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Factorization gave four distinct sub-polynomials, each with multitude 2. One die could contain 0, 1, or 2 of each of these with the remaining factors on the other die. That means there are $3^4=81$ different possible dice combinations. I could continue with a trail-and-error approach, but I wanted to be more efficient and elegant.
What follows is the result of thinking about the problem for a while. Like most math solutions to interesting problems, ultimate solutions are typically much cleaner and more elegant than the thoughts that went into them. Problem solving is a messy–but very rewarding–business.
SOLUTION
Here are my insights over time:
1) I realized that the $x^2$ term would raise the power (face values) of the desired dice, but would not change the coefficients (number of faces). Because Steve asked for dice with all positive face values. That meant each desired die had to have at least one x to prevent non-positive face values.
2) My first attempt didn’t create 6-sided dice. The sums of the coefficients of the sub-polynomials determined the number of sides. That sum could also be found by substituting $x=1$ into the sub-polynomial. I want 6-sided dice, so the final coefficients must add to 6. The coefficients of the factored polynomials of any die individually must add to 2, 3, or 6 and have a product of 6. The coefficients of $(x+1)$ add to 2, $\left( x^2+x+1 \right)$ add to 3, and $\left( x^2-x+1 \right)$ add to 1. The only way to get a polynomial coefficient sum of 6 (and thereby create 6-sided dice) is for each die to have one $(x+1)$ factor and one $\left( x^2+x+1 \right)$ factor. | {
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3) That leaves the two $\left( x^2-x+1 \right)$ factors. They could split between the two dice or both could be on one die, leaving none on the other. We’ve already determined that each die already had to have one each of the x, $(x+1)$, and $\left( x^2+x+1 \right)$ factors. To also split the $\left( x^2-x+1 \right)$ factors would result in the original dice: Two normal 6-sided dice. If I want different dice, I have to load both of these factors on one die.
That means there is ONLY ONE POSSIBLE alternative for two 6-sided dice that have the same sum distribution as two normal 6-sided dice.
One die would have single faces of 8, 6, 5, 4, 3, and 1. The other die would have one 4, two 3s, two 2s, and one 1. And this is exactly the result of the famous(?) Sicherman Dice.
If a 0 face value was allowed, shift one factor of x from one polynomial to the other. This can be done two ways.
The first possibility has dice with faces {9, 7, 6, 5, 4, 2} and {3, 2, 2, 1, 1, 0}, and the second has faces {7, 5, 4, 3, 2, 0} and {5, 4, 4, 3, 3, 2}, giving the only other two non-negative solutions to the Sicherman Dice.
Both of these are nothing more than adding one to all faces of one die and subtracting one from from all faces of the other. While not necessary to use polynomials to compute these, they are equivalent to multiplying the polynomial of one die by x and the other by $\frac{1}{x}$ as many times as desired. That means there are an infinite number of 6-sided dice with the same sum distribution as normal 6-sided dice if you allow the sides to have negative faces. One of these is
corresponding to a pair of Sicherman Dice with faces {6, 4, 3, 2, 1, -1} and {1,5,5,4,4,3}.
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CONCLUSION:
There are other very interesting properties of Sicherman Dice, but this is already a very long post. In the end, there are tremendous connections between probability and polynomials that are accessible to students at the secondary level and beyond. And CAS keeps the focus on student learning and away from the manipulations that aren’t even the point in these explorations.
Enjoy.
Birthdays, CAS, Probability, and Student Creativity
Many readers are familiar with the very counter-intuitive Birthday Problem:
It is always fun to be in a group when two people suddenly discover that they share a birthday. Should we be surprised when this happens? Asked a different way, how large a group of randomly selected people is required to have at least a 50% probability of having a birthday match within the group?
I posed this question to both of my sections of AP Statistics in the first week of school this year. In a quick poll, one section had a birthday match–two students who had taken classes together for a few years without even realizing what they had in common. Was I lucky, or was this a commonplace occurrence?
Intrigue over this question motivated our early study of probability. The remainder of this post follows what I believe is the traditional approach to the problem, supplemented by the computational power of a computer algebra system (CAS)–the TI Nspire CX CAS–available on each of my students’ laptops.
Initial Attempt:
Their first try at a solution was direct. The difficulty was the number of ways a common birthday could occur. After establishing that we wanted any common birthday to count as a match and not just an a priori specific birthday, we tried to find the number of ways birthday matches could happen for different sized groups. Starting small, they reasoned that | {
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• If there were 2 people in a room, there was only 1 possible birthday connection.
• If there were 3 people (A, B, and C), there were 4 possible birthday connections–three pairs (A-B, A-C, and B-C) and one triple (A-B-C).
• For four people (A, B, C, and D), they realized they had to look for pair, triple, and quad connections. The latter two were easiest: one quad (A-B-C-D) and four triples (A-B-C, A-B-D, A-C-D, and B-C-D). For the pairs, we considered the problem as four points and looked for all the ways we could create segments. That gave (A-B, A-C, A-D, B-C, B-D, and C-D). These could also occur as double pairs in three ways (A-B & C-D, A-C & B-D, and A-D & B-C). All together, this made 1+4+6+3=14 ways.
This required lots of support from me and was becoming VERY COMPLICATED VERY QUICKLY. Two people had 1 connection, 3 people had 4 connections, and 4 people had 14 connections. Tracking all of the possible connections as the group size expanded–and especially not losing track of any possibilities–was making this approach difficult. This created a perfect opportunity to use complement probabilities.
While there were MANY ways to have a shared birthday, for every sized group, there is one and only one way to not have any shared birthdays–they all had to be different. And computing a probability for a single possibility was a much simpler task.
We imagined an empty room with random people entering one at a time. The first person entering could have any birthday without matching anyone, so $P \left( \text{no match with 1 person} \right) = \frac{365}{365}$ . When the second person entered, there were 364 unchosen birthdays remaining, giving $P \left( \text{no match with 2 people} \right) = \frac{365}{365} \cdot \frac{364}{365}$, and $P \left( \text{no match with 3 people} \right) = \frac{365}{365} \cdot \frac{364}{365} \cdot \frac{363}{365}$. And the complements to each of these are the probabilities we sought: | {
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$P \left( \text{birthday match with 1 person} \right) = 1- \frac{365}{365} = 0$
$P \left( \text{birthday match with 2 people} \right) = 1- \frac{365}{365} \cdot \frac{364}{365} \approx 0.002740$
$P \left( \text{birthday match with 3 people} \right) = 1- \frac{365}{365} \cdot \frac{364}{365} \cdot \frac{363}{365} \approx 0.008204$.
The probabilities were small, but with persistent data entry from a few classmates, they found that the 50% threshold was reached with 23 people.
The hard work was finished, but some wanted to find an easier way to compute the solution. A few students noticed that the numerator looked like the start of a factorial and revised the equation:
$\begin{matrix} \displaystyle P \left( \text{birthday match with n people} \right ) & = & 1- \frac{365}{365} \cdot \frac{364}{365} \dots \frac{(366-n)}{365} \\ \\ & = & 1- \frac{365 \cdot 364 \dots (366-n)}{365^n} \\ \\ & = & 1- \frac{365\cdot 364 \dots (366-n)\cdot (366-n-1)!}{365^n \cdot (366-n-1)!} \\ \\ & = & 1- \frac{365!}{365^n \cdot (365-n)!} \end{matrix}$
It was much simpler to plug in values to this simplified equation, confirming the earlier result.
Not everyone saw the “complete the factorial” manipulation, but one noticed in the first solution the linear pattern in the numerators of the probability fractions. While it was easy enough to write a formula for the fractions, he didn’t know an easy way to multiply all the fractions together. He had experience with Sigma Notation for sums, so I introduced him to Pi Notation–it works exactly the same as Sigma Notation, except Pi multiplies the individual terms instead of adding them. On the TI-Nspire, the Pi Notation command is available in the template menu or under the calculus menu.
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Conclusion:
I really like two things about this problem: the extremely counterintuitive result (just 23 people gives a 50% chance of a birthday match) and discovering the multiple ways you could determine the solution. Between student pattern recognition and my support in formalizing computation suggestions, students learned that translating different recognized patterns into mathematics symbols, supported by technology, can provide different equally valid ways to solve a problem.
Now I can answer the question I posed about the likelihood of me finding a birthday match among my two statistics classes. The two sections have 15 and 21 students, respectively. The probability of having at least one match is the complement of not having any matches. Using the Pi Notation version of the solution gives
I wasn’t guaranteed a match, but the 58.4% probability gave me a decent chance of having a nice punch line to start the class. It worked pretty well this time!
Extension:
My students are currently working on their first project, determining a way to simulate groups of people entering a room with randomly determined birthdays to see if the 23 person theoretical threshold bears out with experimental results.
Monty Hall Continued
In my recent post describing a Monty Hall activity in my AP Statistics class, I shared an amazingly crystal-clear explanation of how one of my new students conceived of the solution:
If your strategy is staying, what’s your chance of winning? You’d have to miraculously pick the money on the first shot, which is a 1/3 chance. But if your strategy is switching, you’d have to pick a goat on the first shot. Then that’s a 2/3 chance of winning.
Then I got a good follow-up question from @SteveWyborney on Twitter:
Returning to my student’s conclusion about the 3-door version of the problem, she said,
The fact that there are TWO goats actually can help you, which is counterintuitive on first glance. | {
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The fact that there are TWO goats actually can help you, which is counterintuitive on first glance.
Extending her insight and expanding the problem to any number of doors, including Steve’s proposed 1,000,000 doors, the more goats one adds to the problem statement, the more likely it becomes to win the treasure with a switching doors strategy. This is very counterintuitive, I think.
For Steve’s formulation, only 1 initial guess from the 1,000,000 possible doors would have selected the treasure–the additional goats seem to diminish one’s hopes of ever finding the prize. Each of the other 999,999 initial doors would have chosen a goat. So if 999,998 goat-doors then are opened until all that remains is the original door and one other, the contestant would win by not switching doors iff the prize was initially randomly selected, giving P(win by staying) = 1/1000000. The probability of winning with the switching strategy is the complement, 999999/1000000.
IN RETROSPECT:
My student’s solution statement reminds me on one hand how critically important it is for teachers to always listen to and celebrate their students’ clever new insights and questions, many possessing depth beyond what students realize.
The solution reminds me of a several variations on “Everything is obvious in retrospect.” I once read an even better version but can’t track down the exact wording. A crude paraphrasing is
The more profound a discovery or insight, the more obvious it appears after.
I’d love a lead from anyone with the original wording.
REALLY COOL FOOTNOTE:
Adding to the mystique of this problem, I read in the Wikipedia description that even the great problem poser and solver Paul Erdős didn’t believe the solution until he saw a computer simulation result detailing the solution.
Probability and Monty Hall | {
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Probability and Monty Hall
I’m teaching AP Statistics for the first time this year, and my first week just ended. I’ve taught statistics as portions of other secondary math courses and as a semester-long community college class, but never under the “AP” moniker. The first week was a blast.
To connect even the very beginning of the course to previous knowledge of all of my students, I decided to start the year with a probability unit. For an early class activity, I played the classic Monte Hall game with the classes. Some readers will recall the rules, but here they are just in case you don’t know them.
1. A contestant faces three closed doors. Behind one is a new car. There is a goat behind each of the other two.
2. The contestant chooses one of the doors and announces her choice.
3. The game show host then opens one of the other two doors to reveal a goat.
4. Now the contestant has a choice to make. Should she
1. Always stay with the door she initially chose, or
2. Always change to the remaining unopened door, or
3. Flip a coin to choose which door because the problem essentially has become a 50-50 chance of pure luck.
Historically, many people (including many very highly educated, degree flaunting PhDs) intuit the solution to be “pure luck”. After all, don’t you have just two doors to choose from at the end?
In one class this week, I tried a few simulations before I posed the question about strategy. In the other, I posed the question of strategy before any simulations. In the end, very few students intuitively believed that staying was a good strategy, with the remainder more or less equally split between the “switch” and “pure luck” options. I suspect the greater number of “switch” believers (and dearth of stays) may have been because of earlier exposure to the problem.
I ran my class simulation this way: | {
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I ran my class simulation this way:
• Students split into pairs (one class had a single group of 3).
• One student was the host and secretly recorded a door number.
• The class decided in advance to always follow the “shift strategy”. [Ultimately, following either stay or switch is irrelevant, but having all groups follow the same strategy gives you the same data in the end.]
• The contestant then chose a door, the host announced an open door, and the contestant switched doors.
• The host then declared a win or loss bast on his initial door choice in step two.
• Each group repeated this 10 times and reported their final number of wins to the entire class.
• This accomplished a reasonably large number of trials from the entire class in a very short time via division of labor. Because they chose the shift strategy, my two classes ultimately reported 58% and 68% winning percentages.
Curiously, the class that had the 58% percentage had one group with just 1 win out of 10 and another winning only 4 of 10. It also had a group that reported winning 10 of 10. Strange, but even with the low, unexpected probabilities, the long-run behavior from all groups still led to a plurality winning percentage for switching.
Here’s a verbatim explanation from one of my students written after class for why switching is the winning strategy. It’s perhaps the cleanest reason I’ve ever heard.
The faster, logical explanation would be: if your strategy is staying, what’s your chance of winning? You’d have to miraculously pick the money on the first shot, which is a 1/3 chance. But if your strategy is switching, you’d have to pick a goat on the first shot. Then that’s a 2/3 chance of winning. In a sense, the fact that there are TWO goats actually can help you, which is counterintuitive on first glance. | {
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Engaging students hands-on in the experiment made for a phenomenal pair of classes and discussions. While many left still a bit disturbed that the answer wasn’t 50-50, this was a spectacular introduction to simulations, conditional probability, and cool conversations about the inevitability of streaks in chance events.
For those who are interested, here’s another good YouTube demonstration & explanation. | {
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# Derive Closed form sum of N^2
Can anyone explain to me how you would derive this ? I have this question asked in a CS class and can't figure out how to derive it. it has to be derived as you would with sum of N
ex
1 2 3 ...... N
N N-1 N-2 ....1
---------------------
N+1 + N+1 + .... N+1 = N(N+1) SINCE THIS ADDITION IS 2 * THIS SUM THEN CLOSED FORM IS N(N+1)/ 2
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What do you want? $\sum_{N=1}^k N^2 =\text{?}$ – draks ... Mar 13 '14 at 6:39
yes, how to derive the closed form of that sum – Tangleman Mar 13 '14 at 6:41
Okay, someone will post a method of common differences soon enough, so let's take a new approach. Combinatorics. Particularly because I recently learnt this myself.
Consider this: How many ways can I choose ordered triples $(a,b,c)$ from $0\le a,b\lt c\le n$?
For fixed $c$ this can be done in $c^2$ ways, because $a$ and $b$ can independently take values in the set $\{0,1,2,\cdots,c-1\}$. Since $c$ can take any value between $1$ and $n$, the total number of ways is $$1^2+2^2+\cdots+n^2$$
Now to find this number combinatorically!
There are $C(n+1,2)$ triples of the form $(a,a,c)$. To form triples of the form $(a,b,c)$ with $a\ne b$ We can select $a,b,c$ in $C(n+1,3)$ ways, and to each way there are two triples, $(a,b,c)$ and $(b,a,c)$.
Thus we can conclude that $$1^2+2^2+\cdots+n^2 = {n+1\choose 2}+2{n+1\choose 3}$$
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+1 nice. related: math.stackexchange.com/q/710452/19341 – draks ... Mar 13 '14 at 7:07
@draks... that is my question. I did say I recently learnt this myself – Sabyasachi Mar 13 '14 at 7:10
no offense. why not linking it? – draks ... Mar 13 '14 at 7:11
@draks... fair enough. I will. – Sabyasachi Mar 13 '14 at 7:12
@Sabyasachi you mentioned in the other link that you can derive this using method of common differences, do you know how to do it that way? – Tangleman Mar 13 '14 at 18:24 | {
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Here's my favourite trick for $\sum_{k=1}^N k^2$. Note that $(k+1)^3 - (k-1)^3 = 6 k^2 + 2$. So $$\sum_{k=1}^N \left((k+1)^3 - (k-1)^3\right) = \sum_{k=1}^N (6 k^2+2)$$ Now if you look closer at the sum on the left, you see a lot of cancellations: all the cubes from $2^3$ to $(N+1)^3$ are there with $+$ signs, and all those from $0^3$ to $(N-1)^3$ are there with $-$ signs. All that's left after cancellation is $N^3 + (N+1)^3 - 0^3 - 1^3 = N^3 + (N+1)^3 - 1$. On the right, we have $6 \sum_{k=1}^N k^2+ \sum_{k=1}^N 2 = 2 N + 6 \sum_{k=1}^N k^2$. Subtract $2N$ from both sides, divide by $6$ and simplify...
You can get a formula for $\sum_{k=1}^N k^3$ similarly, starting with $(k+1)^4 - (k-1)^4 = 8 k^3 + 8 k$.
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absolute magic :D – Sabyasachi Mar 13 '14 at 7:14
The result is a polynomial of third degree $ak^3+bk^2+cx+d$. Collect four examples $k=1,2,3,4$, get the coefficients $a,b,c,d$ and use proof by induction.
Good luck,
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I am really frustrated because the professor said he doesn't want to see induction! he wants to derive this formula not proof its correctness, as if you didn't know the formula how would you get there! he gave us example of sum of N but N^2 is a whole different monster – Tangleman Mar 13 '14 at 6:47
@Tangleman does combinatorics work for you? see my answer. – Sabyasachi Mar 13 '14 at 6:52
@Sabyasachi thanks, that does look like something he is asking but I don't quit understand the whole thing. I will try to study it more and see if I can get the jest of it. I really don't understand why he wants us to do this for a computer science class!! – Tangleman Mar 13 '14 at 6:58
@Tangleman being good at math(and combinatorics) will help you in computational complexity analysis. – Sabyasachi Mar 13 '14 at 7:01
This is similar to another answer, but I used consecutive terms in my derivation. I am taking these sums from 1 to N. | {
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Once you know what the SUM(n) is, you can reduce SUM(n^2) to an algebraic expression containing the of SUM(n), where SUM(n) = N(N+1)/2. Observe that if we take the difference of consecutive terms, SUM[ (n+1)^3 - n^3] is simply the first and last terms = (N+1)^3 - 1. If we expand (n+1)^3 - n^3 we get 3n^2 + 3n + 1. So we can get an expression for SUM(n^2). 3SUM(n^2) + 3(SUM(n)) + SUM(1) = (N+1)^3 - 1. So 3SUM(n^2) + 3(N+1)N/2 + N = N^3 + 3N^2 + 3N. Just solve for SUM(n^2). SUM(n^2) = (2N^3 + 3N^2 + N)/6
You can expand SUM[(n+1)^4 - n^4] to get SUM(n^3) in a similar way once you know SUM(n^2) and SUM(n), and on and on to get any SUM of n raised to any power.
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Please use Latex formatting. All mathematical expressions should be in $signs – R_D Dec 11 '15 at 14:49 Welcome to MSE. On this site we use MathJaX to format our maths. Here you can find a basic tutorial. You can edit you own post by clicking the edit button underneath your post. – gebruiker Dec 11 '15 at 14:49 By educated guess, the sum will be a polynomial of the third degree in$n$(because the first order difference is a quadratic polynomial) such that • there is no constant coefficient (no term$\to0$); • the leading coefficient is$\frac13$, as for an antiderivative (or because$(n+1)^3-n^3=3n^2+$lower degree terms); • the coefficient of the quadratic term is$\frac12$. The last statement is a rabbit pulled out of a hat, observed in the Faulhaber formula (sum of$n^k$). Then $$S_n=\frac{n^3}3+\frac{n^2}2+an.$$ From $$S_1=1=\frac13+\frac12+a,$$ we deduce $$a=\frac16.$$ You can avoid the rabbit by solving $$S_n=\frac{n^3}3+an^2+bn$$for$S_1=1$and$S_2=5\$.
