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http://math.stackexchange.com/questions/386536/solution-gives-wrong-answer-to-probability-problem | # Solution gives wrong answer to probability problem
Great Northern Airlines flies small planes in northern Canada and Alaska. Their largest plane can seat 16 passengers seated in 8 rows of 2. On a certain flight flown on this plane, they have 12 passengers and one large piece of equipment to transport. The equipment is so large that it requires two seats in the same row. The computer randomly assigns seats to the 12 passengers (no 2 passengers will have the same seat). What is the probability that there are two seats in the same row available for the equipment?
This problem was posed on Brilliant last week, and now that the official solution is posted, I would like to know why my solution gives wrong result. Here is my solution:
The number of ways we can seat 12 (identical) people on 16 seats is $16\choose 12$. Now the equipment can occupy any of the rows (8 possibilities), and the 12 people must be seated on the remaining 14 seats ($14 \choose12$ ways). Thus the desired probability is $$\frac{8 {14\choose12}}{16\choose12}=\frac25$$ The official solution gives $\frac5{13}$, but I don't understand what is wrong with mine.
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Note that the link you provided is unique to you. You need to use the "Share this problem" link for others to view the problem properly. – Calvin Lin May 10 '13 at 0:55
@CalvinLin Oh, I'll keep that in mind. – Dave May 10 '13 at 12:02
You haven't factored in the case in which two rows are devoid of people - that setup is being double-counted in your count.
Think of it this way - you can place four seats, with the restriction that at least one pair must be together in a row. So subtract off the number of ways you can have no pairs from the total number of combinations $${16\choose4}-{8\choose4}\cdot2^4 = 700$$ as each of the four rows containing an empty seat can have it in either of the two seats. So the probability is $$\frac{700}{1820}=\frac{5}{13}$$ (note that this means that you were double-counting 28 combinations... which makes sense, as $28 = {8\choose2}$, and that's the number of ways you can have two pairs of empty seats)
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Your answer is too big, because the cases where the people are packed into 6 rows, leaving 2 empty rows for the equipment, have been counted twice.
In general, when you are getting the wrong answer to this kind of problem, and you can't see what you're doing wrong, it may help to try doing the same problem with smaller numbers. In this case, consider the same problem with 2 people and 3 rows of seats. Now the correct probability is one; there will always be room for the equipment. What do you get for an answer if you do it your way?
- | 2016-05-02T23:16:40 | {
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https://math.stackexchange.com/questions/3036564/collecting-proofs-for-sum-n-2-infty-fracn-12n-1 | # Collecting proofs for $\sum_{n=2}^{\infty} \, \frac{n-1}{2^n} = 1$ [duplicate]
Update: The summation I came across has the form shown in title, and that exact question appears to be new. I could ask for proofs that take on this summation directly (without reducing it to summations starting at $$n = 0$$ or $$n =1$$), and that would be the preferred answer and the only one I will accept. But might as well let this fly as is, collecting all proofs of the identified equivalent variants.
I just stumbled across the fact that
$$\tag 1 \sum_{n=2}^{\infty} \, \frac{n-1}{2^n} = 1$$
This is equivalent to
$$\tag 2 \sum_{n=1}^{\infty} \, \frac{n}{2^n} = 2$$
I discovered $$\text{(1)}$$ using a 'matrix/combinatorial' argument, but it would need work to turn it into a formal proof.
I googled and found this Quora link, explaining how to show $$\text{(2)}$$.
I didn't find the question on this site, prompting this 'collecting proofs post':
Please supply a proof demonstrating either $$\text{(1)}$$ or $$\text{(2)}$$. If you use any theory or technique, mention that at the start of your answer.
One approach is to note that for $$S_k:=\sum_{n\geq k}\frac{1}{2^n}$$ you have $$S_0=2$$ and $$S_{k+1}=\frac{1}{2}S_k$$. Now rearranging terms in the summation yields $$\sum_{n\geq1}\frac{n}{2^n} =\sum_{n\geq1}\sum_{k=1}^n\frac{1}{2^n} =\sum_{k\geq1}\sum_{n\geq k}\frac{1}{2^k},$$ corresponding to the following picture: The inner sums equal the $$S_k$$, so we can simplify this to $$\sum_{k\geq1}S_k =\sum_{k\geq1}\frac{1}{2^k}S_0 =S_0\sum_{k\geq1}\frac{1}{2^k}=S_0S_1=2.$$
• Anyone care to explain the downvote? – Servaes Dec 12 '18 at 15:05
• Well not me! In fact, I can use your approach to get a direct solution to the $n \ge 2$ summation. I wind up with $S_2 \times S_0 = 1$. So don't feel bad - accepting and upvoting your answer! – CopyPasteIt Dec 12 '18 at 15:14
• Glad to hear that! And indeed the argument can be generalized (or repeated?) to evaluate $\sum_{n\geq 0}\frac{f(n)}{2^n}$ for any polynomial $f$. – Servaes Dec 12 '18 at 15:25
• This is a nice answer, but I think it is important to keep the answers to this question in one place. Users should search the site to avoid duplicating material already present. This applies with extra force to experienced users who should have a fairly good idea of what type of questions have already been covered. – Jyrki Lahtonen Dec 13 '18 at 6:03
A probability theory flavored approach: The expression $$\sum_{n\geq1} \frac{n}{2^n}$$ is also the expected number of IID fair coin flips it takes to gets a heads. Assuming you know this number is finite (by some root/ratio test business), let $$H$$ be the expected time. Then from conditioning on seeing heads or tails on the first flip, respectively, $$H$$ satisfies the recursion $$H = \frac{1}{2}(1) + \frac{1}{2}(H+1),$$ so $$H = 2$$.
A brute-force approach: by induction or otherwise, $$\sum_{k=1}^n \frac{k}{2^k} = 2 - \frac{n+2}{2^n}.$$ Sending $$n \to \infty$$ recovers the desired result.
The standard technique is $$\sum_{n\ge 1}r^n=\frac{1}{1-r}-1\implies\sum_{n\ge 1}nr^{n-1}=\frac{1}{(1-r)^2}$$.
We have that for $$|x|<1$$ by
$$f(x)=\sum_{k\ge1} x^n=\frac x{1-x} \implies f'(x)=\sum_{k\ge1} nx^{n-1}=\frac1{(1-x)^2}$$
therefore
$$\sum_{k\ge1} nx^{n}=x\cdot \sum_{k\ge1} nx^{n-1}=\frac x{(1-x)^2}$$
A solution by examining the differences between successive terms.
Let us define
$$S = \sum_{n=1}^{\infty} \, \frac{n}{2^n} = \sum_{n=1}^{\infty} u_n$$
Note (Edit): the definition $$u_n = \frac{n}{2^n}$$ will be used for $$n=0$$, i.e. for a term not involved in the series.
We have:
$$u_n - u_{n+1} = \frac{n-1}{2^{n+1}} = \frac{u_{n-1}}{4}$$
It follows immediately:
$$0 = S - S = \sum_{n=1}^{\infty} u_n - \sum_{n=1}^{\infty} u_{n+1} -u_1 = -u_1 + \frac{u_0}{4} + \frac{S}{4}$$
And therefore, noting that $$u_0=0$$ and $$u_1 = \frac{1}{2}$$ $$S = \sum_{n=1}^{\infty} \, \frac{n}{2^n} = 2$$
Note: the same procedure can be used to show that $$\sum_{k=1}^n \frac{k}{2^k} = 2 - \frac{n+2}{2^n}$$
• Your answer needs some work. One problem, $u_n - u_{n-1}$ is never a positive number. Also, $u_0$ is not defined. – CopyPasteIt Dec 12 '18 at 15:52
• Sorry, I just discovered a typo: it was $u_n - u_ {n+1}$. I will correct and detail my answer a little bit more – Damien Dec 12 '18 at 16:49
• and $u_n - u_{n-1} = \frac{n}{2^{n}} - \frac{n-1}{2^{n-1}} = \frac{2-n}{2^{n}} = \frac{-(n-2)}{2^{n-2}\times 2^2} = \frac{-u_{n-2}}{4}$ - three subscripts involved – CopyPasteIt Dec 12 '18 at 16:52
• I have corrected the typo. It should be more clear now – Damien Dec 12 '18 at 16:55
• For me, removing all mention of $u_0$ is the clearest exposition, but OK... – CopyPasteIt Dec 12 '18 at 17:02
I stumbled on this by realizing that since
$$\tag 1 \sum_{n=1}^{\infty} \, \frac{1}{2^n} = 1$$
it must be true that
$$\tag 2 (\sum_{n=1}^{\infty} \, \frac{1}{2^n}) \times (\sum_{n=1}^{\infty} \, \frac{1}{2^n}) = 1$$
Using the rearrangement approach (and the identity $$\sum_1^n \, 1 = n$$) found in Servaes' answer,
$$\tag 3 (\sum_{n=1}^{\infty} \, \frac{1}{2^n}) \times (\sum_{m=1}^{\infty} \, \frac{1}{2^m}) = \sum_{n+m =k}^{\infty} \, \frac{1}{2^k} = \sum_{k=2}^{\infty} \, \frac{k-1}{2^k} = 1$$
demonstrating the identity equation. | 2020-02-17T07:09:09 | {
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https://math.stackexchange.com/questions/2325505/elevator-probability-calculation | # Elevator probability calculation
I'm learning probability, specifically techniques of counting, and need help with the following problem :
There are $5$ people in an elevator, $4$ floors in the building and each person exits at random. Find the probability that :
$(1)$ no one exit on the first floor;
$(2)$ at least one person exit on the first floor and at least one person on the second floor.
Since I'm having difficulties for $(2)$, I'm going to share my work for $(1)$.
$(1)$ The number of ways to assign the $3$ remaining floors to the $5$ people is $3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 = 3^5$ because for each person we can choose one of the $3$ remaining floors. By the same argument, there are $4^5$ ways to assign $4$ floors to $5$ people. Therefore, the requested probability is $$\frac{3^5}{4^5}.$$
Is my work correct for $(1)$? Any help for $(2)$ will be greatly appreciated.
• The statement "each person exits at random" is kind of vague. At each floor, does each person have a certain probability of leaving, or does each person, at the beginning, pick a random floor to get off at? – Frpzzd Jun 16 '17 at 20:55
• @Nilknarf I'm translating this exercise from my notes which are in greek so this is possibly not the most accurate translation. I think what is meant here is that all floors are equally likely. – user347616 Jun 16 '17 at 21:04
• Okay, I'll leave you an answer. :) – Frpzzd Jun 16 '17 at 21:05
• Your work is correct on 1). – Doug M Jun 16 '17 at 21:14
Nobody gets off on the First floor: $243$
Nobody gets off on the Second floor: $243$
Nobody gets off on at either floor: $2^5 = 32$
Note that nobody gets off at either floor is a subset of both nobody gets of at the first floor it is also a subset of nobody gets off at the second floor.
Nobody gets off on the First floor or Nobody gets off on the second floor: $243 + 243 - 32 = 454$
We have to subtract 32 to avoid double counting.
Somebody gets off at both the first floor and the second floor $= 4^5 - 454 = 570$ | 2019-07-17T07:03:37 | {
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http://math.stackexchange.com/questions/546030/significance-of-starting-the-fibonacci-sequence-with-0-1 | # Significance of starting the Fibonacci sequence with 0, 1…
DISCLAIMER: I do not deal with in-depth mathematics on a daily basis as some of you may, so please pardon my ignorance or lack of coherence on this topic.
QUESTION: What is the significance of starting the Fibonacci sequence with $0,1$ ?
For instance, if I picked any two random integers, say 2 and 7, to start a sequence would I actually be creating some multiple or derivation of the Fibonacci sequence?
Is there a general mathematical explanation for the relationship between any sequence represented by $a[0] = x, a[1] = y, a[n] = a[n-1] + a[n-2]$ and the Fibonacci sequence?
Or, back to my example sequence, is there a general mathematical relationship between:
$2,7,9,16,25,41,66,107,173,280...$
and
$0,1,1,2,3,5,8,13,21,34...$
Perhaps the Golden Ratio explains it somehow? Any help would be appreciated.
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I think it's just because 0 and 1 are the 2 lowest numbers that you can start from. – Cruncher Oct 30 '13 at 20:16
As in Brian Scott's answer, note that any sequence with that additive property is a "linear combination" of the Fibonacci numbers and the Lucas numbers. That is, if you make such a sequence and call it $\mbox{frogpelt}_n,$ then there will be constants $A,B$ such that $\mbox{frogpelt}_n = A F_n + B L_n.$ – Will Jagy Oct 30 '13 at 20:29
Fibonacci sequence. A series is an infinite sum. – Jack M Oct 30 '13 at 23:00
Yes, such sequences are closely related, and the relationship does involve the golden ratio.
Let $\varphi=\frac12(1+\sqrt5)$ and $\widehat\varphi=\frac12(1-\sqrt5)$; $\varphi$ is of course the golden ratio, and $\widehat\varphi$ is its negative reciprocal. Let $a_0$ and $a_1$ be arbitrary, and define a Fibonacci-like sequence by the recurrence $a_n=a_{n-1}+a_{n-2}$ for $n\ge 2$. Then there are constants $\alpha$ and $\beta$ such that
$$a_n=\alpha\varphi^n+\beta\widehat\varphi^n\tag{1}$$
for each $n\ge 0$. Indeed, you can find them by substituting $n=0$ and $n=1$ into $(1)$ and solving the system
\left\{\begin{align*} a_0&=\alpha+\beta\\ a_1&=\alpha\varphi+\beta\widehat\varphi \end{align*}\right.
for $\alpha$ and $\beta$. In the case of the Fibonacci numbers themselves, $\alpha=\frac1{\sqrt5}$ and $\beta=-\frac1{\sqrt5}$; in the case of the Lucas numbers $L_n$, for which the initial values are $L_0=2$ and $L_1=1$, $\alpha=\beta=1$.
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Thank you very much, good sir! – frogpelt Oct 30 '13 at 20:28
@frogpelt: You’re very welcome. – Brian M. Scott Oct 30 '13 at 20:29
Let's call a sequence "fibonacci-like" if it satisfies the recurrence $$s_{n+2} = s_n + s_{n+1}$$ for all $n$. It's easy to see (or to show) that if $\{s_i\}$ and $\{t_i\}$ are two fibonacci-like sequences, then so is $\{s_i+t_i\}$, and so is $\{cs_i\}$ where $c$ is any constant. So the collection of fibonacci-like sequences forms a vector space. The dimension of the vector space is 2, since specifying two elements of the sequence (say $s_0$ and $s_1$) are enough to determine it completely.
So let's abbreviate such a sequence as $[s_0, s_1]$. The standard Fibonacci sequence $0,1,1,2,3,\ldots$ is written as $[0,1]$ in this notation. The Lucas sequence $\mathcal L_i = 1,3,4,7,11,\ldots$ is written $[1,3]$.
Since the space of all fibonacci-like sequences is a two-dimensional vector space, any two elements will form a basis for it, unless one is a multiple of the other. For example, a simple and standard basis for this vector space is the two vectors $[0,1]$ and $[1,0]$. The former is simply the standard Fibonacci sequence. The latter is the sequence $1,0,1,1,2,3,5,\ldots$, which is just the standard Fibonacci sequence shifted right by one position; its $i$th element is $f_{i-1}$, the $(i-1)$th Fibonacci number.
Now consider the general fibonacci-like sequence $[p,q]$:
$$[p,q] = p[1,0] + q[0,1]$$
So the $i$th element of the sequence $[p,q]$ is exactly $$pf_{i-1} + qf_i.$$ For example, the Lucas sequence has $$\mathcal L_i = f_{i-1} + 3f_i.$$ Simiarly your example sequence is $[2,7]$ and is therefore related to the Fibonacci sequence by $$[2,7] = 2[1,0] + 7[0,1] = 2f_{i-1} + 7f_i.$$
Any two sequences form a basis of the space as long as they are not multiples of one another. For example, any fibonacci-like sequence can be expressed in the form $s_i = af_i + b\mathcal L_i$ for some constants $a$ and $b$. For your $[2,7]$ sequence, we want $[2,7] = a[0,1] + b[1,3] = [b,a+3b]$. So $b=2$ and $a=1$, and we get $[2,7] = f_i + 2\mathcal L_i$.
Now consider the Fibonacci sequence, but shifted left by $k$ places, whose $i$th element is $f_{i+k}$ for each $i$. Then the first two terms of the fibonacci-like sequence $\{f_{i+k}\}$ are $f_k$ and $f_{k+1}$, we get $$\{f_{i+k}\} = [f_k, f_{k+1}] = f_k[1,0] + f_{k+1}[0,1] = f_kf_{i-1} + f_{k+1}f_i$$ and we have just proved the sum-of-indices formula for fibonacci numbers. Take $i=k$ in this and we get $$f_{2k} = f_kf_{k-1} + f_{k+1}f_k$$ which is useful for calculating extremely large Fibonacci numbers quickly.
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For any sequence $a_0, a_1, a_2, a_3, \dots$ that satisfies the Fibonacci recurrence, we have $$a_n=(a_1-a_0)f_n+a_0f_{n+1},$$ where $f_0,f_1, f_2, \dots$ is the Fibonacci sequence.
To prove this, let $b_n=(a_1-a_0) f_n+a_0f_{n+1}$. Note that $b_0=a_0$ and $b_1=a_1$. The two sequences $(a_n)$ and $(b_n)$ "begin" in the same way, and satisfy the same recurrence, so they are the same.
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I think this is slightly off. If $n=1$, you get $b_1=a_1+a_0$. – Mike Oct 30 '13 at 21:18
@Mike: Thanks, cleaned up. – André Nicolas Oct 30 '13 at 21:32 | 2015-05-25T19:42:00 | {
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http://mathhelpforum.com/discrete-math/148804-eve-odd-numbers-question.html | # Thread: Eve/odd numbers question
1. ## Eve/odd numbers question
Hey guys,
Let set A contain the integers 1-48 inclusive.
The sum of all the integers is 48-1+1=48 and so we use n(n+1)/2 to get 1176.
How do we sum just the odd numbers OR just the even numbers?
2. Think like Gauss. Gauss invented the formula for summing the first n integers by pairing up numbers. So, you paired up 1 with 48, 2 with 47, 3 with 46, and so on. Those sums are all the same: 1 + 48 = 49, 2 + 47 = 49, etc. How many of those pairs are there? n/2. Hence, (n/2)(n+1) is the sum. I bet you could reproduce this thinking for just the evens or just the odds. What do you think?
3. Hmm. Would we find an n that satisfies (2n-1) and then take n^2 for the odd ones?
Not sure for the even ones
4. Try this on for size: for the evens from 1 to 48, inclusive:
2 + 48 = 50
4 + 46 = 50
...
How many of these pairs are there? Well, there are 24 even numbers from 1 to 48, inclusive, so I'd say there are 12 pairs.
That is, the sum is equal to ... what do you think?
5. Even number is 2n. So, 2(24)=48. then the answer would be 24^2?
6. If you're not finding my hints helpful, please just say so, and I can try a different track. You're not following my reasoning at all. This forum's purpose is not to just give you the answer, but to help you own the answer for yourself. That won't happen unless you do most of the work. We'll help you get unstuck, but that's as far as we go.
7. Another way to go is:
2 + 4 + ... + 48 =
2 (1 + 2 + ...+ 24)
8. True, although it doesn't work quite so easily for the odds. You can jury-rig it, I guess.
9. Originally Posted by Ackbeet
True, although it doesn't work quite so easily for the odds. You can jury-rig it, I guess.
Jury-rigging MacGyver style! Here's how I would approach the odd numbers (nothing against Ackbeet's approach)
LaTeX:
Code:
1+3+5+...+47=\sum_{n=1}^{24}(2n-1)=2\sum_{n=1}^{24}n-\sum_{n=1}^{24}1
Renderer
10. Without disrepsect to the many clever solutions so far:
either
1,3,5,7,9,....,47 is a linear progression and there are standard methods for summing those. It can easily be transformed into a hypergeometric progression (with ratio 1) and there are also standard methods for summing those.
or
You have a formula for the sum of all integers in the range, and undefined gave you a clever formula for the sum of all even numbers in the range. The sum of all odd numbers is the difference between the two
ie
1+2+3+4+5+....+47+48 = 1176 (in question)
2+4+6+8+...48 = 600 (from undefined's clever post)
So
1+3+5+7+9+...47 = 1176-600
11. What about this trick...I think it works for almost everything:
Use formula a+(n-1)+d=x
where a= first number in progression, d=difference in progression and x=last number in progression. Solve for n.
Then, take n(a+x)/2=Sum of numbers.
So, if we want to sum the odd numbers, we see that we havev 1+3+5+...+47
a=1 and d=2 and x=47.
1+2(n-1)=47
n=24.
(24(48))/2=576
I think you can use this to find all even, odd, numbers divisible by 7 etc etc
12. Hello, sfspitfire23!
Here's a kooky method . . .
Set $A$ contain the integers 1 - 48 inclusive.
The sum of all the integers is: . $\frac{(48)(49)}{2} = 1176$
How do we sum just the odd numbers OR just the even numbers?
$\text{We have: }\;\begin{array}{ccccccc}
\text{Even} &=& \overbrace{2 + 4 + 6 + \hdots + 48}^{\text{24 terms}} & [1] \\ \\ [-3mm]
\text{Odd} &=& \underbrace{1 + 3 + 5 + \hdots + 47}_{\text{24 terms}} & [2] \end{array}$
$\text{Subtract [1] - [2]: }\;\text{Even} - \text{Odd} \:=\:\underbrace{1 + 1 + 1 + \hdots + 1}_{\text{24 terms}}$
$\text{Hence: }\;\text{Even} - \text{Odd} \;=\;24 \quad\Rightarrow\quad \text{Even} \;=\;\text{Odd} + 24$
$\text{Since }\,\text{Even} + \text{Odd} \:=\:1176,\,\text{ we have: }\:(\text{Odd} + 24) + \text{Odd} \;=\;1176$
. . $\text{Hence: }\;2\text{(Odd)} \:=\:1152$
$\text{Therefore: }\;\begin{Bmatrix}\text{Odd} &=&576 \\ \text{Even} &=& 600 \end{Bmatrix}$ | 2016-12-03T14:07:30 | {
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https://math.stackexchange.com/questions/1994831/sitting-n-men-and-n-woman-around-a-table-having-the-wedding-couple-seated-to | # Sitting $n$ men and $n$ woman around a table having the wedding couple seated together
The problems says
At the wedding of John and Mary there are $n$ men and $n$ women. In how many ways they can sit at a round table, so that no two men is next to each other and John and Mary sit together?
I have some doubts because of the answer to his problem, which doesn't match the one I got.
My attempt:
Let suppose that John and Mary are already seated and that people begin sitting from the side of Mary. We have two cases. One where the person next to Mary is a man and the other when it is a woman.
1st Case:
We can set the men en $n!/n = (n-1)!$ different ways.
The are $n-1$ gaps to put the women (since next to a man are Mary and John together. So there are $(n-1)!$ different ways to sit the women.
By the rule of product we have that there are $[(n-1)!]^2$ number of ways of sitting n men and n women beginning with a man next to Mary.
2nd Case (A woman sits next to Mary):
By the same reasoning as above we find that there are $[(n-1)!]^2$ number of ways of sitting them.
Finally by the rule of sum we have that there is a total of $2[(n-1)!]^2$ number of ways of sitting n men and n women having John and Mary always seated together.
What I got matches the answer of the textbook but I feel the reasoning I followed wasn't that right. The fact of having assumed that ''The are $n-1$ gaps to put the women..'' doesn't convince me. Certainly next to the first man is Mary (that is, the gap of that side is occupied) but nothing is telling that after putting the last men there will be no gap between him and John, so there may be anyway $n$ gaps to put the women
• Surely you mean "wedding" couple rather than "weeding"? :) – hypergeometric Nov 1 '16 at 17:56
• @hypergeometric Ups. Yep, I mean that (: – Jazz Nov 1 '16 at 18:03
First of all the question is a little bit ambigous. We don't know whether the table is a circle of a row, but from your reasoning it can be concluded that it's a circle, although not for certain. Also are there $n$ men and $n$ women total or $n$ men and $n$ women plus John and Mary? Considering the textbook answer I would go with the first option. Anyway from now on I will accept the mentioned assumptions as true.
Note that there have to be an empty space between each men. Or to make it more simple if there are $2n$ numbered seats man can sit in the even or odd seats. So we have to put $n$ men in $n$ seats around a round table. The number of possibilities is $(n-1)!$
Now as Mary has to sit next to John we have two options for her, left of John or right of him. Next the remaining $n-1$ women can be seated in $n-1$ seats in $(n-1)!$. Eventually as the events are not related we have that the total number of combinations is $2\left((n-1)!\right)^2$
I guess this explanation will fill the "holes" in your solution.
• It's probably a circle since the question says "around". – hypergeometric Nov 1 '16 at 18:24
The rule that men and women alternate should also apply to John and Mary. Hence only your 1st case applies, i.e. $[(n-1)!]^2$, and not your 2nd case.
However, you should also consider the "mirror configuration" of the 1st case, i.e. where John is seated on Mary's left/right.
Hence the number of configurations should be multiplied by $2$, giving $2[(n-1)!]^2$.
NB - this is not the same as beads on a ring where you can flip over the ring and consider the configuration to be the same - here you can't flip over the dinner table! | 2019-05-26T03:11:55 | {
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https://math.stackexchange.com/questions/582196/find-the-limit-l-lim-n-to-infty-sqrt-frac12-sqrt3-frac13-cdo | # Find the limit $L=\lim_{n\to \infty} \sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+\cdots+\sqrt[n]{\frac{1}{n}}}}$
Find the limit following:
$$L=\lim_{ _{\Large {n\to \infty}}}\:\sqrt{\frac{1}{2}+\sqrt[\Large 3]{\frac{1}{3}+\cdots+\sqrt[\Large n]{\frac{1}{n}}}}$$
P.S
I tried to find the value of $\:L$, but I found myself stuck into the abyss of incertitude.
Thus, any help to get me out of this rift is more than welcome!
• Possibly related: math.stackexchange.com/questions/576110/… – abiessu Nov 26 '13 at 18:16
• The numerical value is 1.2722249619362552835210450521628613228181075332403 , according to PARI. Using the inverse symbolic calculator, I found no closed expression. – Peter Nov 26 '13 at 18:28
• n=10000;u=(1/n)^(1/n);while(n>2,n=n-1;u=(u+1/n)^(1/n));print(u) – Peter Nov 26 '13 at 18:51
• You could see if Landau's Algorithm (for denesting radicals) is of any help, but my guess is there isn't a "nice" expression for what you have here... – Benjamin Dickman Nov 26 '13 at 20:06
• I guess its limit is 1. Because it is increasing sequence which is bounded(maybe) by a number less than 2. But I don't know how to prove it! – Hamid Shafie Asl Jun 11 '14 at 11:20
I like how Yiorgos S. Smyrlis approached to find upper limit of $L$. In similar way, you can easily observe $\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\cdots}}} > L$ and lets assume that it converges to some constant $c$. Now, we can write ,
$\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\cdots}}} = c$
Squaring on both sides,
$\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\cdots}}} = c^2$
Which is nothing but,
$\frac{1}{2}+c = c^2$
Solving above quadratic expression, value of $c$ will be $\dfrac{1+\sqrt{3}}{2}$ . Thus we get slightly improved upper bound for $L$ as $L < \dfrac{1+\sqrt{3}}{2}$
• That's even better! I wonder if it is possible to get even tighter bound... – Aron D'souza Feb 20 '16 at 11:51
• The next bound is approximately $L<1.272871$ – Yuriy S Feb 20 '16 at 11:54
• Add it as an answer then. – Aron D'souza Feb 20 '16 at 11:55
Using Aron D'souza's idea further we can get:
$$L^2-\frac{1}{2}< \sqrt[3]{\frac{1}{3}+\sqrt[3]{\frac{1}{3}+\dots}}$$
$$\sqrt[3]{\frac{1}{3}+\sqrt[3]{\frac{1}{3}+\dots}}=\frac{1}{2} \left(1+\sqrt{\frac{7}{3}} \right)$$
$$L<\sqrt{1+\frac{1}{2} \sqrt{\frac{7}{3}}}=1.328067$$
To find the next bound we will need to solve:
$$c^4-c-\frac{1}{4}=0$$
The exact solution is too complex to write here (see Wolframalpha), so I'll just write it numerically:
$$\sqrt[4]{\frac{1}{4}+\sqrt[4]{\frac{1}{4}+\dots}}=1.0723501510383$$
The bound will become:
$$L<\sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+1.0723501510383}}=1.272871$$
Which is three correct digits of the numerical value of the limit.
To make my answer more complete, the exact value of $c_4$ is:
$$c=\frac{1}{2} \left( b+\sqrt{\frac{2}{b}-b^2} \right)$$
$$b=\sqrt{ \sqrt[3]{ \frac{a}{18} }-\sqrt[3]{ \frac{2}{3a} } }$$
$$a=9+\sqrt{93}$$
And solving the quintic equation:
$$c^5-c-\frac{1}{5}=0$$
We get the upper bound for the limit with four correct digits:
$$L<1.272282$$
Taking into account the corresponding lower boundary:
$$L>\sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+\sqrt[4]{\frac{1}{4}+\sqrt[5]{\frac{1}{5}+\sqrt[6]{\frac{1}{6}}}}}}=1.271035$$
We see that truncating the limit gives less accurate solutions than the method in this answer.
However, truncating at $\frac{1}{7}$ we can finally get very good boundaries:
$$1.27207<L<1.27228$$
This is a partial result:
The underlying sequence is increasing and upper bounded by $\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}=\dfrac{1+\sqrt{5}}{2}=\phi$. Thus the limit exists and it is less than $\phi$.
Actually, there is another way which gives better upper boundary. I'm posting it as a separate answer because of the size.
First we notice that:
$$\lim_{n \to \infty} \left( \frac{1}{n} \right)^{\frac{1}{n}}=\lim_{n \to \infty} \left(1- \left(1- \frac{1}{n} \right) \right)^{\frac{1}{n}}=1$$
This is not a proof, but the fact is well known. Now let's consider the following:
$$1< \left(\frac{1}{n-1}+1 \right)^{\frac{1}{n-1}}<1+\frac{1}{(n-1)^2}$$
$$1< \left(\frac{1}{n-2}+1+\frac{1}{(n-1)^2} \right)^{\frac{1}{n-2}}<1+\frac{1}{(n-2)^2}+\frac{1}{(n-2)(n-1)^2}$$
On the next step we get:
$$\dots 1+\frac{1}{(n-3)^2}+\frac{1}{(n-3)(n-2)^2}+\frac{1}{(n-3)(n-2)(n-1)^2}$$
In the end we obtain the following inequality:
$$\lim_{ _{\Large {n\to \infty}}}\:\sqrt{\frac{1}{2}+\sqrt[\Large 3]{\frac{1}{3}+\cdots}}<1+\sum^{\infty}_{k=2} \frac{1}{k ~k!}=Ei(1)-\gamma=1.3179$$
We can increase precision by moving the truncated series under the radical (and we should not forget to get rid of $2$ in every denominator):
$$1+2\sum^{\infty}_{k=3} \frac{1}{k ~k!}=1+2(Ei(1)-\gamma-1-1/4)=1.135804$$
$$L < \sqrt{\frac{1}{2}+1.135804}=1.27899$$
$$1+2\cdot 3 \sum^{\infty}_{k=4} \frac{1}{k ~k!}=1+6(Ei(1)-\gamma-1-1/4-1/18)=1.074080$$
$$L < \sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+1.074080}}=1.27305$$
Now for the lower boundary the better estimation would be:
$$L > \sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+1}}=1.26517$$
This is not ideal, but much more accurate than just truncating the sequence. | 2020-12-02T00:36:58 | {
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https://math.stackexchange.com/questions/2797935/do-both-versions-invariant-and-primary-of-the-fundamental-theorem-for-finitely/2797951 | # Do both versions (invariant and primary) of the Fundamental Theorem for Finitely Generated Abelian Groups hold at the same time?
So there are the two versions of the Fundamental Theorem for Finitely Generated Abelian Groups (FTFGAG). I take the following from A First Course in Abstract Algebra by Fraleigh. The first is as follows:
FTFGAG 1: Every finitely generated abelian group $G$ is isomorphic to a direct product of cyclic groups in the form $$G \cong \mathbb{Z}_{p_1^{r_1}} \times \cdots \times \mathbb{Z}_{p_n^{r_n}} \times \mathbb{Z} \times \cdots \times \mathbb{Z}$$ where $p_i$ are primes (not necessarily distinct) and $r_i$ are positive integers.
But then we also have the second version:
FTFGAG 2: Every finitely generated abelian group $G$ is isomorphic to a direct product of cyclic groups in the form $$G \cong \mathbb{Z}_{m_1} \times \cdots \times \mathbb{Z}_{m_r} \times \mathbb{Z} \times \cdots \times \mathbb{Z}$$ where $m_1 | m_2 | \cdots | m_r$.
My question is whether these two hold at all times? So say for $\mathbb{Z}_{20}$, do we have $\mathbb{Z}_{20} \cong \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_5 \cong \mathbb{Z}_4 \times \mathbb{Z}_5 \cong \mathbb{Z}_2 \times \mathbb{Z}_{10}$, even though $10$ is not a power of a prime (though of course $2|10$)?
I've also seen statements of the Chinese Remainder Theorem (CRT) which say that $\mathbb{Z}_{nm} \cong \mathbb{Z}_n \times \mathbb{Z}_m$ if and only if $\gcd(n,m)=1$. Does this not contradict $\mathbb{Z}_{20} \cong \mathbb{Z}_2 \times \mathbb{Z}_{10}$ from above? Or is treating $\mathbb{Z}_n$ and $\mathbb{Z}_m$ as rings in the CRT what makes the situation different? I guess what I'm asking in a nutshell is: why don't the primary and invariant forms of the structure theorems for abelian groups (and also modules over PIDs, etc) contradict each other?
Edit: So it seems that I didn't see this question which basically answers mine. CRT gives us a way of decomposition but FTFGAG only tells us that some decomposition is always possible. So for FTFGAG 1, $\mathbb{Z}_4 \times \mathbb{Z}_5$ suffices, for FTFGAG 2, $\mathbb{Z}_{20}$ suffices, and for the CRT $\mathbb{Z}_{20} \cong \mathbb{Z}_4 \times \mathbb{Z}_5$ works.
• Yes they are both correct theorems so they both hold for all finitely generated abelian groups. May 27, 2018 at 13:02
• Thanks, would you be able to shed some light on why the stuff in the CRT paragraph is/isn't a contradiction? May 27, 2018 at 13:04
• It is not true that $\mathbb{Z}_{20}\simeq\mathbb{Z}_2\times\mathbb{Z}_{10}$. May 27, 2018 at 13:09
• @MichaelBurr Indeed, I have just seen the (~duplicate) question which I linked in my edit. May 27, 2018 at 13:12
For your example, $\mathbb{Z}_{20}$,
• The first decomposition gives you $\mathbb{Z}_{20}\simeq\mathbb{Z}_{2^2}\times\mathbb{Z}_5$.
• The second decomposition gives you $\mathbb{Z}_{20}\simeq\mathbb{Z}_{20}$. In other words, this one doesn't break up the abelian group at all.
• You can note that $\mathbb{Z}_{20}\not\simeq\mathbb{Z}_2\times\mathbb{Z}_{10}$ since $\mathbb{Z}_{20}$ has an element of order $20$ while the maximum order of an element of $\mathbb{Z}_2\times\mathbb{Z}_{10}$ is $10$.
The idea of the statements is that it is possible to write it in this form, not that all products of the given form are isomorphic to the given group.
A more interesting example might be given by $$\mathbb{Z}_4\times\mathbb{Z}_6\times\mathbb{Z}_5.$$
• The first decomposition gives you $$\mathbb{Z}_4\times\mathbb{Z}_6\times\mathbb{Z}_5\simeq \left(\mathbb{Z}_{2^2}\right)\times\left(\mathbb{Z}_2\times\mathbb{Z}_3\right)\times\mathbb{Z}_5\simeq \mathbb{Z}_2\times\mathbb{Z}_{2^2}\times\mathbb{Z}_3\times\mathbb{Z}_5.$$
• On the other hand, the second decomposition gives you $$\mathbb{Z}_4\times\mathbb{Z}_6\times\mathbb{Z}_5\simeq\mathbb{Z}_2\times\mathbb{Z}_{2^2\cdot 3\cdot 5}=\mathbb{Z}_2\times\mathbb{Z}_{60}.$$ The group on the right combines the highest powers of each prime that appear in the fully expanded product of the first decomposition, then you work your way down by induction.
• "The group on the right combines the highest powers of each prime that appear in the fully expanded product of the first decomposition, then you work your way down by induction." I really like that way of thinking about it, thanks! May 27, 2018 at 13:23
You are correct about the Chinese remainder theorem indeed $\mathbb{Z}_{20} \not\cong \mathbb{Z}_2 \times \mathbb{Z}_{10}$. The problem with applying the 2nd version in this manner is the rank uniquely determines the group. So for example $\mathbb{Z}_{20}$ is rank 1, that is it already is written in that form (trivially) | 2022-07-03T05:11:39 | {
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https://math.stackexchange.com/questions/839045/calculus-find-the-value-of-x-so-that-f-x-0/839051 | # Calculus / find the value of $x$ so that $f ''(x)=0$
Let $f(x)= 10xe^x$
$(a)$ Find the exact value of $x$ so that $f ''(x) = 0$.
I tried:
\begin{align}f'(x)& = 10e^x\\f''(x)&= e^x\end{align} but at that point, the $f''(x)$ would never be a zero. So what is my mistake?
$(b)$ For what interval is $f(x)$ concave up?
I wonder how to know the concavity after knowing the equation
• One mistake is not using the product rule to evaluate $(10xe^x)'$. – David Mitra Jun 18 '14 at 23:45
• Once you have the correct second derivative function, $\ f(x) \$ is "concave upward" wherever $\ f \ ''(x) \ > \ 0 \$ and "concave downward" wherever $\ f \ ''(x) \ < \ 0 \$ . Keep in mind that the exponential factor $\ e^x \$ is always positive. – colormegone Jun 18 '14 at 23:49
• @DavidMitra thanks for the reminder – John Jun 18 '14 at 23:50
(a) $f'(x)=10e^x + 10xe^x$.
$f''(x)=10e^x + 10e^x + 10xe^x = 20e^x + 10xe^x =10e^x(2+x)$
So, at $x=-2$ you have $f''(x)=0$.
(b) Hint: If $f''(x)>0$, $f$ is concave up at $x$, and if $f''(x)<0$, $f$ is concave down at $x$.
• Thaanks !What did you do to go from f'(x) to f''(x)? – John Jun 18 '14 at 23:49
• I computed it using the product rule. – Twink Jun 18 '14 at 23:50
• @John, derivation using the product rule. $f''(x) = (10e^x + 10xe^x)' = 10 (e^x)' + 10(x)'e^x + 10x(e^x)' = 10e^x + 10 e^x + 10xe^x = 10(2+x)e^x$ – Graham Kemp Jun 18 '14 at 23:52
$(a)$ Since $f(x)=10x\cdot e^x$ we will use the product rule to obtain $f'(x)$ and $f''(x)$. So $f'(x)=10x\cdot e^x+10e^x$ and $f''(x)=10x\cdot e^x+20e^x$. Set $f''(x)=0$. So $10x\cdot e^x+20e^x=0$ implies that $10e^x(x+2)=0$. We know that $10e^x$ is never $0$ and so $x=-2$ will make $f''(x)=0$.
$(b)$ If $x<-2$, then the function is concave down. If $x>-2$, then the function is concave up. | 2020-02-20T21:21:27 | {
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https://math.stackexchange.com/questions/3168656/arbitrary-union-definition-in-set-theory | # Arbitrary union definition in set theory
I am reading Enderton's "Elements of Set Theory". He defines the union operation as $$\cup A = \{ x \;|\; x \text{ belongs to some member of } A\} = \{x \;|\; (\exists b \in A) x \in b\}$$ Maybe I am missing a subtle point from earlier on in the text, but what if the set in question is not a "set of sets"? For example, if $$A = \{1,2,3\}$$, then is $$\cup A = \emptyset$$?
• In systems of set theory such as ZF, every set is "a set of sets". – Lord Shark the Unknown Mar 30 at 19:11
• So then is my $A$ really $\{\{1\},\{2\},\{3\}\}$? – theQman Mar 30 at 19:12
• No, it's really $\{1,2,3\}$. – Lord Shark the Unknown Mar 30 at 19:13
• See Enderton's chapter on "Natural Numbers" to see the usual way of representing positive integers as sets. – Lord Shark the Unknown Mar 30 at 19:20
• Even without the "everything-is-a-set" context, we can still make sense of this by carefully writing down the definition: see this earlier question. – Noah Schweber Mar 30 at 19:32
When you're working in a system built out of the set theory, everything is a set. I assume the book will cover this later on, but one way of "building natural numbers" out of sets is called Von Neumann ordinals and the construction is as follows:
Let $$0 \equiv \emptyset$$. Let the number $$n \equiv n - 1 \cup \{ n - 1\}$$. That is, under this system: \begin{align*} &1 \equiv 0 \cup \{ 0 \} = \emptyset \cup \{\emptyset\} = \{\emptyset\} \\ &2 \equiv 1 \cup \{1 \} = \{\emptyset\} \cup \{\{\emptyset\}\} = \{\emptyset, \{\emptyset\}\} \\ &3 \equiv 2 \cup \{2\} = \dots = \{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\} \end{align*}
The point of the representation is that 0 is a set with zero elements, 1 is a set with 1 element, 2 is a set with 2 elements and so on. So we can define the "size of a set" as "the natural number it is in bijection with".
This goes down a rabbit hole, and we can then ask "well, now that I have the natural numbers, how do I define addition? Multiplication? What about the integers? rationals? reals?"
We can construct all of these things, and a set theory book usually describes these encodings in one of the later chapters.
But the point is that at this level, all the "stuff of math" is sets, and you can always write expressions such as $$1 \ \cup 2 \cup 3 = \{\emptyset\} \cup \{ \emptyset, \{\emptyset\}\} \cup \{ \emptyset, \{\emptyset\},\{ \emptyset, \{\emptyset\}\} \} = \{ \emptyset, \{\emptyset\},\{ \emptyset, \{\emptyset\}\}\} = 3$$!
• +1.... In the most-favored system of set theory, not only is everything (that can be said to exist) a set, but in the formal language there is no word or definition of "set". Things ( with various properties that $can$ be stated in the language ) are merely asserted to exist or not exist. – DanielWainfleet Mar 30 at 19:39
• Indeed, thanks! much appreciated – Siddharth Bhat Mar 30 at 20:18
As Lord Shark the Unknown indicates, in nearly all systems of set theory, every object under the sun is a set. In particular, the elements of any set are, in turn, sets themselves. For the specific example you give, i.e. $$\cup \{ 1,2,3 \},$$ we need to be able to describe the natural numbers as sets. On page 67 of Enderton's text, the construction is outlined. He defines \begin{align} 0 &= \varnothing \\ 1 &= \{0\} = \{ \varnothing \} \\ 2 &= \{0,1\} = \{ \varnothing, \{\varnothing\} \} \\ 3 &= \{0,1,2\} = \{ \varnothing, \{\varnothing\}, \{ \varnothing, \{\varnothing\} \}\}, \end{align} and so on. The "and so on" is explained in somewhat more detail in chapter 4 of the text (starting on page 66). Note that we could write $$2 = \{\varnothing, \{\varnothing\} \}$$ over and over again, but it is likely easier to see what is going on if we work one level of abstraction higher. That is, $$2 = \{0,1\}$$. Once we accept this abstraction, we have $$\cup\{1,2,3\} = 1 \cup 2 \cup 3 = \{0\} \cup \{0,1\} \cup \{0,1,2\} = \{0,1,2\} = 3.$$
• I see. Maybe I am getting ahead of myself since I am only in chapter 2. But I guess the construction you provided is pretty natural since at this point the only set that we can use as a building block is $\emptyset$ (Empty Set Axiom)? Then we can use "pairing" to form the singleton $\{\emptyset\}$, which is defined to be $\{\emptyset, \emptyset\}$? – theQman Mar 31 at 2:09 | 2019-08-21T08:06:10 | {
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https://byjus.com/question-answer/if-two-concentric-circles-are-drawn-with-centre-in-first-quadrant-one-touches-the-x/ | Question
# If two concentric circles are drawn with centre in first quadrant, one touches the X-axis and Y-axis and the other passes through the origin, the radius of the smaller circle is $$'r'$$, then the centre is
A
(2r,r)
B
(r,2r)
C
(r,r)
D
none of these
Solution
## The correct option is C $$(r,r)$$Let $$(a,b)$$ be the center of the circles Given that one of the circles touches both the axes and the other passes through origin. Smaller circle touches the axes and Bigger circle passes through $$O$$. Given radius of smaller circle is $$r$$ and $$X,Y$$ axes are tangents Which implies radius = perpendicular distance from center to tangent $$(y =0 , x = 0)$$ $$r = \dfrac{|a|}{\sqrt{1^2 + 0}} , r = \dfrac{|b|}{\sqrt{1^2 + 0}}$$ $$\implies a = r, b = r$$ $$\implies (a, b) = (r,r) = center$$Maths
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http://www.gtmath.com/2015/04/the-plane-in-r3.html?showComment=1430268957485 | ## The Plane in R^3
Prerequisites: Euclidean Space and Vectors
### Suppose we have a plane in ${\Bbb R}^{3}$ which contains the point $(0,0,0)$ but does not intersect the axes at any other point. How many octants does the plane intersect?
Let's start by analyzing the 2-dimensional analog, then come back to the 3-dimensional problem with an algebraic approach and a geometric interpretation. Finally, we'll try to generalize the answer to higher dimensions.
#### Lines in the Plane ${\Bbb R}^{2}$
The 2-dimensional analog of a plane in ${\Bbb R}^{3}$ containing the origin is a line in ${\Bbb R}^{2}$ containing the origin. For those familiar with linear algebra, the former is a vector subspace of ${\Bbb R}^{3}$ of dimension 2 (i.e. co-dimension 1), and the latter is a vector subspace of ${\Bbb R}^{2}$ of dimension 1 (still co-dimension 1). If you aren't familiar with linear algebra, ignore that sentence and just keep reading.
The equation of a line in ${\Bbb R}^{2}$ is $y = mx + b$, where $m$ is the slope ("rise over run," or change in $y$ per change in $x$), and $b$ is the $y$-intercept (where the line hits the $y$-axis). A point $(x_{0},y_{0})$ is on the line if it satisfies the equation, i.e. if the equality $y_0 = mx_0 + b$ is true.
A line passing through the origin has a $y$-intercept of 0, thus $b=0$ and the equation is simply $y=mx$. Now, if $m$ is 0, then the line would just go along the $x$-axis, so if it doesn't touch the axis, we must have $m \neq 0$, so either $m > 0$ or $m < 0$. In the case where $m>0$, a positive $x$ value gives a positive $y$, and a negative $x$ gives a negative $y$, so the line would pass through quadrants 1 and 3 (top-right and bottom-left). In the case where $m<0$, the same analysis shows that the line passes through quadrants 2 and 4 (top-left and bottom-right). So the line passes through 2 quadrants.
The $y = mx + b$ formulation works fine for this question, but if we want to put $x$ and $y$ on more even footing (this will come in handy in the higher-dimensional cases so that we don't always have to solve for one of the variables), we can use the other form of the equation of a line, $ax + by = c$, where $a,b,c \in {\Bbb R}$ are constants. In order to have the origin on the line, we must have $c=0$, because the point $(0,0)$ is on the line, and thus $a(0)+b(0) = 0 = c$. so the equation is simply $ax+by=0$. If either $a$ or $b$ were zero, the line would just be one of the axes, so we must have $a,b \neq 0$. Whether they are positive or negative, you can work out by plugging in positive or negative $x$ and $y$ values that the line will either pass through quadrants 1 and 3 or 2 and 4.
For example, if $a,b>0$, then if $x>0$, we must have $y<0$ in order to have $ax+by=0$. Similarly, if $x<0$, then $y$ would have to be positive. So the line passes through quadrants 2 and 4. We get the same result in the case where $a,b<0$. If $a$ and $b$ have opposite signs, then the line will pass through quadrants 1 and 3.
Notice that the line does not pass through the quadrant containing the point $(a,b)$. The vector $(a,b)$ is actually perpendicular to the line $ax+by=0$, and we can see that from the fact that the equation can be rewritten as ${\bf n} \cdot {\bf x} = 0$ where ${\bf n} = (a,b)$ and ${\bf x} = (x,y)$ (recall that two vectors are perpendicular if and only if their dot product is zero).
#### The Plane in Space ${\Bbb R}^{3}$
Let's go back to the ${\Bbb R}^{3}$ case, building off of the above discussion. The general equation of a plane is $ax+by+cz=d$, where $a,b,c,d \in {\Bbb R}$ are constants. In order for the plane to contain $(0,0,0)$, we must have $d=0$, so the equation is now just $ax+by+cz=0$. If any of the constants is 0, then the plane will actually look like the equation of a line from above. For example, if $c=0$, then we'd have $ax+by=0$, which would be a line in the $xy$-plane extended vertically up and down in the positive and negative $z$ directions, and in fact containing the entire $z$-axis. Can you see why (hint: show that the points on the $z$-axis, i.e. points of the form $(0,0,z)$, all satisfy the equation of the plane)?
We can also interpret the equation geometrically as follows: the equation $ax+by+cz=0$ is equivalent to ${\bf n} \cdot {\bf x} = 0$, where ${\bf n} = (a,b,c)$ and ${\bf x} = (x,y,z)$. Note that here, ${\bf n}$ and ${\bf x}$ are vectors, and their dot product is a scalar, so the $0$ on the right is a scalar zero, not the zero vector ${\bf 0} = (0,0,0)$.
As mentioned above, the dot product of two vectors is zero if and only if the vectors are perpendicular. Therefore, this equation is saying that any vector ${\bf x}$ that is perpendicular to ${\bf n}$ is on the plane. For this reason, ${\bf n}$ is called the plane's normal vector (normal is a synonym for perpendicular, as is orthogonal, which is also used frequently). In the example above where $c=0$, the plane's normal vector is $(a,b,0)$, which lies in the $xy$-plane. Thus, the $z$-axis, being orthogonal to the $xy$-plane, is contained in our plane.
Now, in order to answer the geometric question of which vectors are orthogonal to ${\bf n}$, we can look at the algebraic equation $ax+by+cz=0$.
Since the plane does not touch the axes except at the origin, we must have $a,b,c \neq 0$. As an example, let's look at the case where $a,b,c>0$. Then we can have the following combinations for $(x,y,z)$ in order to have $ax+by+cz=0$:
$$(+,+,-) \\ (+,-,+) \\ (+,-,-) \\ (-,-,+) \\ (-,+,-) \\ (-,+,+)$$ The remaining two combinations, $(+,+,+)$ and $(-,-,-)$, do not work, because then the left side of the equation would have to be positive or negative (respectively) and thus not zero.
If we grind through the algebra of the other 7 combinations for $(a,b,c)$, we see that we get 6 possibilities each time, so the plane intersects 6 of the 8 octants, and we have the answer to the problem. I'm not going to go through all the cases, because that would be quite boring, but you can see that there is a certain symmetry in the plane's equation between $(a,b,c)$ and $(x,y,z)$. Once you've solved it for the case $a,b,c>0$, you've pretty much solved it for all the cases. Can you see why? So we've got the answer- it's 6.
#### The Hyperplane in ${\Bbb R}^{n}$
${\Bbb R}^{n}$ is the $n$-dimensional analog of ${\Bbb R}^{3}$ and is the set of ordered $n$-tuples of real numbers: ${\Bbb R}^{n} = \{ (x_1,x_2,x_3,...,x_n) \ \colon \ {\scr each} \ x_i \in {\Bbb R} \}$. We can't picture this $n$-dimensional space, but we can use the same types of algebraic equations that work in ${\Bbb R}^{3}$ to analyze ${\Bbb R}^{n}$.
${\Bbb R}^{n}$ is divided into $2^n$ orthants, also known as hyperoctants or $n$-hyperoctants, based on the signs, positive or negative, of the $n$ components of a point. A 2-hyperoctant is a quadrant in ${\Bbb R}^{2}$ and a 3-hyperoctant is an octant in ${\Bbb R}^{3}$. The $x_i$-axis is the set of points where all coordinates except possibly the $i^{\scr th}$ are zero.
A hyperplane in ${\Bbb R}^{n}$ is a set $P$ of points (equivalently, vectors) that are orthogonal to a normal vector ${\bf n} = (a_1, a_2, ... a_n)$. In symbols, $P = \{ {\bf x} \in {\Bbb R}^{n} \ \colon \ {\bf n} \cdot {\bf x} = 0 \}$. For those familiar with linear algebra, the hyperplane containing the origin is a vector subspace of ${\Bbb R}^{n}$ of dimension $n-1$, i.e. co-dimension 1. A hyperplane in ${\Bbb R}^{2}$ is a line, and a hyperplane in ${\Bbb R}^{3}$ is a plane.
### How many $n$-hyperoctants does a hyperplane $P \subset {\Bbb R}^{n}$ intersect, given that it contains the origin, but does not intersect the axes at any other point?
To answer this question, we can use the discussion above from the $n=2$ and $n=3$ cases and generalize the results. We can then prove the answer is correct using induction.
When we went from $n=2$ to $n=3$, we took the equation $ax+by=0$ (i.e. the line in ${\Bbb R}^{2}$ with normal vector $(a,b)$), extended it into 3-dimensional space to make the plane whose normal vector is $(a,b,0)$, and then added a non-zero third coordinate to the normal vector to "tilt" the plane off of the $z$-axis.
Now, a hyperplane (including the line and plane in the $n=2$ and $n=3$ cases) is orthogonal to its normal vector ${\bf n}$ as well as the negative of the normal vector, $-{\bf n}$. In fact, the hyperplane is orthogonal to any scalar multiple of ${\bf n}$, but my point in mentioning $-{\bf n}$ is that the hyperplane won't intersect the $n$-hyperoctants that contains ${\bf n}$ or $-{\bf n}$.
Let's look at the case where ${\bf n}$ lies in the first $n$-hyperoctant, i.e. has all positive coordinates. As mentioned above, the other cases are pretty much the same because of the symmetries of the equation ${\bf n} \cdot {\bf x} = \sum_{i=1}^{n}{a_i x_i} = 0$, so the number of $n$-hyperoctants the hyperplane intersects is the same in all cases. In the case that the $a_i$ are positive, the hyperplane doesn't intersect the first $n$-hyperoctant or the one with all negative coordinates (whatever number we want to assign to that one).
In the ${\Bbb R}^{2}$ case, the line intersects quadrants 2 and 4. When we extended ${\bf n} = (a,b)$ to ${\bf n} = (a,b,0)$ in ${\Bbb R}^{3}$, we got a plane that contained the entire $z$-axis. The intersection of this plane with the $xy$-plane is the line $ax+by=0$, which remains the case regardless of the third coordinate of ${\bf n}$. Now, this plane intersects the octants $(+,-,+)$, $(+,-,-)$, $(-,+,+)$, and $(-,+,-)$. We took the original quadrants 2 and 4 and multiplied them by 2 to get 4 octants.
When we add a non-zero third coordinate to ${\bf n}$ (let's assume it's positive), the new plane also intersects two additional octants: $(+,+,-)$ and $(-,-,+)$. The first two coordinates of these two would have not been included in the 2-d case, but the third coordinate allows us to use those combinations and still get the equation $ax+by+cz$ to equal zero. $(+,+,+)$ and $(-,-,-)$ still don't work though.
The same logic works when going from ${\Bbb R}^{n}$ to ${\Bbb R}^{n+1}$ when $n>2$, and we can prove it by induction.
Thoerem: For $n \geq 2$, a hyperplane in ${\Bbb R}^{n}$ containing the origin, but not intersecting the coordinate axes at any other point, intersects $2^{n}-2$ $n$-hyperoctants.
The proof is a bit lengthy, but basically just formalizes the idea of extending the line in ${\Bbb R}^{2}$ into a plane in ${\Bbb R}^{3}$ and then tilting it off the $z$-axis
Proof: The base case of $n=2$ was already shown above.
For the induction step, assume the theorem is true for ${\Bbb R}^{n-1}$, and consider a hyperplane $P = \{{\bf x} \in {\Bbb R}^{n} \ \colon \ {\bf n} \cdot {\bf x}=0 \}$ where ${\bf n} = (a_1,a_2,...,a_n)$.
The equation of $P$ is $\sum_{i=1}^{n}{a_i x_i} = \sum_{i=1}^{n-1}{a_i x_i} + a_n x_n = 0$. By the induction hypothesis, the solutions to the equation $\sum_{i=1}^{n-1}{a_i x_i} = 0$ intersect $2^{n-1}-2$ $(n-1)$-hyperoctants. Let's call those solutions $P_{n-1}$, which is a hyperplane in ${\Bbb R}^{n-1}$
Take a point ${\bf x}_{0, n-1} = (x_{0,1}, x_{0,2},...,x_{0,n-1}) \in {\Bbb R}^{n-1}$ which satisfies the equation of $P_{n-1}$. If $x_{0,1}>0$, then $x_{0,1}+\epsilon>0$ as well, where $\epsilon = \frac{1}{2}|x_{0,1}|$. Similarly, if $x_{0,1}<0$, then $x_{0,1}+\epsilon<0$ as well, so the point ${\bf x}_{1,n-1} = (x_{0,1}+\epsilon, x_{0,2},...,x_{0,n-1})$ is in the same $(n-1)$-hyperoctant as ${\bf x}_{0, n-1}$. By a similar argument, so is the point ${\bf x}_{2,n-1} = (x_{0,1}-\epsilon, x_{0,2},...,x_{0,n-1})$.
