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# Simple Monte Carlo simulation/approximation of 2 pair in a 5 card poker hand

I am very curious about simulation of an event where an estimating/sampling technique is used.

In this example, the goal is to simulate a subset of all the roughly $2.6$ million $5$ card poker hands from a standard $52$ card deck and determine how many are $2$ pair hands.

We already know that the correct count is about $123,552 / 2,598,960$ which is about $4.75$% but I am using this example because it has a reasonable # of outcomes to be simulated in any proportion from $0$ to $100$% and because we can use it to check the accuracy of the partial simulation easily.

I would like if someone could simulate a subset of these hands using some simulation software and perhaps build a small table with 5 columns, namely:

1. Number of hands evaluated
2. % of total possible hands
3. Number of winners found
4. Extrapolated winners expected
5. Actual error

So for example, a good starting point would be to simulate $1$% of the roughly $2.6$ million possible outcomes so that would be about $26,000$. So for our table we would fill in the following:

1. $26,000$
2. $1$%
3. ? (we would expect around $1236$)
4. $100$ * whatever we actually get in # $3$.
5. how far off are we from the expected $1236$ in %?

I'd also be curious of what type of sampling method you used such as random or something else. For example, would you just choose $5$ random cards $26,000$ times and check for a 2 pair? Would that be the best method in this case or would something other than random be more appropriate?

I would like to see the results of some other sample sizes too like $0.01$% ($260$ hands), $0.1$% ($2600$ hands) and $2$% ($52,000$ hands).

Can Mathematica and/or other simulation software handle this problem easily? If so, does it automatically check that the same random hand is not generated multiple times for the same simulation or does that require separate programming or is it not a concern since it would be rare if the number of samples is small (such as $1$% or less of the total # of possible $5$ card hands)?

Assuming your random simulation is indeed random, generating the same hand more than once is not a problem: you do not have to remove duplicates or even have to check whether it has happened. Generating duplicate cards in the same hand would be a problem.

If the probability of an event is $p$ and you sample $n$ times then, using the binomial distribution, you would expect $np$ cases to occur, with a standard deviation of $\sqrt{np(1-p)}$. The ratio of these is $\sqrt{\frac{1-p}{np}}$. Thanks to the central limit theorem, you would expect $95\%$ to within about two standard deviations of the expected value.

For your example this give a mean of about $1236.0$ as you have said and a standard deviation of about $34.3$. The ratio in your example about $0.0278$ suggesting that about $95\%$ of simulations should be within $5.55\%$ of $1236$.

Change the sample size and you change $n$ in the calculations above. If it is small, you should use the binomial distribution directly rather than a Gaussian approximation, especially given the discreteness of the results.

• Can I get some "concrete" data please like someone run a random simulation of some number of trials of a $5$ card poker hand such as $2600$ and report back? I am anxious to see how close to the expected value of $124$ we actually get. Thanks. – David Sep 26 '14 at 10:28
• Here concrete data will tell you less than theory. But if you insist, $260$ attempts gave me $16$ two-pairs (but not full houses or fours-of-a-kind), $2600$ attempts gave me $114$ two-pairs, $26000$ attempts gave me $1263$ two-pairs, and $260000$ attempts gave me $12425$ two-pairs. Consistent with my answer but less informative. – Henry Sep 26 '14 at 12:13
• That is informative thanks. It seems like the Monte Carlo method works when you don't need an exact count/probability but rather just an approximation. However if there were trillions (or more) of possible outcomes to consider, then the % we could actually simulate in a reasonable amount of time would go way down and the possible error would likely go way up. Notice that the $16$ you got here out of $260$ has the greatest % error since it should be a theoretical $12.4$ count so that is about $29$% too high an estimate but I would say it is "ballpark" (although in the "nosebleed" section). – David Sep 26 '14 at 12:49
• The sample error does not really depend on the population size (at least not until they are the same order of magnitude and you are sampling without replacement). My expression $\sqrt{\frac{1-p}{np}}$ indicates how the likely magnitude of the relative error changes with the sample size, and is unaffected by whether there are millions, or billions, or trillions of possibilities in the original population. – Henry Sep 26 '14 at 13:18
• Also your formulas are informative but sometimes we don't know p, that is why we are simulating it. For example, in my original problem of drawing $27$ cards from $54$. I don't even have a feel for what a "ballpark" answer would be on that and because it has about $2$ quadrillion card combinations, even a simulation would likely be considerably off. – David Sep 26 '14 at 13:35


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# Homework Help: Probability: Draw w/o replacement, 3 different kinds of balls, don't understand

1. Oct 18, 2011

### zeion

1. The problem statement, all variables and given/known data

This was an example given in class but I didn't understand how the answer was derived.
Suppose we draw 3 balls without replacement from an urn with 2 red, 2 green, and 2 blue balls. Define the following random variables:

R = number of red balls drawn
G = number of green balls drawn
B = number of blue balls drawn

What is the probability mass function of R?

0 -> 24/120
1 -> 72/120
2 -> 24/120

2. Relevant equations

3. The attempt at a solution

I don't understand how to get those ratios.
For 0 red balls, there will be either 2G 1B or 2B 1G.

2. Oct 18, 2011

### LCKurtz

Why not reduce those fractions?

For example, to get 1 red you would have to choose 1 from the 2 red and 2 from the 4 non-red. That can be done in $\binom2 1 \binom 4 2$ ways. The total number of ways to choose 3 balls from the 6 is $\binom 6 3$ Dividing gives

$$P(2 \hbox{ red}) = \frac{\binom2 1 \binom 4 2}{\binom 6 3} = \frac 3 5$$
ways.

Last edited: Oct 18, 2011
3. Oct 19, 2011

### HallsofIvy

For the purposes of finding the probability of a certain number of red bals, the distinction between green and blue is irrelevant. So just think that there are 2 red and 4 "non-red" balls.

To get 0 red balls you have to get a non-red ball each time. There are 4 non-red out of 6 so the probability of a non-red ball on the first draw is 4/6= 2/3. Since we do not replace it there are 5 balls left, three of which are non-red. The probability of getting a non-red ball is now 3/5. Since we do not replace it, there are now 4 balls, two of which are non-red. The probability that the third ball drawn is also non-red is 2/4= 1/2.

For "one red ball", you could calculate the probability of "RNN" (red, non-red, non-red) in the same way, then use the fact that all three orders, "RNN", "NRN", and "NNR" are all the same.

For "two red balls", look at "RRN", "RNR", and "NRR".

4. Oct 19, 2011

### zeion

Thank you!
That made sense. I copied something like that in my notes but couldn't figure it out.

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# Finding the minimal number of members

I've been working on the following problem

For every issue in the Blue's association, a commission with 10 members (belonging the Blue's) is formed in order to solve the problem. The only condition is

There can't be two commissions having more than one member in common

The Blue's association has formed this year 40 commissions.

What's the minimal amount of members in the Blue's association?

I've only found out the following

For any commission you can form $$\binom{10}{2}=45$$ different pairs and none of them can appear in another commission.

Since 40 different commissions are formed, the minimal number of pairs is $$45\times 40=1800$$.

Denote by $$n$$ the number of members. Thus $$\binom{n}{2}≥1800\Rightarrow n>60$$

The minimal amount of members has to be 100 or less.

You can observe a distribution for 100 members here

My question:

Is 100 the answer or is there an ever smaller possible amount of members? If so, how can I prove it?

• Different but related question, perhaps some of the ideas there are helpful: math.stackexchange.com/questions/2879313/… – Servaes Dec 2 '18 at 18:57
• If there are two members that are in $6$ commisions, then they share at most one commission, and to fill these $11$ commissions then requires another $10\times9+1\times8=98$ distinct members, yielding a total of $100$ members if we count the two members we started with. So any example smaller than the one you found has at most one member in more than $5$ commissions. – Servaes Dec 2 '18 at 20:02
• By a similar (but simpler) argument, no member can be in more than $11$ commisions; to fill $11$ commissions sharing one member requires another $11\times9=99$ distinct members, yielding a total of $100$ members. – Servaes Dec 2 '18 at 20:06
• @Servaes If $a$ and $b$ are two members that are in $6$ commissions, how do you know that those $10\times9+1\times6=98$ members are all distinct? Why can't a commission containing $a$ overlap with a commission containing $b$? – bof Dec 5 '18 at 9:50
• @Servaes Note that, if the points of a projective plane of order $9$ are our "members" and the lines are our "commissions", then we have $91$ members and $91$ commissions, and each member belongs to $10$ commissions. – bof Dec 5 '18 at 11:32

This post exhibits a solution with $$82$$ members. Combined with the excellent answer by @Song, this means $$82$$ is indeed optimal.

Motivation: The excellent answer by @Song and the followup comments by @Servaes make me wonder... perhaps if we look for 41 commissions (not 40) then there is a solution with a great deal of symmetry:

• (a) 82 members (the optimal answer we seek)
• (b) 41 commissions (exceeds OP requirement)
• (c) each member associated with exactly 5 commissions (not part of OP)
• (d) each commission associated with exactly 10 members (equals OP requirement)
• (e) each 2 members have exactly 1 common commission (not part of OP)
• (f) each 2 commissions have exactly 1 common member (exceeds OP requirement)

This would be like a finite projective plane, but with 82 points and 41 lines. However, in a finite projective (respectively: affine) plane, the no. of points and no. of lines are equal (respectively: almost equal), and this is probably why a solution based on FPP only reaches 84. So I decided to look at related structures called Block Designs, Steiner Systems, etc. where there are typically many more "lines" than "points". After quite a bit of digging, I think I found the right structure:

The solution: It is a Steiner $$S(t=2,k=5,n=41)$$ system. A Steiner system is defined by the following properties:

• there are $$n=41$$ objects (these are the commissions)

• there are $$b$$ blocks (these are the members), each block (member) being a subset of objects (i.e. the commissions he/she is associated with)

• each block has $$k=5$$ objects (each member is associated with 5 commissions)

• every $$t=2$$ objects is contained in exactly 1 block (every 2 commissions have exactly 1 common member)

So this already satisfies (b), (c) and (f). Next, quoting from https://en.wikipedia.org/wiki/Steiner_system#Properties we have:

• $$b = {n \choose t} / {k \choose t} = (41 \times 40) / (5 \times 4) = 41 \times 2 = 82$$, satisfying (a)

• $$r = {n-1 \choose t-1} / {k-1 \choose t-1} = 40 / 4 = 10$$, where $$r$$ denotes "the number of blocks containing any given object", i.e. the number of members associated with any given commission, satisfying (d).

Thinking more, I don't think (e) can be satisfied. However, (e) is not needed for the OP, so it doesn't matter.

