrium with maximal sum of expected payoffs to the two players.

## Learning correlated equilibria

One of the advantages of correlated equilibria is that they are computationally less expensive than Nash equilibria. This can be captured by the fact that computing a correlated equilibrium only requires solving a linear program whereas solving a Nash equilibrium requires finding its fixed point completely.[3] Another way of seeing this is that it is possible for two players to respond to each other's historical plays of a game and end up converging to a correlated equilibrium.[4]

## References

1. ^ Aumann, Robert (1974). "Subjectivity and correlation in randomized strategies". Journal of Mathematical Economics. 1 (1): 67–96. CiteSeerX 10.1.1.120.1740. doi:10.1016/0304-4068(74)90037-8.
2. ^ Aumann, Robert (1987). "Correlated Equilibrium as an Expression of Bayesian Rationality". Econometrica. 55 (1): 1–18. CiteSeerX 10.1.1.295.4243. doi:10.2307/1911154. JSTOR 1911154.
3. ^ Papadimitriou, Christos H.; Roughgarden, Tim (2008). "Computing correlated equilibria in multi-player games". J. ACM. 55 (3): 14:1–14:29. CiteSeerX 10.1.1.335.2634. doi:10.1145/1379759.1379762.
4. ^ Foster, Dean P.; Vohra, Rakesh V. (1996). "Calibrated Learning and Correlated Equilibrium". Games and Economic Behavior.


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# Math Help - How to find point of interesection between sin x and sin 0.5x?

1. ## How to find point of interesection between sin x and sin 0.5x?

I'm having trouble manually working out points of intersection between trig graphs. I can do them when its between a cos graph and a sin graph using cos^2(x) + sin^2(x) = 1 but I cant do them when they are the same etc. sin x = sin 2x

How do you manually work these pts of intersection out? Do I have to use the general formula?

2. ## Re: How to find point of interesection between sin x and sin 0.5x?

You can use the fact that sin(2x) - sin(x) = 2sin(x/2)cos(3x/2).

3. ## Re: How to find point of interesection between sin x and sin 0.5x?

Hello, sorkii!

$\text{How to find the intersections of }\,y \:=\:\sin x\,\text{ and }\,y \,=\,\sin\tfrac{x}{2}$

We wish to solve: . $\sin x \:=\:\sin\tfrac{x}{2}$

Recall the identity: . $\sin\theta \:=\:2\sin\tfrac{\theta}{2}\cos\tfrac{\theta}{2}$

So we have: . $2\sin\tfrac{x}{2}\cos\tfrac{x}{2} \:=\:\sin\tfrac{x}{2} \quad\Rightarrow\quad 2\sin\tfrac{x}{2}\cos\tfrac{x}{2} - \sin\tfrac{x}{2} \;=\;0$

. . . . . . $\sin\tfrac{x}{2}\big[2\cos\tfrac{x}{2} - 1\big] \;=\;0$

Hence: . $\sin\tfrac{x}{2} \:=\:0 \quad\Rightarrow\quad \tfrac{x}{2} \:=\:\pi n \quad\Rightarrow\quad \boxed{x \:=\:2\pi n}$

And: . $2\cos\tfrac{x}{2}-1 \:=\:0 \quad\Rightarrow\quad \cos\tfrac{x}{2} \:=\:\tfrac{1}{2}$

. . . . . $\tfrac{x}{2} \:=\:\pm\tfrac{\pi}{3}+ 2\pi n \quad\Rightarrow\quad \boxed{x \:=\: \pm\tfrac{2\pi}{3} + 4\pi n}$


text:
# Root objects and the simplest possible analytic continuation of the Riemann zeta function.