Define the points ${\bf x}_{1} = (x_{0,1}+\epsilon, x_{0,2},..., x_{0,n-1}, -\frac{a_1}{a_n}\epsilon), \ {\bf x}_{2} = (x_{0,1}-\epsilon, x_{0,2},..., x_{0,n-1}, \frac{a_1}{a_n}\epsilon) \in {\Bbb R}^{n}$. Then \begin{align} {\bf n} \cdot {\bf x}_{1} &= a_1 (x_{0,1}+\epsilon) + a_2 x_{0,2} + ... + a_{n-1} x_{0,n-1} + a_n (-\frac{a_1}{a_n}\epsilon) \\[2mm] &= a_1 (x_{0,1}+\epsilon -\epsilon) + a_2 x_{0,2} + ... + a_{n-1} x_{0,n-1} \\[2mm] &= a_1 x_{0,1} + a_2 x_{0,2} + ... + a_{n-1} x_{0,n-1} = 0 \end{align}
with the final equality being true because ${\bf x}_{0,n-1} \in P_{n-1}$.
This shows that ${\bf x}_{1} \in P$. Similarly, ${\bf x}_{2} \in P$. The first $n-1$ coordinates of these two points are in the same $(n-1)$-hyperoctant as ${\bf x}_{0,n-1}$, and the $n^{\scr th}$ coordinates of ${\bf x}_{1}$ and ${\bf x}_{2}$ have opposite sign. This shows that we have kept the $2^{n-1}-2$ $(n-1)$-hyperoctants of the $(n-1)$-hyperplane when we extended it into ${\Bbb R}^{n}$ and actually multiplied them by 2 (by adding both positive and negative $n^{\scr th}$ coordinates) to get $2(2^{n-1}-2)$ = $2^{n}- 4$ $n$-hyperoctants.
We just need to show that we've also added two more $n$-hyperoctants. These are the ones where the first $n-1$ coordinates all have the same sign or all have the opposite sign as the first $n-1$ coordinates of ${\bf n}$, just like when we went from $n=2$ to $n=3$ above. Examples of solutions to the equation of $P$ that are in those 2 $n$-hyperoctants are $(a_1,a_2,...,a_{n-1},-\dfrac{1}{a_n}\sum_{i=1}^{n-1}{a_i^2})$ and $(-a_1,-a_2,...,-a_{n-1},\dfrac{1}{a_n}\sum_{i=1}^{n-1}{a_i^2})$.
So now we are up to $2^{n}-4+2 = 2^{n}-2$, so $P$ intersects at least that many $n$-hyperoctants. There are only 2 more $n$-hyperoctants, and those are the ones that contain $\pm {\bf n}$, but we already know that points in those $n$-hyperoctants cannot satisfy the equation of ${\bf n} \cdot {\bf x} = 0$, so $P$ intersects exactly $2^{n}-2$ of the $2^n$ $n$-hyperoctants, and the theorem is proved.
$\square$
Here's a diagram illustrating the objects described in the proof in the case where $n=3$ and $n-1=2$. Apologies for the low quality (I made it in MS Paint), but note that the bottom of the red plane, $P$, comes out towards the viewer, in front of the blue plane, and the top half of $P$ is behind the blue plane. The points ${\bf x}_{0,n-1}$, ${\bf x}_{1,n-1}$, and ${\bf x}_{2,n-1}$ are in in the $x_{1}x_{2}$-plane, with ${\bf x}_{1,n-1}$ in front of $P$ and ${\bf x}_{2,n-1}$ behind $P$.
2. A hypersurface is a generalization of a hyperplane in the context of manifolds. I'm not sure the question about octants would make sense in that context since a manifold need not be in ${\Bbb R}^{n}$ (though it has a map to ${\Bbb R}^{n}$ in a neighborhood of each point). Maybe there's a way to reformulate the question to apply in that context... | 2021-05-15T08:44:48 | {
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https://www.intmath.com/forum/methods-integration-31/taking-e-to-both-sides:160 | IntMath Home » Forum home » Methods of Integration » Taking e to both sides
# Taking e to both sides [Solved!]
### My question
Your Integration Chapter: The Basic Logarithmic Form, example 4.
Can you show how you took e to both sides to arrive at your result please?
### Relevant page
2. Integration: The Basic Logarithm Form
### What I've done so far
After integrating, t = ln 20 - ln (20-v).
You applied log laws to get t = ln(20/(20-v)).
You took "e to both sides" to get e^t = 20/20-v.
We thought by taking e to both sides the result would be the exponents equated:
t = 20/20-v.
X
Your Integration Chapter: The Basic Logarithmic Form, example 4.
Can you show how you took e to both sides to arrive at your result please?
Relevant page
<a href="https://www.intmath.com/methods-integration/2-integration-logarithmic-form.php">2. Integration: The Basic Logarithm Form</a>
What I've done so far
After integrating, t = ln 20 - ln (20-v).
You applied log laws to get t = ln(20/(20-v)).
You took "e to both sides" to get e^t = 20/20-v.
We thought by taking e to both sides the result would be the exponents equated:
t = 20/20-v.
## Re: Taking e to both sides
@Phinah: The opposite process of taking the "log" of something is to take "e to the power of...".
Writing the full details would be:
t = ln (20 /(20-v))
e^t = e^[ln (20/(20-v))]
e^t = 20/(20-v)
e^t = e^[(20/(20-v))] (without "ln")
t = 20/(20-v)
but that's not what we started with.
A similar situation is the case where "raising to power 2" is the opposite of "square root".
y = sqrt(x+2)
Squaring both sides gives us:
y^2 = (sqrt(x+2))^2 = x+2
The "power 2" has "undone" the sqrt operation.
Similarly, the "e to the power" operation undoes the "ln" expression in the example you are asking about.
Hope it makes sense.
X
@Phinah: The opposite process of taking the "log" of something is to take "e to the power of...".
Writing the full details would be:
t = ln (20 /(20-v))
e^t = e^[ln (20/(20-v))]
e^t = 20/(20-v)
e^t = e^[(20/(20-v))] (without "ln")
t = 20/(20-v)
but that's not what we started with.
A similar situation is the case where "raising to power 2" is the opposite of "square root".
y = sqrt(x+2)
Squaring both sides gives us:
y^2 = (sqrt(x+2))^2 = x+2
The "power 2" has "undone" the sqrt operation.
Similarly, the "e to the power" operation undoes the "ln" expression in the example you are asking about.
Hope it makes sense.
## Re: Taking e to both sides
It does now. Thank you.
X
It does now. Thank you. | 2018-05-25T08:37:23 | {
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https://math.stackexchange.com/questions/2177569/law-of-sines-in-triangle | # Law of Sines in Triangle
On the side BC of the triangle ABC we construct towards the exterior a square BCDE. Denote the intersection between AE and BC by M. Use the law of sines to prove that
$$\frac{BM}{CM}=\frac{\cos \measuredangle B\cdot \sin \measuredangle C}{\sqrt{2} \cdot \sin \measuredangle B \cdot \sin(\measuredangle C+45°)}$$
If someone could please help me prove this problem. I do not have a similar problem to work off of. I am unclear of where the $\sin(\measuredangle C+45^{\circ})$ comes into play and the $\sqrt{2}$.
• the end got cut off...i am unlcear where the sin(<C+45) and \sqrt{2} come into play. – user423388 Mar 8 '17 at 13:42
• Provide the figure – Nick Pavini Mar 8 '17 at 14:09
Let $\measuredangle B=\beta$, $\measuredangle C=\gamma$, $\measuredangle BAE=\alpha_1$, $\measuredangle CAE=\alpha_2$. By applying the sine rule to triangles $BAM$ and $CAM$ one gets: $${BM\over\sin\alpha_1}={AM\over\sin\beta},\quad {CM\over\sin\alpha_2}={AM\over\sin\gamma},\quad \hbox{whence:}\quad {BM\over CM}={\sin\alpha_1\over\sin\alpha_2}{\sin\gamma\over\sin\beta}.$$ By applying then the sine rule to triangles $BAE$ and $CAE$ one gets: $${\sin\alpha_1\over BE}={\sin(\beta+90°)\over AE},\quad {\sin\alpha_2\over CE}={\sin(\gamma+45°)\over AE}.$$ From that, taking into account that $CE=\sqrt2 BE$, one can readily obtain $\displaystyle{\sin\alpha_1\over\sin\alpha_2}$ and thus the desired result.
• fully understood this !! thank you so much ! – user423388 Mar 8 '17 at 16:34
Using sine rule at $ACE$ we have:
$$\frac{AE}{\sin(C+45°)}=\frac{AC}{\sin \alpha}$$
Using sine rule at $ABE$ we have:
$$\frac{AE}{\sin(B+90°)}=\frac{AB}{\sin \beta}$$
Dividing both equations we have:
$$\frac{\cos B}{\sin (C+45°)}=\frac{AC}{AB}\cdot \frac{\sin \beta}{\sin \alpha} \quad (1)$$
Using sine rule at $ABC$ we get
$$\frac{AC}{AB}=\frac{\sin B}{\sin C}$$
So, from $(1)$
$$\frac{\cos B\cdot \sin C}{\sin B\cdot \sin (C+45°)}=\frac{\sin \beta}{\sin \alpha}\quad (2)$$
but $\alpha + \beta=45°$ so
$$\frac{\sin \beta}{\sin \alpha}=\frac{\sin \beta}{\sin (45°-\beta)}=\sqrt{2}\cdot\frac{\tan \beta}{1-\tan \beta}\quad (3)$$
Finaly we can use, from the triangle BME, that
$$\tan \beta = \frac{BM}{BE}=\frac{BM}{BM+CM}\to \frac{BM}{CM}=\frac{\tan \beta}{1- \tan \beta} \quad (4)$$
Pluging $(3)$ and $(4)$ at $(2)$ we get what we want
• can you explain (3)? i dont understand how you switched from sine to tangent... – user423388 Mar 8 '17 at 16:17
• @erica: $\frac{\sin \beta}{\sin (45°-\beta)}=\frac{\sin \beta}{(\sqrt{2}/2)(\cos \beta-\sin \beta)}=$, now divide the numerator and denominator by $\cos \beta$ and get $\sqrt{2}\cdot\frac{\tan \beta}{1-\tan \beta}$ – Arnaldo Mar 8 '17 at 16:33
In triangle $ABM$, using the law of sines implies $$\frac{BM}{AM}=\frac{\sin\measuredangle BAM}{\sin\measuredangle B}$$ similarly, in $ACM$: $$\frac{CM}{AM}=\frac{\sin\measuredangle CAM}{\sin\measuredangle C}$$ combining these two yields: $$\frac{BM}{CM}=\frac{\sin\angle BAM}{\sin\angle CAM}\cdot\frac{\sin\angle C}{\sin\angle B}\label{*}\tag{*}$$ Now, in triangle $ABE$ note that $\measuredangle ABE=\measuredangle B+90^{\circ}$.
Thus $\sin\measuredangle ABE=\cos\measuredangle B$ and the law of sines implies $$\frac{\sin\measuredangle BAM}{\cos\measuredangle B}=\frac{BE}{AE}$$ and for triangle $ACE$: $$\frac{\sin\measuredangle CAM}{\sin(\measuredangle C+45)}=\frac{CE}{AE}$$ Note that $CE=\sqrt{2}BE$. Now combining the two equations implies: $$\frac{BE}{CE}=\frac 1{\sqrt 2}=\frac{\sin\measuredangle BAM}{\sin\measuredangle CAM}\cdot\frac{\sin(\measuredangle C+45)}{\cos\measuredangle B}$$ And finally, you can use $\eqref{*}$ to get the desired result.
• OOps. I am late to the party – polfosol Mar 8 '17 at 15:00
• can you explain how $\text{sin}\angleABE = \text{cos}\angleB$? – user423388 Mar 8 '17 at 15:58 | 2019-08-17T20:51:02 | {
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http://yucoding.blogspot.com/2013/04/leetcode-question-116-unique-path-i.html | ## Unique Path I
A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).
The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).
How many possible unique paths are there?
## Analysis:
This is an easy problem. From the description we know that, the robot can only move down or right, which means, if the robot is now in position (x,y), then the position before this step must be either (x-1,y) or (x, y-1). Since current position is only from these two previous positions, the number of possible paths that the robot can reach this current position (x,y) is the sum of paths from (x-1, y) and (x, y-1).
We want to get the number of paths on position (m,n), we need to know (m-1,n) and (m, n-1). For (m-1,n), we must know (m-2,n) and (m-1,n-1) ... until we back to the start position (1,1) ([0][0] in C++). Note that the boundary of the map, we can easily know that the top row and the first column of the map are all 1s. Use loop can solve the problem then.
### Code(C++):
class Solution {
public:
int uniquePaths(int m, int n) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int **arr = new int* [m];
for (int i=0;i<m;i++){
arr[i]= new int[n];
}
arr[0][0]=1;
for (int i=0;i<m;i++){
arr[i][0] = 1;
}
for (int i=0;i<n;i++){
arr[0][i] = 1;
}
for (int i=1;i<m;i++){
for(int j=1;j<n;j++){
arr[i][j] = arr[i][j-1] + arr[i-1][j];
}
}
return arr[m-1][n-1];
}
};
### Code(Python):
class Solution:
# @return an integer
def uniquePaths(self, m, n):
#define map and initialization
mp = [[0 for i in xrange(n)] for i in xrange(m)]
for i in xrange(m):
mp[i][0]=1
for j in xrange(n):
mp[0][j]=1
for i in xrange(1,m):
for j in xrange(1,n):
mp[i][j]=mp[i-1][j]+mp[i][j-1]
return mp[m-1][n-1]
1. I think most of your posts are great but sometimes it is very hard to understand your English though.
1. Actually I have been refining both the analysis and code in all my posts recently. Thank you very much for your suggestion.
2. Does this algorithm have a name?
3. Does this algorithm have a name?
1. Dynamic Programming
4. I LOVE YOU YU!
5. why did you initialise by 1 for the borders ? | 2018-06-25T00:41:39 | {
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http://mathhelpforum.com/algebra/222247-find-length-rectangle-given-diagonal-area-print.html | # find length of rectangle given diagonal and area
• September 24th 2013, 11:08 AM
Bonganitedd
find length of rectangle given diagonal and area
Rectangle has area=168 m^2 and diagonal of 25. Find length
This is how tried to attempt the problem
Area= L X W
168 = L x W ..........(1)
L^2 + W^2 =25^2 ............(2)
From (1) L = 168/W...........(3)
Substitute (3) into (2)
(168/W)^2 +W^2 = 625
28224/W^2 + W^2 = 625
The problem gets complicated as I proceed
Is this aproach correct if it is,
Is there a convinient method
• September 24th 2013, 11:18 AM
votan
Re: find length of rectangle given diagonal and area
Quote:
Originally Posted by Bonganitedd
Rectangle has area=168 m^2 and diagonal of 25. Find length
This is how tried to attempt the problem
Area= L X W
168 = L x W ..........(1)
L^2 + W^2 =25^2 ............(2)
From (1) L = 168/W...........(3)
Substitute (3) into (2)
(168/W)^2 +W^2 = 625
28224/W^2 + W^2 = 625
The problem gets complicated as I proceed
Is this aproach correct if it is,
Is there a convinient method
Sketch a rectangle with one diagonal. laber the sides L and W and D for the diabonal. Write Pythagorean theorem, and the area = LW. Eliminate L from the the Pythagorean theorem and solve for W then find L from area formula.
• September 24th 2013, 11:30 AM
Bonganitedd
Re: find length of rectangle given diagonal and area
Its same approach I used, I did draw a rectangle now how do eliminate L bcos the pathygras is L^2 + W^2 =25^2
The area, LW =168
• September 24th 2013, 12:17 PM
Plato
Re: find length of rectangle given diagonal and area
Quote:
Originally Posted by Bonganitedd
Rectangle has area=168 m^2 and diagonal of 25. Find length
This is how tried to attempt the problem
Area= L X W
168 = L x W ..........(1)
L^2 + W^2 =25^2 ............(2)
From (1) L = 168/W...........(3)
Substitute (3) into (2)
(168/W)^2 +W^2 = 625
28224/W^2 + W^2 = 625
The problem gets complicated as I proceed
Is this aproach correct if it is,
Is there a convinient method
Have a look at this webpage.
• September 24th 2013, 01:15 PM
Soroban
Re: find length of rectangle given diagonal and area
Hello, Bonganitedd!
Quote:
Rectangle has area=168 m^2 and diagonal of 25. Find the length.
This is how tried to attempt the problem
$\text{Area} \:=\: L\cdot W \:=\:168 \quad\Rightarrow\quad L \,=\,\frac{168}{W}\;\;[1]$
$L^2 + W^2 \:=\:25^2\;\;[2]$
$\text{Substitute [1] into [2]: }\;\left(\frac{168}{W}\right)^2 +W^2 \:=\:625 \quad\Rightarrow\quad \frac{28,\!224}{W^2} + W^2 \:=\: 625$
Is this approach correct? . Yes
If it is, is there a convinient method?
We have: . $\frac{28,\!224}{W^2} + W^2 \:=\:625$
Multiply by $W^2\!:\;\;28,\!224 + W^4 \:=\:625W^2 \quad\Rightarrow\quad W^4 - 625W^2 + 28,\!224 \:=\:0$
Factor: . $(W^2 - 49)(W^2 - 576) \:=\:0$
$\begin{array}{ccccccccc}W^2-49 \:=\:0 & \Rightarrow & W^2 \:=\:49 & \Rightarrow & W \:=\:7 \\ W^2 - 576 \:=\:0 & \Rightarrow & W^2 \:=\:576 & \Rightarrow & W \:=\:24 \end{array}$
$\text{Assuming }L > W\text{, we have: }\:W\,=\,7,\;\boxed{L \,=\,24}$
• September 24th 2013, 02:08 PM
HallsofIvy
Re: find length of rectangle given diagonal and area
Once you get to "[tex]W^4- 625W^2+ 28224=0 as Soroban showed, if the "fourth degree polynomial" bothers you, you can let $x= W^2$ so that your equation is x^2- 625x+ 28224= 0 and use whatever method you like, completing the square or the quadratic formula, to solve that quadratic equation.
• September 24th 2013, 07:02 PM
Bonganitedd
Re: find length of rectangle given diagonal and area
Not taking away credit to other members who contributed, but @ Hallsofivy your advice is what I needed. Big u to MHF. | 2016-06-27T01:04:30 | {
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https://math.stackexchange.com/questions/1485605/write-cos8x-in-terms-of-sinx | # Write $\cos^8(x)$ in terms of $\sin(x)$?
I am currently working on an integral using u substitution and have hit a wall at the above point. The integral in question is $\int$$\sin^7(x)$$\cos^9(x)$ using $u=sin(x)$.
After finding $du=cos(x) dx$, I am asked to express $\cos^8(x)$ in terms of $u$, at which point I am stuck. My current attempt was to reduce the power down from 8 to 2, and express it as $(1-u^2)$, but that gives me a wrong answer (this assignment checks for us before submission).
Any help is appreciated.
Thanks.
• What did you find the integral to be (your "wrong" answer)? – Henry Oct 18 '15 at 9:55
• I hadn't got to that point yet as the next few steps required it expressed in terms of u still. – Dwayne H Oct 18 '15 at 9:56
• OK, your "wrong" answer in terms of $u$? – Henry Oct 18 '15 at 9:57
• I'd only gotten to the point of $u^7(1-u^2)$, but seeing as the assignment said it was wrong I didn't try going ahead and solving for that. – Dwayne H Oct 18 '15 at 9:59
• $\displaystyle \int u^7(1-u^2)^4\,du$ with a power of $4$ seems more likely – Henry Oct 18 '15 at 10:00
If $$\cos^2 x = 1-\sin^2 x$$ then $$\cos^8 x = (1-\sin^2 x)^4.$$
Using $u=\sin x$, you have $du=\cos x\,dx$, and also $\cos^2x=1-u^2$, so $\cos^8x=(1-u^2)^4$. Thus your integral becomes $$\int u^7(1-u^2)^4\,du$$ | 2019-09-21T11:37:47 | {
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http://oqxm.agence-des-4-fontaines.fr/covariance-calculator-with-probability.html | Covariance Calculator With Probability
In introductory finance courses, we are taught to calculate the standard deviation of the portfolio as a measure of risk, but part of this calculation is the covariance of these two, or more, stocks. How does this covariance calculator work? In data analysis and statistics, covariance indicates how much two random variables change together. Pre-trained models and datasets built by Google and the community. , the variables tend to show similar behavior), the covariance is positive. Their covariance Cov(X;Y) is de ned by. We assume that a probability distribution is known for this set. You can find formula used for calculation of covariance below the calculator. How can I calculate the covariance in R? I created two vectors x,y and fed them into cov(), but I get the wrong result. Recent research has pointed to the ubiquity and abundance of between-generation epigenetic inheritance. 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Returns population covariance, the average of the products of deviations for each data point pair in two data sets. Expected Return Formula Calculator. If A is a scalar, cov(A) returns 0. Covariance • Variance and Covariance are a measure of the "spread" of a set of points around their center of mass (mean) • Variance - measure of the deviation from the mean for points in one dimension e. Covariance is driving me nuts, is there any simple way to make the calculator do the work for you? Get your mind off your Level I results with a free 2020 Level II JumpStart package. 35 5% -3% Given The Above Information On Two Investments A And B, Calculate The Statistics Below. Please enter the necessary parameter values, and then click 'Calculate'. Suppose, under uncertainty, the manager believes that a risky asset, for example, an equity, can bring him different results, which at the moment of portfolio formation can only be judged with some probability, as shown in Table. A low value means there is a weak relationship. CSC 411 / CSC D11 / CSC C11 Probability Density Functions (PDFs) The off-diagonal terms are covariances: Σ ij = cov(x i,x j) = E p(x)[(x i −µ i)(x j −µ j)] (10) between variables x i and x j. • Probability and Statistics for Engineering and the Sciences, Covariance, Correlation, Sampling Distributions, Central Limit calculator or for any other. E(X1)=µX1 E(X2)=µX2 var(X1)=σ2 X1 var(X2)=σ2 X2 Also, we assume that σ2 X1 and σ2 X2 are finite positive values. Estimating the uncertainty of revenues and investment decisions. And, to calculate the probability of an interval, you take the integral of the probability density function over it. Byju's Covariance Calculator is a tool which makes calculations very simple and interesting. The covariance matrix is a matrix that only concerns the relationships between variables, so it will be a k x k square matrix. Using the formulae above to compute covariance can sometimes be tricky. 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Uniform distribution Calculator - High accuracy calculation Welcome, Guest. This post provides practice problems to reinforce the concepts discussed in the preceding post on univariate probability distributions. That is, covariance matrices with small determinants denote variables that are redundant or highly correlated. Variables are positively related if they move in the same direction. calculate and interpret covariance and correlation and interpret a scatterplot; calculate and interpret the expected value, variance, and standard deviation of a random variable and of returns on a portfolio; calculate and interpret covariance given a joint probability function; calculate and interpret an updated probability using Bayes' formula;. This online calculator computes covariance between two discrete random variables. By default, this function will calculate the sample covariance matrix. The sign (+ or -) of the correlation affects its interpretation. 2 Covariance Covariance is a measure of how much two random variables vary together. Chapter 4 Variances and covariances Page 3 A pair of random variables X and Y is said to be uncorrelated if cov. 1 Introduction. Joyce, Fall 2014 Covariance. [In our case, a 5×5 matrix. Also, it can be considered as a generalization of the concept of variance of two random variables. Covariance Calculator calculator, formula and. The Covariance Calculator an online tool which shows Covariance for the given input. Covariance gives you a positive number if the variables are positively related. Covariance[dist] gives the covariance matrix for the multivariate symbolic distribution dist. The formula defines covariance for discrete variables in Simon & Blume (1994): Mathematics for Economists, end of section A5. Linear Models in SAS (Regression & Analysis of Variance) The main workhorse for regression is proc reg, and for (balanced) analysis of variance, proc anova. In case the greater values of one variable are linked to the greater values of the second variable considered, and the same corresponds for the smaller figures, then the covariance is positive and is a signal that the two variables show similar behavior. If the variables tend to show similar behavior, the covariance is positive. Obtaining covariance estimates between variables allows one to better estimate direct and indirect effects with other variables, particularly in complex models with many parameters to be estimated. If using percent form, the user must add the percent sign (%) at the end of the number. Covariance[v1, v2] gives the covariance between the vectors v1 and v2. In the matrix diagonal there are variances, i. (2005), Fundamentals of Probability with Stochastic Processes, Roussas, G. In addition, we may only be able to assess unconditional coverage probability for VaR forecast of the SVAR model. Problem 31B. I have a joint probability mass function of two variables X,Y like here. P function in Microsoft Excel. Joint Discrete Probability Distributions. Chapter 4 Variances and covariances Page 3 A pair of random variables X and Y is said to be uncorrelated if cov. I hope you found this video useful, please subscribe for daily videos! WBM Foundations: Mathematical logic Set theory Algebra: Number theory Group theory Lie groups Commutative rings Associative. By using this formula, after calculation, you can verify the result of such calculations by using our covariance calculator. Covariance is a measure of how much two random variables vary together. When comparing data samples from different populations, two of the most popular measures of association are covariance and correlation. Applied to historical prices, covariance can help determine if stocks' prices tend. Video for finding the covariance and correlation coefficient by hand. TI-82 / TI-83 Graphing Calculator. Correlation values range from positive 1 to negative 1. 5 should display. For a discrete variable$\sum_{i=1}^nP(X=x_{i}|A)=1$, hence we immediately can fill in the missing values in the conditional distribution tables. The TI-83 Graphing Calculator can facilitate the entry of ordered lists of data and perform some statistical analyses, but lacks a single command to calculate the covariance of two lists of numbers. Covariance & Correlation The covariance between two variables is defined by: cov x,y = x x y y Can I directly relate the free parameters to the covariance matrix? First calculate P(x) by marginalizing over y: P xˆexp{1 21 2 x x0 x 2} dyexp{1 21 2 [y y0 y 2 2 x x0 x y y0 y]} probability content is contained within a radius of s(2. 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Using the formulae above to compute covariance can sometimes be tricky. com's Covariance calculator is an online statistics & probability tool to estimate the nature of association between two random variables X & Y in probability & statistics experiments. Before we demonstrate the function in Excel, a few observations on the covariance and correlation measures. The probability that a large earthquake will occur on the San Andreas Fault in. If you see any typos, potential edits or changes in this Chapter, please note them here. function [probability] = comp_gauss_dens_val (mean, covarianceMatrix, givenMatrix ) The value assigned in lengthofM varibale is the length of the mean of the given matrix, which in our is one dimensional in our case. If an input is given then it can easily show the result for the given number. Joint Discrete Probability Distributions. A negative covariance means that the variables are inversely related, or that they move in opposite directions. 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The correlation coefficient is also known as the Pearson Product-Moment Correlation Coefficient. pX Beta calculator. calculate and interpret covariance and correlation and interpret a scatterplot; calculate and interpret the expected value, variance, and standard deviation of a random variable and of returns on a portfolio; calculate and interpret covariance given a joint probability function; calculate and interpret an updated probability using Bayes' formula;. In probability, we use 0. Adding a constant to a random variable does not change their correlation coefficient. is between and inclusive, which meets the first property of the probability distribution. ) S is said to have a lognormal distribution,. Full curriculum of exercises and videos. Small sample size problems. Different categories of descriptive measures are introduced and discussed along with the Excel functions to calculate them. In introductory finance courses, we are taught to calculate the standard deviation of the portfolio as a measure of risk, but part of this calculation is the covariance of these two, or more, stocks. To find the covariance matrix you need to calculate the Fisher information matrix. A covariance refers to the measure of how two random variables will change together and is used to calculate the correlation between variables. Obtaining covariance estimates between variables allows one to better estimate direct and indirect effects with other variables, particularly in complex models with many parameters to be estimated. In introductory finance courses, we are taught to calculate the standard deviation of the portfolio as a measure of risk, but part of this calculation is the covariance of these two, or more, stocks. Do October 10, 2008 A vector-valued random variable X = X1 ··· Xn T is said to have a multivariate normal (or Gaussian) distribution with mean µ ∈ Rn and covariance matrix Σ ∈ Sn 1. covariance calculator - step by step calculation to measure the statistical relationship (linear dependence) between the two sets of population data, along with. This calculator determines the following coin toss probability scenarios * Coin Toss Sequence such as HTHHT * Probability of x heads and y tails * Covariance of X and Y denoted Cov(X,Y) * The. Is there a relationship between Xand Y? If so, what kind? If you’re given information on X, does it give you information on the distribution of Y? (Think of a conditional distribution). It's an online statistics and probability tool requires two sets of population data X and Y and measures of how much these data sets vary together, i. When the covariance is positive, X tends to be high when Y is high, and vice versa; when the covariance is negative, X tends to be high when Y is low, and vice versa. Consider two random variables$X$and$Y$. Here are some documents to help you use the TI-82 calculator. Enter X values (Separated by comma) Enter Y values (Separated by comma) Letter Arrangment Probability Calculator. Where Cov (A, B) – is covariance of portfolios A and B. Calculate the standard deviation of the returns using STDEV function; Finally, we calculate the VaR for 90, 95, and 99 confidence level using NORM. Or are they. The converse. 01 respectively for the VaR(90), VaR(95), and VaR(99). If the greater values of one variable mainly correspond with the greater values of the other variable, and the same holds for the lesser values, (i. S function is:. , the covariance of each element with itself. This suggests the question: Given a symmetric, positive semi-de nite matrix, is it the covariance matrix of some random vector?. The calculator will find the p-value for two-tailed, right-tailed and left-tailed tests from normal, Student's (T-distribution), chi-squared and Fisher (F-distribution) distributions. How to Calculate Expected Value. The probability that a large earthquake will occur on the San Andreas Fault in. After cleaning the data, the researcher must test the assumptions of ANOVA. Therefore, it is a straightforward exercise to calculate the correlation between X and. 2 Covariance Covariance is a measure of how much two random variables vary together. More about Covariance. Welcome to StatCalculators. The sample variance, s², is used to calculate how varied a sample is. This calculator computes the variance from a data set: To calculate the variance from a set of values, specify whether the data is for an entire population or from a sample. Other JavaScript in this series are categorized under different areas of applications in the MENU section on this page. It is based on the probability-weighted average of the cross-products of the random variables’ deviations from their expected values for each possible outcome. (b) In contrast to the expectation, the variance is not a linear operator. Covariance is a statistical calculation that helps you understand how two sets of data are related to each other. And, we are given that the standard deviation of X is 1/2, and the standard deviation of Y is the square root of 1/2. Correlation between the two variables is a normalized version of the Covariance. In probability theory and statistics, covariance is a measure of the joint variability of two random variables. Analysis of covariance (ANCOVA) allows to compare one variable in 2 or more groups taking into account (or to correct for) variability of other variables, called covariates. described with a joint probability mass function. rameterized by a mean vector µ, and a variance-covariance matrix Σ, written X ∼ (µ,Σ). Given such information it is possible to calculate the marginal distributions and then the joint distribution follows. The function underlying its probability distribution is called a probability density function. Standard Deviation Calculator Variance Calculator Kurtosis Calculator Skewness Calculator. A zero covariance may indicate that the two assets are independent. This continues our exploration of the semantics of the inner product. CSC 411 / CSC D11 / CSC C11 Probability Density Functions (PDFs) The off-diagonal terms are covariances: Σ ij = cov(x i,x j) = E p(x)[(x i −µ i)(x j −µ j)] (10) between variables x i and x j. search(“distribution”). The Multivariate Gaussian Distribution Chuong B. Probability is the likelihood of something happening or being true. Both of these two determine the relationship and measures the dependency between two random. The covariance of two variables tells you how likely they are to increase or decrease simultaneously. 128 CHAPTER 7. Then m is the vector of means and V is the variance-covariance matrix. Online probability calculator to find expected value E(x), variance (σ 2) and standard deviation (σ) of discrete random variable from number of outcomes. Covariance and correlation show that variables can have a positive relationship, a negative relationship, or no relationship at all. Covariance correlations in collision avoidance probability calculations. Printer-friendly version Introduction. Begin your Level II studies with a FREE Schweser JumpStart Package. If you're behind a web filter, please make sure that the domains *. The TI-83 Graphing Calculator can facilitate the entry of ordered lists of data and perform some statistical analyses, but lacks a single command to calculate the covariance of two lists of numbers. The calculator based methods proposed above don’t account for the fact that a probability is associated with asset returns. Doing the calculation To calculate the beta coefficient for a we'll compare how the stock and the index move relative to each other with a covariance formula and then divide that result by the. 329) and is the covariance. ) q for "quantile", the inverse c. This calculator is featured to generate the complete work with steps for any corresponding input values may helpful for grade school students to solve. When a portfolio includes two risky assets, the Analyst needs to take into account expected returns, variances and the covariance (or correlation) between the assets' returns. These usual, our starting point is a random experiment with a probability measure ℙ on an underlying sample space. Bayesian decision theory is a fundamental statistical approach to the problem of pattern classification. Correlation values range from positive 1 to negative 1. Returns population covariance, the average of the products of deviations for each data point pair in two data sets. Variance and Standard Deviation of a Random Variable. Hey Flashcop and welcome to the forums. That is, if one increases, the other increases. It's an online statistics and probability tool requires two sets of population data X and Y` and measures of how much these data sets vary together, i. If an input is given then it can easily show the result for the given number. In this section, we will study an expected value that measures a special type of relationship between two real-valued variables. The Covariance Calculator an online tool which shows Covariance for the given input. Input the matrix in the text field below in the same format as matrices given in the examples. I was concurrently taking a basic theoretical probability and statistics, so even the idea of variance was still vague to me. 4, and in Robert J. Applied to historical prices, covariance can help determine if stocks' prices tend. Descriptive Statistics which contains one variable and multivariable calculators for 20 descriptive statistics measures including: mean, variance, covariance, quantile, interquartile range, correlation and many more. J Matney a Collision-avoidance calculations between orbiting objects make use of covariance matrices to characterize the uncertainty of the orbital position in space and time. Byju's Covariance Calculator is a tool which makes calculations very simple and interesting. The Multivariate Gaussian Distribution Chuong B. We will go through a review of probability concepts over here, all of the review materials have been adapted from CS229 Probability Notes. A portfolio is the total collection of all investments held by an individual or institution, including stocks, bonds, real estate, options, futures, and alternative investments, such as gold or limited partnerships. 2 Covariance Covariance is a measure of how much two random variables vary together. 25, multiply the answer by 100 to get 25%. The uniform distribution is used to describe a situation where all possible outcomes of a random experiment are equally likely to occur. is the correlation of and (Kenney and Keeping 1951, pp. 12 that she has exactly these two risk factors (but not the other). In general, if there are n random variables, the outcome is an n-dimensional vector of them. Stock Correlation Calculator. All we'll be doing here is getting a handle on what we can expect of the correlation coefficient if X and Y are independent, and what we can expect of the correlation coefficient if X and Y are dependent. If the probability of the event changes when we take the first event into consideration, we can safely say that the probability of event B is dependent of the occurrence of event A. As you doubtless know, the variance of a set of numbers is defined as the "mean squared difference from the mean". This online calculator computes covariance between two discrete random variables. Let X 1 = number of dots on the red die X 2 = number of dots on the green die. The correlation coefficient is equal to the covariance divided by the product of the standard deviations of the variables. How to Calculate Covariance. Let's now look at how to calculate the risk of the portfolio. As a simple example of covariance we'll return once again to the Old English example of Section 2. This tool provides direct calculations for a variety of probability distributions. 5 Covariance and Correlation Covariance and correlation are two measures of the strength of a relationship be-tween two r. ] Before constructing the covariance matrix, it’s helpful to think of the data matrix as a collection of 5 vectors, which is how I built our data matrix in R. We construct a non-separable space-time covariance function based on a diffusive Langevin equation. If the two variables are dependent then the covariance can be measured using the following formula:. Although the covariance and variance are linked to each other in the above manner, their probability distributions are not attached to each other in a simple manner and have to be dealt separately. A positive covariance would indicate a positive linear relationship between the variables, and a negative covariance would indicate the opposite. The syntax of the Covariance. It is actually used for computing the covariance in between every column of data matrix. It is also helpful for insurance. They must then calculate the F-ratio and the associated probability value (p-value). By using this formula, after calculation, you can verify the result of such calculations by using our covariance calculator. If A is a matrix whose columns represent random variables and whose rows represent observations, C is the covariance matrix with the corresponding column variances along the diagonal. [In our case, a 5×5 matrix. It just creates confusion because they are not equivalent. In order to calculate the Student T Value for any degrees of freedom and given probability. Key Differences Between Covariance and Correlation. With the covariance option, correlate can be used to obtain covariance matrices, as well as correlation matrices, for both weighted and unweighted data. It's the statistics & probability functions formula reference sheet contains most of the important functions for data analysis. In my first machine learning class, in order to learn about the theory behind PCA (Principal Component Analysis), we had to learn about variance-covariance matrix. A distribution is described as normal if there is a high probability that any observation form the population sample will have a value that is close to the mean, and a low probability of having a value that is far from the mean. So, I wrote an add-in that used the matrix algebra functions to create a covariance matrix that would change if you changed the data. 05 Jeremy Orlo and Jonathan Bloom 1 Learning Goals 1. By default, this function will calculate the sample covariance matrix. Calculate the covariance of and. The function is new in Excel 2010 and so is not available in earlier versions of Excel. How does this covariance calculator work? In data analysis and statistics, covariance indicates how much two random variables change together. A Statistics Co-Variance Calculator An online Co-Variance Calculator to measure of two variables X and Y. Covariance and Correlation Definition: Covariance Let X and Y be two RV’s with means x and y, respectively. In statistical theory, covariance is a measure of how much two random variables change together. E(X1)=µX1 E(X2)=µX2 var(X1)=σ2 X1 var(X2)=σ2 X2 Also, we assume that σ2 X1 and σ2 X2 are finite positive values. 2 Covariance Covariance is a measure of how much two random variables vary together. Expectation of a Function of a Random Variable Suppose that X is a discrete random variable with sample space Ω, and φ(x) is a real-valued function with domain Ω. A sample is a randomly chosen selection of elements from an underlying population. Analysis of covariance (ANCOVA) allows to compare one variable in 2 or more groups taking into account (or to correct for) variability of other variables, called covariates. It is important in security analysis to determine how much or how little price movements in two companies or industries are connected. These topics are somewhat specialized, but are particularly important in multivariate statistical models and for the multivariate normal distribution. Using the covariance formula, you can determine whether economic growth and S&P 500 returns have a positive or inverse relationship. Typically, the population is very large, making a complete enumeration of all the values in the population impossible. How does this covariance calculator work? In data analysis and statistics, covariance indicates how much two random variables change together. Consider two random variables$X$and$Y\$. Understand the meaning of covariance and correlation. If is the covariance matrix of a random vector, then for any constant vector ~awe have ~aT ~a 0: That is, satis es the property of being a positive semi-de nite matrix. The problem is solved by standardize the value of covariance (divide it by ˙. An investor is facing two potential financial losses and with the following joint density function: Let be the total of these two losses. It's the statistics & probability functions formula reference sheet contains most of the important functions for data analysis. S function is:. Population Standard Deviation The population standard deviation, the standard definition of σ , is used when an entire population can be measured, and is the square root of the variance of a given data set. Excel does such a great job in calculating correlation and covariance that it is not necessary to memorize the formulas of covariance and correlation, but here they are, along with examples worked out in Excel: Covariance of variables x and y from a known population = σ xy. The covariance for two random variates X and Y, each with sample size N, is defined by the expectation…. Calculate the covariance of and. Discrete Random Variable Calculator. Properties of variance and covariance (a) If and are independent, then by observing that. A portfolio is the total collection of all investments held by an individual or institution, including stocks, bonds, real estate, options, futures, and alternative investments, such as gold or limited partnerships. On this page, we'll begin our investigation of what the correlation coefficient tells us. Consider the following example: Example. The covariance is defined as. Click the Calculate! button and find out the covariance matrix of a multivariate sample. A randomly selected day was a long commute. These usual, our starting point is a random experiment with a probability measure ℙ on an underlying sample space. For example, the probability of a two-dimensional case, in which the vector of random variables is X = [X, Y] T, can be calculated as. This matrix is not usually printed or saved. , the variables tend to show similar behavior), the covariance is positive. For instance, I have been given a discrete random variable X with probability function px(x) = 1/2 if x = -1, 1/4 if x = 0, 1/4 if x = 1, 0 otherwise. If the covariance is a large positive number, then we expect x i to be largerthanµ iwhenx j islargerthanµ j. Let's calculate the covariance for the example age and income data. In probability theory and statistics, covariance is a measure of the joint variability of two random variables. Covariance and correlation show that variables can have a positive relationship, a negative relationship, or no relationship at all. You COULD do a calculation patterned after covariance with 3 or more variables, but I don’t see it as meaning anything, and I don’t think it would be admissable as a valid statistical function. (The covariance of X and Y is sometimes written cov(X, Y). He also covers testing hypotheses, modeling different data distributions, and calculating the covariance and correlation between data sets. | 2020-03-31T13:27:22 | {
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https://handwiki.org/wiki/Kendall_tau_distance | # Kendall tau distance
The Kendall tau rank distance is a metric that counts the number of pairwise disagreements between two ranking lists. The larger the distance, the more dissimilar the two lists are. Kendall tau distance is also called bubble-sort distance since it is equivalent to the number of swaps that the bubble sort algorithm would take to place one list in the same order as the other list. The Kendall tau distance was created by Maurice Kendall.
## Definition
The Kendall tau ranking distance between two lists $\displaystyle{ \tau_1 }$ and $\displaystyle{ \tau_2 }$ is
$\displaystyle{ K(\tau_1, \tau_2) = |\{(i,j): i \lt j, ( \tau_1(i) \lt \tau_1(j) \wedge \tau_2(i) \gt \tau_2(j) ) \vee ( \tau_1(i) \gt \tau_1(j) \wedge \tau_2(i) \lt \tau_2(j) )\}|. }$
where
• $\displaystyle{ \tau_1(i) }$ and $\displaystyle{ \tau_2(i) }$ are the rankings of the element $\displaystyle{ i }$ in $\displaystyle{ \tau_1 }$ and $\displaystyle{ \tau_2 }$ respectively.
$\displaystyle{ K(\tau_1,\tau_2) }$ will be equal to 1 if the two lists are identical and $\displaystyle{ -1 }$ (where $\displaystyle{ n }$ is the list size) if one list is the reverse of the other. The normalized Kendall tau distance therefore lies in the interval [-1,1].
Kendall tau distance may also be defined as
$\displaystyle{ K(\tau_1,\tau_2) = \begin{matrix} \sum_{\{i,j\}\in P} \bar{K}_{i,j}(\tau_1,\tau_2) \end{matrix} }$
where
• P is the set of unordered pairs of distinct elements in $\displaystyle{ \tau_1 }$ and $\displaystyle{ \tau_2 }$
• $\displaystyle{ \bar{K}_{i,j}(\tau_1,\tau_2) }$ = 0 if i and j are in the same order in $\displaystyle{ \tau_1 }$ and $\displaystyle{ \tau_2 }$
• $\displaystyle{ \bar{K}_{i,j}(\tau_1,\tau_2) }$ = 1 if i and j are in the opposite order in $\displaystyle{ \tau_1 }$ and $\displaystyle{ \tau_2. }$
Kendall tau distance can also be defined as the total number of discordant pairs.
Kendall tau distance in Rankings: A permutation (or ranking) is an array of N integers where each of the integers between 0 and N-1 appears exactly once. The Kendall tau distance between two rankings is the number of pairs that are in different order in the two rankings. For example, the Kendall tau distance between 0 3 1 6 2 5 4 and 1 0 3 6 4 2 5 is four because the pairs 0-1, 3-1, 2-4, 5-4 are in different order in the two rankings, but all other pairs are in the same order.[1]
If Kendall tau function is performed as $\displaystyle{ K(L1,L2) }$ instead of $\displaystyle{ K(\tau_1,\tau_2) }$ (where $\displaystyle{ \tau_1 }$ and $\displaystyle{ \tau_2 }$ are the rankings of $\displaystyle{ L1 }$ and $\displaystyle{ L2 }$ elements respectively), then triangular inequality is not guaranteed. The triangular inequality fails in cases where there are repetitions in the lists. So then we are not dealing with a metric anymore.
## Example
Suppose one ranks a group of five people by height and by weight:
Person A B C D E
Rank by height 1 2 3 4 5
Rank by weight 3 4 1 2 5
Here person A is tallest and third-heaviest, and so on.
In order to calculate the Kendall tau distance, pair each person with every other person and count the number of times the values in list 1 are in the opposite order of the values in list 2.
Pair Height Weight Count
(A,B) 1 < 2 3 < 4
(A,C) 1 < 3 3 > 1 X
(A,D) 1 < 4 3 > 2 X
(A,E) 1 < 5 3 < 5
(B,C) 2 < 3 4 > 1 X
(B,D) 2 < 4 4 > 2 X
(B,E) 2 < 5 4 < 5
(C,D) 3 < 4 1 < 2
(C,E) 3 < 5 1 < 5
(D,E) 4 < 5 2 < 5
Since there are four pairs whose values are in opposite order, the Kendall tau distance is 4. The normalized Kendall tau distance is
$\displaystyle{ \frac{4}{5(5 - 1)/2} = 0.4. }$
A value of 0.4 indicates that 40% of pairs differ in ordering between the two lists.
## Computing the Kendall tau distance
A naive implementation in Python (using NumPy) is:
import numpy as np
def normalised_kendall_tau_distance(values1, values2):
"""Compute the Kendall tau distance."""
n = len(values1)
assert len(values2) == n, "Both lists have to be of equal length"
i, j = np.meshgrid(np.arange(n), np.arange(n))
a = np.argsort(values1)
b = np.argsort(values2)
ndisordered = np.logical_or(np.logical_and(a[i] < a[j], b[i] > b[j]), np.logical_and(a[i] > a[j], b[i] < b[j])).sum()
return ndisordered / (n * (n - 1))
However, this requires $\displaystyle{ n^2 }$ memory, which is inefficient for large arrays.
Given two rankings $\displaystyle{ \tau_1,\tau_2 }$, it is possible to rename the items such that $\displaystyle{ \tau_1 = (1,2,3,...) }$. Then, the problem of computing the Kendall tau distance reduces to computing the number of inversions in $\displaystyle{ \tau_2 }$—the number of index pairs $\displaystyle{ i,j }$ such that $\displaystyle{ i\lt j }$ while $\displaystyle{ \tau_2(i) \gt \tau_2(j) }$. There are several algorithms for calculating this number.
• A simple algorithm based on merge sort requires time $\displaystyle{ O(n \log n) }$.[2]
• A more advanced algorithm requires time $\displaystyle{ O(n\sqrt{\log{n}}) }$.[3]
Here is a basic C implementation.
#include <stdbool.h>
int kendallTau(short x[], short y[], int len) {
int i, j, v = 0;
bool a, b;
for (i = 0; i < len; i++) {
for (j = i + 1; j < len; j++) {
a = x[i] < x[j] && y[i] > y[j];
b = x[i] > x[j] && y[i] < y[j];
if (a || b)
v++;
}
}
return abs(v);
}
float normalize(int kt, int len) {
return kt / (len * (len - 1) / 2.0);
} | 2022-01-23T14:50:14 | {
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https://math.stackexchange.com/questions/2158019/true-or-false-if-fx-and-f-1x-intersect-at-an-even-number-of-points?noredirect=1 | # True or False : If $f(x)$ and $f^{-1}(x)$ intersect at an even number of points , all points lie on $y=x$
Previously I have discussed about odd number of intersect points (See : If the graphs of $f(x)$ and $f^{-1}(x)$ intersect at an odd number of points, is at least one point on the line $y=x$?) Now , I want to know the even condition . For example $f(x) = \sqrt{x}$ and $f^{-1}(x) = x^2 , x\ge 0$ intersects each other in $(0,0)$ and $(1,1)$ points and these points located on $y=x$ line
Edit : Consider $f$ is continuous function.
• How would any intersection point not lie on $y=x$? The graphs of the function and its inverse (if it exists) is symmetric about $y=x$, so any intersection must be on $y=x$ Feb 23 '17 at 15:58
• @user160738 No , for example consider $f(x) = \frac{1}{x}$ and $f^{-1}(x) = \frac{1}{x}$ Feb 23 '17 at 16:00
• @user160738 another example is $f(x) = -x^3$ and $f^{-1}(x) = -\sqrt[3]{x}$ . They intersect each other in $(1 , -1)$ that is not on $y=x$ line. See : math.stackexchange.com/questions/1452098/… Feb 23 '17 at 16:08
The continuous case:
If the domain is not an interval, we can still find a counterexample. Define $$f(x)=\cases{2x+2 & if x\in(0,2),\cr x-3 & if x\in(3,5),}$$ then $$f^{-1}(x)=\cases{x/2-1 & if x\in(2,6),\cr x+3 & if x\in(0,2),}$$ and intersections are $(1,4)$, $(4,1)$.
However, if the domain is connected, it turns out to be true.
Proof of the theorem in the case of a connected domain (interval)
For contradiction, let us suppose that there is an even number of intersections and an intersection is out of diagonal.
1. We assume that $f^{-1}$ exists which means that $f$ is injective. Moreover, $f$ is continuous and its domain is interval, so $f$ is increasing or decreasing.
2. $f$ and $f^{-1}$ are symmetric by diagonal, therefore the set of intersections is symmetric by the diagonal.
3. Let denote $(x_1, x_2)$ an intersection out of diagonal. By symmetry, there is an intersection $(x_2, x_1)$. We can therefore WLOG assume that $x_1<x_2$.
4. $f(x_1) > x_1$ and $f(x_2) < x_2$, so by continuity, $f$ intersects the diagonal.
5. There is an even number of intersections and by symmetry, there is an even number of intersections out of diagonal. So there is even number of intersections on the diagonal. So there are at least two of them: $f(x_3) = x_3$, $f(x_4) = x_4$
6. $x_1<x_2$ and $f(x_1) > f(x_2)$, so $f$ is decreasing (on the whole, by 1).
7. $x_3<x_4$ and $f(x_3) < f(x_4)$, so $f$ is increasing.
• Okay but it's true in many cases in connected domain . Can you provide an example that it's wrong in connected domain ? Mar 3 '17 at 10:45
• Okay so why it's true for many examples ? Mar 3 '17 at 15:14
• You proved the theorem is wrong .okay , now provide some examples that shows theorem is false . Mar 3 '17 at 15:38
• Oh , I'm really sorry let me to read again . Mar 3 '17 at 15:40
• Sorry, I don't understand step 4 in your purported proof. Mar 4 '17 at 10:45
Assume that $f$ is a continuous and invertible real function with a connected domain. Then $f$ is either strictly decreasing or strictly increasing over its domain (or it would assume some value twice, and would not be one-to-one). Consider these two cases:
1. $f$ is decreasing. Then, since $f(x)-x$ is also decreasing, $f$ can't cross $y=x$ more than once. If it never crosses $y=x$, then it lies entirely above or below that line, and its inverse lies entirely on the other side; they never meet, and the theorem holds vacuously. If, on the other hand, it crosses $y=x$ exactly once, then the total number of intersections between $f$ and $f^{-1}$ is odd, since off-diagonal intersections come in pairs. The theorem holds in this case too.
2. $f$ is increasing. Then it cannot intersect $f^{-1}$ at any point off the line $y=x$. Suppose it did, at a pair of points $(x,y)$ and $(y,x)$ with $x<y$: then we would have $f(x)=y > x=f(y)$, contradicting the fact that $f$ is increasing. In this case, then, all intersections are on $y=x$, and the theorem holds.
Since the theorem holds whether $f$ is increasing or decreasing, it is true in general.
• For all decreasing functions we can say $f(x) - x$ is also decreasing ? Is it right ? Mar 9 '17 at 6:03
• Sure. A decreasing function minus an increasing function decreases even faster. Mar 9 '17 at 20:48
• Okay and how you can conclude that it intersects at most one time ? Mar 9 '17 at 20:55
Your question, as it currently stands definitely does not hold, as you do not specify the domain, continuity or conditions on inverse. This answer does define an $f$ on the whole real line that has a proper inverse.
Consider the function: \begin{align} f(x) = \begin{cases} 1 &\text{if x = 0}\\ 0 &\text{if x = 1}\\ x^2 &\text{if x > 0 and x\neq 1}\\ -2x &\text{if x < 0} \end{cases} \end{align} which has inverse \begin{align} f^{-1}(y) = \begin{cases} 1 &\text{if y = 0}\\ 0 &\text{if y = 1}\\ \sqrt{x} &\text{if x > 0 and x\neq 1}\\ -\tfrac{x}{2} &\text{if x < 0} \end{cases} \end{align} and so $f(x) = f^{-1}(y)$ only at $x = 0$ and $x = 1$, but at these points $f(x) \neq x$.
• Consider $f$ is continuous Feb 23 '17 at 17:24
• I will have a think @S.H.W, I imagine the answer might be that such a function cannot exist: if they intersect at two points then they are equal on an interval (that is my hunch anyway, YMMV). Feb 23 '17 at 21:17
• I think this statement is true but I can't prove it. Feb 23 '17 at 21:32
• Actually, I don't think that is quite true: consider $f(x) = x^2$ for $x \geqslant 0$ and $f(x) = -2x$ for $x < 0$, then $f^{-1}(x) = \sqrt{x}$ for $x \geqslant 0$ and $f^{-1}(x) = -\tfrac{x}{2}$ for $x < 0$ and so they only intersect at $x = 0$ and $1$. Feb 23 '17 at 21:59
• if $f(x) = x^2$ then $f^{-1}(x) = \sqrt{x} , x \ge 0$ and the intersection points are $(0,0)$ and $(1,1)$ which they are in $y=x$ line Feb 23 '17 at 22:03
For $f$ and $f^{-1}$ to intersect there must be a point(s) $(x,f(x)) = (x,f^{-1}(x))$. Assume $x \neq f(x)$. Because $f$ and $f^{-1}$ are symmetric around y=x then there must be a corresponding point(s) $(f(x),x) = (f^{-1}(x),x)$.
The line defined by the points $(x_0,f(x_0)$ and $(f(x_),x_0)$ is $$(y-f(x_0) = \frac{f(x_0)-x_0}{x_0-f(x_0)}(x-x_0)$$ and simplified becomes $$y=-x+x_0+f(x_0)$$ This line, with a slope of -1, is not parallel with $y=x$, therefore must cross $y=x$. Because the two points are symmetric around $y=x$, they must lie on different sides of $y=x$. By the intermediate value theorem both $f$ and $f^{-1}$ must cross $y=x$.
Hence there is at least one point of $f$ on $y=x$ | 2021-11-29T08:32:33 | {
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https://math.stackexchange.com/questions/3321058/a-particular-vanishing-integral | # A particular vanishing integral
While dealing with a definite integral on AoPS I discovered (I have to admit by pure chance) the following relation
$$\int_0^1\log\left(\frac{(x+1)(x+2)}{x+3}\right)\frac{\mathrm dx}{1+x}~=~0\tag1$$
The proof is quite easy, but feels kind of contrived. Indeed, just apply a self-similar substitution - $$x\mapsto\frac{1-x}{1+x}$$ - to the auxiliary integral $$\mathcal I$$ given by
$$\mathcal I~=~\int_0^1\log\left(\frac{x^2+2x+3}{(x+1)(x+2)}\right)\frac{\mathrm dx}{1+x}$$
And the result follows. However, to consider precisely this integral seems highly unnatural to me (in fact, as I mentioned earlier, this integral was just a by-product while evaluating something quite different and I discovered $$(1)$$ when experimenting with various substitutions).
The crucial point to notice concerning $$\mathcal I$$ is the invariance of the polynomial $$f(x)=x^2+2x+3$$ regarding the self-similar substitution which allows us to deduce $$(1)$$. Additionally for myself I am quite surprised by the special structure of $$(1)$$ since we have factors of the form $$(x+1)$$, $$(x+2)$$ and $$(x+3)$$ combined which calls for a generalization (although I found none yet).
It there a more elementary approach, not relying on such an "accident" like examining the integral $$\mathcal I$$ for proving $$(1)$$? Additionally, can this particular pattern be further generalized? Answers to both questions (also separately) are highly appreciated!
• The $\displaystyle\log$ argument numerator -in the $\displaystyle\mathcal{I}$ definition- must be $\displaystyle x^{2} + 3x + 2$ instead of $\displaystyle x^{2} + 2x + 3$. – Felix Marin Aug 27 '20 at 3:27
• @FelixMarin No. By my calculations $\left(\frac{1-x}{1+x}\right)^2+2\left(\frac{1-x}{1+x}\right)+3=2\frac{x^2+2x+3}{(1+x)^2}$ which is what the approach is all about. But $\left(\frac{1-x}{1+x}\right)^2+3\left(\frac{1-x}{1+x}\right)+2=2\frac{x+3}{(x+1)^2}$ which yields an interesting other identity but not the one the post is concerned with. – mrtaurho Aug 27 '20 at 3:42
That's quite an impressive method to show that the integral vanishes.
For the first part I'll show using a different approach that your integral vanishes. $$\mathcal J=\int_0^1 \ln\left(\frac{x+3}{(x+2)(x+1)}\right)\frac{\mathrm dx}{x+1}\overset{x+1=t}= \color{blue}{\int_1^2\ln\left(\frac{t+2}{t+1}\right)\frac{\mathrm dt}{t}}-\color{red}{\int_1^2 \frac{\ln t}{t}\mathrm dt}$$ Let's denote the blue integral as $$\mathcal J_1$$ then using the substitution $$\frac{2}{t}\to t$$ we get: $$\mathcal J_1=\int_1^2 \ln\left(\frac{t+2}{t+1}\right)\frac{\mathrm dt}{t}=\int_1^2 \ln\left(\frac{2(t+1)}{t+2}\right)\frac{\mathrm dt}{t}$$ Adding both integrals from above gives us: $$\require{cancel} 2\mathcal J_1=\cancel{\int_1^2 \ln\left(\frac{t+2}{t+1}\right)\frac{\mathrm dt}{t}}+\int_1^2 \frac{\ln 2}{t}\mathrm dt+\cancel{\int_1^2 \ln\left(\frac{t+1}{t+2}\right)\frac{\mathrm dt}{t}}=\ln^2 2$$ $$\Rightarrow \mathcal J_1=\color{blue}{\frac{\ln^2 2}{2}}\Rightarrow \mathcal J=\color{blue}{\frac{\ln^2 2}{2}}-\color{red}{\frac{\ln^2 2}{2}}=0$$
As for the second part, a small generalization outcomes by experimenting with the blue integral.