So finally we just need to prove that such a Steiner $$S(t=2,k=5,n=41)$$ system exists. This existence is non-trivial, but luckily more digging reveals:

• https://math.ccrwest.org/cover/steiner.html has a list of Steiner systems that are known to exist. $$S(2,5,41)$$ (the webpage sometimes lists the 3 parameters in different order) is not part of any infinite families listed, but if you go further down the page it is listed as a standalone example; clicking on that link goes to...

• https://math.ccrwest.org/cover/show_cover.php?v=41&k=5&t=2 which exhibits the system, created via "Cyclic construction" whatever that means.

I did not check the numbers thoroughly, but if I understand the webpage correctly, there should be 82 rows (members / blocks), each containing 5 numbers (commissions), all the numbers being 1 through 41 inclusive (the 41 commissions), each number (commission) should appear in 10 rows, and every pair-of-numbers should appear in 1 row.

I am not an expert in any of these, so if I had a mistake or misunderstanding above, my apologies. Perhaps someone more expert can check my work?

• +1 Good ideas and good find! I will have a look at the second link. As a side note, indeed (e) is impossible; there are $\binom{82}{2}=3321$ pairs of members. But there are $\binom{10}{2}=45$ pairs of members per commission, so $41\times45=1845$ pairs of members that share a commission. Alternatively, every member is in $5$ commissions, and hence in a commission with $5\times9=45$ members, so $\frac{82\times45}{2}=1845$ pairs of members that share a commission. Either way, there are too many pairs of members to all share a commission. – Servaes Dec 14 '18 at 0:55
• *Another, simpler, way to see that (e) is impossible; this would imply that every member shares a commission with $81$ members. But every member is in only $5$ commissions, and hence shares a commission with only $5\times9=45$ members. – Servaes Dec 14 '18 at 0:57
• @Servaes - yes of course. in fact, more abstractly / vaguely speaking, perhaps insisting on both (e) and (f) is WHY a finite projective plane has the same no. of lines and points. so if we introduce imbalance in (a) vs (b), and hence (c) vs (d), we must necessarily imbalance (e) vs (f) as well. this is all very vague of course. :) – antkam Dec 14 '18 at 1:07
• With duality between lines & points (or equiv: rows & cols of an incidence matrix), there is really no need to translate, but it's up to you. anyway, here's a link I found explaining (I think) the "cyclic construction" of some Steiner systems, although it doesn't mention $S(2,5,41)$ specifically: pitt.edu/~kaveh/Chap7-MATH1050.pdf – antkam Dec 14 '18 at 1:53
• Dan Gordon migrated the La Jolla covering design repository to his new site, so it might be a good idea to update your links. – hardmath Mar 16 at 0:46

Let $$i$$ denote each member of Blue's association and assume that there are $$N$$ members in total, that is, $$i=1,2,\cdots, N.$$ And let $$j,k=1,2,\ldots, 40$$ denote each of 40 commission. We will show that $$N$$ is at least $$82$$.

Consider the set $$S=\{(i,j,k)\;|\;1\leq i\leq N, 1\leq j Let $$d_i$$ denote the number of commissions that $$i$$ joined. We will calculate $$|S|$$ using double counting method. First, note that $$|S|=\sum_{(i,j,k)\in S}1 = \sum_{1\leq j since for each $$j, there is at most one $$i$$ in common. On the other hand, $$|S| = \sum_{1\leq i\leq N} \sum_{(j,k):(i,j,k)\in S}1 = \sum_{1\leq i\leq N} \binom{d_i}{2},$$ since for each $$i$$, the number of pairs $$(j,k)$$ that $$i$$ joined is $$\binom{d_i}{2}$$. We also have $$\sum_{1\leq i\leq N}d_i = 400,$$by the assumption.
Finally, note that the function $$f(x)= \binom{x}{2} = \frac{x^2-x}{2}$$ is convex. Thus by Jensen's inequality we have that $$\binom{40}{2}\geq |S|=\sum_{1\leq i\leq N} \binom{d_i}{2}\geq Nf\left(\frac{\sum_i d_i}{N}\right)=N\binom{\frac{400}{N}}{2}.$$ This gives us the bound $$40\cdot 39 \geq 400\cdot(\frac{400}{N}-1),$$and hence $$N \geq \frac{4000}{49} = 81.63\cdots$$ This establishes $$N\geq 82$$. However, I'm not sure if this bound is tight. I hope this will help.

$$\textbf{Note:}$$ If $$N=82$$ is tight, then above argument implies that $$d_i$$'s distribution is almost concentrated at $$\overline{d} = 400/82 \sim 5$$.
EDIT: @antkam's answer seemingly shows that $$N=82$$ is in fact optimal.

• +1 Fantastic argument! – Servaes Dec 10 '18 at 10:28
• A corollary of this argument is that in the best solution every commission must share a member with some other commission. – Will Fisher Dec 10 '18 at 20:02
• @WillFisher You mean, if there is a solution with $N=82$? In this case, it also follows that there are precisely $72$ members that are in $5$ commissions, and $10$ that are in $4$ commissions. Also, that every commission has precisely $9$ members in $5$ commissions and $1$ member in $4$ commissions. – Servaes Dec 10 '18 at 20:38
• I don't really understand this step: "Finally, note that the function $x\to \binom{x}{2} = \frac{x^2-x}{2}$ is convex. Thus, we have that $$\binom{40}{2}\geq |S|=\sum_{1\leq i\leq N} \binom{d_i}{2}\geq N\binom{\frac{\sum_i d_i}{N}}{2}=N\binom{\frac{400}{N}}{2}.$$" – Dr. Mathva Dec 13 '18 at 23:35
• I should have mentioned that it is by Jensen's inequality. The statement is in en.m.wikipedia.org/wiki/Jensen%27s_inequality. @Dr. Mathva – Song Dec 13 '18 at 23:44

This is just a partial answer. I will show that $$85$$ members suffice; I don't know if $$85$$ is the minimum.

Recall that a projective plane of order $$n$$ exists if $$n$$ is a prime power: it has $$n^2+n+1$$ points and $$n^2+n+1$$ lines; each line has $$n+1$$ points, and there are $$n+1$$ lines through each point; any pair of lines meets in a unique point, and any pair of points determines a unique line.

Consider a projective plane of order $$9$$; it has $$9^2+9+1=91$$ points and $$91$$ lines; there are $$10$$ points on each line and $$10$$ lines through each point. A set of points is in general position if no three of the points are collinear. Note that, if we have a set of $$t$$ points in general position, then the lines determined by those points (taken two at a time) cover a total of at most $$t+8\binom t2$$ points; as long as $$t\le5$$ then the number of covered points is at most $$5+8\binom52=85\lt91$$, so we can add another point to the set and still have them in general position. Thus we can find a set $$S$$ of $$6$$ points in general position.

Let the members of the Blue's association be the $$91-6=85$$ points which are not in $$S$$. The commissions are the lines which do not meet $$S$$; they have $$10$$ members each, and any two have exactly one member in common. Finally, by the in-and-out formula, the number of commissions is $$91-\binom61\cdot10+\binom62\cdot1=46.$$

P.S. Let $$m$$ be the minimum possible number of members. I showed above that $$m\le85$$. On the other hand, I have a small improvement on your lower bound $$m\ge61$$.

Suppose the $$i^\text{th}$$ member belongs to $$d_i$$ commissions; then $$\sum_{i=1}^md_i=400$$ since there are $$40$$ commissions with $$10$$ members each. Moreover $$d_i\le9$$ since $$m\le85\lt91$$. Let $$k=|\{i:d_i\ge5\}|$$. Then $$400=\sum_{i=1}^md_i\le4(m-k)+9k=4m+5k\le340+5k,$$ whence $$k\ge12$$; i.e., there are at least $$12$$ members who are on at least $$5$$ commissions. Choose two members $$i$$ and $$j$$ who are on at least $$5$$ commissions.

Case 1. There is a commission containing both $$i$$ and $$j$$.

First, there are $$10$$ members on the commission which $$i$$ and $$j$$ both belong to. Next $$i$$ belongs to $$4$$ more commissions, with $$36$$ additional members. Finally, $$j$$ belongs to $$4$$ more commissions, each of which contains at most one member of each of the $$5$$ commissions containg $$i$$, and at least $$5$$ members who haven't been counted yet, for a total of $$20$$ new members. This shows that $$m\ge10+36+20=66$$.

Case 2. There is no commission containing both $$i$$ and $$j$$.

In this case a similar argument shows that $$m\ge67$$.

This proves that $$m\ge66$$. Combining this with the upper bound shown earlier, we have $$66\le m\le85.$$

• +1 Great progress and nice ideas! Note that for every prime power $q$ there are arcs of $q+1$ points in $\Bbb{P}(\Bbb{F}_q)$. In particular you can take your set $S$ to contain $7$ points, yielding $42$ commissions and $m\leq84$. With $S$ containing $8$ points we get $39$ commissions and $m\leq83$, perhaps another commission can be fitted in somewhere? – Servaes Dec 5 '18 at 16:53
• Your bottom argument can be extended to show $m\ge 73$. – Will Fisher Dec 5 '18 at 17:04
• @Servaes I don't know much about projective planes. That fact about "arcs of $q+1$ points", is it easy or hard to prove? Is it true for all projective planes of prime power order, or just the field planes? – bof Dec 6 '18 at 10:20
• @WillFisher Nice. How do you prove it? – bof Dec 6 '18 at 10:21
• @bof In any projective plane over a finite field, any nondegenerate conic is an example of an arc of $q+1$ points. I've also checked that the same construction with a set $S$ of $8$ or more points cannot fit any other commission directly (i.e. without adjusting some of the existing $39$ or fewer commissions). – Servaes Dec 6 '18 at 12:46

Allow me to summarize and slightly refine the results in the current answers (if only to straighten out my own thoughts); they show that the minimum number of members $$m$$ satisfies $$82\leq m\leq84$$. They also imply strict conditions on any solution with $$m=82$$.

I also include my result that if $$m=83$$, then no member is in more than $$7$$ commissions. Much more can be said, but I do not have a definitive proof for the cases $$m=82$$ or $$m=83$$.

The upper bound $$m\leq84$$ comes from bof's construction in the projective plane of order $$9$$; the projective plane $$\Bbb{P}^2(\Bbb{F}_9)$$ consists of $$91$$ points on $$91$$ lines, with $$10$$ points on each line and $$10$$ lines through each point. Importantly, each pair of lines meets in precisely one point, and each pair of points is on precisely one line.

For $$7$$ distinct points in general position (no $$3$$ on a line, e.g. points on a smooth conic) there are precisely $$7\times10-\binom{7}{2}\times1=49$$ lines containing these points. Removing these $$7$$ points and the $$49$$ lines containing them leaves $$84$$ points and $$91-49=42$$ lines each containing $$10$$ points, and any pair of lines meets in at most one point. That is, we have $$84$$ members in $$42$$ commissions, with no $$2$$ commissions sharing more than one member, so $$m\leq84$$.