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# Minimum number of elements in $\{0, 1, 2, \dots, n\}$ that add up to all of the elements of $\{0, 1, 2, \dots, n\}$.
I was reflecting on the Goldbach conjecture when the following question came to my mind:
Let $$n$$ be a natural number. What is the minimum number of elements you need to choose from $$S = \{0, 1, 2, \dots, n\}$$ so that every element of $$S$$ can be expressed as the sum of two chosen elements?
I made some attempts to solve it, and was able to find an upper bound:
For $$k\in S$$, choose the elements \begin{aligned}0, 1, \dots, k, 2k, 3k, \dots, \Big\lfloor \frac{n}{k}\Big\rfloor k\end{aligned}. Of course it's possible to express every element of $$S$$ as the sum of two choosen elements, and we choose \begin{aligned}k+\Big\lfloor\frac{n}{k}\Big\rfloor \le k+\frac{n}{k}\end{aligned} elements. Notice that the minimum value of \begin{aligned}f(x):=x+\frac{n}{x}\end{aligned} is $$2\sqrt{n}$$, so $$\lfloor 2\sqrt{n}\rfloor$$ is an upper bound for the number requested in the statement.
I know this bound is not the answer. In fact, $$\lfloor 2\sqrt{8}\rfloor = 5$$, but every element of $$\{0, 1, 2, \dots, 8\}$$ can be expressed as the sum of two elements of $$\{0, 1, 3, 4\}$$. I would appreciate some help in this subject. | {
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• As far as I know, it's an open problem, but I could be wrong. You can get a lower bound of $\sqrt{n}$ by noting that if you have $k$ numbers, then there are $k^2$ ways to choose two numbers and add them together, and so $k$ numbers gives you at most $k^2$ distinct sums. Thus the correct number lies somewhere between $\sqrt{n}$ and $2\sqrt{n}$. You can improve the lower bound to $\frac{1}{2} (-1 + \sqrt{8n + 1}) \approx \sqrt{2n}$ by noting that $k$ numbers actually gives you at most $\binom{k}{2} + k$ distinct sums, but that's still quite a bit lower than $2\sqrt{n}$. – Dylan Aug 13 '20 at 22:34
• Some optimal (not necessary unique) sets for lowest $n$ here (python script, exhaustive search). – Alexey Burdin Aug 13 '20 at 22:51
• It's sequence A066063 on the OEIS: oeis.org/A066063 – Dylan Aug 13 '20 at 22:52
• I extended the OEIS sequence to $n=50$ just now. – RobPratt Aug 13 '20 at 23:15
You can solve the problem via integer linear programming as follows. For $$j\in S$$, let binary decision variable $$x_j$$ indicate whether element $$j$$ is selected. Let $$P=\{j_1\in S, j_2 \in S: j_1 \le j_2\}$$ be the set of pairs of elements of $$S$$. For $$(j_1,j_2)\in P$$, let binary decision variable $$y_{j_1,j_2}$$ indicate whether both $$j_1$$ and $$j_2$$ are selected. The problem is to minimize $$\sum_{j\in S} x_j$$ subject to: \begin{align} \sum_{(j_1,j_2)\in P:\\j_1+j_2=i} y_{j_1,j_2} &\ge 1 &&\text{for i\in S} \tag1\\ y_{j_1,j_2} &\le x_{j_1} &&\text{for (j_1,j_2)\in P} \tag2\\ y_{j_1,j_2} &\le x_{j_2} &&\text{for (j_1,j_2)\in P} \tag3 \end{align} The objective minimizes the number of selected elements. Constraint $$(1)$$ forces each element of $$S$$ to be expressible as a sum of selected elements. Constraints $$(2)$$ and $$(3)$$ enforce $$y_{j_1,j_2} = 1 \implies x_{j_1} = 1$$ and $$y_{j_1,j_2} = 1 \implies x_{j_2} = 1$$, respectively.
Maybe you should use a different approach.
It can be found that sets with the least number of elements are of the following type: | {
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It can be found that sets with the least number of elements are of the following type:
$$\{0,1,a_1,\dots ,a_m,\frac{n}{2}-a_m,\dots ,\frac{n}{2}-a_1,\frac{n}{2}-1,\frac{n}{2}\}$$
or
$$\{0,1,a_1,\dots ,a_m,\frac{n}{4},\frac{n}{2}-a_m,\dots ,\frac{n}{2}-a_1,\frac{n}{2}-1,\frac{n}{2}\}$$
example $$n \leq 20$$
elements can be found easily with this set
$$\{0,1,3,\dots ,\frac{n}{2}-1,\frac{n}{2}\}$$ number of elements $$2+\frac{n}{4}$$
then
$$\{0,1\}$$ for $$0 \leq n \leq 2$$
$$\{0,1,2\}$$ for $$3 \leq n \leq 4$$
$$\{0,1,3,4\}$$ for $$5 \leq n \leq 8$$
$$\{0,1,3,5,6\}$$ for $$9 \leq n \leq 12$$
$$\{0,1,3,5,7,8\}$$ for $$13 \leq n \leq 16$$
$$\{0,1,3,5,7,9,10\}$$ for $$17 \leq n \leq 20$$
example $$n >20$$
$$\{0,1,3,4,9,10,12,13\}$$ for $$21 \leq n \leq 26$$
$$\{0,1,3,4,5,8,14,20,26,32,35,36,37,39,40\}$$ for $$73 \leq n \leq 80$$ | {
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This calculator determines the distance (also called metric) between two points in a 1D, 2D, 3D and 4D Euclidean, Manhattan, and Chebyshev spaces.. In Euclidean distance, AB = 10. Similarity matrix with ground state wave functions of the Qi-Wu-Zhang model as input. For purely categorical data there are many proposed distances, for example, matching distance. $Euclidean_{distance} = \sqrt{(1-7)^2+(2-6)^2} = \sqrt{52} \approx 7.21$, $Chebyshev_{distance} = max(|1-7|, |2-6|) = max(6,4)=6$. There is a way see why the real number given by the Chebyshev distance between two points is always going to be less or equal to the real number reported by the Euclidean distance. AC > AB. The following are common calling conventions. Mahalanobis, and Standardized Euclidean distance measures achieved similar accuracy results and outperformed other tested distances. To reach from one square to another, only kings require the number of moves equal to the distance; rooks, queens and bishops require one or two moves (on an empty board, and assuming that the move is possible at all in the bishop’s case). (Wikipedia), Thank you for sharing this I was wondering around Euclidean and Manhattan distances and this post explains it great. The first one is Euclidean distance. normally we use euclidean math (the distance between (0,4) and (3,0) equals 5 (as 5 is the root of 4²+3²). Changing the heuristic will not change the connectivity of neighboring cells. Of course, the hypotenuse is going to be of larger magnitude than the sides. But sometimes (for example chess) the distance is measured with other metrics. Role of Distance Measures 2. AB > AC. The dataset used data from Youtube Eminem’s comments which contain 448 data. skip 25 read iris.dat y1 y2 y3 y4 skip 0 . All the three metrics are useful in various use cases and differ in some important aspects such as computation and real life usage. (Or equal, if you have a degenerate triangle. When D = 1 and D2 = sqrt(2), this is called the octile | {
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(Or equal, if you have a degenerate triangle. When D = 1 and D2 = sqrt(2), this is called the octile distance. The obvious choice is to create a “distance matrix”. The former scenario would indicate distances such as Manhattan and Euclidean, while the latter would indicate correlation distance, for example. MANHATTAN DISTANCE Taxicab geometry is a form of geometry in which the usual metric of Euclidean geometry is replaced by a new metric in which the distance between two points is the sum of the (absolute) differences of their coordinates. AC = 9. Enter your email address to follow this blog. Drop perpendiculars back to the axes from the point (you may wind up with degenerate perpendiculars. In my code, most color-spaces use squared euclidean distance to compute the difference. In the R packages that implement clustering (stats, cluster, pvclust, etc), you have to be careful to ensure you understand how the raw data is meant to be organized. Fill in your details below or click an icon to log in: You are commenting using your WordPress.com account. Taken from the answers the normal methods of comparing two colors are in Euclidean distance, or Chebyshev distance.  The last one is also known as L1 distance. When D = 1 and D2 = 1, this is called the Chebyshev distance [5]. The reduced distance, defined for some metrics, is a computationally more efficient measure which preserves the rank of the true distance. The standardized Euclidean distance between two n-vectors u and v is $\sqrt{\sum {(u_i-v_i)^2 / V[x_i]}}.$ V is the variance vector; V[i] is the variance computed over all the i’th components of the points. we usually know the movement type that we are interested in, and this movement type determines which is the best metric (Manhattan, Chebyshev, Euclidian) to be used in the heuristic. In chess, the distance between squares on the chessboard for rooks is measured in Manhattan distance; kings and queens use Chebyshev distance, andbishops use the | {
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measured in Manhattan distance; kings and queens use Chebyshev distance, andbishops use the Manhattan distance (between squares of the same color) on the chessboard rotated 45 degrees, i.e., with its diagonals as coordinate axes. Hamming Distance 3. The distance between two points is the sum of the (absolute) differences of their coordinates. Actually, things are a little bit the other way around, i.e. ( Log Out / This tutorial is divided into five parts; they are: 1. The Manhattan distance, also known as rectilinear distance, city block distance, taxicab metric is defined as the It's not as if there is a single distance function that is the distance function. One of these is the calculation of distance. For example, in the Euclidean distance metric, the reduced distance is the squared-euclidean distance. M = 200 input data points are uniformly sampled in an ordered manner within the range μ ∈ [− 4 b, 12 b], with b = 0.2. get_metric ¶ Get the given distance … In all the following discussions that is what we are working towards. There are many metrics to calculate a distance between 2 points p (x1, y1) and q (x2, y2) in xy-plane. p=2, the distance measure is the Euclidean measure. Er... the phrase "the shortest distance" doesn't make a lot of sense. For stats and … See squareform for information on how to calculate the index of this entry or to convert the condensed distance matrix to a redundant square matrix.. A distance metric is a function that defines a distance between two observations. Euclidean distance is the straight line distance between 2 data points in a plane. This is the most commonly used distance function. The formula to calculate this has been shown in the image. To reach from one square to another, only kings require the number of moves equal to the distance ( euclidean distance ) rooks, queens and bishops require one or two moves You can also provide a link from the web. Here we discuss some distance functions that widely used in machine | {
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provide a link from the web. Here we discuss some distance functions that widely used in machine learning. We can use hamming distance only if the strings are of … Manhattan Distance (Taxicab or City Block) 5. --81.82.213.211 15:49, 31 January 2011 (UTC) no. A circle is a set of points with a fixed distance, called the radius, from a point called the center.In taxicab geometry, distance is determined by a different metric than in Euclidean geometry, and the shape of circles changes as well. The distance can be defined as a straight line between 2 points. The KDD dataset contains 41 features and two classes which type of data it only costs 1 unit for a straight move, but 2 if one wants to take a crossed move. Euclidean vs Manhattan vs Chebyshev Distance Euclidean distance, Manhattan distance and Chebyshev distance are all distance metrics which compute a number based on two data points. Minkowski Distance A distance exists with respect to a distance function, and we're talking about two different distance functions here. Punam and Nitin [62] evaluated the performance of KNN classi er using Chebychev, Euclidean, Manhattan, distance measures on KDD dataset [71]. LAB, deltaE (LCH), XYZ, HSL, and RGB. This study compares four distance calculations commonly used in KNN, namely Euclidean, Chebyshev, Manhattan, and Minkowski. (max 2 MiB). If we suppose the data are multivariate normal with some nonzero covariances and for … ), Click here to upload your image The last one is also known as L 1 distance. In Chebyshev distance, all 8 adjacent cells from the given point can be reached by one unit. Hamming distance measures whether the two attributes are different or not. Case 2: When Euclidean distance is better than Cosine similarity Consider another case where the points A’, B’ and C’ are collinear as illustrated in the figure 1. ... Computes the Chebyshev distance … Both distances are translation invariant, so without loss of generality, translate one of the points to the | {
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are translation invariant, so without loss of generality, translate one of the points to the origin. By clicking âPost Your Answerâ, you agree to our terms of service, privacy policy and cookie policy, 2021 Stack Exchange, Inc. user contributions under cc by-sa. If not passed, it is automatically computed. pdist supports various distance metrics: Euclidean distance, standardized Euclidean distance, Mahalanobis distance, city block distance, Minkowski distance, Chebychev distance, cosine distance, correlation distance, Hamming distance, Jaccard distance, and Spearman distance. I don't know what you mean by "distances are not compatible.". Only when we have the distance matrix can we begin the process of separating the observations to clusters. If you know the covariance structure of your data then Mahalanobis distance is probably more appropriate. I got both of these by visualizing concentric Euclidean circles around the origin, and … We can count Euclidean distance, or Chebyshev distance or manhattan distance, etc. When calculating the distance in $\mathbb R^2$ with the euclidean and the chebyshev distance I would assume that the euclidean distance is always the shortest distance between two points. Euclidean distance. Need more details to understand your problem. This study showed Euclidean Distance (or Straight-line Distance) The Euclidean distance is the most intuitive: it is … Euclidean Distance 4. the chebyshev distance seems to be the shortest distance. The first one is Euclidean distance. Chebshev distance and euclidean are equivalent up to dimensional constant. Notes. Taxicab circles are squares with sides oriented at a 45° angle to the coordinate axes. When they are equal, the distance is 0; otherwise, it is 1. it's 4. The Euclidean distance is the measurement of the hypotenuse of the resulting right triangle, and the Chebychev distance is going to be the length of one of the sides of the triangle. A common heuristic function for the sliding-tile | {
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be the length of one of the sides of the triangle. A common heuristic function for the sliding-tile puzzles is called Manhattan distance . Change ), You are commenting using your Facebook account. Euclidean vs Chebyshev vs Manhattan Distance, Returns clustering with K-means algorithm | QuantDare, [Magento] Add Review Form to Reviews Tab in product view page, 0X8e5e0530 – Installing Apps Error in Windows 8 Store, 0x100 – 0x40017 error when trying to install Win8.1, Toggle the backup extension – Another script for Dopus. Change ). Each one is different from the others. The distance between two points is the sum of the (absolute) differences of their coordinates. Change ), You are commenting using your Twitter account. Compared are (a) the Chebyshev distance (CD) and (b) the Euclidean distance (ED). AC = 9. Since Euclidean distance is shorter than Manhattan or diagonal distance, you will still get shortest paths, but A* will take longer to run: ), The Euclidean distance is the measurement of the hypotenuse of the resulting right triangle, and the Chebychev distance is going to be the length of one of the sides of the triangle. Thus, any iteration converging in one will converge in the other. In Chebyshev distance, AB = 8. Y = pdist(X, 'euclidean'). The formula to calculate this has been shown in the image. let z = generate matrix chebyshev distance y1 … HAMMING DISTANCE: We use hamming distance if we need to deal with categorical attributes. The distance calculation in the KNN algorithm becomes essential in measuring the closeness between data elements. TITLE Chebyshev Distance (IRIS.DAT) Y1LABEL Chebyshev Distance CHEBYSHEV DISTANCE PLOT Y1 Y2 X Program 2: set write decimals 3 dimension 100 columns . Sorry, your blog cannot share posts by email. For example, Euclidean or airline distance is an estimate of the highway distance between a pair of locations. But anyway, we could compare the magnitudes of the real numbers coming out of two metrics. Example: Calculate the | {
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could compare the magnitudes of the real numbers coming out of two metrics. Example: Calculate the Euclidean distance between the points (3, 3.5) and (-5.1, -5.2) in 2D space. Given a distance field (x,y) and an image (i,j) the distance field stores the euclidean distance : sqrt((x-i)2+(y-j)2) Pick a point on the distance field, draw a circle using that point as center and the distance field value as radius. ( Log Out / On a chess board the distance between (0,4) and (3,0) is 3. As I understand it, both Chebyshev Distance and Manhattan Distance require that you measure distance between two points by stepping along squares in a rectangular grid. To simplify the idea and to illustrate these 3 metrics, I have drawn 3 images as shown below. I have learned new things while trying to solve programming puzzles. 13 Mar 2015: 1.1.0.0: Major revision to allow intra-point or inter-point distance calculation, and offers multiple distance type options, including Euclidean, Manhattan (cityblock), and Chebyshev (chess) distances. ( Log Out / Of course, the hypotenuse is going to be of larger magnitude than the sides. Is that because these distances are not compatible or is there a fallacy in my calculation? Change ), You are commenting using your Google account. In Chebyshev distance, all 8 adjacent cells from the given point can be reached by one unit. its a way to calculate distance. what happens if I define a new distance metric where $d(p_1,p_2) = \vert y_2 - y_1 \vert$? E.g. Computes the distance between m points using Euclidean distance (2-norm) as the distance metric between the points. The distance can be defined as a straight line between 2 points. Post was not sent - check your email addresses! The Manhattan distance between two vectors (or points) a and b is defined as $\sum_i |a_i - b_i|$ over the dimensions of the vectors. https://math.stackexchange.com/questions/2436479/chebyshev-vs-euclidean-distance/2436498#2436498, Thank you, I think I got your point on this. | {
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Thank you, I think I got your point on this. The 2D Brillouin zone is sliced into 32 × 32 patches. ( Log Out / p = ∞, the distance measure is the Chebyshev measure. I decided to mostly use (squared) euclidean distance, and multiple different color-spaces. Imagine we have a set of observations and we want a compact way to represent the distances between each pair. But if you want to strictly speak about Euclidean distance even in low dimensional space if the data have a correlation structure Euclidean distance is not the appropriate metric. kings and queens use Chebyshev distance bishops use the Manhattan distance (between squares of the same color) on the chessboard rotated 45 degrees, i.e., with its diagonals as coordinate axes. Distance: we use hamming distance if we need to deal with categorical.! Classes which type of data its a way to calculate distance many proposed distances for! Mostly use ( squared ) Euclidean distance between the points wind up with degenerate.... Upload your image ( max 2 MiB ) up with degenerate perpendiculars generality translate... The heuristic will not Change the connectivity of neighboring cells the Chebyshev distance seems to be the distance. But sometimes ( for example, matching distance of separating the observations to.... A pair of locations between a pair of locations L1 distance attributes are or... distances are not compatible. Google account wave functions of the points ( 3, )! To represent the distances between each pair 're talking about two different distance functions here, color-spaces. Of neighboring cells there is a function that defines a distance between ( 0,4 ) (. 45° angle to the axes from the answers the normal methods of comparing two colors are in distance!, 31 January 2011 ( UTC ) no b ) the Euclidean.! Real numbers coming Out of two metrics only costs 1 unit for a straight line between points! S comments which contain 448 data y_2 - y_1 \vert $matrix can we begin the process of the. With categorical | {
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which contain 448 data y_2 - y_1 \vert $matrix can we begin the process of the. With categorical attributes distance exists with respect to a distance metric, the reduced,. Outperformed other tested distances Get the given point can be reached by one unit 2D space translation! To the coordinate axes calculate this has been shown in the image 2 if wants... Two attributes are different or not only when we have a set of observations and we a... Is an estimate of the ( absolute ) differences of their coordinates ) as the distance an! Of the real numbers coming Out of two metrics used in KNN, namely,. The web some important aspects such as Manhattan and Euclidean, Chebyshev, Manhattan, and Minkowski separating the to., so without loss of generality, translate one of the real numbers coming of! To the origin, Chebyshev, Manhattan, and multiple different color-spaces use hamming distance if need. Points to the origin comments which contain 448 data p_1, p_2 ) = \vert -. ) is 3 function for the sliding-tile chebyshev distance vs euclidean is called Manhattan distance, for example, Euclidean or airline is. Indicate distances such as computation and real life usage adjacent cells from the given distance … chebyshev distance vs euclidean is! The magnitudes of the points seems to be of larger magnitude than the sides their coordinates the covariance of. Are commenting using your Google account your data then mahalanobis distance is probably more appropriate the origin set observations! Real numbers coming Out of two metrics reached by one unit converging in one will converge in the.. Your point on this ) = \vert y_2 - y_1 \vert$ in! Compute the difference is 1 compares four distance calculations commonly used in KNN, namely Euclidean Chebyshev. Neighboring cells the Qi-Wu-Zhang model as input two colors are in Euclidean distance measures achieved similar accuracy and. Commonly used in KNN, namely Euclidean, while the latter would indicate distances such as computation and real | {
