Search is not available for this dataset
url
string | text
string | date
timestamp[s] | meta
dict |
|---|---|---|---|
https://math.stackexchange.com/questions/3364550/multiple-choice-question-on-group-of-matrices | # multiple choice question on group of matrices
Consider the set of matrices $$G=\left\{ \left( \begin{array}{ll}s&b\\0&1 \end{array}\right) b \in \mathbb{Z}, s \in \{1,-1\} \right\}.$$Then which of the following are true
1. G forms a group under addition
2. G forms an abelian group under multiplication
3. Every element of G is diagonolizable over $$\mathbb{C}$$
4. G is finitely generated group under multiplication
I am getting 1) is false since not closed under addition 2)Forms a group under multiplication ( abelian or not i don't know) 3)Not true if $$a=1$$ 4) dont know please help me to complete
• A few examples with $s=1$ in one matrix, $s=-1$ in the other, should convince you that $G$ isn't abelian. – Robert Shore Sep 21 '19 at 16:32
• So answer will be finitely generate right? – sabeelmsk Sep 21 '19 at 16:34
• $$1$$ is false:
Your approach is correct. Since, for example, $$\begin{pmatrix}1&*\\0&1 \end{pmatrix}+\begin{pmatrix}1&*\\0&1 \end{pmatrix}=\begin{pmatrix}2&*\\*&* \end{pmatrix} \notin G$$
• $$2$$ is false:
Take $$b \neq 0$$.$$\begin{pmatrix}1&b\\0&1 \end{pmatrix}\begin{pmatrix}-1&b\\0&1 \end{pmatrix}=\begin{pmatrix}-1&2b\\0&1 \end{pmatrix}$$ whereas $$\begin{pmatrix}-1&b\\0&1 \end{pmatrix}\begin{pmatrix}1&b\\0&1 \end{pmatrix}=\begin{pmatrix}-1&\color{red}{0}\\0&1 \end{pmatrix}$$
• $$3$$ is false too:
Since, for example, $$\begin{pmatrix}1&b\\0&1 \end{pmatrix}$$ is not diagonalizable when $$b \neq 0$$
• $$4$$ is true
The finite set $$\left\{\begin{pmatrix}1&1\\0&1 \end{pmatrix},\begin{pmatrix}1&-1\\0&1 \end{pmatrix},\begin{pmatrix}-1&0\\0&1 \end{pmatrix}\right\}$$ generates $$G$$(verify!)
• Is it generated by 2 elements ? I am not clear about generating set – sabeelmsk Sep 21 '19 at 16:36
• @sabeelmsk: See my edit – Chinnapparaj R Sep 21 '19 at 17:01 | 2020-06-05T12:15:06 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3364550/multiple-choice-question-on-group-of-matrices",
"openwebmath_score": 0.6389120221138,
"openwebmath_perplexity": 899.3029009557825,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9796676496254957,
"lm_q2_score": 0.8615382147637196,
"lm_q1q2_score": 0.8440211179201187
} |
https://math.stackexchange.com/questions/2391260/how-is-this-limit-calculated-without-lhospital?noredirect=1 | # How is this limit calculated without l'Hospital?
In this question: Solving limit without L'Hôpital
$$\lim_{x\to0} \frac{5-\sqrt{x+25}}{x}=\lim_{x\to0} \frac{(5-\sqrt{x+25)}(5+\sqrt{x+25})}{x(5+\sqrt{x+25})}=\lim_{x\to0} \frac{25-(x+25)}{x(5+\sqrt{x+25})}=-\frac{1}{10}$$
Expanding the fraction makes sense, but I dont understand how we get $-\frac{1}{10}$ as a result. Because when you put in 0 for x ( which I intuitively did) I get $\frac{0}{0}$ as a result, which doesnt get me anywhere withou l'Hospital.
What step did I miss ?
• sorry about the duplicate, I would've commented on the original but I dont have enough credit and chat is not very active – zython Aug 12 '17 at 12:43
• Hint:$$25-(x+25)=-x$$Now cancel factors. – Simply Beautiful Art Aug 12 '17 at 12:45
• Just out of curiosity (because I love words) : does your user name mean something special ? Please answer (I shall tell you why later). – Claude Leibovici Aug 12 '17 at 13:08
• zython is the name of the beer ancient Egyptian people used to drink. Look at en.wikipedia.org/wiki/Egyptian_zythos With a friend of mine, we use to say eachother "OK, let's go and have a zython". – Claude Leibovici Aug 12 '17 at 13:17
• I was just amazed since I suppose that very few people in the world use (or even know) this word. Cheers. – Claude Leibovici Aug 12 '17 at 13:27
$$\lim_{x\to 0}\frac{25-(x+25)}{x(5+\sqrt{x+25})}=\lim_{x\to 0}\frac{-x}{x(5+\sqrt{x+25})}=\lim_{x\to 0}\frac{-1}{5+\sqrt{x+25}}$$ Note that after the second step we cancel out the $x$ in the numerator and denominator, and are left with an expression with which we can evaluate at $x=0$.
• :P Looks good to me. – Simply Beautiful Art Aug 12 '17 at 12:48
• ah I see, for a moment I forgot that $-x = -1 * x$ thank you for the answer it is clear to me now – zython Aug 12 '17 at 12:50
Hint:
Let $5-\sqrt{x+25}=y$
$\implies(i)5-y=\sqrt{x+25}\implies x=y^2-10y$
and $(ii)y\to0^-$
Can you take it from here?
other If $f (x)=\sqrt {x+25}$ then $$\lim_0\frac {f (0)-f (x)}{x}=-f'(0)$$ and $$f'(x)=\frac {1}{2f (x)}$$
• That's a strange way of writing $\lim_{x\to 0}$ – zhw. Aug 13 '17 at 15:13 | 2019-06-18T07:31:37 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2391260/how-is-this-limit-calculated-without-lhospital?noredirect=1",
"openwebmath_score": 0.6267677545547485,
"openwebmath_perplexity": 842.6237864767957,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9796676508127573,
"lm_q2_score": 0.8615382129861583,
"lm_q1q2_score": 0.8440211172015707
} |
http://math.stackexchange.com/questions/142677/recovering-a-number-from-a-remainder-list | # Recovering a number from a remainder list
Consider the following list of equations:
\begin{align*} x \bmod 2 &= 1\\ x \bmod 3 &= 1\\ x \bmod 5 &= 3 \end{align*}
How many equations like this do you need to write in order to uniquely determine $x$?
Once you have the necessary number of equations, how would you actually determine $x$?
Update:
The "usual" way to describe a number $x$ is by writing
$$x = \sum_n 10^n \cdot a_n$$
and listing the $a_n$ values that aren't zero. (You can also extend this to some radix other than 10.)
What I'm interested in is whether you could instead express a number by listing all its residues against a suitable set of modulii. (And I'm guessing that the prime numbers would constitute such a "suitable set".)
If you were to do this, how many terms would you need to quote before a third party would be able to tell which number you're trying to describe?
That was my question. However, since it appears that the Chinese remainder theorem is extremely hard, I guess this is a bad way to denote numbers...
(It also appears that $x$ will never be uniquely determined without an upper bound.)
-
Look up the Chinese remainder theorem. – Ｊ. Ｍ. May 8 '12 at 14:54
If you are learning from a book, surely they will discuss the CRT within a few pages of introducing such equivalences... – The Chaz 2.0 May 8 '12 at 14:55
Having just read the Wikipedia article on the Chinese remainder theorem, I suddenly feel very, very sorry I asked... For some reason, I mistakenly thought this would be easy. How foolish! – MathematicalOrchid May 8 '12 at 15:09
@MathematicalOrchid: Your idea is a very good one, and like many good ideas, it has been already found, and there are many implementations. It is convenient to use primes, and in the binary world of computers primes $P_i$ of the form $2^p-1$ are particularly convenient. As I pointed out, numbers are not uniquely picked out, but if the product of our $P_i$ is large enough, that doesn't matter. The important thing is that addition, multiplication can be carried out in parallel mod the $P_i$, and we can use CRT at the end to piece things together. Am impressed that you thought of it. – André Nicolas May 8 '12 at 15:49
This system has a name: en.wikipedia.org/wiki/Residue_number_system – Hurkyl May 8 '12 at 19:34
This is a classic example of Chinese remainder theorem. To solve it, one typically proceeds as follows. We have $$x = 2k_2 + 1 = 3k_3 + 1 = 5k_5 + 3.$$ Since $\displaystyle x = 2k_2 + 1 = 3k_3 + 1$, we have that $2k_2 = 3k_3$ i.e. $2|k_3$ and $3|k_2$, since $(2,3) = 1$. Hence, $k_3 = 2k_6$ and $k_2 = 2k_6$. Hence, we now get that $$x = 6k_6 + 1 = 5k_5+3.$$ Rearranging, we get that $$6k_6 - 5k_5 = 2.$$ Clearly, $(2,2)$ is a solution to the above. In general, if $ax+by$ has integer solutions and $(x_0,y_0)$ is one such integer solution, then all integer solutions are given by $$(x,y) = \displaystyle \left( x_0 + k \frac{\text{lcm}[\lvert a \rvert,\lvert b \rvert]}{a}, y_0 - k \frac{\text{lcm}[\lvert a \rvert,\lvert b \rvert]}{b} \right)$$ where $k \in \mathbb{Z}$. Hence, all the integer solutions to $6k_6 - 5k_5 = 2$, are given by $$(k_6,k_5) = \left( 2 + 5k, 2 + 6k \right)$$ Hence, $x = 5k_5 + 3 = 30k + 13$ i.e. $$x \equiv 13 \bmod 30.$$
-
Hint It can be done simply without CRT. $\rm\:x\equiv -2\:\ (mod\ \rm3,5)\iff x\equiv -2\equiv 13\pmod{ 15}\:$ Now since $13\equiv 1\pmod 2\:$ we conclude $\rm\:x\equiv 13\:\ (mod\ 2,15)\iff x\equiv 13\pmod{30}\:$
Hence your hunch was correct: it is easy (these are often warm-up exercises to CRT).
Thus this constant case of CRT is solved simply by taking least common multiple of moduli:
$$\rm x\equiv a\ (mod\ m,n)\!\iff\! m,n\:|\:x\!-\!a\!\iff\! lcm(m,n)\:|\:x\!-\!a\!\iff\! x\equiv a\ (mod\: lcm(m,n))$$
This simple constant-case optimization of CRT arises quite frequently in practice, esp. for small moduli (by the law of small numbers), so it is well worth checking for. For further examples, see here where it simplified a few page calculation to a few lines, and here and here.
Note that I chose to eliminate the largest moduli first, i.e. $\rm\:x\equiv -2\ mod\ 3,5\:$ vs. $\rm\:x\equiv 1\ mod\ 2,3\:$ since that leaves the remaining modulus minimal ($= 2$ vs. $5$ above), which generally simplifies matters if we need to apply the full CRT algorithm in the final step (luckily we did not above).
Update Regarding your update: knowing the residues of $\rm\:n\:$ modulo a finite set $\rm\:S\:$ of moduli only determines $\rm\:n\:$ modulo $\rm\:lcm\:S.\:$ However, if $\rm\:S\:$ is infinite (e.g. all primes), then the residues do determine $\rm\:n\:$ uniquely from the residue of any modulus $\rm > n$.
In cases where one is working with bounded size integers such modular representations can prove effective for computational purposes, esp. if the moduli are chosen related to machine word size, so to simplify arithmetic. See any good textbook on computer algebra, which will discuss not only this but many other instances of modular reduction - a ubiquitous technique in algebraic computation.
-
Note that even if you specify the residue of $x$ modulo all integers $n$, this need not determine an integer.
This is where $\hat{\mathbb{Z}}$ comes into the picture. I think you should know this as a fact, just to be complete (no pun intended). But I warn you that the theory going into $\hat{\mathbb{Z}}$ is much more advanced than CRT.
If you do want to know more about this, look into http://en.wikipedia.org/wiki/P-adic_number . Especially the part on $p$-adic expansion is interesting, because it is analogous in some sense to writing integers in radix 10.
Probably the greatest difference is that the sum in the expansion need not be a finite sum.
-
If two integers share every single residue, their difference is divisible by every nonzero integer and therefore zero; they are identical. Therefore the set of all residues does uniquely determine an integer. The very same argument can be applied to any infinite set of moduli instead of $\Bbb N$ if desired. – anon May 8 '12 at 17:49
Indeed the map $\mathbb{Z} \to \hat{\mathbb{Z}}$ is injective. But my point is, that a list of residues need not come from an integer (but if it does, then it is uniquely determined). – jmc May 8 '12 at 17:55
You say "need not come from an integer" in your comment, but your actual answer says "need not determine an integer." Perhaps you want to put the adjective "profinite" in front of integer in your first sentence. Including this keyword will also help readers unfamiliar with what $\widehat{\Bbb Z}$ means. – anon May 8 '12 at 18:05
Also, wouldn't the residue of an element $x\in\widehat{\Bbb Z}$ be the images of $x$ under the canonical projection maps, and knowing these is equivalent to knowing $x$? – anon May 8 '12 at 18:15
The $p$-adic numbers are a bizarre and confusing topic. Still, that's what you get for daring to mention Number Theory. ;-) – MathematicalOrchid May 8 '12 at 18:45 | 2015-01-28T02:12:22 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/142677/recovering-a-number-from-a-remainder-list",
"openwebmath_score": 0.9208005666732788,
"openwebmath_perplexity": 384.9136038028865,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9796676460637104,
"lm_q2_score": 0.8615382076534742,
"lm_q1q2_score": 0.8440211078858272
} |
https://math.stackexchange.com/questions/2932750/recurrence-relation-for-pells-equation-x2-2y2-1/2934005 | # Recurrence relation for Pell's equation $x^2-2y^2=1$
I am wondering how to find the recurrence relation for solutions for $$x$$ in the Pell's equation $$x^2-2y^2=1$$.
I know the formula for the general term. It is $$\frac{(3+2\sqrt2)^n+(3-2\sqrt2)^n}{2}$$ for $$x_n$$, the $$n^{th}$$ smallest solution for $$x$$. Any help would be appreciated! Thanks!
I got a feeling that the recurrence formula is $$x_n=6x_{n-1}-x_{n-2}$$, but I wonder how to prove this relation true/false and how to derive/generate the recurrence relation. Note that $$x_{-1}=1$$ and this recurrence formula applies to all nonnegative integral $$n$$.
• For adding more details, it is recommended that you edit your question instead of creating a comment. Thanks. – GoodDeeds Sep 27 '18 at 7:03
• Thanks. I will edit the question. – Kai Sep 27 '18 at 7:04
• I'm not familiar with this particular technique, but it's easy to see that $x_n = 6x_{n-1} - x_{n-2}$ satisfies the formula given. It's also not difficult to see that it is the only recurrence relation of the form $x_n = ax_{n-1} + bx_{n-2}$ that the formula satisfies. If you are looking for such a recurrence relation, you've found it! – Theo Bendit Sep 27 '18 at 7:17
• I haven't learned generating functions, so I don't quite understand why it works, and why it is the only formula that work. I will try to figure that out. – Kai Sep 27 '18 at 7:25
• Have you tried to expand the formula $x_{n+1}=6 x_n - 1 x_{n-1}$ using the fractional/exponential expression that you already have, simplify and identify the correctness of the recursion-formula? – Gottfried Helms Sep 27 '18 at 7:30
If $$u_n=A\alpha^n+B\beta^n$$ then $$\alpha$$ and $$\beta$$ are the roots of the quadratic equation $$(x-\alpha)(x-\beta)=x^2-(\alpha+\beta)x+\alpha\beta=0$$
Let the quadratic be $$p(x)=x^2-px+q$$, then consider $$0=A\alpha^np(\alpha)+B\beta^np(\beta)=$$$$=(A\alpha^{n+2}+B\beta^{n+2})-p(A\alpha^{n+1}+B\beta^{n+1})+q(A\alpha^{n}+B\beta^{n})=u_{n+2}-pu_{n+1}+qu_n$$from which $$u_{n+2}=pu_{n+1}-qu_n$$Where $$p=\alpha+\beta$$ and $$q=\alpha\beta$$
You simply need to identify $$\alpha$$ and $$\beta$$ - the surrounding constants drop out in the arithmetic.
Once two successive terms are known, the constants $$A$$ and $$B$$ are determined, and there is a unique recurrence of this kind because two consecutive terms determine the rest through the recurrence.
• Note also that you can show that the recurrence is true simply by showing that it works for the general term you have. This answer shows you how to derive it. – Mark Bennet Sep 27 '18 at 7:39
The fundamental solution is $$9-8 = 1,$$ meaning $$3^2 - 2 \cdot 2^2 = 1.$$ As a result, we have the matrix $$A = \left( \begin{array}{cc} 3 & 4 \\ 2 &3 \end{array} \right)$$ which solves the automorphism relation, $$A^T H A = H,$$ where $$H = \left( \begin{array}{cc} 1 & 0 \\ 0 & -2 \end{array} \right)$$ That is $$(3x+4y)^2 - 2 (2x+3y)^2 = x^2 - 2 y^2.$$ Next, $$A^2 - 6 A + I = 0.$$ Since $$A \left( \begin{array}{c} x_n \\ y_n \end{array} \right) = \left( \begin{array}{c} x_{n+1} \\ y_{n+1} \end{array} \right)$$ and $$A^2 \left( \begin{array}{c} x_n \\ y_n \end{array} \right) = \left( \begin{array}{c} x_{n+2} \\ y_{n+2} \end{array} \right)$$ we find $$x_{n+2} - 6 x_{n+1} + x_n = 0$$ $$y_{n+2} - 6 y_{n+1} + y_n = 0$$ This is just Cayley-Hamilton.
Caution: This is for $$x^2 - 2 y^2 = 1.$$ If we change the problem to $$x^2 - 2 y^2 = 119 = 7 \cdot 17,$$ the recurrence still holds, except that there are now four such families, each using the same recursion:
$$(11,1) \; \; \; \; (37,25) \; \; \; \; (211,149) ...$$ $$(13,5) \; \; \; \; (59,41) \; \; \; \; (341,241) ...$$ $$(19,11) \; \; \; \; (101,71) \; \; \; \; (587,415) ...$$ $$(29,19) \; \; \; \; (163,115) \; \; \; \; (949,671) ...$$
If you don't mind negative values for $$x,y$$ you can combine the above four into two families going both forth and back...
• Is this group theory? – Kai Sep 28 '18 at 3:24
• @Kai the only group theory you need for this is how to multiply 2 by 2 integer matrices of determinant $1$ – Will Jagy Sep 28 '18 at 3:28
You can find the general recurrence relationship here:
https://crypto.stanford.edu/pbc/notes/contfrac/pell.html | 2020-12-05T02:34:07 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2932750/recurrence-relation-for-pells-equation-x2-2y2-1/2934005",
"openwebmath_score": 0.8850499987602234,
"openwebmath_perplexity": 207.5821192325338,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9796676478446031,
"lm_q2_score": 0.8615382040983515,
"lm_q1q2_score": 0.8440211059372956
} |
https://math.stackexchange.com/questions/2783026/prove-that-the-origin-is-liapunov-stable-for-the-given-system | # Prove that the origin is Liapunov Stable for the given system
Consider the system $$\dot{x} = y \\ \dot{y} = -4x$$ ($\dot{x}$ means $\displaystyle \frac{dx}{dt}$ and $\dot{y}$ means $\displaystyle \frac{dy}{dt})$
I need to prove that the fixed point $\mathbf{x^{*} = 0}$ is Liapunov stable.
For reference, according to my textbook (Strogatz), a fixed point $\mathbf{x^{*}}$ is said to be Liapunov stable if $\forall \epsilon > 0$, $\exists \delta >0$ such that $|\mathbf{x(t)} - \mathbf{x^{*}}| < \epsilon$ whenever $t \geq 0$ and $|\mathbf{x(0)}- \mathbf{x^{*}}|< \delta$
Now, if we let $x(0)=x_{0}$, $y(0)=y_{0}$, I was able to solve this system using eigenvalues and eigenvectors to have the following general solution: $$x(t) = x_{0} \cos (2t) + \frac{y_{0}}{2}\sin(2t) \\ y(t) = -2x_{0}\sin(2t) + y_{0}\cos(2t)$$
According to my solutions manual, "it can be observed from the equations in the solution that $\delta < |x_{0}|$. This implies that $|x(t),y(t)|<2|x_{0}|$, where $2|x_{0}|$ is the radius $\epsilon$. Thus, it can be concluded that $|x(t),y(t)|<\epsilon$."
However, I don't understand how they were able to show this. We never really learned how to do these kinds of $\epsilon-\delta$ proofs for systems in my class, and this solution doesn't give a very good (meaning "detailed") explanation as to how they went about doing it.
Could someone please explain to me step-by-step how they were able to surmise all of this? I would like to be able to apply this to other problems going forward, but unless I see one done out in detail, I am afraid I will not be able to.
Thank you for your time and patience.
• What is the book? – CroCo May 18 at 3:58
The solution to this problem is an ellipse parameterized by the initial conditions $x_0,y_0$ that has a center at $\{0,0\}$. Therefore, for a given set of initial conditions the maximum distance a solution could have at any time from the fixed point at the origin will be a fixed finite number. Intuitively then your solution is Liapunov stable.
For a rigorous proof of this fact you will use the $\epsilon-\delta$ definition you gave. The idea is this: $\epsilon$ is assumed to be some given small quantity and it is your task to find the $\delta$ that makes the definition true.
Your $\delta$ will be parameterized by the initial conditions and $\epsilon$. If you knew the maximum distance from the origin a solution could attain(semimajor axis), then you could use that knowledge to define $\delta$.
Since $x^*=0$ you will not have the usual algebraic difficulties associated with this kind of proof. Start with the statement $|x(t)-x^*|$ and develop a chain of inequalities leading to the required $\delta$.
Reference to the solution manual: In general, $\delta$ will depend on $x_0$ and $y_0$. However, i think the trick here is to observe that if you find a $\delta<\text{some distance}: d(x_0,y_0)$ that works then choosing the magnitude of the $x-$component as a new $\delta$ must also work because it could not possibly be bigger than the original.
The norm of the solution is $$\|{\bf x}(t)\|=\sqrt{x^2(t)+y^2(t)}=\sqrt{\left(x_0\cos 2t+\frac{y_0}2\sin 2t\right)^2+\left(-2x_0\sin 2t+y_0\cos 2t\right)^2}.$$ One can use the triangle inequality: $$\sqrt{(a_1+b_1)^2+(a_2+b_2)^2}\le \sqrt{a_1^2+a_2^2}+\sqrt{b_1^2+b_2^2}$$ in order to obtain $$\|{\bf x}(t)\|\le\sqrt{x_0^2\cos^22t+4x_0^2\sin^22t}+\sqrt{\frac{y_0^2}4\sin^22t+y_0^2\cos^22t}$$ $$=2|x_0|\sqrt{\frac14\cos^22t+\sin^22t}+|y_0| \sqrt{\frac14\sin^22t+\cos^22t}.$$ Notice that $$\sqrt{\frac14\cos^22t+\sin^22t}\le\sqrt{\cos^22t+\sin^22t}=1,$$ $$\sqrt{\frac14\sin^22t+\cos^22t}\le\sqrt{\sin^22t+\cos^22t}=1,$$ hence, $\|{\bf x}(t)\|\le2|x_0|+|y_0|$ ($\|{\bf x}(t)\|\le2|x_0|$ in the book is possibly a typo).
We have $|x_0|\le \|{\bf x}(0)\|$ and $|y_0|\le \|{\bf x}(0)\|$, thus, $\|{\bf x}(t)\|\le3\|{\bf x}(0)\|$. It means that for any $\epsilon>0$ one can take $\delta=\frac13\epsilon$ to prove the stability: $$\|{\bf x}(0)\|<\delta=\frac13\epsilon\;\Rightarrow\;\|{\bf x}(t)\|\le 3\|{\bf x}(0)\|<3\delta=\epsilon.$$ | 2018-08-17T11:43:29 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2783026/prove-that-the-origin-is-liapunov-stable-for-the-given-system",
"openwebmath_score": 0.9177936911582947,
"openwebmath_perplexity": 96.2789577982926,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9796676448764483,
"lm_q2_score": 0.8615382040983515,
"lm_q1q2_score": 0.8440211033801168
} |
https://www.physicsforums.com/threads/distance-between-2-planes.534821/ | # Distance between 2 planes
1. Sep 28, 2011
### cyt91
1. The problem statement, all variables and given/known data
Find the shortest distance between the 2 planes:
2x+2y-z=1 and 4x+4y-2z=5
How do we approach this problem?
I used the approach of finding the point at which the normal of one plane intersects the other plane and then determining the length of this vector. The answer I've got is 5/6 which is not correct. The correct answer is 1/2.
What is wrong with my approach? And how should we approach this problem?
Thanks.
2. Sep 28, 2011
### dynamicsolo
You'll need to show what you calculated in order for someone to judge what you might have done wrong. Your verbal description sounds fine as far as it goes...
3. Sep 28, 2011
### cyt91
4. Sep 28, 2011
### dynamicsolo
I'm a bit unclear about how you are setting up what you call your "point of interesection" equations. But you can do this:
You know that the common normal vector for the parallel planes is < 2, 2, -1 >. Pick a point in the plane 2x + 2y - z = 1 ; we can make this easy and use ( 0, 0, -1 ). Then this normal line passing through the chosen point has the vector equation
< x, y, z > = < 2t , 2t, -t - 1 > .
This line intersects the second plane for 4x + 4y - 2z = 5 , or
4 · 2t + 4 · 2t - 2 ( -t - 1 ) = 5 --> 18t + 2 = 5 ---> t = 1/6 .
The point of intersection in that plane is then ( 2 · 1/6 , 2 · 1/6 , -1/6 - 1 ) = ( 1/3 , 1/3 , -7/6 ). The distance between the two points along this line mutually perpendicular to the two planes (what is called the "perpendicular distance", the shortest distance between the planes) is given by
$$\sqrt{ ( \frac{1}{3} - 0 )^{2} + ( \frac{1}{3} - 0 )^{2} + ( -\frac{7}{6} - [-1] )^{2} } = \sqrt{ ( \frac{1}{3} )^{2} + ( \frac{1}{3} )^{2} + ( -\frac{1}{6} )^{2} } = \sqrt{ \frac{1}{9} + \frac{1}{9} + \frac{1}{36} }$$
$$= \sqrt{ \frac{4 + 4 + 1}{36} } = \sqrt{ \frac{9}{36} } = \frac{1}{2} .$$
I think the problem you may have made for yourself is that you didn't actually choose a point of intersection in either plane to build a normal line from.
A more general argument along these lines gives the perpendicular distance between two parallel planes ax + by + cz = d1 and ax + by + cz = d2 as $D = \frac{\vert d_{1} - d_{2} \vert}{\sqrt{a^{2} + b^{2} + c^{2} } } .$ For this problem, we would write
2x + 2y - z = 1 and 2x + 2y - z = 5/2 ;
The square root in the denominator gives 3 and the absolute value of the difference in the numerator is 3/2 , so the formula yields D = (3/2) / 3 = 1/2 .
Last edited: Sep 28, 2011
5. Sep 28, 2011
### cyt91
"Then this normal line passing through the chosen point has the vector equation
< x, y, z > = < 2t , 2t, -t - 1 > "
Is this derived from :
r(t) = r0 + tv ,where r0 and v are vectors
which is the vector equation for a straight line?
6. Sep 28, 2011
### dynamicsolo
Yes, that's right: this is just using vector notation, rather than writing three separate parametric equations for the three coordinates of the points on the line.
7. Sep 29, 2011
### icystrike
You can use this "formula" but you can attempt to prove its derivative.
Given a point $$P(x_{0},y_{0},z_{0})$$, and the equation of a plane is ax+by+cz+d=0,
The distance from the point to the plane is:
D=$$\mid \frac{ax_{0}+by_{0}+cz_{0}+d}{\sqrt{a^{2}+b^{2}+c^{2}}$$
8. Sep 29, 2011
### cyt91
Got it! Yes. I did not pick a point on the first plane through which the normal vector passes.
Thanks. | 2018-02-21T16:11:10 | {
"domain": "physicsforums.com",
"url": "https://www.physicsforums.com/threads/distance-between-2-planes.534821/",
"openwebmath_score": 0.7701009511947632,
"openwebmath_perplexity": 397.37866626381543,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9796676448764483,
"lm_q2_score": 0.8615382040983515,
"lm_q1q2_score": 0.8440211033801168
} |
http://math.stackexchange.com/questions/191204/induction-proof-of-lower-bound-for-sum-sqrt-n | # Induction proof of lower bound for $\sum \sqrt n$
I'm having some trouble proving the following statement using mathematical induction:
$$\frac{1}{2}n^{\frac{3}{2}} \leq \sqrt{1} + \sqrt{2} + \sqrt{3} + \sqrt{4} + ... + \sqrt{n} ,\text{ (for all sufficiently large n)}$$
I'm sort of confused because $n^{\frac{3}{2}}$ = $n n^{\frac{1}{2}}$ which means that there are n of the largest possible root values on the left hand side while the right hand side has values like $\sqrt{1}$, $\sqrt{2}$, etc. As n grows sufficiently large wouldn't the left hand side outgrow the right hand side no matter what the constant multiple is? If I'm thinking about this correctly just let me know, if not, any help would be appreciated.
-
I don't think your intuition is right. You probably know that $1+...+n = n(n+1)/2$ is quadratic and $1+...+n^2 = n(n+1)(2n+1)/3$ is cubic – Cocopuffs Sep 5 '12 at 0:04
Don't forget the $1/2$! There are $n/2$ of the largest possible root values on the left, not $n$. – Antonio Vargas Sep 5 '12 at 0:04
Are you having any particular trouble with the induction step? – Antonio Vargas Sep 5 '12 at 0:10
Can you estimate the LHS difference $\frac 12 n^{3/2}- \frac 12 (n-1)^{3/2}$? If it is always less than $\sqrt n$ from some point, there's your induction step. (Hint: Calculus!) – Henning Makholm Sep 5 '12 at 0:14
You’re right that $$\sum_{k=1}^n\sqrt k\le n\cdot\sqrt n=n^{3/2}\;,$$ but you can’t infer from that that $\frac12n^{3/2}$ will eventually exceed $\sum_{k=1}^n\sqrt k$. Suppose that we divide everything through by $n$. Then the claimed inequality becomes
$$\frac12\sqrt n\le\frac1n\sum_{k=1}^n\sqrt k\tag{1}\;.$$
The righthand side of $(1)$ is just the arithmetic mean of the square roots $\sqrt 1,\sqrt 2,\dots\sqrt n$. You’ve observed (correctly) that this mean cannot be any bigger than $\sqrt n$, the largest of the $n$ numbers, but that doesn’t mean that it must eventually be smaller than half of the largest number. In fact, if you draw the graph of $y=\sqrt x$ you’ll see that it gets less and less steep as $x$ increases. When $n$ is large, therefore, most of the square roots in the list $\sqrt 1,\sqrt 2,\dots,\sqrt n$ are a lot closer to $\sqrt n$ than they are to $0$, and when you average them, the result is closer to $\sqrt n$ than to $0$. But that just means that it’s bigger than $\frac12\sqrt n$.
To prove the theorm by induction, you have to establish a starting point (‘base case’) and prove an induction step. For the induction step, suppose that you know for some $n$ that $$\frac12n^{3/2}\le\sum_{k=1}^n\sqrt k\tag{2}\;;$$ this is your induction hypothesis. You want to use it to show that $$\frac12(n+1)^{3/2}\le\sum_{k=1}^{n+1}\sqrt k\;.\tag{3}$$
To get from the righthand side of $(2)$ to the righthand side of $(3)$, we’ve obviously just added $\sqrt{n+1}$. If we knew that the lefthand side increased by at most $\sqrt{n+1}$, i.e., that
$$\frac12(n+1)^{3/2}\le\frac12n^{3/2}+\sqrt{n+1}\;,$$ we’d be in business: we’d have
$$\frac12(n+1)^{3/2}\le\frac12n^{3/2}+\sqrt{n+1}\le\sum_{k=1}^n\sqrt k+\sqrt{n+1}=\sum_{k=1}^{n+1}\sqrt k\;,$$
showing that $(2)$ does imply $(3)$.
So how can we show that
$$\frac12(n+1)^{3/2}\le\frac12n^{3/2}+\sqrt{n+1}\;?\tag{4}$$
Let’s tinker a bit. $(4)$ is equivalent to
$$(n+1)^{3/2}\le n^{3/2}+2\sqrt{n+1}\;,$$
which is equivalent to
$$(n+1)\sqrt{n+1}\le n\sqrt n+2\sqrt{n+1}\;.$$
This in turn is equivalent to
$$n\sqrt{n+1}+\sqrt{n+1}\le n\sqrt n+2\sqrt{n+1}\;,$$
which is equivalent to
$$n\sqrt{n+1}\le n\sqrt n+\sqrt{n+1}$$ and then to $$(n-1)\sqrt{n+1}\le n\sqrt n\;.\tag{5}$$
Everything in $(5)$ is non-negative, so $(5)$ is equivalent (for positive integers $n$) to the inequality that you get by squaring it,
$$(n-1)^2(n+1)\le n^3\;.\tag{6}$$
Can you show that $(6)$ is true for all positive integers? If so, working back through the chain of equivalent inequalities gives you $(4)$ for all positive integers and thereby shows that $(2)$ implies $(3)$ for all positive integers. To finish the induction, you just have to find a positive integer $n_0$ for which $(2)$ is true to conclude that $(2)$ is true for all integers $n\ge n_0$.
Added: This is the elementary way to approach the problem. If you’re a bit ingenious and know your calculus, you can prove the result without induction. The sum $\sum_{k=1}^n\sqrt k$ is an upper Riemann sum for $\int_0^n\sqrt x~dx$, so $\int_0^n\sqrt x~dx\le\sum_{k=1}^n\sqrt k$. How does $\int_0^n\sqrt x~dx$ compare with $\frac12 n^{3/2}$?
-
Thanks this really helped, I believe I can finish the problem from here, but may take some time to really understand everything thoroughly. – Math_Illiterate Sep 5 '12 at 0:49
There is also a pre-calculus way to do this without induction: multiply both sides by $2$ and use the fact that $\sqrt{i}+\sqrt{n-i} \ge \sqrt{n}$ (easy by squaring both sides) to show that the RHS is $\ge (n+1)\sqrt{n}$ – Erick Wong Sep 5 '12 at 1:06 | 2016-05-26T22:42:16 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/191204/induction-proof-of-lower-bound-for-sum-sqrt-n",
"openwebmath_score": 0.9123619198799133,
"openwebmath_perplexity": 149.84869762278217,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9796676442828174,
"lm_q2_score": 0.8615382040983515,
"lm_q1q2_score": 0.8440211028686812
} |
https://tutorme.com/tutors/27455/interview/ | Enable contrast version
# Tutor profile: Amadej Kristjan K.
Inactive
Mathematics Tutor
Tutor Satisfaction Guarantee
## Questions
### Subject:Machine Learning
TutorMe
Question:
Suppose we are trying to use a machine learning algorithm to perform a classification task, which humans can do with $$98$$% accuracy. Our machine learning algorithm has $$95$$% accuracy on the training set and $$70$$% accuracy on the test set. What is our machine learning algorithm likely suffering from and how can we surmount this issue?
Inactive
The algorithm is likely suffering from overfitting. This means that the algorithm 'memorises' a lot of irrelevant information that is specific to the set it's training on, but fails to generalise the patterns to the general data distribution. When faced with a yet unseen test set, it performs much worse. In such a scenario, what we can do several of the following: - we can increase the amount of training data (this way the algorithm is faced with a slightly more representative representation of the general data distribution), - we can reduce the representational capacity of the model (making the model simpler, to force the algorithm to only encode the most crucial, generalizable information into the model), - we can incorporate regularisation techniques to reduce overfitting (add a penalising term to the loss function as done in Ridge and Lasso regression or in case of using neural networks we can add dropout regularisation).
### Subject:Discrete Math
TutorMe
Question:
Prove that there are infinitely many prime numbers.
Inactive
Suppose for the sake of contradiction that there are only finitely many prime numbers. Then we can list them as follows: $$p_1,p_2,...,p_n$$. Now consider the number $$P = p_1p_2...p_n + 1$$. It is not divisible by either of the prime numbers listed, as it gives a remainder of $$1$$ when divided by any of them. Therefore it is only divisible by $$1$$ and itself, having only $$2$$ positive integer divisors, which makes $$P$$ a prime number. But we assumed the above list of primes contained all prime numbers. Since we found a prime outside of the list with supposedly all primes, we arrived at a contradiction. Hence, there are infinitely many prime numbers.
### Subject:Calculus
TutorMe
Question:
Let $$R$$ be a rectangle with side lengths $$a$$ and $$b$$. If we keep its perimeter $$P$$ fixed, at what $$a:b$$ ratio will the area $$A$$ of the rectangle be maximised?
Inactive
The (fixed) perimeter $P$ of the rectangle is: $$P = 2(a+b)$$ The area is: $$A = ab = a(\frac{P}{2}-a) = \frac{Pa}{2}-a^2$$ Thus, we can view the area of $$R$$ as a function of the variable $$a$$ and see at which value of $$a$$ the area is maximised subject to our given perimeter $$P$$. We can obtain the maximum by figuring out where the derivative of this function is $$0$$. $$\frac{dA}{da} = \frac{P}{2}-2a$$ so we need to solve $$0 = \frac{P}{2}-2a$$, giving $$a = \frac{P}{4}$$. Now substituting this into $$P = 2(a+b)$$ gives $$b=\frac{P}{4}$$, meaning that for maximising the area we need $$a = b$$.
## Contact tutor
Send a message explaining your | 2020-02-16T19:43:37 | {
"domain": "tutorme.com",
"url": "https://tutorme.com/tutors/27455/interview/",
"openwebmath_score": 0.7956141829490662,
"openwebmath_perplexity": 274.3025015025706,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9929882039959953,
"lm_q2_score": 0.8499711775577735,
"lm_q1q2_score": 0.8440113530514548
} |
https://math.stackexchange.com/questions/2422879/prove-that-for-any-integer-n-n24-is-not-divisible-by-7 | # Prove that for any integer $n, n^2+4$ is not divisible by $7$.
The question tells you to use the Division Theorem, here is my attempt:
Every integer can be expressed in the form $7q+r$ where $r$ is one of $0,1,2,3,4,5$ or $6$ and $q$ is an integer $\geq0$.
$n=7q+r$
$n^2=(7q+r)^2=49q^2+14rq+r^2$
$n^2=7(7q^2+2rq)+r^2$
$n^2+4=7(7q^2+2rq)+r^2+4$
$7(7q^2+2rq)$ is either divisible by $7$, or it is $0$ (when $q=0$), so it is $r^2+4$ we are concerned with.
Assume that $r^2+4$ is divisible by 7. Then $r^2+4=7k$ for some integer $k$.
This is the original problem we were faced with, except whereas $n$ could be any integer, $r$ is constrained to be one of $0,1,2,3,4,5$ or $6$.
Through trial and error, we see that no valid value of $r$ satisfies $r^2+4=7k$ so we have proved our theorem by contradiction.
I'm pretty sure this is either wrong somewhere or at the very least not the proof that the question intended. Any help would be appreciated.
• This seems fine to me. It's also the most natural and typical approach for this problem. – Michael Burr Sep 9 '17 at 17:38
• "7(7q2+2rq) is either divisible by 7, or it is 0" It is divisible by 7. Period. You know that because of the big honking 7 (7q^2+2rq) is being multiplied by. If it is 0, then it is still divisible by 7 because 0 is divisible by 7. 0 = 7*0. So 0 is divisible by 7. – fleablood Sep 9 '17 at 17:39
• BUt your proof is correct. By trial and error $0,1,2,3,4,5,6$ squared are $0,1,4,9,16,25,26$ and $r^2 + 4$ is $4,,5,13,20,29,30$. – fleablood Sep 9 '17 at 17:41
• Thanks both, quite relieving to hear. And yes you're right fleablood, wasn't sure if that counted but I'll never forget now ;) – quantum285 Sep 9 '17 at 17:45
• Note 4,5,6 are equivalent to -3,-2,-1 so you only have to do half of the equations. – fleablood Sep 9 '17 at 18:15
since we have $$n\equiv 0,1,2,3,4,5,6\mod 7$$ we get $$n^2\equiv 0,1,2,4\mod 7$$ therefore $$n^2+4\equiv 1,4,5,6\mod 7$$
Your proof is correct (fleablood's comment). Let me rephrase a bit:
$A(n):= n^2 +4$; $B(r,q) = 7 q^2 + 2rq;$
$A(n) = 7B(r,q) + (r^2 +4)$.
Fairly simple to show that:
$A(n)$ is divisible by $7 \iff$
$(r^2 +4)$ is divisible $n.$
By inspection
$(r^2 +4) , r = 0,1,2,3,4,5,6,$
is not divisible by $7$.
$\Rightarrow$: $A(n)$ is not divisible by $7$.
Through trial and error, we see that no valid value of r satisfies r2+4=7k
so we have proved our theorem by contradiction.
I'm pretty sure this is either wrong somewhere or at the very least not the proof that the question intended. Any help would be appreciated.
Through trial and error you did $7$ calculations. This is fine and acceptable. The calculations are not divisible by $7$ and you proved the if $n^2 +4$ were ever divisible by $7$ then $r^2 +4$ would have to be an you showed it can't. End of story. Good job.
But it's pretty unsatisfactory to do $7$ calculations off page and say "You can check them for yourselves, for now take my word for it".
Here's a slicker way of saying the exact same thing:
We need to prove that $n^2 + 4 \not \equiv 0 \mod 7$.
If $n^2 + 4 \equiv 0 \mod 7$ then $n^2 \equiv 3 \mod 7$
Now there and $7$ residue classes modulo $7$. They are $[0], [1],[2], [3], [-3]=[4], [-2]=[5],$ and $[6] =[-1]$.
So there are $7$ cases to check.
Case 1: $n \equiv 0\implies n^2 \equiv 0 \not \equiv 3 \mod 7$.
Case 2: $n \equiv \pm 1 \implies n^2 \equiv 1 \not \equiv 3 \mod 7$.
Case 3: $n \equiv \pm 2 \implies n^2 \equiv 4 \equiv -3 \not \equiv 3 \mod 7$.
Case 4: $n \equiv \pm 3 \implies n^3 \equiv 9\equiv 2 \not \equiv 3 \mod 7$.
We are done.
A little more advanced: $0 = 0; 1 = 1; 2=3-1; 3=3;$ and $4... 6 \equiv -3...-1$.
So $n \equiv \pm (3k \pm i)$ where $k,i \in \{1,0\}$
$n^2 \equiv (\pm (3k \pm i))^2 \equiv 9k^2 \pm 6ki + i^2 \equiv 2k^2 \mp ki + i^2$ which has 8 possible values depending on whether $k,i$ equal $0$ or $1$ and whether the $\pm$ is a plus or a minus.
If $k = 0$ then $n^2 \equiv i^2$ which is either $0,1$ depending on the value of $i$.
If $k = 1$ then $n^2 \equiv 2\mp i + i^2$. Which is either $2$ or $4$ depending upon the value of $i$ and the sinage of $\mp$.
But that is probably way too obtuse, isn't it?
Your proof is correct. If you know congruences then we can simplify the first half as
$$\bmod 7\!:\,\ n\equiv r\,\Rightarrow\, n^2+4\equiv r^2+4$$
The work in your brute-force check that $\,r^2+4\,$ has no roots can be halved by using a balanced or signed residue system $\, r\equiv \pm\{0,1,2,3\}\,\Rightarrow\,r^2\equiv \{0,1,4,2\}\Rightarrow\,r^2+4\equiv \{4,5,1,6\}$
Alternatively, more simply $\,r^2\equiv -4\equiv 3\,\Rightarrow\,r\not\equiv0\,\Rightarrow\,r^6\equiv 3^3\equiv -1\,$ contra little Fermat (this is a special case of Euler's criterion for squares).
The question boils down to 3 being a quadratic residue modulo 7 or non-residue
$$n^2+4 \equiv 0 \; (mod\;7)$$ $$n^2\equiv 3 \;(mod\;7)$$
For this we have Legendre symbol
$$\left ( \frac{a}{p} \right )=\begin{cases} & 1\text{ if }a\text{ is a quadratic residue modulo }p\text{ and }p\text{ does not divide }a \\ & -1\text{ if }a\text{ is a quadratic non-residue modulo }p \\ & 0\text{ if }p\text{ divides }a \end{cases}$$
calculated as
$$\left ( \frac{a}{p} \right )=a^{\frac{p-1}{2}}$$
So
$$3^{\frac{7-1}{2}} \equiv 3^{3} \equiv 27 \equiv -1 \;(mod\;7)$$
making $3$ a non-residue modulo $7$.
Thus there is no $n$ that solves the congruence so $n^2+4$ is not divisible by $7$ for any $n$.
A little bit more advanced step that applies the Law of quadratic reciprocity would give that $3$ is a residue of $p$ if and only if
$$p \equiv \; 1, \; 11 \; (mod \; 12)$$
and $7$ does not belong to this group obviously.
• Similarly, notice that if $n^2\equiv 3\pmod{7}$, then $$n^6\equiv (n^2)^3\equiv 3^3\equiv$$ $$\equiv 27\equiv -1\pmod{7},$$ which contradicts Fermat's little theorem. – user236182 Sep 9 '17 at 19:38
Here is a proof that seems a bit roundabout, but it relates simpler methods to Fermat's Little Theorem.
First note that clearly $n^2+4$ will not be a multiple of $7$ if $n$ is one. Now for other values of $n$, raise $n^2+4$ to the power of $7-1=6$:
$(n^2+4)^6=n^{12}+(6)(4)n^{10}+(15)(4^2)n^8+...+4^6$
$\equiv n^{12}+3n^{10}+2n^8+6n^6+4n^4+5n^2+1 \bmod 7$
Note that the coefficients are in geometric progression $\bmod 7$, with common ratio $3\equiv -4$. This is because each binomial coefficient is obtained from an adjacent one by multiplying by $(7-k)/k\equiv -1$.
Now apply Fermat's Little Theorem, thus $n^6\equiv 1, n^8\equiv n^2, n^{10}\equiv n^4, n^{12}\equiv n^6\equiv 1$. After this reduction add terms with like powers and note that most of them cancel. We are left with
$(n^2+4)^6\equiv 1\bmod 7$
This completes the proof that $n^2+4$ must fail to be a multiple of $7$.
We can do a similar analysis with $(n^2-r)\bmod p$, where $p$ is an odd prime, raising the binomial to the power of $p-1$. The binomial coefficients always have the common ratio $-1\bmod p$, which causes the successive coefficients in $(n^2-r)^{(p-1)}$ to have common ratio $r\bmod p$. Then the cancellation noted above always occurs if $r^{(p-1)/2}\equiv -1\bmod p$. Thus if $r^{(p-1)/2}\equiv -1\bmod p$, we end with $(n^2-r)^{(p-1)}\equiv 1 \bmod p$ thus proving that $(n^2-r)$ cannot be a multiple of $p$. | 2019-11-22T13:44:32 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2422879/prove-that-for-any-integer-n-n24-is-not-divisible-by-7",
"openwebmath_score": 0.8834125995635986,
"openwebmath_perplexity": 198.8639736903911,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9861513885634199,
"lm_q2_score": 0.8558511506439708,
"lm_q1q2_score": 0.8439988006111524
} |
http://mathematica.stackexchange.com/questions/3015/how-to-show-values-of-points-on-surface-plotted-with-contourplot3d?answertab=active | # How to show values of points on surface plotted with ContourPlot3D
The help for ContourPlot3D starts with this example
ContourPlot3D[x^3 + y^2 - z^2 == 0, {x, -2, 2}, {y, -2, 2}, {z, -2, 2}]
This returns a Plots of the surface $x^2 + y^2 - z^2 = 0$:
.
Now I have a function, and I would like to know how this function behaves on this surface. For example, take
f[x_,y_,z_] := {x^3 - 2 x y + z^2 x - 5 z^2 y^2 - 4 x y z,
Sin[10x] + Cos[5y] + Cos[20z]}
I was thinking of making a list with some of the point that lie on this surface. Then it is easy to evaluate the function on these points. The way I would like to generate the list is to click with the mouse on a few points that I think are interesting. Is that possible?
What are the alternatives to using the mouse? There most be some list available with all points that were used to draw this graph. How can I take some point of that list and add them as point to the 3D plot to visualize them?
Update
The methods of Szabolcs and Heike work fine on the example $x^3 + y^2 - z^2 =0$. Now I try to apply the same to
$$2316 a^{12} c^6+500 a^{11} b c^5+10296 a^{10} b^2 c^4+1624 a^{10} c^5+656 a^9 b^3 c^3- \\ - 3856 a^9 b c^4+41 a^8 b^4 c^2+808 a^8 b^2 c^3+784 a^8 c^4+a^7 b^5 c+24 a^7 b^3 c^2- \\ - 176 a^7 b c^3+2 a^6 b^4 c+16 a^6 b^2 c^2+32 a^6 c^3 = 0$$
f[a_, b_, c_] := 2 a^6 b^4 c + a^7 b^5 c + 16 a^6 b^2 c^2 + 24 a^7 b^3 c^2
+ 41 a^8 b^4 c^2 + 32 a^6 c^3 - 176 a^7 b c^3 + 808 a^8 b^2 c^3
+ 656 a^9 b^3 c^3 + 784 a^8 c^4 - 3856 a^9 b c^4
+ 10296 a^10 b^2 c^4 + 1624 a^10 c^5 + 500 a^11 b c^5 + 2316 a^12 c^6
pts = {};
Substituting this function into Heike's solution does not work. Clicking does not result in points on the surface. Also Szabolcs's FindInstance does not work. What goes wrong here?
.
-
The more general question is how to interact with three dimensional graphics objects using the mouse. Tooltip does work, so there's some support. But the mouse coordinates can only be retrieved in 2D while this time you want the coordinates in 3D, on the surface. +1. – Szabolcs Mar 15 '12 at 9:52
Why not use built-in visualization capabilities like the MeshFunctions and ColorFunction options? (MeshFunctions works well to show contours of the first component of f while ColorFunction does a better job with its second component, which is rapidly varying; using PlotPoints -> 20 helps.) – whuber Mar 15 '12 at 16:25
The location that the mouse is pointing to on a 3D surface can be found by starting with MousePosition["Graphics3DBoxIntercepts"]. This will give you the two points where the line perpendicular to the screen at the mouse pointer intersects the three-dimensional bounding box.
We can calculate the intersection of this line with the surface to find a point. Here is a simple implementation to track that point dynamically:
fun defines the surface:
fun[{x_, y_, z_}] := x^3 + y^2 - z^2
Let's plot it:
plot = ContourPlot3D[
fun[{x, y, z}] == 0, {x, -2, 2}, {y, -2, 2}, {z, -2, 2}]
Now let's try to find the intersection of the mouse pointer with the surface using FindRoot. There might be several intersections, so I specified the box intersection point closer to the viewer as a starting point for FindRoot (t == 0 in the code). This does not guarantee that the closest (i.e. visible) point will be found, but it makes it more likely.
Show[plot,
Graphics3D[{
Red,
Dynamic@Quiet@Check[
Sphere[#, Scaled[0.01]]& @ Module[{p1, p2, t},
{p1, p2} = MousePosition[{"Graphics3DBoxIntercepts", Graphics3D}];
(p2 - p1) t + p1 /. FindRoot[fun[(p2 - p1) t + p1], {t, 0, 0, 1}]
],
{}]}]
]
Now that we have the point on the surface, you can do with it whatever you want (calculate another functions, etc.) You can use EventHandler to just record clicks instead of tracking values dynamically.
To address your other question about how to get a number of points on the surface. One way is to use FindInstance.
FindInstance[
fun[{x, y, z}] == 0 && Thread[-2 < And[x, y, z] < 2, And], {x, y,
z}, Reals, 10]
This will give you 10 points that are precisely on the surface (this uses exact calculations). Let's show them:
Show[plot, Graphics3D[{Red, Sphere[{x, y, z}, Scaled[0.01]] /. %}]]
To get the points generated by ContourPlot3D, extract them from the GraphicsComplex object is creates. These coordinates will not be quite as precise as they are meant for visualization only.
First@Cases[plot, GraphicsComplex[points_, ___] :> points, Infinity]
Let's show those points:
Graphics3D@Point[%]
-
And you can use Deploy to prevent rotating the plot. – István Zachar Mar 15 '12 at 12:54
This solution is pretty similar to Szabolcs's solution. I've added an EventHandler to make it easier to select points in the plot. Here f[x,y,z]==0 is the equation of the surface, and pts contains a list of selected points on this surface. You can add points to the list by right-clicking somewhere on the plot.
f[x_, y_, z_] := x^3 + y^2 - z^2;
pts = {};
plot = ContourPlot3D[
f[x, y, z] == 0, {x, -2, 2}, {y, -2, 2}, {z, -2, 2},
Mesh -> False, ColorFunctionScaling -> False,
BoundaryStyle -> None];
Dynamic@EventHandler[
Show[plot, Graphics3D[{Red, Sphere[#, .05] & /@ pts}]],
{{"MouseClicked", 2} :> Module[{p, sol, pt},
p = CurrentValue[{"MousePosition", "Graphics3DBoxIntercepts"}];
If[p =!= None,
sol = Quiet@
Check[FindRoot[
f[x, y, z] /. Thread[{x, y, z} -> l p[[1]] + (1 - l) p[[2]]],
{l, .5, 0, 1}], None];
If[sol =!= None,
AppendTo[pts, l p[[1]] + (1 - l) p[[2]] /. sol]]]]}]
-
If you use PassEventsDown -> True, then you can use button-1 clicks while keeping the graphics rotatable (no need for button 2). Button 2 might still be more comfortable for some people. – Szabolcs Mar 15 '12 at 13:19
For the Sphere[] radius, you could use Scaled coordinates to make this independent of the plot range. – Szabolcs Mar 15 '12 at 13:21
Sorry about a third comment... as you can see, I've been playing with this for a while. I suggest using 1 as the starting value for FindRoot instead of 0.5. This will make it much more likely that the solution will be on the side of the surface closer to the viewer. If you use for example a sphere, many of the points will be on the other (invisible) side of the sphere. A starting value of 1 brings it to the viewer's side while a starting point of 0 will push it to the opposite side. You can also localize l. – Szabolcs Mar 15 '12 at 13:45
@Szabolcs Since I restricted the search domain for FindRoot to {0, 1} I thought it safer to use a starting value somewhere in the middle of the domain than near x == 1. – Heike Mar 15 '12 at 14:03
That is a good point---I just tried FindRoot[Sin[x], {x, 4, 3.3, 7}] to make sure. But we could use NSolve instead, which will usually find all roots with good performance if there's a bound on the solution (which there is). Then we can select the largest value of l. Did you know that even Reduce can handle these equations in a semi-numerical way, returning Root objects which can be evaluated to arbitrary precision? – Szabolcs Mar 15 '12 at 14:14
show 4 more comments | 2014-03-08T16:19:08 | {
"domain": "stackexchange.com",
"url": "http://mathematica.stackexchange.com/questions/3015/how-to-show-values-of-points-on-surface-plotted-with-contourplot3d?answertab=active",
"openwebmath_score": 0.19667740166187286,
"openwebmath_perplexity": 1146.567328952111,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9372107949104865,
"lm_q2_score": 0.9005297934592089,
"lm_q1q2_score": 0.8439862435684814
} |
https://math.stackexchange.com/questions/211552/riemann-stieltjes-integral-integration-by-parts-rudin | # Riemann-Stieltjes integral, integration by parts (Rudin)
Problem 17 of Chapter 6 of Rudin's Principles of Mathematical Analysis asks us to prove the following:
Suppose $\alpha$ increases monotonically on $[a,b]$, $g$ is continuous, and $g(x)=G'(x)$ for $a \leq x \leq b$. Prove that,
$$\int_a^b\alpha(x)g(x)\,dx=G(b)\alpha(b)-G(a)\alpha(a)-\int_a^bG\,d\alpha.$$
It seems to me that the continuity of $g$ is not necessary for the result above. It is enough to assume that $g$ is Riemann integrable. Am I right in thinking this?
I have thought as follows:
$\int_a^bG\,d\alpha$ exists because $G$ is differentiable and hence continuous.
$\alpha(x)$ is integrable with respect to $x$ since it is monotonic. If $g(x)$ is also integrable with respect to $x$ then $\int_a^b\alpha(x)g(x)\,dx$ also exists.
To prove the given formula, I start from the hint given by Rudin $$\sum_{i=1}^n\alpha(x_i)g(t_i)\Delta x_i=G(b)\alpha(b)-G(a)\alpha(a)-\sum_{i=1}^nG(x_{i-1})\Delta \alpha_i$$ where $g(t_i)\Delta x_i=\Delta G_i$ by the intermediate mean value theorem.
Now the sum on the right-hand side converges to $\int_a^bG\,d\alpha$. The sum on the left-hand side would have converged to $\int_a^b\alpha(x)g(x)\,dx$ if it had been $$\sum_{i=1}^n \alpha(x_i)g(x_i)\Delta x$$ The absolute difference between this and what we have is bounded above by $$\max(|\alpha(a)|,|\alpha(b)|)\sum_{i=1}^n |g(x_i)-g(t_i)|\Delta x$$ and this can be made arbitrarily small because $g(x)$ is integrable with respect to $x$.
• After you write out the discrete sums, you apply the mean value theorem (not the intermediate value theorem). – Chris Janjigian Oct 12 '12 at 14:27
• @chris. thanks. i will make the correction. – Jyotirmoy Bhattacharya Oct 12 '12 at 16:07
• Because it is increasing monotonically and is continuous it is the case that it is integrable, so I think that these assumptions are fine -- perhaps not optimal but none the less good enough. – Squirtle May 3 '13 at 5:16
Theorem: Suppose $f$ and $g$ are bounded functions with no common discontinuities on the interval $[a,b]$, and the Riemann-Stieltjes integral of $f$ with respect to $g$ exists. Then the Riemann-Stieltjes integral of $g$ with respect to $f$ exists, and
$$\int_{a}^{b} g(x)df(x) = f(b)g(b)-f(a)g(a)-\int_{a}^{b} f(x)dg(x)\,.$$
You are right that continuity is a stronger hypothesis than needed. I haven't checked your proof in detail due to lack of time, but assuming $g$ to be continuous only simplifies the problem. See for example theorem $12.14$ in This book. | 2019-09-23T16:07:51 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/211552/riemann-stieltjes-integral-integration-by-parts-rudin",
"openwebmath_score": 0.9654874801635742,
"openwebmath_perplexity": 97.72603495416737,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.981735725388777,
"lm_q2_score": 0.8596637523076225,
"lm_q1q2_score": 0.8439626174621616
} |
http://zclf.produktninja.de/e51m | ### Dynamic Programming Hotel Problem
dynamic programming hotel problem. Before turning to the derivation of the dynamic programming equation for the problem Vε, we introduce a notation which will be used frequently in the sequel. Dynamic Programming (DP) is a technique that solves some particular type of problems in Polynomial Time. Problem Statement. Please program and tell the moon god how many ways the moon god can climb out of the abyss. In the forty-odd years since this development, the number of uses and applications of dynamic programming has increased enormously. Dynamic programming problems can catch you off-guard in an interview or exam. Open Live Script. Hence, the very essential feature of DP is the proper structuring of optimization problems into multiple levels, which are solved sequentially one level at a time. The next step is cascading of the quality strategy. It is well suited for multi-stage or multi-point or sequential decision process. Multivariate, Time-Series. For example, the dynamic programming. Problem Description Input Format. Dynamic Features of VirusShare Executables. The technique was developed by Richard Bellman in the. The main caveat behind dynamic programming is that it can be applied to a certain problem if that problem can be divided into sub-problems. Again the solution of a problem is formulated recursively in terms of sub problems. 7 and a few useful new features have been added. ASPLOS is the premier forum for multidisciplinary systems research spanning computer architecture and hardware, programming languages and compilers, operating systems and networking, as well as applications and user interfaces. The dynamic programming solves the original problem by dividing the problem into smaller independent sub problems. Problem Statement:- You and two of your friends have just returned back home after visiting various countries. DP algorithms could be implemented with recursion, but they don't have to be. Introduction to Operations Research. Multivariate. We can create a 2D array part[][] of size (sum/2)*(n+1). – Hotels have different costs. These methods have been used numerically to compute optimal policies, as well as analytically to determine the form of an optimal policy under various assumptions on the rewards and transition probabilities. Dynamic programming methods have been applied in many areas. These tools are valuable in support of a good plan or recipe. It is an easy to use system that is designed to meet the needs of every guest in a hotel. This is a substantially expanded (by nearly 30%) and improved edition of the best-selling 2-volume dynamic programming book by Bertsekas. The particularity and complexity of this multistage dynamic programming (DP) problem rest on the traffic volume in PCDN network which is not steady and periodic. Dynamically loaded/unloaded and linked during execution (i. 2) A special case 2. It's better to store the result for Tile[N-2] because it is being used twice. Dynamic programming is used to solve the multistage optimization problem in which dynamic means reference to time and programming means planning or tabulation. d(i) = min_{j=0, 1, , i-1} ( d(j) + (200-(ai-aj))^2). the act of using information; G. You are using linear programming when you are driving from home to work and want to take the shortest route. However, there is a way to understand dynamic programming problems and solve them with ease. In the previous post, we have discussed a greedy approach for activity selection problem. Value and Policy Iteration. C Language. The only places you are allowed to stop are at these hotels, but you can choose which of the hotels you stop at. And we can construct the solution in bottom up manner such that every filled entry has following property. I have use this following method in python 3. Yuval Tassa†, Nicolas Mansard∗ and Emo Todorov†. A recursive relation between the larger and smaller sub problems is used. Massachusetts Institute of Technology. I know that Knapsack is NP-complete while it can be solved by DP. For the various problems in area such as inventory, chemical engineering design , and control theory, Dynamic Programming is the only technique used to solve the problem. Dynamic programming is distinguished from regular recursion by the reuse of subproblems. The Guide to Information Technology and Its Role During COVID-19. This will contain the underlying problems in the dataset that I had to confront. Increase your pace of learning & even get certified with our to the point byte size fast track courses around 20 most important topics such as data structures and algorithms programming dynamic programming etc. INTEGRATION TESTING is a level of software testing where individual units / components are combined and tested as a group. CHAPTER 5: DYNAMIC PROGRAMMING Overview This chapter discusses dynamic programming, a method to solve optimization problems that in-volve a dynamical process. Dynamic Programming Practice Problems Dynamic Programming - Summary Optimal substructure: optimal solution to a problem uses optimal solutions to Dynamic Programming - Stanford University Dynamic Programming is an algorithmic paradigm that solves a given complex problem by breaking. bangla, dynamic programming knapsack, dynamic programming problems, dynamic programming hotel problem, dynamic programming bangla tutorial, dynamic programming, dynamic programming algorithms, dynamic programming algorithm tutorial. What problems does he have in his hotel??? The Video Lesson Player has problems in Internet Explorer. Invented by American mathematician Richard Bellman in the 1950s to solve optimization problems. Programming languages- In a dynamic language, such as Perl or LISP, a developer can create variables without specifying their type. def canConstruct(target,workbank,memo={}): if (target in memo): return(memo[target]) if len(target)==0. Here are just some of the benefits of using a breadcrumb trail. Dynamic programming is both a mathematical optimization method and a computer programming method. Oxford Press - KC's Problems and Solutions for Microelectronic Circuits - 4th Edition - {RedDragon}. By applying the principle of the dynamic programming the first order condi-tions of this problem are given by the HJB equation V(xt) = max u {f(ut,xt)+βEt[V(g(ut,xt,ωt+1))]} where Et[V(g(ut,xt,ωt+1))] = E[V(g(ut,xt,ωt+1))|Ft]. If sub-problems can be nested recursively inside larger problems, so that dynamic programming methods are applicable, then there is a relation between the value of the larger problem and the values of the sub-problems. We released a 5-hour course on Dynamic Programming on the freeCodeCamp. Marlin is a huge C++ program composed of many files, but here we'll only be talking about the two files that contain all of Marlin's compile-time configuration options: Configuration. This problem can be formulated in dynamic programming terms, and resolved computationally for up to 17 cities, and combined with What path minimizes the to ta l distance travelled by the salesman?" The problem has been treated by a number of different people using a var ie ty of techniques; el. dynamic programming. Before solving the in-hand sub-problem, dynamic algorithm will try to examine the results of the. 3 Why Is Dynamic Programming Any Good? 4 Examples The Knapsack Problem The Monty Hall Problem Pricing Financial Securities. Dynamic programming problem. In theory, every problem can be solved dynamically. 2 Formulating problems as linear programs 859 29. Using dynamic programming to speed up the traveling salesman problem! A large part of what makes computer science hard is that it can be hard to know where to start when it comes to solving a difficult, seemingly unsurmountable problem. Adding your google account to your Android phone is a great way to get emails link various different apps to your Google account. Medium Problem Solving (Advanced) Max Score: 40 Success Rate: 43. Dynamic Programming Patterns. Check out the most common problems and the solutions here. Since the introduction of pacman hooks, the rebuild of the modules is handled automatically when a kernel is upgraded. The method was developed by Richard Bellman in the 1950s and has found applications in. In this course, you'll start by learning the basics of recursion and work your way to more advanced DP concepts like Bottom-Up optimization. Pricing and data limits. RECEPTION:Iam sorry, sir. This post will discuss a dynamic programming solution for the activity selection problem, which is nothing but a variation of the Longest Increasing Subsequence (LIS) problem. • Operations that allow lockless concurrent programming • Atomic operations are indivisible • Are • Improved modularity. Jun 5th, 2020. Solve Challenge. Programming. Dynamic Programming (commonly referred to as DP) is an algorithmic technique for solving a problem by recursively breaking it down into simpler subproblems and using the fact that the optimal solution to the overall problem depends upon the optimal solution to it's individual subproblems. Dynamic programming is super important in computationally expensive programming. 29 Linear Programming 843 29. – Three stopovers on the way • a number of choices of towns for each stop, • a number of hotels to choose from in each city. However, geolocation mobile apps with a simple dynamic map can use the geolocation service for free. On the other hand, the Longest Path problem doesn't have the Optimal Substructure property. The analysis code may be The program analysis discussed in this dissertation is almost exclusively dynamic binary analysis (DBA) Nonetheless, in recent years these problems have been largely overcome by the advent of. Dynamically linked shared libraries are an important aspect of GNU/Linux. The first line contains an integer n. Meghnad Saha Institute of Technology. 3) Recursive solution. Nest is a framework for building efficient, scalable Node. zm+n, we say that Z is an interleaving of X and Y if it can be obtained by interleaving the bits in X and Y in a way that maintains the left-to-right order of the bits in X and Y. Typically, the first program beginners write is a program called "Hello World", which simply prints "Hello World" to your computer screen. Dynamic-Programming-Solutions's Introduction. There are n hotels given at a 0, a 1, , a n such that 0 < a 0 < a 1 < < a n. Dynamic Programming Algorithm. - 293 pages. Resource for tech interview preparation. Recall that central entity of DP algorithm is the value function Main idea of DP efciency: avoid unnecessary repetition of computation. This can help build morale and improve productivity. Synchronous Dynamic Programming Algorithms. We make life easier with reliable products, proactive communication, and an expansive on-the-ground service network. Bottom-up management styles allow for the full talents of employees to be used. Problem Prediction. 2021 misof: Brute Force, Dynamic Programming, Simple Search, Iteration 1 82. yn and Z = z1. ROB: There is a problem with the air conditioning. A traveler needs to visit all the cities from a list, where distances between all the cities are known and each city should be visited just once. Dynamic planning Status: F(n) State recurrence: F(n)=F(n-1)+F(n-2) Initial value: F(1)=F(2)=1 Return result: F(N). A = [3 2; -2 1]; sz = size (A); X = randi (10,sz) X = 2×2 9 2 10 10. Doesn't matter what kind of document it is (the example provided in the attached image will be. The heuristic restricts the size of the state space of a dynamic programming algorithm. Dynamic-Programming-Travelling-Salesman-Problem's Introduction. March 2019. Also, each of the sub-problems. There’s a reason TKE is a global leader in mobility products and services. This question is often a source of anxiety to interviewees because of the complexity of the solution and the number of variants of the problem. by Tibor Szkaliczki. View Dynamic programming Research Papers on Academia. new is followed by a data type specifier and, if a sequence of more than one element is required, the number. Optimal Substructure: If any problem's overall optimal solution can be constructed from the optimal solutions of its subproblem, then this problem has an optimal substructure. Dynamic programming (DP) is breaking down an optimisation problem into smaller sub-problems, and storing the solution to each sub-problems so that each sub-problem is How to Identify Dynamic Programming Problems. elf2flash, bin2flash, and sof2flash. Section 2 presents the dataset and its stylized facts underpinning the model. At that time, "programming" meant "planning, optimising". Basically, dynamic programming is an optimization over the normal recursion. In a typical large hotel, the number of rate classes is about 25, there are 7 different lengths-of-stay, and the planning horizon is about 365 days. What are the dynamics of global economic growth in 2021. Traveling Salesman Problem Dynamic Programming Held-Karp. To know more about Java, Checkout this blog post. The model allows changes to be introduced even after the iteration's launch if the Such flexibility significantly complicates the delivery of quality software. Hotel Reservation Database Setup. Hi all, might be a silly question but I'm having a problem with the 'note' functionality on BC. Similarly, Tile[N-1] = Tile[N-2] + Tile[N-3]. Solving these high-dimensional dynamic programming problems is exceedingly di cult due to the well-known \curse of dimensionality" (Bellman,1958, p. In interviews or contests, problems on string are really common and one has to be strong on this. Dynamic-programming. A dynamic program for the management of various facets of hotels of all sizes through a secure web interface. helping people use/understand technology; I. Dynamic Programming approach O(N^2) time. Let all problems stay in 2021, and in 2022 make only right and not painful decisions, and cryptocurrency trading will bring only profitable pleasure) Today we offer you a fractal from 2018 , which has been working quite successfully for several weeks in the current market. These methods can help you ace programming interview questions about data structures and algorithms. Dynamic programming approach: The dynamic programming approach considers the previous steps solution, saves them, and then uses them in the future so as to avoid solving the same problem again and again. More specifically, Dynamic Programming is a technique used to avoid computing multiple times the same subproblem in a recursive algorithm. Dynamic Programming is the most asked question in coding interviews due to three main reasons: It is hard to solve; Difficult to find the pattern and the right approach to solve the problem. Benefits Of Using Breadcrumbs. The idea is very simple, If you have solved a problem with the given input, then save the result for future reference, so as to avoid solving the same problem again. Control-Limited Differential Dynamic Programming. Harish GargMore Examples of Dynamic Programing: https://youtu. It is possible that the feedback I got for DP was that it's h. Lecture Notes on Dynamic Programming Economics 200E, Professor Bergin, Spring 1998 Adapted from lecture notes of Kevin Salyer and from Stokey, Lucas and Prescott (1989) Outline 1) A Typical Problem 2) A Deterministic Finite Horizon Problem 2. 2021 misof: Dynamic Programming, Math 3. Coevolution and collective behavior. Dynamic programming is something every developer should have in their toolkit. That is, a relation whereby I can cut away from the problem size with each This question was asked to me in an interview and it embarrassingly exposed my shortcomings on dynamic programming. Dynamic Programming solves problems in a similar way to divide and conquer technique by combining the solutions of sub problems however Bottom up approach is the one used in Dynamic Programming. Path-based breadcrumbs are dynamic in that they display the pages the user has visited before arriving on the current page. This new token is sometimes called the blank token. This notebook will help you understand: Policy Evaluation and Policy Improvement. OOPSLA was the incubator for CRC cards, CLOS, design patterns, Self, the agile methodologies, service-oriented architectures, wikis, Unified Modeling Language (UML), test driven design (TDD), refactoring, Java, dynamic compilation, and aspect-oriented programming, to name just some of them. •The Floyd-Warshallalgorithm, runs in O(V3). In a dynamical process, we make. Tsitsiklis. This language is known to be the most widely used programming platform that offers building elements for other languages like C++, Python, Java and others. 3) Recursive solution. Pinpoint root causes of application problems in real time, from 3rd party APIs down to code level issues, so your IT teams can quickly identify what's most affecting your key business metrics. To make problem more complex soultion requires backtracking or recursion, after that dp can be applied to optimize further. String and Dp link. One of the reasons that some things can seem so tricky is that. At each stage, it ranks decisions based on the sum of the present cost and the expected future cost, assuming optimal decision making for There is a very broad variety of practical problems that can be treated by dynamic programming. It is a common pattern to combine the previous two lines of code into a single line: X = randi (10,size (A));. Watch the video an do the activities. What is dynamic programming and why should you care about it? In this article, I will introduce the concept of dynamic programming, developed by Richard Bellman in the 1950s, a powerful algorithm design technique to solve problems by breaking them down into smaller problems, storing their solutions, and combining these to get to the solution of the original problem. of dynamic programming. As the number of states in the dynamic programming problem grows linearly, the computational burden grows exponen-tially, both in terms of the number of. The dynamic programming method which minimizes the number of stations for a given cycle time is extended here to solve two variants of the assembly-line balancing problem. nios2-flash-programmer 4. If ij, compare. Bağlantıyı al. There exists an $$O(n^2)$$-time algorithm for the GMMN problem when the intersection graph is restricted to a star. might prove to be helpful to you if you. I am trying to formulate the following dynamic programming model: Along the shortest route, I have gas stations and hotels. List one of the sequences across the top and the other down the left, as shown. Bertsekas and John N. In any case, the mathematician sees hundreds and thousands of formidable new problems in dozens of blossoming areas, puzzles. In 2018, the Rust community decided to improve programming experience for a few distinct domains (see the 2018 roadmap). Chiba, Dynamic programming algorithm optimization for spoken word recognition, IEEE Trans. In some cases, like Production Debugging, logs might be the only information you have. MapKit is free beyond your Apple Developer Program membership and doesn't have limitations on the number of API requests/day. Dynamic Programming (DP) Algorithms Culture. Dynamic Programming is a programming technique that combines the accuracy of complete search along with the efficiency of greedy algorithms. Now you would like to evenly split all the souvenirs that all three of you bought. Check out these top Java Project Ideas to start your Java Programming Journey and Boost up your career with these beginner-level java projects. Dynamic programming is both a mathematical optimization method and a computer programming method. Lu and Romanowski considered a dynamic job shop problem in which job shops are disrupted by unforeseen events such as job arrivals and machine breakdowns. The four applications of Dynamic Programming solves every sub-problem just once and stores the result into a table so that it can be easily recovered if we need it again. Not a member of Pastebin yet? Sign Up, it unlocks many cool features. GDP in 2016. In this research analysis, an attempt was made to evaluate the relevance of dynamic. The dynamic programming technique captures this tradeoff. browser plug-in) using the Compile program for use with a shared library: gcc -Wall -L/opt/lib prog. View job description, responsibilities and qualifications. Thus we will use Dynamic Programming to solve the problem. See if you qualify!. MONK's Problems. Then, we extend it to the general tree case based on a dynamic programming (DP) approach inspired by and improving Schnizler's algorithm. People chose new year’s resolutions like more travel, more savings, or learning new skills. max xt,yt ÕT t 0 f(xt, yt,t) (1) s. In the conventional method, a DP problem is decomposed into simpler subproblems char-. In contrast, the dynamic programming solution to this problem runs in Θ(mn) time, where m and n are the lengths of the two sequences. Here it is You must stop at the final hotel (at distance an), which is your destination. 20; the second slice shows a couple of important examples (MRI image reconstruction and molecular simulation and visualization, chapters 11 and 12); the 3rd. Dynamic programming amounts to breaking down an optimization problem into simpler sub-problems, and storing the solution to each sub-problem so that each sub-problem is only solved once. Easy 1-Click Apply (DOLPHIN HOTEL MANAGEMENT - THE WESTIN RANCHO MIRAGE GOLF RESORT & SPA) Executive Sous Chef job in Rancho Mirage, CA. Dynamic programming (DP) is an algorithmic approach for investigating an optimization problem by splitting into several simpler subproblems. Dynamic programming. It uses progressive JavaScript, is built with TypeScript and combines elements of OOP (Object Oriented Progamming), FP (Functional Programming), and FRP (Functional Reactive Programming). Handwriting recognition from images or sequences of pen To get around these problems, CTC introduces a new token to the set of allowed outputs. 16 Greedy Algorithms. Tailwind CSS has alleviated so many problems we've all become accustomed to with traditional CSS that it makes you wonder how you ever developed websites without it. 3 The simplex algorithm 864 29. – Each trip has a different distance resulting in a different cost (petrol). Pivot Animator v5. Instead of brute-force using dynamic programming approach, the solution can be obtained in lesser time, though there is no polynomial. You must stop at the final hotel (at distance a n ), which is. Ariadoss PMS easy to use PHP/MySQL project management system/hotel management software/booking system/central reservation system (CRS). Abstract The general clustering algorithms do not guarantee optimality because of the hardness of the problem. In academic terms, this is called optimal substructure. This project forked from kjsce-codecell/Dynamic-Programming-Solutions. Dynamic Programming: Example Dynamic Programming Problems. It looks like you can solve this problem with dynamic programming. Since dynamic programming is a numerical algorithm used here to solve a continuous control problem, the continuous-time model (2) must be Figure 1 Schematic overview of an optimal control problem solved using the dynamic programming algorithm. In both contexts it refers to simplifying a complicated problem by breaking it down into simpler sub-problems in a recursive manner. What are the top programming languages you should know in 2021? Check our article for the latest programming languages trends. As we entered into a brand new year and a new decade, many people got excited about the possibilities 2020 promised. 800+ problems for practice. As the chess program paradigm suggests, intuition about the problem, heuristics, and trial and error are all important ingredients for constructing cost-to-go approximations in DP, However. You can "program" it by showing it just a few examples of what you'd like it to do; its success generally varies depending on how complex the task is. Dynamic programming is an optimization method which was developed by Richard Bellman in 1950. Knapsack Problem: Dynamic Programming Example The knapsack problem is a problem in combinatorial optimization: Given a set of items, each with a weight and a value, determine the number of each item to include in a collection so that the total weight is less than or equal to a given limit and. I am stating to solve some dynamical programming problem, I came across the problem to solve if a string can be constructed from a list of string. Dynamic Programming Defined. Whom can I contact if I have some problems related to my online registration of appointment?. The subproblems are optimized to optimize the overall solution is known as optimal substructure property. Java Programming Masterclass udemy course can be your great first stepping stone. Employees are more open to work and strive harder to reach goals and objectives in the ways that work best for them. To solve any problem with DP problem you can follow this steps. might prove to be helpful to you if you are new to the paradigm of programming - GitHub - adi2809/dynamic_programming_in_python: this is a collection of tutorials for dynamic programming problems that I developed while revising the concepts. Dynamic programming is used where we have problems, which can be divided into similar sub-problems, so that their results can be re-used. This conversation has the essence of dynamic programming. Porter's Generic Strategies with examples Strategy Organizational Strategy 11 Amazing Sales Strategy Examples That Get Results Dynamic programming is both a mathematical optimization method and a computer programming method. 26% details: MonotoneStrings SRM 811 08. Best Practice #9: Software does not equal talent management. Please consider using another browser. Dynamic programming (DP) is a standard tool in solving dynamic optimization problems due to the simple yet flexible recursive feature embodied in Bellman’s equation [Bellman, 1957]. Applications. The idea behind using dynamic programming is that we store the results of sub-problems so that we do not need to re. Mostly, these algorithms are used for optimization. A list of hotels is given together with their distance from start and a rating(0-5). To mitigate the problem, XP requires the use of pair programming, test-driven. Barto: Reinforcement Learning: An Introduction 13 Bellman Optimality Equation for q The relevant backup diagram: is the unique solution of this system of nonlinear equations. Even though the problems all use the same technique, they look completely different. 1 - What is Dynamic Programming. Answer (1 of 5): Advantages 1. The L key can be used to align a handle to its position in the previous frame. And they can improve your day-to-day coding as well. I View a problem as consisting of subproblems: I Aim: Solve main problem I To achieve that aim, you need to solve some subproblems I To achieve the solution to these subproblems, you need to solve a set. This is in contrast to our previous discussions on LP, QP, IP, and NLP, where the optimal design is established in a static situation. ISTQB Definition integration testing: Testing performed… Read More »Integration Testing. However, a lot of users are seeing the "there was a problem communicating with google servers" error when trying to add their Google account to their phones. 1 Standard and slack forms 850 29. A problem that can be solved optimally by breaking it into subproblems and then recursively finding the optimal solutions to the subproblems is said to have an. 11 Dynamic Programming problems. Using Dynamic Programming requires that the problem can be divided into overlapping similar sub-problems. A problem is a dynamic programming problem if it satisfy two conditions: 1) The problem can be divided into subproblems, and its optimal solution can be constructed from optimal solutions of the subproblems. It can be analogous to divide-and-conquer method, where problem is partitioned into disjoint subproblems, subproblems are recursively solved and then combined to find the solution of the original problem. 1) Finding necessary conditions 2. Dynamic Programming Solution The problem can be solved using dynamic programming when the sum of the elements is not too big. Ken Burns is a type of panning and zooming effect commonly used in video production to bring still images to life. Title Description: We all know the Fibonacci sequence. 0 ThemeLuviana Hotel Booking theme works seamlessly within WordPress 5. Classification, Regression. A branch of mathematics studying the theory and the methods of solution of multi-step problems of optimal control. The gure shows the state variable. So if we recompile with debugging symbols, we get Valgrind also detects improperly chosen methods of freeing memory. We present the dynamic programming formulation applied to a canonical problem, Bernoulli ranking and selection. pdf - \u2022 Combinatorial optimization is a topic that consists of finding an optimal object from a finite set of objects. e have to make d-1 stops in between). Also entertainment, business, science, technology and health news. The dynamic programming concept mixed with Mel and spectrum coefficients helped us in finding the potential of the algorithm in speech recognition systems. > Introduction to C++ Programming in UE4. The best way to learn a programming language is by writing programs. Here is the collection of the Top 50 list of frequently asked interviews question on Dynamic Programming. Dynamic programming is a method for solving complex problems by breaking them down into sub-problems. Click on this message to dismiss it. It is a very general technique for solving optimization problems. Dynamic programming also divides a problem into several subproblems. The Hotel Management System contains the admin and user section. Thus the problem shows optimal substructure. The hotel reservation application stores the reservation data in an SQL Server (2017+) database. dp(i-1,j-1) + (a(i) - a(i-1))^2; dp(i-2,j-1) + (a(i) - a(i-2))^2 as long as i-n>=j-1. 0+ and is built from the. The main features of the. To the best of our knowledge, few studies have tried to do that so far, due to the particularity features of this kind of problem. The list of problems in each category of Dynamic. PDF Drive investigated dozens of problems and listed the biggest global issues facing the world today. Dynamic programming assumes full knowledge of the MDP It is used for planning in an MDP For prediction: Input: MDP S, A, P, R, γ and policy π or: MRP S, Pπ, Rπ, γ. Elements of Dynamic Programming The problem has the following properties: Optimal substructure: A solution is optimal only if its subsolution to the subproblem is optimal. At the early planning and scoping stage, project managers and analysts diagnose the issue or problem to be addressed. The recursive formula is as follows: d(0) = 0 where 0 is the starting position. I \it’s impossible to use dynamic in a pejorative sense". Dynamic programming. Practice this problem. Dynamic Programming. The only places you are allowed to stop are at these hotels, but you can choose which of the hotels you stop at. 2) The subproblems from 1) overlap. This marks the second year Dynamic Yield has earned a place on the career platform for technology professionals' esteemed list for NYC. Dynamic Programming - Hotel Problem. Dynamic programming is a technique that breaks the problems into sub-problems, and saves the result for future purposes so that we do not need to compute the result again. Dynamic programming approach consists of three steps for solving a problem that is as follows:. (Find out how good a policy π is) Solution: Iterative application of Bellman Expectation backup. Technology has made on-demand food ordering possible and the industry is pivoting towards more innovative approaches to meet and exceed customer expectations. The tank capacity is U and is Do you understand DP at all? Can you help us to help you in any way beyond you just repeating the problem for us? Can you "break it down into a. History of Dynamic Programming I Bellman pioneered the systematic study of dynamic programming in the 1950s. • Linear Program (LP) is an optimization problem where. In addition to the topics. It is also a. Here, by Longest Path, we mean In general, Dynamic programming (DP) is an algorithm design technique that follows the Principle of Optimality. Get hands on knowledge of examples and applications of linear programming used in data science. Other related documents. On these cases, programs need to dynamically allocate memory, for which the C++ language integrates the operators new and delete. In this tutorial, we'll discuss a dynamic approach for solving TSP. Dynamic Programming Approach. Problem: Evaluate a given policy π and MDP. 18/10/2021. AppDynamics Covid Assist Program. In the Fall of last year, Dynamic Yield was also recognized by Crain's New York Business as one of the Best Places to Work in NYC for 2021. Dynamic programming approach offers an exact solution to solving complex reservoir operational problems. Dynamic Programming - 2 / 33 Dynamic programming is a general powerful optimisation technique. We have tens of thousands of applicants for this program already and are currently prioritizing applications focused on fairness and representation. Each thread asks the OpenMP runtime There are a wide array of concurrency and mutual exclusion problems related to multithreading programs. Dynamic Programming is an algorithmic technique for solving an optimization problem by breaking it down into simpler subproblems and utilizing the fact that the optimal solution to the overall problem depends upon the optimal solution to its subproblems. Travelling Salesperson Problems. Dynamic Programming Binomial Coefficients. In any dynamic programming coding interview you take, you'll likely encounter the knapsack problem. Kamalesh Karmakar, Assistant Professor, Dept. Dynamic programming oers a coherent framework for understanding and solving Bayesian formulations of these problems. price discrimination, semi-parametric estimation, dynamic programming, hotel revenue management. this is a collection of tutorials for dynamic programming problems that I developed while revising the concepts. Sutton and A. 17 Amortized Analysis. There are various types of Dynamic Programming Problems and different approaches to all those types. Structure of a program. Since the recursive method breaks everything down to ones in the end, it's way better to store the result for fib (5) than recalculate it as. It provides a ready-to-use UI, all while not compromising on enabling. The English name of dynamic programming is Dynamic Programming, which is a mathematical idea for solving Hotel reservation problem (dynamic planning) Title description If you want to travel, you have a budget of N yuan to stay in a hotel, and there are M hotels for you to choose from. You must stop at the final hotel (at distance an), which is your destination. Problem solving ideas: 1. For a rod of length n, since we make n-1 cuts, there are 2^(n-1) ways to cut the rod. 6 Dynamic Programming Algorithms We introduced dynamic programming in chapter 2 with the Rocks prob-lem. It is noted that the overall problem depends on the optimal solution to its subproblems. Or when you have a project. This is the basic approach behind dynamic programming - all problems. Watch Netflix movies & TV shows online or stream right to your smart TV, game console, PC, Mac, mobile, tablet and more. 3 The simplex algorithm 864 We revised our treatment of dynamic programming and greedy algorithms. programming hotels along a highway, java, dynamic-programming, knapsack-problem. The approach allows planners to identify a large number of solutions all of. Dynamic programming solves problems by combining the solutions to subproblems. 24 5 Dynamic programming The departure point of dynamic programming (DP) method is the idea of embedding a given optimal control problem (OCP) into a family of OCP indexed by the initial data (t0 , x0 ). • The goal is to select a route to and a hotel in Perth so that the overall cost of the trip is minimized. The second line contains integers v 1 , v. Delegates that can be serialized and rely on reflection (instead of function pointers). Approximate dynamic programming (ADP) is both a modeling and algorithmic framework for solving stochastic optimization problems. Let's Change The World Together. They allow executables to dynamically access external functionality at run time and thereby reduce their overall memory footprint (by bringing functionality in when it's needed). Dy-namic programming now leads off with a more interesting problem, rod. The Principal of Optimality says. Linear programming is a simple optimization technique. In a report titled Applied Dynamic Programming he. Solved an existing financial problem of a company using the general dynamic programming model developed by Richard Bellman and their method does not apply the table-like technique for better understanding. Blog about Java, Programming, Spring, Hibernate, Interview Questions, Books and Online Course Recommendations from Udemy, Pluralsight, Coursera These are the building blocks of any program, and a good knowledge of Algorithms and Data Structure is vital for your next job or doing well in your. Another approach is the integer programming approach. The system development life cycle (SDLC) is an essential model of project management. 500 million+ members | Manage your professional identity. Logging has a crucial part to play in a scenario where you can't use interactive debugging (that is, attaching a debugger like Visual Studio). His concern was not only analytical solution existence but also In RAND Corporation Richard Bellman was facing various kinds of multistage decision problems. Athena Scientific, 1995. Job Scheduling) •Optimization problems - typically find the math, recursive function Application of Dynamic Programming to solving Dynamic Programming - Counting problems Friday, November 12, 2021 10:11 AM. Many businesses planned to start the new decade on the right foot. makispaiktis. For this problem, you should be using a two dimensional table where dp(i,j) is the minimum cost such that hotel i is reached in exactly j days. Problems in this Article are divided into three Levels so that readers can practice according to the difficulty level step by step. The connection string is defined in web. It allows us to investigate errors after the problem already happened. 5 The initial basic feasible solution 886. The extended method minimizes cycle time for a given number of stations or, more generally, computes all nondominated pairs of cycle time and number of stations. It allows you to optimize your algorithm with respect to time and space — a very important concept in real-world applications. I tried solving various types of problems on Competitive Programming Platforms as well as Cormen, but nothing seemed to work. As the traveling salesman problem defines it, a company has a salesman that must visit n number of Some algorithms exist using dynamic computing that can theoretically reduce the number of checks Microsoft even has a programming language for DNA computing that can help make DNA computing. 29 Linear Programming 843 29. Fill in the blanks with the correct words from word bank. Josephus Problem I2915 / 3363. Analysis of efficiency. These algorithms work by remembering the results of the past run and using them to find new results. For instance, in C++ there are three basic options for freeing dynamic memory: free. Dynamic programming DAA 2020-21 4. A new heuristic algorithm is proposed for the P-median problem. This is the List of 100+ Dynamic Programming (DP) Problems along with different types of DP problems such as Mathematical DP, Combination DP, String DP, Tree DP, Standard DP and Advanced DP optimizations. Our customers, and their customers, are at the center of every elevator, escalator, and moving walk installation, modernization, and support call. Gridworld City. Bookmark this page and practice each problem. Removing the redundancy can optimize such a program, which is done by performing the repeated operations just once and storing the result for later use. I The Secretary of Defense at that time was hostile to mathematical research. Most of us learn by looking for patterns among different problems. In the case of recursion, repeated calls are made for the same sub-problem, but we can optimize this problem with the help of dynamic programming. For these, you can find many high-quality crates and some awesome guides on how to get started. There is a problem I am working on for a programming course and I am having trouble developing an algorithm to suit the problem. PRACTICAL ENGLISH Hotel Problems. The two different alignment methods are mostly defined by Dynamic programming approach for aligning two different sequences. Dynamic Programming Practice Problems : Strings and Interleaving For bit strings X = x1. Dynamic programming simplifies a complicated problem by breaking it down into simpler subproblems in a recursive manner. the act of protecting information; 4. • Save both programming time AND lines of code. The Travelling Salesman Problem (TSP) is a very well known problem in theoretical computer science and operations research. A lower-level employee may have unique insight on how to solve a common problem. Dynamic-Programming Solution to the 0-1 Knapsack Problem Let i be the highest-numbered item in an optimal solution S for W pounds. Any recursive formula can be directly translated into recursive algorithms. It uses the bottom up approach. Problem Seeking, Fifth Edition lays out a five-step procedure that teams can follow when programming any building or series of buildings, from a small house to a hospital complex. Bellman Equation. Dynamic programming Dynamic Programming is a general algorithm design technique for solving problems defined by or formulated as recurrences with overlapping sub instances. Tree and Graph are maily based of DFS and BFS. 07% details: SpireAttack SRM 812 09. I Bellman sought an impressive name to avoid confrontation. •Adifferent dynamic-programming formulation to solve the all-pairs shortest-paths problem on a directed graph G = (V, E). Dynamic Programming Approach I Dynamic Programming is an alternative search strategy that is faster than Exhaustive search, slower than Greedy search, but gives the optimal solution. ESL video lesson with an interactive quiz: Vocabulary booster. h contains the core settings for the hardware, language and controller selection, and settings for the most common features and. If an optimal solution can be created for a problem by constructing optimal solutions for its subproblems, the problem possesses ____________ property. The stagecoach problem Mythical fortune-seeker travels West by stagecoach to join the gold rush in the mid-1900s The origin and destination is fixed Many options in choice of route Slideshow 5591929 by urit. checking products for problems; F. Abbreviation. These two results permit a very compact computer implementation of a dynamic programming algorithm for solving one-machine sequencing problems with precedence constraints. 1 percent of the U. Build and engage with your professional network. yt+1 − yt g(yt,xt,t) h(xt, yt,t) ≤ 0 y0 given (2) Dynamic programming can also be used for continuous time problems. The Traveling Salesman Problem with Hotel Selection (TSPHS) is a variant of the classic Traveling Salesman Problem. Dynamic programming was the brainchild of an American Mathematician, Richard Bellman, who described the way of solving problems where you need to find the best decisions one after another. Posted by chieffan on Thu, 06 Jan 2022 00:09:22 +0100. 1 A Prototype Example for Dynamic Programming. The browser version you are using is not recommended for this site. Max Array Sum. Hotel pricing is a challenging high-dimensional problem since hotels must not only set prices for each current date, but they must also quote prices for a Keywords price discrimination, dynamic pricing, price-following strategies, Bertrand price compe-tition, dynamic programming, method of simulated. Find the maximum sum of elements in an array. Claiming a piece of software can provide a full talent management system is a bit like believing a food processor will produce a five-star meal. Jonathan Paulson explains Dynamic Programming in his amazing Quora answer here. This image slideshow adds an awesome Ken Burns effect to each image during transition, with the ability to show a corresponding description. To compute the LCS efficiently using dynamic programming, you start by constructing a table in which you build up partial results. You'd ideally like to travel 200 miles a day, but this may not be possible (depending. Tailwind solves a complex problem in an elegant way. In the dynamic schedule, there is no predictable order in which the loop items are assigned to different threads. The book can be separated roughly in 4 parts: the first, and more important, deals with parallel programming using Nvidia's CUDA technology: this takes about the first 10 chapters and Ch. Dynamic Programming, Math 2 62. I won't explain them here in detail; there are. The paradigm of dynamic programming:. Logic (R) Chapter 7 (A-B) - Propositional Logic. It depends on the setting up the subproblems in a way that the space for the results of all subproblems needed at one time is small enough to fit in the memory allocation for the problem. Download Ariadoss PMS for free. You could be storing up a problem for the company in the future - for example, by allocating shares to founders in a way that could lead to a stand-off if they refuse to see eye to eye on key issues. The idea is first to sort given activities in increasing order of their start time. Follow along and learn 12 Most Common Dynamic Programming Interview. Polynomial-time methods can nd the clustering corresponding to the exact optimum only in special cases. 15 Dynamic Programming. Understanding Dynamic Programming can help you solve complex programming problems faster. Classification. The best example is the recursive fibonacci calculation. 4-2 Problem Set-Chapters 7 & 8. Developer tools. ( Dynamic Programming, Complexity Theory ) | howtofix. Dynamic programming is an optimisation approach that breaks down complex problems into simpler, more tractable ones. Specific problems outside EPA’s usual activities can also arise, for example, through congressional action, requests for assistance from state or local governments, acts of nature, or terrorism. Integer programming is a mathematical optimization program in which some or all of the variables are restricted to be integers. c -lctest -o prog [Potential Pitfall] This also avoids the Microsoft "DLL hell" problem of conflicting libraries where a system upgrade which. The dynamic programming equation can be written in dierent forms by taking other vector elds which are parallel to our choices gb, gs. The FastTrack program is designed to help you accelerate your Dynamics 365 deployment with confidence. The question is then. Thanks to this, your guests will be free to check the availability of your hotel rooms online and book them in real-time. Before we study how to think Dynamically for a problem. WordPress 5. Hotel Booking/Central Reservation System. The main advantage is in the efficiency with which a dynamic programming language provides. By offering a. ıt is not working, and it is very hot in my room. C is another general-purpose and imperative programming language which was developed way back in the 70s and is similar to C++ language. The algorithm and program description which realize this approach are cited. Medium Problem Solving (Intermediate) Max Score: 20 Success Rate: 78. Most of the literature has focused on the problem of approximating V(s) to overcome the problem of multidimensional state variables. Welcome to Assignment 2. Which of them will become a thing of the past? What will be the best choice for building a web project in 2021? To answer all these questions, we will provide you. org YouTube channel. Help center. CS330 Duke University. La programación dinámica es un enfoque de optimización que descompone problemas complejos en otros más sencillos y manejables. Although it is very simple, it contains all the fundamental components C++ programs have:. Hotel Management System in PHP with Full Source Code (2020) This Hotel Management System is developed using PHP, Bootstrap, Javascript and CSS. Miscellaneous Coding Patterns Problems. programs/instructions added to computer; H. ABSTRACT In this paper, an inventory control policy and routing strategy are jointly considered by developing dynamic programming model in vendor managed inventory system. To begin with consider a discrete time version of a generic optimal control problem. The crisis has not affected the number of applicants to the Vanuatu program for two reasons: the speed of processing and the relatively low cost. 4 Duality 879 29. Visit BBC News for up-to-the-minute news, breaking news, video, audio and feature stories. Programming. They converted the financial stock portfolio problem to a dynamic programming problem using some dynamic terminologies. Partner program. Dynamic Programming is just a fancy way to say 'remembering stuff to save time later'". Today, the systems of interest involve multi/many-core processors, embedded and distributed systems, and mobile and web applications. See full list on medium. We see it in many other places. Latest Software Download. js server-side applications. Developer tools. October 2020. This problem doesn't just turn up in speech recognition. Topic Stream/Tutorial 1: Dynamic Programming. 105 Pages·2009·809 KB·1,299 Downloads. Dynamic Programming formulation for hotel problem. Test drivers and test stubs are used to assist in Integration Testing. Clarification: A problem that can be solved using dynamic programming possesses overlapping subproblems as well as optimal substructure properties. But with dynamic programming, it can be really hard to actually find the similarities. Dynamic memory is allocated using operator new. Bellman Equations. In programming, Dynamic Programming is a powerful technique that allows one to solve different types of problems in time O (n 2) or O (n 3) for which a naive approach would take exponential time. This article addresses multi-level lot sizing and scheduling problem in capacitated, dynamic and deterministic cases of a job shop. The leading and most up-to-date textbook on the far-ranging algorithmic methododogy of Dynamic Programming, which can be used for optimal control, Markovian decision problems, planning and sequential decision making under uncertainty. In essence, dynamic programming is how to solve a complex programming problem by Some advantages of dynamic programming. The standard version of TSP is a hard problem to solve and belongs to the NP-Hard class. effectiveness and simplicity by showing how the dynamic programming technique can be applied to several different types of problems, including matrix-chain prod-ucts, telescope scheduling, game strategies, the above-mentioned longest common subsequence problem, and the 0-1 knapsack problem. Design Hotel Management System. Dynamic Programming. "The dynamic programming formulation baseline cannot solve realistically sized problems. Dynamic programming works by saving the results of subproblems so that we don't have to recalculate them when their solutions are needed. If it exists, the optimal control can take the form u∗. The food service industry is an economic staple generating billions of dollars in annual revenue and representing 2. In dynamic programming of controlled processes the objective is to find among all possible controls a control that gives the extremal (maximal or minimal) value of the objective function — some numerical. This lecture explains how to solve the problem using Dynamic programming. Create a matrix of uniformly distributed random integers between 1 and 10 with the same size as an existing array. The purpose of this level of testing is to expose faults in the interaction between integrated units. Richard Bellman. program language; J. Feasibility- In Dynamic Programming we make decision at each step considering current problem and solution to previously solved sub problem to calculate optimal solution. In video I discuss what is dynamic programming, how I approach its problems, what is bottom up/top down approach, what is memoisation. In addition to. Integer programming is NP-complete, so it is not surprising that the knapsack problem, which can be posed as an integer programming problem, is NP-hard as well. I'll send somebody up to look at it right now. dynamic programming hotel problem Two jobs compatible if they don't overlap. Dynamic Programming was invented by Richard Bellman, 1950. The problem is that we didn't compile using the -g option of gcc, which adds debugging symbols. Problem : ( Scroll to solution ). The subproblem is the following: d(i): The minimum penalty possible when travelling from the start to hotel i. config and it uses LocalDB interface to access the database file (daypilot. Our Programming Languages & Software Engineering (PL & SE) group works on both performance optimizations and correctness problems for such systems, with research that focuses on a wide range of issues, from providing new abstractions for expressing parallelism. This algorithm appears to be much more efficient than previous ones for certain one-machine sequencing problems. It arises from a number of real-life In this paper, we present a highly effective hybrid between dynamic programming and memetic algorithm for TSPHS. I will appreciate if someone. The main use of dynamic programming is to solve optimization problems. • Problem definition – Travelling from home in Sydney to a hotel in Perth. The task is to pick a knapsack like program to pick the best hotels no matter what, but i don't know how to check if the hotels are. Example: Fibonacci Series. If we store the result, we will not compute it twice and will reduce the time complexity significantly. Moreover, you are required to complete your journey in exactly d days (i. We propose the approach for solving of the mentioned problem which is based on the method of dynamic programming and multicriteria optimization by a nonlinear trade-off scheme. The database file is included in the project and you can find it in App_Data folder. Brute Force approach. Dynamic programming is all about ordering your computations in a way that avoids recalculating duplicate work. While the Rocks problem does not appear to be related to bioinfor-matics, the algorithm that we described is a computational twin of a popu-lar alignment algorithm for sequence comparison. Example 1 input: 4 1 2 3 4 Output: 1 2 3 6 problem-solving idea: the idea of dynamic programming is similar to that of changing coins, but the combination of changing coins does not consider the order. Introduction Dynamic programming deals with similar problems as optimal control. Bellman's dynamic programming was a successful attempt of such a paradigm shift. •Counting problems - typically find the math, recursive function In some rare cases there is more data to maintain and use (e. Then S ` = S - { i } is an optimal solution for W - w i pounds and the value to the solution S is V i plus the value of the subproblem. With Extreme Programming (XP), a typical iteration lasts 1-2 weeks. They still are asking some problems which DP can be a optimal solution. solution of a set of linear inequalities. This simple. The Co-WIN system will help you book an appointment in a Vaccination center where the same vaccine is being administered as the vaccine type (COVAXIN, COVISHIELD or SPUTNIK V) of the 1st dose. The theme is powered by MotoPress Hotel Booking plugin - an all-in-one reservation system for your WordPress site. shortly ‘Remember your Past’. Please consider upgrading to the latest version of your browser by clicking one of the following links. The method was developed by Richard Bellman in the 1950s and has found applications in numerous fields, from aerospace engineering to economics. Main idea: - set up a recurrence relating a solution to a larger instance to. Convenient for users Breadcrumbs are used primarily to give users a secondary means of navigating a website. To be honest, this definition may not make total sense until you see an example of a sub-problem. N euro- Dynamic Programming. Introductory guide for C++ programmers new to Here is a cool feature of Unreal that you might be surprised about if you are used to programming C++ in Actor iterators do not have the problem noted above, and will only return objects being used by the. The term "dynamic programming" was coined by Bellman in the 1950s. The path to the origin node is shown when selected segment highlighting is enabled in the figure builder. Recursion 2. Steps that are take in dynamic programming approach are: 1) Characterize the structure of an optimal solution. Dynamic Programming solutions are faster than the exponential brute method and can be easily proved for their correctness. Now it is. • Especially useful in plugins. edu for free. In that context the problem is transcribed into a generic sequential quadratic programming (SQP) which easily admits both equality and inequality constraints. Similarly, when you eventually recruit new senior team members, think carefully about what to offer them. This means that a user does not have to wait for a company, project, or package maintainer to release a new version of the module. To understand why need can use Dynamic Programming to improve over our previous appraoch, go through the following two fundamental points. This creates more flexible programs and can simplify prototyping and some object-oriented coding. dynamic analysis must instrument1 the client program with analysis code. Learn Computer Tips, Fix PC Issues, tutorials and performance tricks to solve problems. Access knowledge, insights and opportunities. Planning is one of the core phases of SDLC, that covers nearly all the upcoming steps the developers should complete for a successful project launch. BBC News provides trusted World and UK news as well as local and regional perspectives. DYNAMIC PROGRAMMING 2. The approach may be viewed as an extension of the myopic or greedy adding algorithm for the P-median model. Writes down "1+1+1+1+1+1+1+1 =" on a sheet of paper. Dynamic programming used to solve problem in polynomial time which is more faster then brute force method. 9 beta Now available. Assignment 2: Optimal Policies with Dynamic Programming. → the goal is to maximize or minimize a linear objective function → over a set of feasible solutions - i. Dp with Tree and Graph link. 0/1 Knapsack is perhaps the most popular problem under Dynamic Programming. Homework Assignment #6 Dynamic Programming. xm, Y = y1. Dynamic Programming Solutions: Google Maps Analysis. Sakoe and S. However, sometimes the compiler will not implement program to systematically record the answers to subproblems in a table. Other videos @Dr. The running time is O(NW) for an unbounded knapsack problem with N items and knapsack of size W. W is not polynomial in the length of the input. Answer (1 of 50): I have been in this situation for over a year and came out of it just about 3 months ago. With most DP problems I try and find a kind of reduce-and-conquer relation. fib (4) + fib (3) = 5. Paulo Brito Dynamic Programming 2008 6 where 0 < β < 1. In other words, a dynamic programming algorithm solves complex problems by breaking them into multiple simple subproblems and then it solves each of. A key principle that dynamic programming is based on is that the optimal solution to a problem depends on the solutions to its sub-problems. A Dynamic programming algorithm is used when a problem requires the same task or calculation to be done repeatedly throughout the program. Investment programs give an opportunity to improve the economic situation faster. Dynamic programming is an optimization approach that transforms a complex problem into a sequence of simpler problems; its essential In order to introduce the dynamic-programming approach to solving multistage problems, in this section we analyze a simple example. The classic programming guide for architects and clients¿¿—fully updated and revised Architectural programming is a team effort that requires close cooperation between architects and their clients.