The equation I am trying to solve is:

$$\lim\limits_{k \rightarrow 3} \left( \sum\limits_{n=1}^{n=k} \frac{1}{n^s}+ \frac{1}{k^{s - 1} \cdot (s - 1)}\right)=0 \tag{1}$$

The simplest possible analytic continuation of the Riemann zeta function is: $$\zeta(s)=\lim\limits_{k \rightarrow \infty} \left( \sum\limits_{n=1}^{n=k} \frac{1}{n^s}+ \frac{1}{k^{s - 1} \cdot (s - 1)}\right) \tag{2}$$ $$\mbox{ which appears to be true for }\Re(s)>0$$

So therefore I substituted all $$s$$ with $$x$$ except one of them like this: $$\lim\limits_{k \rightarrow 3} \left( \sum\limits_{n=1}^{n=k} \frac{1}{n^x}+ \frac{1}{k^{x - 1} \cdot (s - 1)}\right)=0 \tag{3}$$

Very crude rational approximations of logarithms are:

$$\log(1) = 0$$
$$\log(2) \approx 7/10$$
$$\log(3) \approx 11/10$$
(That is the level of precision I could afford computationally in this case.)

Notice that: $$\frac{1}{n^x}=\frac{1}{e^{x\log(n)}} \tag{4}$$
and substitute $$\log(n)$$ with the rational approximations for logarithms above.

Solving $$(3)$$ in Mathematica we can then write:

Clear[x, s];
Reduce[1/(E^Round[N[Log[1]], 10^-1])^x +
1/(E^Round[N[Log[2]], 10^-1])^x +
1/(E^Round[N[Log[3]], 10^-1])^x +
1/(E^Round[N[Log[3]], 10^-1])^(x - 1)/(s - 1) == 0, x]

This gives 11 Root objects subject to conditions. Picking the first Root object that Mathematica gives, we have:

x == 10 (2 I \[Pi] C[1] +
Log[Root[-1 + E^(11/10) + s + (-1 + s) #1^4 + (-1 + s) #1^11 &,
1]])

Latexifying it does not help much, but the changes I would do are to replace $$x$$ with $$s$$ and skip the term $$2 i \pi c_1$$ since I have understood that $$c_1$$ is an integer that can be zero.

So the equation that needs to be solved is:

s == 10 (Log[
Root[-1 + E^(11/10) + s + (-1 + s) #1^4 + (-1 + s) #1^11 &, 1]])

Dividing by 10:

s/10 == Log[
Root[-1 + E^(11/10) + s + (-1 + s) #1^4 + (-1 + s) #1^11 &, 1]]

Applying the exponential function we would have:

Exp[s/10] ==
Root[-1 + E^(11/10) + s + (-1 + s) #1^4 + (-1 + s) #1^11 &, 1]

Now this is not solvable in Mathematica so we need a truncated series expansion of $$\exp(s)$$

$$\exp(s/10) \approx 1+s/10+\frac{(s/10)^2}{2}+\frac{(s/10)^3}{6}$$

$$1+\frac{s}{10}+\frac{1}{2} \left(\frac{s}{10}\right)^2+\frac{1}{6} \left(\frac{s}{10}\right)^3=\text{Root}\left[\text{\#1}^{11} (s-1)+\text{\#1}^4 (s-1)+s+e^{11/10}-1\&,1\right] \tag{5}$$

and this Mathematica can solve:

Reduce[(1 + s/10 + (s/10)^2/2 + (s/10)^3/6) ==
Root[-1 + E^(11/10) + s + (-1 + s) #1^4 + (-1 + s) #1^11 &, 1], s]

giving the first Root object starting as:

(-1 + Root[-1088391168000000000000000000000000000000000 +
362797056000000000000000000000000000000000 E^(11/10) +
544195584000000000000000000000000000000000 #1 +
295679600640000000000000000000000000000000 #1^2 + ...

Is this at all true or is it just overly complicated?

I mean we could use the same minimal analytic continuation of the zeta function and the crude rational approximations of the logarithms and only do the series expansion:

Clear[x, s];
Series[1/(E^Round[N[Log[1]], 10^-1])^s +
1/(E^Round[N[Log[2]], 10^-1])^s + 1/(E^Round[N[Log[3]], 10^-1])^s +
1/(E^Round[N[Log[3]], 10^-1])^(s - 1)/(s - 1), {s, 0, 5}]

This then would give:

$$\left(3-e^{11/10}\right)+\frac{1}{10} \left(-18+e^{11/10}\right) s+\left(\frac{17}{20}-\frac{101 e^{11/10}}{200}\right) s^2+\left(-\frac{279}{1000}-\frac{1699 e^{11/10}}{6000}\right) s^3+\left(\frac{8521}{120000}-\frac{82601 e^{11/10}}{240000}\right) s^4+\left(-\frac{29643}{2000000}-\frac{3968999 e^{11/10}}{12000000}\right) s^5+O[s]^6$$ and so on...