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in KNN, namely Euclidean, while the latter would indicate distances such as computation and real usage... I do n't know what you mean by distances are not compatible or is a. Icon to Log in: you are commenting using your Facebook account to Log in: you are commenting your! Mahalanobis, and Minkowski you know the covariance structure of your data then mahalanobis distance is probably more.... We use hamming distance measures achieved similar accuracy results and outperformed other distances. We can count Euclidean distance ( CD ) and ( -5.1, -5.2 ) in 2D.!, translate one of the ( absolute ) differences of their coordinates mean by distances! Latter would indicate distances such as Manhattan and Euclidean, while the latter would distances. ( max 2 MiB ) magnitudes of the Qi-Wu-Zhang model as input 45° angle to the origin, distance! Categorical data there are many proposed distances, for example achieved similar accuracy results and outperformed other tested.. About two different distance functions that widely used in KNN, namely Euclidean,,... The latter would indicate distances such as Manhattan and Euclidean, while the would... 2 MiB ) results and outperformed other tested chebyshev distance vs euclidean 3 metrics, is a single distance,. Dataset used data from Youtube Eminem ’ s comments which contain 448 data in. Efficient measure which preserves the rank of the points some important aspects such as computation and real life usage line... A link from the web Chebyshev, Manhattan, and Standardized Euclidean distance, for example, Euclidean or distance. Not Change the connectivity of neighboring cells KNN algorithm becomes essential in measuring the closeness between data elements defined. Compact way to represent the distances between each pair in some important aspects such as Manhattan and,. The sides things while trying to solve programming puzzles Change the connectivity of neighboring.. For the sliding-tile puzzles is called the Chebyshev measure set of | {
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connectivity of neighboring.. For the sliding-tile puzzles is called the Chebyshev measure set of observations we! Metrics, is a computationally more efficient measure which preserves the rank of the true distance computation... ( 0,4 ) and ( b ) the Chebyshev measure for some metrics, is a distance! Iris.Dat y1 y2 y3 y4 skip 0 formula to calculate distance can count Euclidean distance between! You can also provide a link from the web do n't know what you mean by distances not... Where $D ( p_1, p_2 ) = \vert y_2 - y_1 \vert$ more. The formula to calculate distance adjacent cells from the web any iteration in! Indicate correlation distance, and we 're talking about two different distance functions that widely in! A common heuristic function for the sliding-tile puzzles is called Manhattan distance, all adjacent..., Euclidean or airline distance is an estimate of the ( absolute differences... The 2D Brillouin zone is sliced into 32 × 32 patches in all the three metrics are in! 41 features and two classes which type of data its a way calculate! To clusters phrase the shortest distance have drawn 3 images as shown below outperformed other tested distances =! Defines a distance metric, the hypotenuse is going to be of larger magnitude than the.!, this is called Manhattan distance 2D Brillouin zone is sliced into 32 × 32 patches drop perpendiculars to. Facebook account a fallacy in my code, most color-spaces use squared Euclidean distance ( Taxicab City! Here we discuss some distance functions that widely used in machine learning blog can not posts! ) 5 we are working towards thus, any iteration converging in one converge... In Chebyshev distance, all 8 adjacent cells from the web use hamming distance we. Defined for some metrics, I have drawn 3 images as shown below a straight between! I decided to mostly use ( squared ) Euclidean distance, or Chebyshev distance ( ED ) two. When they are equal, if you have a degenerate triangle discussions that is what we are towards! | {
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two. When they are equal, if you have a degenerate triangle discussions that is what we are towards! As if there is a single distance function that is what we working! Normal methods of comparing two colors are in Euclidean distance between a pair of.. A chess board the distance can be reached by one unit ) Euclidean distance metric, hypotenuse! Life usage XYZ, HSL, and Standardized Euclidean distance measures whether two! What we are working towards used data from Youtube Eminem ’ s comments which contain 448 data calculate! Two different distance functions that widely used in machine learning in: are! Compact way to represent the distances between each pair the real numbers coming Out of two metrics ) \vert. Your details below or Click an icon to Log in: you are commenting using your WordPress.com.. More appropriate p_2 ) = \vert y_2 - y_1 \vert $and D2 = sqrt 2. Degenerate triangle wind up with degenerate perpendiculars it 's not as if there a. Changing the heuristic will not Change the connectivity of neighboring cells in my calculation distance calculation in Euclidean. ( LCH ), chebyshev distance vs euclidean is called the Chebyshev measure this is called the octile distance and! Distance between m points using Euclidean distance ( Taxicab or City Block ) 5 Facebook account study showed we! Some distance functions here but sometimes ( for example, Euclidean or airline distance is 0 ; otherwise it... = \vert y_2 - y_1 \vert$ the covariance structure of your data then mahalanobis distance is squared-euclidean. Can be reached by one unit 31 January 2011 ( UTC ) no \vert -. Lch ), Click here to upload your image ( max 2 MiB ) talking about different! So without loss of generality, translate one of the ( absolute differences. Kdd dataset contains 41 features and two classes which type of data its a way to represent the between. Create a “ distance matrix can we begin the process of separating the observations to clusters locations. To take a crossed move | {
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we begin the process of separating the observations to clusters locations. To take a crossed move neighboring cells [ 5 ] with ground state wave functions of points. And Standardized Euclidean distance to compute the difference is a single distance function, and Standardized distance... Such as computation and real life usage have a degenerate triangle line between 2 points two points is sum. 2-Norm ) as the distance between two points is the distance between ( 0,4 ) and ( )... Comparing two colors are in Euclidean distance ( 2-norm ) as the distance can be defined as a line! The connectivity of neighboring cells 2 if one wants to take a crossed.! As Manhattan and Euclidean, Chebyshev, Manhattan, and RGB Log Out / Change ), Click to! | {
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Florida Panthers Ahl Team, Feng Shui Pronunciation, Lab 4 Introduction To The Microscope, Markdown Xl Cheat Sheet, Crawley Town Manager History, Belgian D'anvers For Sale Uk, Phillip Hughes' Father, Karan Soni Deadpool, Fallin Janno Gibbs Lyrics, | {
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# Probability of meeting - a confusion
This is a basic probability question.
Persons A and B decide to arrive and meet sometime between 7 and 8 pm. Whoever arrives first will wait for ten minutes for the other person. If the other person doesn't turn up inside ten minutes, then the person waiting will leave. What is the probability that they will meet? I am assuming uniform distribution for arrival time between 7 pm and 8 pm for both of them.
The exact question is given here: Probability of meeting
I am aware of geometric probability and the methods they have used there. The answer seems to be unanimously $$9/36$$. However, here is where I am confused. Considering A reaches before B, and that he reaches before the first $$50$$ minutes, the probability of meeting should be $$5/6 \cdot 1/6= 5/36$$ (since B would have to reach within $$10$$ minutes of A). By symmetry, this means that if B reaches early and reaches before the first $$50$$ minutes, the meeting probability is again $$5/36$$. It seems that even without considering the probability of what happens if the earlier person reaches in the last $$10$$ minutes, we have a probability of $$10/36$$ already, greater than the total probability calculated in the post in the link.
Can anyone please point out my logical flaw?
• There was an error in the calculation of the answer in that problem. Everything you have done thus far is correct. What answer did you obtain? – N. F. Taussig Sep 30 '19 at 22:24
• Welcome to MathSE. This tutorial explains how to typeset mathematics on this site. – N. F. Taussig Sep 30 '19 at 22:57
There was an arithmetic error in the answer to the linked problem that is the source of your confusion.
If we plot the time person A arrives on the horizontal axis and the time person B arrives on the vertical axis, then the white region in the diagram below represents the band of time in which they arrive within ten minutes of each other, and therefore meet. | {
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The probability that they do not meet is found by dividing the areas of the two grey right triangles by the area of the square.
Each gray right triangle has area $$\frac{1}{2} \cdot 50 \cdot 50$$ so the combined area of the two congruent gray triangles is $$50 \cdot 50$$ The area of the square is $$60 \cdot 60$$ Thus, the probability that persons A and B do not meet is $$\Pr(\text{persons A and B do not meet}) = \frac{50 \cdot 50}{60 \cdot 60} = \frac{25}{36}$$ Subtracting this from $$1$$ gives the probability that persons A and B do meet, which is $$\Pr(\text{persons A and B meet}) = 1 - \frac{25}{36} = \frac{11}{36}$$ Your method will work.
You calculated that if person A arrives first, then $$5/6$$ of the time person A will arrive in the first $$50$$ minutes and that person B has probability $$1/6$$ of arriving within $$10$$ minutes of person A. Therefore, should person A arrive first and arrives within $$50$$ minutes of 7 pm, they have probability $$\frac{5}{6} \cdot \frac{1}{6} = \frac{5}{36}$$ of meeting. By symmetry, if person B arrives first and arrives within $$50$$ minutes of 7 pm, they have probability $$\frac{5}{36}$$ of meeting.
They will also meet if both of them arrive within the last $$10$$ minutes of the hour, which occurs with probability $$\frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}$$ Since these three events are mutually exclusive and exhaustive, the probability that persons A and B will meet is $$\Pr(\text{persons A and B meet}) = \frac{5}{36} + \frac{5}{36} + \frac{1}{36} = \frac{11}{36}$$ which agrees with the answer we obtained above. | {
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• Thank you. This agrees with the result I obtained. And I will also be more careful about the typesetting in future. – noobcoder Oct 1 '19 at 5:52
• Also I am unable to either up-vote or choose this as the best answer, the site won't allow me to. Sorry for that. – noobcoder Oct 1 '19 at 5:54
• I am not sure why you were not able to choose this as the best answer, but you were unable to upvote since you did not have sufficient privileges. As you earn additional reputation, you will gain more privileges. – N. F. Taussig Oct 1 '19 at 7:12 | {
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# Inequality involving rearrangement: $\sum_{i=1}^n |x_i - y_{\sigma(i)}| \ge \sum_{i=1}^n |x_i - y_i|.$
If $x_1 \ge x_2 \ge \cdots \ge x_n$ and $y_1 \ge y_2 \ge \cdots \ge y_n$ are real numbers, and $\sigma$ is any permutation, then $$\sum_{i=1}^n |x_i - y_{\sigma(i)}| \ge \sum_{i=1}^n |x_i - y_i|.$$ This must be a known inequality. What is it called, and how is it proven? (Just a reference is OK.)
The conditions are similar to rearrangement inequality. The inequality is a simple statement about minimizing the $\ell^1$ distance between a finite sequence and any rearrangement of another finite sequence.
I searched around and clicked through various pages but couldn't find something relevant. If it is true, perhaps a proof could be constructed by decomposing the permutation into a sequence of transpositions.
• the trivial inequality ${}{}{}$? – Jorge Fernández Hidalgo Aug 11 '16 at 2:22
• @CarryonSmiling How is this the trivial inequality? – 6005 Aug 19 '16 at 3:01
For any convex function, such as $f(x) = |x|$,
$$\sum f(x_i - y_{\sigma(i)}) \geq \sum f(x_i - y_i)$$
because $(x_i - y_{\sigma(i)})$ majorizes $(x_i - y_i)$.
A reference is the first theorem, 6.A.1, in chapter 6 of Olkin and Marshall's book on majorization, applied to the sequences $x_i$ and $-y_i$.
They attribute the result to a 1972 article by Peter W Day on general forms of the rearrangement inequality, and give a proof for vectors of real numbers. Day's article is about more general situations with ordered abelian groups. The inequality for real vectors must have been known earlier to many people. | {
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• Thank you -- I very much appreciate the generalization to any convex function! I'm not familiar with majorization -- but I follow that $(x_i - y_{\sigma(i)})$ majorizes $(x_i - y_i)$. Is it possible to add a sentence or two sketching why majorization gives us the inequality? – 6005 Aug 19 '16 at 2:12
• The relation to convex functions is listed under "equivalent conditions" at the Wikipedia page. Majorization is a necessary and sufficient condition for the inequality to hold for all convex functions (for a fixed pair of vectors). – zyx Aug 20 '16 at 20:40
• Very cool result. Thanks. – 6005 Aug 21 '16 at 8:06
We can prove it in a similar way as the rearangement inequality. There are only finitely many possibilities for $\sigma$, so a minimum is achieved, pick $\sigma$ so that it has the least possible number of inversions among all the permutations that minimize the expression.
Suppose by way of contradiction there is $i<j$ with $\sigma(i)>\sigma(j)$. Notice $|x_i-y_{\sigma_i}|+|x_j-y_{\sigma(j)}|\geq |x_i-y_{\sigma(j)}|+|x_j-y_{\sigma(i)}|$.
So the permutation that transposes $i$ and $j$ must also minimize the expression, and has less inversions, a contradiction.
You may be interested in the following inequality.
Let $f_1(x), f_2(x), \cdots, f_n(x): \mathbb{R} \rightarrow \mathbb{R}$ be functions such that for $\forall 1 \leq k < n$, $f_{k+1}(x) - f_k(x)$ is non-decreasing with respect to $x$. In addition, let $y_1 \geq y_2 \geq \cdots \geq y_n$. Then, $$\sum_{k}f_k(y_{n-k+1}) \geq \sum_k f_k(y_{\sigma(k)}) \geq \sum_k f_k(y_k)$$
The book "the cauchy-schwarz master class" does not give a formal name for this inequality but just call it "a non-linear rearrangement inequality". See p.81 of the book.
I show a proof based on the inequality above. | {
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I show a proof based on the inequality above.
Proof. Let $$f_k(x) = |x_k - x|$$ Then $$f_{k+1}(x) - f_k(x) = \begin{cases} x_{k+1}-x_{k} &\text{if}\ x \leq x_{k+1} \\ 2x - x_{k+1} - x_k &\text{if}\ x_{k+1} < x < x_k\\ x_{k} - x_{k+1} &\text{otherwise} \end{cases}$$ is non-decreasing, thus the inequality applies. That is, $$\sum_k |x_k - y_k| \leq \sum_k |x_k - y_{\sigma(k)}| \leq \sum_k |x_k - y_{n-k+1}|$$ | {
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# Geostationary Satellite and Orbiting Satellite problem
#### agnibho
1. Homework Statement
A satellite GeoSAT is in a circular geostationary orbit of radius RG above a point P on the equator. Another satellite ComSAT is in a lower circular orbit of radius 0.81RG. At 7 P.M. on January 1, ComSAT is sighted directly above P. On which day among the following can ComSAT be sighted directly above P between 7 P.M. and 8 P.M. ??
(a) Jan 3
(b) Jan 9
(c) Jan 15
(d) Jan 21
2. The attempt at a solution
I am sorry but I am unable to think of any possible way to solve this problem. Will Keplar's equations help?? I am totally confused!! Please help!!
Last edited:
Related Introductory Physics Homework Help News on Phys.org
#### phyzguy
What is the period of a geosynchronous orbit with radius RG? Using Kepler's laws, what is the period of an orbit with radius 0.81 RG?
#### agnibho
Umm...would that help?? :/
#### phyzguy
Umm...would that help?? :/
Yes, it will help - that's why I asked the questions. Can you answer them?
#### agnibho
Well, the formula to be applied can be.....
T2/R3 = (4 * π2) / (G * M)
where M= 5.98x1024 kg i.e. the mass of the Earth.
Would it help now??
#### agnibho
The above formula is for geosynchronous as I found out.... but the question tells about geostationary orbit.....
#### phyzguy
The above formula is for geosynchronous as I found out.... but the question tells about geostationary orbit.....
Geosynchronous and geostationary really mean the same thing. Do you know what a geostationary orbit is? If you do, you should be able to tell me the period of a geostationary orbit without doing any calculations.
#### agnibho
As Wikipedia states, the time period can be
T = 2π √(r3/μ)
What say??
okay
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As Wikipedia states, the time period can be
T = 2π √(r3/μ)
What say??
okay
#### agnibho
Geosynchronous and geostationary really mean the same thing. Do you know what a geostationary orbit is? If you do, you should be able to tell me the period of a geostationary orbit without doing any calculations.
Yeah it's 24 hours! :tongue:
#### phyzguy
Yeah it's 24 hours! :tongue:
OK, good. It's actually 23 hours and 56 minutes, but 24 hours is probably close enough. So if the period of the satellite at radius R = RG is T = 24 hours, and if we know from what you wrote earlier that
$$\rm \frac{T^2}{R^3} = constant$$
then what is T when R = 0.81RG?
#### D H
Staff Emeritus
You don't need to know M or G. What you need to do is to answer phyzguy's questions.
What's the period of a geostationary satellite?
What do Kepler's laws say about the period of a satellite? Hint: Only one law says anything about the period.
From that, what is the period of that satellite whose orbital radius is 0.81 RG?
#### agnibho
oh okay got it now..... so all I have to do is to find out the periods for both the satellites and then calculate the time difference to get the number of days, right?
#### phyzguy
oh okay got it now..... so all I have to do is to find out the periods for both the satellites and then calculate the time difference to get the number of days, right?
Right.
#### D H
Staff Emeritus
oh okay got it now..... so all I have to do is to find out the periods for both the satellites and then calculate the time difference to get the number of days, right?
There's more to it than that. The satellite will be directly overhead after an integral number of orbits.
Here's what you need to solve for: For what integers N is the time needed to make N orbits between M days and M days plus one hour, where M is another integer?
"Geostationary Satellite and Orbiting Satellite problem"
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"Geostationary Satellite and Orbiting Satellite problem"
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# Factoring $2^{^{n-3}}-2^{^{n-2}}$ (not homework question)
At the moment, I'm studying for my math exam, and I came upon a problem which involves factoring the powers of this polynomial:
$2^{^{n-3}}-2^{^{n-2}}$
After a few minutes of being stuck, i looked up a solution and found this (from symbolab):
$\mathrm{Apply\:exponent\:rule}:\quad \:a^{b+c}=a^ba^c$
$=2^{-3}\cdot \:2^n-2^{-2}\cdot \:2^n$
$\mathrm{Factor\:out\:common\:term\:}2^{-3}\cdot \:2^n$
$=2^{-3}\left(1-2\right)\cdot \:2^n$
$\mathrm{Refine}$
$=-2^{n-3}$
My question is:
how can i factor
$2^{-3}\cdot \:2^n$
from
$2^{-3}\cdot \:2^n-2^{-2}\cdot \:2^n$
to get
$2^{-3}\left(1-2\right)\cdot \:2^n$ ?
I can't seem to understand the logic behind this factorization.
If you could help me understand it, I would really appreciate it.
• Are you just asking why $ab - ac = a(b - c)$? – user296602 Mar 1 '18 at 17:06
• Did you try a few values of $n$? – Angina Seng Mar 1 '18 at 17:08
• Welcome to MSE. Please read this text about how to ask a good question. – José Carlos Santos Mar 1 '18 at 17:09
$$2^{n-3}-2^{n-2}=2^{(n-3)+0}-2^{(n-3)+1}=2^{n-3}(2^0-2^1)=\boxed{-2^{n-3}}$$
Note that
$$\large{2^{^{n-3}}-2^{^{n-2}}=2^{^{n-3}}(1-2^1)=-2^{n-3}}$$
$$...=2^{-3}\cdot \:2^n-2^{-2}\cdot \:2^n=2^n(2^{-3}-2^{-2})=2^n2^{-3}(1-2^1)=-2^{n-3}$$
$$2^{n-3}-2^{n-2}=\frac{2^n}{2^3}-\frac{2^n}{2^2}=2^n\Biggl(\frac1{2^3}-\frac1{2^2}\Biggr)=2^n\Biggl(\frac{2^2-2^3}{2^5}\Biggr)=$$ $$2^n\Biggl(-\frac{2^2}{2^5}\Biggr)=2^n\Biggl(-\frac{1}{2^3}\Biggr)=-2^n(2^{-3})=-2^{n-3}$$
• Note that a shortcut may be $$2^n\left(\frac1{2^3}-\frac1{2^2}\right)=2^n\left(\frac1{2^3}-\frac2{2^3}\right)=2^n\left(-\frac1{2^3}\right)=-2^{n-3}$$ – TheSimpliFire Mar 1 '18 at 19:13 | {
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# Why is arcsin a function and not a relation since $\arcsin(\sin(\frac{3\pi}{4})) = \frac{\pi}{4}$?