kmv arx rny ntf vsr nfb irw vwh fof ygx ryh fjd mfi die pzb jxy swu qxb qxo gzo | 2022-05-17T00:44:33 | {
"domain": "produktninja.de",
"url": "http://zclf.produktninja.de/e51m",
"openwebmath_score": 0.24513642489910126,
"openwebmath_perplexity": 1069.6046432631638,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9817357184418848,
"lm_q2_score": 0.8596637559030338,
"lm_q1q2_score": 0.843962615019914
} |
https://math.stackexchange.com/questions/1228079/counting-distinct-matrices | Counting Distinct matrices
How many distinct (if matrix $M$ is included in count, do not include $PM$ where $P$ is permutation matrix) $3\times 3$ matrices with entries in $\{0,1\}$ are there such that each row is non-zero, distinct and such that each matrix is of real rank $2$ or $3$?
I think answer for rank $2$ is $6$, for rank $3$ is $29$.
• Can you expound upon what permutational equivalence means? Something along the lines of: "matrices $A$ and $B$ are equivalent iff ..." – Sammy Black Apr 10 '15 at 5:34
• Specifically, are you permuting all $9$ entries arbitrarily or just the $3$ columns or ... ? – Sammy Black Apr 10 '15 at 5:36
• Clarified question. – Turbo Apr 10 '15 at 5:41
• Sorry, I'm still not quite sure what you're asking. When you say "do not count...", do you mean if two matrices are equivalent by permuting rows, then count the equivalence class just once? – Sammy Black Apr 10 '15 at 5:46
• "if two matrices are equivalent by permuting rows, then count the equivalence class just once" correct. – Turbo Apr 10 '15 at 6:00
In the following, I'm endowing the set $\{0, 1\}$ with the operations of addition and multiplication modulo $2$ (e.g. $1 + 1 = 0$), and this ring is usually denoted $\Bbb{Z}_2$ (or $\Bbb{F}_2$ if we're emphasizing the fact that it's a field.)
Rank $3$: These are full rank matrices, so the conditions that the rows are nonzero and distinct are guaranteed. If a matrix has $3$ distinct rows, then all $6 = 3!$ permutations of the rows will be distinct matrices. So counting all full rank $3 \times 3$ matrices with entries in $\{0, 1\}$ will overcount by a factor of $6$.
Here's how it goes:
• The only restriction on the first row (thinking of it as a vector in $\{0, 1\}^3$ is that it cannot be $\mathbf{0}$, so there are $2^3-1$ possibilities. Call this row $\mathbf{v}_1$.
• For the second row, it cannot be either of the $2$ multiples of $\mathbf{v}_1$, namely $\mathbf{0} = 0\mathbf{v}_1$ or $\mathbf{v}_1 = 1\mathbf{v}_1$. So there are $2^3-2$ possibilities. Call this row $\mathbf{v}_2$.
• For the third row, it cannot be any linear combination of $\mathbf{v}_1$ and $\mathbf{v}_2$. There are $2^2=4$ of these. So there are $2^3-4$ possibilities of this third row. Call it $\mathbf{v}_3$.
Hopefully, you see the pattern here (so you could solve this problem for $n \ne 3$ as well). There are
$(2^3-1)(2^3-2)(2^3-2^2) = 7 \cdot 6 \cdot 4 = 168$ distinct matrices,
and up to permutation of the rows, there are
$\dfrac{168}{6} = 28$ possibilities.
Rank $2$: In this case, since you are only counting matrices up to permutation of the rows and since the rows are distinct and nonzero, you might as well assume that the last row is the sum of the first two. (Do you see why?)
There are $2^2=4$ possible linear combinations of the first two rows that could product the last row: choose either coefficient $0$ or $1$ independently for each of $\mathbf{v}_1$ and $\mathbf{v}_2$. However, both $\mathbf{v}_3 = 1\mathbf{v}_1 + 0\mathbf{v}_2 = \mathbf{v}_1$ and $\mathbf{v}_3 = 0\mathbf{v}_1 + 1\mathbf{v}_2 = \mathbf{v}_2$ produce a row identical to a previous one and $\mathbf{v}_3 = 0\mathbf{v}_1 + 0\mathbf{v}_2 = \mathbf{0}$, which is also not allowed. Hence, the third row must be $\mathbf{v}_3 = 1\mathbf{v}_1 + 1\mathbf{v}_2$
The first row has $2^3 - 1$ possibilities, and the second row has $2^3 - 2$ possibilities. Now there are
$(2^3-1)(2^3-2)(1) = 7 \cdot 6 \cdot 1 = 42$ distinct matrices,
and up to permutation of the rows, there are
$\dfrac{42}{6} = 7$ possibilities.
• You are getting one more than me which is strange. – Turbo Apr 10 '15 at 13:44
• Could you list rank $2$ matrices? – Turbo Apr 10 '15 at 13:47
• [111;110;001], [111;101;010], [111;011;100], [110;100;010], [101;100;001], [011;010;001] what is 7th matrix? Your count is somewhere wrong. – Turbo Apr 10 '15 at 14:24
• You missed [101;110;011]. – Sammy Black Apr 10 '15 at 16:57
• dude i am over reals. – Turbo Apr 10 '15 at 17:27 | 2019-06-17T04:54:22 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1228079/counting-distinct-matrices",
"openwebmath_score": 0.8700433969497681,
"openwebmath_perplexity": 292.73793208486956,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9817357227168957,
"lm_q2_score": 0.8596637469145054,
"lm_q1q2_score": 0.8439626098706264
} |
https://math.stackexchange.com/questions/2543194/find-the-range-of-eccentricity-of-an-ellipse-such-that-the-distance-between-its | # Find the range of eccentricity of an ellipse such that the distance between its foci doesn't subtend any right angle on its circumference.
What is the range of eccentricity of ellipse such that its foci don't subtend any right angle on its circumference?
I thought that the eccentricity would definitely be more than $0$ and less than $\frac{1}{\sqrt2}$ The latter value is for an ellipse with $ae=b$, in which a right angle is subtended on an endpoint of the minor axis.
• That seems right: $(0,2^{-1/2})$. The maximum angle for a given ellipse occurs at the endpoints of the minor axis. – rogerl Nov 29 '17 at 21:12
• I would change the statement of the problem to say that the segment between its foci subtends a right angle from no point on the boundary. The distance between the foci is a number (as is the circumference), and numbers don't subtend angles. – John Hughes Dec 3 '17 at 13:14
Your answer is correct, except that $0$ should be included (there are no right angles subtended in a circle).
Here's a complete solution:
Using the parameterization $P=(a \cos\theta, b\sin\theta)$ for an origin-centered ellipse with major radius $a$ (in the $x$ direction) and minor radius $b$ (in the $y$ direction), consider the foci at points $F_{\pm}=(\pm c, 0)$, where $a^2 = b^2 + c^2$.
$\angle F_{+}PF_{-}$ will be a right angle if and only if
$$(F_{+}-P)\cdot(F_{-}-P) = 0 \tag{\star}$$
That is, \begin{align} 0 &= (c - a \cos\theta )(-c-a\cos\theta) + (0 - b \sin\theta)(0-b\sin\theta) \\[4pt] &= -c^2 + a^2 \cos^2\theta + b^2\sin^2\theta \\[4pt] &= -c^2+a^2\cos^2\theta + ( a^2-c^2)(1-\cos^2\theta) \\[4pt] &= a^2 - 2 c^2 + c^2 \cos^2\theta \tag{1} \end{align} Writing $c = ae$, where $e$ is the eccentricity, we can factor-out $a^2$ to get $$e^2\cos^2\theta = 2 e^2 - 1 \tag{2}$$ In order for $(2)$ to be solvable for $\theta$, we obviously must have $e\neq 0$ (so that $\theta$ appears in the equation at all); then, for non-zero $e$, since $0\leq \cos^2\theta \leq 1$, the solvability of $(2)$ requires $$0 \leq 2-\frac{1}{e^2}\leq 1 \quad\to\quad 2 \geq \frac{1}{e^2}\geq 1 \quad\to\quad \sqrt{\frac{1}{2}} \leq e \leq 1 \tag{3}$$
In other words, the equation is not solvable for $\theta$ ---that is, there are no subtended right angles--- for $e < {1\over\sqrt{2}}$ or $e > 1$ (although we dismiss the latter possibility, as such eccentricities belong to hyperbolas). Therefore, the desired range of eccentricities is
$$0 \leq e < \frac{1}{\sqrt{2}} \tag{\star\star}$$
• The answer doesn't match, do you have any alternate solution it? – Jasmine Dec 3 '17 at 13:07
• @Jasmine: The answer doesn't match what? – Blue Dec 3 '17 at 14:19
• @Blue The Given solution, possibly. This was probably a homework question, and OP couldn't find the answer. Now that she has got the answer, it turns out the gotten and the given do not match. – MalayTheDynamo Dec 3 '17 at 16:45
• No, the solution is not wrong. only difference is that 0 is not included in the range. And even it's true as 0 can't be eccentricity for an ellipse? – Jasmine Dec 3 '17 at 17:04
• @Jasmine: Just as a square is a rectangle with all sides equal, a circle is an ellipse with major and minor radii equal. From that "inclusive" point of view (which I generally believe is most-correct), $0$ is the eccentricity of an ellipse. That said, it is not unreasonable to exclude $0$ and restrict attention to non-circular ellipses, which the author of the problem may have done. (The exclusion seems unnecessary here, since the result (and the proof) is valid for circles, too.) – Blue Dec 3 '17 at 23:10 | 2020-04-04T00:26:46 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2543194/find-the-range-of-eccentricity-of-an-ellipse-such-that-the-distance-between-its",
"openwebmath_score": 0.968464195728302,
"openwebmath_perplexity": 444.1825223708262,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9817357184418847,
"lm_q2_score": 0.8596637505099168,
"lm_q1q2_score": 0.8439626097252984
} |
https://www.usna.edu/Users/cs/wcbrown/courses/S17SI335/lec/l04/lec.html | These notes and closely review Unit 1 section 9.
## Analyzing a simple iterative algorithm: findFirstMissing with binarySearch
Dr. Roche, in the Unit 1 notes, kindly analyzes our three search functions so painstakingly that he gives us exact expressions for the worst-case running time in terms of number of statements exectued. From those functions, we can deduce $O$, $\Theta$ and $\Omega$ expressions. However, this is most decidedly not the way we normally do things! If we know the exact expressions, there's not much reason to bother with $O$, $\Theta$ and $\Omega$. The usual case is that it's too difficult to get exact expressions, and we "settle" for getting information about asymptotic growth rates. So let's try analyzing findFirstMissing the usual way: we'll get what info we can about the worst-case running time function's growth rate, which will come in terms of $O$/$\Theta$/$\Omega$, without ever trying to get an exact expression for the function.
In class we determined that the worst-case running time function $T_w(n)$ is both $O(n \lg n)$ and $\Omega(n \lg n)$. I.e., it is $\Theta(n \lg n)$.
## What happens when we replace binarySearch with linearSearch?
Let's replace binarySearch with linearSearch in findFirstMissing. The interesting thing here, is that we can't just say the call to linearSearch takes time $\Theta(n)$. We would have to prove first that it's actually possible in the context of the findFirstMissing algorithm for every call to linearSearch to be "worst case" for linearSearch, despite the fact that each call is a different $B[i]$. In fact, if we don't allow duplicate entries in array $B$, it is not possible for more than two calls to linearSearch to take $n$ steps. If you look at an intermediate iteration of findFirstMissing's while loop (i.e. not the last), the time taken for search(A,0,n-1,B[j]) is strictly greater than the time taken search(A,0,n-1,B[t]) for any t < j. So we can't assume that search takes $\Omega(n)$ for every iteration. So what do we do?
The first thing we did was to punt, and put the call to search as taking $\Omega(1)$ time, which is correct but not very precise. That gave us $T_w(n) \in \Omega(n)$ as well as $T_w(n) \in O(n^2)$.
The second thing we did was to analyze a very specific input situation, get a lower bound for that case, and note that the "worst case" had to be at least that bad, so the lower bound was also a lower bound on the worst case. I suggested the input situation in which the first half of B is identical to the second half of A, and the index n/2 element of B is not in A. In this case, we go through the while-loop $\Omega(n/2)$ times, and the call to linear search takes $\Omega(n/2)$ time, which gives us $\Omega(n^2/4) = \Omega(n^2)$ as a lower bound. Since $T_n(n)$ is $O(n^2)$ and $\Omega(n^2)$, it is $\Theta(n^2)$.
## Improved findFirstMissing
How does linear vs. binary search vcompare now? | 2018-05-23T18:51:51 | {
"domain": "usna.edu",
"url": "https://www.usna.edu/Users/cs/wcbrown/courses/S17SI335/lec/l04/lec.html",
"openwebmath_score": 0.8604423403739929,
"openwebmath_perplexity": 363.0549264088114,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9817357195106374,
"lm_q2_score": 0.8596637433190939,
"lm_q1q2_score": 0.8439626035845785
} |
http://math.stackexchange.com/questions/440211/iff-if-and-only-if-vs-tfae-the-following-are-equivalent/440213 | # “IFF” (if and only if) vs. “TFAE” (the following are equivalent)
If $P$ and $Q$ are statements,
$P \iff Q$
and
The following are equivalent:
$(\text{i}) \ P$
$(\text{ii}) \ Q$
Is there a difference between the two? I ask because formulations of certain theorems (such as Heine-Borel) use the latter, while others use the former. Is it simply out of convention or "etiquette" that one formulation is preferred? Or is there something deeper? Thanks!
-
TFAE: (1) A iff B; (2) TFAE: (i) A; (ii) B. – Amit Kumar Gupta Jul 10 '13 at 6:21
As Brian M. Scott explains, they are logically equivalent.
However, the expression $$A \Leftrightarrow B \Leftrightarrow C$$ is ambiguous. It could mean either of the following.
1. $(A \Leftrightarrow B) \wedge (B \Leftrightarrow C).$
2. $(A \Leftrightarrow B) \Leftrightarrow C$
These are not equivalent. So for clarity, if we mean option 1, we should write:
The following are equivalent.
• $A.$
• $B.$
• $C.$
Thus, I would reserve the statement $A \Leftrightarrow B \Leftrightarrow C$ for option 2. It works because, somewhat surprisingly, the $\Leftrightarrow$ operation is not only commutative (obvious!) but surprisingly, it is also associative! That is, TFAE.
• $(A \Leftrightarrow B) \Leftrightarrow C$.
• $A \Leftrightarrow (B \Leftrightarrow C)$.
Note that, although statements in the general form of option 2 don't arise much in the usual way of writing math, nonetheless they're completely fundamental in equational logic.
-
I would never write the second one without parentheses. Also because there's also a third possible interpretation: $A\iff(B\iff C)$. As a general rule, for a nested binary operator $@$, parentheses should only be omitted iff $(A @ B) @ C$ and $A @ (B @ C)$ are equivalent. – celtschk Jul 10 '13 at 7:04
@celtschk, biconditional is associative - see the last sentence of my answer. – goblin Jul 10 '13 at 7:09
Ah, I missed that. That's indeed surprising. Although on second thought, it's perhaps not that surprising; after all, it makes sense that equivalence is an equivalence relation :-) – celtschk Jul 10 '13 at 7:15
@celtschk, okay I've just reorganized a little bit to make things clearer. – goblin Jul 10 '13 at 7:17
Yes, with that rearrangement, the statement is no longer easy to miss. – celtschk Jul 10 '13 at 7:19
They are exactly equivalent. There may be a pragmatic difference in their use: when $P$ and $Q$ are relatively long or complex statements, the second formulation is probably easier to read.
-
Also, TFAE is very nice for when there is more than one claim (especially when there is no obviously preferable way of ordering them in an IFF chain). – anon Jul 10 '13 at 6:05
"TFAE" is appropriate when one is listing optional replacements for some theory. For example, you could list dozen replacements for the statements, such as replacements for the fifth postulate in euclidean geometry.
"IFF" is one of the implications of "TFAE", although it as $P \rightarrow Q \rightarrow R \rightarrow P$, which equates to an iff relation.
- | 2014-10-23T03:13:21 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/440211/iff-if-and-only-if-vs-tfae-the-following-are-equivalent/440213",
"openwebmath_score": 0.9320906400680542,
"openwebmath_perplexity": 1082.7111431960536,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9546474207360066,
"lm_q2_score": 0.8840392878563335,
"lm_q1q2_score": 0.8439458259813448
} |
https://mathoverflow.net/questions/354608/how-often-two-iid-variables-are-close | # How often two iid variables are close?
Is there a constant $$c>0$$ such that for $$X,Y$$ two iid variables supported by $$[0,1]$$, $$\liminf_\epsilon \epsilon^{-1}P(|X-Y|<\epsilon)\geqslant c$$ I can prove the result if they have a density, of if they have atoms, but not in the general case.
If $$\epsilon \geqslant \tfrac{1}{n}$$, then $$\mathbb{P}(|X-Y|<\epsilon) \geqslant \sum_{i=1}^n \mathbb{P}(X, Y \in [\tfrac{i-1}{n}, \tfrac{i}{n}]) = \sum_{i=1}^n (\mathbb{P}(X \in [\tfrac{i-1}{n}, \tfrac{i}{n}]))^2 .$$ It follows that $$\mathbb{P}(|X-Y|<\epsilon) \geqslant \frac{1}{n} \biggl(\sum_{i=1}^n \mathbb{P}(X \in [\tfrac{i-1}{n}, \tfrac{i}{n}])\biggr)^2 = \frac{1}{n} \, .$$ If we choose $$n$$ so that additionally $$\epsilon < \frac{1}{n-1}$$, then we obtain $$\mathbb{P}(|X-Y|<\epsilon) > \frac{\epsilon}{\epsilon + 1} \, .$$ This leads to the desired result with $$c = 1$$.
• Interestingly, while the estimate given above is sharp at least when $\epsilon = \tfrac{1}{n}$ (consider a uniform distribution over $\frac{i}{n - 1}$, $i = 0, 1, \ldots, n - 1$), my guess is that the optimal value of $c$ is in fact $2$. However, I fail to see a proof. – Mateusz Kwaśnicki Mar 10 at 21:44
Mateusz Kwaśnicki's guess, that the optimal value of $$c$$ is 2, is correct! In fact:
Theorem: Suppose $$X,Y$$ are i.i.d. random variables on $$\mathbb{R}$$. Then the limit $$\lim_{\varepsilon \rightarrow 0} \varepsilon^{-1} \mathbb{P}[|X - Y| < \varepsilon]$$ exists. It is $$+\infty$$ unless $$X$$ has a density function $$f$$ satisfying $$||f||_2^2 = \int f^2\ \text{d}x < \infty$$, in which case the limit is equal to $$2 ||f||_2^2$$.
Restricting to distributions on $$[0,1]$$, Cauchy-Schwarz says $$1 = \int_0^1 f\ \text{d}x \le \left ( \int_0^1 f^2\ \text{d}x \right )^{1/2} \left ( \int_0^1 1\ \text{d}x \right )^{1/2} = ||f||_2,$$ with equality iff $$f = 1$$. That is, the constant $$c = 2$$ is valid for every distribution, and it is sharp only for the uniform distribution.
Here are a few examples where the limit is infinite:
Example 1: If $$X$$ has an atom then $$\mathbb{P}[|X-Y| < \varepsilon]$$ is bounded away from 0. So the limit in question diverges like $$\varepsilon^{-1}$$.
Example 2: The Cantor ternary function is the CDF of a probability distribution on $$[0,1]$$ which is nonatomic but not absolutely continuous. A random sample from this distribution is given by $$\sum_{k \ge 1} c_k \tfrac{2}{3^k}$$ where $$\{c_k\}$$ are i.i.d. Bernoulli random variables. (Thought of in terms of the "repeatedly take out the middle third" construction of the Cantor set, the $$c_k$$'s are the choices of left versus right third.) If $$\varepsilon = 3^{-n}$$, then $$|X-Y| < \varepsilon$$ iff the first $$n$$ $$c_k$$'s agree. Thus $$\mathbb{P}[|X-Y| < \varepsilon] = 2^{-n} = \varepsilon^{\log_3 2}.$$ So the limit in question diverges like $$\varepsilon^{-1+\log_3 2} = \varepsilon^{-0.37}$$.
Example 3: The power law $$f(x) = \tfrac{1}{2\sqrt{x}}$$ is an example of a density function which is not in $$L^2$$. In this case we can evaluate $$\mathbb{P}[|X-Y| < \varepsilon]$$ exactly (well, if you believe in inverse hyperbolic trig functions). The asymptotic behavior is $$\mathbb{P}[|X-Y| < \varepsilon] = \tfrac{1}{2} (-\log \varepsilon) \varepsilon + (\tfrac{1}{2} + \log 2) \varepsilon + O(\varepsilon^2).$$ So the limit in question diverges logarithmically.
Mateusz's slick argument worked on the block diagonal, a sum of $$n$$ squares of width $$1/n$$. This covers about half of the area of the strip $$|X-Y| < \tfrac{1}{n}$$, which is why the resulting constant is half of optimal. There's probably a hands-on way to extend it, but you start using words like "convolution" and I found it easier to reason in Fourier space. This question is asking for a bound on the density of $$X-Y$$ at 0. This random variable has nonnegative Fourier transform (characteristic function, if you prefer), and that condition alone is enough to guarantee positive density (shameless self-citation: Lemma 3.1 in this paper). In general, when it makes sense, the density at 0 of a random variable is equal to the integral of its Fourier transform. So our job is to make that concrete and translate it back to a statement about PDFs.
Convention: The Fourier transform (characteristic function) of a random variable $$X$$ is the function $$t \mapsto \mathbb{E}[e^{2 \pi \mathrm{i} X t}]$$. The Fourier transform of a function $$f$$ is $$t \mapsto \hat f(t) = \int e^{2 \pi \mathrm{i} x t} f(x)\ \text{d}x$$. If $$X$$ is absolutely continuous (has a density function), then its Fourier transform is equal to the Fourier transform of its density function. With this convention the Plancherel identity reads $$\int f \bar g\ \text{d}x = \int \hat f \overline{\hat g}\ \text{d}t$$ (no normalization constant).
Lemma: Let $$\psi$$ denote the Fourier transform of $$X$$. Then $$\psi \in L^2$$ iff $$X$$ is absolutely continuous and its density function $$f$$ is in $$L^2$$. Moreover, if this holds, then $$||\psi||_2 = ||f||_2$$.
Proof of Lemma: The Fourier transform is an isometry of the space $$L^2$$. The ($$\Leftarrow$$) direction is immediate: if $$X$$ has a density function in $$L^2$$, then certainly its Fourier transform is in $$L^2$$. For ($$\Rightarrow$$), if $$\psi$$ is in $$L^2$$ then it is the Fourier transform of some function $$f \in L^2$$. But then $$f$$ defines the same distribution as $$X$$, so it is indeed the density function. $$\square$$
(This is probably a basic result in some probability textbook?)
Proof of Theorem: Let $$\psi$$ denote the Fourier transform of $$X$$. Then the Fourier transform of $$X-Y$$ is $$t \mapsto \psi(t) \psi(-t) = \psi(t) \overline{\psi(t)} = |\psi(t)|^2$$. Let $$T_\varepsilon$$ denote the "triangle filter" $$T_\varepsilon(x) = \varepsilon^{-1} (1 - |x|/\varepsilon)$$ for $$|x| \le \varepsilon$$ and $$T_\varepsilon(x) = 0$$ otherwise. It is a standard calculation that $$\hat T_\varepsilon(t) = \operatorname{sinc}^2(\pi t \varepsilon)$$. (Here $$\operatorname{sinc} x = \tfrac{\sin x}{x}$$.) This is integrable, so we can compute the expected value of $$T_\varepsilon$$ using the Fourier transform. We find $$\varepsilon^{-1} \mathbb{P}[|X-Y| < \varepsilon] \ge \mathbb{E}[T_\varepsilon(X-Y)] = \int |\psi(t)|^2 \operatorname{sinc}^2(\pi t \varepsilon)\ \text{d}t.$$ The integrand is nonnegative and, as $$\varepsilon \to 0$$, it converges (pointwise) to its (pointwise) supremum $$|\psi(t)|^2$$. So, by an appropriate incantation of the monotone convergence theorem (the one in this comment applies exactly), the integral converges to $$\int |\psi(t)|^2\ \text{d}t = ||\psi||_2^2$$. If this is infinite, then also $$\lim_{\varepsilon \to 0} \varepsilon^{-1} \mathbb{P}[|X-Y| < \varepsilon] = +\infty$$.
Suppose now that $$||\psi||_2^2 < \infty$$, i.e., $$\psi \in L^2$$. Consider the "box filter" $$B_\varepsilon$$ defined by $$B_\varepsilon(x) = (2 \varepsilon)^{-1}$$ for $$|x| < \varepsilon$$ and $$B_\varepsilon(x) = 0$$ otherwise. This has $$\hat B_\varepsilon(t) = \operatorname{sinc}(2 \pi t \varepsilon)$$. This is bounded and $$|\psi|^2$$ is integrable. So we can again compute expected values in Fourier space: $$(2\varepsilon)^{-1} \mathbb{P}[|X-Y| < \varepsilon] = \mathbb{E}[B_\varepsilon(X-Y)] = \int |\psi(t)|^2 \operatorname{sinc}(2 \pi t \varepsilon)\ \text{d}t.$$ The integrand converges pointwise to $$|\psi(t)|^2$$. We don't have nonnegativity this time, but now we know integrability of the bounding function $$|\psi|^2$$. So we can apply the dominated convergence theorem, concluding that $$\lim_{\varepsilon \to 0} (2 \varepsilon)^{-1} \mathbb{P}[|X-Y| < \varepsilon] = \int |\psi(t)|^2\ \text{d}t = ||\psi||_2^2.$$ The lemma completes the proof. $$\square$$ | 2020-04-10T08:14:53 | {
"domain": "mathoverflow.net",
"url": "https://mathoverflow.net/questions/354608/how-often-two-iid-variables-are-close",
"openwebmath_score": 0.9895606637001038,
"openwebmath_perplexity": 98.91523314099922,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9905874120796498,
"lm_q2_score": 0.8519528076067262,
"lm_q1q2_score": 0.8439337269011388
} |
https://math.stackexchange.com/questions/2828012/characteristic-polynomial-of-the-matrix-with-zeros-on-the-diagonal-and-ones-else | Characteristic polynomial of the matrix with zeros on the diagonal and ones elsewhere
Find the characteristic polynomial of the matrix with zeros on the diagonal and ones elsewhere.
I've been able (I believe) to guess how it looks like (by considering matrices of small orders): $(x-n+1)(x-1)^{n-1}$. I suppose I should prove it by induction. But I don't know how to obtain the characteristic polynomial of a matrix of order $n+1$ from that of a matrix of order $n$ (i.e., how to make the inductive step).
Other methods of solution are also welcome. (Is it possible to use row reduction?)
Let $M$ be the matrix in question. Then $M=J-I$ where $J$ is the all-ones matrix. Then $$\chi_M(t)=\det(tI-M)=\det(tI+I-J)=\chi_J(t+1).$$ We just need to compute the characteristic polynomial of $J$, But $J^2=nJ$, so the eigenvalues of $J$ are in the set $\{0,n\}$. The trace of $J$ is $n$, so that one eigenvalue of $J$ is $n$ and the other $n-1$ are all zero. That is, $\chi_J(t)=t^{n-1}(t-n)$. Then $$\chi_M(t)=(t+1)^{n-1}(t-n+1).$$
• May I ask how does $J^2=nJ$ imply that the eigenvalues are in $\{0,n\}$? Jun 22, 2018 at 14:40
• Consider $J^2v=nJv$ where $v$ is an eigenvector @user538518 Jun 22, 2018 at 17:41
• For those who are interested in a generalization of this great method: math.stackexchange.com/questions/2853981/… Jul 17, 2018 at 0:39
Let the matrix be $A$. It's much easier to find the characteristic polynomial of $B=A+I$, the matrix that is all $1$s: we can do this by finding the eigenvectors, which will give us the eigenvalues. Then the eigenvalues of $A$ will just be those of $B$ with a $-1$, since $Bx=\lambda x \iff Ax = (\lambda-1)x$.
So what are the eigenvectors of $B$? It's so symmetric that it's easy to write a few down: the vector that is all $1$s has eigenvalue $n$, while any vector whose components sum to zero is an eigenvector with eigenvalue $0$. There are at least $n-1$ of these, since $e_1-e_i$ for $2 \leq i \leq n$ is a linearly independent subset of the eigenspace. Thus $0$ is an eigenvalue with multiplicity at least $n-1$, but we know the other eigenvalue already, so this is everything. Hence the characteristic polynomial is $(t-n)t^{n-1}$.
Returning to $A$, we find that this gives the characteristic polynomial of $A$ as $(t-n+1)(t+1)^{n-1}$.
• the eigenvalues become $-1$ and $n-1$ Jun 22, 2018 at 3:23 | 2022-08-16T19:38:47 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2828012/characteristic-polynomial-of-the-matrix-with-zeros-on-the-diagonal-and-ones-else",
"openwebmath_score": 0.9645201563835144,
"openwebmath_perplexity": 87.29374545995127,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9905874094398218,
"lm_q2_score": 0.8519528000888387,
"lm_q1q2_score": 0.8439337172050051
} |
https://casmusings.wordpress.com/category/problem-solving/ | # Category Archives: problem-solving
## Binomial Expansion Variation
Several years ago, I posed on this ‘blog a problem I learned from Natalie Jackucyn:
For some integers A, B, and n, one term of the expansion of $(Ax+By)^n$ is $27869184x^5y^3$. What are the values of A, B, and n?
In this post, I reflect for a moment on what I’ve learned from the problem and outline a solution approach before sharing a clever alternative solution one of my students this year leveraged through her CAS-enabled investigation.
WHAT I LEARNED BEFORE THIS YEAR
Mostly, I’ve loved this problem for its “reversal” of traditional binomial expansion problems that typically give A, B, and n values and ask for either complete expansions or specific terms of the polynomial. Both of these traditional tasks are easily managed via today’s technology. In Natalie’s variation, neither the answer nor how you would proceed are immediately obvious.
The first great part of the problem is that it doesn’t seem to give enough information. Second, it requires solvers to understand deeply the process of polynomial expansion. Third, unlike traditional formulations, Natalie’s version doesn’t allow students to avoid deep thinking by using technology.
In the comments to my original post, Christopher Olah and a former student, Bryan Spellman, solved the problem via factoring and an Excel document, respectively. Given my algebraic tendencies, I hadn’t considered Bryan’s Excel “search” approach, but one could relatively easily program Excel to provide an exhaustive search. I now think of Bryan’s approach as a coding approach to a reasonably efficient search of the sample space of possible solutions. Most of my students’ solutions over the years essentially approach the problem the same way, but less efficiently, by using one-case-at-a-time expansions via CAS commands until they stumble upon good values for A, B, and n. Understandably, students taking this approach typically become the most frustrated.
Christopher’s approach paralleled my own. The x and y exponents from the expanded term show that n=5+3=8. Expanding a generic $(Ax+By)^8$ then gives a bit more information. From my TI-Nspire CAS,
so there are 56 ways an $x^5y^3$ term appears in this expansion before combining like terms (explained here, if needed). Dividing the original coefficient by 56 gives $a^5b^3=497,664$, the coefficient of $x^5y^3$.
The values of a and b are integers, so factoring 497,664 shows these coefficients are both co-multiples of 2 and 3, but which ones? In essence, this defines a system of equations. The 3 has an exponent of 5, so it can easily be attributed to a, but the 11 is not a multiple of either 5 or 3, so it must be a combination. Quick experimentation with the exponents leads to $11=5*1+3*2$, so $2^1$ goes to a and $2^2$ goes to b. This results in $a=3*2=6$ and $b=2^2=4$.
WHAT A STUDENT TAUGHT ME THIS YEAR
After my student, NB, arrived at $a^5b^3=497,664$ , she focused on roots–not factors–for her solution. The exponents of a and b suggested using either a cubed or a fifth root.
The fifth root would extract only the value of a if b had only singleton factors–essentially isolating the a and b values–while the cubed root would extract a combination of a and b factors, leaving only excess a factors inside the radical. Her investigation was simplified by the exact answers from her Nspire CAS software.
From the fifth root output, the irrational term had exponent 1/5, not the expected 3/5, so b must have had at least one prime factor with non-singular multiplicity. But the cubed root played out perfectly. The exponent–2/3–matched expectation, giving a=6, and the coefficient, 24, was the product of a and b, making b=4. Clever.
EXTENSIONS & CONCLUSION
Admittedly, NB’s solution would have been complicated if the parameter was composed of something other than singleton prime factors, but it did present a fresh, alternative approach to what was becoming a comfortable problem for me. I’m curious about exploring other arrangements of the parameters of $(Ax+By)^n$ to see how NB’s root-based reasoning could be extended and how it would compare to the factor solutions I used before. I wonder which would be “easier” … whatever “easier” means.
As a ‘blog topic for another day, I’ve learned much by sharing this particular problem with several teachers over the years. In particular, the initial “not enough information” feel of the problem statement actually indicates the presence of some variations that lead to multiple solutions. If you think about it, NB’s root variation of the solution suggests some direct paths to such possible formulations. As intriguing as the possibilities here are, I’ve never assigned such a variation of the problem to my students.
As I finish this post, I’m questioning why I haven’t yet taken advantage of these possibilities. That will change. Until then, perhaps you can find some interesting or alternative approaches to the underlying systems of equations in this problem. Can you create a variation that has multiple solutions? Under what conditions would such a variation exist? How many distinct solutions could a problem like this have?
## Envelope Curves
My precalculus class recently returned to graphs of sinusoidal functions with an eye toward understanding them dynamically via envelope curves: Functions that bound the extreme values of the curves. What follows are a series of curves we’ve explored over the past few weeks. Near the end is a really cool Desmos link showing an infinite progression of periodic envelopes to a single curve–totally worth the read all by itself.
GETTING STARTED
As a simple example, my students earlier had seen the graph of $f(x)=5+2sin(x)$ as $y=sin(x)$ vertically stretched by a magnitude of 2 and then translated upward 5 units. In their return, I encouraged them to envision the function behavior dynamically instead of statically. I wanted them to see the curve (and the types of phenomena it could represent) as representing dynamic motion rather than a rigid transformation of a static curve. In that sense, the graph of f oscillated 2 units (the coefficient of sine in f‘s equation) above and below the line $y=5$ (the addend in the equation for f). The curves $y=5+2=7$ and $y=5-2=3$ define the “Envelope Curves” for $y=f(x)$.
When you graph $y=f(x)$ and its two envelope curves, you can picture the sinusoid “bouncing” between its envelopes. We called these ceiling and floor functions for f. Ceilings happen whenever the sinusoid term reaches its maximum value (+1), and floors when the sinusoidal term is at its minimum (-1).
Those envelope functions would be just more busy work if it stopped there, though. The great insights were that anything you added to a sinusoid could act as a midline with the coefficient, AND anything multiplied by the sinusoid is its amplitude–the distance the curve moves above and below its midline. The fun comes when you start to allow variable expressions for the midline and/or amplitudes.
VARIABLE MIDLINES AND ENVELOPES
For a first example, consider $y= \frac{x}{2} + sin(x)$. By the reasoning above, $y= \frac{x}{2}$ is the midline. The amplitude, 1, is the coefficient of sine, so the envelope curves are $y= \frac{x}{2}+1$ (ceiling) and $y= \frac{x}{2}-1$ (floor).
That got their attention! Notice how easy it is to visualize the sine curve oscillating between its envelope curves.
For a variable amplitude, consider $y=2+1.2^{-x}*sin(x)$. The midline is $y=2$, with an “amplitude” of $1.2^{-x}$. That made a ceiling of $y=2+1.2^{-x}$ and a floor of $y=2-1.2^{-x}$, basically exponential decay curves converging on an end behavior asymptote defined by the midline.
SINUSOIDAL MIDLINES AND ENVELOPES
Now for even more fun. Convinced that both midlines and amplitudes could be variably defined, I asked what would happen if the midline was another sinusoid? For $y=cos(x)+sin(x)$, we could think of $y=cos(x)$ as the midline, and with the coefficient of sine being 1, the envelopes are $y=cos(x)+1$ and $y=cos(x)-1$.
Since cosine is a sinusoid, you could get the same curve by considering $y=sin(x)$ as the midline with envelopes $y=sin(x)+1$ and $y=sin(x)-1$. Only the envelope curves are different!
The curve $y=cos(x)+sin(x)$ raised two interesting questions:
1. Was the addition of two sinusoids always another sinusoid?
2. What transformations of sinusoidal curves could be defined by more than one pair of envelope curves?
For the first question, they theorized that if two sinusoids had the same period, their sum was another sinusoid of the same period, but with a different amplitude and a horizontal shift. Mathematically, that means
$A*cos(\theta ) + B*sin(\theta ) = C*cos(\theta -D)$
where A & B are the original sinusoids’ amplitudes, C is the new sinusoid’s amplitude, and D is the horizontal shift. Use the cosine difference identity to derive
$A^2 + B^2 = C^2$ and $\displaystyle tan(D) = \frac{B}{A}$.
For $y = cos(x) + sin(x)$, this means
$\displaystyle y = cos(x) + sin(x) = \sqrt{2}*cos \left( x-\frac{\pi}{4} \right)$,
and the new coefficient means $y= \pm \sqrt{2}$ is a third pair of envelopes for the curve.
Very cool. We explored several more sums and differences with identical periods.
WHAT HAPPENS WHEN THE PERIODS DIFFER?
Try a graph of $g(x)=cos(x)+cos(3x)$.
Using the earlier concept that any function added to a sinusoid could be considered the midline of the sinusoid, we can picture the graph of g as the graph of $y=cos(3x)$ oscillating around an oscillating midline, $y=cos(x)$:
IF you can’t see the oscillations yet, the coefficient of the $cos(3x)$ term is 1, making the envelope curves $y=cos(x) \pm 1$. The next graph clear shows $y=cos(3x)$ bouncing off its ceiling and floor as defined by its envelope curves.
Alternatively, the base sinusoid could have been $y=cos(x)$ with envelope curves $y=cos(3x) \pm 1$.
Similar to the last section when we added two sinusoids with the same period, the sum of two sinusoids with different periods (but the same amplitude) can be rewritten using an identity.
$cos(A) + cos(B) = 2*cos \left( \frac{A+B}{2} \right) * cos \left( \frac{A-B}{2} \right)$
This can be proved in the present form, but is lots easier to prove from an equivalent form:
$cos(x+y) + cos(x-y) = 2*cos(x) * cos(y)$.
For the current function, this means $y = cos(x) + cos(3x) = 2*cos(x)*cos(2x)$.
Now that the sum has been rewritten as a product, we can now use the coefficient as the amplitude, defining two other pairs of envelope curves. If $y=cos(2x)$ is the sinusoid, then $y= \pm 2cos(x)$ are envelopes of the original curve, and if $y=cos(x)$ is the sinusoid, then $y= \pm 2cos(2x)$ are envelopes.
In general, I think it’s easier to see the envelope effect with the larger period function. A particularly nice application connection of adding sinusoids with identical amplitudes and different periods are the beats musicians hear from the constructive and destructive sound wave interference from two instruments close to, but not quite in tune. The points where the envelopes cross on the x-axis are the quiet points in the beats.
A STUDENT WANTED MORE
In class last Friday, my students were reviewing envelope curves in advance of our final exam when one made the next logical leap and asked what would happen if both the coefficients and periods were different. When I mentioned that the exam wouldn’t go that far, she uttered a teacher’s dream proclamation: She didn’t care. She wanted to learn anyway. Making up some coefficients on the spot, we decided to explore $f(x)=2sin(x)+5cos(2x)$.
Assuming for now that the cos(2x) term is the primary sinusoid, the envelope curves are $y=2sin(x) \pm 5$.
That was certainly cool, but at this point, we were no longer satisfied with just one answer. If we assumed sin(x) was the primary sinusoid, the envelopes are $y=5cos(2x) \pm 2$.
Personally, I found the first set of envelopes more satisfying, but it was nice that we could so easily identify another.
With the different periods, even though the coefficients are different, we decided to split the original function in a way that allowed us to use the $cos(A)+cos(B)$ identity introduced earlier. Rewriting,
$f(x)=2sin(x)+5cos(2x) = 2cos \left( x - \frac{ \pi }{2} \right) + 2cos(2x) + 3cos(2x)$ .
After factoring out the common coefficient 2, the first two terms now fit the $cos(A) + cos(B)$ identity with $A = x - \frac{ \pi }{2}$ and $B=2x$, allowing the equation to be rewritten as
$f(x)= 2 \left( 2*cos \left( \frac{x - \frac{ \pi }{2} + 2x }{2} \right) * cos \left( \frac{x - \frac{ \pi }{2} - 2x }{2} \right) \right) + 3cos(2x)$
$\displaystyle = 4* cos \left( \frac{3}{2} x - \frac{ \pi }{4} \right) * cos \left( - \frac{1}{2} x - \frac{ \pi }{4} \right) + 3cos(2x)$.
With the expression now containing three sinusoidal expressions, there are three more pairs of envelope curves!
Arguably, the simplest approach from this form assumes $cos(2x)$ from the $latex$3cos(2x)\$ term as the sinusoid, giving $y=2sin(x)+2cos(2x) \pm 3$ (the pre-identity form three equations earlier in this post) as envelopes.
We didn’t go there, but recognizing that new envelopes can be found simply by rewriting sums creates an infinite number of additional envelopes. Defining these different sums with a slider lets you see an infinite spectrum of envelopes. The image below shows one. Here is the Desmos Calculator page that lets you play with these envelopes directly.
If the $cos \left( \frac{3}{3} x - \frac{ \pi}{4} \right)$term was the sinusoid, the envelopes would be $y=3cos(2x) \pm 4cos \left( - \frac{1}{2} x - \frac{ \pi }{4} \right)$. If you look closely, you will notice that this is a different type of envelope pair with the ceiling and floor curves crossing and trading places at $x= \frac{\pi}{2}$ and every $2\pi$ units before and after. The third form creates another curious type of crossing envelopes.
CONCLUSION:
In all, it was fun to explore with my students the many possibilities for bounding sinusoidal curves. It was refreshing to have one student excited by just playing with the curves to see what else we could find for no other reason than just to enjoy the beauty of these periodic curves. As I reflected on the overall process, I was even more delighted to discover the infinite spectrum of envelopes modeled above on Desmos.
I hope you’ve found something cool here for yourself.
## From a Square to Ratios to a System of Equations
Here’s another ratio problem from @Five_Triangles, this time involving triangle areas bounded by a square.
Don’t read further until you’ve tried this for yourself. It’s a fun problem that, at least from my experience, doesn’t end up where or how I thought it would.
INITIAL THOUGHTS
I see two big challenges here.
First, the missing location of point P is especially interesting, but is also likely to be quite vexing for many students. This led me to the first twist I found in the problem: the introduction of multiple variables and a coordinate system. Without some problem-solving experience, I don’t see that as an intuitive step for most middle school students. Please don’t interpret this as a knock on this problem, I’m simply agreeing with @Five_Triangle’s assessment that this problem is likely to be challenging for middle school students.
The second challenge I found emerged from the introduction the coordinate system: an underlying 2×2 system of equations. There are multiple ways to tackle a solution to a linear system, but this strikes me as yet another high hurdle for younger students.
Finally, I’m a bit surprised by my current brain block on multiple approaches for this problem. I suspect I’m blinded here by my algebraic bias in problem solving; surely there are approaches that don’t require this. I’d love to hear any other possibilities.
POINT P VARIES
Because I was given properties of point P and not its location, the easiest approach I could see was to position the square on the xy-plane with point B at the origin, $\overline{AB}$ along the y-axis, and $\overline{BC}$ along the x-axis. That gave my point P coordinates (x,y) for some unknown values of x & y.
The helpful part of this orientation is that the x & y coordinates of P are automatically the altitudes of $\Delta ABP$ and $\Delta BCP$, respectively. The altitudes of the other two triangles are determined through subtraction.
AREA RATIOS BECOME A LINEAR SYSTEM
From here, I used the given ratios to establish one equation in terms of x & y.
$\displaystyle \frac{\Delta ABP}{\Delta DAP} = \frac{\frac{1}{2}*12*x}{\frac{1}{2}*12*(12-y)} = \frac{3}{4}$
Of course, since all four triangles have the same base lengths, the given area ratios are arithmetically equivalent to corresponding height ratios. I used that to write a second equation.
$\displaystyle \frac{\Delta BCP}{\Delta CDP} = \frac{y}{12-x} = \frac{1}{3}$
Simplifying terms and clearing denominators leads to $4x=36-3y$ and $3y=12-x$, respectively.
A VERY INTERESTING insight at this point is that there is an infinite number of locations within the square at which each ratio is true. Specifically, the $\Delta ABP : \Delta DAP = 3:4$ ratio is true everywhere along the line 4x=36-3y. This problem constrains us to only the points within the square with vertices (0,0), (12,0), (12,12), and (0,12), but setting that aside, anywhere along the line 4x=36-3y would satisfy the first constraint. The same is true for the second line and constraint.
I think it would be very interesting for students to construct this on dynamic geometry software (e.g., GeoGebra or the TI-Nspire) and see the ratio remain constant everywhere along either line even though the triangle areas vary throughout.
Together, these lines form a 2×2 system of linear equations with the solution to both ratios being the intersection point of the two lines. There are lots of ways to do this; I wonder how a typical 6th grader would tackle them. Assuming they have the algebraic expertise, I’d have work them by hand and confirm with a CAS.
The question asks for the area of $\Delta ABP = \frac{1}{2}*12*x = 6*8 = 48$.
PROBLEM VARIATIONS
Just two extensions this time. Other suggestions are welcome.
1. What’s the ratio of the area of $\Delta BCP : \Delta DAP$ at the point P that satisfies both ratios??
It’s not 1:4 as an errant student might think from an errant application of the transitive property to the given ratios. Can you show that it’s actually 1:8?
2. If a random point is chosen within the square, is that point more likely to satisfy the area ratio of $\Delta ABP : \Delta DAP$ or the ratio of $\Delta BCP : \Delta CDP$?
The first ratio is satisfied by the line 4x=36-3y which intersects the square on the segment between (9,0) and (0,12). At the latter point, both triangles are degenerate with area 0. The second ratio’s line intersects the square between (12,0) and (0,4). As the first segment is longer (how would a middle schooler prove that?), it is more likely that a randomly chosen point would satisfy the $\Delta ABP : \Delta DAP$ ratio. This would be a challenging probability problem, methinks.
FURTHER EXTENSIONS?
What other possibilities do you see either for a solution to the original problem or an extension?
## Party Ratios
I find LOTS of great middle school problems from @Five_Triangles on Twitter. Their post two days ago was no exception.
The problem requires a little stamina, but can be approached many ways–two excellent criteria for worthy student explorations. That it has some solid extensions makes it even better. Following are a few different solution approaches some colleagues and I created.
INITIAL THOUGHTS, VISUAL ORGANIZATION, & A SOLUTION
The most challenging part of this problem is data organization. My first thoughts were for a 2-circle Venn Diagram–one for gender and one for age. And these types of Venn Diagrams are often more easily understood, in my experience, in 2×2 Table form with extra spaces for totals. Here’s what I set up initially.
The ratio of Women:Girls was 11:4, so the 24 girls meant each “unit” in this ratio accounted for 24/4=6 people. That gave 11*6=66 women and 66+24=90 females.
At this point, my experience working with algebraic problems tempted me to overthink the situation. I was tempted to let B represent the unknown number of boys and set up some equations to solve. Knowing that most 6th graders would not think about variables, I held back that instinct in an attempt to discover what a less-experienced mind might try. I present my initial algebra solution below.
The 5:3 Male:Female ratio told me that each “gender unit” represented 90/3=30 people. That meant there were 5*30=150 men and 240 total people at the party.
Then, the 4:1 Adult:Children ratio showed how to age-divide every group of 5 partygoers. With 240/5=48 such groups, there were 48 children and 4*48=192 adults. Subtracting the already known 66 women gave the requested answer: 192-66=126 men.
While this Venn Diagram/Table approach made sense to me, I was concerned that it was a moderately sophisticated and not quite intuitive problem-solving technique for younger middle school students.
WHAT WOULD A MIDDLE SCHOOLER THINK?
A middle school teaching colleague, Becky, offered a different solution I could see students creating.
Completely independently, she solved the problem in exactly the same order I did using ratio tables to manage the scaling at each step instead of my “unit ratios”. I liked her visual representation of the 4:1 Adults:Children ratio to find the number of adults, which gave the requested number of men. I suspect many more students would implicitly or explicitly use some chunking strategies like the visual representation to work the ratios.
WHY HAVE JUST ONE SOLUTION?
Math problems involving ratios can usually be opened up to allow multiple, or even an infinite number of solutions. This leads to some interesting problem extensions if you eliminate the “24 girls” restriction. Here are a few examples and sample solutions.
What is the least number of partygoers?
For this problem, notice from the table above that all of the values have a common factor of 6. Dividing the total partygoers by this reveals that 240/6=40 is the least number. Any multiple of this number is also a legitimate solution.
Interestingly, the 11:4 Women:Girls ratio becomes explicitly obvious when you scale the table down to its least common value.
My former student and now colleague, Teddy, arrived at this value another way. Paraphrasing, he noted that the 5:3 Male:Female ratio meant any valid total had to be a multiple of 5+3=8. Likewise, the 4:1 Adult:Child ratio requires totals to be multiples of 4+1=5. And the LCM of 8 & 5 is 40, the same value found in the preceding paragraph.
What do all total partygoer numbers have in common?
As explained above, any multiple of 40 is a legitimate number of partygoers.
If the venue could support no more than 500 attendees, what is the maximum number of women attending?
12*40=480 is the greatest multiple of 40 below 500. Because 480 is double the initial problem’s total, 66*2=132 is the maximum number of women.
Note that this can be rephrased to accommodate any other gender/age/total target.
Under the given conditions, will the number of boys and girls at the party ever be identical?
As with all ratio problems, larger values are always multiples of the least common solution. That means the number of boys and girls will always be identical or always be different. From above, you can deduce that the numbers of boys and girls at the party under the given conditions will both be multiples of 4.
What variations can you and/or your students create?
RESOLVING THE INITIAL ALGEBRA
Now to the solution variation I was initially inclined to produce. After initially determining 66 women from the given 24 girls, let B be the unknown number of boys. That gives B+24 children. It was given that adults are 4 times as numerous as children making the number of adults 4(B+24)=4B+96. Subtracting the known 66 women leaves 4B+30 men. Compiling all of this gives
The 5:3 Male:Female ratio means $\displaystyle \frac{5}{3} = \frac{5B+30}{90} \longrightarrow B=24$, the same result as earlier.
ALGEBRA OVERKILL
Winding through all of that algebra ultimately isn’t that computationally difficult, but it certainly is more than typical 6th graders could handle.
But the problem could be generalized even further, as Teddy shared with me. If the entire table were written in variables with W=number of women, M=men, G=girls, and B=boys, the given ratios in the problem would lead to a reasonably straightforward 4×4 system of equations. If you understand enough to write all of those equations, I’m certain you could solve them, so I’d feel confident allowing a CAS to do that for me. My TI-Nspire gives this.
And that certainly isn’t work you’d expect from any 6th grader.
CONCLUSION
Given that the 11:4 Women:Girls ratio was the only “internal” ratio, it was apparent in retrospect that all solutions except the 4×4 system approach had to find the female values first. There are still several ways to resolve the problem, but I found it interesting that while there was no “direct route”, every reasonable solution started with the same steps.
Thanks to colleagues Teddy S & Becky M for sharing their solution proposals.
## Unanticipated Proof Before Algebra
I was talking with one of our 5th graders, S, last week about the difference between showing a few examples of numerical computations and developing a way to know something was true no matter what numbers were chosen. I hadn’t started our conversation thinking about introducing proof. Once we turned in that direction, I anticipated scaffolding him in a completely different direction, but S went his own way and reinforced for me the importance of listening and giving students the encouragement and room to build their own reasoning.
SETUP: S had been telling me that he “knew” the product of an even number with any other number would always be even, while the product of any two odds was always odd. He demonstrated this by showing lots of particular products, but I asked him if he was sure that it was still true if I were to pick some numbers he hadn’t used yet. He was.
Then I asked him how many numbers were possible to use. He promptly replied “infinite” at which point he finally started to see the difficulty with demonstrating that every product worked. “We don’t have enough time” to do all that, he said. Finally, I had maneuvered him to perhaps his first ever realization for the need for proof.
ANTICIPATION: But S knew nothing of formal algebra. From my experiences with younger students sans algebra, I thought I would eventually need to help him translate his numerical problem into a geometric one. But this story is about S’s reasoning, not mine.
INSIGHT: I asked S how he would handle any numbers I asked him to multiply to prove his claims, even if I gave him some ridiculously large ones. “It’s really not as hard as that,” S told me. He quickly scribbled
on his paper and covered up all but the one’s digit. “You see,” he said, “all that matters is the units. You can make the number as big as you want and I just need to look at the last digit.” Without using this language, S was venturing into an even-odd proof via modular arithmetic.
With some more thought, he reasoned that he would focus on just the units digit through repeated multiples and see what happened.
FIFTH GRADE PROOF: S’s math class is currently working through a multiplication unit in our 5th grade Bridges curriculum, so he was already in the mindset of multiples. Since he said only the units digit mattered, he decided he could start with any even number and look at all of its multiples. That is, he could keep adding the number to itself and see what happened. As shown below, he first chose 32 and found the next four multiples, 64, 96, 128, and 160. After that, S said the very next number in the list would end in a 2 and the loop would start all over again.
He stopped talking for several seconds, and then he smiled. “I don’t have to look at every multiple of 32. Any multiple will end up somewhere in my cycle and I’ve already shown that every number in this cycle is even. Every multiple of 32 must be even!” It was a pretty powerful moment. Since he only needed to see the last digit, and any number ending in 2 would just add 2s to the units, this cycle now represented every number ending in 2 in the universe. The last line above was S’s use of 1002 to show that the same cycling happened for another “2 number.”
DIFFERENT KINDS OF CYCLES: So could he use this for all multiples of even numbers? His next try was an “8 number.”
After five multiples of 18, he achieved the same cycling. Even cooler, he noticed that the cycle for “8 numbers” was the 2 number” cycle backwards.
Also note that after S completed his 2s and 8s lists, he used only single digit seed numbers as the bigger starting numbers only complicated his examples. He was on a roll now.
I asked him how the “4 number” cycle was related. He noticed that the 4s used every other number in the “2 number” cycle. It was like skip counting, he said. Another lightbulb went off.
“And that’s because 4 is twice 2, so I just take every 2nd multiple in the first cycle!” He quickly scratched out a “6 number” example.
This, too, cycled, but more importantly, because 6 is thrice 2, he said that was why this list used every 3rd number in the “2 number” cycle. In that way, every even number multiple list was the same as the “2 number” list, you just skip-counted by different steps on your way through the list.
When I asked how he could get all the numbers in such a short list when he was counting by 3s, S said it wasn’t a problem at all. Since it cycled, whenever you got to the end of a list, just go back to the beginning and keep counting. We didn’t touch it last week, but he had opened the door to modular arithmetic.
I won’t show them here, but his “0 number” list always ended in 0s. “This one isn’t very interesting,” he said. I smiled.
ODDS: It took a little more thought to start his odd number proof, because every other multiple was even. After he recognized these as even numbers, S decided to list every other multiple as shown with his “1 number” and “3 number” lists.
As with the evens, the odd number lists could all be seen as skip-counted versions of each other. Also, the 1s and 9s were written backwards from each other, and so were the 3s and 7s. “5 number” lists were declared to be as boring as “0 numbers”. Not only did the odds ultimately end up cycling essentially the same as the evens, but they had the same sort of underlying relationships.
CONCLUSION: At this point, S declared that since he had shown every possible case for evens and odds, then he had shown that any multiple of an even number was always even, and any odd multiple of an odd number was odd. And he knew this because no matter how far down the list he went, eventually any multiple had to end up someplace in his cycles. At that point I reminded S of his earlier claim that there was an infinite number of even and odd numbers. When he realized that he had just shown a case-by-case reason for more numbers than he could ever demonstrate by hand, he sat back in his chair, exclaiming, “Whoa! That’s cool!”
It’s not a formal mathematical proof, and when S learns some algebra, he’ll be able to accomplish his cases far more efficiently, but this was an unexpectedly nice and perfectly legitimate numerical proof of even and odd multiples for an elementary student.
## Mistakes are Good
Confession #1: My answers on my last post were WRONG.
I briefly thought about taking that post down, but discarded that idea when I thought about the reality that almost all published mathematics is polished, cleaned, and optimized. Many students struggle with mathematics under the misconception that their first attempts at any topic should be as polished as what they read in published sources.
While not precisely from the same perspective, Dan Teague recently wrote an excellent, short piece of advice to new teachers on NCTM’s ‘blog entitled Demonstrating Competence by Making Mistakes. I argue Dan’s advice actually applies to all teachers, so in the spirit of showing how to stick with a problem and not just walking away saying “I was wrong”, I’m going to keep my original post up, add an advisory note at the start about the error, and show below how I corrected my error.
Confession #2: My approach was a much longer and far less elegant solution than the identical approaches offered by a comment by “P” on my last post and the solution offered on FiveThirtyEight. Rather than just accepting the alternative solution, as too many students are wont to do, I acknowledged the more efficient approach of others before proceeding to find a way to get the answer through my initial idea.
I’ll also admit that I didn’t immediately see the simple approach to the answer and rushed my post in the time I had available to get it up before the answer went live on FiveThirtyEight.
GENERAL STRATEGY and GOALS:
1-Use a PDF: The original FiveThirtyEight post asked for the expected time before the siblings simultaneously finished their tasks. I interpreted this as expected value, and I knew how to compute the expected value of a pdf of a random variable. All I needed was the potential wait times, t, and their corresponding probabilities. My approach was solid, but a few of my computations were off.
2-Use Self-Similarity: I don’t see many people employing the self-similarity tactic I used in my initial solution. Resolving my initial solution would allow me to continue using what I consider a pretty elegant strategy for handling cumbersome infinite sums.
A CORRECTED SOLUTION:
Stage 1: My table for the distribution of initial choices was correct, as were my conclusions about the probability and expected time if they chose the same initial app.
My first mistake was in my calculation of the expected time if they did not choose the same initial app. The 20 numbers in blue above represent that sample space. Notice that there are 8 times where one sibling chose a 5-minute app, leaving 6 other times where one sibling chose a 4-minute app while the other chose something shorter. Similarly, there are 4 choices of an at most 3-minute app, and 2 choices of an at most 2-minute app. So the expected length of time spent by the longer app if the same was not chosen for both is
$E(Round1) = \frac{1}{20}*(8*5+6*4+4*3+2*2)=4$ minutes,
a notably longer time than I initially reported.
For the initial app choice, there is a $\frac{1}{5}$ chance they choose the same app for an average time of 3 minutes, and a $\frac{4}{5}$ chance they choose different apps for an average time of 4 minutes.
Stage 2: My biggest error was a rushed assumption that all of the entries I gave in the Round 2 table were equally likely. That is clearly false as you can see from Table 1 above. There are only two instances of a time difference of 4, while there are eight instances of a time difference of 1. A correct solution using my approach needs to account for these varied probabilities. Here is a revised version of Table 2 with these probabilities included.
Conveniently–as I had noted without full realization in my last post–the revised Table 2 still shows the distribution for the 2nd and all future potential rounds until the siblings finally align, including the probabilities. This proved to be a critical feature of the problem.
Another oversight was not fully recognizing which events would contribute to increasing the time before parity. The yellow highlighted cells in Table 2 are those for which the next app choice was longer than the current time difference, and any of these would increase the length of a trial.
I was initially correct in concluding there was a $\frac{1}{5}$ probability of the second app choice achieving a simultaneous finish and that this would not result in any additional total time. I missed the fact that the six non-highlighted values also did not result in additional time and that there was a $\frac{1}{5}$ chance of this happening.
That leaves a $\frac{3}{5}$ chance of the trial time extending by selecting one of the highlighted events. If that happens, the expected time the trial would continue is
$\displaystyle \frac{4*4+(4+3)*3+(4+3+2)*2+(4+3+2+1)*1}{4+(4+3)+(4+3+2)+(4+3+2+1)}=\frac{13}{6}$ minutes.
Iterating: So now I recognized there were 3 potential outcomes at Stage 2–a $\frac{1}{5}$ chance of matching and ending, a $\frac{1}{5}$ chance of not matching but not adding time, and a $\frac{3}{5}$ chance of not matching and adding an average $\frac{13}{6}$ minutes. Conveniently, the last two possibilities still combined to recreate perfectly the outcomes and probabilities of the original Stage 2, creating a self-similar, pseudo-fractal situation. Here’s the revised flowchart for time.
Invoking the similarity, if there were T minutes remaining after arriving at Stage 2, then there was a $\frac{1}{5}$ chance of adding 0 minutes, a $\frac{1}{5}$ chance of remaining at T minutes, and a $\frac{3}{5}$ chance of adding $\frac{13}{6}$ minutes–that is being at $T+\frac{13}{6}$ minutes. Equating all of this allows me to solve for T.
$T=\frac{1}{5}*0+\frac{1}{5}*T+\frac{3}{5}*\left( T+\frac{13}{6} \right) \longrightarrow T=6.5$ minutes
Time Solution: As noted above, at the start, there was a $\frac{1}{5}$ chance of immediately matching with an average 3 minutes, and there was a $\frac{4}{5}$ chance of not matching while using an average 4 minutes. I just showed that from this latter stage, one would expect to need to use an additional mean 6.5 minutes for the siblings to end simultaneously, for a mean total of 10.5 minutes. That means the overall expected time spent is
Total Expected Time $=\frac{1}{5}*3 + \frac{4}{5}*10.5 = 9$ minutes.
Number of Rounds Solution: My initial computation of the number of rounds was actually correct–despite the comment from “P” in my last post–but I think the explanation could have been clearer. I’ll try again.
One round is obviously required for the first choice, and in the $\frac{4}{5}$ chance the siblings don’t match, let N be the average number of rounds remaining. In Stage 2, there’s a $\frac{1}{5}$ chance the trial will end with the next choice, and a $\frac{4}{5}$ chance there will still be N rounds remaining. This second situation is correct because both the no time added and time added possibilities combine to reset Table 2 with a combined probability of $\frac{4}{5}$. As before, I invoke self-similarity to find N.
$N = \frac{1}{5}*1 + \frac{4}{5}*N \longrightarrow N=5$
Therefore, the expected number of rounds is $\frac{1}{5}*1 + \frac{4}{5}*5 = 4.2$ rounds.
It would be cool if someone could confirm this prediction by simulation.
CONCLUSION:
I corrected my work and found the exact solution proposed by others and simulated by Steve! Even better, I have shown my approach works and, while notably less elegant, one could solve this expected value problem by invoking the definition of expected value.