The lower bound $$m\geq82$$ comes from Song's answer; the number of pairs of commissions that share a member is at most $$\binom{40}{2}$$, as there are $$40$$ commissions. As every commission shares at most one member, this can also be counted as the number of pairs of commissions that each member is in. If the $$i$$-th member is in $$d_i$$ commissions, then it is in $$\binom{d_i}{2}$$ pairs of commissions and hence $$\sum_{i=1}^m\binom{d_i}{2}\leq\binom{40}{2}.\tag{1}$$ As there are $$40$$ commissions with $$10$$ members each, we also have $$\sum_{i=1}^md_i=400$$. In the inequality above we can bound the left hand side from below using the fact that for all positive integers $$x$$ we have $$\binom{x-1}{2}+\binom{x+1}{2}=2\binom{x}{2}+1.$$ This allows us to even out the $$d_i$$'s to find that $$\sum_{i=1}^m\binom{d_i}{2}\geq(m-n)\binom{x}{2}+n\binom{x+1}{2},\tag{2}$$ for some $$x$$ and $$n$$ with $$0\leq n, where $$(m-n)x+n(x+1)=\sum_{i=1}^md_i=400.$$ The latter simplifies to $$mx+n=400$$, which shows that $$x=\lfloor\frac{400}{m}\rfloor$$ and $$n=400-mx$$. Plugging this back in shows that $$\begin{eqnarray*} \binom{40}{2}&\geq&\sum_{i=1}^m\binom{d_i}{2} \geq(m-n)\binom{x}{2}+n\binom{x+1}{2}\\ &=&(m-(400-m\lfloor\tfrac{400}{m}\rfloor))\binom{\lfloor\frac{400}{m}\rfloor}{2}+(400-m\lfloor\tfrac{400}{m}\rfloor)\binom{\lfloor\frac{400}{m}\rfloor+1}{2}\\ &=&-\frac{m}{2}\lfloor\tfrac{400}{m}\rfloor^2-\frac{m}{2}\lfloor\tfrac{400}{m}\rfloor+400\lfloor\tfrac{400}{m}\rfloor. \end{eqnarray*}$$ The latter is strictly decreasing for $$m$$ in the interval $$[1,84]$$. The inequality is satisfied if and only if $$m\geq82$$, which proves the lower bound.

Let $$S$$ denote the number of times we need to apply the identity $$\binom{x-1}{2}+\binom{x+1}{2}=2\binom{x}{2}+1$$ to reduce the left hand side of $$(2)$$ to the right hand side. We can then write $$(2)$$ more precisely as $$\sum_{i=1}^m\binom{d_i}{2}=(m-n)\binom{x}{2}+n\binom{x+1}{2}+S.$$ Knowing that $$82\leq m\leq84$$ simplifies the above significantly, as then $$x=\lfloor\tfrac{400}{m}\rfloor=4$$ and $$n=400-4m$$. We find that $$780\geq\sum_{i=1}^m\binom{d_i}{2}=1600-10m+S.$$ In particular, for $$m=82$$ we find that $$S=0$$ and hence that there are precisely $$n=400-82\times4=72$$ members that are in $$4$$ commissions and $$10$$ members that are in $$5$$ commissions. We also see that we have equality in $$(1)$$, meaning that every pair of commissions shares a member. This implies $$\sum_{i\in C}(d_i-1)=39$$ for every commission $$C$$, from which it follows that every commission has precisely $$1$$ member that is in $$4$$ commissions, and $$9$$ members that are in $$5$$ commissions.

If $$m=83$$ then $$S\leq10$$, and there are at most $$10$$ pairs of commissions that do not share a member.

Here are a few unincorporated observations that may or may not be helpful. These concern restrictions on minimal examples with $$m<84$$, i.e. $$m=82$$ or $$m=83$$. They are all subsumed by the observations above for $$m=82$$, so I prove them only for $$m=83$$.

Observation 1: For all $$i$$ we have $$d_i\leq9$$.

To fill the commission of member $$i$$ requires $$9d_i+1$$ distinct members, including member $$i$$. We have $$9d_i+1\leq m=83$$ and hence $$d_i\leq9$$.

Observation 2: For all $$i$$ we have $$d_i\leq8$$.

To fill the commission of a member $$i$$ with $$d_i=9$$ requires $$9d_i+1=82$$ distinct members, leaving $$1$$ member remaining as $$m=83$$. Each of the remaining $$40-d_i=31$$ commissions has at most one member from teach of the $$d_i$$ commissions of $$i$$, and hence contains the remaining member. But this member is in at most $$9$$ commissions by observation $$1$$, a contradiction.

Observation 3: For any pair $$i$$, $$j$$ of members in a commission we have $$d_i+d_j\leq14$$.

If the inequality does not hold then without loss of generality $$d_i=8$$ and $$d_j\geq7$$. To fill the shared commission requires another $$8$$ members, and to fill the remaining $$7$$ commissions of member $$i$$ requires another $$9\times7=63$$ members. Each of the $$d_j-1$$ remaining commissions of $$j$$ contains at most $$7$$ members from the $$7$$ commissions of $$i$$, and hence at least $$2$$ new members. Hence we have a total of $$2+8+9\times(d_i-1)+2\times(d_j-1)\geq2+8+63+2\times6=85,$$ members, contradicting $$m=83$$.

Observation 4: For all $$i$$ we have $$d_i\leq7$$.

Suppose toward a contradiction that $$d_i=8$$ for some member $$i$$. To fill these $$d_i=8$$ commissions requires $$9d_i+1=73$$ distinct members, including member $$i$$, leaving $$10$$ members. Each of the remaining $$32$$ commissions has at most $$8$$ members from the $$d_i=8$$ commissions, hence at least $$2$$ members from the remaining $$10$$. Numbering these $$1$$ throught $$10$$ we find that $$\sum_{k=1}^{10}d_k\geq2\times32=64.$$ We distinguish two cases:

Case 1: If $$d_j=8$$ for some $$1\leq j\leq10$$ then $$j$$ shares a commission with at least $$8$$ other of these $$10$$ members, hence they all have $$d_k\leq6$$ by observation $$3$$. To satisfy the inequality there must be one more member $$j'$$ with $$d_{j'}=8$$, and the other $$8$$ have $$d_k=6$$.

We have $$11$$ members, including $$i$$, that together take up $$8+64=72$$ spots in the $$40$$ commissions. The remaining $$83-11=72$$ members then take up $$400-72=328$$ spots. As noted before, the sum $$\sum\binom{d_i}{2}$$ ranging over the remaining $$72$$ members is minimal when the values $$d_i$$ differ by at most $$1$$. This happens precisely when $$d_i=5$$ for $$40$$ members and $$d_i=4$$ for $$432$$ members. Then $$\sum_{k=1}^{83}\binom{d_i}{2}\geq3\binom{8}{2}+8\binom{6}{2}+40\binom{5}{2}+32\binom{4}{2}=796,$$ which exceeds the bound of $$\binom{40}{2}=780$$ we found before, a contradiction.

Case 2: If $$d_j\neq8$$ for all $$10$$ remaining members, then to satisfy $$\sum_{k=1}^{10}d_k\geq64$$ there must be a t least $$4$$ members with $$d_k=7$$. We also have $$\sum_{k=1}^{10}\leq70$$, and we proceed as before.

We have $$5$$ members, including $$i$$, that together take up $$8+28=36$$ spots in the $$40$$ commissions. Hence the remaining $$83-5=78$$ members take up $$400-36=364$$ spots. The sum $$\sum\binom{d_i}{2}$$ over the remaining $$78$$ members is minimized when the $$d_i$$ differ by at most $$1$$. This happens precisely if $$d_i=5$$ for $$52$$ members and $$d_i=4$$ for $$26$$ members, and we $$\sum_{k=1}^{83}\binom{d_k}{2}\geq\binom{8}{2}+4\binom{7}{2}+52\binom{5}{2}+26\binom{4}{2}=788,$$ again contradicting the upper bound of $$\binom{40}{2}=780$$.

Much more can be said, but my computer is already freezing up at this big an answer.

Here's a partial answer that increases the lower bound for any (not necessarily optimal) solution to $$m\ge 74$$.

Suppose there is a solution with $$m$$ members and we know that there are two members each in $$l+1$$ commissions, then $$m\ge 9(l+1)+(8-l)(l+1)+2.$$ This is because if member one is in $$\ge l+1$$ commissions, each commission needs to be filled with $$9$$ new members since these $$l+1$$ commissions already have maximum overlap. For the commissions that member two is in, each needs $$9$$ more members to be accounted for. We can't have any overlap between these commissions because they already have maximum overlap (that being member two). We can at best choose one member from each group with member one in it, giving us $$l+1$$ members, but the other $$9-(l+1)=8-l$$ are new. This gives $$9(l+1)+(8-l)(l+1)$$ members other than the two we started with. (Note that this is the best bound in $$l$$ possible).

Now, suppose $$m$$ members has a solution to the problem. Let $$d_i$$ be the how many commissions the $$i$$-th member is in. First note that $$m\ge 9d_i+1$$ for every $$i$$, so $$d_i\le \lfloor (m-1)/9\rfloor$$. Let $$k_l=|\{i\; :\; d_i>l\}|$$. Then $$400=\sum_{i=1}^md_i\le l(m-k_l)+\lfloor (m-1)/9\rfloor k_l.$$ Hence $$k_l\ge \frac{400-lm}{\lfloor (m-1)/9\rfloor -l}.$$ Since $$k_l$$ is an integer, if $$\frac{400-lm}{\lfloor (m-1)/9\rfloor -l}>1$$ then $$k_l\ge 2$$, meaning that there is at least $$2$$ members in at least $$l+1$$ commissions so by the above $$m\ge 9(l+1)+(8-l)(l+1)+2$$. Note that $$\frac{400-lm}{\lfloor (m-1)/9\rfloor -l}>1$$ exactly when $$\frac{400-\lfloor (m-1)/9\rfloor}{m-1}>l$$. Hence we have that for all $$\frac{400-\lfloor (m-1)/9\rfloor}{m-1}>l$$ we have $$m\ge 9(l+1)+(8-l)(l+1)+2.$$ For this to be satisfied gives $$m\ge 74.$$

• Very nice! I had some trouble understanding the part from "Hence we have that for all...". Perhaps you could make more explicit that if this inequality holds for $l$, then the next inequality must also hold for that $l$. – Servaes Dec 7 '18 at 14:29
• @Servaes I edited the answer. It is basically just solving for the $l$'s such that $k_l\ge 2$, letting the first paragraphs argument follow. – Will Fisher Dec 7 '18 at 16:06


text:
# 5.3 Solve systems of equations by elimination  (Page 4/6)

 Page 4 / 6

The sum of two numbers is 39. Their difference is 9. Find the numbers.