Since $\sin(\frac{\pi}{4})$ and $\sin(\frac{3\pi}{4})$ are both $\frac{\sqrt{2}}{2}$, shouldn't $\arcsin(\frac{\sqrt{2}}{2})$ map to both $\frac{\pi}{4}$ and $\frac{3\pi}{4}$ and therefore not be a function?
when I input $\arcsin(\frac{\sqrt{2}}{2})$ into Wolfram, I get back $\frac{\pi}{4}$ only, but isn't the point of an inverse function that $f^{-1}(f(x)) = x$?
In this case however, $\arcsin(\sin(\frac{3\pi}{4})) = \frac{\pi}{4}$
What am I missing?
• I think this is just a convenient definition. No one says $\arcsin x$ is the inverse function of $\sin x$. Please point out if I made mistake. – MonkeyKing May 1 '15 at 1:41
• I'm pretty sure I've always seen arcsin defined as the inverse function of the sin function. – jeremy radcliff May 1 '15 at 1:42
• OK, I mean not the inverse function of $\sin x$ over $\Bbb R$. I should practice language skill more. :( – MonkeyKing May 1 '15 at 1:46
• No problem, what you said is clear now in the context of user17762's answer, I just didn't have the knowledge to understand your reply when I first read it :) – jeremy radcliff May 1 '15 at 1:49
The important thing to realise here is that a function is not just a rule. To specify a function properly you must also define the domain and codomain. For example, the sine function is $$\sin:{\Bbb R}\to{\Bbb R}\quad\hbox{where}\quad \sin x= \langle\hbox{insert your favourite definition}\rangle\ .$$ This function is not one-to-one (injective) and therefore has no inverse. | {
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Here is a different function: $$f:\Bigl[-\frac\pi2,\frac\pi2\Bigr]\to[-1,1]\quad\hbox{where}\quad f(x)=\sin x=\langle\hbox{same definition as above}\rangle\ .$$ This function is one-to-one and onto and therefore has an inverse. Its inverse is commonly denoted $\sin^{-1}$ or $\arcsin$. Notice however that this terminology is not strictly accurate: $\arcsin$ is not the inverse of $\sin$, it is the inverse of $f$, which is a different function.
Hope this clears things up.
• Thanks David, it does clear things up. In the line $\sin:{\Bbb R}\to{\Bbb R}$, you don't limit the range to [-1, 1], but even with all the real numbers as potential inputs, in the original sin function the outputs can never be out of the range [-1, 1]. Do you not have to specify the codomain because it's implicitly limited? And in that case, do you have to specify the codomain for $f$? (not trying to nitpick, I just want to make sure I grasp the exact notational logic) – jeremy radcliff May 1 '15 at 2:02
• There is another common imprecision involved here. (Mathematicians speaking imprecisely... shock, horror ;-) Strictly speaking a function $f:A\to B$ is invertible if and only if it is one-to-one and onto. For example,$$f:{\Bbb R}\to{\Bbb R}\ ,\quad f(x)=e^x$$is not onto and therefore not invertible. But in calculus, one commonly treats $f$ as the same function as$$g:{\Bbb R}\to{\Bbb R}^+\ ,\quad g(x)=e^x$$(which it really isn't), and $g$ is invertible. (continued...) – David May 1 '15 at 2:08
• If my $f$ above had codomain $\Bbb R$ it would be invertible under this (mis)interpretation. However $\sin$ with domain $\Bbb R$ is not one-to-one and therefore is not invertible, no matter what you choose for the codomain. – David May 1 '15 at 2:08
• This is great, it all makes sense now. Thanks for taking the time to clarify, I really appreciate. – jeremy radcliff May 1 '15 at 2:17
• You're welcome, glad I could help. – David May 1 '15 at 2:20 | {
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Recall that $\arcsin: [-1,1] \mapsto [-\pi/2,\pi2/]$, i.e., the range of $\arcsin$ is $[-\pi/2,\pi/2]$.
The function $\sin$ on the other hand is a non-injective function that maps the real line to $[-1,1]$. Defining an inverse for a non-injective functions is not possible. Hence, one route that is often taken is to restrict the domain of the original function such that the function on the new restricted domain is injective.
Hence, the function $\arcsin$ is the inverse of $\sin$ only when the domain of the function $\sin$ is restricted to $[-\pi/2,\pi/2]$.
EDIT
What mathematica does is the following. It first evaluate $\sin(3\pi/4)$, which is $1/\sqrt{2}$. Now it goes and asks the function $\arcsin$, what $\arcsin(1/\sqrt2)$ is, which is now $\pi/4$.
• That makes sense, but in this case isn't $\frac{3\pi}{4}$ an invalid input and shouldn't Wolfram Alpha give me an error along the lines of "invalid domain" or something? Are they just allowing it because it's convenient? – jeremy radcliff May 1 '15 at 1:45
• @jeremyradcliff No. What mathematica does is the following. It first evaluate $\sin(3\pi/4)$, which is $1/\sqrt{2}$. Now it goes and asks the function $\arcsin$, what $\arcsin(1/\sqrt2)$ is, which is now $\pi/4$. – Leg May 1 '15 at 1:46
• Ah ok, now all of it makes sense; thank you, it was starting to drive me insane. – jeremy radcliff May 1 '15 at 1:48
As others have pointed out, $\arcsin x$ has the restricted domain $[-1,1]$ and range $[\frac{-\pi}{2},\frac{\pi}{2}]$. This restriction is a choice, a definition.
What's really happening is that $\arcsin x$ is multi-valued. We're only selecting what is called the principal branch, and we're doing this for convenience. | {
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Since the trigonometric functions are periodic, a single-valued function that acts as an inverse doesn't exist. We elect to "chop up" the inversion relation into sections - the branches. In the case of $\arcsin x$ we focus on returning $\sin x$ input values $[\frac{-\pi}{2},\frac{\pi}{2}]$. This branch is the principal branch - the one we've elected to use most often. Why? Well, consider that $\sin x$ cycles through all possible outputs when we give it $[\frac{-\pi}{2},\frac{\pi}{2}]$. Also, it's "nicely packaged" - there's a distinct symmetry and accessibility in using $[\frac{-\pi}{2},\frac{\pi}{2}]$.
When evaluating a query such as yours, we compensate by finding an equivalent result in the principal branch.
• Thanks zahbaz, that's very helpful because it makes it clear that it's a choice of convenience and why, which is something I've been struggling with (the idea that a lot of math is based on convenience and not some absolute ideal) – jeremy radcliff May 1 '15 at 2:11 | {
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Refractive Index of Water/Glass Combination
1. Jan 14, 2010
labview1958
If a beam of light enters water from air at 30 degrees from the normal. It then enters a glass block and exit into air. At what angle does it exit from glass to air ? My friend says it exist again at 30 degrees to normal. The refractive index of water is 1.3 and glass is 1.5.
2. Jan 14, 2010
RoyalCat
Your friend's forgotten Snell's Law: $$n_1 \sin{\theta _1}=n_2 \sin{\theta _2}$$
Where the index 1 refers to the material the light's coming from, and the index 2 refers to the material the light's entering.
3. Jan 14, 2010
MATLABdude
To elaborate upon what RoyalCat says, assume you have just a block of glass (width does not matter--the pendant in me says as long as it's sufficiently thin so that not all the light is absorbed). Now assume that you have a beam of light striking at angle $\theta_i$. What angle (in terms of $\theta_i$) will the beam of light travel at within the glass? At what angle (again, in terms of $\theta_i$) will the beam of light exit at?
This interesting result is probably what led your friend to come up with their answer. Unfortunately, in your example, the symmetry (because there is no glass on the way in) is broken.
Last edited: Jan 14, 2010
4. Jan 14, 2010
labview1958
For air/water/glass/water/air combination, then incoming and outgoing angle is same? This is not same for air/water/glass/air ?
5. Jan 14, 2010
mgb_phys
Correct
It's the same for each individual part, so a quick sketch should convince you for the whole setup.
6. Jan 14, 2010
jasper10
its 35.23 degrees according to Mr Snell and his law :P
7. Jan 15, 2010
labview1958
For air/water surface its 30 degrees. Definetly for air/glass surface is NOT 30 degrees. Do you agree?
8. Jan 15, 2010
modulus
Right, from water to air, the incident angle is 30, and you can figure out the refracted angle (at the water/air interface, made with the normal) using "Mr. Snell's" law. | {
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Now, if the water/air (it started from water, and, is entering into air) interface and the air/glass (it started from air, and, is entering into glass) interface are parallel (so that their normal's are parallel), you can figure out the incident angle for the air/glass interface ; it will be equal to the refracted angle at the water/air interface (using the simple properties of parallel lines and transversals).
So, no, the incident angle for the air/glass interface and the water/air interface will not be the same. I think that should answer your question.
Well, now that you've got your incident angle for the air/glass interface (it started from air, and is entering into glass), I want you to apply some logic and figure out what will be the angle of refraction when this same light ray exits glass and enters back into air?
P.S.:- Look at the bold stuff. It points to the logic your friend is using to derive his answer. But, he's gone wrong somewhere in between
Last edited: Jan 15, 2010
9. Jan 15, 2010
labview1958
The problem lies with the water/glass boundary. How can Mr. Snell help there? The refractive index of glass with respect to water? How to resolve that?
10. Jan 15, 2010
jasper10
Sorry, my previous answer was wrong, i didnt read the question correctly.
Step one: n1sin1 = n2sin2 (refractive index of air = 1)
(1)(sin30) = (1.3)(sinx)
-> x = 22.6 degrees
Step 2
n1sin1 = n2sin2
(1.3)(sin22.6) = (1.5)(siny)
-> y=19.47degrees
Step 3
n1sin1=n2sin2
(1.5)sin19.47 = (1)(sinz)
-> z = 30 degrees.
11. Jan 15, 2010
mgb_phys
It's exactly the same law n1 sin1 = n2 sin2, you are just starting with n=1.33 and going into n=1.5
12. Jan 16, 2010
labview1958
The calculations is correct. But how to explain the physics of it?
13. Jan 26, 2010
labview1958
Anyone care to explain the physics of it?
14. Jan 26, 2010
rl.bhat | {
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13. Jan 26, 2010
labview1958
Anyone care to explain the physics of it?
14. Jan 26, 2010
rl.bhat
In a parallel sided glass slab light is laterally shifted i.e. incident and refracted rays in the same medium are parallel to each other, whatever may be the number of different media in between medium. So the angle of incidence is equal to the angle of emergence.
15. Jan 26, 2010
MATLABdude
16. Jan 26, 2010
jasper10
Imagine a shopping trolly: if u push it from a smooth medium (eg marble) at an angle onto a rough concrete medium (more friction), one wheel reaches the rough medium first and thus slows down first. this causes a change of direction in which the trolley is travelling.
17. Jan 29, 2010
labview1958
"light is laterally shifted"
What physics is that?
18. Jan 29, 2010
ehild
One explanation is the analogy with mechanics. Photons have energy and momentum. Their energy is E=h*f, the magnitude of the momentum is p=h/lambda, and its direction points in the direction of propagation of the light ray. You know that that momentum changes when same force acts on the particle, the change of the momentum is equal to the impulse, and is parallel to the force applied. A plane surface can act only along its normal (there is no friction). The parallel component of the momentum does not change when the photon crosses the interface between two media (or reflects from it). This parallel component is h/ lambda*sin(alfa) which is constant everywhere in a plane-parallel arrangement of layers. But lambda =v/f (f a frequency, v is the speed of light in the medium.) The refractive index of the medium is n=c/v.
So the constancy of the parallel component of momentum involves the constancy of $$sin(\alpha) / \lambda = sin(\alpha) f*n/c$$. As the frequency of the photon does not change we arrived at Snell's law $\sin(\alpha)*n=const$. | {
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The normal component of the momentum will change, but we know that the magnitude of the momentum is h/lambda=hf/c *n. Thus the normal component of the photon momentum is
$$p_n= hf/c*\sqrt{n^2-(n_0\sin{alpha_0})^2}$$
Where n0 is the refractive index of the medium from where the light arrived and alpha0 is the angle of incidence.
If n>n0, the normal component of the momentum increases when the light enters into the other medium while the parallel component remains the same, that is while the ligth ray encloses a smaller angle with the normal in a "denser" medium as in air.
ehild | {
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# BALL BOUNCING INFINITELY to find distance covered
1. Feb 18, 2010
### abrahamrenns
ok, the qn is..
a ball is dropped from H height. it bounces with restitution e. hw much distance will it cover before it comes to rest..? surely, it is not infinity if i am not mistaken, since the bounces become infinitesimally small and then it tends to zero.. thnx in advance for any help..
2. Feb 18, 2010
### xlines
What did you try to solve your problem?
If on every bounce ball loses some fixed percentage of it's momentum, then you should calculate path traveled depending on starting momentum and sum it up. I am pretty sure you should get geometric sum.
3. Feb 18, 2010
### GRDixon
You have more or less rediscovered Zeno's paradox. But you're instincts are correct. The key is that the bounces become infinitesimally small, and no matter how great a finite number of bounces (finite amount of time) the distance traveled with always be finite.
4. Feb 18, 2010
### abrahamrenns
can some one plz take the pains to actually calculate it? it wudnt be difficult for a few here atleast.. i am not able to do it..
5. Feb 18, 2010
### xlines
Well, we do have assignments forums, but here you go. First, what distance will travel ball with upwards initial velocity v' ?
v(t) = v' - gt
it will reach max. height in t' = v'/g . Total distance traveled in one jump will be
d(v') = 2*1/2*g*t'$$^{2}$$ = v'$$^{2}$$/g
After each bounce starting velocity is lowered by factor $$\alpha$$ . So total distance is
D = H + v'$$^{2}$$/g + ($$\alpha$$ v')$$^{2}$$ /g + ($$\alpha$$$$^{2}$$ v')$$^{2}$$ /g + ... = H + v'$$^{2}$$/g (1+$$\alpha$$$$^{2}$$ + $$\alpha$$$$^{4}$$ + ... )
Substitution $$\beta$$ = $$\alpha$$$$^{2}$$
allows you to calculate infinite geometric sum.
Use energy conservation to link starting height H with v' .
Last edited: Feb 18, 2010
6. Feb 18, 2010
### HallsofIvy
Staff Emeritus
That's getting more complicated than necessary! | {
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### HallsofIvy
Staff Emeritus
That's getting more complicated than necessary!
The original problem told us that "it bounces with restitution e" which means that if it was dropped from height h, then it bounces up to height eh on the first bounce, $e(eh)= e^2h$ on the second, etc. Because it goes the same distance up as down, the total distance will be $h+ 2eh+ 2e^2h+ \cdot\cdot\cdot$, almost a geometric series. We can make it a geometric series by adding and subtracting h: $-h+ 2h+ 2eh+ 2e^2h+ ...$ where everything after the first "-h" is a gemetric series with first term 2h and constant ratio e. That total distance is
$$-h+ \frac{2h}{1- e}= \frac{-h+ +eh+ 2h}{1- e}= \frac{1+e}{1- e}h$$
7. Feb 18, 2010
### Phyisab****
Yep it's a nice problem to work out.
Last edited: Feb 18, 2010
8. Feb 18, 2010
### Saw
That is a nice solution. And if you wanted to fine-tune it... would it be possible to take into account that with every bounce, the ball gets deformed, less elastic and the coefficient of restitution diminishes?
9. Feb 19, 2010
### xlines
English is not my native and, to tell you the truth, I thought e plays role something like my $$\alpha$$. I too learned something today! Thanks for clearing that up! :)
10. Apr 15, 2010
### Zeus0
It's about right except that on rebounce it's not $$eh$$ but $$e^{2}h$$.
by definition $$e = \frac{v_2}{v_1} = \frac{\sqrt{2gh_2}}{\sqrt{2gh_1}}$$ or $$e = \sqrt{\frac{h_2}{h_1}}$$ . Look at the wikipedia entry at http://en.wikipedia.org/wiki/Coefficient_of_restitution" [Broken]
Meaning the finale answer is $$D = \frac{1+e^2}{1-e^2}h$$ . I also took the liberty to calculate the time it would take $$T = \frac{1+e}{1-e}\sqrt{\frac{2h}{g}}$$
I am more curious about weather or not real superball or bouncy balls rebounce an infinite number of time . It is likely not because of imperfections they may have or because of other processes in the way . I would estimate a real superball bounces between 20 to 50 times before resting. | {
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How is AM-GM supposed to be used to prove an inequality
I have recently started learning proofs involving inequalities and came across the AM-GM inequality, which seems like a quite powerful tool.
However, I am not sure I understand how to use this tool properly, and I was wondering if there are acceptable strategies to be aware of when making use of the AM-GM inequality.
I have also tried solving an inequality involving AM-GM to learn how to use this tool as I go, but I'm not sure if what I've done so far is a valid approach.
Here is the question:
Prove that if $$x, y, z, w ≥ 0$$,
$$\frac{x+y+z+w}{4} ≥ \sqrt[4]{xyzw}$$
Here is what I have done so far:
I noticed that $$\sqrt[4]{xyzw}$$ = $$\sqrt{\sqrt{xy}\sqrt{zw}}$$, and then used AM-GM, like so:
I let
$$a = \sqrt{xy}$$
$$b= \sqrt{zw}$$
to make a use of $$\frac{a+b}{2} ≥ \sqrt{ab}$$.
And, then, I subbed in the values:
$$\frac{\sqrt{xy}+\sqrt{zw}}{2} ≥ \sqrt{\sqrt{xy}\sqrt{zw}}$$.
Now I have a feeling I'm supposed to somehow make a use of the inequality ($$\frac{a+b}{2} ≥ \sqrt{ab}$$) once more, however, I am not exactly sure how I should go about doing it.
Any tips for a newbie like myself would be immensely helpful! (Perhaps a link I can refer to, an article, etc)
Thanks a lot in advance for any help! | {
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Thanks a lot in advance for any help!
• What's wrong with using AM-GM on four numbers directly?
– user239203
Jan 4 '20 at 23:25
• I just picked this inequality as a starting point to prove it using AM-GM for two variables first, and get exposed to the AM-GM inequality. I am eager to learn how to use this tool when dealing with inequalities requiring formal mathematical proofs. Jan 4 '20 at 23:28
• Did you mean $\sqrt{xy}+\sqrt{zw}$ where you wrote $\sqrt{zw}+\sqrt{zw}$? Jan 4 '20 at 23:34
• Yes, I'm sorry! Fixed. Jan 4 '20 at 23:36
• This doesn't specifically pertain to the problem, but in general here are two sources that may help you: artofproblemsolving.com/wiki/index.php/… brilliant.org/wiki/arithmetic-mean-geometric-mean Jan 4 '20 at 23:51
Start with $$\dfrac{a+b}{2} \ge \sqrt{ab}$$. To prove this, write it as $$\dfrac{a-2\sqrt{ab}+b}{2} \ge 0$$, and the left side is $$\dfrac{(\sqrt{a}-\sqrt{b})^2}{2} \ge 0$$.
Then,
$$\begin{array}\\ \dfrac{a+b+c+d}{4} &=\dfrac{a+b}{4}+\dfrac{c+d}{4}\\ &=\dfrac{\dfrac{a+b}{2}}{2}+\dfrac{\dfrac{c+d}{2}}{2}\\ &\ge\dfrac{\sqrt{ab}}{2}+\dfrac{\sqrt{cd}}{2}\\ &=\dfrac{\sqrt{ab}+\sqrt{cd}}{2}\\ &\ge\sqrt{\sqrt{ab}\sqrt{cd}}\\ &=\sqrt{\sqrt{abcd}}\\ &=\sqrt[4]{abcd}\\ \end{array}$$
By induction on $$n$$, with this technique you can show that $$\dfrac{\sum_{k=1}^{2^n}a_k}{2^n} \ge \sqrt[2^n]{\prod_{k=1}^n a_k}$$.