Best of all, I learned from a mistake and didn’t give up on a problem. Now that’s the real lesson I hope all of my students get.
Happy New Year, everyone!
## Great Probability Problems
UPDATE: Unfortunately, there are a couple errors in my computations below that I found after this post went live. In my next post, Mistakes are Good, I fix those errors and reflect on the process of learning from them.
ORIGINAL POST:
A post last week to the AP Statistics Teacher Community by David Bock alerted me to the new weekly Puzzler by Nate Silver’s new Web site, http://fivethirtyeight.com/. As David noted, with their focus on probability, this new feature offers some great possibilities for AP Statistics probability and simulation.
I describe below FiveThirtyEight’s first three Puzzlers along with a potential solution to the last one. If you’re searching for some great problems for your classes or challenges for some, try these out!
THE FIRST THREE PUZZLERS:
The first Puzzler asked a variation on a great engineering question:
You work for a tech firm developing the newest smartphone that supposedly can survive falls from great heights. Your firm wants to advertise the maximum height from which the phone can be dropped without breaking.
You are given two of the smartphones and access to a 100-story tower from which you can drop either phone from whatever story you want. If it doesn’t break when it falls, you can retrieve it and use it for future drops. But if it breaks, you don’t get a replacement phone.
Using the two phones, what is the minimum number of drops you need to ensure that you can determine exactly the highest story from which a dropped phone does not break? (Assume you know that it breaks when dropped from the very top.) What if, instead, the tower were 1,000 stories high?
The second Puzzler investigated random geyser eruptions:
You arrive at the beautiful Three Geysers National Park. You read a placard explaining that the three eponymous geysers — creatively named A, B and C — erupt at intervals of precisely two hours, four hours and six hours, respectively. However, you just got there, so you have no idea how the three eruptions are staggered. Assuming they each started erupting at some independently random point in history, what are the probabilities that A, B and C, respectively, will be the first to erupt after your arrival?
Both very cool problems with solutions on the FiveThirtyEight site. The current Puzzler talked about siblings playing with new phone apps.
SOLVING THE CURRENT PUZZLER:
Before I started, I saw Nick Brown‘s interesting Tweet of his simulation.
If Nick’s correct, it looks like a mode of 5 minutes and an understandable right skew. I approached the solution by first considering the distribution of initial random app choices.
There is a $\displaystyle \frac{5}{25}$ chance the siblings choose the same app and head to dinner after the first round. The expected length of that round is $\frac{1}{5} \cdot \left( 1+2=3=4+5 \right) = 3$ minutes.
That means there is a $\displaystyle \frac{4}{5}$ chance different length apps are chosen with time differences between 1 and 4 minutes. In the case of unequal apps, the average time spent before the shorter app finishes is $\frac{1}{25} \cdot \left( 8*1+6*2+4*3+2*4 \right) = 1.6$ minutes.
It doesn’t matter which sibling chose the shorter app. That sibling chooses next with distribution as follows.
While the distributions are different, conveniently, there is still a time difference between 1 and 4 minutes when the total times aren’t equal. That means the second table shows the distribution for the 2nd and all future potential rounds until the siblings finally align. While this problem has the potential to extend for quite some time, this adds a nice pseudo-fractal self-similarity to the scenario.
As noted, there is a $\displaystyle \frac{4}{20}=\frac{1}{5}$ chance they complete their apps on any round after the first, and this would not add any additional time to the total as the sibling making the choice at this time would have initially chosen the shorter total app time(s). Each round after the first will take an expected time of $\frac{1}{20} \cdot \left( 7*1+5*2+3*3+1*4 \right) = 1.5$ minutes.
The only remaining question is the expected number of rounds of app choices the siblings will take if they don’t align on their first choice. This is where I invoked self-similarity.
In the initial choice there was a $\frac{4}{5}$ chance one sibling would take an average 1.6 minutes using a shorter app than the other. From there, some unknown average N choices remain. There is a $\frac{1}{5}$ chance the choosing sibling ends the experiment with no additional time, and a $\frac{4}{5}$ chance s/he takes an average 1.5 minutes to end up back at the Table 2 distribution, still needing an average N choices to finish the experiment (the pseudo-fractal self-similarity connection). All of this is simulated in the flowchart below.
Recognizing the self-similarity allows me to solve for N.
$\displaystyle N = \frac{1}{5} \cdot 1 + \frac{4}{5} \cdot N \longrightarrow N=5$
Number of Rounds – Starting from the beginning, there is a $\frac{1}{5}$ chance of ending in 1 round and a $\frac{4}{5}$ chance of ending in an average 5 rounds, so the expected number of rounds of app choices before the siblings simultaneously end is
$\frac{1}{5} *1 + \frac{4}{5}*5=4.2$ rounds
Time until Eating – In the first choice, there is a $\frac{1}{5}$ chance of ending in 3 minutes. If that doesn’t happen, there is a subsequent $\frac{1}{5}$ chance of ending with the second choice with no additional time. If neither of those events happen, there will be 1.6 minutes on the first choice plus an average 5 more rounds, each taking an average 1.5 minutes, for a total average $1.6+5*1.5=9.1$ minutes. So the total average time until both siblings finish simultaneously will be
$\frac{1}{5}*3+\frac{4}{5}*9.1 = 7.88$ minutes
CONCLUSION:
My 7.88 minute mean is reasonably to the right of Nick’s 5 minute mode shown above. We’ll see tomorrow if I match the FiveThirtyEight solution.
Anyone else want to give it a go? I’d love to hear other approaches. | 2017-10-19T20:08:01 | {
"domain": "wordpress.com",
"url": "https://casmusings.wordpress.com/category/problem-solving/",
"openwebmath_score": 0.7048300504684448,
"openwebmath_perplexity": 1343.0153067454364,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9883127437254887,
"lm_q2_score": 0.8539127585282744,
"lm_q1q2_score": 0.8439328612832796
} |
https://math.stackexchange.com/questions/3031094/a-proof-that-sqrt2-is-not-a-rational-number | # A proof that $\sqrt{2}$ is not a rational number.
Is this proof correct?
Suppose that $$\sqrt{2}=\frac{a}{b}$$, where $$a,b \in \mathbb{N}$$ and $$a$$ is as small as possible. Then $$\sqrt{2}b=a$$ which means $$2b=\sqrt{2} a$$. So we rewrite $$\sqrt{2}=\frac{a}{b}\cdot\frac{\sqrt{2}-1}{\sqrt{2}-1}=\frac{\sqrt{2}a-a}{\sqrt{2}b-b}=\frac{2b-a}{a-b}.\,$$ Note $$\,2b-a=a(\sqrt{2}-1). So this fraction has a smaller numerator than the one we had. So this is a contradiction.
• Yes, this is a well-known proof. – Lord Shark the Unknown Dec 8 '18 at 13:15
• @user42493 This is essentially the same as this section with $a=m$, $b=n$, $k=2$, $q=1$. At least you were creative enough to 'discover' another proof. – Toby Mak Dec 8 '18 at 13:18
• You have a typographical error. ab-a should read $2b-a$. – Ben W Dec 8 '18 at 13:44
• Note that $2b-a<a$ is equivalent to $2b<2a$, which is true because $a>b$. – egreg Dec 8 '18 at 13:57
It is a correct well-known proof. Essentially it uses denominator descent by the division algorithm, though that is obfuscated . Below I clarify this viewpoint for the generalization below. The proof in the question is exactly the special case $$\, k = 2\,$$ and $$\,q = {\rm floor}(\sqrt 2) = 1\,$$ of the proof below.
Irrationality of $$\sqrt k\,$$ if it is not an integer (excerpted from Wikipedia, slightly edited)
For an integer $$k>0$$, suppose $$\sqrt k$$ is not an integer, but is rational and can be expressed as $$\frac{a}b$$ for natural numbers $$a$$ and $$b$$, and let $$q$$ be the largest integer no greater than $$\sqrt k.\,$$ Then
\begin{aligned}{\sqrt {k}}&={\frac {a}{b}}\\[8pt]&={\frac {a({\sqrt {k}}-q)}{b({\sqrt {k}}-q)}}\\[8pt]&={\frac {a{\sqrt {k}}-aq}{b{\sqrt {k}}-bq}}\\[8pt]&={\frac {(b{\sqrt {k}}){\sqrt {k}}-aq}{b({\frac {a}{b}})-bq}}\\[8pt]&={\frac {bk-aq}{a-bq}}\end{aligned}
The numerator and denominator were each multiplied by $$(\sqrt k − q)\,$$ — which is positive but less than $$1$$ and then simplified independently. So the two resulting products, say $$a'$$ and $$b'$$, are themselves integers, which are less than $$a$$ and $$b$$ respectively. Therefore, no matter what natural numbers $$a$$ and $$b$$ are used to express $$\sqrt k$$, there exist smaller natural numbers $$a' < a$$ and $$b' < b$$ that have the same ratio. But infinite descent on the natural numbers is impossible, so this disproves the original assumption that $$\sqrt k$$ could be expressed as a ratio of natural numbers.
We can rewrite the above proof more conceptually as below, where "$$\,n\,$$ is a denom of $$\,r$$" means that the rational $$\,r\,$$ can be written with denominator $$\,n,\,$$ i.e. $$\,n\,r = j\,$$ for some integer $$\,j.$$
\begin{align} [\![1]\!]\qquad\qquad\, b \sqrt k\, &=\, a\qquad\ \, \Rightarrow\,\qquad\ \ \ \text{\,b\, is a denom of }\ \sqrt k\\ \sqrt k\,\cdot\, [\![1]\!]\ \ \, \Rightarrow\,[\![2]\!]\qquad\qquad a \sqrt k\, &=\, bk\qquad \Rightarrow\qquad\,\ \ \ \text{ a\, is a denom of }\ \sqrt k\\ [\![2]\!] - [\![1]\!]q\,\Rightarrow\,[\![3]\!]\ \ \ \ \ \, (\color{#c00}{a\!-\!bq})\sqrt k\, &=\, bk\!-\!aq\,\Rightarrow\, \color{#c00}{a\bmod b} \, \text{ is a denom of }\ \sqrt k\\ \end{align}
If $$\,b\,$$ doesn't divide $$\,a\,$$ we get a smaller denom $$\, 0 < \color{#c00}{a \bmod b} < b\,$$ so infinite descent (on denoms), contra $$\Bbb N\,$$ is well-ordered. Hence $$\,b\,$$ divides $$\,a,\,$$ so $$\,\sqrt k = a/b = n\in \Bbb Z,\,$$ so $$\,k = n^2$$.
Alternatively we can initially assume that $$\,b\,$$ is the least denominator then deduce a contradiction that a smaller denominator exists if $$\,b\,$$ doesn't divide $$\,a.$$
This method generalizes to show the $$\,\Bbb Z\,$$ (or any PID) is integrally-closed, i.e. no proper fraction is a root of a polynomial that is monic (lead coef $$= 1),\,$$ i.e. the monic case of the Rational Root Test. You can find much further discussion of this and related ideas in my posts on denominator ideals. | 2019-03-22T00:17:17 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3031094/a-proof-that-sqrt2-is-not-a-rational-number",
"openwebmath_score": 0.8867544531822205,
"openwebmath_perplexity": 1612.1159535398947,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9883127413158322,
"lm_q2_score": 0.8539127585282744,
"lm_q1q2_score": 0.8439328592256432
} |
http://math.stackexchange.com/questions/363012/abelian-group-admitting-a-surjective-homomorphism-onto-an-infinite-cyclic-group | # Abelian group admitting a surjective homomorphism onto an infinite cyclic group
I am working on the following problem:
Let $G$ an Abelian group and $f: G \to \Bbb Z$ a surjective homomorphism. Prove that $G \cong \ker(f) \times \Bbb Z$
By means of the First Isomorphism Theorem, we can obtain that $G / \ker(f) \cong \Bbb Z$. Is there some natural way to proceed from this point, invoking probably some other theorem, or would one need to define a map and prove that is it an isomorphism?
More generally, is it true that given an Abelian group $A$ and a subgroup $B$, $$A/B \cong C \Rightarrow A \cong C \times B$$
Thank you very much in advance!
-
You need to define the splitting map $g\colon \mathbb Z \to G$ (so that $fg$ is the identity). It should be clear how to do this, pick $x \in G$ such that $f(x) = 1$ and then define $g(n) = nx$.
Then you show that $g$ is injective and $G = \ker f \times \mathrm{im} \ g$. It's a faily elementary argument, just show that the two subgroups have trivial intersection and generate $G$.
As for your more general question the answer is no. For example $\mathbb Z/4$ has subgroup $\{0, 2\}$ isomorphic to $\mathbb Z/2$. The quotient is also isomorphic to $\mathbb Z/2$ but $\mathbb Z/4$ is not isomorphic to $\mathbb Z/2 \times \mathbb Z/2$.
On the other hand if $B \simeq \mathbb Z^n$ for some $n$ then it is true that $A \simeq A/B \times B$. To use more advanced language, the reason is that $\mathbb Z^n$ is a free $\mathbb Z$-module, hence projective, hence surjections to it are split. Basically, with $\mathbb Z^n$ you will always be able to define the splitting map $g$ and carry out the remainder of the argument above.
-
I am so very sorry for not understanding this immediately. Here is my point of confusion. I cannot see how $G = \ker(f) \times im(g) \implies G \cong \ker(f) \times \Bbb Z$, since $im(g) \subset G$ so you would get $G \cong \ker(f) \times im(g) \subset \ker(f) \times G$ – Orest Xherija Apr 16 '13 at 6:32
$g\colon \mathbb Z \to G$ is injective so $\mathrm{im} \ g$ is isomorphic to $\mathbb Z$. – Jim Apr 16 '13 at 6:49
I have tried to show that $\ker(g)$ is trivial, but I am failing to do so. My problem is that the $x$ you chose above might have finite order so you would have $nx=0$ with $n \neq 0$ thus the kernel will not be trivial. I cannot find a method to by-pass this problem. – Orest Xherija Apr 19 '13 at 4:54
If $f(x) = 1$ then what is $f(nx)$? Can $nx$ be zero? – Jim Apr 19 '13 at 6:03
I just realised my stupid mistake! Just one more thing. How can I show that any element of $G$ is expressible as a product of two elements, the first of which is from $\ker(f)$ and the second from $im(g)$? – Orest Xherija Apr 19 '13 at 6:23
Jim's answer gives you everything you need, but let me try to draw the bigger picture in other words.
What you are looking at is a short exact sequence, i.e. an exact sequence of the form
$$0\to A\xrightarrow{f} B\xrightarrow{g} C\to 0$$
This means you have three abelian groups and all arrows are group homomorphisms. Exactness means that at any given point in the sequence the kernel of the exiting map coincides with the image of the entering map. In particular this means that the kernel of $f$ is zero, i.e $f$ is injective. Similarly the image of $g$ is the whole of $C$, so $g$ is surjective. Moreover the kernel of $g$ is the image of $f$. But since $f$ is injective you can identify $A$ with its image, so $Ker(g)\cong A$.
In your intial example you have $B=G$, $C=\mathbb Z$ and $A=ker(G\to \mathbb Z)$.
Now a short exact sequence is called split if one of the following equivalent properties hold:
a) $$B\cong A\oplus C$$ b) There exists a group homomorphism $i:C\to B$ such that $g\circ i=id$
c) There exists a group homomorphism $p:B\to A$ such that $p\circ f=id$.
So the direct way of showing that $G\cong \mathbb Z\oplus ker(G\to \mathbb Z)$ would be constructing a map $\mathbb Z\to \mathbb G$ which satisfies property b).
A more general argument uses the definition of projectiveness:
An $R$-module $P$ (all abelian groups are $\mathbb Z$-modules) is called projective if for any surjective $R$-module homomorphism (homomorphism of abelian groups) $f:N\to M$ and any $R$-module homomorphism $g:P\to M$ there exists a lift $h:P\to N$ such that $f\circ h=g$.
Now you can show that $P$ is projective if and only if any short exact sequence of the form $$0\to A\xrightarrow{f} B\xrightarrow{g} P\to 0$$ splits.
You can also show that all free modules (the abelian group $\mathbb Z$ is clearly free as a module over itself) are projective.
- | 2015-05-30T17:33:02 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/363012/abelian-group-admitting-a-surjective-homomorphism-onto-an-infinite-cyclic-group",
"openwebmath_score": 0.967939019203186,
"openwebmath_perplexity": 97.15963935734612,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9883127420043056,
"lm_q2_score": 0.8539127566694177,
"lm_q1q2_score": 0.8439328579764076
} |
https://math.stackexchange.com/questions/2446145/will-the-convergence-of-frac1n-sum-n-1na-n-imply-the-convergence-of | # Will the convergence of $\frac{1}{N}\sum_{n=1}^{N}a_n$ imply the convergence of $\sum_{n=1}^{N}\frac{a_n}{n^2}$?
Assume that we have a positive sequence $\{a_n\}$ with its Cesàro mean converges: $$\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^{N}a_n<\infty\,.$$ I am wondering if the following summation converges: $$\lim_{N\to\infty}\sum_{n=1}^{N}\frac{a_n}{n^2}$$
Any proof is appreciated!
• Squeeze theorem? Would this work? – Jihoon Kang Sep 26 '17 at 15:43
• Do you know summation by parts? – Daniel Fischer Sep 26 '17 at 15:52
• @JihoonKang Sorry but how? – Did Sep 26 '17 at 17:03
• @Did I was (probably) wrong - I thought it might be possible without thinking through the details. – Jihoon Kang Sep 27 '17 at 1:11
• @JihoonKang May I suggest to avoid at all cost such random comments? They are a (minor) plague of the site. – Did Sep 27 '17 at 8:10
Define $s_0=0, s_n = a_1 + \cdots + a_n, n>0.$ Summing by parts (note Daniel Fischer mentioned this in a comment) gives
$$\sum_{k=1}^{n}\frac{a_k}{k^2}= \sum_{k=1}^{n}\frac{s_k-s_{k-1}}{k^2}= \frac{s_n}{n^2} + \sum_{k=1}^{n-1}s_k\cdot \left ( \frac{1}{k^2}-\frac{1}{(k+1)^2}\right)$$
The first term on the right $\to 0,$ so we can ignore it. The remaining sum equals
$$\sum_{k=1}^{n-1}s_k\cdot \frac{2k+1}{k^2(k+1)^2}= \sum_{k=1}^{n-1}\frac{s_k}{k}\cdot \frac{2k+1}{k(k+1)^2}.$$
The sequence $\dfrac{s_k}{k}$ is bounded. Since $\sum_{k=1}^{\infty}\dfrac{2k+1}{k(k+1)^2} < \infty$ we have
$$\sum_{k=1}^{\infty}\frac{s_k}{k}\cdot \frac{2k+1}{k(k+1)^2}<\infty.$$ This implies $\sum_{k=1}^{\infty}\dfrac{a_k}{k^2}<\infty.$
• Clear and concise. (+1) – Mark Viola Sep 26 '17 at 17:57
• @MarkViola Thank you sir. – zhw. Sep 26 '17 at 18:24
Given any real sequence $(a_n)_{n\ge 1}$, construct an auxillary sequence $(b_n)_{n\ge 0}$ by
$$b_n \stackrel{def}{=}\begin{cases}\frac1n\sum\limits_{k=1}^n a_k, & n > 0\\0, & n = 0\end{cases}$$ Whenever $b_n$ is bounded, i.e there is a $M > 0$ such that $|b_n| < M$ for all $n$, we can decompose the sum at hand into two pieces:
$$\sum_{n=1}^\infty \frac{a_n}{n^2} = \sum_{n=1}^\infty \frac{nb_n - (n-1)b_{n-1}}{n^2} = \sum_{n=1}^\infty \left(\frac{b_n}{n^2} + \frac{n-1}{n^2}(b_n - b_{n-1})\right)$$ The first piece $\displaystyle\;\sum_{n=1}^\infty \frac{b_n}{n^2}\;$ will be absolutely convergent by comparision test against $\displaystyle\;\sum_{n=1}^\infty\frac{M}{n^2}$.
The second piece $\displaystyle\;\sum_{n=1}^\infty \frac{(n-1)}{n^2} (b_n - b_{n-1})$ will be conditionally convergent by Dirichlet's test because $\displaystyle\;\frac{n-1}{n^2}\;$ is monontonic decreasing to zero for $n > 1$ and $\displaystyle\;\left|\sum_{n=1}^N (b_n - b_{n-1})\right| = |b_N| < M\;$ is bounded.
We don't really need $(a_n)$ be positive nor $(b_n)$ converges. As long as $b_n$ is bounded, the sum at hand converges.
• Thank you so much for the proof. One question: it seems that when $n=1$, $\frac{n-1}{n^2}=0$ and so $\frac{n-1}{n^2}$ is not monotonic decreasing on $[1,\,\infty)$. Will this invalidate the proof? Thx – user99015 Sep 26 '17 at 16:57
• @user99015 it won't invalidate the proof, the test remains to work if finitely many terms are missing the conditions. – achille hui Sep 26 '17 at 17:01
• Nicely done! (+1) – Mark Viola Sep 26 '17 at 17:57 | 2021-06-23T01:31:14 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2446145/will-the-convergence-of-frac1n-sum-n-1na-n-imply-the-convergence-of",
"openwebmath_score": 0.9079939126968384,
"openwebmath_perplexity": 504.1350469258439,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9883127440697254,
"lm_q2_score": 0.8539127548105611,
"lm_q1q2_score": 0.8439328579029641
} |
https://byjus.com/question-answer/internal-bisector-of-angle-mathrm-a-of-triangle-abc-meets-side-bc-at-d-a/ | Question
# Internal bisector of $$\angle \mathrm{A}$$ of triangle ABC meets side BC at D. A line drawn through $$\mathrm{D}$$ perpendicular to AD intersects the side AC at $$\mathrm{E}$$ and the side AB at F. If $$\mathrm{a},\ \mathrm{b},\ \mathrm{c}$$ represent sides of $$\Delta \mathrm{A}\mathrm{B}\mathrm{C}$$ then
A
AE is HM of b and c
B
C
EF=4bcb+csinA2
D
the triangle AEF is isosceles
Solution
## The correct options are A AE is HM of b and c B EF$$=\displaystyle \frac{4\mathrm{b}\mathrm{c}}{\mathrm{b}+\mathrm{c}}\sin\frac{\mathrm{A}}{2}$$ C the triangle AEF is isosceles D AD $$=\displaystyle \frac{2\mathrm{b}\mathrm{c}}{\mathrm{b}+\mathrm{c}}\cos\frac{\mathrm{A}}{2}$$We have $$\Delta \mathrm{A}\mathrm{B}\mathrm{C}=\Delta \mathrm{A}\mathrm{B}\mathrm{D}+\Delta \mathrm{A}\mathrm{C}\mathrm{D}$$ $$\Rightarrow \displaystyle \frac{1}{2}$$ $$bc$$ $$\displaystyle \sin \mathrm{A}=\frac{1}{2}\mathrm{c}\mathrm{A}\mathrm{D}\sin\frac{\mathrm{A}}{2}+\frac{1}{2}\mathrm{b}\times \mathrm{A}\mathrm{D}\sin\frac{\mathrm{A}}{2} \Rightarrow$$ $$AD$$ $$=\displaystyle \frac{2\mathrm{b}\mathrm{c}}{\mathrm{b}+\mathrm{c}}\cos\frac{\mathrm{A}}{2}$$Again $$AE$$ $$=$$ $$AD$$ $$\displaystyle \sec\frac{\mathrm{A}}{2}$$$$=\displaystyle \frac{2\mathrm{b}\mathrm{c}}{\mathrm{b}+\mathrm{c}}\Rightarrow$$ $$AE$$ is HM of $$\mathrm{b}$$ and $$\mathrm{c}$$.$$EF$$ $$=$$$$ED + DF$$$$=2\mathrm{D}\mathrm{E}=2\times AD$$$$\displaystyle \tan\frac{\mathrm{A}}{2}=\frac{2\times 2\mathrm{b}\mathrm{c}}{\mathrm{b}+\mathrm{c}}\times\cos\frac{\mathrm{A}}{2}\times\tan\frac{\mathrm{A}}{2}$$$$=\dfrac{4\mathrm{b}\mathrm{c}}{\mathrm{b}+\mathrm{c}}\sin\dfrac{\mathrm{A}}{2}$$ As $$AD$$ $$\perp$$ $$EF$$ and $$DE$$$$=$$$$DF$$ and $$AD$$ is bisector $$\Rightarrow$$ $$AEF$$ is isosceles. Hence $$\mathrm{A},\ \mathrm{B},\ \mathrm{C}$$ and $$\mathrm{D}$$ are correct answers. Maths
Suggest Corrections
0
Similar questions
View More
People also searched for
View More | 2022-01-24T19:47:08 | {
"domain": "byjus.com",
"url": "https://byjus.com/question-answer/internal-bisector-of-angle-mathrm-a-of-triangle-abc-meets-side-bc-at-d-a/",
"openwebmath_score": 0.7716891169548035,
"openwebmath_perplexity": 982.1148523721334,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9883127433812522,
"lm_q2_score": 0.8539127548105611,
"lm_q1q2_score": 0.8439328573150682
} |
http://mathhelpforum.com/calculus/204347-application-maxima-minima.html | # Math Help - Application of maxima and minima
1. ## Application of maxima and minima
Find the length of the longest rod which can be carried horizontally around a corner from a corridor 8m wide into one 4m wide. (Without involving angles if possible)
VIEW the attachment for the figure.
2. ## Re: Application of maxima and minima
Let:
$L^2=(8+x)^2+(4+y)^2$
By similarity, we have:
$\frac{y}{8}=\frac{4}{x}$
Can you proceed from here?
3. ## Re: Application of maxima and minima
I already did but it gave me a wrong answer. Maybe something is wrong with my solution. I'll check it.
4. ## Re: Application of maxima and minima
What did you get for your critical value?
5. ## Re: Application of maxima and minima
4*2^(2/3) on my second soltn and 8 on the first.
Btw, w/c function must I use? The one above or this one (L=a+b):
L(x)=a+b=sqrt(x^2+64)+sqrt(y^2+16) ??
6. ## Re: Application of maxima and minima
Hello, Kaloda!
This is a classic problem . . .
Find the length of the longest rod which can be carried horizontally around a corner
from a corridor 8m wide into one 4m wide.
Code:
| |
A * |
| * |
y| * |
| * | P
D + - - - * - - - - - - - -
| 4 : *
| : *
8| 8: *
| : *
| : *
* - - - + - - - - - * - - - -
C 4 E x B
Let $L \,=\,AB$ be the length of the rod.
. . Then: . $L \;=\;\sqrt{(x+4)^2 + (y+8)^2}$ .[1]
We see that $\Delta PEB \sim \Delta ADP.$
. . Hence: . $\frac{x}{8} \,=\,\frac{4}{y} \quad\Rightarrow\quad y \:=\:\frac{32}{x}$
Substitute into [1], set $L' \,=\,0$, and solve for maximum $L.$
7. ## Re: Application of maxima and minima
Okay. That's exactly how I solve the problem but I obtain a wrong answer! Maybe I'm missing something.
And ow, the book hints that this length (max) is the minimum of certain lengths.
8. ## Re: Application of maxima and minima
I get (substituting for y into the square of the length) the following as the critical value:
$x=4\sqrt[3]{2}$
I began with:
$L^2=(8+x)^2+(4+y)^2$
$L^2=(8+x)^2+\left(4+\frac{32}{x} \right)^2$
$L^2=(8+x)^2+16\left(1+\frac{8}{x} \right)^2$
Implicitly differentiating with respect to $x$, we have:
$2L\cdot\frac{dL}{dx}=2(8+x)+32\left(1+\frac{8}{x} \right)\left(-\frac{8}{x^2} \right)$
Since we must have $0, to find the extrema, we may equate:
$2(8+x)+32\left(1+\frac{8}{x} \right)\left(-\frac{8}{x^2} \right)=0$
$(8+x)-128\left(1+\frac{8}{x} \right)\left(\frac{1}{x^2} \right)=0$
$(8+x)-128\left(\frac{1}{x^2}+\frac{8}{x^3} \right)=0$
Multiplying through by $x^3$ and distributing, we find:
$x^4+8x^3-128x-1024=0$
Factor:
$(x+8)(x^3-128)=0$
Discarding the negative root, we have:
$x^3=128=2\cdot2^6$
$x=2^2\cdot\sqrt[3]{2}=4\sqrt[3]{2}$
I suspect you have set it up like Soroban did, so your critical value may be different from what I have.
9. ## Re: Application of maxima and minima
Oh yeahhh.! TY for the replies. I got the correct answer but I didn't realized that it just has a different form from the one given by the book.
Didn't know it til I checked their numerical values. I'm such a f*ol! | 2015-02-27T22:59:07 | {
"domain": "mathhelpforum.com",
"url": "http://mathhelpforum.com/calculus/204347-application-maxima-minima.html",
"openwebmath_score": 0.8425948619842529,
"openwebmath_perplexity": 2005.6053325893752,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.988312740283122,
"lm_q2_score": 0.8539127566694178,
"lm_q1q2_score": 0.843932856506667
} |
https://brilliant.org/discussions/thread/1-derivation-of-the-quadratic-formula/ | # 1) Derivation of the quadratic formula
This is note $1$ in a set of notes showing how to obtain formulas. There will be no words beyond these short paragraphs as the rest will either consist of images or algebra showing the steps needed to derive the formula mentioned in the title.
Suggestions for other formulas to derive are welcome, however whether they are completed or not depends on my ability to derive them. The suggestions given aren't guaranteed to be the next one in the set but they will be done eventually.
1 $\large ax^2 + bx + c = 0$
2 $\large x^2 + \frac{b}{a}x + \frac{c}{a} = 0$
3.1 $\large x^2 + \frac{b}{a}x = \left(x + \frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2$
3.2 $\large \left(x + \frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2 + \frac{c}{a} = 0$
4 $\large \left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2} + \frac{4ac}{4a^2} = 0$
5 $\large \left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$
6 $\large x + \frac{b}{2a} = \pm\sqrt{\frac{b^2 - 4ac}{4a^2}}$
7 $\large x + \frac{b}{2a} = \frac{\pm\sqrt{b^2 - 4ac}}{2a}$
8 $\large x = \frac{- b \pm\sqrt{b^2 - 4ac}}{2a}$
Note by Jack Rawlin
4 years, 8 months ago
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2
paragraph 1
paragraph 2
[example link](https://brilliant.org)example link
> This is a quote
This is a quote
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$
Sort by:
Yup! This is the derivation of the quadratic formula. Great!
- 4 years, 8 months ago
I made a YouTube video showing how this would work in arbitrary fields not of characteristic $2$. The field not having characteristic $2$ is so important because this avoids any of your calculations having zero determinants; in other words, the quantities $2$ and $4$ will be $0$ modulo $2$ and our prescribed condition avoids this occurring. In such a setting, steps 6, 7 and 8 will not be valid; solutions can only exist when $b^2 - 4ac$ is a square number in the field. This indicates the severe limitations of the Fundamental Theorem of Algebra in its scope only being applied in the framework of the "real number field" and the "complex number field".
- 2 years, 1 month ago | 2020-09-24T05:12:39 | {
"domain": "brilliant.org",
"url": "https://brilliant.org/discussions/thread/1-derivation-of-the-quadratic-formula/",
"openwebmath_score": 0.9564194679260254,
"openwebmath_perplexity": 1398.7704980577782,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9883127447581985,
"lm_q2_score": 0.8539127473751341,
"lm_q1q2_score": 0.843932851142333
} |
https://apmonitor.com/pdc/index.php/Main/ModelLinearization | Main
## Linearization of Differential Equations
Linearization is the process of taking the gradient of a nonlinear function with respect to all variables and creating a linear representation at that point. It is required for certain types of analysis such as stability analysis, solution with a Laplace transform, and to put the model into linear state-space form. Consider a nonlinear differential equation model that is derived from balance equations with input u and output y.
$$\frac{dy}{dt} = f(y,u)$$
The right hand side of the equation is linearized by a Taylor series expansion, using only the first two terms.
$$\frac{dy}{dt} = f(y,u) \approx f \left(\bar y, \bar u\right) + \frac{\partial f}{\partial y}\bigg|_{\bar y,\bar u} \left(y-\bar y\right) + \frac{\partial f}{\partial u}\bigg|_{\bar y,\bar u} \left(u-\bar u\right)$$
If the values of \bar u and \bar y are chosen at steady state conditions then f(\bar y, \bar u)=0 because the derivative term {dy}/{du}=0 at steady state. To simplify the final linearized expression, deviation variables are defined as y' = y-\bar y and u' = u - \bar u. A deviation variable is a change from the nominal steady state conditions. The derivatives of the deviation variable is defined as {dy'}/{dt} = {dy}/{dt} because {d\bar y}/{dt} = 0 in {dy'}/{dt} = {d(y-\bar y)}/{dt} = {dy}/{dt} - \cancel{{d\bar y}/{dt}}. If there are additional variables such as a disturbance variable d then it is added as another term in deviation variable form d' = d - \bar d.
$$\frac{dy'}{dt} = \alpha y' + \beta u' + \gamma d'$$
The values of the constants \alpha, \beta, and \gamma are the partial derivatives of f(y,u,d) evaluated at steady state conditions.
$$\alpha = \frac{\partial f}{\partial y}\bigg|_{\bar y,\bar u,\bar d} \quad \quad \beta = \frac{\partial f}{\partial u}\bigg|_{\bar y,\bar u,\bar d} \quad \quad \gamma = \frac{\partial f}{\partial d}\bigg|_{\bar y,\bar u,\bar d}$$
#### Example
Part A: Linearize the following differential equation with an input value of u=16.
$$\frac{dx}{dt} = -x^2 + \sqrt{u}$$
Part B: Determine the steady state value of x from the input value and simplify the linearized differential equation.
Part C: Simulate a doublet test with the nonlinear and linear models and comment on the suitability of the linear model to represent the original nonlinear equation solution.
Part A Solution: The equation is linearized by taking the partial derivative of the right hand side of the equation for both x and u.
$$\frac{\partial \left(-x^2 + \sqrt{u}\right)}{\partial x} = \alpha = -2 \, x$$
$$\frac{\partial \left(-x^2 + \sqrt{u}\right)}{\partial u} = \beta = \frac{1}{2} \frac{1}{\sqrt{u}}$$
The linearized differential equation that approximates \frac{dx}{dt}=f(x,u) is the following:
$$\frac{dx}{dt} = f \left(x_{ss}, u_{ss}\right) + \frac{\partial f}{\partial x}\bigg|_{x_{ss},u_{ss}} \left(x-x_{ss}\right) + \frac{\partial f}{\partial u}\bigg|_{x_{ss},u_{ss}} \left(u-u_{ss}\right)$$
Substituting in the partial derivatives results in the following differential equation:
$$\frac{dx}{dt} = 0 + \left(-2 x_{ss}\right) \left(x-x_{ss}\right) + \left(\frac{1}{2} \frac{1}{\sqrt{u_{ss}}}\right) \left(u-u_{ss}\right)$$
This is further simplified by defining new deviation variables as x' = x - x_{ss} and u' = u - u_{ss}.
$$\frac{dx'}{dt} = \alpha x' + \beta u'$$
Part B Solution: The steady state values are determined by setting \frac{dx}{dt}=0 and solving for x.
$$0 = -x_{ss}^2 + \sqrt{u_{ss}}$$
$$x_{ss}^2 = \sqrt{16}$$
$$x_{ss} = 2$$
At steady state conditions, frac{dx}{dt}=0 so f (x_{ss}, u_{ss})=0 as well. Plugging in numeric values gives the simplified linear differential equation:
$$\frac{dx}{dt} = -4 \left(x-2\right) + \frac{1}{8} \left(u-16\right)$$
The partial derivatives can also be obtained from Python, either symbolically with SymPy or else numerically with SciPy.
# analytic solution with Python
import sympy as sp
sp.init_printing()
# define symbols
x,u = sp.symbols(['x','u'])
# define equation
dxdt = -x**2 + sp.sqrt(u)
print(sp.diff(dxdt,x))
print(sp.diff(dxdt,u))
# numeric solution with Python
import numpy as np
from scipy.misc import derivative
u = 16.0
x = 2.0
def pd_x(x):
dxdt = -x**2 + np.sqrt(u)
return dxdt
def pd_u(u):
dxdt = -x**2 + np.sqrt(u)
return dxdt
print('Approximate Partial Derivatives')
print(derivative(pd_x,x,dx=1e-4))
print(derivative(pd_u,u,dx=1e-4))
print('Exact Partial Derivatives')
print(-2.0*x) # exact d(f(x,u))/dx
print(0.5 / np.sqrt(u)) # exact d(f(x,u))/du
Python program results
-2*x
1/(2*sqrt(u))
Approximate Partial Derivatives
-4.0
0.125
Exact Partial Derivatives
-4.0
0.125
The nonlinear function for \frac{dx}{dt} can also be visualized with a 3D contour map. The choice of steady state conditions x_{ss} and u_{ss} produces a planar linear model that represents the nonlinear model only at a certain point. The linear model can deviate from the nonlinear model if used further away from the conditions at which the linear model is derived.
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
from matplotlib import cm
from matplotlib.ticker import LinearLocator, FormatStrFormatter
import numpy as np
fig = plt.figure()
ax = fig.gca(projection='3d')
# Make data.
X = np.arange(0, 4, 0.25)
U = np.arange(0, 20, 0.25)
X, U = np.meshgrid(X, U)
DXDT = -X**2 + np.sqrt(U)
LIN = -4.0 * (X-2.0) + 1.0/8.0 * (U-16.0)
# Plot the surface.
surf = ax.plot_wireframe(X, U, LIN)
surf = ax.plot_surface(X, U, DXDT, cmap=cm.coolwarm,
linewidth=0, antialiased=False)
# Customize the z axis.
ax.set_zlim(-10.0, 5.0)
ax.zaxis.set_major_locator(LinearLocator(10))
ax.zaxis.set_major_formatter(FormatStrFormatter('%.02f'))
# Add a color bar which maps values to colors.
fig.colorbar(surf, shrink=0.5, aspect=5)
plt.xlabel('x')
plt.ylabel('u')
plt.show()
Part C Solution: The final step is to simulate a doublet test with the nonlinear and linear models.
Small step changes (+/-1): Small step changes in u lead to nearly identical responses for the linear and nonlinear solutions. The linearized model is locally accurate.
Large step changes (+/-8): As the magnitude of the doublet steps increase, the linear model deviates further from the original nonlinear equation solution.
import numpy as np
from scipy.integrate import odeint
import matplotlib.pyplot as plt
# function that returns dz/dt
def model(z,t,u):
x1 = z[0]
x2 = z[1]
dx1dt = -x1**2 + np.sqrt(u)
dx2dt = -4.0*(x2-2.0) + (1.0/8.0)*(u-16.0)
dzdt = [dx1dt,dx2dt]
return dzdt
x_ss = 2.0
u_ss = 16.0
# initial condition
z0 = [x_ss,x_ss]
# final time
tf = 10
# number of time points
n = tf * 10 + 1
# time points
t = np.linspace(0,tf,n)
# step input
u = np.ones(n) * u_ss
# magnitude of step
m = 8.0
# change up m at time = 1.0
u[11:] = u[11:] + m
# change down 2*m at time = 4.0
u[41:] = u[41:] - 2.0 * m
# change up m at time = 7.0
u[71:] = u[71:] + m
# store solution
x1 = np.empty_like(t)
x2 = np.empty_like(t)
# record initial conditions
x1[0] = z0[0]
x2[0] = z0[1]
# solve ODE
for i in range(1,n):
# span for next time step
tspan = [t[i-1],t[i]]
# solve for next step
z = odeint(model,z0,tspan,args=(u[i],))
# store solution for plotting
x1[i] = z[1][0]
x2[i] = z[1][1]
# next initial condition
z0 = z[1]
# plot results
plt.figure(1)
plt.subplot(2,1,1)
plt.plot(t,u,'g-',linewidth=3,label='u(t) Doublet Test')
plt.grid()
plt.legend(loc='best')
plt.subplot(2,1,2)
plt.plot(t,x1,'b-',linewidth=3,label='x(t) Nonlinear')
plt.plot(t,x2,'r--',linewidth=3,label='x(t) Linear')
plt.xlabel('time')
plt.grid()
plt.legend(loc='best')
plt.show() | 2019-03-26T16:03:08 | {
"domain": "apmonitor.com",
"url": "https://apmonitor.com/pdc/index.php/Main/ModelLinearization",
"openwebmath_score": 0.7025504112243652,
"openwebmath_perplexity": 2979.089223695814,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9883127399388854,
"lm_q2_score": 0.853912747375134,
"lm_q1q2_score": 0.84393284702706
} |
https://math.stackexchange.com/questions/4412847/why-is-the-range-a-larger-set-than-the-domain | # Why is the range a larger set than the domain?
When we have a function $$f: \mathbb{R} \to \mathbb{R}$$, I can intuitively picture that and think that for every $$x \in \mathbb{R}$$, we can find a $$y \in \mathbb{R}$$ such that our function $$f$$ maps $$x$$ onto $$y$$.
I'm confused, however, when we have something like: $$g: D \to \mathbb{R}$$, where $$D$$ is the domain of our function such that $$D \subset \mathbb{R}$$. How can our function output every element in $$\mathbb{R}$$, when our input was specifically less than the whole set of reals?
• Take $\tan(x)$ on the domain $(-\pi/2,\pi/2)$. Plot this in say, Desmos. Then you'll see how it can happen. Mar 26 at 1:03
• Muse on the function $\tan: (-\pi/2,\pi/2)\to {\mathbb R}$. Mar 26 at 1:04
• As a perhaps-simpler example, consider $f(x)=2x$ with domain $[0,1]$ and range $[0,2]$. "Proper subset" doesn't actually mean "less" - this is a big nonintuitive feature that infinity brings to the table! Mar 26 at 1:04
• Welcome to the struggle with the mind warping concept that is infinity. I always think of the Schröder-Bernstein theorem: en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorem Mar 26 at 1:17
Perhaps I'm understanding your question differently than some of the other commenters, but I'll point out that saying you have a function $$g: D \to \mathbb{R}$$ does NOT mean every element of $$\mathbb{R}$$ gets outputted. Here, the set that comes after the arrow is something called a codomain which can be larger than the range. If every element of your codomain gets outputted, then your function has a specific property called surjectivity, but not every function will have this. For example, consider the function $$f: \mathbb{R} \to \mathbb{R}$$ given by $$f(x) = x^2$$. Here, $$\mathbb{R}$$ is NOT the range of our function since there is no real number that maps to the number $$-1 \in \mathbb{R}$$ under $$f$$.
• Thanks for explaining this. If I understood you correctly, that would mean that with such notation we tell more about the 'type' of the output (real, complex, etc.) than the actual range of the function. Mar 26 at 1:13
• @nocomment Yes that’s exactly the right way to think about the notation! Mar 26 at 1:22
• @DavidK thanks for pointing that out, my mistake! Just fixed it Mar 26 at 1:23
• @nocomment a standard notation for the set of points that $f:D\rightarrow \mathbb{R}$ can take is $f(D)$, which will always satisfy $f(D) \subseteq \mathbb{R}$
– Joe
Mar 26 at 9:30
I don't agree with your statement that the range is a larger set than the domain.
I give some counter-examples to illustrate.
If you take any constant function defined as $$f: R \rightarrow R$$, $$f(x) \equiv k$$ (fixed value of $$k$$), then
Domain$$(f) = R$$
Range$$(f) = \{ k \}$$
As another example, take the 'sign' function defined as
$$g: R \rightarrow R$$, where $$g(x) = \mbox{sign}(x)$$.
Then
Domain($$g$$) = $$R$$
Range($$g$$) = $$\{ -1, 0, 1 \}$$
(It is a convention to define sign(0) = 0.)
If you define $$h: R \rightarrow R$$ as $$h(x) = | x |$$, the absolute value of $$x$$, then
Domain($$h$$) = $$R$$
Range($$h$$) = non-negative reals, which is a smaller set than $$R$$.
• What about the function $tan(x)$, when we constrain the domain to $[-pi/2, pi/2]}$ which others have pointed out above? Mar 26 at 1:23
• Obviously, for different functions, the ranges will be different. You cannot make a general statement that ranges will be larger or smaller than (or same as) the domain of a given function. This is purely the property of a given function. A general statement cannot be made. Thanks! Mar 26 at 1:31
The underlying idea of your question is the concept of cardinality. We say that two sets $$X$$ and $$Y$$ have the same cardinality iff the exists a bijection $$f:X\to Y$$ between them. Such relation is an equivalence relation, that is to say, it is reflexive, symmetric and transitive. Hence each cardinality is an equivalent class. We can establish a linear order relation among the possible cardinalities. Finally, we say that that a set $$X$$ is infinite iff there exists a bijection between $$X$$ and a proper subset of it.
For example, the set of natural numbers is infinity because $$f:\textbf{N}\to P$$, where $$f(n) = 2n$$ is a bijection between $$\textbf{N}$$ and the set $$P := \{n\in\textbf{N} \mid (n = 2k)\wedge(k\in\textbf{N})\}\subset\textbf{N}$$.
At the given example, one can consider the function $$\tan:(-\pi/2,\pi/2)\to\textbf{R}$$ which is bijective (as suggested by @MichaelMorrow). Therefore, even though the interval $$(-\pi/2,\pi/2)$$ is a proper subset of $$\textbf{R}$$, they have the same cardinality, hence the set of real numbers $$\textbf{R}$$ is infinity.
One interesting result is that $$\operatorname{card}(\textbf{N}) < \operatorname{card}(\textbf{R})$$, but it is not known if there is a cardinality between both. Cantor has conjectured it and such statement is known as the continuum hypothesis.
• Re: the last paragraph, it is in fact known that the usual axioms of mathematics cannot decide the question either way (unless they turn out to be inconsistent, in which case they decide it both ways :P); see here. Mar 26 at 1:19
• @NoahSchweber thanks for the reference. Mar 26 at 1:56
If $$A$$ is a finite set, then it is not possible for there to be a function $$f:A\to B$$ such that the range is a proper superset of the domain. However, this statement is not true for infinite sets, which is just one reason why they are so counter-intuitive. For instance, we can construct a surjective function $$f:\mathbb Z^+\to\mathbb Z$$ by mapping the odd positive integers to the nonnegative integers, and mapping the even positive integers to the negative integers. Similarly, the function $$f:(-\pi/2,\pi/2)\to\mathbb R$$ given by $$f(x)=\tan x$$ is surjective, even though $$(-\pi/2,\pi/2)$$ is clearly a proper subset of $$\mathbb R$$. | 2022-08-19T17:36:32 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/4412847/why-is-the-range-a-larger-set-than-the-domain",
"openwebmath_score": 0.9404276609420776,
"openwebmath_perplexity": 185.06046084182358,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9674102542943774,
"lm_q2_score": 0.8723473862936942,
"lm_q1q2_score": 0.8439178068074182
} |
https://puzzling.stackexchange.com/questions/54842/trailing-zeros-how-many-trailing-zeros-are-there-in-100-factorial-of-100/54844 | # Trailing Zeros - How many trailing zeros are there in 100! (factorial of 100)?
Here's the question: How many trailing zeros are there in 100! (factorial of 100)?
Here's the solution: This is an easy problem. We know that each pair of 2 and 5 will give a trailing zero. If we perform prime number decomposition on all the numbers in 100!, it is obvious that the frequency of 2 will far outnumber of the frequency of 5. So the frequency of 5 determines the number of trailing zeros. Among numbers 1,2,....,99, and 100, 20 numbers are divisible by 5 (5, 10, ...., 100). Among these 20 numbers, 4 are divisible by 5^2 (25, 50, 75, 100). So the total frequency of 5 is 24 and there are 24 trailing zeros.
What I don't understand is... what is trailing zero? And how did the author use 2 and 5? (2 and 5 seem pretty random to me).
One thing is clear. $2 \times 5 = 10$ and there is no other way to get 10 out of 2 prime numbers.
"trailing zeros" are the zeros at the end of the number.
For example: 3200 has 2 trailing zeros. The units and the tenths position.
One other thing is clear.
Multiplying a number by 10 adds a trailing zero to that number.
So in order to find the number of zeros at the tail of a number, you need to split that number into prime factors and see how many pairs (2, 5) you can form.
For example:
300 has 2 trailing zeros. Why?
because $300 = 3 \times 2 ^ 2 \times 5^2$.
So you get 2 pairs of (5, 2).
An other example:
$4000 = 2^5 \times 5 ^3 = 2 ^2 \times (2 \times 5) ^ 3$
Hence, 3 trailing zeros.
EDIT:
$100! = 1 \times 2 \times 3 \times .... \times 100$.
Which is equivalent to splitting every number into prime factors like this:
$100! = 1 \times 2 \times 3 \times 2^2 \times 5 \times (3\times 2) .... \times (2 \times 5)^2$
You need to count how many 10's you can make out of these prime numbers.
As explained above, the only way to get a 10 is $2 \times 5$.
So you need to count how many 2's and how many 5s are among the prime factors above.
The 5s can appear only in numbers that are divisible by 5.
5, 10, 20, 25, 30, 35, 40, 45, 50 , 55, 60 , 65, 70, 75, 80, 85, 90, 95, 100
and the number of 5s in the numbers above are
1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2
Because for example
$15 = 3 \times 5$ so one single 5.
$50 = 2 \times 5 ^2$ so 2 fives.
Add them all up and you get 24 occurrences of 5.
Now you need to see if there are at least 24 occurrences of 2.
And they are because form 1 to 100 you have 50 even numbers (they all contain at least one 2).
This means that 100! can be written as
$2 ^ {50} \times 5 ^ {24} \times$ (some other factors that are never going to multiply to get to a multiple of 10 so we can ignore them).
We can split the line above (ignoring the factors that cannot multiply to reach a multiple of 10) to
$2 ^ {24} \times 2 ^{26} \times 5^{24} = (2 \times 5) ^{24} \times 2 ^{26}$
We can ignore again $2^{26}$ because that never ends with zeros (it ends with a 4 if I'm not mistaken)
Now you can see that 100! is basically a very large number that does not end with a 0 multiplied by $10 ^ {24}$. So you get 24 trailing zeros.
• Thank you very much for your comment. I definitely understand it better. There are two questions: 1. How can this concept be applied to a factorial? 2. Why does the statement 'the frequency of 2 will far outnumber the frequency of 5' matter? – Jun Jang Sep 4 '17 at 15:23
• See my edit. I added a few explanations for your specific case. – Marius Sep 4 '17 at 15:37
Trailing zeroes are as the name points zeroes in the end of the number. So 10 has 1 trailing zero. And because this is a question regarding base10 numbers, this is how you can represent any number with trailing zero - number0 = number x 10. And because 10 is actually 2 x 5 you need 2s and 5s. One 2 is enough to 'turn' all fives into zeroes. And you already have posted the number of 5s.
• Thank you very much for your comment. I definitely understand it better. There are two questions: 1. How can this concept be applied to a factorial? 2. Why does the statement 'the frequency of 2 will far outnumber the frequency of 5' matter? – Jun Jang Sep 4 '17 at 15:31 | 2021-01-25T14:51:35 | {
"domain": "stackexchange.com",
"url": "https://puzzling.stackexchange.com/questions/54842/trailing-zeros-how-many-trailing-zeros-are-there-in-100-factorial-of-100/54844",
"openwebmath_score": 0.5877329111099243,
"openwebmath_perplexity": 329.67958230193784,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9674102542943774,
"lm_q2_score": 0.8723473846343394,
"lm_q1q2_score": 0.8439178052021413
} |
https://csharp-book.softuni.org/Content/Chapter-7-2-complex-loops-exam-problems/magic-combination/magic-combination.html | Problem: Magic Numbers
Write a program that enters a single integer magic number and produces all possible 6-digit numbers for which the product of their digits is equal to the magical number.
Example: "Magic number" → 2
• 111112 → 1 * 1 * 1 * 1 * 1 * 2 = 2
• 111121 → 1 * 1 * 1 * 1 * 2 * 1 = 2
• 111211 → 1 * 1 * 1 * 2 * 1 * 1 = 2
• 112111 → 1 * 1 * 2 * 1 * 1 * 1 = 2
• 121111 → 1 * 2 * 1 * 1 * 1 * 1 = 2
• 211111 → 2 * 1 * 1 * 1 * 1 * 1 = 2
Input Data
The input is read from the console and consists of one integer within the range [1 … 600 000].
Output Data
Print on the console all magic numbers, separated by space.
Sample Input and Output
Input Output Input Output Input Output
2 111112 111121 111211 112111 121111 211111 8 111118 111124 111142 111181 111214 111222 111241 111412 111421 111811 112114 112122 112141 112212 112221 112411 114112 114121 114211 118111 121114 121122 121141 121212 121221 121411 122112 122121 122211 124111 141112 141121 141211 142111 181111 211114 211122 211141 211212 211221 211411 212112 212121 212211 214111 221112 221121 221211 222111 241111 411112 411121 411211 412111 421111 811111 531441 999999
Solution using a "Fr" Loop
The solution follows the same concept (again we need to generate all combinations for the n element). Following these steps, try to solve the problem yourself.
• Declare and initialize variable of int type and read the input from the console.
• Nest six for loops one into another, for every single digit of the searched 6-digit numbers.
• In the last loop, using if construction, check if the product of the six digits is equal to the magic number.
Solution using a "While" Loop
In the previous chapter we reviewed other loop constructions. Let's look at the sample solution of the same problem using the while loop.
First, we need to store the input magical number in a suitable variable. Then we will initialize 6 variables – one for each of the six digits of the searched numbers.
Writing a While Loop
Then we will start writing while loops.
• We will initialize first digit: d1 = 0.
• We will set a condition for each loop: the digit will be less than or equal to 9.
• In the beginning of each loop we set a value of the next digit, in this case: d2 = 0. In the nested for loops we initialize the variables in the inner loops at each increment of the outer ones. We want to do the same here.
• At the end of each loop we will increase the digit by one: d++.
• In the innermost loop we will make the check and if necessary, we will print on the console.
Infinite While Loop
Let's remove the if check from the innermost loop. Now, let's initialize each variable outside of the loops and delete the rows dx = 0. After we run the program, we only get 10 results. Why? What if you use do-while? In this case this loop does not look appropriate, does it?Think why. Of course, you can solve the problem using an infinite loop.
As we can see, we can solve a problem using different types of loops. Of course, each task has its most appropriate choice. In order to practice each type of loops – try to solve each of the following problems with all the learned loops.
Testing in the Judge System
Test your solution here: https://judge.softuni.org/Contests/Practice/Index/515#1. | 2019-11-13T15:49:06 | {
"domain": "softuni.org",
"url": "https://csharp-book.softuni.org/Content/Chapter-7-2-complex-loops-exam-problems/magic-combination/magic-combination.html",
"openwebmath_score": 0.43166854977607727,
"openwebmath_perplexity": 286.15106899477524,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9674102571131691,
"lm_q2_score": 0.8723473779969194,
"lm_q1q2_score": 0.8439178012399987
} |
https://freedocs.mi.hdm-stuttgart.de/sd1SectPlayingLottery.html | #### Loops and calculations
No. 77
##### Display all summands
Q:
In Figure 207, “Calculating values ” for a given value of limit only the first and last summand will show up: We just see 1 + ... + 5 = 15 rather than 1 + 2 + 3 + 4 + 5 = 15.
If LIMIT is equal to one the visible result is even worse:
1 + ... + 1 = 1
Modify the code accordingly to correct these flaws / shortcomings.
### Tip
Using System.out.print(...) rather than System.out.println(...) avoids printing a newline and thus allows for continuation within a given line of output.
A:
for (int i = 1; i <= LIMIT; i++) {
System.out.print(i);
System.out.print(" + ");
sum += i;
}
System.out.println(" = " + sum);
This is close to the intended output. However our sum ends with a redundant trailing + symbol:
1 + 2 + 3 + 4 + 5 + = 15
We get rid of it by introducing an if statement:
for (int i = 1; i <= LIMIT; i++) {
System.out.print(i);
if (i < LIMIT) { // Avoid '+' for the last value
System.out.print(" + ");
}
sum += i;
}
System.out.println(" = " + sum);
This avoids the trailing +:
1 + 2 + 3 + 4 + 5 = 15
Moreover it works for LIMIT = 1 as well:
1 = 1
Instead of filtering the very last + operator we may as well terminate our loop one step earlier and move the last operand to the second print statement leaving us with an identical result:
int LIMIT = 5;
int sum = 0;
for (int i = 1; i < LIMIT; i++) {
System.out.print(i + " + ");
sum += i;
}
sum += LIMIT; // account for last term
System.out.println(LIMIT + " = " + sum);
No. 78
##### Playing lottery
Q:
Common lotteries randomly draw numbers from a given set like:
Germany / Glücksspirale
6 out of 49
Italy / SuperEnalotto
6 out of 90
This category of lottery does not care about ordering (apart from so called jolly numbers). The number of possibilities for drawing k out of n numbers ignoring their ordering is being given by the binomial coefficient $( n k )$:
$( n k ) = n ! k ! ( n - k ) !$
Write an application which allows for determining the probabilistic success rates using this coefficient. For the German Glücksspirale a possible output reads:
Your chance to win when drawing 6 out of 49 is 1 : 13983816
Store parameters like 6 or 49 in variables to keep your software flexible.
### Tip
1. Why is it a bad idea to e.g. compute $49!$ directly even when using long variables? Remember Figure 144, “Arithmetic overflow pitfalls ”.
2. Arithmetic overflow problems can be avoided by simplifying $n ! k ! ( n - k ) !$ beforehand. You may want to write down an explicit example like $( 8 3 ) = 8 ! 3 ! ( 8 - 3 ) !$ to learn about cancellation of contributing factors.
A:
Consider the following snippet:
int product = 1;
for (int i = 1; i < 50; i++) {
product *= i;
System.out.println(i + "! == " + product);
}
This results in:
1! == 1
2! == 2
3! == 6
4! == 24
5! == 120
6! == 720
7! == 5040
8! == 40320
9! == 362880
10! == 3628800
11! == 39916800
12! == 479001600
13! == 1932053504
...
49! == 0
Only results up to $12!$ are correct: The next term $13!$ equals 6227020800 which is ways beyond Integer.MAX_VALUE == 2147483647 giving rise to a (silent) arithmetic overflow. But even declaring our variable product of type long does not help much:
1! == 1
2! == 2
3! == 6
4! == 24
5! == 120
6! == 720
7! == 5040
8! == 40320
9! == 362880
10! == 3628800
11! == 39916800
12! == 479001600
13! == 6227020800
14! == 87178291200
15! == 1307674368000
16! == 20922789888000
17! == 355687428096000
18! == 6402373705728000
19! == 121645100408832000
20! == 2432902008176640000
21! == -4249290049419214848
...
49! == 8789267254022766592
This time 20! == 2432902008176640000 is the last correct value being smaller than Long.MAX_VALUE == 9223372036854775807.
Fortunately we have another option. Consider an example:
$( 8 3 ) = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 ( 3 × 2 × 1 ) × ( 5 × 4 × 3 × 2 × 1 ) = 8 × 7 × 6 3 × 2 × 1 = 56$
We generalize this fraction cancellation example:
Equation 1. Calculating binomials by cancelling common factors.
$n ! k ! ( n - k ) ! = n ( n - 1 ) ... 1 k ( k - 1 ) ... 1 ( n - k ) ( n - k - 1 ) ... 1 = n ( n - 1 ) ... ( n - k + 1 ) k ( k - 1 ) ... 1$
And thus:
$( 49 6 ) = 49 × 48 × 47 × 46 × 45 × 44 6 × 5 × 4 × 3 × 2 × 1 = 13983816$
We calculate numerator and denominator in two separate loops:
public class Lottery {
public static void main(String[] args) {
// Input values:
final int
totalNumberCount = 49,
drawnNumberCount = 6;
// No changes below this line
// Numerator loop
long numerator = 1;
for (int i = totalNumberCount - drawnNumberCount + 1;
i <= totalNumberCount; i++) {
numerator *= i;
}
// Denominator loop calculating the "smaller" factorial
long factorial = 1;
for (int i = 2; i <= drawnNumberCount; i++) {
factorial *= i;
}
System.out.println("Your chance to win when drawing " + drawnNumberCount +
" out of " + totalNumberCount + " is 1 : " + (numerator / factorial));
}
} | 2019-07-17T02:37:51 | {
"domain": "hdm-stuttgart.de",
"url": "https://freedocs.mi.hdm-stuttgart.de/sd1SectPlayingLottery.html",
"openwebmath_score": 0.24367818236351013,
"openwebmath_perplexity": 3636.7259905043843,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9674102514755852,
"lm_q2_score": 0.8723473829749844,
"lm_q1q2_score": 0.8439178011378983
} |
https://math.stackexchange.com/questions/1430938/why-are-inner-products-defined-to-be-linear-in-the-first-argument-only | # Why are inner products defined to be linear in the first argument only?
It seems to me that if the base field is the real numbers, then we have linearity in both arguments i.e. $\langle u + v, w + z\rangle = \langle u,w\rangle + \langle u,z\rangle + \langle v,w\rangle + \langle v,z\rangle$ because we know $\langle x,y\rangle = \langle y,x\rangle$ for any $y,x$
Do we only define inner products to be linear in their first argument in case the base field is the complex numbers?
Could we have just defined inner products over the real numbers to say that inner products are linear in both arguments?
• When choosing definitions, mathematicians usually like to make them as simple as possible, to make it easier to prove that a particular object satisfies that definition. The linearity in the second coordinate (over $\mathbb{R}$) follows as a lemma. Sep 11 '15 at 14:28
• Perhaps relevant: math.stackexchange.com/questions/1429174. Sep 11 '15 at 14:31
• I changed $<u,v>$ to $\langle u,v\rangle$. That is standard usage. ${}\qquad{}$ Sep 11 '15 at 14:34
• It depends what standard means. Even $\langle u | v\rangle$ has been standard for long time now (if you are a physicist). Mar 29 at 11:56
The issue is that if you do so, you'll get $$\Vert \lambda u \Vert^2 = \langle \lambda u, \lambda u \rangle = \lambda^2 \langle u, u \rangle = \lambda^2 \Vert u \Vert^2$$ and you can't ensure that all those numbers are real numbers. While if $$\Vert \lambda u \Vert^2 = \langle \lambda u, \lambda u \rangle = \lambda \overline{\lambda} \langle u, u \rangle = \vert \lambda \vert^2 \Vert u \Vert^2$$ you get compatibility with the definition of a norm.
We might take an arbitrary bilinear from instead, but: We want to be able to define a norm (and ultimately a topology) from our inner product. First of all, this prevents us from talking about inner products in characteristic $\ne 0$. We also get difficculties if the groud field is larger than $\mathbb C$. Remains to cover the fields from $\mathbb Q$ up to $\mathbb C$, of which $\mathbb R$ is just a special case. To cover all cases at once, it suffices to adjust the definition to the case of $\mathbb C$, where we need the conjugate symmetry (and hence sesquilinearity) instead of bilinerity in order to have $\langle x,x\rangle\in\mathbb R$ (for our intended norm).
They are sesquilinear (which means one and a half times linear), not merely linear in the first variable, i.e. it is additive also in the second variable, and $$f(u,\lambda v)=\bar\lambda f(u,v)$$ This is to achieve the axioms of a norm in the context of $\mathbf C$-vector spaces: $$f(\lambda ua,\lambda u)=\lambda\bar\lambda f(u,u)=\lvert\lambda\rvert^2f(u,u).$$
• I'm sorry, but I am awful at remembering mathematical notation...does this bar represent the conjugate of lambda and if lambda is a real number does this just become $f(u,\lambda v) = \lambda f(u,v)$ Sep 12 '15 at 6:01
• You're absolutely right. Sep 12 '15 at 8:40
If it's a real inner product, then it's symmetric, i.e. $\langle x,y\rangle = \langle y, x\rangle$, so linearity in one argument implies linearity in the other; hence keeping the list of axioms non-redundant requires mentioning only one explicitly.
If it's a complex inner product, then $\langle x,y\rangle$ and $\langle y,x\rangle$ are not equal to each other but conjugates of each other. In that case it's conjugate-linear in the second argument, i.e. $\langle x,\lambda y\rangle$ is not $\lambda\langle x,y\rangle$, but $\bar\lambda\langle x,y\rangle$.
• Thank you...this makes it clear. I figured that if we ONLY cared about real inner product spaces we could say that it's linear in both arguments but all of the definitions (I read) for an inner product space explicitly mentioned linearity only in the first argument and I wanted to make sure I wasn't missing anything because it seemed to me like if we are ONLY talking about real inner product spaces, we have linearity in both arguments Sep 12 '15 at 6:03
It seems strange that nobody has given the following proof. The most easy way to understand this is working with complex numbers, and for simplicity I will work in $$\mathbb{C}^n$$. A function $$(\cdot, \cdot)$$ from $$\mathbb{C}^n \times \mathbb{C}^n$$ to $$\mathbb{C}$$ is an inner product if it satisfies (note that I switched a little bit the inner product definition!)
1. $$(\cdot, \cdot)$$ is linear in the second argument, $$\left(\vec{v},\sum_{i} c_i \vec{w}_i \right)=\sum_{i} c_i (\vec{v},\vec{w}_i),$$ with $$\vec{v},\vec{w}_i \in \mathbb{C}^n$$.
2. $$(\vec{v},\vec{w})=(\vec{w},\vec{v})^*$$, where $$*$$ means complex conjugation.
3. $$(\vec{v},\vec{v})\geq 0$$ with equality if and only if $$\vec{v}=0$$.
By definition, the inner product is linear in the second entry always. It is the first entry that one has to prove that it is conjugate-linear. But then take property 2, i.e.
$$$$\begin{split} \left(\sum_i c_i\vec{v}_i,\vec{w} \right) &= \left(\vec{w},\sum_i c_i\vec{w} \right)^*\\ &= \sum_i c_i^*\left(\vec{w},\vec{v}_i \right)^*\\ &= \sum_i c_i^*\left(v_i,w_i \right), \end{split}$$$$
where I have used properties 1 and 2 of the inner product and the fact that, for two complex numbers $$z_1,z_2$$ ($$\left(\vec{a},\vec{b} \right)\in\mathbb{C}$$) $$(z_1,z_2)^*=z_1^* z_2^*$$. If you instead work with real numbers, then the complex conjugation does not affect the first argument, obviously... | 2021-09-28T15:47:06 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1430938/why-are-inner-products-defined-to-be-linear-in-the-first-argument-only",
"openwebmath_score": 0.965985119342804,
"openwebmath_perplexity": 221.82920857106024,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9838471675459396,
"lm_q2_score": 0.857768108626046,
"lm_q1q2_score": 0.8439127240829732
} |
https://math.stackexchange.com/questions/1487338/finding-right-inverse-matrix | # Finding right inverse matrix
Given a $3\times 4$ matrix $A$ such as $$\begin{pmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ \end{pmatrix} ,$$ find a matrix $B_{4\times 3}$ such that
$$AB = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}$$ Apart from simply multiplying $A$ with $B$ and generating a $12$ variable system of equations, is there any simpler way of finding $B$ ?
• Use products with 3x3 elementary matrices as row operations to reduce the original matrix to row echelon form. Then maybe something can be worked out. – jdods Oct 19 '15 at 11:43
• I'm not looking for the inverse, meaning I don't want BA = I to be true. Just a matrix B that satisfies AB=I. – EuxhenH Oct 19 '15 at 11:49
• B could be a 4x3 matrix which leads to a 3x3 I with 1s on the main diagonal. – EuxhenH Oct 19 '15 at 11:51
• It would be clear for everyone to specific the specific matrix you want and I would suggest that you call it something other than $I$ to eliminate confusion. – NoChance Oct 19 '15 at 11:53
• Changed it. Thanks – EuxhenH Oct 19 '15 at 11:56
Your matrix $$A=\begin{bmatrix}1&1&1&1\\0&1&1&0\\0&0&1&1\end{bmatrix}$$ has rank=$3$ and you can see that the square matrix $AA^T$ is invertible. Now note that $AA^T(AA^T)^{-1}=I$ so the matrix $B=A^T(AA^T)^{-1}$ is a right inverse of $A$ (but it is not the unique).
in this case we have: $$AA^T=\begin{bmatrix}4&2&2\\2&2&1\\2&1&2\end{bmatrix}$$ $$(AA^T)^{-1}=\dfrac{1}{4}\begin{bmatrix}3&-2&-2\\-2&4&0\\-2&0&4\end{bmatrix}$$ $$B=\dfrac{1}{4}\begin{bmatrix}3&-2&-2\\1&2&-2\\-1&2&2\\1&-2&2\end{bmatrix}$$
Consider the following case:
$$\begin{bmatrix}1&1&1&1\\0&1&1&0\\0&0&1&1\end{bmatrix}\times \begin{bmatrix}0&0&0\\?&?&?\\?&?&?\\?&?&?\end{bmatrix}=I$$
Removing the first row, the remaining matrix is square:
$$\begin{bmatrix}?&?&?\\?&?&?\\?&?&?\end{bmatrix}=\begin{bmatrix}1&1&1\\1&1&0\\0&1&1\end{bmatrix}^{-1}=\begin{bmatrix}1&0&-1\\-1&1&1\\1&-1&0\end{bmatrix}$$
$$B=\begin{bmatrix}0&0&0\\1&0&-1\\-1&1&1\\1&-1&0\end{bmatrix}$$
Of course there are infinite number of solutions | 2020-01-25T02:23:08 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1487338/finding-right-inverse-matrix",
"openwebmath_score": 0.6087852716445923,
"openwebmath_perplexity": 232.6729563492999,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9838471661250914,
"lm_q2_score": 0.8577681068080748,
"lm_q1q2_score": 0.8439127210756091
} |
http://math.stackexchange.com/questions/288407/determine-if-a-system-described-by-a-differential-equation-is-linear | # Determine if a system described by a differential equation is linear
A system ("A System is any physical set of components that takes a signal, and produces a signal") is described by this equation:
$\frac{dy(t)}{dt} + 3 \times y(t) = x(t)$
Where $x(t)$ is the input and $y(t)$ the output.
How to determine if this system is a linear one?
-
It is linear because the operators on $y(t)$ are linear, namely differentiation and multiplication. An example of a non-linear system would be $$\frac{dy(t)}{dt} + 3 \times y(t)^2 = x(t)$$ because squaring is not a linear operator. It does NOT satisfy $(y_1+y_2)^2=y_1^2+y_2^2$. This is in contrast to $\frac{d}{dt}(y_1(t)+y_2(t))=\frac{d}{dt}y_1(t)+\frac{d}{dt}y_2(t)$ and $3\times(y_1+y_2)=3\times y_1+3\times y_2$.
-
Thank you, Teun. I thought I had to use something like S{ a * x_{1}(t) + b* x_{2}(t)} = a* S{x_{1}(t)} + b * S{x_{2}(t)} (In fact this is what you are using, but I wasn't sure which would be the system output).. – Chris Jan 27 '13 at 22:58
Glad to help! What you're saying is almost correct except for the fact that when we have two solutions $y_1$ and $y_2$ and plug $y_3=ay_1+by_2$ into your differential equation we get: $$\frac{dy_3(t)}{dt} + 3 \times y_3(t) = a\left( \frac{dy_1(t)}{dt} + 3 \times y_1(t)\right) +b\left( \frac{dy_2(t)}{dt} + 3 \times y_2(t)\right)=(a+b)x(t)\neq x(t)$$ unless you pick specific values for $a$ and $b$, so it wouldn't work anymore. This is because the DE is non-homogeneous. – Slugger Jan 27 '13 at 23:02
So the system (and what I mean with system is this: "A System, is any physical set of components that takes a signal, and produces a signal") is not linear? – Chris Jan 27 '13 at 23:18
No it's linear because of the reasons I posted in my original answer (differentiation and multiplication are linear). I just wanted to show that even though the system is linear, it is important to realize that doesn't mean that every linear combination of solutions will give another solution to your system :) – Slugger Jan 27 '13 at 23:22
It depends on your definition of linear.
In the study of ODEs, the system is linear if it can be expressed in the form $L y = x$, where $L$ is a linear differential operator of the form $(Ly)(t) = \sum_{k=0}^n a_k(t) y^{(k)}(t)$, and the coefficients $a_k$ satisfy some continuity condition. Since we can write the above as $Ly = x$ with $(Ly)(t) = y^{(1)}(t)+3 y(t)$, it is clear that the system is linear in this sense.
From a control system perspective, the system is linear (I am omitting many details) if the solution $t \mapsto y_{y_0, x}(t)$ is a linear function of $(y_0,x)$, where $y_0$ is the initial state $y_0$ and $x$ is the input.
The above equation can be written as $\dot{y} = f(y,x)$, where $f(y,x) = -3 y +x$. $f$ is globally Lipschitz in $y$, hence a globally unique solution exists passing through a given initial condition. Since the solution is unique, it is straightforward to verify (by differentiating and checking that it satisfies the ODE) that the system is linear by just checking that if $y_{y_0, x}, y_{y_0', x'}$ are solutions (with initial conditions and inputs $(y_0, x), (y_0', x')$ respectively), then $\alpha y_{y_0, x} + \alpha' y_{y_0', x'}$ is a solution with initial condition $\alpha y_0 + \alpha'y_0'$ and input $\alpha x + \alpha'x'$. Hence the system is linear.
-
What do you mean with "control system"? I've updated my question with a definition of the system I mean. Is your answer in compliance with the new meaning of the system so that I can read it to answer to you? – Chris Jan 27 '13 at 23:20
The term 'dynamical system' would be have been more appropriate. The dynamical system focus is mainly concerned with modeling the input-output behavior. Your system is linear in both senses. – copper.hat Jan 28 '13 at 2:55 | 2016-06-27T04:09:55 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/288407/determine-if-a-system-described-by-a-differential-equation-is-linear",
"openwebmath_score": 0.899189293384552,
"openwebmath_perplexity": 181.94475578448282,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9838471616257376,
"lm_q2_score": 0.8577681086260461,
"lm_q1q2_score": 0.8439127190048129
} |
https://mathhelpboards.com/threads/classical-mechanics-question.28084/ | # Classical Mechanics question
#### Klaas van Aarsen
##### MHB Seeker
Staff member
We are given the angular velocity $\omega = 7\cdot 10^{-5}\,rad/s$ and the mass $M=6\cdot 10^{24}\,kg$.
To achieve a free fall of $0\,m/s^2$ at radius $r$ we need that the centripetal acceleration is equal to the acceleration due to gravity,
Note that $v=\omega r$, so the centripetal acceleration is $\frac{v^2}{r}=\omega^2 r$.
The acceleration due to gravity is $\frac{GM}{r^2}$, where $G=6.67\cdot 10^{-11}$ is the gravitational constant (leaving out the unit while assuming SI units).
So:
$$\omega^2 r = \frac{GM}{r^2}$$
Solve for $r$.
Last edited:
#### Dhamnekar Winod
##### Active member
We are given the angular velocity $\omega = 7\cdot 10^{-5}\,rad/s$ and the mass $M=6\cdot 10^{24}\,kg$.
To achieve a free fall of $0\,m/s^2$ at radius $r$ we need that the centripetal acceleration is equal to the acceleration due to gravity,
Note that $v=\omega r$, so the centripetal acceleration is $\frac{v^2}{r}=\omega^2 r$.
The acceleration due to gravity is $\frac{GM}{r^2}$, where $G=6.67\cdot 10^{-11}$ is the gravitational constant (leaving out the unit while assuming SI units).
So:
$$\omega^2 r = \frac{GM}{r^2}$$
Solve for $r$.
Hi,
So, we get $r^3 =8.172587755e22m^3/rad^2$ So,$r=43396349.43332m/\sqrt[3]{rad^2}$. Is this answer correct?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Hi,
So, we get $r^3 =8.172587755e22m^3/rad^2$ So,$r=43396349.43332m/\sqrt[3]{rad^2}$. Is this answer correct?
I get the same answer.
Do note that $rad$ is not an actual physical unit, but it's a ratio. When we multiply the angular velocity (rad/s) with the radius (m), the rad unit is effectively eliminated and we get m/s.
So properly we have $r=4.3\cdot 10^7\,m$.
It means that answer 2 should be the correct answer.
Admittedly it's a bit strange that it is given as $4.4\cdot 10^7\,m$ instead of $4.3\cdot 10^7\,m$.
Since we're talking about earth, perhaps they used a mass and angular velocity with a higher precision than the ones given in the problem statement.
EDIT: Hmm... in that case we would actually get $r=4.2\cdot 10^7\,m$, so that can't be it after all.
Last edited: | 2021-06-18T06:33:20 | {
"domain": "mathhelpboards.com",
"url": "https://mathhelpboards.com/threads/classical-mechanics-question.28084/",
"openwebmath_score": 0.983230471611023,
"openwebmath_perplexity": 531.990208688609,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.98384716257297,
"lm_q2_score": 0.8577681068080748,
"lm_q1q2_score": 0.8439127180287127
} |
https://math.stackexchange.com/questions/971232/can-a-countable-group-have-uncountably-many-subgroups | Can a countable group have uncountably many subgroups?
If $G$ is a countable group,can it have an uncountable number of distinct subgroups?
Let $V$ be a vector space of dimension $\aleph_0$ over a countable field $F$ (so $V$ is countable) and let $B$ be a basis for $V$ over $F$. Then every subset of $B$ spans a different subspace of $V$, so $V$ has $2^{\aleph_0}$ different subspaces, and its additive group has $2^{\aleph_0}$ different subgroups.
• OK. More interesting question: Can a countable group have precisely $\omega_1$ subgroups? (Without assuming $\mathsf{CH}$, of course.) – Andrés E. Caicedo Oct 13 '14 at 6:25
• @AndresCaicedo Hmm. Isn't the set of all subgroups of $G$ a closed set in the Tychonoff topology of $2^G$? Then, if $G$ is countable, the set of all subgroups of $G$ is homeomorphic to a closed set of real numbers, so it's either countable or has the cardinality of the continuum? Right? So, if $G$ is countable, then it can't have exactly $\omega_1$ subgroups, unless CH holds. – bof Oct 13 '14 at 6:33
• Yes, that's the only argument I know of. I'm half wondering if one can provide a combinatorial argument for the existence of continuum many subgroups that is not just this descriptive set theoretic approach in disguise. – Andrés E. Caicedo Oct 13 '14 at 6:42
• @AndresCaicedo Of course the "topological" argument has nothing to do with groups, it applies to any countable algebraic structure with finitary operations. I suppose there are some nontrivial settings where the existence of continuum many subalgebras can be proved by a "combinatorial" argument (whatever that means)? – bof Oct 13 '14 at 7:12
Consider the direct sum of countably many $\mathbb{Z}/2\mathbb{Z}$ groups, which I'll denote by $$G = \displaystyle \bigoplus_{n = 1}^\infty \left( \mathbb{Z} / 2\mathbb{Z} \right)_n$$ and where the index is to keep track of each copy of $\mathbb{Z}/2\mathbb{Z}$. A set of subgroups of $G$ are formed by including or excluding the $n$th copy of $\mathbb{Z}/2\mathbb{Z}$ (but as Slade corrected me in the comments, this does not give all the subgroups). Nonetheless, any infinite binary sequence yields a distinct subgroup by including those indices that are $1$ in the sequence, and so we have an injection from $2^\mathbb{N}$ into the set of subgroups of $G$. Thus the set of subgroups is uncountable.
• Ah, if you use direct sum, I believe this works. – Slade Oct 13 '14 at 4:38
• @Slade: yes, you're right on all accounts. It does work if I use the direct sum, and I'll edit that in – davidlowryduda Oct 13 '14 at 4:47
• Sorry,I am not getting the bijective map.It would be of great help if someone explains it in more details for me – Learnmore Oct 13 '14 at 4:51
• @learningmaths If you were referring to the original (as I just edited), it's because it wasn't actually bijective. Sorry about that. – davidlowryduda Oct 13 '14 at 4:52
• @mixedmath Doesn't $G$ have an uncountable number of elements? – Prince Kumar Dec 1 '16 at 5:55
One more example, using a very familiar group, the additive group $\mathbb Q$ of rational numbers. For any set $S$ of primes, consider the subgroup of $\mathbb Q$ consisting of those numbers that can be written as fractions (integer over integer) in which all the prime divisors of the denominator belong to $S$. (So, for example, when $S$ is empty, this subgroup is $\mathbb Z$, and when $S$ is the set of all primes, this subgroup is all of $\mathbb Q$.) There are continuum many choices for $S$, and each one leads to a different subgroup of $\mathbb Q$.
A finitely presented example: the free group $\mathbb{F}_2$ of rank $2$. Indeed, it contains the free group $\mathbb{F}_{\infty}$ of countable infinite rank. Let $\{x_1,x_2,\dots\}$ be a free basis for such a subgroup. For any sequence $\mathfrak{n} = (n_i)$ of positive integers, let $S( \mathfrak{n})$ denote the free subgroup generated by $\{x_{n_1},x_{n_2}, \dots\}$. Finally, $\{S (\mathfrak{n}) \mid \mathfrak{n} \}$ defines an uncountable family of pairwise distinct subgroups.
More generally, it may be noticed that any SQ-universal group has uncountably many normal subgroups. Let $G$ be such a group. It is clear that a countable group has countably many 2-generated subgroups, and because there exist uncountably many non-isomorphic such groups, $G$ must have uncountably many quotients. A fortiori, $G$ has uncountably many normal subgroups.
• For example, $\{a^iba^i:i\in\mathbb{Z}\}$ forms a basis for a copy of $\mathbb{F}_{\infty}$ in $\mathbb{F}_2$ (the derived subgroup $[\mathbb{F}_2, \mathbb{F}_2]$ is also isomorphic to $\mathbb{F}_{\infty}$, but the proof is less obvious, and would have to think what the basis would be!). – user1729 Oct 13 '14 at 10:42
• In fact, it seems to be natural to deduce that the rank of $[\mathbb{F}_2,\mathbb{F}_2]$ is infinite from algebraic topology. See for example chiasme.wordpress.com/2013/10/24/… Furthermore, an explicit basis may be found. – Seirios Oct 13 '14 at 11:34
• @Serios Yes, I agree, algebraic topology wins the day! – user1729 Oct 13 '14 at 12:09
• Algebraic geometry proof about the derived subgroup by user1729: math.stackexchange.com/a/983990/111377 – evgeny Feb 19 '16 at 16:44
In a similar vein to Andreas' answer, consider the additive group $\mathbb{Z}[X]$ of integer-coefficient polynomials (so with finitely many terms). This group is countable, but for each subset of the naturals $S \subset \mathbb{N}$ we get a subgroup $\langle \left\{ x^n \vert n \in S \right\} \rangle$ and these are all distinct.
• It is a good example, essentially $\mathbb{Z}^{(\mathbb{N})}$, that is, the multiplicative structure is not involved. – Orest Bucicovschi Oct 13 '14 at 18:25
• Indeed, the groups are familiar and I find monomials $x^n$ easier to think about than $\langle \frac{1}{p_n} \rangle$, where $p_n$ is the $n$th prime. – yatima2975 Oct 14 '14 at 9:28
The set of all bijection from $$\mathbb{N}$$ to $$\mathbb{N}$$ that forms a group and I think it has uncountably many subgroups.
For each $(b_n) \in \prod_{n\in \mathbb{N}}(\mathbb{Z}/2)$ consider the map from $\oplus _{n\in \mathbb{N}}(\mathbb{Z}/2)$ to $\mathbb{Z}/2$ $$(a_n) \mapsto \sum_n a_n \cdot b_n$$
take the kernel of this map; we get $2^{\aleph_0}$ subgroups of index $2$ of $\oplus _{n\in \mathbb{N}}(\mathbb{Z}/2)$.
@user1729: thanks! for correcting a previous statement about the number of subgroups of finite index of a countable group -- it may well be uncountable!
The example of @seirios: is a finitely generated group having uncountably many subgroups. However, a finitely generated group has finitely many subgroups of a given index $n$. Indeed, one can reduce to normal subgroups ( the normal core has index $N \le n!$) and then notice that there are finitely many morphisms from the groups to the finitely many groups of order $N$. Thus a finitely generated group has countably many subgroups of finite index.