• Your first line doesn't make sense. Did you mean finding the zeros of $F(s)=\sum_{n=1}^3 n^{-s}+ 3^{1-s}/(s-1)$ ? If so the first step is to find $a,b$ such that for $\Re(s) \not \in [-a,b]$ then $F(s) \ne 0$ – reuns Mar 24 at 13:30

For $$k=3$$:

$$\sum _{n=1}^k \frac{1}{n^s}+\frac{1}{k^{s-1} (s-1)}=1+2^{-s}+3^{-s}+\frac{1}{3^{s-1} (s-1)}$$

For $$s=0$$:

$$1+2^{-s}+3^{-s}+\frac{1}{3^{s-1} (s-1)}=0$$

Note for $$s=0$$, $$\sum\limits_{n=1}^k \frac{1}{n^s}+\frac{1}{k^{s-1} (s-1)}=0$$ for all values of $$k$$.

For $$F(s)=1+2^{-s}+3^{-s}+\frac{1}{3^{s-1} (s-1)}$$, the contour plots of $$\Re(F(s))=0$$ (blue curve) and $$\Im(F(s))=0$$ (orange curve) in Figure (1) below suggest the location of the complex roots of $$F(s)$$ are in a vertical strip centered near the origin. In Figure (1) below, the horizontal axis is the real axis and the vertical axis is the imaginary axis.

Figure (1): Contour Plots of $$\Re(F(s))=0$$ and $$\Im(F(s))=0$$ in blue and orange respectively

For $$G(s)=\sum _{n=1}^k \frac{1}{n^s}+\frac{1}{k^{s-1} (s-1)}$$, the contour plots of $$\Re(G(s))=0$$ (blue curve) and $$\Im(G(s))=0$$ (orange curve) in Figures (2) and (3) below illustrate how four of the zeros of $$G(s)$$ move closer to the first four non-trivial zeta zeros in the upper-half plane (red discrete portion of the plots) as the evaluation limit is increased from $$k=3$$ in Figure (2) to $$k=100$$ in Figure (3).

Figure (2): Contour Plots of $$\Re(G(s))=0$$ and $$\Im(G(s))=0$$ in blue and orange respectively with $$k=3$$

Figure (3): Contour Plots of $$\Re(G(s))=0$$ and $$\Im(G(s))=0$$ in blue and orange respectively with $$k=100$$

The following four figures illustrate how the four zeros of $$G(s)$$ associated with the first four non-trivial zeta zeros in the upper-half plane ($$\rho_1$$ to $$\rho_4$$) migrate from their initial values at $$k=3$$ to their values at $$k=100$$ in steps of $$\Delta_k=1$$.

Figure (4): Evaluation of $$G(s)$$ zero associated with $$\rho_4$$ from $$k=3$$ to $$k=100$$

Figure (5): Evaluation of $$G(s)$$ zero associated with $$\rho_3$$ from $$k=3$$ to $$k=100$$

Figure (6): Evaluation of $$G(s)$$ zero associated with $$\rho_2$$ from $$k=3$$ to $$k=100$$

Figure (7): Evaluation of $$G(s)$$ zero associated with $$\rho_1$$ from $$k=3$$ to $$k=100$$

Another fairly simple formula I've found useful is the following (see Riemann zeta function - Dirichlet series and Chapter 2 of Part II of "Theory of Functions" by Konrad Knopp):

$$\zeta(s)=\frac{1}{s-1}\sum_{k=1}^\infty\left(\frac{k}{(k+1)^s}-\frac{k-s}{k^s}\right),\quad \Re(s)>0$$

• The proof of $\zeta(s)=\lim\limits_{k \rightarrow \infty} \left( \sum\limits_{n=1}^k \frac{1}{n^s}+ \frac{1}{k^{s - 1} \cdot (s - 1)}\right)$ takes 1 line, the proof of your's takes 4 – reuns Jul 2 at 21:28


text:
Lemma 77.8.3. Situation as in Definition 77.8.2, The action $a$ is free if and only if

$G \times _ B X \to X \times _ B X, \quad (g, x) \mapsto (a(g, x), x)$

is a monomorphism of algebraic spaces.