## Solution

$\begin{array}{ccc}\mathbf{\text{Step 1. Read}}\phantom{\rule{0.2em}{0ex}}\text{the problem}\hfill & & \\ \mathbf{\text{Step 2. Identify}}\phantom{\rule{0.2em}{0ex}}\text{what we are looking for.}\hfill & & \hfill \text{We are looking for two numbers.}\hfill \\ \mathbf{\text{Step 3. Name}}\phantom{\rule{0.2em}{0ex}}\text{what we are looking for.}\hfill & & \begin{array}{c}\hfill \text{Let}\phantom{\rule{0.2em}{0ex}}n=\phantom{\rule{0.2em}{0ex}}\text{the first number.}\hfill \\ \hfill \phantom{\rule{2.3em}{0ex}}m=\text{the second number}\hfill \end{array}\hfill \\ \begin{array}{c}\mathbf{\text{Step 4. Translate}}\phantom{\rule{0.2em}{0ex}}\text{into a system of equations.}\hfill \\ \\ \\ \\ \\ \\ \\ \end{array}\hfill & & \hfill \begin{array}{}\\ \text{The sum of two numbers is 39.}\hfill \\ \hfill \phantom{\rule{0.3em}{0ex}}n+m=39\hfill \\ \hfill \text{Their difference is 9.}\hfill \\ \hfill n-m=9\hfill \end{array}\hfill \\ \text{The system is:}\hfill & & \hfill \left\{\begin{array}{c}n+m=39\hfill \\ n-m=9\hfill \end{array}\hfill \\ \begin{array}{c}\mathbf{\text{Step 5. Solve}}\phantom{\rule{0.2em}{0ex}}\text{the system of equations.}\hfill \\ \text{To solve the system of equations, use}\hfill \\ \text{elimination. The equations are in standard}\hfill \\ \text{form and the coefficients of}\phantom{\rule{0.2em}{0ex}}m\phantom{\rule{0.2em}{0ex}}\text{are}\hfill \\ \text{opposites. Add.}\hfill \\ \\ \text{Solve for}\phantom{\rule{0.2em}{0ex}}n.\hfill \\ \\ \\ \\ \\ \end{array}\hfill & & \hfill \begin{array}{c}\hfill \underset{\text{____________}}{\left\{\begin{array}{c}n+m=39\hfill \\ n-m=9\hfill \end{array}}\hfill \\ \hfill 2n\phantom{\rule{1.8em}{0ex}}=48\hfill \\ \\ \hfill \phantom{\rule{2.21em}{0ex}}n=24\hfill \end{array}\hfill \\ \begin{array}{c}\text{Substitute}\phantom{\rule{0.2em}{0ex}}n=24\phantom{\rule{0.2em}{0ex}}\text{into one of the original}\hfill \\ \text{equations and solve for}\phantom{\rule{0.2em}{0ex}}m.\hfill \end{array}\hfill & & \hfill \begin{array}{c}\hfill \phantom{\rule{0.2em}{0ex}}\begin{array}{c}\hfill n+m=39\\ \hfill 24+m=39\\ \hfill m=15\end{array}\hfill \end{array}\hfill \\ \mathbf{\text{Step 6. Check}}\phantom{\rule{0.2em}{0ex}}\text{the answer.}\hfill & & \phantom{\rule{1.55em}{0ex}}\text{Since}\phantom{\rule{0.2em}{0ex}}24+15=39\phantom{\rule{0.2em}{0ex}}\text{and}\hfill \\ & & \hfill 24-15=9,\phantom{\rule{0.2em}{0ex}}\text{the answers check.}\hfill \\ \mathbf{\text{Step 7. Answer}}\phantom{\rule{0.2em}{0ex}}\text{the question.}\hfill & & \hfill \text{The numbers are 24 and 15.}\hfill \end{array}$

The sum of two numbers is 42. Their difference is 8. Find the numbers.

The numbers are 25 and 17.

The sum of two numbers is −15. Their difference is −35. Find the numbers.

The numbers are −25 and 10.

Joe stops at a burger restaurant every day on his way to work. Monday he had one order of medium fries and two small sodas, which had a total of 620 calories. Tuesday he had two orders of medium fries and one small soda, for a total of 820 calories. How many calories are there in one order of medium fries? How many calories in one small soda?

## Solution

 Step 1. Read the problem. Step 2. Identify what we are looking for. We are looking for the number of calories in one order of medium fries and in one small soda. Step 3. Name what we are looking for. Let f = the number of calories in 1 order of medium fries.     s = the number of calories in 1 small soda. Step 4. Translate into a system of equations: one medium fries and two small sodas had a total of 620 calories two medium fries and one small soda had a total of 820 calories. Our system is: Step 5. Solve the system of equations. To solve the system of equations, use elimination. The equations are in standard form. To get opposite coefficients of f , multiply the top equation by −2. Simplify and add. Solve for s . Substitute s = 140 into one of the original equations and then solve for f . Step 6. Check the answer. Verify that these numbers make sense in the problem and that they are solutions to both equations. We leave this to you! Step 7. Answer the question. The small soda has 140 calories and the fries have 340 calories.

Malik stops at the grocery store to buy a bag of diapers and 2 cans of formula. He spends a total of $37. The next week he stops and buys 2 bags of diapers and 5 cans of formula for a total of$87. How much does a bag of diapers cost? How much is one can of formula?

The bag of diapers costs $11 and the can of formula costs$13.

To get her daily intake of fruit for the day, Sasha eats a banana and 8 strawberries on Wednesday for a calorie count of 145. On the following Wednesday, she eats two bananas and 5 strawberries for a total of 235 calories for the fruit. How many calories are there in a banana? How many calories are in a strawberry?

There are 105 calories in a banana and 5 calories in a strawberry.

write in this form a/b answer should be in the simplest form 5%
convert to decimal 9/11
August
Equation in the form of a pending point y+2=1/6(×-4)
write in simplest form 3 4/2
August
From Google: The quadratic formula, , is used in algebra to solve quadratic equations (polynomial equations of the second degree). The general form of a quadratic equation is , where x represents a variable, and a, b, and c are constants, with . A quadratic equation has two solutions, called roots.
Melissa
what is the answer of w-2.6=7.55
10.15
Michael
w = 10.15 You add 2.6 to both sides and then solve for w (-2.6 zeros out on the left and leaves you with w= 7.55 + 2.6)
Korin
Nataly is considering two job offers. The first job would pay her $83,000 per year. The second would pay her$66,500 plus 15% of her total sales. What would her total sales need to be for her salary on the second offer be higher than the first?
x > $110,000 bruce greater than$110,000
Michael
Estelle is making 30 pounds of fruit salad from strawberries and blueberries. Strawberries cost $1.80 per pound, and blueberries cost$4.50 per pound. If Estelle wants the fruit salad to cost her $2.52 per pound, how many pounds of each berry should she use? nawal Reply$1.38 worth of strawberries + $1.14 worth of blueberries which=$2.52
Leitha
how
Zaione
is it right😊
Leitha
lol maybe
Robinson
8 pound of blueberries and 22 pounds of strawberries
Melissa
8 pounds x 4.5 = 36 22 pounds x 1.80 = 39.60 36 + 39.60 = 75.60 75.60 / 30 = average 2.52 per pound
Melissa
8 pounds x 4.5 equal 36 22 pounds x 1.80 equal 39.60 36 + 39.60 equal 75.60 75.60 / 30 equal average 2.52 per pound
Melissa
hmmmm...... ?
Robinson
8 pounds x 4.5 = 36 22 pounds x 1.80 = 39.60 36 + 39.60 = 75.60 75.60 / 30 = average 2.52 per pound
Melissa
The question asks how many pounds of each in order for her to have an average cost of $2.52. She needs 30 lb in all so 30 pounds times$2.52 equals $75.60. that's how much money she is spending on the fruit. That means she would need 8 pounds of blueberries and 22 lbs of strawberries to equal 75.60 Melissa good Robinson 👍 Leitha thanks Melissa. Leitha nawal let's do another😊 Leitha we can't use emojis...I see now Leitha Sorry for the multi post. My phone glitches. Melissa Vina has$4.70 in quarters, dimes and nickels in her purse. She has eight more dimes than quarters and six more nickels than quarters. How many of each coin does she have?
10 quarters 16 dimes 12 nickels
Leitha
A private jet can fly 1,210 miles against a 25 mph headwind in the same amount of time it can fly 1,694 miles with a 25 mph tailwind. Find the speed of the jet.
wtf. is a tail wind or headwind?
Robert
48 miles per hour with headwind and 68 miles per hour with tailwind
Leitha
average speed is 58 mph
Leitha
Into the wind (headwind), 125 mph; with wind (tailwind), 175 mph. Use time (t) = distance (d) ÷ rate (r). since t is equal both problems, then 1210/(x-25) = 1694/(×+25). solve for x gives x=150.
bruce
the jet will fly 9.68 hours to cover either distance
bruce
Riley is planning to plant a lawn in his yard. He will need 9 pounds of grass seed. He wants to mix Bermuda seed that costs $4.80 per pound with Fescue seed that costs$3.50 per pound. How much of each seed should he buy so that the overall cost will be $4.02 per pound? Vonna Reply 33.336 Robinson Amber wants to put tiles on the backsplash of her kitchen counters. She will need 36 square feet of tiles. She will use basic tiles that cost$8 per square foot and decorator tiles that cost $20 per square foot. How many square feet of each tile should she use so that the overall cost of the backsplash will be$10 per square foot?
Ivan has $8.75 in nickels and quarters in his desk drawer. The number of nickels is twice the number of quarters. How many coins of each type does he have? mikayla Reply 2q=n ((2q).05) + ((q).25) = 8.75 .1q + .25q = 8.75 .35q = 8.75 q = 25 quarters 2(q) 2 (25) = 50 nickles Answer check 25 x .25 = 6.25 50 x .05 = 2.50 6.25 + 2.50 = 8.75 Melissa John has$175 in $5 and$10 bills in his drawer. The number of $5 bills is three times the number of$10 bills. How many of each are in the drawer?
7-$10 21-$5
Robert
Enrique borrowed $23,500 to buy a car. He pays his uncle 2% interest on the$4,500 he borrowed from him, and he pays the bank 11.5% interest on the rest. What average interest rate does he pay on the total \$23,500? (Round your answer to the nearest tenth of a percent.)
Two sisters like to compete on their bike rides. Tamara can go 4 mph faster than her sister, Samantha. If it takes Samantha 1 hour longer than Tamara to go 80 miles, how fast can Samantha ride her bike?
8mph
michele
16mph
Robert
3.8 mph
Ped
16 goes into 80 5times while 20 goes into 80 4times and is 4mph faster
Robert
what is the answer for this 3×9+28÷4-8
315
lashonna
how do you do xsquard+7x+10=0
What
(x + 2)(x + 5), then set each factor to zero and solve for x. so, x = -2 and x = -5.
bruce
I skipped it
What


text:
# A mathematically illogical argument in the derivation of Hamilton's equation in Goldstein

In the book of Goldstein, at page 337, while deriving the Hamilton's equations (canonical equations), he argues that

The canonical momentum was defined in Eq. (2.44) as $$p_i = \partial L / \partial \dot q_i$$; substituting this into the Lagrange equation (8.1), we obtain

$$\dot p_i= \frac{\partial L}{\partial q_i} \tag{8.14}$$

so Eq. (8.13) can be written as

$$dL = \dot p_i dq_i + p_id \dot q_i + \frac{\partial L}{\partial t}dt \tag{8.13′}$$

The Hamiltonian $$H(q,p,t)$$ is generated by the Legendre transformation

$$H(q,p,t) = \dot q_i p_i - L(q, \dot q, t), \tag{8.15}$$

which has the differential

$$dH = \dot q_i d p_i - \dot p_i d q_i - \frac {\partial L}{\partial t}, \tag{8.16}$$

where the term $$p_i d \dot q_i$$ is removed by the Legendre transformation. Since $$dH$$ can also be written as

$$dH = \frac{\partial H}{\partial q_i}d q_i + \frac{\partial H}{\partial p_i}d p_i + \frac{\partial H}{\partial t}d t, \tag{8.17}$$

However, if $$H$$ is defined to be a function of $$q,p,t$$, then how can we define $$H(q,p,t) = \dot q *p - L(q,\dot q,t)$$, i.e $$\dot q$$ is not an argument of $$H$$ whereas it is in its definition.