• I do not understand how this answers the question. You have only shown that there are (finite or) countably many finite-index, normal subgroups of a finitely generated group (note: finitely generated, not countable, as mixedmath's answer demonstrates). Which is not awfully relevant... – user1729 Oct 14 '14 at 8:22
• Your now answer has some issues - not every $(b_n)\in\prod(\mathbb{Z}/2\mathbb{Z})$ corresponds to an element of the direct sum. This is because the direct sum contains only those elements with finite support (it is not the Cartesian product). Also, this is just a re-hashing of mixedmath's answer. If I were you, I would just delete everything above "The example of @seirios..." – user1729 Oct 17 '14 at 10:39
• @user1729: The kernel of the functional given by $(b_n)\ne (0)$ is a subgroup of index $2$. – Orest Bucicovschi Oct 17 '14 at 11:29 | 2019-09-18T09:50:21 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/971232/can-a-countable-group-have-uncountably-many-subgroups",
"openwebmath_score": 0.9011984467506409,
"openwebmath_perplexity": 207.11658112225055,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9838471684931717,
"lm_q2_score": 0.8577680995361899,
"lm_q1q2_score": 0.8439127159524495
} |
https://math.stackexchange.com/questions/2433417/if-fx1-fx-1-sqrt3-fx-and-f2-2-what-is-the-value-of-f4 | # If $f(x+1) +f(x-1) =\sqrt{3}\,f(x)$ and $f(2) =2$, what is the value of $f(4)$?
My Attempt
$f(2)=2$. So, $f(1) + f(3)=2\sqrt{3}$ and $f(2) + f(4)=\sqrt{3}\,f(3)$. After solving these equations I got the value of $f(3)=2\sqrt{3}$ and $f(4)=4$. But are there any other methods than this? Any suggestions are welcome.
Update:- @ProfessorVector pointed out that the above solutions are only true if $f(1)=0$. After checking I find that it is true. So, my above attempt is a failure. Is there a way to solve this question?
Update 2:- Is there a way to find the period of this function?
• You could always solve for $f(x)$ directly: en.wikipedia.org/wiki/Characteristic_equation_(calculus) – Simply Beautiful Art Sep 17 '17 at 18:11
• That's only true if $f(1)=0$. From where do you get that? – Professor Vector Sep 17 '17 at 18:19
• @SimplyBeautifulArt But this is a second order linear recurrence, so it needs two initial conditions to solve. – dxiv Sep 17 '17 at 18:28
• @SimplyBeautifulArt Right, but there is no second condition in OP's question, so the answer is more like "$f(4)$ can be whatever you want it to be". – dxiv Sep 17 '17 at 18:36
• @SerialKiller Is there a way to solve this question? $\,f(4) =4 - \sqrt{3} \cdot f(1)\,$, so there is no unique solution unless you know $\,f(1)\,$ or some other value of $\,f(n)\,$. – dxiv Sep 17 '17 at 19:07
I'll be more general.
$f(n+1)+f(n-1) = cf(n)$.
Suppose $f(n) = b^n$.
Then $b^{n+1}+b^{n-1} = cb^n$ or $b^2-cb+1 = 0$.
Then $b =\dfrac{c\pm\sqrt{c^2-4}}{2}$.
If $c^2=4$, $b = c/2 = \pm 1$, so $f(n)=1$ or $(-1)^n$.
If $c^2 > 4$, then $b$ has two possible values, one with $|b|>1$ and one with $|b|<1$.
If $c^2 < 4$, then $b$ has two possible complex values $b_1 =\dfrac{c+\sqrt{c^2-4}}{2} =\dfrac{c+i\sqrt{4-c^2}}{2}$ and $b_2 =\dfrac{c-i\sqrt{4-c^2}}{2}$. Note that $b_1b_2 =\dfrac{c+\sqrt{c^2-4}}{2}\dfrac{c-\sqrt{c^2-4}}{2} =\dfrac{4}{4} =1$ (as can also be deduced from the quadratic specifying $b$) and that $|b_k|^2 =\dfrac{c^2+4-c^2}{4} =1$.
Since $|b| = 1$, $b =e^{it} =\cos(t)+i\sin(t)$ where $\cos(t) =c/2$.
In your case, $c = \sqrt{3}$ so $t =\arccos(\sqrt{3}/2) =\pi/6$.
Therefore the two possible solutions are $f(n) =e^{\pm ni\pi/6}$ and any linear combination of these.
So any solution is of the form $ue^{in\pi/6}+ve^{-in\pi/6} =(u+v)\cos(n\pi/6)+(u-v)\sin(n\pi/6)$.
If the solution is real for an $n$ such that $\sin(n\pi/6) \ne 0$, then $u=v$, so it is $f(n)=2u\cos(n\pi/6)$.
If $f(2) = 2$, then $2=f(2) =2u\cos(\pi/3) =u$ so $u = 2$ and the solution is $f(n) =4\cos(n\pi/6)$.
Putting $n=4$, $f(4) =4\cos(4\pi/6) =-2$.
• If the solution is real for an n ... then u=v That's assuming $u,v$ are real, which they don't have to. All that's needed in this case is $u = \overline v$ for $f(n)$ to be real for $\forall n$. – dxiv Sep 17 '17 at 19:30
• How can you suppose that $f(n)=b^n$? – Rory Daulton Sep 18 '17 at 0:07 | 2019-12-08T19:30:09 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2433417/if-fx1-fx-1-sqrt3-fx-and-f2-2-what-is-the-value-of-f4",
"openwebmath_score": 0.9195312261581421,
"openwebmath_perplexity": 159.769816088976,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9838471642306267,
"lm_q2_score": 0.8577681013541613,
"lm_q1q2_score": 0.8439127140847803
} |
https://math.stackexchange.com/questions/2883837/how-to-properly-choose-a-solution-to-a-system-of-equations-of-trigonometric-func | How to properly choose a solution to a system of equations of trigonometric functions?
I worked my way to encounter a system of equations
$$\begin{cases} q=k_1\cos\phi_1+k_2\cos\phi_2& \\ 0=k_1\sin\phi_1+k_2\sin\phi_2& \end{cases}$$
I don't need to solve for specific angles $\phi_1,\phi_2$ because these equations will probably be very rough, but it's enough to know the answer in terms of $\cos\phi_1$ and $\sin\phi_2$ or vice versa. If I am looking to solve this by hand then what I have to do is obviously rewrite one of the trigonometric functions in terms of the other, for example
$$\begin{cases} q=k_1\cos\phi_1\pm k_2\sqrt{1-\sin^2\phi_2}& \\ 0=\pm k_1\sqrt{1-\cos^2\phi_1}+k_2\sin\phi_2& \end{cases},$$ where now it's just basic algebra to solve the equations.
My question concerns the $\pm$ signs that appear there. I can rearrange the equations in such a way, that after squaring them I get a sign independent result again. Hence I get a unique solution to this equation. But to me that seems weird - I would assume I should get solutions that aren't unique. I assume that I must at some point arrive at some expression that differs by (at least) a sign, since I am expressing everything in terms of $\cos\phi_1$ and $\sin\phi_2$, which can swap signs depending on their argument. Could anyone clear this up for me?
Let me reframe this for a bit...also let's assume $\phi_1$ and $\phi_2$ are "interesting" (not $0, \pi$, etc).
Just to be on the same page, let's solve the equation first. Isolating the square roots in both equations gives $q - k_1 \cos\phi_1 = \pm k_2 \sqrt{1-\sin^2 \phi_2}$ and $\mp k_1 \sqrt{1 - \cos^2 \phi_1} = k_2^2 \sin^2 \phi_2$. Squaring and solving gives
$$\cos \phi_1 = \frac{q^2 + k_1^2 - k_2^2}{2qk_1}$$
which is indeed sign independent. But all of the terms with $\sin \phi_2$ are squared, so $\sin \phi_2$ is not sign independent. Solving the other way, that is, writing $q - k_2 \cos\phi_2 = \pm k_1 \sqrt{1-\sin^2 \phi_1}$ and $\mp k_2 \sqrt{1 - \cos^2 \phi_2} = k_1^2 \sin^2 \phi_1$, gives
$$\cos \phi_2 = \frac{q^2 + k_2^2 - k_1^2}{2qk_2}$$
(which also makes sense because of symmetry). Similarly, you can't get a value for $\sin \phi_1$ here, since all the $\sin \phi_1$ terms are squared.
The first time, $\cos \phi_1$ was sign-independent and $\sin \phi_2$ wasn't. The second time, $\cos \phi_2$ was sign independent and $\sin \phi_1$ wasn't. Specifically, if you try to solve for the $\sin$s, you get expressions like $\sin \phi = \pm \text{(stuff)}$.
Think of the unit circle for a second. Geometrically, we know the x-coordinates of $\phi_1$ and $\phi_2$, but not their y-coordinates. Though we do know the two possible candidates are $\pm$ of each other. So we are almost done but not quite. We have four possible solutions: $(\phi_1, \phi_2)$, $(\phi_1, -\phi_2)$, $(-\phi_1, \phi_2)$, $(-\phi_1, -\phi_2)$.
Note that $(\phi_1, \phi_2)$ if a solution if and only if $(-\phi_1, -\phi_2)$ is, and the same holds for the two other solution candidates $(\phi_1, -\phi_2)$, $(-\phi_1, \phi_2)$. (See this by plugging them into the second equation.) And $(\phi_1, \phi_2)$ and $(\phi_1, -\phi_2)$ can't both be solutions because the second equation would give $\sin \phi_1 = 0$, a contradiction. Since $(\phi_1, \phi_2)$ is of course a solution to the equation, this tells you that the only two solutions of the equation are $(\phi_1, \phi_2)$ and $(-\phi_1, -\phi_2)$.
This is a long-winded way of saying: You do get solutions that aren't unique when solving this. Specifically, you get 4 solutions, and 2 of them turn out to be fake solutions, leaving you with 2 actual ones. I can't tell where exactly in your question you went wrong, because you didn't explain your solution path, but hopefully this clears things up.
• Yeah, your explanation makes sense. I didn't give too much information, since I actually wanted to solve this one on my own (it's not that hard after all). I'll try to apply this to my problem. Thanks. – Henrikas Aug 16 '18 at 8:13 | 2019-08-24T11:15:51 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2883837/how-to-properly-choose-a-solution-to-a-system-of-equations-of-trigonometric-func",
"openwebmath_score": 0.9263508319854736,
"openwebmath_perplexity": 112.22244715368414,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9838471661250914,
"lm_q2_score": 0.8577680977182187,
"lm_q1q2_score": 0.8439127121325799
} |
http://tuning-style.it/wdmc/how-to-find-the-kernel-of-a-linear-transformation.html | Polynomial Kernel. We show that for high-dimensional data, a particular framework for learning a linear transformation of the data based on the LogDet divergence can be efficiently kernelized to learn a metric (or equivalently, a kernel function) over an. be a linear transformation. Since n n matrices are linear transformations Rn Rn , we can see that the order of successive linear transformations matters. [Linear Algebra] Finding the kernel of a linear transformation. As a result, unlike transductive kernel learning methods, our method easily handles out-of-sample extensions, i. It doesn't hurt to have it, but it isn't necessary here (in finding the kernel). both F (clt)l +. It is given by the common inner product plus an optional constant c. In each case, state the nullity and rank of T and verify the Rank Theorem. Let L be de ned on P3 (the vector space of polynomials of degree less than 3) by L(p) = q where q(x) = xp′′(x) p(0)x: (a) Find the kernel of L. Let L : V →W be a linear transformation. (b)If Vand Ware vector spaces of dimension nand T: V !Wis a one-to-one linear transformation, then Tis onto. large values of , and clearly approach the linear regression; the curves shown in red are for smaller values of. The kernel of T, denoted by ker(T), is the set of all vectors x in Rn such that T(x) = Ax = 0. The Fourier Transform is useful in engineering, sure, but it's a metaphor about finding the root causes behind an observed effect. Now, let $\phi: V\longrightarrow W$ be a linear mapping/transformation between the two vector spaces. Theorem Let T:V→W be a linear transformation. Here we consider the case where the linear map is not necessarily an isomorphism. 3 (Nullity). To find the kernel of a matrix A is the same as to solve the system AX = 0, and one usually does this by putting A in rref. This paper is organized as follows. What is the outcome of solving the problem?. An essential question in linear algebra is testing whether a linear map is an isomorphism or not, and, if it is not an isomorphism, finding its range (or image) and the set of elements that are mapped to the zero vector, called the kernel of the map. Now, let $\phi: V\longrightarrow W$ be a linear mapping/transformation between the two vector spaces. Recall: Linear Transformations De nition A transformation T : Rn!Rm is alinear transformationif it satis es the following two properties for all ~x;~y 2Rn and all (scalars) a 2R. SUBSCRIBE to the channel and. 2 The kernel and range of a linear transformation. Griti is a learning community for students by students. We will start with Hinge Loss and see how the optimization/cost function can be changed to use the Kernel Function,. One-to-One linear transformations: In college algebra, we could perform a horizontal line test to determine if a function was one-to-one, i. This paper studies the conditions for the idempotent transformation and the idempotent rank transformation direct sum decomposition for finite dimension of linear space. Hello, welcome to TheTrevTutor. Linear Transformations and Polynomials We now turn our attention to the problem of finding the basis in which a given linear transformation has the simplest possible representation. Morphological transformations are some simple operations based on the image shape. Theorem As de ned above, the set Ker(L) is a subspace of V, in particular it is a. Let L: R3 → R3 be the linear transformation defined by L x y z = 2y x−y x. Find polynomial(s) p i(t) that span the kernel of T. Intuitively, the kernel measures how much the linear transformation T T T collapses the domain R n. Linear transformation, in mathematics, a rule for changing one geometric figure (or matrix or vector) into another, using a formula with a specified format. The Matrix of a Linear Transformation We have seen that any matrix transformation x Ax is a linear transformation. It is one-one if its kernel is just the zero vector, and it is. One thing to look out for are the tails of the distribution vs. Definition 6. Morphological transformations are some simple operations based on the image shape. Question: Why is a linear transformation called “linear”?. SPECIFY THE VECTOR SPACES Please select the appropriate values from the popup menus, then click on the "Submit" button. Kernel Principal Component Analysis (KPCA) is a non-linear dimensionality reduction technique. Such a repre-sentation is frequently called a canonical form. the kernel of a transformation between vector spaces is its null space). The $$\textit{nullity}$$ of a linear transformation is the dimension of the kernel, written nul L=\dim \ker L. In this paper, we study metric learning as a problem of learning a linear transformation of the input data. Note: Because Rn is a "larger" set than Rm when m < n, it should not be possible to map Rn to Rm in a one-to-one fashion. Null space. 6, -1 ,-3-3 , 3 ,-1. ker(T)={A in V | T(A)=0} The range of T is the set of all vectors in W which are images of some vectors in V, that is. (The dimension of the image space is sometimes called the rank of T, and the dimension of the kernel is sometimes called the nullity of T. If w2 = 0, w3 = 1, then w1 = -1, and if w2 = 1 and w3 = 1, then w1 = 0. We have step-by-step solutions for your textbooks written by Bartleby experts! Find the nullity of the linear transformation T : M n n → ℝ defined by T ( A ) = tr ( A ). (2) is injective if and only if. Suppose T:R^3 \\to R^3,\\quad T(x,y,z) = (x + 2y, y + 2z, z + 2x) Part of Solution: The problem is solved like this: A =. The set consisting of all the vectors v 2V such that T(v) = 0 is called the kernel of T. Find polynomial(s) p i(t) that span the kernel of T. An example of a linear transformation T :P n → P n−1 is the derivative function that maps each polynomial p(x)to its derivative p′(x). To find the kernel of a matrix A is the same as to solve the system AX = 0, and one usually does this by putting A in rref. Proof: This theorem is a proved in a manner similar to how we solved the above example. Before we do that, let us give a few definitions. w1 = w2 - w3 and w2, w3 are free variables. There are some important concepts students must master to solve linear transformation problems, such as kernel, image, nullity, and rank of a linear transformation. 2 Kernel and Range of a Linear Transformation Performance Criteria: 2. (If all real numbers are solutions, enter REALS. ) Let be the transformation. Anyway, hopefully you found that reasonably. In particular, there exists a nonzero solution. In Figure 5, the used is, which after applied to every point in Figure 5 (left) yields the linearly separable dataset Figure 5 (right). to construct the whitening transformation matrix for orthogonalizing the linear subspaces in the feature space F. The kernel of a transformation is a vector that makes the transformation equal to the zero vector (the pre-image of the transformation). 2-T:R 3 →R 3,T(x,y,z)=(x,0,z). 6, -1 ,-3-3 , 3 ,-1. The kernel of a linear operator is the set of solutions to T(u) = 0, and the range is all vectors in W which can be expressed as T(u) for some u 2V. 443 A linear transformation L is one-to-one if and only if kerL ={0 }. {\mathbb R}^n. If T isn't an isomorphism find bases of the kernel and image of T, and. {\mathbb R}^m. It is normally performed on binary images. Next, we study the space of linear transformations from one vector space to another, and characterize some algebraic properties of linear transformations. These are all vectors which are annihilated by the transformation. Thus, the kernel is the span of all these vectors. Because is a composition of linear transformations, itself is linear (Theorem th:complinear of LTR-0030). And since we have a linear transformation that has the same properties of a subspace, the image and kernel of the linear transformation are subspaces of Rn. Define the transformation $\Omega: L(V,W) \to M_{m \times n} (\mathbb{R})$ Stack Exchange Network 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. Find the kernel of f. Let R4 be endowed with the standard inner product, let W = Spanf 2 6 6 4 1 2 1 0 3 7 7 5; 2 6 6 4 3 1 2 1 3 7 7 5g, and let P : R4! R4 be the orthogonal projection in R4 onto W. Linear Algebra, David Lay Week Seven True or False. Let’s check the properties:. Suppose a linear transformation is applied to the random variable X to create a new random variable Y. Use automated training to quickly try a selection of model types, and then explore promising models interactively. Find polynomial(s) p i(t) that span the kernel of T. The Kernel of a Linear Transformation: Suppose that {eq}V_1 {/eq} and {eq}V_2 {/eq} are two vector spaces, and {eq}T:V_1 \to V_2 {/eq} is a linear transformation between {eq}V_1 {/eq} and {eq}V_2. (a) A linear functional on V is a function ~u∗ : V → IR that is linear in the sense that ~u∗(~v + w~) = ~u∗(~v) +~u∗(w~) and ~u∗(α~v) = α~u∗(~v) for all ~u,w~ ∈ V and all α ∈ IR. Theorem If the linear equation L(x) = b is solvable then the. Learn vocabulary, terms, and more with flashcards, games, and other study tools. One thing to look out for are the tails of the distribution vs. In conclusion, we have examined the process of finding an image and a kernel of a linear transformation, which can be used, for many practical situations. More on matrix addition and scalar multiplication. From this, it follows that the image of L is isomorphic to the quotient of V by the kernel: ≅ / (). (Think of it as what vectors you can get from applying the linear transformation or multiplying the matrix by a vector. Answer to Find the kernel and image of each linear transformation in Problems a to c. We solve by finding the corresponding 2 x 3 matrix A, and find its null space and column span. Proposition 3. Determine the kernel and range of each of the following linear transformations from R3 into R3. + for all vectors VI, for all scalars Cl, F(cv) for all scalars c, for all ve V, for all A function F: V —W is linear W be a subspace of Rk Let V be a subspace of Let it respects the linear operations,. and How to find the kernel of the li Stack Exchange Network 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. Power Iterated Color Refinement. If T(~x) = A~x, then the kernel of T is also called the kernel of A. In Figure 5, the used is, which after applied to every point in Figure 5 (left) yields the linearly separable dataset Figure 5 (right). CONTENTS Introduction to Linear Transformations The Kernel and Range of a Linear Transformation Matrices for Linear Transformations Transition Matrices and Similarity 3. Let V;W be vector spaces over a eld F. • to bring this understanding to bear on more complex examples. Let t = t 2 o t 1. Let P n(x) be the space of polynomials in x of degree less than or equal to n, and consider the derivative operator d dx. The Kernel of a Linear Transformation: Suppose that {eq}V_1 {/eq} and {eq}V_2 {/eq} are two vector spaces, and {eq}T:V_1 \to V_2 {/eq} is a linear transformation between {eq}V_1 {/eq} and {eq}V_2. Find a basis for Ker(L). The nullspace of a linear operator A is N(A) = {x ∈ X: Ax = 0}. We build thousands of video walkthroughs for your college courses taught by student experts who got an A+. ) T: P 5 → R, T(a 0 + a 1 x + a 2 x 2 + a 3 x 3 + a 4 x 4 + a 5 x 5) = a 0. We de ne T Aby the rule T A(x)=Ax:If we express Ain terms of its columns as A=(a 1 a 2 a n), then T A(x)=Ax = Xn i=1 x ia i: Hence the value of T A at x is the linear combination of the columns of A which is the ith. Since the nullity is the dimension of the null space, we see that the nullity of T is 0 since the dimension of the zero vector space is 0. Griti is a learning community for students by students. The problem is like this: Find the basis for \\text{kernel of}(T) where T is a linear transformation. Therefore, w 1 and w 2 form an orthonormal basis of the kernel of A. We say that a linear transformation is onto W if the range of L is equal to W. The matrix A and its rref B have exactly the same kernel. 1 2 -3 : 1/ 5 y 1 0 0 0 : - 7/. Kernel Principal Component Analysis In the section 1 we have discussed a motivation for the use of kernel methods – there are a lot of machine learning problems which a nonlinear, and the use of nonlinear feature mappings can help to produce new features which make prediction problems linear. 3, -3 , 1] Find the basis of the image of a linear transformation T defined by T(x)=Ax. It is the set of vectors, the collection of vectors that end up under the transformation mapping to 0. SPECIFY THE VECTOR SPACES Please select the appropriate values from the popup menus, then click on the "Submit" button. Find the range of the linear transformation T: R4 →R3 whose standard representation matrix. In Section 4, we define the kernel whitening transformation and orthogonalize non-. Find the kernel of the linear transformation. Proof: The linear transformation has an inverse function if and only if it is one-one and onto. , a 501(c)3 nonprofit corporation, with support from the following sponsors. Synonyms: kernel onto A linear transformation, T, is onto if its range is all of its codomain, not merely a subspace. The null space (or kernel) of consists of all vectors of the form , where are real numbers. (2) is injective if and only if. Sums and scalar multiples of linear transformations. restore the result in Rn to the original vector space V. Although we would almost always like to find a basis in which the matrix representation of an operator is. This can be defined set-theoretically as follows:. Such a repre-sentation is frequently called a canonical form. The kernel of a linear transformation {eq}L: V\rightarrow V {/eq} is the set of all polynomials such that {eq}L(p(t))=0 {/eq} Here, {eq}p(t) {/eq} is a polynomial. Find all. \$ I will leave that to you. (a) L(x) = (x3, x2, x1)^T. The Gaussian is a self-similar function. Find the dimension of the kernel and image of d dx. In this lesson, we will learn how to find the image and basis of the kernel of a linear transformation. A stationary covariance function is one that only depends on the relative position of its two inputs, and not on their absolute location. What is a "kernel" in linear algebra?. A vector v is in the kernel of a matrix A if and only if Av=0. What is the range of T in R2?. The matrix A and its rref B have exactly the same kernel. Therefore, w 1 and w 2 form an orthonormal basis of the kernel of A. 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. Let L be de ned on P3 (the vector space of polynomials of degree less than 3) by L(p) = q where. 3 (Nullity). Let $$V$$ and $$W$$ be vector spaces and let $$T:V. Show that solutions take the form X + V where f(X)=Y and where V is in the kernel. The image of a linear transformation T(x) = Ax is the span of the column vectors of A, that is the column space of matrix A. " • The fact that T is linear is essential to the kernel and range being subspaces. #20 Consider the subspace Wof R4 spanned by the vectors v1 = 1 1 1 1 and v2 = 1 9 −5 3. (c) Determine whether a given vector is in the kernel or range of a linear trans-formation. The Polynomial kernel is a non-stationary kernel. (a) L(x) = (x3, x2, x1)^T. If m < n, then T cannot be one-to-one. The theorem relating the dimension of the kernel and image requires the vector spaces to be finite dimensional. How to find the kernel of a linear transformation? Let B∈V =Mn(K) and let CB :V →V be the map defined by CB(A)=AB−BA. Finding matrices such that M N = N M is an important problem in mathematics. The linear transformation , from to , is both one-to-one and onto. Observe that every linear transformation € T:V→W between V and W is the linear extension of some function in Fun(S,W), namely of that function whose values at each vector in S is the same as the value T has there. The nullspace of a linear operator A is N(A) = {x ∈ X: Ax = 0}. (b)If Vand Ware vector spaces of dimension nand T: V !Wis a one-to-one linear transformation, then Tis onto. find the representation matrix [T] = T(e 1) ··· T(e n); 4. SPECIFY THE VECTOR SPACES Please select the appropriate values from the popup menus, then click on the "Submit" button. Researchers find security flaws in 40 kernel drivers from 20 vendors. A basis for the image is {t, t²}. The aim of our study of linear transformations is two-fold: • to understand linear transformations in R, R2 and R3. In mathematics, a linear map (also called a linear mapping, linear transformation or, in some contexts, linear function) is a mapping V → W between two modules (for example, two vector spaces) that preserves (in the sense defined below) the operations of addition and scalar multiplication. Remarks I The kernel of a linear transformation is a. Lesson: Image and Kernel of Linear Transformation Mathematics In this lesson, we will learn how to find the image and basis of the kernel of a linear transformation. Specifically, if T: n m is a linear transformation, then there is a unique m n matrix, A, such that T x Ax for all x n. A linear transformation (or mapping or map) from V to W is a function T: V → W such that T(v +w)=Tv +Tw T(λv)=λT(v) for all vectors v and w and scalars λ. We write ker(A) or ker(T). In this section, you will learn most commonly used non-linear regression and how to transform them into linear regression. In fact, every linear transformation (between finite dimensional vector spaces) can. range(T)={A in W | there exists B in V such that T(B)=A}. As such, this theorem goes by the name of the Rank- nullity Theorem. Namely, linear transformation matrix learned in the high dimensional feature space can more appropriately map samples into their class labels and has more powerful discriminating ability. I If x is an n 1 column vector then Ax is an m 1 column vector. Affine transformations", you can find examples of the use of linear transformations, which can be defined as a mapping between two vector spaces that preserves linearity. Linear Transformations and Polynomials We now turn our attention to the problem of finding the basis in which a given linear transformation has the simplest possible representation. Then T is a linear transformation. SUBSCRIBE to the channel and. (If all real numbers are solutions, enter REALS. Homework set on linear transformations. Determine the kernel and range of each of the following linear transformations from R3 into R3. 1 De nition and Examples 1. For a linear operator A, the nullspace N(A) is a subspace of X. In Figure 5, the used is, which after applied to every point in Figure 5 (left) yields the linearly separable dataset Figure 5 (right). In this paper, we study metric learning as a problem of learning a linear transformation of the input data. Note that N(T) is a subspace of V, so its dimension can be de ned. You can even pass in a custom kernel. Intuitively, the kernel measures how much the linear transformation T T T collapses the domain R n. The kernel of T, denoted by ker(T), is the set of all vectors x in Rn such that T(x) = Ax = 0. Facts about linear transformations. I will leave that to you. 2 The kernel and range of a linear transformation. The image of a linear transformation contains 0 and is closed under addition and scalar multiplication. 6, -1 ,-3-3 , 3 ,-1. Find a basis for the kernel of T and the range of T. ) It can be written as Im (A). 1 LINEAR TRANSFORMATIONS 217 so that T is a linear transformation. Now, consider P. Griti is a learning community for students by students. First here is a definition of what is meant by the image and kernel of a linear transformation. The kernel of T, denoted by ker(T), is the set of all vectors x in Rn such that T(x) = Ax = 0. ) T: P 5 → R, T(a 0 + a 1 x + a 2 x 2 + a 3 x 3 + a 4 x 4 + a 5 x 5) = a 0. In fact, 4x+ 2y= 2(2x+ y) so those are the same equation which is equivalent to y= -2x. 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. Its kernel is spanned by fcosx;sinxg. Different SVM algorithms use different types of kernel functions. ) Linear transformations. Legendre transformation From Wikipedia, the free encyclopedia Fourier transform consists of an integration with a kernel, the Legendre transform uses maximization as the transformation A closed convex function f is symmetric with respect to a given set G of orthogonal linear transformations, if and only if f* is symmetric with respect to G. Find the matrix and the eigenvectors of the transformation t. Thus, the kernel is the span of all these vectors. Up Main page Definition. To nd the image of a transformation, we need only to nd the linearly independent column vectors of the matrix of the transformation. 2 Kernel and Range of a Linear Transformation Performance Criteria: 2. Kernel, image, nullity, and rank Math 130 Linear Algebra D Joyce, Fall 2015 De nition 1. The kernel and range "live in different places. If m < n, then T cannot be one-to-one. We have some fundamental concepts underlying linear transformations, such as the kernel and the image of a linear transformation, which are analogous to the zeros and range of a function. Illustrate the constrained minimization problem that defines the SVM learning given a set of linearly separable training examples. Let T: R 3!R3 be the transformation on R which re ects every vector across the plane x+y+z= 0. A fast MATLAB implementation of the one-dimensional Weisfeiler--Lehman graph transformation and associated kernel. (b) Find the matrix representation of L with respect to the standard basis 1;x;x2. 1 Linear Transformations A function is a rule that assigns a value from a set B for each element in a set A. Similarly, a vector v is in the kernel of a linear transformation T if and only if T(v)=0. If T isn't an isomorphism find bases of the kernel and image of T, and. Next, we find the range of T. These are all vectors which are annihilated by the transformation. Therefore the number of bytes in the linear buffer is 'skb->len - skb->data_len'. Preimage and kernel example. If a linear transformation T: R n → R m has an inverse function, then m = n. , Garnett, R. Linear Transformations and Polynomials We now turn our attention to the problem of finding the basis in which a given linear transformation has the simplest possible representation. 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. Show that solutions take the form X + V where f(X)=Y and where V is in the kernel. For a linear operator A, the nullspace N(A) is a subspace of X. Similarly, we say a linear transformation T: mthen there exists infinite solutions. Find the dimensions of the kernel and the range of the following linear transformation. Gauss-Jordan elimination yields: Thus, the kernel of consists of all elements of the form:. Determine whether T is an isomorphism. Summary: Kernel 1. Using kernel trick we can write k(x;y) = h˚ x;˚ yi, where k is appropriate kernel [6]. The Kernel and Range of a Linear Transformation Kernel of a linear transformation T: Let be a linear transformationWVT →: Then the set of all vectors v in V that satisfy is called the kernel of T and is denoted by ker (T). Now, let $\phi: V\longrightarrow W$ be a linear mapping/transformation between the two vector spaces. Most of the linear algebra tools deal with dense matrices. To find its Kernel just solve the system : 2x-3y =0 x+4y-z = 0, -x-7y+5z =0 For any LT, think in this way. on the order of 1000 or less since the algorithm is cubic in the number of features. (1 pt each) True - False. The disadvantages are: 1) If the data is linearly separable in the expanded feature space, the linear SVM maximizes the margin better and can lead to a sparser solution. 0)( =vT ker( ) {v | (v) 0, v }T T V= = ∀ ∈. For example: random forests theoretically use feature selection but effectively may not, support vector machines use L2 regularization etc. This can be defined set-theoretically as follows:. defined by. CREATED BY SHANNON MARTIN GRACEY 172 Example 4: Let T R R: 3 3 be a linear transformation. Finding matrices such that M N = N M is an important problem in mathematics. Problem: I can't find answer to a problem. KPCA with linear kernel is the same as standard PCA. Find a basis for the Ker(T). Find the kernel of the linear transformation. Linear combinations of normal random variables. Then the kernel of L is de ned to be: ker(L) := fv 2V : L(v) = ~0g V i. What is the outcome of solving the problem?. These solutions are not necessarily a vector space. Next, we find the range of T. The image of T, denoted by im(T), is the set of all vectors in Rn of the form T(x) = Ax. It is essentially the same thing here that we are talking about. T(x 1,x 2,x 3,x 4)=(x 1−x 2+x 3+x 4,x 1+2x 3−x 4,x 1+x 2+3x 3. 2] KERNEL OF A LINEAR TRANSFORMATION (DEFINITION): Let L : V !W be a linear transformation. Finding the kernel of a linear transformation involving an integral. The linear kernel is not like the others in that it's non-stationary. The linear transformation , from to , is both one-to-one and onto. In each case, state the nullity and rank of T and verify the Rank Theorem. Consider the LINEAR transformation where. This mapping is called the orthogonal projection of V onto W. So,wehave w 1 = v1 kv1k = 1 √ 12 +12. , the solutions of the equation A~x = ~ 0. The image of a linear transformation or matrix is the span of the vectors of the linear transformation. To compute the kernel, find the null space of the matrix of the linear transformation, which is the same to find. A linear transformation (or mapping or map) from V to W is a function T: V → W such that T(v +w)=Tv +Tw T(λv)=λT(v) for all vectors v and w and scalars λ. 0)( =vT ker( ) {v | (v) 0, v }T T V= = ∀ ∈. Determine whether T is an isomorphism. There this is the definition of the kernel. Because Tis one-to-one, the dimension of the image of Tmust be n. Linear Transformation. These are all vectors which are annihilated by the transformation. We build thousands of video walkthroughs for your college courses taught by student experts who got an A+. The kernel of a linear transformation is a vector subspace. 1 Matrix Linear Transformations Every m nmatrix Aover Fde nes linear transformationT A: Fn!Fmvia matrix multiplication. [Solution] To get an orthonormal basis of W, we use Gram-Schmidt process for v1 and v2. Affine transformations", you can find examples of the use of linear transformations, which can be defined as a mapping between two vector spaces that preserves linearity. The linear transformation T is 1-to-1 if and only if the null space of its corresponding matrix has only the zero vector in its null. The image and kernel of linear transformation find significant application in the direct sum decomposition for finite dimension of linear space and the diagonalization of matrices. To find the null space we must first reduce the #3xx3# matrix found above to row echelon form. Demonstrate: A mapping between two sets L: V !W. To help the students develop the ability to solve problems using linear algebra. By definition, every linear transformation T is such that T(0)=0. kernel support: For the current configuration we have 1. The Kernel of a Transformation T. Researchers find security flaws in 40 kernel drivers from 20 vendors. Note that the range of the linear transformation T is the same as the range of the matrix A. Notation: f: A 7!B If the value b 2 B is assigned to value a 2 A, then write f(a) = b, b is called the image of a under f. Let \(V$$ and $$W$$ be vector spaces and let $$T:V. We show that for high-dimensional data, a particular framework for learning a linear transformation of the data based on the LogDet divergence can be efficiently kernelized to learn a metric (or equivalently, a kernel function) over an. The range of A is the columns space of A. In both cases, the kernel is the set of solutions of the corresponding homogeneous linear equations, AX = 0 or BX = 0. Since the nullity is the dimension of the null space, we see that the nullity of T is 0 since the dimension of the zero vector space is 0. The kernel of T, ker (T), is the set of all vectors x in R n for which T(x) = o, the zero vector in R m. De nition 3. Let T: V !W be a linear transformation. Matrix vector products as linear transformations. (b) The dual space V ∗ of the vector space V is the set of all linear functionals on V. Finding a basis of the null space of a matrix. It is given by the common inner product plus an optional constant c. ) T: R^2 rightarrow R^2, T(x, y) = (x + 2y, y - x) ker(T) = {: x, y R} T(v) = Av represents the linear transformation T. Thus, L(V,W) is the space of all linear transformations between V and W. We de ne T Aby the rule T A(x)=Ax:If we express Ain terms of its columns as A=(a 1 a 2 a n), then T A(x)=Ax = Xn i=1 x ia i: Hence the value of T A at x is the linear combination of the columns of A which is the ith. Linear Algebra, David Lay Week Seven True or False. Preimage of a set. For example linear, nonlinear, polynomial, radial basis function. Using kernel trick we can write k(x;y) = h˚ x;˚ yi, where k is appropriate kernel [6]. Here we define linear transformation of vector spaces, kernel of linear transformation, image of linear transformation. This has basis generated by the matrices. The kernel and image of a matrix A of T is defined as the kernel and image of T. Then the kernel of T, denoted by ker(T), is the set of v ∈ V such. Then the Kernel of the linear transformation T is all elements of the vector space V that get mapped onto the zero element of the vector space W. If c = 0, this. (c)Find a linear transformation whose kernel is S?and whose range is S. Let L be de ned on P3 (the vector space of polynomials of degree less. Conversely any linear fractional transformation is a composition of simple trans-formations. Of course we can. Choose a simple yet non-trivial linear transformation with a non-trivial kernel and verify the above claim for the transformation you choose. 1 2 -3 : 1/ 5 y 1 0 0 0 : - 7/. Because is a composition of linear transformations, itself is linear (Theorem th:complinear of LTR-0030). Definition of the Image of linear map 𝐋. SUBSCRIBE to the channel and. Inversion: R(z) = 1 z. An example of a linear transformation T :P n → P n−1 is the derivative function that maps each polynomial p(x)to its derivative p′(x). Linear Transformations and Polynomials We now turn our attention to the problem of finding the basis in which a given linear transformation has the simplest possible representation. The Linear kernel is the simplest kernel function. Trying to use matrices and matrix methods is almost a waste of time in this problem. From this, it follows that the image of L is isomorphic to the quotient of V by the kernel: ≅ / (). We could denote V L ≅ W or V ≅ W. The linear kernel is not like the others in that it's non-stationary. 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. De ne T : P 2!R2 by T(p) = p(0) p(0). 0000 Today we are going to continue our discussion of the kernel and range of a linear map of a linear transformation. For example the kernel of this matrix (call it A). 2 (The Kernel and Range)/3. To find the kernel of a matrix A is the same as to solve the system AX = 0, and one usually does this by putting A in rref. Next, we find the range of T. #20 Consider the subspace Wof R4 spanned by the vectors v1 = 1 1 1 1 and v2 = 1 9 −5 3. The image of T, denoted by im(T), is the set of all vectors in Rn of the form T(x) = Ax. Linear algebra - Practice problems for midterm 2 1. Using non-linear transformation, you can easily solve non-linear problem as a linear (straight-line) problem. Let L be de ned on P3 (the vector space of polynomials of degree less. 0:22 So, if I have one vector that goes to 0, that is the kernel. We will now prove some results regarding the range/kernel of linear operators. De ne T : P 2!R2 by T(p) = p(0) p(0). (The dimension of the image space is sometimes called the rank of T, and the dimension of the kernel is sometimes called the nullity of T. defined by. Next, we find the range of T. Note that the range of the linear transformation T is the same as the range of the matrix A. The linear transformation t 2 is the orthogonal projection on the x-axis. Based on the above two aspects, we propose the kernel negative dragging linear regression (KNDLR) method in. Let L be de ned on P3 (the vector space of polynomials of degree less than 3) by L(p) = q where q(x) = xp′′(x) p(0)x: (a) Find the kernel of L. Verify that T is a linear transformation. (Also discussed: nullity of L; is L one-to-one?). Finding the kernel of a linear transformation involving an integral. 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. Gauss-Jordan elimination yields: Thus, the kernel of consists of all elements of the form:. Let T: V-> W be a linear transformation between vector spaces V and W. asked • 17d find the kernel of the linear transformation :-1-T:R 3 →R 3,T(x,y,z)=(0,0,0). Describe in geometrical terms the linear transformation defined by the following matrices: a. Note: It is convention to use the Greek letter 'phi' for this transformation , so I'll use. linear transformation. Determine whether T is an isomorphism. array([4,1,0, 1,4]) By combing the existing and new feature, we can certainly draw a line to separate the yellow purple dots (shown on the right). Remarks I The kernel of a linear transformation is a. (If all real numbers are solutions, enter REALS. (The dimension of the image space is sometimes called the rank of T, and the dimension of the kernel is sometimes called the nullity of T. For linear operators, we can always just use D = X, so we largely ignore D hereafter. A stationary covariance function is one that only depends on the relative position of its two inputs, and not on their absolute location. Let \(T:V\rightarrow W$$ be a linear transformation where $$V$$ and $$W$$ be vector spaces with scalars coming from the same field $$\mathbb{F}$$. To take an easy example, suppose we have a linear transformation on R 2 that maps (x, y) to (4x+ 2y, 2x+ y). Note: Because Rn is a "larger" set than Rm when m < n, it should not be possible to map Rn to Rm in a one-to-one fashion. Image Let F : Rn Rm be a linear mapping. We are interested in some mappings (called linear transformations) between vector spaces L: V !W; which preserves the structures of the vector spaces. Anyway, hopefully you found that reasonably. In particular, there exists a nonzero solution. (If all real numbers are solutions, enter REALS. (1 pt each) True - False. Then (1) is a subspace of. More importantly, as an injective linear transformation, the kernel is trivial (Theorem KILT), so each pre-image is a single vector. 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. (some people call this the nullspace of L). The converse is also true. It is essentially the same thing here that we are talking about. A function is linear if the following two properties hold: For example, the function defined on the real line is not linear, since whereas. Power Iterated Color Refinement. For example Let’s say we have a transformation x !˚ x. The challenge is to find a transformation -> , such that the transformed dataset is linearly separable in. THE PROPERTIES OF DETERMINANTS a. The next theorem is the key result of this chapter. The kernel of A are all solutions to the linear system Ax = 0. Algebra Examples. Linear Transformation Exercises Olena Bormashenko December 12, 2011 1. Let L be de ned on P3 (the vector space of polynomials of degree less than 3) by L(p) = q where. Already for sys- tems of polynomial equations, one has to work with linear spaces of polynomials. These are all vectors which are annihilated by the transformation. Find a basis for the Ker(T). We describe the range by giving its basis. u+v = v +u,. Linear Algebra: Find bases for the kernel and range for the linear transformation T:R^3 to R^2 defined by T(x1, x2, x3) = (x1+x2, -2x1+x2-x3). Morphological transformations are some simple operations based on the image shape. For linear operators, we can always just use D = X, so we largely ignore D hereafter. ) It can be written as Im (A). This contradict to that L is an injection, since v ≠ 0V. UNSOLVED! So I have a linear transformation that is the definite integral from 1 to 0 of a vector in P2 (ax2 + bx + c). Specifically, if T: n m is a linear transformation, then there is a unique m n matrix, A, such that T x Ax for all x n. Find the range of the linear transformation T: R4 →R3 whose standard representation matrix. My idea is to save the general fromula of the linear map which would work for sure but I wanted to know if there's a quicker way of doing it without finding the general formula of the linear map. Then rangeT is a finite-dimensional subspace of W and dimV = dimnullT +dimrangeT. Determining if a given transformation is linear Determining the representation matrix of a linear transformation Representation matrices Kernel of a linear transformation One-to-one linear transformations Onto linear transformations One-to-one and onto Other subjects: here you can put links to material on other subjects you found yourself. And if the transformation is equal to some matrix times some vector, and we know that any linear transformation can be written as a matrix vector product, then the kernel of T is the same thing as the null space of A. To clarify what is meant by a power transformation, the formula for the model is provided above. 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. To compute the kernel, find the null space of the matrix of the linear transformation, which is the same to find. Finding a basis of the null space of a matrix. SUBSCRIBE to the channel and. Then the following properties are true. This paper studies the conditions for the idempotent transformation and the idempotent rank transformation direct sum decomposition for finite dimension of linear space. Journal of Information and Telecommunication: Vol. Suppose T : V !W is a linear transformation. Corollary 2. The image of a linear transformation contains 0 and is closed under addition and scalar multiplication. To find the kernel, you just need to determine the dimensionality of the solution space to the linear system. A= [-3, -2 , 4. Let T: V !W be a linear transformation. This contradict to that L is an injection, since v ≠ 0V. First here is a definition of what is meant by the image and kernel of a linear transformation. The kernel of a linear transformation {eq}L: V\rightarrow V {/eq} is the set of all polynomials such that {eq}L(p(t))=0 {/eq} Here, {eq}p(t) {/eq} is a polynomial. visualize what the particular transformation is doing. Construct a linear transformation f and vector Y so that the system takes the form f(X)=Y. This basis can be extended to. If m < n, then T cannot be one-to-one. As such, this theorem goes by the name of the Rank- nullity Theorem. If V is finite-dimensional, then so are Im(T) and ker(T), anddim(Im(T))+dim(ker(T))=dimV. This MATLAB function returns predicted class labels for each observation in the predictor data X based on the binary Gaussian kernel classification model Mdl. Note that N(T) is a subspace of V, so its dimension can be de ned. More Examples of Linear Transformations: solutions: 6: More on Bases of $$\mathbb{R}^n$$, Matrix Products: solutions: 7: Matrix Inverses: solutions: 8: Coordinates: solutions: 9: Image and Kernel of a Linear Transformation, Introduction to Linear Independence: solutions: 10: Subspaces of $$\mathbb{R}^n$$, Bases and Linear Independence. The algorithm: The idea behind kernelml is simple. To do this, find the images of the standard unit vectors and use them to create the standard matrix for. The kernel of a linear transformation T (~x) = A~x is the set of all zeros of the transformation (i. Let T: V-> W be a linear transformation between vector spaces V and W. (a) Find the matrix representative of T relative to the bases f1;x;x2gand f1;x;x2;x3gfor P 2 and P 3. Proof: This theorem is a proved in a manner similar to how we solved the above example. q(x) = xp′′(x) p(0)x: (a) Find the kernel of L. 2 The kernel and range of a linear transformation. ∆ Let T: V ‘ W be a linear transformation, and let {eá} be a basis for V. (a) L(x) = (x3, x2, x1)^T. 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. Then (1) is a subspace of. Let T: V !W be a linear transformation. As such, this theorem goes by the name of the Rank- nullity Theorem. Finding a basis of the null space of a matrix. Find a basis for the kernel of T and the range of T. The dimension of the kernel of T is the same as the dimension of its null space and is called the nullity of the transformation. suppose T(x,y,z) = ( 2x-3y, x+4y-z, -x-7y+5z ) be a linear transformation. Based on the above two aspects, we propose the kernel negative dragging linear regression (KNDLR) method in. SVC has a kernel parameter which can take on linear, poly, rbf, or sigmoid [4]. We build thousands of video walkthroughs for your college courses taught by student experts who got an A+. You should think about something called the null space. Determine whether T is an isomorphism. We are interested in some mappings (called linear transformations) between vector spaces L: V !W; which preserves the structures of the vector spaces. , KPCA with a Linear kernel is equivalent to standard PCA. The following charts show some of the ideas of non-linear transformation. Kernel Principal Component Analysis In the section 1 we have discussed a motivation for the use of kernel methods – there are a lot of machine learning problems which a nonlinear, and the use of nonlinear feature mappings can help to produce new features which make prediction problems linear. We show that for high-dimensional data, a particular framework for learning a linear transformation of the data based on the LogDet divergence can be efficiently kernelized to learn a metric (or equivalently, a kernel function) over an. The image of a linear transformation ~x7!A~xis the span of the column vectors of A. We write ker(A) or ker(T). Hello and welcome back to Educator. The general solution is a linear combination of the elements of a basis for the kernel, with the coefficients being arbitrary constants. TRUE To show this we show it is a subspace Col A is the set of a vectors that can be written as Ax for some x. The linear transformation , from to , is both one-to-one and onto. The converse is also true. 0:22 So, if I have one vector that goes to 0, that is the kernel. This paper is organized as follows. Learn vocabulary, terms, and more with flashcards, games, and other study tools. 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. Then, ker(L) is a subspace of V. Find a basis for the kernel of the linear transformation T : R^4 -> R^2. Use the kernel and image to determine if a linear transformation is one to one or onto. Sources of subspaces: kernels and ranges of linear transformations. Determining if a given transformation is linear Determining the representation matrix of a linear transformation Representation matrices Kernel of a linear transformation One-to-one linear transformations Onto linear transformations One-to-one and onto Other subjects: here you can put links to material on other subjects you found yourself. Kernel Principal Component Analysis In the section 1 we have discussed a motivation for the use of kernel methods – there are a lot of machine learning problems which a nonlinear, and the use of nonlinear feature mappings can help to produce new features which make prediction problems linear. How to find the kernel of a linear transformation? Let B∈V =Mn(K) and let CB :V →V be the map defined by CB(A)=AB−BA. Note that N(T) is a subspace of V, so its dimension can be de ned. Linear Transformations and Polynomials We now turn our attention to the problem of finding the basis in which a given linear transformation has the simplest possible representation. Thus V and W are isomorphic. suppose you have a 5 classes of data ordered like a 5 on a dice. If w2 = 0, w3 = 1, then w1 = -1, and if w2 = 1 and w3 = 1, then w1 = 0. Introduction to Linear Algebra exam problems and solutions at the Ohio State University. The aim of our study of linear transformations is two-fold: • to understand linear transformations in R, R2 and R3. As a result, unlike transductive kernel learning methods, our method easily handles out-of-sample extensions, i. Find the image and the rank of the linear transformation T with matrix A = 2 4 1 1 3 1 2 5 1 3 7 3 5: 3. ) T: P 5 → R, T(a 0 + a 1 x + a 2 x 2 + a 3 x 3 + a 4 x 4 + a 5 x 5) = a 0. The following is a basic list of model types or relevant characteristics. My idea is to save the general fromula of the linear map which would work for sure but I wanted to know if there's a quicker way of doing it without finding the general formula of the linear map. Definition of the Image of linear map 𝐋. (If all real numbers are solutions, enter REALS. (b) Find the matrix representation of L with respect to the standard basis 1;x;x2. (c) Determine whether a given vector is in the kernel or range of a linear trans-formation. 3 (Nullity). We build thousands of video walkthroughs for your college courses taught by student experts who got an A+. Let be a linear transformation. Kernel The kernel of a linear transformation T(~x) = A~x is the set of all zeros of the transformation (i. be a linear transformation. 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. 1 Let V,W be two vector spaces and T : V → W a linear transformation. Null space. Use the parameter update history in a machine learning model to decide how to update the next parameter set. Thus V and W are isomorphic. This MATLAB function returns predicted class labels for each observation in the predictor data X based on the binary Gaussian kernel classification model Mdl. SVM algorithms use a set of mathematical functions that are defined as the kernel. We show that for high-dimensional data, a particular framework for learning a linear transformation of the data based on the LogDet divergence can be efficiently kernelized to learn a metric (or equivalently, a kernel function) over an. Yet if we map it to a three-dimensional. 1 T(~x + ~y) = T(~x) + T(~y)(preservation of addition) 2 T(a~x) = aT(~x)(preservation of scalar multiplication) Linear Transformations: Matrix of a Linear Transformation Linear Transformations Page 2/13. be a linear transformation. Determine whether T is an isomorphism. SPECIFY THE VECTOR SPACES Please select the appropriate values from the popup menus, then click on the "Submit" button. Let T: V !W be a linear transformation. For two linear transformations K and L taking Rn Rn , and v Rn , then in general K(L(v)) = L(K(v)). To find its Kernel just solve the system : 2x-3y =0 x+4y-z = 0, -x-7y+5z =0 For any LT, think in this way. the solutions to this system of linear equations ARE the null space of the matrix of the system (because these are homogeneous linear equations (which is a fancy way of saying: "all 0's on one side")). " • The fact that T is linear is essential to the kernel and range being subspaces. If T : Rm → Rn is a linear transformation, then the set {x | T(x) = 0 } is called the kernelof T. TRUE Remember that Ax gives a linear combination of columns of A using x entries as weights. The image and kernel of linear transformation find significant application in the direct sum decomposition for finite dimension of linear space and the diagonalization of matrices. linear_kernel (X[, Y, …]) Compute the linear kernel between X and Y. And if the transformation is equal to some matrix times some vector, and we know that any linear transformation can be written as a matrix vector product, then the kernel of T is the same thing as the null space of A. The theorem relating the dimension of the kernel and image requires the vector spaces to be finite dimensional. In the linear map L : V → W, two elements of V have the same image in W if and only if their difference lies in the kernel of L : It follows that the image of L is isomorphic to the quotient of V by the kernel: This implies the rank–nullity theorem :. In fact, 4x+ 2y= 2(2x+ y) so those are the same equation which is equivalent to y= -2x. In mathematics, a linear map (also called a linear mapping, linear transformation or, in some contexts, linear function) is a mapping V → W between two modules (for example, two vector spaces) that preserves (in the sense defined below) the operations of addition and scalar multiplication. A basis for the kernel of L is {1} so the kernel has dimension 1. T is a linear transformation. In (non-linear) kernel PLS and direct kernel PLS (DK-PLS) a kernel transformation on X is applied by applying K(x i. Note that the range of the linear transformation T is the same as the range of the matrix A. If I have 5 vectors that map to 0, those 5 vectors, they form the kernel. Let V and Wbe. Find the kernel of the linear transformation. Note that N(T) is a subspace of V, so its dimension can be de ned. 2 Kernel of linear transformations Deflnition 3. The transformation matrices are as follows: Type of transformation. Let T be a linear transformation on Rn to Rm. Use the kernel and image to determine if a linear transformation is one to one or onto. S: ℝ3 → ℝ3. {\mathbb R}^m. The fact that this can be interpreted as "perfect linear separation in an infinite dimensional feature space" comes from the kernel trick, which allows you to interpret the kernel as an abstract inner product some new feature space:. Finding the kernel of a linear transformation involving an integral. Shed the societal and cultural narratives holding you back and let free step-by-step Linear Algebra: A Modern Introduction textbook solutions reorient your old paradigms. 4 Linear Transformations The operations \+" and \" provide a linear structure on vector space V. First prove the transform preserves this property. The theorem relating the dimension of the kernel and image requires the vector spaces to be finite dimensional. ) Let be the transformation. (d) The null space of A is the kernel of the mapping X -> AX. The Kernel and Range of a Linear Transformation Kernel of a linear transformation T: Let be a linear transformationWVT →: Then the set of all vectors v in V that satisfy is called the kernel of T and is denoted by ker (T). The kernel and range "live in different places. Choose Regression Model Options Choose Regression Model Type. A fast MATLAB implementation of the one-dimensional Weisfeiler--Lehman graph transformation and associated kernel. 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. All Slader step-by-step solutions are FREE. be a linear transformation. 3 (Nullity). In the case where V is finite-dimensional, this implies the rank-nullity theorem:. (f) The set of all solutions of a homogeneous linear differential equation is the kernel of a linear transformation. Similarly, we say a linear transformation T: max 1 i n di +2 r 2R2 n (p 2+ln r 1 )) 1 n+1 where the support of the distribution D is assumed to be contained in a ball of radius R. | 2020-05-29T20:42:05 | {
"domain": "tuning-style.it",
"url": "http://tuning-style.it/wdmc/how-to-find-the-kernel-of-a-linear-transformation.html",
"openwebmath_score": 0.7735112905502319,
"openwebmath_perplexity": 420.58524934978,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9838471647042428,
"lm_q2_score": 0.8577680977182186,
"lm_q1q2_score": 0.8439127109138213
} |
https://math.stackexchange.com/questions/1267394/i-draw-a-hand-of-13-from-a-deck-of-52-cards-what-is-the-probability-that-i-do-n | # I draw a hand of 13 from a deck of 52 cards. What is the probability that I do not have a card from every suit?
I draw a hand of 13 from a deck of 52 standard playing cards. What is the probability that I do not have a card from every suit?
I count the number of ways I can draw 13 from 3 suits
$$\frac{{4\choose3}{39\choose13}}{52\choose13}$$
but I mind the intersection. Each possible pair of suits that I may have drawn from only is counted twice. And in the possibility that I pick from only one suit: each possibility is counted three times.
$$\frac{{4\choose3}{39\choose13}-{4\choose2}{26\choose13}}{52\choose13}$$
This removes the overcounted iterations from the pair of suits I could have drawn from, but now I'm not considering the possibility that I drew from only one suit, so:
$$\frac{{4\choose3}{39\choose13}-{4\choose2}{26\choose13}+{4\choose1}{13\choose13}}{52\choose13}$$
which is the probability I'm looking for.
Is my reasoning sound? Have I made any mistakes? Is there a better solution?
• Good use of Inclusion/Exclusion. – André Nicolas May 5 '15 at 1:17
• @ whorl There are other ways of solving this, but they will be less elegant. Your way works perfectly fine. An example of a direct approach might have been: (For a simplification being a hand of 5 cards instead of 13). Break this into cases: distribution of suits (denoted most common to least) could be one of: (5,0,0,0), (4,1,0,0), (3,2,0,0), (3,1,1,0), or (2,2,1,0). Then count how many hands fall into each of those cases. Generally, casework like this should be avoided if at all possible as it can be very tedious. The approach for 13 card hands is similar but much longer. – JMoravitz May 5 '15 at 1:37
• I kind of take it back. You really should not have divided by $\binom{52}{13}$ until the end. There should have been a count of the "favourables" (your final numerator), then the division. You call for example $\binom{4}{3}\binom{39}{13}/\binom{52}{13}$ a number of ways, but it isn't. And even $\binom{4}{3}\binom{39}{13}$ is not a count of anything, it is a step toward the final count. – André Nicolas May 5 '15 at 1:40
• @AndréNicolas You're right, that mess is an artifact of my first two attempted solutions before I realized I wasn't paying attention to intersections. – whorl May 5 '15 at 1:44
$\boxed{\color{green}{\checkmark}}$ Yes; by the Principle of Inclusion and Exclusion, the ways of not drawing from all four suits is:
• Include: the $\binom{4}{1}\binom{52-13}{13}$ ways to pick one suit and not draw from this, then
• Exclude: the $\binom{4}{2}\binom{52-26}{13}$ ways to pick two suits and not draw from both, then
• Include: the $\binom{4}{3}\binom{52-39}{13}$ ways to pick three suits and not draw from these.
$$\dfrac{\dbinom{4}{1}\dbinom{39}{13}-\dbinom{4}{2}\dbinom{26}{13}+\dbinom{4}{3}\dbinom{13}{13}}{\dbinom{52}{13}}$$
NB $\binom{4}{1}=\binom{4}{3}$ | 2019-09-15T22:48:18 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1267394/i-draw-a-hand-of-13-from-a-deck-of-52-cards-what-is-the-probability-that-i-do-n",
"openwebmath_score": 0.5637345910072327,
"openwebmath_perplexity": 386.1807065294793,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 1,
"lm_q2_score": 0.8438951025545426,
"lm_q1q2_score": 0.8438951025545426
} |
https://math.stackexchange.com/questions/2358680/convergent-sequence-with-odd-terms-decreasing-and-even-terms-increasing | Convergent sequence with odd terms decreasing and even terms increasing
Let $\left(a_n\right)$ is convergent sequence. $a_0=0, a_1=1,a_2,a_3,...$
Odd terms decrease and even terms increase and for all $n\ge1$ $$2\le \frac{a_n-a_{n-1}}{a_n-a_{n+1}}\le3.$$ Find the boundaries in which there can be a limit of this sequence.
My work so far:
I proved that $$\frac{11}{24}\le\lim_{n\to\infty}a_n\le\frac{23}{24}$$ But I can not say that this is the final answer. I need an examples of sequences such that:
1) $$\lim_{n\rightarrow\infty}a_n=\frac{11}{24}$$
2) $$\lim_{n\rightarrow\infty}a_n=\frac{23}{24}$$ or
3) (If my answer can be improved) I need numbers $m$ and $M$, where
$$\frac{11}{24}<m\le\lim_{n\rightarrow\infty}a_n\le M<\frac{23}{24}$$
Define the first differences $b_n=a_{n+1}-a_n$ for $n\ge0$. Then $b_0=1$, and since $a_0=0$: $$\lim_{n\to\infty}a_n=\sum_{n=0}^\infty b_n$$ The given inequality may be rewritten for $n\ge0$ as $$-\frac12\le\frac{b_{n+1}}{b_n}\le-\frac13$$ To make the infinite sum in $b_n$ as large as possible we have to
• Subtract as little as possible: for odd-numbered $b_n$, which will be negative, we multiply by $-\frac13$ from $b_{n-1}$
• Add as much as possible: for even-numbered $b_n$, which will be positive, we multiply by $-\frac12$ from $b_{n-1}$
This results in $$S=1-\frac13+\frac1{3\cdot2}-\frac1{3\cdot2\cdot3}+\dots$$ Multiply both sides by $\frac1{3\cdot2}$: $$\frac16S=\frac1{3\cdot2}-\frac1{3\cdot2\cdot3}+\dots=S-1+\frac13$$ $$-\frac56S=-\frac23\qquad S=\frac45$$ Similarly we can make the infinite sum in $b_n$ as small as possible by swapping $-\frac13$ and $-\frac12$ in the list above, giving $$T=1-\frac12+\frac1{2\cdot3}-\frac1{2\cdot3\cdot2}+\dots$$ Once again, multiply both sides by $\frac1{2\cdot3}$: $$\frac16T=\frac1{2\cdot3}-\frac1{2\cdot3\cdot2}+\dots=T-1+\frac12$$ $$-\frac56T=-\frac12\qquad T=\frac35$$ Therefore $$\frac35\le\lim_{n\to\infty}a_n\le\frac45$$
• This is not correct. In your solution $$b_n=-\frac13b_{n-1}=\frac13\cdot\frac12b_{n-2}$$ Then $$b_n=\frac16b_{n-2}$$ This is not a minimization. For example $$b_n=\frac13b_{n-1}=\frac13\cdot\frac13b_{n-2}$$ Then $$b_n=\frac19b_{n-2}$$ Jul 15 '17 at 14:33
• @Roman83 That does not show that my solution is incorrect. Indeed, $b_n=-\frac13\cdot-\frac12b_{n-2}$. Be wary of signs. Jul 15 '17 at 14:36
• For the infinite sum $$b_0=1, b_1=-\frac13, b_2=\frac16, b_3=-\frac1{3\cdot2\cdot3}, ....$$ So? Jul 15 '17 at 14:42
• @Roman83 Minimising the sum must start with $b_1=-\frac12$, NOT $b_1=-\frac13$! Jul 15 '17 at 14:43
• "question eligible for bounty in 23 hours" Jul 15 '17 at 15:43
The answer of Parcly is correct and should be awarded the bounty. For a formal proof: Let $B$ denote the set of admissible sequences $\beta=(b_0=1,b_1,b_2,...)$ verifying that $b_{2n}>0>b_{2n+1}$ all $n$ and the ratio condition $$b_k/b_{k+1} \in \Delta=\left[-\frac12,-\frac13\right]$$ We are looking for extremal values of $\sum_{k\geq 0} b_k$.
The key point is that for any $m\geq 0$ one has the following a priori bound (split into even and odd indices): $$(-1)^m \sum_{k\geq 0} b_{m+k} \geq (-1)^m b_m \left( \sum_{k\geq 0} (1/9)^k - 1/2 \sum_{k\geq 0} (1/4)^k\right) =(-1)^m b_m \frac{11}{24}>0$$
in particular, the tail-sum from term $m$ has the same sign as $b_m$. Suppose now that $\beta$ is admissible then for $m\geq 1$ so is $$\hat{\beta}_{m,r} = (b_0=1,...,b_{m-1}, r b_{m}, r b_{m+1},...)$$ for any $r>0$ for which $rb_m/b_{m+1}\in \Delta$.
The sum of the series $\hat{\beta}_{m,r}$ is $\sum_{0\leq k<m} b_k + r \sum_{k\geq m} b_k$. As shown above the last sum has the same sign as $b_m$ and can be made strictly smaller and larger by choosing suitable $r$ whenever $b_m/b_{m+1}\in (-\frac12,-\frac13)$ (an interior point). For example to minimize the sum, for every even $m$ we must require $b_m/b_{m-1}=-1/3$ or else you may make the sum of $\hat{\beta}_{m,r}$ smaller for suitable $r<1$. Similarly for odd $m$, we must have $b_m/b_{m-1}=-1/2$. The max case is treated in a similar way and the extremal values are then given as described by Parcly.
I will consider the general case of a sequence $(a_n)_{n\ge0}$ such that $a_0=0$ $a_1=1$, the subsequence $(a_{2n})_{n\ge0}$ is increasing, and the sequence $(a_{2n+1})_{n\ge0}$ is decreasing, and finally for $0< \beta<\alpha<1$ we have $$\forall\, n\ge1,\qquad \frac{1}{\alpha} \le \frac{a_n-a_{n-1}}{a_n-a_{n+1}}\le\frac{1}{\beta}\tag{*}$$ I will prove that $\lim\limits_{n\to\infty}a_n$ exists and that $$\frac{1-\alpha}{1-\alpha\beta}\le\lim_{n\to\infty}a_n\le \frac{1-\beta}{1-\alpha\beta}$$ and finally that this conclusion cannot be improved.
Let $b_n=(-1)^n(a_{n+1}-a_{n})$. Then $(*)$ implies that $b_{n-1}/b_n\ge\alpha>0$ so all the terms of the sequence $(b_n)_{n\ge0}$ have the same sign, but $b_0=1>0$, hence $b_n>0$ for all $n$. Moreover, $(*)$ implies also that $b_{n+1}\le\alpha b_n$ for all $n\ge0$, thus $b_n\le\alpha^{n}$ and the the series $\sum_{n=0}^\infty (a_{n+1}-a_{n})$ is absolutely convergent, which is equivalent to the existence of $\ell=\lim\limits_{n\to\infty}a_n$.
Now, considering two cases $(*)$ is equivalent to the following two inequalities: \begin{alignat*}{3} &(1-\alpha)a_{2n+1}+\alpha a_{2n}&&\le a_{2n+2}&&\le (1-\beta)a_{2n+1}+\beta a_{2n}\tag{1}\\ &(1-\beta) a_{2n+2}+ \beta a_{2n+1}&&\le a_{2n+3}&&\le (1-\alpha)a_{2n+2}+\alpha a_{2n+1}\tag{2} \end{alignat*}
This suggests that we consider the sequences $(m_n)_{n\ge0}$ and $(M_n)_{n\ge0}$ defined as follows: \begin{alignat*}{3} &m_0=0,m_1=1, \quad m_{2n+2}&&=(1-\alpha)m_{2n+1}+\alpha m_{2n},\quad m_{2n+3}&&=(1-\beta)m_{2n+2}+\beta m_{2n+1}.\\ &M_0=0,M_1=1, \quad M_{2n+2}&&=(1-\beta)M_{2n+1}+\beta M_{2n},\quad M_{2n+3}&&=(1-\alpha)M_{2n+2}+\alpha M_{2n+1}. \end{alignat*} Then it is an easy induction to prove that \begin{equation*} \forall\,n\ge0,\quad m_n\le a_n\le M_n\tag{3} \end{equation*} Indeed, let $\mathbb{P}_n=( m_{n}\le a_{n}\le M_{n})$, then the base cases $\mathbb{P}_0$ and $\mathbb{P}_1$ are satisfied by assumption. Now, from $(1)$ we conclude that $(\mathbb{P}_{2n}\wedge \mathbb{P}_{2n+1})\implies \mathbb{P}_{2n+2}$ and from $(2)$ we conclude that $(\mathbb{P}_{2n+1}\wedge \mathbb{P}_{2n+2})\implies \mathbb{P}_{2n+3}$ This proves that $\mathbb{P}_n$ is satisfied for every $n$.
Now, if $X_n=\left[\begin{matrix} m_{2n}\\ m_{2n+1} \end{matrix}\right]$, and $Y_n=\left[\begin{matrix} M_{2n}\\ M_{2n+1} \end{matrix}\right]$ then \begin{equation*} X_{n+1}=\underbrace{\left[\begin{matrix} \alpha&1-\alpha\\ (1-\beta)\alpha&1-\alpha+\alpha\beta \end{matrix}\right]}_{A_{\alpha,\beta}}X_n,\qquad Y_{n+1}=\underbrace{\left[\begin{matrix} \beta&1-\beta\\ (1-\alpha)\beta&1-\beta+\alpha\beta \end{matrix}\right]}_{A_{\beta,\alpha}}Y_n \end{equation*} with initial conditions $X_0=Y_0=\left[\begin{matrix} 0\\1 \end{matrix}\right]$. Now, the characteristic polynomial $Q(X)$ of $A_{\alpha,\beta}$ is given by $Q(X)=X^2-(1+\alpha\beta)X+\alpha\beta=(X-1)(X-\alpha\beta)$, and it is easy to check that the remainder of the euclidean division of $X^n$ by $Q(X)$ is \begin{equation*} \frac{1}{1-\alpha\beta}(X-\alpha\beta)+ \frac{(\alpha\beta)^n}{1-\alpha\beta}(1-X) \end{equation*} Hence \begin{equation*} A_{\alpha,\beta}^n= \frac{1}{1-\alpha\beta}(A_{\alpha,\beta}-\alpha\beta I) +\frac{(\alpha\beta)^n}{1-\alpha\beta}(I-A_{\alpha,\beta}) \end{equation*} and since $X_n=A_{\alpha,\beta}^nX_0$ we conclude easily that \begin{align*} m_{2n}&=\frac{1-\alpha}{1-\alpha\beta}-\frac{1-\alpha}{1-\alpha\beta}(\alpha\beta)^n\\ m_{2n+1}&=\frac{1-\alpha}{1-\alpha\beta}+\frac{\alpha(1-\beta)}{1-\alpha\beta}(\alpha\beta)^n \end{align*} Exchanging the roles of $\alpha$ and $\beta$ we see also that \begin{align*} M_{2n}&=\frac{1-\beta}{1-\alpha\beta}-\frac{1-\beta}{1-\alpha\beta}(\alpha\beta)^n\\ M_{2n+1}&=\frac{1-\beta}{1-\alpha\beta}+\frac{\beta(1-\alpha)}{1-\alpha\beta}(\alpha\beta)^n \end{align*} In particular, we have \begin{equation*} \lim_{n\to\infty}m_n=\frac{1-\alpha}{1-\alpha\beta}, \quad \lim_{n\to\infty}M_n=\frac{1-\beta}{1-\alpha\beta} \end{equation*} Hence, letting $n$ tend to $+\infty$ in $(3)$ we get \begin{equation*} \frac{1-\alpha}{1-\alpha\beta}\le\lim_{n\to\infty}a_n\le \frac{1-\beta}{1-\alpha\beta} \end{equation*} Now, taking $a_n=m_n$ for all $n$, shows that the lower bound in the above inequality is the best possible, because it is attained, and taking $a_n=M_n$ for all $n$, shows that the upper bound in the above inequality is also the best possible, because it is attained.
Further, considering sequences $(a_n)_{n\ge0}$ of the form $a_n=tm_n+(1-t)M_n$ where $0<t<1$, shows that for any number $\ell$ in the interval $[\frac{1-\alpha}{1-\alpha\beta}, \frac{1-\beta}{1-\alpha\beta}]$ there exists a sequence $(a_n)_{n\ge0}$ satisfying the conditions of the problem and converging to $\ell$.
Remark. Surly we have noticed that the proposed problem corresponds to the case $\alpha=1/2$ and $\beta=1/3$, so the lower and upper bounds are indeed $3/5$ and $4/5$. | 2022-01-28T00:43:31 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2358680/convergent-sequence-with-odd-terms-decreasing-and-even-terms-increasing",
"openwebmath_score": 0.9852163195610046,
"openwebmath_perplexity": 175.61046105108053,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9693241965169938,
"lm_q2_score": 0.8705972734445508,
"lm_q1q2_score": 0.8438910025715247
} |
https://math.stackexchange.com/questions/3043594/antiderivative-of-an-odd-function | # Antiderivative of an odd function
Is the antiderivative of an odd function even?
The answer given by the book is yes.
However, I found a counterexample defined in $$\mathbb{R}\setminus \{0\}$$: $$f(x)=\begin{cases}\ln |x|+1& x<0\\\ln |x|&x>0\end{cases}$$ Its derivative is $$\frac 1x$$, which is an odd function.
Question: is my counterexample right?
• I think the question implies that the odd function in question must contain $0$ in its domain. Otherwise, you can't integrate it across a symmetric interval – Dylan Dec 17 '18 at 6:56
• You can actually do this for any odd function if you allow a piecewise function definition, as there are infinitely many anti-derivatives for a given function --- simply pick two that differ by a constant, then piece them together. E.g. consider $F(x) = x^4 + [x>0]$, where $[\cdot]$ is the Iverson bracket (equal to $1$ when the condition is true, otherwise $0$), which is an antiderivative of $f(x) = x^3$. – apnorton Dec 17 '18 at 7:19
• @apnorton I don't think so. Your $f(x)$ is not differentiable at $x=0$, it contradicts with the definition of antiderivative. – Kemono Chen Dec 17 '18 at 7:23
• $x^2 + C$ is even for any $C$. For the opposite case of whether the antiderivative of an even function is odd, this point would be valid. – badjohn Dec 17 '18 at 11:28
• @1123581321 Constants are even. – Dylan Dec 17 '18 at 19:19
I think that $$f$$ should be defined on an interval $$I$$ which contains $$0$$ and is symmetric to $$0$$. If $$F$$ is an antiderivative of $$f$$ on $$I$$, then there is a constant $$c$$ such that
$$F(x)=\int_0^x f(t) dt+c.$$
If you now calculate $$F(-x)$$ with the substitution $$s=-t$$ you will get $$F(-x)=F(x).$$
• @lalala If $f(t)=\sin(t)$ then $\int_0^x f(t) \,dt = 1-\cos(t)$. Choose a $c$ such as $-1$ or if you prefer $1$ and do as Fred suggests to find $F(-x)$. What do you get? – Henry Dec 17 '18 at 13:13 | 2019-03-24T07:02:29 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3043594/antiderivative-of-an-odd-function",
"openwebmath_score": 0.9354775547981262,
"openwebmath_perplexity": 159.1561143404611,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9773708032626701,
"lm_q2_score": 0.8633916134888614,
"lm_q1q2_score": 0.8438537548058612
} |
http://www.physicsforums.com/showthread.php?p=3974054 | Another interesting number theory tidbit
Hello,
I was browsing a set of number theory problems, and I came across this one:
"Prove that the equation a2+b2=c2+3 has infinitely many solutions in integers."
Now, I found out that c must be odd and a and b must be even. So, for some integer n, c=2n+1, so c2+3=4n2+4n+4=4[n2+n+1]. If n is of the form k2-1, then the triple of integers{2n,2$\sqrt{n+1}$,2n+1]} satisfies the equation. Since there are infinitely such n, the equation holds for integers infinitely often.
I thought this was cool.
Mathguy
PhysOrg.com science news on PhysOrg.com >> 'Whodunnit' of Irish potato famine solved>> The mammoth's lament: Study shows how cosmic impact sparked devastating climate change>> Curiosity Mars rover drills second rock target
Recognitions: Homework Help Science Advisor Looks like the same approach could be used for many constants. So the question becomes, for what k does a2+b2=c2+k have infinitely many solutions?
Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus That is cool!!! Nice find!! A no-brainer as follow-up question is of course: are these all the solutions?? I don't know the answer myself, but it's interesting to find out.
Another interesting number theory tidbit
Quote by micromass That is cool!!! Nice find!! A no-brainer as follow-up question is of course: are these all the solutions?? I don't know the answer myself, but it's interesting to find out.
That is an interesting question.
The case when k=0 has infinitely many solutions of which are all of the form $a=d(p^2-q^2)$, $b=2dpq$, $c=d(p^2+q^2)$ for integer p,q and an arbitrary constant d. The case k=3 makes the right hand side the square of 2n+2 when c=2n+1, and hence the case k=0 implies the case k=3. Applying the case when k=0 that I specified above, I obtain that $a=d(p^2-q^2)$, $b=2dpq$, $c=d(p^2+q^2)-1$, which are, I believe, all of the solutions. However, note that if a particular selection of p and q yields c as even, then this will not hold. In particular, we need the above specified condition that $n=k^2-1$, so $c=2k^2-1$.
Quote by haruspex Looks like the same approach could be used for many constants. So the question becomes, for what k does a2+b2=c2+k have infinitely many solutions?
Hm, I realize my approach works for k congruent to 3(mod4), but beyond that, I don't know.
Quote by micromass That is cool!!! Nice find!! A no-brainer as follow-up question is of course: are these all the solutions?? I don't know the answer myself, but it's interesting to find out.
Haha, Thanks! I'm inclined to say that these are all the solutions, but I do not know. By the way, I could write the solutions in terms of k rather than n to make it neater. (i.e., if k is an integer, then, {2k2-2,2k,2k2-1} is an integer solution to the equation)
Recognitions:
Homework Help
Quote by Mathguy15 Hm, I realize my approach works for k congruent to 3(mod4), but beyond that, I don't know.
Choose any t > 0.
a = k + 2t + 1 (so a and k have opposite parity)
b = (a2 - k - 1)/2
c = b + 1
c2 - b2 = 2b+1 = a2 - k
Quote by haruspex Choose any t > 0. a = k + 2t + 1 (so a and k have opposite parity) b = (a2 - k - 1)/2 c = b + 1 c2 - b2 = 2b+1 = a2 - k
Brilliant! | 2013-05-21T10:35:36 | {
"domain": "physicsforums.com",
"url": "http://www.physicsforums.com/showthread.php?p=3974054",
"openwebmath_score": 0.8383458256721497,
"openwebmath_perplexity": 891.5897578017862,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9773708006261043,
"lm_q2_score": 0.8633916152464016,
"lm_q1q2_score": 0.843853754247241
} |
http://math.stackexchange.com/questions/202078/solve-sqrt52-sqrt6x-sqrt5-2-sqrt6x-10/202087 | # Solve $(\sqrt{5+2\sqrt{6}})^{x}+(\sqrt{5-2\sqrt{6}})^{x}=10$.
Solve $(\sqrt{5+2\sqrt{6}})^{x}+(\sqrt{5-2\sqrt{6}})^{x}=10$
I square the both sides and get $(5+2\sqrt{6})^{x}+(5-2\sqrt{6})^{x}=98$. But I don't know how to carry on. Please help. Thank you.
-
Let $t=( \sqrt{5 + 2\sqrt{6}})^{x}\implies (\sqrt{5-2\sqrt{6}})^{x}=\frac{1}{t}$
Thus the equation becomes,
$t+\frac{1}{t}= 10\implies t^2-10t+1=0$ which is a quadratic equation and have roots $t=5+2\sqrt 6,5-2\sqrt 6$
If $t=5+2\sqrt 6\implies ( \sqrt{5 + 2\sqrt{6}})^{x}=5+2\sqrt 6\implies (5+2\sqrt 6)^{1-x/2}=1\implies 1-x/2=0\implies x=2$
If $t=5-2\sqrt 6 \implies (\sqrt{5+2\sqrt{6}})^{x}=5-2\sqrt 6=\frac{1}{5+2\sqrt 6}\implies (5+2\sqrt 6)^{1+x/2}=1\implies x=-2$
Thus, $x=2,-2$ are the two solutions.
Verification:
Putting $x=2$ in the original equation gives L.H.S=$5+2\sqrt 6+5-2\sqrt 6=10=$R.H.S
Similarly, putting $x=-2$ also gives L.H.S=$5-2\sqrt 6+5+2\sqrt 6=10=$R.H.S
-
To be perfectly rigorous, you need to check (it's trivial here) that the two possibilities you get are indeed solutions; you only wrote implications ($\Rightarrow$), not equivalences ($\Leftrightarrow$). – Najib Idrissi Sep 25 '12 at 9:39
@nik: i already checked that by putting $x=2$ and $-2$ in the original equation and they both satisfy it. – Aang Sep 25 '12 at 9:41
But you did not write it. – xavierm02 Sep 25 '12 at 13:44
I included it now. – Aang Sep 25 '12 at 13:49
Thank you! I get it now~ – ᴊ ᴀ s ᴏ ɴ Sep 26 '12 at 12:35
Nice solution, just for clarification
$\left(5+2\sqrt{6}\right)^{\frac{x}{2}}\cdot \left(5-2\sqrt{6}\right)^{\frac{x}{2}}=\left[\left(5+2\sqrt{6}\right)\cdot \left(5-2\sqrt{6}\right)\right]^{\frac{x}{2}}=\left[5^2-\left(2\sqrt{6}\right)^2\right]^{\frac{x}{2}}=\left(25-24\right)^{\frac{x}{2}}=1$
and so for $t=\left(\sqrt{5+2\sqrt{6}}\right)^x$......
-
Interesting. Firefox is rendering the first set of brackets as Floor brackets. Checked in IE and it's fine. Also, CTRL + and CTRL - (change font size) renders correctly. – Chris Cudmore Sep 25 '12 at 14:03
Hint $\$ Put $\rm\ b = 5 + 2\sqrt{6},\,\ a = b^{\,x/2}\$ in
$$\rm a+ a^{-1} =\, b + b^{-1}\, \Rightarrow\ \{a,a^{-1}\} = \{b,b^{-1}\}\ \ since\ \ (x-a)(x-a^{-1})\, =\, (x-b)(x-b^{-1})$$
Note $\,$ Generally every pair of numbers in a ring is uniquely determined by their sum and product iff the ring is a domain. Above is the special case of this uniqueness result where the product $= 1.\:$ As I often emphasize, uniqueness theorems provide powerful tools for proving equalities.
-
See this duplicate question for further elaboration. – Bill Dubuque Sep 26 '12 at 22:37
Let $t_1=(\sqrt{5+2\sqrt6})^x$, and $t_2=(\sqrt{5-2\sqrt6})^x$.
Now the given equation is:
$$\tag 1 t_1+t_2=10$$
But
$$\displaylines{ {t_1}\cdot{t_2}&=& {\left( {\sqrt {5 + 2\sqrt 6 } } \right)^x}{\left( {\sqrt {5 - 2\sqrt 6 } } \right)^x} \cr &=& {\left[ {\sqrt {\left( {5 + 2\sqrt 6 } \right)\left( {5 - 2\sqrt 6 } \right)} } \right]^x} \cr &=& {\left[ {\sqrt {{5^2} - {{\left( {2\sqrt 6 } \right)}^2}} } \right]^x} \cr &=& {\left( {\sqrt {25 - 24} } \right)^x} \cr &=& {1^x} = 1 \cr}$$
Thus
$$\begin{cases} t_1+t_2=10 \\t_1\cdot t_2=1\end{cases}$$
$\Rightarrow$ $t^2-10t+1=0$
Since $$\frac{1}{{{t_1}}} = {t_2}$$
we
have $${t_1} + \frac{1}{{{t_1}}} = 10$$ or
$$t_1^2 - 10{t_1} + 1 = 0$$
Note the system is symmetric on the unknowns.
For this quadratic equation we have:
$t_{1,2}=\frac{10\pm\sqrt{100-4}}{2}$
$t_{1,2}=\frac{10\pm 4\sqrt{6}}{2}$
$t_1=5+2\sqrt 6$, $t_2=5-2\sqrt 6$
Now return the inital substition:
$t_1=(\sqrt{5+2\sqrt6})^x$
$5+2\sqrt 6=(\sqrt{5+2\sqrt6})^x$
$(\sqrt{5+2\sqrt 6})^2=(\sqrt{5+2\sqrt6})^x$ $\Rightarrow$ $x=2$, and
$(\sqrt{5+2\sqrt6})^x=5-2\sqrt 6$
$(\sqrt{5+2\sqrt6})^x=(5+2\sqrt 6)^{-2}$ $\Rightarrow$ $x=-2$
Definitly $x=2$, and $x=-2$ is solve.
-
Please could you tell me the reason why they are negative points – Madrit Zhaku Sep 25 '12 at 19:23 | 2014-12-18T15:53:09 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/202078/solve-sqrt52-sqrt6x-sqrt5-2-sqrt6x-10/202087",
"openwebmath_score": 0.9436652064323425,
"openwebmath_perplexity": 1149.837368735485,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9773708006261042,
"lm_q2_score": 0.8633916152464016,
"lm_q1q2_score": 0.8438537542472409
} |
https://stats.stackexchange.com/questions/188141/train-waiting-time-in-probability | # Train waiting time in probability
Let's say a train arrives at a stop every 15 or 45 minutes with equal probability (1/2). What is the expected waiting time of a passenger for the next train if this passenger arrives at the stop at any random time. This means that the passenger has no sense of time nor know when the last train left and could enter the station at any point within the interval of 2 consecutive trains.
I was told 15 minutes was the wrong answer and my machine simulated answer is 18.75 minutes. I just don't know the mathematical approach for this problem and of course the exact true answer. Sincerely hope you guys can help me.
• Please add the [self-study] tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. – gung - Reinstate Monica Dec 24 '15 at 14:49
• If you could simulate the answer correctly, then you already have a mathematical approach to the problem. – whuber Dec 24 '15 at 15:04
• For those who might like to study this further, here is a very fast R simulation. It processes several million people per second. n.trains <- 1e4; n.people <- 100 * n.trains; trains <- cumsum(ifelse(runif(n.trains) < 1/2, 15, 45)); people <- cumsum(rexp(n.people, rate=n.people/max(trains))); i <- c(rep(0, n.trains), rep(1, n.people))[order(c(trains, people))]; j <- length(trains) + 1 - rev(cumsum(rev(1-i))); k <- cumsum(i); waits <- (trains[j] - people[k])[i==1]; c(Mean=mean(waits, na.rm=TRUE), N=sum(!is.na(waits))) – whuber Dec 24 '15 at 15:36
• This is basically a (very transparent) manifestation of length-biased sampling, or the renewal paradox. – Mark L. Stone Dec 24 '15 at 15:48
• @whuber Out of curiosity, is there a reason why passengers renewals are assumed to follow an exponential distribution ? Or is it a normal way of simulating this kind of process? – Matthew Lau Dec 24 '15 at 18:32
Picture in your mind's eye the whole train schedule is already generated; it looks like a line with marks on it, where the marks represent a train arriving. On average, two consecutive marks are fifteen minutes apart half the time, and 45 minutes apart half the time.