Proof. Immediate from the definitions. $\square$

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text:
# 2-Wasserstein (optimal transport) and extension to the set of all signed measures

Consider the 2-Wasserstein distance between probability measures $\mu$ and $\nu$ (on $\mathbb{R}^d$), defined as $$d_{W_2}(\mu,\nu) = \inf_{\gamma} \Big[\int \|x-y\|^2 d\gamma(x,y)\Big]^{1/2}$$ where the $\inf$ is over all couplings $\gamma$ of $\mu$ and $\nu$. Can we define a norm (or something norm-like) on the space of signed measures (or a linear subspace of it containing the cone of probability measures) which gives rise to $W_2$ for probability measures. (I suppose not, but why?)

If not, can we approximate $d_{W_2}$ by a norm?

-
There is a dual formulation of Wasserstein distance which makes perfect sense for signed measures, although I don't know what pathological behaviors it might have in that generality. For comparison, this paper discusses the fact that if bounded-Lipschitz distance is extended in the obvious way to signed measures then it fails to be a complete metric: worldscientific.com/doi/abs/10.1142/S0219493712003584 – Mark Meckes Feb 12 '13 at 3:40
Thanks for the reference. I will think more about the dual version. A more direct approach is also welcome. – passerby51 Feb 12 '13 at 4:21
Related question: mathoverflow.net/questions/120291/… – Dirk Feb 12 '13 at 7:15
@Dirk: Thanks for the link. – passerby51 Feb 12 '13 at 16:37

(I guess you missed a square in your definition.)

2-Wasserstein distance doesn't respect the convex structure on measures. Consider two points $x_1 \ne x_2$ and Dirac measures $\delta(x_1), \delta(x_2)$. The measure $\frac{\delta(x_1)+\delta(x_2)}{2}$ is not a midpoint between $\delta(x_1)$ and $\delta(x_2)$.

-
You are right, I missed a square. I guess you are arguing why it can't be approximated by a norm? If I remember correctly, 2-Wasserstein is geodesically convex. So maybe it is possible to approximate it locally by a norm? – passerby51 Oct 25 '15 at 17:55

This paper has several links to relevant literature by Kantorovich & Rubinstein who define an OT inspired norm for signed measures.

https://hal.inria.fr/inria-00072186/en

-
Thanks for the pointer. – passerby51 Feb 2 at 5:36


text:
## Tuesday, February 28, 2017

### Reversing a segment

One of the standard transformations of a tour used in local optimization is reversing a segment. In the new tour all the cities of the segment occur in reverse order.

pos.    0 1 2 3 4 5 6 7 ...
before  A-B-C-D-E-F-G-H-...
after   A-B-F-E-D-C-G-H-...

We use the following algorithm: swap the first and the last city of the segment, then the second and one before last, and so on, to the middle of the segment. This way number of swaps is equal half of the segment size.

Beware

We should not forget that a tour is cyclic, and therefore after tour[N-1] city (in ascending order of indices) goes tour[0], to close the tour. Therefore segment from tour[N-2] to tour[2] is 5 cities long — it constists of: tour[N-2], tour[N-1], tour[0], tour[1] and tour[2], in this order. Note that reversing the complementary segment, consisting of the cities on positions from 2 to (N-2), is not the exactly same and in some cases may lead to erroneous results. (But see later reversing a segment given by city numbers, not by positions, more complex and using additional structure, but where this problem does not occur).

proc Reverse_Segment(tour: var Tour_Array;
startIndex, endIndex: Tour_Index) =
## Reverses order of elements in segment [startIndex, endIndex]
## of tour.
## NOTE: Final and initial part of tour make one segment.
## NOTE: While the order of arguments can be safely changed when
## reversing segment in 2-optimization, it makes difference for
## 3-opt move - be careful and do not reverse (x,z) instead of (z,x).
var
left, right: Tour_Index

let inversionSize = ((N + endIndex - startIndex + 1) mod N) div 2

left  = startIndex
right = endIndex

for counter in 1 .. inversionSize:
swap(tour[left], tour[right])
left  = (left + 1) mod N
right = (N + right - 1) mod N