Moreover, when he is taking the differential of $$H$$, he argues that $$pd\dot q$$ is removed, but he does not say why.

I mean mathematically speaking this whole argument is plan wrong, as far as I can see, so assuming that it is not the case, what am I missing in here ?

$$\boldsymbol{\S\:}\textbf{A. In General}$$

Consider a real function $$\:f\left(x\right)\:$$ of a real variable $$x \in \left[\alpha,\beta\right]$$ with continuous 1st and 2nd derivatives. Suppose that its 2nd derivative is everywhere negative so that its graph in the $$\:xy-$$plane is as in Figure-01. From every point of the graph, we have a tangent line.

Now, the graph of the function could be sketched by the family of the tangent lines, see Figure-02. We say that this curve (graph) is the envelope of the family of the tangent lines. From this fact we note that we could define the function $$\:f\left(x\right)\:$$ by the family of its tangent lines. Indeed, as shown in Figure-03, if from the angle $$\:\theta\:$$ of any tangent line we know the point where this line intersects the $$\:y-$$axis, let $$\:\boldsymbol{-}\omega\:$$ (the minus sign used for future purposes), then we would have an equivalent definition of the function $$\:f\left(x\right)$$. So, we must have the function $$\:\omega\left(\theta\right)$$. For the domain of angle $$\:\theta\:$$ we have from Figure-03 as example

$$$$\theta \in \left[\theta_1,\theta_2\right] \quad \text{where} \quad \theta_1\boldsymbol{=}\min{(\theta_\alpha,\theta_\beta)}\quad \text{and} \quad \theta_2\boldsymbol{=}\max{(\theta_\alpha,\theta_\beta)} \tag{A-01}\label{A-01}$$$$

Instead of using the angle $$\:\theta\:$$ we equally well use the variable $$\:u\boldsymbol{=}\tan\theta\boldsymbol{=}\dfrac{\mathrm df}{\mathrm dx}$$. For the domain of $$\:u\:$$ we have
$$$$u \in \left[u_1,u_2\right] \quad \text{where} \quad u_1\boldsymbol{=}\min{(\tan\theta_\alpha,\tan\theta_\beta)}\quad \text{and} \quad u_2\boldsymbol{=}\max{(\tan\theta_\alpha,\tan\theta_\beta)} \tag{A-02}\label{A-02}$$$$

From Figure-03 we have $$$$y\boldsymbol{+}\omega\boldsymbol{=}\tan\theta \cdot x\boldsymbol{=}u \cdot x \tag{A-03}\label{A-03}$$$$ so $$$$\boxed{\:\:\omega\left(u\right)\boldsymbol{=}u \cdot x\boldsymbol{-}f\left(x\right)\vphantom{\dfrac{a}{b}}\:\:} \tag{A-04}\label{A-04}$$$$ Now looking in above equation it seems mathematically illogical the argument that the function $$\:\omega\:$$ doesn't depend on the variable $$\:x\:$$ and must we write $$$$\omega\left(u,x\right)\stackrel{???}{\boldsymbol{=}}u \cdot x\boldsymbol{-}f\left(x\right) \tag{A-05}\label{A-05}$$$$ But this is not this case here because from \eqref{A-04} $$$$\dfrac{\partial\omega}{\partial x}\boldsymbol{=}u \boldsymbol{-}\dfrac{\partial f}{\partial x}\boldsymbol{=}\dfrac{\mathrm df}{\mathrm dx} \boldsymbol{-}\dfrac{\mathrm df}{\mathrm dx}\boldsymbol{=}0 \tag{A-06}\label{A-06}$$$$ that is $$\:\omega\:$$ is independent of $$\:x$$. It depends only on $$\:u\:$$ that's why we write $$\:\omega\left(u\right)$$.

In Figure-04 this fact is explained graphically : Suppose that a value $$\:u\in \left[u_1,u_2\right]\:$$ is given. This is like to give a direction, that is a line $$\:\varepsilon\:$$ at an angle $$\:\phi\boldsymbol{=}\arctan(u)$$. We find a unique line $$\:\varepsilon_t\:$$ tangent to the curve-graph of $$\:f\left(x\right)\:$$ and parallel to $$\:\varepsilon\:$$ which intersects the $$\:y-$$axis at $$\:\boldsymbol{-}\omega(u)$$. Beyond the value of the independent variable $$\:u\:$$ there is no need of any value of $$\:x$$. To the contrary, this value of $$\:x\:$$ is determined underground automatically from the contact point of the tangent line $$\:\varepsilon_t\:$$ with the graph.

We call the function $$\:\omega\left(u\right)\:$$ the Legendre transform of the function $$\:f\left(x\right)\:$$ with respect to the variable $$\:x$$.

Note that differentiating \eqref{A-04} with respect to $$\:u\:$$ we have $$$$x\boldsymbol{=}\dfrac{\mathrm d\omega\left(u\right)}{\mathrm du} \tag{A-07}\label{A-07}$$$$ So, the function $$\:f\left(x\right)\:$$ and its Legendre transform with respect to $$\:x\:$$, that is the function $$\:\omega\left(u\right)$$, fulfill the following set of equations \begin{align} f\left(x\right) \boldsymbol{+}\omega\left(u\right) & \boldsymbol{=}u \cdot x \tag{A-08a}\label{A-08a}\\ u & \boldsymbol{=}\dfrac{\mathrm df\left(x\right)}{\mathrm dx} \tag{A-08b}\label{A-08b}\\ x & \boldsymbol{=}\dfrac{\mathrm d\omega\left(u\right)}{\mathrm du} \tag{A-08c}\label{A-08c} \end{align}

If in above equations we interchange the roles as follows \begin{align} f & \boldsymbol{\rightleftarrows} \omega \tag{A-09a}\label{A-09a}\\ x & \boldsymbol{\rightleftarrows} u \tag{A-09b}\label{A-09b} \end{align} then equations \eqref{A-08a},\eqref{A-08b} and \eqref{A-08c} give respectively \begin{align} \omega\left(u\right)\boldsymbol{+} f\left(x\right)& \boldsymbol{=}x \cdot u \tag{A-10a}\label{A-10a}\\ x & \boldsymbol{=}\dfrac{\mathrm d\omega\left(u\right)}{\mathrm du} \tag{A-10b}\label{A-10b}\\ u & \boldsymbol{=}\dfrac{\mathrm df\left(x\right)}{\mathrm dx} \tag{A-10c}\label{A-10c} \end{align}

But this set of equations is identical to that of (A-08) : The function $$\:f\left(x\right)\:$$ is the Legendre transform of $$\:\omega\left(u\right)$$ with respect to $$\:u$$. That is application of two successive Legendre transformations returns the initial function.

$$\boldsymbol{\S\:}\textbf{B. Classical Mechanics - Lagrange and Hamilton functions}$$

In Classical Mechanics the Euler-Lagrange equation of motion for one degree of freedom is $$$$\dfrac{\mathrm d}{\mathrm d t}\left(\dfrac{\partial L}{\partial\dot q}\right)\boldsymbol{-}\dfrac{\partial L}{\partial q}\boldsymbol{=}0 \tag{B-01}\label{B-01}$$$$ where \begin{align} L\left(q,\dot q,t\right) & \boldsymbol{\equiv}\text{the Lagrange function} \tag{B-02a}\label{B-02a}\\ q & \boldsymbol{\equiv}\text{the generalized coordinate} \tag{B-02b}\label{B-02b}\\ \dot q & \boldsymbol{\equiv}\dfrac{\mathrm d q}{\mathrm d t} \tag{B-02c}\label{B-02c} \end{align} For the Legendre transform of the Lagrange function $$\:L\left(q,\dot q,t\right)\:$$ with respect to the independent variable $$\:\dot q\:$$ we replace all Variables, Functions and Differential Operators in $$\:\boldsymbol{\S\:}\textbf{A}\:$$ as follows \begin{align} \text{Variables}\:\:\: : \:\:\:& \left. \begin{cases} x\!\!\! &\!\!\! \boldsymbol{-\!\!\!-\!\!\!-\!\!\!\rightarrow} \dot q\\ u\!\!\! &\!\!\! \boldsymbol{-\!\!\!-\!\!\!-\!\!\!\rightarrow} p \end{cases}\right\} \tag{B-03a}\label{B-03a}\\ \text{Functions}\:\:\: : \:\:\:& \left. \begin{cases} f\!\!\! &\!\!\! \boldsymbol{-\!\!\!-\!\!\!-\!\!\!\rightarrow} L\\ \omega\!\!\! &\!\!\! \boldsymbol{-\!\!\!-\!\!\!-\!\!\!\rightarrow} H \end{cases}\right\} \tag{B-03b}\label{B-03b}\\ \text{Operators}\:\:\: : \:\:\:& \left. \begin{cases} \dfrac{\mathrm d \hphantom{x}}{\mathrm d x}\!\!\! &\!\!\! \boldsymbol{-\!\!\!-\!\!\!-\!\!\!\rightarrow} \dfrac{\partial \hphantom{x}}{\partial \dot q}\vphantom{\dfrac{a}{\dfrac{a}{b}}}\\ \dfrac{\mathrm d \hphantom{u}}{\mathrm d u}\!\!\! &\!\!\! \boldsymbol{-\!\!\!-\!\!\!-\!\!\!\rightarrow} \dfrac{\partial \hphantom{p}}{\partial p} \end{cases}\right\} \tag{B-03c}\label{B-03c} \end{align} Equations \eqref{A-08a},\eqref{A-08b} and \eqref{A-08c} give respectively \begin{align} H\left(q,p,t\right)\boldsymbol{+} L\left(q,\dot q,t\right) & \boldsymbol{=}p\,\dot q \tag{B-04a}\label{B-04a}\\ p & \boldsymbol{=}\dfrac{\partial L\left(q,\dot q,t\right)}{\partial \dot q} \tag{B-04b}\label{B-04b}\\ \dot q & \boldsymbol{=}\dfrac{\partial H\left(q,p,t\right)}{\partial p} \tag{B-04c}\label{B-04c} \end{align} So the Legendre transform of the Lagrange function $$\:L\left(q,\dot q,t\right)\:$$ with respect to the independent variable $$\:\dot q\:$$ is the Hamilton function $$\:H\left(q,p,t\right)\:$$, where from \eqref{B-04a} $$$$H\left(q,p,t\right) \boldsymbol{=}p\,\dot q\boldsymbol{-} L\left(q,\dot q,t\right) \tag{B-05}\label{B-05}$$$$ In the spirit of the discussion in $$\:\boldsymbol{\S\:}\textbf{A}\:$$ the Hamilton function $$\:H\left(q,p,t\right)\:$$ is independent of the variable $$\:\dot q$$, it depends on the independent variable $$\:p\boldsymbol{\equiv}\text{the generalized momentum}$$.