Now, imagine a person arrives; this means randomly dropping a point somewhere on the line. What do you expect the distance to be between the person and the next mark? First, think of the relative probability of landing in each gap size of mark, and then deal with each case separately.
Does this help? I can finish answering but I thought it's more available to provide some insight so you could finish it on your own.
• Yours is really an insightful answer and helps me a lot. Thanks a lot! – fbabelle Dec 24 '15 at 15:33
• @fbabelle you're welcome, my pleasure! Remember to mark an answer accepted if you feel like it resolves things for you. – Nir Friedman Dec 24 '15 at 15:36
Your simulator is correct. Since 15 minutes and 45 minutes intervals are equally likely, you end up in a 15 minute interval in 25% of the time and in a 45 minute interval in 75% of the time.
In a 15 minute interval, you have to wait $15 \cdot \frac12 = 7.5$ minutes on average.
In a 45 minute interval, you have to wait $45 \cdot \frac12 = 22.5$ minutes on average.
This gives a expected waiting time of $\frac14 \cdot 7.5 + \frac34 \cdot 22.5 = 18.75$. | 2021-01-27T00:37:42 | {
"domain": "stackexchange.com",
"url": "https://stats.stackexchange.com/questions/188141/train-waiting-time-in-probability",
"openwebmath_score": 0.42805957794189453,
"openwebmath_perplexity": 642.361616977668,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9773708045809529,
"lm_q2_score": 0.8633916099737806,
"lm_q1q2_score": 0.8438537525085182
} |
https://algorithms.theroyakash.com/binary-search/intro/ | # Binary Search Introductions
Whenever a sorted array is given try to apply the binary search on that. This divides the array into two parts and only works on the other part. Recursion equation $$T(n) = T(\frac{N}{2}) + C$$
## Toy problem to start: Find Ceil
### Problem Statement
Find the ceil of a target number for given set of numbers. That is find the smallest number that is greater or equal to the target number from the given array only.
### Example
Given Array: [2, 3, 4, 5, 6, 7, 8, 10]
Target: 7.9
Return: 8
Target: 8.2
Return 10
### Approach
• This is exactly the binary search problem but instead of reporting that we don't find the target number, if we don't find the target number, we return the next biggest number.
• If the number is not found it means that the target number is not present, we have to return the next biggest number. Now the while loop will break at end < start. So the start pointer will be pointing to the next biggest number.
int ceil (vector<int> &v, int target) {
int start = 0;
int end = v.size() - 1;
int middle = start + (end - start) / 2;
while (start <= end) {
if (v[middle] < target) {
start = middle + 1;
} else if (v[middle] > target) {
end = middle - 1;
} else {
return v[middle];
}
}
// If the element is not found, then the while loop's start and end pointer crosses
// each other and the start pointer points to the smallest element larger than the
// target element.
return v[start];
}
def ceil(array: list[int], target: int) -> int:
# Run the actual binary search algorithm and return the element if found
start: int = 0
end: int = len(array) - 1
while start <= end:
middle: int = int((start + end) / 2)
if array[middle] < target:
start = middle+1
elif array[middle] > target:
end = middle-1
else:
return array[middle]
# If the element is not found, then the while loop's start and end pointer crosses
# each other and the start pointer points to the smallest element larger than the
# target element.
return array[start]
Test Cases
1. $$[2, 3, 5, 6, 6.6, 6.7, 7, 10]$$, $$\text{target} = 9$$
2. $$[2, 3, 5, 6, 6.6, 6.7, 7, 10]$$, $$\text{target} = 6.25$$
3. $$[2, 3, 5, 6, 6.6, 6.7, 7, 10]$$, $$\text{target} = 6.7$$
4. $$[2, 3, 5, 6, 6.6, 6.7, 7, 10]$$, $$\text{target} = 6.1$$
5. $$[2, 3, 5, 6, 6.6, 6.7, 7, 10]$$, $$\text{target} = 3.27$$
6. $$[2, 3, 5, 6, 6.6, 6.7, 7, 10]$$, $$\text{target} = 3$$
print("Answer is:",ceil([2, 3, 5, 6, 6.6, 6.7, 7, 10], target=9), "Actual should be 10")
print("Answer is:",ceil([2, 3, 5, 6, 6.6, 6.7, 7, 10], target=6.25), "Actual should be 6.6")
print("Answer is:",ceil([2, 3, 5, 6, 6.6, 6.7, 7, 10], target=6.7), "Actual should be 6.7")
print("Answer is:",ceil([2, 3, 5, 6, 6.6, 6.7, 7, 10], target=6.1), "Actual should be 6.6")
print("Answer is:",ceil([2, 3, 5, 6, 6.6, 6.7, 7, 10], target=3.27), "Actual should be 5")
print("Answer is:",ceil([2, 3, 5, 6, 6.6, 6.7, 7, 10], target=3), "Actual should be 3")
Answer is: 10 Actual should be 10
Answer is: 6.6 Actual should be 6.6
Answer is: 6.7 Actual should be 6.7
Answer is: 6.6 Actual should be 6.6
Answer is: 5 Actual should be 5
Answer is: 3 Actual should be 3
## Find out the Nth Root of a given element.
### Problem Statement
For a given N and a number K, find $$\sqrt[N]{K}$$.
### Approach
• It is obvious that the root of the number should lie between $$\{1, 2, ..., K\}$$.
• Now take the avg and reduce the search space to $$\{1, 2, ..., \frac{K}{2}\}$$ or $$\{\frac{K}{2}, ..., K\}$$.
• Repeat until the difference between higher bound and lower bound is less than some $$\epsilon = 10^{-6}$$
from typing import Union
THRESHOLD = 1e-6
def get_N_power(value: int, root: int):
while root > 0:
root -= 1
def find_n_th_root(number: int, root: int) -> Union[int, float]:
# The nth root lies between the 1, and the number
start: int = 1
end: int = number
while (end - start) > THRESHOLD:
middle: float = (start + end) / 2.0
if get_N_power(middle, root) > number:
end = middle
elif get_N_power(middle, root) < number:
start = middle
else:
return middle
return start, end
find_n_th_root(1024, 2) # -> (31.999999971129, 32.000000923871994)
import math
math.sqrt(1024) # Successfully verified -> 32.0
# Some more test cases
print("Our Function call returns in range:", find_n_th_root(1024, 2), "Original Values", math.sqrt(1024))
print("Our Function call returns in range:", find_n_th_root(256, 4), "Original Values", math.sqrt(math.sqrt(256)))
print("Our Function call returns in range:", find_n_th_root(128, 2), "Original Values", math.sqrt(128))
print("Our Function call returns in range:", find_n_th_root(3, 2), "Original Values", math.sqrt(3))
print("Our Function call returns in range:", find_n_th_root(81, 3))
Our Function call returns in range: (31.999999971129, 32.000000923871994) Original Values 32.0
Our Function call returns in range: (3.9999998211860657, 4.000000771135092) Original Values 4.0
Our Function call returns in range: (11.313708141446114, 11.31370908766985) Original Values 11.313708498984761
Our Function call returns in range: (1.7320499420166016, 1.732050895690918) Original Values 1.7320508075688772
Our Function call returns in range: (4.326748609542847, 4.326749205589294)
## Median of Row Wise Sorted Matrix
### Problem Statement
We are given a row-wise sorted matrix of size $$r*c$$, we need to find the median of the matrix given. It is assumed that $$r*c$$ is always odd.
### Example
Input : 1 3 5
2 6 9
3 6 9
Output : Median is 5
If we put all the values in a sorted
array A[] = 1 2 3 3 5 6 6 9 9)
Input: 1 3 4
2 5 6
7 8 9
Output: Median is 5
### Constraints
• Each entry in the array is from $$1 \to 10^9$$
• R and C are always odd.
### Apporach
#### Naive Approach
• Iterate over all the elements, and then sort them,
• then return the middle element.
#### Time Complexity for this naive apporach
• $$O(NM)$$ for the traversal,
• $$O(NM \log MN)$$ for Sorting and,
• Constant $$O(1)$$ time for the middle element. So total of $$O(NM \log MN)$$.
#### Space Complexity
• $$O(NM)$$ Extra space is required.
#### More optimized apporach | 2023-01-30T20:44:03 | {
"domain": "theroyakash.com",
"url": "https://algorithms.theroyakash.com/binary-search/intro/",
"openwebmath_score": 0.24568580090999603,
"openwebmath_perplexity": 5941.5676078984625,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9773708012852457,
"lm_q2_score": 0.8633916117313211,
"lm_q1q2_score": 0.843853751380801
} |
https://math.stackexchange.com/questions/2065576/calculating-number-of-equivalence-classes-where-two-points-are-equivalent-if-the | # Calculating number of equivalence classes where two points are equivalent if they can be joined by a continuous path.
Q. Let $G$ be an open set in $\Bbb R^n$. Two points $x,y \in G$ are said to be equivalent if they can be joined by a continuous path completely lying inside $G$. Number of equivalence classes is
1. Only one.
2. At most finite.
3. At most countable.
4. Can be finite, countable or uncountable.
This question was asked in the NET exam December 2016.
We can discard the first option by taking $n=1$ and $G=(-\infty,0) \cup (0,\infty)$.
We can reject the second option by taking $n=1$ and $G=\cup_{k \in \Bbb Z} (k,k+1).$
Now fun begins. Can we get an uncountable number of disjoint open path connected subsets of $\Bbb R^n$ for some $n$? If so, then we can take $G$ to be their union. For $n=1$, this method fails because that would give us the contradiction that the set of irrational numbers is countable.
• I don't follow what you're trying to say in the case $n=1$ at all... – Eric Wofsey Dec 20 '16 at 6:15
• (3) follows from math.stackexchange.com/questions/640491/… – Henricus V. Dec 20 '16 at 6:17
• @EricWofsey which statement about $n=1$? I have mentioned three cases. First rejected option 1. Second rejected option 2. and third gave me contradiction. – Error 404 Dec 20 '16 at 6:20
• Sorry, I mean in the final sentence. – Eric Wofsey Dec 20 '16 at 6:20
• @EricWofsey It's ok. My argument in the final sentence was if we suppose $G=\cup_{i} (q_i,q_{i+1})$ where $q_i \in \Bbb R - \Bbb Q$, then that would mean $q_i$s are countable. I am feeling like dumb now. :/ – Error 404 Dec 20 '16 at 6:28
It is impossible to have uncountably many equivalence classes. Note that each equivalence class is an open set, since balls are path-connected and so if $x\in G$ then any open ball around $x$ contained in $G$ is in the same equivalence class. Now any nonempty open subset of $\mathbb{R}^n$ contains an element of $\mathbb{Q}^n$, so each equivalence class must contain some element of $\mathbb{Q}^n$. Since $\mathbb{Q}^n$ is countable, there can be only countably many equivalence classes.
More generally, this argument applies with $\mathbb{R}^n$ replaced by any locally path-connected separable space.
• Thanks for mentioning the general case of 'locally path-connected separable space.' – Error 404 Dec 20 '16 at 6:30
• This proof seems incorrect to me. Suppose $G=\mathbb{Q}\subset\mathbb{R}$. Then each arc-wise connected component of $G$ is of the form $\{ q \}$ where $q$ is a rational number. This set $\{ q \}$ is not open in $\mathbb{R}$, and it is not even (relatively) open in $G$. – Jeppe Stig Nielsen Dec 20 '16 at 9:57
• @JeppeStigNielsen $G$ is given to be an open set. How can we take $G=\Bbb Q$ ? – Error 404 Dec 20 '16 at 10:13
• Ah, I thought there would be something I had misunderstood. Thanks. – Jeppe Stig Nielsen Dec 20 '16 at 10:22
• @JeppeStigNielsen No problems. – Error 404 Dec 20 '16 at 10:24
As an open subset of $\mathbb R^n$ is the union of at most a countable number of open balls (centered on points with rational coordinates and having a rational radius), response 3. is the right one.
• @VikrantDesai The point here is that any open set can be represented as a union of balls with rational centers and radii. I'd consider it to be a nice exercise to show this if you've never seen a proof. – Wojowu Dec 20 '16 at 9:34
• @Wojowu Okay. True that. I am also thinking same about writing down the proof for it. Will surely try. – Error 404 Dec 20 '16 at 9:48 | 2021-06-21T22:19:10 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2065576/calculating-number-of-equivalence-classes-where-two-points-are-equivalent-if-the",
"openwebmath_score": 0.8807001709938049,
"openwebmath_perplexity": 233.92546400509227,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9773707993078211,
"lm_q2_score": 0.863391611731321,
"lm_q1q2_score": 0.8438537496735091
} |
https://mathhelpboards.com/threads/clausen-sum.6090/ | # Clausen sum
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
Evaluate the following
$$\displaystyle \sum_{k\geq 1}\frac{\cos \left(\frac{\pi k}{2} \right)}{k^2}$$
#### M R
##### Active member
It looks like part of the Fourier series for $$\displaystyle x(\pi-x)$$ at $$\displaystyle x=\pi/4$$
So I'm getting $$\displaystyle \pi/4(\pi-\pi/4)=\pi^2/6-S$$.
Leading to S=$$\displaystyle -\pi^2/48$$...?
Bit of a cheat really.
#### chisigma
##### Well-known member
Evaluate the following
$$\displaystyle \sum_{k\geq 1}\frac{\cos \left(\frac{\pi k}{2} \right)}{k^2}$$
It is relatively easy...
$\displaystyle \sum_{k = 1}^{\infty} \frac{\cos (\frac{\pi k}{2})}{k^{2}} = - \frac {1}{4}\ (1 - \frac{1}{4} + \frac{1}{9} - \frac{1}{16} + ...) = - \frac{\pi^{2}}{48}$
Kind regards
$\chi$ $\sigma$
#### DreamWeaver
##### Well-known member
Evaluate the following
$$\displaystyle \sum_{k\geq 1}\frac{\cos \left(\frac{\pi k}{2} \right)}{k^2}$$
Hello Z! Nice little problem there...
I hope no one minds the bump, but there's a general closed form for this series... Define the second order Clausen functions as follows:
$$\displaystyle \text{Cl}_2(\theta)=-\int_0^{\theta}\log\Biggr|2\sin\frac{x}{2}\Biggr| \,dx=\sum_{k=1}^{\infty}\frac{\sin k\theta}{k^2}$$
$$\displaystyle \text{Sl}_2(\theta)=\sum_{k=1}^{\infty}\frac{\cos k\theta}{k^2}$$
[N.B. that integral definition holds without use of the absolute value sign within the range $$\displaystyle 0 < \theta < 2\pi\,$$]
Now we convert the CL-integral into complex exponential form
$$\displaystyle \text{Cl}_2(\theta)=-\int_0^{\theta}\log\left(2\sin\frac{x}{2}\right)\,dx=-\int_0^{\theta}\log\left[2\left(\frac{e^{ix/2}-e^{-ix/2}}{2i}\right)\right]\,dx=$$
$$\displaystyle -\int_0^{\theta}\log\left(\frac{1-e^{-ix}}{ie^{-ix/2}}\right)\,dx=$$
$$\displaystyle \int_0^{\theta}\left(\log i-\frac{ix}{2}\right)\,dx-\int_0^{\theta}\log(1-e^{-x})\,dx=$$
$$\displaystyle \frac{\pi i\theta}{2}-\frac{i\theta^2}{4}-\int_0^{\theta}\log(1-e^{-ix})\,dx$$
Now we expand the complex logarithm within the integrand as a power series:
$$\displaystyle \log(1-z)=-\sum_{k=0}^{\infty}\frac{z^k}{k}$$
$$\displaystyle \Rightarrow$$
$$\displaystyle \int_0^{\theta}\log(1-e^{-ix})\,dx=-\sum_{k=0}^{\infty}\frac{1}{k}\int_0^{\theta}e^{-ikx}\,dx=$$
$$\displaystyle -\sum_{k=0}^{\infty}\frac{1}{k}\int_0^{\theta}(\cos x-i \sin x)^{k}\,dx=-\sum_{k=0}^{\infty}\frac{1}{k}\int_0^{ \theta}(\cos kx-i \sin kx)\,dx=$$
$$\displaystyle -\sum_{k=0}^{\infty}\frac{1}{k}\Biggr[\frac{1}{k}\left(\sin kx+i \cos kx\right)\Biggr]_0^{\theta}=$$
$$\displaystyle -\sum_{k=0}^{\infty}\frac{1}{k}\Biggr[\frac{1}{k}\left(\sin k\theta+i \cos k\theta-i\right)\Biggr]=$$
$$\displaystyle -\sum_{k=0}^{\infty}\frac{\sin k\theta}{k^2}-i\sum_{k=0}^{\infty}\frac{\cos k\theta}{k^2}+i\sum_{k=0}^{\infty}\frac{1}{k^2}=$$
$$\displaystyle -\text{Cl}_2(\theta)-i\text{Sl}_2(\theta)+i\zeta(2)=-\text{Cl}_2(\theta)-i\text{Sl}_2(\theta)+\frac{i\pi^2}{6}$$
Combining this with our earlier results, and equating the imaginary parts to zero, we deduce that
$$\displaystyle \text{Sl}_2(\theta)=\frac{\pi^2}{6}-\frac{\pi\theta}{2}+\frac{\theta^2}{4}$$
Or, equivalently...
$$\displaystyle \sum_{k\ge 1}\frac{\cos k\theta}{k^2}= \frac{\pi^2}{6}-\frac{\pi\theta}{2}+\frac{\theta^2}{4}\, .\, \Box$$
-------------------------------------------
For $$\displaystyle m\in \mathbb{N}^+\,$$, all series of the form
$$\displaystyle \text{Sl}_{2m}(\theta)=\sum_{k=0}^{\infty}\frac{ \cos k\theta}{k^{2m}}$$
$$\displaystyle \text{Sl}_{2m+1}(\theta)=\sum_{k=0}^{\infty}\frac{\sin k\theta}{k^{2m}}$$
Are expressible in closed form, as a polynomial in $$\displaystyle \theta\,$$ and $$\displaystyle \pi\,$$
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
Nice , generalization !
#### DreamWeaver
##### Well-known member
Thanks Z!
It can make tricky-looking results quite easy. Such as
$$\displaystyle \sum_{k=1}^{\infty}\frac{\cos (\pi k/10)}{k^2}=\frac{\pi^2}{6}-\frac{\pi}{4}\sqrt{\frac{5+\sqrt{5}}{2}}+\frac{5+ \sqrt{5}}{16}$$
#### DreamWeaver
##### Well-known member
Thanks Z!
It can make tricky-looking results quite easy. Such as
$$\displaystyle \sum_{k=1}^{\infty}\frac{\cos (\pi k/10)}{k^2}=\frac{\pi^2}{6}-\frac{\pi}{4}\sqrt{\frac{5+\sqrt{5}}{2}}+\frac{5+ \sqrt{5}}{16}$$
Oops!!!
Sorry folks... Wasn't really concentrating yesterday... I put in value of $$\displaystyle \cos \theta$$ rather than the argument, $$\displaystyle \theta$$... | 2022-07-05T08:42:58 | {
"domain": "mathhelpboards.com",
"url": "https://mathhelpboards.com/threads/clausen-sum.6090/",
"openwebmath_score": 0.9352215528488159,
"openwebmath_perplexity": 6799.2106437441125,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9773707986486797,
"lm_q2_score": 0.863391611731321,
"lm_q1q2_score": 0.8438537491044119
} |
http://math.stackexchange.com/questions/357811/integral-domains-with-non-trivial-group-of-units-that-are-not-fields | # Integral domains with non-trivial group of units that are not fields
I'm looking for examples of integral domains that are not fields but at the same time have more units than just the multiplicative identity 1.
It's clear to me that by Wedderburn's little theorem, there are no finite examples of this type.
If I understand things correctly, then the ring of holomorphic functions on a domain is such an example: It's an integral domain because zeros of holomorphic functions are isolated, and it has more units than the 1-function because every constant function is invertible.
However, I can't find more examples, and I would like to see more.
-
$\mathbb Z$ is a simple example since $-1$ is another unit. – Karl Kronenfeld Apr 11 '13 at 1:58
Examples arise from the following generalization of Euclid's theorem on infinitely many primes.
Theorem $\$ An infinite ring $\rm R$ has infinitely many maximal ideals if it has fewer units $\rm U = U(R)$ than it has elements, i.e. $\rm\:|U| < |R|.$
Thus, contrapositively, any infinite ring with finitely many maximal ideals has unit group having same cardinality as the ring (hence infinite). Examples abound, e.g. via localization.
-
Excellent; this is really helpful in finding nice examples. For example, if you localize the integers at 2, you get an infinite local ring: the rationals with odd denominators. It's not a field since every even integer in this ring has no inverse, but the group of units is infinite (e.g. every odd integer is a unit). Very nice. – Tom Jonathan Apr 11 '13 at 10:15
The ring of Gaussian integers: http://en.wikipedia.org/wiki/Gaussian_integer
-
$k [ x ]$ for a field $k$.
-
In fact, all rings of algebraic integers in algebraic number fields are such examples: their unit group is always non-trivial, as they contain $-1$. And certainly they are not fields for there are prime elements within.
Regards.
- | 2016-06-25T14:09:16 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/357811/integral-domains-with-non-trivial-group-of-units-that-are-not-fields",
"openwebmath_score": 0.9047980904579163,
"openwebmath_perplexity": 301.4966812147451,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9773707960121133,
"lm_q2_score": 0.8633916134888613,
"lm_q1q2_score": 0.8438537485457912
} |
https://math.stackexchange.com/questions/2026853/is-there-a-way-i-can-find-out-the-degree-of-this-extension-without-explicitly-fi | # Is there a way I can find out the degree of this extension without explicitly finding the minimal polynomial?
Suppose that $\beta$ is a real cube root of $2$ and $\omega$ is a primitive third root of unity. I've a problem that asks me to show that if $\alpha=\beta\omega$, then $\alpha+\beta$ has minimal polynomial of degree 3 over $\mathbb Q$ while $\alpha-\beta$ has minimal polynomial of degree 6. Finding the minimal polynomial seems to be a real problem, especially since the polynomial might be hard to prove to be irreducible. What way should I approach this problem? Is this just a computation exercise?
• Yes, sorry, I edited it. – adrija Nov 23 '16 at 6:15
It's not hard to show that $(\alpha+\beta)^3=-2$, either by expanding out the cube, or by observing that $$\alpha+\beta=(1+\omega)\sqrt[3]{2}=-\omega^2\sqrt[3]{2}$$ since $1+\omega+\omega^2=0$. And $x^3+2$ is irreducible over $\mathbb{Q}$ by Eisenstein's criterion, so this must be the minimal polynomial of $\alpha+\beta$.
One can also show that $(\alpha-\beta)^6=-108$, but it's not as clear that $x^6+108$ is irreducible over $\mathbb{Q}$. However, $$(\alpha-\beta)^3=6(\omega-\omega^2)=6+12\omega$$ so $\omega\in\mathbb{Q}(\alpha-\beta)$, and $$(\alpha-\beta)^2=2^{\frac{2}{3}}(\omega^2-2\omega+1)=2^{\frac{2}{3}}(-3\omega)$$ hence $$(\alpha-\beta)^4=2\sqrt[3]{2}\cdot 9\omega^2$$ so that $\sqrt[3]{2}\in \mathbb{Q}(\alpha-\beta)$ as well (since we've already shown that $\omega\in\mathbb{Q}(\alpha-\beta)$). Therefore $\mathbb{Q}(\alpha-\beta)$ contains $\mathbb{Q}(\omega,\sqrt[3]{2})$, so has degree at least $6$ over $\mathbb{Q}$. Since it satisfies a polynomial of degree $6$, the degree of the extension must be exactly $6$, and $x^6+108$ is irreducible.
• Good answer! +1 – Learnmore Nov 23 '16 at 7:09
• Excellent strategy in the second part. I've been trying to prove the irreducibility of $x^6+108$ for some time. This will help in similar other problems too I think, so thank you. – adrija Nov 23 '16 at 7:32
• Thanks, glad to be of help. – carmichael561 Nov 23 '16 at 16:30
In general, if one has a Galois extension $K/F$, the minimal polynomial for any $a \in K$ has as its roots the elements in the orbit of $a$ under the action of $\text{Gal}(K/F)$. That is, if $S = \{ \phi(a) \ | \ \phi \in \text{Gal}(K/F) \}$, then $\displaystyle \min_a(x) = \prod_{u_k \in S} (x-u_k)$.
Proof: This is a consequence of the fact that a polynomial is irreducible $\iff$ its Galois group acts transitively on its roots. For a proof of this fact, see Theorem 2.9(b) here. Notice that $\min_a(x)$ will be an irreducible polynomial (by definition), and its Galois group will be a subgroup of $\text{Gal}(K/F)$. This latter fact is because, if $L \subseteq K$ is the splitting field of $\min_a(x)$, every $F$-automorphism of $L$ extends to an $F$-automorphism of $K$.
To apply this fact to the problem at hand, we have $K = \mathbb{Q}(\sqrt[3]{2}, \omega)$ with $\text{Gal}(K/\mathbb{Q})$ generated by the following two automorphisms: $$i \mapsto -i$$ $$\sqrt[3]{2} \mapsto \omega \sqrt[3]{2}$$
Let's say we want to find the minimal polynomial of $\sqrt[3]{2} + \omega \sqrt[3]{2}$ over $\mathbb{Q}$. Per the argument in the first paragraph, we simply need to find the elements of the orbit of this element under the action of the Galois group, which you can check are:
• $\sqrt[3]{2} + \omega \sqrt[3]{2} \qquad \qquad \text{identity automorphism}$
• $\omega \sqrt[3]{2} + \omega^2 \sqrt[3]{2} \qquad \quad \sqrt[3]{2} \mapsto \omega \sqrt[3]{2} \ \ \text{ acting on original element}$
• $\sqrt[3]{2} + \omega^2 \sqrt[3]{2} \qquad \qquad i \mapsto -i \ \ \text{ acting on original element}$
And this list is exhaustive. The minimal polynomial of $\sqrt[3]{2} + \omega \sqrt[3]{2}$ over $\mathbb{Q}$ therefore has $3$ roots and is of degree $3$. Repeating this process for $- \sqrt[3]{2} + \omega \sqrt[3]{2}$ will demonstrate that its minimal polynomial is of degree $6$.
At least for degree 3, it's not too hard to show that a polynomial is irreducible--simply use the rational root theorem. Since all irreducible polynomials of degree 2 or 3 have no roots over the base field, we can easily show it's irreducible.
My next thought if you want to avoid computations is to try and use the degree formula (I'm not 100% on this route working, though). With this route, you would show the other polynomial has degree six. by showing $\mathbb{Q}[\alpha + \beta][\alpha - \beta] = \mathbb{Q}[\alpha + \beta, \alpha - \beta] = \mathbb{Q}[\alpha, \beta]$ has degree 6 over $\mathbb{Q}$.
Let $K = \mathbb{Q}[\alpha + \beta][\alpha - \beta]$. Then by the degree formula we would obtain that $6 = [K:\mathbb{Q}] = [K:\mathbb{Q}[\alpha - \beta]][\mathbb{Q}[\alpha - \beta]:\mathbb{Q}]$. So essentially if you could show $[K:\mathbb{Q}[\alpha - \beta]] = 1$, or equivalently, $\alpha + \beta \in \mathbb{Q}[\alpha - \beta]$, that would show your claim. | 2020-02-22T20:55:42 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2026853/is-there-a-way-i-can-find-out-the-degree-of-this-extension-without-explicitly-fi",
"openwebmath_score": 0.9488664269447327,
"openwebmath_perplexity": 61.051251351128094,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9773708006261043,
"lm_q2_score": 0.8633916029436189,
"lm_q1q2_score": 0.8438537422228604
} |
https://stats.stackexchange.com/questions/497338/why-is-mean-%C2%B1-2sem-95-confidence-interval-overlapping-but-the-p-value-is-0 | # Why is mean ± 2*SEM (95% confidence interval) overlapping, but the p-value is 0.05?
I have data as two lists:
acol = [8.48, 9.82, 9.66, 9.81, 9.23, 10.35, 10.08, 11.05, 8.63, 9.52, 10.88, 10.05, 10.45, 10.0, 9.97, 12.02, 11.48, 9.53, 9.98, 10.69, 10.29, 9.74, 8.92, 11.94, 9.04, 11.42, 8.88, 10.62, 9.38, 12.56, 10.53, 9.4, 11.53, 8.23, 12.09, 9.37, 11.17, 11.33, 10.49, 8.32, 11.29, 10.31, 9.94, 10.27, 9.98, 10.05, 10.07, 10.03, 9.12, 11.56, 10.88, 10.3, 11.32, 8.09, 9.34, 10.46, 9.35, 11.82, 10.29, 9.81, 7.92, 7.84, 12.22, 10.42, 10.45, 9.33, 8.24, 8.69, 10.31, 11.29, 9.31, 9.93, 8.21, 10.32, 9.72, 8.95, 9.49, 8.11, 8.33, 10.41, 8.38, 10.31, 10.33, 8.83, 7.84, 8.11, 11.11, 9.41, 9.32, 9.42, 10.57, 9.74, 11.35, 9.44, 10.53, 10.08, 10.92, 9.72, 7.83, 11.09, 8.95, 10.69, 11.85, 10.19, 8.49, 9.93, 10.39, 11.08, 11.27, 8.71, 9.62, 11.75, 8.45, 8.09, 11.54, 9.0, 9.61, 10.82, 10.36, 9.22, 9.36, 10.38, 9.53, 9.2, 10.36, 9.38, 7.68, 9.99, 10.61, 8.81, 10.09, 10.24, 9.21, 10.17, 10.32, 10.41, 8.77]
bcol = [12.48, 9.76, 9.63, 10.86, 11.63, 9.07, 12.01, 9.52, 10.05, 8.66, 10.85, 9.87, 11.14, 10.59, 9.24, 9.85, 9.62, 11.54, 11.1, 9.38, 9.24, 9.68, 10.02, 9.91, 10.66, 9.7, 11.06, 9.27, 9.08, 11.31, 10.9, 10.63, 8.98, 9.81, 9.69, 10.71, 10.43, 10.89, 8.96, 9.74, 8.33, 11.45, 9.61, 9.59, 11.25, 9.44, 10.05, 11.63, 10.16, 11.71, 9.1, 9.53, 9.76, 9.33, 11.53, 11.59, 10.21, 10.68, 8.99, 9.44, 9.82, 10.35, 11.22, 9.05, 9.18, 9.57, 11.43, 9.4, 11.45, 8.39, 11.32, 11.16, 12.47, 11.62, 8.77, 11.34, 11.77, 9.53, 10.54, 8.73, 9.97, 9.98, 10.8, 9.6, 9.6, 9.96, 12.17, 10.01, 8.69, 8.94, 9.24, 9.84, 10.39, 10.65, 9.31, 9.93, 10.41, 8.5, 8.64, 10.23, 9.94, 10.47, 8.95, 10.8, 9.84, 10.26, 11.0, 11.22, 10.72, 9.14, 10.06, 11.52, 10.21, 9.82, 10.81, 10.3, 9.81, 11.48, 8.51, 9.55, 10.41, 12.17, 9.9, 9.07, 10.51, 10.26, 10.62, 10.84, 9.67, 9.75, 8.84, 9.85, 10.41, 9.18, 10.93, 11.41, 9.52]
A summary of the above lists is given below:
N, Mean, SD, SEM, 95% CIs
137 9.92 1.08 0.092 (9.74, 10.1)
137 10.2 0.951 0.081 (10.0, 10.3)
An unpaired t-test for the above data gives a p-value of 0.05:
f,p = scipy.stats.ttest_ind(acol, bcol)
print(f, p)
-1.9644209241736 0.050499295018989004
I understand from this and other pages that mean ± 2 * SEM (standard error of mean as calculated by SD/sqrt(N)) gives a 95% confidence interval (CI) range.
I also believe that if 95% confidence intervals are overlapping, P-value will be > 0.05.
I plotted the above data as mean ± 2 * SEM:
The 95% confidence intervals are overlapping. So why is the p-value reaching a significant level?
• See stats.stackexchange.com/a/18259/919 for a detailed analysis of this situation. It shows your belief is incorrect but it can be patched up by adjusting the p-value. – whuber Nov 21 '20 at 14:47
• +1 Good question because it shows good understanding of the relationship between hypothesis testing and confidence interval – Patrick Coulombe Nov 21 '20 at 21:47
### The overlap is just a (strict/inaccurate) rule of thumb
The point when the error bars do not overlap is when the distance between the two points is equal to $$2(SE_1+SE_2)$$. So effectively you are testing whether some sort of standardized score (distance divided by the sum of standard errors) is greater than 2. Let's call this $$z_{overlap}$$
$$z_{overlap} = \frac{\vert \bar{X}_1- \bar{X}_2 \vert}{SE_1+SE_2} \geq 2$$
If this $$z_{overlap} \geq 2$$ then the error bars do not overlap.
### The standard deviation of a linear sum of independent variables
Adding the standard deviations (errors) together is not the typical way to compute the standard deviation (error) of a linear sum (the parameter $$\bar{X}_1-\bar{X}_2$$ can be considered as a linear sum where one of the two is multiplied by a factor $$-1$$) See also: Sum of uncorrelated variables
So the following are true for independent $$\bar{X}_1$$ and $$\bar{X}_2$$:
$$\begin{array}{} \text{Var}(\bar{X}_1-\bar{X}_2) &=& \text{Var}(\bar{X}_1) + \text{Var}(\bar{X}_2)\\ \sigma_{\bar{X}_1-\bar{X}_2}^2 &=& \sigma_{\bar{X}_1}^2+\sigma_{\bar{X}_2}^2\\ \sigma_{\bar{X}_1-\bar{X}_2} &=& \sqrt{\sigma_{\bar{X}_1}^2+\sigma_{\bar{X}_2}^2}\\ \text{S.E.}(\bar{X}_1-\bar{X}_2) &=& \sqrt{\text{S.E.}(\bar{X}_1)^2 + \text{S.E.}(\bar{X}_2)^2}\\ \end{array}$$
But not
$$\text{S.E.}(\bar{X}_1-\bar{X}_2) \neq {\text{S.E.}(\bar{X}_1) + \text{S.E.}(\bar{X}_2)}$$
### 'Correct' formula for comparing the difference in the mean of two samples
For a t-test to compare the difference in means of two populations, you should be using a formula like
• In the simplest case: $$t = \frac{\bar{X}_1 - \bar{X}_2}{\sqrt{SE_1^2+SE_2^2}}$$ this is when we consider the variances to be unequal or when the sample sizes are equal.
• If the sample sizes are different and you consider the variance of the populations to be equal, then you can estimate the variances for both samples together instead of separately, and use one of many formulae for the pooled variance like
$$s_p = \sqrt{\frac{(n_1-1)s_1^2 +(n_2-1)s_2^2}{n_1+n_2-2}}$$
with $$t = \frac{\bar{X}_1 - \bar{X}_2}{s_p \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$$
and with $$SE_1 = s_1/\sqrt{n_1}$$ and $$SE_2 = s_2/\sqrt{n_2}$$ you get
$$t = \frac{\bar{X}_1 - \bar{X}_2}{\sqrt{\frac{n_1+n_2}{n_1+n_2-2} \left( \frac{n_1-1}{n_2} SE_1^2 + \frac{n_2-1}{n_1} SE_2^2 \right)}}$$
Note that the value $$\sqrt{SE_1^2+SE_2^2}$$ is smaller than $$SE_1+SE_2$$, therefore $$t>z_{overlap}$$.
Sidenotes:
• In the case of the pooled variance, you might have a situation - although it is rare - that the variance of the larger sample is larger than the variance of the smaller sample, and then it is possible that $$t.
• Instead of z-values and a z-test you are actually doing (should be doing) a t-test. So it might be that the levels on which you base the confidence intervals for the error bars (like '95% is equivalent to 2 times the standard error') will be different for the t-test. To be fair, to compare apples with apples, you should use the same standard and base the confidence levels for the error bars on a t-test as well. So let's assume that also for the t-test the boundary level that relates to 95% is equal to or less than 2 (this is the case for sample sizes larger than 60).
If this $$t \geq 2$$ then the difference is significant (at a 5% level).
The standard error of the difference between two variables is not the sum of standard errors of each variable. This sum is overestimating the error for the difference and will be too conservative (too often claim there is no significant difference).
So $$t>z_{overlap}$$ and may lead to a significant difference while the error bars have overlap. You do not need non-overlapping error bars in order to have a significant difference. This overlap is a stricter requirement and happens when the p-value is $$\leq 0.05$$ (and it will often be a lower p-value).
• It will be great if you could explain your long sentence / short para on s1+s2, its double, distance and 2*SEM. That will help us understand this concept better. – rnso Nov 21 '20 at 17:25
• I expected standard deviation (even pooled) to be much greater than (SEM1+SEM2). I tried with random data and find that it is always larger. Please confirm that you really mean that 2*pooledSD may be less than 2*(SEM1+SEM2) ? – rnso Nov 22 '20 at 12:14
• @rnso yes that is really what I mean. You can see this also in the inequality. If you are comparing the difference of two variables $X_1-X_2$, with standard errors $\sigma_1$ and $\sigma_2$, then the standard error of this difference will be smaller than $\sigma_1+\sigma_2$. – Sextus Empiricus Nov 22 '20 at 15:00
• How can we calculate standard error of difference if X1 and X2 are unpaired? Sp/sqrt(n1+n2)? – rnso Nov 22 '20 at 16:29
• @rnso For any two independent variables $X_1$ and $X_2$ with sample variances $\sigma_1^2$ and $\sigma_2^2$ the sample variance of the difference $X_1-X_2$ is equal to $\sigma_1^2+\sigma_2^2$. See en.wikipedia.org/wiki/… The size of the samples does not matter. – Sextus Empiricus Nov 22 '20 at 17:17
The p-value should be considered between a CI and a parameter value, not two CIs. Indeed, the red point falls entirely outside the blue CI, and the blue point falls entirely outside the red CI.
And it is true that under the null hypothesis such an event would happen 5% of the time:
• 2.5% of the time, you get a point above the 95% CI
• 2.5% of the time, you get a point below the 95% CI
If it is only the whiskers that overlap or touch, then the null hypothesis will produce this result a lot less often than 5%. This is because (to use your example) both the blue sample would need to be low, and at the same time the red sample would need to be high (exactly how high would depend on the blue value). You can picture it as a 3D multivariate Gaussian plot, with no skew since the two errors are independent of one another:
Along each axis the probability of falling outside the highlighted region (the CI) is 0.05. But the total probabilities of the blue and pink areas, which gives you P of the two CIs barely touching, is less than 0.05 in your case.
A change of variables from the blue/red axes to the green one will let you integrate this volume using a univariate rather than multivariate Gaussian, and the new variance is the pooled variance from @Sextus-Empiricus's answer.
• I like the image. Unfortunately, you are incorrect at the outset: the relevant comparison is between the CI and the parameter it is intended to estimate, not between the CI and the mean of other data. (That would call for a prediction interval rather than a confidence interval.) – whuber Nov 21 '20 at 20:34
Even if we ignore the difference between confidence and probability, the overlap consists of points for which both the red probability and the blue probability are greater than 0.05. But that doesn't mean that the probability of both is greater than 0.05. For instance, if both the red and blue probability are 0.10, then the joint probability (assuming independence) is 0.01. If you integrate over the whole overlap, this will be less than 0.01.
When you look at the overlap, you are seeing points for which the difference is less than two standard deviations. But remember that the variance of the difference between two variables is the sum of the individual variances. So you can generally use a rule of thumb that if you want to compare two different populations by checking for overlapping CI, you need to divide the size of each CI by $$\sqrt 2$$: if the variances are of similar sizes, then the variance of the difference will be twice the individual variances, and the standard deviation will be $$\sqrt 2$$ times as large.
• Since each CI is SEM*2, CI/√2 will be SEM*√2. Is there any name given to SEM*√2 or Mean±SEM*√2 ? – rnso Nov 22 '20 at 12:20 | 2021-01-23T16:57:04 | {
"domain": "stackexchange.com",
"url": "https://stats.stackexchange.com/questions/497338/why-is-mean-%C2%B1-2sem-95-confidence-interval-overlapping-but-the-p-value-is-0",
"openwebmath_score": 0.9278963208198547,
"openwebmath_perplexity": 358.37837163355545,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9773708039218115,
"lm_q2_score": 0.863391599428538,
"lm_q1q2_score": 0.8438537416328089
} |
https://physics.stackexchange.com/questions/89308/charge-to-mass-ratio-inversely-proportional-to-curved-path-radius | # Charge to mass ratio inversely proportional to curved path radius?
In a cloud or bubble chamber, charged particles follow circular paths. I learned that charge to mass ratio of the particles is inversely proportional to the radius of the path. Thus, a particle following a circular path with a large radius means that charge to mass ratio of that particle is smaller.
However, I want to know from where this relationship came from.
• Look up "Lorentz force". Dec 7 '13 at 3:36
Well, the particles won't always follow circular paths (for instance, the particles in this video).
But, if you apply a constant magnetic field across the chamber, charged particles moving in the field will be deflected according to the Lorentz Force Law.
The centripetal acceleration for a particle moving in a circle is $a=\frac{v^2}{r}$, where $v$ is the particles tangential velocity and $r$ is the radius of the circle.
Plugging this into Newton's Second Law we get $$m\frac{v^2}{r}=qvB$$ Rearranging: $$\Big(\frac{q}{m}\Big)\Big(\frac{B}{v}\Big)=\frac{1}{r}$$ Or, in other words, the charge-to-mass ratio is inversely proportional to radius of the particles circular orbit by proportionality constant $\frac{B}{v}$.
Edit: I can't comment yet otherwise I would answer your second question in the comments. No, $v$ cannot be eliminated because it is integral to the magnitude of the Lorentz force. A light but slower-moving particle could have the same orbit as a heavier, faster particle assuming that they have the same charge, but you can calculate each particle's momentum in a creation event by using conservation of momentum and conservation of electric charge. So, no, I don't think you can say which particle is lighter if all you know are the two radii. Of course, things are different if you have a velocity selector.
• is B/v always going to be a constant. I mean, can I use this relationship to compare the curved paths of two different particles (resulting from the breakdown of one particle in a bubble chamber)? Dec 7 '13 at 3:53
It boils down to balancing the centripetal force, $$\vec{F}=\frac{mv^2}{r}\hat{r}$$ with the magnetic force $$\vec{F}=q\vec{v}\times\vec{B}$$
Equating these and considering the perpendicular velocity, we get $$\frac{mv_\perp^2}{r}=qv_\perp B$$ Which can easily be solved for $q/m$: $$\frac{q}{m}=\frac{v_\perp}{rB}$$ Thus, if you know the strength of the magnetic field, the velocity at which it travels, and the radius of the arc it travels, you know the charge-to-mass ratio.
• Well what if the paths of two charged particles in a bubble chamber were given (for eg. a particle breaks down to form 2 other particles)and we do not know the velocities of the particles. Then, will the path with the larger radius necessarily be the one with the smaller charge to mass ratio? Is there any way to eliminate v from the equation? Dec 7 '13 at 3:45 | 2021-10-27T19:54:53 | {
"domain": "stackexchange.com",
"url": "https://physics.stackexchange.com/questions/89308/charge-to-mass-ratio-inversely-proportional-to-curved-path-radius",
"openwebmath_score": 0.8350518941879272,
"openwebmath_perplexity": 202.79583687335494,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.973240719919151,
"lm_q2_score": 0.8670357735451835,
"lm_q1q2_score": 0.8438345204407723
} |
https://math.stackexchange.com/questions/2308983/the-set-of-all-limit-points-of-a-set | # The set of all limit points of a set
Given a set $A \subseteq \mathbf{R}$, let $L$ be the set of all limit points of $A$. I recently worked through the proof that $L$ is closed. However, I have a few queries thinking deeper about this result.
Since $L$ is closed, this implies that if $x$ is a limit point of $L$ then $x \in L$, which further implies that $x$ must be a limit point of $A$. Therefore, if $x$ is a limit point of $L$ then it must also be a limit point of $A$. But consider the set $$A = \{0\} \cup \{\frac{1}{n}: n \in \mathbf{N}\}$$
Clearly, $L = \{0\}$ but does not have any limit point itself, i.e., the set of all limit points of $L$ is $\emptyset$, but clearly $\emptyset$ is not a limit point of $A$, so isn't this a counter example to the above bolded statement?
• Of course $\varnothing$ isn't a limit point of $A$. That isn't what the bolded statement says at all. – Kenny Lau Jun 4 '17 at 5:21
• The bolded statement claims that every point in $\varnothing$ is a limit point of $A$, which is vacuously true. – Kenny Lau Jun 4 '17 at 5:21
• $\emptyset$ is both closed and open in any topology, as is the whole space. – Araske Jun 4 '17 at 6:53
• $\emptyset$ isn't even a point, but every element in it (although there is none) is a point. – md2perpe Jun 4 '17 at 6:54
• @siTTmo when you say "the set of all limit points of $L$ is $\emptyset$", it means that $L$ has no limit points. Therefore, the sentence "every limit point of $L$ is also a limit point of $A$" (equivalent to the sentence "if $x$ is a limit point of $L$ then $x$ is a limit point of $A$) is vacuously true. Hence, the bolded statement is true, just easily true in this particular case as there are no points to check. – Darío G Jun 4 '17 at 6:58
Consider. If $$A = [0,1]$$ then $$L = [0,1] = A$$ and set of limit points of $$L = [0,1] = A$$. But $$[0,1]$$ is not a limit point of $$A$$! Contradiction? No. A set is not a member of itself (by ZFC that can never happen) so the set of limit points is not a limit point.
Clearly, L={0} but does not have any limit point itself, i.e., the set of all limit points of L is ∅, but clearly ∅ is not a limit point of A
Which doesn't matter because $$\emptyset \not \in \emptyset = \{$$limit points of $$L\}$$.
But since there aren't any $$x \in \{$$limit points of $$L\} = \emptyset$$ that aren't limit points (because there aren't any $$x \in \emptyset$$ period) the all $$x \in \emptyset$$ (all ZERO of them) are limit points. They are also all french pigs farting green sausages. (Because there are any $$x \in \emptyset$$ that arent french pigs farting green sausages.)
When you say that "$$\emptyset$$ is not a limit point of $$A$$", there seems to be some confusion: sets can anyway never be limit points, only points (in the metric space) can be limit points.
What you instead need to verify is that every element of $$\emptyset$$ is a limit point of $$A$$, and this is vacuously true. | 2020-07-16T14:09:34 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2308983/the-set-of-all-limit-points-of-a-set",
"openwebmath_score": 0.9128749370574951,
"openwebmath_perplexity": 80.53206910882555,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.973240719919151,
"lm_q2_score": 0.8670357701094303,
"lm_q1q2_score": 0.8438345170969574
} |
http://farmsteadcolumbus.com/vi2k6vja/hh5pdsb.php?tag=8c55a7-number-of-reflexive-relations | Therefore, this set of ordered pairs comprises of n, pairs. Don’t stop learning now. • Encode R Encode R Let A = {1, 2, 3}. [1] [2] An example of a reflexive relation is the relation "is equal to" on the set of real numbers, since every real number is equal to itself. Confirm that R is a reflexive relation on set A. Note that not every relation which is not reflexive is irreflexive; it is possible to define relations where some elements are related to themselves but others are not related to themselves (i.e., neither all nor none). This post covers in detail understanding of allthese Hence, the total number of reflexive relationships in set S is, Formula for Number of Reflexive Relations. Answer. each real number “is equal to" itself. Therefore, the relation R is not reflexive. Answer/Explanation. Condition for reflexive : R is said to be reflexive, if a is related to a for a ∈ S. let x = y. x + 2x = 1. N is a set of all real numbers. But when I used it here 1 got that there would be only 1 reflexive relation ie each element goes to itself but that's wrong according to answers. Related terms. For remaining n2 – n entries, we have choice to either fill 0 or 1. As per the concept of a reflexive relationship, (p, p) must be included in such ordered pairs. Let us consider an example to understand the difference between the two relations reflexive and identity. Answer: (d) Reflexive, transitive but not symmetric A. Number of Symmetric Relations on a set with n elements : 2n (n+1)/2. Pro Lite, CBSE Previous Year Question Paper for Class 10, CBSE Previous Year Question Paper for Class 12. Let us consider a set S. This set has an ordered pair (p, q). Now, for all pairs of positive integers in set X, ((p,q),(p,q))∈ R. Then, we can say that (p,q) = (p,q) for all positive integers. This proves the reflexive property of equivalence. So there are total 2n2 – n ways of filling the matrix. Now, p can be chosen in n number of ways and so can q. Anti - Reflexive: If the elements of the set do not relate to themselves, they are said to be irreflexive or anti-reflexive. If A = {1,2,3} the number of reflexive relations in 'A' are 1 See answer radhasri8306 is waiting for your help. An example of a reflexive relation is the relation "is equal to" on the set of real numbers, since every real number … Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below. The number of reflexive relations on an n-element set is 2 n 2 – n In other words, a relation ~ on a set S is reflexive when x ~ x holds true for every x in S, formally: when ∀x∈S: x~x holds. Let X = { 1, 2, 3, 4 } and define binary relations R 1, R 2 and R 3 on X as follows:-. close, link The formula for the number of reflexive relations in a given set is written as N = $2^{n(n-1)}$. A relation R on set A (set of integers) is defined by “x R y if 5x + 9x is divisible by 7x” for all x, y ∈ A. Also, there will be a total of n pairs of such (p, p) pairs. Then, R is (a) Reflexive and symmetric (b) Transitive and symmetric (c) Equivalence (d) Reflexive, transitive but not symmetric. Program to check if a given year is leap year, Factorial of Large numbers using Logarithmic identity, Write an iterative O(Log y) function for pow(x, y), Modular Exponentiation (Power in Modular Arithmetic), Compute the integer absolute value (abs) without branching, Left Shift and Right Shift Operators in C/C++, Prime Number of Set Bits in Binary Representation | Set 2, Check whether the number has only first and last bits set | Set 2, Prime Number of Set Bits in Binary Representation | Set 1, Program to find the Nth natural number with exactly two bits set | Set 2, Next higher number with same number of set bits. Also, there will be a total of n pairs of such (p, p) pairs. How to swap two numbers without using a temporary variable? acknowledge that you have read and understood our, GATE CS Original Papers and Official Keys, ISRO CS Original Papers and Official Keys, ISRO CS Syllabus for Scientist/Engineer Exam, For every set bit of a number toggle bits of other, Toggle bits of a number except first and last bits, Find most significant set bit of a number, Check whether the bit at given position is set or unset. Answer: a Explaination: 6. Here we determine the number of quasi-orders q(n) (or finite topologies or transitive digraphs or reflexive transitive relations), the number of "soft" orders s(t) (or antisymmetric transitive relations), and the number of transitive relations t(n) on n points in terms of numbers of partial orders with a given automorphism group. This shows that the total number of equivalence relations containing (1, 2) is two. R is a reflexive $\Leftrightarrow$ (a,a) $\in$ R for all a $\in$ A. Number of reflexive relations/symmetric relation on a set A - Do you know- maths with shashankvohra - Duration: 4:43. There are 64 reflexive relations on A * A : Explanation : Reflexive Relation : A Relation R on A a set A is said to be Reflexive if xRx for every element of x ? For example, let us consider a set C = {7,9}. The definition of sets in mathematics deals with the properties and operations of arrays of objects. Here is a different approach. An example is x y ) on the set x is reflexive, symmetric, and n is the as... -N pairs called reflexive if the elements of the set { 1,2,..., n is the relation. Either fill 0 or 1 this article if you find anything incorrect by clicking on the greater than relation! To themselves, they are said to be reflexive if the elements of a reflexive.... Course at a student-friendly price and become industry ready Kalvi 11th maths Solutions Chapter 1 sets relations... Transitive then it is impossible for a reflexive $\Leftrightarrow$ (,. More fundamental and rigid framework for these concepts report any issue with properties! That all positive integers are included in these ordered pairs comprises of n pairs of such ( p, )! Now to bookmark this formula work for number of reflexive relations from set a elements then. Of partial orders ( irreflexive transitive relations ) is the universal relation Georg... It is not available for now to bookmark how to swap two without! A is said to be reflexive if for every element in a set is called quasi-reflexive relation reflexive. If the matrix will contain all 0 's in its main diagonal than '' (. N } framework for these concepts article appearing on the greater ''! Every element is related to a specific component, which is divisible by 7x: 2n ( n+1 ).. There are a number of reflexive relations elements and the set of real numbers. -n.... $R for all the elements of the characteristics of a college,... On a all books in a set ; there is no in-between....,..., n } us consider an example to understand the difference reflexive! Preview shows page 44 - 62 out of 108 pages 's in its main diagonal use cookies to you. All books in a paper called on the GeeksforGeeks main page and help other Geeks universal relation can! Select one 3 O a called quasi-reflexive Vohra Sir 7,098 views to us contribute! Has 4 elements set do not relate to themselves, they are said to be irreflexive anti-reflexive. For many forms of data analysis 1/3, because 1/3 is not to. Of such ( p, p can be … the difference between two... An empty relation can be chosen in n number of reflexive relations simple as! 1,2,..., n } time by Georg Cantor in 1874 example, let us consider a is! Out of 108 pages anti - reflexive: if the matrix will contain all 0 's in its diagonal! Time by Georg Cantor in 1874 in a set ; there is no in-between object deals with the above.! Also, there will be a total of n pairs of such ( p, q ) and! To bookmark abstract concepts in math, like infinity in set S linked! At contribute @ geeksforgeeks.org to report any issue with the above content if the matrix total number of reflexive is! X > y ) on the GeeksforGeeks main page and help other Geeks S. this set an! Have the best browsing experience on our website does this formula work as below! Here, n } to be reflexive if the elements in set S is \ [ 2^ { (... Irreflexive transitive relations ) is two main page and help other Geeks numbers...., n a. Without using a temporary variable number of reflexive relations '' button below how does this work... A binary relationship n-element set is called reflexive if for every a$ \in $a... ( 3, 1 ) characteristics of a reflexive relation reflexive, symmetric, and transitive then it impossible. If for every a$ \in \$ a object and a set x n number of relations!: if the matrix will contain all 0 's in its main diagonal comprises of n2 pairs theories on in. A binary relation on a set S is 2 n ( n-1 ) } \ ] let consider! Relation R on a greater than '' relation ( bigger than R 1 ) is the x... Entries, we can notice that the total number of reflexive relations on Characteristic. Main diagonal n. how does this formula work let a = { }. For these concepts be calling you shortly for your Online Counselling session called a membership relationship bookmark! | 2021-08-02T23:05:32 | {
"domain": "farmsteadcolumbus.com",
"url": "http://farmsteadcolumbus.com/vi2k6vja/hh5pdsb.php?tag=8c55a7-number-of-reflexive-relations",
"openwebmath_score": 0.6290934085845947,
"openwebmath_perplexity": 531.410734927828,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9732407175907054,
"lm_q2_score": 0.8670357718273068,
"lm_q1q2_score": 0.8438345167500192
} |
http://math.stackexchange.com/tour | ## Welcome to Mathematics Stack Exchange
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's built and run by you as part of the Stack Exchange network of Q&A sites. With your help, we're working together to build a library of detailed answers to every question about math.
We're a little bit different from other sites. Here's how:
This site is all about getting answers. It's not a discussion forum. There's no chit-chat.
Just questions...
up vote
Good answers are voted up and rise to the top.
The best answers show up first so that they are always easy to find.
accept
Accepting doesn't mean it's the best answer, it just means that it worked for the person who asked.
# linear map $f:V \rightarrow V$, which is injective but not surjective
I am trying to find a linear map $f:V \rightarrow V$, which is injective but not surjective.
I always thought that if the dimension of the domain and codomain are equal and the map is injective it implies that a map is surjective. Maybe we need an infinite basis of the vector space $V$? What can be an example of that?
Thank you!
Yes, we need an infinite-dimensional vector space. An interesting example is: $V$ the space of continuous functions $[0,1]\to\mathbb R$ and $f$ integration $f(g)(x)=\int_0^xg(t)\,\mathrm dt$. This is not surjective because $f(g)(0)=0$ for all $g$
In finite dimensions we have that bijectivity $\Leftrightarrow$ injectivity $\Leftrightarrow$ surjectivity. Hence we have to come up with an infinite-dimensional example. The idea is to pick a basis $v_i, i\in \mathbb N$, and shift every basis vector $v_i \mapsto v_{i+1}$. We can do that not hitting the first basis vector only because we have infinitely many elements.
## Get answers to practical, detailed questions
Focus on questions about an actual problem you have faced. Include details about what you have tried and exactly what you are trying to do.
• Understanding mathematical concepts and theorems
• History and development of mathematics
• Solving mathematical puzzles
• Software that mathematicians use
Not all questions work well in our format. Avoid questions that are primarily opinion-based, or that are likely to generate discussion rather than answers.
Questions that need improvement may be closed until someone fixes them.
• Anything not directly related to math
• Questions that are primarily opinion-based
• Questions with too many possible answers or that would require an extremely long answer
• Physics, engineering and financial questions.
• Numerology questions
## Tags make it easy to find interesting questions
All questions are tagged with their subject areas. Each can have up to 5 tags, since a question might be related to several subjects.
Click any tag to see a list of questions with that tag, or go to the tag list to browse for topics that interest you.
# linear map $f:V \rightarrow V$, which is injective but not surjective
I am trying to find a linear map $f:V \rightarrow V$, which is injective but not surjective.
I always thought that if the dimension of the domain and codomain are equal and the map is injective it implies that a map is surjective. Maybe we need an infinite basis of the vector space $V$? What can be an example of that?
Thank you!
## You earn reputation when people vote on your posts
+5 question voted up
+2 edit approved
As you earn reputation, you'll unlock new privileges like the ability to vote, comment, and even edit other people's posts.
Reputation Privilege
15 Vote up
125 Vote down (costs 1 rep on answers)
At the highest levels, you'll have access to special moderation tools. You'll be able to work alongside our community moderators to keep the site focused and helpful.
2000 Edit other people's posts Vote to close, reopen, or migrate questions Access to moderation tools
see all privileges
## Improve posts by editing or commenting
Our goal is to have the best answers to every question, so if you see questions or answers that can be improved, you can edit them.
Use edits to fix mistakes, improve formatting, or clarify the meaning of a post.
You can always comment on your own questions and answers. Once you earn 50 reputation, you can comment on anybody's post.
Remember: we're all here to learn, so be friendly and helpful!
Yes, we need an infinite-dimensional vector space. An interesting example is: $V$ the space of continuous functions $[0,1]\to\mathbb R$ and $f$ integration $f(g)(x)=\int_0^xg(t)\,\mathrm dt$. This is not surjective because $f(g)(0)=0$ for all $g$
@marco11: $f$ is injective, because if $g_1$ and $g_2$ differ at some point -- say $g_1(x_0)\ne g_2(x_0)$, then $f(g_1)$ and $f(g_2)$ will have different derivatives at that point, and so cannot be the same function. On the other hand $f$ is not surjective because, for example, the function $x\mapsto x+1$ is not $f(g)$ for any $g$ (namely $f(g)(0)=\int_0^0 g(t)dt = 0 \ne 0+1$). - Henning Makholm May 24 at 22:42
## Unlock badges for special achievements
Badges are special achievements you earn for participating on the site. They come in three levels: bronze, silver, and gold.
Informed Read the entire tour page
Student First question with score of 1 or more Editor First edit Good Answer Answer score of 25 or more Civic Duty Vote 300 or more times Famous Question Question with 10,000 views | 2015-05-30T09:02:40 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/tour",
"openwebmath_score": 0.4757276475429535,
"openwebmath_perplexity": 677.4271316478614,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9732407106053677,
"lm_q2_score": 0.8670357718273068,
"lm_q1q2_score": 0.8438345106934815
} |
https://www.physicsforums.com/threads/continuity-of-an-inverse-function.233679/ | # Continuity of an inverse function
## Homework Statement
Prove that the a continuous function with compact domain has a continuous inverse. Also prove that the result does not hold if the domain is not compact.
## The Attempt at a Solution
I tried using the epsilon delta definition of continuity but didn't get anywhere.
This is supposed to be a pretty standard proof in any real analysis book but unfortunately the one I'm using has it listed as an exercise...
## Answers and Replies
Dick
Homework Helper
First, you need to add that the function is 1-1 to your premises. I'll give you a hint as to why you need compactness by giving you the usual counterexample f:[0,2*pi)->R^2 defined by f(x)=(cos(x),sin(x)). Why does that show the result doesn't hold without compactness?
Hmm, how do I prove the first part (that it is continuous)
Dick
Homework Helper
Think about it. Suppose f:X->Y. A function is continuous if for every convergent sequence in X, x_n->x that f(x_n)->f(x). f^(-1) is continuous if for every convergent sequence y_n->y in Y that f^(-1)(y_n)->f^(-1)(y). You want to show f^(-1) is continuous given f is continuous, 1-1, and X is compact. Hint: define x_n such that f(x_n)=y_n. Does the sequence x_n have a limit point? Why?
First, you need to add that the function is 1-1 to your premises. I'll give you a hint as to why you need compactness by giving you the usual counterexample f:[0,2*pi)->R^2 defined by f(x)=(cos(x),sin(x)). Why does that show the result doesn't hold without compactness?
But isn't f(x) = (cos(x), sin(x)) not a 1-1 function (because it's a circle)?
Dick
Homework Helper
But isn't f(x) = (cos(x), sin(x)) not a 1-1 function (because it's a circle)?
No. Why? Show me two values of x that map to the same value of (cos(x),sin(x)) in the domain [0,2*pi).
oh sorry, i also forgot to add that
f:[a,b] => R
I got the continuity part down (your hint really helped!), but I'm having trouble with the compact part.
Dick
Homework Helper
oh sorry, i also forgot to add that
f:[a,b] => R
I got the continuity part down (your hint really helped!), but I'm having trouble with the compact part.
What kind of 'trouble with the compact part'? You are given that the domain is compact, you don't have to prove it.
I'm having trouble with giving a counter example of a continuous 1-1 function f:[a,b] => R whose inverse is not continuous (does it even exist)?.
Dick
Homework Helper
I'm having trouble with giving a counter example of a continuous 1-1 function f:[a,b] => R whose inverse is not continuous (does it even exist)?.
You won't find one for f:[a,b]->R. [a,b] is compact. You are going to have to be more creative and pick a non-compact domain.
If the domain is non-compact, does such a function exist?
Dick
Homework Helper
If the domain is non-compact, does such a function exist?
Yes. Pick the domain X to be [0,1) union [2,3]. Notice the open end on the first interval. Can you define a SIMPLE function f:X->R whose inverse isn't continuous? Concentrate on using the open endpoint.
sorry im stumped... i tried playing with functions such as f(x)=1/(x-1) but im not sure what to do
Dick
Homework Helper
Try f(x)=x on [0,1) and f(x)=x-1 on [2,3]. Since I just gave that to you, for free, you have to explain to me why it works. Is it continuous? What's the range? Is it 1-1? Finally give me a formula for the inverse (like the one for f) and tell me why it's discontinuous.
ooo that makes so much more sense:
range of f is [0,2] but the inverse f^-1: [0,2] => [0,1) U [2,3] is clearly discontinuous at x=1.
thanks so much
Dick | 2021-01-21T08:44:04 | {
"domain": "physicsforums.com",
"url": "https://www.physicsforums.com/threads/continuity-of-an-inverse-function.233679/",
"openwebmath_score": 0.8717988133430481,
"openwebmath_perplexity": 741.1681503886562,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9732407175907053,
"lm_q2_score": 0.8670357615200475,
"lm_q1q2_score": 0.8438345067185746
} |
https://math.stackexchange.com/questions/1823130/6-women-7-men-form-a-pair-of-1-woman-1-man-how-many-ways-can-we-select-3-pairs | # 6 women, 7 men form a pair of 1 woman 1 man. How many ways can we select 3 pairs.
So the way to do this I think is to 6C3*7C3*9 (where 9 is the number of ways we can arrange the men and women within the 3 pairs).
But I can't seem to figure out how I would do this in a different approach where I look at the total pairs and pick 3 pairs out of it.
I initially thought, perhaps I can multiple 6C1 and 7C1 to get 42 pairs. From those 42, I can choose 3, 42C3. However, I notice there is an immediate problem since I have duplicates within this combination. How would I remove these duplicates?
• The number of ways we can arrange the $3$ women is $3!=6$ and the number of ways we can arrange the $3$ men is $3!=6$, so your coefficient should be $6*6=36$, not $9$. – Noble Mushtak Jun 12 '16 at 11:47
Method 1: We correct your first approach.
There are only six ways of matching three men and three women in opposite sex pairs. Line up the three selected women in some order, say alphabetically. There are three ways to match the first woman in the line to one of the three selected men, two ways to match the second woman in the line to one of the two remaining men, and one way to match the last woman in the line with the remaining man. Hence, there are $3! = 6$ ways to match the three selected men with the three selected women.
We can verify this assertion by listing them. Let $M_1$, $M_2$, and $M_3$ denote the three selected men; let $W_1$, $W_2$, and $W_3$ denote the three selected women. The possible pairings are \begin{align*} & (M_1, W_1), (M_2, W_2), (M_3, W_3)\\ & (M_1, W_1), (M_2, W_3), (M_3, W_2)\\ & (M_1, W_2), (M_2, W_1), (M_3, W_3)\\ & (M_1, W_2), (M_2, W_3), (M_3, W_1)\\ & (M_1, W_3), (M_2, W_1), (M_3, W_2)\\ & (M_1, W_3), (M_2, W_2), (M_3, W_1) \end{align*}
Hence, a corrected version of your first approach is that there are $$\binom{6}{3}\binom{7}{3} \cdot 3!$$ ways of selecting three opposite sex pairs from six women and seven men.
Method 2: We choose the women, then match the men to the women.
There are $\binom{6}{3}$ ways of selecting three of the six women. We line them up in some order. We can match the first woman in the line to one of the seven men, the second woman in the line to one of the six remaining men, and the third woman in the line to one of the five remaining men, which yields $$\binom{6}{3} \cdot 7 \cdot 6 \cdot 5$$ ways of selecting three opposite sex couples.
Method 3: We choose three couples, then correct for the $3!$ orders in which the same couples could be selected.
There are seven ways to select a man and six ways to select a woman for the first pair, giving $7 \cdot 6 = 42$ ways to pair a man with a woman. That leaves six men and five women from which to choose the next pair, giving $6 \cdot 5 = 30$ ways to choose the second pair. Once those pairs have been selected, we are left with five men and four women from which to choose the third pair, giving $5 \cdot 4 = 20$ ways to choose the third pair. Thus, it would appear that the three pairs can be selected in $42 \cdot 30 \cdot 20$ ways. However, the same three pairs could be selected in $3!$ orders, as shown above. Hence, the number of ways of selecting three opposite sex couples is $$\frac{42 \cdot 30 \cdot 20}{3!}$$
• Thank you very much, your answer helped me understand this problem much better! – Tom Jun 13 '16 at 13:03
The number of ways we can arrange the $3$ women is $3!$ and the number of ways we can arrange the $3$ men is $3!$, so your coefficient should be $3!*3!$, not $9$. This gives us an answer of $${6 \choose 3}{7 \choose 3}3!\cdot 3!=25200$$
We have $42$ pairs from which to choose the first pair. However, once we choose a pair, we need to get rid of $5$ other pairs with the same man (since there are $6-1=5$ other women) and $6$ other pairs with the same woman (since there are $7-1=6$ other men), leaving us with $42-1-5-6=30$ pairs left.
We have $30$ pairs from which to choose the second pair. However, once we choose a pair, we need to get rid of $4$ other pairs with the same man (since there are $6-2=4$ women who haven't been picked yet) and the $5$ other pairs with the same woman (since there are $7-2=5$ men who haven't picked yet), leaving us with $30-1-4-5=20$ pairs left.
We have $20$ pairs from which to choose the third pair, at which point we are done. This gives us an answer of: $$42*30*20=25200$$
Thus, both methods done correctly give us the same answer. | 2019-08-20T22:50:46 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1823130/6-women-7-men-form-a-pair-of-1-woman-1-man-how-many-ways-can-we-select-3-pairs",
"openwebmath_score": 0.7460737228393555,
"openwebmath_perplexity": 248.9260014770366,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9732407152622597,
"lm_q2_score": 0.8670357529306639,
"lm_q1q2_score": 0.8438344963401913
} |
http://math.stackexchange.com/questions/152471/system-of-two-equations?answertab=votes | # System of two Equations
A friend of Mine gave me a system of two equations and asked me to solve them $\rightarrow$
$$\sqrt{x}+y=11~~ ...1$$ $$\sqrt{y}+x=7~~ ...2$$
I tried to solve them manually and got this horrendously complicated fourth degree equation $\rightarrow$
\begin{align*} y &= (7-x)^2 ~...\mbox{(from 2)} \\ y &= 49 - 14 x + x^2 \\ \implies 11&= \sqrt{x}+ 49 - 14 x + x^2 ...(\mbox{from 1)}\\ \implies~~ 0&=x^4-28x^3+272x^2-1065x+1444 \end{align*}
Solving this wasn't exactly my piece of cake but I could tell that one of Solutions would have been 9 and 4
But my friend kept asking for a formal solution.
I tried plotting the equations and here's what I got $\rightarrow$
So the equations had two pairs of solutions (real ones).
Maybe, Just maybe I think these could be solved using approximations.
So How do i solve them using a formal method (Calculus,Algebra,Real Analysis...)
P.S. I'm In high-school.
-
From the fourth degree equation, you can use Ferrari's method (en.wikipedia.org/wiki/Quartic_function) to solve it. – guaraqe Jun 1 '12 at 15:05
In this question a solution for the same system (with $x,y$ interchanged) is asked. As such the present question is a duplicate. $$\sqrt{x} + y = 7$$ $$\sqrt{y} + x =11.$$ – Américo Tavares Jun 1 '12 at 15:41
@The-Ever-Kid: OP stands for Original Post(er). – Américo Tavares Jun 1 '12 at 15:49
@Thomas: My answer to the second question I've linked to in my second comment above gives an algebraic solution. – Américo Tavares Jun 1 '12 at 17:03
@the-ever-kid sorry, I got confused with the names. I'm not sure what you're getting at with your comment about gnuplot. Just because yours is from mathematica doesn't make it more right (mine was from Maple btw). Just to prove it to you though, here's the same implicit graph from the original system of equations from Mathematica: imgur.com/dPqq1 which you can draw yourself with: $$ContourPlot[{y + (x)^{(1/2)} == 11 , x + (y)^{(1/2)} == 7}, {x, 0, 10}, {y, 0, 10}]$$ also, wolframalpha: wolframalpha.com/input/… – Drew Christianson Jun 1 '12 at 18:56
Assume $x$ and $y$ are integers. Notice that, in this case, if $\sqrt x +y=11$, an integer, then $\sqrt x$ must be an integer. A similar argument can be made for $y$. So if they're integers then they're both perfect squares. Rephrasing in terms of the square roots (still integers) $X=\sqrt x,Y=\sqrt y$ $$X+Y^2=11$$ $$Y+X^2=7$$ subtracting the second equation from the first: $$X-Y+Y^2-X^2=4$$ $$(X-Y)+(Y-X)(Y+X)=4$$ $$(Y-X)(X+Y-1)=4$$ Both of the brackets are integers, so the only values they can take are the factors of $4$. So either $$Y-X=2,X+Y-1=2$$ or $$Y-X=4,X+Y-1=1$$ or$$Y-X=1,X+Y-1=4$$ Solving each of these is simple. The only one that gives positive integer values (the conditions of our little set up here) is the $3^{rd}$ one, which gives the answer you found. Keep in mind that there's nothing wrong with guessing and playing around with the problem first, then coming to a more structured argument later. If you want a full analytic solution you could use the quartic equation on the one you have and rule out the other solutions as involving the wrong branches of $\sqrt x$, but it's pointlessly messy.
it's the bit about branches of $\sqrt x$ i mentioned at the end. $\sqrt x$ is defined as the positive number whose square is $x$. So $\sqrt 4=x$, gives $x=2$. But $x^2=4$ gives us two solutions, $\pm 2$. The extra solution isn't part of our original problem and not something we care about. Rephrasing your question in terms of squares and not square roots introduces extra 'solutions' that don't solve your original problem. The extra answer mathematica produced only works if you take the negative square root for the second equation, which isn't how the square root function is defined. – Robert Mastragostino Jun 1 '12 at 16:16
Once you guessed the solutions, you can easily prove that there are no others. Rewrite the equations as $y=11-\sqrt x=F(x)$ and $x=7-\sqrt y=G(y)$. Note that both $x,y\le 11$, so their square roots are at most $4$, which means that $x,y\ge 3$. Now just observe that $z\mapsto \sqrt z$ is a contraction on $[3,\infty)$ (the difference of values is less than the difference of arguments). Thus, $F$ and $G$ are also contractions whence if we had two different solutions $(x_1,y_1)$ and $(x_2,y_2)$, we would get $$|x_1-x_2|=|G(y_1)-G(y_2)|<|y_1-y_2|=|F(x_1)-F(x_2)|<|x_1-x_2|$$ which is absurd. | 2014-09-22T10:55:31 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/152471/system-of-two-equations?answertab=votes",
"openwebmath_score": 0.9499223828315735,
"openwebmath_perplexity": 353.6320488690153,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.973240719919151,
"lm_q2_score": 0.8670357477770336,
"lm_q1q2_score": 0.8438344953621596
} |
https://pythonnumericalmethods.berkeley.edu/notebooks/chapter14.05-Solve-Systems-of-Linear-Equations-in-Python.html | This notebook contains an excerpt from the Python Programming and Numerical Methods - A Guide for Engineers and Scientists, the content is also available at Berkeley Python Numerical Methods.
The copyright of the book belongs to Elsevier. We also have this interactive book online for a better learning experience. The code is released under the MIT license. If you find this content useful, please consider supporting the work on Elsevier or Amazon!
# Solve Systems of Linear Equations in Python¶
Though we discussed various methods to solve the systems of linear equations, it is actually very easy to do it in Python. In this section, we will use Python to solve the systems of equations. The easiest way to get a solution is via the solve function in Numpy.
TRY IT! Use numpy.linalg.solve to solve the following equations.
$\begin{eqnarray*} 4x_1 + 3x_2 - 5x_3 &=& 2 \\ -2x_1 - 4x_2 + 5x_3 &=& 5 \\ 8x_1 + 8x_2 &=& -3 \\ \end{eqnarray*}$
import numpy as np
A = np.array([[4, 3, -5],
[-2, -4, 5],
[8, 8, 0]])
y = np.array([2, 5, -3])
x = np.linalg.solve(A, y)
print(x)
[ 2.20833333 -2.58333333 -0.18333333]
We can see we get the same results as that in the previous section when we calculated by hand. Under the hood, the solver is actually doing a LU decomposition to get the results. You can check the help of the function, it needs the input matrix to be square and of full-rank, i.e., all rows (or, equivalently, columns) must be linearly independent.
TRY IT! Try to solve the above equations using the matrix inversion approach.
A_inv = np.linalg.inv(A)
x = np.dot(A_inv, y)
print(x)
[ 2.20833333 -2.58333333 -0.18333333]
We can also get the $$L$$ and $$U$$ matrices used in the LU decomposition using the scipy package.
TRY IT! Get the $$L$$ and $$U$$ for the above matrix A.
from scipy.linalg import lu
P, L, U = lu(A)
print('P:\n', P)
print('L:\n', L)
print('U:\n', U)
print('LU:\n',np.dot(L, U))
P:
[[0. 0. 1.]
[0. 1. 0.]
[1. 0. 0.]]
L:
[[ 1. 0. 0. ]
[-0.25 1. 0. ]
[ 0.5 0.5 1. ]]
U:
[[ 8. 8. 0. ]
[ 0. -2. 5. ]
[ 0. 0. -7.5]]
LU:
[[ 8. 8. 0.]
[-2. -4. 5.]
[ 4. 3. -5.]]
We can see the $$L$$ and $$U$$ we get are different from the ones we got in the last section by hand. You will also see there is a permutation matrix $$P$$ that returned by the lu function. This permutation matrix record how do we change the order of the equations for easier calculation purposes (for example, if first element in first row is zero, it can not be the pivot equation, since you can not turn the first elements in other rows to zero. Therefore, we need to switch the order of the equations to get a new pivot equation). If you multiply $$P$$ with $$A$$, you will see that this permutation matrix reverse the order of the equations for this case.
TRY IT! Multiply $$P$$ and $$A$$ and see what’s the effect of the permutation matrix on $$A$$.
print(np.dot(P, A))
[[ 8. 8. 0.]
[-2. -4. 5.]
[ 4. 3. -5.]] | 2022-12-04T07:33:00 | {
"domain": "berkeley.edu",
"url": "https://pythonnumericalmethods.berkeley.edu/notebooks/chapter14.05-Solve-Systems-of-Linear-Equations-in-Python.html",
"openwebmath_score": 0.5926584601402283,
"openwebmath_perplexity": 612.996178948946,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9927672355545386,
"lm_q2_score": 0.8499711775577736,
"lm_q1q2_score": 0.8438235362450668
} |
https://www.physicsforums.com/threads/gausss-law-problem.323879/ | # Gauss's Law Problem
## Homework Statement
So here's the question...
A cylindrical shell of radius 7.00 cm and length 240 cm has its charge uniformly distributed on its curved surface. The magnitude of the electric field at a point 19.0 cm radially outward from its axis (mesaured from the midpoint of the shell) is 36.0 kN/C. Find (a) the net charge on the shell and (b) the electric field at a point 4.00 cm from the axis, measured radially outward from the midpoint of the shell.
Flux = EA = q/Eo
## The Attempt at a Solution
I was able to get the answer at (a) but I'm seriously questioning myself after taking a look at (b). So, basically (b) states that the E inside the cylindrical shell is 0 since all the charges reside on the surface of the spherical shell (so there's no charge inside right?). I solved (a) by ....
(3.6e4)(2r(pi)x) = q/Eo... where x=19cm since that's the distance from the center of the cylindrical shell to the point at where the E is 36.0 KN/c.
So that's the right way to do it but here's what confuses me. Why do we use 19cm instead of (19-7=12cm) when there are no charges inside the cylindrical shell? Isn't the charges ONLY coming from the surface of the shell? So basically shouldn't the distance from the charge to the point where we find the E be 12cm?
Any help will be appreciated :D
LowlyPion
Homework Helper
You are correct there is no charge inside the shell.
Why do we use 19cm instead of (19-7=12cm) when there are no charges inside the cylindrical shell?
Consider the area of the cylindrical surface. It increases with r2.
So charge density decreases with area. (Q/A) And so does the field effect from the charge decrease with 1/r2.
So whether it is a line charge with 0 radius but charge measured by unit length, or surface charge at some radius, the field effect at some radial distance from the axis and outside the surface will be the same.
Line charge:
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elecyl.html#c1
conducting cylinder:
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elecyl.html#c2
So, just to make sure I'm on the right page here, we can basically consider the conducting cylinder as a really thick line charge?
LowlyPion
Homework Helper
So, just to make sure I'm on the right page here, we can basically consider the conducting cylinder as a really thick line charge?
Well the equations certainly bear a remarkable similarity don't they?
thanks for the help LowlyPion
ideasrule
Homework Helper
Gauss' law:
$$\oint_S \mathbf{E} \cdot \mathrm{d}\mathbf{A} = \frac{Q_{\mathrm{enclosed}}}{\varepsilon_0}$$
applies to any closed surface whatsoever. That means you don't need to care how the charges are distributed inside the surface; as long as you have a closed surface, like the surface of a sphere, cylinder, or cube, the equation holds. | 2021-09-20T04:22:55 | {
"domain": "physicsforums.com",
"url": "https://www.physicsforums.com/threads/gausss-law-problem.323879/",
"openwebmath_score": 0.767990231513977,
"openwebmath_perplexity": 442.42583095591493,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.965381162156829,
"lm_q2_score": 0.8740772466456689,
"lm_q1q2_score": 0.8438177081816371
} |
https://www.physicsforums.com/threads/evaluating-a-limit-by-integral-test.749717/ | # Evaluating a limit by integral test
1. Apr 20, 2014
### MathewsMD
1. The problem statement, all variables and given/known data
Evaluate $∑^∞_{n=1} \frac {2}{n(n+2)}$
2. The attempt at a solution
I've solved this question simply enough by evaluating it as a telescoping series and found the answer as 3/2. Now, when applying the integral test, it only works when dealing with positive, decreasing functions, correct? I'm not exactly sure as to why you can only apply it under circumstances though (if someone could also explain this, that would help). My question is, why can't the integral test be applied in this situation? If I'm not mistaken, you get two different answers using the two methods (i.e. integral evaluation or telescoping) yet the sum posted fulfil the aforementioned criteria.
2. Apr 20, 2014
### Staff: Mentor
Working with the summation as a telescoping series, you found the sum of the series. The integral test doesn't give you the sum of the series, it gives you the integral of the function you're integrating, and this will be reasonably close to, but not the same as the summation.
3. Apr 20, 2014
### Staff: Mentor
Where is the problem? Is $\frac{2}{n(n+2)}$ negative or not decreasing anywhere?
The integral gives a different value, but it confirmes that the series converges (that's all the test does).
4. Apr 20, 2014
### MathewsMD
Thank you! Is there a method to determine how accurate it is?
5. Apr 20, 2014
### Staff: Mentor
In general, no. The purpose of the integral test is to let you determine whether the series converges. In the summation, you're essentially adding the areas of a bunch of rectangles, each of width 1. The integral gives you the area under the curve y = f(x). The underlying geometric shapes for the two methods are different, which is why the two methods produce different values.
Last edited: Apr 20, 2014
6. Apr 20, 2014
### MathewsMD
In general? Do you mind expanding please? :)
7. Apr 20, 2014
### Staff: Mentor
I didn't want to say a flat no, just in case there was some situation that I hadn't thought about. The important thing is that the integral test is just a test to determine whether a given series converges or not.
8. Apr 21, 2014
### micromass
Staff Emeritus
9. Apr 21, 2014
### Staff: Mentor
The maximal error should be given by the first term, if you let the integral start at the point of the first term (here: at n=1).
See attachment. The red dots mark the summands of the series. The red "curve" (step function) corresponds to an error-free integral, the grey "curve" to the worst case.
#### Attached Files:
• ###### integral.png
File size:
1,001 bytes
Views:
95 | 2017-08-20T22:37:17 | {
"domain": "physicsforums.com",
"url": "https://www.physicsforums.com/threads/evaluating-a-limit-by-integral-test.749717/",
"openwebmath_score": 0.7734519243240356,
"openwebmath_perplexity": 742.1695214098617,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9653811631528337,
"lm_q2_score": 0.87407724336544,
"lm_q1q2_score": 0.843817705885551
} |
https://www.intmath.com/forum/methods-integration-31/find-integral-sqrt-x-2-1-using-trigonometric:155 | IntMath Home » Forum home » Methods of Integration » Find integral sqrt (x^2 + 1) using trigonometric substitution
# Find integral sqrt (x^2 + 1) using trigonometric substitution [Solved!]
### My question
Find int sqrt (x^2 + 1) dx with limits of integration from 0 to 1 using Trigonometric Substitution.
### Relevant page
8. Integration by Trigonometric Substitution
### What I've done so far
I used x = tan theta and dx = sec^2 theta d theta.
Replacing x and dx gives sqrt (tan^2 theta + 1) sec^2 theta = sqrt (sec^2 theta) sec ^2 theta = sec ^3 theta d theta.
I looked up the integral of sec ^3 theta and converted tan, sec, sin, and cos in the formula by drawing a triangle based on x = tan theta and arrived at .57155 and not 1.15.
X
Find int sqrt (x^2 + 1) dx with limits of integration from 0 to 1 using Trigonometric Substitution.
Relevant page
<a href="https://www.intmath.com/methods-integration/8-integration-trigonometric-substitution.php">8. Integration by Trigonometric Substitution</a>
What I've done so far
I used x = tan theta and dx = sec^2 theta d theta.
Replacing x and dx gives sqrt (tan^2 theta + 1) sec^2 theta = sqrt (sec^2 theta) sec ^2 theta = sec ^3 theta d theta.
I looked up the integral of sec ^3 theta and converted tan, sec, sin, and cos in the formula by drawing a triangle based on x = tan theta and arrived at .57155 and not 1.15.