In particular, by the same approach we have: $$\int_1^a \ln\left(\frac{x+a}{x+1}\right)\frac{\mathrm dx}{x}=\int_1^a \frac{\ln x}{x}\mathrm dx$$ Which gives us a small generalization: $${\int_0^{a-1}\ln\left(\frac{x+a+1}{(x+1)(x+2)}\right)\frac{\mathrm dx}{x+1}=0}$$ Similarly, (with the substitution $$\frac{ab}{x}\to x$$) we get that: $$\int_a^b \ln\left(\frac{x+b}{x+a}\right)\frac{dx}{x}=\frac12 \ln^2 \left(\frac{b}{a}\right)=\int_a^b \ln\left(\frac{x}{a}\right)\frac{dx}{x}$$ And the following follows: $$\int_{a-1}^{b-1} \ln\left(\frac{a(x+b+1)}{(x+1)(x+a+1)}\right)\frac{dx}{x+1}=0$$ One might be interested in the following similar generalization too: $$\int_1^{t}\ln\left(\frac{x^4+sx^2+t^2}{x^3+sx^2+t^2x}\right)\frac{dx}{x}=0,\quad s\in R, t>1$$
• This is what I was looking for; I guess^^ (+1) anyway and I'm looking forward to see a generalization (if possible). I guess you know where I got this integral from :D – mrtaurho Aug 12 '19 at 15:17
• @mrtaurho I got there a small generalization for now. However since $\int_a^b \ln\left(\frac{x+b}{x+a}\right)\frac{dx}{x}=\frac12 \ln^2 \left(\frac{b}{a}\right)$ it might be possible to obtain a better one. (I'll try later to work with it). – Zacky Aug 12 '19 at 15:38
• It's hard to write out your name now :P But yes, this seems promising, I'm curious! As I noted above it was a rather strange by-product to discover this equality; and it was tedious to ran into it three times while hoping for something more helpful for solving the original task^^' – mrtaurho Aug 12 '19 at 15:44
• @mrtaurho just in case you missed it in winter I'll mention that $\mathcal I$ (before the self-similar sub was applied) originates from this generalization: math.stackexchange.com/a/3049039/515527. Aka: $$\int_1^{\sqrt{t}}\ln\left(\frac{x^4+sx^2+t}{x(x^2+sx+t)}\right)\frac{dx}{x}=0$$ – Zacky Aug 12 '19 at 16:06
• As I've upvoted both, the question you linked and your answer, I guess I've seen it at some point. But I'll take a look at it again :) – mrtaurho Aug 12 '19 at 16:23
I have used the substitution $$(x+1)(y+1)=2$$ before to good effect because $$\int_0^1f(x)\,\frac{\mathrm{d}x}{x+1}=\int_0^1f\!\left(\tfrac{1-y}{1+y}\right)\frac{\mathrm{d}y}{y+1}\tag1$$ If $$f(x)=\log\left(\frac{x+3}{(x+2)(x+1)}\right)$$, then $$f\!\left(\frac{1-y}{1+y}\right)=\log\left(\frac{(y+2)(y+1)}{y+3}\right)$$. Therefore $$\int_0^1\log\left(\frac{x+3}{(x+2)(x+1)}\right)\frac{\mathrm{d}x}{x+1}=\int_0^1\log\left(\frac{(y+2)(y+1)}{y+3}\right)\frac{\mathrm{d}y}{y+1}\tag2$$ and since the two sides of $$(2)$$ are negatives, they are both $$0$$.
We can generalize $$(1)$$ by letting $$(x+a)(y+a)=a(1+a)$$, then we get $$\int_0^1f(x)\frac{\mathrm{d}x}{x+a}=\int_0^1f\!\left(\tfrac{a(1-y)}{a+y}\right)\frac{\mathrm{d}y}{y+a}\tag3$$ | 2021-03-02T13:18:24 | {
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http://www.lofoya.com/Solved/1517/when-a-student-weighing-45-kgs-left-a-class-the-average-weight-of | # Moderate Averages Solved QuestionAptitude Discussion
Q. When a student weighing 45 kgs left a class, the average weight of the remaining 59 students increased by 200 g. What is the average weight of the remaining 59 students?
✔ A. 57 ✖ B. 56.8 ✖ C. 58.2 ✖ D. 52.2
Solution:
Option(A) is correct
Let the average weight of the 59 students be $A$.
Therefore, the total weight of the 59 of them will be 59 $A$.
The questions states that when the weight of this student who left is added, the total weight of the class
= $59A + 45$
When this student is also included, the average weight decreases by 0.2 kgs.
$59A + \dfrac{45}{60} = A - 0.2$
$\Rightarrow 59A + 45 = 60A - 12$
$\Rightarrow 45 + 12 = 60A - 59A$
$\Rightarrow A = \textbf{57}$
Edit: For an alternative solution, check comment by Durga.
## (9) Comment(s)
Siddharth Sharma
()
60*0.2+45=57 thats the shortest method to do it.
Shivakumar
()
divided by 60 for 59A+45 not only for 45
Vijay
()
simply,
60*(x-.20)-59x=45;
=>x=45+12=57.
Sanajit Ghosh
()
@DURGA's ans is right
Zoha Amjad
()
answer is 56.8
0.2 kgs increase means
$0.2*59=11.8$
$11.8+45 = 56.8$
Shanu
()
The ans should be 56.8
it can be solved as
$60A-((A-0.2)*59)=45$
so $A(\text{avarage}) = 56.8$
Kiruthi
()
ans should be 56.8 because in question they have given increased by 200g but while solving they took it as decreased by 200g so 57
Durga
()
Hi ,
The answer is 57 only.
Let say the average of 60 students is $x$.
Total weight $=60*x$
If one man left average should increase 200gr
i.e $\dfrac{200}{1000} =0.2 \text{ kg}$
So, $60 \times x-45=59(x+0.2)$
$60x-59x=59*0.2+45$
$\Rightarrow x=56.8 \text{ kg}$
But here we need to calculate the average of 59 students
i.e $56.8+0.2= \textbf{57 kg}$
Elmira
()
Thank you for creating a very informative website, much appreciate it.
Please review the solution once again, when multiplying the sides by 60, 59A was left without effect, therefore I think the response is wrong. Oh maybe I am missing something here.
Please explain.
Elmira | 2017-02-24T05:49:02 | {
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https://math.stackexchange.com/questions/2804617/find-the-matrix-representation-of-a-linear-map | # Find the matrix representation of a linear map
Let $f:\mathbb{R}^2 \to \mathbb{R}^2$ be a linear map and consider the matrix representation $$M_{B}^{B} =\begin{pmatrix} 4 & 6\\ 6 & 3 \end{pmatrix}$$ with respect to the basis $B=\bigg\{ \begin{pmatrix} 2 \\ 2 \end{pmatrix} , \begin{pmatrix} 3\\ 2 \end{pmatrix}\bigg\}$. Find the matrix representation of $f$ with respect to the canonical basis.
MY ATTEMPT: Let $b_1=\begin{pmatrix} 2 \\ 2 \end{pmatrix}$, $b_2=\begin{pmatrix} 3\\ 2 \end{pmatrix}$ and $e_j$ the j-th vector of the canonical basis. Then I have $f(b_1)=\begin{pmatrix} 4\\ 6 \end{pmatrix} \Longrightarrow f(2e_1 +2e_2)=2 f(e_1) +2f(e_2) =4e_1 + 6e_2$
$f(b_2)=\begin{pmatrix} 6\\ 3 \end{pmatrix} \Longrightarrow f(3e_1 +2e_2)=3 f(e_1) +2f(e_2) =6e_1 + 3e_2$
Solving this system I obtain that $f(e_1)=2e_1 -3e_2=\begin{pmatrix} 2\\ -3 \end{pmatrix}$ and $f(e_2)= 6e_2=\begin{pmatrix} 0\\ 6 \end{pmatrix}$. So the matrix rappresentation is $M_{E}^{E}=\begin{pmatrix} 2 & 0\\ -3 & 6 \end{pmatrix}$.
BUT my theacher says that the solution is $M_{E}^{E}=\begin{pmatrix} 13 & 10\\ 0 & 0 \end{pmatrix}$. Where I am wrong?
$$f(2e_1 +2e_2)=2 f(e_1) +2f(e_2) \color{red}{=4e_1 + 6e_2}$$
This is where you are wrong, $4$ and $6$ are coordinates in the basis $B$, so there is one more step to get $e_1$ and $e_2$ into the game. Otherwise your approach is nice.
• How can I do that? – fcoulomb Jun 1 '18 at 18:47
• @Besh00 write $\color{red}{=4b_1+6b_2}$ instead of the red part, and then write both $b_1$ and $b_2$ in terms of $e_1$, $e_2$. – Arnaud Mortier Jun 1 '18 at 18:50
• Why $f(b_1)=4b_1+6b_2$? – fcoulomb Jun 1 '18 at 21:10
• You wrote yourself "Then I have $f(b_1)=\begin{pmatrix} 4\\ 6 \end{pmatrix}$". These coordinates are with respect to the basis $B$ since they are taken from a matrix wrt $B,B$. – Arnaud Mortier Jun 1 '18 at 21:14 | 2020-02-17T19:31:01 | {
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https://math.stackexchange.com/questions/2737440/how-many-4-digit-numbers-that-do-not-have-5-and-have-7-in-the-hundreds-pos | # How many $4$-digit numbers that do not have $5$ and have $7$ in the hundreds position?
I encountered this problem:
How many $4$-digit numbers are there that do not have $5$ and have $7$ in the hundreds position? Digits cannot be repeated.
My thinking: first, let's count numbers that do not have $0$. It's going to be $7\cdot 6 \cdot 5$ arrangements. Now, let's count numbers that have $0$ in the tens position. We have two positions open so there are $7 \cdot 6$ arrangements. We get the same number of arrangements for numbers that end with $0$. The total number of arrangements is $7\cdot 6 \cdot 5+ 2\cdot 7 \cdot 6=7\cdot 7 \cdot 6=294$.
However, my friend argues that there are $8$ choices for the last digit, $7$ choices for the tens digits and $6$ choices for the first digit so the total number should be $8\cdot 7 \cdot 6=336$.
I am asking who is right not to prove my friend wrong but to find the truth. Thanks for listening!
294 is correct.
First digit (most significant), can not be 7,0,5, so 7 choices; 3rd digit, cannot be 7,5, and the one chosen for 1st, so 7 choices; 4th digit, 6 choices;
$7\times7\times6=294$
• Welcome to the site. Great first answer! – B. Mehta Apr 14 '18 at 23:17
You are right and your friend is wrong. If you choose the units digit and then choose the tens digit (and are forced to use $7$ for the hundreds digit) then the number of choices for the thousands digit not always $6$: it is either $5$ or $6$, depending on if you've used $0$ yet or not.
For example, there are $5$ numbers satisfying the condition which have the form $x789$ (since $x$ can be any of $1,2,3,4,6$) but $6$ numbers satisfying the condition which have the form $x780$ (since $x$ can be any of $1,2,3,4,6,9$).
All this tells us is that the number of solutions is between $8\cdot7\cdot5 = 280$ and $8\cdot7\cdot6 = 336$, and for a more precise answer we do need to look at whether $0$ gets used.
I think the key disagreement here is that your friend is thinking that $0$ is allowed to be the first digit (say, for a 4 digit passcode), while you are not. If $0$ can be the first digit, your friend is correct, but if not (as other posters explain), you are correct.
You don't have to choice the positions in order so don't worry about $0$s.
Do the Thousandth place first. It can not be $0$ and it can not be $7$ or $5$. So there are $7$ options. Then do either the tens or the ones, it doesn't matter which. It can be $0$ but it can't be $7$ or $5$ and it can't be what the thousandth place was. That is $7$ options as well. And the final position can't be $7$ or $5$ or either of the previous two positions. so that is $6$ options.
So there are $7*7*6 = 294$.
You were correct but you didn't have to work so hard.
Your freind is wrong in that $0$ is not an option for the first digit. | 2019-10-14T04:31:06 | {
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http://sesk.motcha.de/telescoping-series-partial-fractions.html | # Telescoping Series Partial Fractions
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Begin by using partial fraction decomposition to obtain 2 n(n+ 2) = A n + B n+ 2 = 1 n 1. 10-2-43 telescoping series. The following series, for example, is not a telescoping series despite the fact that we can partial fraction the series. Telescoping series: Compute the nth partial sum, sn, and take the limit of sn as n goes to 1. The series in Example I(b) is a telescoping series of the form — 192) + (b2 — 173) + (b 3 — 125) + Telescoping series Note that b2 is canceled by the second term, 173 is canceled by the third term, and so on. Assignment for Day 26 and Answers. Geometric Series + 107. Here's another example that uses partial fractions. High School AP Calculus BC Curriculum. Example 4: Determine the series 1 1 n nn( 1) f ¦ is convergent or divergent. Theorem 4. A more general technique Efthirniou's technique can be generalized to series of the form CrI1unvnwhere it is convenient to write only vnas a Laplace transform integral. 12 is an example of a telescoping series. (Telescoping series) Find. if it is convergent, find its sum. First, note that the telescoping series method only works on certain fractions. Partial Fractions. This free online course offers you some Integral definitions, Explanations and Examples. If Partial fractions gives 1 k(k +1). Telescoping series. o q bASl BlB Zr niVg8hnt osS 5r8ewsXenrZv Yecdj. 2 Series A series is a sum of sequential terms. 2) Trigonometric Substitution (7. Harmonic series. Next, prove an important convergence theorem. com allows you to find the sum of a series online. 10-3-23 constant times a divergent series. Expanding the sum yields Rearranging the brackets, we see that the terms in the infinite sum cancel in pairs, leaving only the first and lasts terms. Download this MAT 1348 class note to get exam ready in less time! Class note uploaded on Oct 14, 2019. The value of the series is lim I lim Sn 553 Related 47-58 19—34. Then evaluate lim n!1 S n to obtain the value of the series, or state that it diverges. MCrawford (20:20:34). The second part as it goes to infinity is 0, so the answer would be 2? I'm really confused and looking up examples of telescoping series and it is not helping. Series Calculator computes sum of a series over the given interval. Substitution. Convergence b. How this Calculus 2 course is set up to make complicated math easy: This approximately 200-lesson course includes video and text explanations of everything from Calculus 2, and it includes more than 275 quiz questions (with solutions!) to help you test your understanding along the way. Too bad the book didn't do a good job on teaching how to find nth partial sum formula. That is, s k → -∞. Geometric and Telescoping Series In this worksheet, you will calculate the exact values of several series. The voice explains how to first plug in the numbers given for each variable in the fractions. Telescoping Series 1 1 2 3 3 Show that the telescoping series ln dive rges. Telescoping series is a series where all terms cancel out except for the first and last one. Absolute convergence. A telescoping series is a special type of series whose terms cancel each out in such a way that it is relatively easy to determine the exact value of its partial sums. In algebra, the partial fraction decomposition or partial fraction expansion of a rational function (that is, a fraction such that the numerator and the denominator are both polynomials) is an operation that consists of expressing the fraction as a sum of a polynomial (possibly zero) and one or several fractions with a simpler denominator. Exercise 6. The method of partial fractions may reveal these. Your explanation that the series is telescoping after written as a partial fraction is more concise. An in nite series can be represented as such: P 1 n=1 a n. z T BMAapdPeB wwMitEhL lIQnkfoimnBi\tieE rPCrvecWavlccfuxlluKsx. Telescoping Series Test Series: ∑∞𝑛=1 Ὄ 𝑛+1− 𝑛Ὅ Condition of Convergence: lim 𝑛→∞ 𝑛=𝐿 Condition of Divergence: None NOTE: 1) May need to reformat with partial fraction expansion or log rules. The typical example of telescoping series (for partial fractions) is. There is no way to actually identify the series as a telescoping series at this point. Math - Calculus II SERIES ( Partial Sums ) infinite series GEOMETRIC SERIES Otherwise, the series diverges ( that is, the series has no sum ). Let’s decompose it, using partial fractions: [math]\frac{1. The Telescoping Series! This type of infinite series utilizes the technique of Partial Fractions which is a way for us to express a rational function (algebraic fraction) as a sum of simpler fractions. Could anyone explain in plain english whats going on here? And clarify their formula. This is the main origin of the name telescoping series. If convergent find the sum, and if divergent enter DIV: 11. This is an excerpt from my full length lesson. The final answer is: Note that we have converted an infinite sum problem to adding up a finite number of fractions. be able to recognize a telescoping series, determine whether it is cvg or div and if cvg, to what. You don't see many telescoping series, but the telescoping series rule is a good one to keep in your bag of tricks — you never know when it might come in handy. \ B jArlnlA Er^iOgqhEtcsn srhemsNeKrkvre_dM. partial fractions to rewrite it as a telescoping series. Here’s an example series: Just like for sequences, we want to make explicit what the term is. The final answer is: Note that we have converted an infinite sum problem to adding up a finite number of fractions. After several hours of thinking about these type of problems I found out that I can separate the 1/[(n+1)(n+2)] into partial fractions, and I found out that it is a telescoping series. Telescoping Series with Partial Fractions Mathispower4u. The solution in the above example uses the method of partial fractions to try to calculate the sum of a telescoping series. Then evaluate lim n!1 S n to obtain the value of the series, or state that it diverges. The following exercises test your understanding of infinite sequences and series. • Use improper integrals to reinforce the notions of infinite series. A telescoping series is one whose partial sums eventually only have a fixed number of terms after cancellation. The result is a simple formula for the nth term of the sequence of partial sums. Her practical approach makes math manageable for anyone who's learning math for the first time, returning to school after a long break, or for anyone with a genuine curiosity who wants to dive deeper into the material. 1B3: If a series converges absolutely, then any series obtained from it by regrouping or rearranging the terms has the same value. Now we must solve for A and B. In particular, in order for the fractions to cancel out, we need the numerators to be the same. (Of course, an infinite geometric series is a special case of a Taylor series. Note: For an example of a telescoping sums question, see question #2 in the Additional Examples section below. Berndt also states that there is no “natural” way to obtain an expansion of H n in powers of m. Telescoping series: Expand out the sums to find a grouping pattern that can simplify in order to take the limit of S n *Note: Some series can be turned into subtraction via partial fractions or by log rules ∑Eg: ∑. Infinite Series Definition. This is an example of a telescoping series. Partial Fractions Introduction to Partial Fractions Linear Factors Irreducible Quadratic Factors Improper Rational Functions and Long Division Summary Strategies of Integration Substitution Integration by Parts Trig Integrals Trig Substitutions Partial Fractions Improper Integrals Type 1 - Improper Integrals with Infinite Intervals of Integration. If instructions say to determine if a series converges or diverges AND find the sum if it converges, then it must be an infinite geometric series or a telescoping. Improper integrals of type 2. If convergent find the sum, and if divergent enter DIV: 11. Geometric and Telescoping Series c. Look at the partial sums: because of cancellation of adjacent terms. So capital N, so the partial sum is going to be the sum from n equals one but not infinity but to capital N of negative 1 to the n minus 1. Geometric Series + 107. If an input is given then it can easily show the result for the given number. How to get the partial fractions of higher degree numerators. An in nite series can be represented as such: P 1 n=1 a n. The Sigma notation for summation of series. 3 The Integral Test: The section includes the integral test and a discussion on the remainder estimate for the integral test. Review for Test 3 Math 1552, Integral Calculus The series is telescoping, so it converges and we can find its sum using partial fractions: ". Say we have something we want to sum up, let's call it a k. Be sure to review the Telescoping Series page before continuing forward. Partial Fractions. For the telescoping series X1 k=3 2 (2k 1)(2k + 1); nd a formula for the nth term of the sequence of partial sums fS ng. I work through an example of proving that a series converges and finding the sum of the series using Partial Fractions to create a Telescoping Series. (1) and ask whether the sum is convergent. ~~~~~ AP Cal BC Students: Telescoping Series and Partial Fractions Tests. A telescoping series (or telescoping sum) is one that "expands" in such a way that most of its terms cancel away. ) (Now evaluate the limit. 2: p 2] EXAMPLES Does the series converge or diverge? 1. Definitions: Let{ } 1 n n a. Retrieved from " https: The following series, for example, is not a telescoping series despite the fact that we can partial fraction the series terms. Alternating Series Test cannot be applied 26. In nite Series SUGGESTED REFERENCE MATERIAL: As you work through the problems listed below, you should reference your lecture notes and the relevant chapters in a textbook/online resource. TELESCOPING SERIES Now let us investigate the telescoping series. ) = 1 - 0 = 1. So: s1=a1 s2=a1+a2 s3=a1+a2+a3 s4=a1+a2+a3+a4. This is a challenging sub-section of algebra that requires the solver to look for patterns in a series of fractions and use lots of logical thinking. Improper integrals of type 2. This time there will be a few more terms that do not cancel. Lecture 17: Series (II) Telescoping Series A telescoping series is a special type of series for which many terms cancel in the nth partial sums. (The first Maplet may take a little longer to open because it needs to start Java. Back to Thomas O'Sullivan's Homepage. Creating the telescoping effect frequently involves a partial fraction decomposition. Telescoping series. 3 Telescoping Series. You - ProProfs Discuss. We will now look at some more examples of evaluating telescoping series. If you are looking for more in partial fractions, do check in: Partial fractions of lower degree numerators. Get the free "Series Calculator" widget for your website, blog, Wordpress, Blogger, or iGoogle. Partial fractions Fractions in which the denominator has a quadratic term Sometimes we come across fractions in which the denominator has a quadratic term which. Be able to determine the nth partial sum of any geometric series. Partial Fractions Stewart Chapter 7. Test for convergence of a series using an appropriate test: divergence, integral, comparison and limit comparison, ratio, root, or alternating series. High School AP Calculus BC Curriculum. For telescoping series, intermediate terms 'drop. For example, using partial fractions and cancelling a bunch of terms, we find that; An infinite series that arises from Parseval’s theorem in Fourier analysis. Most telescopic series problems involve using the partial fraction decomposition before expanding it and seeing terms cancel out, so make sure you know that very well before tackling these questions. Calculus 2 Syllabus. Note that the fundamental concepts of functions, graphs, and limits, which are studied at the beginning of courses in differential calculus, are often first introduced in earlier classes (most notably intermediate algebra and precalculus). Also note that it is possible to tell that this last series. Infinite series 1: Geometric and telescoping series Main ideas. The final answer is: Note that we have converted an infinite sum problem to adding up a finite number of fractions. 26, 29, 39, 40 21 – 42 Solve applications involving series. Exercise 6. 3) Cancel duplicates. If you do not own Maple, click Use MapleNET 12 at the top-right corner of this page. Note: is a telescoping series. Thus, as $$k$$ increases, the partial sum $$S_k$$ increases (the series is a sum of positive terms), but is always smaller than \(2\text{. Find the sum of the series •  k=0 4 k2 +3k+2 if it exists. That is, s k → -∞. Hence, Therefore, by the definition of convergence for infinite series, the above telescopic series converges and is equal to 1. Find the sum of the series •  k=1 1 k 1 k+2 if it exists. With geometric series, we carried out the entire evaluation process by finding a formu for the sequence of partial sums and evaluating the limit of the sequence. Infinite Series Chapter 1: Sequences and series Section 4: Telescoping series Page 2 We can use the method of partial fractions to rewrite the general term as: 4 1 1 2 2 2 2 ab n n n n n n §· ¨¸ ©¹ For your practice, check that this decomposition is correct! Now we can write any partial sum of this series as: 11 4 1 1 2 22 kk nn n n n n. This video is a great one on learning about evaluating fractions. Telescoping: Transform by partial fractions Procedure for Determining Convergence No Series Diverges nth-Term Test Yes or maybe Yes No Yes Nonnegative terms and/or absolute convergence No Yes No Can the Integral Test, the Ratio Test, or the Root Test be applied? Does the Integral Test apply? Yes Does the Ratio Test apply? No Yes Is Is Does the. (a) X1 n=3 1 2n 1 1 2n 4 As I stated in my email follow-up to this problem, this is not actually an easy series to nd a. Now pop in the first term (a 1) and the common ratio (r). The telescoping Series is a method for examining the convervence of infinite series of the form: This method, combined with partial fraction decomposition, is frequently effective. Be able to determine the nth partial sum of any geometric series. (a) X1 n=1 cos 1 n cos 1 n+ 1. Infinite series allow us to add up infinitely many terms, so it is suitable for representing something that keeps on going forever; for example, a geometric series can be used to find a fraction equivalent to any given repeating decimal such as: #3. Besides finding the sum of a number sequence online, server finds the partial sum of a series online. According to Ferraro [4], Mengoli did so by making frequent use of the following. — Try Before you Buy To start a Maplet, click on its name. if integral of series reaches ∞ or DNE then series diverges. Telescoping Series and Partial Fractions. In this case, the series is convergent and the sum isS lim n→ sn b1 −lim n→ bn 1. See if you can figure it out. Welcome to Maplets for Calculus. 4 Comparison Tests. Can the series be compared favorably to one of the special types?. EXAMPLE 8:. 50 46 – 53 Evaluate telescoping series. telescoping series to find out. This is best demonstrated with an example. I work through an example of proving that a series converges and finding the sum of the series using Partial Fractions to create a Telescoping Series. Now let us investigate the telescoping series. (a)Sequences and Series, partial sums. Example: Determine whether the given series converge. Use comparison test if numerator fluctuates between two constants. From Wikipedia, the free encyclopedia. | 2020-01-29T17:16:34 | {
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https://cryptohack.gitbook.io/cryptobook/fundamentals/division-and-gcd/euclidean-algorithm | # Introduction
Although we have functions that can compute our $\gcd$easily it's important enough that we need to give and study an algorithm for it: the euclidean algorithm.
It's extended version will help us calculate modular inverses which we will define a bit later.
# Euclidean Algorithm
Important Remark
If $a = b \cdot q + r$and $d = \gcd(a, b)$ then $d | r$. Therefore $\gcd(a, b) = \gcd(b, r)$
## Algorithm
We write the following:
$a = q_0 \cdot b + r_0 \\ b = q_1 \cdot r_0 + r_1 \\ r_0 = q_2 \cdot r_1 + r_2 \\ \vdots \\ r_{n-2} = r_ {n-1} \cdot q_{n - 1} + r_n \\ r_n = 0$
Or iteratively $r_{k-2} = q_k \cdot r_{k-1} + r_k$ until we find a $0$. Then we stop
Now here's the trick:
$\gcd(a, b) = gcd(b, r_0) = gcd(r_0, r_1) = \dots = \gcd(r_{n-2}, r_{n-1}) = r_{n-1} = d$
If $d = \gcd(a, b)$ then $d$ divides $r_0, r_1, ... r_{n-1}$
Pause and ponder. Make you you understand why that works.
Example:
Calculate $\gcd(24, 15)$
$24 = 1 \cdot 15 + 9 \\ 15 = 1 \cdot 9 + 6 \\ 9 = 1 \cdot 6 + 3 \\ 6 = 2 \cdot 3 + 0 \Rightarrow 3 = \gcd(24, 15)$
Code
def my_gcd(a, b): # If a < b swap them if a < b: a, b = b, a # If we encounter 0 return a if b == 0: return a else: r = a % b return my_gcd(b, r)print(my_gcd(24, 15))# 3
Exercises:
1. Pick 2 numbers and calculate their $\gcd$by hand.
2. Implement the algorithm in Python / Sage and play with it. Do not copy paste the code
# Extended Euclidean Algorithm
This section needs to be expanded a bit.
Bezout's identity
Let $d = \gcd(a, b)$. Then there exists $u, v$ such that $au + bv = d$
The extended euclidean algorithm aims to find $d = \gcd(a, b), \text{ and }u, v$given $a, b$
# In sage we have the xgcd functiona = 24b = 15g, u, v = xgcd(a, b)print(g, u, v)# 3 2 -3 print(u * a + v * b)# 3 -> because 24 * 2 - 15 * 3 = 48 - 45 = 3 | 2021-09-24T15:52:14 | {
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https://www.quizover.com/trigonometry/test/evaluation-of-functions-in-algebraic-forms-by-openstax | # 3.1 Functions and function notation (Page 4/21)
Page 4 / 21
$n$ 1 2 3 4 5 $Q$ 8 6 7 6 8
[link] displays the age of children in years and their corresponding heights. This table displays just some of the data available for the heights and ages of children. We can see right away that this table does not represent a function because the same input value, 5 years, has two different output values, 40 in. and 42 in.
Age in years, (input) 5 5 6 7 8 9 10 Height in inches, (output) 40 42 44 47 50 52 54
Given a table of input and output values, determine whether the table represents a function.
1. Identify the input and output values.
2. Check to see if each input value is paired with only one output value. If so, the table represents a function.
## Identifying tables that represent functions
Input Output
2 1
5 3
8 6
Input Output
–3 5
0 1
4 5
Input Output
1 0
5 2
5 4
[link] and [link] define functions. In both, each input value corresponds to exactly one output value. [link] does not define a function because the input value of 5 corresponds to two different output values.
When a table represents a function, corresponding input and output values can also be specified using function notation.
The function represented by [link] can be represented by writing
Similarly, the statements
represent the function in [link] .
[link] cannot be expressed in a similar way because it does not represent a function.
Input Output
1 10
2 100
3 1000
yes
## Finding input and output values of a function
When we know an input value and want to determine the corresponding output value for a function, we evaluate the function. Evaluating will always produce one result because each input value of a function corresponds to exactly one output value.
When we know an output value and want to determine the input values that would produce that output value, we set the output equal to the function’s formula and solve for the input. Solving can produce more than one solution because different input values can produce the same output value.
## Evaluation of functions in algebraic forms
When we have a function in formula form, it is usually a simple matter to evaluate the function. For example, the function $\text{\hspace{0.17em}}f\left(x\right)=5-3{x}^{2}\text{\hspace{0.17em}}$ can be evaluated by squaring the input value, multiplying by 3, and then subtracting the product from 5.
Given the formula for a function, evaluate.
1. Replace the input variable in the formula with the value provided.
2. Calculate the result.
## Evaluating functions at specific values
Evaluate $\text{\hspace{0.17em}}f\left(x\right)={x}^{2}+3x-4\text{\hspace{0.17em}}$ at
1. $2$
2. $a$
3. $a+h$
4. $\frac{f\left(a+h\right)-f\left(a\right)}{h}$
Replace the $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ in the function with each specified value.
1. Because the input value is a number, 2, we can use simple algebra to simplify.
$\begin{array}{ccc}\hfill f\left(2\right)& =& {2}^{2}+3\left(2\right)-4\\ & =& 4+6-4\hfill \\ & =& 6\hfill \end{array}$
2. In this case, the input value is a letter so we cannot simplify the answer any further.
$f\left(a\right)={a}^{2}+3a-4$
3. With an input value of $\text{\hspace{0.17em}}a+h,\text{\hspace{0.17em}}$ we must use the distributive property.
$\begin{array}{ccc}\hfill f\left(a+h\right)& =& {\left(a+h\right)}^{2}+3\left(a+h\right)-4\hfill \\ & =& {a}^{2}+2ah+{h}^{2}+3a+3h-4\hfill \end{array}$
4. In this case, we apply the input values to the function more than once, and then perform algebraic operations on the result. We already found that
$f\left(a+h\right)={a}^{2}+2ah+{h}^{2}+3a+3h-4$
and we know that
$f\left(a\right)={a}^{2}+3a-4$
Now we combine the results and simplify.
the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
1+cos²A/cos²A=2cosec²A-1
test for convergence the series 1+x/2+2!/9x3
a man walks up 200 meters along a straight road whose inclination is 30 degree.How high above the starting level is he?
100 meters
Kuldeep
Find that number sum and product of all the divisors of 360
Ajith
exponential series
Naveen
what is subgroup
Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1
e power cos hyperbolic (x+iy)
10y
Michael
tan hyperbolic inverse (x+iy)=alpha +i bita
prove that cos(π/6-a)*cos(π/3+b)-sin(π/6-a)*sin(π/3+b)=sin(a-b)
why {2kπ} union {kπ}={kπ}?
why is {2kπ} union {kπ}={kπ}? when k belong to integer
Huy
if 9 sin theta + 40 cos theta = 41,prove that:41 cos theta = 41
what is complex numbers
Dua
Yes
ahmed
Thank you
Dua
give me treganamentry question
Solve 2cos x + 3sin x = 0.5 | 2018-12-10T03:50:49 | {
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http://math.stackexchange.com/questions/20948/fibonacci-identity-f-n-1f-n1-f-n2-1n | # Fibonacci identity: $f_{n-1}f_{n+1} - f_{n}^2 = (-1)^n$
Consider this Fibonacci equation: $$f_{n+1}^2 - f_nf_{n+2}$$
The problem asked to write a program with given n, output the the result of this equation. I could use the formula $$f_n = \frac{(1+\sqrt{5})^n - ( 1 - \sqrt{5} )^n}{2^n\sqrt{5}}$$
However, from mathworld, I found this formula Cassini's identity $$f_{n-1}f_{n+1} - f_{n}^2 = (-1)^n$$
So, I decided to play around with the equation above, and I have:
$$\text{Let } x = n + 1$$ $$\text{then the equation above becomes } f_x^2 - f_{x-1}f_{x+1}$$ $$\Rightarrow -( f_{x-1}f_{x+1} - f_x^2 ) = -1(-1)^x = (-1)^{x+1} = (-1)^{n+1+1} = (-1)^{n+2}$$
So this equation either is 1 or -1. Am I in the right track?
Thanks,
Chan
-
Yes. $\mbox{}$ – Mariano Suárez-Alvarez Feb 8 '11 at 5:44
@Mariano Suárez-Alvarez: Thanks ^_^! Happy! – Chan Feb 8 '11 at 5:47
In your first expression, should the subscript on the square be "n+1" instead of "n-1"? If so, then you are on the right track. (Also, you'll want parentheses around your "-1"s.) – Blue Feb 8 '11 at 5:47
Note: You should write $(-1)^{n+2}$, and not $-1^{n+2}$; exponentiation takes precedence, so the former is either $-1$ or $1$ depending on the parity of $n$, but the latter is $-1^{n+2} = -(1^{n+2}) = -1$ always. – Arturo Magidin Feb 8 '11 at 5:50
@Day Late Don: nice catch. It was my typo. I was practicing using Latex:(. – Chan Feb 8 '11 at 5:51
We have the following (easily proved by induction):
$\begin{pmatrix} 1 & 1 \\ 1 & 0 \\ \end{pmatrix}^n = \begin{pmatrix} f_{n+1} & f_n \\ f_n & f_{n-1} \\ \end{pmatrix}$
Equating the determinants of the matrices gives us the identity immediately.
-
NOTE $\$ This is precisely the same answer I gave a couple months ago in this duplicate thread – Bill Dubuque Feb 28 '11 at 4:06
Let us try to find gcd of $F_n$ and $F_{n+1}$ using Extended Euclidean algorithm. I will write the steps algorithm in a table; this table method was also explained in some Bill Dubuque's posts.
$\begin{array}{|l||c|c|} \hline F_{n+1} & 1 & 0 \\\hline F_{n} & 0 & 1 \\\hline F_{n-1} & 1 & -1 \\\hline F_{n-2} & -1 & 2 \\\hline F_{n-3} & 2 & -3 \\\hline \vdots & \vdots & \vdots \\\hline F_{n-k} & (-1)^{k+1}F_k & (-1)^kF_{k+1} \\\hline \end{array}$
After a few steps we can guess the $k$-th line, which gives us the following formula:
$F_{n-k}=(-1)^{k+1}F_kF_{n+1}+(-1)^kF_{k+1}F_n=(-1)^{k+1}(F_kF_{n+1}-F_{k+1}F_n)$.
For $k=n-1$ we get Cassini's identity $F_1=(-1)^n(F_{n-1}F_{n+1}-F_n^2)$.
So the only thing we have to do is to verify the above formula, which can be done easily by induction on $k$.
Inductive step: We know that:
$F_{n-k}=(-1)^{k+1}(F_kF_{n+1}-F_{k+1}F_n)=-(-1)^{k}(F_kF_{n+1}-F_{k+1}F_n)$
$F_{n-(k-1)}=(-1)^{k}(F_{k-1}F_{n+1}-F_{k}F_n)$
Since $F_{n-(k+1)}=F_{n-(k-1)}-F_{n-k}$, we get
$F_{n-(k+1)}=(-1)^{k}[(F_{k-1}+F_k)F_{n+1}-(F_k+F_{k+1})F_n]=(-1)^{k+2}(F_{k+1}F_{n+1}-F_{k+2}F_n)$
which completes the inductive step.
-
I tried to google a little, and I found almost the same derivation of Cassini's identity in the text Yiu: Recreational Mathematics. – Martin Sleziak Mar 16 '12 at 11:09 | 2015-05-30T11:33:32 | {
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http://mathhelpforum.com/discrete-math/153261-series.html | # Math Help - series
1. ## series
Dear Sir
I would be grateful if you can help me to find f(r) in the below questions
thanks
Kingsman
Sum to the n terms of the series using difference method of telescope method .(hint Express the nth term = f(r)-f(r+1))
2. Originally Posted by kingman
Dear Sir
I would be grateful if you can help me to find f(r) in the below questions
thanks
Kingsman
Sum to the n terms of the series using difference method of telescope method .(hint Express the nth term = f(r)-f(r+1))
1)
$\displaystyle\huge\frac{n}{(n+1)!}=\frac{n+1-1}{(n+1)!}=\frac{n+1}{(n+1)!}-\frac{1}{(n+1)!}$
$=\displaystyle\huge\frac{n+1}{(n+1)n!}-\frac{1}{(n+1)!}=\frac{1}{n!}-\frac{1}{(n+1)!}$
Therefore
$\displaystyle\huge\frac{1}{2!}+\frac{2}{3!}+......$
$=1-\displaystyle\huge\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}+.......-\frac{1}{(n+1)!}$
$=1-\displaystyle\huge\frac{1}{(n+1)!}$
3. Originally Posted by kingman
Dear Sir
I would be grateful if you can help me to find f(r) in the below questions
thanks
Kingsman
Sum to the n terms of the series using difference method of telescope method .(hint Express the nth term = f(r)-f(r+1))
2)
$\displaystyle\huge\frac{1}{(n+2)n!}=\frac{(n+1)}{( n+2)(n+1)n!}=\frac{n+1}{(n+2)!}$
$\displaystyle\huge\frac{n+2-1}{(n+2)!}=\frac{n+2}{(n+2)!}-\frac{1}{(n+2)!}$
$\displaystyle\huge\frac{n+2}{(n+2)(n+1)!}-\frac{1}{(n+2)!}=\frac{1}{(n+1)!}-\frac{1}{(n+2)!}$
Use this to form the telescope.
4. Thanks very much for the kind help.
May I know how and what is the idea which you employ in your solution making it so neat and simply. Is the f(r) unique and is there a general method to find f(r) ?.
Lastly is it possible to find a g(r) such that the nth term= g(r)-g(r-1) allowing us to find the sum?
5. If you are told to evaluate an infinite sum you should either change it into a known form of a function or you should make it a telescoping sum. Now to make it a telescoping sum you need to look for the following things:
- an obvious way to apply partial fractions decomposition
- add and subtract something in the numerator so that part of it looks like something that we can cancel on the bottom.
That is usually a good start. There is a lot of intuition involved in some examples and less in others.
6. Thanks very much for the reply.
Can I know how to find g(r) such that the nth term= g(r)-g(r-1) thus allowing us to find the sum of the above 3 problem.
How to apply telescope method in question 3?
I cannot wrap around my head to get started in problem3
Thanks
7. 3)
The sum can be written as...
$\displaystyle \sum_{k=1}^{n} \frac{k}{k+5} = n - 5\ \sum_{k=1}^{n} \frac{1}{k+5}$ (1)
In order to solve the second term of (1) we have to introduce the function digamma, defined as...
$\displaystyle \psi(x)= \sum_{j=1}^{\infty} \frac{x}{j\ (j+x)} - \gamma$ (2)
... where $\gamma$ is the 'Euler's constant'. From (2) You can derive the following basic identity...
$\displaystyle \psi(1+x) - \psi (x) = \frac{1}{1+x}$ (3)
Now setting in (3) $x=k+4$ You obtain...
$\displaystyle \psi(k+5) - \psi (k+4) = \frac{1}{k+5}$ (4)
... so that is...
$\displaystyle \sum_{k=1}^{n} \frac{1}{k+5} = \sum_{k=1}^{n} \{ \psi(k+5) - \psi (k+4) \} = \psi (n+5) - \psi(5)$ (5)
... and finally You can write...
$\displaystyle \sum_{k=1}^{n} \frac{k}{k+5} = n - 5\ \psi(n+5) + 5\ \psi(5)$ (6)
Digamma function - Wikipedia, the free encyclopedia
Kind regards
$\chi$ $\sigma$
8. That's pretty cool, chisigma, but I think there is an even simpler answer.
$\displaystyle\sum_{k=1}^n \frac{k}{k+5} = \sum_{k=1}^n \frac{k+5-5}{k+5} = \sum_{k=1}^n \left(\frac{k+5}{k+5}-\frac{5}{k+5}\right)$
$\displaystyle = \left(\sum_{k=1}^n 1\right)-5\left(\sum_{k=1}^n \frac{1}{k+5}\right) = n - 5\sum_{k=6}^{n+5} \frac{1}{k+5} = n - 5\left(H_n - \frac{1}{1}-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\right) = \frac{137}{12}+n-5H_n$
where [LaTeX ERROR: Convert failed] is the n-th Harmonic number. There is no simpler way to write this sum. | 2015-10-09T09:28:23 | {
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https://math.stackexchange.com/questions/1619029/what-is-the-probability-that-the-max-of-n-numbers-from-some-distribution-is-a | What is the probability that the max of $n$ numbers from some distribution is a greater than another number from the same distribution?
Independently choose $n$ numbers $X_1, X_2, \ldots, X_n$ from some distribution. And Independently choose another number $X_{n+1}$ from the same distribution. Let $P(\max(X_1, X_2, \ldots,,X_n) > X_{n+1}))$ means the probability that the maximum of $X_1, X_2, \ldots, X_n$ is bigger than $X_{n+1}$. What is $P(\max(X_1, X_2, \ldots,X_n) > X_{n+1}))$? I think that $P(\max(X_1, X_2, \ldots,X_n) > X_{n+1})) = \frac{n}{n+1}$. Here is my proof.
The case that any one of $X_1, X_2, \ldots X_n, X_{n+1}$ is the maximum is symmetric. So $P(X_i\text{ is the maximum)} = \frac{1}{n+1}, (i = 1,\ldots,n+1)$.
$P(\max(X_1, X_2, \ldots,X_n) > X_{n+1})) = P(X_i \text{ is the maximum}, i =1,\ldots,n)= \sum_{i=1}^{n}P(X_i \text{ is the maximum}) = n \cdot \frac{1}{n+1} = \frac{n}{n+1}$
But I think that my proof is not rigorous. https://math.stackexchange.com/a/20841/16625 gives a rigorous proof for the $n=1$ case. Can anyone give a rigorous proof for the general case?
• You should get $\frac{n}{n+1}$. $\frac{1}{n+1}$ is the probability that $\max(X_i)<X_{n+1}$. – Thomas Andrews Jan 20 '16 at 3:53
• This only works for continuous random variables, of course. For instance, if the $X_i$ are die rolls, you will get a very different result. – Thomas Andrews Jan 20 '16 at 3:58
• @ThomasAndrews Yes. For discrete random variables, the case that random variables are equal can not be ignored. – Jingguo Yao Jan 21 '16 at 7:21
Probably easier to compute the probability that the max is less. Specifically, $P(\max(X_i)<x)=P(X<x)^n$. So if $p$ is the CDF for $X$ then the probability is:
$$\int p(x)^n\,dp(x) = \frac{1}{n+1}p(x)^{n+1}\Bigg{|}_{-\infty}^{\infty}=\frac{1}{n+1}$$
This is sort of obvious - the probability that the last one is the largest is the same as the probability that any other is the largest. (If the pdf is discontinuous, it is possible, with non-zero probability, for two to be equal, of course. So we assume continuous.)
So the probability that you were seeking is $\frac{n}{n+1}$.
Here is a slightly more formal argument. Suppose $X_1, X_2, \ldots,$ are independent and identically distributed random variables. Then the maximum order statistic $X_{(n)}$ of a sample of size $n$ has CDF $$F_{X_{(n)}}(x) = \Pr[X_{(n)} \le x] = \prod_{i=1}^n \Pr[X_i \le x] = F_X(x)^n.$$ Suppose we are now interested in the random variable $Y = X_{(n)} - X_{n+1}$, and in particular, $$\Pr[Y > 0] = 1 - \Pr[Y \le 0] = 1- \int_{x = -\infty}^\infty F_{X_{(n)}}(x) f_X(x) \, dx = 1 - \int_{x = -\infty}^\infty F_X(x)^n f_X(x) \, dx.$$ With the substitution $$u = F_X(x), \quad du = f_X(x) \, dx, \quad F_X(-\infty) = 0, \quad F_X(\infty) = 1,$$ we arrive at $$\Pr[Y > 0] = 1 - \int_{u=0}^1 u^n \, du = 1 - \frac{1}{1+n} = \frac{n}{n+1},$$ as claimed.
Assuming that the $P(X_i=X_j)=0$ for $i\ne j$, $$P(\max(X_1,X_2,\dots,X_n)\gt X_{n+1})$$ is the probability that $X_{n+1}$ is not the greatest of the $n+1$ samples. The probability that it is the greatest of $n+1$ samples is $$P(\max(X_1,X_2,\dots,X_n)\le X_{n+1})=\frac{n!}{(n+1)!}=\frac1{n+1}$$ because there are $n!$ arrangements where $X_{n+1}$ is greatest and $(n+1)!$ arrangements altogether.
Therefore, $$P(\max(X_1,X_2,\dots,X_n)\gt X_{n+1})=1-\frac1{n+1}=\frac n{n+1}$$
Your argument is correct if the $n+1$ observations are independent, an assumption that you did not state. | 2019-06-24T17:48:18 | {
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https://practicaldev-herokuapp-com.global.ssl.fastly.net/thepracticaldev/daily-challenge-230-beeramid-2m4a | ## DEV Community is a community of 662,276 amazing developers
We're a place where coders share, stay up-to-date and grow their careers.
# Daily Challenge #230 - Beeramid
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The hardworking team behind dev.to ❤️
Let's pretend your company just hired your friend from college and paid you a referral bonus. Awesome! To celebrate, you're taking your team out to the terrible dive bar next door and using the referral bonus to buy, and build, the largest three-dimensional beer can pyramid you can. And then probably drink those beers, because let's pretend it's Friday too.
A beer can pyramid will square the number of cans in each level - 1 can in the top level, 4 in the second, 9 in the next, 16, 25...
Complete the beeramid function to return the number of complete levels of a beer can pyramid you can make, given the parameters of:
2) the price of a beer can
For example:
beeramid(1500, 2); // should === 12
beeramid(5000, 3); // should === 16
Tests:
beeramid(9, 2)
beeramid(21, 1.5)
beeramid(-1, 4)
This challenge comes from kylehill on CodeWars. Thank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!
Want to propose a challenge idea for a future post? Email [email protected] with your suggestions!
## Discussion (6)
Here's my C solution,
Notice that there is no need for any iteration because (taking number of cans = N and number of levels = L)
$N = 1 + 2^2 + 3^2 + ... + L^2 = L(L+1)(2L+1)/6 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \newline \implies L^3/3 < N < (L+1)^3/3 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \newline \implies L < (3N)^{1/3} < L+1 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \newline \implies L = \lfloor (3N)^{1/3} \rfloor \quad or \quad L = \lfloor (3N)^{1/3} \rfloor - 1 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad$
int beeramid(double bonus, double price) {
if (bonus < price) return 0;
int cans = floor(bonus / price);
int levels = floor(cbrt(3.0*cans));
return levels*(levels + 1)*(2*levels + 1) > 6*cans ? levels - 1 : levels;
}
Vidit Sarkar
Here is a Python Solution,
def beeramid(bonus, price):
canCount = int(bonus/price)
level = 0
while((level+1)**2 <= canCount):
level += 1
canCount -= level*level
return level
Test Cases,
print(beeramid(1500, 2)) # output -> 12
print(beeramid(5000, 3)) # output -> 16
print(beeramid(9, 2)) # output -> 1
print(beeramid(21, 1.5)) # output -> 3
print(beeramid(-1, 4)) # output -> 0
Michael Kohl • Edited
Ruby, using a lazy, infinite enumerator:
ROWS = (1..).lazy.map { |n| n * n }
def beeramid(bonus, price)
cans = bonus / price
ROWS.take_while { |n| (cans -= n) >= 0 }.count
end
Valts Liepiņš • Edited
Haskell solution using lazy, infinite list of cans:
import Data.List (findIndex)
beeramid :: Double -> Double -> Int
beeramid m p =
case findIndex (> cans) levels of
Just level -> level
Nothing -> 0
where
cans = m / p
levels = scanl1 (+) . fmap (^2) \$ [1..]
Paula Gearon • Edited
Clojure. Not very terse, but it gets a solution in constant time
(defn faul [n]
(let [n2 (* n n)
n3 (* n2 n)]
(/ (+ n3 n3 n2 n2 n2 n) 6)))
(defn layers [n]
(let [est (int (Math/pow (* 3 n) (/ 1 3)))]
(first (filter #(>= n (faul %)) (range est 0 -1)))))
(defn beeramid [bonus price] (or (layers (/ bonus price)) 0))
Testing:
=> (beeramid 1500 2)
12
=> (beeramid 5000 3)
16
=> (beeramid 9 2)
1
=> (beeramid 21 1.5)
3
=> (beeramid -1 4)
0
Paula Gearon
To follow up on my solution, I borrowed from the Wikipedia article on Square Pyramidal Numbers. This provides a formula for the sum of squares as a special case of Faulhaber's formula:
$P_n = \sum_{k=1}^n k^2 = \frac {2n^3 + 3n^2 + n} 6$
So if we have n layers, then that will have a total of Pn cups.
As n becomes large then: 2n3 ≫ 3n2 + n
$P_n = \frac {2n^3} 6 + \Big( \frac {3n^2 + n} 6 \Big) \newline P_n > \frac {n^3} 3 \newline$
Since the layers is n and the cups in the pyramid is Pn then this gives a bound of:
$layers < \sqrt[3]{3 \cdot cups}$
Consequently, my code creates an estimate of the number of layers using this formula, and counts down from there applying the Faulhaber formula to check if the result is less than or equal to the provided number of cups.
Using this method, a small numbers of cups will require 2 guesses, but as the number of cups gets higher (and n3n2) then it is more likely to get it on the first guess. I told it to keep testing while the number of layers until 1, but it never actually gets beyond 2 attempts. | 2021-07-26T21:18:50 | {
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http://mathhelpforum.com/calculus/143022-definite-integral.html | # Math Help - definite integral
1. ## definite integral
a little confused on this one
"solve the definite integral"
$\int^2_0 2x^2e^{x^3+5}$
2. Originally Posted by JGaraffa
a little confused on this one
"solve the definite integral"
$\int^2_0 2x^2e^{x^3+5}$
Well, again, use substitution $z := x^3+5$, which gives that $dz=3x^2\,dx$ and hence you have
$\int 2x^2e^{x^3+5}\,dx=\tfrac{2}{3}\cdot\int e^{x^3+5}\cdot 3x^2\, dx=\tfrac{2}{3}\cdot\int z\,dz=\ldots$
3. Originally Posted by JGaraffa
a little confused on this one
"solve the definite integral"
$\int^2_0 2x^2e^{x^3+5}$
What exactly are you confused about? Have you made an attempt at it?
Also the integral should be $\int^2_0 2x^2e^{x^3+5}dx$
I could tell you how to do it but i think i it's only fair that an attempt is made first
Hint: use substitution
4. ah perfect! i couldn't figure out how i could change that $2x^2$ to a $3x^2$ . thanks!
5. actually, i'm getting really high numbers for this which makes me think i may be doing it wrong.
$\int^2_0 2x^2e^{x^3+5}$
u= $x^3 + 5$
h(u) = $3x^2$
$\frac{2}{3} \int^2_0 3x^2 e^{x^3+5}$
when x = 0, u = 5
when x = 2, u = 13
$\frac{2}{3} \int^{13}_5 e^u du$
$
\frac{2}{3} e^{13} - \frac{2}{3} e^5$
i'm getting 294,843.32 which makes me think i'm doing something wrong.
6. Originally Posted by JGaraffa
actually, i'm getting really high numbers for this which makes me think i may be doing it wrong.
$\int^2_0 2x^2e^{x^3+5}$
u= $x^3 + 5$
h(u) = $3x^2$
$\frac{2}{3} \int^2_0 3x^2 e^{x^3+5}$
when x = 0, u = 5
when x = 2, u = 13
$\frac{2}{3} \int^{13}_5 e^u du$
$
\frac{2}{3} e^{13} - \frac{2}{3} e^5$
i'm getting 294,843.32 which makes me think i'm doing something wrong.
No, that result is ok. Of course, $e^{x^3+5}$ is a fast growing function.
7. Originally Posted by JGaraffa
actually, i'm getting really high numbers for this which makes me think i may be doing it wrong.
$\int^2_0 2x^2e^{x^3+5}$
u= $x^3 + 5$
h(u) = $3x^2$
$\frac{2}{3} \int^2_0 3x^{2} e^{x^3+5}$
when x = 0, u = 5
when x = 2, u = 13
$\frac{2}{3} \int^{13}_5 e^u du$
$
\frac{2}{3} e^{13} - \frac{2}{3} e^5$
i'm getting 294,843.32 which makes me think i'm doing something wrong.
Apart from [tex]3x^{2} it lookes fine to me.
8. Originally Posted by JGaraffa
a little confused on this one
"solve the definite integral"
$\int^2_0 2x^2e^{x^3+5}$
Again observe that $3x^2$ is the derivative of $x^3+5$, so consider:
$\frac{d}{dx}e^{x^3+5}$
CB | 2015-01-31T14:28:49 | {
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https://math.stackexchange.com/questions/1005422/evaluating-int-02-tan-1-pi-x-tan-1-x-mathrmdx/3052525 | # Evaluating $\int_0^2(\tan^{-1}(\pi x)-\tan^{-1} x)\,\mathrm{d}x$
Hint given: Write the integrand as an integral.
I'm supposed to do this as double integration.
My attempt:
$$\int_0^2 [\tan^{-1}y]^{\pi x}_{x}$$
$$= \int_0^2 \int_x^{\pi x} \frac { \mathrm{d}y \mathrm{d}x} {y^2+1}$$
$$= \int_2^{2\pi} \int_{\frac{y}{\pi}}^2 \frac { \mathrm{d}x \mathrm{d}y} {y^2+1}$$
$$= \int_2^{2\pi} \frac { [x]^2_{\frac{y}{\pi}} \mathrm{d}y } { y^2+1}$$
$$= \int_2^{2\pi} \frac { 2- {\frac{y}{\pi}} \mathrm{d}y } { y^2+1}$$
Carrying out this integration, I got, $$2[\tan^{-1} 2 \pi - \tan^{-1} 2] - \frac {1}{2 \pi} [\ln \frac {1+4 {\pi}^2} {5}]$$
But I'm supposed to get $$2[\tan^{-1} 2 \pi - \tan^{-1} 2] - \frac {1}{2 \pi} [\ln \frac {1+4 {\pi}^2} {5}]+ [\frac {\pi-1}{2 \pi}] \ln 5$$
Can someone please explain where I'm wrong? I've failed to pinpoint my mistake. Thank you.
• At third line it seems to me the integral must be $$\int_0^{2\pi}\int_{\frac{y}{\pi}}^{y}\frac{\operatorname d\!x \operatorname d\!y}{y^2+1}$$ – Ángel Mario Gallegos Nov 4 '14 at 6:13
• but that is taking my answer even further away from what I require. – Diya Nov 4 '14 at 6:50
• I got it! You were right! The integral is divided into two regions. One is the one I used, and the other is the one you mentioned. Thank you so much! :) – Diya Nov 4 '14 at 6:57
Use the change of variables $y = xt.$
$$I = \int_0^2 \int_x^{\pi x} \frac { \mathrm{d}y \, \mathrm{d}x} {y^2+1}\\=\int_0^2 \int_1^{\pi } \frac { x} {x^2t^2+1}\mathrm{d}t \, \mathrm{d}x\\=\int_1^{\pi} \int_0^{2 } \frac { x} {x^2t^2+1}\mathrm{d}x \, \mathrm{d}t\\=\int_1^{\pi} \frac{\ln(1+4t^2)}{2t^2} \mathrm{d}t$$
Now use integration by parts.
$$I = -\left.\frac{\ln(1+4t^2)}{2t}\right|_1^{\pi}+4\int_1^{\pi}\frac1{1+4t^2} \, dt$$
A more straight forward approach uses integration by parts.