Equation \eqref{B-05} yields $$$$\dfrac{\partial H\left(q,p,t\right)}{\partial q}\boldsymbol{=}\boldsymbol{-}\dfrac{\partial L\left(q,\dot q,t\right)}{\partial q} \tag{B-06}\label{B-06}$$$$ From this equation and the definition of $$\:p$$, see equation \eqref{B-04b}, the Euler-Lagrange equation of motion \eqref{B-01} gives $$$$\dot p \boldsymbol{=}\boldsymbol{-}\dfrac{\partial H\left(q,p,t\right)}{\partial q} \tag{B-07}\label{B-07}$$$$ Equations \eqref{B-04c} and \eqref{B-07} together constitute the Hamilton equations of motion $$$$\text{Hamilton equations of motion}\:\:\: : \:\:\: \left. \begin{cases} \dot q & \!\!\boldsymbol{=}\boldsymbol{+}\dfrac{\partial H\left(q,p,t\right)}{\partial p}\vphantom{\dfrac{a}{\dfrac{a}{b}}}\\ \dot p & \!\!\boldsymbol{=}\boldsymbol{-}\dfrac{\partial H\left(q,p,t\right)}{\partial q} \end{cases}\right\} \tag{B-08}\label{B-08}$$$$

• Great and clarifying answer! I just made some minor modifications which don't do any harm to your reasoning. A pity though you didn't continue with $\boldsymbol{\S\:} B$ – descheleschilder May 4 '19 at 18:07
• @descheleschilder : Thanks for your attention and editing. – Frobenius May 4 '19 at 21:07
• @descheleschilder I totally agree with you. – Sebastiano Jun 26 '19 at 12:04

However, if $$H$$ is defined to be a function of $$q,p,t$$, then how can we define $$H(q,p,t) = \dot q *p - L(q,\dot q,t)$$, i.e $$\dot q$$ is not an argument of $$H$$ whereas it is in its definition.

As usual in a Legendre transform, the above expression for $$H$$ should be understood as a shortened notation for $$H(q,p,t) = \dot q(q,p,t) \cdot p-L(q,\dot q(q,p,t),t)$$ where $$\dot q(q,p,t)$$ is obtained by inverting the definition of $$p$$ $$p = \frac{\partial L}{\partial \dot q}(q, \dot q, t)$$ to obtain the function $$\dot q(q,p,t)$$.

• @onurcanbektas For nice Lagrangian because those are convex functions of $\dot q$. This is the typical situation. For not nice Lagrangians this is a separate issue. – GiorgioP May 1 '19 at 7:35
• @onurcanbektas I do not disagree. If you want a mathematically proper presentation, you should try Arnold. Goldstein is mediocre in many ways. – knzhou May 1 '19 at 10:55
• @onurcanbektas I agree that a better discussion (maybe in an appendix) of the issues elted to Legende transform could have been better than nothing. However I would not be too critic against Goldstein's choices. Nowadays, his book (in all its editions) can be seen as an excellent attempt to present in a contained way the core elements of Classical Mechanics as immediately useful for practical purposes. So, no differential geometry methods and no extensive discussion of many mathematical-physics issue. For them, there are tons of specialized literature or more comprehensive treatises. – GiorgioP May 1 '19 at 11:32
• "So, you are basically saying that the author is making assumptions without even saying (or maybe noticing), again." Welcome to physics @onurcanbektas – Jannik Pitt May 1 '19 at 12:58
• So, you are basically saying that the author is making assumptions without even saying - No, the author is writing for Physicists, not Mathematicians. The former are expected to understand the shorthands, notations, and formulations used in their field. The latter, naturally, are generalists. Physics often reads like mathematics with an accent and some of its own patois vocabulary. If you're in that field you just understand what is meant. If you're trying to parse it as a native speaker of Mathematics it can seem, at first glance, to be missing things. – J... May 1 '19 at 19:51

The Formula Goldstein has given (8.15) is not a definition of the Hamiltonian (because you are right in that, the Formula depends on $$\dot{q}$$, which is not an argument of the Hamiltonian. However, the formla can be understood as an equation we want $$H$$ to satisfy if the variable $$p$$ satisfies \begin{align} p = \frac{\partial L}{\partial \dot{q}}(q, \dot{q}, t) \end{align}

Unlike suggested in the previous version of this answer, $$p$$, $$q$$ and $$\dot{q}$$ can be independent variables in these equations.

By that now is also clear why the $$p \dot{q}$$ vanishes here:The differential of the term $$\dot{q} p$$ in formula 8.15 cancels with the one from the differential of $$L$$ in 8.13.

Written down: \begin{align} dH = d \dot{q} p - \dot{q} dp - dL \end{align}
With $$dL$$ from 8.13, you arrive at the same formula Goldstein arrives at.

Important Note from my side: Goldstein argues with the Legendre Transform here when talking about why the differential vanishes. In fact, the way he "defined" $$H$$ is a Legendre Transformation. However, since he started to define $$H$$ without making use of the term "Legendre-Transform", he could have argued without it later on as well when talking about the differentials. As I did, you can perfectly understand why $$d \dot{q} p$$ vanishes without making use of the term "Legendre-Transformation". Conversely, when Goldstein writes that $$d \dot{q} p$$ vanishes because of the "Legendre-Transformation", he implicitly means exactly what I wrote down.

• Time dependence of $q, \dot q$ and $p$ is simply misleading when discussing the relation between Hamiltonian and Lagrangian. They are intended to be functions defined for any value of their arguments, not only on the solution of the equations of motion. In the whole argument, the only role played by time is via the possible explicit dependence of $L$ and $H$ On $t$. – GiorgioP May 1 '19 at 7:31
• If I understand @GiorgioP, I agree; the time dependence gen. coordinates is just a way of saying that a solution of those equations is a curve in the phase space. In other words, when we obtain the solutions, we have a parametrisation of that particular solution wrt $t$, but when we are doing our analysis, $p,q,\dot q$ are just variables just like $x,y,z$. – onurcanbektas May 1 '19 at 7:37
• You are right, I thought too mutch. The only thing important is that $p$ satisfies 2.44, if that is the case, $p$, $q$ and $\dot{q}$ need not be thought of as a trajectory. I never wrote, however, that the trajectory needs to be a solution of the equations of motion. I'll edit my answer accordingly! – Quantumwhisp May 1 '19 at 10:13

Right, $$\dot{q}$$ is not an agrument of $$H$$ and you will only see it afterwards, but it is a function of time so you must make $$dH$$ from the definition (8.15) keeping it in mind and you must use $$dL$$ from (8.13'). Then you arrive at the right Hamiltonian differential. Note, eq. (8.16) misses a factor $$dt$$ at the last term (a typo).

• do you mean to subtract 8.17 from the differential of 8.15 by using 8.13', and then equation the coefficient of each $dq, dp, dt$'s to zero ? – onurcanbektas May 1 '19 at 7:17

First lets try out the legendre transformation on a particular example.

$$L = \frac12 m \dot{q}^2 - V(q),$$

according to Goldstein the hamiltonian for this system is,

$$H = \dot{q} p - L,$$

initially we think of $$p$$ and $$\dot{q}$$ as being independent variables. If we take $$\partial H / \partial \dot{q}$$ we will get,

$$\frac{\partial H}{\partial \dot{q}} = p - \frac{\partial L}{\partial \dot{q}},$$

if we now constrain ourselves to the surface $$p=\frac{\partial L}{\partial \dot{q}}$$ we find that the partial derivative of $$H$$ with respect to $$\dot{q}$$ vanishes.

For the purposes of calculating the dynamics then we would constrain our results,

$$H\Big|_{p=m\dot{q}} = \Big( p\dot{q} - L(\dot{q},q)\Big) \Big|_{p=m\dot{q}}$$

$$H\Big|_{p=m\dot{q}} = \Big( \frac{p^2}{m} - L(p/m,q)\Big) \Big|_{p=m\dot{q}}$$

$$H\Big|_{p=m\dot{q}} = \Big( \frac{p^2}{m} - \frac{p^2}{2m} + V(q) \Big) \Big|_{p=m\dot{q}}$$

$$H\Big|_{p=m\dot{q}} = \Big( \frac{p^2}{2m} + V(q) \Big) \Big|_{p=m\dot{q}}$$

This kind of "constraining our variables after the fact" practice is very common in Classical Mechanics.


text:
### MiniSymposium-14: Nonlocal Models: Analysis Aspects

Abstract: Nonlocal theories have been successfully employed in many settings, such as fracture modeling, image processing, biology, sand pile formation, and more. Nonlocal models use integral operators which take into account the cumulative behavior in the neighborhood of a point, and thus they allow solutions or domains to be discontinuous. On the other hand, nonlocal models based on integro-differential equations have fundamentally different mathematical structure and properties compared to their local PDE-based counterparts, requiring new and novel analytical and numerical techniques to analyze and utilize them. In this minisymposium we will focus on recent advances in nonlocal theories, including theoretical results (well-posedness, regularity, and asymptotic behavior of solutions), numerical methods and analysis (such as convergence and conditioning), and practical application (modeling, simulation, engineering analysis) related to nonlocal theories that have appeared in fracture modeling (peridynamics), nonlocal diffusion, biology, image processing, and more.