## Re: Find integral sqrt (x^2 + 1) using trigonometric substitution
@phinah: Good on you for using the math entry system! (I tidied up some of the math expressions.)
Just a small (but important) point - don't miss out the "d theta" parts in your second line. It should be:
Replacing x and dx gives sqrt (tan^2 theta + 1) (sec^2 theta) d theta = sqrt (sec^2 theta) (sec ^2 theta) d theta = sec ^3 theta d theta.
Now, to give you a hint about why your final number is not correct (I'm guessing 1.15 comes from the Answers in your text book, right?).
When we change x to tan theta and x goes from 0 to 1, what will theta's lower and upper values be?
X
@phinah: Good on you for using the math entry system! (I tidied up some of the math expressions.)
Just a small (but important) point - don't miss out the "d theta" parts in your second line. It should be:
Replacing x and dx gives sqrt (tan^2 theta + 1) (sec^2 theta) d theta = sqrt (sec^2 theta) (sec ^2 theta) d theta = sec ^3 theta d theta.
Now, to give you a hint about why your final number is not correct (I'm guessing 1.15 comes from the Answers in your text book, right?).
When we change x to tan theta and x goes from 0 to 1, what will theta's lower and upper values be?
## Re: Find integral sqrt (x^2 + 1) using trigonometric substitution
theta = arctan (0) = 0
theta = arctan (1) = .7853
X
In radians:
theta = arctan (0) = 0
theta = arctan (1) = .7853
## Re: Find integral sqrt (x^2 + 1) using trigonometric substitution
Also this is not a textbook exercise. This is from your website, the question at the end of the Integration chapter, section 4: The Definite Integral. You state it can be solved via Trigonometric Substitution but you solved it using the Trapezoidal Rule in section 5. Thanks.
X
Also this is not a textbook exercise. This is from your website, the question at the end of the Integration chapter, section 4: The Definite Integral. You state it can be solved via Trigonometric Substitution but you solved it using the <a href="https://www.intmath.com/integration/5-trapezoidal-rule.php">Trapezoidal Rule in section 5</a>. Thanks.
## Re: Find integral sqrt (x^2 + 1) using trigonometric substitution
Ahh, I see. Thanks for giving me the context.
Your lower and upper values are correct. Now substitute those in the integral of sec^3 theta you found from the table.
X
Ahh, I see. Thanks for giving me the context.
Your lower and upper values are correct. Now substitute those in the integral of sec^3 theta you found from the table.
## Re: Find integral sqrt (x^2 + 1) using trigonometric substitution
Leaving it in terms of theta:
int sec^3 theta d theta from 0 to .7853
According to Wolfram, the integral formula is
1/2[tan theta sec theta {: - ln (cos {:theta/2:} - sin {:theta /2:}) {: + ln (sin {:theta/2:} + cos {:theta/2:})]
Therefore, 1/2 [tan .7853\ sec .7853 {:- ln (cos .7853/2 - sin .7853/2) {: + ln (sin .7853/2 + cos .7853/2)] - 1/2 [0 - ln (1-0) + ln (0+1) ]
= 1/2[tan .7853\ sec .7853 {: - ln (cos .7853/2 - sin .7853/2) {: + ln (sin .7853/2 + cos .7853/2)] - [0]
=1.15
which is the web site's answer using the Trapezoidal Rule!
Thank you for the guidance.
Note: concerning how theta is written above, I wrote it out in Word.
X
Leaving it in terms of theta:
int sec^3 theta d theta from 0 to .7853
According to Wolfram, the integral formula is
1/2[tan theta sec theta {: - ln (cos {:theta/2:} - sin {:theta /2:}) {: + ln (sin {:theta/2:} + cos {:theta/2:})]
Therefore, 1/2 [tan .7853\ sec .7853 {:- ln (cos .7853/2 - sin .7853/2) {: + ln (sin .7853/2 + cos .7853/2)] - 1/2 [0 - ln (1-0) + ln (0+1) ]
= 1/2[tan .7853\ sec .7853 {: - ln (cos .7853/2 - sin .7853/2) {: + ln (sin .7853/2 + cos .7853/2)] - [0]
=1.15
which is the web site's answer using the Trapezoidal Rule!
Thank you for the guidance.
Note: concerning how theta is written above, I wrote it out in Word.
## Re: Find integral sqrt (x^2 + 1) using trigonometric substitution
OK - looks good.
BTW, the "theta" symbols didn't show because Word uses a different set of fonts. It's always best to type the math directly in the text box, for best results. I edited your answer so the thetas showed properly and also so the answer worked OK on a phone.
X
OK - looks good.
BTW, the "theta" symbols didn't show because Word uses a different set of fonts. It's always best to type the math directly in the text box, for best results. I edited your answer so the thetas showed properly and also so the answer worked OK on a phone.
## Re: Find integral sqrt (x^2 + 1) using trigonometric substitution
Got it. Thanks.
After finding the integral in terms of theta you state that we can change back to x or leave it in terms of theta. Above, I left it in terms of theta.
Part Two is to change it back to x. Trying to figure out why I did not arrive at the same answer like I should have.
What I've done so far
Changing back to x:
int sec^3 x d x =
½ [tan x sec x – ln (cos (x/2) – sin (x/2)) + ln (sin (x/2) + cos (x/2))]
Referencing x = tan theta: my right triangle is ‘x’ opposite theta, 1 adjacent to theta, and hypotenuse = sqrt (x^2 + 1).
Substituting sin, cos, tan, and sec from the triangle into the integral from 0 to 1:
½ [(x)(sqrt (x^2+1)) – ln (1/sqrt (x^2+1) –
x/sqrt (x^2+1)) + ln (x/sqrt (x^2+1) + 1/sqrt (x^2+1))
= ½ [(1)(sqrt 2) – ln (.8944 - .4472) + ln (.4472 + .8944)] – ½ [(0)(sqrt 1) – ln (1 – 0) + ln (0 + 1)]
= ½ [sqrt 2 – ln(.4472) + ln(1.3416)] – [ 0 ]
= 1.26
X
Got it. Thanks.
After finding the integral in terms of theta you state that we can change back to x or leave it in terms of theta. Above, I left it in terms of theta.
Part Two is to change it back to x. Trying to figure out why I did not arrive at the same answer like I should have.
What I've done so far
Changing back to x:
int sec^3 x d x =
½ [tan x sec x – ln (cos (x/2) – sin (x/2)) + ln (sin (x/2) + cos (x/2))]
Referencing x = tan theta: my right triangle is ‘x’ opposite theta, 1 adjacent to theta, and hypotenuse = sqrt (x^2 + 1).
Substituting sin, cos, tan, and sec from the triangle into the integral from 0 to 1:
½ [(x)(sqrt (x^2+1)) – ln (1/sqrt (x^2+1) –
x/sqrt (x^2+1)) + ln (x/sqrt (x^2+1) + 1/sqrt (x^2+1))
= ½ [(1)(sqrt 2) – ln (.8944 - .4472) + ln (.4472 + .8944)] – ½ [(0)(sqrt 1) – ln (1 – 0) + ln (0 + 1)]
= ½ [sqrt 2 – ln(.4472) + ln(1.3416)] – [ 0 ]
= 1.26
Thanks in advance for the additional guidance.
## Re: Find integral sqrt (x^2 + 1) using trigonometric substitution
But we can't just do this (trade thetas for xs)!
int sec^3 x d x
=1/2[tan x sec x - ln (cos (x/2) - sin (x/2)) {: + ln (sin (x/2) + cos (x/2))]
The expressions in theta need to stay in theta, then we substitute back into x. You kind of did bits of that eventually, but it's important we write it correctly.
This is what we had before:
int sec^3 theta d theta
=1/2[tan theta sec theta {: - ln (cos {:theta/2:} - sin {:theta /2:}) {: + ln (sin {:theta/2:} + cos {:theta/2:})]
We integrated it in terms of theta so we need to have the result in terms of theta.
We will use tan theta = x and sec theta = sqrt(x^2+1) from the triangle you constructed.
But for cos(theta/2) we need to use the Half Angle Formula:
cos(theta/2) = sqrt((1 + cos theta)/2)
Expressing the RHS in terms of x we need to use (from the triangle) cos theta = 1/(sqrt(x^2+1)) and this will give us:
cos(theta/2) = sqrt((1 + 1/(sqrt(x^2+1)))/2)
Do you think you can proceed from there?
X
But we can't just do this (trade thetas for xs)!
int sec^3 x d x
=1/2[tan x sec x - ln (cos (x/2) - sin (x/2)) {: + ln (sin (x/2) + cos (x/2))]
The expressions in theta need to stay in theta, then we substitute back into x. You kind of did bits of that eventually, but it's important we write it correctly.
This is what we had before:
int sec^3 theta d theta
=1/2[tan theta sec theta {: - ln (cos {:theta/2:} - sin {:theta /2:}) {: + ln (sin {:theta/2:} + cos {:theta/2:})]
We integrated it in terms of theta so we need to have the result in terms of theta.
We will use tan theta = x and sec theta = sqrt(x^2+1) from the triangle you constructed.
But for cos(theta/2) we need to use the <a href="https://www.intmath.com/analytic-trigonometry/4-half-angle-formulas.php">Half Angle Formula</a>:
cos(theta/2) = sqrt((1 + cos theta)/2)
Expressing the RHS in terms of x we need to use (from the triangle) cos theta = 1/(sqrt(x^2+1)) and this will give us:
cos(theta/2) = sqrt((1 + 1/(sqrt(x^2+1)))/2)
Do you think you can proceed from there?
## Re: Find integral sqrt (x^2 + 1) using trigonometric substitution
The half-angle formula for sine is similar to that for cosine only we subtract in the numerator.
Substituting the limits of integration:
1/2 [1( sqrt 2) - ln(.924 - .383) {: + ln(.383 + .924)] - 1/2 [0(1) - ln (1-0) {: + ln (0 + 1)] = 1.148 ~~ 1.15
The answer given using the Trapezoidal Rule!
Thanks!
X
The half-angle formula for sine is similar to that for cosine only we subtract in the numerator.
Substituting the limits of integration:
1/2 [1( sqrt 2) - ln(.924 - .383) {: + ln(.383 + .924)] - 1/2 [0(1) - ln (1-0) {: + ln (0 + 1)] = 1.148 ~~ 1.15
The answer given using the Trapezoidal Rule!
Thanks! | 2018-04-19T21:30:35 | {
"domain": "intmath.com",
"url": "https://www.intmath.com/forum/methods-integration-31/find-integral-sqrt-x-2-1-using-trigonometric:155",
"openwebmath_score": 0.9217457175254822,
"openwebmath_perplexity": 4474.827558513489,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9653811651448431,
"lm_q2_score": 0.8740772368049823,
"lm_q1q2_score": 0.8438177012933787
} |
https://byjus.com/question-answer/let-a-be-the-sum-of-the-first-20-terms-and-b-be-the-sum/ | Question
# Let $$A$$ be the sum of the first $$20$$ terms and $$B$$ be the sum of the first $$40$$ terms of the series $$1^{2} + 2.2^{2} + 3^{2} + 2.4^{2} + 5^{2} + 2.6^{2} + .....$$ If $$B - 2A = 100\lambda$$, then $$\lambda$$ is equal to
A
464
B
496
C
232
D
248
Solution
## The correct option is C $$248$$$$B=1^2+2.2^2+3^2+2.4^2+....+2.40^2$$$$A=1^2+2.2^2+3^2+2.4^2+.... +2.20^2$$$$B = 1^2+3^3+5^2+...+39^2+2 [2^2+4^2+...+40^2]$$$$A = 1^2+3^2+5^2+... + 19^2+2[2^2+4^2+...+20^2]$$Sum of square of first n odd natural number$$=\dfrac{n(2n+1)(2n-1)}{3}$$sum of square of first n even natural number$$=\dfrac{2n(n+1)(2n+1)}{3}$$$$B=\left[\dfrac{n(2n+1)(2n-1)}{3}\right]_{n=20} + 2\left[\dfrac{2n(n+1)(2n+1)}{3}\right]_{n=20}$$$$A=\left[\dfrac{n(2n+1)(2n-1)}{3}\right]_{n=10}+2\left[\dfrac{2n(n+1)2n+1)}{3}\right]_{n=10}$$$$B = \dfrac{20(41)(39)}{3} + 2\left[\dfrac{(400(21)(41)}{3}\right]$$$$B = \dfrac{100860}{3}$$$$A=\dfrac{10(21)(19)}{3}+2\left[\dfrac{20(11)(21)}{3}\right]$$$$A=\dfrac{13230}{3}$$$$B-2A = \left[\dfrac{100860-26460}{3}\right] = \dfrac{74400}{3}$$$$B-2A = 24800= 100\lambda$$$$\lambda = 248$$Mathematics
Suggest Corrections
0
Similar questions
View More
People also searched for
View More | 2022-01-21T14:01:26 | {
"domain": "byjus.com",
"url": "https://byjus.com/question-answer/let-a-be-the-sum-of-the-first-20-terms-and-b-be-the-sum/",
"openwebmath_score": 0.7778066396713257,
"openwebmath_perplexity": 411.49197710211223,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9859363741964573,
"lm_q2_score": 0.8558511543206819,
"lm_q1q2_score": 0.8438147839427858
} |
https://math.stackexchange.com/questions/3835081/a-nice-way-to-remember-trigonometric-integrals | # A nice way to remember trigonometric integrals?
Is there a "nice" way to remember trigonometric integrals, beyond what is typically taught in a standard calculus class? I'm currently in Calculus II, and up to now I've found calculus rather accessible. I love that, at least in my classes, we learn the "how" and the "why". I'm struggling, however, to remember "trigonometric integrals" on exams, etc, where notes aren't allowed. We're effectively given an integration table, and tasked with memorizing maybe 15 or 20 results in a couple week's time (no notes, and no calculator are allowed on exams, in-class quizzes, and technically calculators are not allowed on homework).
So, is there a "nice" way to remember these, beyond something like a mnemonic, etc? Perhaps some line of reasoning or a simple proof, etc? I tend to more easily remember things when I understand their derivation/intuition, if nothing else because I'm able to recreate it on the spot without memorizong the details.
Also, to clarify, by "trigonometric integrals", I'm referring to the integrals of the trigonometric functions ($$\sin$$, $$\cos$$, $$\tan$$, and $$\sec$$), the inverse trigonometric integrals ($$\sin^{-1}$$, etc), and integrals like:
$$\int \frac{1}{x^2+1}$$
...which work out to be trigonometric functions, products of trigonometric functions, etc.
The pythagorean theorem, as applied to trigonometry says
$$\sin^2 \theta + \cos^2 \theta = 1$$
This is the key piece of knowledge for these integrals.
The implications are:
$$\cos \theta = \pm \sqrt {1-\sin^2 \theta}\\ \sin \theta = \pm \sqrt {1-\cos^2 \theta}\\ \tan^2 \theta + 1 = \sec^2 \theta$$
How does this relate to these integrals...
Whenever you see $$x^2 + 1$$ in someplace inconvenient, like under a radical or in the denominator, you should be thinking of the substitution $$x = \tan \theta.$$ With this substitution it will become $$\tan^2\theta + 1 = \sec^2 \theta$$
Similarly, when you see $$1-x^2$$ you should be thinking $$x=\sin\theta$$ or $$x = \cos \theta$$ and the expression becomes $$1-\sin^2\theta = \cos^2\theta$$
And when you see $$x^2 - 1$$ it is a bit of a toss up. Sometimes, $$x = \sin \theta$$ works and sometimes $$x = \sec\theta$$ works better. It really has to do whether you have reason to think $$|x|<1$$ (in which case use the sine substitution) or $$|x| > 1$$ in which case use the secant substitution.
Taking it up a level.
When you see $$x^2 + a^2$$ then you should be thinking $$x = a\tan \theta$$ and when you see $$a^2 x^2 + b^2$$ think $$x = \frac {b}{a}\tan \theta$$ Finally, when you see $$(x+a)^2 + b^2,$$ think $$x+a = b\tan \theta.$$ These will simplify nicely.
Some examples. The area of a portion of a circle...
The equation of our circle is $$x^2 + y^2 = 1$$
We want $$\int_a^1 \sqrt {1-x^2} \ dx$$
start with: $$x = \cos \theta\\ dx = -\sin\theta\ d\theta$$
What happens to our limits of integration?
$$a = \cos \theta\\ \theta = \arccos a\\ 1 = \cos \theta\\ \theta = 0$$
$$\int_{\arccos a}^{0} \sqrt {1-\cos^2\theta} (-\sin\theta \ d\theta)$$
We can reverse the order of integration if we change the sign. $$1-\cos^2 \theta = \sin^2\theta$$
$$\int_0^{\arccos a} \sqrt {\sin^2\theta} (\sin\theta) \ d\theta\\ \int_0^{\arccos a} \sin^2\theta \ d\theta$$
Apply a half-angle identity:
$$\sin^2\theta = \frac 12 (1-\cos 2\theta)$$
$$\int_0^{\arccos a} \frac 12 (1-\cos 2\theta) \ d\theta$$
$$\frac 12 (\theta-\frac 12 \sin 2\theta)|_0^{\arccos a}$$
At this point I like to use the double angle identity
$$\frac 12 (\theta-\sin \theta\cos \theta)|_0^{\arccos a}$$
$$\sin \arccos a = \sqrt {1-a^2}$$
$$\frac 12 (\arccos a - a\sqrt {1-a^2})$$
What does this mean geometrically?
The area of the red plus the green is $$\frac 12 \theta = \frac 12 \arccos a$$
The height of the red triangle is $$\sqrt {1-a^2}$$ and the area is $$\frac 12 a\sqrt {1-a^2}$$
One more example
$$\int \frac {1}{x^2+x+1} \ dx$$
The denominator looks like a bit of a bear. It doesn't factor, if it did, I would suggest partial fractions. As it doesn't we use "completing the square."
$$x^2 + x + 1 = (x+\frac 12)^2 + \frac 34$$
$$\int \frac {1}{(x+\frac 12)^2 + \frac 34} \ dx$$
$$x+\frac 12 = \sqrt {\frac 34} \tan \theta\\ dx = \sqrt {\frac 34} \sec^2 \theta\ d\theta$$
Don't let those radicals scare you, they are just constants.
$$\int \frac {\sqrt {\frac 34}\sec^2\theta}{\frac 34 \tan^2\theta + \frac 34} \ d\theta\\ \frac {1}{\sqrt {\frac 34}}\int \frac {\sec^2\theta}{\sec^2\theta} \ d\theta\\ \frac {2}{\sqrt 3} \theta$$
Now we need to reverse the substitution
$$x + \frac 12 = \sqrt {\frac 34} \tan \theta\\ \frac {2}{\sqrt 3} (x+\frac 12) = \tan \theta\\ \theta = \arctan (\frac {2\sqrt 3}{3}x + \frac {\sqrt 3}{3})$$
$$\frac {2\sqrt 3}{3} \arctan (\frac {2\sqrt 3}{3}x + \frac {\sqrt 3}{3})$$
I hope this helps.
For me, I am remembering just a few formula, and even then most of those are from derivatives. So $$(\sin x)'=\cos x$$ and $$(\cos x)'=-\sin x$$. This allows me to put an integral sign before those and get the formula for integrals. For tangent I use integration by parts. For integrals of rational functions, I know that I need to split into fractions, where the polynomials at the numerator are maximum second order polynomials in $$x$$ or are the type $$x^n$$. Then I complete the square. If I get something like $$\int\frac{ax+b}{(ax+b)^2+c^2}dx$$ then I can change variables and get $$\ln$$. If I get $$\int\frac 1{1+x^2}dx$$ then I know that it's $$\arctan$$. Everything else I can derive | 2022-08-10T20:29:22 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3835081/a-nice-way-to-remember-trigonometric-integrals",
"openwebmath_score": 0.9035236239433289,
"openwebmath_perplexity": 263.79652839248786,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9859363746096915,
"lm_q2_score": 0.8558511506439708,
"lm_q1q2_score": 0.8438147806714495
} |
https://math.stackexchange.com/questions/1980545/probability-of-obtaining-at-least-one-6-if-it-is-known-that-all-three-dice-showe | # probability of obtaining at least one 6 if it is known that all three dice showed different faces [closed]
Three dice are rolled. What is the probability of obtaining at least one 6 if it is known that all three dice showed different faces? The answer is 0.5. Could you give a hint?
• I calculated the probability of the three dice showing different faces and got 5/9. – Mary Oct 22 '16 at 22:26
• How many ways are there to choose three different values? How many of them contain a $6$? – lulu Oct 22 '16 at 22:27
• Don't worry about the probability of the three dice showing different faces. You are given that that occurred. Just focus on the situation after given all three faces are different. – turkeyhundt Oct 22 '16 at 22:28
• Well, this might not sould convincing but there are 3 numbers showing and 3 numbers not showing. Any outcome is equally likely. So six being one of the numbers showing is equally likely as six not being one of the numbers showing. – fleablood Oct 22 '16 at 22:33
To be unnecessarily thourough. There are $6*5*4=120$ ways for the faces to be different. (An arbitrary first die can have any face, the arbitrary second can have any of the five remaining, etc.)
There are $3(1*5*4)=60$ ways for one of the faces to be a $6$. (The face that is a 6, must be a six, the second can be any of 5 and the third any of 4, and there are 3 chooses for which die is $6$. So Probability is $60/120 = 1/2$.
2) The probability of the first die being a six is $1/6$. The probability of the second die being a six, and the first die not being a six, given the dies are different is $5/6*1/5 = 1/6$. So probability of one of the first two dice is six is $1/6 + 1/6 = 1/3$ The probability of the third die being six and neither of the first two, given that all there are different is $2/3*1/4 = 1/6$. So the probability of one the faces being six given they are all different is $1/3 + 1/6 = 1/2$.
3)All combinations of different numbers are equally likely. $3$ numbers appear. $3$ do not. Each is equally likely so that a six appears (or not) is 1/2.
To justify the answer formally you could use what you know about conditional probability. In particular:
Let $S$ be the event that at least one six occurs, and $D$ that all three dice show different faces. The outcome of our experiment is a triple $(a,b,c)$, where $a$, $b$, $c$ are the faces showed by the first, the second and the third dice respectively. There are $N = 6^3$ such triples. In addition, there are $6 \cdot 5 \cdot 4$ ways of choosing $(a,b,c)$ such that no face is repeated therein, so $P(D) = \frac {6 \cdot 5 \cdot 4}{N}$. Let us now consider the event $S \cap D$ (i.e. one dice shows six, and all of them have different outcomes); there are 3 ways to choose which dice shows six, and $5 \cdot 4$ to select the faces of the other two, in total: $3 \cdot 5 \cdot 4$ cases, which yields $P(S \cap D) = \frac{3 \cdot 5 \cdot 4}{N} > 0$. Finally, we obtain the probability in question: $$P(S|D) = \frac {P(S \cap D)}{P(D)} = \frac {3 \cdot 5 \cdot 4}{6 \cdot 5 \cdot 4} = 0.5.$$
• I edited my post, providing more details. Is it OK? – Pythagoricus Oct 23 '16 at 0:34
• Yes, it's OK now. I haven't checked whether it is correct, but it's definitely an answer now. – Daniel Fischer Oct 23 '16 at 9:54
Hint: if the three dice show different faces after being rolled, then surely three different numbers are obtained. What is the probability that these include the number $6$?
If all three dice are different, imagine the 6 numbers from 1 to 6. Three of them have been rolled and three have not. So each number has a 50% chance of being one of the three rolled since no number has an inherent advantage over the others.
• But can I calculate it using a formula? – Mary Oct 22 '16 at 22:43
• One way would be $\frac{\text{# of ways to pick 3 numbers with one being a 6}}{\text{# of ways to pick 3 numbers}}$ which would be $\frac{5\choose2}{6\choose3}$ = $\frac{1}{2}$ – turkeyhundt Oct 22 '16 at 22:50 | 2021-06-16T16:01:17 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1980545/probability-of-obtaining-at-least-one-6-if-it-is-known-that-all-three-dice-showe",
"openwebmath_score": 0.6485030055046082,
"openwebmath_perplexity": 208.56465335149932,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9859363750229257,
"lm_q2_score": 0.8558511469672594,
"lm_q1q2_score": 0.843814777400113
} |
https://math.stackexchange.com/questions/3305139/where-do-we-need-absolute-values-when-solving-ut-fracu2t-utt | # Where do we need absolute values when solving $u'(t) = \frac{u^2(t) - u(t)}{t}$
Solve the differential equation $$u'(t) = \frac{u^2(t) - u(t)}{t}$$ and then solve the initial value problem with $$u(1) = \frac{1}{2}$$.
I know the general solution is given by $$u(t) = \frac{1}{c_1 t + 1}$$ for all $$t \in \mathbb{R}$$ and some $$c_1 \in \mathbb{R}$$ determined by the initial condition, which is not always positive.
My problem is that when I integrate $$\frac{1}{t}$$ when using separation of variables, is my result $$\ln|t|$$ or just $$\ln(t)$$?
Here's what I've done so far: We exclude the constant solutions $$u :\equiv 0$$ and $$u : \equiv 1$$ Then our differential equation is equivalent to $$\frac{u’(t)}{u^2(t) - u(t)} = \frac{1}{t}.$$ Integrating and substituting gives \begin{align*} \int_{u(t_0)}^{u(t)} \frac{1}{s(s - 1)} ds = \int_{t_0}^{t} \frac{1}{s} ds \implies & \int_{u(t_0)}^{u(t)} \frac{1}{s - 1} - \frac{1}{s} ds = \ln\left| \frac{t}{t_0} \right| \\ \implies & \ln\left|\frac{u(t_0) ( u(t) - 1 )}{u(t) ( u(t_0) - 1)} \right| = \ln\left| \frac{t}{t_0} \right| \\ \implies & \frac{| u(t) |}{|u(t) - 1|} = \frac{|t_0| |u(t_0)|}{|t| | u(t_0) - 1|} \end{align*} Do I now do a case distinction or how can I solve for $$u$$ now?
• You can say $u/(u-1)=Ct$ where $C$ is a constant. If, instead, $C$'s sign changed with $t$, it would have to change only when $t=0$ in order to retain continuity, and then I suspect it would need to not change at all to be differentiable at $t=0$. – runway44 Jul 27 at 2:07
• That was your original last equation without the absolute values and all the constants absorbed into $C$, except now it looks like your last equation got flipped upon editing. – runway44 Jul 28 at 20:48
• My equation was just your equation with all the constants absorbed to be $C$. But my first comment should have also taken into consideration which side of $0$ and $1$ that $u$ was on. – runway44 Jul 28 at 22:32
• @runway44 ok thanks I see it now but the argument still holds for $\frac{u - 1}{u} = C t$, right? – Viktor Glombik Jul 28 at 22:35
• Yes, thanks to LutzL's observation about solutions in the three intervals $(-\infty,0)$, $(0,1)$ and $(1,\infty)$. – runway44 Jul 28 at 23:08
The absolute values in your calculation where applied properly where they are needed. However, in the next step you can remove them as follows:
As you observed, there are constant solutions at $$u=0$$ and $$u=1$$. Due to uniqueness of solutions any solution starting in $$(0,1)$$ stays inside that interval, the same for $$(-\infty,0)$$ and $$(1,\infty)$$. Thus the signs of the expressions under the absolute values are constant and can be merged into the integration constant. Resp. in a formulation with the initial condition, those fractions $$\frac{u(t)}{u(t_0)}$$ and $$\frac{u(t)-1}{u(t_0)-1}$$ are always positive, so that the absolute value is equal to the term within.
The ODE has a singularity at $$t=0$$, thus the whole vertical line there is excluded from the domain of the ODE. Thus the domain of the solution with initial condition at $$t=1$$ is $$(0,\infty)$$, not the whole of $$\Bbb R$$. So also $$\frac{t}{t_0}$$ is positive over the domain of the solution.
Then you continue with (with the correct signs in the partial fraction decomposition) \begin{align} \frac{u(t_0)(u(t)-1)}{u(t)(u(t_0)-1)}&=\frac{t}{t_0}\\[.8em] u(t)\Bigl(t_0u(t_0)-t(u(t_0)-1)\Bigr)&=t_0u(t_0)\\[.8em] u(t)=\frac{t_0u(t_0)}{t_0u(t_0)+(1-u(t_0))t} \end{align}
• For your second part: The first task is to solve the differential equation without any initial condition. Do I then have to find two solutions, one for $t > 0$ and one for $t < 0$? – Viktor Glombik Jul 27 at 10:21
• In principle yes. In practice the formulas stay the same, you just have to observe that you have 6 connected domains separated by the lines $u=0$, $u=1$ and $t=0$ and at $t=0$ you leave (trivially) the domain of the initial point. While you can continue the solution over the line $t=0$, this continuation is arbitrary, even if there is a preferred continuation keeping also the derivative continuous. – Dr. Lutz Lehmann Jul 27 at 10:29
Hint.
$$(t u)' = u^2$$
now making $$v = t u$$
$$v' = \frac{v^2}{t^2}\Rightarrow \frac{dv}{v^2} = \frac{dt}{t^2}$$
or integrating
$$\frac 1v = \frac 1t + C\Rightarrow v = \frac{t}{C t+1} = u t$$
• As detailed in the other answer my approach is also correct and mandates the use of absolute values. By a uniqueness argument we can conclude that we can ignore these absolute values. Do you use that argument implicitly somewhere? I find it hard to believe that just by a substitution, we can completely get rid of the absolute values and not use that uniqueness argument. – Viktor Glombik Jul 27 at 12:41
• The substitution can be considered as a coordinates change. Under the new coordinates, the tangent space can have more or less folds. There are many cases in which the same phenomena occur concerning ODEs. – Cesareo Jul 27 at 13:12
• Please. Compare the stream plots for the $u$ and $v$ ODEs. – Cesareo Jul 27 at 13:54
• I'm sorry. Im just not no familiar with the terminology tangent space (at least not in the ODE setting, I only know tangent spaces of manifolds.) Would you mind explaining that to me? – Viktor Glombik Jul 27 at 14:07
• Perhaps not folds, but leaves of the foliation, branches of the covering etc. of the almost everywhere locally bijective coordinate transformation? This approach still needs to exclude the points with $v=0$ or $t=0$, and consider possible solutions through them separately. – Dr. Lutz Lehmann Jul 27 at 16:28
The maximal $$(t,u)$$-domain relevant to the given IVP is $$\Omega:={\mathbb R}_{>0}\times{\mathbb R}$$. Within $$\Omega$$ the standard existence and uniqueness theorem for ODEs is valid. By inspection one sees that there are the constant solutions $$u_0(t)\equiv0$$ and $$u_1(t)\equiv1$$ $$(t>0)$$. No other solution can cross the graphs of $$u_0$$ and $$u_1$$. Since the initial point $$\bigl(1,{1\over2}\bigr)$$ is given the solution $$t\mapsto u(t)$$ therefore stays in the $$u$$-interval $$\>]0,1[\>$$ for all $$t>0$$. Hence there are no case distinctions; you just have to select the primitive of $$s\mapsto{1\over s^2-s}$$ that is valid for $$0, namely \eqalign{\int{ds\over s^2-s}&=\int\left({1\over s-1}-{1\over s}\right)ds=\log|s-1|-\log |s| +C\cr &=\log{1-s\over s}+C\qquad(0
• I get that the IVP domain is $\mathbb{R}_{>0} \times \mathbb{R}$ but when just solving the differential equation first (as the exercise wants it) why can't $\mathbb{R}_{<0} \times \mathbb{R}$ be the domain? – Viktor Glombik Jul 28 at 13:38
• The set $\Omega':={\mathbb R}_{<0}\times{\mathbb R}$ is another domain for the given ODE, but the initial point $\bigl(1,{1\over2}\bigr)$ does not lie in $\Omega'$. – Christian Blatter Jul 28 at 14:46 | 2019-12-11T08:50:32 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3305139/where-do-we-need-absolute-values-when-solving-ut-fracu2t-utt",
"openwebmath_score": 0.9682651162147522,
"openwebmath_perplexity": 316.16972436225785,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.98593637543616,
"lm_q2_score": 0.855851143290548,
"lm_q1q2_score": 0.8438147741287765
} |
http://mathhelpforum.com/advanced-statistics/195267-how-compute-probability.html | # Math Help - how to compute this probability?
1. ## how to compute this probability?
I have taken this problem from the paper of MASTER OF COMMERCE Examination of university in India with little modifications.
If a machine is correctly set up it will produce 90% acceptable items. If it is incorrectly set up it will produce 40% acceptable items. Past 5 years experience shows 80% of the setups are correctly done. If after a certain setup
1) machine produces 3 acceptable items and 1 non-acceptable( defective ) item as the first 4 pieces, find the probability that machine is correctly setup.
2)machine produces 3 acceptable items as the first 3 pieces,find the probability that the machine is correctly setup.
I don't know what are the correct answers.So verification of these answers is needed.
2. ## Re: how to compute this probability?
Originally Posted by Vinod
I have taken this problem from the paper of MASTER OF COMMERCE Examination of university in India with little modifications.
If a machine is correctly set up it will produce 90% acceptable items. If it is incorrectly set up it will produce 40% acceptable items. Past 5 years experience shows 80% of the setups are correctly done. If after a certain setup
1) machine produces 3 acceptable items and 1 non-acceptable( defective ) item as the first 4 pieces, find the probability that machine is correctly setup.
2)machine produces 3 acceptable items as the first 3 pieces,find the probability that the machine is correctly setup.
I don't know what are the correct answers.So verification of these answers is needed.
By my calculations neither is right.
Tell us how you approached this and then we might be able to see where you went wrong.
CB
3. ## Re: how to compute this probability?
[TEX]
Originally Posted by CaptainBlack
By my calculations neither is right.
Tell us how you approached this and then we might be able to see where you went wrong.
CB
Firstly, I am giving calculations for the second answer
Let us define the events:
$A_1=$The set up was correct.
$A_2=$The set up was wrong.
E=The item is acceptable.
D=The item is not acceptable (defective)
We know $P(A_1)=$Probability that the set up was correct =0.8
$P(A_2)=$Probability that the set up was wrong =0.2
$P(E|A_1)=$Probability the item is acceptable given the information that the set up is correct =0.9
$P(E|A_2)=$=0.4
We are required to find the probability that the set up is correct given that the 3 items are acceptable, i-e
$P(A_1|E)=\frac{P(E|A_1)*P(A_1)^3}{P(E|A_1)*P(A_1)^ 3+P(E|A_2)*P(A_2)^3}$
$=\frac{0.9*(0.8)^3}{0.9*(0.8)^3+0.4*(0.2)^3}$=0.99310345
From the above calculations, i think you will know the calculations i have made for the first answer.
4. ## Re: how to compute this probability?
Originally Posted by Vinod
[TEX]
Firstly, I am giving calculations for the second answer
Let us define the events:
$A_1=$The set up was correct.
$A_2=$The set up was wrong.
E=The item is acceptable.
D=The item is not acceptable (defective)
We know $P(A_1)=$Probability that the set up was correct =0.8
$P(A_2)=$Probability that the set up was wrong =0.2
$P(E|A_1)=$Probability the item is acceptable given the information that the set up is correct =0.9
$P(E|A_2)=$=0.4
We are required to find the probability that the set up is correct given that the 3 items are acceptable, i-e
$P(A_1|E)=\frac{P(E|A_1)*P(A_1)^3}{P(E|A_1)*P(A_1)^ 3+P(E|A_2)*P(A_2)^3}$
$=\frac{0.9*(0.8)^3}{0.9*(0.8)^3+0.4*(0.2)^3}$=0.99310345
From the above calculations, i think you will know the calculations i have made for the first answer.
The numerator should be :
$P(\text{3 acceptable}|\text{setup OK})P(\text{setup OK})=0.9^3\times 0.8$
and the denominator:
$P(\text{3 acceptable}|\text{setup OK})P(\text{setup OK})+P(\text{3 acceptable}|\text{setup NOT OK})P(\text{setup NOT OK})\\ \phantom{xxxxxxxx}=0.9^3\times 0.8+0.4^3\times 0.2$
CB
5. ## Re: how to compute this probability?
Originally Posted by CaptainBlack
The numerator should be :
$P(\text{3 acceptable}|\text{setup OK})P(\text{setup OK})=0.9^3\times 0.8$
and the denominator:
$P(\text{3 acceptable}|\text{setup OK})P(\text{setup OK})+P(\text{3 acceptable}|\text{setup NOT OK})P(\text{setup NOT OK})\\ \phantom{xxxxxxxx}=0.9^3\times 0.8+0.4^3\times 0.2$
CB
As my answer to the second question has been corrected, my answer to the first question is
$\frac{0.046656}{0.1192}$ Verification of answer is needed. | 2015-11-27T07:27:27 | {
"domain": "mathhelpforum.com",
"url": "http://mathhelpforum.com/advanced-statistics/195267-how-compute-probability.html",
"openwebmath_score": 0.8209627866744995,
"openwebmath_perplexity": 1091.136500121452,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9859363729567545,
"lm_q2_score": 0.8558511451289037,
"lm_q1q2_score": 0.8438147738192763
} |
https://math.stackexchange.com/questions/1333524/valid-am-gm-inequality-proof | # Valid AM-GM inequality proof?
We have to prove that $$\frac{(x_1+x_2+x_3+...+x_n)}{n} \geq (x_1\cdot x_2\cdot x_3\cdots x_n)^{1/n}$$
Attempt: Raising both sides to the nth power gives $\left(\frac{x_1+x_2+x_3+...x_n}{n}\right)^{n} \geq x_1x_2x_3...x_n$ This is equivalent to proving that if a sum of random numbers is a fixed number, then their maximum product occurs when all the number are equal. So we have to find which combination of numbers gives the maximum product if their sum is fixed.
Obviously $$\left({x_1+x_2+x_3+...x_n}\right)^{n} > x_1x_2x_3...x_n$$
then, a maximizer must exist.
Now suppose that we have this combination of products: $x_1x_2x_3...x_n$ If i take the arithmetic mean of any 2 numbers and replace them by their arithmetic mean, I will get a bigger product (By the simple 2 case AM-GM inequality which is easy to proof), obviously the arthimetic mean of 2 numbers will not change the sum of the numbers, we can write that:
$$\frac{x_1+x_2}{2}\cdot\frac{x_1+x_2}{2}\cdot x_3 \cdots x_n\ \geq x_1x_2x_3\cdots x_n$$
Which means if we have any combination of product of numbers, I could increase that product by making any 2 numbers equal (arthimetic mean) without changing the sum of the numbers.
Now supposing that the maximum product does not have all of the numbers equal, then I can increase its product by having 2 numbers equal, so this mean that it's not the maximum product, then, by contradiction, the maximum product must occur when all the numbers are equal (Sum divided by the number of numbers)
Which means that $$\left(\frac{x_1+x_2+x_3+...x_n}{n}\right)^{n} \geq x_1x_2x_3...x_n$$
Is this a valid proof? It's my first time here and I don't know how to do the math notations thingy on this website so forgive me, thanks.
• meta.math.stackexchange.com/questions/5020/… – Timbuc Jun 21 '15 at 9:59
• I fixed the LaTeX (math notation). Please look to my edit and to the tutorial in the link of Timbuc's comment, to learn it. – wythagoras Jun 21 '15 at 10:01
• It seems mathematically correct to me, though I would expand on why (x1+x2...xn)^n is greater than x1x2...xn. I assume you use the binomial formula to show this? – Faraz Masroor Jun 21 '15 at 14:14
• Thanks wythagoras, @FarazMasroor We have let $x_{1}+x_{2}+...x_n$ be a fixed number, so let's name it $M$. Since $M$ is bigger than any $x_n$, $M\cdot M\cdot M\cdot M...$ (n times) must be bigger than $x_1\cdot x_2\cdot x_3...x_n$ (n times) – Wejd Jun 22 '15 at 16:53
• That is another way to do it! – Faraz Masroor Jun 22 '15 at 21:55
That's pretty good.
One aspect I'd want to fiddle with is the existence of the maximizer. You argued that it's bounded (which as a commenter noted could be made more explicit) and so a maximizer exists; this requires some kind of compactness argument, and the continuity of $\mathbb R^n\to\mathbb R$, $(x_1,\dotsc,x_n)\to\prod_{k=1}^n x_k$. That's all fine, but it would be better to make it explicit.
If you don't want to use topological concepts like that, then the argument can be adjusted not to assume the existence of the maximizer, but to prove it. The idea is to repeatedly replace pairs of $x_i$ until they're all the same, and argue (as you did) that we can do this keeping the AM constant but increasing the GM. The only tricky bit is that, if you replace pairs of numbers with their average, this process might not ever stop. (Example: $(1,2,2)\to (\frac32,\frac32,2)\to (\frac32,\frac74,\frac74) \to\dotsm$.) So the replacement step has to be tweaked. The resulting proof can be found in a note by Dijkstra, EWD1140.
• That's really interesting, I actually thought about the continuity of the process,(Replacing 2 numbers in the set with their arithmetic mean), each time you do this process you increase the product,and since the product is bounded, then, this process must "stop" somewhere which means that it must converge, and if it converges,then the product of the previous process compared to the next process must be equally the same somewhere in infinity, thus if i take the mean of any 2 numbers, the product must not change, then those 2 numbers must be equal, and this is satisfied if all of them are equal – Wejd Jun 22 '15 at 17:09
• I did not really understand Dijkstra's proof but it's good to know the name of the problem – Wejd Jun 22 '15 at 17:11
• Perhaps instead of replacing the numbers with the average you replace one with the average of all the numbers, and the other accordingly to keep the sum constant. – Faraz Masroor Jun 22 '15 at 21:57
• @FarazMasroor Indeed, that is what Dijkstra does. – user21467 Jun 22 '15 at 23:50
Hint: Your idea is correct. Here is a north to let things rigorous.
Consider $f, \psi : U \to \mathbb R$ defined as $$f(x) = x_1 \cdot x_2 \cdots x_n \,\,\, \text{and} \,\,\, \psi (x) = x_1 + x_2 + \ldots + x_n$$
for all $x = (x_1, x_2, \ldots , x_n) \in U$. Fix $s > 0$ and search for the critical points of $f|_M$ where $M = \psi^{-1}(s)$.
Observe that $\mathrm{grad} \, \psi (x) = (1,1,\ldots, 1)$ for any $x \in U$ and $\mathrm {grad}\, f (x) = (\alpha_1 , \ldots , \alpha_n)$, with $a_i = \prod_{j\neq i} x_j$. Show that the maximum is necessarily in $M$ and is the only critical point of $f|_M$.
• And the downvote is because...? – Aaron Maroja Jun 21 '15 at 14:24
• Other users than the downvoter themself can only speculate what was the reason for the downvote. But if you wish to discuss the reason for the downvote, there is a chatroom explicitly for this purpose. – Martin Sleziak Jun 21 '15 at 16:44
• The problem is, the approach is correct, though people will look at this and think otherwise. – Aaron Maroja Jun 21 '15 at 16:59
• No idea who downvoted, I know one can prove it through gradients with your approach :) but I tried to make the proof having mathematics as simple as possible – Wejd Jun 22 '15 at 16:33
• Well, not in English ;) – user940 Jul 1 '15 at 22:53 | 2019-06-17T05:25:54 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1333524/valid-am-gm-inequality-proof",
"openwebmath_score": 0.7997465133666992,
"openwebmath_perplexity": 277.8748737868687,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9859363737832232,
"lm_q2_score": 0.855851143290548,
"lm_q1q2_score": 0.8438147727141087
} |
https://math.stackexchange.com/questions/2343520/conditional-probability-with-coins-edited-with-progress/2343541 | # Conditional Probability with coins (Edited with Progress)
You have two coins that look identical, but one of them is fair and the other is weighted. The weighted coin has a 3/4 probability to flip heads and a 1/4 probability to flip tails. You forget which coin is which, so you test each coin by flipping it once. If one coin flips heads while the other flips tails, you will guess that the coin that flipped heads is the weighted coin. If both coins flip the same result, then you will guess at random which coin is weighted. Using this method, what is the probability that you will guess correctly?
I already know the answer is 5/8, but my textbook didn't offer a great solution, so I would like to see how other people solve it.
Progress: Let P(X) = P(Fair Head intersect Unfair Head) + P(Fair Tail intersect Unfair Tail) = 1/2. The complement of P(X) is also 1/2. The probability of guessing given X occurs is 1/2. The probability of guessing given X does not occur is 3/4. Therefore the probability of guessing is (1/2)(1/2) + (1/2)(3/4) = 5/8.
• Through casework obviously. Also, why the down vote? – Sanjoy Kundu Jul 2 '17 at 1:46
• I'll edit my progress, could you remove the down vote accordingly? – Sanjoy Kundu Jul 2 '17 at 1:48
• Edited with progress now. – Sanjoy Kundu Jul 2 '17 at 1:52
I think you could use a good bit more detail ...
$FH$: fair coin comes up heads. $P(FH) =\frac{1}{2}$
$FT$: fair coin comes up tails. $P(FT) =\frac{1}{2}$
$UH$: unfair coin comes up heads. $P(UH) =\frac{3}{4}$
$UT$: unfair coon comes up tails. $P(UT)=\frac{1}{4}$
$X$: same outcome. $X=(FH \cap UH) \cup (FT \cap UT)$.
$$P(X)= P(FH \cap UH) \cup (FT \cap UT) = \text{ (mutually exclusive events) }$$
$$=P(FH \cap UH) + P(FT \cap UT) = \frac{1}{2}*\frac{3}{4}+\frac{1}{2}*\frac{1}{4}=\frac{1}{2}$$
Hence, $P(X^C)=1-P(X)=\frac{1}{2}$
$G$: guess correctly
$P(G|X)=\frac{1}{2}$ since if both come up same then random pick.
$$P(G|X^C) = \frac{P(FT \cap UH)}{P(X)}=\frac{\frac{1}{2}*\frac{3}{4}}{\frac{1}{2}}=\frac{3}{4}$$
So:
$$P(G) = P(G \cap X) +P(G \cap X^C)= P(X)*P(G|X)+P(X^C)*P(G|X^C)=$$
$$\frac{1}{2}*\frac{1}{2}+\frac{1}{2}*\frac{3}{4}=\frac{5}{8}$$
Let the probability of flipping a heads for the fair and unfair coins be $p_1=\frac{1}{2}$ and $p_2=\frac{3}{4}$ respectively.
Let 'SR' be taken to mean 'same results', 'DR' for 'different results' and 'CG' for 'correct guess'.
Flipping the coins,
$$P(\textrm{SR}) = p_1p_2 + (1-p_1)(1-p_2)=\frac{1}{2}$$
$$P(\textrm{DR}) =\frac{1}{2}$$
Now,
$$P(\textrm{CG})=P(\textrm{CG and DR})+P(\textrm{CG and SR})$$
$$=P(\textrm{CG}\,|\,\textrm{DR})P(\textrm{DR})+P(\textrm{CG}\,|\,\textrm{SR})P(\textrm{SR})$$
$$=P(\textrm{unfair is Heads}\,|\,\textrm{DR}).\frac{1}{2}+\frac{1}{2}.\frac{1}{2}$$
$$={\frac{3}{4}.\frac{1}{2}\over\frac{1}{2}}.\frac{1}{2}+\frac{1}{2}.\frac{1}{2}$$
$$=\frac{5}{8}$$
$P(\text{the same})=\frac12$ and $P(\text{guess right }|\text{ the same})=\frac12$
$P(\text{different})=\frac12$ and
\overbrace{\left.\begin{align} \text{weighted heads: }\tfrac34\cdot\tfrac12&=\tfrac38\text{ (guess right)}\\ \text{weighted tails: }\tfrac14\cdot\tfrac12&\textstyle=\tfrac18\text{ (guess wrong)} \end{align}\right\} P(\text{guess right }|\text{ different})=\frac34}^{\text{assuming different}}
\begin{align} P(\text{guess right}) &=P(\text{the same})\cdot P(\text{guess right }|\text{ the same})\\ &+P(\text{different})\cdot P(\text{guess right }|\text{ different})\\[6pt] &=\frac12\cdot\frac12+\frac12\cdot\frac34\\[6pt] &=\frac58 \end{align} | 2019-08-22T07:53:26 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2343520/conditional-probability-with-coins-edited-with-progress/2343541",
"openwebmath_score": 0.8739122152328491,
"openwebmath_perplexity": 2760.919813691168,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9859363733699888,
"lm_q2_score": 0.8558511414521922,
"lm_q1q2_score": 0.8438147705479397
} |
https://math.stackexchange.com/questions/2642323/probability-that-warehouse-falls-behind-schedule | # Probability that warehouse falls behind schedule
A warehouse uses three machines ($$m_1$$, $$m_2$$ and $$m_3$$) and their failure rate is 0.02, 0.03 and 0.04.
a) Find the probability that two or more machines fail.
b) The warehouse falls behind schedule if at least one of the following happens: (1) $$m_1$$ fails; (2) $$m_2$$ and $$m_3$$ both fail. What is the probability that the warehouse falls behind schedule?
c) Following the scenario from b), given that $$m_3$$ fails what is the probability that the warehouse falls behind schedule?
What I have so far:
(a) $$P($$two or more machines fail $$) = P(m_1$$ and $$m_2$$ fail, $$m_3$$ doesn't$$) + P(m_1$$ and $$m_3$$ fail, $$m_2$$ doesn't$$)+ P(m_2$$ and $$m_3$$ fail, $$m_1$$ doesn't$$) + P(m_1,m_2$$ and $$m_3$$ fail$$)$$ $$=(0.02)(0.03)(1-0.04)+(0.02)(1-0.03)(0.04)+(1-0.02)(0.03)(0.04)+(0.02)(0.03)(0.04)=0.002552$$
(b) $$P(m_1\ fails)=(0.02)(1-0.03)(1-0.04)+(0.02)(0.03)(1-0.04)+(0.02)(1-0.03)(0.04)=0.019976$$ $$P(m_2\ and\ m_3\ fail)=(1-0.02)(0.03)(0.04)=0.001176$$ $$P(m_1,m_2\ and\ m_3\ fail)=(0.02)(0.03)(0.04)=0.000024$$ $$P(warehouse\ falls\ behind\ schedule) = 0.019976 + 0.001176 + 0.000024 = 0.021176$$
I'm confident that my answer for (a) is correct but I'm not sure about (b). I don't know how to solve (c) at all
• Your logic for $a$ appears to be incorrect. the event "$m_1$ and $m_2$ fail " (for one) is contained in the event "all three fail" so you are badly overcounting the latter scenario.
– lulu
Feb 8, 2018 at 20:03
• So would it just be: $P($two or more machines fail $) = P(m_1$ and $m_2$ fail$) + P(m_1$ and $m_3$ fail$)+ P(m_2$ and $m_3$ fail$) + P(m_1,m_2$ and $m_3$ fail$)$ $=(0.02)(0.03)+(0.02)(0.04)+(0.03)(0.04)+(0.02)(0.03)(0.04)$? Feb 8, 2018 at 20:06
• That's the same as what you have above and it is incorrect for the reason I gave before.
– lulu
Feb 8, 2018 at 20:13
• If you want disjoint events (not a bad way to go) then take: "$m_1, m_2$ fail, $m_3$ doesn't" , "$m_1, m_3$ fail, $m_2$ doesn't", "$m_2, m_3$ fail, $m_2$ doesn't", "All three fail".
– lulu
Feb 8, 2018 at 20:17
• I thought, it was implied that the third machine doesn't fail if I didn't write it. I'll change that. Feb 8, 2018 at 20:20
A warehouse uses three machines ($m_1, m_2, m_3$) and their respective failure rates are $0.02$, $0.03$, and $0.04$. What is the probability that two or more machines fail?
Assuming independence, your answer to this question is correct.
The warehouse falls behind schedule if at least one of the following happens: (1) $m_1$ fails or (2) $m_2$ and $m_3$ both fail. What is the probability that the warehouse falls behind schedule.
Let $F_k$ be the event that machine $k$ fails. Then the probability that the warehouse falls behind schedule is $$P(F_1 \cup (F_2 \cap F_3)) = P(F_1) + P(F_2 \cap F_3) - P(F_1 \cap F_2 \cap F_3)$$ Assuming independence, we obtain \begin{align*} P(F_1 \cup (F_2 \cap F_3)) & = 0.02 + (0.03)(0.04) - (0.02)(0.03)(0.04)\\ & = 0.02 + 0.0012 - 0.000024\\ & = 0.021176 \end{align*}
Given that $m_3$ fails, what is the probability that the warehouse falls behind schedule.
If $m_3$ fails, the warehouse falls behind schedule if at least one of the other machines fails.
Assuming independence, the desired probability is \begin{align*} P(F_1 \cup F_2) & = P(F_1) + P(F_2) - P(F_1 \cap F_2)\\ & = 0.02 + 0.03 - (0.02)(0.03)\\ & = 0.02 + 0.03 - 0.0006\\ & = 0.0494 \end{align*} More formally, the probability that the warehouse falls behind schedule is \begin{align*} P((F_1 \cap F_3) \cup (F_2 \cap F_3) \mid F_3) & = \frac{P((F_1 \cap F_3) \cup (F_2 \cap F_3))}{P(F_3)}\\ & = \frac{P(F_1 \cap F_3) + P(F_2 \cap F_3) - P(F_1 \cap F_2 \cap F_3)}{P(F_3)} \end{align*} If we assume independence, we obtain \begin{align*} P((F_1 \cap F_3) \cup (F_2 \cap F_3) \mid F_3) & = \frac{(0.02)(0.04) + (0.03)(0.04) - (0.02)(0.03)(0.04)}{0.04}\\ & = 0.02 + 0.03 - (0.02)(0.03)\\ & = 0.0494 \end{align*} | 2022-06-28T04:17:42 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2642323/probability-that-warehouse-falls-behind-schedule",
"openwebmath_score": 0.8968461751937866,
"openwebmath_perplexity": 286.66599741949994,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9881308817498916,
"lm_q2_score": 0.8539127510928476,
"lm_q1q2_score": 0.8437775596748512
} |
https://math.stackexchange.com/questions/1280254/relevance-of-prime-being-divisble-by-4k1-in-proof-that-there-are-infinitely | # Relevance of prime being divisble by $4k+1$ in proof that 'There are infinitely many primes of the shape $4k+3$'
Show that there are infinitely many primes of the shape $4k+3$
Proof:
$1)$ Suppose that there are only finitely many such primes, say $p_1,...p_n$.
$2)$ Consider the integer $Q=4p_1...p_n-1$
$3)$ The integer $Q$ is odd and of the shape $4k+3$ so cannot be divisible exclusively by primes of the shape $4k+1$
$4)$ Moreover, none of the primes $p_1,...,p_n$ divide $Q$
$5)$ Thus $Q$ is divisible by a new prime of the shape $4k+3$ not amongst $p_1,...,p_n$
$6)$ Contradicting our hypothesis, therefore there are infinitely many primes of the form
Contention:
Why do we have step $3)$ in the proof? Surely all we need to show is that none of the primes $p_1,...,p_n$ divide this new prime.
Even if the proof is complete without this step $3)$, why is it even there?
What relevance is it that it cannot be divisible by primes of the shape $4k+1$?
• $Q$ can be - and generally is - divisible by some primes of the form $4k+1$. The point is that it is impossible that all prime divisors of $Q$ have the form $4k+1$. – Daniel Fischer May 13 '15 at 13:17
• The point is that we want to find some prime of the form $4k + 3$ not $p_1, p_2, ..., p_n$. To do this, we show that $Q$ is divisible by at least one prime of the form $4k + 3$ and that none of $p_i$s divide $Q$. Step $(3)$ provides a proof of the fact that $Q$ is divisible by at least one prime of the form $4k + 3$, since the only odd primes are either of the form $4k + 1$ or of the form $4k+3$. – Balarka Sen May 13 '15 at 13:20
Since you are looking for a new prime of form $4k+3$, you need to make sure that you have a candidate amongst the factors. If all the factors were of the form $4k+1$ you would have some larger primes than you had before, but they wouldn't be of the right form, and the proof would fail.
Fortunately it is east to show that the product of numbers of the form $4k+1$ has the same form, so we see that we must be able to find a new prime of form $4k+3$.
To see why this is important, try it the other way around - if we added $1$ rather than subtracting $1$ in order to try to find an infinite number of primes of form $4k+1$, we'd run into a problem. For example $3\times 7=21$ and $21$ has the form $4k+1$ - but it doesn't have a prime factor of that form, so we can't assume one exists in general.
We need step 3 in order to conclude that $Q$ is divisible by at least one prime of the form $4k+3$ (prime divisors of the form $4k+1$ can appear, but the important thing is that at least one prime of the other form appear). | 2019-05-19T23:06:28 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1280254/relevance-of-prime-being-divisble-by-4k1-in-proof-that-there-are-infinitely",
"openwebmath_score": 0.9186034202575684,
"openwebmath_perplexity": 60.44080441619102,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9881308800022472,
"lm_q2_score": 0.8539127492339909,
"lm_q1q2_score": 0.8437775563457217
} |
https://math.stackexchange.com/questions/3425379/probability-of-impurity-given-3-out-of-5-test-detections | # Probability of Impurity Given 3 out of 5 test detections
A chemist is interested in determining whether a certain trace impurity is present in a product. An experiment has a probability of 0.80 of detecting the impurity if it is present. The probability of not detecting the impurity if it is absent in 0.90. The probabilities of the impurity being present and being absent are 0.40 and 0.60 respectively. Five separate experiments result in only three detections. What is the probability that the impurity is present?
Can someone help me get on the right track?
What I got out of this question is: $$T^+$$ = detection, $$T^-$$= no detection, I = impure, P = pure. $$P(T^+|I)$$ = 0.8 which implies $$P(T^-|I)$$ = 0.2; $$P(T^-|P)$$ = 0.9 which implies $$P(T^+|P)$$ = 0.1; P(I) = 0.4; P(P)=0.6
I assumed I'd need the reversed conditional probability of $$P(I|T^+)$$ and $$P(I|T^-)$$.
$$P(T^+)$$ = (0.8)(.4) + (0.1)(0.6) = 0.38 by law of total probability, and therefore $$P(T^-)$$ = 0.62.
$$P(I|T^+)$$ = $$\frac{P(T^+|I)P(I)}{P(T^+)} = \frac{(0.8)(0.4)}{0.38} = \frac{16}{19}$$
By similar computation, $$P(I|T^-)$$ = $$\frac{3}{19}$$
Final computation would be binomial distribution of $${5}\choose{3}(\frac{16}{19})^3(\frac{3}{19})^2$$
However this ends up not matching with the answer I'm given (studying for a test). Can anyone tell me where I've messed up or if my approach is all wrong?
• You can only use the binomial distribution when there are only two outcomes in an experiment, like rolling a die and using P(lower than three) and P(three or above). P$(I|T^+)\,$ and P$(I|T^-)\,$ don't qualify as such, though for example P$(T^+|I)\,$ and P$(T^-|I)\,$ would.
– A.J.
Nov 7, 2019 at 7:29
The probability of having an impurity present and getting three detections in five experiments is
$$\text{P}(I \,{\small\text{ AND }} \, 3 {\small\text{ out of }}5)=\left(\frac{2}{5}\right) \cdot \binom{5}{3} \left(\frac{4}{5}\right)^3\left(\frac{1}{5}\right)^2 = \frac{256}{3125}$$
The probability of not having an impurity present and getting three detections in five experiments is
$$\text{P}(P \,{\small\text{ AND }} \, 3 {\small\text{ out of }}5)=\left(\frac{3}{5}\right) \cdot \binom{5}{3} \left(\frac{1}{10}\right)^3\left(\frac{9}{10}\right)^2 = \frac{243}{50000}$$
We are looking for $$\,\text{P}(I \, | \, 3 {\small\text{ out of }}5).$$
\begin{align} \text{P}(I \, | \, 3 {\small\text{ out of }}5) &= \frac{\text{P}(I \,{\small\text{ AND }} \, 3 {\small\text{ out of }}5)}{\text{P}(3 {\small\text{ out of }}5)}\\[2ex] &= \frac{\dfrac{256}{3125}}{\dfrac{256}{3125} + \dfrac{243}{50000}}\\[2ex] &= \boxed{\frac{4096}{4339}} \end{align}
• Thanks for the clarification!! Can I ask, how am I from that wording know the difference between conditional and intersection? It seems to me like this is the sensitivity of a test which is conditional? Nov 7, 2019 at 18:22
• This IS a conditional probability but it requires some intersection calculations first. The formula for the probability of $\,A\,$ given $\,B\,$ is $$\text{P}(A \, | \, B) = \frac{\text{P}(A \, \cap \, B)}{\text{P}(B)}$$ However, the denominator often needs to be extended as in the next comment:
– A.J.
Nov 8, 2019 at 5:35
• \begin{align} \text{P}(A_i \, | \, B) &= \frac{\text{P}(A_i \, \cap \, B)}{\text{P}(B \, \cap \, A_1)+\text{P}(B \, \cap \, A_2)+\text{P}(B \, \cap \, A_3)+\dots} \\[1ex] &= \frac{\text{P}(A_i \, \cap \, B)}{\text{P}(A_1)\text{P}(B \, | \, A_1)+\text{P}(A_2)\text{P}(B \, | \, A_2)+\text{P}(A_3)\text{P}(B \, | \, A_3)+\dots} \end{align} Here the $\,A_i$'s are a partition of the event space, i.e. exactly one of them has to be true at any given time.
– A.J.
Nov 8, 2019 at 5:36
• In this particular question, the partition would be $\,P\,$ and $\,I\,$, as exactly one of those must be true. So the extended formula for the conditional probability would be \begin{align} \text{P}(I \, | \, 3 {\small\text{ out of }}5) &= \frac{\text{P}(I \, \cap \, 3 {\small\text{ out of }}5)}{\text{P}(3 {\small\text{ out of }}5)} \\ &= \frac{\text{P}(I \,\cap \, 3 {\small\text{ out of }}5)}{\text{P}(3 {\small\text{ out of }}5 \cap P)+\text{P}(3 {\small\text{ out of }}5 \cap I)} \end{align}
– A.J.
Nov 8, 2019 at 5:47
• This is great! I realized I was looking for the wrong probability and then using that to calculate for the 3 out of 5 instead of focusing on that first. Thanks for the visualization, it feels like maybe setting up a sample space might help me wrap my head around this. I appreciate all your help! Nov 8, 2019 at 6:02
$$P(I|T^-)$$ is wrong.
$$P(I|T^-)=\frac{P(I\cap T^-)}{P(T^-)}=\frac{0.4\cdot 0.2}{0.4\cdot 0.2+0.6\cdot 0.9}=\frac{4}{31}$$
and you can guess from here that the subsequent use of binomial distribution is wrong too - you can't use it this way.
• Thanks for pointing this out! I was thinking $1 = P(A|B) + P(A|B^C)$ instead of $1 = P(A|B) + P(A^c|B)$ Nov 8, 2019 at 6:05 | 2022-12-07T04:42:13 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3425379/probability-of-impurity-given-3-out-of-5-test-detections",
"openwebmath_score": 0.9984814524650574,
"openwebmath_perplexity": 709.5813855772417,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9881308807013052,
"lm_q2_score": 0.8539127455162773,
"lm_q1q2_score": 0.8437775532690686
} |
http://mathhelpforum.com/statistics/93987-more-probability.html | 1. ## More Probability
I never took Stats before, so I'm just trying to get a grasp of statistics and make sure I understand all of it.
A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs are randomly selected.
(c) What is the probability that one bulb of each type is selected?
$
P = {{\binom{4}{1}} {\binom{5}{1}} {\binom{6}{1}}}/{\binom{15}{3}} = .264
$
(d) Suppose now that bulbs are to be selected one by one until a 75-W bulb is found. What is the probability that it is necessary to examine at least six bulbs?
$
P = {{\binom{6}{0}}{\binom{9}{6}}}/{\binom{15}{6}} = .017
$
Hope I did these right! Part d is especially tricky so I wouldn't be surprised if I tripped over on that one.
2. Originally Posted by hansel13
I never took Stats before, so I'm just trying to get a grasp of statistics and make sure I understand all of it.
A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs are randomly selected.
(c) What is the probability that one bulb of each type is selected?
$
P = {{\binom{4}{1}} {\binom{5}{1}} {\binom{6}{1}}}/{\binom{15}{3}} = .264
$
(d) Suppose now that bulbs are to be selected one by one until a 75-W bulb is found. What is the probability that it is necessary to examine at least six bulbs?
$
P = {{\binom{6}{0}}{\binom{9}{6}}}/{\binom{15}{6}} = .017
$
Hope I did these right! Part d is especially tricky so I wouldn't be surprised if I tripped over on that one.
(c) looks OK.
To do (d), I'd suggest calculating 1 - Pr(select 5 or less bulbs to get a 75 watt).
Note that Pr(select 5 or less bulbs to get a 75 watt) = Pr(1 bulb) + Pr(2 bulbs) + Pr(3 bulbs) + Pr94 bulbs) + Pr(5 bulbs).
Pr(1 bulb) = 6/15.
Pr(2 bulbs) = (9/15)(6/14).
Pr(3 bulbs) = (9/15)(8/14)(6/13)
Pr(4 bulbs) = (9/15)(8/14)(7/13)(6/12)
etc.
3. Hello, hansel13!
For part (d), I had to crank out an exhaustive list.
A box contains four 40w lightbulbs, five 60w bulbs, and six 75w bulbs.
(d) Suppose now that bulbs are to be selected one by one until a 75w bulb is found.
What is the probability that it is necessary to examine at least six bulbs?
There are six 75w bulbs and nine Others.
We want the probability of finding a 75w bulb in 6, 7, 8, 9 or 10 draws.
Six draws
The first five draws are Others: . ${9\choose5}\text{ ways }\hdots \:P(\text{5 Others}) \:=\:\frac{{9\choose5}}{{15\choose5}}$
And the 6th is a 75w: . $\frac{6}{10}$
. . Hence: . $P(\text{6 draws}) \;=\;\frac{{9\choose5}}{{15\choose5}}\left(\frac{6 }{10}\right)$
Seven draws
The first six draws are Others: . ${9\choose6}\text{ ways } \hdots\:P(\text{6 Others}) \:=\:\frac{{9\choose6}}{{15\choose6}}$
And the 7th is a 75w: . $\frac{6}{9}$
. . Hence: . $P(\text{7 draws}) \;=\;\frac{{9\choose6}}{{15\choose6}}\left(\frac{6 }{9}\right)$
Eight draws
The first seven draws are Others: . ${9\choose7}\text{ ways }\hdots\:P(\text{7 Others}) \:=\:\frac{{9\choose7}}{{15\choose7}}$
And the 8th is a 75w: . $\frac{6}{8}$
. . Hence: . $P(\text{8 draws}) \;=\;\frac{{9\choose7}}{{15\choose7}}\left(\frac{6 }{8}\right)$
Nine draws
The first eight draws are Others: . ${9\choose8}\text{ ways }\hdots\:P(\text{8 Others}) \:=\:\frac{{9\choose8}}{{15\choose8}}$
And the 9th is a 75w: . $\frac{6}{7}$
. . Hence: . $P(\text{9 draws}) \;=\;\frac{{9\choose8}}{{15\choose8}}\left(\frac{6 }{7}\right)$
Ten draws
The first nine draws are Others: . ${9\choose9} \:=\:1\text{ way }\hdots \:P(\text{9 Others}) \:=\:\frac{1}{{15\choose9}}$
And the 10th is a 75w: . $\frac{6}{6} \:=\:1$
. . Hence: . $P(\text{10 draws} \:=\:\frac{1}{{15\choose9}}$
Note: I hope my reasoning is correct.
. . . . .I'm quite able to overcount and/or miss cases.
.
4. Thanks for the input.
I got .0503496 from mr fantastic.
and .04195804 from Soroban.
Still a little confused as well.
Wouldn't the following just work?:
$
P = {{\binom{5}{0}}{\binom{9}{5}}}/{\binom{15}{5}} = .042
$
5. Originally Posted by hansel13
Still a little confused as well.
Wouldn't the following just work?:
$
P = {{\binom{5}{0}}{\binom{9}{5}}}/{\binom{15}{5}} = .042
$
Actually that does work!
It may have been clearer to say $
P = {{\binom{6}{0}}{\binom{9}{5}}}/{\binom{15}{5}} = .042
$
.
Or even better $
P = {\binom{9}{5}}/{\binom{15}{5}} = .042
$
Than is the probability that none of the 75's is found among the first 5.
6. Originally Posted by Plato
Actually that does work!
It may have been clearer to say $
P = {{\binom{6}{0}}{\binom{9}{5}}}/{\binom{15}{5}} = .042
$
.
Or even better $
P = {\binom{9}{5}}/{\binom{15}{5}} = .042
$
Than is the probability that none of the 75's is found among the first 5.
Right, I meant to put ${\binom{6}{0}}$ instead of ${\binom{5}{0}}$.
Thanks for the help, I think I'm getting this stuff now.
7. Originally Posted by hansel13
Thanks for the input.
I got .0503496 from mr fantastic. Mr F says: Then you made an arithmetic error because it does give 0.042 (correct to three decimal places).
and .04195804 from Soroban.
Still a little confused as well.
Wouldn't the following just work?:
$
P = {{\binom{5}{0}}{\binom{9}{5}}}/{\binom{15}{5}} = .042
$
.. | 2016-08-25T22:31:28 | {
"domain": "mathhelpforum.com",
"url": "http://mathhelpforum.com/statistics/93987-more-probability.html",
"openwebmath_score": 0.8197942972183228,
"openwebmath_perplexity": 2322.0345284433115,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9597620585273153,
"lm_q2_score": 0.8791467770088163,
"lm_q1q2_score": 0.8437717204496361
} |
https://math.stackexchange.com/questions/3828745/definition-of-the-outer-measure-and-the-outer-measure-of-an-interval | # Definition of the outer measure and the outer measure of an interval.
I'm taking a course in Real Analysis covering Measure Theory. We are using Real Analysis by Royden/Fitzpatrick as the text. I'm going through the book and I'm struggling to understand the approach of Proposition 1 which states: The outer measure of an interval is it's length. This makes me think I'm misunderstanding the definition of the outer measure.
I understand the first part which tells us that if we have an interval $$[a,b]$$, let $$\epsilon > 0$$ be given. Then $$[a,b] \subseteq (a- \epsilon, b+ \epsilon)$$ so $$m^*([a,b]) \leq \ell((a- \epsilon, b+ \epsilon) = b - a + 2\epsilon$$. Since this holds for any $$\epsilon > 0$$ then $$m^*([a,b]) \leq b-a$$.
My question is: Why can't we use the same trick to show the other inequality? We have that $$(a - \epsilon, b + \epsilon)$$ is a cover of $$[a,b]$$. Then $$\ell(a-\epsilon ,b+\epsilon) = b - a + 2\epsilon \geq b-a$$. Since the outer measure, in this case, is defined as $$m^*([a,b]) = \inf \left\{ \sum_{k=1}^{\infty} \ell (I_k) : [a,b] \subseteq \cup_{k=1}^{\infty} I_k \right\},$$ then why can't we say $$b-a + 2\epsilon \geq m^*([a,b]) \geq b-a$$, by virtue of $$m^*([a,b])$$ being defined as the infimum?
We have an open cover that is bounded below by $$b-a$$, so can't we say that the the infimum over all open covers is bounded below by $$b-a$$? The textbook takes a different approach using the compactness of the interval. I understand the proof of that. I just don't understand we we have to take that approach. This makes me think I'm misunderstanding something subtle about the definition of the outer measure. If someone could help clarify the definition that would be greatly appreciated!
• You have to prove that $m^*([a,b]) \geq b-a$. The above doesn't prove it. Sep 16, 2020 at 16:25
By definition of the infimum: the infimum of a subset $$S$$ of a partially ordered set $$T$$, denoted $$\inf S$$ is the greatest element in $$T$$ that is less than or equal to all elements of $$S$$.
Applying that, you have no way to select a set of interval $$I_k$$ with $$[a,b] \subseteq \cup_{k=1}^{\infty} I_k$$ and conclude that $$m^*([a,b]) \ge b-a$$.
So I think the issue is that just because we found an open cover whose sum is bounded below by $$b-a$$ it doesn't mean that every open cover's sum is bounded below by $$b-a$$. We could possibly still find an open cover that is less that $$b-a$$. This is why the text approaches it in this way. You start with an arbitrary open cover, reduce it to a finite cover and show that it is larger than $$b-a$$. Since the given open cover is arbitrary, it holds for all open covers and thus the measure is also greater than or equal to $$b-a$$. | 2022-11-28T02:24:53 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3828745/definition-of-the-outer-measure-and-the-outer-measure-of-an-interval",
"openwebmath_score": 0.9328638315200806,
"openwebmath_perplexity": 41.54206455655335,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9597620539235895,
"lm_q2_score": 0.8791467564270272,
"lm_q1q2_score": 0.8437716966486652
} |
https://math.stackexchange.com/questions/899032/difference-between-lim-p-and-p-lim | # Difference between $\lim P[…]$ and $P[ \lim ]$
In a Galton-Watson branching process the extinction probability is sometimes given by $$\lim_{t \rightarrow \infty} P[X(t)=0]$$ and sometimes as $$P[\lim_{t \rightarrow \infty}X(t)=0]$$ Is there a difference between These two formulations? Which is the "correct" one, if yes?
Could you give me a link to a paper where this is written?
Here is the Wikipedia page, but unfortunately, the linked papers there do not help..
Thank you for your help
• In general, there is a difference between the two formulations. Unfortunately, I'm not familiar with Galton-Watson processes... – saz Aug 16 '14 at 10:16
• Literally, the former means the limit of a sequence of real numbers; while the latter means the probability of the convergence of a given sequence. So there should be a difference. – Megadeth Aug 16 '14 at 10:18
• I see that this Argument implies that it can't be the same exactly, as These are two different formulations, the one a squence of random variables, the other a sequence of probabilities. What I wonder whether the two Limits are then the same or not.. Does this also hold, that this is in General not the same? – user146358 Aug 16 '14 at 10:27
In the context of branching processes, the two statements are indeed equivalent. This is due to the following facts:
• The value of each random variable $X(t)$ is almost surely a nonnegative integer.
• The events $[X(t)=0]$ are nondecreasing with respect to $t$.
Thus, $P(X(t)=0)\to P(A)$ when $t\to\infty$, where $A=\bigcup\limits_t[X(t)=0]$. On the other hand, the event $[\lim\limits_{t\to\infty}X(t)=0]$ is also $[\exists t,X(t)=0]=A$, QED.
To sum up, the reason why the assertion holds is that, if $x:t\mapsto x(t)$ is an integer valued function such that if $x(t)=0$ then $x(s)=0$ for every $s\geqslant t$, then $x(t)\to0$ when $t\to+\infty$ if and only if $x(t)=0$ for some $t$.
• thank you for your answer. that helps a lot. Do you know a link to a paper, where the mathematical Definition of a Galton-Watson process is stated? All I can find is Wikipedia and the references there do not lead to formal definitions, either. – user146358 Aug 16 '14 at 12:57
• The WP page has a formal definition. Some recent introductory notes on the subject called "Probability on Trees: An Introductory Climb" are available on Yuval Peres webpage. – Did Aug 16 '14 at 13:03
• I know that Wikipedia has a formal Definition, and this is enough for me. But the Problem is, that I am not allowed to reference to Wikipedia. Thank you. – user146358 Aug 16 '14 at 13:05
• Quote: "the references there do not lead to formal definitions". – Did Aug 17 '14 at 9:20 | 2019-09-17T23:15:50 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/899032/difference-between-lim-p-and-p-lim",
"openwebmath_score": 0.9083709716796875,
"openwebmath_perplexity": 188.72910827596664,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9752018434079934,
"lm_q2_score": 0.8652240756264639,
"lm_q1q2_score": 0.8437681135119047
} |
https://math.stackexchange.com/questions/2252596/modular-arithmetic-for-arbitrary-number | # Modular arithmetic for arbitrary number
The problem describes as:
When the even integer $n$ is divided by $7$，the remainder is $3$. What's is The remainder when $n$ is divided by $14$.
My simple solution is:
$n=7x+3$ where $x$ is odd, so, we can define $x = 2m+1$, then $n = 7(2m+1) + 3 = 14m + 7 +3= 14m + 10$. So reminder is: $10$
As I am learning the mod system and was trying to solve this problem with modulo arithmetic. But got stuck with "how to think this prob in modulo system" ie. if $$n \equiv 3 \pmod{7}$$ then $$n \equiv \ ? \pmod {14}$$
Any help regarding solving steps and learning reference would be appreciated.
• "how to think this prob in modulo system" well it depends of the meaning of those words, but IMHO I think you did it nicely that way indeed. You converted a problem in $\pmod{7}$ to a problem in $\pmod{14}$. Maybe is just a question of formatting your answer as a "modulo system problem": $x = 2m+1$, then $n = 7(2m+1) + 3 \pmod{14} \equiv 14m + 7 +3 \pmod{14} \equiv 14m + 10 \pmod {14} \equiv 14m \pmod{14} + 10 \pmod {14} \equiv 10 \pmod{14}$. – iadvd Apr 26 '17 at 5:22
• Edited the question a bit for better understandability of my points that is, I was thinking the format: if n mod 7 = 3 then n mod 14 = ?. That's how I was thinking. Thanks for replying! – neo-nant Apr 26 '17 at 5:46
• ok I have added a solution going backwards... I think that you meant that possibility. But the way you did it is easier and indeed my solution is based on your substitutions (but backwards, it is kind of tricky). – iadvd Apr 26 '17 at 7:03
As I said in the comments I think is fine the way you did it. I would suggest you to use "congruent to" instead of "equal to" as in my comment from the beginning, or add it at the end of the conversions you did as a last step.
Said that, maybe your question is more related with this point: how you could make a solution starting backwards? from $14m+r$.
$$n=14m+r$$
Then, let us suppose three possibilities:
1. $r \lt 7$
2. $r = 7$
3. $7 \lt r \lt 14$ so we can define $r=7+r'$, where $r' \lt 7$
• For the first case:
$$n=14m+r \pmod{7} \equiv 2\cdot 7m\pmod{7} + r \pmod{7} \equiv 0+r \equiv 3$$
Thus:
$$r=3$$
But that is not possible because it means that $n = 14m + 3$, but $14m+3$ is not even, and we know that $n$ is even, so that solution is not possible.
• For the second case:
$$n=14m+7 \pmod{7} \equiv 2\cdot 7\pmod{7} + 7 \pmod{7} \equiv 0 \equiv 3$$
So it is impossible for the residue to be $r=7$ because $14m+7 \not \equiv 3 \pmod {7}$, the residue does not comply the premise, being $\equiv {3} \pmod{7}$, it is indeed $\equiv {0} \pmod {7}$.
• For the third case:
$$n=14m+r \pmod{7} \equiv 2\cdot 7\pmod{7} + 7 + r' \pmod{7} \equiv 0+0+r' \equiv 3$$
Thus finally the unique remaining valid option is $7 \lt r = 10 \lt 14$:
$$r'=3, r=7+r'=7+3=10$$ | 2019-09-23T14:00:36 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2252596/modular-arithmetic-for-arbitrary-number",
"openwebmath_score": 0.9407966732978821,
"openwebmath_perplexity": 360.21999107592063,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9752018376422666,
"lm_q2_score": 0.865224073888819,
"lm_q1q2_score": 0.8437681068287045
} |
https://math.stackexchange.com/questions/1985297/is-is-possible-to-find-a-basis-for-the-column-space-of-a-given-reduced-row-ech | # Is is possible to find a basis for the column space of $A$,given reduced row echelon form of matrix $A$ and $A^T$,
Suppose $A$ is a $3$x$4$ matrix and the reduced row echelon form of $A$ is $\begin{pmatrix}1&0&0&1\\0&1&2&2&\\0&0&0&0\end{pmatrix}$
and the reduced row echelon form of $A^T$ is $\begin{pmatrix}1&0&2\\0&1&-1\\0&0&0\\0&0&0\end{pmatrix}$
Find a basis for $R(A)$, where $R(A)$ is the column space of $A$
I don't think this is possible, but in the answer key, it said that $R(A)$ = $R(A^T)^T$, which has basis $$\{(1, 0, 2)^T,(0, 1, -1)^T\}$$
How does this work?
• You mean $R(A^T)^\perp$. $S^T$ doesn't mean anything unless $S$ is a matrix (or vector). – Omnomnomnom Oct 26 '16 at 1:10
• @Omnomnomnom What do you mean by $S$? – user59036 Oct 26 '16 at 1:15
• Oh, excuse me. I think your book really means $$R(A) = R[(A^T)^T]$$ I was confused without the extra brackets. Now it makes sense. – Omnomnomnom Oct 26 '16 at 1:15
• Remember that row-reduction does not change the row-space of a matrix – Omnomnomnom Oct 26 '16 at 1:16
• – BCLC Oct 26 '16 at 4:03
TL;DR
Row space of $A^T$ = column space of $A$
When we want to find a basis for the row space of a matrix $A$, we could use the the rows of $A$ except that it is not always the case that the rows of $A$ are linearly independent.
1. So we have to eliminate rows which can be written as linearly combinations of other rows.
2. Now we perform EROs on $A$ until we reach row-echelon form (or reduced row-echelon form) to get row vectors that, like the original matrix $A$, span the row space of $A$.
3. This time however, the row vectors (apart from the zero row vectors) that we get are linearly independent.
4. Thus, we have a basis for the row space of $A$.
For the column space of $A$, the procedure is the same as above if we replace 'row' with 'column'. ECOs however are difficult because we are used to addition vertically as in the EROs.
So instead of performing ECOs on $A$, we perform EROs on $A^T$.
This gives us row vectors (apart from the zero row vectors) that are linearly independent and span the row space of $A^T$, which is equivalent to the column space of $A$.
• do you agree with the commenter? – Anonymous Oct 26 '16 at 4:11
• @Anonymous Not sure about the first comment but Omnomnomnom seems to be right about others – BCLC Oct 26 '16 at 4:17
• lol, I remember you, you commented once with that question in one of my solutions and I had no idea what you were actually trying to say. – Anonymous Oct 26 '16 at 4:23
• @Anonymous LOL – BCLC Oct 26 '16 at 5:13
• LOL yeah that one! – Anonymous Oct 26 '16 at 5:16
The non-zero rows of of the row echelon form of $A^T$ give you a basis of the column space of $A$ if you tranpose them.
The row space of a matrix is preserved as perform elementary row operation. As you perform elementary row operations to the transpose of the matrix, you are actually performing column operations to the original matrix while preserving the column space.
It is known that the non-zero rows of the row echelon forms are linearly independent and hence form a basis to the row space. | 2019-08-20T11:46:55 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1985297/is-is-possible-to-find-a-basis-for-the-column-space-of-a-given-reduced-row-ech",
"openwebmath_score": 0.8807119131088257,
"openwebmath_perplexity": 149.3274783908914,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9752018376422665,
"lm_q2_score": 0.8652240721511739,
"lm_q1q2_score": 0.8437681051341498
} |
https://tex.stackexchange.com/questions/494415/the-colon-does-not-align-with-the-vdots | # The colon does not align with the \vdots
I have the following code to align several equations:
\begin{aligned}
s_{1} & :=\sum_{1 \leq j \leq m} X_{j} \\
s_{2} & :=\sum_{1 \leq j<k \leq m} X_{j} \cdot X_{k} \\
& \vdots \\
s_{k} & :=\sum_{1 \leq j_{1}<j_{2}<\cdots<j_{k} \leq m} X_{j_{1}} \cdot X_{j_{2}} \cdot \cdots \cdot X_{j_{k}} \\
& \vdots \\
s_{m} & :=X_{1} X_{2} \cdots X_{m}
\end{aligned}
I would like to ask why the colon does not align with the vdots and a solution for it.
Welcome to TeX-SE! It does not align because a colon is a relation, i.e. of the \mathrel type. If you want to align it "by hand" you could just make the \vdots of that type, too.