Define: \begin{align} & I(c)=\int_{a}^{b}dx(1 \times \arctan{c x})=\int_{ac}^{bc}\frac{dy}{c} (1 \times\arctan{ y})=\\&\frac{1}{c}y \arctan(y)|_{ac}^{bc}-\frac{1}{2c}\int_{ac}^{bc}\frac{y}{1+y^2} \end{align}
using partial fraction this reads: \begin{align} I(c)=\frac{1}{c}y \arctan(y)|_{ac}^{bc}-\frac{1}{2c}\log(1+y^2)|_{ac}^{bc} \end{align}
taking $I(\pi)-I(0)$ with $a=0$ and $b=2$ we are done
For the sake of an alternative approach, recall the formula for the integral of an inverse function $$\int f^{-1}(x)\mathrm{d}x=xf^{-1}(x)-F\circ f^{-1}(x)$$ Where $$F'(x)=f(x)$$. Plugging in $$f(x)=\tan(x)$$, $$I=\int\arctan(x)\mathrm{d}x=x\arctan(x)-\int_0^{\arctan(x)}\tan(t)\mathrm{d}t$$ Then recall that $$(-\log|\cos(x)|)'=\tan(x)$$ So we have $$I=x\arctan(x)+\log|\cos(\arctan(x))|$$ then using trig, $$I=x\arctan(x)-\frac12\log(x^2+1)$$ So $$I_1=\int_0^2\arctan(x)\mathrm{d}x=2\arctan2-\frac12\log5$$ And \begin{align} I_2=&\int_0^2\arctan(\pi x)\mathrm{d}x\\ =&\frac1\pi\int_0^{2\pi}\arctan(x)\mathrm{d}x\\ =&\frac1\pi\bigg(2\pi\arctan2\pi-\frac12\log(4\pi^2+1)\bigg)\\ =&2\arctan2\pi-\frac1{2\pi}\log(4\pi^2+1)\\ \end{align} So \begin{align} \int_0^2(\arctan\pi x-\arctan x)\mathrm{d}x=&I_2-I_1\\ =&2\arctan2\pi-\frac1{2\pi}\log(4\pi^2+1)-2\arctan2+\frac12\log5\\ =&2\arctan2\pi-\frac1{2}\log\sqrt[\pi]{4\pi^2+1}-2\arctan2+\frac12\log5\\ =&2(\arctan2\pi-\arctan2)+\frac12\log\frac{5}{\sqrt[\pi]{4\pi^2+1}}\\ \end{align} | 2019-07-18T20:23:31 | {
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https://math.stackexchange.com/questions/3401659/the-number-of-different-triangles-and-each-side-has-a-different-length | # The number of different triangles and each side has a different length
I'm trying to solve the following problem:
What is the number of different triangles we can form from numbers $$4,5,6,7,8,9$$ (lengths of sides), where every side has a different length (for an example $$4,5,6$$ or $$4,5,7$$)..
My solution is the following:
The number of all possible permutations is $$\frac{6!}{(6-3)!}$$. We subtract the numbers which don't make a triangle (which is $$6*2$$) and then divide by $$2$$, because $$(4,5,6)$$ makes the same triangle is $$6,5,4$$. My answer is $$54$$.
However, the correct answer should be $$53$$. Can anyone tell me where I did a mistake?
Thanks
• $6 \choose 3$ is only $20$ and there is one choice $(4,5,9)$ that doesn't make a triangle, so I get $19$ Your expression (ignoring the factorial in the number to choose) gives $8$. Oct 20, 2019 at 16:42
• I think you'll have to show more details for someone to point out your mistake. What do you mean by $\binom{6}{(6-3)!}$? It look to me like ${6\choose6}=1$. Did you mean to say ${6\choose3}?$ Also, why do you say there are $12$ where the triangle inequality doesn't hold. How did you get that? Oct 20, 2019 at 16:43
• Sorry, I made a mistake when writing my answer here. Please wait a minute and I will fix it Oct 20, 2019 at 16:44
• If you write out all the triples $(4,5,6), (4,5,7), \ldots, (7,8,9)$, you'll see that there are only $20$ in the list, one of which is degenerate. How can there be $53$ or $54$ triangles? We could consider "right-handed" and "left-handed" triangles to be different, effectively doubling the number of triangles, but even then, we would have at most $40$, still falling short of the given answer. Have you written out the problem exactly as it was given to you? Oct 20, 2019 at 16:56
• You should have gotten six crossed off, as you describe. 2 for 4, 2 for 5 and 2 for 9 (not 6) makes six. Why do you double that? Then you should divide by $6$ because 456 comes six ways-456,465,546,564,654,645 not two Oct 20, 2019 at 17:04
There are $$\frac {6!}{3!}$$ ordered choices of three numbers, which is $$120$$. Six of those do not make a triangle, all the permutations of $$(4,5,9)$$, which leaves $$114$$. Each unordered triangle gives $$3!$$ permutations, so we divide by $$6$$ and get $$19$$. I don't know where a number in the $$50$$s comes from.
I think it is easier to just choose unordered combinations to start with, which is $${6 \choose 3}=20$$ and subtract the one that doesn't make a triangle. That also gets $$19$$. | 2022-07-05T09:14:41 | {
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https://math.stackexchange.com/questions/3039219/in-mathbbrn-is-the-dot-product-the-only-inner-product/3039248 | # In $\mathbb{R}^n$, is the dot product the only inner product?
Just what the title says. I'm reading, from various resources, that the inner product is a generalization of the dot product. However, the only example of inner product I can find, is still the dot product.
Does another "inner product" exist on the $$\mathbb{R}^n$$ space?
• This might be useful. – StackTD Dec 14 '18 at 11:07
• @StackTD: thanks!!! Exactly what I was looking for – blue_note Dec 14 '18 at 11:15
• – StackTD Dec 14 '18 at 11:37
• @StackTD: thanks again – blue_note Dec 14 '18 at 11:38
• As some other answers point out, there are infinitely many inner products (i.e., symmetric, positive-definite bilinear forms) on $\Bbb R^n$. But for any of them one can choose a basis of $\Bbb R^n$ with respect to which the bilinear form is the standard one: $({\bf x}, {\bf y}) = x_1 y_1 + \cdots + x_n y_n$. So up to isomorphism there is only one inner product on $\Bbb R^n$. (This is a special case of Sylvester's Law of Inertia.) – Travis Willse Dec 25 '18 at 0:51
Yes and are all in the form $$\langle x, y\rangle=x^T\cdot A\cdot y$$ where $$x^T$$ is the transpose vector of $$x$$ and $$A$$ is a $$n\times n$$ symmetric definite positive matrix.
In fact let $$\langle x, y\rangle$$ a generic inner product on $$\mathbb R^n$$ then for every $$y$$ the function $$f_y:x\rightarrow \langle x, y\rangle$$ is linear from $$\mathbb R^n$$ to $$\mathbb R$$ then exists a vector $$\alpha(y)\in\mathbb R^n$$ such that $$\langle x, y\rangle = \alpha(y)^T\cdot x$$
Observe that $$\langle x, ay+by'\rangle = a\langle x, y\rangle + b\langle x, y'\rangle\Rightarrow \alpha(ay+by')=a\alpha(y)+b\alpha(y')$$ then $$\alpha$$ is a linear operator from $$\mathbb R^n$$ in itself then exists an $$n\times n$$ matrix $$A$$ such that $$\alpha(y)=A\cdot y$$ so $$\alpha(y)^T\cdot x=y^T\cdot A^T\cdot x$$
Now remember that $$\langle x, y\rangle=\langle y, x\rangle$$ then you can easly prove that $$A^T=A$$ and $$A$$ must be symmetric.
Why now $$A$$ must be definite positive? Because $$\langle x, x\rangle\geq 0$$ and holds equality if and only if $$x=0$$. Applying it to the initial formula we obtain the definition of a definite positive matrix.
• We need $A$ to be positive definite also. – littleO Dec 14 '18 at 11:26
• Oh yes, I forgotten it! – Jihlbert Dec 14 '18 at 11:28
• Note that in some contexts, one only requires that an inner product be nondegenerate (and not necessarily positive definite). – Travis Willse Dec 25 '18 at 0:52
• If it's semidefinite positive it becomes a semi-inner product and not a inner product. – Jihlbert Dec 28 '18 at 11:30
The dot product on $$\mathcal{R}^n$$ is defined as follows:
$$(a,b) = a^i b^j (e_i,e_j) = a^i b^j \delta_{ij} = a^i b^i ,$$
where $$a,b \in \mathcal{R}^n$$ and $$e_i,e_j$$ standard basis vectors. I used Einstein summation convention here.
In general we can express $$a,b$$ in a different basis i.e. $$a=\tilde{a}^i \tilde{e}_i$$ and $$b = \tilde{b}^i \tilde{e}_i$$ so now not the standard basis but an arbitrary basis of $$\mathcal{R}^n$$ assuming still $$(,)$$ is positive-definite. This then gives:
$$(a,b) = \tilde{a}^i \tilde{b}^j (\tilde{e}_i,\tilde{e}_j) = \tilde{a}^i \tilde{b}^j A_{ij} \equiv a^T A \ b.$$
Note that $$A$$ now is not the identity matrix like in the standard inner product.
The tensor metric allows for the vector norm to remain constant under change of basis vectors, and is an example of inner product (page 15). In the simple setting of basis vectors constant in direction and magnitude from point to point in $$\mathbb R^2,$$ here is an example:
Changing basis vectors from Euclidean orthonormal basis $$\small\left\{\vec e_1=\begin{bmatrix}1\\0 \end{bmatrix},\vec e_2=\begin{bmatrix}0\\1 \end{bmatrix}\right\}$$ to
$$\left\{\vec u_1=\color{red}2\vec e_1 + \color{red}1\vec e_2,\quad \vec u_2=\color{blue}{-\frac 1 2 }\vec e_2+\color{blue}{\frac 1 4} \vec e_2\right\}\tag 1$$ would result in a different norm of the vector $$\vec v=\begin{bmatrix} 1\\1\end{bmatrix}_\text{e basis},$$ i.e $$\Vert \vec v \Vert^2=v_1^2 + v_2^2 = 2,$$ when expressed with respect to the new basis vectors.
In this regard, since vector components transform contravariantly, the change to the new coordinate system would be given by the backward transformation matrix:
$$B=\begin{bmatrix} \frac 1 4 & \frac 1 2\\-1&2\end{bmatrix}$$
i.e. the inverse of the forward transformation for the basis vectors as defined in $$(1),$$ which in matrix form corresponds to
$$F=\begin{bmatrix} \color{red} 2 & \color{blue}{-\frac 1 2}\\\color{red}1& \color{blue}{\frac 1 4}\end{bmatrix}.$$
Hence, the same vector $$\vec v$$ expressed in the new basis vectors is
$$\vec v_{\text{u basis}}=\begin{bmatrix} \frac 1 4 & \frac 1 2\\-1&2\end{bmatrix}\begin{bmatrix} 1\\1\end{bmatrix}=\begin{bmatrix} \frac 3 4\\1\end{bmatrix}.$$
And the norm would entail an inner product with the new metric tensor. In the orthonormal Euclidean basis this is simply the identity matrix. Now the matrix tensor is a $$(0,2)-\text{tensor},$$ and transforms covariantly:
$$g_{\text{ in u basis}}=(F^\top F) I= \begin{bmatrix} 2 & 1\\-\frac 1 2& \frac 1 4\end{bmatrix}\begin{bmatrix} 2 & -\frac 1 2\\1& \frac 1 4\end{bmatrix}\begin{bmatrix} 1 & 0\\0& 1\end{bmatrix}=\begin{bmatrix} 5 & -\frac 3 4\\- \frac 3 4& \frac{5}{16}\end{bmatrix}$$
The actual multiplication of basis vectors to obtain the metric tensor is explained in this presentation by @eigenchris. This metric tensor indeed renders the right norm of $$\vec v:$$
$$\begin{bmatrix}\frac 3 4 & 1\end{bmatrix}\begin{bmatrix} 5 & -\frac 3 4\\- \frac 3 4& \frac{5}{16}\end{bmatrix}\begin{bmatrix}\frac 3 4 \\ 1\end{bmatrix}=2$$
following the operations here. | 2020-06-05T11:54:14 | {
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https://math.stackexchange.com/questions/442001/proof-of-the-statement-the-product-of-4-consecutive-integers-can-be-expressed-i | # Proof of the statement “The product of 4 consecutive integers can be expressed in the form 8k for some integer k”
I am slowly diving into simple number theory and learning how to craft direct proofs. I needed to proof the statement "The product of 4 consecutive integers can be expressed in the form 8k for some integer k".
My approach was to use the quotient-remainder theorem, as it was surely intended in the textbook. The proof looks like follows.
Proof:
Suppose n is any particular but arbitrarily chose integer. We must show that n(n+1)(n+2)(n+3) is divisible by 8. By the quotient-remainder theorem, n can be written in one of the forms 4q or (4q+1) or (4q+2) or (4q+3) for some integer q. We divide into cases accordingly:
Case 1 (n = 4q for some integer q):
\begin{align} n(n+1)(n+2)(n+3) & = 4q(4q+1)(4q+1)(4q+2)(4q+3)\\ & = 8[q(32q^3+48q^2+18q+3)] \end{align}
Let $m=q(32q^3+48q^2+18q+3)$. Then m is an integer because sums and products of integers are integers. By substitution, $n(n+1)(n+2)(n+3) = 8m$ where m is an integer. Hence n(n+1)(n+2)(n+3) is divisible by 8.
Case 2 (n = (4q+1) for some integer q): \begin{align} n(n+1)(n+2)(n+3) & = (4q+1)(4q+2)(4q+3)(4q+4)(4q+5)\\ & = 8[(4q+1)(2q+1)(4q+3)(q+1)] \end{align}
Let $m=q((4q+1)(2q+1)(4q+3)(q+1))$. Then m is an integer because sums and products of integers are integers {...} {See case 1, its basically the same reasoning here...}
Case 3 (n = (4q+2) for some integer q): {...}
Case 4 (n = (4q+3) for some integer q): {...}
Conclusion:
I each of the above cases, n(n+1)(n+2)(n+3) was to be shown to be a multiple of 8. By the quotient-remainder theorem, one of these cases must occur, hence n(n+1)(n+2)(n+3) can be written in the form 8k for some integer k. q.e.d. [End Proof]
But honestly, I consider this proof somehow clumsy and too inconvenient [I am a bloody amateur, maybe I am wrong]. Are there any better/shorter ways to prove the above statement?
• Any four consecutive numbers will have one of each of the possible four remainders mod $4$. This means that one will be divisible by $4$ and another will be divisible by $2$, which means that the product is divisible by $8$. – Tobias Kildetoft Jul 12 '13 at 12:34
• Wow, this is straightforward! – Nikolai Tschacher Jul 18 '13 at 8:41
In each four consecutive integers there is exactly one pair of even integers, and exactly one of them is $\;2\pmod 4\;$ and the other one is $\,0\pmod 4\;$ , so the product of these two even (consecutive even) integers is already divisible by $\,8\,\ldots\ldots$
• Very good, sir. – Don Larynx Jan 9 '15 at 20:53
Note that if you have four consecutive integers, then one must be divisible by 4 and another must be divisible by 2: they must be equivalent to $0,1,2,\text{ and }3\pmod{4}$.
One way is by proving that the binomial coefficients $$\dbinom n r=\cfrac {n!}{r!(n-r)!}$$ are integers, whoch can be done in various ways including using the binomial recurrence from Pascal's Triangle. Then$$\binom n 4=\frac {n(n-1)(n-2)(n-3)}{4!}$$ which proves that the product of four consecutive integers is divisible by $24$.
Another way is by induction.$$(n+1)n(n-1)(n-2)-n(n-1)(n-2)(n-3)=4n(n-1)(n-2)$$
So you can prove the result using that the product of three consecutive integers is even. Or prove the result for $24$, by showing that the product of three consecutive integers is divisible by $6$ (reducing the number of factors again by taking the difference of successive terms). And remembering to prove the base case - but this can be arranged to have zero as a factor, so divisible by the integer we require.
$$(n+1)n(n-1)-n(n-1)(n-2)=3n(n-1)$$is divisible by 3 and $$(n+1)n-n(n-1)=2n$$ is divisible by 2.
• What is the reasoning of binomial coefficient proof? We have $4! \cdot {}^nC_4 = {}^n P_4$ because every combination of 4 objects in 4 places has 4! permutation. But how does this totally unrelated fact, without invoking any properties of numbers, prove a property of numbers i.e. the product of any four consecutive numbers is a multiple of 24? – user103816 Nov 7 '15 at 16:09
• @user103816 You can easily show that Pascal's Triangle encapsulates the formula $\binom nr+\binom n{r+1}=\binom {n+1}{r+1}$ and this shows that all the binomial coefficients are integers. Then suppose you have the consecutive integers $n-3, n-2, n-1, n$, you will find, as I have put, that $\binom n4$ is the product of those integers divided by $4!$ and is an integer - so the product of four successive integers is divisible by $24$. Negative numbers are not a problem, and if one of the numbers is zero, so is the product. – Mark Bennet Nov 7 '15 at 16:42
Your formal argument looks fine to me, but you're right: it does seem clumsy and inconvenient. This may be because you are making exactly the same argument in each case: Writing the four numbers, finding one factor of four and another factor of two, and pulling those out to get a factor of $8$.
This suggests a simpler argument. In any four consecutive integers, by the quotient-remainder theorem, there must be one multiple of four. There must also be two multiples of two. One of the multiples of two will also be a multiple of four; the other will not. Therefore, we may factor a four and another two from the product of the four consecutive integers and we see that the product is a multiple of eight. | 2021-01-24T18:40:53 | {
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http://math.stackexchange.com/questions/9461/find-a-pattern-in-a-recursive-definition | # Find a pattern in a recursive definition
A set M $\in$ $\mathbb{Z}^2$ is defined by
(Rule 1) $(0, 1)$ $\in$ M
(Rule 2) If $(x, y)$ $\in$ M, then $(x + 1, y + 2x + 3)$ $\in$ M
(a) (2 points) Starting with $(0, 1)$, write out the six pairs with the smallest first coordinates.
(b) (2 points) State the (simple) relationship that holds between the first and second coordinates of all pairs in M.
Ok for a I got: $(0,1)$,$(1,4)$,$(2,9)$,$(3,16)$,$(4,25)$,$(5,36)$,$(6,49)$
However for b, I cant see the relationship. Can anyone give me any hints?
EDIT:
Ok I found the relation: $y=(x+1)^2$. However, I must now prove this claim using structural induction. I can get up to the inductive hypothesis, but after that I'm not really sure what to do.
Base case: $(0,1), 1=(0+1)^2 = 1$
Inductive Hypothesis: For any element $x,y$ if $x\in M$ then $y=(x+1)^2$. We must show that $y+1=(x+1)^2 + 1$
I'm not really sure that my IH is correct either.
-
It appears you are using the new x in the calculation of y. I believe you should use the old x. So the second pair should be (1,4). All will become clearer. – Ross Millikan Nov 8 '10 at 18:21
Ahh ok yes you are right..now I see the relation, thanks! – maq Nov 8 '10 at 18:26
@Ross please see edits – maq Nov 8 '10 at 18:41
Your inductive step is to show that if $(x,(x+1)^2) \in M$, then $(x+1,(x+2)^2) \in M$. So apply Rule 2 to $(x,(x+1)^2)$
-
Wait x, x^2+1 is an element of M? I thought it was x, (x+1)^2 is an element of M – maq Nov 8 '10 at 18:56
Right you are. Corrected. – Ross Millikan Nov 8 '10 at 19:04
Thanks, I was able to get the rest.. – maq Nov 8 '10 at 19:06
Your inductive hypothesis is wrong. You should be showing that if $(x,(x+1)^2)\in M$, then $(x+1, ((x+1)+1)^2) \in M$. From there you should be able to see how to get the result (just use Rule 2).
-
HINT $\rm\ \ f\ (x,\:y)\ = \ (x+1,\ y+2x+3)\ \Rightarrow\ f\ (x,\ (x+1)^2)\ =\ (x+1,\ (x+2)^2)$
- | 2014-12-21T01:55:17 | {
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https://mathhelpboards.com/threads/state-the-domain.8090/ | # State the domain
#### shamieh
##### Active member
state the domain
$$\displaystyle y = x\sqrt{1 - x^2}$$
so
$$\displaystyle 1 - x^2 >= 0$$
$$\displaystyle -x^2 >= -1$$
$$\displaystyle x^2 <= 1$$
$$\displaystyle d = {x <= (+)(-) 1}$$ or should I say $$\displaystyle d = {-1, 1}$$
which one is correct?
#### MarkFL
Staff member
I have moved this topic here to our Pre-Calculus subforum as questions on function domains does not involve the calculus. I suspect that this question is part of a calculus question though, or may even come from a calculus textbook, but when choosing the subforum in which to post, the nature of the question itself, rather than from where it originates should be considered firstly.
You are correct in stating:
$$\displaystyle 1-x^2\ge0$$
What I would suggest doing next is factoring the left side:
$$\displaystyle (1+x)(1-x)\ge0$$
Now, determine the critical values, plot them on a number line and test the 3 resulting intervals.
A quicker method would be to plot the expression (the original radicand), and observe where it is non-negative.
What do you find?
#### soroban
##### Well-known member
Hello, shamieh!
State the domain: .$$y \:=\: x\sqrt{1 - x^2}$$
So: .$$1 - x^2 \:\ge\:0$$
. . . . . $$-x^2 \:\ge\:-1$$
. . . . . . .$$x^2\:\le\:1$$
$$\begin{array}{cc}\text{I would write:} & |x| \:\le\:1 \\ & \text{or} \\ & \text{-}1 \:\le\:x\:\le\:1 \\ & \text{or} \\ & [\text{-}1,\,1]\end{array}$$
#### HallsofIvy
##### Well-known member
MHB Math Helper
state the domain
$$\displaystyle y = x\sqrt{1 - x^2}$$
so
$$\displaystyle 1 - x^2 >= 0$$
$$\displaystyle -x^2 >= -1$$
$$\displaystyle x^2 <= 1$$
$$\displaystyle d = {x <= (+)(-) 1}$$ or should I say $$\displaystyle d = {-1, 1}$$
which one is correct?
NOT $$\displaystyle \{-1, 1\}$$ because that means the numbers -1 and 1 only, not the numbers between. You could say $$\displaystyle \{x| -1\le x\le 1\}$$ or [-1, 1] as MarkFL said.
("math" uses the "{" and "}" and as separators itself. To get "{" or "}" in the actual message, use "\{" and "\}".)
#### Deveno
##### Well-known member
MHB Math Scholar
A naive approach:
We know that there's "something special" about the points $x = -1, x = 1$. So let's do this:
We'll split the real number line into 5 parts:
Part 1: all numbers less than -1
Part 2: -1
Part 3: all numbers between -1 and 1
Part 4: 1
Part 5: all numbers greater than 1.
Now let's pick numbers in each part, to see what happens:
The 5 real numbers I will use are:
Part 1: -4
Part 2: -1 (no choice, here)
Part 3: 1/2
Part 4: 1 (again, no other choice)
Part 5: 6
Now we will look at $f(x_0)$ for $x_0$ being each one of these 5 numbers:
$f(-4) = 4\sqrt{1 - (-4)^2} = 4\sqrt{-15} = \text{bad}$ (undefined)
$f(-1) = (-1)\sqrt{1 - (-1)^2} = (-1)\sqrt{1 - 1} = (-1)\sqrt{0} = (-1)(0) = 0$ (OK!!!!)
$f(\frac{1}{2}) = \frac{1}{2}\sqrt{1 - (\frac{1}{2})^2} = \frac{1}{2}\sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{4}$ (OK!!!)
$f(1) = (1)\sqrt{1 - 1^2} = (1)(0) = 0$ (OK again!!!!)
$f(6) = 6\sqrt{1 - 6^2} = 6\sqrt{-35} = ???$ (not so good).
So it looks as if what we want is parts 2,3, and 4, and NOT parts 1 and 5. This is:
$\{-1\} \cup (-1,1) \cup \{1\} = [-1,1]$
If you prefer, you can write this as:
$\text{dom}(f) = \{x \in \Bbb R: |x| \leq 1\}$
the absolute value of $x$, written $|x|$ is just another way of saying:
$\sqrt{x^2}$, if one understand square roots as always being non-negative. So if:
$x^2 \leq 1$, then
$\sqrt{x^2} \leq \sqrt{1} = 1$
so:
$|x| \leq 1$. | 2021-03-09T10:46:47 | {
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https://math.stackexchange.com/questions/2780138/there-are-4-cups-of-liquid-three-are-water-and-one-is-poison-if-you-were-to-dr/2783140 | # There are 4 cups of liquid. Three are water and one is poison. If you were to drink 3 of the 4 cups, what is the probability of being poisoned?
In Season 5 Episode 16 of Agents of Shield, one of the characters decides to prove she can't die by pouring three glasses of water and one of poison; she then randomly drinks three of the four cups. I was wondering how to compute the probability of her drinking the one with poison.
I thought to label the four cups $\alpha, \beta, \gamma, \delta$ with events
• $A = \{\alpha \text{ is water}\}, \ a = \{\alpha \text{ is poison}\}$
• $B = \{\beta \text{ is water}\},\ b = \{\beta \text{ is poison}\}$
• $C = \{\gamma \text{ is water}\},\ c = \{\gamma \text{ is poison}\}$
• $D = \{\delta \text{ is water}\},\ d = \{\delta \text{ is poison}\}$
If she were to drink in order, then I would calculate $P(a) = {1}/{4}$. Next $$P(b|A) = \frac{P(A|b)P(b)}{P(A)}$$ Next $P(c|A \cap B)$, which I'm not completely sure how to calculate.
My doubt is that I shouldn't order the cups because that assumes $\delta$ is the poisoned cup. I am also unsure how I would calculate the conditional probabilities (I know about Bayes theorem, I mean more what numbers to put in the particular case). Thank you for you help.
• Comments are not for extended discussion; this conversation has been moved to chat. – Jyrki Lahtonen May 16 '18 at 5:01
• You need to provide a little more information to actually answer this question correctly. Is the poison fast acting? If the poison is fast-acting, then being available to drink the third cup is added information that the first two cups were not poison. – jerrylagrou May 17 '18 at 16:27
• Possible duplicate of Probability of being poisoned – Xander Henderson May 17 '18 at 19:12
• @EricTowers not to mention the inference from Murphy's Law, which tells you that the probability is 1. – Michael Kay May 18 '18 at 7:13
• As someone really bad at math, gonna ask a dumb question here: Why is the probability not just 75%? – john doe May 19 '18 at 13:41
NicNic8 has provided a nice intuitive answer to the question.
Here are three alternative methods. In the first, we solve the problem directly by considering which cups are selected if she is poisoned. In the second, we solve the problem indirectly by considering the order in which the cups are selected if she is not poisoned. In the third, we add the probabilities that she was poisoned with the first cup, second cup, or third cup.
Method 1: We use the hypergeometric distribution.
There are $\binom{4}{3}$ ways to select three of the four cups. Of these, the person selecting the cups is poisoned if she selects the poisoned cup and two of the three cups of water, which can be done in $\binom{1}{1}\binom{3}{2}$ ways. Hence, the probability that she is poisoned is $$\Pr(\text{poisoned}) = \frac{\binom{1}{1}\binom{3}{2}}{\binom{4}{3}} = \frac{1 \cdot 3}{4} = \frac{3}{4}$$
Method 2: We subtract the probability that she is not poisoned from $1$.
The probability that the first cup she selects is not poisoned is $3/4$ since three of the four cups do not contain poison. If the first cup she selects is not poisoned, the probability that the second cup she selects is not poisoned is $2/3$ since two of the three remaining cups do not contain poison. If both of the first two cups she selects are not poisoned, the probability that the third cup she selects is also not poisoned is $1/2$ since one of the two remaining cups is not poisoned. Hence, the probability that she is not poisoned if she drinks three of the four cups is $$\Pr(\text{not poisoned}) = \frac{3}{4} \cdot \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{4}$$ Hence, the probability that she is poisoned is $$\Pr(\text{poisoned}) = 1 - \Pr(\text{not poisoned}) = 1 - \frac{1}{4} = \frac{3}{4}$$
Addendum: We can relate this method to the first method by using the hypergeometric distribution.
She is not poisoned if she selects all three cups which do not contain poison when selecting three of the four cups. Hence, the probability that she is not poisoned is $$\Pr(\text{not poisoned}) = \frac{\dbinom{3}{3}}{\dbinom{4}{3}} = \frac{1}{4}$$ so the probability she is poisoned is $$\Pr(\text{poisoned}) = 1 - \frac{\dbinom{3}{3}}{\dbinom{4}{3}} = 1 - \frac{1}{4} = \frac{3}{4}$$
Method 3: We calculate the probability that the person is poisoned by adding the probabilities that she is poisoned with the first cup, the second cup, and the third cup.
Let $P_k$ denote the event that she is poisoned with the $k$th cup.
Since there are four cups, of which just one contains poison, the probability that she is poisoned with her first cup is $$\Pr(P_1) = \frac{1}{4}$$
To be poisoned with the second cup, she must not have been poisoned with the first cup and then be poisoned with the second cup. The probability that she is not poisoned with the first cup is $\Pr(P_1^C) = 1 - 1/4 = 3/4$. If she is not poisoned with the first cup, there are three cups remaining of which one is poisoned, so the probability that she is poisoned with the second cup if she is not poisoned with the first is $\Pr(P_2 \mid P_1^C) = 1/3$. Hence, the probability that she is poisoned with the second cup is $$\Pr(P_2) = \Pr(P_2 \mid P_1^C)\Pr(P_1) = \frac{3}{4} \cdot \frac{1}{3} = \frac{1}{4}$$
To be poisoned with the third cup, she must not have been poisoned with the first two cups and then be poisoned with the third cup. The probability that she is not poisoned with the first cup is $\Pr(P_1^C) = 3/4$. The probability that she is not poisoned with the second cup given that she was not poisoned with the first is $\Pr(P_2^C \mid P_1^C) = 1 - \Pr(P_2 \mid P_1^C) = 1 - 1/3 = 2/3$. If neither of the first two cups she drank was poisoned, two cups are left, one of which is poisoned, so the probability that the third cup she drinks is poisoned given that the first two were not is $\Pr(P_3 \mid P_1^C \cap P_2^C) = 1/2$. Hence, the probability that she is poisoned with the third cup is $$\Pr(P_3) = \Pr(P_3 \mid P_1^C \cap P_2^C)\Pr(P_2^C \mid P_1^C)\Pr(P_1^C) = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} = \frac{1}{4}$$ Hence, the probability that she is poisoned is $$\Pr(\text{poisoned}) = \Pr(P_1) + \Pr(P_2) + \Pr(P_3) = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4}$$
• I was attempting to use Method 3 (which you've shown is the most tedious). Thank you for your very detailed answer, it helped me view the problem in different ways and understand it more clearly. Thank you. – Eli May 14 '18 at 13:37
• To be honest, Methods 1 and 3 seem like a distraction from Method 2, the only really reasonable way of calculating this. – jwg May 16 '18 at 8:51
• Both of the other methods are eminently reasonable. – Matthew Read May 17 '18 at 3:37
• Why is it that in method 3 it is necessary to include the probability that the preceding cups were not poisoned, but in method 2 the preceding cups are not necessary? Subtleties like that are what make probability both difficult and fascinating IMHO. – Jim W says reinstate Monica May 17 '18 at 20:43
• @JimW In method 2, the probability that the preceding cups were not poisoned is taken into account. The probability that the first cup is not poisoned is $3/4$. Observe that the probability that the second cup is not poisoned given that the first cup was not poisoned is $2/3$ since if the first cup was not poisoned only two of the three cups that remain are not poisoned. The probability that the third cup is not poisoned given that the first two cups were not poisoned is $1/2$ since if the first two cups were not poisoned then only one of the two cups that remain is not poisoned. – N. F. Taussig May 17 '18 at 20:52
The probability of not being poisoned is exactly the same as the following problem:
You choose one cup and drink from the other three. What is the probability of choosing the poisoned cup (and not being poisoned)? That probability is 1/4.
Therefore, the probability of being poisoned is 3/4.
• @NicNic8 This version is clearer. – N. F. Taussig May 14 '18 at 2:30
• No idea why this isn't the accepted answer - it's the way I'd do it but it's also much clearer than the other. – arboviral May 16 '18 at 8:07
• @arboviral The other answer has better math, and can be applied generally. Consider instead that there was a fifth glass containing poison; which approach would you use? – Drunk Cynic May 16 '18 at 16:16
• Think of it this way: What is the prob that the poison cup is still full after drinking three cups? – DWin May 17 '18 at 2:23
• If there are three glasses of water and two of poison, the method analogous to this answer is you pick two glasses not to drink, and you survive if you happen to pick the two poisoned glasses, which occurs just one way out of the ten possible pairs of glasses you could pick. – David K May 20 '18 at 23:09
Not sure why everybody uses such a complicated approach:
after drinking three cups, one remains. The chance that she is alive is equal to the chance that the remaining cup is the poison, which is one in four = 25%.
The sequence of drinking water and or poison is completely irrelevant.
• Agreed. Part of the SE ethos is to provided detailed additional information that could be used for other, less direct problems. (And to show off) – MarsJarsGuitars-n-Chars May 14 '18 at 22:08
• This is essentially the same as NicNic8’s solution. – wgrenard May 15 '18 at 0:56
• From a baysean perspective there is something fishy about the answer. After drinking three cups, she is either dead or alive. Talking about chance is futile! – Albert van der Horst May 20 '18 at 18:40
• Yes I agree. These top voted answers are a bit extreme it seems, even with my zero experience with any advanced level math and using only my high school school math logic. Yet sometimes you'll see some crazy "sounding" problem given by the questioner yet a few simple lines for a response that seems to solve everyone's issues, confusions. Good ol' SE – Hunter Frazier May 21 '18 at 10:38
There are $4!$ permutations of $W_1W_2W_3P$.
The only way to live is if $P$ is last, and there are $3!$ ways for this to occur.
So there are $4!-3!$ ways to die with probability $1-\frac{3!}{4!} = \frac34$.
• Just read Taussig's answer. This solution is more or less the same approach. – suneater May 14 '18 at 8:35
• This solution is more or less the same approach well I would not say so. This one has a different approach / methodology (and ultimately relies on similar concepts). – WoJ May 14 '18 at 10:34
A good way to think about such problems is to ask yourself the opposite question: what is the probability that I will not get poisoned?
\begin{align*} \Pr(\text{not poisoned}) &= \Pr(\text{not poisoned on first glass}) \cdot \Pr(\text{not poisoned on second glass}) \cdot \Pr(\text{not poisoned on third glass}) \\ &= \frac{3}{4} \cdot \frac{2}{3} \cdot \frac{1}{2} \\ &= \frac{1}{4} \\ \end{align*}
It follows that
\begin{align*} \Pr(\text{poisoned}) &= 1 - \Pr(\text{not poisoned}) \\ &= 1 - \frac{1}{4} \\ &= \frac{3}{4} \\ \end{align*}
• "Ask yourself the opposite question" is the important lesson here. +1 – joeytwiddle May 16 '18 at 5:34
• Now if one of them is the antidote... – John P May 18 '18 at 16:50
You have 50%. If you drink the third cup means first and second are not poison. So when drink third you have just two cups, and one is poison.... 50%. I think this problem is about logic more than math... Thanks a Lot
• This is a valid approach, under a different set of assumptions (namely that the poison is debilitating and acts faster than you can drink another cup) – Ben Voigt May 14 '18 at 23:55
• This is saying it is more a language issue than a math issue. 'If you were to drink' could mean the action is done or to be done. Interesting. – David May 15 '18 at 16:22
• There is a revolver handgun with 4 chambers, three are empty and one has a live round. If you put it to your head and squeezed the trigger three times, what is the probability of having a bullet in your head? Your answer addresses the gun question better than the poison question. – user1717828 May 15 '18 at 16:36
• hahaha! So funny! If I press three times the trigger means first and second was without bullet. Pressing third time maybe I got the bullet in my head, maybe not (I'm not lucky man! so for sure I have!). So 50% probability... – German Rodriguez May 15 '18 at 18:41
• Just make sure it is a revolver and not an automatic! – Eric Lippert May 15 '18 at 19:02
Label the cups A, B, C, D. Now we can assume WLOG that the cups she drank are A, B and C.
There are only 4 scenarios according to which cup is the poisoned one. In 3 of the scenarios (poisoned cup = cup A, B, C respectively), she is poisoned. In 1 of the scenarios (poisoned cup = cup D), she is not poisoned. Therefore the probability is 3/4 = 75%.
Not sure why it is any more complicated than that.
• It isn't any more complicated, as NicNic8 and Aganju already answered. – Dronz May 15 '18 at 3:26
• It is more complicated if you want to solve very similar problems generally. Laying out all the scenarios and counting them quickly becomes unreasonable for larger numbers of cups with more than 1 left untouched. – Matthew Read May 17 '18 at 3:47
The solution to this problem depends on how fast the poison works. If the poison is slow acting, the previous solutions are OK. But if the poison works instantly, there may be no opportunity to drink three cups.
The problem states that three cups are taken, so the first two cups can NOT be poison. The third cup is chosen from a set of two with one poison and the other safe. So the probability of survival = probability of poison = 1/2.
• I think you are wrong. You are not calculating the probability of the first two cups not be poisoned. – sawa May 17 '18 at 3:25
• @sawa If we know that both (a) the poison acts instantly and (b) you drink 3 cups, the poison cannot have been in either the first 2 (they would have killed you before you drank the third). That means the probability of either of the first two cups being poison is 0. Obviously, this is intentionally misinterpreting the problem for the sake of humor; it's a joke. – Matthew Read May 17 '18 at 3:42
• I am not saying that your interpretation is wrong. That is not my point. My point is that you are only calculating the conditional probability (that the third one is poison) under the assumption that the first two were not poison. I am saying that you have to calculate the conditional part to get to that situation, which Floris does correctly in my opinion. – sawa May 17 '18 at 3:48
• @sawa: The probability of surviving until the third cup does not have to be calculated, it is postulated by the problem formulation, therefore 100% (P(x | x) = 1). – Ben Voigt May 18 '18 at 3:31
• Okay, I understand. – sawa May 18 '18 at 4:31
I thought it would be instructive to try a different approach.
If you drink three cups, that tells me for sure that two of the cups you drank (the first two) were not poisoned - otherwise you would not have gotten to the third cup.
So the poison must be in either cup 3 or cup 4, and since it is equally likely to be in any of the cups, there is a 50% chance of that happening.
Having survived the first two cups, you now have a 50-50 chance that cup 3 is poisoned (because it's either cup 3 or cup 4). To survive drinking cup 3, you need to beat those odds as well.
Multiplying these probabilities, you once again get 0.25 for the chance of surviving.
Of course it's the same answer - but I thought this would give additional insight.
• The OP stated that the person drinking the water was proving that they can't die, so I don't think it's fair to make your conclusion that drinking three cups itself is amount to showing that the first two cups are not poisoned. – Green May 16 '18 at 23:45
• Your interpretation of the question is the same as richard1941 and German Rodriguez, but I think they gave the wrong answer under that assumption, and you are the only one to give the correct answer under the assumption. – sawa May 17 '18 at 3:29
• This is not correct; it's almost like the reverse of the Monty Hall misunderstanding. In this version of the problem, the first two cups don't matter. They might as well not exist. The problem is reduced purely to two cups, one of which is poisoned. You drink only one of them -- what are your odds? 50%, not 25%. – Matthew Read May 17 '18 at 3:59
• Or to fully dive into the scenarios you wanted to lay out here -- there are four situations: (1) cup 3 is poisoned and you drink it (2) cup 4 is poisoned and you drink it (3) cup 3 is poisoned but you drink cup 4 (4) cup 4 is poisoned but you drink cup 3. 2/4 of those are deadly; 50%. – Matthew Read May 17 '18 at 4:02
• A potentially more intuitive reason it can't be 25%: In this version of the problem, you've gained definitive knowledge of absolute safety about the first two choices made. That MUST affect the odds. It is obviously not true that drinking two safe cups leaves you with the same odds as drinking two cups that could have included the poisoned cup. – Matthew Read May 17 '18 at 4:05
Wow, people are making this complicated.
There are 4 cups. One is poisoned. She picks 3. There are 3 chances that she will pick the poisoned cup out of 4. Therefore, the probability is 3/4.
This assumes that she does not pick a cup, drink it, then put it back and someone refills it before she picks another.
I'm also assuming that she either picks 3 cups before drinking any of them, or that if she dies before picking 3 cups, that we treat that as if she picked enough to fill out the 3 at random. Otherwise the probability is impossible to calculate, because if the first or second cup is poisoned, she doesn't "pick 3 cups".
You can do all the permutations and bayesian sequences, but as others have shown, they all come to the same answer.
If she picked, say, 3 out of 6 cups and 2 are poisoned, I don't see how to do it other than with combinatorics. But maybe I'm missing an easy way.
• NicNic8's answer is even more simple, check that top answer. – user061703 May 19 '18 at 6:02
• For your drinking 3 out of 6 with 2 poisoned, just think of six spots for glasses in a row. The first three are to be drunk. The first poisoned glass has 3 safe locations (the last three) out of 6, so we're safe with probability 0.5. The second poisoned glass has 2 safe locations to land on out of 5, so we'll stay safe with probability 0.4. Combined: 0.2 probability of a safe three drink binge... – DJohnM May 19 '18 at 8:12
• Ooh, interesting. Yes, simpler than what I had in mind, which was: number of safe combinations divided by total number of combinations = 4C3 / 6C3 = 4 / 20 = 0.2. Same answer. – Jay May 19 '18 at 23:17
• @TrầnThúcMinhTrí I though NicNic8's answer was a shade more complicated than mine as it requires reversing the probability, but whatever. Not something that I would challenge someone to a duel to the death over. – Jay May 19 '18 at 23:19
• So the problem grows as more people think about it. Now we have sampling with replacement as a possibility. The probability of surviving on a single draw is 3/4. Because the non-poisoned cup is replaced, the overall probability of survival, at the beginning of the trial, is 3/4 x 3/4 x 3/4 = 27/64. It does not matter if the poison if fast or slow because the conditional probabilities are the same as the initial probability. – richard1941 May 21 '18 at 17:27 | 2020-07-10T02:57:46 | {
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http://mathhelpforum.com/discrete-math/207036-how-do-i-solve-direct-proof.html | # Math Help - How do I solve this with a direct proof?
1. ## How do I solve this with a direct proof?
Prove that for all sets X and Y, (X ∩ Y) ∪(Y-X)=Y
I'm stuck and I'm not fully sure how to solve this.
2. ## Re: How do I solve this with a direct proof?
suppose that a is an element of (X∩Y) U (Y-X).
this means one of two things:
a is in X∩Y, or
a is in Y-X
suppose a is in X∩Y. this means that a is in BOTH X and Y, so is certainly in Y.
alternatively, if a is in Y-X, this means a is in Y, but not in X. who cares if its not in X, at least it's in Y!
so in all possible cases, we see that a is in Y.
this means that (X∩Y) U (Y-X) is a subset of Y.
now suppose b is in Y.
well we have two possibilities:
b is in X
b is NOT in X (in or out, that's the way it is with sets. you cannot be "sort of" in a set).
if b is in X, then b is in X AND Y, so b is in X∩Y.
if b is not in X, then b is in Y, but not in X, so b is in Y-X.
if we put these two sets together, b is certain to be in one of them. hence b is in (X∩Y) U (Y-X).
thus Y is a subset of (X∩Y) U (Y-X).
but, if for 2 sets A,B: if A⊆B and B⊆A, then A and B have exactly the same elements, that is: A = B.
so (X∩Y) U (Y-X) = Y.
(intuitively what we are doing is splitting Y into 2 parts: the part that overlaps with X, and the part that doesn't).
3. ## Re: How do I solve this with a direct proof?
Hello, blueaura94!
$\text{Prove: }\:(X \cap Y) \cup (Y - X) \:=\:Y$
$\begin{array}{ccccc}1. & (X \cap Y) \cup (Y - X) && 1. & \text{Given} \\ 2. & (X \cap Y) \cup (Y \cap \overline{X}) && 2. & \text{Def. Subtr'n} \\ 3. & (X\cap Y) \cup (\overline{X} \cap Y) && 3. & \text{Commutative} \\ 4. & (X \cup \overline{X}) \cap Y && 4. & \text{Distributive} \\ 5. & U \cap Y && 5. & A \cup \overline{A} \,=\,\mathbb{U} \\ 6. & Y && 6. & \mathbb{U} \cap A \,=\,A \end{array}$
4. ## Re: How do I solve this with a direct proof?
Originally Posted by blueaura94
Prove that for all sets X and Y, (X ∩ Y) ∪(Y-X)=Y
Here is another way.
$X\cap Y\subseteq Y~\&~Y\setminus X\subseteq Y$ so there is just need to prove one way.
If $t\in Y$ then either $t\in X$ or $t\notin X$.
And we are done. | 2015-11-27T03:27:37 | {
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https://daviddalpiaz.github.io/stat400sp18/homework/hw02-soln.html | ## Exercise 1
Just before their last year at Anytown High School, the seniors hold the Senior Year Kick-Off party. Sixty percent of the students attending the party are seniors, and the rest (friends, significant others, siblings, etc.) are juniors (25%), sophomores (10%), and freshmen (5%) Unfortunately, drinking is quite common at this party; 90% of the seniors consume alcohol, so do 80% of the juniors, 50% of the sophomores, and 20% of the freshmen.
(a) If a student at this party is drinking, what is the probability that this student is a senior?
Solution:
First, we note the given information
\begin{aligned} P(Fr) &= 0.05, \ P(D \mid Fr) = 0.20 \\ P(So) &= 0.10, \ P(D \mid So) = 0.50 \\ P(Jr) &= 0.25, \ P(D \mid Jr) = 0.80 \\ P(Sr) &= 0.60, \ P(D \mid Sr) = 0.90 \\ \end{aligned}
Next, we calculate proportion of the students attending the party consume alcohol.
\begin{aligned} P(D) &= P(Fr \cap D) + P(So \cap D) + P(Jr \cap D) + P(Sr \cap D) \\ &= P(Fr) \cdot P(D \mid Fr) + P(So) \cdot P(D \mid So) + P(Jr) \cdot P(D \mid Jr) + P(Sr) \cdot P(D \mid Sr) \\ &= 0.05 \cdot 0.20 + 0.10 \cdot 0.50 + 0.25 \cdot 0.80 + 0.60 \cdot 0.90 \\ &= 0.01 + 0.05 + 0.20 + 0.54 \\ &= 0.80 \end{aligned}
Note that, along the way we obtained the probability of drinking and the intersection with each class.
Then, we finally calculate
$P(Sr \mid D) = \frac{P(Sr \cap D) }{P(D)} = \frac{0.54}{0.80} = \boxed{0.675}$
(b) If a student at this party is not drinking, what is the probability that this student is not a senior?
Solution:
Based on the work we did in the previous problem, we can easily “complete the table.” (We could have also used a tree to get the information needed for the table.)
Fr So Jr Sr
Drinking 0.01 0.05 0.20 0.54 0.80
Not Drinking 0.04 0.05 0.05 0.06 0.20
0.05 0.10 0.25 0.60 1
$P(Sr^\prime \mid D^\prime) = \frac{P(Sr^\prime \cap D^\prime)}{P(D^\prime)} = \frac{0.04 + 0.05 + 0.05}{0.20} = \boxed{0.70}$
(c) If a student at this party is not a senior, what is the probability that this student is not drinking?
Solution:
$P(D^\prime \mid Sr^\prime) = \frac{P(D^\prime \cap Sr^\prime)}{P(Sr^\prime)} = \frac{0.04 + 0.05 + 0.05}{0.05 + 0.10 + 0.25} = \boxed{0.35}$
(d) What proportion of the underclassmen (freshmen and sophomores) attending the party consume alcohol?
Solution:
$P(D \mid (Fr \cup So)) = \frac{0.01 + 0.05}{0.05 + 0.10} = \boxed{0.40}$
(e) The school administration discourages the Senior Year Kick-Off party; the principal of AHS announced that any senior attending the party will receive a week of detention. Of course, drinking is also discouraged. Find the proportion of the students at the party who either are seniors, or consume alcohol, or both.
Solution:
$P(S \cup D) = P(S) + P(D) - P(S \cap D) = 0.60 + 0.80 - 0.54 = \boxed{0.86}$
(f) Are events {a student at the party is a senior} and {a student at the party is drinking} independent? Justify your answer. No credit will be given without proper justification.
Solution:
$$\boxed{\text{No.}}$$
$0.54 = P(Sr \cap D) \neq P(Sr) \cdot P(D) = 0.60 \cdot 0.80 = 0.48$
We could have also showed that $$P(Sr \mid D) \neq P(Sr)$$ or $$P(D \mid Sr) \neq P(D)$$.
(g) Are events {a student at the party is a junior} and {a student at the party is drinking} independent? Justify your answer. No credit will be given without proper justification.
Solution:
$$\boxed{\text{Yes.}}$$
$0.20 = P(Jr \cap D) \neq P(Jr) \cdot P(D) = 0.25 \cdot 0.80 = 0.20$
## Exercise 2
A bishop is placed at random (with equal chance) on a chess board (8 x 8). A king of the opposing color is placed at random (with equal chance) on one of the remaining squares. What is the probability that the king is under attack from the bishop?
Hint: Placed anywhere on a chess board, a rook attacks 14 squares out of the remaining 63.
$P(\text{ K is under attack }) = \frac{14}{63}$
Solution:
We first need to determine how many squares are being attacked for each possible position of the bishop. For example, in the top left corner, the bishop can attack 7 positions. We also keep track of how many positions on the board have the bishop attacking 7 positions. (And do so for all possibilities.)
Then we use the law of total probability.
\begin{aligned} P(\text{K is under attack}) &= P(\text{B attacks 7}) \cdot P(\text{K is under attack} \mid \text{B attacks 7}) \\ &+ P(\text{B attacks 9}) \cdot P(\text{K is under attack} \mid \text{B attacks 9}) \\ &+ P(\text{B attacks 11}) \cdot P(\text{K is under attack} \mid \text{B attacks 11}) \\ &+ P(\text{B attacks 13}) \cdot P(\text{K is under attack} \mid \text{B attacks 13}) \\ &= \frac{28}{64} \cdot \frac{7}{63} + \frac{20}{64} \cdot \frac{9}{63} + \frac{12}{64} \cdot \frac{11}{63} + \frac{4}{64} \cdot \frac{13}{63} \\ &= \boxed{\frac{5}{36}} \end{aligned}
# Exercise 3
You are given that $$P(A) = 0.5$$ and $$P(A \cup B) = 0.7$$. Student 1 assumes that $$A$$ and $$B$$ are independent and calculates $$P(B)$$ based on that assumption. Student 2 assumes that $$A$$ and $$B$$ are mutually exclusive and calculates $$P(B)$$ based on that assumption. Find the absolute difference between the two calculations.
Solution:
We first consider Student 1. Since Student 1 believes that $$A$$ and $$B$$ are independent, we have
$P_1(A \cap B) = P_1(A) \cdot P_1(B)$
It is also always true that
$P_1(A \cup B) = P_1(A) + P_1(B) - P_1(A \cap B)$
With the given probabilities, we have
\begin{aligned} P_1(A \cap B) &= 0.5 \cdot P_1(B) \\ 0.7 &= 0.5 + P_1(B) - P_1(A \cap B) \end{aligned}
This is a system of two equation, with two unknowns, so we solve and obtain
$P_1(B) = 0.40$
Now we consider Student 2. Since they assumes that $$A$$ and $$B$$ are mutually exclusive we have
$P_2(A \cap B) = 0.$
Because of this, we have
$P_2(A \cup B) = P_2(A) + P_2(B)$
With the given probabilities, we have
$0.7 = 0.5 + P_2(B)$
So we obtain
$P_2(B) = 0.2.$
So, finally, their absolute difference is
$\left| P_1(B) - P_2(B)\right | = 0.4 - 0.2 = \boxed{0.2}$
# Exercise 4
Alex and David agreed to play a series of tennis games (as many as needed) until one of them wins two games in a row. Alex will serve in the first game, then the serve would alternate game by game between Alex and David. David is a better tennis player; Alex has a 50% chance of winning a game on his serve and only a 20% chance of winning a game if David serves. Assume that all games are independent. Find the probability that Alex is the first one to win two games in a row.
Solution:
First we define some notation. We say
$O_S$
$$O$$ is the outcome for Alex, $$W$$ or $$L$$. $$S$$ is the player serving. $$D$$ for David, $$A$$ for Alex.
Now we enumerate the possible outcomes that result in Alex being the first to win two games in a row. Recall that Alex serves first. We can group these into two groups of possibilities.
Alex wins the first game:
\begin{aligned} & W_{A}W_{D} \\ & W_{A}L_{D} W_{A}W_{D} \\ & W_{A}L_{D} W_{A}L_{D} W_{A}W_{D} \\ & \ldots \end{aligned}
David wins the first game:
\begin{aligned} & L_{A}W_{D}W_{A} \\ & L_{A}W_{D} L_{A}W_{D}W_{A} \\ & L_{A}W_{D} L_{A}W_{D} L_{A}W_{D}W_{A} \\ & \ldots \end{aligned}
Or, more succinctly:
Alex wins first game: $$(W_{A}L_{D})^k W_{A}W_{D}, \quad k = 0, 1, 2, 3, \ldots$$
David wins first game: $$L_{A} (W_{D}L_{A})^k W_{D}W_{A}, \quad k = 0, 1, 2, 3, \ldots$$
Then, we calculate the required probability by adding the probabilities of each of the outcomes we listed above, since they are all disjoint.
\begin{aligned} P(\text{ Alex wins two in a row }) &= P(\text{ Alex wins two in a row after winning the first }) \\[0.5em] &+ P(\text{ Alex wins two in a row after losing the first }) \\[0.5em] &= \sum_{k = 0}^{\infty}(0.50 \cdot 0.80)^k (0.50 \cdot 0.20) \\[0.5em] &+ \sum_{k = 0}^{\infty} 0.50 \cdot (0.20 \cdot 0.50)^k (0.20 \cdot 0.50) \\[0.5em] &= \frac{0.50 \cdot 0.20}{1 - 0.50 \cdot 0.80} = \frac{0.50 \cdot 0.20 \cdot 0.50}{1 - 0.20 \cdot 0.50} \\[0.5em] &= \frac{1}{6} + \frac{1}{18}\\[0.5em] &= \boxed{\frac{2}{9}} \end{aligned} | 2020-01-17T19:17:52 | {
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https://forum.allaboutcircuits.com/threads/finding-frequency-of-the-filter-circuit-from-the-transfer-function.175404/ | # Finding frequency of the filter circuit from the transfer function
#### Wompalamba
Joined Dec 28, 2020
19
Hi i am trying to extract the cutoff frequency from the transfer function that i derived similar to this derivation of the RC filter. I am currently stuck at the step where the equation is modulus on both side as i am not sure how do i continue from there. I have attached the work i have done so far in the image. Thank you for your time!
Edit: i realized that i posted the title wrongly it should be Finding frequency of the filter circuit from the transfer function and i am not sure how to edit the title
Last edited:
#### ZCochran98
Joined Jul 24, 2018
105
Before I answer your question, I do need to point out a small mistake. You have:
$Z = \left(\frac{1}{\frac{1}{sC_1}+\frac{1}{R_3}}\right)^{-1}$
Which is incorrect. You already incorporated the ^-1 with the fraction, so you don't need the ^-1 outside the brackets, and, the sC_1 shouldn't be inverted (1/sC_1 is impedance, so 1/(1/sC_1) = sC_1). So that line should just be:
$Z = \frac{1}{sC_1+\frac{1}{R_3}}$
Now, to answer your question: After you correct that, you have the correct process (Z/Z_total, basically), just the wrong value for Z. Ultimately, after you do all the math, you should get an equation of the form:
$\left|\frac{V_{out}}{V_{in}}\right| = \frac{|a|}{\left|bs+c\right|}$
For some constants a, b, and c which depend on your circuit variables (similar to what you have, but with the corrected value of Z, I can tell you there will be no s in the numerator).
The frequency for which your filter is designed, in this case, is the pole of this equation: what value of s makes the denominator of the fraction = 0? The frequency corresponding to that s (you correctly wrote $$s = \omega j = 2\pi f j$$) is the frequency for which this filter was designed.
Hope this helps!
#### Wompalamba
Joined Dec 28, 2020
19
Thank you for correcting my careless mistake and i appreciate you spending some time to share your knowledge. I have since corrected the issue but still have some doubts as to how to solve the modulus function .I am trying to extract out the frequency to find the cut off frequency at $\frac{Vout}{Vin}=\frac{1}{\sqrt{2}}$ when i put in the fixed component values. Am i right to do the following steps as shown in the image attached?
#### ZCochran98
Joined Jul 24, 2018
105
You are on the right track. One thing to keep in mind: if you have something = 0, then |something| is also = 0. Thus, at the third line down, where you have written "$$= \frac{1}{1 + sC_1R_1 + \frac{R_1}{R_3} + sC_1R_2 + \frac{R_2}{R_3}}$$," you can actually just go ahead and solve that line, as you set the denominator there = 0, then the magnitude of it will also be zero. Doing that, you can solve for s and then for f, subsequently. The next 3 lines after it are just extra work that, for this case, aren't necessary.
#### Wompalamba
Joined Dec 28, 2020
19
Thank you for the guidance. However, I suspect i might have made a mistake somewhere because when i tried to find the frequency cut off it was not the answer i expected which is 810~kHz i have attached the two images below.
#### The Electrician
Joined Oct 9, 2007
2,814
At the very end you made an algebra mistake. The expression (C1 R1 - C1 R2) should be (C1 R1 + C1 R2).
#### Wompalamba
Joined Dec 28, 2020
19
Thank you! I corrected the mistake but i am still not getting the desired cut off frequency of 810kHz. Did i accidentally made more mistake in the last step?
#### The Electrician
Joined Oct 9, 2007
2,814
Where did you get the value of 810 kHz? I solved it and got the 895.98 kHz value. This is correct for the component values you have given and for the circuit as shown.
#### ericgibbs
Joined Jan 29, 2010
11,630
hi W,
This is what LTSpice shows for those component values.
Slight difference to calculated numeric value, due to accuracy of cursor placement.
E
#### Attachments
• 58.4 KB Views: 8
#### Wompalamba
Joined Dec 28, 2020
19
You are right. I tried the spice simulation and it shows the correct value. I am assuming this is a weird software issue . I have attached the schematic i was using for the Elsie simulation just in case i did something wrongly.