Organizers:

Saturday, October 19th, 2019 at 10:20 - 11:40 (268 Carver Hall)
10:20 - 10:40 Pablo Raúl Stinga,
Iowa State University
The Obstacle Problem for a Fractional Monge-Ampère Equation

We study the obstacle problem for a degenerate fractional Monge-Ampère equation. This nonlocal operator was introduced by Caffarelli and Charro. We show the existence of a unique, globally Lipschitz and semiconcave classical solution, at which the fractional Monge-Ampère equation becomes locally uniformly elliptic. The uniform ellipticity is used to deduce local regularity of the solution and the free boundary. This is joint work with Yash Jhaveri (IAS).

10:40 - 11:00 Mikil Foss,
University of Nebraska - Lincoln
Poincare Inequalities for Nonhomogeneous Nonlocal Operators

Convolution operators with integrable kernels have become an important component for nonlocal models of phenomenon with discontinuities, such as fracture and tearing. For these models, existence and stability results often rely on coercivity properties of the operator in an appropriate space. I will present some new Poincare type inequalities for nonhomogeneous operators that provide coercivity in Lebesgue spaces.

11:00 - 11:20 Simon Bortz,
University of Washington - Seattle
Some Soft Tools for PDE Regularity

Complex analysis and functional analysis come together in some very interesting and powerful ways. Perhaps the most useful fusion of these fields is the method of complex interpolation, which will be the primary tool in this talk. We will show how to exploit features of complex interpolation in a variety of situations, (local and non-local elliptic and parabolic equations) to show improved integrability and/or differentiability properties of solutions to differential equations. Time permitting we will discuss some future directions.

11:20 - 11:40 Hatim Khudhair,
Missouri University of Science and Technology
Turing Patterns in the Schnakenberg Equations: From Normal to Anomalous Diffusion

In recent years, anomalous diffusion has been observed in many biological experiments. Instead of classical Laplace operator, the anomalous diffusion is described by the fractional Laplacian, leading to nonlocal models. In this talk, we will discuss the Turing patterns of the Schnakenberg equation and compare the pattern selection under normal and anomalous diffusion. Our analysis shows that the wave number of the Turing instability increases with the exponent of the fractional Laplacian. The interplay of the nonlinearity and long-range diffusion are studied numerically, and especially comparisons are provided to understand the nonlocal effects of the fractional Laplacian.

Saturday, October 19th, 2019 at 14:40 - 16:00 (268 Carver Hall)
14:40 - 15:00 Cory Wright,
Lehigh University
Nonlocal Trace Theorems and Their Applications

The classical trace theory demonstrates that the trace operator on continuous functions can be extended continuously as a map on sobolev spaces $$H^1$$ to $$H^{1/2}$$. This theory has a myriad of useful applications throughout different aspects of analysis. It is then natural to study similar theories for the nonlocal framework, more specifically peridynamics. We will show two different versions of a nonlocal trace theorem involving peridynamic kernels. We define a nonlocal energy space motivated by a nonlocal gradient operator and use this as a basis to create a trace theorem involving $$L^2$$. Viewing this energy space as a nonlocal analogue to $$H^1$$ also motivates a nonlocal trace theorem incorporating a nonlocal version of the space $$H^{1/2}$$ as in the classical trace theory. We then will demonstrate some useful applications of these theorems to different problems in nonlocal analysis.

15:00 - 15:20 Mary Vaughan,
Iowa State University
Fractional Derivatives in One-sided Weighted Sobolev Spaces

Fractional derivatives can be defined in the most general way in the distributional sense. We start from this point and then show that the Marchaud pointwise formulas for these operators indeed hold for functions in more general classes. The classes we consider are weighted Sobolev spaces with one-sided Sawyer weights. These classes capture the one-sided nature of fractional derivatives. The pointwise and norm limits as the orders of the derivatives converge to an integer are also analyzed.

15:20 - 15:40 Animesh Biswas,
Iowa State University
Regularity Theory for Nonlocal Space-time Master Equations

We analyze recent novel regularity theory for fractional power of parabolic operators in divergence form. These equations are fundamental in continuous time random walk models and appear as generalized Master equation. These equations are non-local in nature and were studied by Luis Caffarelli and Luis Silvestre. We developed a parabolic method of semigroups that allows us to prove a local extension problem. As a consequence we obtain interior and boundary Harnack inequalities and sharp interior and global parabolic Schauder estimates. For the latter, we also prove a characterization of the correct intermediate parabolic Hölder spaces in the spirit of Sergio Campanato. This is joint work with Marta de León-Contreras (Universidad Autonoma de Madrid, Spain) and Pablo Raúl Stinga (Iowa State University).

15:40 - 16:00 Evan Camrud,
Iowa State University
Fractional Operators via Analytic Interpolation of Integer Powers

Although the study of functional calculus has already established necessary and sufficient conditions for operators to be fractionalized, this paper aims to use our well-conceived notion of integer powers of operators to construct non-integer powers of operators. In doing so, we not only provide a more intuitive understanding of fractional theories, but also provide a framework for producing new fractional theories. Such interpolations allow one to approximate fractional powers by finite sums of integer powers of operators, and thus may find much use in numerical analysis of fractional PDEs and the time-frequency analysis of the fractional Fourier transform. Further, these results provide an interpolation procedure for the Riemann zeta function.

Coordinated by
Iowa State University


text:
# Contributions to a conference proceedings

 2019-05-2020:52 [PUBDB-2019-02319] Journal Article/Contribution to a conference proceedings HERMES Collaboration Medium-indused modification of kaons spectra measured in SIDIS at HERMES 24th International Baldin Seminar on High Energy Physics Problems, ISHEPP 2018, DubnaDubna, Russia, 17 Sep 2018 - 22 Sep 2018   The predicted high sensitivity of the nuclear modification factor for K$^−$ in SIDIS due to the QCD based effect of medium-induced flavour conversion in the fragmentation function is studied at HERMES experiment. Unlike π+, π− and K$^+$ nuclear modification factor for K$^−$ is assumed to increase at high value of Bjorken variable $x_B$ and a hadron fraction energy $z$. [...] OpenAccess: PDF PDF (PDFA); External link: Fulltext 2019-05-1612:47 [PUBDB-2019-02297] Journal Article/Contribution to a conference proceedings et al Strong-field physics in the molecular frame XXI International Conference on Ultrafast Phenomena 2018, UP 2018, HamburgHamburg, Germany, 15 Jul 2018 - 20 Jul 2018   Laser-aligned carbonyl-sulfide molecules were strong-field ionized using mid-infrared light. Investigating the strong-field effects in the molecular frame allowed to add novel facets to the understanding of the intrinsic nature of strong-field physics.. Fulltext: GIF; OpenAccess: PDF; 2019-05-1609:36 [PUBDB-2019-02294] Journal Article/Contribution to a conference proceedings et al Generation and complete temporal characterization of 5-fs EUV pulses XXI International Conference on Ultrafast Phenomena 2018, UP 2018, HamburgHamburg, Germany, 15 Jul 2018 - 20 Jul 2018   Extreme-ultraviolet pulses as short as 5 fs, produced by high- order harmonic generation and selected by a time-delay compensated monochromator, were fully reconstructed thanks to an advanced ptychographic technique, while most commonly used reconstruction methods fail. These pulses constitute the shortest extreme ultraviolet pulses ever achieved via a time-delay compensated monochromator.. Fulltext: PNG; OpenAccess: PDF PDF (PDFA); External link: Fulltext 2019-05-1510:26 [PUBDB-2019-02285] Journal Article/Report/Contribution to a conference proceedings Sala, F. A light dimuon resonance in B decays? [arXiv:1809.11061; DESY-18-172] 7th Workshop on Theory, Phenomenology and Experiments in Flavour Physics : The Future of BSM Physics., FPCapri 2018, Anacapri, CapriAnacapri, Capri, Italy, 8 Jun 2018 - 10 Jun 2018Contributions / Info Nuclear and particle physics proceedings 303-305, 14 - 19 (2018) [10.1016/j.nuclphysbps.2019.03.004]   The observed deviations from the Standard Model in several $b \to s\mu\mu$ processes can be explained in terms of a new vector boson produced on-shell in B meson decays. A mass of 2.5-3 GeV and a total width of 10-20% allow to hide the associated dimuon bump in the poorly known charmonium region, and the large invisible decay width can be interpreted in terms of Dark Matter. [...] External link: Fulltext 2019-04-1711:55 [PUBDB-2019-01990] Journal Article/Contribution to a conference proceedings STEFANIAK, T. A Higgs Boson at 96 GeV?! Prospects for Charged Higgs Discovery at Colliders, CHARGED2018, UppsalaUppsala, Sweden, 25 Sep 2018 - 28 Sep 2018 Proceedings of Science / International School for Advanced Studies (CHARGED2018), 016 (2019) [10.22323/1.339.0016]   We briefly summarize some searches for Higgs bosons with a mass of $m_ϕ ≤ 110$~GeV at LEP and the LHC. We discuss a possible signal in the diphoton decay mode at $m_ϕ∼96~$GeV as reported by CMS, together with a $∼2σ$ hint in the $b\bar{b}$ final state at LEP [...] OpenAccess: PDF PDF (PDFA); External link: Fulltext 2019-04-0819:48 [PUBDB-2019-01909] Contribution to a conference proceedings Sarler, B. Computational Modelling of Gas Focused thin Liquid Sheets VIII International Conference on Computational Methods for Coupled Problems in Science and Engineering, COUPLED PROBLEMS 2019, BarcelonaBarcelona, Spain, 3 Jun 2019 - 5 Jun 2019 8 pp. (2019) [10.3204/PUBDB-2019-01909]   OpenAccess: PDF PDF (PDFA); 2019-04-0819:44 [PUBDB-2019-01908] Contribution to a conference proceedings et al Validation of a Computational Model for a Coupled Liquid and Gas Flow in Micro-Nozzles VIII International Conference on Computational Methods for Coupled Problems in Science and Engineering, COUPLED PROBLEMS 2019, BarcellonaBarcellona, Spain, 3 Jun 2019 - 5 Jun 2019 7 pp. (2019) [10.3204/PUBDB-2019-01908]   The work presents verification of a numerical model for micro-jet focusing, where a coupled liquid and gas flow occurs in a gas dynamic virtual nozzle (GDVN). Nozzlesof this type are usedinserial femtosecond crystallographyexperimentsto deliver samplesintoX-ray beam. [...] OpenAccess: PDF PDF (PDFA); 2019-04-0808:36 [PUBDB-2019-01881] Journal Article/Contribution to a conference proceedings et al A Microfluidic Device for the Investigation of Rapid Gold Nanoparticle Formation in Continuous Turbulent Flow 16th International Conference on X-ray Absorption Fine Structure , XAFS16, KarlsruheKarlsruhe, Germany, 23 Aug 2015 - 28 Aug 2015   A new setup with an integrated microfluidic chip with small dead time, high time resolution and compatibility with in situ X-ray absorption (XAS) measurements is presented. It can also be combined with a free liquid jet [...] OpenAccess: PDF PDF (PDFA); 2019-04-0315:00 [PUBDB-2019-01848] Journal Article/Contribution to a conference proceedings/Contribution to a book et al High-flux XAFS-beamline P64 at PETRA III Proceedings of the 13th International Conference on Synchrotron Radiation Instrumentation, SRI2018, TaipeiTaipei, Taiwan, 10 Jun 2018 - 15 Jun 2018 AIP conference proceedings 2054, 060031 (2019) [10.1063/1.5084662]   Beamline P64 at PETRA III is dedicated to x-ray absorption spectroscopy experiments which require high flux, like quick extended x-ray absorption fine structure spectroscopy on the sub-second time scale, high-resolution resonant emission spectroscopy, and x-ray absorption fine-structure spectroscopy of highly diluted systems. The beamline is installed at the high-energy storage ring PETRA III. [...] Published on 2019-01-16. Available in OpenAccess from 2020-01-16.: PDF PDF (PDFA); External link: Fulltext 2019-04-0309:52 [PUBDB-2019-01842] Journal Article/Contribution to a conference proceedings/Contribution to a book et al Simulation framework SYRIS tested for microtomography applications at the imaging beamline P05/PETRA III Proceedings of the 13th International Conference on Synchrotron Radiation Instrumentation, SRI2018, TaipeiTaipei, Taiwan, 10 Jun 2018 - 14 Jun 2018 AIP conference proceedings 2054, 060084 (2019) [10.1063/1.5084715]   The Helmholtz-Zentrum Geesthacht, Germany, is operating the user experiments for microtomography at the beamlines P05 and P07 using synchrotron radiation produced in the storage ring PETRA III at DESY, Hamburg, Germany. In recent years the software pipeline and sample changing hardware for performing high throughput experiments were developed. [...] Published on 2019-01-16. Available in OpenAccess from 2020-01-16.: PDF PDF (PDFA);