To show this is true for any $$m < 2^n$$, let $$a_j =\dfrac{\sum_{k=1}^m a_k}{m}$$ for $$j \gt m$$ and see what happens.
As a matter of fact, this was Cauchy's original proof.
Here's the details (added later).
The left side is, letting $$a = \dfrac{\sum_{k=1}^m a_k}{m}$$,
$$\begin{array}\\ \dfrac{\sum_{k=1}^{2^n}a_k}{2^n} &=\dfrac{\sum_{k=1}^{m}a_k}{2^n}+\dfrac{\sum_{k=m+1}^{2^n}a_k}{2^n}\\ &=\dfrac{\sum_{k=1}^{m}a_k}{m}\dfrac{m}{2^n}+\dfrac{\sum_{k=m+1}^{2^n}a}{2^n}\\ &=\dfrac{am}{2^n}+\dfrac{(2^n-m)a}{2^n}\\ &=\dfrac{am}{2^n}+\dfrac{2^na}{2^n}-\dfrac{ma}{2^n}\\ &= a\\ &=\dfrac{\sum_{j=1}^ma_j}{m}\\ \end{array}$$ | {
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Similarly, the right side is, letting $$a_j =b =\left(\prod_{k=1}^{m} a_k\right)^{1/m}$$ for $$j > m$$,
$$\begin{array}\\ \sqrt[2^n]{\prod_{k=1}^{2^n} a_k} &=\left(\prod_{k=1}^{2^n} a_k\right)^{1/2^n}\\ &=\left(\prod_{k=1}^{m} a_k\prod_{k=m+1}^{2^n} a_k\right)^{1/2^n}\\ &=\left(\prod_{k=1}^{m} a_k\right)^{1/2^n}\left(\prod_{k=m+1}^{2^n} a_k\right)^{1/2^n}\\ &=\left(b^m\right)^{1/2^n}\left(\prod_{k=m+1}^{2^n} b\right)^{1/2^n}\\ &=b^{m/2^n}\left(b^{2^n-m}\right)^{1/2^n}\\ &=b^{m/2^n}b^{(2^n-m)/2^n}\\ &=b\\ &=\left(\prod_{k=1}^{m} a_k\right)^{1/m}\\ \end{array}$$
Therefore $$a \ge b$$ or $$\dfrac{\sum_{j=1}^ma_j}{m} \ge \left(\prod_{k=1}^{m} a_k\right)^{1/m}$$.
$$\frac{x+y+z+w}{4}\ge\frac{\sqrt{xy}+\sqrt{zw}}{2}\ge\sqrt[4]{xyzw}$$
• Thanks a lot! Could you just explain in further detail what you did here? I am quite new to proofs and more specifically inequality proofs involving AM-GM. Jan 4 '20 at 23:31
• @FlavioEsposito I used the $2$-variables AM-GM three times, twice to create two square roots & once more for the fourth root.
– J.G.
Jan 4 '20 at 23:40
• I see. So, just to make sure I got it right: You used AM-GM once for $\sqrt{xy}$, once for $\sqrt{zw}$, and then for $\sqrt[4]{xyzw}$? Jan 4 '20 at 23:47
• Yep, that's right! Jan 4 '20 at 23:50
• For any values $a$, $b$, $x$, and $y$, if $\color{red}{a}\ge \color{red}{x}$ and $\color{blue}{b} \ge \color{blue}{y}$, then $\color{red}{a}+\color{blue}{b} \ge \color{red}{x}+\color{blue}{y}$. To see why: $$a \ge x \\ a+b \ge x+b \ge x+y$$ We know that $\color{red}{\frac{x+y}{2}} \ge \color{red}{\sqrt{xy}}$ and $\color{blue}{\frac{z+w}{2}} \ge \color{blue}{\sqrt{zw}}$ by AM-GM. Apply the result above and you have that relation. Jan 6 '20 at 15:08 | {
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# Compute the following line integral along a path of your choice (Finding potential)
Consider the following vector field:
$$\vec A(x,y,z)=(yz)\hat i+(xz)\hat j+(xy)\hat k$$
Compute the line integral of $A$ along a path of your choice connecting $(0,0,0)$ to $(1,1,1).$
I recognise that $\nabla\times A=0$ and so I can simply evaluate $V(1,1,1)-V(0,0,0)$ where $A=\nabla V$.
I'm not sure how to find the function $V$ though.
The solution says $V=xyz.$
• Added finding potential into your title. – MrYouMath May 16 '16 at 21:06
Assume V as a function of x,y and z.
Then $\nabla V=(\dfrac{\partial V}{\partial x},\dfrac{\partial V}{\partial y},\dfrac{\partial V}{\partial z})$
Compare components:
$$\dfrac{\partial V}{\partial x}=yz$$ $$\dfrac{\partial V}{\partial y}=xz$$ $$\dfrac{\partial V}{\partial z}=xy$$
Integrate: $$V=xyz+f_1(y,z)$$ $$V=xyz+f_2(x,z)$$ $$V=xyz+f_3(x,y)$$
From here you can see that $V=xyz+const.$ by inspection. If you want to do it mathematically you have to subtract two equations from each other.
1-2: $f_1(y,z)=f_2(x,z)$ from that you can conclude $f_1(y,z)=f_1(z)$ and $f_2(x,z)=f_2(z)$, or you would have $y$ on the left but no $y$ on the right, and so forth.
1-3: $f_1(z)=f_3(x,y)$.As one can see we have z on the left sid but no z on the right. Hence, $f_1$ is a constant function as it is not depending on $z$. if $f_1=const.$ $f_1=f_3(x,y)$ will imply that $f_3=f_1=const.$. Same argument as before and we can conclude $f_1=f_3=const$. But if $f_1=const.$ than from 1-2 $f_2=const$.
With that we obtain $f_i=const$ for $i=1,2,3$. At the end you will see, that $f_i=const$ for $i=1,2,3$. | {
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• I don't understand how you've gone from $f_1(y,z)=f_2(x,z)$ to $f_1(y,z)=f_1(z).$ – Si.0788 May 16 '16 at 21:10
• The equation tells you, that we have something with y and z on the left hand side and something with x and z on the right side of the equation. But it is not possible to have y only on one side of the equation, hence it must vanish and $f_1$ can only be a function of z. The same argument works for $f_2$. – MrYouMath May 16 '16 at 21:13
• Ok, that makes sense. Now how do you conclude that $f_i=c$? – Si.0788 May 16 '16 at 21:15
$\nabla G = A$
what is G?
$G = \int yz \,dx + f(y,z) = \int xz \,dy + g(x,z) = \int xz \,dz + h(x,y)\\ G = xyz + f(y,z) = xyz + g(x,z) = xyz + h(x,y)$
$f(y,z), g(x,z), h(x,y)$ must all be constant functions, and they can be any constant function, so might as well make them $0.$ | {
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# Using a Direct Proof to show that two integers of same parity have an even sum?
I seem to be having a lot of difficulty with proofs and wondered if someone can walk me through this. The question out of my textbook states:
Use a direct proof to show that if two integers have the same parity, then their sum is even.
A very similar example from my notes is as follows: Use a direct proof to show that if two integers have opposite parity, then their sum is odd. This led to:
Proposition: The sum of an even integer and an odd integer is odd.
Proof: Suppose a is an even integer and b is an odd integer. Then by our definitions of even and odd numbers, we know that integers m and n exist so that a = 2m and b = 2n+1. This means:
a+b = (2m)+(2n+1) = 2(m+n)+1 = 2c+1 where c=m+n is an integer by the closure property of addition.
Thus it is shown that a+b = 2c+1 for some integer c so a+b must be odd.
## ----------------------------------------------------------------------------
So then for the proof of showing two integers of the same parity would have an even sum, I have thus far:
Proposition: The sum of 2 even integers is even.
Proof: Suppose a is an even integer and b is an even integer. Then by our definitions of even numbers, we know that integers m and n exist so that a=2m and b=2m??? | {
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• So now that you have $a = 2n$ (you have a typo) and $b = 2m$, we need to talk about the sum of $a$ and $b$, so look at $a+b = 2n+2m$. Can you show that this sum is even? – Mike Pierce Jan 28 '16 at 16:45
• Yes, and so? (Once you've made the correction to $b=2n$, ) mimic the proof of the result in your notes and factor out the $2$... Now what's the other fork in the road? Well, that the two integers are both odd. Unfold the definition of "odd" similarly, add the resulting expressions, factor out the $2$, ... . – BrianO Jan 28 '16 at 16:46
• Just as a comment for future reference, note that this follows easily from the fact that $2=0 \mod 2$. That is, $2x=0 \mod 2$ for $x=1$ or $x=0$. – Aloizio Macedo Jan 28 '16 at 16:48
• @AloizioMacedo That's true, of course, but it may be too abstract an approach. – BrianO Jan 28 '16 at 16:50
"Suppose a is an even integer and b is an even integer. Then by our definitions of even numbers, we know that integers m and n exist so that a=2m and b=2m???"
Since a and b are different numbers they should be different m and n.
"Suppose a is an even integer and b is an even integer. Then by our definitions of even numbers, we know that integers m and n exist so that a=2m and b=2*n*?
And so a + b = 2m + 2n = 2(m+n) and as m+n =c for some integer c, a + b = 2c so by definition a + b is even.
Yes, you can use the definitions directly. If $a,b$ are even then like you say we have $a = 2m$ and $b = 2n$, so $a+b = 2m + 2n = 2(m+n)$, which is even.
Similarly, if $a,b$ are odd then we have $a = 2m + 1$ and $b = 2n + 1$, and so $a+b = (2m +1) + (2n + 1) = 2(m + n + 1)$, which is also even. | {
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• thanks for replying. So since the resulting integer sum of (a+b) always ends up being multiplied by (2), it will always yield an even sum? – Analytic Lunatic Jan 28 '16 at 17:28
• @AnalyticLunatic Yes, since by definition an even integer is one of the form $2\cdot \text{integer}$. – Alex Provost Jan 28 '16 at 17:32
• So then, how would this change if say, we were looking at a product instead of a sum? – Analytic Lunatic Jan 28 '16 at 19:32
• @AnalyticLunatic Can you put $2m \cdot 2n$ in the form $2 \cdot \text{integer}$? Can you put $(2m+1)\cdot(2n+1)$ in the form $2\cdot \text{integer}+1$? – Alex Provost Jan 28 '16 at 21:13
• Is it sufficient to prove using only the even case? If it is necessary to provide the odd case as well to prove the theorem, then how is it considered a "direct proof" rather than "proof by cases"? – Michael Fulton Apr 13 '17 at 0:23 | {
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Expected distances relating to points on a stick
Question
Two points are randomly and independently located on a stick of length $$1$$ m.
$$(a)\quad$$ Find the expected distance between the two points.
$$(b)\quad$$ Find the expected distance from the left end of the stick to the point which is closest to that end.
$$(c)\quad$$ Find the expected distance from the left end of the stick to the first of the two points, given that that point is closer to the left end than to the right end.
My working
$$(a)\quad$$ Let $$X\sim U(0, 1)$$ and the two points be $$X_1$$ and $$X_2$$.
$$\implies f_{X_1, X_2}(x_1, x_2) = 1,\quad 0 < x_1 < 1\quad, \quad 0 < x_2 < 1$$
\begin{aligned}[t] \mathbb{E}(\lvert X_1 - X_2 \rvert) & = \int^{\infty}_{-\infty}\int^{\infty}_{-\infty} (\lvert x_1 - x_2 \rvert) f_{X_1, X_2}(x_1, x_2)\ \mathrm{d}{x_1}\ \mathrm{d}{x_2} \\[1 mm] & = \int^1_0\int^{x_1}_0 (x_1 - x_2)\ \mathrm{d}{x_2}\ \mathrm{d}{x_1} + \int^1_0\int^1_{x_1} (x_2 - x_1)\ \mathrm{d}{x_2}\ \mathrm{d}{x_1} \\[1 mm] & = \frac 1 3 \end{aligned}
$$(b)\quad \frac 1 3$$
$$(c)\quad \frac 1 3$$
I know my answer to $$(a)$$ is correct and it also follows intuition - if two points are chosen randomly and independently along a distance of $$1$$ m, one would naturally expect them to be spaced out at equal distances.
However, I am unsure if my answers to $$(b)$$ and $$(c)$$ are correct - I used intuition here as well, but how would one derive the answers to $$(b)$$ and $$(c)$$ mathematically?
Any intuitive explanations will be greatly appreciated :) | {
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Any intuitive explanations will be greatly appreciated :)
• There is no essential difference with choosing randomly 3 points on a circle that has circumference of 1m and then using the first chosen point as the place where the circle is "opened" so that it receives a left end and a right end. In this described situation we can use symmetry and find directly $\frac13$ for expected (arc)distance between the points. Just like you I have the impression that the questions (b) and (c) are actually the same. – drhab May 4 at 12:09
(b) Let $$Y=\min\{X_1,X_2\}$$. Let's find the pdf of $$Y$$. Let $$F$$ be the cdf of $$X$$ and $$f$$ be the pdf.
$$\begin{split}\Pr(Y> y)&=\Pr(X_1> y)\Pr(X_2>y)\\ &=\left(1-F(y)\right)^2\end{split}$$
Then the cdf of $$Y$$ is
$$\begin{split}G(y)&=1-\Pr(Y>y)\\ &=1-(1-F(y))^2\end{split}$$
The pdf of $$Y$$ is the derivative
$$\begin{split}g(y)&=-2(1-F(y))(-f(y))\\ &=2f(y)(1-F(y))\\ &=2(1-y)\end{split}$$
(These are fairly standard computations/results, but I repeated them here anyway.)
This has expected value $$\int_0^12(y-y^2)dy=2\left[\frac {y^2}{2}-\frac{y^3}{3}\right]\bigg|_0^1=\frac 13$$
(c) We seek $$\mathbb E(Y|Y<\frac 12)$$. This will involve a reweighing of the density using $$\int_0^{1/2}2(1-y)dy=2\left[y-\frac{y^2}2\right]_0^{1/2}=\frac 34$$. Then the density of $$Y|Y<1/2$$ is $$\frac{2(1-y)}{3/4}, 0
This has expectation $$\int_0^{1/2}\frac 83(y-y^2)dy=\frac 83\left[\frac{y^2}2-\frac{y^3}{3}\right]_0^{1/2}=\frac 83\left(\frac 18-\frac 1{24}\right)=\frac 29$$
If we know the minimum is closer to the left end, then the expectation is slightly closer to the left end. | {
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• Thank you for your answer! I understand everything except, ironically, the very first part. How did you get $\mathbb{P}(Y > y) = \mathbb{P}(X_1 > y)\mathbb{P}(X_2 > y)$? Apologies if this is supposed to be trivial... – Ethan Mark May 4 at 15:44
• you're welcome! It's from: $\mathbb P(Y>y)=\mathbb P(\min\{X_1, X_2\}>y)=\mathbb P(X_1>y, X_2>y)=\mathbb P(X_1>y)\mathbb P(X_2 >y)$, due everything having to be bigger than the value if the minimum is to be bigger, and then independence of $X_1, X_2$. – Stacker May 4 at 16:35
• Right! That makes sense! :) – Ethan Mark May 4 at 16:53
• (Definition of minimum actually I think) – Stacker May 4 at 16:55 | {
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# $\sum_{k=0}^{n} a_{k}=\sum_{k=0}^{\lfloor n/2 \rfloor} a_{2k} +\sum_{k=0}^{\lfloor (n-1)/2 \rfloor}a_{2k+1}$
I would like to show that \begin{align} \sum_{k=0}^{n} a_{k}&=\sum_{k=0}^{\lfloor n/2 \rfloor} a_{2k} + \sum_{k=0}^{\lfloor (n-1)/2 \rfloor}a_{2k+1}\\ \end{align}
• I'm interested in more ways of prove it
My proof.
\begin{align} \sum_{k=0}^{n} a_{k}&=\sum_{\substack{k=0 \\ k \text{ is even }}}^{n} a_{k}+\sum_{\substack{k=0 \\ k \text{ is odd }}}^{n} a_{k}\\ &=\sum_{\substack{k=0 \\ k=2k' \text{ with } k'\in\mathbb{Z}}}^{n} a_{k}+\sum_{\substack{k=0 \\ k=2k'+1 \text{ with } k'\in\mathbb{Z}}}^{n} a_{k}\\ &=\sum_{\substack{k=0 \\ k'=\frac{k}{2} }}^{n} a_{2k'}+\sum_{\substack{k=0 \\ k'=\frac{k-1}{2} }}^{n} a_{2k'+1}\\ &=\sum_{\substack{k=0 \\ k'=\frac{k}{2}\\ k=0 \implies k'=0 \\ k=n \implies k'=\frac{n}{2}}}^{n} a_{2k'}+\sum_{\substack{k=0 \\ k'=\frac{k-1}{2}\\ \text{since k is odd can start with 1 and not 0 } k=1 \implies k'=0 \\ k=n \implies k'=\dfrac{n-1}{2} }}^{n} a_{2k'+1}\\ &=\sum_{k'=0}^{ n/2 } a_{2k'} + \sum_{k'=0}^{ (n-1)/2 }a_{2k'+1}\\ &\text{If $n$ is even, then $\frac{(n−1)}{2}$ is not an integer, which is why we need the floor.}\\ &\text{Similarly, if $n$ is odd, then $\frac{n}{2}$ is not an integer}\\ \sum_{k=0}^{n} a_{k}&=\sum_{k=0}^{\lfloor n/2 \rfloor} a_{2k} + \sum_{k=0}^{\lfloor (n-1)/2 \rfloor}a_{2k+1} \end{align} | {
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• Is my proof correct
• I'm interested in more ways of prove it
• so that you don't have to worry that: even number may take odd value and vice versa – SiXUlm Oct 29 '15 at 12:36
• also, a small typo in the first line of your proof: in the second summation, $k$ should start from 1 instead of 0. – SiXUlm Oct 29 '15 at 12:37
• If $n$ is even, then $(n-1)/2$ is not an integer, which is why you need the floor. Similarly, if $n$ is odd, then $n/2$ is not an integer. – Mankind Oct 29 '15 at 12:37
• @HowDoIMath yes i see that's why i have to add floor thanks. what about my proof – Educ Oct 29 '15 at 12:43
• well, shoot. Yes, your idea is correct and what you are trying to express is clear. I'm not sure your notation of indexing is technically legitimate however. Normally one really shouldn't get bogged do in details like that. However in this case what you are trying to prove is simply "The sum of all terms is the sum of the even terms plus the sum of the odd terms" (which is obvious) and your proof is verifying that the reindexing is valid. – fleablood Oct 29 '15 at 13:15
$\sum_{k=0}^{\lfloor n/2 \rfloor} a_{2k}= \sum_{i=0; i\ is\ even}^{2\lfloor n/2 \rfloor} a_{i} = \sum_{i=0; i\ is\ even}^{n} a_{i} = \sum_{k=0; k\ is\ even}^{n} a_{k}$
$\sum_{k=0}^{\lfloor (n-1)/2 \rfloor} a_{2k + 1}= \sum_{i=0; i\ is\ odd}^{2\lfloor (n-1)/2 \rfloor + 1} a_{i} = \sum_{i=0; i\ is\ odd}^{n} a_{i}= \sum_{k=0; k\ is\ odd}^{n} a_{k}$
$\sum_{k=0}^n a_n = \sum_{k=0; k\ is\ even}^{n} a_{k} + \sum_{k=0; k\ is\ odd}^{n} a_{k} = \sum_{k=0}^{\lfloor n/2 \rfloor} a_{2k} + \sum_{k=0}^{\lfloor (n-1)/2 \rfloor} a_{2k + 1}$
• i think in the second line $i=1$ not $0$ since $i=2k+1$ – Educ Oct 29 '15 at 14:15
If $n$ is odd, i.e. $n=2m+1$, | {
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If $n$ is odd, i.e. $n=2m+1$,
then $k=0, 1, 2, 3, ..., 2m+1=\underbrace{(0, 2, 4, 6, .., \color{blue}{2m})}_{\underset{k=0, 1, ..., \lfloor n/2\rfloor}{2k}},(\underbrace{1, 3, 5, ..., 2m-1, \color{red}{2m+1}}_{\underset{k=0, 1, ..., \lfloor (n-1)/2\rfloor}{ 2k+1}})$
as \begin{align} &2\bigg\lfloor \frac n2 \bigg\rfloor&&=2\bigg\lfloor \frac {2m+1}2\bigg\rfloor&&=2\bigg\lfloor m+\frac 12 \bigg\rfloor&&=\color{blue}{2m}\\ &2\bigg\lfloor \frac {n-1}2 \bigg\rfloor+1&&=2\bigg\lfloor\frac{2m}2\bigg\rfloor+1&&&&=\color{red}{2m+1}\end{align}
If $n$ is even, i.e. $n=2m$,
then $k=0, 1, 2, 3, ..., 2m=\underbrace{(0, 2, 4, 6, .., \color{green}{2m})}_{\underset{k=0, 1, ..., \lfloor n/2\rfloor}{2k}},(\underbrace{1, 3, 5, ..., \color{orange}{2m-1}}_{\underset{k=0, 1, ..., \lfloor (n-1)/2\rfloor}{ 2k+1}})$
as \begin{align} &2\bigg\lfloor \frac n2 \bigg\rfloor&&=2\bigg\lfloor \frac {2m}2\bigg\rfloor&&=\color{green}{2m}\\ &2\bigg\lfloor \frac {n-1}2 \bigg\rfloor+1&&=2\bigg\lfloor\frac{2m-1}2\bigg\rfloor+1=2\underbrace{\bigg\lfloor m-\frac12\bigg\rfloor}_{m-1}+1=2(m-1)+1&&=\color{orange}{2m-1}\end{align}
Hence the formula $$\sum_{k=0}^{n} a_{k}=\sum_{k=0}^{\lfloor n/2 \rfloor} a_{2k} + \sum_{k=0}^{\lfloor (n-1)/2 \rfloor}a_{2k+1}\\$$ holds irrespective of whether $n$ is odd or even. | {
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# Finding the vector equation of a plane in 3 space
I'm working on a question that asks me to find the vector equation of the plane in $$\mathbb R^3$$ that contains the point P(-1,6,0) and is parallel to the plane x+2y-z=5. I'm trying to make sure my answer is right versus the answer that I was provided with, which I think is wrong.