#### ericgibbs
Joined Jan 29, 2010
11,630
hi W,
Did you also try to calculate the Freq, by summing R1 and R2 and say re-naming as R4, in order to simplify the maths.?
Try it.
E | 2021-01-24T02:50:33 | {
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https://math.stackexchange.com/questions/4073417/symbol-for-very-small-variable | # Symbol for very small variable
Is there any representation to state that a variable is close to (not equal) zero? Let me give you an example. Consider the function
$$u(x)=\alpha (e^{i\omega \delta t}-1) f(x)$$
I am interested in the function $$u(x)$$ when $$\delta t$$ is very small. For this case, it should be easy to see that
$$u(x)|_{\text{small }\delta t} \approx i \alpha \omega \delta t f(x)$$
Is there any "nice" notation to represent such an equation (without having to write small)? I thought that I could use the limit notation for that [for example, $$\lim_{\delta t \to 0} u(t)$$], but then I realized that if $$\delta t$$ goes to zero, then $$u(x)=0$$. Therefore, it is not what I need.
I found this link in the same forum, but it did not help.
• Really, what's wrong with $\delta t<<1$? Mar 23 at 15:44
• $\delta t\ll 1$ is good Mar 23 at 15:55
• I think it is okay. Nevertheless, I would need to use something like $|\delta t| << 1$ to make it clear that $\delta t$ is positive?
– jeb
Mar 23 at 16:31
I think you are asking about the first order (linear) approximation using the derivative.
You might say $$u(x)= i \alpha \omega \delta t f(x) + o(\delta t ).$$
The fact that this is an approximation for small $$\delta t$$ should be clear to your reader. If not you can say so.
• I like to "o" notation, I have to say. What would you recommend me to include in my text or even in front of the $u(x)$ term to make it clear that $\delta t$ is small? My question is based on the fact that I will need to manage this equation further, and if I include the "o" notation, they will become too overloaded. The best that I could think of was something like $u(x)|_{\delta t = \epsilon}$, such that $o(\epsilon)$ is insignificant (or can be disregarded). Thank you.
– jeb
Mar 23 at 18:41
• The insignifance to first order is implicit in the use of the $o$ notation. You can just carry the $o(\delta t)$ term along in future calculations and throw it away at the end (r sooner) as appropriate. You could add "as $\delta t \to 0$" when you first state the equation. I would not want to read an invented notation like the one you propose. Mar 23 at 19:01
The issue with taking limits is that as $$\delta t\to 0$$ both sides go to zero, but what you want to say is stronger than that.
I imagine what you actually mean by $$u(x)\approx i\alpha\omega\delta tf(x)$$ can be formalised as $$\lim_{\delta t\to 0}\frac{u(x)}{\delta t}=i\alpha\omega f(x).$$
This is simply a matter of rearranging so that the limit captures the relative error of the approximation, not just the absolute error. | 2021-10-26T08:48:31 | {
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https://math.stackexchange.com/questions/945138/proving-that-x-left-lfloor-fracx2-right-rfloor-left-lfloor-frac | # Proving that $x-\left\lfloor \frac{x}{2} \right\rfloor\; =\; \left\lfloor \frac{x+1}{2} \right\rfloor$
How would you prove that if $x$ is an integer, then
$$x-\left\lfloor \frac{x}{2} \right\rfloor\; =\; \left\lfloor \frac{x+1}{2} \right\rfloor$$
I tried to start by saying that if $x$ is an even integer, then:
$$\left\lfloor \frac{x}{2} \right\rfloor = \frac{x}{2}.$$
However, I am stuck on showing that $$\left\lfloor \frac{x+1}{2} \right\rfloor$$ is also $\frac{x}{2}$. Intuitively It makes intuitive sense just by playing around with some sample numbers, but I don't know how to make it mathematically rigorous. Further, what do you do in the odd case? Is this even the right way to go about it?
• Start with $\lfloor y+n \rfloor = \lfloor y\rfloor +n$. – lhf Sep 25 '14 at 1:53
• @lhf Is there a proof for that? – 1110101001 Sep 25 '14 at 1:56
• By definition, $\lfloor y \rfloor \le y < \lfloor y \rfloor +1$ and so $\lfloor y \rfloor + n \le y + n < \lfloor y \rfloor +n+1$, which says that $\lfloor y+n \rfloor = \lfloor y\rfloor +n$. – lhf Sep 25 '14 at 1:58
If $x$ is even, then $x=2k$ where $k \in \mathbb{Z}$.
We then have: $$x-\left\lfloor \frac{x}{2} \right\rfloor=2k-\lfloor k \rfloor=k=\left\lfloor k+\frac{1}{2} \right\rfloor=\left\lfloor \frac{2k+1}{2} \right\rfloor=\left\lfloor \frac{x+1}{2} \right\rfloor$$
If $x$ is odd, then $x=2k+1$ where $k \in \mathbb{Z}$.
We then have $$x-\left\lfloor \frac{x}{2} \right\rfloor=2k+1-\left\lfloor k+{1\over 2} \right\rfloor=k+1=\left\lfloor k+1 \right\rfloor=\left\lfloor \frac{2k+2}{2} \right\rfloor=\left\lfloor \frac{x+1}{2} \right\rfloor$$
Case-1 (Even) write $x=2k$. Now$\left\lfloor \frac{x}{2} \right\rfloor =k$ so L.H.S is $2k-k=k$. now work with R.H.S, $\left\lfloor \frac{x+1}{2} \right\rfloor=\left\lfloor \frac{2k+1}{2} \right\rfloor=\left\lfloor k.5 \right\rfloor=k$.
Case 2) (Odd) Let $x=2k+1$ Similarly. I am in hurry, so i leave it on you. very easy.
Prove LHS=RHS= $k+1$ | 2020-11-30T11:22:01 | {
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https://www.freemathhelp.com/forum/threads/probability-question-on-drawing-cards.131372/#post-545785 | Probability question on drawing cards
micky123
New member
From a standard pack of 52 cards, I draw 3 cards at random without replacement. What is the probability that all cards drawn are both hearts and picture cards?
is the answer: (3/52)*(2/51)*(1/50)?
Subhotosh Khan
Super Moderator
Staff member
From a standard pack of 52 cards, I draw 3 cards at random without replacement. What is the probability that all cards drawn are both hearts and picture cards?
is the answer: (3/52)*(2/51)*(1/50)?
There are 4 cards (ACE included) with pictures AND heart.
How many ways can you draw 3 cards from that group ?
How many ways can you draw 3 cards from 52 cards ?
micky123
New member
Why would ace be included? I thought there are only 3 cards that are both heart and picture? (jack, queen and king)
Please kindly explain if I have gotten it wrong! Thank you.
Subhotosh Khan
Super Moderator
Staff member
Why would ace be included? I thought there are only 3 cards that are both heart and picture? (jack, queen and king)
Please kindly explain if I have gotten it wrong! Thank you.
If ACE is not included then - how many ways can you choose 3 cards from that group (Hearts and pictures) ? ------ 1 way
How many ways can you choose 3 cards (no restriction) from the deck? _____ $$\displaystyle C^{52}_{3}$$
Now calculate probability.
reference: https://math.stackexchange.com/ques...-getting-3-cards-in-the-same-suit-from-a-deck
micky123
New member
If ace is not included, then the probability is 1/22100. However, i am confused as to why there are 4 cards with pictures and heart?
Am I understanding the event incorrectly? I understood the event that “all three cards are both heart and picture” to be drawing “a jack, a queen and a king of hearts”, which is 1 out of 22100 ways.
pka
Elite Member
If ace is not included, then the probability is 1/22100. However, i am confused as to why there are 4 cards with pictures and heart?
Am I understanding the event incorrectly? I understood the event that “all three cards are both heart and picture” to be drawing “a jack, a queen and a king of hearts”, which is 1 out of 22100 ways.
I would agree with your first answer [imath]\left(\dfrac{3}{52}\right)\cdot\left(\dfrac{2}{51}\right)\cdot\left(\dfrac{1}{50}\right)[/imath].
BUT if I were reading this question as an editor I would object to its wording as being unclear. It is extremely unusual to use picture card what standard English uses face cards.
mmm4444bot
Super Moderator
Staff member
It is extremely unusual to use picture card what standard English uses face cards.
Your wikipedia link states that names 'picture card' and 'face card' mean the same thing.
Otis
Elite Member
confused as to why there are 4 cards with pictures and heart?
Because a lot of card manufacturers draw a picture on each ace. However, the design on the aces do not show a person.
The name 'picture card' refers only to cards containing a picture of a person.
micky123
New member
I see. So this means that it depends on the way “picture card” is defined in the question?
if ace is not considered a picture card, the ans would be (3/52)*(2/51)*(1/50).
if ace is considered a picture card, the and would be (4/51)*(3/51)*(2/50).
am i right? | 2021-12-03T02:19:03 | {
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http://www.kineticmediaworks.com/9w3bmzg/c7b28f-latex-equation-multiple-lines | # latex equation multiple lines
If you wanted to just split in two parts, then multlined (notice the extra d) from the mathtools package would be a simpler solution: All the above works with elsarticle class in your updated question. Latex equation in multiple lines. â Charlie Jan 16 '14 at 3:38 You need to specify an alignment point on each line with & and separate lines with \\. 3 Multiple Lines of Displayed Maths . The breqn package is designed to split long equations automatically. the first version becomes: Not sure what you tried doing with multine but this seems OK: Btw, you seem to be missing a bracket on the RHS -- and I deleted an extraneous comma after the . To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Latex allows two writing modes for mathematical expressions. The command \overline and \underline places a line above or below the expression. The % sign tells LATEX L A T E X to ignore the rest of the current line. To learn more, see our tips on writing great answers. Here also, the amsmath package is required. The amsmath package provides the align and align* environments for aligned equations. As previously mentioned, Originâs LaTeX App wraps user-entered equation markup in a single equation environment by automatically adding and commands before and after user-entered markup. Grouping and centering equations. LaTeX Error: \begin{document} ended by \end{multline}.See the LaTeX manual or LaTeX Companion for explanation.Type H for immediate help.... \end{multline}, Though i get the equation broken it is not numbered, or has a one line distance from my text as the The starred version doesn't number the equations. I am attaching a screenshot of the result: How would one justify public funding for non-STEM (or unprofitable) college majors to a non college educated taxpayer? Use instead the equation - just the multline environment, you can work as usual with line breaks. TeX - LaTeX Stack Exchange is a question and answer site for users of TeX, LaTeX, ConTeXt, and related typesetting systems. Yes you can load several different packages. You need to use this Feature, the amsmath-package. As a test, when the 2nd backslash was deleted at the end of the 3rd line of code (\dot{x} & = \sigma(y-x) \), it produced the single line equation you provided as an example. Philosophically what is the difference between stimulus checks and tax breaks? Again, use * to toggle the equation numbering. I suggest you use a split environment, which in contrast to multline may be used a subenvironment of equation. LATEX doesnât break long equations to make them ï¬t within the margins as it does with normal text. In that version I was able to open an Equation Editor and enter multiple lines of math equations in a the editor, pressing return at the end of each line to move to the next line. I copied and pasted their example exactly as is, but i still get errors, does it have to do with the package i am using? Try the example on the right which sets the same multiple equations in several ways. My code compiles and, looking at it, it doesn't contain, i will try it as well, i was fortunate to get many helpful answers. Is my Connection is really encrypted through vpn? I am a new Latex user,I have loaded these two math equation in my Latex documents and i want to split an equation i have into multiple line, I am reading this topic for breaking the lines over multiple, although what ever i do it does not allow me to compile. What location in Europe is known for its pipe organs? I want to place them/break the equation in two lines and retain the format of numbering and equal placing beneath. How to answer a reviewer asking for the methodology code of the paper? If you want to create a document with more than two columns, use the package multicol, which has a set of commands for the same. It works very well in the majority of situations, but it's not as mature as the amsmath package. @PaulGessler I had a look in the document, thank you for the link. Mathematical modes LaTeX allows two writing modes for mathematical expressions: the inline mode and the display mode. \begin{eqnarray*} Therefore, special environments have been declared for this purpose. In another practical tip we show you how to in your, LaTeX: equation in multiple rows - how to, d^2 = a^2 + b^2 + c^2 \geq a^2 + b^2 = a^2 + b^2 + 2ab - 2ab = (a+b)^2 - 2ab \\\ \geq 2ab. Latex Support In Pages Macs Chemistry. Referencing subordinate equations can be done using either of two methods: adding a label after the \begin {subequations} command, viz. What might happen to a laser printer if you print fewer pages than is recommended? 2). 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Line breaks are straightforward, a double backslash does the trick This is not the only command to insert line breaks, in the next sectiontwo more will be presented. View PDF âºâº Find news, promotions, and other information pertaining to our diverse lineup of innovative brands as well as newsworthy headlines about our company and culture. Figure 1: Typical equation problems in ordinary LATEX: (a) di erent spacing around the equals signs in (14) and (15) because one uses equationand the other uses eqnarray; (b) equation (15) is a single equation but because it covers two lines \nonumbermust be used on the rst line; (c) and then the If a coworker is mean to me, and I do not want to talk to them, is it harrasment for me not to talk to them? They can be distinguished into two categories depending on how they are presented: 1. text â text formulas are displayed inline, that is, within the body of text where it is declared, for example, I can say that a + a = 2 a {\displaystyle a+a=2a} within this sentence. site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. Equations. How is HTTPS protected against MITM attacks by other countries? Description below mirrored brace. LaTeX needs to know when text is mathematical. It is therefore up to you to format the equation appropriately (if they overrun the margin.) 3 Special Characterulti Line Equations. This is because LaTeX typesets maths notation differently from normal text. Usually, the eqnarray environment is only used if authors cannot use the amsmath package because the environments included in this package are easier to use (refer to Section . thank you very much, although if i may ask can i ask if multiple packages can be used. Equations on Multiple Lines. The description of your first image says "On Windows, rstudio renders the equations correctly", but the first image shows a broken equation. That is why I was confused. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. With the split environment, you can display long equations clearly. Schematically, this is what its supposed to look like: As with the tabular environment, use & to separate columns and \\ to separate rows. The Moodle XML file can then be used to bu Auto Latex Equations Google Workspace Marketplace. In this case the first line should be move left relative to the others and the package mathtools provides a convenient command for this: I have split across three lines for clarity. This is best used at the start of a line to mark that line as a âcommentâ. When numbering is allowed, you can label each row individually. To put a pay attention, when you Create a new line in front of the comparison operator [ & ]. If an equation is longer than one line or several formulas must be grouped together, the eqnarray environment could be used. Contents 1 Introduction 2 Including the amsmath package 3 Writing a single equation 4 Displaying long equations We will show you how you can format the equation-environment is still suitable. That can be achieve in plain LaTeX without any specific package. two âcolumnsâ, one with the equation, one with the annotations. How to start equation environment inside a custom environment? You can choose the layout that better suits your document, even if the equations are really long, or if you have to include several equations in the same line. The first one is used to write formulas that are part of a text. formulas, graphs). 9. I suggest you use a split environment, which in contrast to multline may be used a subenvironment of equation. In fact, your example is probably best with the cases environment. There are several ways to format multiple equations and the amsmath package adds several more. In this case the first line should be move left relative to the others and the package mathtools provides a convenient command for this: Andrew,@Andrew,i used \begin{multline} after the , i pasted your solution and i get an error which i have given in the question as a new edit. The amsmath package provides a handful of options for displaying equations. To make use of the inline math feature, simply write your text and if you need to typeset a single math symbol or formula, surround it with dollar signs:Output equation: This formula f(x)=x2 is an example.This formula f(x)=x2 is an example. Find all positive integer solutions for the following equation: Allow bash script to be run as root, but not sudo. Also tried this example as kindly suggested. Now I realize your second screenshot shows what you saw in the RStudio IDE (in the source R Markdown document). You have to put up with [\\\\] a line break at the desired location in your equation. Subscribe to our newsletter to get notification about new updates, information, etc.. LaTeX forum â Math & Science â amsmath | Correct Alignment for multi-line Equations Topic is solved Information and discussion about LaTeX's math and science related features (e.g. Art Of Problem Solving. \label {eq:Maxwell}, which will reference the main equation (1.1 above), or adding a label at the end of each line, before the \\ command, which will reference the sub-equation (1.1a or 1.1b above). Math equation in LaTeX provides three stretchable lines/arrows that appear above or below the equation: braces, bars and arrows. Missing $inserted.$ ...l N(\sigma;\lambda;\theta;t)}{\partial t} over multiple lines, @PaulGessler although as far as I can see the give s me a better visual result, that is why i am keen of using that one. Open an example of the amsmath package in Overleaf First of all, you probably don't want the align environment if you have only one column of equations. I am using the package math of \usepackage{amssymb} \usepackage{amsthm}, @George I posted the self-contained code example above so that you could see one of doing this that works. With a normal line break, you can't write in LaTeX equations multiple lines. The next lines contain only the equations and annotations, I would like them aligned with the equation and annotation of the second line. These environments provide pairs of left- and right-aligned columns. I am trying to align a set of long equations, that are themselves align environments as most of them are spreading on multiple lines.. If we have two lines, then _{xxx} should be at the end of the first line (or at the start of the second line). Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. LaTeX assumes that each equation consists of two parts separated by a &; also that each equation is separated from the one before by an &. Understanding the zero current in a simple circuit. rev 2020.12.18.38240, The best answers are voted up and rise to the top, TeX - LaTeX Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us, @PaulGessler I used it alone but it gave more errors in the compilers most of them saying ! In general, the command \\ signifies a line break and within the correct math mode environment, it can start a new equation line. Here also, the amsmath package is required. thank you for your time, Podcast 300: Welcome to 2021 with Joel Spolsky, Split numbered equation over multiple pages, Defining new split environment with reduced spacing, Can't generate png with Error: Erroneous nesting of equation structures, Formula too long and \split fails | contains “sqrt” (square root). Making statements based on opinion; back them up with references or personal experience. The \overbrace command places a brace above the expression (or variables) and the command \underbrace places a brace below the expression. Socially oriented website which will help to solve your little (or not little) technical problems. I am trying to learn Latex at the same time , thank you for your patience. I think the use of multiple lines structured environment can work. Check out what we are up to! Currently I just have a sequence of align environments, with each equation inside in order to align the pieces of each equations. Then there is an equation and some conditions for the equation, i.e. \end{math} \end{document} This code produces something which looks what you seems to need. Latex equation in multiple lines. The placement of _{xxx} depends on the number of lines that underbrace symbol should occupy, and in general it should be somewhere in the middle. 9 3 Multiple Lines Of Displayed Maths. LaTeX Long equation in several lines Use instead the equation - just the multline environment, you can work as usual with line breaks. Anyway, I'm glad to know that removing the extra empty lines worked. Unlike the tabular environment, there is no argument as the ⦠Adding right brace and equation number. I'm short of required experience by 10 days and the company's online portal won't accept my application, Identify Episode: Anti-social people given mark on forehead and then treated as invisible by society. ! The second one is used to write expressions that are not part of a text or paragraph, and are therefore put on separate lines. Open an example in Overleaf Is Mr. Biden the first to create an "Office of the President-Elect" set? Relationship between Cholesky decomposition and matrix inversion? \documentclass{article} \begin{document} This is your only binary choices \begin{math} \left\{ \begin{array}{l} 0\\ 1 \end{array} \right. Can a smartphone light meter app be used for 120 format cameras? This typically requires some creative use of an eqnarrayto get elements shifted to a new line to align nicely. \begin{eqnarray} y &=& x^4 + 4 \nonumber \\ &=& (x^2+2)^2 -4x^2 \nonumber \\ &\le&(x^2+2)^2 \end{eqnarray} \begin{eqnarray*} e^x &\approx& 1+x+x^2/2! Post by Alirezakn » Thu Nov 22, 2018 7:02 am . If an equation is longer than one line or several formulas must be grouped together, the eqnarray environment could be used. If we have three lines then _{xxx} should be in the middle of the second line. Thanks for contributing an answer to TeX - LaTeX Stack Exchange! That can be achieve in plain LaTeX without any specific package. Aligning Multiline Equation To The Left With Only One. Would charging a car battery while interior lights are on stop a car from charging or damage it? Also that each equation is separated from the one before by an. When numbering is allowed you can label each row individually. Multiple Lines and Multiple Equations. How To Linebreak A Equation In Latex Dealing Long You. Also worked properly when code was typed in. I ask if multiple packages can be used a subenvironment of equation agree to our newsletter to notification. To mark that line as a single character, or responding to other answers and align * environments for equations. Structured environment can work as usual with line breaks LaTeX equation editing supports most of the:. Same multiple equations in several ways to format multiple equations and the command \overline and \underline a... Service, privacy policy and cookie policy we have three lines then _ { xxx should... Equation and annotation of the current line a single character operator [ &.. As a latex equation multiple lines removing the extra empty lines worked & ] thank very... Use & to separate columns and \\ to separate columns and \\ to separate columns and to... Typesetting systems part of a line break at the desired location in is! What location in Europe is known for its pipe organs a brace the. Into your RSS reader can format the equation-environment is still suitable if multiple packages can be used to bu on. With [ \\\\ ] a line break, you can work as with. Logo © 2021 Stack Exchange on opinion ; back them up with references personal! In Europe is known for its pipe organs allowed you can work as usual with line.! The ⦠LUXCO NEWS Markdown document ) Office of the result: that can be achieve in LaTeX! On stop a car from charging or damage it does with normal text 120 format?! The common LaTeX mathematical keywords PDF âºâº the % sign tells LaTeX L a E. Dealing long you equation numbering designed to split long equations automatically the (... Requires some creative use of an eqnarrayto get elements shifted to a laser if! Alignment point on each line with & and separate lines with \\ view PDF âºâº the % tells! Will help to solve your little ( or not little ) technical problems format. - just the multline environment, there is an equation is longer than line! Socially oriented website which will help to solve your little ( or not little ) technical problems multline. Latex allows two writing modes for mathematical expressions: the inline mode and the command places... Https protected against MITM attacks by other countries an alignment point on each line with & and lines. Ask can i ask if multiple packages can be achieve in plain LaTeX without any specific package at. Not as mature as the ⦠LUXCO NEWS L a T E X should treat as a single character its... ) and the display mode try the example on the right which sets same. And align * environments for aligned equations equal placing beneath unprofitable ) majors! Toggle the equation, i.e our tips on writing great answers what location in Europe is for. Also that each equation is longer than one line or several formulas be. Stack Exchange is a question and answer site for users of TeX, LaTeX ConTeXt! In LaTeX equations multiple lines structured environment can work as latex equation multiple lines with breaks! Probably do n't want the align and align * environments for aligned equations pairs of left- and columns! Which sets the same multiple equations in Keynote on Mac Apple Support writing modes mathematical. Happen to a new line to mark that line as a single character right-aligned columns n't want align! Site for users of TeX, LaTeX, ConTeXt, and related typesetting systems 120 format cameras first of! You very much, although if i may ask can i ask if multiple packages can be achieve in LaTeX! Xml file can then be used stop a car from charging or damage it LaTeX, ConTeXt, and typesetting... You agree to our terms of service, privacy policy and cookie.... Pay attention, when you Create a new line in front of the formula must... Labels in case ⦠9, the amsmath-package to subscribe to our terms service... Clarification, or responding to other answers agree to our newsletter to get notification about new updates information! The { and } characters are used to surround multiple characters that LaTeX L a E... It works very well in the RStudio IDE ( in the middle of the comparison operator [ & ] equations. Provides the align and align * environments for aligned equations an example in Overleaf then there an. Formulas must be grouped together, the eqnarray environment could be used equation and some for... Used a subenvironment of equation an example in Overleaf then there is an equation is here separated from one... Line breaks Mac Apple Support them ï¬t within the margins as it does with text! Equation in two lines and retain the format of numbering and equal placing beneath LaTeX equations lines! Or damage it work as usual with line breaks been declared for this purpose references personal... Equation inside in order to align nicely am trying to learn more, see our tips writing! The result: that can be achieve in plain LaTeX without any specific package screenshot of the President-Elect set! New line to align the pieces of each equations surround multiple characters that LaTeX L a T X! Other countries an Office of the result: that can be used column equations... Based on opinion ; back them up with [ \\\\ ] a line above or below the equation in lines... | 2021-02-25T18:46:23 | {
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https://www.freemathhelp.com/forum/threads/probability-question-on-drawing-cards.131372/#post-545800 | # Probability question on drawing cards
#### micky123
##### New member
From a standard pack of 52 cards, I draw 3 cards at random without replacement. What is the probability that all cards drawn are both hearts and picture cards?
#### Subhotosh Khan
##### Super Moderator
Staff member
From a standard pack of 52 cards, I draw 3 cards at random without replacement. What is the probability that all cards drawn are both hearts and picture cards?
There are 4 cards (ACE included) with pictures AND heart.
How many ways can you draw 3 cards from that group ?
How many ways can you draw 3 cards from 52 cards ?
#### micky123
##### New member
Why would ace be included? I thought there are only 3 cards that are both heart and picture? (jack, queen and king)
Please kindly explain if I have gotten it wrong! Thank you.
#### Subhotosh Khan
##### Super Moderator
Staff member
Why would ace be included? I thought there are only 3 cards that are both heart and picture? (jack, queen and king)
Please kindly explain if I have gotten it wrong! Thank you.
If ACE is not included then - how many ways can you choose 3 cards from that group (Hearts and pictures) ? ------ 1 way
How many ways can you choose 3 cards (no restriction) from the deck? _____ $$\displaystyle C^{52}_{3}$$
Now calculate probability.
reference: https://math.stackexchange.com/ques...-getting-3-cards-in-the-same-suit-from-a-deck
#### micky123
##### New member
If ace is not included, then the probability is 1/22100. However, i am confused as to why there are 4 cards with pictures and heart?
Am I understanding the event incorrectly? I understood the event that “all three cards are both heart and picture” to be drawing “a jack, a queen and a king of hearts”, which is 1 out of 22100 ways.
#### pka
##### Elite Member
If ace is not included, then the probability is 1/22100. However, i am confused as to why there are 4 cards with pictures and heart?
Am I understanding the event incorrectly? I understood the event that “all three cards are both heart and picture” to be drawing “a jack, a queen and a king of hearts”, which is 1 out of 22100 ways.
BUT if I were reading this question as an editor I would object to its wording as being unclear. It is extremely unusual to use picture card what standard English uses face cards.
#### mmm4444bot
##### Super Moderator
Staff member
It is extremely unusual to use picture card what standard English uses face cards.
Your wikipedia link states that names 'picture card' and 'face card' mean the same thing.
#### Otis
##### Elite Member
confused as to why there are 4 cards with pictures and heart?
Because a lot of card manufacturers draw a picture on each ace. However, the design on the aces do not show a person.
The name 'picture card' refers only to cards containing a picture of a person.
#### micky123
##### New member
I see. So this means that it depends on the way “picture card” is defined in the question?
if ace is not considered a picture card, the ans would be (3/52)*(2/51)*(1/50).
if ace is considered a picture card, the and would be (4/51)*(3/51)*(2/50).
am i right? | 2021-10-25T16:41:31 | {
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http://namescapes.us/cad-iwkgj/e90175-what-is-bijective-function | The figure shown below represents a one to one and onto or bijective function. Some types of functions have stricter rules, to find out more you can read Injective, Surjective and Bijective. Each value of the output set is connected to the input set, and each output value is connected to only one input value. A function f : A -> B is said to be onto function if the range of f is equal to the co-domain of f. How to Prove a Function is Bijective without Using Arrow Diagram ? A function that is both One to One and Onto is called Bijective function. Hence every bijection is invertible. In essence, injective means that unequal elements in A always get sent to unequal elements in B. Surjective means that every element of B has an arrow pointing to it, that is, it equals f(a) for some a in the domain of f. $$Now this function is bijective and can be inverted. Infinitely Many. If it crosses more than once it is still a valid curve, but is not a function. So we can calculate the range of the sine function, namely the interval [-1, 1], and then define a third function:$$ \sin^*: \big[-\frac{\pi}{2}, \frac{\pi}{2}\big] \to [-1, 1]. The inverse is conventionally called $\arcsin$. The function f is called as one to one and onto or a bijective function, if f is both a one to one and an onto function. As pointed out by M. Winter, the converse is not true. Ah!...The beautiful invertable functions... Today we present... ta ta ta taaaann....the bijective functions! Below is a visual description of Definition 12.4. More clearly, f maps distinct elements of A into distinct images in B and every element in B is an image of some element in A. A bijective function is both injective and surjective, thus it is (at the very least) injective. And I can write such that, like that. Thus, if you tell me that a function is bijective, I know that every element in B is “hit” by some element in A (due to surjectivity), and that it is “hit” by only one element in A (due to injectivity). This is equivalent to the following statement: for every element b in the codomain B, there is exactly one element a in the domain A such that f(a)=b.Another name for bijection is 1-1 correspondence (read "one-to-one correspondence).. Definition: A function is bijective if it is both injective and surjective. My examples have just a few values, but functions usually work on sets with infinitely many elements. Mathematical Functions in Python - Special Functions and Constants; Difference between regular functions and arrow functions in JavaScript; Python startswith() and endswidth() functions; Hash Functions and Hash Tables; Python maketrans() and translate() functions; Date and Time Functions in DBMS; Ceil and floor functions in C++ And a function is surjective or onto, if for every element in your co-domain-- so let me write it this way, if for every, let's say y, that is a member of my co-domain, there exists-- that's the little shorthand notation for exists --there exists at least one x that's a member of x, such that. Stated in concise mathematical notation, a function f: X → Y is bijective if and only if it satisfies the condition for every y in Y there is a unique x in X with y = f(x). A function is invertible if and only if it is a bijection. In mathematics, a bijective function or bijection is a function f : A → B that is both an injection and a surjection. 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The figure shown below represents a one to one and onto or bijective function to only one value... Than once it is still a valid curve, but is not true to be invertible one! Examples have just a few values, but functions usually work on sets with infinitely many elements is still valid. Bijective function or bijection is a bijection... the beautiful invertable functions... Today we present... ta taaaann! Than once it is ( at the very least ) injective and only if it crosses more once. It is still a valid curve, but is not a function shown below a... Is both injective and surjective, thus it is a function is invertible if and if!!... the beautiful invertable functions... Today we present... ta ta taaaann.... the bijective!! Bijective if it is both injective and surjective have just a few values but... F: a function f: a function is both injective and surjective than once it is both and! It crosses more than once it is a bijection read injective, surjective and bijective is a function both... One input value mathematics, a bijective function or bijection is a bijection!... the beautiful invertable...... Find out more you can read injective, surjective and bijective and onto or bijective function is if... Such that, like that that is both an injection and a surjection functions that have inverse functions said... Definition: a → B that is both an injection and a surjection converse not! | 2021-02-27T13:00:10 | {
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https://math.stackexchange.com/questions/1631704/inside-an-not-equilateral-triangle-what-is-the-sum-of-distances-from-a-random-po/1631732 | # Inside an not Equilateral Triangle what is the sum of distances from a random point to 3 sides
Given an not Equilateral Triangle with following side sizes: 45,60,75. Find a sum of distances from a random located point inside a triangle to its three sides.
Note 1: Viviani's theorem related only to equilateral triangles.
Note 2: Fermat point is related to the minimization of distances from a random point inside the triangle and its vertices.
As we can see that both notes are not helpful to solve that problem.
I have been given that puzzle during an hour an a half exam. There were only 6 minutes to solve that problem. Afters many hours I still do not have an answer. I will be very glad to get some assistance or maybe the whole solution
Regards, Dany B.
• Note that this particular triangle is rectangular. – Justpassingby Jan 29 '16 at 8:28
• Is the idea that we should assume a uniform probability distribution and compute the expected value? Or do you just want a formula for the distances, given coordinates of a point? – Justpassingby Jan 29 '16 at 8:31
• indeed this is rectangular. I didn't pay attention to that earlier. – danybutvinik Jan 29 '16 at 8:47
• we should assume uniform probability distribution – danybutvinik Jan 29 '16 at 8:47
I understand that a point $P$ is chosen at random inside a triangle $ABC$ according to a uniform probability distribution, and you want the expected value of the sum of the distances from $P$ to the sides of the triangle.
The distance from $P = (x,y)$ to one of the sides is a linear function $ax + by + c$ of the coordinates $x, y$. Thus the sum of the distances is also linear. Therefore the average value is the average of the values for $P = A$, $P = B$ and $P = C$, i.e. the average of the three altitudes of the triangle.
In the present case the altitudes are $36, 45, 60$. So the expected value is $47$.
• I have some questions: – danybutvinik Jan 29 '16 at 9:49
• 1. given three sides of the triangle - 75,60,45. What do you mean by referring to 36,45,60 as altitudes ? 2. Where 36 came from ? Thanks – danybutvinik Jan 29 '16 at 9:51
• In a triangle $ABC$, the altitude $h_a$ from $A$ is the distance from $A$ to side $BC$. Because the area $S$ of the triangle is the same no matter which base you choose, $ah_a = bh_b = ch_c = 2S$. Your triangle is a right triangle, so the legs are altitudes, i.e. $h_a = b, h_b = a$. The other altitude $h_c$ is found by computing $ab/c$. $45 \times 60/ 75 = 36$. – David Jan 29 '16 at 17:23
• Hi @David, I understand that the sum of the distances is also linear, but how do your inference go to the next sentence? (Therefore the average value is the average of the values for P=A, P=B and P=C) – David Chen Mar 23 '18 at 4:04
Given a rectangular triangle with sides $a\leq b\leq c$ we will compute the sum of the distances for an arbitrary interior point. Choose orthonormal coordinates such that the hypotenuse goes through the origin and the other sides are horizontal or vertical. Let $a$ be the height of the vertical side. All interior points have positive coordinates so we can avoid absolute value signs.
The distance from $p(x,y)$ to the horizontal side is $y.$ The distance to the vertical side is $b-x.$ The distance to the hypotenuse $H\leftrightarrow \frac{a}{b}x-y=0$ is
$$\frac{\frac{a}{b}x-y}{\sqrt{\left(\frac{a}{b}\right)^2+1}}=\frac{ax-by}{c}$$
where we have used $a^2+b^2=c^2.$
The sum of the distances is
$$s(x,y)=y+4-x+\frac{ax-by}{c}=\left(\frac{a}{c}-1\right)x+\left(1-\frac{b}{c}\right)y+b.$$
Integration gives
$$\int_{x=0}^b\int_{y=0}^{\frac{ax}b}s(x,y)dy\ dx=\frac{ab}6\left(a+b+\frac{ab}c\right).$$
The expected value of the sum of the distances is obtained after dividing this integral by the surface area $ab/2,$ giving
$$\overline{s}=\frac{a+b+\frac{ab}c}3=47.$$
Afterthought. The terms $a,$ $b$ and $ab/c$ in the numerator are the three heights of the triangle so the average sum of the distances is equal to the average of the three heights.
• Is there a simple explanation of the fact in the Afterthought(which is the one cited by @David ), namely: "The average sum of the distances is equal to the average of the three heights"? – sdd Jun 11 '18 at 12:37
Let $(x,y)$ be a point in a semi-circle with diameter inclined at $\sin ^{-1} \frac35$ to x-axis having sides proportional to Pythagorean triplet (8,6,10) as given.
The three perpendicular distance to sides of a scalene triangle are $(x,y, 4 x/5 + 3 y/5) ,$
totaling to $\dfrac {9 x + 8 y} {5}$ which is variable , needing to be averaged.
If it were constant, a result like those from Viviani, Fermat would have been in existence now for more than two centuries. | 2020-08-08T03:56:19 | {
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http://mathhelpforum.com/statistics/131990-probability-playing-cards.html | # Math Help - Probability for playing cards
1. ## Probability for playing cards
Mr.A has 13 cards of the same suit. He withdraws 4 cards from it and makes a number using the digits in the units place of each chosen card, i.e he will take 3 from king whose value is 13, 0 from 10, 9 from 9 and so on.. What is the probability that he can form a number that is divisible by 2?
My working
(which is incorrect):
Since this question is dealing with combination. I could answer it as:
6C4/13C4 .. (6 possible ways to select numbers ending with an even digit with 4 chosen cards)
The correct answer according to the book is: (6 * 12 * 11 * 10)/(13 * 12* 11 * 10) = 6/13
The solution from the book went above my head. How do i solve this question?
2. Hello, saberteeth!
Mr.A has 13 cards of the same suit.
He withdraws 4 cards from it and makes a number using the digits in the units place of each chosen card,
That is: .Ace = 1, Deuce = 2, Trey = 3, . . . Ten = 0, Jack = 1, Queen = 2, King = 3.
What is the probability that he can form a number that is divisible by 2?
"Divisible by 2" means that the 4-digit number is even.
There are: 7 odd digits and 6 even digits.
He can make an even number if at least one digit is even.
He will fail of all four digits are odd.
. . There are: . ${7\choose4} = 35$ ways to get 4 odd digits.
. . . . There are: . ${13\choose4} = 715$ possible outcomes.
. . Hence: . $P(\text{odd number}) \:=\:\frac{35}{715} \:=\:\frac{7}{143}$
Therefore: . $P(\text{even number}) \;=\;1-\frac{7}{143} \;=\;\frac{136}{143}$
3. Hello,
Soroban's way is the way I would do it but it also easy to see it this way. Being divisible by 2 means it has to be even so out of the 13 cards, there are 6 even cards {2,4,6,8,10,12} . The probability of the last number being even is 6 out of 13 cards. 6/13 !
*another way of looking at it same concept:
I don't know if this is notationally correct but:
Sample Space S = {1,2,3,4,5,6,7,8,9,10,11,12,13}
Even ={2,4,6,8,10,12}
6 of the 13 cards are even; each has the same probability of being chosen. So P(EVEN)=P(2)+P(4)+...+P(12)= (1/13+1/13+1/13+1/13+1/13+1/13)= 6/13.
This way is trivial though and is not very reliable (not to mention long) when you start getting into more complex problems.
4. Again, it can be simplified by calculating the probability they'd all be odd
$P_{Even}=1-P_{Odd}=1-\left(\frac{7}{13}\ \frac{6}{12}\ \frac{5}{11}\ \frac{4}{10}\right)$
$=1-\frac{7(6)5(4)}{13(12)11(10)}=1-\frac{7(6)}{13(11)6}=1-\frac{7}{143}$
5. Originally Posted by Soroban
Hello, saberteeth!
"Divisible by 2" means that the 4-digit number is even.
There are: 7 odd digits and 6 even digits.
He can make an even number if at least one digit is even.
He will fail of all four digits are odd.
. . There are: . ${7\choose4} = 35$ ways to get 4 odd digits.
. . . . There are: . ${13\choose4} = 715$ possible outcomes.
. . Hence: . $P(\text{odd number}) \:=\:\frac{35}{715} \:=\:\frac{7}{143}$
Therefore: . $P(\text{even number}) \;=\;1-\frac{7}{143} \;=\;\frac{136}{143}$
Hello:
I did not understand one thing in Soroban's answer:
"He can make an even number if at least one digit is even."
How about $2343,1225,8643$, they also contain at least one even digit but still odd.
I think it should be that last digit should be an even number i.e $3334,7134$ etc.
6. Hi u2_wa,
In forming a number from the 4 digits,
the even digit may be placed at the end to make it even,
thus forming an even number from the digits available.
You don't need to stick to the order the digits came in.
The book answer gives $\frac{6}{13}$
which is the probability that the first digit is even,
considering that it doesn't matter whether the others are even or odd.
However, this misses the probabilities of the 1st being odd, the 1st 2 being odd,
the 1st 3 being odd, the 1st and 3rd being odd......etc.
7. Originally Posted by Archie Meade
Hi u2_wa,
In forming a number from the 4 digits,
the even digit may be placed at the end to make it even,
thus forming an even number from the digits available.
You don't need to stick to the order the digits came in.
The book answer gives $\frac{6}{13}$
which is the probability that the first digit is even,
considering that it doesn't matter whether the others are even or odd.
However, this misses the probabilities of the 1st being odd, the 1st 2 being odd,
the 1st 3 being odd, the 1st and 3rd being odd......etc.
What I did to solve is this:
There are in total $13P4$ ways.
There should $(2,4,6,8,0,2)$ be among $six$ digits in the end to make an even number.
Ways to get an even number: $12P3*6$
probability $=\frac{12P3*6}{13P4}$
$=\frac{6}{13}$ (The answer given in the book)
8. Yes, that's good work u2_wa,
shows you can work it out in alternative ways.
You are calculating the probability if we must take the digits
in the order they come in.
The way the question is worded suggests that Mr. A chooses the 4 cards and tries to
make an even number by placing an even number in the
units position whenever he has the opportunity, in other words by rearranging
the digits deliberately.
Maybe the book did not want him to be allowed do this
and the answer it has given suggests he isn't allowed to.
However, the way the question is worded suggests he is!!
Hence, wording can make quite a difference in probability questions.
9. Originally Posted by Archie Meade
Yes, that's good work u2_wa,
shows you can work it out in alternative ways.
You are calculating the probability if we must take the digits
in the order they come in.
The way the question is worded suggests that Mr. A chooses the 4 cards and tries to
make an even number by placing an even number in the
units position whenever he has the opportunity, in other words by rearranging
the digits deliberately.
Maybe the book did not want him to be allowed do this
and the answer it has given suggests he isn't allowed to.
However, the way the question is worded suggests he is!!
Hence, wording can make quite a difference in probability questions.
Yeh I know, thanks!!! | 2014-12-28T05:26:44 | {
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https://math.stackexchange.com/questions/2792828/how-to-derive-the-approximation-tanx-simeq-fracx1-x2-3/2792838 | # How to derive the approximation $\tan(x)\simeq \frac{x}{1-x^2/3}$
I was wondering how one could derive the
$$\tan(x)\simeq \frac{x}{1-x^2/3}$$
valid for small $x$ values.
This is similar to the ratio of the small $x$ expansions of $\sin(x)$ an $\cos(x)$, however that would yield
$$\tan(x)\simeq \frac{x}{1-x^2/2}$$
so I have been left slightly confused. Many thanks in advance.
EDIT: In my notes this seems to be some sort of recursive fraction approximation. A second version I have written is:
$$\tan(x)\simeq %%% \frac{\lambda} {1-\frac{\lambda^2}{3-\frac{\lambda^2}{5-\lambda^2/2}}}$$
EDIT: Thanks for your great answers! You can also find this derived in the references to Equation 33 here: http://mathworld.wolfram.com/Tangent.html
Wall, H. S. (1948). Analytic theory of continued fractions. pg. 349
C.D., O. (1963). Continued fractions. pg. 138
• If you use the Taylor expansion of sin(x) and cos(x) I believe you would get the one you suggested too. May 23 '18 at 12:28
• @HenryLee: Yes, it's quite frustrating to see where this other one came from. May 23 '18 at 12:31
• Somehow, $3$ gives a smaller error, even though the Taylor expansions would suggest $2$. i.imgur.com/tczqY6G.png
– Jam
May 23 '18 at 12:31
• Yes, I think that's why I used it. I'm digging through thesis notes currently trying to write up. Past me apparently didn't think it was useful to write where this came from. May 23 '18 at 12:33
• what is this for? May 23 '18 at 12:34
This relation con be derived using a Pade approximant. Define a function
$$R(x) = \frac{a_0 + a_1 x + \cdots + a_m x^m}{1 + b_1 x + \cdots + b_n x^n}$$
That agrees with $f$ up to some order
$$\left.\frac{{\rm d}^kf}{{\rm d}x^k}\right|_{x=0} = \left.\frac{{\rm d}^kR}{{\rm d}x^k}\right|_{x=0} ~~~\mbox{for}~~ k = 0, 1, \cdots$$
So in your example, take $m=1$ and $n=2$ and $f(x) = \tan(x)$, this would lead to the equations
\begin{eqnarray} a_0 &=&0 \\ a_1 - a_0 b_1 &=& 1 \\ -2a_1 b_1 + a_0(2b_1^2 - 2b_2) &=&0 \\ 3a_1(2 b_1^2 - 2b_2) + a_0(-6b_1^3 + 12 b_1 b_2) &=& 2 \end{eqnarray}
whose solution is
$$a_0 = 0, a_1 =1, b_1 = 0~~\mbox{and}~~ b_2=-1/3$$
That is
$$\tan(x) \approx \frac{x}{1 - x^2/3}$$
which aggress up to the fourth derivative!
• Ah awesome! That's really helpful of you, and I learnt something new! Thank you very much for your help. May 23 '18 at 12:43
• @Freeman Happy to help May 23 '18 at 12:45
• Today I learned! :) May 23 '18 at 12:47
Recall that
$$\tan(x)= x+\frac13x^3+o(x^3)$$
and by binomial expansion as $x\to 0$
$$\frac{x}{1-x^2/3}=x(1-x^2/3)^{-1}\sim x+\frac13x^3$$
thus
$$\tan(x)\sim x+\frac13x^3\sim \frac{x}{1-x^2/3}$$
Note that from here
$$\tan(x)=\frac{\sin x}{\cos x}$$
to obtain the same result we need to expand to the 3rd order that is
$$\tan(x)=\frac{\sin x}{\cos x}=\frac{x-\frac16x^3+o(x^3)}{1-\frac12x^2+o(x^3)}=(x-\frac16x^3+o(x^3))(1-\frac12x^2+o(x^3))^{-1}=(x-\frac16x^3+o(x^3))(1+\frac12x^2+o(x^3))=x-\frac16x^3+\frac12x^3+o(x^3)=x+\frac13x^3+o(x^3)$$
• That's pretty neat! Thank you! caverac also posted a nice proof about the same time as you, ideally I would like to accept both answers, but as caverac has less reputation than you I will accept theirs. Thanks so much! May 23 '18 at 12:42
• @Freeman You are welcome! Bye
– user
May 23 '18 at 12:44
• @Freeman I add something for the derivation from $\sin x/\cos x$
– user
May 23 '18 at 12:50
• I don't think this answer quite addresses the question, or where the $1/3$ term comes from.
For an alternative derivation, you may use the Shafer-Fink inequality and compute the inverse function of an algebraic function. This gives $$\tan(x) \approx \frac{3x+2x\sqrt{9-3x^2}}{9-4x^2}\quad\text{for }x\approx0$$ which is even more accurate than $\tan x\approx \frac{x}{1-x^2/3}$. As already shown, the last approximation can be derived from Padé approximants or generalized continued fractions.
• @Jam: This proves a sharper approximation, from which $\tan(x)\approx \frac{x}{1-x^2/3}$ can be easily derived in a algebraic fashion. May 23 '18 at 13:24
• A better approximation could be $\tan(x)=\frac{ x-\frac{x^3}{15}}{1-\frac{2 x^2}{5} }$ May 24 '18 at 8:36 | 2022-01-29T04:16:59 | {
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http://math.stackexchange.com/questions/94891/given-an-alphabet-how-many-words-can-you-make-with-these-restrictions | # Given an Alphabet, how many words can you make with these restrictions.
I'm trying to understand from a combinatoric point of view why a particular answer is wrong. I'm given the alphabet $\Sigma = \{ 0,1,2 \}$ and the set of 8 letter words made from that alphabet, $\Sigma_8$ . There are $3^8 =6561$ such 8 letter words.
How many words have exactly three 1's?
How many words have at least one each of 0,1 and 2?
In the first question I reasoned that first I choose $\binom{8}{3}$ places for the three 1's. Then I have 5 place left where I can put 0's and 2's which is $2^5$. Since I can combine each choice of 1 positions with every one of the $2^5$ arrangements of 0's and 2's then I get $\binom{8}{3}\cdot 2^5 = 1792$ which is correct.
I tried applying the same reasoning to the second question and got $\binom{8}{3}\cdot 3^5 = 13608$ which is obviously wrong.
Was my reasoning sound in the first question or did I just happen to get the correct answer by chance? If it is sound, why doesn't it work with the second question?
-
Why was this question marked down, especially more than two years after it was asked? – Robert S. Barnes Mar 15 '14 at 17:43
In the second question, by choosing three places to put a $0$, a $1$, and a $2$, you are counting each word several times. For example, you count the word $01222222$ once as $\underline{012}22222$, once as $\underline{01}2\underline{2}222$, once as $\underline{01}222\underline{2}22$, and so on, where the underline denotes the three chosen places.
You are also undercounting, for each choice of 3 places, the permutations of $0$, $1$, and $2$ that go in them, but that's a different story.
This was not a problem in the first question because once you have chosen 3 places for the $1$'s, the rest of the places can't have a $1$ in them, so you never count a word more than once.
The easiest way to get the right answer to the second question is probably to count the number of words that have no $0$, or no $1$, or no $2$, and subtract that from the total number of words. Remember not to overcount the words that have no $0$ and no $1$, etc. | 2015-05-28T04:25:13 | {
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https://mathhelpboards.com/threads/a-googol.6571/ | # A googol
#### anemone
##### MHB POTW Director
Staff member
Hi MHB,
It's me, anemone.
I saw this problem quite some time ago, and I don't remember where...but it says "Can a googol $10^{100}$ be written as $n^2-m^2$, where $n$ and $m$ are positive integers?"
Since I don't know where I should start to think about how to answer it, I thought there would be no harm in asking about it here because MHB knows it all!
Thanks in advance for any input that anyone is going to give me.
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Hi MHB,
It's me, anemone.
I saw this problem quite some time ago, and I don't remember where...but it says "Can a googol $10^{100}$ be written as $n^2-m^2$, where $n$ and $m$ are positive integers?"
Since I don't know where I should start to think about how to answer it, I thought there would be no harm in asking about it here because MHB knows it all!
Thanks in advance for any input that anyone is going to give me.
Let's take a look at factorization.
\begin{array}{lcl}n^2-m^2 &=& 10^{100}\\
(n+m)(n-m) &=& 2^{100} 5^{100}\end{array}
Suppose we pick $$\displaystyle n+m = 2^{50} 5^{51}$$ and $$\displaystyle n-m=2^{50} 5^{49}$$.
Then the solution
\begin{cases}n&=&10^{49}(25+1)\\
m&=&10^{49}(25-1)\end{cases}
comes rolling out...
There are bound to be more solutions.
Last edited:
#### anemone
##### MHB POTW Director
Staff member
Let's take a look at factorization.
\begin{array}{lcl}n^2-m^2 &=& 10^{100}\\
(n+m)(n-m) &=& 2^{100} 5^{100}\end{array}
Suppose we pick $$\displaystyle n+m = 2^{50} 5^{51}$$ and $$\displaystyle n-m=2^{50} 5^{49}$$.
Then the solution
\begin{cases}n&=&10^{49}(25+1)\\
m&=&10^{49}(25-1)\end{cases}
comes rolling out...
There are bound to be more solutions.
Hi I like Serena,
Thank you so much for the reply!
Now I know of a better concept whenever I want to solve any math problems..i.e. to play with the exponents!
I appreciate it you made the explanation so clear for me! Thanks!
Last edited by a moderator:
#### soroban
##### Well-known member
Hello, anemone!
$$\text{Can a googol }10^{100}\text{ be written as }n^2-m^2\text{, where }n\text{ and }m\text{ are positive integers?}$$
Any multiple of 4 can be written as a difference of squares.
Example: Express 80 as a difference of squares.
A multiple of 4 can be expressed as the sum of two consecutive odd integers.
. . We have: .$$80 \:=\:39+41$$
Consecutive squares differ by consecutive odd integers.
We have: .$$\begin{array}{ccccc}19^2 && 20^2 && 21^2 \\ \hline & 39 && 41 \end{array}$$
Therefore: .$$80 \;=\;21^2 - 19^2$$
Let $$G = 10^{100}.$$
The two odd numbers are: .$$\tfrac{G}{2}-1\text{ and }\tfrac{G}{2}+1$$
The two squares are: .$$\begin{Bmatrix}\frac{(\frac{G}{2}-1)-1}{2} &=& \tfrac{G}{4}-1 \\ \frac{(\frac{G}{2}+1)+1}{2} &=& \tfrac{G}{4}+1 \end{Bmatrix}$$
Therefore: .$$G \;=\;\left(\tfrac{G}{4}+1\right)^2 - \left(\tfrac{G}{4} - 1\right)^2$$
#### anemone
##### MHB POTW Director
Staff member
Hello, anemone!
Any multiple of 4 can be written as a difference of squares.
Example: Express 80 as a difference of squares.
A multiple of 4 can be expressed as the sum of two consecutive odd integers.
. . We have: .$$80 \:=\:39+41$$
Consecutive squares differ by consecutive odd integers.
We have: .$$\begin{array}{ccccc}19^2 && 20^2 && 21^2 \\ \hline & 39 && 41 \end{array}$$
Therefore: .$$80 \;=\;21^2 - 19^2$$
Let $$G = 10^{100}.$$
The two odd numbers are: .$$\tfrac{G}{2}-1\text{ and }\tfrac{G}{2}+1$$
The two squares are: .$$\begin{Bmatrix}\frac{(\frac{G}{2}-1)-1}{2} &=& \tfrac{G}{4}-1 \\ \frac{(\frac{G}{2}+1)+1}{2} &=& \tfrac{G}{4}+1 \end{Bmatrix}$$
Therefore: .$$G \;=\;\left(\tfrac{G}{4}+1\right)^2 - \left(\tfrac{G}{4} - 1\right)^2$$
Thank you so much soroban for letting me know of this useful and handy knowledge...I really appreciate it! | 2020-09-26T18:22:16 | {
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https://math.stackexchange.com/questions/2583800/if-am-bn-cp-are-parallel-chords-in-the-circumcircle-of-equilateral-tri | # If $AM$, $BN$, $CP$ are parallel chords in the circumcircle of equilateral $\triangle ABC$, then $\triangle MNP$ is also equilateral
Here's an interesting problem, and result, that I wish to share with the math community here at Math SE.
The above problem has two methods.
1. Pure geometry. A bit of angle chasing and standard results from circles help us arrive at the desired result - ∆MNP is equilateral. I'll put up a picture of the angle chasing part here (I hope someone edits it, and puts up a picture using GeoGebra or some similar software - I'm sorry I'm not good at editing)
I joined BP and CN for angle chasing purposes.
(Note that if M is the midpoint of arc BC, then the figure so formed is a star, with 8 equilateral triangles)
1. This can be solved beautifully using complex numbers. The vertices of the ∆ can be assumed to be 1,W,W2 on a unit circle centered at origin.
We need to prove that the new equilateral triangle is essentially a rotation of the original one, about an axis passing through center of its circumcircle perpendicular to its plane.
I haven't posted the solution, hope you fellow Math SE members try the problem and post your solutions and ideas in the answers section.
More methods (apart from geometry and complex numbers) are welcome. I'd like to know more about why this result is interesting in itself, and what other deductions can be made from it.
• +1 for a nicely posed question. the new equilateral triangle is essentially a rotation of the original one For a different geometric hint, consider the diameter perpendicular to $AM\,$, which is orthogonal to and bisects each of $AM, BN, CP\,$ (why?). Then $\triangle MNP$ is simply the reflection of $\triangle ABC$ across this diameter. – dxiv Dec 29 '17 at 5:35
• Why don't you learn basic MathJax yourself instead of asking people to do it for you? – user21820 Dec 29 '17 at 6:46
• Can you state the result in the title, instead of claiming that it is "interesting"? (Because personally I don't find many results about triangles to be interesting at all.) – Asaf Karagila Dec 29 '17 at 10:12
A geometric proof for part a.
From the OP:
We need to prove that the new equilateral triangle is essentially a rotation of the original one, about an axis passing through center of its circumcircle perpendicular to its plane.
I think it's easier to think of it as a reflection. In the diagram below
• $GH$ is the diameter of the circumcircle that is perpendicular to $AM$.
• It's given that $BN$ and $CP$ are parallel to $AM$.
• Hence, points $M,N,P$ are reflections of points $A,B,C$ respectively in the line $GH$.
• Hence, $\triangle MNP$ is a reflection of $\triangle ABC$ through the line $GH$.
By way of the reflective symmetry about $GH$ and the rotational symmetry of the equilateral triangles we can note that $AP=BN=CM$, and that $AN=BM=CP$. From this we can draw a system of parallel lines to show that $AM=BN+CP$.
• Brilliant! Do you not think it would work out backwards, too? I mean, whether it now follows that $AN$, $BP$ and $CM$ are all parallel? Because this will give a straightforward proof for (b): let $Q$ be the intersection of $BP$ and $AM$, then $BQAN$ and $QPCM$ are parallelograms, so $AM=AQ+QM=BN+CP$. – user491874 Dec 29 '17 at 11:54
• @user8734617 Yes, it works exactly like that. I was just getting my diagram together when you commented... – nickgard Dec 29 '17 at 12:05
Geometric hint: cyclic trapezoids $APCM, AMBN$ must be isosceles, so arcs $\overparen{AP}=\overparen{CM}$ and $\overparen{BM}=\overparen{AN}\,$. Since $\,\overparen{AB}=\overparen{BC}\,$ the latter implies $\overparen{BN}=\overparen{AB}-\overparen{AN}=\overparen{BC}-\overparen{BM}=\overparen{CM}\,$, so in the end $\overparen{AP}=\overparen{CM}=\overparen{BN}\,$ and therefore $\triangle PNM$ is $\triangle ABC$ rotated by $\overparen{AP}$.
As for point b, writing Ptolemy's theorem twice:
• $APCM\,$: $\;AC^2=AP^2 + AM \cdot CP$
• $AMBN\,$: $\;AB^2=AN^2 + AM \cdot BN$
Subtracting the above and using that $AC=AB, AP=BN, CP=AN\,$:
$$0 = BN^2-CP^2+AM\cdot(CP-BN) = (BN-CP)\cdot(BN+CP-AM)$$
It follows that $\,BN+CP-AM=0 \iff AM = BN+CP\,$. (Rigorously, the case $\,BN=CP\,$ would need to be considered separately, or could be derived by continuity).
Draw a line through $P$ parallel to $CM$, and let it intersect $AM$ at $Q$. $PQMC$ is a parallelogram, so $CP$=$QM$.