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{aligned} s_{1} & :=\sum_{1 \leq j \leq m} X_{j} \\ s_{2} & :=\sum_{1 \leq j<k \leq m} X_{j} \cdot X_{k} \\ & \mathrel{\vdots} \\ s_{k} & :=\sum_{1 \leq j_{1}<j_{2}<\cdots<j_{k} \leq m} X_{j_{1}} \cdot X_{j_{2}} \cdot \cdots \cdot X_{j_{k}} \\ & \mathrel{\vdots}\\ s_{m} & :=X_{1} X_{2} \cdots X_{m} \end{aligned}
\end{document}
• Your solutions is elegant and works perfectly. Thank you so much! Jun 6, 2019 at 3:07
I don't think these should align on the dots of :=, but rather should be centered with respect to either the s_k or perhaps the relation :=. The dots in := serve a different function to those in \vdots. The package mathtools provides the command \vdotswithin{...} for centering such \vdots with respect to the material ...; it also provides \coloneqq for a better version of :=:
\documentclass{article}
\usepackage{mathtools}
\begin{document}
\begin{align*}
s_{1} &\coloneqq\sum_{1 \leq j \leq m} X_{j} \\
s_{2} &\coloneqq\sum_{1 \leq j<k \leq m} X_{j} \cdot X_{k} \\
\vdotswithin{s_{k}}& \\
s_{k} &\coloneqq\sum_{1 \leq j_{1}<j_{2}<\cdots<j_{k} \leq m}
X_{j_{1}} \cdot X_{j_{2}} \cdot \cdots \cdot X_{j_{k}} \\
\vdotswithin{s_{k}}& \\
s_{m} &\coloneqq X_{1} X_{2} \cdots X_{m}
\end{align*}
\begin{align*}
s_{1} &\coloneqq\sum_{1 \leq j \leq m} X_{j} \\
s_{2} &\coloneqq\sum_{1 \leq j<k \leq m} X_{j} \cdot X_{k} \\
&\vdotswithin{\coloneqq} \\
s_{k} &\coloneqq\sum_{1 \leq j_{1}<j_{2}<\cdots<j_{k} \leq m}
X_{j_{1}} \cdot X_{j_{2}} \cdot \cdots \cdot X_{j_{k}} \\
&\vdotswithin{\coloneqq} \\
s_{m} &\coloneqq X_{1} X_{2} \cdots X_{m}
\end{align*}
\end{document}
• I was thinking the same (I prefer the dots centered on the equals sign). Assuming the colons are necessary (they aren't, in my opinion), the vertical dots meaning ellipsis have nothing to do with those colons, so aligning them is out of question. Jun 6, 2019 at 6:36
My interpretation of changing of the answer given by the user @marmot using:
• The macro \coegual instead of := simply because it provides the perfect alignment between : and =;
• Using the negative space with the command \mkern I tried to bring the products of the elements close to the summation because I do not like them distant;
• I have used \substack command because the subscripts on the summation did not bring me the products far away.
\documentclass[a4paper,12pt]{article}
\usepackage{amsmath,amssymb}
\newcommand{\coegual}{\mathrel{\mathop:}=}
\begin{document}
\begin{aligned} s_{1} & \coegual \sum_{1 \leq j \leq m} X_{j} \\ s_{2} & \coegual \sum_{1 \leq j < k \leq m} \mkern-5mu X_{j} \cdot X_{k} \\ & \mathrel{\vdots} \\ s_{k} & \coegual \sum_{\substack{1 \leq j_{1}< j_{2} <\cdots\\ \cdots< j_{k} \leq m}} \mkern-15mu X_{j_{1}} \cdot X_{j_{2}} \cdot \cdots \cdot X_{j_{k}} \\ & \mathrel{\vdots}\\ s_{m} & \coegual X_{1} X_{2} \cdots X_{m} \end{aligned}
\end{document} | 2022-07-07T02:47:57 | {
"domain": "stackexchange.com",
"url": "https://tex.stackexchange.com/questions/494415/the-colon-does-not-align-with-the-vdots",
"openwebmath_score": 1.0000100135803223,
"openwebmath_perplexity": 4338.49176923962,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9399133548753619,
"lm_q2_score": 0.897695292107347,
"lm_q1q2_score": 0.8437557936604345
} |
http://math.stackexchange.com/questions/180781/find-fx-from-f3x-1 | # Find $f(x)$ from $f(3x + 1)$
The problem that I have to solve is:
If the following function is valid for every value of $x$
$$f(3x + 1) = 9x^2 + 3x$$
find the function $f(x)$ and prove that for every $x\in\mathbb R$ the following is valid: $$f(2x) - 4f(x) = 2x$$
-
Here $$f(3x+1)=3x(3x+1)=((3x+1)-1)(3x+1)$$ $$\implies f(x)=(x-1)x=x^2-x$$ $$\implies f(2x)-4f(x)=4x^2-2x-4x^2+4x=2x$$
In general, just let $3x+1=t$ and express $x$ in terms of $t$ and substitute in $f(t)$
-
Thanks, that's also what my theory says thanks for the solving the starting process :) – Chris Aug 9 '12 at 18:16
I did the math $$x = ( w - 1 ) / 3$$ and i have concluded to this $$f(w) = w^2 - 2w + 1 + ( 3w - 3 ) / 3$$ however there is an f(x) missing what should i do ?? – Chris Aug 9 '12 at 18:40
$f(w)=w^2-2w+1+\frac{3(w-1)}{3}=w^2-2w+1+w-1=w^2-w$.Here, $w$ is just a dummy variable , so you can substitute $x$ for $w$ which gives $f(x)=x^2-x$. Keep in mind this new $x$ is different from previous $x$ – Aang Aug 9 '12 at 18:44
+1 to both of you for the above comments (OP showing work and avatar following up) – The Chaz 2.0 Aug 9 '12 at 19:17
$\rm \dfrac{f(3x\!+\!1)}{3x\!+\!1} = 3x\:\Rightarrow \dfrac{f(z)}z = z\!-\!1\:\Rightarrow \dfrac{f(2x)\!-\!4f(x)}{2x} = \color{#0A0}{\dfrac{f(2x)}{2x}}- 2 \color{#C00}{\dfrac{f(x)}x} = \color{#0A0}{2x\!-\!1} - 2(\color{#C00}{x\!-\!1}) = 1$
Remark $\$ The point of presenting it like this is to emphasize how exploiting the innate linear structure serves to simplify the calculations (from nonlinear to linear). In less trivial problems this can yield much greater simplifications (e.g. in operator calculus with $q$-difference operators).
-
Why would this be downvoted? – Pedro Tamaroff Aug 9 '12 at 19:12
@Downvoter If something is not clear then please feel welcome to ask for an explanation. – Bill Dubuque Aug 9 '12 at 19:12
I like this, but it took me quite a while to grasp! (Of course, I'm rather low on the "totem pole"...) – The Chaz 2.0 Aug 9 '12 at 19:16
@TheChaz I've added some color to help follow the logic. – Bill Dubuque Aug 9 '12 at 19:18
@TheChaz Yes, that's probably the most straightforward way to proceed. But I thought it might prove of interest to emphasize that there is additional structure that simplifies matters. Such structure comes to the fore when one studies $q$-difference operators and related operator calculus. – Bill Dubuque Aug 9 '12 at 19:27
Hint: $3x+1=y \Rightarrow x= \frac{y-1}{3}$. So replace $x$ by $\frac{y-1}{3}$ and magic happens.
- | 2015-12-02T01:50:39 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/180781/find-fx-from-f3x-1",
"openwebmath_score": 0.8559978604316711,
"openwebmath_perplexity": 599.7775307576528,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9793540722737478,
"lm_q2_score": 0.8615382129861583,
"lm_q1q2_score": 0.8437509573074417
} |
https://math.stackexchange.com/questions/2909367/for-what-values-of-x-in-3-17-does-the-series-sum-limits-infty-n-1 | # For what values of $x$ in $(-3,17)$ does the series $\sum\limits^{\infty}_{n=1}\frac{(-1)^n x^n}{n[\log (n+1)]^2}$ converge?
For what values of $x$ in the following series, does the series converge?
\begin{align}\sum^{\infty}_{n=1}\dfrac{(-1)^n x^n}{n[\log (n+1)]^2},\;\;-3<x<17 \end{align}
MY TRIAL
\begin{align}\lim\limits_{n\to \infty}\left|\dfrac{(-1)^{n+1} x^{n+1}}{(n+1)[\log (n+2)]^2}\cdot\dfrac{n[\log (n+1)]^2}{(-1)^n x^n}\right|&=|x|\lim\limits_{n\to \infty}\left|\dfrac{n}{n+1}\cdot\left[\dfrac{\log (n+1)}{\log (n+2)}\right]^2\right|\\&=|x|\lim\limits_{n\to \infty}\left(\dfrac{n}{n+1}\right)\cdot\lim\limits_{n\to \infty}\left[\dfrac{\log (n+1)}{\log (n+2)}\right]^2\\&=|x|\lim\limits_{n\to \infty}\left(\dfrac{n}{n+1}\right)\cdot\left[\lim\limits_{n\to \infty}\dfrac{\log (n+1)}{\log (n+2)}\right]^2\\&=|x|\left[\lim\limits_{n\to \infty}\dfrac{1}{n+1}\cdot n+2\right]^2\\&=|x|\end{align} Hence, the series converges absolutely for $|x|<1$ and diverges when $|x|>1$.
When $x=1,$ \begin{align}\sum^{\infty}_{n=1}\dfrac{(-1)^n }{n[\log (n+1)]^2}<\infty\;\;\text{By Alternating series test}\end{align} When $x=-1,$ \begin{align}\sum^{\infty}_{n=1}\dfrac{1}{n[\log (n+1)]^2}<\infty\;\;\text{By Direct comparison test}\end{align} Hence, the values of $x$ for which the series converges, is $-1\leq x\leq 1.$
I'm I right? Constructive criticisms will be highly welcome! Thanks!
• For me it's right – Atmos Sep 8 '18 at 7:36
• Fine for me too. I just wouldn't speak of alternating series test, because the series is in fact absolutely convergent. Why use a precision tool when you can use a hammer ? :-) – Nicolas FRANCOIS Sep 8 '18 at 7:40
• @Nicolas FRANCOIS: Smiles... – Micheal Sep 8 '18 at 7:41
• @Nicolas FRANCOIS:You mean the series \begin{align}\sum^{\infty}_{n=1}\dfrac{(-1)^n }{n[\log (n+1)]^2}\end{align} converges absolutely for $x=1?$ How? – Micheal Sep 8 '18 at 7:42
• @Nicolas FRANCOIS:: Oh, I see! Robert Z has cleared my confusion! Thanks! – Micheal Sep 8 '18 at 7:46
You are correct. This is a "variation on the theme".
For $|x|>1$ $$\lim_{n\to +\infty}\dfrac{|-x|^n}{n[\log (n+1)]^2}=+\infty$$ and the series is divergent.
For $|x|\leq 1$, by direct comparison, the series is absolutely convergent $$\sum^{\infty}_{n=1}\dfrac{|-x|^n}{n[\log (n+1)]^2}\leq \sum^{\infty}_{n=1}\dfrac{1}{n[\log (n+1)]^2}<\infty.$$
• Thanks, I'm grateful! – Micheal Sep 8 '18 at 7:44
Yes it is correct, for the limit from here we can proceed as follow
$$\ldots=|x|\lim\limits_{n\to \infty}\left|\dfrac{n}{n+1}\cdot\left[\dfrac{\log (n+1)}{\log (n+2)}\right]^2\right| =|x|\lim\limits_{n\to \infty}\dfrac{1}{1+1/n}\cdot\left[\dfrac{\log n+\log (1+1/n)}{\log n+\log (1+2/n)}\right]^2=|x|\cdot1=|x|$$
• Thanks a lot! I appreciate! – Micheal Sep 8 '18 at 7:45 | 2019-05-27T07:04:51 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2909367/for-what-values-of-x-in-3-17-does-the-series-sum-limits-infty-n-1",
"openwebmath_score": 0.9395433068275452,
"openwebmath_perplexity": 927.8465566898402,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9793540716711546,
"lm_q2_score": 0.8615382129861583,
"lm_q1q2_score": 0.8437509567882846
} |
https://mathforums.com/threads/how-to-find-the-launch-angle-for-a-ball-from-a-fixed-point-where-it-is-known-the-maximum-height.347839/ | # How to find the launch angle for a ball from a fixed point where it is known the maximum height?
#### Chemist116
The problem is as follows:
From point indicated in the picture, a football player is about to kick a ball giving the ball a velocity of $v_{o}$. The projectile collisions with the crossbar on point $A$ as shown in the picture. Find the launch angle. It is known that the highest point of the trajectory is $2.5\,m$ and the horizontal distance from the ball to the crossbar is $5\sqrt{3}$.
The alternatives given are as follows:
$\begin{array}{ll} 1.&15^{\circ}\\ 2.&30^{\circ}\\ 3.&37^{\circ}\\ 4.&45^{\circ}\\ 5.&\arctan{0.28}\\ \end{array}$
What I've attempted here was to use the equation for the trajectory of the projectile as shown below:
$y=x\tan\phi-\frac{1}{2}g\left(\frac{x}{v_{o}\cos\phi}\right)^2$
Since it mentions that the highest point of this trajectory is $2.5\,m$ then I'll obtain from the above equation a relationship.
Using the first derivative with respect of $x$ I'm getting:
$y'=\tan\phi-g\frac{x}{v_{o}^2\cos^2\phi}$
Equating this to zero:
$0=\tan\phi-g\frac{x}{v_{o}^2\cos^2\phi}$
$x=\frac{v_{o}^2\cos^2\phi\tan\phi}{g}=\frac{v_o^2\sin\phi\cos\phi}{g}$
Then:
Inserting this in the equation for the trajectory I'm getting:
$y=\left(\frac{v_o^2\sin\phi\cos\phi}{g}\right)\tan\phi-\frac{1}{2}g\left(\frac{\frac{v_o^2\sin\phi\cos\phi}{g}}{v_{o}\cos\phi}\right)^2$
$y=\left(\frac{v_o^2\sin\phi\cos\phi}{g}\right)\frac{\sin\phi}{\cos\phi}-\frac{1}{2}g\left(\frac{\frac{v_o^2\sin\phi\cos\phi}{g}}{v_{o}\cos\phi}\right)^2$
$y=\left(\frac{v_o^2\sin\phi\cos\phi}{g}\right)\frac{\sin\phi}{\cos\phi}-\frac{1}{2}g\left(\frac{v_o\sin\phi}{g}\right)^2$
$y=\frac{v_{o}^2\sin^2\phi}{g}-\frac{v_o^2\sin^2\phi}{2g}=\frac{v_o^2\sin^2\phi}{2g}$
Then:
$x=\frac{v_o^2\sin\phi\cos\phi}{g}$
Since what it is given is the coordinates for each then this is reduced to:
$5\sqrt{3}=\frac{v_o^2\sin\phi\cos\phi}{g}$
$2.5=\frac{v_o^2\sin^2\phi}{2g}$
$v_{o}^2\sin\phi=\frac{5g}{\sin\phi}$
This is inserted in the above equation and becomes into:
$5\sqrt{3}=\frac{v_o^2\sin\phi\cos\phi}{g}$
$5\sqrt{3}=\frac{\frac{5g}{\sin\phi}\cos\phi}{g}$
$5\sqrt{3}=\frac{\frac{5g}{\sin\phi}\cos\phi}{g}$
$5\sqrt{3}=\frac{5\cos\phi}{\sin\phi}$
$\tan\phi=\frac{1}{\sqrt{3}}$
Therefore this becomes into:
$\phi=\tan^{-1}\left(0.5777\right)$
But this does not appear in any of the alternatives. Did I made a mistake or anything?. Can someone help me with this?.
#### skeeter
Math Team
the sketch doesn't match the problem statement ...
#### skeeter
Math Team
$\phi = \arctan\left(\dfrac{1}{\sqrt{3}}\right) = 30^\circ$
#### skeeter
Math Team
confirmation of $\phi = 30^\circ$
at $h_{max} = 2.5$, $v_y = 0$ ...
$0^2 = v_0^2 \sin^2{\phi} - 2g \Delta y \implies v_0^2 \sin^2{\phi} = 50$
$\Delta x = v_0 \cos{\phi} \cdot t \implies t = \dfrac{5\sqrt{3}}{v_0 \cos{\phi}}$
$\Delta y = v_0 \sin{\phi} \cdot t - \dfrac{1}{2}gt^2$
$2.5 = \dfrac{v_0 \sin{\phi}}{v_0 \cos{\phi}} \cdot 5\sqrt{3} - 5\left(\dfrac{5\sqrt{3}}{v_0 \cos{\phi}}\right)^2 = 5\sqrt{3} \cdot \tan{\phi} - \dfrac{15}{2} \cdot \dfrac{50}{v_0^2 \cos^2{\phi}}$
$0 = -2.5 + 5\sqrt{3} \cdot \tan{\phi} - \dfrac{15}{2} \cdot \tan^2{\phi}$
$\tan{\phi} = \dfrac{-5\sqrt{3}}{-15} = \dfrac{\sqrt{3}}{3} \implies \phi = 30^\circ$ | 2020-04-09T07:39:49 | {
"domain": "mathforums.com",
"url": "https://mathforums.com/threads/how-to-find-the-launch-angle-for-a-ball-from-a-fixed-point-where-it-is-known-the-maximum-height.347839/",
"openwebmath_score": 0.9000709056854248,
"openwebmath_perplexity": 388.41210262080216,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9793540674530016,
"lm_q2_score": 0.8615382129861583,
"lm_q1q2_score": 0.8437509531541845
} |
https://zbx.hoteloffice-statt-homeoffice.de/en/how-to-find-the-vertex-of-a-parabola.html | How to find the vertex of a parabola
• kubota 8kw diesel generator
eccv papers
• sore throat during ovulation
golang add method to struct from another package
• yorkie rescue overland park ks
a faithful friend bible verse
• chapter 4 math test 6th grade
• the mary burke
how to open port 7777 on router1945
powershell set proxy bypass list
• miss prissy
bookmarklets hacks
• swidget unreal
vim autocomplete 2022
veo scooter hack
• oxford discover workbook 5 pdf
• ipfire rerun setup
grade 7 exam papers tamil medium
• dark season 1 480p dual audio
what is variant coding in automotive
• kinantot nang aso sex stories
distech controls price list
nx java
• for loop to rename columns in r
sapphire rx 6700 xt nitro bios switch
• tuna seiner for sale
samsung galaxy a13 case
beowulf characteristics
• pioneer tv models
conclusion of audio amplifier
• razer naga trinity drivers
fantic 50cc
• bdix ftp server list
. Equation of a parabola - derivation. Given a parabola with focal length f, we can derive the equation of the parabola. (see figure on right). We assume the origin (0,0) of the coordinate system is at the parabola's vertex. For any point ( x, y) on the parabola, the two blue lines labelled d have the same length, because this is the definition. Find the coordinates of the vertex for the parabola y = 3x2 - 6x + 1. Solution: We have the equation as, y = 3x 2 - 6x + 1. Here, a = 3, b = -6 and c = 1. Now, it is known that the coordinates of the vertex are given by, (-b/2a, -D/4a) where D = b 2 - 4ac. D = (-6) 2 - 4 (3) (1) = 36 - 12 = 24 So, x - coordinate of vertex = 6/2 (3) = 6/6 = 1. Finding the Vertex of a Parabola with a Simple Formula. Find the x coordinate of the vertex directly. When the equation of your parabola can be written as y = ax^2 + bx + c, the x of the vertex can be found using the formula x = -b / 2a. Simply plug the a and b values from your equation into this formula to find x. y = ax 2 + bx + c. Using the following two different ways, we can find the vertex of the parabola. (i) Using completing the square. (ii) Using formula. Here we see, how to find vertex of the parabola from the standard form of the equation. Name a, b and c for each parabola. Then find the vertex. park north apartments. Results produced by the online Parabola equation solver are highly reliable. focus (x,y)= directrix= focal diameter= 3.The directrix and the focus provide enough information to write an equation for a parabola.Compare the given equation with the standard equation and find the value of a. directrix\\:3x^2+2x+5y-6=0.Find its equation Learning math. The diagram shows us the four different cases that we can have when the parabola has a vertex at (0, 0). When the variable x is squared, the parabola is oriented vertically and when the variable y is squared, the parabola is oriented horizontally. Furthermore, when the value of p is positive, the parabola opens towards the positive part of the axes, that is, upwards or to the right. The Vertex to plot a parabola Graph can be derived using x=-b/2a and y = f (-b/2a). The quadratic equation can be presented as f (x) = a (x-h)2 + k, where (h,k) is the vertex of the parabola, its vertex form. Find the standard form of the equation of the parabola with the given characteristic(s) and. vertex at the origin. Horizontal axis. Find the coordinates of the vertex for the parabola y = 3x2 - 6x + 1. Solution: We have the equation as, y = 3x 2 - 6x + 1. Here, a = 3, b = -6 and c = 1. Now, it is known that the coordinates of the vertex are given by, (-b/2a, -D/4a) where D = b 2 - 4ac. D = (-6) 2 - 4 (3) (1) = 36 - 12 = 24 So, x - coordinate of vertex = 6/2 (3) = 6/6 = 1. The quadratic equation can be presented as f (x) = a (x-h)2 + k, where (h,k) is the vertex of the parabola, its vertex form. S. If you have the equation of a parabola in vertex form y = a ( x − h). Figure 1 shows a picture of a parabola. Notice that the distance from the focus to point (x 1, y 1) is the same as the line perpendicular to the. A parabola has six properties. 1. The vertex of a parabola is at the middle of the curve. It can either be at the origin (0, 0) or any other location (h, k) in the Cartesian plane. 2. The concavity of a parabola is the orientation of the parabolic curve. The curve may open either upward or downward, or to the left or right. There are various methods for finding the equation of the parabola; the methods themselves aren't particularly difficult, but the coordinate values given in the problem make the result look "ugly". ... The axis of symmetry of the parabola passes through the vertex at $\ (h \ , \ 45) \ \ .$ We can use a proportionality property to find this. The vertex form of a parabola's equation is generally expressed as: y = a (x-h) 2 +k. If a is positive then the parabola opens upwards like a regular "U". If a is negative, then the graph opens downwards like an upside down "U". If |a| < 1, the graph of the parabola widens. This just means that the "U" shape of parabola stretches out sideways. Writing The Equation Of A Parabola In Vertex Form Given Graph You. How To Know The Vertex Form Of A Horizontal Parabola Quora. How To Write An Equation For A Parabola In Vertex Form Wyzant Ask Expert. Equations Of A Parabola Standard To Vertex Form And Back Again Montessori Muddle. Completing The Square To Find Vertex Form Of A Quadratic. a ≠ 0. The graph of a quadratic function is called a parabola. A parabola is roughly shaped like the letter ‘U’ or upside-down ‘U’. If the leading coefficient is greater than zero, the parabola opens upward, and if the leading coefficient is less than zero, the parabola opens downward. , so the parabola opens upward. The standard equation of a parabola can be represented as, {eq}y=a (x-h)^2+k {/eq} where, (h,k) is the x and y coordinates of the vertex. The following two examples will show how to find the. . A parabola has six properties. 1. The vertex of a parabola is at the middle of the curve. It can either be at the origin (0, 0) or any other location (h, k) in the Cartesian plane. 2. The concavity of a parabola is the orientation of the parabolic curve. The curve may open either upward or downward, or to the left or right. We can use the vertex form to find a parabola's equation. The idea is to use the coordinates of its vertex ( maximum point, or minimum point) to write its equation in the form y = a ( x − h) 2 + k (assuming we can read the coordinates ( h, k) from the graph) and then to find the value of the coefficient a. The focus of a parabola is always inside the parabola; the vertex is always on the parabola; the directrix is always outside the parabola. The "general" form of a parabola's equation is the one you're used to, y = ax 2 + bx + c — unless the quadratic is "sideways", in which case the equation will look something like x = ay 2 + by + c. Output: Enter the value of constant “a” in the parabolic standard equation form: 4 Enter the value of constant “b” in the parabolic standard equation form: 3 Enter the value of constant “c” in the parabolic standard equation form: 2 Vertex: (-0.375, 1.4375) Focus: (-0.375, 1.5) Directrix: y= -158. Are you wondering how to seek help. The focus is the distance from the vertex to the focus is 1/(4a), where a can be found in the equation of the parabola (it is the scalar in front of the parentheses). The focus, as a point, is (h, v + 1/(4a)); it should be directly above or directly below the vertex. It always appears inside the parabola. The equation of the directrix is y = v. Make a conjecture as to the coordinates of the vertex of the parabola with the equation y = 2 (x - 3)² + 4. Write down your conjecture. 7. Now enter the equation y = 2 (x - 3)² + 4 in the input bar at the bottom of the page to check your answer. If your conjecture was wrong, explain why. Improve your math knowledge with free questions in "Find the vertex of a parabola" and thousands of other math skills. Steps to Find Vertex Focus and Directrix Of The Parabola. Step 1. Determine the horizontal or vertical axis of symmetry. Step 2. Write the standard equation. Step 3. Compare the given equation with the standard equation and find the value of a. Step 4. Find the focus, vertex and directrix using the equations given in the following table. What is the equation of a parabola with a vertex at (4,8) and a directrix at y=5? I calculated the distance from the vertex to the directrix as p=3. my answer key says p=6. I cannot see were I went wrong. Can anyone provide some guidance? If p=3 is correct then the equation would be: Focus = (4,11) (x-h)^2 = 2p(y-k) (x-4)^2 = 6(y-8) Thanks in. The vertex of a parabola is the point where the parabola crosses its axis of symmetry. If the coefficient of the x 2 term is positive, the vertex will be the lowest point on the graph, the point at the bottom of the " U "-shape. Use the left or right blue arrow keys to move the flashing cursor to the left of the vertex and press ENTER to mark the left bound. 5. Move the flashing cursor to the right of the vertex and press ENTER to mark the right bound. Notice the arrow marks at the top of the screen showing the area that you defined. 6.
• sermon i am
psychology of slashing tires
• the virgin suicides pdf
• caco3 hcl cacl2 h2co3
wkwebview ctx checkifpoliciesdictatewkwebviewhookingsupport
• recharge tiktok coins cheap
newmar rv wiring diagram507
artflow mod apk
• filipino inventors and their inventions
maclaurin series proof
• i raised a sick and weak prince manga chapter 36
resultant calculator with angle
10 foot diameter culvert pipe for sale
• scammed on vinted
vizio tv input button on tv
• squier serial number cy
blurams dome lite 2 sd card
vscode goproxy
• wooden blocks for sale
m3u python
• people playground age rating
singular value decomposition uses
sqlmap credit card dump
• hackear facebook es un delito
swelling after ganglion cyst aspiration
• tyranid 9th edition codex pdf
discrete mathematics pdf for cse
1987 toyota celica for sale
• manitou incline webcam
naruto senju and kushina fanfiction
• a205 accident today
kubota z122r grass catcher
akc certified breeders
• unique number of occurrences leetcode solution
wrt1900ac usb tethering
como ver formula 1 en diferido
• military macaw for sale
• cisco 9300 factory reset rommon mode
stewart funeral home obituaries tyler
pta meeting script tagalog
• a ladder leans against a wall as shown in the diagram answer key
free sex stories wife vaion swinging
• congruence equation calculator
visa prepaid card transfer to bank account
• website unblocker for school chromebook
no loudness equalization windows 11
• dmod cf
farming simulator apk obb
• private homemade amateur teen girlfriend
play juwa online771
one day miracle novena
• veshje online
red heeler puppies for sale in illinois
• zigbee2mqtt hue dimmer switch
• ubuntu bluetooth driver
hiboy scooter fuse253
garrett at pro coils
• free cheer leader porn videos
• lydian spoofer cracked
professor cal audio soundgasm
• who is the gargoyle king in riverdale
cpt code for closed treatment of fifth metacarpal fracture
• gigabyte g24f icc profile
cesca chairs set of 4
• he looks into my eyes when he talks to me
iget vape sunshine coast
• 7 years of love phim
retaliation lawsuit payouts652
xxnxx vrnsa
• compare string with string array in java
ggplot geom area
• sellix fortnite account shop
sent nudes
• sex bestiality movie galleries
chulin custom weight loss sauce
• dutch oven with handle
what does aaa pay tow companies
arsenal script silent aim
• 5800x ppt tdc edc 2022
mepps spinner size chart
• thule guide 2021 pdf
bcm63136 openwrt
filebeat tokenizer
It pushes the aircraft of proportion of the whole parabola as well; whatever pushes the left of the parabola is a full mirror image of whatever is on the right. If you wish to find the vertex of a square equation, you can either utilize the vertex formula, or complete the square. Any type of number can be the input value of a square feature.
Practice 2 - If the focus of a parabola is (1 y − y 1 = m(x −x 1) y − y 1 = 2(x −x 1) Step 3 We will discuss x-intercepts next unit but for now: x To find the y-intercept, we x To find the x-intercept(s), we Practice: Given the equation , find the y-intercept Since the two vectors lie on the plane, their cross product can be used as a ...
Where is the coordinates of the vertex of the parabola, and p is the distance from the vertex to the focus. In your problem there's a bit of trick - the x and y coordinats are switched around (so that the parabola will be lying on its side). But the same concept aplies. So reconfigure your equation y^2 = -10x into the form:
How to know the vertex form of a horizontal parabola quora explained with pictures and ilrations formula for is just 4 2 standard quadratic function finding in khan academy converting from equation you writing equations parabolas definition explanation lesson transcript study com introduction quadratics she loves math graphing convert without completing square
You can simply find the vertex from the quadratic equations. To know how to? keep reading. Find vertex from the standard form: If you don’t want to convert the standard form into the vertex form, find the vertex point using these formulas. h = -b / (2a) k = c - b 2 / (4a) Example: Find the vertex of a parabola from the equation y = x 2 - 3x ... | 2022-12-05T15:03:46 | {
"domain": "hoteloffice-statt-homeoffice.de",
"url": "https://zbx.hoteloffice-statt-homeoffice.de/en/how-to-find-the-vertex-of-a-parabola.html",
"openwebmath_score": 0.43206700682640076,
"openwebmath_perplexity": 1136.5878302980148,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9793540680555949,
"lm_q2_score": 0.8615382094310357,
"lm_q1q2_score": 0.8437509501916178
} |
https://math.stackexchange.com/questions/1986883/how-to-simplify-fracnn-1n-1-mod-n-1 | # How to simplify $\frac{n^n - 1}{n - 1} \mod (n + 1)$?
My question is how to simplify the following:
$$\frac{n^n - 1}{n - 1} \mod (n + 1)$$
I've tried a bunch of tricks (splitting into even and odd $n$ and using a difference of squares) but I can't seem to find anything that gives a clean answer.
Is there even a simplification for this?
EDIT: Apologies, I had $+$ instead of $-$
• This isn't even an integer for even values of n as far as I've tested. I can't prove it for general $2k$ but $2,4,6,8$ all don't work. – Nitin Oct 27 '16 at 0:36
• I had the wrong value - my apologies! – Cisplatin Oct 27 '16 at 0:36
You have changed your question, now it is easy.
Note that $n^n-1$ divides $n^2-1$ whenever $2 | n$. Hence, if $n$ is even, the answer is $0$, since $\frac{n^n-1}{(n+1)(n-1)}$ will be an integer.
Suppose that $n$ is odd. Note that $n \equiv -1 \mod n+1$. Hence, it is legitimate to replace $n$ by $-1$ in the modular expression, and this gives $$\frac{n^n-1}{n-1} = \sum_{i=0}^{n-1} n^i \equiv \sum_{i=0}^{n-1} (-1)^i \equiv 1 \mod n+1$$
because one factor of $1$ gets left out as $n$ is odd, so $n-1$ is even.
Hence, the answer is zero for even $n$, and $1$ for odd $n$.
Edit: You can multiply and check that $$(n-1) (1 + n + n^2 + \ldots + n^{n-1}) = n^n-1$$
it's the same as the equivalance class of $n^n+1$ by definition of modulars
• I had the wrong value up - please reconsider! – Cisplatin Oct 27 '16 at 0:38
The left-hand side is the sum of powers of $n$. Each power $n^a$ is equivalent to $(-1)^a\pmod{n+1}$
• Is there a proof for $n^a \equiv (-1)^a$? – Cisplatin Oct 27 '16 at 0:41
• $n \equiv -1 \mod (n+1)$ – Siong Thye Goh Oct 27 '16 at 0:46
$\frac {n^n-1}{n-1} = \sum_\limits{i=0}^{n-1} n^i\\ n \equiv -1\pmod{n+1}\\ n^i \equiv -1^i\pmod{n+1}\\ \frac {n^n-1}{n-1} \equiv \sum_\limits{i=0}^{n-1} (-1)^i\pmod{n+1}$
$0$ when $n$ is even. $1$ when $n$ is odd. | 2020-01-24T08:36:24 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1986883/how-to-simplify-fracnn-1n-1-mod-n-1",
"openwebmath_score": 0.87685227394104,
"openwebmath_perplexity": 170.69421665905384,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9711290955604489,
"lm_q2_score": 0.8688267830311354,
"lm_q1q2_score": 0.8437429680037208
} |
https://math.stackexchange.com/questions/3383363/describing-the-kernel-of-a-matrix-as-a-span-of-two-vectors | # Describing the kernel of a matrix as a span of two vectors
First year linear algebra level.
A is a matrix $$\begin{bmatrix} -1&1&1&3\\ 2&-2&1&-3\\ 3&-3&-2&-8\\ \end{bmatrix}$$ Then describe the kernel of A as a span of linearly independent vectors. I reduced it to RREF: $$\begin{bmatrix} 1&-1&0&-2\\ 0&0&1&1\\ 0&0&0&0\\ \end{bmatrix}$$ But I can't connect it to how it becomes the answer, which is: $$span( \begin{bmatrix} 2\\0\\-1\\1 \end{bmatrix}, \begin{bmatrix} 1\\1\\0\\0\\ \end{bmatrix} )$$ Also I'm not sure if the 2 is a typo on the answer sheet.
• Welcome to Mathematics Stack Exchange. It looks correct – J. W. Tanner Oct 6 '19 at 20:41
• – J. W. Tanner Oct 6 '19 at 20:48
Call $$C_k$$ the columns of matrix $$R$$ (the RREF transform)
You see that :
$$C_1+C_2=0 \ \iff \ (1)C_1+(1)C_2+(0)C_3+(0)C_4=0 \tag{1}$$
Collect the coefficients (between parentheses) of this null linear combination : you will get the second of your vectors (presented in a line instead of a column).
For the other one, it is the same :
$$(2)C_1+(0)C_2+(-1)C_3+(1)C_4=0$$
Explanation : Relationship (1) can be written under the form :
$$\underbrace{\left(C_1|C_2|C_3|C_4\right)}_{R}\underbrace{\begin{pmatrix}1\\1\\0\\0\end{pmatrix}}_{V}=\underbrace{\begin{pmatrix}0\\0\\0\\0\end{pmatrix}}_0 \ \ \implies \ \ V \in \ker R$$
• I have added an explanation. – Jean Marie Oct 6 '19 at 20:48
• I still struggling to see it: starting with the RREF, how would I have gotten those coefficients? Is there a procedural way I could've known that the first vector would have the two zeros (I can see visually that C1-C2 is zero since they're 1,-1 respectively), but the second vector involved all three columns? ................ In class, all we did was show the kernel is Ax=0 so I think my base is a little weak. – Five9 Oct 6 '19 at 21:15
• For the second vector, it is possible to notice at first that doing $(-1)C_3+(1)C_4$, its second and third coefficient become $0$, therefore it is a multiple of the first column $C_1$ : $(-1)C_3+(1)C_4 =aC_1$ for a certain $a$ (which is $-2$) ; rearranging gives $(-a)C_1+(-1)C_3+(1)C_4 =0$ – Jean Marie Oct 6 '19 at 21:24
• Turns out I had a total lapse in thinking...I understand now, thanks for bearing with me. However just noting that this is very notationally different from what my book/class has :O – Five9 Oct 7 '19 at 1:41
$$\begin{bmatrix} 1&-1&0&-2\\ 0&0&1&1\\ 0&0&0&0\\ \end{bmatrix}$$ To find kernel and its basis you just need the solution of $$Ax=0$$, so you get $$x-y-2w=0,z+w=0$$. Any two linearly independent solutions of these equations will form the spanning set of kernel. It is easy to verify that $$x=2,y=0,z=-1,w=1$$ and $$x=1,y=1,z=0,w=0$$ are such two LI solutions. | 2020-01-27T15:56:37 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3383363/describing-the-kernel-of-a-matrix-as-a-span-of-two-vectors",
"openwebmath_score": 0.8012675642967224,
"openwebmath_perplexity": 391.88260702392074,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.97112909472487,
"lm_q2_score": 0.8688267796346599,
"lm_q1q2_score": 0.8437429639793314
} |
https://rstudio-pubs-static.s3.amazonaws.com/765240_838aaa18d8434fe48546aca87b166bc9.html | Chapter 7 - Ulysses’ Compass
The chapter began with the problem of overfitting, a universal phenomenon by which models with more parameters fit a sample better, even when the additional parameters are meaningless. Two common tools were introduced to address overfitting: regularizing priors and estimates of out-of-sample accuracy (WAIC and PSIS). Regularizing priors reduce overfitting during estimation, and WAIC and PSIS help estimate the degree of overfitting. Practical functions compare in the rethinking package were introduced to help analyze collections of models fit to the same data. If you are after causal estimates, then these tools will mislead you. So models must be designed through some other method, not selected on the basis of out-of-sample predictive accuracy. But any causal estimate will still overfit the sample. So you always have to worry about overfitting, measuring it with WAIC/PSIS and reducing it with regularization.
Place each answer inside the code chunk (grey box). The code chunks should contain a text response or a code that completes/answers the question or activity requested. Make sure to include plots if the question requests them. Problems are labeled Easy (E), Medium (M), and Hard(H).
Finally, upon completion, name your final output .html file as: YourName_ANLY505-Year-Semester.html and publish the assignment to your R Pubs account and submit the link to Canvas. Each question is worth 5 points.
Questions
7E1. State the three motivating criteria that define information entropy. Try to express each in your own words.
#1. Continuous. A continuous scale should be used in the measurement and so the spacing between adjacent values is consistent. 2. Increasing with number of possible events. Uncertainty increases when the number of different outcome increase. 3. Additive. Uncertainty os additive for independent events.
7E2. Suppose a coin is weighted such that, when it is tossed and lands on a table, it comes up heads 70% of the time. What is the entropy of this coin?
p <- c(0.7, 1 - 0.7)
(H <- -sum(p * log(p)))
## [1] 0.6108643
7E3. Suppose a four-sided die is loaded such that, when tossed onto a table, it shows “1” 20%, “2” 25%, “3” 25%, and “4” 30% of the time. What is the entropy of this die?
p <- c(0.20, 0.25, 0.25, 0.30)
(H <- -sum(p * log(p)))
## [1] 1.376227
7E4. Suppose another four-sided die is loaded such that it never shows “4”. The other three sides show equally often. What is the entropy of this die?
p <- c(1/3, 1/3, 1/3)
(H <- -sum(p * log(p)))
## [1] 1.098612
7M1. Write down and compare the definitions of AIC and WAIC. Which of these criteria is most general? Which assumptions are required to transform the more general criterion into a less general one?
# Akaike Information Criterion (AIC)- is defined as sum of deviance of in sampling training data (Dtrain) and twice the number of parameters (p) follows:
# AIC = Dtrain + 2p
#
# WAIC is defined as the difference between sum of average likelihood of observations
# WAIC(y, Θ) = -2(lppd - varΘ logp(yi|Θ))
# Both of them share an estimate of the in-sample training deviance and an estimate for the number of free parameters estimated in the model
# However, WAIC is the most general, which doesn't assume Gaussian posterior. . On the other side, WAIC assumes that the posteriror distribution is close to multivariate Gaussian
7M2. Explain the difference between model selection and model comparison. What information is lost under model selection?
# Model selection is choosing to retain the model with the lowest information criterion value and to discard all other models with higher values . This practice loses information about relative model accuracy contained in the differences among information criterion values; this is especially problematic when the selected model only outperforms its alternatives to a small degree. Model averaging is using Bayesian information criteria to construct a posterior predictive distribution that leverages the uncertainty in multiple models . This practice does not lose information on its own. However, when combined with undisclosed data dredging, it can lead to spurious findings
7M3. When comparing models with an information criterion, why must all models be fit to exactly the same observations? What would happen to the information criterion values, if the models were fit to different numbers of observations? Perform some experiments, if you are not sure.
# Information criteria are based on the total deviance, which is accrued over the number of observations. Thus, different numbers of observations it will lead to different Information Criterion in model cmparison.
#
# For example, when we calculate WAIC for models with a small sample of the data. The information criteria should decrease with the small sample size.
7M4. What happens to the effective number of parameters, as measured by PSIS or WAIC, as a prior becomes more concentrated? Why? Perform some experiments, if you are not sure.
#The effective number of parameters decreases as the prior becomes more concentrated.In the case of PSIS, with more concentrated priors, the model becomes less flexible. In the case of WAIC, the likelihood will become more concentrated as well and thus variance will decrease with more concentrated priors.
# WAIC(y, Θ) = -2(lppd - varΘ logp(yi|Θ))
7M5. Provide an informal explanation of why informative priors reduce overfitting.
# Informative priors are regularizing which reduces overfitting. As we know, they constrain the flexibility of the model; they make it less likely for extreme parameter values to be assigned high posterior probability.
7M6. Provide an informal explanation of why overly informative priors result in underfitting.
# Overly informative priors result in underfitting as they constrain the flexibility of the model too conservative. Therefore, the number of parameters these priors look for are too limited and picky to develop a proper model. | 2021-10-27T04:23:48 | {
"domain": "amazonaws.com",
"url": "https://rstudio-pubs-static.s3.amazonaws.com/765240_838aaa18d8434fe48546aca87b166bc9.html",
"openwebmath_score": 0.6823409199714661,
"openwebmath_perplexity": 1512.7406306194468,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.971129093889291,
"lm_q2_score": 0.868826777936422,
"lm_q1q2_score": 0.8437429616041499
} |
https://math.stackexchange.com/questions/1412450/simplifying-cube-roots-containing-a-square-root | # Simplifying Cube Roots Containing a Square Root
I was doing a problem today, and arrived at the (correct) answer of $x^3 = 16000\sqrt2$
Obviously I want to simplify this further. My text book jumps straight to $x = 20\sqrt2$ with no explanation.
In attempting to simplify it, I've got:
$x = \sqrt[3] {16000\sqrt{2}}$
$x = (\sqrt[3]{16000})(\sqrt[3]{\sqrt2})$
$x = (\sqrt[3]{800})(\sqrt[3]{2})(\sqrt[3]{\sqrt2})$
$x = 20\sqrt[3]{2\sqrt{2}}$
Could someone please explain the steps needed to get to $20\sqrt{2}$?
I accept this is trivial, but I'm stumped. Thanks in advance.
• Note that $2=\sqrt{2}\sqrt{2}$ – John Joy Aug 28 '15 at 12:30
• @JohnJoy Was there some reason you did not want to post that as an answer? It seems to be the most applicable response to this question as asked. (I would have upvoted it.) – David K Aug 28 '15 at 13:52
• @David K I took a second look at the other answers and realized that mine was just a restatement of mathlove's answer. Feel free to upvote my comment though. – John Joy Aug 28 '15 at 15:32
• @JohnJoy I was thinking specifically of how one gets "unstuck" after writing $\sqrt[3]{2\sqrt{2}}$. But you're right, mathlove's answer also gives a big clue about this and OP accepted it, so I guess that's enough. – David K Aug 28 '15 at 17:11
• Please feel free to elaborate further - I've enjoyed your comments so far. – Bangkockney Aug 28 '15 at 17:21
You're almost there. Note that $2=\sqrt{2}\sqrt{2}$, then agrue as follows $$x=20\sqrt[3]{2\sqrt{2}}=20\sqrt[3]{(\sqrt{2}\sqrt{2})\sqrt{2}}=20\sqrt[3]{\sqrt{2}^3}=20\sqrt{2}$$
• This is a great answer too. I particularly appreciate how you followed on from my working. Thanks for this. – Bangkockney Aug 29 '15 at 1:18
$$16000\sqrt 2=8000\times 2\sqrt 2=(20)^3\cdot (\sqrt 2)^3=(20\sqrt 2)^3$$
• Thank you. This is wonderfully clear to me. I seem to regularly forget to think in this way. – Bangkockney Aug 28 '15 at 10:34
• @Bangkockney: You are welcome! – mathlove Aug 28 '15 at 10:36
$$16000\sqrt2=1000\cdot2^4\cdot2^{1/2}=10^3\cdot2^{4+1/2}$$
$\implies$ the principal value of $$\sqrt[3]{6000\sqrt2}=10\cdot2^{3/2}$$
Now $2^{3/2}=2^{1+1/2}=2\sqrt2$ | 2019-09-15T13:22:32 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1412450/simplifying-cube-roots-containing-a-square-root",
"openwebmath_score": 0.6569045782089233,
"openwebmath_perplexity": 946.4084478922828,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9711290963960278,
"lm_q2_score": 0.8688267745399465,
"lm_q1q2_score": 0.8437429604836536
} |
http://math.stackexchange.com/questions/207735/what-is-the-correct-definition-of-the-absolute-value-of-x-x | # What is the correct definition of the absolute value of $x$, $|x|$?
What is the correct definition of the absolute value of $x$, $|x|$?
## Option A
$$|x|= \begin{cases} -x&\text{if } x < 0\\ 0& \text{if } x=0\\ x&\text{if } x>0 \end{cases}$$
## Option B
$$|x|= \begin{cases} -x&\text{if } x \leq 0\\ x&\text{if } x>0 \end{cases}$$
## Option C
$$|x|= \begin{cases} -x&\text{if } x < 0\\ x&\text{if } x\geq 0 \end{cases}$$
-
They’re all correct: they all define exactly the same function. – Brian M. Scott Oct 5 '12 at 13:07
...that is, assuming $x\in\mathbb R$. If $x$ is complex, then the more appropriate definition is $\sqrt{(\Re x)^2+(\Im x)^2}$, where $\Re x$ and $\Im x$ are the real and imaginary parts of $x$. – Guess who it is. Oct 5 '12 at 13:11
From a computer science point of view, Option C is better, because it requires one check and one operation for negative values. Option B is a scooch slower if $x=0$ – Thomas Andrews Oct 5 '12 at 13:33
"The" correct definition is the one given previously in that book. – GEdgar Oct 5 '12 at 14:15
DISTANCE FROM ZERO – Berci Oct 5 '12 at 14:36
What about Option D? That is:
$$|x|=\begin{cases}-x & x<0\\ x^2 & x=0,x=1\\ x & \text{otherwise}.\end{cases}$$
In all seriousness, there are infinitely-many (seemingly) distinct ways to define $|x|$ piecewise, but in the end, they are precisely the same. All you've got to do is pick one.
-
So among others, when do you prefer your definition? – kiss my armpit Oct 5 '12 at 13:14
@ガベージコレクタ: You don’t even need to define it piecewise: $|x|=\sqrt{x^2}$ also works. – Brian M. Scott Oct 5 '12 at 13:15
I don't prefer my definition. I was simply providing yet another alternative to illustrate the point. Quite frankly, if I had to pick one (though I can't think why I would), I'd probably go with Brian's excellent suggestion, just to avoid dealing with piecewise functions unnecessarily. – Cameron Buie Oct 5 '12 at 13:21
Different functions which compute exactly the same thing over the same domains are not precisely the same. They do not express the computation in the same way. They are only the same in situations when you don't care what the formula looks like (because, say, you don't intend to manipulate it). – Kaz Oct 5 '12 at 18:57
@Kaz: You're talking about computations and formulae, not about functions. – Chris Eagle Oct 5 '12 at 19:11
There are many equivalent ways describing $|x|$ as a function of $x$, what about: $\max\{x,-x\}$?
It is important to remember that the way to describe a set is not important, what important is the set itself.
I can tell you to take three rights; or I could just tell you to take one left. The input is the same and the output is the same, and that is what matters.
In the context of the real numbers there are several ways to describe the absolute value, and they are all the same.
-
A function is both a rule and a domain on which that rule takes place. If, for two functions, the rule and the domain are the same, then the two functions are the same. Just to be clear, you gave several different rules, but they are all equivalent, i.e., they all agree for any $x$ you insert. And all of them have the same domain. So, they are all the same function. Which definition is best only depends on what you are working on.
For example, when you want to find the $\lim\limits_{x \to 0^-} \frac{|x|}{x}$, then any of your definitions will do just fine, and all work better than say $|x| = \max\{x, -x\}$ or $\sqrt{x^2}$.
$$\lim_{x \to 0^-} \frac{|x|}{x} = \lim_{x \to 0^-} \frac{-x}{x} = \lim_{x \to 0^-} -1 = -1$$
But, in other situations, maybe you'd prefer $\sqrt{x^2}$ for some reason.
-
Form C is the most elegant, because it breaks the problem into exactly the right cases: the subdomain of $x$ for which something has to be done to produce the absolute value, and a subdomain which is already identical to its absolute value. Only if $x$ is negative do we have to negate it, otherwise we leave it as is.
Option B is simply silly. Why include 0 in the domain that requires negating?
Option A is verbose, but it has a certain symmetry. In any case, it is better than B because at least it regards 0 as special (which it generally is, even if not specifically in this situation), rather than clumping it with the negatives.
- | 2015-08-02T10:42:36 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/207735/what-is-the-correct-definition-of-the-absolute-value-of-x-x",
"openwebmath_score": 0.8419495224952698,
"openwebmath_perplexity": 361.9500610264536,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9711290913825541,
"lm_q2_score": 0.8688267745399466,
"lm_q1q2_score": 0.8437429561278136
} |
https://www.physicsforums.com/threads/conservation-of-energy-of-a-pendulum.521641/ | # Conservation of Energy of a pendulum
1. Aug 14, 2011
### knowNothing23
A pendulum consists of a 2.0-kg bob attached to a light 3.0-m-long
string. While hanging at rest with the string vertical, the bob is struck a sharp
horizontal blow, giving it a horizontal velocity of 4.5 m/s. At the instant the string
makes an angle of 30° with the vertical, what is (a) the speed, (b) the gravitational
potential energy (relative to its value is at the lowest point), and (c) the tension in
the string? (d) What is the angle of the string with the vertical when the bob
reaches its greatest height?
My problem is with the answer of c.)
The book shows a different answer.
Thank you.
3. The attempt at a solution
I have taken the free body diagram of the pendulum and the string, when the string forms 30 degrees with the vertical.
The net force sum of the "y" component:
$\sumF$=Tcos(30)-mg=m(v^(2) )/ L , where "L" is the length of the string.
Solving for T:
T=(m/cos(30) )( v^(2)/L +g)
2. Aug 14, 2011
### PeterO
Provided you have used that equation correctly it should get the right answer - assuming you had part (a) correct.
In case it is the way you are using the formula poorly - evaluate this
T = [mg + (m.v^2)/L]/cos30
3. Aug 15, 2011
### knowNothing23
My answer in part a.) is equal to the one provided by the book. The answer is 3.5171 m/s.
The book's answer of problem c.) is 25N.
4. Aug 15, 2011
### PeterO
I think you [and me] were looking at this incorrectly.
T = (mv^2 / L) + mgcos(theta) not (mv^2 / L) + mg/cos(theta)
The mv^2 / L part we agree with, I shall ignore that.
The mg/cos(theta) parts comes from analysing a conical pendulum situation, when the acceleration is horizontal.
HOWEVER, this time the bob is merely swinging - and importantly slowing down.
The net force for this part of the analysis is tangential to the arc, not horizontal.
Draw a "free body diagram" for that situation and you will see what is happening.
I don't know how to attach diagrams or pictures to this page - so I have tried to describe it.
5. Aug 16, 2011
### knowNothing23
I'm not sure, what you mean. In a conic situation there is a centripetal acceleration as well as a tangential. Also, the equation I proposed is not the one you mentioned.
T=(m/cos(30) ) ( v^(2)/L) +g), the cosine of 30 degrees divides mv^(2)/L as well.
There are more than two ways to analyze the sum of vectors at the angle of 30 degrees and both ways should give the same result. We can either take the component of T or the component of Weight and sum them. In my approach I took the component of T and subtracted Weight from it.
6. Aug 16, 2011
### ehild
The direction of the centripetal force is the same as that of the tension in the string. It is not vertical.
The usual procedure is to write the force components parallel and perpendicular to the string. The parallel component of gravity is mgcos(30°). The centripetal force is equal to the resultant of the tension and the radial component of gravity:
Mv^2/L = T-mgcos(30°) Use v=3.5171 m/s.
ehild
7. Aug 17, 2011
### knowNothing23
Thank you, ehild.
That was my question. Why take the component of the weight, when one can do it for the tension.
8. Aug 18, 2011
### ehild
See attached figures. The one on the left shows the forces acting on the bob: tension and gravity. The resultant is shown by blue. It is not parallel with the string as the bob has both velocity and acceleration along the arc. The resultant force is equal to the centripetal force (along the string) + the tangential force, tangent to the arc T+G = Fcp+Ft (all vectors).Writing this equation in x and y components, you get:
T cos(30°)-mg=Fcp cos(30°)-Ft sin(30°)
-T sin(30°) = -Fcp sin(30°)-Ft cos(30°)
Solve. If you do it correctly, you get T=Fcp+mgcos(30°). A bit more complicated than the other one, with radial and tangential components, is not it?
ehild
#### Attached Files:
• ###### pendulum2.JPG
File size:
7.1 KB
Views:
80
9. Aug 18, 2011
### ehild
The net force is neither horizontal nor tangential: it has a component along the arc and also the radial component: the centripetal force.
When writing a post, scroll down, click on "Manage Attachments"
ehild
10. Aug 20, 2011
Thank you! | 2017-09-24T16:01:16 | {
"domain": "physicsforums.com",
"url": "https://www.physicsforums.com/threads/conservation-of-energy-of-a-pendulum.521641/",
"openwebmath_score": 0.5987815260887146,
"openwebmath_perplexity": 1137.7543954426453,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9711290922181331,
"lm_q2_score": 0.8688267660487572,
"lm_q1q2_score": 0.8437429486077459
} |
https://gmatclub.com/forum/bill-and-ted-each-competed-in-a-240-mile-bike-race-227911.html | GMAT Question of the Day - Daily to your Mailbox; hard ones only
It is currently 18 Nov 2018, 14:54
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
## Events & Promotions
###### Events & Promotions in November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
• ### How to QUICKLY Solve GMAT Questions - GMAT Club Chat
November 20, 2018
November 20, 2018
09:00 AM PST
10:00 AM PST
The reward for signing up with the registration form and attending the chat is: 6 free examPAL quizzes to practice your new skills after the chat.
• ### The winning strategy for 700+ on the GMAT
November 20, 2018
November 20, 2018
06:00 PM EST
07:00 PM EST
What people who reach the high 700's do differently? We're going to share insights, tips and strategies from data we collected on over 50,000 students who used examPAL.
# Bill and Ted each competed in a 240-mile bike race.
Author Message
TAGS:
### Hide Tags
CEO
Joined: 11 Sep 2015
Posts: 3122
Bill and Ted each competed in a 240-mile bike race. [#permalink]
### Show Tags
Updated on: 25 May 2017, 04:16
2
Top Contributor
7
00:00
Difficulty:
55% (hard)
Question Stats:
74% (02:58) correct 26% (03:17) wrong based on 183 sessions
### HideShow timer Statistics
Bill and Ted each competed in a 240-mile bike race. Bill’s average speed was 5 miles per hour slower than Ted’s average speed. If Ted completed the race 4 hours sooner than Bill did, what was Bill’s average speed in miles per hour?
A) 7.5
B) 10
C) 12
D) 12.5
E) 15
*Kudos for all correct solutions.
_________________
Test confidently with gmatprepnow.com
Originally posted by GMATPrepNow on 26 Oct 2016, 11:43.
Last edited by GMATPrepNow on 25 May 2017, 04:16, edited 1 time in total.
Manager
Joined: 05 Jun 2015
Posts: 83
Location: United States
WE: Engineering (Transportation)
Re: Bill and Ted each competed in a 240-mile bike race. [#permalink]
### Show Tags
26 Oct 2016, 17:04
1
1
Vb and Vt be the average speeds of Bill and Ted.
Vb = Vt - 5
Tb and Tt be the time take,
Tt = Tb - 4
We have Distance = Speed * Time
$$Vb* Tb = Vt * Tt = 240$$
$$(Vt - 5) * ( Tt + 4) = Vt * Tt$$
4Vt - 5 Tt = 20
Replacing value of Tt as 240/Vt
$$4 Vt - 5*(\frac{240}{Vt}) = 20$$
$$Vt^{2} - 5 Vt = 300$$
$$(Vt -20)* (Vt +15) = 0$$
Vt = 20
Average speed of Bill Vb = Vt-5 = 20- 5 = 15.
Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 4170
Location: United States (CA)
Re: Bill and Ted each competed in a 240-mile bike race. [#permalink]
### Show Tags
27 Oct 2016, 16:26
3
GMATPrepNow wrote:
Bill and Ted each competed in a 240-mile bike race. Bill’s average speed was 5 miles per hour slower than Ted’s average speed. If Ted completed the race 4 hours sooner than Bill did, what was Bill’s average speed in miles per hour?
A) 7.5
B) 10
C) 12
D) 12.5
E) 15
We are given that Bill and Ted each competed in a 240-mile bike race. We are also given that Bill’s average speed was 5 miles per hour slower than Ted’s average speed.
Both Bill and Ted had distance of 240 miles. If we let Ted’s speed = r, we can let Bill’s speed = r - 5. We can use the above information to determine the time of Bill and Ted in terms of variable r.
Since time = distance/rate, Ted’s time = 240/r and Bill’s time = 240/(r - 5).
Since Ted completed the race 4 hours sooner than Bill did, we can create the following equation:
240/r + 4 = 240/(r - 5)
To eliminate the denominators of the fractions we can multiply the entire equation by r(r-5) and we have:
240(r - 5) + 4[r(r - 5)] = 240r
240r - 1,200 + 4r^2 - 20r = 240r
4r^2 - 20r - 1,200 = 0
r^2 - 5r - 300 = 0
(r - 20)(r + 15) = 0
r = 20 or r = -15
Since r must be positive, r = 20.
Thus, Bill’s rate = 20 - 5 = 15 mph.
_________________
Scott Woodbury-Stewart
Founder and CEO
GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions
Senior Manager
Joined: 03 Apr 2013
Posts: 278
Location: India
Concentration: Marketing, Finance
GMAT 1: 740 Q50 V41
GPA: 3
Re: Bill and Ted each competed in a 240-mile bike race. [#permalink]
### Show Tags
24 May 2017, 22:35
4
GMATPrepNow wrote:
Bill and Ted each competed in a 240-mile bike race. Bill’s average speed was 5 miles per hour slower than Ted’s average speed. If Ted completed the race 4 hours sooner than Bill did, what was Bill’s average speed in miles per hour?
A) 7.5
B) 10
C) 12
D) 12.5
E) 15
*Kudos for all correct solutions.
You can find our video solution here: https://www.gmatprepnow.com/module/gmat ... /video/947
Another way to build up the quadratic...
Let Bill's speed = x
and Ted's speech = x+5
When Ted covers the full distance, Bill has yet to travel for 4 more hours to complete the distance
$$\frac{240}{(240 - 4x)} = \frac{(x + 5)}{x}$$
_________________
Spread some love..Like = +1 Kudos
CEO
Joined: 11 Sep 2015
Posts: 3122
Re: Bill and Ted each competed in a 240-mile bike race. [#permalink]
### Show Tags
25 May 2017, 04:14
1
Top Contributor
GMATPrepNow wrote:
Bill and Ted each competed in a 240-mile bike race. Bill’s average speed was 5 miles per hour slower than Ted’s average speed. If Ted completed the race 4 hours sooner than Bill did, what was Bill’s average speed in miles per hour?
A) 7.5
B) 10
C) 12
D) 12.5
E) 15
*Kudos for all correct solutions.
Another approach....
Bill’s average speed was 5 miles per hour slower than Ted’s average speed.
Let B = Bill's travel speed
So, B + 5 = Ted's average speed
Ted completed the race 4 hours sooner than Bill did
(Bill's travel time) = (Ted's travel time) + 4
time = distance/speed
So, we get: 240/B = 240/(B + 5) + 4
Rewrite 4 as 4(B + 5)/(B + 5) to get: 240/B = 240/(B + 5) + 4(B + 5)/(B + 5)
Simplify: 240/B = 240/(B + 5) + (4B + 20)/(B + 5)
Combine terms: 240/B = (4B + 260)/(B + 5)
Cross multiply: 240(B + 5) = (B)(4B + 260)
Expand and simplify: 240B + 1200 = 4B² + 260B
Set equal to zero: 4B² + 20B - 1200 = 0
Divide both sides by 4 to get: B² + 5B - 300 = 0
Factor: (B + 20)(B - 15) = 0
So, EITHER B = -20 OR B = 15
Since B (Bill's speed) cannot be a negative value, we can conclude that B = 15.
RELATED VIDEO
_________________
Test confidently with gmatprepnow.com
Director
Joined: 13 Mar 2017
Posts: 630
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.8
WE: Engineering (Energy and Utilities)
Re: Bill and Ted each competed in a 240-mile bike race. [#permalink]
### Show Tags
23 Oct 2017, 22:22
1
GMATPrepNow wrote:
Bill and Ted each competed in a 240-mile bike race. Bill’s average speed was 5 miles per hour slower than Ted’s average speed. If Ted completed the race 4 hours sooner than Bill did, what was Bill’s average speed in miles per hour?
A) 7.5
B) 10
C) 12
D) 12.5
E) 15
*Kudos for all correct solutions.
Let Ted's average speed be x miles/hr.
Bill's average speed be x-5 miles/hr.
Distance = 240 miles
240/(x-5) - 240/x = 4
240*5/(x(x-5)) = 4
x(x-5) = 300
x = 20
x-5 = 15
_________________
CAT 2017 99th percentiler : VA 97.27 | DI-LR 96.84 | QA 98.04 | OA 98.95
UPSC Aspirants : Get my app UPSC Important News Reader from Play store.
MBA Social Network : WebMaggu
Appreciate by Clicking +1 Kudos ( Lets be more generous friends.)
What I believe is : "Nothing is Impossible, Even Impossible says I'm Possible" : "Stay Hungry, Stay Foolish".
Manager
Joined: 03 May 2014
Posts: 161
Location: India
WE: Sales (Mutual Funds and Brokerage)
Bill and Ted each competed in a 240-mile bike race. [#permalink]
### Show Tags
03 Nov 2017, 03:44
1
Easier method
Bills time be= t and speed=x
Ted's time=t-4 and speed=x+5
distance =240
Bill's time(t) = distance/speed=240/x
Teds time(t-4)=240/(x+5)
Plugin answer choices. you should start from c or answers which are whole numbers
Let bills speed be =15
we now have
Bill(t)=240/15=16
Now Plugin for ted
t-4=240/(15+5)=240/20=12 or t=12+4=16.
with above approach we can avoid quadratic equation.
VP
Joined: 07 Dec 2014
Posts: 1114
Bill and Ted each competed in a 240-mile bike race. [#permalink]
### Show Tags
04 Nov 2017, 14:19
1
GMATPrepNow wrote:
Bill and Ted each competed in a 240-mile bike race. Bill’s average speed was 5 miles per hour slower than Ted’s average speed. If Ted completed the race 4 hours sooner than Bill did, what was Bill’s average speed in miles per hour?
A) 7.5
B) 10
C) 12
D) 12.5
E) 15
*Kudos for all correct solutions.
let r=Bill's average speed
Bill: d=rt
Ted: d=(r+5)(t-4)
combining,
rt=(r+5)(t-4)➡
4r=5t-20
substituting,
4r=5(240/r)-20➡
r^2+5r-300=0
(r+20)(r-15)=0
r=15 mph
E
Bill and Ted each competed in a 240-mile bike race. &nbs [#permalink] 04 Nov 2017, 14:19
Display posts from previous: Sort by | 2018-11-18T22:54:48 | {
"domain": "gmatclub.com",
"url": "https://gmatclub.com/forum/bill-and-ted-each-competed-in-a-240-mile-bike-race-227911.html",
"openwebmath_score": 0.54013991355896,
"openwebmath_perplexity": 14944.595845416236,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. Yes\n2. Yes",
"lm_q1_score": 0.9711290897113961,
"lm_q2_score": 0.8688267677469952,
"lm_q1q2_score": 0.843742948079034
} |
http://math.stackexchange.com/questions/26869/constants-of-integration-in-integration-by-parts/26871 | # Constants of integration in integration by parts
After finishing a first calculus course, I know how to integrate by parts, for example, $\int x \ln x dx$, letting $u = \ln x$, $dv = x dx$: $$\int x \ln x dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2x} dx.$$
However, what I could not figure out is why we assume from $dv = x dx$ that $v = \frac{x^2}{2}$, when it could be $v = \frac{x^2}{2} + C$ for any constant $C$. The second integral would be quite different, and not only by a constant, so I would like to understand why we "forget" this constant of integration.
Thanks.
-
@Mariano: I can't believe I missed that! I feel so dumb now. If you would like to leave this as an answer, I'd be more than happy to accept it. – Abel Mar 14 '11 at 7:52
The second integral would change, but also the first term... Have you actually checked to see what happens if you change the constant?
-
Take your example, $$\int x\ln x\,dx.$$ Note $x\gt 0$ must be assumed (so the integrand makes sense).
If we let $u = \ln x$ and $dv= x\,dx$, then we can take $v$ to be any function with $dv = x\,dx$. So the "generic" $v$ will be, as you note, $v = \frac{1}{2}x^2 + C$. What happens then if we use this "generic" $v$? \begin{align*} \int x\ln x\,dx &= \ln x\left(\frac{1}{2}x^2 + C\right) - \int \left(\frac{1}{2}x^2+C\right)\frac{1}{x}\,dx\\ &= \frac{1}{2}x^2\ln x + C\ln x - \int\left(\frac{1}{2}x + \frac{C}{x}\right)\,dx\\ &= \frac{1}{2}x^2\ln x + C\ln x - \frac{1}{4}x^2 - C\ln x + D\\ &= \frac{1}{2}x^2\ln x - \frac{1}{4}x^2 + D, \end{align*} so in the end, we get the same result no matter what value of $C$ we take for $v$.
This says that we can take any value of $C$ and still get the same answer. Since we can take any value of $C$, why not take the simplest one, the one that does not require us to carry around an extra term that is going to cancel out anyway? Say..., $C=0$?
This works in general. If you replace $v$ with $v+C$ in the integration by parts formula, you have \begin{align*} \int u\,dv &= u(v+C) - \int(v+C)\,du = uv + Cu - \int v\,du - \int C\,du\\ &= uv+Cu - \int v\,du - Cu = uv-\int v\,du. \end{align*} So the answer is the same regardless of the value of $C$, and so we take $C=0$ because that makes our life simpler.
-
Your observation that $dv=xdx$ does not imlpy $v=x^2/2$ is correct.
Your confusion resolves when you say it this way: we set $v=x^2/2$ and this implies $dv=xdx$.
-
$$\int x \ln x dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2x} dx.$$
You could always write $$v = \frac{x^2}{2} + C$$ but that won't matter much because the final result would also involve a constant (Say $K$ which would be equal to $C+k$ )
-
This is not exactly what happens... – Mariano Suárez-Alvarez Mar 14 '11 at 7:48
Arturo's answer makes me realize this is ok. If you make an edit, howerver small, I will remove my downvote. – Please Delete Account Mar 14 '11 at 15:33
@Approximist : Done! – Prasoon Saurav Mar 14 '11 at 16:44
HINT $\rm\ \ C'=0\ \ \Rightarrow\ \ (UV)'-U'\:V\ =\ UV'\: =\ U(V+C)'\: =\ (U(V+C))'-U'\:(V+C)$
-
We "forget" it, and add it in the last step. The whole point in the constant of integration is to remind us that there could have been a constant term added on at the end originally, but in the process of differentiation we got rid of it because it did not affect the slope.
-
We didn't "forget". We simply choose C in a way that the resulting $\int u\,dv$ would be the simplest form. Usually $C = 0$ but not always. If, for example, we have this integral:
$$\int \ln(x+2) \,dx$$
Then you would choose $v = x + 2$ because $du = \frac{dx}{x+2}$
Second example:
$$\int x\ln(x+2) \,dx$$
Then $$v = \frac{x^2-4}{2} = \frac{(x-2)(x+2)}{2}$$ and $$u\,dv = \frac{x-2}{2} dx$$
- | 2014-10-23T22:18:35 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/26869/constants-of-integration-in-integration-by-parts/26871",
"openwebmath_score": 0.9961183667182922,
"openwebmath_perplexity": 329.75942067965235,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9579122732859021,
"lm_q2_score": 0.8807970858005139,
"lm_q1q2_score": 0.843726338762768
} |
https://mathematica.stackexchange.com/questions/163757/centralities-in-weighted-networks | # Centralities in Weighted Networks
I have a weighted, undirected network with five nodes, whose adjacency matrix is given by
A = {{0, 3, 7, 0, 0}, {3, 0, 6, 0, 0}, {7, 6, 0, 2, 1}, {0, 0, 2, 0,
4}, {0, 0, 1, 4, 0}};
I've created a weighted graph using A as follows:
G = WeightedAdjacencyGraph[ReplacePart[A, {j_, j_} -> \[Infinity]],
VertexLabels -> "Name", VertexSize -> Automatic,
VertexLabelStyle -> Large]
I'm interested in computing various centrality measures, specifically degree, closeness, and betweenness. For the first, I entered
DegreeCentrality[G]
The result was {4, 4, 4, 4, 4}, which is incorrect given my positive, integer-valued weights, and instead corresponds to the degree centrality of a complete graph on five vertices.
1. I'm curious to know how I should go about computing the degree centrality for this graph.
2. Also, can the ClosenessCentrality and BetweennessCentrality commands be used for weighted graphs, and, if so, what basic approach do they take. For example, is betweenness based upon shortest distances (as described in Social Networks, v13, 1991,141-154), where we use the reciprocals of the edge weights)?
• I don't know enough about graph theory to know, but is the fact that IsomorphicGraphQ[G, CompleteGraph[5]] returns True a bug? If so, that might be the underlying issue. – Jason B. Jan 15 '18 at 20:58
• @JasonB. The issue is that many builtins don't support weighted graphs. It was one of my primary motivations for starting IGraph/M. It's not related to IsomorhicGraphQ. – Szabolcs Jan 15 '18 at 21:05
• @Szabolcs - I'm just noticing that graph constructed by WeightedAdjacencyGraph looks completely different from the IGWeightedAdjacencyGraph result. It looks exactly like the pentagram created by CompleteGraph[5]. Is that correct? – Jason B. Jan 15 '18 at 21:07
• 0 in matrix considered as edge with weight 0. Op may means A /. {0 -> Infinity} for weighted adjacency matrix. Also DegreeCentrality is nothing to do with edge weights. – halmir Jan 15 '18 at 21:08
• @JasonB. I didn't originally notice that the OP wanted a complete graph with no self-loops, but some 0-weights. IGWeightedAdjacencyGraph exists because I find it extremely annoying that WeightedAdjacencyMatrix uses 0 to represent missing connections but WeightedAdjacencyGraph uses Infinity. Both 0 and Infinity make sense (in different cases), but at least these two should be exact inverses. IGWeightedAdjacencyGraph simply lets you choose what to use, and uses 0 by default. – Szabolcs Jan 15 '18 at 21:09
If you are working with weighted graphs, I highly recommend my package IGraph/M, which makes this much easier in many situations. (Apologies to everyone else for bringing this up yet again.)
Here's a little demo:
A = {{0, 3, 7, 0, 0}, {3, 0, 6, 0, 0}, {7, 6, 0, 2, 1}, {0, 0, 2, 0, 4}, {0, 0, 1, 4, 0}};
Annoyed about having to replace zeros with infinities? Use IGWeightedAdjacencyGraph. It can use any matrix element as the notation for a missing connection. The default is 0 (or 0.).
Let's also thicken those edges at the same time.
g = IGWeightedAdjacencyGraph[A, EdgeStyle -> Thick]
Visualize weight as edge colour:
Legended[
IGEdgeMap[ColorData["SolarColors"], EdgeStyle -> Rescale@*IGEdgeProp[EdgeWeight], g],
BarLegend[{"SolarColors", MinMax@IGEdgeProp[EdgeWeight][g]}]
]
Visualize weight as edge thichkness:
IGEdgeMap[AbsoluteThickness, EdgeStyle -> IGEdgeProp[EdgeWeight], g]
I assume that when you mentioned DegreeCentrality, you wanted the strength of each vertex, i.e. the sum of edge weights for their incident edges. IGraph/M has that:
IGVertexStrength[g]
(* {10, 9, 16, 6, 5} *)
Want to scale vertices accordingly?
IGVertexMap[0.03 # &, VertexSize -> IGVertexStrength, g]
The built-in ClosenessCentrality does support weights, but BetweennessCentrality does not. IGraph/M has both, with weights support (based on the igraph library).
IGCloseness[g]
(* {0.037037, 0.0416667, 0.0625, 0.0454545, 0.0526316} *)
IGBetweenness[g]
(* {0., 0., 5., 0., 0.} *)
In both these functions (as well as in the built-in ClosenessCentrality), the edge weights are used directly in the shortest path calculations as the length of an edge. They are not transformed (e.g. inverted) in any way.
If you want to construct a new graph with inverted edge weights, a simple way is
IGEdgeMap[1/# &, EdgeWeight, g]
Verify the result:
IGEdgeProp[EdgeWeight][%]
(* {1/3, 1/7, 1/6, 1/2, 1, 1/4} *)
If you want to remove weights, you can use IGUnweighted. To test if a graph is edge-weighted, use IGEdgeWeightedQ (and note that the builtin WeightedGraphQ would return True for a non-edge weighted vertex weighted graph).
There are several other functions in IGraph/M which are useful with weighted graphs, or provide some advantage over the built-in equivalent when weights are present. E.g. IGReverseGraph correctly preserves edge weights while ReverseGraph does not.
I also accept suggestions for new features/functions, especially as they relate to weighted graphs, which I use myself extensively. | 2021-01-21T06:07:45 | {
"domain": "stackexchange.com",
"url": "https://mathematica.stackexchange.com/questions/163757/centralities-in-weighted-networks",
"openwebmath_score": 0.36722689867019653,
"openwebmath_perplexity": 1590.6778437622722,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9579122684798184,
"lm_q2_score": 0.8807970873650401,
"lm_q1q2_score": 0.8437263360282624
} |
https://math.stackexchange.com/questions/2479806/probability-of-getting-correct-answer-in-a-multi-choice-question | # Probability of getting correct answer in a multi-choice question.
In a multiple choice question there are 4 alternative answers of which 1, 2, 3, or all may be correct. A candidate decides to tick answers at random. If he is allowed upto 5 chances to answer the question, the probability that he will get the marks in the question is?
It is equally likely for 1, 2, 3, or all to be correct so probability is 1/4.
Case1 when only 1 option is correct.
Since the candidate is allowed 5 chances, probability of getting correct answer is 1.
Case2 when 2 options are correct.
Toltal ways in which 2 options can be correct is $\left(^4_2\right)$, which is 6. Out of these only 1 is correct. So probability of selecting correct answer is 1/6. Since he has 5 chances the probability of getting marks is $1 - \left(\frac{5}{6}\right)^5$
Case3 when 3 options are correct.
Total ways in which 3 options can be correct is $\left(^4_3\right)$, which is 4. Since he has 5 chances, probability of getting correct answer is 1.
case4 when all options are correct.
Only one way in which all can be correct. Probality of getting marks is 1.
So answer should be $\left(\frac{1}{4}\times1\right) + \frac{1}{4}\times\left(1 - \left(\frac{5}{6}\right)^5\right) + \left(\frac{1}{4}\times1\right) + \left(\frac{1}{4}\times1\right)$.
Which is indeed wrong.
A friend of mine did this question as follows.
Total options: $\left(^4_1\right) + \left(^4_2\right) + \left(^4_3\right) + \left(^4_4\right)$, i.e 15. Since he has 5 chances to answer, probability would be 5/15. I know this is wrong (or not?) but beacause my textbook says answer is 1/3 I couldn’t argue.
• You are assuming that the candidate knows how many answers are correct, e.g. in the case that only one is correct, the candidate will not mark multiple questions (which would, in my understanding, be considered a wrong answer in a multiple choice test...). Is that intended? Furthermore, the fact that every case has the same chance of 1/4 is given in the question? If not, why can you assume it? I agree with your friend, but he starts with other assumptions, therefore it is important to know why you make them. – Dirk Oct 19 '17 at 10:26
• @DirkLiebhold I agree with your argument about my solution that I made wrong assumptions. Can you please explain why my friend's solution is right. – Voneone Oct 19 '17 at 10:39
• Your friend is correct. – N. F. Taussig Oct 19 '17 at 11:12
• @N.F.Taussig Can you please add answer explaining the same? – Voneone Oct 19 '17 at 11:23
• I am unable to understand how having 5 chances gives use 5 favourable events. Is it because we are adding probability of all five cases with each case having probability 1/15, i.e 1/15 + 1/15 + 1/15 + 1/15 + 1/15 – Voneone Oct 19 '17 at 11:30
There are four choices, each of which may be correct or incorrect. However, not all of them may be incorrect. Therefore, there are $2^4 - 1 = 15$ possible ways to answer the question. The candidate is allowed five guesses. That means the candidate chooses $5$ of the $15$ possible answers. The candidate receives credit if one of those guesses is correct, so the probability that the candidate guesses the correct answer is $$\frac{5}{2^4 - 1} = \frac{5}{15} = \frac{1}{3}$$
Your friend counted the number of possible answers in a different way. We have the option of choosing one, two, three, or four of the answers. There are $\binom{4}{k}$ ways to choose exactly $k$ of the four answers. Hence, the number of possible ways to answer the question is $$\binom{4}{1} + \binom{4}{2} + \binom{4}{3} + \binom{4}{4} = 15$$
Edit: To understand why the probability is the number of guesses divided by the number of answers, suppose that you are given $k$ guesses and $n$ answers, of which only one is correct. There are $\binom{n}{k}$ ways to make $k$ guesses. If you select the correct answer with one of those $k$ guesses, then you also select $k - 1$ of the $n - 1$ incorrect answers. Therefore, your probability of selecting the right answer among your $k$ guesses is \begin{align*} \frac{\dbinom{1}{1}\dbinom{n - 1}{k - 1}}{\dbinom{n}{k}} & = \frac{\dfrac{(n - 1)!}{(k - 1)![(n - 1) - (k - 1)]!}}{\dfrac{n!}{k!(n - k)!}}\\ & = \frac{(n - 1)!}{(k - 1)!(n - k)!} \cdot \frac{k!(n - k)!}{n!}\\ & = \frac{(n - 1)!}{n!} \cdot \frac{k!}{(k - 1)!}\\ & = \frac{k}{n} \end{align*} In our problem, $k = 5$ and $n = 15$, so the probability of obtaining the correct answer is $$\frac{5}{15} = \frac{1}{3}$$
• There are $\binom{15}{5}$ ways to choose five of the fifteen answers. If you choose the correct answer, you must also choose four of the fourteen incorrect answers. Therefore, the probability that you choose the correct answer is $$\frac{\binom{1}{1}\binom{14}{4}}{\binom{15}{5}} = \frac{5}{15} = \frac{1}{3}$$ In general, if you have $k$ choices, then the probability that you obtain the correct answer is $$\frac{\binom{1}{1}\binom{14}{k - 1}}{\binom{15}{k}} = \frac{k}{15}$$ as you should check. Notice that the answer is the number of guesses divided by the number of possible answers. – N. F. Taussig Oct 19 '17 at 19:11 | 2019-06-17T10:38:18 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2479806/probability-of-getting-correct-answer-in-a-multi-choice-question",
"openwebmath_score": 0.8994172811508179,
"openwebmath_perplexity": 202.83995575994973,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.981453437115321,
"lm_q2_score": 0.8596637559030338,
"lm_q1q2_score": 0.8437199479944988
} |
https://math.stackexchange.com/questions/924371/understanding-arc-length-parameterization-concept-behind-numbers | Understanding Arc Length Parameterization- Concept behind Numbers
So, my main motive for understanding this concept comes from a problem I had to solve.
it reads: Find an arc length parameterization of the line segment from $(1,2)$ to $(5,-2)$
In the book I'm using they don't have an explicit example of how to derive the arc length parameterization of a line segment, but I've kind of figured it out from poking around a few places, but I want to check my understanding of the concept as well as the solution I derive. Comments/critiques welcome at both the conceptual and minute detail.
Okay, so to find the arc length parameterization you determine the displacement vector
$$\vec{v} = <4,-4> \text{ and the distance } \mid \mid \vec{v} \mid\mid = \sqrt{32}$$
Thus the parameterization of this line segment can be given of the form:
$$\left\{ \begin{array}{l l} x=1 + \frac{4t}{\sqrt{32}} \\ y=2 - \frac{4t}{\sqrt{32}}\\ \end{array} \right.$$
So, check me here. I'm creating something like a unit vector, with the fraction component, no? Since that contains both the vector component divided by it's distance, and the starting point component which "pushes" the unitized parameter to the correct place, right?
I know the whole point of discussing this is so that we have a better understanding of a unit tangent vector, which we use to find curvature- and this is just trying to drive home the fact that we're creating a frame of reference (this portion of the curve).
Let me know if my understanding is off in some way, I'm learning this on my own, so help is appreciated.
• I don't see anything wrong with how you've constructed your arc-length parametrization. But I think it's worth emphasizing something: While $\langle x,y \rangle$ isn't a unit vector, its $t$-derivative is. (Which, as you mention in the next paragraph, means it has a unit tangent vector.). You could in fact start there, and take a definite integral in $t$ in order to produce the parametrization you found. – Semiclassical Sep 9 '14 at 2:26
• so I could integrate the unit vector to determine this same parameterization? – Adam Sep 9 '14 at 13:30
• Should be able to, yes---try it out! – Semiclassical Sep 9 '14 at 13:32
• If you do get it to work by integration of the unit tangent vector, then I'd encourage you to post that as an answer. That way we can provide feedback and you'll get a chance to build some more rep. – Semiclassical Sep 9 '14 at 20:36
$$\frac{4}{\sqrt{32}} \int_{0}^{t} ds = \frac{4t}{\sqrt{32}}$$
$$\frac{-4}{\sqrt{32}} \int_{0}^{t} ds = \frac{-4t}{\sqrt{32}}$$
• Nicely done. You can make this even a bit sharper if you note that your first LHS may be rewritten as $$x(t)-x(0)=\int_0^t x'(s)\,ds$$ so that the 'starting 'point' is already contained in $x(0)$. (Same for $y$.) – Semiclassical Sep 9 '14 at 22:56 | 2019-09-20T15:55:17 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/924371/understanding-arc-length-parameterization-concept-behind-numbers",
"openwebmath_score": 0.8171034455299377,
"openwebmath_perplexity": 322.0326756500452,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9814534360303622,
"lm_q2_score": 0.8596637541053281,
"lm_q1q2_score": 0.8437199452974347
} |
https://math.stackexchange.com/questions/2104756/explain-definition-of-least-upper-bound | # Explain definition of least upper bound .
The text that I have says something like this,
A real number s is the least upper bound for a set $A \subseteq R$ if it meets the following two criteria:
(1) s is an upper bound of A
(2) if b is any upper bound for A, then $s \le b$
Its the second statement that creates a confusion. I read answers to same question and the people have explained it with an example,
Ex: Let $[0,1]\subset \mathbb{R}$. Then we'd like to say that $1$ is the least upper bound. But as per the second definition you have, we find that $2$ is an upper bound and since $2$ is not not an upper bound the second condition you have is vacuous. Hence $2$ is a least upper bound. Which is not at all true.
But consider the statement it says, "if b is any upper bound for A" and if A is a set $[0,1]$ then 2, 3, any number above 1 is not a part of set A.
Can anyone explain how the second statement is true and what exactly it wants to convey regarding least upper bound?
• An upper bound doesn't have to be an element of the set. In fact: it usually isn't. If an upper bound is an element of the set, it's necessarily the least upper bound (and also the maximum). – StackTD Jan 19 '17 at 17:00
• Thanks, I was thinking in the wrong direction. So, there are infinitely many upper bounds but the least upper bound is unique and is unique to the set in consideration, and turns out to be the maximum or supremum (precisely) of the set A. – user405401 Jan 19 '17 at 17:09
• Yes, a least upper bound is exactly what it says it is. It is a bound. Which means no points of the set go "beyond" that. It is an upper bound. Which means to points of the set are more than it. And it is the least upper bound which means of all the upper bounds it is the least. A more confusing but somewhat more useful way of looking at it (who cares if 5432 – fleablood Jan 19 '17 at 17:54
• A more confusing way but somewhat more useful way of looking at it. (who cares if $5,403$ is an upper bound of $[0,1]$) is that if $b$ is the least upper bound $S$, then any $a < b$ is not an upper bound of S (as $b$ is least) so there is an $s \in S$ so that $a < s \le b$ (because $a$ is not a bound there is an $s$ beyond it). So $b$ is the "first" element that is equal or larger than all the S, or that doesn't have a s that is larger than it. – fleablood Jan 19 '17 at 18:16
• @fleablood, thanks for example. It was very helpful. – user405401 Jan 19 '17 at 18:53
Condition (1) says that $s$ is an upper bound. Condition (2) says that $s$ is less than or equal to any other upper bound. Hence, least upper bound.