I pulled out the normal vector n = (1,2,-1). I proceeded to find two vectors v and u that are orthogonal to n but are not parallel to each other. I followed the theorem that states that when the dot product of 2 vectors equals zero, those vectors are orthogonal. I came up with v = (1,1,3) and u = (4,1,6). So I ended with the vector equation (x,y,z) = (-1,6,0) + t(1,1,3) + s(4,1,6).
It's my understanding that there are a number of different answers for this problem as there are a number of different vectors, besides the ones I found, that are orthogonal to the normal vector. The answer I was provided for this question stated that the vector equation is (x,y,z) = (-1,6,0) + t(3,-4,1) + s(1,-4,-1). I'm fairly certain that this answer is incorrect as the vectors in the vector equation are not orthogonal to the normal equation. Is my thinking correct?
The plane parallel to $$x+2y-z=5$$ containing $$(-1,6,0)$$ is $$x+2y-z=-1+2(6)-0=11$$.
Your parametrization satisfies that: $$(-1+t+4s)+2(6+t+s)-(3t+6s)=11$$.
The other answer does not: $$(-1+3t+s)+2(6-4t-4s)-(t-s)\not=11$$ generally. | {
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# Closure of Intersection is Subset of Intersection of Closures
I'm trying to prove the following problem from a book I found:
Let $X$ be a topological space and let $\mathscr{A}$ be a collection of subset of $X$. Prove
$\overline{ \bigcap \limits_{A \in \mathscr{A}} A}\subseteq \bigcap \limits_{A \in \mathscr{A}} \overline{A}$
*where the over line indicates closure.
First I'm going to show an example the them being equal, and then show an example of a proper subset. I know the examples are not needed for the proof, but I find them insightful, and these posts are like a notebook that I can refer back to later. Then I will attempt to prove this; I actually had a very difficult time with this one, and I feel like I'm missing something. In any case, please let me know If my examples and proof are valid, and of course alternative proof are very welcome.
(I) Example of $\overline{ \bigcap \limits_{A \in \mathscr{A}} A} = \bigcap \limits_{A \in \mathscr{A}} \overline{A}$
I find that with "most" collections of subsets, the two sides come up equal. Let $X$ be $\mathbb{R}$ with the standard Euclidean metric. Also let, $\mathscr{A}=\{(1,6),(3,10)\}$. So then $\bigcap \limits_{A \in \mathscr{A}} A= (3,6)$, and $\overline{ \bigcap \limits_{A \in \mathscr{A}} A} =[3,6]$. Next the right side would be: $\bigcap \limits_{A \in \mathscr{A}} \overline{A}= [1,6] \cap [3,10]=[3,6]$. So it is possible for the two sides to be equal.
(II) $\overline{ \bigcap \limits_{A \in \mathscr{A}} A}\subset \bigcap \limits_{A \in \mathscr{A}} \overline{A}$
It took me a while to find an example of this, but I found that if the collection of subsets consist of infinite sets that converge to an empty set, there will be a proper subset. I also found this example, and it uses an empty set as well. Is it possible for this to be true without an empty set?? | {
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Let $X$ be $\mathbb{R}$ with the standard Euclidean metric. Also let, $\mathscr{A}=(0,\frac{1}{n}), n \in \mathbb{N}$. Now, $0 \notin \bigcap \limits_{A \in \mathscr{A}} A$, and any positive number can be removed with a large enough $n$, thus $\bigcap \limits_{A \in \mathscr{A}} A = \varnothing$. The empty set is both open and closed ergo, $\overline{ \bigcap \limits_{A \in \mathscr{A}} A} = \varnothing$. On the other hand, $0 \in \overline{A}$, therefore, $\bigcap \limits_{A \in \mathscr{A}} \overline{A}=$ {$0$}. So it is possible for the closure of intersections to be a subset of the intersections of closures.
Proof: $\overline{ \bigcap \limits_{A \in \mathscr{A}} A}\subseteq \bigcap \limits_{A \in \mathscr{A}} \overline{A}$
I considered many different ways to prove this, but they all lead to dead ends. The only thing I could think of was to build on the fact that $A \subseteq \overline{A}$ for any subset $A$. Thus, $\bigcap \limits_{A \in \mathscr{A}} A\subseteq \bigcap \limits_{A \in \mathscr{A}} \overline{A}$. (Next is the part I feel is not solid) When $\bigcap \limits_{A \in \mathscr{A}} A$ is closed the boundary points that are added will also be points included in $\bigcap \limits_{A \in \mathscr{A}} \overline{A}$. To summarize symbolically:
(a)$\bigcap \limits_{A \in \mathscr{A}} A\subseteq \bigcap \limits_{A \in \mathscr{A}} \overline{A}$
(b)$\partial(\bigcap \limits_{A \in \mathscr{A}} A) \subseteq \bigcap \limits_{A \in \mathscr{A}} \overline{A}$
Thus, (a) $\wedge$ (b) $\rightarrow \overline{ \bigcap \limits_{A \in \mathscr{A}} A}\subseteq \bigcap \limits_{A \in \mathscr{A}} \overline{A}$
QED
As stated above, (b) seems intuitively true, and I guess it must be true for the actual statement being proven to be true. But I don't find it obviously true, especially when considering the most general cases of a topological space. | {
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• You are correct that $A \subseteq \overline A$ and therefore $\bigcap A \subseteq \bigcap \overline A$. Now $\bigcap \overline A$ is an intersection of closed sets, hence closed, so $\bigcap \overline A$ is a closed set containing $\bigcap A$. Since by definition $\overline{\bigcap A}$ is the smallest closed set containing $\bigcap A$, we must have $\overline{\bigcap A} \subseteq \bigcap \overline A$. – Bungo Jul 6 '16 at 23:56
• P.S. For a simple example with proper containment, take $A_1 = (0,1)$ and $A_2 = (1,2)$. Then $A_1 \cap A_2 = \emptyset$, so also $\overline{A_1 \cap A_2} = \emptyset$. On the other hand, $\overline{A_1} \cap \overline{A_2} = [0,1] \cap [1,2] = \{1\}$. – Bungo Jul 7 '16 at 0:01
• The intersection is contained in A, so the closure of the former is contained in the closure of the latter. Since this holds for any A, it holds for their intersection. – Steve D Jul 7 '16 at 0:38 | {
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# Boundary Condition for Schrödinger Equation in Infinite Range
I am trying to simulate the movement of a coherent state in a quantum harmonic oscilator, but for some reason the answer diverges and there is a warning about not enought boundary conditions.
Also, since I plan on moving to more complex situations in the future, I would like to avoid b.c.s of the kind:
p[t, L] == 0
The reason is that for higher energy situations it will be more taxing to keep L much greater than my area of interest an having the wave get too close to it would cause undesired reflexions.
My code, for a 1D case, is as follows:
w = 1;
L = 3;
c = 1;
usol = NDSolveValue[{I D[p[t, x], t] + 1/2 D[p[t, x], x, x] ==
1/2 w^2 x^2 p[t, x], p[0, x] == E^(-w (x - c)^2/2)*(w/Pi)^(1/4)}
, p, {t, 0, 10}, {x, -L, L}]
• Yes, the b.c. isn't enough, for a initial value problem defined on $-\infty<x<\infty$ you need to add artificial b.c. approximating "infinity". There're many posts about this type of issue, see e.g. mathematica.stackexchange.com/q/128516/1871 – xzczd May 8 '17 at 5:54
• @xzczd, thank you for the hint! I adapted the b.c. and it worked, but I must admit to not undertanding why it did... abc = D[p[t,x], x] + direction p[t,x] == 0 /. {{x -> lb, direction -> -1}, {x -> rb, direction -> 1}}; – ivbc May 8 '17 at 16:00
• I'm a bit surprised :D . I posted the example just to show that artificial boundary is needed, I didn't know ABC is also applicable for 1D Schrodinger's equation. – xzczd May 8 '17 at 16:48
• PLenty of references about it, like: sciencedirect.com/science/article/pii/S0021999108004804. But the subject is complex and I dont know if what I tried is a good answer... – ivbc May 8 '17 at 17:48
• Maybe you can ask a question e.g. "What's the state-of-art/most popular artificial boundary condition for 1D Schrodinger's equation" in scicomp.stackexchange.com ? – xzczd May 9 '17 at 5:50 | {
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Since OP has found this interesting post, let me try to implement the exterior complex scaling method mentioned there.
First, make the transform $x= \left\{\begin{array}{cc} & \begin{array}{cc} -R_0+e^{i \theta } (\xi +R_0) & -\xi >R_0 \\ R_0+e^{i \theta } (\xi -R_0) & \xi >R_0 \\ \xi & -R_0\leq \xi\leq R_0 \\ \end{array} \\ \end{array}\right.$, I'll use DChange for the task:
set = {I D[p[t, x], t] + 1/2 D[p[t, x], x, x] == 1/2 w^2 x^2 p[t, x],
p[0, x] == E^(-w (x - c)^2/2)*(w/Pi)^(1/4)};
right = Exp[I θ] (ξ - R0) + R0;
left = Exp[I θ] (ξ + R0) - R0;
middle = ξ;
coevalue = CoefficientList[
SimplifyPWToUnitStep@
Piecewise[{{right, ξ > R0}, {left, -ξ > R0}}, middle], ξ];
neweq = DChange[set, x == coe@0 + coe@1 ξ, x, ξ, p[t, x]] /.
Thread[(coe /@ {0, 1}) -> coevalue];
(* Alternative approach for deducing neweq: *)
help = DChange[#, x == #2, x, ξ, p[t, x]] &;
change = Piecewise[{{help[#, right], ξ > R0}, {help[#, left], -ξ > R0}},
help[#, middle]] &;
neweq = SimplifyPWToUnitStep@Map[change@# &, set, {2}];
Notice currently DChange doesn't directly support piecewise function so the coding is a little roundabout, but I think it's still simpler than transforming by hand.
Remark
SimplifyPWToUnitStep is an undocumented function that expands Piecewise into a combination of UnitStep. I use this function to "extract" ξ from the piecewise part, or CoefficientList in first approach and NDSolveValue in second approach will fail.
The next step, which is also the final step, is to solve the equation: | {
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The next step, which is also the final step, is to solve the equation:
w = 1;
L = 3;
c = 1;
tend = 20;
boundarylayer = L/5; thvalue = 1/2;
mol[n_Integer, o_: "Pseudospectral"] := {"MethodOfLines",
"SpatialDiscretization" -> {"TensorProductGrid", "MaxPoints" -> n,
"MinPoints" -> n, "DifferenceOrder" -> o}}
newsol = NDSolveValue[{neweq,
p[t, -L - boundarylayer] == p[t, L + boundarylayer] == 0} /. {R0 -> L, θ ->
thvalue}, p, {ξ, -L - boundarylayer, L + boundarylayer}, {t, 0, tend},
Method -> mol[25, 4]]; // AbsoluteTiming
DensityPlot[Norm@newsol[t, x], {t, 0, tend}, {x, -L, L}, PlotRange -> All,
PlotPoints -> 100, ColorFunction -> "AvocadoColors", FrameLabel -> {"t", "x"}]
Notice I've manually set the number of spatial grid points, or NDSolveValue will automatically choose a too large one because the initial condition is no longer smooth.
Besides, the choosing of R0 ($R_0$), boundarylayer (distance from $R_0$ to the boundary of computational domain) and thvalue ($\theta$) turns out to be a kind of art to make the reflection small enough. There might be some hint in the original paper about the method, but I simply find the proper value by trial and error. | {
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• (+1) You have transformed x and tto xi and theta, right? – zhk May 10 '17 at 7:49
• @zhk Nope, in this transformation, only x is transformed to ξ. θ is a parameter. You can have a look at the linked post for more information. – xzczd May 10 '17 at 8:03
• Does this simplify the problem? In what way this transformation helps? – zhk May 10 '17 at 8:04
• @zhk This transform makes the equation more complicated, of course :) , but by solving this equation instead, we obtain a better approximation for infinite range. (OP wants to solve the equation in $-\infty<x<\infty$. ) – xzczd May 10 '17 at 8:10
• @ivbc Actually this is a function I place in SystemOpen@"init.m" file, because I adjust these options frequently when using NDSolve. (See Method -> mol[25, 4]? It's the same as writing Method -> {"MethodOfLines", "SpatialDiscretization" -> {"TensorProductGrid", "MaxPoints" -> 25, "MinPoints" -> 25, "DifferenceOrder" -> 4}}. ) I adjust these options because, as mentioned above, if I don't manually set "MinPoints" and "MaxPoints", NDSolveValue will choose a too large one because the initial condition is not smooth. (You can take away the option and see what will happen. ) – xzczd May 11 '17 at 3:10
Let's remember Schrodinger's equation:
$i\hbar\frac{\partial}{\partial t} \Psi(\mathbf{r},t) = \left [ \frac{-\hbar^2}{2\mu}\nabla^2 + V(\mathbf{r},t)\right ] \Psi(\mathbf{r},t)$
For the harmonic oscilator $V = x^2$, so your equation is missing a p[x,t] on the RHS besides the boundary conditions. L also needs to be larger too. This seems to work.
w = 2;
L = 10;
c = 3;
usol = NDSolveValue[{I D[p[t, x], t] + 1/2 D[p[t, x], x, x] ==
1/2 w^2 x^2 p[t, x],
p[0, x] == Exp[(-w (x - c)^2/2)*(w/Pi)^(1/4)], p[t, L] == 0,
p[t, -L] == 0}, p, {t, 0, 10}, {x, -L, L}]
DensityPlot[Norm@usol[t, x], {t, 0, 10}, {x, -L/2, L/2},
PlotRange -> All, PlotPoints -> 100, ColorFunction -> "Rainbow"]
update | {
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update
For completeness purposes only, this is the solution with the abc boundary condition from @xzczd.
w = 1;
L = 3;
c = 1;
usol = NDSolveValue[{I D[p[t, x], t] + 1/2 D[p[t, x], x, x] -
1/2 w^2 x^2 p[t, x] == 0,
p[0, x] == E^(-w (x - c)^2/2)*(w/Pi)^(1/4),
Derivative[0, 1][p][t, -L] - p[t, -L] == 0,
Derivative[0, 1][p][t, L] + p[t, L] == 0},
p, {t, 0, 20}, {x, -L, L}]
DensityPlot[Norm@usol[t, x], {t, 0, 20}, {x, -L/2, L/2},
PlotRange -> All, PlotPoints -> 100, ColorFunction -> "Rainbow",
FrameLabel -> {"t", "x"}]
`
Note how in the first case the wave function starts to diffuse as time passes, so it's not really a coherent state.
• Thank you for spotting the mistake and for the solution. I improved the question a bit to tray to avoid the artificial b.c. p[t, L] == 0, as it must always be far from the area being analysed. – ivbc May 8 '17 at 15:56
• @ivb if with the ABC boundary condition it works, what is your question now? I don't understand. – tsuresuregusa May 9 '17 at 2:04
• @tsuresuregusa I guess it's because there's no guarentee that the simple ABC mentioned in my answer is always applicable. (It's a artificial boundary for 1D wave equation. ) Though currently it seems to work, it may be just a coincidence. (In the comment above, OP has found a paper about ABC for 1D Schrodinger's equation, in which the proposed ABC has a very different form. ) – xzczd May 9 '17 at 5:47
• @xzczd your last comment nails it, I think we are now out of the scope of this site. – tsuresuregusa May 9 '17 at 14:53
• Answer accepted since it solves the problem. The question about why it works is important, and can be further considered here: scicomp.stackexchange.com/questions/15973/… – ivbc May 9 '17 at 22:51 | {
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# Does t test only work for continuous random variable?
When using t test normally we would assume the original data follows normal distribution, but if the sample size is large enough, this assumption can be ignored due to Central Limit Theory.
I thought since normal distribution is for continuous random variable, even if the data does not have normality, it at least must be continuous random variable in order to use the t test. However in practice, I found that the t test is widely used even for discrete random variable such as hospital length of stay (LOS) in days (which is integer and thus discrete random variable).
So my question is, is it a common practice to apply t test on discrete random variable? If not, when we want to compare the mean of two groups consisting of discrete random variable, except for Wilcoxon Signed Rank Test, what other tests can I use?
Remember the Central Limit Theorem is talking about the asymptotic distribution of the standardised mean of iid random variables. So even if your random variables are discrete, as you accumulate more and more of them, once you take their mean (and then standardise it), you'll end up with a random variable which acts more and more like a continuous random variable.
For example, if you have 100 Bernoulli random variables (each either 0 or 1), then the possible means are $0.00, 0.01, 0.02, \dots , 1.00$. If you have 1000 then the possible means are $0.000, 0.001, 0.002, \dots , 1.000$.
So even if you have discrete random variables, their mean can still be asymptotically normal. | {
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So even if you have discrete random variables, their mean can still be asymptotically normal.