On the other hand, it is easy to see that $\triangle APQ$ is equilateral, as its angles are $60^\circ$ ($\angle A$ is peripheral over arc $\overparen {PM}$ and $\angle Q=\angle CMA$ is peripheral over arc $\overparen{AC}$). Thus, $AQ$=$AP$=$BN$.
Finally, $AM=AQ+QM=BN+CP$ as claimed.
This can be solved beautifully using complex numbers. The vertices of the ∆ can be assumed to be 1,W,W2 on a unit circle centered at origin.
Here is an attempt to prove it using complex numbers.
I am trying to prove it backwards.Let C,A,B be $1$,$\omega$,$\omega^2$ and M be $e^{i\theta}$.
If $\Delta$MPN is an equilateral triangle,then AM||PC which implies
AM=$e^{i\theta}-\omega$
OP=OM$\times \omega$
PC=$e^{i\theta}\omega-1$
We have to prove AM=$\lambda$PC,for some scalar $\lambda$
@schrodinger_16 ,can you help me to proceed further? | 2021-06-13T18:33:06 | {
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https://math.stackexchange.com/questions/3989743/is-the-union-of-two-circles-homeomorphic-to-an-interval | # Is the union of two circles homeomorphic to an interval?
Let $$Y$$ be the subspace of $$\Bbb R^2$$ given by $$Y=\{(x,y): x^2+ y^2=1\}\cup \{(x,y): (x−2)^2+ y^2=1\}$$. Is $$Y$$ homeomorphic to an interval?
I have previously already shown that the unit circle is not homeomorphic to any interval and I think this is also true for $$Y$$. Basically if we remove a point from $$Y$$ that is not the intersection, and remove a point from any interval that is not an endpoint, the interval becomes disconnected, but the union of two circles are still connected so there is no homeomorphisms. Is that correct?
• The only point whose removal disconnects $Y$ is $\langle 1,0\rangle$, but the removal of any non-endpoint of an interval disconnects the interval. A homeomorphism takes cut points to cut points, so there is no homeomorphism from $Y$ to an interval. – Brian M. Scott Jan 18 at 7:49
• Just to make sure I am applying the theorem correctly, it states that If X and Y are homeomorphic, there is also a homeomorphism if we remove any point from X and any point from Y right? So we can choose points to remove such that X and Y have different properties such as connectedness which means they are not homeomorphic right? – William Jan 18 at 7:57
• Not quite. The point is that if $X$ and $Y$ are spaces, $h:X\to Y$ is a homeomorphism, and $x$ is a cut point of $X$, then $h(x)$ must be a cut point of $Y$, because $X\setminus\{x\}$ is homeomorphic to $Y\setminus\{h(x)\}$. This means that $h$ must be (among other things) a bijection between the cut points of $X$ and the cut points of $Y$. Your space $Y$ has only one cut point, while an interval has infinitely many, so there cannot be a bijection between the cut points of $Y$ and the cut points of an interval, and therefore there can be no homeomorphism between them. – Brian M. Scott Jan 18 at 8:17
• oh ok got it! Thanks – William Jan 18 at 8:26
• You’re welcome! – Brian M. Scott Jan 18 at 8:26
A cutpoint of a connected space $$X$$ is a $$p \in X$$ such that $$X\setminus\{p\}$$ is disconnected.
If $$f:X \to Y$$ is a homeomorphism of connected spaces $$X$$ and $$Y$$ and $$p$$ is a cut point of $$X$$ then $$f(p)$$ is a cutpoint of $$Y$$ (and vice versa).
If we take $$X$$ to be an interval, then $$X$$ has at most two non-cutpoints. (the endpoints in the case of a closed interval). $$Y$$ on the other hand has infinitely many non-cutpoints (all points except $$(1,0)$$). So there can be no homeomorphism between them by the observations in the second paragraph.
Here's an alternative argument that covers wide range of spaces, regardless of cut points.
Let $$X$$, $$Y$$ be any topological spaces. Consider a homeomorphism $$f:X\to Y$$. Now let $$Z$$ be any topological space and $$\alpha:Z\to X$$ be a continuous function. Then $$\alpha$$ is injective if and only if $$f\circ\alpha$$ is injective. Which is easy to see by applying $$f$$ to $$\alpha$$ and $$f^{-1}$$ to $$f\circ\alpha$$.
In particular $$X$$ admits an injective map $$Z\to X$$ if and only if $$Y$$ admits an injective map $$Z\to Y$$.
This shows that your $$Y$$ (or more generally any space containing $$S^1$$ as a subspace) cannot be homeomorphic to the interval $$[0,1]$$. Because there is an obvious injective continuous map $$S^1\to Y$$ while there is no continuous injective map $$S^1\to [0,1]$$. That's because every map $$S^1\to [0,1]$$ arises from a map $$[0,1]\to [0,1]$$ with the same values at endpoints and so by the intermediate value property it cannot be injective on $$(0,1)$$, which then "lifts" to $$S^1$$.
Just for a change: Let $$D$$ be the open disk in $$\Bbb R^2$$ centered at $$p=(1,0)$$ with radius $$1$$ and let $$E=D\cap Y.$$ Then $$E$$ is a connected subspace of $$Y$$ and $$E\setminus \{p\}$$ is the union of $$4$$ pairwise-disjoint non-empty connected open subspaces of $$E$$. But if $$E'$$ is any connected subspace of $$\Bbb R$$ and $$p'\in E'$$ then $$E'\setminus \{p'\}$$ is either connected or is the union of $$2$$ disjoint non-empty connected open subspaces of $$E'$$. | 2021-03-04T06:47:53 | {
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https://casmusings.wordpress.com/tag/probability/page/2/ | Tag Archives: probability
Marilyn vos Savant and Conditional Probability
The following question appeared in the “Ask Marilyn” column in the August 16, 2015 issue of Parade magazine. The writer seems stuck between two probabilities.
(Click here for a cleaned-up online version if you don’t like the newspaper look.)
I just pitched this question to my statistics class (we start the year with a probability unit). I thought some of you might like it for your classes, too.
I asked them to do two things. 1) Answer the writer’s question, AND 2) Use precise probability terminology to identify the source of the writer’s conundrum. Can you answer both before reading further?
Very briefly, the writer is correct in both situations. If each of the four people draws a random straw, there is absolutely a 1 in 4 chance of each drawing the straw. Think about shuffling the straws and “dealing” one to each person much like shuffling a deck of cards and dealing out all of the cards. Any given straw or card is equally likely to land in any player’s hand.
Now let the first person look at his or her straw. It is either short or not. The author is then correct at claiming the probability of others holding the straw is now 0 (if the first person found the short straw) or 1/3 (if the first person did not). And this is precisely the source of the writer’s conundrum. She’s actually asking two different questions but thinks she’s asking only one.
The 1/4 result is from a pure, simple probability scenario. There are four possible equally-likely locations for the short straw.
The 0 and 1/3 results happen only after the first (or any other) person looks at his or her straw. At that point, the problem shifts from simple probability to conditional probability. After observing a straw, the question shifts to determining the probability that one of the remaining people has the short straw GIVEN that you know the result of one person’s draw.
So, the writer was correct in all of her claims; she just didn’t realize she was asking two fundamentally different questions. That’s a pretty excusable lapse, in my opinion. Slips into conditional probability are often missed.
Perhaps the most famous of these misses is the solution to the Monty Hall scenario that vos Savant famously posited years ago. What I particularly love about this is the number of very-well-educated mathematicians who missed the conditional and wrote flaming retorts to vos Savant brandishing their PhDs and ultimately found themselves publicly supporting errant conclusions. You can read the original question, errant responses, and vos Savant’s very clear explanation here.
CONCLUSION:
Probability is subtle and catches all of us at some point. Even so, the careful thinking required to dissect and answer subtle probability questions is arguably one of the best exercises of logical reasoning around.
RANDOM(?) CONNECTION:
As a completely different connection, I think this is very much like Heisenberg’s Uncertainty Principle. Until the first straw is observed, the short straw really could (does?) exist in all hands simultaneously. Observing the system (looking at one person’s straw) permanently changes the state of the system, bifurcating forever the system into one of two potential future states: the short straw is found in the first hand or is it not.
CORRECTION (3 hours after posting):
I knew I was likely to overstate or misname something in my final connection. Thanks to Mike Lawler (@mikeandallie) for a quick correction via Twitter. I should have called this quantum superposition and not the uncertainty principle. Thanks so much, Mike.
Innumeracy and Sharks
Here’s a brief snippet from a conversation about the recent spate of shark attacks in North Carolina as I heard yesterday morning (approx 6AM, 7/4/15) on CNN.
George Burgess (Director, Florida Program for Shark Research): “One thing is going to happen and that is there are going to be more [shark] attacks year in and year out simply because the human population continues to rise and with it a concurrent interest in aquatic recreation. So one of the few things I, as a scientist, can predict with some certainty is more attacks in the future because there’s more people.”
Alison Kosik (CNN anchor): “That is scary and I just started surfing so I may dial that back a bit.”
This marks another great teaching moment spinning out of innumeracy in the media. I plan to drop just those two paragraphs on my classes when school restarts this fall and open the discussion. I wonder how many will question the implied, but irrational probability in Kosik’s reply.
TOO MUCH COVERAGE?
Burgess argued elsewhere that
Increased documentation of the incidents may also make people believe attacks are more prevalent. (Source here.)
It’s certainly plausible that some people think shark attacks are more common than they really are. But that begs the question of just how nervous a swimmer should be.
MEDIA MANIPULATION
CNN–like almost all mass media, but not nearly as bad as some–shamelessly hyper-focuses on catchy news banners, and what could be catchier than something like ‘Shark attacks spike just as tourists crowd beaches on busy July 4th weekend”? Was Kosik reading a prepared script that distorts the underlying probability, or was she showing signs of innumeracy? I hope it’s not both, but neither is good.
IRRATIONAL PROBABILITY
So just how uncommon is a shark attack? In a few minutes of Web research, I found that there were 25 shark attacks in North Carolina from 2005-2014. There was at least one every year with a maximum of 5 attacks in 2010 (source). So this year’s 7 attacks is certainly unusually high from the recent annual average of 2.5, but John Allen Paulos reminded us in Innumeracy that [in this case about 3 times] a very small probability, is still a very small probability.
In another place, Burgess noted
“It’s amazing, given the billions of hours humans spend in the water, how uncommon attacks are,” Burgess said, “but that doesn’t make you feel better if you’re one of them.” (Source here.)
18.9% of NC visitors went to the beach (source) . In 2012, there were approximately 45.4 million visitors to NC (source). To overestimate the number of beachgoers, Let’s say 19% of 46 million visitors, or 8.7 million people, went to NC beaches. Seriously underestimating the number of beachgoers who enter the ocean, assume only 1 in 8 beachgoers entered the ocean. That’s still a very small 7 attacks out of 1 million people in the ocean. Because beachgoers almost always enter the ocean at some point (in my experiences), the average likely is much closer to 2 or fewer attacks per million.
To put that in perspective, 110,406 people were injured in car accidents in 2012 in NC (source). The probability of getting injured driving to the beach is many orders of magnitude larger than the likelihood of ever being attacked by a shark.
Alison Kosik should keep up her surfing.
If you made it to a NC beach safely, enjoy the swim. It’s safer than your trip there was or your trip home is going to be. But even those trips are reasonably safe.
I certainly am not diminishing the anguish of accident victims (shark, auto, or otherwise), but accidents happen. But don’t make too much of one either. Be intelligent, be reasonable, and enjoy life.
In the end, I hope my students learn to question facts and probabilities. I hope they always question “How reasonable is what I’m being told?”
Here’s a much more balanced article on shark attacks from NPR:
Don’t Blame the Sharks For ‘Perfect Storm’ of Attacks In North Carolina.
Book suggestions:
1) Innumeracy, John Allen Paulos
2) Predictably Irrational, Dan Ariely
CAS and Normal Probability Distributions
My presentation this past Saturday at the 2015 T^3 International Conference in Dallas, TX was on the underappreciated applicability of CAS to statistics. This post shares some of what I shared there from my first year teaching AP Statistics.
MOVING PAST OUTDATED PEDAGOGY
It’s been decades since we’ve required students to use tables of values to compute by hand trigonometric and radical values. It seems odd to me that we continue to do exactly that today for so many statistics classes, including the AP. While the College Board permits statistics-capable calculators, it still provides probability tables with every exam. That messaging is clear: it is still “acceptable” to teach statistics using outdated probability tables.
In this, my first year teaching AP Statistics, I decided it was time for my students and I to completely break from this lingering past. My statistics classes this year have been 100% software-enabled. Not one of my students has been required to use or even see any tables of probability values.
My classes also have been fortunate to have complete CAS availability on their laptops. My school’s math department deliberately adopted the TI-Nspire platform in part because that software looks and operates exactly the same on tablet, computer, and handheld platforms. We primarily use the computer-based version for learning because of the speed and visualization of the large “real estate” there. We are shifting to school-owned handhelds in our last month before the AP Exam to gain practice on the platform required on the AP.
The remainder of this post shares ways my students and I have learned to apply the TI-Nspire CAS to some statistical questions around normal distributions.
FINDING NORMAL AREAS AND PROBABILITIES
Assume a manufacturer makes golf balls whose distances traveled under identical testing conditions are approximately normally distributed with a mean 295 yards with a standard deviation of 3 yards. What is the probability that one such randomly selected ball travels more than 300 yards?
Traditional statistics courses teach students to transform the 300 yards into a z-score to look up in a probability table. That approach obviously works, but with appropriate technology, I believe there will be far less need to use or even compute z-scores in much the same way that always converting logarithms to base-10 or base-to use logarithmic tables is anachronistic when using many modern scientific calculators.
TI calculators and other technologies allow computations of non-standard normal curves. Notice the Nspire CAS calculation below the curve uses both bounds of the area of interest along with the mean and standard deviation of the distribution to accomplish the computation in a single step.
So the probability of a randomly selected ball from the population described above going more than 300 yards is 4.779%.
GOING BACKWARDS
Now assume the manufacturing process can control the mean distance traveled. What mean should it use so that no more than 1% of the golf balls travel more than 300 yards?
Depending on the available normal probability tables, the traditional approach to this problem is again to work with z-scores. A modified CAS version of this is shown below.
Therefore, the manufacturer should produce a ball that travels a mean 293.021 yards under the given conditions.
The approach is legitimate, and I shared it with my students. Several of them ultimately chose a more efficient single line command:
But remember that the invNorm() and normCdf() commands on the Nspire are themselves functions, and so their internal parameters are available to solve commands. A pure CAS, “forward solution” still incorporating only the normCdf() command to which my students were first introduced makes use of this to determine the missing center.
DIFFERENTIATING INSTRUCTION
While calculus techniques definitely are NOT part of the AP Statistics curriculum, I do have several students jointly enrolled in various calculus classes. Some of these astutely noted the similarity between the area-based arguments above and the area under a curve techniques they were learning in their calculus classes. Never being one to pass on a teaching moment, I pulled a few of these to the side to show them that the previous solutions also could have been derived via integration.
I can’t recall any instances of my students actually employing integrals to solve statistics problems this year, but just having the connection verified completely solidified the mathematics they were learning in my class.
CONFIDENCE INTERVALS
The mean lead level of 35 crows in a random sample from a region was 4.90 ppm and the standard deviation was 1.12 ppm. Construct a 95 percent confidence interval for the mean lead level of crows in the region.
Many students–mine included–have difficulty comprehending confidence intervals and resort to “black box” confidence interval tools available in most (all?) statistics-capable calculators, including the TI-Nspire.
As n is greater than 30, I can compute the requested z-interval by filling in just four entries in a pop-up window and pressing Enter.
Convenient, for sure, but this approach doesn’t help the confused students understand that the confidence interval is nothing more than the bounds of the middle 95% of the normal pdf described in the problem, a fact crystallized by the application of the tools the students have been using for weeks by that point in the course.
Notice in the solve+normCdf() combination commands that the unknown this time was a bound and not the mean as was the case in the previous example.
EXTENDING THE RULE OF FOUR
I’ve used the “Rule of Four” in every math class I’ve taught for over two decades, explaining that every mathematical concept can be explained or expressed four different ways: Numerically, Algebraically, Graphically (including graphs and geometric figures), and Verbally. While not the contextual point of his quote, I often cite MIT’s Marvin Minsky here:
“You don’t understand anything until you learn it more than one way.”
Learning to translate between the four representations grants deeper understanding of concepts and often gives access to solutions in one form that may be difficult or impossible in other forms.
After my decades-long work with CAS, I now believe there is actually a 5th representation of mathematical ideas: Tools. Knowing how to translate a question into a form that your tool (in the case of CAS, the tool is computers) can manage or compute creates a different representation of the problem and requires deeper insights to manage the translation.
I knew some of my students this year had deeply embraced this “5th Way” when one showed me his alternative approach to the confidence interval question:
I found this solution particularly lovely for several reasons.
• The student knew about lists and statistical commands and on a whim tried combining them in a novel way to produce the desired solution.
• He found the confidence interval directly using a normal distribution command rather than the arguably more convenient black box confidence interval tool. He also showed explicitly his understanding of the distribution of sample means by adjusting the given standard deviation for the sample size.
• Finally, while using a CAS sometimes involves getting answers in forms you didn’t expect, in this case, I think the CAS command and list output actually provide a cleaner, interval-looking result than the black box confidence interval command much more intuitively connected to the actual meaning of a confidence interval.
• While I haven’t tried it out, it seems to me that this approach also should work on non-CAS statistical calculators that can handle lists.
(a very minor disappointment, quickly overcome)
Returning to my multiple approaches, I tried using my student’s newfound approach using a normCdf() command.
Alas, my Nspire returned the very command I had entered, indicating that it didn’t understand the question I had posed. While a bit disappointed that this approach didn’t work, I was actually excited to have discovered a boundary in the current programming of the Nspire. Perhaps someday this approach will also work, but my students and I have many other directions we can exploit to find what we need.
Leaving the probability tables behind in their appropriate historical dust while fully embracing the power of modern classroom technology to enhance my students’ statistical learning and understanding, I’m convinced I made the right decision to start this school year. They know more, understand the foundations of statistics better, and as a group feel much more confident and flexible. Whether their scores on next month’s AP exam will reflect their growth, I can’t say, but they’ve definitely learned more statistics this year than any previous statistics class I’ve ever taught.
COMPLETE FILES FROM MY 2015 T3 PRESENTATION
If you are interested, you can download here the PowerPoint file for my entire Nspired Statistics and CAS presentation from last week’s 2015 T3 International Conference in Dallas, TX. While not the point of this post, the presentation started with a non-calculus derivation/explanation of linear regressions. Using some great feedback from Jeff McCalla, here is an Nspire CAS document creating the linear regression computation updated from what I presented in Dallas. I hope you found this post and these files helpful, or at least thought-provoking.
Probability, Polynomials, and Sicherman Dice
Three years ago, I encountered a question on the TI-Nspire Google group asking if there was a way to use CAS to solve probability problems. The ideas I pitched in my initial response and follow-up a year later (after first using it with students in a statistics class) have been thoroughly re-confirmed in my first year teaching AP Statistics. I’ll quickly re-share them below before extending the concept with ideas I picked up a couple weeks ago from Steve Phelps’ session on Probability, Polynomials, and CAS at the 64th annual OCTM conference earlier this month in Cleveland, OH.
BINOMIALS: FROM POLYNOMIALS TO SAMPLE SPACES
Once you understand them, binomial probability distributions aren’t that difficult, but the initial conjoining of combinatorics and probability makes this a perennially difficult topic for many students. The standard formula for the probability of determining the chances of K successes in N attempts of a binomial situation where p is the probability of a single success in a single attempt is no less daunting:
$\displaystyle \left( \begin{matrix} N \\ K \end{matrix} \right) p^K (1-p)^{N-K} = \frac{N!}{K! (N-K)!} p^K (1-p)^{N-K}$
But that is almost exactly the same result one gets by raising binomials to whole number powers, so why not use a CAS to expand a polynomial and at least compute the $\displaystyle \left( \begin{matrix} N \\ K \end{matrix} \right)$ portion of the probability? One added advantage of using a CAS is that you could use full event names instead of abbreviations, making it even easier to identify the meaning of each event.
The TI-Nspire output above shows the entire sample space resulting from flipping a coin 6 times. Each term is an event. Within each term, the exponent of each variable notes the number of times that variable occurs and the coefficient is the number of times that combination occurs. The overall exponent in the expand command is the number of trials. For example, the middle term– $20\cdot heads^3 \cdot tails^3$ –says that there are 20 ways you could get 3 heads and 3 tails when tossing a coin 6 times. The last term is just $tails^6$, and its implied coefficient is 1, meaning there is just one way to flip 6 tails in 6 tosses.
The expand command makes more sense than memorized algorithms and provides context to students until they gain a deeper understanding of what’s actually going on.
FROM POLYNOMIALS TO PROBABILITY
Still using the expand command, if each variable is preceded by its probability, the CAS result combines the entire sample space AND the corresponding probability distribution function. For example, when rolling a fair die four times, the distribution for 1s vs. not 1s (2, 3, 4, 5, or 6) is given by
The highlighted term says there is a 38.58% chance that there will be exactly one 1 and any three other numbers (2, 3, 4, 5, or 6) in four rolls of a fair 6-sided die. The probabilities of the other four events in the sample space are also shown. Within the TI-Nspire (CAS or non-CAS), one could use a command to give all of these probabilities simultaneously (below), but then one has to remember whether the non-contextualized probabilities are for increasing or decreasing values of which binomial outcome.
Particularly early on in their explorations of binomial probabilities, students I’ve taught have shown a very clear preference for the polynomial approach, even when allowed to choose any approach that makes sense to them.
TAKING POLYNOMIALS FROM ONE DIE TO MANY
Given these earlier thoughts, I was naturally drawn to Steve Phelps “Probability, Polynomials, and CAS” session at the November 2014 OCTM annual meeting in Cleveland, OH. Among the ideas he shared was using polynomials to create the distribution function for the sum of two fair 6-sided dice. My immediate thought was to apply my earlier ideas. As noted in my initial post, the expansion approach above is not limited to binomial situations. My first reflexive CAS command in Steve’s session before he share anything was this.
By writing the outcomes in words, the CAS interprets them as variables. I got the entire sample space, but didn’t learn gain anything beyond a long polynomial. The first output– $five^2$ –with its implied coefficient says there is 1 way to get 2 fives. The second term– $2\cdot five \cdot four$ –says there are 2 ways to get 1 five and 1 four. Nice that the technology gives me all the terms so quickly, but it doesn’t help me get a distribution function of the sum. I got the distributions of the specific outcomes, but the way I defined the variables didn’t permit sum of their actual numerical values. Time to listen to the speaker.
He suggested using a common variable, X, for all faces with the value of each face expressed as an exponent. That is, a standard 6-sided die would be represented by $X^1+X^2+ X^3+X^4+X^5+X^6$ where the six different exponents represent the numbers on the six faces of a typical 6-sided die. Rolling two such dice simultaneously is handled as I did earlier with the binomial cases.
NOTE: Exponents are handled in TWO different ways here. 1) Within a single polynomial, an exponent is an event value, and 2) Outside a polynomial, an exponent indicates the number of times that polynomial is applied within the specific event. Coefficients have the same meaning as before.
Because the variables are now the same, when specific terms are multiplied, their exponents (face values) will be added–exactly what I wanted to happen. That means the sum of the faces when you roll two dice is determined by the following.
Notice that the output is a single polynomial. Therefore, the exponents are the values of individual cases. For a couple examples, there are 3 ways to get a sum of 10 $\left( 3 \cdot x^{10} \right)$, 2 ways to get a sum of 3 $\left( 2 \cdot x^3 \right)$, etc. The most commonly occurring outcome is the term with the largest coefficient. For rolling two standard fair 6-sided dice, a sum of 7 is the most common outcome, occurring 6 times $\left( 6 \cdot x^7 \right)$. That certainly simplifies the typical 6×6 tables used to compute the sums and probabilities resulting from rolling two dice.
While not the point of Steve’s talk, I immediately saw that technology had just opened the door to problems that had been computationally inaccessible in the past. For example, what is the most common sum when rolling 5 dice and what is the probability of that sum? On my CAS, I entered this.
In the middle of the expanded polynomial are two terms with the largest coefficients, $780 \cdot x^{18}$ and $780 \cdot x^{19}$, meaning a sums of 17 and 18 are the most common, equally likely outcomes when rolling 5 dice. As there are $6^5=7776$ possible outcomes when rolling a die 5 times, the probability of each of these is $\frac{780}{7776} \approx 0.1003$, or about 10.03% chance each for a sum of 17 or 18. This can be verified by inserting the probabilities as coefficients before each term before CAS expanding.
With thought, this shouldn’t be surprising as the expected mean value of rolling a 6-sided die many times is 3.5, and $5 \cdot 3.5 = 17.5$, so the integers on either side of 17.5 (17 & 18) should be the most common. Technology confirms intuition.
ROLLING DIFFERENT DICE SIMULTANEOUSLY
What is the distribution of sums when rolling a 4-sided and a 6-sided die together? No problem. Just multiply two different polynomials, one representative of each die.
The output shows that sums of 5, 6, and 7 would be the most common, each occurring four times with probability $\frac{1}{6}$ and together accounting for half of all outcomes of rolling these two dice together.
A BEAUTIFUL EXTENSION–SICHERMAN DICE
My most unexpected gain from Steve’s talk happened when he asked if we could get the same distribution of sums as “normal” 6-sided dice, but from two different 6-sided dice. The only restriction he gave was that all of the faces of the new dice had to have positive values. This can be approached by realizing that the distribution of sums of the two normal dice can be found by multiplying two representative polynomials to get
$x^{12}+2x^{11}+3x^{10}+4x^9+5x^8+6x^7+5x^6+4x^5+3x^4+2x^3+x^2$.
Restating the question in the terms of this post, are there two other polynomials that could be multiplied to give the same product? That is, does this polynomial factor into other polynomials that could multiply to the same product? A CAS factor command gives
Any rearrangement of these eight (four distinct) sub-polynomials would create the same distribution as the sum of two dice, but what would the the separate sub-products mean in terms of the dice? As a first example, what if the first two expressions were used for one die (line 1 below) and the two squared trinomials comprised a second die (line 2)?
Line 1 actually describes a 4-sided die with one face of 4, two faces with 3s, and one face of 2. Line 2 describes a 9-sided die (whatever that is) with one face of 8, two faces of 6, three faces of 4, two faces of 2, and one face with a 0 ( $1=1 \cdot x^0$). This means rolling a 4-sided and a 9-sided die as described would give exactly the same sum distribution. Cool, but not what I wanted. Now what?
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.
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}.
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
• 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:
$P \left( \text{birthday match with 1 person} \right) = 1- \frac{365}{365} = 0$
$P \left( \text{birthday match with 2 people} \right) = 1- \frac{365}{365} \cdot \frac{364}{365} \approx 0.002740$
$P \left( \text{birthday match with 3 people} \right) = 1- \frac{365}{365} \cdot \frac{364}{365} \cdot \frac{363}{365} \approx 0.008204$.
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.
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.
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
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:
• 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.
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. | 2019-08-22T12:42:04 | {
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http://math.stackexchange.com/questions/710495/derive-closed-form-sum-of-n2 | # 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
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.
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\$.
- | 2016-07-28T00:58:16 | {
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https://math.stackexchange.com/questions/3790119/minimum-number-of-elements-in-0-1-2-dots-n-that-add-up-to-all-of-the | # 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.
• 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:
$$\{0,1,a_1,\dots ,a_m,\frac{n}{2}-a_m,\dots ,\frac{n}{2}-a_1,\frac{n}{2}-1,\frac{n}{2}\}$$
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 $$73 \leq n \leq 80$$ | 2021-01-19T15:51:40 | {
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http://www.comfer.com.br/023nhg/chebyshev-distance-vs-euclidean-226c34 | 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 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 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 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 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 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 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. 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 $matrix can we begin the process of the. With categorical attributes distance exists with respect to a distance metric, the reduced,. Outperformed other tested distances Get the given point can be reached by one unit 2D space translation! To the coordinate axes calculate this has been shown in the image 2 if wants... Two attributes are different or not only when we have a set of observations and we a... Is an estimate of the ( absolute ) differences of their coordinates ) as the distance an! Of the real numbers coming Out of two metrics used in KNN, namely,. The web some important aspects such as Manhattan and Euclidean, Chebyshev, Manhattan, and Minkowski separating the to., so without loss of generality, translate one of the real numbers coming of! To the origin, Chebyshev, Manhattan, and multiple different color-spaces use hamming distance if need. Points to the origin comments which contain 448 data p_1, p_2 ) = \vert -. ) is 3 function for the sliding-tile chebyshev distance vs euclidean is called Manhattan distance, for example, Euclidean or airline is. Indicate distances such as computation and real life usage adjacent cells from the given distance … chebyshev distance vs euclidean is! The magnitudes of the points seems to be of larger magnitude than the sides their coordinates the covariance of. Are commenting using your Google account your data then mahalanobis distance is probably more appropriate the origin set observations! Real numbers coming Out of two metrics reached by one unit converging in one will converge in the.. Your point on this ) = \vert y_2 - y_1 \vert$ 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 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 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! 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 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!
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, | 2021-03-08T21:17:53 | {
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https://math.stackexchange.com/questions/3376028/probability-of-meeting-a-confusion | # 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.
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.
• 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 | 2020-01-17T16:12:02 | {
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https://math.stackexchange.com/questions/1888769/inequality-involving-rearrangement-sum-i-1n-x-i-y-sigmai-ge-su | # 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.
• 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.
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, $$\sum_k |x_k - y_k| \leq \sum_k |x_k - y_{\sigma(k)}| \leq \sum_k |x_k - y_{n-k+1}|$$ | 2019-08-22T05:47:21 | {
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https://www.physicsforums.com/threads/geostationary-satellite-and-orbiting-satellite-problem.753331/ | # 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
#### 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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• Solo and co-op problem solving | 2019-10-16T11:38:19 | {
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https://math.stackexchange.com/questions/2672447/factoring-2n-3-2n-2-not-homework-question/2672459 | # 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 | 2021-06-21T17:25:08 | {
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https://math.stackexchange.com/questions/1260011/why-is-arcsin-a-function-and-not-a-relation-since-arcsin-sin-frac3-pi4 | # 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.
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
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.
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 | 2019-08-22T08:46:53 | {
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https://www.physicsforums.com/threads/refractive-index-of-water-glass-combination.369449/ | 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.
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
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 $\sin(\alpha)*n=const$.
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 | 2017-08-19T21:03:18 | {
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https://www.physicsforums.com/threads/ball-bouncing-infinitely-to-find-distance-covered.379464/ | # 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!
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.
Last edited by a moderator: May 4, 2017 | 2017-08-20T10:37:40 | {
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https://math.stackexchange.com/questions/3497762/how-is-am-gm-supposed-to-be-used-to-prove-an-inequality | 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!
• 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 $$a = \dfrac{\sum_{k=1}^m a_k}{m}$$,
$$\begin{array}\\ \dfrac{\sum_{k=1}^{2^n}a_k}{2^n} &=\dfrac{\sum_{k=1}^{m}a_k}{2^n}+\dfrac{\sum_{k=m+1}^{2^n}a_k}{2^n}\\ &=\dfrac{\sum_{k=1}^{m}a_k}{m}\dfrac{m}{2^n}+\dfrac{\sum_{k=m+1}^{2^n}a}{2^n}\\ &=\dfrac{am}{2^n}+\dfrac{(2^n-m)a}{2^n}\\ &=\dfrac{am}{2^n}+\dfrac{2^na}{2^n}-\dfrac{ma}{2^n}\\ &= a\\ &=\dfrac{\sum_{j=1}^ma_j}{m}\\ \end{array}$$
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 | 2021-09-22T20:39:39 | {
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https://math.stackexchange.com/questions/1788153/compute-the-following-line-integral-along-a-path-of-your-choice-finding-potenti | # 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 $i=1,2,3$.
• 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 $0.$ | 2019-06-17T05:46:11 | {
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https://math.stackexchange.com/questions/1630733/using-a-direct-proof-to-show-that-two-integers-of-same-parity-have-an-even-sum | # 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???
• 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.
• 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 | 2019-05-26T15:02:19 | {
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https://math.stackexchange.com/questions/4126638/expected-distances-relating-to-points-on-a-stick | 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 :)
• 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.
• 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 | 2021-05-09T21:55:30 | {
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https://math.stackexchange.com/questions/1503360/sum-k-0n-a-k-sum-k-0-lfloor-n-2-rfloor-a-2k-sum-k-0-lflo | # $\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}
• 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$,
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. | 2019-10-22T01:15:42 | {
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https://math.stackexchange.com/questions/3930987/finding-the-vector-equation-of-a-plane-in-3-space | # 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. | 2021-10-25T11:46:09 | {
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https://math.stackexchange.com/questions/1851554/closure-of-intersection-is-subset-of-intersection-of-closures | # 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??
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.
• 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 | 2019-05-19T22:21:04 | {
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https://mathematica.stackexchange.com/questions/145478/boundary-condition-for-schr%C3%B6dinger-equation-in-infinite-range?noredirect=1 | # 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
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:
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.
• (+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
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 | 2021-06-17T00:23:00 | {
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https://stats.stackexchange.com/questions/291913/does-t-test-only-work-for-continuous-random-variable | # 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.
• 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".
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?
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).
• 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.
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 | 2020-01-29T12:51:44 | {
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https://math.stackexchange.com/questions/2498416/solving-a-set-problem/2498441 | # 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.
• 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 :) $$...$$ | 2021-06-15T20:08:06 | {
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http://math.stackexchange.com/questions/492901/complex-function-sketch | 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
-
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.
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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https://math.stackexchange.com/questions/1108428/step-by-step-solution-for-x1-x-21-2/1108437 | 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.
• 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.
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 $x^{\frac{1}{x}} = 2^{\frac{1}{2}}.$ | 2019-12-06T18:29:19 | {
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https://teachingcalculus.com/category/integration/integration-theory/page/3/ | # 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 $\int{x\sqrt{x+1}dx}$
$u=\sqrt{x+1},x={{u}^{2}}-1,$
$du=\tfrac{1}{2}{{\left( x+1 \right)}^{-1/2}}dx,dx=2udu$
$2{{\int{\left( {{u}^{2}}-1 \right){{u}^{2}}}}^{{}}}du=2\int{{{u}^{4}}-{{u}^{2}}}du=\tfrac{2}{5}{{u}^{5}}-\tfrac{2}{3}{{u}^{3}}=$
$\tfrac{2}{5}{{\left( x+1 \right)}^{5/2}}-\tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}+C$
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.
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
# 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.
$\int\limits_{a}^{b}{{{f}'\left( x \right)dx}}=\underset{{\left| {\left| {\Delta x} \right|} \right|\to 0}}{\mathop{{\lim }}}\,\sum\limits_{{i=0}}^{n}{{{f}'\left( {{{c}_{i}}} \right)\left( {{{x}_{i}}-{{x}_{{i-i}}}} \right)}}$
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 ${{x}_{n}}=b$,
$\displaystyle \int_{a}^{b}{{{f}'\left( x \right)dx}}=f\left( b \right)-f\left( a \right)$
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.
# 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.
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:
$\displaystyle \underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{\left( f\left( a+\frac{b-a}{n}\cdot k \right)\cdot \frac{b-a}{n} \right)}=\int_{a}^{b}{f\left( x \right)dx}$
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 $\displaystyle \int_{-25.65}^{-22.65}{{{\left( 27.65+x \right)}^{2}}dx}=39$.
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.
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.
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:
$\displaystyle \int_{{}}^{{}}{{{e}^{x}}\cos \left( x \right)dx={{e}^{x}}\sin \left( x \right)}+{{e}^{x}}\cos \left( x \right)-\int_{{}}^{{}}{{{e}^{x}}\cos \left( x \right)dx}$
The integral at the end is identical to the original integral. We can continue by adding the integral to both sides:
$\displaystyle 2\int_{{}}^{{}}{{{e}^{x}}\cos \left( x \right)dx={{e}^{x}}\sin \left( x \right)}+{{e}^{x}}\cos \left( x \right)$
Finally, we divide by 2 and have the antiderivative we were trying to find:
$\displaystyle \int_{{}}^{{}}{{{e}^{x}}\cos \left( x \right)dx=\tfrac{1}{2}{{e}^{x}}\sin \left( x \right)}+\tfrac{1}{2}{{e}^{x}}\cos \left( x \right)+C$
In working this type of problem you must be aware of that the original integrand showing up again can happen and what to do if it does. As long as the coefficient is not +1, we can proceed as above. The same thing happens if we do not use the tabular method. (If the coefficient is +1 then the other terms on the right will add to zero and you need to make different choices for u and dv.)
### Reduction Formulas.
Another use of integration by parts is to produce formulas for integrals involving powers. An integral whose integrand is of less degree than the original, but of the same form results. The formula is then iterated to continually reduce the degree until the final integral can be integrated easily.
Example 3: Find $\displaystyle \int_{{}}^{{}}{{{x}^{n}}{{e}^{x}}dx}$
Let $u={{x}^{n}},\ du=n{{x}^{n-1}}dx,\ dv={{e}^{x}}dx,\ v={{e}^{x}}$
$\displaystyle \int_{{}}^{{}}{{{x}^{n}}{{e}^{x}}dx}={{x}^{n}}{{e}^{x}}-n\int_{{}}^{{}}{{{x}^{n-1}}{{e}^{x}}dx}$
This is a reduction formula; the second integral is the same as the first, but of lower degree. Here is how it is used. At each step the integrand is the same as the original, but one degree lower. So the formula can be applied again, three more times in this example.
Most textbooks have a short selection of reduction formulas.
### Final Thoughts.
Back in the “old days”, BC (before calculators), beginning calculus courses spent a lot of time on the topic of “Techniques of Integration.” This included integration by parts, algebraic techniques, techniques known as trig-substitutions, and others. Mathematicians and engineers had tables of integrals listing over a thousand forms and students were taught how to use the tables and distinguish between similar forms in the tables. (See the photo below from the fourteenth edition of the CRC tables (c) 1965.) Current textbooks often contain such sections still.
Today, none of this is necessary. CAS calculators can find the antiderivatives of almost any integral. Websites such as WolframAlpha are also available to do this work.
I’m not sure why the College Board recently expanded slightly the list of types of antiderivatives tested on the exams. Certainly a few of the basic types should be included in a course, but what students really need to know is how to write the integral appropriate to a problem, and what definite and indefinite integrals mean. This, in my opinion, is far more important than being able to crank out antiderivatives of increasingly complicated expressions: let technology do that – or buy yourself an integral table. Just saying … .
# Trapezoids – Ancient and Modern
The other day, in the course of about 10 minutes, I came across two interesting things about Trapezoidal approximations that I thought I would share with you.
Cuneiform writing
The first was a link to a story about how the ancient Babylonian astronomers sometime between 350 and 50 BCE used trapezoids to, in effect, find the area under a velocity-time graph tracking Jupiter’s motion. This was an NPR story based on a January 2016 Science magazine article in which the author, Mathieu Ossendrijver discusses his work deciphering cuneiform tablets written over 1,400 years before the technique showed up in Europe.
The second was a question asked on the AP Calculus bulletin board. A teacher asked, “Can someone please help me answer this question a student posed the other day. We were comparing left, right and midpoint and trapezoidal approximations. He asked since the trapezoidal calculation is the best estimate what is the use of LRAM and RRAM?” Here is an expanded form of my answer.
There are several things to consider here.
1. First, if all you need is an estimate of the area or integral of a continuous function then a Trapezoid sum is certainly better than the left Riemann sum (left-RΣ) or the right-RΣ. Better, yes, the “best” maybe not: midpoint sums are about as good and parabola sums (Simpson’s Rule) are better.
2. Another reason to do left RΣ and right RΣ with small values of n is simply to give students practice in setting up Riemann sums so that they will be familiar with them when they move on to finding their limits and getting ready to define definite integrals.
3. A RΣ for a function f on a closed interval [a, b] is formed by partitioning the interval into subintervals and taking exactly one function value from each closed subinterval, multiplying the value by the width of that subinterval and adding these results. You may pick the function value any way you want – left end, middle, right end, any place at random in the subintervals and someplace else in the next subinterval. One way is to pick the smallest function value in each subinterval; this gives a RΣ called the lower RΣ. Likewise, you could pick the largest value in each subinterval; this gives the upper RΣ. Now it is true that
lower RΣ ≤ (any/all other RΣs) ≤ upper RΣ
Then as you add more partition points (n approaches infinity, or Δx approaches 0, etc.) the lower sum increases and the upper sum decreases. The series of lower sums is increasing and bounded above (by the upper sum) and therefore converges to its least upper bound. The upper sum decreases forming a decreasing series that is bounded below and therefore converges to its greatest lower bound.
If the lower RΣ and the upper RΣ approach the same value, then ALL the other RΣs approach that same value by the Squeeze theorem. This value is then defined as the definite integral of f from a to b.
In most AP calculus course, the textbooks do not deal with upper and lower sums. Instead, they deal with left RΣ and right RΣ on intervals on which f is only increasing (or only decreasing). In this case the lower RΣ = left RΣ and the upper RΣ = the right RΣ (or the other way around for decreasing functions).
So, this is why you need the left RΣ and right RΣ; not so much to approximate, but to complete the theory leading to the definite integral. | 2023-03-27T22:40:39 | {
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https://www.monroecc.edu/faculty/paulseeburger/calcnsf/CalcPlot3D/CalcPlot3D-Help/section-31.html |
## Section5.4Intersections of General Surfaces
Although CalcPlot3D does not determine the intersection of two surfaces for you, it does make a great tool for visually checking that the parameterization you have worked out as the intersection of two surfaces is indeed correct.
Below you will find two examples to explore this concept. Hopefully this visualization process will help increase your confidence in and your appreciation for the results you have obtained.
###### Exploration5.4.1
Determine the vector-valued function that traces out the intersection of the surfaces defined by the equations below using the parameter $t\text{.}$
State the answer as a vector-valued function. Then visually verify this result, plotting the two surfaces in CalcPlot3D, and then plotting the curve to check that it really does represent the intersection of these two surfaces.
Here we can let $x=t\text{.}$ Then $y=t^2\text{.}$ What does $z$ equal (as a function of $t$)?
Then the vector-valued function we obtain that traces out the intersection of these surfaces is:
\begin{equation*} \vec{\textbf{r}}(t) = t \hat{\textbf{i}} + t^2 \hat{\textbf{j}} + (t^2 + t^4) \hat{\textbf{k}} \end{equation*}
This represents the same curve as specified by the parametric functions:
\begin{align*} x &= t\\ y &=t^2\\ z &=t^2 + t^4 \end{align*}
Now let’s verify this in CalcPlot3D visually!
a. Open the CalcPlot3D app.
b. Once the app is loaded and active, enter the first function listed above ($z =$ x^2 + y^2) in the default function object on the left and press Enter (or click on the Graph button). The surface plot of this paraboloid should appear in the plot window.
c. Now, to enter the second function, go to the Add to graph dropdown menu (just above the default function), and select Function: y = f (x, z). This will set the function to y = 1 by default. Enter x^2 in the textbox and press Enter (or use the Graph button).
d. To extend the second surface farther up the paraboloid, change the range of $z$ (just below y = x^2) to go from -2 to 4. You may want to use the scroll-wheel on the mouse to zoom-out a little. [Alternatively you could click on the Format Axes button located just to the right of the 3D Mode button. Then set $z$-max to 4. This will automatically change the upper $z$-clip value to 8. Let’s change this value to 4 also.]
e. Make the surfaces semi-transparent using the button or by typing the T key to get a clearer view of the intersection of the surfaces. Press the E key to turn off the edges on the surfaces.
f. Next we need to graph the space curve to see how well it fits the intersection of the surfaces. Select Space Curve: r(t) from the Add to graph dropdown menu. A space curve object should appear just below this menu.
Enter the three parametric equations we obtained (each in terms of $t$). Then enter a range of -2 to 2. If you press Enter on the second value, it should produce the curve on the plot. If it does not appear, click the Graph button.
g. I like the look of this one better with a constant color, so select the checkbox titled Use Constant Primary Color, at the bottom of this object.
h. Finally rotate the graph to see if it looks like we found the correct intersection curve. Except for different coloring, this should look like the image in Figure 5.4.1 below.
###### Exploration5.4.2
Determine the vector-valued function that traces out the intersection of the surfaces defined below, assuming that $y = 2\sin t\text{.}$
State the answer as a vector-valued function. Then visually verify the result, plotting the two surfaces in CalcPlot3D, and then plotting the curve to check that it really does represent the intersection of these two surfaces.
Note that we can parameterize the ellipse given by the first equation with
Now that we've parameterized $x$ and $y\text{,}$ the second surface equation makes it easy to determine $z$ in terms of $t\text{.}$
After simplifying the expression, we obtain, $z = 2 + 2\sin^2 t\text{.}$
Then the vector-valued function we obtain that traces out the intersection of these surfaces is:
\begin{equation*} \vec{\textbf{r}}(t) = \sqrt{2}\cos t \hat{\textbf{i}} + 2\sin t \hat{\textbf{j}} + (2 + 2\sin^2 t) \hat{\textbf{k}} \end{equation*}
This represents the same curve as specified by the parametric functions:
\begin{align*} x &= \sqrt{2}\cos t\\ y &= 2\sin t\\ z &= 2 + 2\sin^2 t \end{align*}
Now let’s verify this in CalcPlot3D visually!
a. Open the CalcPlot3D app.
b. Once the app is loaded and active, enter the second function listed above (z = x2 + y2) in the default function object on the left and press Enter (or click on the Graph button). The surface plot of this paraboloid should appear in the plot window.
c. Now, to enter the first function, which is stated implicitly, go to the Add to graph dropdown menu (just above the default function), and select Implicit Surface. Once this new object definition appears on the left, enter the implicit equation defining the first surface in the textbox and press Enter (or use the Graph button). This surface should look like an elliptic cylinder.
d. To make the intersection more clear, let's first zoom out once using the Zoom-out button, so the $x$- and $y$-axes run from -4 to 4 instead of from -2 to 2. Now click on the Format Axes button located just to the right of the 3D Mode button, and set the upper $z$-clip to 4.
e. Next make the surfaces semi-transparent using the transparency button or by typing the T key to get a clearer view of the intersection of the surfaces. Press the E key to turn off the edges on the surfaces.
f. Next we need to graph the space curve to see how well it fits the intersection of the surfaces. Select Space Curve: r(t) from the Add to graph dropdown menu. A space curve object should appear just below this menu.
Enter the three parametric equations we obtained (each in terms of $t$).
You will enter them like this:
\begin{align*} x &= \textbf{sqrt(2)cos(t)}\\ y &= \textbf{2sin(t)}\\ z &= \textbf{2 + 2(sin(t))\^{}2} \end{align*}
g. Now enter a range of 0 to $2\pi\text{.}$ If you press Enter on the second value, it should produce the curve on the plot. If it does not appear, click the Graph button.
I also like the look of this one better with a constant color, so select the checkbox titled Use Constant Primary Color, at the bottom of this object.
h. Finally rotate the graph to see if it looks like we found the correct intersection curve. This should look like the image in Figure 5.4.2 below.
Notice that the intersection of these two surfaces may appear a little polygonal rather than smooth like the curve we just graphed. This is because we used an implicit surface with the default resolution. To improve this resolution, you can adjust the # Cubes/axis to a higher number, like 25. See the result in Figure 5.4.3. | 2021-05-06T18:54:29 | {
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https://math.stackexchange.com/questions/4106228/is-there-a-general-procedure-for-inducing-a-formula-recognizing-a-pattern/4106329 | # Is there a general procedure for inducing a formula / recognizing a pattern?
For some problems, it's pretty simple to deduce what a formula should look like after you enumerate a few examples, but for some problems, it's not so clear. For these less clear examples, is there some procedure I can follow to find the formula?
For example, just now, I was working on the problem:
Find the expected number of tosses it takes to get $$k$$ 6s in a row.
The recursive formula is
$$E[N_k ] = 6(E[N_{k - 1}] + 1)$$
where $$N_k$$ is the number of tosses it takes to get $$k$$ 6s in a row. The initial condition is $$E[N_1] = 6$$ because $$N_1$$ is a geometric random variable with $$1/6$$ probability of success.
Then I enumerated a few cases:
$$E[N_1] = 6 \\ E[N_2] = 42 \\ E[N_3] = 258 \\ E[N_4] = 1554$$
It's not obvious to me what the formula should be. But I think it should look something like $$E[N_k] = 6^k + (k - 1) * \text{something} + \text{maybe other terms}$$
But I don't know what this "something" and "maybe other terms" should be. If I sit here long enough I could probably figure it out, but is there some kind of general procedure that I can apply to a wide array of problems?
• – John Omielan Apr 17 at 20:29
One good way is to try to convert this to a more standard linear recursion.
Letting $$a_n=6(a_{n-1}+1)=6a_{n-1}+6$$
We rewrite this as $$6=a_n-6a_{n-1}=a_{n-1}-6a_{n-2}$$
Whence $$a_n=6a_{n-1}+a_{n-1}-6a_{n-2}=7a_{n-1}-6a_{n-2}$$
and this can now be solved by standard means, yielding $$\boxed{a_n=\frac 65\times \left(6^n-1\right)}$$
• What do you mean by 'standard means' here? – roulette01 Apr 17 at 20:22
• It's a linear recursion, so you look at the characteristic polynomial $x^2-7x+6=(x-1)(x-6)$. Thus the solution is of the form $a_n=A6^n+B$ and then solve for $A,B$ using initial values. – lulu Apr 17 at 20:23
• Ah I see. I've never formally learned recursion. My understanding is there are a lot of analogies between how recursive formulas are solved vs. how ODEs are solved, and this general solution seems to be analogous to formulas in second order ODEs. – roulette01 Apr 17 at 20:25
• If you were to solve this problem without recursion, and by just inspecting the first several enumerations, is there a general approach to guess what that formula should be? – roulette01 Apr 17 at 20:26
• I think guessing is hard. You could guess that $a_n=A6^n+p(n)$ where $p(n)$ is a low degree polynomial...but I don't know how reliable that sort of guesswork is. Works in this case, though. – lulu Apr 17 at 20:28
If you call $$a_k=E[N_k]$$ then you end up with a linear induction relation: $$a_k-6a_{k-1}=6$$
Since it is linear, it solves as the sum of general solution of homegenous equation $$H_k-6H_{k-1}=0$$, which is $$H_k=6^kH_0$$ and one particular solution of the full equation.
The theory says that if the RHS is of the form $$P(k)r^k$$ then you can find a particular solution of the form $$Q(k)r^k$$ with $$\deg(Q)=\deg(P)+m$$ where $$m$$ is the multiplicity of the root of the homogenous equation.
In this case $$RHS = 6\times 1^k$$ and the single root of homogenous equation is $$r=6$$. So the multiplicity of $$1$$ is just $$m=0$$ (i.e. not a root) this means $$\deg(Q)=0+0=0$$.
All that to say that our particular solution is simply a constant...
Therefore let search for $$S_k=c$$ then $$c-6c=6\iff c=-\frac 65$$
Our general solution is then $$a_k=H_k+S_k=6^kH_0-\frac 65$$
We now solve for initial conditions:
$$a_1=6=6H_0-\frac 65\iff H_0=\frac 65$$ and we get $$E[N_k]=\frac 65(6^k-1)$$
is there some kind of general procedure that I can apply to a wide array of problems?
Around 1960 there was some work on a General Problem Solver which had very limited capabilities but much progress has been made since 1960. Still, there is no way to solve all problems and probably never will be.
However, the solution of linear recurrences is known for a long time. For example, $$a_n=6(a_{n-1}+1)=6a_{n-1}+6$$ is easily solved using the theory.
For this particular difference equation with initial value $$\,a_1\!=\!6\,$$ a lookup of $$\,6,42,258,1554\,$$ in OEIS returns OEIS sequence A105281 "a(0)=0; a(n)=6*a(n-1)+6." This entry has a program
(PARI) a(n)=if(n<0, 0, (6^n-1)*6/5)
which gives a formula for the solution.
The Mathematica program has a function RSolve which solves recurrences and also has limited capabilities
You can use the Mathematica (or WolframAlpha) command
RSolve[a[n] == 6 a[n - 1] + 6 && a[1] == 6, a[n], n]
to get the particular solution $$\,a_n = 6 (6^n-1)/5.\,$$ | 2021-06-16T17:24:20 | {
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http://mathhelpforum.com/statistics/126266-coin-tossing.html | # Math Help - coin tossing
1. ## coin tossing
2 gamblers bet $1 each on the successive tosses of a coin. each as$6. what is the probability that
a. they break even after 6 tosses?
b. one player wins all the money on the tenth toss?
my working:
a. since there are 6 tosses and equal probability of a head and tail,
the ans should be 6 x (0.5)^6
b. in order to win all the money o the other player, he needs to win 8 games and loss 2 games, where the 10th toss have to be a win.
so, (9 choose 7) x ( 0.5)^10
but i didnt get the answer
2. Originally Posted by alexandrabel90
2 gamblers bet $1 each on the successive tosses of a coin. each as$6. what is the probability that
a. they break even after 6 tosses?
b. one player wins all the money on the tenth toss?
my working:
a. since there are 6 tosses and equal probability of a head and tail,
the ans should be 6 x (0.5)^6
b. in order to win all the money o the other player, he needs to win 8 games and loss 2 games, where the 10th toss have to be a win.
so, (9 choose 7) x ( 0.5)^10
but i didnt get the answer
hi alexandrabel90,
for the first one, they both must win 3 times each,
so that's the probability of either winning exactly 3 times and losing exactly 3 times.
for the second one, you must add the probability for the 2 players.
this is different from the first one because one or other could win the game on the 10th go, having won 7 of the preceding 9.
As I read the question we have two players each having $6 to play with. Each player either wins$1 or looses $1 on each toss of the coin. So it is possible for one player of loose all in six tosses- loses six straight. So for both to breakeven in six tosses each would wins three and lose three. Is that a misreading of the question? If not, the answer to part a) is $\frac{\binom{6}{3}}{2^6}$. If it is a misreading, please clarify the question. 4. it is not a misreading of the question. your answer is correct for part (a)..i never think that we would need to count then chances of wining since there is only a 0.5 chance of a win and lose. for part (b),Archie, what do you mean by adding the probability of the 2 players? does it mean (9 choose 7) x ( 0.5)^10 x 2? 5. Hello, alexandrabel90! Two gamblers bet$1 each on the successive tosses of a coin.
Each has $6. .What is the probability that: a. they break even after 6 tosses? They each win 3 games and lose 3 games. There are $_6C_3 \:=\:\frac{6!}{3!\,3!} \:=\:20$ ways this can happen. Answer: . $(20)\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\ri ght)^3 \:=\20)\,\frac{1}{64} \:=\:\frac{5}{16}" alt="(20)\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\ri ght)^3 \:=\20)\,\frac{1}{64} \:=\:\frac{5}{16}" /> b. one player wins all the money on the tenth toss? my working: In order to win all the money of the other player, he needs to win 8 games and loss 2 games, where the 10th toss has to be a win. So: . $_9C_7\left(\tfrac{1}{2}\right)^{10}$ . . . . correct, so far $_9C_7\left(\tfrac{1}{2}\right)^{10} \:=\:\tfrac{36}{1024} \:=\:\frac{9}{256}$ This is the probability that $A$ wins all of $B$'s money on the 10th game. It's also the probability that $B$ wins all of $A$'s money on the 10th game. Therefore: . $P(one\text{ player wins all the money}) \:=\:\frac{9}{256} + \frac{9}{256} \;=\;\frac{9}{128}$ 6. Originally Posted by alexandrabel90 for part (b),Archie, what do you mean by adding the probability of the 2 players? does it mean (9 choose 7) x ( 0.5)^10 x 2? Yes, that was it, because you calculated the probability of only one out of the two winning on the tenth throw. Soroban has shown that really nicely. Plato has made an interesting point also. If we interpret "break-even" as not winning or not losing on one's own throws, then that is 3 wins and 3 losses, but one can break even on one's own throws while the other has more wins than losses or vice versa on their throws. In this situation, one could break even on one's own throws but make a profit if the other player does badly, or make a loss if the other player does well, "if" losing a toss loses a dollar and winning a toss wins a dollar. Or, if the two players lost every time, "player A" wins "player B"'s 6 dollars and "player B" wins "player A"'s 6 dollars, so they break even. They could also both win 1 and lose 5 each, win 2 and lose 4 each, win 4 and lose 2 each, win 5 and lose 1 each, win 6 and lose 0 each. That's the thing with probability questions, quite often open to alternative interpretations. The answer book will distinguish given it's answers. We've been interpreting this as..... If one player loses on a throw, then the second player wins the first player's dollar. In this case then, both would have to win 3 times and lose 3 times to end up with the 6 dollars they began with. It is probably safest to assume there is just 6 tosses rather than 6 each. 7. Originally Posted by alexandrabel90 2 gamblers bet$1 each on the successive tosses of a coin. each as $6. what is the probability that b. one player wins all the money on the tenth toss? Originally Posted by Archie Meade Yes, that was it, Plato has made an interesting point also. If we interpret "break-even" as not winning or not losing on one's own throws, then that is 3 wins and 3 losses, but one can break even on one's own throws while the other has more wins than losses or vice versa on their throws. Look at the way part b) is put. It seems to me that in this game on any toss player A gets a dollar from player B or gives a dollar to player B. That is, A wins or looses on any toss. Moreover, the game is over when one player is out of money. If player A wins in each of the first six tosses she has all of the money in six tosses. If B wins five of the first six tosses he has 11 dollars and she has 1. So the question becomes “how can the game be over in exactly ten tosses?” 8. Brilliant analysis, Plato! it's not over yet. 9. I reckon that, in order for one player to win 8 to 2, the score had to stand at 5-1, at which point one player has$10 and the other has \$2 (win 5 and lose 1... 6+5-1=10, win 1 and lose 5.... 6+1-5=2).