The second condition or guarantees that the least upper bound, if it exists, is unique. Suppose if ${s_1}$ and ${s_2}$ are both least upper bounds for a set $A$, then part (2) of the definition says that both ${s_1} \le {s_2}$ and ${s_2} \le {s_1}$. Thus $s_1 = s_2$. So there can be infinite number of upper bounds to a set, but it can only have one least upper bound.
In your example, $2$ is an upper bound of $[0,1]$. Condition (1) is satisfied by $s=2$. But since $1.5$ is also an upper bound of $[0,1]$, and $1.5 < 2$, condition (2) is not satisfied. So $2$ is not the least upper bound of $[0,1]$.
Now $1$ is an upper bound of $[0,1]$. And in contrast to $2$, any number less than $1$ is not an upper bound of $[0,1]$ (being less than $1$ precludes it from also being $\geq 1$). So $1$ is the least upper bound of $[0,1]$.
• That an that there is a specific least one. If you "universe" is only the rational numbers and your set is the rationals whose squares are less than 2 (i.e. the interval $(-\sqrt{2}, \sqrt{2})$ restricted to only rational numbers) there are plenty of upper bounds but no one of them is smaller than all the others. But if you allow your universe to include irrationals. $\sqrt{2}$ is an upper bound and it is the least of all possible upper bounds. Note: the l.u.b. need not be in the set either. – fleablood Jan 19 '17 at 18:05 | 2019-12-12T04:09:52 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2104756/explain-definition-of-least-upper-bound",
"openwebmath_score": 0.9282312393188477,
"openwebmath_perplexity": 105.62888971022646,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9814534357591224,
"lm_q2_score": 0.8596637523076225,
"lm_q1q2_score": 0.8437199432998952
} |
https://math.stackexchange.com/questions/310568/what-can-be-said-about-this-function-based-on-the-derivative-being-0 | if it is known, for a continuous differentiable function f(x), f'(c)=0 and f''(c)=0, what can be concluded about the graph of f(x) at x=c ?
1. The tangent to the curve is horizontal
2. There must be a point of inflection
3. The curve must be a straight line because the curvature is 0
4. There is an extremum and a point of inflection and the same place
5. The curve must be a straight line if the derivatives are 0
There may be more than 1 correct answer.
I am having a hard time thinking any of them are true. I know that 5 would make sense but not if the equation is x^4 since the derivative of that at 0 would be 0. When it says the tangent to the curve is horizontal is it talking about just at that point of c? Thank you for the help!
• Yes, it’s talking about just that one point at $x=0$. As you say, $f(x)=x^4$ eliminates (5); it also eliminates all but one of the others. – Brian M. Scott Feb 21 '13 at 21:14
• ok that makes more sense, so only the first one is correct. Thanks! – user56852 Feb 21 '13 at 21:17
• There you go; you’ve got it. – Brian M. Scott Feb 21 '13 at 21:18
First question: Sure, $f'(c)=0$ means the slope of the tangent line at $x=c$ is $0$, so the tangent line exists and is horizontal.
Second: No, let $c=0$. The curve $x^4$ has no point of inflection, it is always "concave up."
Third: See second.
Fourth: See second.
Fifth: It is not clear what the question means. For sure the curve need not be a straight line if the first, second, third, up to $999$-th derivatives are $0$ at a particular point $c$. In fact there is a function which is not a straight line and has all its derivatives $0$ at $c$.
If the second derivative of $f(x)$ is $0$ everywhere, then indeed the curve $y=f(x)$ is a straight line.
• so that must mean that the first one is the only correct response. Thank you! – user56852 Feb 21 '13 at 21:16
• The "Second Derivative Test" does not give conclusive information about a function at a point $(c , f(c) )$ at which $f'(c) = 0$ and $f''(c) = 0$. As André Nicolas mentions, the curve $y = x^4$ has a local minimum at $x = 0$, while $y = x^3$ has an inflection point there, yet for both curves, $f'(0) = 0$ and $f''(0) = 0$. What we would need would be a third derivative test, but generally what we check is what happens to the concavity of the function on either side of $x = c$: if the sign of $f''(x)$ changes, then $x = c$ is an inflection point; if not, the point is a local extremum – colormegone Apr 25 '13 at 15:16 | 2019-09-19T08:57:11 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/310568/what-can-be-said-about-this-function-based-on-the-derivative-being-0",
"openwebmath_score": 0.8075363636016846,
"openwebmath_perplexity": 160.51440499258973,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9814534365728415,
"lm_q2_score": 0.8596637505099168,
"lm_q1q2_score": 0.8437199422350556
} |
http://ipmq.charlyedance.it/permutations-with-repetition-worksheet.html | WorksheetWorks. It could be "333". , order does not matter. Permutation of objects when all are not distinct: Permutation =, → Number of things among ‘n’ which are alike of rth type. Use permutations with repeated objects to solve the problem. To use the sine rule to find a missing angle, just flip the whole thing over. An inversion of a permutation σ is a pair (i,j) of positions where the entries of a permutation are in the opposite order: i < j and σ_i > σ_j. Excel provides functions that help you with factorials, permutations, and combinations. Total […]. So, number of three-digit odd numbers using the digits 0, 3, 5 and 7 (repetition not allowed) = 18 - 6 = 12. I teach all these topics in CS 317 (Discrete Information Structures), a required course for computer science majors at my university. ) Easy Combinations and Permutations. Example of Repetition. with repetition ♦ T will go over permutation worksheets and assign homework Permutation without repetition ♦ S will do Getting Started activity sheet ♦ T will model how to solve permutations without repetition (using proper notation),,in reading problem form, and using calculators to solve ♦ S will do worksheets on notation, writing out. Dim vAllItems As Variant Dim Buffer() As String Dim BufferPtr As Long Dim Results As Worksheet ' ' Posted by Myrna Larson ' July 25, 2000 ' Microsoft. At first this section may seem difficult but after some practicing some online problems and going through the detailed solution one can gain confidence. They react and respond to objects and experience measurement attributes in practical situations. Displaying top 8 worksheets found for - Repetition. b) The number of permutations of distinct objects taken r at a time is nr !! n P nr Example 5. 35 Permutations, Combinations and Proba- bility. To score well in Quantitative aptitude one should be thoroughly familiar with Permutation and Combination. Hm, if the term is "combinations 5 out of 10", then in my opinion 9,9,9,8,0 and 0,8,9,9,9 and 8,0,9,9,9 and 9,8,0,9,9 and are exactly the same combination: 3 times 9, one 8 and one 0. How to use the FACT function. In both cases we start with a set containing a a total of n elements. Permutations and combinations. A lock contains 3 dials, each with ten digits. Each number in the code can be chosen from the digits 0 through 9. Hint, hint, wink, wink. This chapter talk about selection and arrangement of things which could be any numbers, persons,letters,alphabets,colors etc. Permutationa method (Excel) 05/24/2019; 2 minutes to read +1; In this article. This Permutations with Repetition: Permutations and Repetition Interactive is suitable for 9th - 12th Grade. NOTES: Section 10. (permutations) are presented, consists of simple examples and problems. Permutation worksheets cover the topics such as listing possible permutations, finding the number of permutations using the formula, evaluating the expressions, solving equations involving. The search committee will choose four of them, and rank the chosen four from strongest to weakest. How many different codes can you have? n = 10, r = 5 105 = 100,000 codes Permutation without. PERMUTATIONS AND COMBINATIONS WORKSHEET CTQR 150 1. The two-by-two table, pictured in Table 1 below, is a component of most discrete mathematics or combinatorics textbooks (e. One permutation is. This form allows you to generate randomized sequences of integers. 1 — Probabilitv. A permutation is an arrangement of a set of objects in an ordered way. Okay, so "combinations and permutations" sounds like the name of a class you would take at wizards' college, but these are actually topics that you would cover in a statistics class. 4) Derek shu"ed a pack of 52 playing cards and asked his friend, Ian to choose any three cards. This type of activity is required in a mathematics discipline that is known as combinatorics; i. Description. Permutations and Combinations problems with solutions or questions covered for all Bank Exams, Competitive Exams, Interviews and Entrance tests. 7 digits are selected (without repetition) to from a telephone number. How many ways can 4 students from a group of 15 be lined up for a photograph? There are 15 P 4 possible permutations of 4 students from a group of 15. Example of permutation? Wiki User 2010-07-11 04:57:31. Due to the rapidly-changing and large number of permutations of the text available, please be very careful to make sure you are buying exactly the product(s) you want. For an in-depth explanation of the formulas please visit Combinations and Permutations. Name: Categorizing Counting Problems Activity Permutations and Combinations 1. Seperate lists with a blank line. Suppose we are given a total of n distinct objects and want to select r of them. In this chapter, you will learn about : • Permutation of r objects from n different objects. Create your lucky lotto numbers for game draws such as Mega Millions, Powerball, EuroMillions, OZ lotto, Lotto 6/49, Lotto Max, and practically any other Lottery/Keno system currently in existence around the world!. 2 n! a!b!c! Permutations with identical objects. Some of the worksheets displayed are Combinationspermutations work indicate whether each, Permutations vs combinations, Tree diagrams and the fundamental counting principle, Accelerated 67 mathematics curriculum guide, The fundamental counting principle and permutations, Ds80 introduction and planning guide. Sound Reading Comprehension Worksheets. Permutation With Repetition Problems With Solutions : Distributive property of multiplication worksheet - II. (a) (b) Solution. TRIGONOMETRY. Grade level skilltopic search. Combination. 1 - Apply the Counting Principle and Permutations. 1 048 576 2. ICS 141: Discrete Mathematics I 6. Most commonly, the restriction is that only a small number of objects are to be considered, meaning that not all the objects need to be ordered. In writing, repetition can occur at many levels: with individual letters and sounds, single words, phrases, or even ideas. How many probabilities are there. - Axel Richter Sep 7 '16 at 12:35. Also, give an example of a problem whose solution is a permutation and an example of a problem whose solution is a combination. Candidates will be expected to be familiar with the scientific notation for the expression of compound units, e. If repetition is allowed then how many different three digits numbers can be formed using the digits from 1 to 5? A. Each selection can go with any other selection, so each number is multiplied together. Multiplication Rule. The number of ways to do this is C(70+5-1, 5-1) = C(74,4) = C(74, 70). Practice Questions for Permutations - Concept - Formula - Problems with step by step solutions. Math Analysis Honors – Worksheet 53 Basic Combinatorics – Combinations Tell whether permutations or combinations are being described. and then take \cos^ {-1} of the whole thing once it’s in your calculator. These two topics are very similar and are easy to get confused. Let this code be universal (not only 3 but cover N unique arguments in a data set) Let's this macro to do it in a new sheet and list an output in columns. So, number of three-digit odd numbers using the digits 0, 3, 5 and 7 (repetition not allowed) = 18 - 6 = 12. A permutation is an arrangement of objects, without repetition, and order being important. You will be quizzed on probability and permutation topics. FACT FACT, which computes factorials, is surprisingly not categorized as Statistical. Two examples which are worked out in three different ways each: Using FCP, Formula, Calculator 4. 3) Vanessa chooses a combination of four digits from 0 to 9 without repetition for her school locker. Use this simple online tool to generate random letters. A local restaurant is offering a 3 item lunch special. The Iliad, The Odyssey. Step 5: Divide 360 by 24. We calculated that there are 630 ways of rearranging the non-P letters and 45 ways of inserting P's, so to find the total number of desired permutations use the basic principle of counting, i. Hence it is a permutation problem. Remember that you can always check your answers to "solving" problems by plugging them back in to the original question. 4 P 3 = 4! / (4 - 3)! = 24. In some cases, repetition of the same element is allowed in the permutation. But now, all the ordered permutations of any three people (and there are 3! = 6 of them, by FACT 1), will “collapse” into one single unordered combination, e. Permutations with Restrictions. I've got an exam on prob. Mississippi. Explanation: Number of ways of selecting 3 consonants from 7. Showing top 8 worksheets in the category - Repetition. A permutation pays attention to the order that we select our objects. If she chooses one type of fish types of permutation: Repetition is Allowed: such as the lock above. ) ! P(20,9)= 20! (20"9)! = 20! 11! = 60,949,324,800 EXAMPLE 1. Include Answer Key. If you will eventually take Math 2D/E, however, it is possible that your instructors in those classes will assign homework directly from the textbook, so please bear this in mind. Combinations are arrangements of objects without regard to order and without repetition. Counting permutations (Section 4. A formula for permutations Using the factorial, we can rewrite 𝑃𝑛,𝑘=𝑛𝑛−1 𝑛−2⋯𝑛−𝑘+1 as 𝑃𝑛,𝑘= 𝑛! 𝑛−𝑘! This formula is theoretically useful, for proving formulas involving permutations (and combinations), but it is of no computational relevance. Permutations with repetition. The fundamental difference between permutation and combination is the order of objects, in permutation the order of objects is very important, i. Tons of Free Math Worksheets at: Permutations and Combinations - Independent Practice Worksheet Permutations and Combinations Independent Practice Worksheet. In how many ways can you create a five-letter password if letters may be repeated? If letters cannot be repeated? The prix fixe menu at a local restaurant offers three choices of appetizer: soup, Caesar salad or mixed greens, three mains: chicken, pork or fish, and two desserts: crème brule or sherbert. Apr 27, 2018 - In this packet, you will find the following: 1. Created: Jan 16, 2017 | Updated: Feb 22, 2018. So P(H or 4) is $\frac{7}{12}$ Again, we can work this out from the tree diagram, by selecting every branch which includes a Head or a 4: Each of the ticked branches shows a way of achieving the desired outcome. Answers and Solutions for CAT Permutation and Combination Questions PDF (Set -2): Solutions: 1) Answer (D) The nth term is a + (n-1)d 1000 = 1 + (n-1)d So, (n-1)d = 999 999 = 3^3 * 37 So, the number of factors is 4*2 = 8 Since there should be at least 3 terms in the series, d cannot be 999. They will find the value (evaluate) of expressions using permutations. This is combinatorics. 3-digit permutations, repetition allowed: 9 x 9 x 9 = 729 3-digit permutations, no repetition: 9 x 8 x 7 = 504 3-digit. Counting problems are presented along with their detailed solutions and detailed explanations. Worksheets: I have attached to the email I sent out for your student's class. At first this section may seem difficult but after some practicing some online problems and going through the detailed solution one can gain confidence. 4) Derek shu"ed a pack of 52 playing cards and asked his friend, Ian to choose any three cards. Page 1 of 2 The number of permutations of r objects taken from a group of n distinct objects is denoted by nP r and is given by: nP r = (n n º! r)! PERMUTATIONS OF n OBJECTS TAKEN r AT A TIME USING PERMUTATIONS An ordering of n objects is a of the objects. Circular permutation 1. In how many different ways can the individuals finish the competition?. Suppose I have 4 letters and I want to arrange them in 3 places (repetition allowed), so I would have 43=64 possible permutations. #3 - I can use permutations with repetition to count. Then solve the problem. permutation formula, letting n = 20 and r = 9. Identify the following as Permutations, Combinations or Counting Principle problems. Permutations with Repetition These are the easiest to calculate. Description. If there are 25 cars of this. Repetition can be problematic in writing if it leads to dull work, but it can also be an effective poetic or rhetorical strategy to strengthen your message, as our examples of repetition in writing demonstrate. Permutations with repetition. This list was not organized by years of schooling but thematically. If you're seeing this message, it means we're having trouble loading external resources on our website. Recall: Five people can line up in a row in 5 x 4 x 3 x 2 x 1 = 5! = 120 ways The total number of permutations is denoted by P(n, r) By the Fundamental Counting Principle, P(n, r) = n!. A permutation is an arrangement in which order matters. A worksheet is provided for student. ©P DO NOT COPY. 8 26 customer reviews. Since the letter o appears twice we need to divide by 2! The letters a and p are the ones that repeat. Permutations with Reruns 1 - Cool Math has free online cool math lessons, cool math games and fun math activities. 1 Permutations when all the objects are distinct. When a thing has n different types we have n choices each time!. What is the probability that you listened to exactly four country songs? 8) Asanji is carrying ten pages of math homework and five pages of English homework. pdf), Text File (. Putting 5 books in order. counting principle to find the number of different faces. After entering the characters, and then click OK button, all the possible permutations are displayed in column A of active worksheet. com is an online resource used every day by thousands of teachers, students and parents. It doesn't matter in what order we add our ingredients but if we have a combination to our padlock that is 4-5-6 then the. In both cases we start with a set. The number of permutations of n objects, taken r at a time, when repetition of objects is allowed, is nr. Permutations combinations worksheets free members only. These samplings are given as follows: PERMUTATIONS WITH REPETITION/REPLACEMENT. , f ( a 1) ∈ B = { b 1, b 2, b 3,, b n }. The same set of objects, but taken in a different order will give us different permutations. Sample Space. ) r)! arrangement of all n objects arrangement of n objects taken r at a time arrangement of n objects with some like objects. (a)Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is to be always included in each arrangement. Create different worksheets. The Iliad, The Odyssey. List permutations with repetition and how many to choose. Meritnation. Circular permutation is a very interesting case. Permutation formula is used to find the number of ways an object can be arranged without taking the order into consideration. The Best Office Productivity Tools. Name: Categorizing Counting Problems Activity Permutations and Combinations 1. ): The number of r-permutations from a set of n objects with repetition allowed is nr. How many different ways can you arrange the letters in the word number?. A third item C can be first, second, or third for each order above. Combination & Permutation Generator. Repetition can be problematic in writing if it leads to dull work, but it can also be an effective poetic or rhetorical strategy to strengthen your message, as our examples of repetition in writing demonstrate. Permutations A permutation of length r is a list (x1; ;xr) with distinct components (no repetition); that is, xi 6= xj when i 6= j. After entering the characters, and then click OK button, all the possible permutations are displayed in column A of active worksheet. Permutation formula is used to find the number of ways an object can be arranged without taking the order into consideration. Clarifying What You Heard, Asking for Repetition, and Confirming Your Understanding in Business English Even if you have been speaking English for years, you are going to find yourself in a situation where someone uses a word or a phrase you don’t know, speaks a little too fast, or mumbles (speaks unclearly). Later on, in Section 3, we suggest more complicated problems which can be used for later lessons or in lessons for. You will be quizzed on probability and permutation topics. Revising notes in exam days is on of the best tips recommended by teachers during exam days. Permutations. With random shuffle and no repetition, you listen to seven songs. 2 3 letter words, without letter repetition. If somebody could help me with these problems, that would be great. Example of permutation A club has 20 members The members want to elect a president and a vice-president to be in charge of the club. Worksheets are Reading on the move, Permutations, Using proverbs to illustrate grammar points, Poetic devices work 5, Contents, Parallel structure work, Transitional words and phrasesrevised815, Work a2 fundamental counting principle factorials. C program to print all natural numbers from 1 to n. I know there are 10,000 possibilities (10 x 10 x 10 x 10), but I need to show them in an Excel spreadsheet. An example of an ordinary combination is a choice of 6 numbers from 1 to 49 for a lottery draw: you must pick 6 different numbers out of 49, you are not allowed any repeat numbers and, of course the order you select the numbers in doesn't matter s. From n objects, nr = n n (r factors) lists of. PERMUT function. A lock contains 3 dials, each with ten digits. Permut w/o repetition. If we have n things of which x number of things are of same kind, y number of things are of same type and similarly z number of things are of the same type. Another definition of permutation is the number of such arrangements that are possible. 6 Counting Principles, Permutations, and Combinations 1021 We use the Fundamental Counting Principle to find the number of three-course schedules. Letter X Tracing Worksheets. Permutation formula is used to find the number of ways an object can be arranged without taking the order into consideration. Name: Categorizing Counting Problems Activity Permutations and Combinations 1. 4 Responses to "List permutations with repetition [UDF]". How many probabilities are there. , is a professor of mathematics at Anderson University and the author of "An Introduction to Abstract Algebra. To cover the answer again, click "Refresh" ("Reload"). 4) Derek shu"ed a pack of 52 playing cards and asked his friend, Ian to choose any three cards. Other common types of restrictions include restricting the type of objects. The probability distribution of a hypergeometric random variable is called a hypergeometric distribution. Solution: There are 4 letters in the word love and making making 3 letter words is similar to arranging these 3 letters and order is important since LOV and VOL are different words because of the order of the same letters L, O and V. There is one way to arrange one item A. 5-digit numbers are to be formed from the digit 1, 2, 3, 4, 5, 6, 7, 8. A 5-member team and a captain will be selected out of these 10 players. (A true "combination lock" would accept both 10-17-23 and 23-17-10 as correct. The fundamental difference between permutation and combination is the order of objects, in permutation the order of objects is very important, i. A permutation with repetition is included. A permutation is an arrangement of objects, without repetition, and order being important. Permutations A permutation of length r is a list (x1; ;xr) with distinct components (no repetition); that is, xi 6= xj when i 6= j. They react and respond to objects and experience measurement attributes in practical situations. The Odyssey, The Iliad. How many different ways can the letters in the word "micro" be arranged if it always has to start with a vowel? Exercise 3. ( n – r)! where n = number of objects r = number of positions. Circular Permutations Before diving into circular permutation let us discuss Permutation of n things not all different taken all together. 3 Permutations and Combinations 6. The order you put the numbers in matters. times such that repetition is allowed and ordering does not matter. Let this code be universal (not only 3 but cover N unique arguments in a data set) Let's this macro to do it in a new sheet and list an output in columns. Counting problems are presented along with their detailed solutions and detailed explanations. of codes that can be created with no repetition of any letter or digit?. Stage C: Random Identification by Quantity. The order you put the numbers in matters. Permutations 12 13 10 1 10 8 14 16 11 15 7 16 11 10 1 3 letter words, with repetition allowed. A lock has a 5 digit code. indd 1 07/11/13 10:42 AM. Displaying top 8 worksheets found for - Repetition. C)How many arrangements are possible if repetition is not allowed? Do the number of choices stay the same or are they reduced? D) In which case is there a greater number of permutations possible? Homework:Page 93 – 95: Questions 1 – 8, 11. 1 — Probabilitv. Name: Categorizing Counting Problems Activity Permutations and Combinations 1. • Circular Permutation C. Example of Repetition. It does not cover all rules of inference in propositional logic. The formula is given below. Visual Basic for Applications code The code below supports 4 public entry points (createSubset, createPowerSet, createPermutations, and createCombinations) for use by other code and 4 user defined functions or UDFs (UDFSubset. Definition: A permutation of “n” objects is an arrangement of the “n” objects, with regard to order. Algebra 1 prentice hall quiz, addition properties-commutative worksheet 3rd grade, solving radical expression calculator, objectives of college algebra repetition, help with algebra division, free answers for a math equation, matthematics for school+ free download book. These combination problems are sometimes called 'less than' problems. 2 3 letter words, without letter repetition. For example a true “combination lock” would accept both 17–01–24 and. Videos, worksheets, games and activities to help Algebra II students learn about permutations. In other words: "My fruit salad is a combination of apples, grapes and bananas" We don't care what order the fruits are in, they could also be "bananas, grapes and apples" or "grapes, apples and bananas", its the same. Your dog has 8 puppies, 3 male & 5 female. EffortlessMath. 10 Chapter 8: Permutations and Combinations DO NOT COPY. How many different ways can you arrange the letters in the word number?. Thanks jan It was very helpfull to know UINT8 reduce the memory print by a factor of 8. Algorithms for Generating Permutations and Combinations Section 6. Hyperbole Worksheet Author: lainastapleton Created Date: 7/21/2011 9:56:45 PM. Properties of Permutation. A pemutation is a sequence containing each element from a finite set of n elements once, and only once. How many ways can those offices be filled? A-2 The company Sea Esta has ten members on its board of directors. 2 Permutations and Factorials Worksheet B 3 13. An example of an ordinary combination is a choice of 6 numbers from 1 to 49 for a lottery draw: you must pick 6 different numbers out of 49, you are not allowed any repeat numbers and, of course the order you select the numbers in doesn't matter s. The permutations with repetition are denoted by PR(n,k). number of ways 10 semi—finalists can finish in Miss America (5 places: 1st, 2nd, 3rd, 4th, 5th). 1 — Probabilitv. When some of those objects are identical, the situation is transformed into a problem about permutations with repetition. Now arranging words in dictionary first letter is C, Fixing the position of C remaining 6 words can be arranged in 6! Ways But as S repeats thrice we can i. Next Post Permutations with repetition using Excel 2 thoughts on " Using named ranges and worksheet functions in Excel VBA " Pingback: Combinations and Permutations with Excel | The Universe Divided. Change a pic. Worksheets based on US Common Core standards curricuulum are listed under the respective concepts of ELA & Math. Examples of permutations with repetition include passcodes, phone numbers, and even three-scoop ice cream cones if you care about the scoop order and can get multiple scoops of the same flavor. Permutations. **Important note: The formulas below are only appropriate for problems involving selection from a single source with no repetition. Multiple worksheets. A permutation with repetition is included. Most commonly, the restriction is that only a small number of objects are to be considered, meaning that not all the objects need to be ordered. #3 - I can use permutations with repetition to count. When considering the differences between combinations and permutations, we are essentially concerned with the concept of order. Suppose we are given a total of n distinct objects and want to select r of them. At first this section may seem difficult but after some practicing some online problems and going through the detailed solution one can gain confidence. Permutation. Homework: Lesson 10. Permutation formula is used to find the number of ways an object can be arranged without taking the order into consideration. This is a permutation and repeats are not allowed. A multiple-choice test has 10 questions. I need a Permutation with repetition for all possible 4-digit numbers using values of 0 through 9. Counting ordered selections (Section 4. Math Worksheets High School Math based on the topics required for the Regents Exam conducted by NYSED. Permutation can be done in two ways, Permutation with repetition: This method is used when we are asked to make. Using the formula, we see that there are: $$\dfrac{15!}{5!5!5!}=756756$$ ways in which 15 pigs can be assigned to the 3 diets. Permutations with Repetition. Worksheet A2 : Fundamental Counting Principle, Factorials, Permutations Intro. The Iliad, The Odyssey. Am currently working through a probability worksheet for revision purposes. Print Permutation & Combination: Problems & Practice Worksheet 1. Generating a line up for the 9 players on a ball team if the pitcher bats first and the catcher bats last. 2 Solve counting problems using permutations. ©P Multiple-Choice Questions 1. We want to order 3 books on a shelf, so we must have 3 spaces on the shelf:. As the number of things (letters) increases, their permutations grow astronomically. Homework: Lesson 10. Number of permutations = = = 360 (ii) Number of letters used = 6. The number of words is given by. Sal explains the permutation formula and how to use it. Home › Math › Navigate a Grid Using Combinations And Permutations Puzzles can help develop your intuition -- figuring how to navigate a grid helped me understand combinations and permutations. With a combination, we still select r objects from a total of n , but the order is no longer considered. One permutation is. From n objects, nr = n n (r factors) lists of. Thanks in advance in column "A", i have the numbers, suppose it is 0 to 99 (1) Now i want the list of generated numbers through combination with no repeats in the set of 5, please provide any vba code (2) I want the list of generated numbers through permutation with no repeats in the set of. 1) There are 50 applicants for three jobs:. There are methods for calculating permutations, and it's important to understand the difference between a set with and without repetition. Then find the number of possibilities. A student has to take one course of physics, one of science and one of mathematics. Work out permutations and combinations for a set of items, numbers, letters separated by commas with or without repetition, with or without order. P (n, r) denotes the number of permutations of n objects taken r at a time. Worksheets based on US Common Core standards curricuulum are listed under the respective concepts of ELA & Math. Repetitions are not allowed. 35 Permutations, Combinations and Proba- bility. The concept of permutation is used for the arrangement of objects in a specific order i. As the number of things (letters) increases, their permutations grow astronomically. 5! = 120 Permutations rep Sl!s ! n!=n n — (6-2)! Co total possibilities when order is important (arrange, order, rank, line up, etc. Combinations (Unordered Selections) A combination of n objects taken r at a time is a selection which does not take into account the arrangement of the objects. This is a combinations with repetition question without any special circumstances. Random Name Picker No Repeats. Students insist that the die on the table doesn't alter the die under the. Mathematics and statistics disciplines require us to count. Selecting a 4-digit pin number. Permutation with repetition choose (Use permutation formulas when order matters in the problem. What if I want a random permutation such that each item is selected once?. ( total number of letters)! ( number of repeats)! A! ⋅ B! ⋅ C!! How many ways can you arrange the letters of the word 'loose'? There is only one letter that repeats. In order to answer the question, we will use the combinations formula, where n = the total number of items (10) and k = the number of items. Some of the worksheets for this concept are Reading on the move, Permutations, Using proverbs to illustrate grammar points, Poetic devices work 5, Contents, Parallel structure work, Transitional words and phrasesrevised815, Work a2 fundamental counting principle factorials. Permutation Combination - Practice Questions A collection of questions that typically appear from the topic of Permutation and Combination. Detailed visual description of the Standard Deck of Cards 2. Multiplication Rule If one event can occur in m ways, a second event in n ways and a third event in r, then the three events can occur in m × n × r ways. Using 100,000 permutations reduces the uncertainty near p = 0:05 to 0:1% and allows p-values as small as 0. If repetition is allowed then how many different three digits numbers can be formed using the digits from 1 to 5? A. There are 7 color patterns available. It is very useful and interesting as a topic. , is a professor of mathematics at Anderson University and the author of "An Introduction to Abstract Algebra. we do not allow repetition: AAB is not a valid permutation of ABC. O A iARlpl` Qr]ilgWhstAsc lroeNseefrRvMendR. Permutation with repetition choose (Use permutation formulas when order matters in the problem. Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 2 Permutations Name_____ Date_____ Period____ ©Y V2h0U1j6b WKSuCtpae LSeoOfRtEwRaYrse\ JLfLACe. txt) or read online for free. PERMUT function. It also does not cover some counting problems like combinations with repetition and permutations with indistinguishable objects. 5 When repetition of objects is allowed The number of permutations of n things taken all at a time, when repetion of objects is allowed is nn. Important Results on’Permutation. Permut w/o repetition. Re: Can VBA be used to generate Permutations / Combinations / Iterations/ Variations. extension allows students to use the formula for permutations with repetition. Permutations with Restricted Position By Frank Harary In his book on combinatorial analysis, Riordan [4, p. To use the sine rule to find a missing angle, just flip the whole thing over. with repetition ♦ T will go over permutation worksheets and assign homework Permutation without repetition ♦ S will do Getting Started activity sheet ♦ T will model how to solve permutations without repetition (using proper notation),,in reading problem form, and using calculators to solve ♦ S will do worksheets on notation, writing out. You will be quizzed on probability and permutation topics. Use this simple online tool to generate random letters. I will also be sending the worksheets in an email. Solve problems about arrangements of objects in a line, including those involving:. CliffsNotes study guides are written by real teachers and professors, so no matter what you're studying, CliffsNotes can ease your homework headaches and help you score high on exams. Permutations, Combinations, and the Counting Principle Task Cards Students will practice solving problems using the fundamental counting principle, permutations, and combinations by working through these 20 task cards. Create your lucky lotto numbers for game draws such as Mega Millions, Powerball, EuroMillions, OZ lotto, Lotto 6/49, Lotto Max, and practically any other Lottery/Keno system currently in existence around the world!. If there are objects, with objects being non-distinct and of a. A permutation is an ordering of a set of objects. Hint: Line up the 36 students and then determine how. # of permutations of k = 3 from n = 5 is equal to 5! 2! = 60. This type of activity is required in a mathematics discipline that is known as combinatorics; i. Permutations are for lists (order matters) and combinations are for groups (order doesn’t matter). 3 4 letter words, without letter repetition. Print Permutation & Combination: Problems & Practice Worksheet 1. Permutation With Repetition and Circular Permutations During an activity at school, 10 children are asked to sit in a circle Is the arrangement of children a linear or circular permutation? Explain. Teachers may use Activity 15. PERMUT function. Then give the possible number of outcomes. Combined sets can have a prefix and/or suffix added via the prefix/suffix fields. Multiplication Rule If one event can occur in m ways, a second event in n ways and a third event in r, then the three events can occur in m × n × r ways. Which of the following represents the number of codes that can be created with no repetition of any letter or digit? A. Permutations And binations Study Material for IIT JEE from Permutations And Combinations Worksheet, source: askiitians. Permutations and binations ppt from Permutations And Combinations Worksheet, source: slideplayer. 6 Counting Principles, Permutations, and Combinations 1021 We use the Fundamental Counting Principle to find the number of three-course schedules. Explain the di˜erence between permutations and combinations. 3 Probability and Odds 13-3 Practice Answers Probabiility and Odds. , f ( a 1) ∈ B = { b 1, b 2, b 3,, b n }. Permutation is the arrangement of a given set of numbers or things in a certain order. The number of permutations if objects, where n are the same,a b are the same, c are the same etc. 1 Worksheet Example #1: For a wedding dinner, you have a choice between soup or salad. Permutations and combinations worksheet ctqr 150 1. If we have a set of 𝑛 items, the total number of permutations is given by 𝑛 × ( 𝑛 − 1 ) × ( 𝑛 − 2 ) × ⋯ × 2 × 1. You can extend the idea to any number of choices. Worksheet A2 : Fundamental Counting Principle, Factorials, Permutations Intro. 13) Heather and Bill are planning trips to three countries this year. At first this section may seem difficult but after some practicing some online problems and going through the detailed solution one can gain confidence. An example of Permutation with Repetition 5. Topics for the Final Exam: ===== From Test 1 (note, all of cardinality is included): ----- Logic: - Proving logical formulas are equivalent: both by truth tables and logical equivalences - Translating from English to logic, and logic to English - De Morgan's law for negation - Quantifiers, negation - Translating to and from English with quantifiers and negation of quantifiers - Boolean. List permutations with repetition and how many to choose from Noel asks: Is there a way where i can predict all possible outcomes in excel in the below example. In these worksheets, your students will solve permutations. Worksheets are Reading on the move, Permutations, Using proverbs to illustrate grammar points, Poetic devices work 5, Contents, Parallel structure work, Transitional words and phrasesrevised815, Work a2 fundamental counting principle factorials. If these letters are written down in a row, there are six different possible arrangements: PQR or PRQ or QPR or QRP or RPQ or RQP. The number of 3-digit decimal numbers with repetition (and possible leading zeros) allowed is simply 103 = 1000. Number of possible permutations: Permutations with repetition. It does not cover all rules of inference in propositional logic. You can't be first and second. 6 people get on a bus. I start this class by reviewing conditional probability (from the previous lesson). No Repetition: for. These are the easiest to calculate. Videos, worksheets, games and activities to help Algebra II students learn about permutations. Example of permutation A club has 20 members The members want to elect a president and a vice-president to be in charge of the club. Arithmetic Ability provides you all type of quantitative and competitive aptitude mcq questions on Permutation And Combination with easy and logical explanations. That's a lot of ways!. 163-164] discusses permu-tations with restricted position and mentions an open question : "Any restrictions of position may be represented on a square, with the elements to be permuted as column heads and the positions as row heads, by. Worksheet B2 : Permutations 1. How to use the FACT function. Permutation vs. 12) • Teachers may remind students to draw a diagram to illustrate. 2 Special Permutations Worksheet B 4 13. To use the sine rule to find a missing angle, just flip the whole thing over. How many possible sequences of numbers exist? 2. It is very useful and interesting as a topic. There are two types of arrangements: Use permutations with. Permutations with Repetition. 13 cards selected from a deck of 52 for a hand in a game of “13”. Repetition is widely used in both poetry and prose; throughout all genres and forms of literature and oral tradition. This means that for the example of the combination lock above, this calculator does not compute the case where the combination lock can have repeated values, for example 3-3-3. Using the formula, we see that there are: $$\dfrac{15!}{5!5!5!}=756756$$ ways in which 15 pigs can be assigned to the 3 diets. In this case, I would check whether nine pennies plus nine nickels plus nine dimes totalled $1. Remember, verbal irony is when someone says the opposite of what they really mean. Combinations & Permutations (hints) Date: RHHS Mathematics Department 28. Sound Reading Comprehension Worksheets. I teach all these topics in CS 317 (Discrete Information Structures), a required course for computer science majors at my university. Create your lucky lotto numbers for game draws such as Mega Millions, Powerball, EuroMillions, OZ lotto, Lotto 6/49, Lotto Max, and practically any other Lottery/Keno system currently in existence around the world!. The order you put the numbers in matters. This Permutations with Repetition: Permutations and Repetition Interactive is suitable for 9th - 12th Grade. 3: Permutations When All Objects Are Distinguishable is the notation used to represent the number of permutations that can be made from a set of n different objects, where only r of them are used in each arrangement. You can't be first and second. No Repetition: for example the first three people in a running race. Page 1 of 2 The number of permutations of r objects taken from a group of n distinct objects is denoted by nP r and is given by: nP r = (n n º! r)! PERMUTATIONS OF n OBJECTS TAKEN r AT A TIME USING PERMUTATIONS An ordering of n objects is a of the objects. Identifies how language use in their own writing differs according to their purpose, audience and subject matter. Example 1: A college offers 3 different English courses, 5 different math course, 2 different art courses, and 4 different history courses. This article describes the formula syntax and usage of the PERMUT function in Microsoft Excel. Permutation and combination? In your own words, explain the difference between a permutation and a combination. With a combination, we still select r objects from a total of n , but the order is no longer considered. notebook Stat day 1 notes December 09, 2014 Apple has come out with two new products recently, the iPhone 6 and the iPhone 6+. Category Questions section with detailed description, explanation will help you to master the topic. Hint: Line up the 36 students and then determine how. Combination: the order of the events doesn’t matter and it does not matter which item is placed first. Solution: The word ‘REMAINS’ has 7 letters. There are different types of permutations and combinations, but the calculator above only considers the case without replacement, also referred to as without repetition. ( n – r)! where n = number of objects r = number of positions. Homework: Lesson 10. Permutation: A permutation of n differenct elements is an ordering of the elements such that one element is first, one is second, one is third, and so on. How many ways can those offices be filled? A-2 The company Sea Esta has ten members on its board of directors. A permutation is an arrangement of all or part of a set of objects. Another definition of permutation is the number of such arrangements that are possible. Worksheet B2 : Permutations 1. Permutation Combination Worksheet - Free download as PDF File (. Remember that you can always check your answers to "solving" problems by plugging them back in to the original question. It does not cover all rules of inference in propositional logic. The basic difference between permutation and combination is of order Permutation is basically called as a arrangement. You can extend the idea to any number of choices. times such that repetition is allowed and ordering does not matter. So a descent is just an inversion at two adjacent positions. So, if we have 3 tin cans to give away, there are 3! or 6 variations for every choice we pick. Permutations - Definitions Counting Formula: Permutations with repetition The number of ways to arrange robjects from a set of n objects, in order, with repetition allowed is: nr Example: (a) What are the permutations of the letters in the word COMPUTER? (b) If only 5 letters are used from the above, what is the number of permutations?. Permutation and combination problems with solutions pdf and combination hard problems with solutions, permutation and combination problems with answers, combination with repetition problems, solving permutations and combinations, basic permutation and combination problems, permutation and combination gre problems, permutation and. Worksheet #1-12,14,15ab. Permutations and Combinations BUILDING ON listing outcomes of probability experiments solving equations BIG IDEAS Counting strategies can be used to determine the number of ways to choose objects from a set or to arrange a set of objects. Selecting 3 officers (Pres, VP and Sec) from a club with 9 members. If there are m ways to do one thing, and n ways to do another, then there are m × n ways of doing both. This lesson on Properties of Real Numbers is one that gets covered at the beginning of every Algebra course. 2) Rob and Mary are planning trips to nine. A worksheet is provided for student. A permutation is an arrangement or sequence of selections of objects from a single set. You will be quizzed on probability and permutation topics. How many ways can a 10 member jury be selected form a pool of 25 people? Without Repetition 7. Whether you've loved the book or not, if you give your honest and detailed thoughts then people will find new books that are right for them. ) with full confidence. - Axel Richter Sep 7 '16 at 12:35. Honestly, it caused my son to melt into a puddle of tears every day and I had to find something that caused less of a fight and moved along with less repetition. No Repetition: for. How many 5 digit numbers can be named using the digits 5, 6, 7, 8, and 9 without repetition?, with repetition?. So, the number of possibilities is 7. They'll give your presentations a professional, memorable appearance - the kind of sophisticated look that today's audiences expect. Permutation can be done in two ways, Permutation with repetition: This method is used when we are asked to make. There are methods for calculating permutations, and it's important to understand the difference between a set with and without repetition. The symbol for this number is P(n;k). Type in lists of items, one item per line. Letter X Tracing Worksheets. Permutations are studied in almost every branch of mathematics, and in many other fields of science. In order to answer the question, we will use the combinations formula, where n = the total number of items (10) and k = the number of items. An example of Permutation with Repetition 5. Solution: There are 4 letters in the word love and making making 3 letter words is similar to arranging these 3 letters and order is important since LOV and VOL are different words because of the order of the same letters L, O and V. (no need to solve): The model of the car you are thinking of buying is available in nine different colors and three different styles (hatchback, sedan, or station wagon). (a) (b) Solution. There are 7 countries they would like to visit. Permutations A permutation is an ordered sequence of k elements selected from a given finite set of n numbers, without repetitions, and not necessarily using all n elements of the given set. Random Name Picker No Repeats. 1 Permutations and Combinations The Fundamental Counting Principle allow us to count large numbers of possibilities quickly. Introducing Repetition - Week 2 Worksheet Arrangements An arrangement is a grouping of objects. Permutations with repetition. Example of permutation A club has 20 members The members want to elect a president and a vice-president to be in charge of the club. Consider arranging 3 letters: A, B, C. FUNDAMENTAL COUNTING PRINCIPLE: Use the fundamental counting principle to determine the number of items in the sample space for the following events: 1a. PERMUTATIONS Recall Example 5:. Permutation If n is the number of distinct things and r things are chosen at a time. 5 When repetition of objects is allowed The number of permutations of n things taken all at a time, when repetion of objects is allowed is nn. a) The number of permutations of n distinct objects is n!. Similarly we have n options for f ( a 2), and so on. Worksheet MATHS11WK20143. Also, give an example of a problem whose solution is a permutation and an example of a problem whose solution is a combination. Use it only when the order of things is important. PERMUTATIONS AND COMBINATIONS WORKSHEET CTQR 150 1. He may choose one of 3 physics courses (P1, P2, P3), one of 2 science courses (S1, S2) and one of. permutation of 20 elements leads to huge number. 7: Permutations and Combinations Permutations In this section, we will develop an even faster way to solve some of the problems we have already learned to solve by other means. Tracing Upper And Lowercase Letters. FUNDAMENTAL COUNTING PRINCIPLE: Use the fundamental counting principle to determine the number of items in the sample space for the following events: 1a. All the above formulas are defined for Number of Permutations or Combinations of r objects chosen from n objects. 5) Stephanie rearranged the letters in the word # TOGETHER# and formed new words beginning with R and ending with T. In how many ways can this be accomplished? 3. How many ways can 5 paintings be line up on a wall? 3. As the number of things (letters) increases, their permutations grow astronomically. Selecting a 4-digit pin number. replacement or repetition, where the number of elements is greater than or equal to the number of slots. Permutation is an ordered arrangement of items that occurs when a. Permutations are for lists (order matters) and combinations are for groups (order doesn’t matter). Now, if we were to have objects, not all distinct, then this is a different matter, and in fact there does exist a formula for such a case. Excel provides functions that help you with factorials, permutations, and combinations. When we don't allow repetition of letters, there are: 26 × 25 × 24 × 23 = 358,800. If we are taking an r-permutation of an n-set with repetition allowed, the number of such arrangements is nr. An addition of some restrictions gives rise to a situation of permutations with restrictions. Sal explains the permutation formula and how to use it. Candidates will be expected to be familiar with the scientific notation for the expression of compound units, e. This lesson on Properties of Real Numbers is one that gets covered at the beginning of every Algebra course. Order doesn’t matter. (permutations) are presented, consists of simple examples and problems. This calculation. Permut w/o repetition. It could be "333". Permutation or Combination? Ans= 132600 ways 4. Combination Formulas. In a certain state's lottery, 48 balls numbered 1 through 48 are placed in a machine and six of them are drawn at random. 1 This is pronounced 'n factorial', and written n!. How many possible combinations of pizza with one topping are there? 2. Sal explains the permutation formula and how to use it. Each number in the code can be chosen from the digits 0 through 9. As the number of things (letters) increases, their permutations grow astronomically. How to use the FACT function. Brett Berry. 1, 3, or 5 Minute Drill Multiplication Worksheets A multiplication math drill is a worksheet with all of the single digit problems for multiplication on one page. A permutation with repetition is included. Grade level skilltopic search. com Subscribe to My Channel: https://. List outcomes of Sample Space. Combinations. A permutation is an arrangement of objects, without repetition, and order being important. For example, if you are trying to come up with ways to arrange. Permutations. (a) (b) Solution. It can be used as a worksheet function (WS) in Excel. Permutations, Combinations, and the Counting Principle Task Cards Students will practice solving problems using the fundamental counting principle, permutations, and combinations by working through these 20 task cards. Google shows. To refer to combinations in which repetition is allowed, the terms k-selection, k-multiset, or k-combination with repetition are often used. Avoiding duplicate permutations % Progress. In this case, I would check whether nine pennies plus nine nickels plus nine dimes totalled$1. Permutation and Combination are two closely related concepts. When a collection of units includes some members that appear the same, the number of permutations is reduced by the number of arrangements that result only from exchanging the identical members. Practice Questions for Permutations - Concept - Formula - Problems with step by step solutions. Permutations A permutation is an ordered sequence of k elements selected from a given finite set of n numbers, without repetitions, and not necessarily using all n elements of the given set. Worksheet 15-3 B Date_____ Period_____ Combinations – total possibilities when order is NOT important (choose, pick, group, select) C n n C n,1 nn number of groups formed when all objects are used nr !,!! n C n r C n r r number of groups formed from objects taken r at a time (no repetition / order doesn’t matter). How many ways can 5 paintings be line up on a wall? 3. A permutation is an arrangement of all or part of a set of objects. Consider the selection of a set of 4 different letters from the English alphabet. Permutation Problem 1. The number of permutations of n objects taken r at a time is given by: nP r = n!. The symbol for this number is P(n;k). Evaluate the following expressions. To see the answer, pass your mouse over the colored area. Permutation is the arrangement of a given set of numbers or things in a certain order. In a race of 15 horses you beleive that you know the best 4 horses and that 3 of them will finish in the top spots: win, place and show (1st, 2nd and 3rd). Before beginning any strength-training program, it is advisable that a lifter determine their one-repetition-maximum (1RM) to gauge their current strength level.
1gg0qf5zr0lu 8vnn66l474zjbt mlzi4faqnan 604007i0yl gbx1xi8zwo l3wum2je1g5yhkw 3f69qiw5v61kmet dmfx6tov8i3iqt kxl7b9n95qg6a abtu4di24x bpfrzik9qsmo0 8re4fi6wuoy8jzm 1fsfzr8ey7 s94zkwow63yl ffyrhkeix2o0em ecpggokxuvgm37f b1ys795h43a 0jbm2niiegjytw 91o7tpniyd ls2vxmy1epzbdsf 7xofbb2h8i3fu 4q3m5n25se2ardp fvk9hl8ydn11 1r1qjsdcczl9ae3 3c15dmx8e0 qwmjgddrbff4 jl6c9i1lez cushopwoc5y0l61 | 2020-07-08T01:36:10 | {
"domain": "charlyedance.it",
"url": "http://ipmq.charlyedance.it/permutations-with-repetition-worksheet.html",
"openwebmath_score": 0.5433921813964844,
"openwebmath_perplexity": 971.650366932591,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9814534398277176,
"lm_q2_score": 0.8596637451167997,
"lm_q1q2_score": 0.8437199397400613
} |
http://openstudy.com/updates/55e4d1ebe4b0819646d7fe13 | ## Trisaba one year ago the principal of a school plans a school picnic for June 2. a few days before the event, the weather forecast predicts rain for June 2, so the principle decides to cancel the picnic. consider the following information. in the school's town, the probability that it rains on any day in June is 3%. when it rains, the forecast correctly predicts rain 90% of the time. when it does not rain, the forecast incorrectly predicts rain 5% of the time. let event a be the event that it rains on a day in June. let event b be the event that the forecast predicts rain. do you think the principal...
1. Trisaba
made a good decision? why or why not?
2. Trisaba
this is a series of questions
3. Trisaba
@dan815
4. Trisaba
what is probability of b given a
5. Trisaba
P(B l A)
6. dan815
okay what is B and A
7. Trisaba
in the question
8. Trisaba
last two lines
9. dan815
P(B|A) means A given B or B given A?
10. Trisaba
b given a
11. dan815
okay so chance the forecast predicts right given it rains then
12. dan815
when it rains, the forecast correctly predicts rain 90% of the time.
13. dan815
P(B|A)=0.9=90%
14. Trisaba
a formula for P(B l A) is that is P(A intersection B) times P(A) all divided by P(B)
15. Trisaba
?
16. Trisaba
no not times P(A) formula above
17. Trisaba
how did you figure out that the second thing was that
18. dan815
P(B|A)=P(a n b)/P(b) like that?
19. Trisaba
yeah
20. dan815
because the statement is already telling us what we want
21. Trisaba
no
22. Trisaba
P(A) at bottom
23. Trisaba
how did you know it was it
24. dan815
oh sry ya p(a) bottom
25. dan815
P(B|A) is asking the probability of B given A probability of rain prediction when it rains
26. dan815
and this statement is giving you the answer "when it rains, the forecast correctly predicts rain 90% of the time. "
27. Trisaba
oh
28. Trisaba
now what is P(A)
29. Trisaba
now what is P(A)
30. dan815
3% chance it rains on a day in june
31. Trisaba
is a?
32. dan815
ya
33. Trisaba
$P \left( a ^{c} \right)$ is what
34. Trisaba
.97?
35. dan815
|dw:1441061445919:dw|
36. dan815
or100%-3%= 1-0.03
37. dan815
yes compliment raining is not raining, which is 97%
38. Trisaba
P(B l A complement) is?
39. dan815
that means chance of prediction when there is no rain
40. dan815
|dw:1441061765626:dw|
41. Trisaba
5%?
42. dan815
yeah
43. Trisaba
$P(A) \times P(B l A) + P(A ^{c}) \times P(B l A ^{c})=P(B)$
44. Trisaba
|dw:1441062094509:dw|
45. Trisaba
a complement we said was .97
46. dan815
yeah
47. Trisaba
so what are the question marks
48. dan815
when it does not rain, the forecast predits 95% of the time it wont rain and 5% it will rain
49. dan815
that what it means by 5% wrong prediction when it does not rain
50. dan815
|dw:1441062757048:dw|
51. Trisaba
thanks for the help
52. dan815
|dw:1441062793507:dw|
53. dan815
this is the full break down
54. dan815
they will probably ask u a final question like the chance it will rain, when there is a prediction of rain
55. Trisaba
60%
56. Trisaba
bayes' theorem
57. Trisaba
thanks for helping dan
58. dan815
wait i got 36% lemme see
59. dan815
okay think about it like this out of every 97 days, 5% of those days say it will rain, and it wont rain out of every 3 days, 90% of those days say it will rain, and it will rain! what is the total number of days it will rain to not raining then?
60. Trisaba
61. dan815
0.03*0.9 : 0.05*0.97 this means that (0.03*0.9)/((0.03*0.9)+(0.05*0.97)) = chance it rains when it say it rains
62. Trisaba
60% is what i got
63. dan815
what did u say p(b) was
64. Trisaba
P(B) is .0755
65. Trisaba
look at my sheet posted pic
66. dan815
hard to read, what did u say P(B|A) was then
67. Trisaba
.9
68. Trisaba
$\frac{ 90 }{ 151 }$
69. Trisaba
it rounds to 60%
70. dan815
hmm i am not seeing anything wrong with the way im finding the answer, ill think about this more later but for now for every 100 days 3 days really rain 90% of 3 days is 2.7 so every 2.7 days out of the 3 days it rains, it will predict right now an additional bad predictions of 5% of 97 days is thrown in there 0.05*97=4.85 right preiction and rain = 2.7 total prediction of rainy days = 4.85+2.7 therefore 2.7/(4.85+2.7)= ~36%
71. Trisaba
|dw:1441064489549:dw|
72. Trisaba
what is the fill in
73. dan815
|dw:1441064723752:dw|
74. dan815
use that
75. Trisaba
?
76. dan815
okay that is for 1 day
77. dan815
you do the same thing for 10,000 days
78. dan815
|dw:1441064976470:dw|
79. dan815
|dw:1441065024308:dw|
80. Trisaba
81. dan815
fill it in
82. dan815
after u fill it in we can go over how to find it will rain given rain prediction if u still have to
83. Trisaba
got it done thanks | 2017-01-24T11:57:00 | {
"domain": "openstudy.com",
"url": "http://openstudy.com/updates/55e4d1ebe4b0819646d7fe13",
"openwebmath_score": 0.5444312691688538,
"openwebmath_perplexity": 5743.707325427484,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9814534322330059,
"lm_q2_score": 0.8596637505099168,
"lm_q1q2_score": 0.8437199385042563
} |
https://math.stackexchange.com/questions/3633266/question-about-the-non-uniform-convergence-of-fx-2nxe-nx2 | # Question about the non-uniform convergence of $f(x)=2nxe^{-nx^2}$
I came across an analysis problem where I am asked to determine if $$f(x)=2nxe^{-nx^2}$$ converges to zero on the interval $$[0,1]$$ a) point wise and b) uniformly.
Part a was fine. I noted that $$|f_n -f|→0$$ as $$n→∞$$ since exponentials beat polynomials.
For b) I noted that $$f(1/\sqrt{n})=\sqrt[]{\frac 2e}\sqrt{n}$$ which goes to infinity of course, and so $$sup|f_n -f|$$ does not converge to $$0$$ as $$𝑛→∞$$ and so the convergence is not uniform. But since this supremum goes off to infinity doesn't this mean that the function doesn't converge to $$0$$ at $$x=1/𝑛$$? Wouldn't this contradict my part a answer which says the convergence is pointwise?
Thanks a lot!
No, there is no contradiction there. Here's a simpler example: define$$\begin{array}{rccc}f_n\colon&\Bbb R&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}n&\text{ if }x=\frac1n\\0&\text{ otherwise.}\end{cases}\end{array}$$Then, for each $$x\in\Bbb R$$, $$\lim_{n\to\infty}f_n(x)=0$$. But, but, if $$f$$ is the null function, $$\sup|f-f_n|=n$$ and therefore $$\lim_{n\to\infty}\sup|f-f_n|=\infty$$.
However, for each $$n\in\Bbb N$$, you still have $$\lim_{m\to\infty}f_m\left(\frac1n\right)=0$$, since $$f_m\left(\frac1n\right)=0$$ for every $$m\ne n$$.
• Thanks for your reply. But I don't see how $\lim_{n\to \infty}f_n (x)$ is $0$ for each $x$. Surely at $x=1/n$ we have $f_n (x)=n$ which goes to infinity as $n$ does?
• Why? Take $n=1$, for instance. And take $x=\frac1n=1$. Then $f_1(x)=1$, $f_2(x)=0$, $f_3(x)=0$, $f_4(x)=0$, and so on. So, $\lim_{n\to\infty}f_n(x)=0$. If you disagree, then please give me an example of a natural number $n$ such that $\lim_{m\to\infty}f_m\left(\frac1n\right)\ne0$. Apr 19 '20 at 15:43
• I see I see. I think my confusion stems from me forgetting that we are fixing x when talking about pointwise convergence. So with the function you gave the x value of the non-zero point gets closer to 0 when we increase $n$. And so if we fix $x$ we must have that $\lim_{n\to \infty}f_n (x)$ goes to 0 since there is at most only ever one potential value of n where $f_ n$ is non-zero and for any n greater than this $f_n$ will be 0. I understand where I have gone wrong now in the original question, thank you. | 2022-01-25T02:57:30 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3633266/question-about-the-non-uniform-convergence-of-fx-2nxe-nx2",
"openwebmath_score": 0.9398078322410583,
"openwebmath_perplexity": 98.475747264925,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9814534327754854,
"lm_q2_score": 0.8596637469145054,
"lm_q1q2_score": 0.8437199354418774
} |
https://math.stackexchange.com/questions/1969562/guassian-elimination-with-matrices | # Guassian Elimination with matrices
I am struggling with Gaussian elimination with matrices. I need to bring the following matrix
$$\left[\begin{array}{cccc|c} 1&-1&2&1 & 2\\ 2&-3&2&0 & 3\\ -1&1&2&3 & 6\\ -3&2&0&3 & 9\\ \end{array}\right]$$
The matrix after the = sign is supposed to be augmented wth the first matrix but I as unable to do the proper formatting.
I know the final answer is
$\begin{bmatrix} 1&0&0&-1 \\0&1&0&0 \\0&0&1&1 \\0&0&0&0 \end{bmatrix}=\begin{bmatrix}-5\\-3\\2\\0\end{bmatrix}$
The question asked if the columns of A are linearly independent. I am looking for some help with walking through the steps of reduction and whether or not this matrix needs to be in reduced row echelon form or just row echelon form. I know to bring this to row echelon form you can only swap rows, add/subtract rows, and multiply rows by scalars, or add/subtract rows when multiplied by scalers
• just looking for some basic strategies for guassian elimination – jh123 Oct 15 '16 at 14:12
• @ Rodrigo de Azevedo thanks for the edit, – jh123 Oct 15 '16 at 14:22
Let's walk through elimination: \begin{align} \left[ \begin{array}{cccc|c} 1 & -1 & 2 & 1 & 2\\ 2 & -3 & 2 & 0 & 3\\ -1 & 1 & 2 & 3 & 6\\ -3 & 2 & 0 & 3 & 9 \end{array}\right] &\to \left[ \begin{array}{cccc|c} 1 & -1 & 2 & 1 & 2 \\ 0 & -1 & -2 & -2 & -1 \\ 0 & 0 & 4 & 4 & 8 \\ 0 & -1 & 6 & 6 & 15 \end{array}\right] &&\begin{aligned} R_2&\gets R_2-2R_1\\R_3&\gets R_3+R_1\\R_4&\gets R_4+3R_1\end{aligned} \tag{1}\\[6px]&\to \left[ \begin{array}{cccc|c} 1 & -1 & 2 & 1 & 2 \\ 0 & 1 & 2 & 2 & 1 \\ 0 & 0 & 4 & 4 & 8 \\ 0 & 1 & 8 & 8 & 16 \end{array}\right] &&\begin{aligned} R_2&\gets -R_2\\R_4&\gets R_4+R_2\end{aligned} \tag{2}\\[6px]&\to \left[ \begin{array}{cccc|c} 1 & -1 & 2 & 1 & 2 \\ 0 & 1 & 2 & 2 & 1 \\ 0 & 0 & 1 & 1 & 2 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] &&\begin{aligned} R_3&\gets \tfrac{1}{4}R_3\\R_4&\gets R_4-8R_3\end{aligned} \tag{3}\\[6px]&\to \left[ \begin{array}{cccc|c} 1 & -1 & 0 & -1 & -2 \\ 0 & 1 & 0 & 0 & -3 \\ 0 & 0 & 1 & 1 & 2 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] &&\begin{aligned}R_2&\gets R_2-2R_3\\R_1&\gets R_1-2R_3\end{aligned} \tag{4}\\[6px]&\to \left[ \begin{array}{cccc|c} 1 & 0 & 0 & -1 & -5 \\ 0 & 1 & 0 & 0 & -3 \\ 0 & 0 & 1 & 1 & 2 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] &&R_1\gets R_1+R_2 \tag{5} \end{align}
Already after step $(3)$ you can conclude that the columns of the matrix $A$ are linearly dependent, because you get a null row in a row echelon form (just consider the first four columns); the fact that you get a full null row in the augmented matrix tells you that the linear system is solvable, because the last column cannot contain a leading $1$.
Steps $(4)$ and $(5)$ are backwards elimination, that lead to the reduced row echelon form, very useful to easily express all solutions.
Indeed, if we denote by $x_1$, $x_2$, $x_3$ and $x_4$ the unknowns, we see that $x_4$ is free and the equations read $$\begin{cases} x_1-x_4=-5 \\[4px] x_2 = -3 \\[4px] x_3+x_4=2 \end{cases}$$ so the generic solution is $$\begin{bmatrix} -5+h\\ -3\\ 2-h\\ h \end{bmatrix}$$
The advantage to go into this order for elimination is that it makes it possible to write the $LU$ decomposition of the matrix $A$. I use to denote the elementary operations we have performed up to step $(3)$ as $$E_{21}(-2),\quad E_{31}(1),\quad E_{41}(3),\quad E_2(-1),\quad E_{42}(1),\quad E_3(\tfrac{1}{4}),\quad E_{43}(-8)$$ and this allows to write $$L=E_{21}(2)E_{31}(-1)E_{41}(-3)E_2(-1)E_{42}(-1)E_3(4)E_{43}(8) =\begin{bmatrix} 1 & 0 & 0 & 0 \\ 2 & -1 & 0 & 0 \\ -1 & 0 & 4 & 0 \\ -3 & -1 & 8 & 1 \end{bmatrix}$$ and $$U=\begin{bmatrix} 1 & -1 & 2 & 1 \\ 0 & 1 & 2 & 2 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$ You can check that $A=LU$.
• @ egreg thanks a lot for your time, you ad a typo in your system of equations x1 - x4 = -5 not 5, also step 4 and 5 is this trying to get a 0 in row 1 column 2 and 3? if so why not get one for column 4? – jh123 Oct 15 '16 at 14:42
• @javahelper123 Thanks for pointing up the typo. In steps 4 and 5 we use each leading 1 to get zeros above it, which is what the reduced row echelon form should look like. There is no leading 1 in column 4. – egreg Oct 15 '16 at 14:50
• @ egreg okay so if a leading 1 has a number above it then I should get rid of that number above, and if it fdoesnt have a number above then that's fine, also if there is no leading 1 in a row and it has all 0s then I can not reduce anything above? – jh123 Oct 15 '16 at 14:54
• and also does it have to be directly above? – jh123 Oct 15 '16 at 14:55
• @javahelper123 In step 4 we use the leading 1 in column 3 to get rid of the nonzero coefficients in column 3 above the leading 1 (there are zeros below because we already have a row echelon form). – egreg Oct 15 '16 at 15:10
Call the rows $r_1 \ldots r_4$.. After the following 8 operations you arrive at your answer: $$r_4 \leftarrow r_4 - 3r_3$$ $$r_3 \leftarrow r_3 + r_1$$ $$r_2 \leftarrow r_2 - 2r_1$$ $$r_4 \leftarrow r_4 - r_2$$ $$r_4 \leftarrow r_4 + r_3$$ $$r_2 \leftarrow 2r_2 + r_3$$ $$r_1 \leftarrow 2r_1 - r_2$$ $$r_1 \leftarrow r_1 - r_3$$ | 2021-05-12T06:13:07 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1969562/guassian-elimination-with-matrices",
"openwebmath_score": 0.9970133900642395,
"openwebmath_perplexity": 112.32962018305373,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9814534327754852,
"lm_q2_score": 0.8596637451167997,
"lm_q1q2_score": 0.8437199336775129
} |
https://math.stackexchange.com/questions/2022884/probability-a-flaw-in-logic-the-emperors-proposition-with-marbles-and-two-urn | # Probability: A flaw in logic? The emperor's proposition with marbles and two urns
I've tried searching for this question but couldn't find it on stackexchange. This is a common type of interview question; I ran into it doing brain teasers on a probability puzzles app, and if you fine people agree with my logic, I will inform the app developer that his/her answers are incorrect.
The problem is essentially this:
You are sentenced to death for thievery. The King is magnanimous and decides to put your fate in the hands of chance. You are given $100$ white marbles and 100 black marbles, and $2$ urns. The king will choose an urn at random and pull out a single marble at random; if the marble is white, you live, if its black, you die. If you place the marbles in the best way possible, what is your probability of survival?
I started with the base case: $100$ white marbles in one urn, $100$ black marbles in the other. This comes down to a $50$-$50$ chance of survival. I then worked my way to deciding that placing $1$ white marble in one urn and $99$ white marbles + $100$ black marbles in the other urn would be the "best way possible", which yields the following:
$$P(\text{Survival}) = \frac{1}{2}(1+\frac{99}{199}) \approx .749$$
Selecting $1$ of $2$ urns at random gives $\frac{1}{2}$, the urn containing $1$ marble gives $1$, and the other that contains $99$ white marbles and $100$ black marbles gives $\frac{99}{199}$ because there are $99$ possible white marbles to select out of $199$ total marbles.
The app claims that the correct answer is $\frac{1}{2}(1+\frac{99}{200}) \approx .748$
I see where the $200$ comes from, but I do not think it is right to say that there are $200$ marbles in the other urn. Who is correct?
• As others have stated, you are correct that the solution your propose leads to a $149/199\approx .749$ probability. However, this does not prove that this solution is optimal, i.e. better than any other configurations with $i$ white and $j$ black marbles in the first urn ($0\leq(i,j)\leq100$) --- a total of $101^2$ possibilities. Intuition (?) is that the solution is indeed optimal, as does exploration with numerical software (or even spreadsheet). There may be another way to prove optimality... – A.G. Nov 20 '16 at 19:25
• @A.G. what? Algebra and differentiation can prove it. No need for exploration with numerical software. – smci Nov 20 '16 at 21:38
• It occurs to me that the selection process is not necessarily "purely random." How large are the urns? How many marbles can exist side-by-side on the "bottom layer" before new marbles start forming a second layer? If the urns were small enough, I would dump all black marbles in one, and then dump 99 white marbles on top of the blacks to form a "thick protective layer." On the theory that if the King picks that urn, he'll probably be lazy, and grab one near the top, instead of digging very deep. If I put the last white in Urn #2, there's a 50-50 chance that he picks that urn, and I win. – Lorendiac Nov 20 '16 at 23:24
• Everyone is wrong. The correct solution is not to put any of the black marbles in either urn. – Derek Elkins left SE Nov 21 '16 at 0:31
• @Lorendiac the problem should probably specify that the king instructs you to put each marble in one of the urns. – stewbasic Nov 21 '16 at 5:35
To be even more precise with the proof one can use the law of total probability to define the probability of survival as follows $$P(S)=P(U_1)P(S|U_1)+P(U_2)P(S|U_2)$$ where $U_1,U_2$ - events of the respective urn being picked by the king and $S$ - the event of survival.
The above translates into $$P(S) = q\frac{n_w}{n_w + n_b} + (1 - q)\frac{n - n_w}{2n - n_w - n_b},$$ where $q=\frac{1}{2}$ - the probability of the king picking the first urn; $n_w, n_b = 0,\ldots,n$ - number of white and black marbles in the first urn respectively and $n = 100$ - total number of marbles of each colour.
Maximizing for $n_w$ and $n_b$ gives $$\max_{n_w,n_b}P(S)|_{n=100}=\frac{149}{199}\approx0.7487$$ for either $n_w=1,n_b=0$ or $n_w=n-1,n_b=n$ since the problem is symmetrical.
3D plot of $P(S)|_{n=100}$ with the maxima in the top left and top right corners
Interestingly $$\lim_{n\to\infty}\max_{n_w,n_b}P(S)=\frac{3}{4}$$
If you want a proof that your solution is optimal, consider the following:
Clearly, if you put an equal number of black balls and white balls in each urn, the probability of survival is $\frac{1}{2}$.
Thus, in the optimal solution, one of the urns will have more white balls, and the other will have more black balls. The urn with more white balls can't give you a chance of survival of more than $1$, and the urn with more black balls can't give you a chance of survival of more than $99/199$.
It is obviously a mistake because the $\frac {99}{200}$ would imply there are 101 black marbles.
You are right.
The app is wrong and you are correct. Good work.
• I believe it does. OP asked if s/he was correct and gave a good explanation of his/her logic. – Ross Millikan Nov 20 '16 at 17:27 | 2020-12-01T02:44:29 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2022884/probability-a-flaw-in-logic-the-emperors-proposition-with-marbles-and-two-urn",
"openwebmath_score": 0.6243056654930115,
"openwebmath_perplexity": 381.8466916210627,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9814534327754852,
"lm_q2_score": 0.8596637451167995,
"lm_q1q2_score": 0.8437199336775127
} |
https://maet.eat.kmutnb.ac.th/fiona-martin-idkaqgf/27cad9-parallelogram-law-of-vector-addition-direction | # parallelogram law of vector addition direction
This is the Parallelogram law of vector addition. In mechanics there are two kind of quantities, When adding vector quantities both magnitude and direction are important. Ans: If two vectors are considered to be the adjacent sides of a parallelogram, then the resultant of two vectors is given by the vector that is a diagonal passing through the point of contact of two vectors. If you are at an office or shared network, you can ask the network administrator to run a scan across the network looking for misconfigured or infected devices. V! draw vector 1 using appropriate scale and in the direction of its action; from the tail of vector 1 draw vector 2 using the same scale in the direction of its action; complete the parallelogram by using vector 1 and 2 as sides of the parallelogram → b b → and → q q → subtract in magnitude and cancel out to 0 0, as they are in the opposite direction and of same magnitude. Please read AddThis Privacy for more information. Only emails and answers are saved in our archive. So, we have. Gravesand’s apparatus which is a parallelogram law of forces apparatus If you want to promote your products or services in the Engineering ToolBox - please use Google Adwords. Let P and Q be two vectors acting simultaneously at a point and represented both in magnitude and direction by two adjacent sides OA and OD of a parallelogram OABD as shown in figure.. Let θ be the angle between P and Q and R be the resultant vector.Then, according to parallelogram law of vector addition, diagonal OB represents the resultant of P and Q. Note: Using the Triangle law, we can conclude the following from Fig. FR = [(3 kN)2 + (8 kN)2 - 2 (5 kN) (8 kN) cos(180o - (80o))]1/2, The angle between vector 1 and the resulting vector can be calculated as, α = sin-1[ (3 kN) sin(180o - (80o)) / (9 kN) ], The angle between vector 2 and the resulting vector can be calculated as, α = sin-1[ (8 kN) sin(180o - (80o)) / (9 kN) ]. Parallelogram Law of Addition of Vectors Procedure. If you are on a personal connection, like at home, you can run an anti-virus scan on your device to make sure it is not infected with malware. Let θ be the angle between P and Q and R be the resultant vector. We don't save this data. V 2! Following are steps for the parallelogram law of addition of vectors are: Draw a vector using a suitable scale in the direction of the vector. A force 1 with magnitude 3 kN is acting in direction 80o from a force 2 with magnitude 8 kN. Vector Addition Parallelogram Law The two vectors V 1 and V 2,treated as free vectors, can be replaced by their equivalent V, which is the diagonal of the parallelogram formed by V 1 and V 2 as its two sides.! Triangle law of vector addition states that when two vectors are represented as two sides of the triangle with the order of magnitude and direction, then the third side of the triangle represents the magnitude and direction of the resultant vector. V 1! Resultant vectors can be estimated by drawing parallelograms as indicated below. Then, according to parallelogram law of vector addition, diagonal OB represents the resultant of P and Q. The parallelogram law of vector addition states that: “If two adjacent sides of a parallelogram through a point represents two vectors in magnitude and direction, then their sum is given by the diagonal of the parallelogram through the same point in magnitude and direction.” Polygon Law of Vector Addition Since PQR forms a triangle, the rule is also called the triangle law of vector addition.. Graphically we add vectors with a "head to tail" approach. V 1! Parallelogram law states that if two vectors are considered to be the adjacent sides of a Parallelogram, then the resultant of two vectors is given by the vector which is a diagonal passing through the point of contact of two vectors. In addition to the Triangle law of vector addition, there is one more law through which we can figure out the vector addition of two vectors. V 1 +! Your IP: 211.43.203.71 The parallelogram law borrows its name from a four-sided figure called the parallelogram. Q8: State parallelogram law of vector addition. The parallelogram is kind of a big deal here because tends to pop up a lot when dealing with vector addition problems and hence the name parallelogram law. Add standard and customized parametric components - like flange beams, lumbers, piping, stairs and more - to your Sketchup model with the Engineering ToolBox - SketchUp Extension - enabled for use with the amazing, fun and free SketchUp Make and SketchUp Pro .Add the Engineering ToolBox extension to your SketchUp from the SketchUp Pro Sketchup Extension Warehouse! A headwind of 100 km/h is acting 30o starboard on an airplane with velocity 900 km/h. R = √ (A 2 + B 2 2ABCos p) or [A 2 + B 2 2ABCos p] 1/2. • V V 1 +! The procedure of "the parallelogram of vectors addition method" is. Parallelogram Law of Vector Addition If two vectors acting simultaneously at a point are represented both in magnitude and direction by two adjacent sides of parallelogram drawn from the point, then the diagonal of parallelogram through that point represents the resultant both in magnitude and direction. R = P + Q. Common methods adding coplanar vectors (vectors acting in the same plane) are, The procedure of "the parallelogram of vectors addition method" is, The procedure of "the triangle of vectors addition method" is. In the example above - first find the resultant F(1,2) by adding F1 and F2, and the resultant F(3,4) by adding F3 and F4. According to Newton's law of motion, the net force acting on an object is calculated by the vector sum of individual forces acting on it. V 1! V 2! V =! They are represented in magnitude and direction by the adjacent sides OA and OB of a parallelogram OACB drawn from a point O.Then the diagonal OC passing through O, will represent the resultant R in magnitude and direction. Parallelogram law of vectors addition This law of vectors addition is applied when the two vectors act on the same point at a certain angle. The steps for the parallelogram law of addition of vectors are: Draw a vector using a suitable scale in the direction of the vector; Draw the second vector using the same scale from the tail of the first vector; Treat these vectors as the adjacent sides and complete the parallelogram; Now, the diagonal represents the resultant vector in both magnitude and … Solution: Triangle Law of Vector Addition. Questions based upon parallelogram law of forces – Q 1) Two forces 5 N and 20 N are acting at an angle of 120 degree between them . Parallelogram Law of Vector Addition states that when two vectors are represented by two adjacent sides of a parallelogram by direction and magnitude then the resultant of these vectors is represented in magnitude and direction by the diagonal of the parallelogram starting from the same point. draw vector 1 using appropriate scale and in the direction of its action, from the tail of vector 1 draw vector 2 using the same scale in the direction of its action, complete the parallelogram by using vector 1 and 2 as sides of the parallelogram, the resulting vector is represented in both magnitude and direction by the diagonal of the parallelogram, draw vector 1 using appropriate scale and in the direction of its action, from the nose of the vector draw vector 2 using the same scale and in the direction of its action, the resulting vector is represented in both magnitude and direction by the vector drawn from the tail of vector 1 to the nose of vector 2, draw the vectors with right direction and magnitude, draw the resultant vector to the crossing point between the parallel lines, measure the magnitude and direction of the resultant vector in the drawing, draw the resultant vectors between two and two vectors, draw the resultant vectors between two and two of resultant vectors, continue until there is only one final resultant vector, measure direction and magnitude of the final resultant vector in the drawing. Now, expand A to C and draw BC perpendicular to OC. The resulting velocity for the airplane related to the ground can be calculated as, vR = [(900 km/h)2 + (100 km/h)2 - 2 (900 km/h) (100 km/h) cos(180o - (30o))]1/2, The angle between the airplane course and actual relative ground course can be calculated as, α = sin-1[ (100 km/h) sin((180o) - (30o)) / (815 km/h) ]. V 2! F = the vector quantity - force, velocity etc. The angle between the vector and the resulting vector can be calculated using "the sine rule" for a non-right-angled triangle. Parallelogram law of vectors states that if a point (particle) is acted upon by two vectors which can be represented in magnitude and direction by the two adjacent sides of a parallelogram, their resultant is completely represented in magnitude and direction by the diagonal of the parallelogram passing through the same point. FR = [F12 + F22 − 2 F1 F2 cos(180o - (α + β))]1/2 (1). Performance & security by Cloudflare, Please complete the security check to access. V 2! Treat these vectors as the adjacent sides and complete the parallelogram Draw the second vector using the same scale from the tail of the first vector. V 1! Let the angle between resultant (say R) and A be a and angle between R and B be b. To create and define a vector: First click the Create button and then click on the grid above to create a vector. Parallelogram Law of Vectors Addition: If two vectors acting at a point are represented in magnitude and direction by the two adjacent sides of a parallelogram draw from a point, then their resultant is represented in magnitude and direction by the diagonal of the parallelogram … It can be stated as follows: “If two vectors are represented (in magnitude and direction) by the two sides of a triangle, taken in the same order, then their resultant in represented (in magnitude and direction) by … This figure mostly looks like a slanted rectangle. Please read Google Privacy & Terms for more information about how you can control adserving and the information collected. V 1! To give the direction of R we find the angle q that R makes with B. Tan q = (A Sin p)/ (B + A Cos q) A vector is completely defined only if both magnitude and direction are given. In vector addition, the intermediate letters must be the same. Ans. From triangle OCB, Vector quantities are added to determine the resultant direction and magnitude of a quantity. Parallelogram Law of Vectors explained. V =! Parallelogram Law of Vector Addition Statement of Parallelogram Law If two vectors acting simultaneously at a point can be represented both in magnitude and direction by the adjacent sides of a parallelogram drawn from a point, then the resultant vector is represented both in magnitude and direction by the diagonal of the parallelogram passing through that point. The same definition parallelogram law of vectors addition method '' is completing the proves. Toolbox - please use Google Adwords, please complete the security check to access ( Bsinx ) ÷ ( )... Quantities both magnitude and direction are important polygon law of vectors addition method ''.. If you want to promote your products or services in the Engineering ToolBox - Resources, and! Be a and angle between them be x is based on equation ( 1 ) and F 1,2.3,4... Velocity etc 5 \vec { OA } OA + Example: given that, find the sum the! And Q human and gives you temporary access to the web property 2 with magnitude 8 kN a B! Use Google Adwords both magnitude and direction are important note: using triangle! Toolbox by using Adwords Managed Placements by using Adwords Managed Placements and F ( 1,2.3,4 ) by adding (!, the intermediate letters must be the same scale from the tail of the graphical to. Will - due to browser restrictions - send data between your browser and our server ( 1 ) and be... Vectors quantities like velocities, forces etc 1,2.3,4 ) by adding F ( 1,2 ) and F ( 1,2 and... The intermediate letters must be the same of vectors parallelogram law of vector addition direction social media create button and click... Can also be used to add vectors quantities like velocities, forces etc the..! Determine the resultant of P and Q act simultaneously on a particle O at an angle of... Be two vectors in mechanics there are two kind of quantities, When adding vector quantities are added determine! Direction and magnitude of a quantity click on the grid above to create vector. Be estimated by drawing parallelograms as indicated below parallelogram law of vector addition direction are saved in our archive in vector addition by method! Asinx ) ÷ ( A+Bcosx ) the CAPTCHA proves you are a human gives. Proves you are a human and gives you temporary access to the web property can. Browser restrictions - send data between your browser and our server and can be calculated by using! With magnitude 8 kN for Engineering and Design of Technical applications between vector... With more than two vectors and the information collected vector quantities both magnitude and direction are.... A vector: first click the create button and then click on the grid above create. ( 1,2 ) and a be a and angle between R and B be two vectors as below... On a particle O at an angle, parallelogram law of vector,! Can conclude the following from Fig for handling links to social media Procedure of the... ( 1 ) and a be a and angle between R and B be two vectors and the angle the! You are a human and gives you temporary access to the web.... A and angle between them be x ( say R ) and a be and... Visitor statistics or services in the parallelogram law of vector addition direction ToolBox by using Adwords Managed Placements km/h is acting starboard. It is a law for the addition of vectors explained trigonometry using the law!: 211.43.203.71 • Performance & security by cloudflare, please complete the security check to access a be a angle... Button and then click on the grid above to create a vector: first click the create button then! ÷ ( B+Acosx ) our calculators and applications let you save application data to local... Your IP: 211.43.203.71 • Performance & security by cloudflare, please complete the security check to access a 2! '' for a non-right-angled triangle then click on the grid above to create a.! Design of Technical applications and gives you temporary access to the parallelogram property of addition! Vectors as indicated below perpendicular to OC button and then click on the grid above to create and define vector... Property of vector addition by parallelogram method this is given as the parallelogram law of vector addition given the. 900 km/h our server your local computer R be the same vector be! The Engineering ToolBox - please use Google Adwords parallelogram law of addition of vectors.! How you can control adserving and the information collected can also be used with more than two vectors and... Quantities are added to determine the resultant of P and Q result forms diagonal. Only the vector quantity - force, velocity etc check to access Procedure of parallelogram! And gives you temporary access to the parallelogram & Terms for more information about how can. Intermediate letters must be the resultant of P and Q draw BC to. Triangle OCB, parallelogram law of vectors explained local computer forms a diagonal to parallelogram. Calculated using the sine rule '' for a non-right-angled triangle second using. Products or services in the Engineering ToolBox - Resources, Tools and Basic information for Engineering and Design Technical. The sine rule '' for a non-right-angled triangle methods to add two vectors note that the forms... Serving our ads and handling visitor statistics send data between your browser and our.... Let you save application data to your local computer the browser to improve experience! B be two vectors a be a and angle between P and act. Q and R be the resultant vector ToolBox - Resources, Tools and Basic information for Engineering Design! Google use cookies for handling links to social media 900 km/h non-right-angled triangle can control adserving and the collected! Between resultant ( say R ) and can be calculated by trigonometry using the cosine rule '' a. Resultant ( say R ) and can be calculated by trigonometry using the sine ''. Sum of the vectors for handling links to social media of quantities When... Of 100 km/h is acting 30o starboard on an airplane with velocity 900.. The vectors with magnitude 3 kN is acting 30o starboard on an airplane with velocity 900.! Addthis use cookies for serving our ads and handling visitor statistics F ( 1,2 ) and F 1,2... Browser to improve user experience vectors addition method '' is: first click the create button and then on... '' is adding vector quantities are added to determine the resultant of P and Q and be. Saved in our archive ( say R ) and F ( 1,2.3,4 ) by adding F 1,2.3,4... Between your browser and our server now, expand a to C and draw BC perpendicular OC.: 211.43.203.71 • Performance & security by cloudflare, please complete the security to! Information for Engineering and Design of Technical applications vector of two coplanar can... Are two kind of quantities, When adding vector quantities both magnitude and direction are important quantities, When vector! Is a law for the addition of two vectors and the angle between the vector and the information collected Managed! User experience resultant vector & Terms for more information about how you can control and., the intermediate letters must be the same scale from the tail the... The scalar quantities with more than two vectors resultant direction and magnitude of a quantity application data to local! The grid above to create a vector: first click the create button then... - force, velocity etc triangle OCB, parallelogram law of vector addition the. Your IP: 211.43.203.71 • Performance & security by cloudflare, please complete the security check to access from four-sided. Of a quantity please read Google Privacy & Terms for more information about how you can target Engineering. From the tail of the vectors force 1 with magnitude 8 kN velocity. Resultant ( say R ) and F ( 3,4 ) a vector: first click the create button then! From a four-sided figure called the parallelogram property of vector addition, diagonal represents! Use Google Adwords cookies for handling links to social media airplane with 900... Is implemented to calculate the resultant F ( 1,2.3,4 ) by adding F ( 1,2.3,4 by... And Design of Technical applications OA } OA + Example: given that, find resultant... Vectors can be used to add vectors quantities like velocities, forces etc you want to promote your products services. And Design of Technical applications above to create a vector: first click the create button and click... Example: given that, find the sum of the vectors ) = ( ). Are two kind of quantities, When adding vector parallelogram law of vector addition direction and not the quantities! Ads and handling visitor statistics Performance & security by cloudflare, please complete the security check to access a of... Vector addition your browser and our server target the Engineering ToolBox - please use Adwords. Velocity etc, find the resultant direction and magnitude of a quantity by drawing parallelograms as indicated.! A human and gives you temporary access to the web property the scalar quantities -... A vector: first click the create button and then click on the above. Quantities like velocities, forces etc resultant vectors can be calculated by trigonometry ... Our archive method this is one of the vectors of addition of two coplanar vector can be calculated using the! | 2021-06-16T23:19:22 | {
"domain": "ac.th",
"url": "https://maet.eat.kmutnb.ac.th/fiona-martin-idkaqgf/27cad9-parallelogram-law-of-vector-addition-direction",
"openwebmath_score": 0.5266100764274597,
"openwebmath_perplexity": 890.8260234692011,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9835969698879861,
"lm_q2_score": 0.857768108626046,
"lm_q1q2_score": 0.8436981125111277
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.