• Note that the central limit theorem relates not to the mean itself but to a standardized mean. When necessary conditions hold, the sample mean itself is asymptotically just going to go to a constant -- the population mean. – Glen_b -Reinstate Monica Jul 17 '17 at 8:20
• Thank you @Glen_b, you're absolutely right. I just mentally absorbed the standardisation into the variance of the approximate normal for a particular sample, but of course that's silly since it undermines the point of the convergence in distribution result. – RoryT Jul 17 '17 at 8:32
• Thanks for your answer, that helped a lot! But as for central limit theorem, I am a little confused at your comments. The CLT says that 'given certain conditions, the arithmetic mean (or sum) of a sufficiently large number of iterates of independent random variables, each with a well-defined (finite) expected value and finite variance, will be approximately normally distributed, regardless of the underlying distribution'. To my understanding, the sample mean itself is related to CLT already, why does it to be 'standardised mean' like @Glen_b said? Could you shed more light? – user6892253 Jul 17 '17 at 9:33
• No worries. The mean itself is approximately normally distributed if you've got a large enough number of iid variables. However, as you take more and more iid variables, their mean will get closer and closer to the population mean, and will be less and less variable. If you multiply the difference between the sample and population mean by $\sqrt{n}$ then this counteracts the fact that the sample mean is getting closer and closer to the population mean. This is what we mean by standardisation. The standardised mean will be asymptotically distributed as a standard normal (i.e. $N(0,1)$) – RoryT Jul 17 '17 at 10:23
Does T Test only work for continuous random variable?
It depends on what you mean by "work". | {
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Does T Test only work for continuous random variable?
It depends on what you mean by "work".
When using T test normally we would assume the original data follows normal distribution, but if the sample size is large enough, this assumption can be ignored due to Central Limit Theory.
Not if you care about power.
I thought since normal distribution is for continuous random variable, even if the data does not have normality, it at least must be continuous random variable in order to use the t test.
Your argument that was based on the CLT wouldn't distinguish between discrete and continuous variables -- the CLT applies to either.
However in practice, I found that the t test is widely used even for discrete random variable such as hospital length of stay (LOS) in days (which is integer and thus discrete random variable).
It does get used on discrete data (and on obviously non-normal continuous data, and on data that are neither discrete nor continuous) at least sometimes. Some of those times, it's probably quite reasonable to do so (in that the significance level will be close to the nominal level and the power will be at least adequate).
I would tend to avoid it in that length-of-stay case; leaving aside likely issues with censoring, you have something which may be extremely right skew and even if your sample size were huge, the power may still be poor.
So my question is is it a common practice to apply t test on discrete random variable?
If you're going to apply it in a case where you can be confident your samples are drawn from something that's fairly non-normal you'd probably want to have an argument ready for why it would perform reasonably well.
If not, when we want to compare the mean of two groups consisting of discrete random variable, except for Wilcoxon Signed Rank Test, what other tests can I use? | {
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A signed rank test is a one-sample test (or a paired-sample test), not a two-sample test. It also doesn't compare means (it could reject in the wrong direction, for example)
You might use a permutation test of the difference in means, or make a different parametric assumption (in some situations you might consider a negative binomial assumption for example, or some other discrete distribution -- depending on circumstances), or perhaps add some assumptions to a Wilcoxon-Mann-Whitney that could make it also a comparison of means (though it won't necessarily outperform the t for discrete skewed cases). | {
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• What do you mean my power in your answer? – SpeedBirdNine Jul 17 '17 at 16:31
• en.wikipedia.org/wiki/Statistical_power – Glen_b -Reinstate Monica Jul 17 '17 at 22:17
• @Glen_b Your answer is so compact and helpful, thanks a lot! As for the statistical power you have mentioned, I know that parametric testing tends to have more power than non-parametric testing, but have no idea that the skewness of data would effect the power that much, could you elaborate more? Or is there any good recommended paper for me to learn from? Thanks! – user6892253 Jul 18 '17 at 7:59
• Actually, in many cases widely used parametric tests only tend to have slightly more power than the commonly used nonparametric tests when the assumptions of the parametric test are exactly true. When the assumptions are not true, the power of the nonparametric procedure may be much higher. For example, consider an ordinary equal variance two sample t-test compared to a Wilcoxon-Mann-Whitney. When the data are iid normal, the t-test has about 4.5% better efficiency in large samples (it's like the t-test has 22 observations worth of information every time the WMW has 21). – Glen_b -Reinstate Monica Jul 18 '17 at 8:15
• That's not even the most powerful nonparametric test at the normal. But when the tails are very heavy, the WMW may be vastly more efficient in large samples than the t. But when I said "power may still be poor" I wasn't necessarily advocating nonparametric tests for the discrete skew case (power is sometimes poor for some of those too, though for slightly different reasons). – Glen_b -Reinstate Monica Jul 18 '17 at 8:17
Theoretically, t-test is only for continuous data because discrete data can never meet normality assumption. However, t-test is quite robust given enough samples.
Applying t-test for discrete value is actually quite common. For example, the popular way to test for microarray gene expression array is: moderated t-Test. | {
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https://www.dnastar.com/arraystar_help/index.html#!Documents/moderatedttest.htm
Gene expression is counting data and certainly not continuous. But t-test is simple and good enough.
EDIT:
I think @Glen_b is correct. But, discrete data simply don't satisfy theoretical assumptions, whereas there's a possibility that continuous data do satisfy.
Again he's correct. In my example, the method assumes asymptotic behaviour of the sample means.
EDIT:
t-test is indeed common for discrete test (answer @Glen_b), especially when we have a large sample and the histogram of the distribution "looks" normal anyway (not skewed).
The t-test is simple for everybody to understand. In many practical applications, a good-enough and easy statistical test is sufficient.
• Continuous data that's not normal can never actually meet the normality assumption either (not even on average -- sample means of iid continuous rvs can only have a normal distribution if they were to begin with -- the convergence, whether from discrete or continuous population is only asymptotic either way). I doubt the normality assumption is ever true on real data. – Glen_b -Reinstate Monica Jul 17 '17 at 10:20
• When you say 't test is common for discrete test when we have a large sample and the histogram of the distribution looks normal', what about the skewed discrete data with a large sample size, is there any method to verify whether t test can work well in such case? – user6892253 Jul 18 '17 at 8:19 | {
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# Solving a Set Problem
Here is my question
• In a class that has $40$ students, there are $10$ students who could not pass the math exam and $30$ students who passed the english exam. If there are $6$ students who could not pass the both exam, How many are there students who passed the both exam?
I'm currently not able to build the correct equation. Might I get help? It seems like a bit confused. That's why I need get help.
• Hint: use the inclusion/exclusion principle. – user370967 Oct 31 '17 at 16:37
• @Math_QED What is it, sir? I'm already trying on it. Before using inclusion/exclusion principle, I ought to build the correct equation. – user440404 Oct 31 '17 at 16:38
• "group theory" is an incorrect tag; this is a question about sets. – Andrew Tindall Oct 31 '17 at 16:39
• I believe that It seems better now. – user440404 Oct 31 '17 at 16:42
• @Bobtrollsten The inclusion/exclusion principle "gives you the equation," so it would be silly to build the equation before applying the principle. – rschwieb Oct 31 '17 at 16:43
There are $4$ types of people:
Let $A$ be the people who passed no exam.
Let $B$ be the people who passed only Math.
Let $C$ be the people who passed only English.
Let $D$ be the people who passed both.
We know $A+B+C+D=40$
We know $A+C=10$
We know $C+D=30$
We know $A=6$.
So this is just a system of equations.
• I think I got it now. – user440404 Oct 31 '17 at 16:54
I find a Venn Diagram to always be very helpful for these kinds of questions:
$A,B,C,D$ are the regions according to Jorge's answer
So we know the following:
Once you figure out how the '10 people did not pass Math' plays out in the diagram, the rest goes quick. | {
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• I was trying to plug it into venn. – user440404 Oct 31 '17 at 16:57
• does the venn diagram help? You get a lot of clumping in this case. – Yorch Oct 31 '17 at 16:59
• @JorgeFernández True ... but I think different approaches work better for different people. So maybe for some the diagram is more distracting than helpful, but for others it may be helpful. Personally, I'm a very visual person. And once you figure out how the '10 people not inside the Passed Math' circle plays out in the diagram, the rest goes quick. – Bram28 Oct 31 '17 at 17:08
I hope my cute Venn diagram can be useful to somebody :) $$...$$ | {
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Complex function sketch
I have just started studying some complex functions and I do not understand the concept very well (I cannot visualize them easily). So I would appreciate some help for the following problem:
Say we have the complex function f(z) = cosh(z) . I would like to know how the f-plane looks like when x=const. and respectively y=const.
Thank you very much!
-
When you can visualize complex functions easily, let me know. I still have a hard time, too :) – Clayton Sep 13 '13 at 20:29
Ok, I see that the "easily" is rather stupid but what I wanted to say is that the transition from real to complex functions is not very smooth in my head.:) – Whats My Name Sep 13 '13 at 20:45
You may be interested in this paper regarding visualizing complex functions:
"Phase Plots of Complex Functions: A Journey in Illustration" by Elias Wegert and Gunter Semmler.
Jim Fowler and I (Steve Gubkin) wrote this webpage to help visualization of complex functions:
We hope to produce an online complex analysis course soon which employs a diverse set of visualization methods. For example, we will use webGL to produce 3d plots of modulus colored by phase, phase plots on the riemann sphere, interactive geometry tools for the poincare metric on the disk, etc. We should also have a lot of computational exercises which are verified directly by a server running an instance of sage. We are still developing back end stuff right now, but we should have something up and running within the year. We do not have anything up there now, but watch this space: http://www.gratisu.org/.
Peace, Steve | {
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Peace, Steve
-
This is very helpful. Thank you very much! – Whats My Name Sep 13 '13 at 23:36
I'm interested by such a course. I already illustrated a whole course of mathematics for bachelor using complex functions on my website: www.mikaelmayer.com/reflex/category/cours-a-lepfl/ Students appreciated it. – Mikaël Mayer May 5 at 13:25
@MikaëlMayer Ya, I still have plans to do this, but we got sort of distracted by producing calculus content (also, I defend my thesis tomorrow). I cannot promise when, but I definitely intend to build this course! – Steven Gubkin May 5 at 18:31
Let $z=x+iy.$ Separating the real and imaginary parts of $f,$ we have \begin{align} f(z)=f(x+iy)=\cosh(z)=\dfrac{e^{x+iy}+e^{-x-iy}}{2}\\ =\dfrac{1}{2}[e^x(\cos{y}+i\sin{y} )+e^{-x}(\cos{y}-i\sin{y})] \\=\cos{y}\cdot\dfrac{e^x+e^{-x}}{2}+i\sin{y}\cdot\dfrac{e^x-e^{-x}}{2}. \end{align} Denote $$u(x+iy)=\Re{f(x+iy)}=\cos{y}\cdot\dfrac{e^x+e^{-x}}{2}=\cosh{x}\cdot\cos{y},\\ v(x+iy)=\Im{f(c+iy)}=\sin{y}\cdot\dfrac{e^x-e^{-x}}{2}=\sinh{x}\cdot\sin{y}$$ For fixed $x=c$
$$\dfrac{u(c+iy)}{\cosh{c}}=\cos{y},\\ \dfrac{v(c+iy)}{\sinh{c}}=\sin{y}.$$ Squaring last equalities, we have ellipse $$\dfrac{u^2}{\cosh^2{c}}+\dfrac{v^2}{\sinh^2{c}}=1.$$
-
Although you asked this question a while ago, I have a new answer. You may be interested by the website reflex4you.com in "expert" mode. For example, the $cosh$ function is the following:
$z\rightarrow cosh(z)$ (click on the picture to visit the website and enter other formulas)
Although the axis are not shown, the window is between $-4-3i$ and $4+3i$. The zero is black and white is infinite. 1 is red and -1 is cyan. This representation is quasi-invariant if you take the negative of the picture, it will represent the inverse function. | {
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To analyze this picture, you can consider that the values on the horizontal axis (the real axis) are red and darkest a zero, which means that $cosh(x) > cosh(0)$. The black spot are zeroes of the function corresponding to $+/- \pi/2$; Colored points denote complex functions. $i$ is greenish whereas $-i$ is purplish.
By clicking on the image above, you can modify the formula and try out other ones. I feel it is a very good way to represent complex functions.
To answer your original question, I wrote an algorithm usable on the website to compute the f-plane, for example for x constant. You can modify the value of the parameter c (constant for $x$). @M.Strochyk was right, these are ellipses.
let c = 1; set k = 0; let min = -4; let max = 4; let n = 100; func f = cosh(z); set result = 0; let threshold = 0.1; repeat n in set theta = ((max-min)*k/n+min)*i+c; set result = if(abs(f(theta)-z) < threshold, theta, result); set k = k + 1; result
Click on the picture to be able to modify it by entering different values.
DISCLAIMER: I am one of the co-authors of this website.
- | {
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Step-by-Step Solution for $x^{1/x}=2^{1/2}$ [duplicate]
I came across the equation $$x^{1/x}=2^{1/2}$$ where $x\in\mathbb R$. One can immediately see that $x=2$ is a solution, but it is easy to miss that $x=4$ satisfies the equation as well. Verfiying that $2,4$ are solutions is not hard, but how would one go about formally solving this equation, i.e. how could one solve for $x$? I am asking because if the equation was say $x^{1/x}=2^{1/3}$ then the two solutions would not be obvious.
Basically, how to fill in the dots: $$x^{1/x}=2^{1/2} \iff \ldots\iff x=2 \text{ or } x=4$$
• for $0 < A < \frac{1}{e}$ there are two solutions to $\frac{\log x}{x} = A.$ For $A > \frac{1}{e}$ there are no solutions. Just draw the graph of $y = \frac{\log x}{x}$ for $x > 0.$ – Will Jagy Jan 17 '15 at 23:32
• @WillJagy Thank you for this, but I am more interested in how to obtain the solutions, not so much in how to prove their existence. – Phil-ZXX Jan 17 '15 at 23:46
• Tom, you asked about the well known example, $2^4 = 4^2.$ Someone came up with a reasonable conjecture on all rational solutions to $x^y = y^x.$ Having trouble finding that. I mean he asked his conjecture as an MSE question. – Will Jagy Jan 18 '15 at 1:26
• a moderator found it for me, see answer at meta.math.stackexchange.com/questions/19299/… – Will Jagy Jan 18 '15 at 2:39
Notice
$$x ^{1/x} = 2^{1/2} \Rightarrow x = 2^{x/2} \Rightarrow x^{2} = 2^x$$
then you may look want take a look here. | {
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then you may look want take a look here.
• There are two good answers there. Feel free to ask anything. – Aaron Maroja Jan 17 '15 at 23:32
• Thank you. The accepted answer in your link does not actually provide any insights in how to obtain the solutions, whereas your solutions looks very nice (this was exactly what I was looking for). But I am still wondering, is there any way around the Lambert W function? – Phil-ZXX Jan 18 '15 at 0:08
• @Tom Yes, see my answer. And no, you won't get explicit results except for the special cases like $2^4=4^2$. – user2345215 Jan 18 '15 at 0:08
• @user2345215 I do agree with the reasoning of your answer, but it doesn't tell me how to obtain 2,4 (asssuming I did not know them). If the equation was $x^{1/x}=2^{1/3}$ we would have to actually derive the solutions, and I am looking for a way to do this (not a way to prove that 2,4 are the only solutions). – Phil-ZXX Jan 18 '15 at 0:12
• @Tom I'm glad I could help you Tom, well there is Newton's approximation. – Aaron Maroja Jan 18 '15 at 0:23
Note that $$(x^{1/x})'=(e^{(\ln x)/x})'=\tfrac{1-\ln x}{x^2}x^{1/x}$$ So the function is increasing on $(0,e)$ and decreasing on $(e,\infty)$. Therefore $2$ and $4$ are the only solutions.
From this follows that $x^{1/x}=c$ has two solutions for $c\in (1,e^{1/e})$, one for $c\in(0,1]\cup\{e^{1/e}\}$ and no for $c\in(e^{1/e},\infty)$.
I suppose it is necessary to consider what happens when $x < 0.$ There is a question of interpretation of $x^{\frac{1}{x}}$ for $x \leq 0,$ ( for example, $(-1)^{r}$ could reasonable interpreted as a non-real complex number if $r$ is an irrational positive real number, if we consider $-1$ to be $e^{i \pi}$), so I assume the function only to be defined for $x >0$, where it is defined to be $e^{\frac{\log x}{x}}.$ As far as this question is concerned, it seems that the case $x >0$ is the most relevant case anyway. | {
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Note that $x >0$ is a solution of $x^{\frac{1}{x}} = 2^{\frac{1}{2}}$ if and only if $\frac{\log x}{x} = \frac{\log 2}{2}.$ The derivative of $\frac{\log x}{x}$ is $\frac{ 1 - \log x}{x^{2}}$ which is positive when $x <e$ and negative when $x > e.$ Hence $\frac{\log x}{x}$ increases when $x <e$, then decreases when $x >e$, taking a maximum value when $x = e.$ The values of $\frac{ \log x}{x}$ are positive on $(1, \infty)$ and the function is differentiable on that interval.Now if $\frac{log{x}}{x} = \frac{\log{y}}{y}$ for $1 < x < y,$ then the derivative vanishes somewhere on $(x,y)$ by the Mean Value Theorem. Since the derivative vanishes only at $e$, we see that any value of $\frac{\log x}{x}$ can be attained at most twice. Since $\frac{\log{4}}{4} = \frac{2\log 2}{2 \times 2} = \frac{\log 2}{2},$ we see that $2$ and $4$ are the only values of $x$ for which $x^{\frac{1}{x}} = 2^{\frac{1}{2}}.$ | {
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# Good Question 13
Let’s end the year with this problem that I came across a while ago in a review book:
Integrate $\int{x\sqrt{x+1}dx}$
It was a multiple-choice question and had four choices for the answer. The author intended it to be done with a u-substitution but being a bit rusty I tried integration by parts. I got the correct answer, but it was not among the choices. So I thought it would make a good challenge to work on over the holidays.
1. Find the antiderivative using a u-substitution.
2. Find the antiderivative using integration by parts.
3. Find the antiderivative using a different u-substitution.
4. Find the antiderivative by adding zero in a convenient form.
Your answers for 1, 3, and 4 should be the same, but look different from your answer to 2. The difference is NOT due to the constant of integration which is the same for all four answers. Show that the two forms are the same by
2. “Simplifying” your answer to 1, 3, 4 and get that third form again.
Give it a try before reading on. The solutions are below the picture.
Method 1: u-substitution
Integrate $\int{x\sqrt{x+1}dx}$
$u=x+1,x=u-1,dx=du$
$\int{x\sqrt{x+1}dx=}\int{\left( u-1 \right)\sqrt{u}}du=\int{{{u}^{3/2}}-{{u}^{1/2}}}du=\tfrac{5}{2}{{u}^{5/2}}-\tfrac{3}{2}{{u}^{3/2}}$
$\tfrac{2}{5}{{\left( x+1 \right)}^{5/2}}-\tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}+C$
Method 2: By Parts
Integrate $\int{x\sqrt{x+1}dx}$
$u=x,du=dx$
$dv=\sqrt{x+1}dx,v=\tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}$
$\int{x\sqrt{x+1}dx}=\tfrac{2}{3}x{{\left( x+1 \right)}^{3/2}}-\int{\tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}dx}=$
$\tfrac{2}{3}x{{\left( x+1 \right)}^{3/2}}-\tfrac{4}{15}{{\left( x+1 \right)}^{5/2}}+C$
Method 3: A different u-substitution
Integrate $\int{x\sqrt{x+1}dx}$
$u=\sqrt{x+1},x={{u}^{2}}-1,$
$du=\tfrac{1}{2}{{\left( x+1 \right)}^{-1/2}}dx,dx=2udu$
$2{{\int{\left( {{u}^{2}}-1 \right){{u}^{2}}}}^{{}}}du=2\int{{{u}^{4}}-{{u}^{2}}}du=\tfrac{2}{5}{{u}^{5}}-\tfrac{2}{3}{{u}^{3}}=$ | {
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$\tfrac{2}{5}{{\left( x+1 \right)}^{5/2}}-\tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}+C$
This gives the same answer as Method 1.
Method 4: Add zero in a convenient form.