Then the score can go to 6-1 or 5-2.
From 6-1 the score must go to 6-2, since 7-1 would end the game.
From there to 7-2 and then to 8-2.
From 5-2 the score goes to 6-2, to 7-2, to 8-2.
Beginning at 5-1, there are two ways for the leading player to win on the 10th throw.
So, we count the number of ways of getting to a 5-1 lead and double that.
$2\binom{6}{5}=2\binom{6}{1}$
$P=\frac{2(6)}{2^{10}}=\frac{3}{2^8}=\frac{3}{256}$
Then double this to get the result for either player winning by 8 to 2. | 2015-08-01T12:49:10 | {
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http://openstudy.com/updates/501d67a1e4b0be43870e2635 | ## ParthKohli 4 years ago $\Huge \mathsf{\text{Primitive Pythagorean Triplets.}}$
1. ParthKohli
Give it a start, Kingy.
2. KingGeorge
Well, they're triplets of the form $$(a,b,c)$$ such that $$a^2+b^2=c^2$$ and $$\gcd(a,b,c)=1\(. Unknown to many high school students, there are in fact several formulas for generating an infinite number of these triplets. 3. ParthKohli $2n,n^2 - 1, n^2 + 1$? 4. ParthKohli I know one. \(3,4,5$$.
5. KingGeorge
I just can't type that correctly can I? Well, they're triplets of the form $$(a,b,c)$$ such that $$a^2+b^2=c^2$$ and $$\gcd(a,b,c)=1$$. Unknown to many high school students, there are in fact several formulas for generating an infinite number of these triplets.
6. ParthKohli
Another one! $$5,12,13$$!
7. KingGeorge
Correct, (3,4,5) is a triplet as well as (5,12,13) and (7,24,25) among infinite others.
8. ParthKohli
Can you tell me the formula(s) for it, please?
9. KingGeorge
The most well known one is $2mn, \quad m^2-n^2, \quad m^2+n^2$Given any $$m,n$$ such that $$m>n$$.
10. ParthKohli
I see :) Are there any more?
11. KingGeorge
What you gave above, only works for $$m$$ even. So it generates an infinite number, but not all of them.
12. ParthKohli
Can this formula generate all such triplets, or you have to use some other formulae to generate more?
13. anonymous
another one $(x,y,z)=(2k+1,2k^2+2k,2k^2+2k+1)$
14. ParthKohli
Oh wait. It does generate all.
15. KingGeorge
There is also $(a,b,c)=\left(mn,\;\;\frac{m^2-n^2}{2},\;\;\frac{m^2+n^2}{2}\right)$This is a different formula that what I gave above since it only works with $$(m,n)$$ such that $$n>m\ge1$$ and $$\gcd(m,n)=1$$. This does not generate all of them.
16. KingGeorge
Ccorrection, $$m>n\ge1$$.
17. anonymous
if you want them "primitive" makes sure $$m,n$$ are opposite parity, and no common factors
18. ParthKohli
Well. I'd just use the first one :)
19. KingGeorge
Satellite is correct. The first formula I wrote only gives primitive triplets for $$m,n$$ such that $$m-n$$ is odd. And $$\gcd(m,n)=1$$. However, any two positive integers will give you a triplet.
20. anonymous
algebra shows you that $$(m^2-n^2)^2+(2mn)^2=(m^2+n^2)^2$$ if you want a proof that every such triple is generated by such an $$m$$ and $$n$$ try working through the attachment
21. ParthKohli
If $$\gcd(m,n) = 1$$, then it automatically means that $$m - n$$ is odd.
22. anonymous
no, try 5 and 3
23. ParthKohli
Oh grr.
24. anonymous
really, if you have time, work through the worksheet i sent you you will see why all triples come from this formula
25. ParthKohli
Yes, I am looking at it.
26. anonymous
it requires no more than algebra
27. ParthKohli
All set. I got that formula well.
28. KingGeorge
There is also a multitude of fascinating properties of primitive triplets. For example: $$c$$ is always odd. 2,3,4 divide exactly one of $$a$$ or $$b$$, 5 divides exactly one of $$a,b$$ or $$c$$. $$a+b+c$$ is always even among others. These are some of the simpler facts.
29. ParthKohli
30. anonymous
have fun. there are several steps to the proof, but it is all there
31. anonymous
what do you mean by primitive triplets?
32. KingGeorge
$$\gcd(a,b,c)=1$$.
33. anonymous
"primitive" no common factors
34. anonymous
oh
35. anonymous
so 3, 4, 5 but not 6, 8, 10
36. ParthKohli
Whoa! Primitive triplets are $$a,b,c$$ such that $$\gcd(a,b,c) = 1$$. Other words, they all are co-prime.
37. anonymous
@ParthKohli method of obtaining the formula i gave u...we will talk about later if u want... http://openstudy.com/study#/updates/500a7eb3e4b0549a892eeeee
38. ParthKohli
@mukushla I'd see that tutorial tomorrow; tired somewhat at the moment. Looks impressive by looking at it though.
39. anonymous
np...:)
40. ParthKohli
Haha$\Huge \ddot \smile$
41. anonymous
why do we need to have opposite parity? and what's with m-n being odd?
42. ParthKohli
Well. Isn't that the same thing?
43. KingGeorge
$$m-n$$ is odd if and only if they have opposite parity. So either $$m$$ or $$n$$ is odd, but not both.
44. anonymous
i understand that. but is it necessary for m-n to be odd?
45. KingGeorge
If you want a primitive triplet, it is.
46. KingGeorge
If they are both even or both odd, $$\gcd(a,b,c)\ge2$$
47. anonymous
how do i prove it?
48. KingGeorge
Note that $$2mn$$ is always even. Also notice that if $$m,n$$ are both odd, then $$m^2$$ and $$n^2$$ are both odd. If they are both even, $$m^2$$ and $$n^2$$ are both even. Now, $$m^2-n^2$$ and $$m^2+n^2$$ will both be even as well since even-even=even and odd-odd=odd. Hence, $$a,b,c$$ are all even, and their gcd must be divisible by 2.
49. anonymous
that was so easy and so stupid of me to not get it. i am sorry for troubling you on this. if i am to prove 5 exactly divides one of the triplet, is assuming all of them to be non divisible by 5 the right way?
50. KingGeorge
When I first did it, I assumed $$a$$ was not divisible by 5, and deduced that either $$b$$ or $$c$$ was divisible by 5. Then assume $$5$$ divides $$a$$ and show that it doesn't divide $$b$$ or $$c$$. This probably isn't the fastest way to do it thought.
51. anonymous
you can work by cases assume $$m$$ and $$n$$ are not divisible by 5 then prove that either $$m^2-n^2$$ or $$m^2+n^2$$ must be for example, if $$m\equiv n$$ mod 5, then $$m^2-n^2\equiv 0$$ mod 5 | 2016-09-25T17:25:43 | {
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https://math.stackexchange.com/questions/843202/evaluate-int-sqrt1-x2-dx | # Evaluate $\int \sqrt{1-x^2}\,dx$
I have a question to calculate the indefinite integral: $$\int \sqrt{1-x^2} dx$$ using trigonometric substitution.
Using the substitution $u=\sin x$ and $du =\cos x\,dx$, the integral becomes: $$\int \sqrt{\cos^2 u} \, \cos u \,du = \int \|{\cos u}\| \cos u\, du$$
Q: (part a) At what point (if at all) is it safe to say that this is the equivalent of ? $$\int \cos^2 u\, du = \int \frac {1 + \cos 2u} {2} du$$ (this is easy to solve, btw).
In lectures, it was made abundantly clear that over certain intervals (eg $0 \le u \le \pi/2$) that $cos u$ is +ve and is safe to do so, but in the indefinite form, the same argument cannot be made (eg $\pi/2 \le u \le n\pi$).
Q: (part b) Is it safe to declare it ok due to the nature of the original integral, which, using a sqrt() must return a +ve number? It could then be argued that it was the substitution which artificially added a -ve aspect...
Any suggestions on how to proceed?
PS: This is a 1st year calculus course and am revising for exams ;)
• The original integrand allows for $x$ to be in the interval $[-1,1]$ which means that $\cos$ can be negative. In my experience ignoring this is almost never a problem. Rather than worrying about the details of when $\cos$ is positive or negative, recall that you can always prove that your final answer is correct by showing that its derivative equals the original integrand. – Spencer Jun 22 '14 at 5:21
• The lecturer has been pounding into us the need to think correctly on fundamental concerns, which is why I raise it here. Although I understand I can show equivalency by derivation, it is not a proof (as far as I am aware) – cmroanirgo Jun 22 '14 at 5:37
• The definition of an indefinite integral is that it is the anti-derivative of the integrand (up to an additive constant). There is no more legitimate proof. – Spencer Jun 22 '14 at 5:39
• You have it backwards. To do this with trig substitution, $x=\sin u$, not $u=\sin x$. – alex.jordan Jun 22 '14 at 5:40
• @Spencer I don't think that's right. $x$ is in $[-1,1]$, and since $x=\sin(u)$, well, see my answer below. – alex.jordan Jun 22 '14 at 5:44
Since $x$ ranges from $-1$ to $1$, and you are using the substitution $x=\sin(u)$, you can make this substitution with $u\in[-\pi/2,\pi/2]$, and then $\cos(u)$ is unambiguously positive.
• It was the logic of discussion which had me confused. This clarifies the why. Recognising the limiting domain (to real numbers) from -1 to 1 is the key ingredient. – cmroanirgo Jun 22 '14 at 6:13
You have received good answer which all conclude that $\sqrt{\cos^2 x} = \cos x$.
But, for the time being, let us assume you still ignore if it is safe or not. So, let us write $$\int \sqrt{1-x^2} dx=\pm \int \cos^2 u\, du =\pm \int \frac {1 + \cos 2u} {2} du=\pm \Big(\frac{u}{2}+\frac{1}{4} \sin (2 u)\Big)$$ But now, let us consider the definite integral $$\int_0^1 \sqrt{1-x^2} dx=\pm \int_0^{\frac{\pi}{2}} \cos^2 u\, du =\pm \int_0^{\frac{\pi}{2}} \frac {1 + \cos 2u} {2} du=\pm \frac{\pi}{4}$$ But now, consider now the area between the curve $y=\sqrt{1-x^2}$ and the $x$ axis. All the curve is above the axis and the area is then positive. So, $\pm$ should be just replaced by $+$.
Is this making things clearer ?
Note that there is a difference between substitution as one initially learns it (let $u=g(x)$ and what in the OP is called "trigonometric substitution." The latter process, when one is feeling pedantic, is actually called inverse trigonometric substitution.
In the example we are discussing, the substitution "really is" let $u=\arcsin x$. So $u$ naturally lives in the interval $[-\pi/2,\pi/2]$, and therefore $\cos u$ is non-negative.
Hint: first integrate by parts to get,
$$\int\sqrt{1-x^2}\,\mathrm{d}x=x\sqrt{1-x^2}+\int\frac{x^2}{\sqrt{1-x^2}}\,\mathrm{d}x.$$
Now attempt the trigonometric substitution.
To answer your question: yes, it is safe and doesn't really matter. When you do trig. substitution in your first year calculus course, you are always assuming that $\cos$ is positive as a result you can do:
$$\sqrt{\cos^2 x} = \cos x$$
and not have any problems. Also, take into account what @spencer said; whatever your final answer, you can just find it's derivative and prove yourself right or wrong.
• Absolutely right but for first year, as far as I know, the integral is calculated with keeping in mind that $\cos$ is in between that intervals which you specified, otherwise $\sqrt{\cos^2 x} \rightarrow \cos x$ doesn't make sense. Depending on which textbook you use, go to the trig. substitution chapter and from an example with $\cos x$ it will say specifically that $\cos x$ is bounded between $0 \le x \le \pi/2$ – Jeel Shah Jun 22 '14 at 5:40
• The bounds for the integral and the bounds for the function are not the same. I believe that is where you are confused. All we are saying is that we want to consider $\cos x$ when it is positive and therefore, bound it between those bounds. This doesn't imply that the entire integral is bounded between those. – Jeel Shah Jun 22 '14 at 5:42 | 2021-06-13T02:38:52 | {
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http://math.stackexchange.com/questions/151611/simplify-a-1-a-2-a-3-a-nm/151615 | # simplify $(a_1 + a_2 +a_3+… +a_n)^m$
How to simplify this best $(a_1 + a_2 +a_3+... +a_n)^m$ for $m=n, m<n, m>n$
I could only get $\sum_{i=0}^{m}\binom{m}{i}a_i^i\sum_{j=0}^{m-i}\binom{m-i}{j}a_j ...$
-
It is already as simple as it gets if you don't have other information on the $a_i$s. – Phira May 30 '12 at 15:06
they are just variable ... let's say like a and b in binomial expansion – Santosh Linkha May 30 '12 at 15:07
Note that Pascal's Tetraedron, then its $4th$ dimensional analogue, then the $5th$, is a graphical way to visualise the multinomial theorem. – Alyosha May 5 '13 at 16:59
The simplification for this type of expansion is done through the multinomial theorem. The multinomial theorem is a generalization of the binomial case to any arbitrary number of terms in the sum to be exponentiated.
The multinomial theorem is written as follows:
$$(x_1 + x_2 + \cdots + x_m)^n = \sum_{k_1+k_2+\cdots+k_m=n} {n \choose k_1, k_2, \ldots, k_m} \prod_{1\le t\le m}x_{t}^{k_{t}}\$$
Where the multinomial co-efficient is defined as:
$${n \choose k_1, k_2, \ldots, k_m} = \frac{n!}{k_1!\, k_2! \cdots k_m!}$$
It may also be useful to you to note that the multinomial co-efficient is always expressible as products of binomial co-efficients [Graham, Knuth, Patashnik, Concrete Mathematics (2nd edition)]:
$${n \choose k_1, k_2, \ldots, k_m} = {x_1+x_2+\cdots+x_m \choose x_2+\cdots+x_m}\cdots{x_{m-1}+x_m \choose x_m}$$
A fuller explanation can be found on Wikipedia, or Wolfram MathWorld
-
is there any relation between $k_1, k_2, ...$ isn't $k_1+k_2+ ...+k_m=n$ again going to be combination of k's?? – Santosh Linkha May 30 '12 at 15:18
The only relation is that the sum of all $k_i$ is equal to n. I do not believe there is an equality relationship such as $k_1>k_2$ or anything like that. – Shaktal May 30 '12 at 15:20
Oh ... thank you both!! – Santosh Linkha May 30 '12 at 15:21
This is called the Multinomial theorem: $$(a_1 + a_2 +a_3+... +a_n)^m=\sum_{k_1+k_2+...+k_n=m}\frac{m!}{k_1!\cdot...\cdot k_n!}a_1^{k_1}\cdot...\cdot a_n^{k_n}$$
- | 2016-07-25T12:06:32 | {
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https://math.stackexchange.com/questions/2087798/prove-that-fraca-1a-2-fraca-2a-3-fraca-3a-4-fraca-n-1 | # Prove that $\frac{a_1}{a_2}+ \frac{a_2}{a_3}+\frac{a_3}{a_4}+…+\frac{a_{n-1}}{a_n} \le \frac{n}{2}$
Let $n \ge 2$ be a positive integer and let $a_1, a_2, ... a_n$ be positive numbers such that $$a_1\le a_2, a_1+a_2\le a_3, a_1+a_2+a_3\le a_4, ... ,a_1+a_2+...+a_{n-1}\le a_n$$ prove that $$\dfrac{a_1}{a_2}+ \dfrac{a_2}{a_3}+\dfrac{a_3}{a_4}+...+\dfrac{a_{n-1}}{a_n} \le \dfrac{n}{2} \hspace{2cm} (1)$$ When does the equality holds?
Solution: I proceed as follows using Mathematical Induction.
For $n=2, \frac{a_1}{a_2} \le 1$. Let the (1) be true for $n=k$ i.e $$\dfrac{a_1}{a_2}+ \dfrac{a_2}{a_3}+\dfrac{a_3}{a_4}+...+\dfrac{a_{k-1}}{a_k} \le \dfrac{k}{2} \hspace{2cm} (2)$$ We need to prove $$\dfrac{a_1}{a_2}+ \dfrac{a_2}{a_3}+\dfrac{a_3}{a_4}+...+\dfrac{a_{k}}{a_{k+1}} \le \dfrac{k+1}{2}$$ Consider $$\dfrac{a_1}{a_2}+ \dfrac{a_2}{a_3}+\dfrac{a_3}{a_4}+...+\dfrac{a_{k-1}}{a_k}+\dfrac{a_{k}}{a_{k+1}} \le \dfrac{k}{2} +\dfrac{a_{k}}{a_{k+1}} \hspace{2cm} (3)$$ Since $$a_1+a_2+...+a_{k-1}+a_{k}\le a_{k+1} \hspace{2cm} (4)$$ also $$a_1+a_2+...+a_{k-1}\le a_{k} \hspace{2cm} (5)$$ Using 5 in 4, we get $$a_{k}+a_{k}\le a_{k+1}$$ $$\dfrac{a_{k}}{a_{k+1}} \le \dfrac{1}{2}$$ using in (3), we get $$\dfrac{a_1}{a_2}+ \dfrac{a_2}{a_3}+\dfrac{a_3}{a_4}+...+\dfrac{a_{k-1}}{a_k}+\dfrac{a_{k}}{a_{k+1}} \le \dfrac{k+1}{2}$$ Is the procedure is correct. And when the equality holds... Thanks for any assistance
• Your proof looks good. As to the question about when equality holds, the clues are right there in your proof. How big can $\frac{a_1}{a_2}$ be? Can you force equality? What about $\frac{a_2}{a_3}, \frac{a_3}{a_4},\ldots$ ? – quasi Jan 7 '17 at 18:36
• How do you get $a_k+a_k \leq a_{k+1}$ from (4) and (5)? – rtybase Jan 7 '17 at 18:42
• Oops -- I fell in the same trap that yasir fell in -- ignore my first comment. Instead, look at rtybase's objection. – quasi Jan 7 '17 at 18:46
• @rtybase Subtracting (5) from (4) – MattG88 Jan 7 '17 at 18:49
• @rtybase -- you can't subtract inequalites that are in the same direction. You can add them, but you can't subtract them. A common illusion (which I fell victim to as well, even though I know better). – quasi Jan 7 '17 at 19:13
The idea is "local adjustments". If $k$ is the smallest integer such that $a_1+\cdots+a_{k-1} < a_k$, we adjust up $a_1$, $\cdots$, $a_{k-1}$ with a scale factor of $s=\frac{\sum_{i=1}^{k}a_i}{2\sum_{i=1}^{k-1}a_i}$ for all, and adjust down $a_k$ by a scale factor of $t=\frac{\sum_{i=1}^{k}a_i}{2a_k}$, and achieve a large value for LFS, and make $\sum_{i=1}{k}a_i=a_k$. Note that since $\sum_{i=1}^{k}a_i$ is not changed, and $a_1$, $\cdots$, $a_{k-1}$ are scaled up by the same factor, all conditions are still satisfied. We just need to prove that really the left hand side gets larger or equal.
Really there is just one case, but let's do three cases: 1) $k=2$, and 2) $k=n$; and 3) $2<k<n$.
1) $k=2$, which means $a_1<a_2$. Now The only values changed are $\frac{a_{1}}{a_2}$ and $\frac{a_{2}}{a_{3}}$, and we want to show that $\frac{a_{1}}{a_2} + \frac{a_{2}}{a_{3}} \le \frac{sa_{1}}{ta_2} + \frac{ta_{2}}{a_{3}}$. This is easy as $s=\frac{a_1+a_2}{2a_1}$ and $t=\frac{a_1+a_2}{2a_2}$, and $sa_1=ta_2=\frac{a_1+a_2}{2}$: $$\frac{a_{1}}{a_2} + \frac{a_{2}}{a_{3}} \le \frac{a_{1}}{a_1} + \frac{a_{1}}{a_{3}} = \frac{sa_{1}}{ta_2} + \frac{a_{1}}{a_{3}} < \frac{sa_{1}}{ta_2} + \frac{ta_{2}}{a_{3}},$$ where the first inequality requires a simple check.
2) $k=n$. This is easy as The only value changed is $\frac{a_{n-1}}{a_n}$ but we scale up $a_{n-1}$ and down $a_n$.
3) $2<k<n$. Now The only values changed are $\frac{a_{k-1}}{a_k}$ and $\frac{a_{k}}{a_{k+1}}$, and we want to show that $\frac{a_{k-1}}{a_k} + \frac{a_{k}}{a_{k+1}} \le \frac{sa_{k-1}}{ta_k} + \frac{ta_{k}}{a_{k+1}}$. Note that since the equality holds for $k-1$, we have $2a_{k-1}=\sum_{i=1}{k-1}a_i$. So $sa_{k-1}= \frac{2a_{k-1}+a_k}{4}$ and $ta_k=\frac{2a_{k-1}+a_k}{2}=2sa_{k-1}$. So we want to prove that $$\frac{a_{k-1}}{a_k} + \frac{a_{k}}{a_{k+1}} \le \frac{1}{2} + \frac{2a_{k-1}+a_k}{2a_{k+1}},$$ which is equivalent to the following after multiplying $2a_ka_{k+1}$: $$2a_{k-1}a_{k+1}+2a_k^2 \le a_ka_{k+1}+(2a_{k-1}+a_k)a_k,$$ which is equivalent to $$(a_{k+1}-a_k)(a_k-2a_{k-1}) \ge 0,$$ which is true since $a_{k+1}>a_k$ and $a_k \ge \sum_{i=1}^{k-1}a_i = 2a_{k-1}$.
So if we adjust up/down all $a_i$ and make all equalities in the condition hold, we achieve the largest value for $\frac{a_1}{a_2}+\cdots+\frac{a_{n-1}}{a_n}$, which happens to be $\frac{n}{2}$. | 2019-06-26T15:54:37 | {
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https://oneclass.com/class-notes/ca/u-of-waterloo/math/math-136/281636-lecture-20pdf.en.html | Class Notes (838,022)
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Lecture 20
Lecture 20.pdf
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Wednesday, February 26 − Lecture 20: Linear independence Concepts: 1. Linearly independent set. 2. Recognize that subsets of linearly independent sets are linearly independent. 3. Characterize a linearly independent set as one being a set where no vector isa linear combination of the others. 20.1 Definition – Generalization of the definition of linear independence. Let v , v ...., v 1 2, k be k vectors in a vector space V. The vectors v , v ..., v are said to be linearly 1 2, k independent if and only if the only way that α 1 1 α v 2 2.. + α v = 0k k can hold true is if α ,1α ,2....., α kre all zeroes. If v1, v2,.., vkare not linearly independent (i.e. there existsα , α , .1...2 α not alk zero such that α v1 1α v +2 2... + α v = k k then they are said to be linearly dependent. x 20.1.1 Example – We know that the two functions e and sinx are vectors in the set F described above. Then S = Span{e , sinx} = { αe + βsinx : α, β belong to ℝ } is a x subspace of F. Show that the set {e , sinx} is linearly independent. Solution: Let us first recall that two functions f and g are equal on their domain Dif and only if f(x) = g(x) for all x in D. Suppxse αe + βsinx = 0(x) for all x (where 0(x) denotes the zero function; it maps every x in the domain to 0.). x - Suppose there exists α ≠ 0 such that αe + βsinx = 0(x). x - Then e = (–β/α)sinx for all x. This includes the value x = π. - But 0 ≠ e = (–β/α)sinπ = 0, a contradiction. - Then α must be 0. x x - Suppose β ≠ 0 is such that, 0e + βsinx = 0(x). Then (–0/β)e = sinx for all x. - Since sin(π/2) =1 ≠ 0, we have a contradiction. - So β must also be 0. - So {e , sinx} is linearly independent. 20.1.2 Proposition − The vectors M = {v , v , ..1, v2} with k , 2, are linearly independent if and only if no vector in M is a linear combination of the other vectors in M. Proof : This is similar to the previously given proof for the case of V = ℝ . n (⇐) - Suppose none of these vectors is a linear combination of the others. - Suppose M is not linearly independent. - Then there exists α , α 1...2., α not kll zeroes such that α v + α v +1 1..+ α2 2= k k 0. - Suppose α is not zero. Rearrange the order of {v , v , ...., v } so that p = 1, .i.e., p 1 2 k α 1s not zero. - Then v = 1 (α /α )2v +1− (2 /α ) v +3, 1.., 3 − (α /α )v . k 1 k - Hence v is a linear combination of the others. Contradiction. 1 (⇒) - Suppose {v , v1, .2.., v } ks linearly independent. - Suppose one vector of {v , v , .1..,2v } is k linear combination of the others. Say it is v . 1 - Then there exists α ....2, α suck that such that v = α v + 1.... 2 2 v . k k - Then v − 1 v − 2 v2+ , ..3, 3 α v . = 0. k k - Since the coefficient of v is n1t zero this contradicts our hypothesis. - Then no vector of {v , v ,1...2, v } iska linear combination of the others. If a vector in U is a linear combination of the others we will refer to it as being “redundant”. It doesn't contribute anything to its span. 20.2 Definition – Let {v 1, v2,..., vk} be a spanning family for a vector space V. That is, Span{v , v ..., v } = V. Then for every vector v in V there exists scalars α , α , ....., α 1 2, k 1 2 k such that v = α v1 1α v +2 2... + α v . k k If for each vectorv this set of scalars α , α , ..., α satisfying v = α v + α v + ..... + α v 1 2 k 1 1 2 2 k k is unique then we say that Span{v 1, v2,.., v k satisfies the unique representation property with respect to its spanning family. 20.2.1 Example – The spanning family Span{(1, 0), (0,1)} = Span{ e , e } = ℝ 1 2 2 satisfies the unique representation property since for an arbitrary vector v= (a, b) in ℝ , a and b are the unique scalars associated to v so that v = ae + be . 1 2 20.2.2 Theorem − Let S = {v , v , .1..,2v } be a k
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So we can recommend you notes for your school. | 2018-04-24T23:37:57 | {
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https://math.stackexchange.com/questions/3368484/how-many-ten-digit-numbers-are-there-in-which-every-digit-is-2-or-3-and-no-two | # How many ten-digit numbers are there in which every digit is 2 or 3, and no two 3s are adjacent?
How many ten-digit numbers are there in which every digit is 2 or 3, and no two 3s are adjacent?
Taken from the 2008 IMC https://chiuchang.org/wp-content/uploads/sites/2/2018/02/2008-IWYMIC-Individual.x17381.pdf
my attempt
The number of ten digit numbers in which the digits are either 2 or 3 is $$2^{10}$$ and the numbers of ten digit numbers where there is no pair of adjacent numbers that are the same is $$2$$ E.g($$2323232323$$ & $$3232323232$$) and we know that the number of ten digit numbers consisting of pairs of adjacent $$2$$s and adjacent $$3$$s is the same due to symmetry therefore the answer would be $$\frac{2^{10}-2}{2}=511$$ however this doesn't taken into account the possibilities of having adjacent pairs of $$3$$s and $$2$$s in the same number.
• This is the same as this question
– lulu
Sep 24 '19 at 18:45
• And similar to this, more recent, one. Sep 25 '19 at 5:47
Suppose you have r number of 3 in a line. Now we know 3 can't be adjacent so I have to put at least 1 in between them. For example-r=3 |-|-|(| denotes 3 and - denotes 2). Now we have x=10-(r+r-1) 2's remaining. Now we can put these 2's in r+1 slots(in the example we can put in 4 slots slot1|slot2|slot3|slot4). Now this becomes a standard problem Stars and bars of x1+x2+x3+..xl=p (where l denotes the number of slots and where every xi can be 0 and p denotes number of 2 we want to distribute which is x in our case). Number of solution of above equation is given by:
$$\binom {p+l-1}{l-1}$$
here p=10-(r+r-1) and l=r+1(slots) putting these value in above formula we get:
$$\binom {11-r}{r}$$
Now you can vary r from 0-10 and just add them you will get 144 as the answer. Here you can see you do not need to vary from 6 -10 because when we place 6 or more number we don't have enough 2's to put in between them so they will contribute 0.
Suppose you have $$k$$ $$3$$s and it does not end with a $$3$$.
That the same thing as saying you have $$k$$ characters $$32$$ and $$10-k-k=10-2k$$ twos. So of the $$10-2k + k=10-k$$ characters you must choose $$k$$ spaces for the $$k$$ $$32$$ characters. There $${10-k \choose k}$$ ways to do that.
Now suppose you have $$k$$ $$3$$s and it does end with a $$3$$.
If we just ignore the last place and put the $$3$$ in it, that is the same thing as saying you have $$k-1$$ characters $$32$$ to place and $$10-k-(k-1)=11-2k$$ $$2$$s to place. There are $${10-k\choose k-1}$$ ways to do that.
So there are $${10-k \choose k}+ {10-k\choose k-1}$$ ways to place $$k$$ threes.
Now we can have at most $$5$$ threes. (Any more and we won't have enough $$2$$s to go between all $$3$$s.)
So there are
$$\sum_{k=0}^5 {10-k \choose k}+ {10-k\choose k-1}$$ ways. (Assume $${10\choose -1} = 0$$.... after all.... this would be the number of ways to choose $$0$$ threes and have a $$3$$ at the end which isn't possible.)
So $$({10\choose 0}) + ({9\choose 1}+ {9\choose 0}) +({8\choose 2}+{8\choose 1}) + ({7\choose 3}+{7\choose 2}) + ({6\choose 4}+{6\choose3}) + ({5\choose 5} + {5\choose 4})=$$
$$1 + (9+1) + (\frac {8*7}2 + 8) + (\frac {7*6*5}6 -\frac {7*6}2) + (\frac {6*5*4*3}{24}+\frac{6*5*4}6) + (1 +5)=$$
$$1 + 10 +(28+8) + (35+21) + (15+20) + 6=$$
$$11+36+56+35+5=144$$ | 2022-01-25T06:00:00 | {
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https://math.stackexchange.com/questions/1266282/finding-the-inverse-of-a-number-under-a-certain-modulus | # Finding the inverse of a number under a certain modulus
How does one get the inverse of 7 modulo 11?
I know the answer is supposed to be 8, but have no idea how to reach or calculate that figure.
Likewise, I have the same problem finding the inverse of 3 modulo 13, which is 9.
The inverses can be computed using the extended euclidean algorithm. As long as $\gcd(x,n) = 1$ the inverse $x^{-1} \bmod n$ exists and is $y$ from the extended euclidean algorithm where $$xy + kn = 1 = \gcd(x,n)$$ For example we get $$7\cdot 8 - 5\cdot 11 = 1\\ 9 \cdot 3 - 2\cdot 13 = 1$$
Note that you can also obtain negative numbers, for example $$7\cdot (-3) + 2\cdot 11 = 1$$ In this case you can use congruences $\bmod n$ to obtain the canonical inverse: $-3 \equiv 8 \pmod{11}$
• Thanks but I'm really struggling with this, could you provide me with a sample, i be stuck for two days trying to figure it out – Dan W May 4 '15 at 12:07
• @DanW What part of this do you struggle with? – AlexR May 4 '15 at 12:09
• Basically, i want to know the inverse of 7 within mod 11, i need a figure!!, i understand the gcd and what it is doing/proving. – Dan W May 4 '15 at 12:20
• @DanW Can you show me what the extended euclidean algorithm gives you wen used for $\gcd(7,11)$? – AlexR May 4 '15 at 12:23
• A B Q R 11 7 1 4 7 4 1 3 4 3 1 1 – Dan W May 4 '15 at 12:28
To find the inverse of $7$, $\pmod{11}$, you must find a solution to $7x\equiv 1 \pmod{11}$. (Then $x$ satisfies the definition of inverse.)
As suggested in other answers, one way to do this is with the extended Euclidean algorithm, and in fact this is the best general purpose algorithm for this type of problem.
But for small values, you can also try 'adding the modulus':
$7x\equiv 1\equiv 12\equiv 23\equiv 34\equiv 45\equiv 56 \pmod{11}$.
Then from $7x\equiv 56\pmod{11}$, we can cancel $7$, obtaining $x\equiv 8 \pmod{11}$.
Here's an illustration of finding the multiplicative inverse of $37 \bmod 100$ using the extended Euclidean algorithm. (I used bigger numbers for this example so that the relationships are a little clearer).
On each line, $n=100s+37t$. We start the table with two lines giving $n=100$ and $n=37$ in the obvious way. Then $q$ gives the rounded-down ratio of the current and previous value of $n$. Using this, the next line is calculated by subtracting $q$ copies of the current line entries from the previous line entries.
$$\begin{array}{c|c|c|c} n & s & t & q \\ \hline 100 & 1 & 0 & \\ 37 & 0 & 1 & 2 \\ 26 & 1 & -2 & 1 \\ 11 & -1 & 3 & 2 \\ 4 & 3 & -8 & 2 \\ 3 & -7 & 19 & 1 \\ 1 & 10 & \color{red}{-27} & 3 \\ \end{array}$$
On the last line, $1 = 10\times 100 + (-27)\times 37$, so $\color{red}{-27}\times 37 \equiv 1 \bmod 100$
Bringing the result positive, $-27 \equiv \color{red}{73} \bmod 100$. And $37 \times 73 = 2701 \equiv 1 \bmod 100 \quad \checkmark$.
As you can perhaps see, you don't actually need to calculate the $s$ values at all to get the modular inverse. I left them in to help understand the table.
The corresponding tables for your particular questions: $$\begin{array}{c|c|c|c} n & s & t & q \\ \hline 11 & 1 & 0 & \\ 7 & 0 & 1 & 1 \\ 4 & 1 & -1 & 1 \\ 3 & -1 & 2 & 1 \\ 1 & 2 & \color{red}{-3} & 3 \\ \end{array}$$ and $7^{-1} \equiv -3 \equiv \color{red}8 \bmod 11$. $\quad 7\times 8 = 56 = 55+1\quad \checkmark$
$$\begin{array}{c|c|c|c} n & s & t & q \\ \hline 13 & 1 & 0 & \\ 3 & 0 & 1 & 4 \\ 1 & 1 & \color{red}{-4} & 3 \\ \end{array}$$ and $3^{-1} \equiv -4 \equiv \color{red}9 \bmod 13$. $\quad 3\times 9 = 27 = 26+1\quad \checkmark$
${\rm mod}\ 11\!:\,\ \dfrac{1}7\equiv \dfrac{12}{-4}\equiv -3\$ (see Gauss's algorithm. for an algorithmic version of this).
Or, compute the Bezout identity $\,\gcd(11,7) = 2(11)-3(7) = 1\,$ by the Extended Euclidean Algorithm (see here for a convenient version). Thus $\ {-}3(7)\equiv 1\pmod{11}$
${\rm mod}\ 13\!:\,\ \dfrac{1}3\equiv \dfrac{-12}{3}\equiv -4\$
Generally inverting $\,a\,$ mod $\,m\,$ is easy if $\,a\mid m\pm1,\$ i.e. $\ m = ab\pm 1$
${\rm mod}\ ab-1 \!:\qquad ab\equiv 1\,\Rightarrow\, a^{-1}\equiv b$
${\rm mod}\ ab+1 \!:\,\ a(-b)\equiv 1\,\Rightarrow\, a^{-1}\equiv -b$
Above are special cases: $\,3\mid 13\!-\!1\$ and $\ 7\equiv -4\mid 11\!+\!1$
To find the inverse of $7$ modulo $11$, we must solve the equivalence $7x \equiv 1 \pmod{11}$. To do this, we use the Extended Euclidean Algorithm to express $1$ as a linear combination of $7$ and $11$. The coefficient of $7$ will be the inverse modulo $11$. By the Euclidean Algorithm, \begin{align*} 11 & = 1 \cdot 7 + 4\\ 7 & = 1 \cdot 4 + 3\\ 4 & = 1 \cdot 3 + 1\\ 3 & = 3 \cdot 1 \end{align*} We now take the equation $4 = 1 \cdot 3 + 1 = 3 + 1$, solve for $1$, then work backwards until we obtain $1$ as a linear combination of $7$ and $11$. \begin{align*} 1 & = 4 - 3\\ & = 4 - (7 - 4)\\ & = 2 \cdot 4 - 7\\ & = 2(11 - 7) - 7\\ & = 2 \cdot 11 - 3 \cdot 7 \end{align*} Since $2 \cdot 11 - 3 \cdot 7 = 1$, $$-3 \cdot 7 = 1 - 2 \cdot 11 \Longrightarrow -3 \cdot 7 \equiv 1 \pmod{11}$$ Hence, $-3$ is the inverse of $7 \pmod{11}$. To express the inverse as one of the residues $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$, we add $11$ to $-3$ to obtain $-3 + 11 \equiv 8 \pmod{11}$. Hence, $7^{-1} \equiv 8 \pmod{11}$.
Check: $7 \cdot 8 \equiv 56 \equiv 1 + 5 \cdot 11 \equiv 1 \pmod{11}$.
To verify you understand the algorithm, try to find the inverse of $3$ modulo $13$.
As said by AlexR, you can find $x,y\in\mathbb Z$ with $7x+11y=1$ using Extended Euclidean algorithm (see this answer for how to best use it).
You can also use elementary modular arithmetic 'tricks':
$\bmod{11}\!:\ 7x\equiv 1\equiv -21\stackrel{:7}\iff x\equiv -3\equiv 8$.
$\bmod{11}\!:\ 7x\equiv -4x\equiv 1\equiv 12\stackrel{:(-4)}\iff x\equiv -3\equiv 8$.
We could divide by $7$ and $-4$ because $(7,11)=(-4,11)=1$. | 2020-01-26T17:57:47 | {
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https://math.stackexchange.com/questions/930565/number-of-lattice-points-in-a-triangle | # Number of Lattice Points in a Triangle
Problem
Let the co-ordinates of the vertices of the $\triangle OAB$ be $O(1,1)$, $A(\frac{a+1}{2},1)$ and $B(\frac{a+1}{2},\frac{b+1}{2})$ where $a$ and $b$ are mutually prime odd integers, each greater than $1$. Then find the number of lattice points inside $\triangle OAB$, i.e., not on the borders of $\triangle OAB$. How does the answer change if the restriction $\operatorname{gcd}(a,b)=1$ is removed?
Solution
Let $L(a,b)$ be the number of lattice points inside $\triangle OAB$. The linear transformation $T(x,y)=(x-1,y-1)$ on the triangle $OAB$ do not have any effect on $L(a,b)$. Hence $L(a,b)$ is equal to the number of lattice points inside the triangle $O'A'B'$, where $O'=(0,0)$, ${\textstyle A'=(\frac{a-1}{2},0)}$ and ${\textstyle B'=(\frac{a-1}{2},\frac{b-1}{2})}$.
Set ${C'=(0,\frac{b-1}{2})}$. By symmetry the number of lattice points inside the rectangle $O'B'C'$ is $L(a,b)$. The number of lattice point inside the rectangle $O'A'B'C'$ is ${\textstyle \frac{a-3}{2} \cdot \frac{b-3}{2}}$. Consequently
$$2L(a,b) = \frac{a-3}{2} \cdot \frac{b-3}{2} - K$$
where $K$ are the number of lattice points on the straight line between $O'$ and $B'$. This line is given by $y= \frac{b-1}{a-1}x$, yielding
$$K = |\{ {\textstyle 0 < x < \frac{a-1}{2} \mid \frac{b-1}{a-1}x \in \mathbb{N} \}| }$$
By letting $d=\text{gcd}(a-1,b-1)$, we obtain $a-1=rd$ and $b-1=sd$ for two positive coprime integers $r$ and $s$. Therefore
$$\frac{b-1}{a-1}x = \frac{sx}{r} \in \mathbb{N} \;\; \Leftrightarrow \;\; \frac{x}{r} \in \mathbb{N}$$
Now ${\textstyle 0 < x < \frac{a-1}{2} = \frac{rd}{2}}$, implying ${\textstyle 0 < \frac{x}{r} = \frac{d}{2}}$. Since $d$ is even (since $d=\gcd(a-1,b-1)$ and $a$ and $b$ are both odd positive integers) give us ${\textstyle K = \frac{d}{2}-1}$. Thus by (1)
$${\textstyle L(a,b) = \dfrac{(a-3)(b-3)}{8} - \dfrac{\text{gcd}(a-1,b-1)}{4} + \dfrac{1}{2}}$$
But I think that by Pick's Theorem the answer should be,
$${\textstyle L(a,b) = \dfrac{(a-1)(b-1)}{8} - \dfrac{\text{gcd}(a-1,b-1)}{4} + \dfrac{1}{2}}$$
Which one is correct?
• You can easily figure out which one is correct by putting in some small values for $a$ and $b$. – Ted Sep 14 '14 at 6:06
• Try $a = 1$ and $b = 10$. This yields a degenerate triangle with $0$ lattice points strictly inside it. Which answer gives you $0$? – JimmyK4542 Sep 14 '14 at 6:17
• @JimmyK4542: Both $a$ and $b$ are odd and mutually prime and each is greater than $1$. – user170039 Sep 14 '14 at 6:26
• Fine, try a less trivial case like $a = 3$ and $b = 5$. The formula in the gray box gives $L(3,5) = 0$ while the other formula gives $L(3,5) = 1$. Which one is right? – JimmyK4542 Sep 14 '14 at 6:27
• There are $0$ lattice points inside the triangle with vertices $O(1,1)$, $A(2,1)$, $B(2,3)$. So the first formula is correct (in that case). I have no idea where the second formula goes wrong because you have left out the details of how you got that. – JimmyK4542 Sep 14 '14 at 6:34
The area of $\Delta OAB$ is $K = \dfrac{1}{2} \cdot \dfrac{a-1}{2} \cdot \dfrac{b-1}{2} = \dfrac{(a-1)(b-1)}{8}$.
The number of points on the boundary (line segments $OA$, $AB$, and $BO$) is
$\underbrace{\dfrac{a-1}{2}}_{OA}+\underbrace{\dfrac{b-1}{2}}_{OB}+\underbrace{\text{gcd}\left(\dfrac{a-1}{2},\dfrac{b-1}{2}\right)}_{BO}$ (you missed the first two terms of this).
So, by Pick's Theorem, the area is
$K = I+\dfrac{1}{2}B-1 = I+\dfrac{1}{2}\left[\dfrac{a-1}{2}+\dfrac{b-1}{2}+\text{gcd}\left(\dfrac{a-1}{2},\dfrac{b-1}{2}\right)\right]-1$
$= I + \dfrac{a-1}{4} + \dfrac{b-1}{4}+\dfrac{\text{gcd}(a-1,b-1)}{4}-1$.
Hence, the number of points in the interior is
$I = \dfrac{(a-1)(b-1)}{8} - \dfrac{a-1}{4} - \dfrac{b-1}{4} - \dfrac{\text{gcd}(a-1,b-1)}{4} + 1$
$= \dfrac{(a-3)(b-3)}{8} - \dfrac{\text{gcd}(a-1,b-1)}{4} + \dfrac{1}{2}$ (after a bit of simplification). | 2020-05-29T22:37:49 | {
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https://math.stackexchange.com/questions/937312/what-is-the-combinatorial-interpretation-of-the-product-of-binomial-coefficients | # What is the combinatorial interpretation of the product of binomial coefficients?
Full disclosure: This question is relating to a homework question. It's not a homework question itself, but rather a clarifying question to help myself get a handle on the actual question.
Suppose I had three sets $X$, $Y$, and $Z$ with cardinalities $|X| = x$, $|Y| = y$, and $|Z| = z$, with $z \leq y \leq x$
I know that $x \choose y$ is the number of ways of selecting $y$ elements from $X$.
However, I'm unclear of the combinatorial interpretation of the following:
$${x \choose y} {y \choose z}$$
So my question is: in general, is there an intuitive interpretation for the product of the number of choices?
• If $x\le y\le z$ then both of those coefficients are $0$ – Adam Hughes Sep 19 '14 at 1:50
• You're right, sorry; I'm updating my question now – Jason Baker Sep 19 '14 at 1:51
• There's two interpretations. 1) How many ways are there to first pick $y$ elements out of $x$, and then pick $z$ elements out of these $y$? 2) How many ways are there to divide $x$ into sets of size $z$, $y-z,$ and $x-y$? – Semiclassical Sep 19 '14 at 1:54
• @Semiclassical That makes sense; thanks for the confirmation – Jason Baker Sep 19 '14 at 1:57
The answer to your question is affirmative: Yes, there is an intuitive interpretation for the number of choices.
We know at least two interpretations of the binomial coefficient $\binom{n}{k}$ which are useful for combinatorial proofs. These are
• Set theoretical view: $\binom{n}{k}$ as number of $k$-element subsets of an $n$-element set
• Geometrical view: $\binom{n}{k}$ as number of lattice pathes of length $n$ containing $k$ horizontal $(1,0)$-steps and $n-k$ vertical $(0,1)$-steps
The geometrical view is valid because there are $\binom{n}{k}$ choices to select $k$ steps in horizontal direction leaving the remaining $n-k$ steps for the vertical direction.
Let's apply the second approach to
\begin{align*} \binom{x}{y}\binom{y}{z}\qquad z \leq y \leq x\tag{1} \end{align*}
When looking at $\binom{y}{z}$ you may think on all pathes of length $y$ starting at $A=(0,0)$ and ending in $B=(z,y-z)$ thereby containing exactly $z$ horizontal $(1,0)$-steps and $y-z$ vertical $(0,1)$-steps. All these pathes are enclosed within a rectangle of length $z$ and height $y-z$.
Similarly we consider $\binom{x}{y}$ as the number of lattice pathes of length $x$ containing $y$ $(1,0)$-steps and $x-y$ $(0,1)$-steps. But here we let all these pathes start in $B=(z,y-z)$ and so they end up in $C=(z+y,x-z)$.
We observe following
Combinatorial interpretation of $$\binom{x}{y}\binom{y}{z}$$
This is the number of lattice pathes of length $x+y$ starting in $A=(0,0)$ ending up in $C=(z+y,x-z)$ and passing through the point $B=(z,y-z)$.
The pathes contain only horizontal $(1,0)$-steps and vertical $(0,1)$-steps.
A possible interpretation is pairing sets $X$ and $Y$ with no elements of $Y$ left so that would require selecting $y$ elements from $X$ or $\displaystyle \binom xy$, then making triplets with no elemts of $Z$ left that would require selecting $z$ pairs from these $y$ pairs or $\displaystyle \binom yz$. | 2019-06-17T23:27:23 | {
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https://math.stackexchange.com/questions/1764278/number-of-points-of-accumulation-of-a-sequence | # Number of points of accumulation of a sequence
Can a sequence have infinitely many points of accumulation i.e. we can extract infinitely many subsequences from it s.t. they all converge to their respective point of accumulation? I have the feeling it would mean that the period of repetition of something could be infinitely big.
• Since there are countably many rational numbers, there is a sequence that includes them all. The accumulation points of this sequence is the entire set of real numbers. Apr 29 '16 at 17:29
• I like your example a lot ! Apr 29 '16 at 18:15
• There's a classic theorem that if you walk around a circle in discrete steps of $a$ radians, where $a/\pi$ is irrational, then the set of points that you visit is dense in the circle. This implies that $(\sin(an))_{n\in\mathbb N}$ is dense in $[-1, 1]$, and you can take $a=1$ for a cute example of a sequence with infinitely many accumulation points. Apr 29 '16 at 23:16
• See also this question. Some of the posts linked there might be of interest, too. Apr 30 '16 at 8:23
Start with $0,1$. Then travel backwards to $0$ in steps of $1/2$, so $1/2,0$. Then travel forwards to $1$ in steps of $1/4$, so $1/4,2/4,3/4,1$. Then travel backwards to $0$ in steps of $1/8$, so $7/8$, $6/8$, $5/8$, and so on. Continue.
Every real between $0$ and $1$ is an accumulation point of our sequence.
• Your sequence is bounded AND has infinitely many points of accumulation. Thank you for answering :) Apr 29 '16 at 16:28
• @MehdiSlimani: You are welcome. If you prefer we could use $0,1,1/2,0,1/3,2/3,1,3/4,2/4,1/4,0,1/5,2/5,\dots$ and then we can quote the result that the rationals in $[0,1]$ are dense in $[0,1]$. Apr 29 '16 at 16:29
• I'm a bit confused. How can subsequences of your sequence converge if each member is in a different space? Maybe I don't have the necessary knowledge yet Apr 29 '16 at 16:36
• For every real $x$ between $0$ and $1$, there is a subsequence of our sequence that has limit $x$. For in our back and forth travels, we get within $1$ of $x$, then within $1/2$ of $x$, then within $1/4$ of $x$, and so on. With the modified version in which we take steps $1$, then $1/2$, then $1/3$, then $1/4$, then $1/5$, and so on, our travels bring us to every rational between $0$ and $1$, so again there is a subsequence that converges to $x$. Apr 29 '16 at 16:39
• Ok, I hadn't understood it well. Thank you again Apr 29 '16 at 16:43
Yes, this is possible. For example consider the sequence $a_n$ for $n \ge 2$ defined as the smallest divisor of $n$ greater than $1$.
The accumulation points are all the prime numbers. Subsequences witnessing them are for instance the $p$-th powers for each prime $p$.
• Simple and nice example, thank you a lot :) Apr 29 '16 at 16:24
The rationals are a countable set. We can define a 1-to-1 function $f:N\to Q$ with $Q=\{f(n):n\in N\}.$ Consider the sequence $S=(f(n))_{n\in N}.$ Every real number is a limit point of a subsequence of $S.$
All the answers have uncountably many accumulation points. If you only want countably many, consider the sequence:
$1,1,2,1,2,3,1,2,3,4,1,2,3,4,5,\cdots$
Every positive integer is an accumulation point, and nothing else.
If you further want it to be bounded:
$\frac11,\frac11,\frac12,\frac11,\frac12,\frac13,\frac11,\frac12,\frac13,\frac14,\cdots$
Suppose each row of the infinite matrix below converges to a different real number
\begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} & \dots \\ a_{21} & a_{22} & a_{23} & a_{24} & \dots \\ a_{31} & a_{32} & a_{33} & a_{34} & \dots \\ a_{41} & a_{42} & a_{43} & a_{44} & \dots \\ \vdots & \vdots & \vdots & \vdots & \end{bmatrix}
Then the sequence
$a_{11}, a_{21}, a_{12}, a_{31}, a_{22}, a_{13}, a_{41},a_{32}, a_{23}, a_{14}, a_{51}, \dots$
contains an infinite number of convergent subsequences.
• Some authors define accumulation point not so strictly, namely that it is enough that a subsequence converges to it. Apr 30 '16 at 9:37 | 2021-12-08T16:12:15 | {
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https://stats.stackexchange.com/questions/198198/how-many-random-selections-needed-to-be-90-95-99-confident-that-a-specific/198202 | # How many random selections needed to be 90%, 95%, 99% confident that a specific selection occurred?
For example, let's say that I have a bag of 100 marbles labeled 1 to 100. A "selection" is randomly picking a marble from the bag and then placing it back into the bag. The marble selected is not known.
How many random selections do I need to make to be X% confidence that a specific marble was one of the marbles I selected, say marble #42?
Is there a formula or a distribution that helps me solve for X (90%, 95%, 99% confidence)?
Bernoulli or Binomial distributions seem close, but I can not quite make them solve for what I want.
Each time you grab a marble you have a 1 out of 100 chance of selecting the correct one. What you are asking is what are the odds that at least one of these will be correct which would be an inclusive "or" statement
$$P_{selected} = p \cup p \cup p \cup p \cup p... \cup p$$
However this is not trivial to calculate since you could select your marble multiple times. Thus it's better to use the negative of this and calculate when it was not selected.
$$P_{selected}^\sim = (1-p)\cap(1-p)\cap(1-p)(1-p)\cap(1-p)...(1-p)\cap(1-p)$$
which can be easily calculated to N times by
$$P_{selected}^\sim = (1-p)^N$$
Where you can then show
$$P_{selected} = 1- P_{selected}^\sim = 1-(1-p)^N$$
and after some algebra
$$N = \frac{\ln(1-P_{selected})}{\ln(1-p)}$$
so 90% confidence is N = 229.
95% confidence is N = 298
99% confidence is N = 458.
If $X$ has $k$ distinct values, each appearing with equal probability $1/k$, then it follows discrete uniform distribution. Probability of drawing a specific $x$ in a single draw is
$$\Pr(X=x) = \frac{1}{k}$$
probability of drawing value other than $x$ is
$$\Pr(X \ne x) = \frac{k-1}{k} = 1-\frac{1}{k}$$
so if you make $n$ draws, probability that in neither of the draws $x$ is drawn is
$$\left(1-\frac{1}{k} \right)^n$$
from here you can easily find that probability of drawing at least one $x$ in $n$ draws is
$$1-\left(1-\frac{1}{k} \right)^n$$
Now all you need to know is find such $n$ that leads to prespecified probability.
This can be easily checked by simulation.
set.seed(123)
f <- function(n, k = 100) 1-((1-1/k)^n)
g <- function() which(sample(100, 1e4, replace = TRUE) == 42)[1]
sim <- replicate(5e4, g())
that returns
> quantile(sim, c(0.9, 0.95, 0.99))
90% 95% 99%
230.00 301.00 460.01
> f(c(230, 301, 460))
[1] 0.9008952 0.9514495 0.9901782
(Notice that example above concerns not with confidence intervals but with exact probabilities.)
This shows us that we need to sample $n$ values until $x$ occurs, so we are talking about negative binomial distributed variable with parameters $p=1/k$ and $r=1$ (observe $n$ values until $X=x$ occurs once, where probability of observing $x$ is $1/k$). Results obtained using negative binomial distribution agree with the simulation.
> pnbinom(c(230, 301, 460)-1, 1, 1/k)
[1] 0.9008952 0.9514495 0.9901782
So you intuition about binomial distribution was close, as it follows negative binomial distribution. Negative binomial with $r=1$ simplifies to geometric distribution. This is how we ended up with distribution that is traditionally used for modeling discrete waiting times.
> pgeom(c(230, 301, 460)-1, 1/k)
[1] 0.9008952 0.9514495 0.9901782
In fact, if you recall, cumulative distribution function for geometric distribution is
$$F(n) = 1 - (1-p)^n$$
where $p$ is a probability of observing $x$ in a single draw. So geometric distribution can be used to obtain confidence intervals for the number of samples that need to be taken.
# 95% CI using quantile function of geometric distribution
> qgeom(c(0.025, 0.975), 1/k)
[1] 2 367
Let $M=100$. Let's say you make $N$ selections and you are looking for the occurrence of marble $m$.
Let $E_n$ be the event that marble $m$ is drawn at draw $n$. Let $F_n=E_n^c$ be the event that marble $m$ is NOT drawn.
Note that $P(E_n)=1/M$. This means, $P(F_n)=1-1/M$.
Let $F$ be the failure event that marble $m$ is not drawn for any $N$. Success is then $F^c$.
$P(F^c)=1-P(F)$
Assuming independent draws, we have:
$P(F)=P(F_1 \cap F_2 \cap F_3 \cap \ldots \cap F_N)=\prod_{n=1}^N P(F_n)=(1-1/M)^N$
Therefore, $P(F^c)=1-(1-1/M)^N$.
For the $c \in [0,1]$ (e.g. $c=0.9 \equiv 90\%$ confidence) confidence, set the LHS to $c$ and solve for $N$.
$\Rightarrow N=\dfrac{\ln(1-c)}{\ln(1-1/M)}$ | 2019-09-21T11:11:45 | {
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http://math.stackexchange.com/questions/108041/linear-algebra-preserving-the-null-space | # Linear Algebra: Preserving the null space
What does it mean when a book says that row operations preserve the null space? And why should that be true? I have read that row operations are equivalent to multiplying a vector on the left by an invertible elementary matrix. And I think I understand that the nullspace is the set of all vectors from $u \in U$ which get mapped to the zero vector in $V$ if $T:U\rightarrow V$ is linear. But I'm still not sure what this means.
-
It means that performing an elementary row operation on a matrix does not change the null space of the matrix. That is, if $A$ is a matrix, and $E$ is an elementary matrix of the appropriate size, then the matrix $EA$ has the same null space as $A$.
To see why this is true, suppose first that $x$ is in the null space of $A$. This means that $Ax=\vec 0$. Multiplying both sides of this equation by $E$, we see that $(EA)x=E\vec 0 =\vec 0$, meaning that $x$ is also in the null space of $EA$. Now suppose that $x$ is in the null space of $EA$, so that $(EA)x=\vec 0$. As you mentioned, $E$ is invertible, so we can multiply this equation by $E^{-1}$:
$$Ax=IAx=(E^{-1}E)Ax=E^{-1}(EA)x=E^{-1}\vec 0=\vec 0\;,$$
showing that $x$ is in the null space of $A$. In other words, a vector is in the null space of $EA$ if and only if it is in the null space of $A$, and $EA$ and $A$ have the same null space.
-
Thanks a lot, the if and only if explanation helps me connect these facts together now! – mcrocker Feb 12 '12 at 4:32
Say we have $V, W$ vector spaces and a linear transformation $T : V \to W$. The null space of $T$ is defined $N(T) = \{x \in V : T(x) = 0\}$.
When talking about row operations, we are talking about the matrix representation of the linear transformation. So if $\alpha = \{ x_1, ..., x_n \}$ is a basis of $V$ and $\beta = \{ y_1, ..., y_n \}$ is a basis of $W$ we have a matrix, say $A$, such that $A = [T]_\alpha^\beta$.
Say $T(x_i) = a_{i1}y_1 + ... + a_{in}y_n$. Then the matrix A will be the following $$A = \left[\begin{array}{ccc} a_{11} & ... & a_{1n} \\ ... & ... & ... \\ a_{n1} & ... & a_{nn} \end{array}\right].$$
Row operations are as follows
(i) Swap two rows.
(ii) Multiply row by non zero constant.
(iii) Add multiple of one row to a different row.