text:
# Does Cauchy continuity imply uniform continuity? [No.] [closed]

It is well known that if $X$ is a first countable topological space and $Y$ is a topological space, then $f : X \rightarrow Y$ is continuous iff $$\forall x \in {\rm map}(\mathbb{N},X),\forall p \in X \quad x_{n} \rightarrow p \Rightarrow f(x_{n}) \rightarrow f(p)$$

It is also well known that if $X$ and $Y$ are metric spaces and $f : X \rightarrow Y$ is uniformly continuous, then $f$ maps Cauchy sequences to Cauchy sequences.

By analogy it seems plausible that if a function between metric spaces maps Cauchy sequences to Cauchy sequences then it must be uniformly continuous. However mimicking the proof of the analogous result for continuous maps doesn't work, which makes me think the result if false. Does anyone know any counterexamples?

Also on the uniform continuity wikipedia page, it says that the result is true if $X$ and $Y$ are subsets of $\mathbb{R}^{n}$. EDIT: It actually doesn't say this, I misread the page.

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## closed as off-topic by Ricardo Andrade, Willie Wong, Karl Schwede, Ryan Budney, David RobertsApr 16 '14 at 1:51

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – Ricardo Andrade, Karl Schwede, Ryan Budney, David Roberts
If this question can be reworded to fit the rules in the help center, please edit the question.

I wrote the portion of the wikipedia page en.wikipedia.org/wiki/Uniform_continuity dealing with the extension problem and the condition of Cauchy continuity. Formerly the page only implied (in the usual mathematical sense, but also in the conventional sense of suggested, rather than said explicitly) that Cauchy continuity does not imply uniform continuity, since any continuous, non-uniformly continuous function on a complete space gives a counterexample, and such a function, $x \mapsto x^2$, was provided. I just edited the article to be more explicit on this point. –  Pete L. Clark Jun 12 '10 at 17:08
@Daniel Barter: you write "it says that the result is true if $X$ and $Y$ are subsets of $\mathbb{R}^n$". Please be more specific, so we can tell if there is an error and fix it if necessary: which page says that what result is true? –  Pete L. Clark Jun 12 '10 at 17:09
Summary of the question: "A implies B", by analogy it's plausible that "B implies A". Is it true? A="function is uniformly continuous", B="function is Cauchy continuous". –  Victor Protsak Jun 12 '10 at 21:51
@Pete: I was under the impression that the uniform continuity wikipedia page said that if a function between two subsets of euclidean space mapped Cauchy sequences to Cauchy sequences then it was uniformly continuous. However i misread. –  Daniel Barter Jun 12 '10 at 23:50

No it's not true.

f(x) = x^2 on whole real line.

It maps Cauchy sequences to Cauchy sequences but it's not uniformly continuous on the whole real line.

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Take for $X$ the disjoint union of a sequence $A_n$ of two point sets. Set $d(p,q)=1/n$ if $p$ and $q$ are different points in $A_n$ and $d(p,q)=1$ if $p\in A_n$ and $q\in A_m$ with $m\not=n$. Every Cauchy sequence in $X$ is eventually constant. Consider the identity mapping from $X$ with this metric to $X$ with the discrete metric.

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X in this metric is also a discrete space, as the ball around x with radius 1/n, when x is in A_n, is {x}. Any function from a discrete space is continuous. –  Henno Brandsma Jun 12 '10 at 6:51
It is a counterexample to the statement "a map that preserves Cauchy sequences needs to be uniformly continuous", as required by the OP. The above counterexample shows that the statement fails to be true even with the identity map on the same topological space, with respect to 2 different metrics. –  Pietro Majer Jun 12 '10 at 7:11

Several people have already given examples to the effect that preservation of Cauchyness is not enough to prove that a map is uniformly continuous. It is still possible however, to characterize uniform continuity in terms of sequences. In case you are interested here goes the result (for metric spaces only, for uniform spaces you would need nets (or filters)).

Theor: Let $f:X\to Y$ be a map between metric spaces (both metrics denoted by d). Then $f$ is uniformly continuous iff for every pair of sequences $(x_n)$ and $(z_n)$ in $X$ such that $d(x_n, z_n)$ converges to $0$, then $d(f(x_n), f(z_n))$ converges to zero.

Proof: exercise to the reader.

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Well, here is [most probably another] Wikipedia page: http://en.wikipedia.org/wiki/Cauchy-continuous_function. HTH.

Alternatively, you may use the function $f:\mathbb{R\rightarrow\mathbb{R}}$ , expressed by $f\left(x\right)=x^{2}$ to show that it is possible for a continuous function to send Cauchy sequences to Cauchy sequences without being uniformly continuous.

As for the continuity question: the function must be continuous. For, embed $Y$ into its completion, say $Y^{\sim}$, let $p\in X$ , and let $\left(x_{n}\right)$ be a sequence in $X$ converging to $p$ . Then the sequence $\left(x_{1}, p, x_{2}, p,...\right)$ is Cauchy in $X$ , isn't it ? And, therefore, its image by $f$ should be a Cauchy sequence in $Y$ , hence convergent in $Y^{\sim}$ . Yet, that image contains a constant subsequence... isn't it ?

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I am a bit sceptical about this page, for two reasons. No link to a proof is given and secondly, they have hyper linked the term "if and only if"... –  Daniel Barter Jun 12 '10 at 5:49
@Daniel: It is certainly a good idea to be somewhat skeptical about wikipedia articles. But this one seems basically sound. They do give a reference, to a book even. Regarding the hyperlink, well, the style and quality of the writing on wikipedia varies, to put it mildly. –  Harald Hanche-Olsen Jun 12 '10 at 11:54
An answer which, in fact, is not a truly answer, is over voted, due to the name of the author. Some annoying comments about this there were discreetly erased. This is due to the "young wolves" mentality: "Let’s catch a big name for our site". Another answer, which is correct, but partial, and following my answer and being included in this, is also fully voted. Still, my answer is not yet voted. In conclusion, I'll give myself a "-1", just as a protest, not because I'm frustated, but since I'm sickened of hypocrisy. –  Ady Jun 12 '10 at 12:09
OK, since I can't, I'll ask someone downvoting me. –  Ady Jun 12 '10 at 12:11
You are right to be critical of my answer, Ady, as $x^2$ is quicker and more elementary. I was thinking that the implication is true for totally bounded spaces and wanted a more or less minimal and simple counterexample for bounded (and complete) spaces. –  Bill Johnson Jun 12 '10 at 15:49


text:
+0

0
51
1

What is the largest value of $x$ such that the expression

$$$\dfrac{x+1}{8x^2-65x+8}$$$

is not defined?

Guest Dec 29, 2017

#1
+5931
+1

The expression  $$\frac{x+1}{8x^2-65x+8}$$   is not defined when the denominator equals zero.

So to find all the  x  values that cause the expression to be undefined, set the denominator = 0 .

8x2 - 65x + 8  =  0            Now we need to solve this equation for  x .

We can factor the left side like this....

8x2 - 64x - x + 8  =  0

Factor  8x  out of the first two terms, factor  -1  out of the last two terms.

8x(x - 8) - 1(x - 8)  =  0

Factor  x-8  out of both terms.

(x - 8)(8x - 1)  =  0

Set each factor equal to zero and solve for  x .

x - 8  =  0          or          8x - 1  =  0

x  =  8               or          8x  =  1

x  =  1/8

So the expression is not defined when  x = 8  and when  x = 1/8  .

The largest of these values is  8 .

hectictar  Dec 29, 2017
Sort:

#1
+5931
+1

The expression  $$\frac{x+1}{8x^2-65x+8}$$   is not defined when the denominator equals zero.

So to find all the  x  values that cause the expression to be undefined, set the denominator = 0 .

8x2 - 65x + 8  =  0            Now we need to solve this equation for  x .

We can factor the left side like this....

8x2 - 64x - x + 8  =  0

Factor  8x  out of the first two terms, factor  -1  out of the last two terms.

8x(x - 8) - 1(x - 8)  =  0

Factor  x-8  out of both terms.

(x - 8)(8x - 1)  =  0

Set each factor equal to zero and solve for  x .

x - 8  =  0          or          8x - 1  =  0

x  =  8               or          8x  =  1

x  =  1/8

So the expression is not defined when  x = 8  and when  x = 1/8  .

The largest of these values is  8 .

hectictar  Dec 29, 2017

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# How to calculate return on investment for an adjustment to a complex options position?

Say I currently hold a set of options positions with the same symbol/expiry that collectively have a net present value based on the estimated value at expiration of +10. I could also liquidate the positions now at a value of +6.

I am considering a set of multiple transactions with the same symbol/expiry. If I executed these transactions, I would net +8 including commissions. The estimated present value at expiration for the resulting positions would decline by -3 to 7. I could also liquidate the entire position after the transactions at a value of +5.

How to I determine the ROI of executing these transactions?