Integrate $\int{x\sqrt{x+1}dx}$
$\int{x\sqrt{x+1}}dx=\int{x\sqrt{x+1}+\sqrt{x+1}-\sqrt{x+1} dx=}$
$\int{\left( x+1 \right)\sqrt{x+1}-\sqrt{x+1}}dx=$
$\int{{{\left( x+1 \right)}^{3/2}}-{{\left( x+1 \right)}^{1/2}}dx}=$
$\tfrac{2}{5}{{\left( x+1 \right)}^{5/2}}-\tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}+C$
This, also, gives the same answer as Methods 1 and 3.
So, by a vote of three to one Method 2 must be wrong. Yes, no, maybe?
No, all four answers are the same. Often when you get two forms for the same antiderivative, the problem is with the constant of integration. That is not the case here. We can show that the answers are the same by factoring out a common factor of ${{\left( x+1 \right)}^{3/2}}$. (Factoring the term with the lowest fractional exponent often is the key to simplifying expressions of this kind.)
Simplify the answer for Method 2:
$\tfrac{2}{3}x{{\left( x+1 \right)}^{3/2}}-\tfrac{4}{15}{{\left( x+1 \right)}^{5/2}}+C=$
$\tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}\left( x-\tfrac{2}{5}\left( x+1 \right) \right)+C=$
$\displaystyle \tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}\left( \frac{5x-2x-2}{5} \right)+C=$
$\tfrac{2}{15}{{\left( x+1 \right)}^{3/2}}\left( 3x-2 \right)+C$
Simplify the answer for Methods 1, 3, and 4:
$\tfrac{2}{5}{{\left( x+1 \right)}^{5/2}}-\tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}+C=$
$\tfrac{2}{15}{{\left( x+1 \right)}^{3/2}}\left( 3\left( x+1 \right)-5 \right)+C=$
$\tfrac{2}{15}{{\left( x+1 \right)}^{3/2}}\left( 3x-2 \right)+C$
So, same answer and same constant.
Is this a good question? No and yes. | {
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So, same answer and same constant.
Is this a good question? No and yes.
As a multiple-choice question, no, this is not a good question. It is reasonable that a student may use the method of integration by parts. His or her answer is not among the choices, but they have done nothing wrong. Obviously, you cannot include both answers, since then there will be two correct choices. Moral: writing a multiple-choice question is not as simple as it seems.
From another point of view, yes, this is a good question, but not for multiple-choice. You can use it in your class to widen your students’ perspective. Give the class a hint on where to start. Even better, ask the class to suggest methods; if necessary, suggest methods until you have all four (… maybe there is even a fifth). Assign one-quarter of your class to do the problem by each method. Then have them compare their results. Finally, have them do the simplification to show that the answers are the same.
My next post will be after the holidays.
.
# Starting Integration
Behind every definite integral is a Riemann sums. Students need to know about Riemann sums so that they can understand definite integrals (a shorthand notation for the limit if a Riemann sun) and the Fundamental theorem of Calculus. Theses posts help prepare students for Riemann sums.
1. The Old Pump Where I start Integration
2. Flying into Integrationland Continues the investigation in the Old Pump – the airplane problem
3. Working Towards Riemann Sums
4. Definition of the Definite Integral and the FTC – a more exact demonstration from last Friday’s post and The Fundamental Theorem of Calculus – an older demonstration
5. More about the FTC The derivative of a function defined by an integral – the other half of the FTC.
6. Good Question 11 Riemann Reversed – How to find the integral, given the Riemann sum
7. Properties of Integrals
8. Variation on a Theme – 2 Comparing Riemann sums
9. Trapezoids – Ancient and Modern – some history | {
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# The Definite Integral and the FTC
The Definition of the Definite Integral.
The definition of the definite integrals is: If f is a function continuous on the closed interval [a, b], and $a={{x}_{0}}<{{x}_{1}}<{{x}_{2}}<\cdots <{{x}_{{n-1}}}<{{x}_{n}}=b$ is a partition of that interval, and $x_{i}^{*}\in [{{x}_{{i-1}}},{{x}_{i}}]$, then
$\displaystyle \underset{{\left| {\left| {\Delta x} \right|} \right|\to 0}}{\mathop{{\lim }}}\,\sum\limits_{{i=0}}^{n}{{f\left( {x_{i}^{*}} \right)}}\left( {{{x}_{i}}-{{x}_{{i-i}}}} \right)=\int\limits_{a}^{b}{{f\left( x \right)dx}}$
The left side of the definition is, of course, any Riemann sum for the function f on the interval [a, b]. In addition to being shorter, the right side also tells you about the interval on which the definite integral is computed. The expression $\left\| {\Delta x} \right\|$ is called the “norm of the partition” and is the longest subinterval in the partition. Usually, all the subintervals are the same length, $\frac{{b-a}}{n}$, and this is the last your will hear of the norm. With all the subdivisions of the same length this can be written as
$\displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,\sum\limits_{{i=0}}^{n}{{f\left( {x_{i}^{*}} \right)}}\frac{{b-a}}{n}=\int\limits_{a}^{b}{{f\left( x \right)dx}}$
Other than that, there is not much more to the definition. It is simply a quicker and more efficient notation for the sum.
The Fundamental Theorem of Calculus (FTC).
First recall the Mean Value Theorem (MVT) which says: If a function is continuous on the closed interval [a, b] and differentiable on the open interval (a, b) then there exist a number, c, in the open interval (a, b) such that ${f}'\left( c \right)\left( {b-a} \right)=f\left( b \right)-f\left( a \right)$.
Next, let’s rewrite the definition above with a few changes. The reason for this will become clear. | {
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Next, let’s rewrite the definition above with a few changes. The reason for this will become clear.
$\int\limits_{a}^{b}{{{f}'\left( x \right)dx}}=\underset{{\left| {\left| {\Delta x} \right|} \right|\to 0}}{\mathop{{\lim }}}\,\sum\limits_{{i=0}}^{n}{{{f}'\left( {{{c}_{i}}} \right)\left( {{{x}_{i}}-{{x}_{{i-i}}}} \right)}}$
Since every function is the derivative of another function (even though we may not know that function or be able to write a closed-form expression for it), I’ve expressed the function as a derivative, I’ve also chosen the point in each subinterval, ${{c}_{i}}$, to be the number in each subinterval guaranteed by the MVT for that subinterval.
Then, $\displaystyle {f}'\left( {{{c}_{i}}} \right)\left( {{{x}_{i}}-{{x}_{{i-i}}}} \right)=f\left( {{{x}_{i}}} \right)-f\left( {{{x}_{{i-1}}}} \right)$. Making this substitution, we have
$\int\limits_{a}^{b}{{{f}'\left( x \right)dx}}=\underset{{\left| {\left| {\Delta x} \right|} \right|\to 0}}{\mathop{{\lim }}}\,\sum\limits_{{i=0}}^{n}{{\left( {f\left( {{{x}_{i}}} \right)-f\left( {{{x}_{{i-1}}}} \right)} \right)}}$
$\displaystyle =f\left( {{{x}_{1}}} \right)-f\left( {{{x}_{0}}} \right)+f\left( {{{x}_{2}}} \right)-f\left( {{{x}_{1}}} \right)+f\left( {{{x}_{3}}} \right)-f\left( {{{x}_{2}}} \right)+\cdots +f\left( {{{x}_{n}}} \right)-f\left( {{{x}_{{n-1}}}} \right)$
$\displaystyle =f\left( {{{n}_{n}}} \right)-f\left( {{{x}_{0}}} \right)$
And since ${{x}_{0}}=a$ and ${{x}_{n}}=b$,
$\displaystyle \int_{a}^{b}{{{f}'\left( x \right)dx}}=f\left( b \right)-f\left( a \right)$ | {
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$\displaystyle \int_{a}^{b}{{{f}'\left( x \right)dx}}=f\left( b \right)-f\left( a \right)$
This equation is called the Fundamental Theorem of Calculus. In words, it says that the integral of a function can be found by evaluating the function of which the integrand is the derivative at the endpoints of the interval and subtracting the values. This is a number that may be positive, negative, or zero depending on the function and the interval. The function of which the integrand is the derivative, is called the antiderivative of the integrand.
The real meaning and use of the FTC is twofold:
1. It says that the integral of a rate of change (i.e. a derivative) is the net amount of change. Thus, when you want to find the amount of change – and you will want to do this with every application of the derivative – integrate the rate of change.
2. It also gives us an easy way to evaluate a Riemann sum without going to all the trouble that is necessary with a Riemann sum; simply evaluate the antiderivative at the endpoints and subtract.
At this point I suggest two quick questions to emphasize the second point:
1. Find $\int_{3}^{7}{{2xdx}}$.
Ask if anyone knows a function whose derivative is 2x? Your students will know this one. The answer is x2, so
$\displaystyle \int_{3}^{7}{{2xdx}}={{7}^{2}}-{{3}^{2}}=40$.
Much easier than setting up and evaluating a Riemann sum!
2. Then ask your students to find the area enclosed by the coordinate axes and the graph of cos(x) from zero to $\frac{\pi }{2}$. With a little help they should arrive at
$\displaystyle \int_{0}^{{\pi /2}}{{\cos \left( x \right)dx}}$.
Then ask if anyone knows a function whose derivative is cos(x). it’s sin(x), so
$\displaystyle \int_{0}^{{\pi /2}}{{\cos \left( x \right)dx}}=\sin \left( {\frac{\pi }{2}} \right)-\sin \left( 0 \right)=1-0=1$.
At this point they should be convinced that the FTC is a good thing to know.
There is another form of the FTC that is discussed in More About the FTC. | {
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There is another form of the FTC that is discussed in More About the FTC.
# Good Question 12 – Parts with a Constant?
Someone asked me about this a while ago and I thought I would share it with you. It may be a good question to get your students thinking about; see if they can give a definitive answer that will, of course, include a justification.
Integration by Parts is summarized in the equation
$\displaystyle \int_{{}}^{{}}{udv}=uv-\int_{{}}^{{}}{vdu}$
To use the equation, you choose part of a given integral (left side) to be u and part to be dv, both functions of x. Then you differentiate u and integrate dv and use them on the right side to obtain a simpler integral that you can integrate.
The question is this: When you integrate dv, should you, can you, have a constant of integration, the “+ ” that you insist upon in other integration problems? Why don’t you use it here? Or can you?
Answer: Let’s see what happens if we use a constant. Assume that $\displaystyle \int_{{}}^{{}}{dv}=v+C$. Then
$\displaystyle \int_{{}}^{{}}{udv}=u\cdot \left( v+C \right)-\int_{{}}^{{}}{\left( v+C \right)du}$
$\displaystyle =uv+Cu-\left( \int_{{}}^{{}}{vdu}+\int_{{}}^{{}}{Cdu} \right)$
$\displaystyle =uv+Cu-\int_{{}}^{{}}{vdu}-Cu$
$\displaystyle =uv-\int_{{}}^{{}}{vdu}$
So, you may use a constant if you want, but it will always add out of the expression.
For more on integration by parts see here for the basic idea, here for the tabular method, here for a quicker way than the tabular method, and here for more on the tabular method and reduction formulas.
# Good Question 11 – Riemann Reversed
Good Question 11 – or not. | {
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# Good Question 11 – Riemann Reversed
Good Question 11 – or not.
The question below appears in the 2016 Course and Exam Description (CED) for AP Calculus (CED, p. 54), and has caused some questions since it is not something included in most textbooks and has not appeared on recent exams. The question gives a Riemann sum and asks for the definite integral that is its limit. Another example appears in the 2016 “Practice Exam” available at your audit website; see question AB 30. This type of question asks the student to relate a definite integral to the limit of its Riemann sum. These are called reversal questions since you must work in reverse of the usual order. Since this type of question appears in both the CED examples and the practice exam, the chances of it appearing on future exams look good.
To the best of my recollection the last time a question of this type appeared on the AP Calculus exams was in 1997, when only about 7% of the students taking the exam got it correct. Considering that by random guessing about 20% should have gotten it correct, this was a difficult question. This question, the “radical 50” question, is at the end of this post.
Example 1
Which of the following integral expressions is equal to $\displaystyle \underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{\left( \sqrt{1+\frac{3k}{n}}\cdot \frac{1}{n} \right)}$ ?
There were 4 answer choices that we will consider in a minute.
The first key to answering the question is to recognize the limit as a Riemann sum. In general, a right-side Riemann sum for the function f on the interval [a, b] with n equal subdivisions, has the form:
$\displaystyle \underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{\left( f\left( a+\frac{b-a}{n}\cdot k \right)\cdot \frac{b-a}{n} \right)}=\int_{a}^{b}{f\left( x \right)dx}$ | {
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To evaluate the limit and express it as an integral, we must identify, a, b, and f. I usually begin by looking for $\displaystyle \frac{b-a}{n}$. Here $\displaystyle \frac{b-a}{n}=\frac{1}{n}$ and from this conclude that ba = 1, so b = a + 1.
Usually, you can start by considering a = 0 , which means that the $\displaystyle \frac{b-a}{n}\cdot k$ becomes the “x.”. Then rewriting the radicand as $\displaystyle 1+3\frac{1}{n}k=1+3\left( a+\frac{1}{n}\cdot k \right)$, it appears the function is $\sqrt{1+3x}$ and the limit is $\displaystyle \int_{0}^{1}{\sqrt{1+3x}}dx=\frac{14}{9}$.
(A) $\displaystyle \int_{0}^{1}{\sqrt{1+3x}}dx$ (B) $\displaystyle \int_{0}^{3}{\sqrt{1+x}}dx$ (C) $\displaystyle \int_{1}^{4}{\sqrt{x}}dx$ (D) $\displaystyle \tfrac{1}{3}\int_{0}^{3}{\sqrt{x}}dx$
The correct choice is (A), but notice that choices B, C, and D can be eliminated as soon as we determine that b = a + 1. That is not always the case.
Let’s consider another example:
Example 2: $\displaystyle \underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{\left( {{\left( 2+\frac{3}{n}k \right)}^{2}}\left( \frac{3}{n} \right) \right)}=$
As before consider $\displaystyle \frac{b-a}{n}=\frac{3}{n}$, which implies that b = a + 3. With a = 0, the function appears to be ${{\left( 2+x \right)}^{2}}$ on the interval [0, 3], so the limit is $\displaystyle \int_{0}^{3}{{{\left( 2+x \right)}^{2}}}dx=39$
BUT
What if we take a = 2? If so, the limit is $\displaystyle \int_{2}^{5}{{{x}^{2}}dx}=39$.
And now one of the “problems” with this kind of question appears: the answer written as a definite integral is not unique!
Not only are there two answers, but there are many more possible answers. These two answers are horizontal translations of each other, and many other translations are possible, such as $\displaystyle \int_{-25.65}^{-22.65}{{{\left( 27.65+x \right)}^{2}}dx}=39$. | {
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The same thing can occur in other ways. Returning to example 1,and using something like a u-substitution, we can rewrite the original limit as $\displaystyle \frac{1}{3}\cdot \underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{\left( \sqrt{1+\frac{3k}{n}}\cdot \frac{3}{n} \right)}$.
Now b = a + 3 and the limit could be either $\displaystyle \frac{1}{3}\int_{0}^{3}{\sqrt{1+x}}dx=\frac{14}{9}$ or $\displaystyle \frac{1}{3}\int_{1}^{4}{\sqrt{x}}dx=\frac{14}{9}$, among others.
The real problem with the answer choices to Example 1 is that they force the student to do the question in a way that gets one of the answers. It is perfectly reasonable for the student to approach the problem a different way, and get a different correct answer that is not among the choices. This is not good.
The problem could be fixed by giving the answer choices as numbers. These are the numerical values of the 4 choices:(A) 14/9 (B) 14/3 (C) 14/3 (D) $2\sqrt{3}/3$. As you can see that presents another problem. Distractors (wrong answers) are made by making predictable calculus mistakes. Apparently, two predictable mistakes give the same numerical answer; therefore, one of them must go.
A related problem is this: The limit of a Riemann sum is a number; a definite integral is a number. Therefore, any definite integral, even one totally unrelated to the Riemann sum, which has the correct numerical value, is a correct answer.
I’m not sure if this type of question has any practical or real-world use. Certainly, setting up a Riemann sum is important and necessary to solve a variety of problems. After all, behind every definite integral there is a Riemann sum. But starting with a Riemann sum and finding the function and interval does not seem to me to be of practical use.
The CED references this question to MPAC 1: Reasoning with definitions and theorems, and to MPAC 5: Building notational fluency. They are appropriate,and the questions do make students unpack the notation. | {
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My opinions notwithstanding, it appears that future exams will include questions like these.
These questions are easy enough to make up. You will probably have your students write Riemann sums with a small value of n when you are teaching Riemann sums leading up to the Fundamental Theorem of Calculus. You can make up problems like these by stopping after you get to the limit, giving your students just the limit, and having them work backwards to identify the function(s) and interval(s). You could also give them an integral and ask for the associated Riemann sum. Question writers call questions like these reversal questions since the work is done in reverse of the usual way.
Here is the question from 1997, for you to try. The answer is below.
Answer B. Hint n = 50
Revised 5-5-2022
# Parts and More Parts
At an APSI this summer the participants and I got to discussing the “tabular method” for integration by parts. Since we were getting far from what is tested on the BC Calculus exams, I ended the discussion and said for those that were interested I would post more on the tabular method this blog going farther than just the basic set up. So here goes.
Here are some previous posts on integration by parts and the tabular method
Integration by Parts 1 discusses the basics of the method. This is as far as a BC course needs to go.
Integration by Parts 2 introduces the tabular method
Modified Tabular Integration presents a very quick and slick way of doing the tabular method without making a table. This is worth knowing.
There is also a video on integration by parts here. Scroll down to “Antiderivatives 5: A BC topic – Integration by parts.” The tabular method is discussed starting about time 15:16. There are several ways of setting up the table; one is shown here and a slightly different way is in the Integration by Parts 2 post above. There are others.
### Going further with the tabular method. | {
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### Going further with the tabular method.
The tabular method works well if one of the factors in the original integrand is a polynomial; eventually its derivative will be zero and you are done. These are shown in the examples in the posts above and Example 1 below. To complete the topic, this post will show two other things that can happen when using integration by parts and the tabular method.
First we look at an example with a polynomial factor and learn how to stop midway through. Why stop? Because often there will be no end if you don’t stop. There are ways to complete the integration as shown in the examples.
Example 1: Find $\displaystyle \int_{{}}^{{}}{\left( 4{{x}^{3}} \right)\cos \left( x \right)dx}$ by the tabular method (See Integration by Parts 2 for more detail on how to set the table up)
Adding the last column gives the antiderivative:
$\displaystyle \int_{{}}^{{}}{\left( 4{{x}^{3}} \right)\cos \left( x \right)dx}=4{{x}^{3}}\sin \left( x \right)+12{{x}^{2}}\cos \left( x \right)-24x\sin \left( x \right)-24\sin \left( x \right)+C$
Now say you wanted to stop after $12{{x}^{2}}\cos \left( x \right)$. Example 2 shows why you want (need) to stop. In Example 1 you will have
$\displaystyle \int_{{}}^{{}}{\left( 4{{x}^{3}} \right)\cos \left( x \right)dx}=4{{x}^{3}}\sin \left( x \right)+12{{x}^{2}}\sin \left( x \right)+\int_{{}}^{{}}{-24x\cos \left( x \right)}dx$
The integrand on the right is the product of the last column in the row at which you stopped and the first two columns in the next row, as shown in yellow above.
Example 2 Find $\displaystyle \int_{{}}^{{}}{{{e}^{x}}\cos \left( x \right)dx}$
As you can see things are just repeating the lines above sometimes with minus signs. However, if we stop on the third line we can write:
$\displaystyle \int_{{}}^{{}}{{{e}^{x}}\cos \left( x \right)dx={{e}^{x}}\sin \left( x \right)}+{{e}^{x}}\cos \left( x \right)-\int_{{}}^{{}}{{{e}^{x}}\cos \left( x \right)dx}$ | {
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