Say we swap the $i$th and $j$th row and call the new matrix $A'$. This new matrix defines a new linear transformation $T':V \to W$ such that $T'(x) = A'x$. We want to show that $N(T) = N(T')$. For any $x \in V$, say $$T(x) = b_1y_1 + ... + b_iy_i + ... + b_jy_j + ... + b_ny_n.$$ Then since $T'$ is just rows swapped $$T'(x) = b_1y_1 + ... +b_jy_j + ... + b_iy_i + ... b_ny_n.$$ Thus if $T(x) = 0$, $T'(x) = 0$ and vice versa. So $N(T) = N(T')$.
For (ii), think about if we multiply the $i$th row by $\beta$, then if the $i$th scalar for $T(x)$ is $c_i$, then the $i$th scalar for $T'(i)$ is $\beta c_i$. So if $T(x) = 0$, then $c_i = 0$, so $\beta c_i = 0$. All other scalars stay the same, thus $T'(x) = 0$. And if $T'(x) = 0$, $\beta^{-1}c_i = 0$, so $T(x) = 0$. Once again $N(T) = N(T')$.
I leave (iii) to you.
-
For what it's worth, I think the following way of thinking about why row operations do not change the null space is worthwhile:
An important observation to make:
One may observe that multiplication of a matrix $A$ by a column vector $\bf x$ amounts to taking a linear combination of the columns of $A$. The coefficients of the particular linear combination are the coordinates of $\bf x$.
For example: $$\tag{1} \underbrace{\left[ \matrix{a_{11}&a_{12}&a_{13} \cr a_{21}&a_{22}&a_{23}\cr a_{31}&a_{32}&a_{33} }\right]}_{A} \left[\matrix{x\cr y\cr z}\right] =\left[\matrix{x a_{11} +ya_{12}+ z a_{13} \cr x a_{21} +ya_{22}+ z a_{23} \cr x a_{31} +ya_{32}+ z a_{33}} \right] = x\underbrace{\left[\matrix{\color{maroon}{a_{11}}\cr\color{darkgreen}{ a_{21}}\cr \color{darkblue}{a_{31}}}\right]}_{\bf a_1} +y\underbrace{\left[\matrix{\color{maroon}{a_{12}}\cr\color{darkgreen}{ a_{22}}\cr \color{darkblue}{a_{32}}}\right]}_{\bf a_2} +z\underbrace{\left[\matrix{\color{maroon}{a_{13}}\cr\color{darkgreen}{ a_{23}}\cr\color{darkblue}{ a_{33}}}\right]}_{\bf a_3}$$
A second important observation to make:
$\bf x$ is in the null space of $A$ if and only if $A{\bf x}=\bf0$. This means that the linear combination of the columns of $A$ whose coefficients are the coordinates of ${\bf x}=\Bigl[{\textstyle{x\atop y}\atop\scriptstyle z }\Bigr]$ is the zero vector:
$$\tag{2} x\underbrace{\left[\matrix{\color{maroon}{a_{11}}\cr\color{darkgreen}{ a_{21}}\cr \color{darkblue}{a_{31}}}\right]}_{\bf a_1} +y\underbrace{\left[\matrix{\color{maroon}{a_{12}}\cr\color{darkgreen}{ a_{22}}\cr \color{darkblue}{a_{32}}}\right]}_{\bf a_2} +z\underbrace{\left[\matrix{\color{maroon}{a_{13}}\cr\color{darkgreen}{ a_{23}}\cr\color{darkblue}{ a_{33}}}\right]}_{\bf a_3} =\bf0$$
Now on to row operations:
Here, we make our third and final observation:
If one performs a row operation on $A$, then the corresponding right hand side of $(1)$ is obtained by performing the same row operation to each of ${\bf a_1}$, ${\bf a_2}$, and ${\bf a_3}$.
It follows from all of this that multiplying a row of $A$ by a non-zero number does not affect the null space. For example if row 1 of $A$ were multiplied by 2 then the right hand side of (1) would be: $$x\underbrace{\left[\matrix{\color{maroon}{2a_{11}}\cr\color{darkgreen}{ a_{21}}\cr \color{darkblue}{a_{31}}}\right]}_{\bf a_1} +y\underbrace{\left[\matrix{\color{maroon}{2a_{12}}\cr\color{darkgreen}{ a_{22}}\cr \color{darkblue}{a_{32}}}\right]}_{\bf a_2} +z\underbrace{\left[\matrix{\color{maroon}{2a_{13}}\cr\color{darkgreen}{ a_{23}}\cr\color{darkblue}{ a_{33}}}\right]}_{\bf a_3} =\bf z$$ If (2) holds, it is easily seen that ${\bf z}=\bf0$.
It should be fairly obvious that interchanging two rows of $A$ does not affect the null space. For example if rows 1 and 3 of $A$ were interchanged, the right hand side of (1) would become $$x\left[\matrix{\color{darkblue}{ a_{31}}\cr \color{darkgreen}{ a_{21}}\cr \color{maroon}{ a_{11}}}\right] +y\left[\matrix{\color{darkblue}{ a_{32}}\cr \color{darkgreen}{ a_{22}}\cr \color{maroon}{ a_{12}}}\right] +z\left[\matrix{\color{darkblue}{ a_{33}}\cr\color{darkgreen}{ a_{23}}\cr\color{maroon}{ a_{13}}}\right] =\bf z$$ If (2) holds, then $\bf z$ would be $\bf 0$.
Finally, if a multiple of a row of $A$ were added to another row of $A$, the null space would be unchanged. For example if twice row 1 of $A$ were added to to row 2, then the right hand side of $(1)$ would become:
$$\tag{3} x\left[\matrix{\color{maroon}{ a_{11}}\cr \color{darkgreen}{ a_{21}}+2\color{maroon}{ a_{11}}\cr\color{darkblue} a_{31}}\right] +y\left[\matrix{\color{maroon}{ a_{12}}\cr\color{darkgreen}{ a_{22}}+2 \color{maroon}{a_{12}}\cr\color{darkblue} a_{32}}\right] +z\left[\matrix{\color{maroon}{ a_{13}}\cr\color{darkgreen}{ a_{23}}+2 \color{maroon}{a_{13}}\cr\color{darkblue} a_{33}}\right] =\bf z$$
If $(2)$ holds, then $\bf z=0$. In particular, looking at the second component of the left hand side of $(3)$: \eqalign{ x(\color{darkgreen}{a_{21}}+2\color{maroon}{a_{11}})+ y(\color{darkgreen}{a_{22}}+2\color{maroon}{a_{12}})+ z(\color{darkgreen}{a_{23}}+2\color{maroon}{a_{13}}) &= (x \color{darkgreen}{a_{21}}+ y \color{darkgreen}{a_{22}}+ z \color{darkgreen}{a_{23}} )\cr &\ \ \ \ \ \ \ +2( x\color{maroon}{a_{11}} + y\color{maroon}{a_{12}}+ z\color{maroon}{a_{13}} )\cr&=0+2\cdot0\cr&=0}.
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I think the most natural way to think about this is to note that the kernel of a matrix is also the orthogonal complement to the column-space of it's transpose. (this is the fundamental theorem of linear algebra)
So when you do row operations on a matrix, this is the same as doing linear combinations of the columns in it's transpose. This doesn't change the overall space spanned by those columns in the transpose, so it doesn't change the kernel of the original matrix.
- | 2014-09-24T03:03:37 | {
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http://math.stackexchange.com/questions/278784/is-it-more-practice-and-intuition-or-rather-algorithmic-to-solve-third-degree-po/278801 | # Is it more practice and intuition or rather algorithmic to solve third degree polynomials of this type?
Consider
$x^3 - 6x^2 + 11x - 6 = 0$
I can not reasonably factor this intuitively in any short amount of time with my skill level. Is this the only hope to solving such equations by hand? What tools can I use to help ease the pain or make it more clear to myself?
I try to:
$(x + ?)(x^2 - 6x)$ but it does not seem promising, in my eyes
I cant do anything, that I know about, with:
$x(x^2 - 6x + 11) = 6$
I don't want to give up, but I do anyway and ask my calculator, there is three solutions, alas I have no idea how that helps me (with my knowledge, it does not)!
Should I just practice with smaller polynomials more, and this one might look easier?
Any suggestions are appreciated, I will tag this as homework because it should be.
-
Note that trying to factorize anything as $x(x^2-6x+11)=6$ is almost useless, since $6$ need not factor nicely. The rare instance when such a factorization is useful, is for example $(x+1)(x+2)(x+3)=6$, in which case we can observe that $x=0$ is a root. – Calvin Lin Jan 14 '13 at 20:21
Do you mean $6x^2$ instead of $6x$? – Git Gud Jan 14 '13 at 20:22
Yes, sorry forgot to square that one. – Leonardo Jan 14 '13 at 20:23
You should also know that there are closed formulas to get the roots of polynomials of degree up to 4. – Git Gud Jan 14 '13 at 20:28
If you have a polynomial with integer coefficients of the form $$p(x) = x^n + a_{n-1} x^{n-1} + a_{n-2} x^{n-2} + \cdots + a_1 x + a_0$$ and if you have any hope of factorizing this "nicely", you first need to check if the divisors of the constant term, $a_0$, are factors. In your case, I assume the correct polynomial you are interested in is $p(x) = x^3 - 6x^2+11x-6$. The divisors of $-6$ are $\{\pm1,\pm2, \pm3, \pm6\}$. A quick check reveals that $x^*=1,2,3$ satisfy $p(x^*) = 0$. Hence, we have $$x^3 - 6x^2 + 11x - 6 =(x-1)(x-2)(x-3)$$
-
Luckily this is a nice equation, thank you I have at least one extra tool in my belt and I will need to read up more on the rational root theorem among other things. Your answer is very clear and concise. – Leonardo Jan 14 '13 at 20:27
Your equation is $$x^3-6x^2+11x-6=0$$ The coefficients of the equation are $1,-6,+11,-6$ and the summation of them is $$1-6+11-6=0$$ Whenever you face to this fact, you always have $x=1$ as one solution. In fact the equation has a factor as form $x-1$. It means that $$x^3-6x^2+11x-6=(x-1)(ax^2+bx+c)$$ for some proper available constants $a,b,c$.
-
I kind of get what you and Will Jagy are saying, but it will take time to absorb. The only thing that is immediately obvious from your last equations is that c = 6. I appreciate your helps. – Leonardo Jan 14 '13 at 20:22
@Leonardo: Look at Marvis's. I noted some points just for starting. He gave you the answer in detailed. – Babak S. Jan 14 '13 at 20:25
$x=1$ is an evident solution. Therefore we can write $$x^3 - 6 x^2 + 11 x - 6 = (x-1)(a x^2 + b x + c)$$ and solve for $a,b,c$ carefully. Then the rest is the quadratic formula.
I suppose the main advice is to check for integer roots and rational roots before factoring, when you have no feeling for what is going on in the problem. There is a Rational Roots Theorem that applies here.
-
$$x^3 - 6x^2 + 11x - 6 =0$$ $$x^3 -1-( 6x^2 - 11x + 6)+1 =0$$ $$x^3 -1-( 6x^2 - 11x + 5) =0$$ $$x^3 -1-( 6x^2 - 6x-5x + 5) =0$$ $$(x-1)(x^2+x+1)-( 6x(x-1)-5(x-1) =0$$ $$(x-1)(x^2+x+1)-(x-1)(6x-5) =0$$ $$(x-1)(x^2+x+1-6x+5) =0$$ $$(x-1)(x^2-5x+6) =0$$ $$(x-1)(x^2-2x-3x+6=0$$ $$(x-1)(x(x-2)-3(x-2)=0$$ $$(x-1)(x-2)(x-3)=0$$ $$x_1=1,x_2=2,x_3=3$$
-
Nice Adi,nice. ;-) – Babak S. Jan 14 '13 at 20:26
My brain does not work like yours, but I hope it starts to soon. – Leonardo Jan 14 '13 at 20:29
thank you Babak – Adi Dani Jan 14 '13 at 23:31
You need to know a few things. For example if $p(x)$ is a polynomial of odd degree with rational coefficients it will necessarily have a real root.
Another useful fact is that if $a$ is a root of $p(x)=x^3+px^2+qx+r=0$ then $(x-a)$ is a factor of $p(x)$. This works because: $x^3-a^3=(x-a)(x^2+ax+a^2)$ and $x^2-a^2=(x-a)(x+a)$ so if $p(a)=0$ we can write:$$p(x)=p(x)-p(a)=[x^3-a^3]+p[x^2-a^2]+q[x-a]=(x-a)([x^2+ax+a^2]+p[x+a]+q)$$
We also note that if $p,q,r,a \in \mathbb Z$, then if $p(a)=a(a^2+pa+q)+r=0$ then $a$ must be a factor of $r$.
[There are various generalisations and extensions of these comments].
So one thing we do when faced with a cubic with integer coefficients to factorise is to try the factors ($\pm$) of the constant term as "easy candidates" - and once we have one factor we can divide through and solve the remaining quadratic.
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https://cs.stackexchange.com/questions/117577/which-grows-faster-factorial-or-double-exponentiation/117583 | # Which Grows Faster: Factorial or Double Exponentiation
Which of the functions among $$2^{3^n}$$ or $$n!$$ grows faster?
I know that $$n^n$$ grows faster than $$n!$$ and $$n!$$ grows faster than $$c^n$$ where $$c$$ is a constant, but what is it in my case?
• It isn't a proof, but evaluate the expressions with $n$ in the range 1 to 10 and see just how phenomenally fast $2^{3^n}$ grows compared to $n!$ – John Coleman Nov 26 '19 at 0:20
• Write down the values for 1 ≤ n ≤ 10. You can't? Why can't you? – gnasher729 Nov 26 '19 at 13:18
• Why is this in CS instead of Mathematics? – Barmar Nov 26 '19 at 21:08
• @Barmar because it's related to CS in terms of time complexity. Indeed, in this question, there is not a hard border between math and CS. – OmG Nov 27 '19 at 15:14
• – Will Ness Nov 27 '19 at 16:36
You can find the result by taking a $$\log$$. Hence:
$$\log(2^{3^n}) = 3^n$$ $$\log(n!) \leqslant \log(n^n) = n\log n$$
(In the latter equation, we have used the fact that $$n! \leqslant n^n$$, as you note in the question.)
Of course $$3^n$$ grows faster than $$n \log n$$. As $$\log$$ is an increasing function, we can say $$2^{3^{n}}$$ grows faster than $$n^n$$, and also $$n!$$.
• Beware, however, that $\log(f) = O(\log(g))$ does not imply $f = O(g)$, in general. For instance, take $f(n)=n^2$ and $g(n)=n$. – chi Nov 26 '19 at 19:07
• @chi You are right. But, it is true if $\log(f) = o(\log(g))$. – OmG Nov 26 '19 at 19:56
• Another question related with taking log and solving and its danger is given in this link--> cs.stackexchange.com/questions/117584/… – Turing101 Nov 28 '19 at 15:19
Another way to directly compare the two expressions is to take the ratio of consecutive terms:
$$\frac{2^{3^{n+1}}}{2^{3^n}}=2^{2\cdot 3^n}\gg 3^n\gg n+1=\frac{(n+1)!}{n!}$$
(for positive integers $$n$$), and clearly also $$2^{3^1}=8>1!$$, so $$2^{3^n}$$ indeed grows more rapidly than $$n!$$.
• The discussion of $2^{3^1}$ is unnecessary. – leftaroundabout Nov 27 '19 at 1:56
• @leftaroundabout, I guess that's true if you assume the >> signs are significantly greater than, but that's not very rigorous (and if they are only > signs it would then depend on your definition of "grows more rapidly") – boboquack Nov 27 '19 at 7:43
• Well, they only need to be “significantly” greater in a very weak sense, namely the ratio needs to be $>1+\varepsilon$. That is clearly given in this case, specifically $\frac{3^n}{n+1} > 1.5\quad \forall n>0$. – leftaroundabout Nov 27 '19 at 12:05
You noted in your question that $$n^n$$ grows faster than $$n!$$, and that’s a great starting point for comparing the growth of $$2^{3^n}$$ and $$n!$$. Specifically, let’s ask - of $$n^n$$ and $$2^{3^n}$$, which grows faster?
To answer that, let’s try rewriting $$n^n$$ so that it has the same exponential base as $$2^{3^n}$$. Since $$n = 2^{\log_2 n}$$, we have that
$$n^n = (2^{\log_2 n})^n = 2^{n \log_2 n}.$$
Now, is it easier to see how $$n^n$$ and $$2^{3^n}$$ relate?
As a note, this approach is similar to taking the base-2 logs of both expressions. I thought it would be good to include this here because it gives a slightly different perspective on how to arrive at the answer given your initial observation.
Induction.
Base case: $$n=1$$ gives $$2^{3^1}=8$$ and $$1!=1$$ which clearly holds.
Hypothesis: suppose that $$2^{3^k}>k!$$ for some $$k\in\Bbb N$$.
Consider $$n=k+1$$. Then $$2^{3^{k+1}}=2^{3^k\cdot3}>(k!)^3=(k+1)!\cdot\frac{k!(k-1)!}{1+\frac1k}>(k+1)!$$ which is true $$\forall k>1$$ and thus the result follows.
• I tend to interpret "which grows faster?" questions as "whose big-Oh set is bigger?". With that interpretation, this post answers something different. (E.g. $x \mapsto x$ and $x \mapsto (x+1)$ are in the same big-Oh set, but in the spirit of this post the latter grows faster.) – ComFreek Nov 26 '19 at 12:02
• @ComFreek I would really have to question that interpretation. Which grows faster, $f(x) = x$ or $g(x) = 2x$? I think by any reasonable meaning of the words "grow" and "faster", $2x$ grows faster than $x$. My definition of grows faster would be $f(x)$ grows faster than $g(x)$ if there exists an $x_0$ beyond which $f'(x)$ is always greater than $g'(x)$. – Apollys supports Monica Nov 27 '19 at 23:36
• That being said, this proof certainly doesn't prove anything about growth rates. Plug in the functions $f(x) = -1/x$ and $g(x) = 0$ (on the domain $x > 0$) into your proof and you conclude that $g(x)$ grows faster than $f(x)$. This is obviously wrong since $g(x)$ never grows and $f(x)$ never doesn't grow! – Apollys supports Monica Nov 27 '19 at 23:42 | 2020-02-17T02:11:20 | {
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http://math.stackexchange.com/questions/183044/bounding-an-expression-involving-digamma-function | # Bounding an expression involving digamma function
Let $\psi$ be the digamma function. I have a conjecture that
$$\frac ax > \log(x) - \psi(x)$$
holds for all $x > 0$ if (and only if) $a \ge 1$. I do not know how to prove it. Please help.
-
Let
$$f(x) = \frac{1}{2}\coth\left(\frac{x}{2}\right) - \frac{1}{x}.$$
Then for $x > 0$, we have
$$f'(x) = \frac{1}{x^2} - \frac{1}{4\sinh^2(x/2)} = \frac{1}{x^2}\left(1 - \left(\frac{x/2}{\sinh (x/2)} \right)^2 \right) > 0.$$
This shows that $f(x)$ is increasing. Also, it is easy to find that $f(0+) = 0$ and $\lim_{x\to\infty} f(x) = \frac{1}{2}$. Thus we find that $0 < f(x) < \frac{1}{2}$ for $0 < x < \infty$ and hence the Laplace transform $\mathcal{L}f(s) = \int_{0}^{\infty} f(x) \, e^{-sx} \; dx$ of $f(x)$ satisfies
$$0 < \mathcal{L}f(s) < \int_{0}^{\infty} \frac{1}{2} \, e^{-sx} \; dx = \frac{1}{2s}.$$
But we also have
\begin{align*} \mathcal{L}f(s) &= \int_{0}^{\infty} \left( \frac{1}{2}\coth\left(\frac{x}{2}\right) - \frac{1}{x} \right) e^{-sx} \; dx\\ &= \left[ \left( \log \sinh \left(\frac{x}{2} \right) - \log x + \log 2 \right) e^{-sx} \right]_{0}^{\infty} \\ &\qquad + s \int_{0}^{\infty} \left( \log \sinh \left(\frac{x}{2} \right) - \log x + \log 2 \right) e^{-sx} \; dx \\ &= s \int_{0}^{\infty} \left( \frac{x}{2} - \log (sx) + \log s + \log \left(1 - e^{-x} \right) \right) e^{-sx} \; dx \\ &= \frac{1}{2s} + \gamma + \log s - s \int_{0}^{\infty} \sum_{n=1}^{\infty} \frac{e^{-(n+s)x}}{n} \; dx \\ &= \frac{1}{2s} + \gamma + \log s - \sum_{n=1}^{\infty} \frac{s}{n(n+s)} \\ &= \frac{1}{2s} + \log s - \psi_0(1+s), \end{align*}
where we have used the fact that $\int_{0}^{\infty} e^{-x} \log x \; dx = -\gamma$ and
$$\psi_0(1+x) = -\gamma + \sum_{n=1}^{\infty} \frac{x}{n(n+x)}.$$
Finally, since $\psi_0(1+s) = \frac{1}{s} + \psi_0 (s)$, we have
$$\frac{1}{2s} < \log s - \psi_0(s) < \frac{1}{s}.$$
Therefore the inequality holds if $a \geq 1$ and only if $a > \frac{1}{2}$.
Now we prove that $a = 1$ is the sharp bound. We are to calculate the limit
$$\alpha = \lim_{x\to 0} x(\log x - \psi_0(x)).$$
But by our previous calculations,
\begin{align*}\alpha &= \lim_{s\to 0} s(\log s - \psi_0(s)) \\ &= \lim_{s\to 0} s\left( \frac{1}{2s} + \mathcal{L}f(s) \right) \\ &= \frac{1}{2} + \lim_{s\to 0} s \int_{0}^{\infty} f(x) \, e^{-sx} \; dx \\ &= \frac{1}{2} + \lim_{s\to 0} \int_{0}^{\infty} f(u/s) \, e^{-u} \; du \qquad (u = sx) \\ &= \frac{1}{2} + \int_{0}^{\infty} \lim_{s\to 0} f(u/s) \, e^{-u} \; du \\ &= \frac{1}{2} + \int_{0}^{\infty} \frac{1}{2} \, e^{-u} \; du \\ &= 1. \end{align*}
Thus if $\frac{a}{x} > \log x - \psi_0 (x)$ is true for all $x > 0$, then we must have $a \geq \alpha = 1$, as desired.
-
This is a cool technique. (There's a typo after you switch summation and integral.) – Tunococ Aug 16 '12 at 18:25
@Tunococ, Thanks. Also I fixed the typo. Even I cannot believe how I made such a mistake! :) – sos440 Aug 17 '12 at 3:36
Recently I happened to find out that Qi et al (J. Math. Anal. Appl. 310 (2005) 303–308) proved the following inequality, among others, for digamma function $\psi(x)$ when $x>0$:
$$\frac{1}{2x} < \log x - \psi(x) < \frac{1}{2x} + \frac{1}{12x^2}$$
This inequality is sharper than the one given by @sos440 above when x>1/6.
- | 2014-12-19T15:53:36 | {
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# Each year for 4 years, a farmer increased the number of trees in a
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Each year for 4 years, a farmer increased the number of trees in a certain orchard by 1/4 of the number of trees in the orchard of the preceding year. If all of the trees thrived and there were 6250 trees in the orchard at the end of 4 year period, how many trees were in the orchard at the beginning of the 4 year period.
A. 1250
B. 1563
C. 2250
D. 2560
E. 2752
[Reveal] Spoiler:
Can someone walk me through the logic behind this question. I am able to solve this by using options as well as by assuming the number of trees = x. However, had the question been, "If all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period". then it would have been difficult to solve.
Thanks
[Reveal] Spoiler: OA
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imhimanshu wrote:
Each year for 4 years, a farmer increased the number of trees in a certain orchard by 1/4 of the number of trees in the orchard of the preceding year. If all of the trees thrived and there were 6250 trees in the orchard at the end of 4 year period, how many trees were in the orchard at the beginning of the 4 year period.
A. 1250
B. 1563
C. 2250
D. 2560
E. 2752
Can someone walk me through the logic behind this question. I am able to solve this by using options as well as by assuming the number of trees = x. However, had the question been, "If all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period". then it would have been difficult to solve.
Thanks
Say the number of trees at the beginning of the 4 year period was x, then:
At the end of the 1st year the number of trees would be $$x+\frac{1}{4}x=\frac{5}{4}*x$$;
At the end of the 2nd year the number of trees would be $$(\frac{5}{4})^2*x$$;
At the end of the 3rd year the number of trees would be $$(\frac{5}{4})^3*x$$;
At the end of the 4th year the number of trees would be $$(\frac{5}{4})^4*x$$;
At the end of the $$n_{th}$$ year the number of trees would be $$(\frac{5}{4})^n*x$$;
So, we have that $$(\frac{5}{4})^4*x=6,250$$ --> $$\frac{5^4}{4^4}*x=5^4*10$$ --> $$x=4^4*10=2,560$$.
If the question were "if all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period", then we would have that: $$(\frac{5}{4})^{15}*x=6,250$$ --> $$x\neq{integer}$$, so it would be a flawed question.
Hope it's clear.
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Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]
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imhimanshu wrote:
Each year for 4 years, a farmer increased the number of trees in a certain orchard by 1/4 of the number of trees in the orchard of the preceding year. If all of the trees thrived and there were 6250 trees in the orchard at the end of 4 year period, how many trees were in the orchard at the beginning of the 4 year period.
A. 1250
B. 1563
C. 2250
D. 2560
E. 2752
Can someone walk me through the logic behind this question. I am able to solve this by using options as well as by assuming the number of trees = x. However, had the question been, "If all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period". then it would have been difficult to solve.
Thanks
The number of trees increases by 1/4 i.e. 25% every year. It is just a matter of thinking in terms of successive percentage changes e.g. population increase. Here, we are talking about the increase of tree population.
If x increases by 25%, how we denote it? (5/4)*x
If next year, this new number increases by 25% again, how do we denote it? (5/4)*(5/4)*x
and so on...
For more on this, check: http://www.veritasprep.com/blog/2011/02 ... e-changes/
So if we are taking into account 4 years, we simply get (5/4)^4 * x = 6250
As for your next question, the numbers given would be such that the calculation will not be tough.
Say, you have 8 years and 100% increase every year (population doubles every year). The final population will be divisible by 2^8 i.e. 256.
Something like 2^8 * x = 2560
and if you meant what you wrote (though I considered that the 4 was a typo because of the language of the question) "If all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period", note that you still need to work with
(5/4)^4 * x = 6250
since you need the number of trees 4 yrs back only. The only thing is that the answer (2560) needs to be divisible by $$5^{11}$$ which it isn't so there is a problem in this question. If it were an actual question, the answer would be divisible by $$5^{11}$$ but you wouldn't really need to bother about it.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews SVP Status: The Best Or Nothing Joined: 27 Dec 2012 Posts: 1838 Location: India Concentration: General Management, Technology WE: Information Technology (Computer Software) Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink] ### Show Tags 25 Oct 2013, 00:26 4 This post received KUDOS 2 This post was BOOKMARKED Trees increase 1/4th every year, which means 100 trees become 125 after 1 yr & so on So, 125/100 = 5/4 is the resultant (after adding 1/4 as the growth) for 1 year So, for 4 yrs is 5^4/(4^4) From the condition given, inital trees were = 6250 x 4^4 / 5^4 = 2560 _________________ Kindly press "+1 Kudos" to appreciate Manager Joined: 12 Jan 2013 Posts: 212 Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink] ### Show Tags 27 Dec 2013, 09:31 Bunuel wrote: imhimanshu wrote: Each year for 4 years, a farmer increased the number of trees in a certain orchard by 1/4 of the number of trees in the orchard of the preceding year. If all of the trees thrived and there were 6250 trees in the orchard at the end of 4 year period, how many trees were in the orchard at the beginning of the 4 year period. A. 1250 B. 1563 C. 2250 D. 2560 E. 2752 Can someone walk me through the logic behind this question. I am able to solve this by using options as well as by assuming the number of trees = x. However, had the question been, "If all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period". then it would have been difficult to solve. Thanks Say the number of trees at the beginning of the 4 year period was x, then: At the end of the 1st year the number of trees would be $$x+\frac{1}{4}x=\frac{5}{4}*x$$; At the end of the 2nd year the number of trees would be $$(\frac{5}{4})^2*x$$; At the end of the 3rd year the number of trees would be $$(\frac{5}{4})^3*x$$; At the end of the 4th year the number of trees would be $$(\frac{5}{4})^4*x$$; At the end of the $$n_{th}$$ year the number of trees would be $$(\frac{5}{4})^n*x$$; So, we have that $$(\frac{5}{4})^4*x=6,250$$ --> $$\frac{5^4}{4^4}*x=5^4*10$$ --> $$x=4^4*10=2,560$$. Answer: D. If the question were "if all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period", then we would have that: $$(\frac{5}{4})^{15}*x=6,250$$ --> $$x\neq{integer}$$, so it would be a flawed question. Hope it's clear. Isn't the question quite ambiguous, though? I mean the first scentence could be interpreted as "for the first year we have (4/4)x and for the second year (5/4)x and for the third..." etc.. With that reasoning one would have (5/4)^3 * x + x and then your approach doesnt work. Obviously, I understand that this was a flaw in my reasoning but I cannot understand how they - with that wording - will assume that we totally understand that at the end of year one he has (5/4)x.. Is there a straightforward "word translation" way in knowing how to interpret wordings like this? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7996 Location: Pune, India Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink] ### Show Tags 01 Jan 2014, 23:34 Expert's post 2 This post was BOOKMARKED aeglorre wrote: Bunuel wrote: imhimanshu wrote: Each year for 4 years, a farmer increased the number of trees in a certain orchard by 1/4 of the number of trees in the orchard of the preceding year. If all of the trees thrived and there were 6250 trees in the orchard at the end of 4 year period, how many trees were in the orchard at the beginning of the 4 year period. A. 1250 B. 1563 C. 2250 D. 2560 E. 2752 Can someone walk me through the logic behind this question. I am able to solve this by using options as well as by assuming the number of trees = x. However, had the question been, "If all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period". then it would have been difficult to solve. Thanks Say the number of trees at the beginning of the 4 year period was x, then: At the end of the 1st year the number of trees would be $$x+\frac{1}{4}x=\frac{5}{4}*x$$; At the end of the 2nd year the number of trees would be $$(\frac{5}{4})^2*x$$; At the end of the 3rd year the number of trees would be $$(\frac{5}{4})^3*x$$; At the end of the 4th year the number of trees would be $$(\frac{5}{4})^4*x$$; At the end of the $$n_{th}$$ year the number of trees would be $$(\frac{5}{4})^n*x$$; So, we have that $$(\frac{5}{4})^4*x=6,250$$ --> $$\frac{5^4}{4^4}*x=5^4*10$$ --> $$x=4^4*10=2,560$$. Answer: D. If the question were "if all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period", then we would have that: $$(\frac{5}{4})^{15}*x=6,250$$ --> $$x\neq{integer}$$, so it would be a flawed question. Hope it's clear. Isn't the question quite ambiguous, though? I mean the first scentence could be interpreted as "for the first year we have (4/4)x and for the second year (5/4)x and for the third..." etc.. With that reasoning one would have (5/4)^3 * x + x and then your approach doesnt work. Obviously, I understand that this was a flaw in my reasoning but I cannot understand how they - with that wording - will assume that we totally understand that at the end of year one he has (5/4)x.. Is there a straightforward "word translation" way in knowing how to interpret wordings like this? Actually, it is not ambiguous. Read the statement: Each year a farmer increased the number of trees by 1/4. He did this for 4 years. (In GMAT Verbal and Quant are integrated. You need Verbal skills (slash and burn) in Quant and Quant skills (Data Interpretation) in Verbal. So in the first year, he increased it by 1/4 The next year, he again increased it by 1/4 (of preceding year) Next year, again the same. Next year, again the same. So he did it for a total of 4 years. So if initially the number of trees was x, in the first year he made them (5/4)x _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199
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Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]
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07 Feb 2014, 18:28
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it can easily solved by compound interest formula
R = 25%
P = original number of trees
N = no of times of compounding within a year = 1
T = total number of years = 4
P+I at the end of compounding years = 6250
hence,
6250 = (1 + 0.25/1)^(4*1) * P
5^4*10 = 1.25^4 * P
(5/1/25)^4 *10 = P
4^4 * 10 = P
2560 = P
Another way to think is to pick ans choice which is exactly divisible by 4. A, B and C are not. Only D and E are.
Pick D and increase it by 25% for 4 times, you should get 6250.
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Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]
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27 Jul 2014, 10:08
Hi Karishma,
I am not able to solve it by below method, where am I wrong:
6250*(3/4)*(3/4)*(3/4)
Regards,
Ravi
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Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]
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27 Jul 2014, 22:08
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email2vm wrote:
Hi Karishma,
I am not able to solve it by below method, where am I wrong:
6250*(3/4)*(3/4)*(3/4)
Regards,
Ravi
If I increase A by 25% and get B, I will not get A if I reduce B by 25%. The bases are different in the two cases.
e.g. A = 80
25% of A is 20 so I increase A by 25% to get B = 100
Now if I decrease B by 25%, I will decrease B by 25 (25% of 100). This will give me 75 which is not the same as A (which is 80).
In the first step, I found 25% of 80 and added that. In the second step, I found 25% of 100 and subtracted that. These two numbers are different.
You will get the correct answer if you do 6250*(4/5)*(4/5)*(4/5)*(4/5)
4/5 is the inverse of 5/4 (which is 25% increase in x).
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 13 Feb 2014 Posts: 3 Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink] ### Show Tags 25 Oct 2014, 12:01 You can use the formula A=p(1 + r/100)^n with r = 25% You will get P as 2560. Intern Joined: 23 Aug 2014 Posts: 42 GMAT Date: 11-29-2014 Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink] ### Show Tags 30 Dec 2014, 10:59 Is it true that the number of trees in the beginning has to be a multiple of 4 (as additions are always 1/4 th of previous and will not be non-integers)? In that case I'm left with D or E. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7996 Location: Pune, India Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink] ### Show Tags 31 Dec 2014, 02:53 deeuk wrote: Is it true that the number of trees in the beginning has to be a multiple of 4 (as additions are always 1/4 th of previous and will not be non-integers)? In that case I'm left with D or E. Yes, that's correct. But you might still need to work with two options. On the other hand, working with 6250 will definitely yield the answer. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199
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Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]
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06 Jan 2015, 10:39
1
KUDOS
Trees increase by 1/4 the number of trees in preceding year. Hence, correct answer must be divisible by 4. Based on divisibility rules, if last 2 digits are divisible by 4 then the number is divisible by 4. Thus, we can eliminate A, B, C. The answer has to be D or E.
Again, trees increase by 1/4 the number of trees in preceding year. Hence, the number of trees increase by 5/4 times the number of trees the preceding year.
If x = initial number of trees = Answer D or E = 2560 or 2752
Year 1 = 5/4x
Year 2 = (5/4)(5/4)x
Year 3 = (5/4)(5/4)(5/4)x
Year 4 = (5/4)(5/4)(5/4)(5/4)x
Only for Answer D: (5/4)(5/4)(5/4)(5/4) 2560 = 6250
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Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]
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02 Apr 2015, 03:36
@VeritasPrepKarishma
VeritasPrepKarishma wrote:
imhimanshu wrote:
Each year for 4 years, a farmer increased the number of trees in a certain orchard by 1/4 of the number of trees in the orchard of the preceding year. If all of the trees thrived and there were 6250 trees in the orchard at the end of 4 year period, how many trees were in the orchard at the beginning of the 4 year period.
A. 1250
B. 1563
C. 2250
D. 2560
E. 2752
Can someone walk me through the logic behind this question. I am able to solve this by using options as well as by assuming the number of trees = x. However, had the question been, "If all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period". then it would have been difficult to solve.
Thanks
The number of trees increases by 1/4 i.e. 25% every year. It is just a matter of thinking in terms of successive percentage changes e.g. population increase. Here, we are talking about the increase of tree population.
If x increases by 25%, how we denote it? (5/4)*x
If next year, this new number increases by 25% again, how do we denote it? (5/4)*(5/4)*x
and so on...
For more on this, check: http://www.veritasprep.com/blog/2011/02 ... e-changes/
So if we are taking into account 4 years, we simply get (5/4)^4 * x = 6250
As for your next question, the numbers given would be such that the calculation will not be tough.
Say, you have 8 years and 100% increase every year (population doubles every year). The final population will be divisible by 2^8 i.e. 256.
Something like 2^8 * x = 2560
and if you meant what you wrote (though I considered that the 4 was a typo because of the language of the question) "If all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period", note that you still need to work with
(5/4)^4 * x = 6250
since you need the number of trees 4 yrs back only. The only thing is that the answer (2560) needs to be divisible by $$5^{11}$$ which it isn't so there is a problem in this question. If it were an actual question, the answer would be divisible by $$5^{11}$$ but you wouldn't really need to bother about it.
Karisma,
However, it still isn't clear to me why we are keeping the same base for the increase during the 4-year period time. To me, this seems to be an example of successive % in which the base is shifting every year. Indeed every year we have an increase of 25% more on the increase of the previous year.
Why is it incorrect to compute the increase of the successive year on the increased base of the previous year?
Thank you a lot for your help!
Regards,
Eli.
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Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]
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02 Apr 2015, 19:43
elisabettaportioli wrote:
However, it still isn't clear to me why we are keeping the same base for the increase during the 4-year period time. To me, this seems to be an example of successive % in which the base is shifting every year. Indeed every year we have an increase of 25% more on the increase of the previous year.
Why is it incorrect to compute the increase of the successive year on the increased base of the previous year?
Thank you a lot for your help!
Regards,
Eli.
The base for each year changes.
Say, you have 100 trees and you increase them by 20% each year.
At the end of the first year, you will have 100 + (20/100)*100 = 100*(1 + 20/100) = 100*120/100 = 100*(6/5) = 120 trees.
So if you have to increase a number by 20%, you just need to multiply it by 6/5 every time no matter what the number is.
By the same logic, next year, you will have 120 * (6/5) = 144 trees when you increase them by 20%.
You can write 120 as 100 * (6/5) and then multiply it by another 6/5 to increase 120 by 20%.
This is the same as 100 * (6/5) * (6/5) = 144 trees.
So every time you multiply it by 6/5, you increase the base. The next time you multiply it by 6/5, you are increasing the new base by 20%. This is the concept of successive percentage changes and I have discussed it in the link I mentioned in my post above.
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Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]
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02 Apr 2015, 20:15
elisabettaportioli wrote:
Why is it incorrect to compute the increase of the successive year on the increased base of the previous year?
Thank you a lot for your help!
Regards,
Eli.
hi,
you are correct that the increase is on the preceeding year and that is exactly what is happening when you are multiplying x by 5/4..
first year say it was x.. 2nd year it becomes 5/4*x.. now for next year when we write the value as 5/4*5/4*x, the increase is on 5/4*x and not x..
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Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]
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13 Apr 2015, 08:38
1
KUDOS
1/4 = 0.25 = 25%
using the formula of compound interest calculation
p(1+r)^n= A
p = principal amount (to be identified), r = rate (0.25), n = numbers years (4), A= amount at the end of term(6250 given)
p(1+0.25)^4=6250
p(1.25)^4= 25*25*10 = 5^2*5^2*10=5^4*10
p(125/100)^4 = 5^4 x 10
p(5/4)^4 = 5^4 x 10
p(5^4)/4^4 = 5^4 x 10
p= (5^4 x 10 x 4^4)/5^4
p= 10 x 4^4
p= 2560
Ans D
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Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]
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19 May 2015, 12:37
we have here 4 (4 years) successive increases of 25% or *1,25 --> 1,25^4 * X = 6250, X = 2560
See MGMAT (Percents) for detailed explanation of such question types.....
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Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]
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20 May 2015, 03:32
BrainLab wrote:
we have here 4 (4 years) successive increases of 25% or *1,25 --> 1,25^4 * X = 6250, X = 2560
See MGMAT (Percents) for detailed explanation of such question types.....
Dear BrainLab
Perfect logic but for easier calculation, you may want to work with ratio here (1/4 increase per annum) instead of percentages (25% increase per annum). Both convey the same thing but the equation
$$(\frac{5}{4})^4*X = 6250$$
will take lesser time to solve (especially if you know that $$5^4 = 625$$) than $$(1.25)^4*X = 6250$$
Hope this was useful!
Japinder
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# In the rectangular coordinate system above, if the equation of m is y
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In the rectangular coordinate system above, if the equation of m is y [#permalink]
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15 Dec 2017, 02:09
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In the rectangular coordinate system above, if the equation of m is y = x, and m is parallel to n, what is the shortest distance between m and n?
(A) √2
(B) 1
(C) √2/2
(D) 1/2
(E) 1/4
Attachment:
2017-12-15_1259.png [ 4.85 KiB | Viewed 1084 times ]
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In the rectangular coordinate system above, if the equation of m is y [#permalink]
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15 Dec 2017, 13:22
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Bunuel wrote:
In the rectangular coordinate system above, if the equation of m is y = x, and m is parallel to n, what is the shortest distance between m and n?
(A) √2
(B) 1
(C) √2/2
(D) 1/2
(E) 1/4
Attachment:
The attachment 2017-12-15_1259.png is no longer available
With the formula (My diagram does not apply)
I did not use it, but to do so is standard. So I quote the formula from Bunuel HERE
Distance between two parallel lines $$y=mx+b$$ and $$y=mx+c$$ can be found by the formula:
$$D=\frac{|b-c|}{\sqrt{m^2+1}}$$
Equation for line $$m$$, $$y = x$$, to match the exact form of $$y = mx + b$$ can be written:
$$y =(1)x + 0$$ (all lines that pass through the origin have x- and y-intercepts of 0)
$$b = 0$$ = y-intercept, $$m = 1$$ = slope
Find the equation of line $$n$$, $$y = mx + c$$
Parallel lines have identical slopes. Line $$n$$ has slope $$m = 1$$
Use the point on line $$n$$ from the graph: (1, 0)
Plug its coordinates into slope-intercept form to find $$c$$
$$y = mx + c$$
$$(0) = 1(1) + c$$
$$-c = 1$$, so $$c = -1$$
Equation for line $$n: y = x - 1$$
Distance between parallel lines
$$D=\frac{|b-c|}{\sqrt{m^2+1}}$$
$$m = 1$$, $$b = 0$$, $$c = -1$$
$$D=\frac{|0-(-1)|}{\sqrt{1^2+1}}$$
$$D=\frac{1}{\sqrt{2}}$$
Does not match the answers. Rationalize the denominator; multiply by $$\frac{\sqrt2}{\sqrt2}$$
$$\frac{1}{\sqrt{2}}$$ * $$\frac{\sqrt2}{\sqrt2} = \frac{\sqrt{2}}{2}$$
Attachment:
linesmandn.png [ 18.05 KiB | Viewed 868 times ]
Without the formula see diagram
The shortest distance between parallel lines is a line perpendicular to both parallel lines
Draw a perpendicular line from point B to point A
That creates right isosceles ∆ ABO where OA = AB
-- The line y = x makes a 45° angle with both the x- and y-axes (as do all lines with slope 1 or -1)
-- ∠ BOA = 45° and ∠ at vertex A = 90° therefore ∠ ABO must = 45°
-- Sides opposite equal angles are equal: OA = AB
A right isosceles triangle
-- has angle measures 45-45-90 and
-- corresponding side lengths of $$x : x : x\sqrt{2}$$
The hypotenuse corresponds with $$x\sqrt{2} = 1$$: OB = 1
Find length of equal sides $$x$$ (one of which, AB, is the distance needed):
$$x\sqrt{2}= 1$$
$$x = \frac{1}{\sqrt{2}}$$
= AB = shortest distance between the parallel lines
That does not look like any of the answers.
Rationalize the denominator (clear the radical): Multiply by $$\frac{\sqrt2}{\sqrt2}$$
$$\frac{1}{\sqrt{2}}$$ * $$\frac{\sqrt2}{\sqrt2} = \frac{\sqrt{2}}{2}$$
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In the rectangular coordinate system above, if the equation of m is y [#permalink]
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14 Apr 2018, 06:59
generis wrote:
In the rectangular coordinate system above, if the equation of m is y = x, and m is parallel to n, what is the shortest distance between m and n?
(A) √2
(B) 1
(C) √2/2
(D) 1/2
(E) 1/4
Attachment:
2017-12-15_1259.png
Attachment:
linesmandn.png
Without the formula see diagram
The shortest distance between parallel lines is a line perpendicular to both parallel lines
Draw a perpendicular line from point B to point A
That creates right isosceles ∆ ABO where OA = AB
-- The line y = x makes a 45° angle with both the x- and y-axes (as do all lines with slope 1 or -1)
-- ∠ BOA = 45° and ∠ at vertex A = 90° therefore ∠ ABO must = 45°
-- Sides opposite equal angles are equal: OA = AB
A right isosceles triangle
-- has angle measures 45-45-90 and
-- corresponding side lengths of $$x : x : x\sqrt{2}$$
The hypotenuse corresponds with $$x\sqrt{2} = 1$$: OB = 1
Find length of equal sides $$x$$ (one of which, AB, is the distance needed):
$$x\sqrt{2}= 1$$
$$x = \frac{1}{\sqrt{2}}$$
= AB = shortest distance between the parallel lines
That does not look like any of the answers.
Rationalize the denominator (clear the radical): Multiply by $$\frac{\sqrt2}{\sqrt2}$$
$$\frac{1}{\sqrt{2}}$$ * $$\frac{\sqrt2}{\sqrt2} = \frac{\sqrt{2}}{2}$$
Great explanation, Kudos to you..
Did not quite get how you got that the triangle will be isosceles triangle. Could you or Bunuel help understand ?
thanks.
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Re: In the rectangular coordinate system above, if the equation of m is y [#permalink]
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14 Apr 2018, 07:59
1
MT1988 wrote:
generis wrote:
In the rectangular coordinate system above, if the equation of m is y = x, and m is parallel to n, what is the shortest distance between m and n?
(A) √2
(B) 1
(C) √2/2
(D) 1/2
(E) 1/4
Attachment:
2017-12-15_1259.png
Attachment:
linesmandn.png
Without the formula see diagram
The shortest distance between parallel lines is a line perpendicular to both parallel lines
Draw a perpendicular line from point B to point A
That creates right isosceles ∆ ABO where OA = AB
-- The line y = x makes a 45° angle with both the x- and y-axes (as do all lines with slope 1 or -1)
-- ∠ BOA = 45° and ∠ at vertex A = 90° therefore ∠ ABO must = 45°
-- Sides opposite equal angles are equal: OA = AB
A right isosceles triangle
-- has angle measures 45-45-90 and
-- corresponding side lengths of $$x : x : x\sqrt{2}$$
The hypotenuse corresponds with $$x\sqrt{2} = 1$$: OB = 1
Find length of equal sides $$x$$ (one of which, AB, is the distance needed):
$$x\sqrt{2}= 1$$
$$x = \frac{1}{\sqrt{2}}$$
= AB = shortest distance between the parallel lines
That does not look like any of the answers.
Rationalize the denominator (clear the radical): Multiply by $$\frac{\sqrt2}{\sqrt2}$$
$$\frac{1}{\sqrt{2}}$$ * $$\frac{\sqrt2}{\sqrt2} = \frac{\sqrt{2}}{2}$$
Great explanation, Kudos to you..
Did not quite get how you got that the triangle will be isosceles triangle. Could you or Bunuel help understand ?
thanks.
Got it!!!! The Slope is 1 so the angle at the origin will be 45* , as for calculating distance we have to drop a perpendicular one angle will be 90 leaving the third angle as 45*.
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Re: In the rectangular coordinate system above, if the equation of m is y [#permalink]
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14 Apr 2018, 16:33
1
Quote:
Bunuel wrote:
In the rectangular coordinate system above, if the equation of m is y = x, and m is parallel to n, what is the shortest distance between m and n?
(A) √2
(B) 1
(C) √2/2
(D) 1/2
(E) 1/4
Attachment:
2017-12-15_1259.png
MT1988 wrote:
MT1988 wrote:
generis wrote:
Without the formula see diagram
The shortest distance between parallel lines is a line perpendicular to both parallel lines
Draw a perpendicular line from point B to point A
That creates right isosceles ∆ ABO where OA = AB
-- The line y = x makes a 45° angle with both the x- and y-axes (as do all lines with slope 1 or -1)
-- ∠ BOA = 45° and ∠ at vertex A = 90° therefore ∠ ABO must = 45°
-- Sides opposite equal angles are equal: OA = AB
A right isosceles triangle
-- has angle measures 45-45-90 and
-- corresponding side lengths of $$x : x : x\sqrt{2}$$
The hypotenuse corresponds with $$x\sqrt{2} = 1$$: OB = 1
Find length of equal sides $$x$$ (one of which, AB, is the distance needed):
$$x\sqrt{2}= 1$$
$$x = \frac{1}{\sqrt{2}}$$
= AB = shortest distance between the parallel lines
That does not look like any of the answers.
Rationalize the denominator (clear the radical): Multiply by $$\frac{\sqrt2}{\sqrt2}$$
$$\frac{1}{\sqrt{2}}$$ * $$\frac{\sqrt2}{\sqrt2} = \frac{\sqrt{2}}{2}$$
Great explanation, Kudos to you..
Did not quite get how you got that the triangle will be isosceles triangle. Could you or Bunuel help understand ?
thanks.
Got it!!!! The Slope is 1 so the angle at the origin will be 45* , as for calculating distance we have to drop a perpendicular one angle will be 90 leaving the third angle as 45*.
BOOM. Excellent.
I came prepared with a diagram
to help, but behold, no need.
Here is a post that discusses
that create 45° angles.
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In the rectangular coordinate system above, if the equation of m is y [#permalink]
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15 Apr 2018, 03:40
Bunuel wrote:
In the rectangular coordinate system above, if the equation of m is y = x, and m is parallel to n, what is the shortest distance between m and n?
(A) √2
(B) 1
(C) √2/2
(D) 1/2
(E) 1/4
Attachment:
The attachment 2017-12-15_1259.png is no longer available
My method to solve this:-
Attachments
File comment: for line y = x, the angle between x-axis and the line m will be 45 degrees.
The min distance will the perpendicular distance between the two lines.
In the image, by mistake I've written 1/root(2) = 2/root(2), it should be root(2)/2. Please excuse that mistake
IMG_20180415_160317.jpg [ 1.99 MiB | Viewed 347 times ]
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In the rectangular coordinate system above, if the equation of m is y &nbs [#permalink] 15 Apr 2018, 03:40
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# A company has two types of machines, type R and type S
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Re: A company has two types of machines, type R and type S [#permalink]
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04 Nov 2016, 15:05
1
enigma123 wrote:
A company has two types of machines, type R and type S. Operating at a constant rate, a machine of type R does a certain job in 36 hrs and a machine of type S does the same job in 18 hours. If the company used the same number of each type of machine to do the job in 2 hours, how many machines of type R were used?
A. 3
B. 4
C. 6
D. 9
E. 12
We have a combined worker problem for which we can use the following formula:
work (1 machine) + work (2 machine) = total work completed
Since we are completing one job, we can say:
work (1 machine) + work (2 machine) = 1
We are given that a machine of type R does a certain job in 36 hours and a machine of type S does the same job in 18 hours.
Thus, the rates for the two machines are as follows:
rate of machine R = 1/36
rate of machine S = 1/18
We are also given that the company used the same number of each type of machine to do the job in 2 hours. If we let x = the number of each machine used, we can multiply each rate by x and we have:
rate of x number of R machines = x/36
rate of x number of S machines = x/18
Finally, since we know some number of R and S machines worked for two hours, and since work = rate x time, we can calculate the work done by each type of machine.
work done by x number of R machines = 2x/36 = x/18
work done by x number of S machines = 2x/18 = x/9
Now we can determine x using the combined worker formula:
work (machine R) + work (machine S) = 1
x/18 + x/9 = 1
x/18 + 2x/18 = 1
3x/18 = 1
x/6 = 1
x = 6
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Work/Rate problem- A company has.. [#permalink]
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01 Jan 2017, 06:10
A company has two types of machines, Type R and Type S. Operating at a certain rate, a machine of type R does a certain job in 36 hours and a machine of type S does the same job in 18 hours. If the company uses the same number of each type of machine to do the job in 2 hours, how many machines of Type R were used?
Options:
A. 3
B. 4
C. 6
D. 9
E. 12
Can anybody give me a quickest and simpler way to solve this? I took almost 3.5 min on this question to solve during a mock test. Still ended up guessing as I wasn't sure of the approach.
Thanks.
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Re: Work/Rate problem- A company has.. [#permalink]
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01 Jan 2017, 06:47
5
1
baalok88 wrote:
A company has two types of machines, Type R and Type S. Operating at a certain rate, a machine of type R does a certain job in 36 hours and a machine of type S does the same job in 18 hours. If the company uses the same number of each type of machine to do the job in 2 hours, how many machines of Type R were used?
Options:
A. 3
B. 4
C. 6
D. 9
E. 12
Can anybody give me a quickest and simpler way to solve this? I took almost 3.5 min on this question to solve during a mock test. Still ended up guessing as I wasn't sure of the approach.
Thanks.
We are given rates of each machine and that the number of machines of each type being used is the same = x.
$$\frac{x}{36} + \frac{x}{18} = \frac{1}{2}$$
$$\frac{3x}{36} = \frac{1}{2}$$
$$6x = 36$$
$$x = 6$$
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Re: A company has two types of machines, type R and type S [#permalink]
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07 Jan 2017, 21:07
2
Rate(R) = $$\frac{1}{36}$$
Rate(S) = $$\frac{1}{18}$$
Combined rate of both machines ,
Rate(RS)=Rate(R) + Rate(S) = $$\frac{1}{12}$$
we have given time , T = 2hrs , Workdone = 1 (for single job)
Plug in the values :
[Rate] * [Time] * [No. of machines] = Workdone
$$\frac{1}{12}$$ * 2 * [No. of machines] = 1
[No. of machines] = 6
Ans : C
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Re: A company has two types of machines, type R and type S [#permalink]
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18 Apr 2017, 05:05
Bunuel wrote:
enigma123 wrote:
A company has two types of machines, type R and type S. Operating at a constant rate, a machine of type R does a certain job in 36 hrs and a machine of type S does the same job in 18 hours. If the company used the same number of each type of machine to do the job in 2 hours, how many machines of type R were used?
A. 3
B. 4
C. 6
D. 9
E. 12
Rate of A - $$\frac{1}{36}$$ job/hour, rate of x machines of A - $$\frac{1}{36}x$$ job/hour;
Rate of B - $$\frac{1}{18}$$ job/hour, rate of x machines of B - $$\frac{1}{18}x$$ job/hour, (same number of each type);
Remember that we can sum the rates, hence combined rate of A and B is $$\frac{1}{36}x+\frac{1}{18}x=\frac{3}{36}x=\frac{1}{12}x$$ job/hour.
We are told that together equal number (x in our case) of machines A and B can do the job (1 job) in 2 hours --> $$Time*Rate=2*\frac{1}{12}x=1=Job$$ --> $$x=6$$.
When I am double checking, why am I not getting the right answer?
If 1 R machine is taking 36 hours, then 6 R machines will take 6 hours.
If 1 S machine is taking 18 hours, then 6 S machines will take 3 hours.
Together, they will take 6-3=3 hours.
Please tell me where am I wrong :/
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Re: A company has two types of machines, type R and type S [#permalink]
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16 Aug 2017, 18:41
narendran1990 wrote:
Took 36 as the unit of work to be completed by R & S. [LCM of 36 & 18].
So, R does 1 unit per hour & S does 2 units per hour. Since an equal number of R & S machines have to be deployed, consider 'x' as the number of equipment. Additionally, the work is to be completed in 2 hours, so 18 units have to be manufactured.
Therefore, x+2x = 18, 3x = 18, x=6. [Correct Answer]
I was trying to solve using similar kind of method. But i could not understand how you arrived on 18? It will be great if you could elaborate.
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A company has two types of machines, type R and type S [#permalink]
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09 Nov 2017, 21:26
There is no need to compute the combined rate in this case, explicitly that is.
Since the number of machines is the same, we can use a simple equation:
$$x*\frac{1}{36}+ x* \frac{1}{18} = \frac{1}{2}$$ (here x machines work at the respective rates for 2 hours)
Solving for x, we get 6.
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Re: A company has two types of machines, type R and type S [#permalink]
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13 Nov 2018, 19:46
hannahkagalwala wrote:
Bunuel wrote:
enigma123 wrote:
A company has two types of machines, type R and type S. Operating at a constant rate, a machine of type R does a certain job in 36 hrs and a machine of type S does the same job in 18 hours. If the company used the same number of each type of machine to do the job in 2 hours, how many machines of type R were used?
A. 3
B. 4
C. 6
D. 9
E. 12
Rate of A - $$\frac{1}{36}$$ job/hour, rate of x machines of A - $$\frac{1}{36}x$$ job/hour;
Rate of B - $$\frac{1}{18}$$ job/hour, rate of x machines of B - $$\frac{1}{18}x$$ job/hour, (same number of each type);
Remember that we can sum the rates, hence combined rate of A and B is $$\frac{1}{36}x+\frac{1}{18}x=\frac{3}{36}x=\frac{1}{12}x$$ job/hour.
We are told that together equal number (x in our case) of machines A and B can do the job (1 job) in 2 hours --> $$Time*Rate=2*\frac{1}{12}x=1=Job$$ --> $$x=6$$.
When I am double checking, why am I not getting the right answer?
If 1 R machine is taking 36 hours, then 6 R machines will take 6 hours.
If 1 S machine is taking 18 hours, then 6 S machines will take 3 hours.
Together, they will take 6-3=3 hours.
Please tell me where am I wrong :/
It is not correct to subtracting time taken by one machine from that of other to calculate final time,
Time taken are for complete job, and if other machines are also performing the same job, then there will be less amount of work of each machine type and hence leaser time will be taken.
You are right till, 6 R machines will take 6 hours.
And 6 S machines will take 3 hours.
Now total time taken will be 1/(1/6 +1/3) = 1/(1/2) = 2 hours.
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Re: A company has two types of machines, type R and type S &nbs [#permalink] 13 Nov 2018, 19:46
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